Numerical Treatment for Solving Linear Fractional-Order Volterra Integro-Differential Equations with Constant Time-Delay of Retarded Type A thesis Submitted to the Council of Faculty of Science and Science Education School of Science at the University of Sulaimani In partial fulfillment of the requirements for the degree of Master of Science in Mathematics (Numerical Analysis)
By Miran Bayan Mohammed Amin B.Sc. in Mathematics (2008), University of Sulaimani
Supervised by Dr. Shazad Shawki Ahmed Assistant professor
June
2016
Pwshpar
2716
� ِْﺴ ِﻢ � ِ ا� �اﻟﺮ ْ َﲪ ِﻦ �اﻟﺮ ِﺣ ِﲓ
�َﺮﻓَ ْﻊ ٔأ ُ� ٔأ � ِ� َ�ﻦ َءا َﻣنُﻮ ْا ِﻣنﲂُ َو ٔأ � ِ� َ�ﻦ �أوﺗُﻮ �أ ٔأﻟ ِﻌ َﲅ َد َر َﺟ ٍ �ﺎت ُ ﻮن َﺧب ٌِﲑ َو ٔأ ُ� ِﺑ َﻤﺎ ﺗَ ْﻌ َﻤﻠ َ ﺳﻮرة ا�ﺎد� ا�ٓﻳﺔ 11
Supervisor's Certification I certify that the preparation of thesis titled “Numerical Treatment for Solving Linear
Fractional-Order
Volterra Integro-Differential Equations with
Constant Time-Delay of Retarded Type” accomplished by (Miran Bayan Mohammed Amin) was prepared under my supervision at the School of Science, Faculty of Science and Science Education at the University of Sulaimani, as partial fulfillment of the requirements for the degree of Master of Science in (Mathematics).
Signature: Name: Dr. Shazad Shawki Ahmed Title: Assistant Professor Date:
/
/ 2016
In view of the available recommendation, I forward this thesis for debate by the examining committee.
Signature: Name: Dr. Karwan Hama Faraj Jwamer Title: Professor Date:
/
/ 2016
Linguistic Evaluation Certification I herby certify that this thesis titled “Numerical Treatment for Solving Linear Fractional-Order Volterra Integro-Differential Equations with Constant Time-Delay of Retarded Type” prepared by (Miran Bayan Mohammed Amin), has been read and checked and after indicating all the grammatical and spelling mistakes; the thesis was given again to the candidate to make the adequate corrections. After the second reading, I found that the candidate corrected the indicated mistakes. Therefore, I certify that this thesis is free from mistakes.
Signature: Name: Sawen Salih Aziz Position: English Department, School of Languages, University of Sulaimani Date: 17 / 4 / 2016
Examination Committee Certification We certify that we have read this thesis entitled “Numerical Treatment for Solving Linear Fractional-Order Volterra Integro-Differential Equations with Constant Time-Delay of Retarded Type” prepared by (Miran Bayan Mohammed Amin), As the Examining Committee, examined the student in its content and in what is connected with it, and in our opinion it meets the basic requirements toward the degree of Master of Science in Mathematics (Numerical Analysis).
Signature:
Signature:
Name: Dr. Karwan H.F Jwamer
Name: Dr. Burhan F. Jumaa
Title: Professor
Title: Assistant Professor
Date: 2 / 6 / 2016
Date: 2 / 6 / 2016 (Member)
(Chairman)
Signature:
Signature:
Name: Dr. Ayad M. Ramadan
Name: Dr. Shazad Sh. Ahmad
Title: Assistant Professor
Title: Assistant Professor
Date: 2 / 6 / 2016
Date: 2 / 6 / 2016
(Member)
(Supervisor and Member)
Approved by the Dean of the Faculty of Science and Science Education.
Signature: Name: Dr. Bakhtiar Q. Aziz Title: Professor Date:
/
/ 2016
Dedications
Dedicated To: My Father and My Mother My Lovely Wife “Rozha” My Lovely Son “Mazan” My Brothers and Sisters All My Friends
Miran Bayan Mohammed Amin 2016
Acknowledgements
Alhamdulillah, all praises belong to Allah. (S.W.T). who has given the writer the health and the strength to finish this thesis.
I am grateful for the cooperation and constant encouragement from my honorable assistant professor, Dr. Shazad Shawki Ahmed, whose regular suggestions made my work easy and proficient My deep thanks to Professor Dr. Karwan Hama Faraj Jwamer, the Head of the Department of Mathematics and all the staff of the Department of Mathematics, Faculty of Science and Science Education, University of Sulaimani; And all our teachers for their invaluable instructions and comments on the study. Finally, special thanks to my Lovely wife, Rozha, for helping me throughout the years of my study; without her encouragement I could not complete this work.
Miran Bayan Mohammed amin 2016
Abstract
In this thesis, the main goal is to study and modify some approximate methods as well as new algorithms of numerical solutions that have been proposed for the first time to treat the higher order Linear Volterra Integro-Fractional Differential equations (LVIFDEs) in Caputo sense of constant multi-time Retarded delay. Two numerical methods for solving LVIFDEs of constant multi-time Retarded delay have been concluded by using Block-by-Block methods and Least-square orthogonal methods: Three different blocks of Block-by-Block methods (two, three and four) have also been presented and modified successfully to employ with finite difference techniques to approximate the solution for this equation numerically. Moreover, the least-square techniques with aid of orthogonal polynomials (Legendre and Chebyshev) associated with the delays idea have been applied in one method for the first time to find the approximate solution for LVIFDEs of constant multi-time Retarded delay with variable coefficients. Finally, programs in general procedure are written for the proposed algorithms in MatLab (V.8) and applied for several illustrative examples to show the effectiveness and accuracy of the presented method.
i
List of Abbreviations Symbols
Descriptions
VIE FIE IDE FIDE ODE LFIDDE LFFIDDE LVIDDE FVIDDE VIDDE VIFDE VIDE FC DDE IFDE R-L C[a, b] C m [a, b] Cγ [a, b] Cγm [a, b]
Volterra Integral Equation Fredholm Integral Equation Integro-Differential Equation Fredholm Integro-Differential Equation Ordinary Differential Equation Linear Fractional Integro Delay Differential Equation Linear Fredholm Fractional Integro Delay Differential Equation LinearVolterra Integro Delay Differential Equation Fractional Volterra Integro-Delay Differential Equation Volterra Integro-Delay Differential Equation Volterra Integro-Fractional Differential Equation Volterra Integro-Differential Equation Fractional Calculus Delay Differential Equation Integro-Fractional Differential Equation Riemann-Liouville Space of Continuous Function on [a, b] Collection of all m-times Continuously Differentiable Function on [a, b] Space of γ-Functions Collection of all Functions the m-Th Derivative in the Space of γ-Functions Caputo Fractional Derivative of Order α Riemann-Liouville Fractional Derivative of Order α Riemann-Liouville Fractional integral operator Integer Ceiling Function Integer Floor Function Laplace Transform. Inverse Laplace Transform. Laplace Adomain Decomposition Method Adomain Decomposition Method Algorithm of Two Block Method, Algorithm of Three Block Method ALgorithm of Four Block Method Block-by-Block Method Program of Two Block Method Program of Three Block Method Program of Four Block Method Two Block Method, Three Block Method, Four Block Method Least Square Error Least Square Error function Running Time of any Program per Second Least Square Orthogonal Method Algorithm Delay Chebyshev polynomial Algorithm Delay Legendre polynomial Delay Closed Chebyshev polynomial method Delay Open Chebyshev polynomial method Delay Legendre polynomial method
C α a Dt R α a Dt α a Jt
⌈∗⌉ ⌊∗⌋ ℒ{u(t)} ℒ −1 {u(t)} LADM ADM A2BM, A3BM, A4BM BBM P2BM, P3BM, P4BM 2BM, 3BM, 4BM L.S.E L. S. Eu R.Time/Sec LSOM ADCH ADLP DCH DOH DLE
ii
Contents
Subjects General introduction
Page no.
Chapter One : Basic Concepts 1.1 Introduction ……………………………………………………………. 1.2 Some Special Functions of the Fractional Calculus…………………… 1.2.1 The Gamma Function ……………………………………………. 1.2.2 The Beta Function………………………………………………… 1.2.3 Mittag-Leffler Function…………………………………………... 1.3 Fractional Integration and Differentiation……………………………... 1.3.1 Riemann-Liouville Fractional Operators…………………………. 1.3.2 Properties of R-L Fractional Operators…………………………... 1.3.3 Composition of Riemann-Liouville operators……………............. 1.3.4 Caputo Fractional Derivative…………………………………....... 1.3.5 Properties of Caputo Fractional Operator……………………........ 1.4 Classification of integral equation………………………………........... 1.5 Delay Differential Equations………………………………………....... 1.6 Fractional Integro-Delay Differential equations………………………. 1.7 Occurrence of Linear VIFDEs of Retarded constant time delays........... 1.8 The Contribution of the Thesis………………………………………. 1.9 Test Examples………………………………………………………… Chapter Two : Block-by-Block Method 2.1 Introduction ……………………………………………………………. 2.2 Basic Propositions………………………………………………............ 2.3 Basic Quadrature Formulas……………………………………………. 2.4 A Parameter Estimation Procedure…………………………………….. 2.5 Solution Method for Higher-Fractional Order Linear VIDEs of Constant Multi-Time Retarded Delay …………........................................... 2.5.1 Two Block Method…………………………………………........... 2.5.2 Three Block Method………………………………………………. 2.5.3 Four Block Method……………………………………………....... 2.6 Implementation of the method…………………………………………. 2.7 Discussion…………………………………………………………........
iii
1 2 2 3 3 4 4 5 6 8 8 11 14 15 17 26 27 30 30 31 34 36 39 43 50 56 81
Chapter Three : Least Square Orthogonal Method 3.1 Introduction ………………………………………………………......... 3.2 Orthogonal polynomials……………………………………………….. 3.2.1 Shifted Chebyshev Polynomials……………………………........... 3.2.2 Shifted Legendre Polynomials…………………………………….. 3.3 Gauss and Clenshaw-Curtis Formulas…………………………………. 3.3.1 Gauss-Chebyshev Quadrature Formula……………………........... 3.3.2 Gauss-Legendre Quadrature Formula…………………………….. 3.3.3 Cleanshaw-Curtis Integration Formula……………………............ 3.4 Solution Technique for VIFDEs with multi-time Delay………………. 3.4.1 Using Chebyshev Polynomials……………………………………. 3.4.2 Using Legendre Polynomials………………………………............ 3.5 Implementation of the method…………………………………………. 3.6 Discussion…………………………………………………………........ Chapter Four : Conclusions and Recommendations 4.1 Conclusions…………………………………………………………….. 4.2 Recommendations………………………………………………............ Appendix…………………………………………………………………... References………………………………………………………………….
iv
82 83 83 84 85 86 86 86 87 103 108 112 125 126 133 134 184
General Introduction
General Introduction Fractional Calculus (FC) is a natural generalization of integration and derivation to non-integer order operators, one of the major advantages of fractional calculus is that it can be considered as a super set of integer-order calculus. Thus, fractional calculus has the potential to accomplish what integer order calculus cannot. The idea of FC has been known as a science since the development of the regular calculus [22, 69]. Presently there exist several versions for defining the fractional operator, such as the Riemann's, the Liouville's, the Gruwald-Letnikov's, the Riemann- Liouville's, and the Caputo's fractional integrals and derivatives [37, 42]. The first book published on FC is the book of Oldham and Spanmier [37] published in 1974. The remarkably comprehensive encyclopedic-type monograph by Samko, Kilbas and Marichev [64], which was published in English in 1993. One of the most recent works on the subject of FC is the book of Podlubny [69], published in 1999. Some of the latest works especially on fractional physics models of anomalous kinetics of complex processes were edited by Hilfer [57] in 2000. During the last decades FC has blossomed and grown in pure mathematics as well as in scientific applications. We can see applications in many fields of science and engineering, including fluid mechanic, theology, hydrogeology, control theory of Dynamical systems, signal processing, mechanical engineering and finance [30 , 42 , 57, 69]. Furthermore, the fractional order integro-differential field is a rapidly growing field on both theory and applications [64], it is natural to study the analytical and numerical computations of such problem related to the fractional calculus.
This thesis is organized as follows: In chapter one an introduction of the fractional calculus was presented which includes Mathematical formulation of special functions (the gamma and beta v
General Introduction
function and Mittag-Leffler function) which play the most important role in the theory of fractional derivatives and integrals. It also, contains the definition and some potentially useful properties of the Riemann-Liouville fractional operators (integrals and derivatives) and also Caputo derivatives, classification of integral equations, and occurrence of Linear Volterra Integro-Fractional Differential Equation (LVIFDEs) of constant multi-time Retarded delay with variable coefficients. In addition, the mixed multi-term linear fractional orders DE’s reduced to a higher order LVIFDE of delay type with variable coefficients was presented.
In chapter two and four Numerical method are presented: In chapter two Block-by-Block method of type two, three and four Blocks with aid of finite difference techniques are applied to treat linear VIFDE of constant retarded delays type with variable coefficient numerically. While chapter three considers the numerical solution of linear VIFDE of constant retarded delays type with variable coefficients using least-square techniques with aid of orthogonal (Legendre and Chebyshev) polynomials which reduced our problem to an algebraic equations, by solving it, the constants of approximate expression were obtained. Finally, chapter four demonstrates the conclusions and recommendations for future work, and some basic points and comparisons between the methods are shown in tables and figures.
For each numerical method an algorithm had been built, a program was written, examples were solved, and results have been tabulated. Comparisons were made between the exact and numerical solutions depending on the least square errors. All the computer programs were written in MatLab (V.8).
vi
CHAPTER ONE
Basic Concepts
Chapter One
Basic Concepts
1.1 Introduction Fractional Calculus (FC) is the brunch of calculus that generalizes the derivative of a function to non-integer order, allowing calculations such as deriving a function to ½ orders. The name “fractional” is used for denoting this kind of derivative [27]. The fractional calculus may be considered an old and yet novel topic, which is starting from some speculations of Leibniz (1695, 1697) and L. Euler (1730); it has been developed up to nowadays. The idea of generalizing the notion of derivative, meaning of 𝑑𝑑 𝑛𝑛 𝑢𝑢⁄𝑑𝑑𝑡𝑡 𝑛𝑛 = 𝐷𝐷𝑛𝑛 𝑢𝑢(𝑡𝑡), to non-integer order (𝑛𝑛), in particular to the order ½, is contained in the correspondence of Leibniz with Bernoulli, L’Hopital and Wallis. Euler took the first step by observing that the result of the evaluation of the derivative of the power function has a meaning for non-integer thanks to his Gamma Function. One of the major advantages of fractional calculus is that it can be considered as a super set of integer-order calculus. Thus, fractional calculus has the potential to accomplish what integer order calculus cannot. For more details in historical development of fractional calculus see [9, 27, 37, 39]. The last decades, FC (integral and derivative) have gained considerable importance recently due to their wide range of applications in many contexts of science. Nowadays it is impossible to describe a viscoelastic process without using a fractional integral and derivative. FC has also been used in engineering; physics, finance and hydrology, and other applications are which polymer physics, thermodynamics, Electrochemistry of corrosion, optics, Electrical Networks, biophysics and the behavior of viscoplastic materials [22, 27, 30, 37, 57]. Presently, there exist several versions for defining the fractional operator, such as the Riemann's, the Liouville's, the Gruwald-Letnikov's, the Riemann- Liouville's, and the Caputo's fractional integrals and derivatives [37, 39, 46]. Delay Differential Equation (DDE’s) is an important class of a functional differential equation in which the derivatives of some unknown functions at present time are dependent on the values of the functions at previous times. One reason for the importance is that they describe processes with “after-effects”, that is, time lag (if the independent variable denotes time). Naturally, therefore, delay differential equations had found many applications in mechanics, physics, engineering, economics and biology and especially in the theory of automatic control [26]. However, in practice, these functional differential equations arise in many areas of mathematical modeling: for example, control system [23], lasers, traffic models [15], metal cutting, epidemiology, population dynamics [41], chemical kinetics [28], etc.There are now a collection of books that indicate application areas for 1
Chapter One
Basic Concepts
DDE’s and we cite, in particular, Bellman and Cooke [53], Hale [31], Driver [18], and El’sgol’ts and Norkin [19]. Integro-Fractional Differential Equations (IFDE's) for both types Volterra Fredholm have also gained a lot of interest in many applications, such as biological, physical and engineering problems [40]. The concept of higher-order Volterra integro-fractional differential equations with multi-time delay combine three important subjects: Fractional differentiation, integro-differential equations of Volterra type and delay functional equations which have motivated a huge amount of research work in the last thirty years. 1.2 Some Special Functions of the Fractional Calculus: In this section we present the definitions and properties of the Euler gamma function and some basic theory of the special functions connected with this function which are the most common and important properties in the study of the theory of the fractional calculus. Some information are given here on: 1.2.1 The Gamma Function: [42, 59] Let ℝ denote the set of real numbers. In terms of the familiar Gamma function Γ(𝛼𝛼) , Γ ∶ ℝ ∖ ℤ− For 𝛼𝛼 ∈ ℝ+ , the integral 0 ⟶ ℝ , is defined as: definition is formed ∞
Γ(𝛼𝛼 ) = � 𝑡𝑡 𝛼𝛼−1 𝑒𝑒 −𝑡𝑡 𝑑𝑑𝑑𝑑 0
For 𝛼𝛼 ∈ ℝ− ∖ ℤ− 0 , there exist 𝑛𝑛 ∈ ℕ such that 𝛼𝛼 + 𝑛𝑛 ∈ (0,1) the gamma function is formed Γ(𝛼𝛼 ) =
Г(𝛼𝛼 + 𝑛𝑛) ∏𝑗𝑗𝑛𝑛−1 =0 (𝛼𝛼 + 𝑗𝑗 )
The Gamma Function is simply the generalization of the factorial and binomial coefficient for all real numbers. The following recursion relation on gamma is true: • Г(𝛼𝛼 + 1) = 𝛼𝛼! , for all 𝛼𝛼 ∈ ℕ • A generalized binomial coefficient �𝜇𝜇𝜆𝜆 � may be defined, for all 𝜆𝜆, 𝜇𝜇 ∈ ℝ , by 𝜆𝜆 Г(𝜆𝜆 + 1) 𝜆𝜆 � �= � =� 𝜇𝜇 Г(𝜇𝜇 + 1) Г(𝜆𝜆 − 𝜇𝜇 + 1) 𝜆𝜆 − 𝜇𝜇
• Г(𝛼𝛼 + 1) = 𝛼𝛼 Г(𝛼𝛼) , for all 𝛼𝛼 ∈ ℝ ∖ ℤ− 0 . • Г(𝛼𝛼) it never reaches Zero, for all 𝛼𝛼 . 2
Chapter One
Basic Concepts
1.2.2 The Beta Function: [42, 59] The Beta Function is usually denoted by ℬ(𝛼𝛼, 𝛽𝛽) , which is a special function in two variables, defined as: 1
ℬ (𝛼𝛼, 𝛽𝛽) = � 𝑡𝑡 𝛼𝛼 −1 (1 − 𝑡𝑡 )𝛽𝛽 −1 𝑑𝑑𝑑𝑑 , 0
Also, equivalent to
1
ℬ (𝛼𝛼, 𝛽𝛽) = � 0
𝛼𝛼, 𝛽𝛽 ∈ ℝ+
𝑡𝑡 𝛼𝛼−1 𝑑𝑑𝑑𝑑 , (1 + 𝑡𝑡 )𝛼𝛼+𝛽𝛽
𝛼𝛼, 𝛽𝛽 ∈ ℝ+
Another expression of beta function can be connected with gamma function in a direct way: ℬ(𝛼𝛼, 𝛽𝛽) =
Moreover:
• ℬ(𝛼𝛼, 𝛽𝛽 + 𝑛𝑛) = (𝛼𝛼+𝛽𝛽
• ℬ(𝛼𝛼 + 𝑛𝑛, 𝛽𝛽) = (𝛼𝛼+𝛽𝛽 •
Γ(𝛼𝛼 )Γ(𝛽𝛽) = ℬ (𝛽𝛽, 𝛼𝛼 ) Γ(𝛼𝛼 + 𝛽𝛽 + 1)
(𝛽𝛽 +𝑛𝑛 −1)(𝛽𝛽 +𝑛𝑛 −2)…(𝛽𝛽 +1)𝛽𝛽
+𝑛𝑛 −1)(𝛼𝛼 +𝛽𝛽 +𝑛𝑛−2)…(𝛼𝛼 +𝛽𝛽 +1)(𝛼𝛼+𝛽𝛽 ) (𝛼𝛼+𝑛𝑛 −1)(𝛼𝛼 +𝑛𝑛 −2)…(𝛼𝛼+1)𝛼𝛼
+𝑛𝑛 −1)(𝛼𝛼 +𝛽𝛽 +𝑛𝑛−2)…(𝛼𝛼 +𝛽𝛽 +1)(𝛼𝛼+𝛽𝛽 )
Farther more, for −∞ < 𝑠𝑠 < 𝑥𝑥 < ∞ 𝑥𝑥
ℬ(𝛼𝛼, 𝛽𝛽) for all 𝑛𝑛 ∈ ℕ ℬ(𝛼𝛼, 𝛽𝛽) for all 𝑛𝑛 ∈ ℕ
�(𝑥𝑥 − 𝑡𝑡)𝛼𝛼−1 (𝑡𝑡 − 𝑠𝑠)𝛽𝛽 −1 𝑑𝑑𝑑𝑑 = (𝑥𝑥 − 𝑠𝑠)𝛼𝛼+𝛽𝛽 −1 ℬ(𝛼𝛼, 𝛽𝛽) . 𝑠𝑠
1.2.3 The Mittag-Leffler Function: [27, 59] The Mittag-Leffler Function plays an important role in the theory of calculus for arbitrary orders. The generalization of The Mittag-Leffler Function to several parameters can be defined in terms of a power series as: ∞
𝑡𝑡 𝑘𝑘 𝐸𝐸𝛼𝛼 1 ,𝛼𝛼 2 ,…,𝛼𝛼 𝑚𝑚 ,𝛽𝛽 (𝑡𝑡 ) = � , Γ(𝛽𝛽 + ∑𝑚𝑚 𝑖𝑖=1 𝛼𝛼𝑖𝑖 ℓ𝑖𝑖 ) 𝑘𝑘=0
(𝛼𝛼1 , 𝛼𝛼2 , … , 𝛼𝛼𝑚𝑚 , 𝛽𝛽 > 0)
Where ℓ1 + ℓ2 + ℓ3 + ⋯ + ℓ𝑚𝑚 = 𝑘𝑘 and ℓ1 > 0, ℓ2 > 0, … , ℓ𝑚𝑚 > 0 1: 𝑚𝑚), 𝛽𝛽 we obtain wellFor special choices of the values of the parameters 𝛼𝛼𝑖𝑖 (𝑖𝑖 = ������ known classical function, e.g.: 𝐸𝐸1,1 (𝑡𝑡 ) = 𝑒𝑒 𝑡𝑡
𝐸𝐸𝛼𝛼 ,𝛽𝛽 (0) = 1
𝐸𝐸2,1 (𝑡𝑡 ) = cosh�√𝑡𝑡�
;
;
3
𝐸𝐸2,2 (𝑡𝑡 ) =
sinh�√𝑡𝑡� 𝑡𝑡
Chapter One
Basic Concepts
1.3 Fractional Integration and Differentiation: There are several types of the fractional integrals and fractional derivatives on bounded interval of the real line. This section presents some formula definitions, properties, fractional integration and differentiation operators. The most common ones are: 1.3.1 Riemann-Liouville Fractional Operators: [4, 6] Riemann-Liouville Fractional integral operator is direct generalization of Cauchy’s formula for n-fold integral of the function 𝑢𝑢(𝑡𝑡) thus: 𝑛𝑛 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡 )
𝑡𝑡
=� � 𝑎𝑎
𝜇𝜇 𝑛𝑛 −1
𝑎𝑎
�
𝜇𝜇 𝑛𝑛 −2
𝑎𝑎
𝜇𝜇 1
⋯ � 𝑢𝑢(𝜇𝜇)𝑑𝑑𝑑𝑑𝑑𝑑𝜇𝜇1 ⋯ 𝑑𝑑𝜇𝜇𝑛𝑛−1 𝑎𝑎
𝑡𝑡
1 �(𝑡𝑡 − 𝜇𝜇)𝑛𝑛−1 𝑢𝑢(𝜇𝜇)𝑑𝑑𝑑𝑑 , 𝑛𝑛 ∈ ℕ = (𝑛𝑛 − 1)! 𝑎𝑎
𝑛𝑛 𝑎𝑎 𝐽𝐽𝑡𝑡
… (1.1)
Where is the 𝑛𝑛-fold integral operator with 𝑎𝑎 𝐽𝐽𝑡𝑡0 𝑢𝑢(𝑡𝑡) = 𝑢𝑢(𝑡𝑡), easily proved by induction. Replacing the discrete factorial (𝑛𝑛 − 1)! with continuous gamma function Γ(𝑛𝑛), which satisfies (𝑛𝑛 − 1)! = Γ(𝑛𝑛) for 𝑛𝑛 ∈ ℕ, we observe that the right hand side of equation (1.1) may have a meaning for non-integer values of 𝑛𝑛 , which is the generalization of it.
Definition (1.1): [70]
A real valued function 𝑢𝑢 defined on [𝑎𝑎, 𝑏𝑏] is in the space of 𝛾𝛾-functions 𝐶𝐶𝛾𝛾 [𝑎𝑎, 𝑏𝑏] 𝛾𝛾 ∈ ℝ, if there exists a real number 𝑟𝑟 > 𝛾𝛾, such that 𝑢𝑢(𝑡𝑡) = (𝑡𝑡 − 𝑎𝑎)𝑟𝑟 𝑢𝑢�(𝑡𝑡), where 𝑢𝑢� ∈ 𝐶𝐶[𝑎𝑎, 𝑏𝑏], and it is said to be in the space 𝐶𝐶𝛾𝛾𝑛𝑛 [𝑎𝑎, 𝑏𝑏] if and only if 𝑢𝑢(𝑛𝑛 ) ∈ 𝐶𝐶𝛾𝛾 [𝑎𝑎, 𝑏𝑏], 𝑛𝑛 ∈ ℕ0 . Definition (1.2): [9, 27]
The Reimann-Liouville fractional integral operator 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 of order 𝛼𝛼 ≥ 0 of a function 𝑢𝑢 ∈ 𝐶𝐶𝛾𝛾 [𝑎𝑎, 𝑏𝑏], 𝛾𝛾 ≥ −1 is defined as: 𝑡𝑡
1 �(𝑡𝑡 − 𝑥𝑥)𝛼𝛼−1 𝑢𝑢(𝑥𝑥)𝑑𝑑𝑥𝑥 𝛼𝛼 ( ) 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢 𝑡𝑡 = �Γ(α) 𝑢𝑢(𝑡𝑡 )
𝑎𝑎
,
,
Moreover, the integral operator 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 has a linear operator that is 𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡
𝛼𝛼 > 0
𝛼𝛼 = 0
[𝑐𝑐1 𝑢𝑢(𝑡𝑡) + 𝑐𝑐2 𝑣𝑣(𝑡𝑡)] = 𝑐𝑐1 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡) + 𝑐𝑐2 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 𝑣𝑣(𝑡𝑡)
For any 𝑐𝑐1 , 𝑐𝑐2 ∈ ℝ and 𝛼𝛼 ∈ ℝ+ .
4
… (1.2)
Chapter One
Basic Concepts
Definition (1.3): [9, 27] The Reimann-Liouville fractional derivative 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 of order 𝛼𝛼 ≥ 0 of 𝑚𝑚 [𝑎𝑎, 𝑏𝑏], 𝑚𝑚 ∈ ℕ is defined as: 𝑢𝑢(𝑡𝑡) ∈ 𝐶𝐶−1 𝑑𝑑 𝑚𝑚 𝑚𝑚 −𝛼𝛼 𝐽𝐽 𝑢𝑢(𝑡𝑡 ) 𝑚𝑚 𝑎𝑎 𝑡𝑡 𝑑𝑑𝑡𝑡 𝑅𝑅 𝛼𝛼 ( ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡 = � 𝑚𝑚 𝑑𝑑 𝑢𝑢(𝑡𝑡 ) 𝑑𝑑𝑡𝑡 𝑚𝑚
,
𝑚𝑚 − 1 < 𝛼𝛼 < 𝑚𝑚
,
𝛼𝛼 = 𝑚𝑚
… (1.3)
Moreover, the derivative operator 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 has a linear operator that is 𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
[𝑐𝑐1 𝑢𝑢(𝑡𝑡) + 𝑐𝑐2 𝑣𝑣(𝑡𝑡)] = 𝑐𝑐1 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡) + 𝑐𝑐2 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑣𝑣(𝑡𝑡)
For any 𝑐𝑐1 , 𝑐𝑐2 ∈ ℝ and 𝛼𝛼 ∈ ℝ+ .
1.3.2 Properties of R-L Fractional Operators: [39, 42]
The three important properties of 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 and follows:
𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
operators can be pointed as
1. An interesting property of the R-L fractional operator is that, in general, the R-L fractional derivative of a constant is not equal to zero, namely: 𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝒞𝒞
(𝑡𝑡−𝑎𝑎 )−𝛼𝛼
,
𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡 )
=
𝑅𝑅 𝛼𝛼 ( ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡
=
= 𝒞𝒞
Γ(1−𝛼𝛼)
for any constant 𝒞𝒞 and all 𝛼𝛼 ≥ 0 , (𝛼𝛼 ∉ ℕ), and it is
zero when 𝛼𝛼 is integer order. 2. Let 𝛼𝛼 ≥ 0 , and for 𝑢𝑢(𝑡𝑡) = (𝑡𝑡 − 𝑎𝑎)𝛽𝛽 for some 𝛽𝛽 > −1 . Then Г(𝛽𝛽 + 1) (𝑡𝑡 − 𝑎𝑎)𝛽𝛽 +𝛼𝛼 Г(𝛽𝛽 + 𝛼𝛼 + 1)
Г(𝛽𝛽 + 1) (𝑡𝑡 − 𝑎𝑎)𝛽𝛽 −𝛼𝛼 Г(𝛽𝛽 − 𝛼𝛼 + 1)
… (1.4) … (1.5)
3. The value [ 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡)]𝑡𝑡=𝑎𝑎 is also used occasionally. The existence of the one sided limit, lim 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡) depending on behavior from 𝑢𝑢 in the near point 𝑡𝑡→𝑎𝑎
𝑡𝑡 = 𝑎𝑎, if 𝑢𝑢 is bounded and integrable, the value is zero, and it is not zero if 𝑢𝑢 is not continuous. In fact, we can see function 𝑢𝑢(𝑡𝑡) = (𝑡𝑡 − 𝑎𝑎)𝛽𝛽 −1 , 𝑡𝑡 > 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 > 0, then: lim 𝑡𝑡→𝑎𝑎
𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡)
0 = � Г(𝛽𝛽) ∞
5
𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖
𝛼𝛼 + 𝛽𝛽 > 1 𝛼𝛼 + 𝛽𝛽 = 1 𝛼𝛼 + 𝛽𝛽 < 1
Chapter One
Basic Concepts
1.3.3 Composition of Riemann-Liouville Operators: We need some composition relations of R-L fractional integral and fractional derivatives for the solution of fractional differential and integral equations; it is described in the following lemmas: Lemma (1.1): (Composition of fractional integration) [42, 69] The operator’s 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 commute (is known as the semi-group property with identity operator 𝑎𝑎 𝐽𝐽𝑡𝑡0 ). This algebraically formulated result implies directly 𝛽𝛽 𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝛼𝛼+𝛽𝛽
= 𝑎𝑎 𝐽𝐽𝑡𝑡
𝛽𝛽 +𝛼𝛼
𝑢𝑢(𝑡𝑡) = 𝑎𝑎 𝐽𝐽𝑡𝑡
𝛽𝛽
𝑢𝑢(𝑡𝑡) = 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡)
… (1.6)
If 𝑢𝑢 ∈ 𝐶𝐶𝛾𝛾 [𝑎𝑎, 𝑏𝑏], 𝛾𝛾 ≥ −1 and 𝛼𝛼, 𝛽𝛽 ∈ ℝ+which is a well known result in the integer case. As a special case, from equation (1.6) we conclude that: 𝛼𝛼 ( 𝑎𝑎 𝐽𝐽� … � 𝐽𝐽𝑡𝑡𝛼𝛼 ) 𝑢𝑢(𝑡𝑡 ) = 𝑎𝑎 𝐽𝐽𝑡𝑡𝑘𝑘𝑘𝑘 𝑢𝑢(𝑡𝑡 ) , 𝑢𝑢 ∈ 𝐶𝐶𝛾𝛾 [𝑎𝑎, 𝑏𝑏], 𝛾𝛾 ≥ −1 , 𝛼𝛼 ≥ 0 , 𝑘𝑘 ∈ ℕ �� 𝑎𝑎�� 𝑡𝑡��� 𝑘𝑘−𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
Lemma (1.2): (Composition of R-L fractional differentiation) [27, 46]
Let 𝛼𝛼, 𝛽𝛽 > 0 then the composition of two R-L fractional derivatives operator
𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 , (𝑛𝑛
− 1 < 𝛼𝛼 ≤ 𝑛𝑛) and
𝑅𝑅 𝛽𝛽 (𝑚𝑚 𝑎𝑎 𝐷𝐷𝑡𝑡 ,
𝛼 + 𝛽𝛽 < 𝑛𝑛 we can state it as follows: 𝑅𝑅 𝛼𝛼 𝑅𝑅 𝛽𝛽 𝑎𝑎 𝐷𝐷𝑡𝑡 � 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)�
=
𝑅𝑅 𝛼𝛼+𝛽𝛽 𝑢𝑢(𝑡𝑡) 𝑎𝑎 𝐷𝐷𝑡𝑡
Remark (1.1): [46]
−
− 1 < 𝛽𝛽 ≤ 𝑚𝑚); (𝑛𝑛, 𝑚𝑚 ∈ ℕ) be such that
𝑚𝑚
𝛽𝛽 −𝑗𝑗 � � 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)� 𝑡𝑡=𝑎𝑎 𝑗𝑗 =1
(𝑡𝑡 − 𝑎𝑎)−𝑗𝑗 −𝛼𝛼 Г(1 − 𝑗𝑗 − 𝛼𝛼)
Generally, the R-L fractional derivative operators commute, with only one exception: For 𝛼𝛼 ≠ 𝛽𝛽 we have 𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
𝛽𝛽
𝛽𝛽
𝛼𝛼+𝛽𝛽
� 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡 )� = 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡 � 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡 )� = 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡
If and only if satisfies the following two conditions 𝛽𝛽 −𝑗𝑗
[ 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡
𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
𝑢𝑢(𝑡𝑡 )]𝑡𝑡=𝑎𝑎 = 0,
and
𝑢𝑢(𝑡𝑡 )
𝑗𝑗 = 1,2, … , 𝑚𝑚 ; 𝑚𝑚 = ⌈𝛽𝛽⌉
And the condition [ 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼−𝑗𝑗 𝑢𝑢(𝑡𝑡)]𝑡𝑡=𝑎𝑎 = 0 ,
𝑗𝑗 = 1,2, … , 𝑛𝑛 ; 𝑛𝑛 = ⌈𝛼𝛼 ⌉
6
… (1.7) 𝑅𝑅 𝛽𝛽 𝑎𝑎 𝐷𝐷𝑡𝑡
do not
Chapter One
Basic Concepts
Lemma (1.3): (Mixed fractional integration and differentiation) [27, 66] i.
ii.
Let 𝛼𝛼 ≥ 𝛽𝛽 ≥ 0 , then for 𝑢𝑢(𝑡𝑡) ∈ 𝐶𝐶[𝑎𝑎, 𝑏𝑏] , the relation is valid at every point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏]: 𝛼𝛼−𝛽𝛽
𝛼𝛼 𝑅𝑅 𝛽𝛽 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡)
= 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡) … (1.8) In particular, where 𝛽𝛽 = 𝑘𝑘 ∈ ℕ and 𝛼𝛼 > 𝑘𝑘, then 𝐷𝐷𝑡𝑡𝑘𝑘 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡) = 𝑎𝑎 𝐽𝐽𝑡𝑡𝛼𝛼−𝑘𝑘 𝑢𝑢(𝑡𝑡) 𝛽𝛽
Let 𝛼𝛼 ≥ 𝛽𝛽 ≥ 0 , if the fractional derivative 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡 , (𝑚𝑚 − 1 < 𝛽𝛽 ≤ 𝑚𝑚), of a 𝑚𝑚 −𝛽𝛽
function 𝑢𝑢(𝑡𝑡) is integrable,(or, if 𝑢𝑢(𝑡𝑡) ∈ 𝐶𝐶[𝑎𝑎, 𝑏𝑏] and 𝑎𝑎 𝐽𝐽𝑡𝑡 then: 𝛼𝛼 𝑅𝑅 𝛽𝛽 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
=
𝛼𝛼−𝛽𝛽 𝑢𝑢(𝑡𝑡) − 𝑎𝑎 𝐽𝐽𝑡𝑡
𝑚𝑚
𝛽𝛽 −𝑗𝑗 � � 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)� 𝑗𝑗 =1
𝑢𝑢(𝑡𝑡) ∈ 𝐶𝐶 𝑚𝑚 [𝑎𝑎, 𝑏𝑏]),
(𝑡𝑡 − 𝑎𝑎)𝛼𝛼−𝑗𝑗 … (1.9) 𝑡𝑡=𝑎𝑎 Г(1 + 𝛼𝛼 − 𝑗𝑗)
Lemma (1.4) :( Mixed fractional and integer-order differentiation) [46, 66] Let 𝛼𝛼 ≥ 0 , 𝑛𝑛 ∈ ℕ and 𝑢𝑢 ∈ 𝐶𝐶 𝑛𝑛+𝑚𝑚 [𝑎𝑎, 𝑏𝑏] ; 𝑚𝑚 = ⌈𝛼𝛼⌉, then 𝑑𝑑 𝑛𝑛 𝑅𝑅 𝛼𝛼 𝐢𝐢. � 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)� = 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝑛𝑛 +𝛼𝛼 𝑢𝑢(𝑡𝑡) 𝑛𝑛 𝑑𝑑𝑑𝑑 𝑛𝑛 −1 (𝑗𝑗 ) 𝑛𝑛 𝑢𝑢(𝑡𝑡) 𝑑𝑑 𝑢𝑢 (𝑎𝑎)(𝑡𝑡 − 𝑎𝑎)𝑗𝑗 −𝛼𝛼−𝑛𝑛 𝑅𝑅 𝛼𝛼 𝑅𝑅 𝛼𝛼+𝑛𝑛 𝐢𝐢𝐢𝐢. 𝑎𝑎 𝐷𝐷𝑡𝑡 � � = 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) − � Г(1 + 𝑗𝑗 − 𝛼𝛼 − 𝑛𝑛) 𝑑𝑑𝑑𝑑 𝑛𝑛 𝑗𝑗 =0
… (1.10)
… (1.11)
Remark (1.2): We see that, from Lemma (1.4), the R-L fractional derivative operator commutes with
𝑑𝑑 𝑛𝑛
𝑑𝑑𝑑𝑑 𝑛𝑛
, i.e. 𝑑𝑑 𝑛𝑛 𝑢𝑢(𝑡𝑡 ) 𝑑𝑑 𝑛𝑛 𝑅𝑅 𝛼𝛼 𝛼𝛼 𝑅𝑅 � 𝐷𝐷 𝑢𝑢(𝑡𝑡 )� = 𝑎𝑎 𝐷𝐷𝑡𝑡 � � = 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼+𝑛𝑛 𝑢𝑢(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝑛𝑛 𝑑𝑑𝑑𝑑 𝑛𝑛 𝑎𝑎 𝑡𝑡
𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
If and only if at the lower terminal 𝑡𝑡 = 𝑎𝑎 of the function 𝑢𝑢(𝑡𝑡) satisfies with the conditions 𝑢𝑢(𝑗𝑗 ) (𝑎𝑎) = 0 , (𝑗𝑗 = 0,1, … , 𝑛𝑛 − 1). Lemma (1.5): [61] Let 𝑓𝑓 be continuous function on [𝑎𝑎, 𝑏𝑏] × [𝑎𝑎, 𝑏𝑏]. Then for 𝛼𝛼 ≥ 0
Lemma (1.6): [56]
𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡
𝑡𝑡
𝑡𝑡
� 𝑓𝑓(𝑡𝑡, 𝑠𝑠)𝑑𝑑𝑑𝑑 = � 𝑠𝑠 𝐽𝐽𝑡𝑡𝛼𝛼 𝑓𝑓(𝑡𝑡, 𝑠𝑠)𝑑𝑑𝑑𝑑 , 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏]
𝑎𝑎
𝑎𝑎
Let 𝛼𝛼 ∈ ℝ+, and 𝑚𝑚 − 1 < 𝛼𝛼 ≤ 𝑚𝑚, 𝑚𝑚 ∈ ℕ, for 𝒞𝒞𝑗𝑗 ∈ ℝ, then: 𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑚𝑚
… (1.12)
= 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑣𝑣(𝑡𝑡) if and only if 𝑢𝑢(𝑡𝑡) = 𝑣𝑣(𝑡𝑡) + � 𝒞𝒞𝑗𝑗 (𝑡𝑡 − 𝑎𝑎)𝛼𝛼−𝑗𝑗 … (1.13) 7
𝑗𝑗 =1
Chapter One
Basic Concepts
1.3.4 Caputo Fractional Derivative: [33] The Caputo fractional derivative was introduced to alleviate some of the difficulties associated with the R-L approach to fractional differential equations when applied to the solution of physical problems. M. Caputo in 1967 introduced an alternative definition, today called Caputo fractional derivative, which is modification of the Riemman-Liouville definition and has advantage of dealing with property with initial value problems in which the initial conditions are given. Here we present the definition and some properties of new Caputo derivative operator. Definition (1.4): [27, 42] 𝑚𝑚 [𝑎𝑎, 𝑏𝑏], 𝑚𝑚 ∈ ℕ is The Caputo fractional derivative of order 𝛼𝛼 ≥ 0 of 𝑢𝑢(𝑡𝑡) ∈ 𝐶𝐶−1 defined as: 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
=
𝑚𝑚 −𝛼𝛼 (𝑚𝑚 ) (𝑡𝑡) 𝑢𝑢 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑚𝑚 �
𝑑𝑑 𝑢𝑢(𝑡𝑡) 𝑑𝑑𝑡𝑡 𝑚𝑚
,
,
𝑚𝑚 − 1 < 𝛼𝛼 < 𝑚𝑚 𝛼𝛼 = 𝑚𝑚
… (1.14)
Remark (1.3): [56] We note that the definition of the Caputo differential operator also utilizes the RL-integral operator, but compared with the RL-derivative the sequence, in which integer order differentiation and fractional integration are applied, is interchanged. i.e. 𝑚𝑚 𝑚𝑚 −𝛼𝛼 𝑅𝑅 𝛼𝛼 𝑢𝑢(𝑡𝑡) ≠ 𝑎𝑎 𝐽𝐽𝑡𝑡𝑚𝑚 −𝛼𝛼 𝐷𝐷𝑡𝑡𝑚𝑚 𝑢𝑢(𝑡𝑡) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝐷𝐷𝑡𝑡 𝑎𝑎 𝐽𝐽𝑡𝑡 1.3.5 Properties of Caputo Fractional Operator: [4, 33] Three important properties of Caputo fractional operators can be pointed as follows: 1. The Caputo differential operator is a linear operator, i.e. 𝑐𝑐 𝛼𝛼 𝑐𝑐 𝛼𝛼 𝑐𝑐 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 [𝑐𝑐1 𝑢𝑢(𝑡𝑡) + 𝑐𝑐2 𝑣𝑣(𝑡𝑡)] = 𝑐𝑐1 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) + 𝑐𝑐2 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑣𝑣(𝑡𝑡) For any 𝑐𝑐1 , 𝑐𝑐2 ∈ ℝ and 𝛼𝛼 ∈ ℝ+ .
2. The 𝛼𝛼-Caputo fractional derivative of a constant function is equal to zero; i.e. 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝒞𝒞 = 0 , for any constant 𝒞𝒞 and all 𝛼𝛼 > 0 . 3. Let 𝛼𝛼 > 0 ; 𝑚𝑚 = ⌈𝛼𝛼⌉ and for 𝑢𝑢(𝑡𝑡) = (𝑡𝑡 − 𝑎𝑎)𝛽𝛽 for some 𝛽𝛽 ≥ 0. Then: 0 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = � Γ(𝛽𝛽 + 1) (𝑡𝑡 − 𝑎𝑎)𝛽𝛽 −𝛼𝛼 Γ(𝛽𝛽 + 1 − 𝛼𝛼) 8
𝑖𝑖𝑖𝑖 𝛽𝛽 ∈ {0,1,2, ⋯ , 𝑚𝑚 − 1} … (1.15) 𝑖𝑖𝑖𝑖 𝛽𝛽 ∈ ℕ 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 ≥ 𝑚𝑚 𝑜𝑜𝑜𝑜 𝛽𝛽 ∉ ℕ 𝑎𝑎𝑎𝑎𝑎𝑎 𝛽𝛽 > 𝑚𝑚 − 1
Chapter One
Basic Concepts
Lemma (1.7): (Relations between R-L and Caputo Operators) [33, 42] Let 𝛼𝛼 ≥ 0 and 𝑚𝑚 = ⌈𝛼𝛼⌉. Moreover, assume that 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑢𝑢 exists and 𝑢𝑢 possesses (𝑚𝑚 − 1) derivative at 𝑎𝑎 then: 𝐶𝐶 𝛼𝛼 𝑅𝑅 𝛼𝛼 (𝑚𝑚 − 1 < 𝛼𝛼 ≤ 𝑚𝑚) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝑎𝑎 𝐷𝐷𝑡𝑡 �𝑢𝑢 (𝑡𝑡) − 𝑇𝑇𝑚𝑚 −1 [𝑢𝑢; 𝑎𝑎]�, Where 𝑇𝑇𝑚𝑚 −1 [𝑢𝑢; 𝑎𝑎] denotes the Taylor polynomial of degree 𝑚𝑚 − 1 for the function 𝑢𝑢, centered at 𝑎𝑎. Also, we can express this as follows: 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) (𝑘𝑘)
=
𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
𝑛𝑛−1
𝑢𝑢(𝑘𝑘) (𝑎𝑎) 𝑢𝑢(𝑡𝑡) − � (𝑡𝑡 − 𝑎𝑎)𝑘𝑘−𝛼𝛼 , 𝑡𝑡 > 𝑎𝑎 Γ(𝑘𝑘 − 𝛼𝛼 + 1) 𝑘𝑘=0
… (1.16)
Moreover, if 𝑢𝑢 (𝑎𝑎) = 0 for 𝑘𝑘 = 0,1, … , 𝑛𝑛 − 1 ; i.e. we assume 𝑢𝑢 to have an 𝑛𝑛fold zero at 𝑎𝑎 . Then: 𝐶𝐶 𝛼𝛼 𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) The following two lemmas express that the Caputo derivative is left inverse of the R-L integral while it has no right inverse. Lemma (1.8): (Caputo derivative is left inverse of the RL-integral) [33, 42] i. ii.
If 𝑢𝑢 is continuous and 𝛼𝛼 ≥ 0 with 𝑛𝑛 = ⌈𝛼𝛼⌉ , then 𝐶𝐶 𝛼𝛼 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝑢𝑢(𝑡𝑡), 𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏 Assume that 𝛼𝛼 ≥ 0 , 𝑚𝑚 = ⌈𝛼𝛼⌉ , 𝑎𝑎𝑎𝑎𝑎𝑎 𝑢𝑢 ∈ 𝐶𝐶 𝑚𝑚 [𝑎𝑎, 𝑏𝑏]. Then 𝛼𝛼 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑚𝑚 −1
𝑢𝑢(𝑘𝑘) (𝑎𝑎) (𝑡𝑡 − 𝑎𝑎)𝑘𝑘 = 𝑢𝑢(𝑡𝑡) − � 𝑘𝑘! 𝑘𝑘=0
… (1.17) … (1.18)
We note that the classic 𝑛𝑛-fold integral (𝑛𝑛 ∈ ℕ) and differential operators of 𝑛𝑛-order satisfy like formula: 𝐷𝐷𝑡𝑡𝑛𝑛 𝑎𝑎 𝐽𝐽𝑡𝑡𝑛𝑛 𝑢𝑢(𝑡𝑡)
= 𝑢𝑢(𝑡𝑡) ;
Lemma (1.9): [61]
𝑛𝑛 𝑛𝑛 𝑎𝑎 𝐽𝐽𝑡𝑡 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑛𝑛 −1
𝑢𝑢(𝑘𝑘) (𝑎𝑎) (𝑡𝑡 − 𝑎𝑎)𝑘𝑘 = 𝑢𝑢(𝑡𝑡) − � 𝑘𝑘! 𝑘𝑘=0
… (1.19)
Let 𝛼𝛼 > 𝛽𝛽 ≥ 0, 𝑚𝑚𝛼𝛼 − 1 < 𝛼𝛼 ≤ 𝑚𝑚𝛼𝛼 and 𝑚𝑚𝛽𝛽 − 1 < 𝛽𝛽 ≤ 𝑚𝑚𝛽𝛽 (𝑚𝑚𝛼𝛼 , 𝑚𝑚𝛽𝛽 ∈ ℕ) be such that 𝑢𝑢(𝑡𝑡) ∈ 𝐶𝐶 𝑚𝑚 𝛽𝛽 [𝑎𝑎, 𝑏𝑏]. Then 𝛼𝛼 𝑎𝑎 𝐽𝐽𝑡𝑡
𝛽𝛽 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)�
=
𝛼𝛼−𝛽𝛽 𝑢𝑢(𝑡𝑡) 𝑎𝑎 𝐽𝐽𝑡𝑡
Lemma (1.10): [33, 42]
𝑚𝑚 𝛽𝛽 −1
𝑢𝑢(𝑘𝑘) (𝑎𝑎) (𝑡𝑡 − 𝑎𝑎)𝑘𝑘+𝛼𝛼−𝛽𝛽 … (1.20) − � Γ(𝑘𝑘 + 𝛼𝛼 − 𝛽𝛽 + 1) 𝑘𝑘=0
Let 𝛼𝛼 ≥ 0, (𝛼𝛼 ∉ ℕ) and 𝑚𝑚 = ⌈𝛼𝛼⌉. Moreover, assume that 𝑢𝑢 ∈ 𝐶𝐶 𝑚𝑚 [𝑎𝑎, 𝑏𝑏]. Then 𝐶𝐶 𝛼𝛼 the Caputo fractional derivative 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) is continuous on [𝑎𝑎, 𝑏𝑏] and 𝐶𝐶 𝛼𝛼 lim𝑡𝑡⟶𝑎𝑎 [ 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)] = 0. 9
Chapter One
Basic Concepts
Lemma (1.11): [27, 46] Let 𝛼𝛼 ≥ 0 and 𝑛𝑛 ∈ ℕ. Assume that 𝑢𝑢 is such that both 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼+𝑛𝑛 𝑢𝑢 and 𝐶𝐶 𝛼𝛼 (𝐷𝐷 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑡𝑡 𝑢𝑢) exist.Then: 𝑛𝑛 𝐶𝐶 𝛼𝛼 𝐶𝐶 𝛼𝛼+𝑛𝑛 𝑢𝑢(𝑡𝑡), 𝑚𝑚 − 1 < 𝛼𝛼 ≤ 𝑚𝑚 … (1.21) 𝑎𝑎 𝐷𝐷𝑡𝑡 �𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)� = 𝑎𝑎 𝐷𝐷𝑡𝑡
Moreover, the Caputo differential operator with differential of integer-order is commute, i.e., whenever, 𝑢𝑢(𝑘𝑘) (𝑎𝑎) = 0, for all 𝑘𝑘 = 𝑚𝑚 , 𝑚𝑚 + 1, … , 𝑛𝑛 ∶ 𝐶𝐶 𝛼𝛼 ( 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡))
Lemma (1.12): [22, 56]
= 𝐷𝐷𝑡𝑡𝑛𝑛 ( 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡)) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼+𝑛𝑛 𝑢𝑢(𝑡𝑡)
Let 𝛼𝛼 ∈ ℝ+, and 𝑚𝑚 − 1 < 𝛼𝛼 ≤ 𝑚𝑚 , 𝑚𝑚 ∈ ℕ, then, for 𝒞𝒞𝑗𝑗 ∈ ℝ :
𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑚𝑚
= 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑣𝑣(𝑡𝑡) if and only if 𝑢𝑢(𝑡𝑡) = 𝑣𝑣(𝑡𝑡) + � 𝒞𝒞𝑗𝑗 (𝑡𝑡 − 𝑎𝑎)𝑚𝑚 −𝑗𝑗 𝑗𝑗 =1
Lemma (1.13): [61]
… (1.22)
If 𝛼𝛼 > 0 , 𝛼𝛼 ∉ ℕ , 𝑚𝑚 = ⌈𝛼𝛼⌉ and 𝑛𝑛 ∈ ℤ+ with 𝑛𝑛 ≥ 𝑚𝑚 and for any arbitrary 𝑡𝑡0 ≥ 𝑎𝑎 ≥ 0. Then, for all 𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏 𝐶𝐶 𝛼𝛼 (𝑡𝑡 𝑎𝑎 𝐷𝐷𝑡𝑡
− 𝑡𝑡0
)𝑛𝑛
𝑛𝑛−𝑚𝑚
(−1)𝑘𝑘 𝑡𝑡0 − 𝑎𝑎 𝑘𝑘 = 𝑛𝑛! � � � � (𝑡𝑡 − 𝑎𝑎)𝑛𝑛−𝛼𝛼 � 𝑘𝑘! Г(𝑛𝑛 − 𝑘𝑘 − 𝛼𝛼 + 1) 𝑡𝑡 − 𝑎𝑎 𝑘𝑘=0
… (1.23)
Lemma (1.14): [66] Let 𝜑𝜑 be continuously differentiable function on [𝑎𝑎, 𝑏𝑏] × [𝑎𝑎, 𝑏𝑏]. Then for 0 < 𝛼𝛼 < 1, and 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] we have: 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
𝑡𝑡
𝑡𝑡
� 𝜑𝜑(𝑡𝑡, 𝑠𝑠)𝑑𝑑𝑑𝑑 = � 𝐶𝐶𝑠𝑠 𝐷𝐷𝑡𝑡𝛼𝛼 𝜑𝜑(𝑡𝑡, 𝑠𝑠)𝑑𝑑𝑑𝑑 + 𝑎𝑎 𝐽𝐽𝑡𝑡1−𝛼𝛼 � lim 𝜑𝜑(𝑡𝑡, 𝑠𝑠))�
𝑎𝑎
Remark (1.4): [46, 62]
𝑠𝑠→𝑡𝑡−𝑎𝑎
𝑎𝑎
… (1.24)
The most relevant differences between R-L and Caputo fractional derivatives in equation (1.3) and (1.14) may be summarized in the following points: 1. In general, the R-L fractional derivative operator of a constant is not equal to zero, whereas the Caputo fractional derivative of a constant is equal to zero, i.e. 𝑅𝑅 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝒞𝒞
= 𝒞𝒞
(𝑡𝑡−𝑎𝑎)−𝛼𝛼 Γ(1−𝛼𝛼)
;
𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝒞𝒞
= 0, for any constant 𝒞𝒞 and all 𝛼𝛼 > 0 . 10
Chapter One
Basic Concepts
2. Application problems require definitions of fractional derivatives allowing utilization of physically interpretable initial conditions, which contain 𝑢𝑢(𝑎𝑎), 𝑢𝑢′ (𝑎𝑎), 𝑒𝑒𝑒𝑒𝑒𝑒. the main advantage of Caputo's approach is that the initial conditions for FDE's with Caputo derivatives take on the same form as for integer-order differential equations, i.e. it contains the limit values of integerorder derivatives of unknown functions at the lower terminal 𝑡𝑡 = 𝑎𝑎 . Unfortunately, the R-L approach leads to initial conditions containing the limit values of the R-L fractional derivatives at the lower terminal 𝑡𝑡 = 𝑎𝑎, for example: 𝛼𝛼−𝑚𝑚 𝛼𝛼−1 𝛼𝛼−2 lim 𝑅𝑅𝑎𝑎𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝜇𝜇1 , lim 𝑅𝑅𝑎𝑎𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝜇𝜇2 , … lim 𝑅𝑅𝑎𝑎𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) = 𝜇𝜇𝑚𝑚 𝑡𝑡→𝑎𝑎
𝑡𝑡→𝑎𝑎
𝑡𝑡→𝑎𝑎
… (1.25)
where 𝜇𝜇ℓ , ℓ = 1,2, … , 𝑚𝑚, are given constants. 3. Laplace transform of the R-L fractional derivative for solving applied problems allows utilization of initial condition of the type (1.23), which causes polynomial with their physical interpolation .On the contrary, the Laplace transform of the Caputo derivative allows utilization of initial values of classical integer order derivatives with known physical interpolations. The above mentioned reasons clarify why Caputo fractional derivative is adopted in these works which are instead of R-L derivative.
1.4 Classification of Integral Equation: [2, 3, 17, 40, 47, 52, 60, 61] The functional equation in which the unknown function appears under one or more integral sign is called integral equations. Whenever the unknown function appears in the equation in a nonlinear manner then it is called nonlinear integral equation, and can be present in the form. 𝑣𝑣(𝑡𝑡)
𝑄𝑄(𝑡𝑡 )𝑢𝑢(𝑡𝑡 ) = 𝑓𝑓 (𝑡𝑡 ) + 𝜆𝜆 � 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 ; 𝑡𝑡 ∈ 𝐼𝐼 = [𝑎𝑎, 𝑏𝑏] 𝑎𝑎
… (1.26)
Where the known functions 𝑄𝑄, 𝑓𝑓: 𝐼𝐼 → ℝ and 𝐾𝐾: 𝒮𝒮 × ℝ → ℝ; with 𝒮𝒮 = {(𝑡𝑡, 𝑥𝑥): 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑡𝑡 ≤ 𝑏𝑏}, and 𝑢𝑢(𝑡𝑡) is the unknown function and 𝜆𝜆 is a scalar parameter. The integral equation (1.26) is said to be linear integral equation if 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)� = 𝒦𝒦(𝑡𝑡, 𝑥𝑥)𝑢𝑢(𝑥𝑥).
11
Chapter One
Basic Concepts
Definition (1.5): The integral equation (1.26) is said to be:
First kind Integral equation; if 𝑄𝑄(𝑡𝑡) = 0, ∀ 𝑡𝑡 ∈ 𝐼𝐼, then the equation (1.26) becomes:
i.
𝑣𝑣(𝑡𝑡)
𝑓𝑓 (𝑡𝑡 ) = 𝜆𝜆∗ � 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 , 𝑎𝑎
𝜆𝜆∗ = −𝜆𝜆
Second kind Integral equation; if 𝑄𝑄(𝑡𝑡) ≠ 0, ∀ 𝑡𝑡 ∈ 𝐼𝐼, then the equation (1.26) becomes:
ii.
𝑣𝑣(𝑡𝑡)
iii.
𝑢𝑢(𝑡𝑡 ) = 𝑔𝑔(𝑡𝑡 ) + 𝜆𝜆 � 𝒦𝒦�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 , 𝑎𝑎
𝑔𝑔 = 𝑓𝑓 ⁄𝑄𝑄 ; 𝒦𝒦 = 𝐾𝐾 ⁄𝑄𝑄
Third Kind Integral equation; if 𝑄𝑄(𝑡𝑡) is a continuous function possessing a finite number of zeros in the interval 𝐼𝐼.
Definition (1.6):
The integral equation (1.26) is said to be Homogenous integral equation if 𝑓𝑓(𝑡𝑡) = 0, otherwise it is called Non-Homogeneous integral equation. Definition (1.7): The integral equation (1.26) is said to be: i.
Volterra Integral Equation’s; if 𝑣𝑣(𝑡𝑡) = 𝑡𝑡, where 𝑎𝑎 < 𝑡𝑡 < 𝑏𝑏. Hence the integral equations 𝑡𝑡
𝑡𝑡
𝑓𝑓 (𝑡𝑡 ) = 𝜆𝜆 � 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 ; 𝑢𝑢(𝑡𝑡 ) = 𝑓𝑓 (𝑡𝑡 ) + 𝜆𝜆 � 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 𝑎𝑎
𝑎𝑎
Represent VIE’s of the First and Second Kind respectively. ii.
Fredholm Integral Equation’s; if 𝑣𝑣(𝑡𝑡) = 𝑏𝑏, where 𝑏𝑏 is a constant and 𝑏𝑏 > 𝑎𝑎. Hence the integral equations 𝑏𝑏
𝑏𝑏
𝑓𝑓(𝑡𝑡 ) = 𝜆𝜆 � 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 ; 𝑢𝑢(𝑡𝑡 ) = 𝑓𝑓 (𝑡𝑡 ) + 𝜆𝜆 � 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)�𝑑𝑑𝑑𝑑 𝑎𝑎
𝑎𝑎
Represent FIE’s of the First and Second kind respectively.
Definition (1.8): The integral equation (1.26) is called Convolution type of integral equation if the kernel depends only on the difference (𝑡𝑡 − 𝑥𝑥), i.e. 𝐾𝐾�𝑡𝑡, 𝑥𝑥, 𝑢𝑢(𝑥𝑥)� = 𝒦𝒦(𝑡𝑡 − 𝑥𝑥, 𝑢𝑢(𝑥𝑥)); such a kernel is called difference kernel. 12
Chapter One
Basic Concepts
Definition (1.9): An Integro-Differential Equation (IDE) is an equation involving one (or more) unknown functions 𝑢𝑢(𝑡𝑡), together with both differential and integral operations on 𝑢𝑢. Definition (1.10): [61, 47] The Linear Fredholm Integro-ordinary Differential Equation (FIDE) of order (n ∈ ℕ), can be represented as: 𝑏𝑏
𝑛𝑛 −1
�𝐷𝐷𝑛𝑛 + � 𝒫𝒫𝑖𝑖 (𝑡𝑡)𝐷𝐷𝑖𝑖 � 𝑢𝑢(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � 𝒦𝒦(𝑡𝑡, 𝑥𝑥)𝑢𝑢(𝑥𝑥)𝑑𝑑𝑑𝑑 ; 𝑡𝑡 ∈ 𝐼𝐼 = [𝑎𝑎, 𝑏𝑏] … (1.27) 𝑖𝑖=0
𝑎𝑎
With the two-point boundary conditions: 𝑛𝑛 −1
��𝑟𝑟𝑗𝑗 ,𝑖𝑖 𝑢𝑢(𝑗𝑗 ) (𝑎𝑎) + 𝑟𝑟𝑗𝑗 ,𝑛𝑛+𝑖𝑖 𝑢𝑢(𝑗𝑗 ) (𝑏𝑏)� = 𝑐𝑐𝑗𝑗 𝑖𝑖=0
, 𝑗𝑗 = 0,1,2, … , 𝑛𝑛 − 1
Where 𝑓𝑓 and 𝒫𝒫i (i = 0,1, … , n − 1) are assumed to be continuous and bounded realvalued functions on 𝐼𝐼 and 𝒦𝒦 : 𝑆𝑆 × ℝ → ℝ , (𝑆𝑆 ≔ {(𝑡𝑡, 𝑥𝑥): 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑡𝑡 ≤ 𝑏𝑏}) denote a continuity function, while 𝑢𝑢(𝑥𝑥) is to be determined and λ is a scalar parameter. Definition (1.11): [61, 47] The Linear Volterra Integro- ordinary Differential Equation (VIDE) of higher order 𝑛𝑛(𝑛𝑛 ∈ 𝑁𝑁)if the only integrals of Volterra type appear (i.e. if 𝒦𝒦 (𝑡𝑡, 𝑥𝑥) = 0 𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡 < 𝑥𝑥) and defines initial conditions (𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟𝑗𝑗 ,𝑖𝑖 = 0; 𝑓𝑓𝑓𝑓𝑓𝑓 𝑖𝑖 ≥ 𝑛𝑛) , which has the following form : 𝑡𝑡
𝑛𝑛 −1
�𝐷𝐷𝑛𝑛 + � 𝒫𝒫𝑖𝑖 (𝑡𝑡)𝐷𝐷𝑖𝑖 � 𝑢𝑢(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � 𝒦𝒦(𝑡𝑡, 𝑥𝑥)𝑢𝑢(𝑥𝑥)𝑑𝑑𝑑𝑑 ; 𝑡𝑡 ∈ 𝐼𝐼 = [𝑎𝑎, 𝑏𝑏] … (1.28) 𝑖𝑖=0
With the initial conditions: 𝑛𝑛−1
𝑎𝑎
� 𝑟𝑟𝑗𝑗 ,𝑖𝑖 𝐷𝐷 𝑖𝑖 𝑢𝑢(𝑎𝑎) = 𝜂𝜂𝑗𝑗 𝑖𝑖=0
, 𝑗𝑗 = 0,1,2, … , 𝑛𝑛 − 1
where 𝒦𝒦, f and 𝒫𝒫i (i = 0,1, … , n − 1) are continuous functions, λ is a scalar parameter and 𝑢𝑢 is the unknown function. 13
Chapter One
Basic Concepts
1.5 Delay Differential Equations: [18, 53, 63] Delay Differential Equations (DDEs) are Differential Equations in which the derivatives of some unknown functions at present time are dependent on the values of the functions at the previous times. Mathematically, a general (DDE) of the 𝑚𝑚-th order for unknown smoothness functions 𝑢𝑢(𝑡𝑡) with 𝑛𝑛-th multiple time delays: 𝐹𝐹�𝑡𝑡, 𝑢𝑢(𝑡𝑡), 𝑢𝑢′ (𝑡𝑡), … , 𝑢𝑢(𝑚𝑚 ) (𝑡𝑡), 𝑢𝑢(𝑡𝑡 − 𝜏𝜏1 ), 𝑢𝑢′ (𝑡𝑡 − 𝜏𝜏1 ), … , 𝑢𝑢(𝑚𝑚 ) (𝑡𝑡 − 𝜏𝜏1 ), …,
𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑛𝑛 ), 𝑢𝑢′ (𝑡𝑡 − 𝜏𝜏𝑛𝑛 ), … , 𝑢𝑢(𝑚𝑚 ) (𝑡𝑡 − 𝜏𝜏𝑛𝑛 )� = 0
… (1.29)
Where the quantities 𝜏𝜏𝑖𝑖 ≥ 0 , 𝑖𝑖 = 1,2, … , 𝑛𝑛 are called the delays or time lags and 𝐹𝐹: [𝑎𝑎, 𝑏𝑏] × ℝ𝑤𝑤 ⟶ ℝ is nonlinear smoothness function, such that 𝑤𝑤 = (𝑚𝑚 + 1)(𝑛𝑛 + 1) .
A fundamental technique for solving DDEs (1.29) depending on the kind of delays, we distinguish: Constant delay: all time-delays are constants (𝜏𝜏𝑖𝑖 = 𝐶𝐶𝑖𝑖 ∈ ℝ+) for all 𝑖𝑖 = 1,2, … , 𝑛𝑛.
Time-dependent delay: the time-delay is a function of the time 𝑡𝑡 such that 𝜏𝜏𝑖𝑖 = 𝜏𝜏𝑖𝑖 (𝑡𝑡) , for some 𝑖𝑖. State-dependent delay: the time-delay has the form 𝜏𝜏𝑖𝑖 = 𝜏𝜏𝑖𝑖 �𝑡𝑡, 𝑢𝑢(𝑡𝑡)� for some 𝑖𝑖 .
Neutral delay: the time delays has the form, for some 𝑖𝑖 .
𝜏𝜏𝑖𝑖 = 𝜏𝜏𝑖𝑖 �𝑡𝑡, 𝑢𝑢(𝑡𝑡), 𝑢𝑢′ (𝑡𝑡), … , 𝑢𝑢(𝑚𝑚 ) (𝑡𝑡)�
The most obvious difference between ODEs and DDEs is the initial data. The solution of an ODE is determined by its value at the initial point 𝑡𝑡 = 𝑎𝑎. In
evaluating the DDEs (1.29) for 𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏, a term like 𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑖𝑖 ) may represent
values of the solution at points prior to the initial point. For example, at 𝑡𝑡 = 𝑎𝑎 we
must have the solution at 𝑎𝑎 − 𝜏𝜏𝑖𝑖 . It is easy to see that if 𝜏𝜏∗ is the longest ����� 𝑛𝑛} , the equations generally require us to provide the delay; 𝜏𝜏∗ = max{𝜏𝜏𝑖𝑖 : 𝑖𝑖 = 1:
solution 𝜑𝜑(𝑡𝑡) for 𝑎𝑎 − 𝜏𝜏∗ ≤ 𝑡𝑡 ≤ 𝑎𝑎 . For DDEs we must provide not just the value of
the solution at the initial point, but also the “history”, the solution at times prior to the initial point.
14
Chapter One
Basic Concepts
For one thing, such discontinuities are not unusual for ODEs, but they are almost always present for DDEs, because the methods for both of them are intended for problems with solutions that have several continuous derivatives. From these considerations we will find that the most natural answer appears to be that one should specify an initial function on some interval of length 𝜏𝜏∗ , say [𝑎𝑎 − 𝜏𝜏∗ , 𝑎𝑎], and they try to satisfy equation(1.29) from 𝑡𝑡 ≥ 𝑎𝑎. Thus we set 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡),
𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎 − 𝜏𝜏∗ ≤ 𝑡𝑡 ≤ 𝑎𝑎
… (1.30)
We assume that 𝜑𝜑 ∈ 𝐶𝐶 𝑚𝑚 ([𝑎𝑎 − 𝜏𝜏∗ , 𝑎𝑎], ℝ) given function. See the figure below:
𝑡𝑡0 − 𝜏𝜏∗
𝑡𝑡 − 𝜏𝜏∗
𝑡𝑡0 = 𝑎𝑎
𝑡𝑡
𝑡𝑡0 + 𝜏𝜏∗
𝑡𝑡0 + 2𝜏𝜏∗
The action of the evaluation operator 𝜑𝜑 for a delay differential equation of the form (1.29) is to take a function defined over a time interval of length τ∗ and to map it into another function defined over a similar time interval. For instance, the initial function defined over the interval [𝑎𝑎 − 𝜏𝜏∗ , 𝑎𝑎] is mapped into a solution curve on the interval [𝑎𝑎, 𝑎𝑎 + 𝜏𝜏∗ ] .A fundamental technique for solving DDEs is to reduce it to a sequence of ODE. 1.6 Fractional Integro-Delay Differential Equations: [61] The Fractional integro-delay differential Equation is a delay functional equation which involves both integral and fractional derivative operators of the unknown. In classifying fractional integro-delay lags differential equation, we have the same category used in integro-differential equations. Mathematically, the general form of Linear Fractional Integro-Delay Differential Equations (LFIDDEs) of Caputo-type can be formulated as: 𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝑢𝑢(𝑡𝑡 − 𝜎𝜎𝑖𝑖 ) + 𝑃𝑃0 (𝑡𝑡)𝑢𝑢(𝑡𝑡 − 𝜎𝜎𝑛𝑛 ) 𝑖𝑖=1
𝑚𝑚 𝑣𝑣(𝑡𝑡)
= 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝑢𝑢(𝑥𝑥 − 𝜏𝜏𝑗𝑗 )𝑑𝑑𝑑𝑑 , 𝑗𝑗 =1 𝑎𝑎
15
𝑡𝑡 ∈ 𝐼𝐼 = [𝑎𝑎, 𝑏𝑏]
… (1.31)
Chapter One
Basic Concepts
Connected with 𝜇𝜇-conditions with 𝜇𝜇-historical functions, that is 𝜇𝜇 = max{𝛼𝛼1 , 𝛼𝛼2 , … , 𝛼𝛼𝑛𝑛 }, where 𝑢𝑢(𝑡𝑡) is the unknown function which is the solution of equation (1.31) and 𝒦𝒦𝑗𝑗 : 𝑆𝑆 × ℝ → ℝ (with 𝑆𝑆 = {(𝑡𝑡, 𝑥𝑥): 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑡𝑡 ≤ 𝑏𝑏}) for all 𝑗𝑗 = 1 , ⋯ , 𝑚𝑚 Given continuous functions and 𝑓𝑓, 𝑃𝑃𝑖𝑖 ∈ 𝐶𝐶(𝐼𝐼, ℝ ) , for all 𝛼𝛼 𝑖𝑖 = 0,1, … , 𝑛𝑛 − 1 where 𝑢𝑢(𝑡𝑡) ∈ ℝ , 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑖𝑖 𝑢𝑢(𝑡𝑡) is the 𝛼𝛼𝑖𝑖 -fractional order Caputoderivative of 𝑢𝑢 and all 𝛼𝛼𝑖𝑖 , ∈ ℝ+ for (𝑖𝑖 ≠ 0), 𝑛𝑛𝛼𝛼 𝑖𝑖 −1 < 𝛼𝛼𝑖𝑖 ≤ 𝑛𝑛𝛼𝛼 𝑖𝑖 , 𝑛𝑛𝛼𝛼 𝑖𝑖 = ⌈𝛼𝛼𝑖𝑖 ⌉, for all ����� 𝑖𝑖 = 1: 𝑛𝑛 . Furthermore, the quantities 𝜏𝜏𝑗𝑗 ≥ 0 , 𝑗𝑗 = 1,2, … , 𝑚𝑚 and 𝜌𝜌 ≥ 0, 𝑖𝑖 = 1,2, … , 𝑛𝑛 𝑖𝑖
are called the delay or time-lags.
The equation (1.31) is called Linear Fredholm Fractional Integro-Delay Differential Equations (LFFIDDEs) where 𝑣𝑣(𝑡𝑡) is fixed (𝑣𝑣(𝑡𝑡) = 𝑏𝑏) , while it is Linear Volterra Fractional Integro-Delay Differential Equations (LVFIDDEs) when 𝑣𝑣(𝑡𝑡) = 𝑡𝑡 . Fractional integro-delay differential equ. (1.31) are classified into three types:
First Type: Equation (1.31) of Retarded type if, max{𝛼𝛼0 , 𝛼𝛼1 , … , 𝛼𝛼𝑛𝑛 −1 } < 𝛼𝛼𝑛𝑛
or max�𝑛𝑛𝛼𝛼 0 , 𝑛𝑛𝛼𝛼 1 , … , 𝑛𝑛𝛼𝛼 𝑛𝑛 −1 � < 𝑛𝑛𝛼𝛼 𝑛𝑛 where 𝑛𝑛𝛼𝛼 𝑖𝑖 = ⌈𝛼𝛼𝑖𝑖 ⌉ , (i.e., the highest-fractional
order derivative of unknown function appears without difference arguments).
Second Types: Equation (1.31) of Neutral type if, max{𝛼𝛼0 , 𝛼𝛼1 , … , 𝛼𝛼𝑛𝑛 −1 } = 𝛼𝛼𝑛𝑛 or where 𝑛𝑛𝛼𝛼 𝑖𝑖 = ⌈𝛼𝛼𝑖𝑖 ⌉ , (i.e., the highest-fractional
max�𝑛𝑛𝛼𝛼 0 , 𝑛𝑛𝛼𝛼 1 , … , 𝑛𝑛𝛼𝛼 𝑛𝑛 −1 � = 𝑛𝑛𝛼𝛼 𝑛𝑛
order derivative of unknown function appears both with and without different arguments). Third
Types:
Equation
max{𝛼𝛼0 , 𝛼𝛼1 , … , 𝛼𝛼𝑛𝑛 −1 } > 𝛼𝛼𝑛𝑛
(1.31) or
of
Advanced
(Mixed)
max�𝑛𝑛𝛼𝛼 0 , 𝑛𝑛𝛼𝛼 1 , … , 𝑛𝑛𝛼𝛼 𝑛𝑛 −1 � > 𝑛𝑛𝛼𝛼 𝑛𝑛
type
if,
which
are
combinations of the previous two types.
For the purpose of clarity, assume that all 𝜎𝜎𝑖𝑖 = 0 for 𝑖𝑖 = 1,2, … , 𝑛𝑛 − 1 and
𝛼𝛼𝑛𝑛 > 𝛼𝛼𝑛𝑛−1 > 𝛼𝛼𝑛𝑛−2 > ⋯ > 𝛼𝛼1 > 𝛼𝛼0 = 0 with 𝜎𝜎𝑛𝑛 = 𝜏𝜏 (for simply), we have the
following linear Integro-Retarded constant delay fractional differential equation of the Volterra type.
16
Chapter One
Basic Concepts
Definition (1.12): The higher order Linear Volterra Integro-Retarded constant delay fractional differential equation with variable coefficients: 𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝑢𝑢(𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡)𝑢𝑢(𝑡𝑡 − 𝜏𝜏) 𝑖𝑖=1
𝑚𝑚
𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝑢𝑢(𝑥𝑥 − 𝜏𝜏𝑗𝑗 )𝑑𝑑𝑑𝑑 , 𝑗𝑗 =1 𝑎𝑎
𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏
… (1.32)
Together with initial conditions and historical functions: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻)
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑢𝑢(𝑡𝑡 = 𝑎𝑎) = 𝑢𝑢0 𝑢𝑢′ (𝑡𝑡 = 𝑎𝑎) = 𝑢𝑢1 ⋮ (𝜇𝜇 −1) (𝑡𝑡 = 𝑎𝑎) = 𝑢𝑢𝜇𝜇 −1 𝑢𝑢
𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) 𝑢𝑢′ (𝑡𝑡) = 𝜑𝜑 ′ (𝑡𝑡) ⋮ (𝜇𝜇 −1) (𝑡𝑡) = 𝜑𝜑 (𝜇𝜇 −1) (𝑡𝑡) 𝑢𝑢
𝜇𝜇 = ⌈𝛼𝛼𝑛𝑛 ⌉
for all, 𝑎𝑎� ≤ 𝑡𝑡 ≤ 𝑎𝑎
𝑎𝑎 − max�𝜏𝜏, 𝜏𝜏𝑗𝑗 ∶ 𝑗𝑗 = 1: 𝑚𝑚 �
Where 𝑢𝑢(𝑡𝑡) is the unknown function which is the solution of equation (1.32) the functions 𝒦𝒦𝑗𝑗 : 𝑆𝑆 × ℝ ⟶ ℝ , (𝑆𝑆 = {(𝑡𝑡, 𝑥𝑥) ∶ 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑡𝑡 ≤ 𝑏𝑏}) , 𝑗𝑗 = 1,2, … , 𝑚𝑚 and 𝑓𝑓, 𝑃𝑃𝑖𝑖 ∶ [𝑎𝑎, 𝑏𝑏] ⟶ ℝ ; 𝑖𝑖 = 0,1,2, … , 𝑛𝑛 − 1 for all continuous functions. In addition 𝛼𝛼𝑖𝑖 ∈ ℝ+ for 𝑖𝑖 ≠ 0 , 𝑛𝑛𝛼𝛼 𝑖𝑖 − 1 < 𝛼𝛼𝑖𝑖 ≤ 𝑛𝑛𝛼𝛼 𝑖𝑖 , 𝑛𝑛𝛼𝛼 𝑖𝑖 = ⌈𝛼𝛼𝑖𝑖 ⌉ for all 𝑖𝑖 = 1,2, … , 𝑛𝑛 with property 𝛼𝛼𝑛𝑛 > 𝛼𝛼𝑛𝑛 −1 > ⋯ > 𝛼𝛼1 > 𝛼𝛼0 = 0 and positive constant time-lags (delay), 𝜏𝜏𝑗𝑗 , 𝜏𝜏 .for all 𝑗𝑗 = 1,2, … , 𝑚𝑚 .
1.7 Occurrence of Linear VIFDEs of Retarded constant time delays:
In this part some problems are discussed, whose direct representation is in terms of multi-time fractional orders delay differential equations, reduced to a higher fractional order Integro-Retarded delay Differential equations with Variable Coefficients: Consider the multi-term fractional order delay differential equations in general form:
17
Chapter One 𝐶𝐶 𝛼𝛼 2 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
Basic Concepts
+ 𝑃𝑃1 (𝑡𝑡)
𝐶𝐶 𝛼𝛼 1 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑑𝑑 𝑟𝑟 𝑢𝑢(𝑡𝑡 − 𝜏𝜏) + 𝑃𝑃0 (𝑡𝑡) = 𝑑𝑑𝑡𝑡 𝑟𝑟
𝑚𝑚
� 𝜓𝜓𝑗𝑗 (𝑡𝑡) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 ) … (1.33) 𝑗𝑗 =1
Where 𝑃𝑃1 (𝑡𝑡), 𝑃𝑃0 (𝑡𝑡) and 𝜓𝜓𝑗𝑗 (𝑡𝑡) ; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚 are given continuous functions for all 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] and 𝑚𝑚𝛼𝛼 ≥ 𝑟𝑟 (𝑟𝑟 ∈ ℕ0 ) with property that 𝛼𝛼2 > 𝛼𝛼1 > 𝛼𝛼0 = 0, 𝜇𝜇 = ⌈𝛼𝛼2 ⌉ ; (𝜇𝜇 − 1 < 𝛼𝛼2 , 𝛼𝛼1 ≤ 𝜇𝜇) . with the initial conditions and historical functions: 𝑢𝑢(𝑘𝑘) (𝑎𝑎) = 𝑢𝑢𝑘𝑘 ; 𝑘𝑘 = 0,1,2, … , 𝜇𝜇 − 1 and 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) for 𝑡𝑡 ∈ [𝑎𝑎�, 𝑎𝑎] where 𝑎𝑎� = 𝑎𝑎 − max�𝜏𝜏, 𝜏𝜏𝑗𝑗 ∶ 𝑗𝑗 = 1: 𝑚𝑚 � and 𝜑𝜑 ∈ 𝐶𝐶𝜇𝜇 ([𝑎𝑎�, 𝑎𝑎], ℝ) .
By the following new technique we convert equation (1.33) to our problem. For 𝑟𝑟 = 0 and assume that min {𝛼𝛼𝑖𝑖 } ≥ 1, the equation (1.33) with the same initial condition and initial function becomes: 𝐶𝐶 𝛼𝛼 2 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
+
𝛼𝛼 𝑃𝑃1 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡)
𝑚𝑚
+ 𝑃𝑃0 (𝑡𝑡) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏) = � 𝜓𝜓𝑗𝑗 (𝑡𝑡) 𝑢𝑢�𝑡𝑡 − 𝜏𝜏𝑗𝑗 � 𝑗𝑗 =1
… (1.34)
Take 𝜂𝜂 ≥ 1 then using Lemmas (1.7) and (1.4(i)) with 𝑛𝑛 = 1, the following is obtained: 𝐶𝐶 𝜂𝜂 ( ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡
𝑑𝑑−1
𝑢𝑢(𝑘𝑘) (𝑎𝑎) 𝜕𝜕 𝜂𝜂 −1 = 𝑅𝑅𝑎𝑎 𝐷𝐷𝑡𝑡 �𝑢𝑢(𝑡𝑡 ) − � (𝑡𝑡 − 𝑎𝑎)𝑘𝑘 � ; (𝑑𝑑 − 1 < 𝜂𝜂 ≤ 𝑑𝑑) 𝜕𝜕𝜕𝜕 𝑘𝑘! 𝑘𝑘=0
𝑑𝑑−2
𝜕𝜕 𝑅𝑅 𝜂𝜂−1 𝑢𝑢(𝑘𝑘) (𝑎𝑎) 𝑢𝑢(𝑑𝑑−1) (𝑎𝑎) 𝑅𝑅 𝜂𝜂 −1 𝑘𝑘 ( ) ( ) (𝑡𝑡 − 𝑎𝑎)𝑑𝑑−1 � � �𝑢𝑢 � � = 𝑡𝑡 − − 𝑡𝑡 − 𝑎𝑎 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑎𝑎 𝐷𝐷𝑡𝑡 (𝑑𝑑 − 1)! 𝜕𝜕𝜕𝜕 𝑘𝑘! 𝑘𝑘=0
Using lemma (1.7) with equation (1.5) we get: 𝐶𝐶 𝜂𝜂 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
=
𝜕𝜕 𝐶𝐶 𝜂𝜂 −1 𝑢𝑢𝑑𝑑−1 (𝑡𝑡 − 𝑎𝑎)𝑑𝑑−𝜂𝜂 � � 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) − 𝜕𝜕𝜕𝜕 Γ(𝑑𝑑 − 𝜂𝜂 + 1)
… (1.35)
The Equation (1.35) to be true for all 𝜂𝜂 = 𝛼𝛼2 , 𝛼𝛼1 , applying equation (1.35) to each part of equation (1.34) the following can be obtained:
18
Chapter One
Basic Concepts
𝑢𝑢𝜇𝜇 𝛼𝛼 −1 𝜕𝜕 𝐶𝐶 𝛼𝛼 2 −1 𝛼𝛼 −1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 2 𝑢𝑢(𝑡𝑡) + 𝑃𝑃1 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡) − � 𝑎𝑎 𝐷𝐷𝑡𝑡 𝜕𝜕𝜕𝜕 Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼2 + 1) 𝑢𝑢𝜇𝜇 𝛼𝛼 −1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 � − 𝑃𝑃1 (𝑡𝑡) Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 1) 𝑢𝑢𝜇𝜇 𝛼𝛼 −1 𝛼𝛼 −1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 � − 𝑃𝑃1′ (𝑡𝑡) � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡) − Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 1) 𝑚𝑚
+ 𝑃𝑃0 (𝑡𝑡)𝑢𝑢(𝑡𝑡 − 𝜏𝜏) = � 𝜓𝜓𝑗𝑗 (𝑡𝑡) 𝑢𝑢�𝑡𝑡 − 𝜏𝜏𝑗𝑗 � 𝑗𝑗 =1
Integrating both sides on an interval (𝑎𝑎, 𝑡𝑡), ∀𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], and let: 𝛼𝛼�2 = 𝛼𝛼2 − 1, 𝛼𝛼�1 = 𝛼𝛼1 − 1, 𝜇𝜇̅𝛼𝛼 = 𝜇𝜇𝛼𝛼 − 1 Thus 𝜇𝜇̅𝛼𝛼 − 1 < 𝛼𝛼�2 , 𝛼𝛼�1 < 𝜇𝜇̅𝛼𝛼 . So �2 �1 𝐶𝐶 𝛼𝛼 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) + 𝑃𝑃1 (𝑡𝑡) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑚𝑚 +2 𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
With the initial conditions 𝑢𝑢(𝑘𝑘) (𝑎𝑎) = 𝑢𝑢𝑘𝑘 ; 𝑘𝑘 = 0,1,2, … , 𝜇𝜇̅ − 1 . Where
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 𝑓𝑓(𝑡𝑡) = 𝑢𝑢𝜇𝜇�𝛼𝛼 � + 𝑃𝑃1 (𝑡𝑡) − ℋ(𝑡𝑡)� Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) 𝑡𝑡
1 ℋ(𝑡𝑡) = � 𝑃𝑃1′ (𝑥𝑥) (𝑥𝑥 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 𝑑𝑑𝑑𝑑 𝛤𝛤(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1)
𝜓𝜓𝑗𝑗 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = �−𝑃𝑃0 (𝑥𝑥) 𝑃𝑃1′ (𝑥𝑥)
And
As
a
𝑎𝑎
; 𝑗𝑗 = 1, ⋯ , 𝑚𝑚 ; ℓ = 𝑚𝑚 + 1 ; ℓ = 𝑚𝑚 + 2
𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = � 𝑢𝑢(𝑥𝑥 − 𝜏𝜏) �1 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑥𝑥 𝑢𝑢(𝑥𝑥)
special
case
if
𝑃𝑃1 (𝑡𝑡)=𝒞𝒞,
where
; 𝑗𝑗 = 1, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 𝒞𝒞
; 𝑗𝑗 = 𝑚𝑚 + 2 is
a
constant
and
𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ; 𝑗𝑗 = 1,2, … , 𝑚𝑚 and 𝑢𝑢(𝑥𝑥 − 𝜏𝜏) ; for 𝑗𝑗 = 𝑚𝑚 + 1, so we
obtain:
�2 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) +
� 𝛼𝛼 𝒞𝒞 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡)
𝑚𝑚 +1 𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
19
Chapter One
Basic Concepts
Where (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 + 𝒞𝒞 𝑓𝑓(𝑡𝑡) = 𝑢𝑢𝜇𝜇�𝛼𝛼 � � 𝛤𝛤(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) 𝛤𝛤(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) 𝜓𝜓𝑗𝑗 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = � −𝑃𝑃0 (𝑥𝑥)
; 𝑗𝑗 = 1, … , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1
This means that a higher fractional order LVIDDEs with variable and constant coefficients has been obtained respectively. For 𝑟𝑟 = 1 , the equation (1.33) with the same initial conditions and initial functions becomes: 𝑚𝑚
𝑑𝑑𝑑𝑑 (𝑡𝑡 − 𝜏𝜏) 𝐶𝐶 𝛼𝛼 2 ( ) 𝐶𝐶 𝛼𝛼 1 = � 𝜓𝜓𝑗𝑗 (𝑡𝑡) 𝑢𝑢�𝑡𝑡 − 𝜏𝜏𝑗𝑗 � 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡 + 𝑃𝑃1 (𝑡𝑡 ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡 ) + 𝑃𝑃0 (𝑡𝑡 ) 𝑑𝑑𝑑𝑑 𝑗𝑗 =1
… (1.36)
After the same procedure, we obtain the higher fractional order LVIDDEs with variable coefficient of the form: �2 �1 𝐶𝐶 𝛼𝛼 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) + 𝑃𝑃1 (𝑡𝑡) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
+ 𝑃𝑃0 (𝑡𝑡) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏)
𝑚𝑚 +2 𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
Where
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 + 𝑃𝑃1 (𝑡𝑡) − ℋ(𝑡𝑡)� 𝑓𝑓(𝑡𝑡) = 𝑢𝑢𝜇𝜇�𝛼𝛼 � Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) 𝑡𝑡
And
With
1 � 𝑃𝑃1′ (𝑥𝑥) (𝑥𝑥 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 𝑑𝑑𝑑𝑑 ℋ(𝑡𝑡) = Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) 𝑎𝑎
𝜓𝜓𝑗𝑗 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = �−𝑃𝑃0′ (𝑥𝑥) 𝑃𝑃1′ (𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = �𝑢𝑢(𝑥𝑥 − 𝜏𝜏) �1 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑥𝑥 𝑢𝑢(𝑥𝑥) 20
; 𝑗𝑗 = 1, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 ; 𝑗𝑗 = 𝑚𝑚 + 2 ; 𝑗𝑗 = 1, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 ; 𝑗𝑗 = 𝑚𝑚 + 2
Chapter One
Basic Concepts
As a special case if 𝑃𝑃1 (𝑡𝑡)=𝒞𝒞 and if, where 𝒞𝒞 is a constant and 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ; 𝑗𝑗 = 1,2, … , 𝑚𝑚 and 𝑢𝑢(𝑥𝑥 − 𝜏𝜏) ; for 𝑗𝑗 = 𝑚𝑚 + 1, so we obtain: �2 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) +
Where
� 𝛼𝛼
𝒞𝒞 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏)
𝑚𝑚 +1 𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 𝑓𝑓(𝑡𝑡) = 𝑢𝑢𝜇𝜇�𝛼𝛼 � + 𝒞𝒞 � 𝛤𝛤(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) 𝛤𝛤(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) 𝜓𝜓𝑗𝑗 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = � ′ −𝑃𝑃0 (𝑠𝑠)
; 𝑗𝑗 = 1,2, … , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1
For 𝜂𝜂 ≥ 2 then using Lemmas (1.7) and (1.4(i)) with𝑛𝑛 = 2, we get: 𝐶𝐶 𝜂𝜂 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝜕𝜕 2 𝐶𝐶 𝜂𝜂 −2 𝑢𝑢𝑑𝑑−2 (𝑡𝑡 − 𝑎𝑎)𝑑𝑑−𝜂𝜂 = 2 � 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) − 𝜕𝜕𝑡𝑡 Γ(𝑑𝑑 − 𝜂𝜂 + 1) −
𝑢𝑢𝑑𝑑−1 (𝑡𝑡 − 𝑎𝑎)𝑑𝑑−𝜂𝜂 +1 � Γ(𝑑𝑑 − 𝜂𝜂 + 2)
… (1.37)
Assume that 𝑚𝑚𝑚𝑚 𝑛𝑛{𝛼𝛼𝑖𝑖 } ≥ 2. And if (𝑟𝑟 = 0) so the equation (1.37) is true for all 𝜂𝜂 = 𝛼𝛼2 , 𝛼𝛼1 , applying equation (1.37) to each part of equation (1.34), the following can be obtained: 𝜕𝜕 2 𝐶𝐶 𝛼𝛼 2 −2 𝛼𝛼 −2 𝑢𝑢(𝑡𝑡) + 𝑃𝑃1 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡) � 𝑎𝑎 𝐷𝐷𝑡𝑡 2 𝜕𝜕𝑡𝑡 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 2 𝑃𝑃1 (𝑡𝑡)(𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 + − 𝑢𝑢𝜇𝜇 𝛼𝛼 −2 � � Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼2 + 1) Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 1) (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 2 +1 𝑃𝑃1 (𝑡𝑡)(𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 +1 + − 𝑢𝑢𝜇𝜇 𝛼𝛼 −1 � �� Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼2 + 2) Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 2) −2𝑃𝑃1′ (𝑡𝑡)
𝑢𝑢𝜇𝜇 𝛼𝛼 −2 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 𝑢𝑢𝜇𝜇 𝛼𝛼 −1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 +1 𝜕𝜕 𝐶𝐶 𝛼𝛼 1 −2 𝑢𝑢(𝑡𝑡) − − � 𝐷𝐷 � Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 1) Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 2) 𝜕𝜕𝜕𝜕 𝑎𝑎 𝑡𝑡
𝛼𝛼 −2 −𝑃𝑃1′′ (𝑡𝑡) � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡)
𝑢𝑢𝜇𝜇 𝛼𝛼 −2 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 𝑢𝑢𝜇𝜇 𝛼𝛼 −1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇 𝛼𝛼 −𝛼𝛼 1 +1 − − � Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 1) Γ(𝜇𝜇𝛼𝛼 − 𝛼𝛼1 + 2) 𝑚𝑚
21
+𝑃𝑃0 (𝑡𝑡) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏) = � 𝜓𝜓𝑗𝑗 (𝑡𝑡) 𝑢𝑢�𝑡𝑡 − 𝜏𝜏𝑗𝑗 � 𝑗𝑗 =1
Chapter One
Basic Concepts
By integrating both sides twice on an interval (𝑎𝑎, 𝑡𝑡) and letting 𝜇𝜇̅ = 𝜇𝜇 − 2 𝛼𝛼�1 = 𝛼𝛼1 − 2 , 𝛼𝛼�2 = 𝛼𝛼2 − 2 the following is obtained: �2 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡 )
𝑚𝑚 +3 𝑡𝑡
� 𝛼𝛼
+ 𝑃𝑃1 (𝑡𝑡 ) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡 ) = 𝑓𝑓(𝑡𝑡 ) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
Where 𝑓𝑓 (𝑡𝑡 ) = 𝑢𝑢𝜇𝜇�𝛼𝛼 �
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 ( ) + 𝑃𝑃1 𝑡𝑡 − ℋ1 (𝑡𝑡 )� Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 +1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 +1 + 𝑢𝑢𝜇𝜇� 𝛼𝛼 +1 � + 𝑃𝑃1 (𝑡𝑡 ) − ℋ2 (𝑡𝑡 )� Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 2) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 2) 𝑡𝑡
1 �(𝑡𝑡 − 𝑥𝑥)[2(𝜇𝜇̅ − 𝛼𝛼�1 )𝑃𝑃′ 1 (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 ̅−𝛼𝛼� 1 −1 ℋ1 (𝑡𝑡 ) = ) ( 𝛤𝛤 𝜇𝜇̅ − 𝛼𝛼�1 + 2 +
𝑎𝑎 ′′ ( )( 𝑃𝑃1 𝑥𝑥 𝑥𝑥 −
𝑡𝑡
𝑎𝑎)𝜇𝜇 −𝛼𝛼� 1 ]𝑑𝑑𝑑𝑑
1 �(𝑡𝑡 − 𝑥𝑥)[2(𝜇𝜇̅ − 𝛼𝛼�1 )𝑃𝑃′ 1 (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 ̅−𝛼𝛼� 1 ℋ2 (𝑡𝑡 ) = ) ( 𝛤𝛤 𝜇𝜇̅ − 𝛼𝛼�1 + 2 +
And
As
a
𝑎𝑎 ′′ ( )( 𝑃𝑃1 𝑥𝑥 𝑥𝑥 −
𝑎𝑎)𝜇𝜇 −𝛼𝛼� 1 +1 ]𝑑𝑑𝑑𝑑
(𝑡𝑡 − 𝑥𝑥)𝜓𝜓𝑗𝑗 (𝑥𝑥) ⎧ −(𝑡𝑡 − 𝑥𝑥)𝑃𝑃0 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = ′′ ⎨ (𝑡𝑡 − 𝑥𝑥)𝑃𝑃1 (𝑥𝑥) ⎩ 2(𝑡𝑡 − 𝑎𝑎)𝑃𝑃1′ (𝑥𝑥)
𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ⎧ ⎪𝑢𝑢(𝑥𝑥 − 𝜏𝜏) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑥𝑥𝛼𝛼� 1 𝑢𝑢(𝑥𝑥) ⎨ ⎪ 𝜕𝜕 𝐶𝐶 𝛼𝛼� 1 ( ) ⎩ 𝜕𝜕𝜕𝜕 𝑎𝑎 𝐷𝐷𝑥𝑥 𝑢𝑢 𝑥𝑥
special
case
if
𝑃𝑃1 (𝑡𝑡)
=𝒞𝒞,
; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 ; 𝑗𝑗 = 𝑚𝑚 + 2 ; 𝑗𝑗 = 𝑚𝑚 + 3
; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 ; 𝑗𝑗 = 𝑚𝑚 + 2
where
; 𝑗𝑗 = 𝑚𝑚 + 3
𝒞𝒞
is
a
constant
𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ; 𝑗𝑗 = 1,2, … , 𝑚𝑚 and 𝑢𝑢(𝑥𝑥 − 𝜏𝜏) ; for 𝑗𝑗 = 𝑚𝑚 + 1,
so
obtain:
�2 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
+
� 𝛼𝛼 𝒞𝒞 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡)
𝑚𝑚 +1 𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑
Where ℋ1 (𝑡𝑡) = 0 , ℋ2 (𝑡𝑡) = 0 and
22
𝑗𝑗 =1 𝑎𝑎
and we
Chapter One
Basic Concepts
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 𝑓𝑓(𝑡𝑡) = 𝑢𝑢𝜇𝜇� 𝛼𝛼 � + 𝒞𝒞 � Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) Γ(𝜇𝜇̅ 𝛼𝛼 − 𝛼𝛼�1 + 1) (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 +1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 +1 + 𝑢𝑢𝜇𝜇�𝛼𝛼 +1 � + 𝒞𝒞 � Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 2) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 2) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = �
(𝑡𝑡 − 𝑥𝑥) 𝜓𝜓𝑗𝑗 (𝑥𝑥) −(𝑡𝑡 − 𝑥𝑥)𝑃𝑃0 (𝑥𝑥)
; 𝑗𝑗 = 1,2, … , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1
This means that a higher fractional order LVIDDEs with variable and constant coefficients has been obtained respectively. By the same previous technique with 𝑟𝑟 = 1 from equation (1.33) we obtain: �2 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
Where
𝑓𝑓 (𝑡𝑡 ) = 𝑢𝑢𝜇𝜇�𝛼𝛼 �
𝑚𝑚 +3 𝑡𝑡
� 𝛼𝛼
+ 𝑃𝑃1 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 + 𝑃𝑃1 (𝑡𝑡 ) − ℋ1 (𝑡𝑡 )� 𝛤𝛤 (𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) 𝛤𝛤 (𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 +1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 +1 + 𝑢𝑢𝜇𝜇� 𝛼𝛼 +1 � + 𝑃𝑃1 (𝑡𝑡 ) − ℋ2 (𝑡𝑡 )� 𝛤𝛤 (𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 2) 𝛤𝛤 (𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 2) 𝑡𝑡
1 �(𝑡𝑡 − 𝑥𝑥)[2(𝜇𝜇̅ − 𝛼𝛼�1 )𝑃𝑃′ 1 (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 ̅−𝛼𝛼� 1 −1 ℋ1 (𝑡𝑡 ) = 𝛤𝛤 (𝜇𝜇̅ − 𝛼𝛼�1 + 2) 𝑎𝑎
+ 𝑃𝑃1′′ (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 −𝛼𝛼� 1 ]𝑑𝑑𝑑𝑑 𝑡𝑡
1 �(𝑡𝑡 − 𝑥𝑥)[2(𝜇𝜇̅ − 𝛼𝛼�1 )𝑃𝑃′ 1 (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 ̅−𝛼𝛼� 1 ℋ2 (𝑡𝑡 ) = 𝛤𝛤 (𝜇𝜇̅ − 𝛼𝛼�1 + 2) 𝑎𝑎
+ 𝑃𝑃1′′ (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 −𝛼𝛼� 1 +1 ]𝑑𝑑𝑑𝑑
And
(𝑡𝑡 − 𝑥𝑥)𝜓𝜓𝑗𝑗 (𝑥𝑥) ⎧ −(𝑡𝑡 − 𝑥𝑥)𝑃𝑃0 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = ′′ ⎨ (𝑡𝑡 − 𝑥𝑥)𝑃𝑃1 (𝑥𝑥) ⎩ 2(𝑡𝑡 − 𝑎𝑎)𝑃𝑃1′ (𝑥𝑥)
𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ⎧ 𝜕𝜕 ⎪ ⎪ 𝑢𝑢(𝑥𝑥 − 𝜏𝜏) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝜕𝜕𝜕𝜕 � 𝛼𝛼 ⎨ 𝐶𝐶𝑎𝑎 𝐷𝐷𝑥𝑥 1 𝑢𝑢(𝑥𝑥) ⎪ ⎪ 𝜕𝜕 𝐶𝐶 𝛼𝛼� 1 ( ) ⎩ 𝜕𝜕𝜕𝜕 𝑎𝑎 𝐷𝐷𝑥𝑥 𝑢𝑢 𝑥𝑥
23
; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 ; 𝑗𝑗 = 𝑚𝑚 + 2 ; 𝑗𝑗 = 𝑚𝑚 + 3
; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚
; 𝑗𝑗 = 𝑚𝑚 + 1
; 𝑗𝑗 = 𝑚𝑚 + 2
; 𝑗𝑗 = 𝑚𝑚 + 3
Chapter One
As
a
Basic Concepts
special
case
if
𝑝𝑝1 (𝑡𝑡) =𝒞𝒞,
where
𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ; 𝑗𝑗 = 1,2, … , 𝑚𝑚 and obtain:
�2 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 (𝑡𝑡 )
� 𝛼𝛼 𝒞𝒞 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 1 𝑢𝑢(𝑡𝑡 )
+
𝜕𝜕𝜕𝜕
is
a
constant
and
𝑢𝑢(𝑥𝑥 − 𝜏𝜏) ; for 𝑗𝑗 = 𝑚𝑚 + 1, so we
𝑚𝑚 +3 𝑡𝑡
= 𝑓𝑓 (𝑡𝑡 ) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
where ℋ1 (𝑡𝑡) = 0 , ℋ2 (𝑡𝑡) = 0 and 𝑓𝑓(𝑡𝑡 ) = 𝑢𝑢𝜇𝜇�𝛼𝛼
𝜕𝜕
𝒞𝒞
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 � � + 𝒞𝒞 Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 +1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 +1 � + 𝑢𝑢𝜇𝜇�𝛼𝛼 +1 � + 𝒞𝒞 Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 2) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 2)
(𝑡𝑡 − 𝑥𝑥) 𝜓𝜓𝑗𝑗 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = � −(𝑡𝑡 − 𝑥𝑥)𝑃𝑃0 (𝑥𝑥)
; 𝑗𝑗 = 1,2, … , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1
This means that a higher fractional order LVIDDEs type with variable and constant coefficients has been obtained respectively. Finally, for 𝑟𝑟 = 2 the following FVIDDE with variable coefficients can be concluded: �2 �1 𝐶𝐶 𝛼𝛼 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) + 𝑃𝑃1 (𝑡𝑡) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
+ 𝑃𝑃0 (𝑡𝑡) 𝑢𝑢(𝑥𝑥 − 𝜏𝜏)
𝑚𝑚 +4 𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
where
(𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 2 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇�𝛼𝛼 −𝛼𝛼� 1 𝑓𝑓 (𝑡𝑡 ) = 𝑢𝑢𝜇𝜇�𝛼𝛼 � + 𝑃𝑃1 (𝑡𝑡 ) − ℋ1 (𝑡𝑡 )� Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 1) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 1) (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 1 +1 (𝑡𝑡 − 𝑎𝑎)𝜇𝜇� 𝛼𝛼 −𝛼𝛼� 2 +1 ( ) + 𝑢𝑢𝜇𝜇� 𝛼𝛼 +1 � + 𝑃𝑃1 𝑡𝑡 − ℋ2 (𝑡𝑡 )� Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�2 + 2) Γ(𝜇𝜇̅𝛼𝛼 − 𝛼𝛼�1 + 2) 𝑡𝑡
1 �(𝑡𝑡 − 𝑥𝑥)[2(𝜇𝜇̅ − 𝛼𝛼�1 )𝑃𝑃′ 1 (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 ̅−𝛼𝛼� 1 −1 ℋ1 (𝑡𝑡 ) = ) ( 𝛤𝛤 𝜇𝜇̅ − 𝛼𝛼�1 + 2 𝑎𝑎
+ 𝑃𝑃1′′ (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 −𝛼𝛼� 1 ]𝑑𝑑𝑑𝑑
24
Chapter One
Basic Concepts 𝑡𝑡
1 �(𝑡𝑡 − 𝑥𝑥)[2(𝜇𝜇̅ − 𝛼𝛼�1 )𝑃𝑃′ 1 (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 ̅−𝛼𝛼� 1 ℋ2 (𝑡𝑡 ) = ) ( 𝛤𝛤 𝜇𝜇̅ − 𝛼𝛼�1 + 2 𝑎𝑎
+ 𝑃𝑃1′′ (𝑥𝑥)(𝑥𝑥 − 𝑎𝑎)𝜇𝜇 −𝛼𝛼� 1 +1 ]𝑑𝑑𝑑𝑑
And
(𝑡𝑡 − 𝑥𝑥)𝜓𝜓𝑗𝑗 (𝑥𝑥) ⎧ ′′ ⎪ (𝑡𝑡 − 𝑥𝑥)𝑃𝑃0 (𝑥𝑥) 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = 2(𝑡𝑡 − 𝑥𝑥)𝑃𝑃1′ (𝑥𝑥) ⎨ (𝑡𝑡 − 𝑥𝑥)𝑃𝑃 ′′ (𝑥𝑥) 1 ⎪ ⎩ 2(𝑡𝑡 − 𝑎𝑎)𝑃𝑃1′ (𝑥𝑥)
𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � ⎧ ⎪ 𝑢𝑢(𝑥𝑥 − 𝜏𝜏) ⎪ 𝜕𝜕 ( ) 𝐺𝐺𝑗𝑗 �𝑢𝑢�𝑥𝑥, 𝜏𝜏𝑗𝑗 �� = 𝜕𝜕𝜕𝜕 𝑢𝑢 𝑥𝑥 − 𝜏𝜏 ⎨ 𝐶𝐶 𝛼𝛼� 1 ( ) ⎪ 𝑎𝑎 𝐷𝐷𝑥𝑥 𝑢𝑢 𝑥𝑥 ⎪ 𝜕𝜕 𝐶𝐶 𝛼𝛼� 1 ( ) ⎩𝜕𝜕𝜕𝜕 𝑎𝑎 𝐷𝐷𝑥𝑥 𝑢𝑢 𝑥𝑥
; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1 ; 𝑗𝑗 = 𝑚𝑚 + 2 ; 𝑗𝑗 = 𝑚𝑚 + 3 ; 𝑗𝑗 = 𝑚𝑚 + 4
; 𝑗𝑗 = 1,2, ⋯ , 𝑚𝑚 ; 𝑗𝑗 = 𝑚𝑚 + 1
; 𝑗𝑗 = 𝑚𝑚 + 2 ; 𝑗𝑗 = 𝑚𝑚 + 3
; 𝑗𝑗 = 𝑚𝑚 + 4
By the same procedure for all 𝑚𝑚𝑚𝑚𝑚𝑚{𝛼𝛼𝑖𝑖 } ≥ {3,4,5, ⋯ } corresponding values of 𝑟𝑟 ∈ ℕ0 , to get the higher order FVIDDE with variable (constant) coefficients. Lemma (1.15): Consider the linear VIEs of form
𝑡𝑡
𝑚𝑚
𝑢𝑢(𝑡𝑡 ) = 𝑓𝑓(𝑡𝑡 ) + � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 ) 𝑑𝑑𝑑𝑑 𝑗𝑗 =1 𝑎𝑎
… (1.38)
With one continuous historical function 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) ; 𝑎𝑎� ≤ 𝑡𝑡 ≤ 𝑎𝑎 where 𝑓𝑓 and 𝒦𝒦𝑗𝑗 (𝑗𝑗 = 1,2, … , 𝑚𝑚) are given continuous functions. Therefore, taking 𝛼𝛼-Caputo differentiation for both sides of equation (1.38), we obtain: 𝐶𝐶 𝛼𝛼 ( ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡
𝑚𝑚
𝑡𝑡
= 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑓𝑓(𝑡𝑡 ) + 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 �� � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 ) 𝑑𝑑𝑑𝑑� 𝑗𝑗 =1 𝑎𝑎
First, if 0 < 𝛼𝛼 ≤ 1 , using linearity property of 𝛼𝛼-Caputo derivative with lemma (1.14), to have: 𝐶𝐶 𝛼𝛼 ( ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡
= 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑓𝑓(𝑡𝑡 )
By assumption:
𝑡𝑡
𝑚𝑚
+ � �� 𝐶𝐶𝑥𝑥 𝐷𝐷𝑡𝑡𝛼𝛼 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝑢𝑢�𝑡𝑡 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 + 𝑎𝑎 𝐽𝐽𝑡𝑡1−𝛼𝛼 � lim 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 )�� 𝑥𝑥→𝑡𝑡−𝑎𝑎
𝑗𝑗 =1 𝑎𝑎
�𝑗𝑗 (𝑡𝑡, 𝑥𝑥) = 𝐶𝐶𝑥𝑥 𝐷𝐷𝑡𝑡𝛼𝛼 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) ; 𝒦𝒦
𝑓𝑓 (̅ 𝑡𝑡) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑓𝑓(𝑡𝑡)
𝑃𝑃𝑗𝑗 (𝑡𝑡) = − 𝐽𝐽1−𝛼𝛼 � lim 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 )� 𝑡𝑡 𝑎𝑎
𝑥𝑥→𝑡𝑡−𝑎𝑎
25
Chapter One
Basic Concepts
Thus 𝐶𝐶 𝛼𝛼 ( ) 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢 𝑡𝑡
𝑚𝑚
𝑡𝑡
�𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 )𝑑𝑑𝑑𝑑 + 𝑃𝑃𝑗𝑗 (𝑡𝑡 )𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑗𝑗 0 ) = 𝑓𝑓 (̅ 𝑡𝑡 ) + � � 𝒦𝒦 𝑗𝑗 =1 𝑎𝑎
With initial condition 𝑢𝑢(𝑎𝑎) = 𝑓𝑓(𝑎𝑎) and one continuous historical function 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) ; 𝑎𝑎� ≤ 𝑡𝑡 ≤ 𝑎𝑎 we can repeat this procedure on the last equation to obtain another formula of higher fractional order of linear VIDDEs with variable coefficients.
1.8 The Contribution of the Thesis: The main goal of this thesis is to introduce a general problem called linear higher fractional-order Volterra integro-differential equations with constant timedelay of Retarded type of variable coefficients in the sense of Caputo-fractional derivative, and it modifies some numerical methods that have been used for the first time to treat such problem: First of all, it discusses general definitions, properties and lemmas in fractional calculus, integral with integro-differential equations and timedelay differential equations which combine these three important subjects in a more general formula. The mixed multi-term linear fractional delay differential equations and integral delay differential equations are reduced to higher fractional-order Volterra integro-differential equations with constant time-delay of Retarded type of variable coefficients in the sense of Caputo-fractional derivative, and it is illustrated by more details. The study explains three algorithms 2BM,3BM and 4BM which present numerical methods to compute integro-fractional differential equation of Volterra type and Retarded delays depending on a finite difference approximation combined with Block-by-Block methods of second, third and fourth blocks respectively. The study also describes a least-square technique with basic orthogonal polynomials including Chebyshev and Legendre functions to solve LVIFDDE's. For each numerical method, an algorithm has been built, a program was written, examples were solved, and results have been tabulated. Comparison were made between the exact and numerical solutions depending on the least square errors and running time of all computer programs which were written in MatLab.
26
Chapter One
Basic Concepts
1.9 Test Examples: This thesis takes the following higher fractional linear Volterra integrodifferential equations with constant multi-time Retarded delay with variable coefficients as test examples: Test example (1): Consider a higher fractional linear VIDE's with constant multi-time Retarded delay on the closed interval [0,1]: 𝑐𝑐 2𝛼𝛼 0𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
− cos(𝑡𝑡) 𝐶𝐶0𝐷𝐷𝑡𝑡𝛼𝛼 𝑢𝑢(𝑡𝑡) + 𝑡𝑡 3 𝑢𝑢(𝑡𝑡 − 0.8 ) = 𝑓𝑓(𝑡𝑡) 𝑡𝑡
+ �[−𝑥𝑥 sin(𝑡𝑡) 𝑢𝑢(𝑥𝑥 − 0.2) +𝑒𝑒 𝑥𝑥−𝑡𝑡 𝑢𝑢(𝑥𝑥 − 0.4) + (𝑡𝑡 + 𝑥𝑥 2 )𝑢𝑢(𝑥𝑥 − 0.6)]𝑑𝑑𝑑𝑑 𝑎𝑎
where 𝑓𝑓(𝑡𝑡) =
4 4 cos(𝑡𝑡) 2−𝛼𝛼 4 23 1 𝑡𝑡 2−2𝛼𝛼 − 𝑡𝑡 + sin(𝑡𝑡) � 𝑡𝑡 4 − 𝑡𝑡 3 − 𝑡𝑡 2 � Γ(3 − 2𝛼𝛼) 15 50 Γ(3 − 𝛼𝛼) 2 +
With the initial conditions:
123 −𝑡𝑡 123 28 43 118 3 49 4 8 5 𝑒𝑒 − + 𝑡𝑡 − 𝑡𝑡 2 + 𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 25 25 5 25 75 15 5 �
if 0 < 𝛼𝛼 < 0.5; 𝑢𝑢(0) = −1 if 0 < 𝛼𝛼 ≤ 1; 𝑢𝑢(0) = −1 , 𝑢𝑢′ (0) = 0
And the historical function 𝜑𝜑(𝑡𝑡) = 2𝑡𝑡 2 − 1 , 𝑎𝑎� ≤ 𝑡𝑡 ≤ 0 While the exact solution is 𝑢𝑢(𝑡𝑡) = 2𝑡𝑡 2 − 1
Test example (2): Consider a linear Volterra integro-differential equations with constant multi-time Retarded delay for fractional order lies in (0,1) : 𝐶𝐶 0.9 0𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
− 𝑡𝑡
1 𝐶𝐶 0.6 𝐷𝐷 𝑢𝑢(𝑡𝑡) + 𝑡𝑡 𝐶𝐶0𝐷𝐷𝑡𝑡0.2 𝑢𝑢(𝑡𝑡) + cos(𝑡𝑡) 𝑢𝑢(𝑡𝑡 − 0.33) = 𝑓𝑓(𝑡𝑡) 6 0 𝑡𝑡
+ �[5 𝑡𝑡 𝑥𝑥 𝑢𝑢(𝑥𝑥 − 0.2) + (𝑥𝑥 2 + 𝑡𝑡 2 ) 𝑢𝑢(𝑥𝑥 − 0.72) − 𝑒𝑒 −2𝑥𝑥−𝑡𝑡 𝑢𝑢(𝑥𝑥 − 0.6)] 𝑑𝑑𝑑𝑑 𝑎𝑎
27
Chapter One
Basic Concepts
where 𝑓𝑓(𝑡𝑡) =
2 6 1 1 2 𝑡𝑡 0.1 − 𝑡𝑡 2.1 − 𝑡𝑡 0.4 + 𝑡𝑡 2.4 + 𝑡𝑡 1.8 Γ(1.1) Γ(3.1) 3 Γ(2.6) Γ(3.4) Γ(1.8) 6 − 𝑡𝑡 3.8 + cos(𝑡𝑡) [0.375937 + 1.6733 𝑡𝑡 + 0.99 𝑡𝑡 2 − 𝑡𝑡 3 ] Γ(3.8) 67078 3 13001 4 3 61 313 1 + 𝑒𝑒 −3𝑡𝑡 � 𝑡𝑡 3 − 𝑡𝑡 2 − 𝑡𝑡 − 𝑡𝑡 − 𝑡𝑡 �− 46875 20 100 1000 3750 2 951 5 17 6 313 −𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 + 𝑒𝑒 500 12 1000
With initial condition and historical function: 𝑢𝑢(0) = 1; 𝜑𝜑(𝑡𝑡) = 1 + 2𝑡𝑡 − 𝑡𝑡 3 . While the exact solution is 𝑢𝑢(𝑡𝑡) = 1 + 2𝑡𝑡 − 𝑡𝑡 3 .
Test example (3):
Consider a linear VIFDE’s with constant multi-time Retarded delay on 0 ≤ 𝑡𝑡 ≤ 1 for fractional order lies in (0,1): 𝐶𝐶 0.7 0𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) +
1
1
cosh(𝑡𝑡) 𝐶𝐶0𝐷𝐷𝑡𝑡0.6 𝑢𝑢(𝑡𝑡) + 𝑡𝑡 𝐶𝐶0𝐷𝐷𝑡𝑡0.3 𝑢𝑢(𝑡𝑡) − 𝑒𝑒 𝑡𝑡 𝑢𝑢(𝑡𝑡 − 0.1) = 𝑓𝑓(𝑡𝑡) 𝑡𝑡
2
5
+ �[−𝑥𝑥 sin (𝑡𝑡) 𝑢𝑢(𝑥𝑥 − 0.3) + (𝑡𝑡 + 𝑥𝑥 2 ) 𝑢𝑢(𝑥𝑥 − 0.9)] 𝑑𝑑𝑑𝑑 𝑎𝑎
Where: 𝑓𝑓(𝑡𝑡) = +
2 2 24 24 3.4 𝑡𝑡 3.3 − 𝑡𝑡 0.3 + cosh(𝑡𝑡) � 𝑡𝑡 − 𝑡𝑡 0.4 � Γ(1.3) Γ(1.4) Γ(4.3) Γ(4.4)
12 4.7 1 27 4 6 1 6081 2 527 3 𝑡𝑡 − 𝑡𝑡 1.7 + sin(𝑡𝑡) � 𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 5 + 𝑡𝑡 6 � Γ(4.7) Γ(1.7) 750 200 25 6 20000 −
24561 2 16393 3 391 4 9 5 2 6 1 7 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 10000 10000 1000 125 5 7 +𝑒𝑒 𝑡𝑡 �−
2001 501 3 2 2 1 + 𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 3 − 𝑡𝑡 4 � 50000 1250 250 25 5
With initial condition and the initial function : 𝑢𝑢(0) = 0 , 𝜑𝜑(𝑡𝑡) = 𝑡𝑡 4 − 2𝑡𝑡 . While the exact solution is 𝑢𝑢(𝑡𝑡) = 𝑡𝑡 4 − 2𝑡𝑡 . 28
Chapter One
Basic Concepts
Test example (4): Consider a linear VIFDE’s with constant multi-time Retarded delay on [0,1] for fractional order lies in (0,1): 𝑐𝑐 0.6 0𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡) −
3𝑡𝑡 0𝑐𝑐 𝐷𝐷𝑡𝑡0.3 𝑢𝑢(𝑡𝑡) + 𝑒𝑒 𝑡𝑡 0𝑐𝑐 𝐷𝐷𝑡𝑡0.2 𝑢𝑢(𝑡𝑡) + 𝑡𝑡 2 𝑢𝑢(𝑡𝑡 − 1.6) = 𝑓𝑓(𝑡𝑡) 𝑡𝑡
(𝑥𝑥 − 4𝑡𝑡) +�� 𝑢𝑢(𝑥𝑥 − 0.5) + (1 − 𝑥𝑥 2 ) 𝑢𝑢(𝑥𝑥 − 0.63)� 𝑑𝑑𝑑𝑑 2 𝑎𝑎
where
36 3.7 12 𝑒𝑒 𝑡𝑡 2.8 1 6 511 5 222593 4 12 2.4 𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 + 𝑡𝑡 + 𝑡𝑡 − 𝑡𝑡 𝑓𝑓(𝑡𝑡) = Γ(3.7) 3 250 20000 Γ(3.4) Γ(3.8) 27054953 3 35351 2 249953 + 𝑡𝑡 − 𝑡𝑡 − 𝑡𝑡 1500000 5000 500000
With initial condition: 𝑢𝑢(0) = 1, and the historical function 𝜑𝜑(𝑡𝑡) = 2𝑡𝑡 3 + 1 While the exact solution is 𝑢𝑢(𝑡𝑡) = 2𝑡𝑡 3 + 1
Test example (5):
Consider a higher fractional linear VIDE's with constant multi-time Retarded delay on the closed interval [0,1]: 𝐶𝐶 1.6 0 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
+ t 𝐶𝐶0𝐷𝐷𝑡𝑡1.1 𝑢𝑢(𝑡𝑡) − 3 𝐶𝐶0𝐷𝐷𝑡𝑡0.3 𝑢𝑢(𝑡𝑡) + (𝑡𝑡 + 2) 𝑢𝑢(𝑡𝑡 − 0.45) = 𝑓𝑓(𝑡𝑡)
+ � [(𝑥𝑥 − cos(𝑡𝑡)) 𝑢𝑢(𝑥𝑥 − 0.7) − 𝑒𝑒1−2𝑡𝑡 𝑢𝑢(𝑥𝑥 − 0.45)] 𝑑𝑑𝑑𝑑 𝑎𝑎
where 𝑓𝑓(𝑡𝑡) =
𝑡𝑡
6 6 6 18 1.7 𝑡𝑡 0.4 + 𝑡𝑡 1.9 + 𝑡𝑡 0.7 − 𝑡𝑡 Γ(1.4) Γ(1.9) Γ(1.7) Γ(2.7) 47 2 31 387 3 2 1003 + 𝑒𝑒 1−2𝑡𝑡 � 𝑡𝑡 − 𝑡𝑡 + 𝑡𝑡 3 � + cos(𝑡𝑡) �𝑡𝑡 3 − 𝑡𝑡 2 + 𝑡𝑡� − 𝑡𝑡 4 20 10 100 4 400 76 3 127 2 2757 1003 + 𝑡𝑡 − 𝑡𝑡 − 𝑡𝑡 + 15 200 400 200
With initial condition: 𝜑𝜑(𝑡𝑡) = 1 − 2𝑡𝑡 + 3𝑡𝑡 2 .
𝑢𝑢(0) = 1 , 𝑢𝑢′ (0) = −2 and the historical function
While the exact solution is 𝑢𝑢(𝑡𝑡) = 1 − 2𝑡𝑡 + 3𝑡𝑡 2 29
CHAPTER TWO
Block By Block Method
Chapter Two
Block By Block Method
2.1 Introduction A Block-by-Block method is based on interpolatory quadrature rules, these form class of good self-starting methods which produce a block of value at a time that is effective for the large interval [16], and thus the methods are simple switching stepsize procedures. We can see that the most higher-fractional linear Volterra integro-differential equations in Caputo sense of constant multi-time Retarded delay with variable coefficients do not have exact analytic solutions, therefore approximate and numerical procedures must be used. One of the good techniques for solving differential and integral equation is Block methods. The Block-by-Block approach was first suggested by Young [10] in connection with product integration techniques. A similar concept for differential equations was given by Miline [68]. Here we recall some published works on this subject: In [49, 50] Linz has presented second order method, with modified, for solving nonlinear Volterra integral equations of the second kind. While Borhan F.Jumaa [14], Katani and shahmorad [34,35] and Attia [13] were used only the fourth order of it for treating system of nonlinear Volterra integral equations. Moreover, in [48] Rasdi, Majid and Radzi present a two point and three point one-step block method for computing second order delay differential equations. And Makroglou [12], Muna with Latiff [45], Hassan [29] and Atheer [11] had applied it to solve numerically delay integro-differential equations of Volterra and Fredholm types. In this chapter, a new technique for computing numerically the linear Volterra Integro-Fractional Differential Equations (VIFDE's) with constant multi-time Retarded delay of positive arbitrary order which are less than or equal to one is applied for the first time combining with the aid of finite difference and Block-byBlock methods including two, three and four blocks. Three new algorithms for solving linear VIFDE's with constant multi-time Retarded delay using this procedure have been developed. Further, each algorithm program was written in MatLab (V.8) with several illustrative examples presented to show the effectiveness and accuracy of this method. A comparison between the algorithms has been made depending on the least square error and running time. 2.2 Basic Propositions: Before starting the investigations, it is necessary to identify two important basic propositions in our works: 30
Chapter Two
Block By Block Method
Proposition (2.1): [61] The finite difference approximation of Caputo derivative for 0 < 𝛼𝛼 ≤ 1 at given points 𝑡𝑡 = 𝑡𝑡𝑟𝑟+1 ; 𝑟𝑟 = 0,1, … , 𝑁𝑁 − 1 and ℎ = (𝑏𝑏 − 𝑎𝑎)⁄𝑁𝑁 , (𝑁𝑁 ∈ ℤ+) is formed as: 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)|𝑡𝑡=𝑡𝑡 𝑟𝑟+1
𝑟𝑟
ℎ−𝛼𝛼 = �[𝑢𝑢(𝑡𝑡𝑟𝑟−𝑠𝑠+1 ) − 𝑢𝑢(𝑡𝑡𝑟𝑟−𝑠𝑠 )] 𝑏𝑏𝑠𝑠𝛼𝛼 Γ(2 − α)
where 𝑏𝑏𝑠𝑠𝛼𝛼 = (𝑠𝑠 + 1)1−𝛼𝛼 − 𝑠𝑠1−𝛼𝛼 .
𝑠𝑠=0
… (2.1)
Proposition (2.2)
For any p-orders of Block-by-Block methods the Caputo fractional derivative of order 0 < 𝛼𝛼 ≤ 1 for any smooth function 𝑢𝑢(𝑡𝑡) on [𝑎𝑎, 𝑏𝑏], can evaluate it at any points 𝑡𝑡 = 𝑡𝑡𝑟𝑟𝑟𝑟 +𝜗𝜗 for each 𝑟𝑟 = 0,1, … , 𝑁𝑁 − 1 (𝑁𝑁 ∈ ℤ+) with 𝜗𝜗 = 1,2, … , 𝑝𝑝 by the following formula: 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)| 𝑡𝑡=𝑡𝑡 𝑟𝑟𝑟𝑟 +𝜗𝜗
−𝛼𝛼 �ℎ�� = ��𝑢𝑢�𝑡𝑡𝑟𝑟𝑟𝑟 +𝜗𝜗 � − 𝑢𝑢(𝑡𝑡𝑟𝑟𝑟𝑟 +𝜗𝜗 −1 )] Γ(2 − α)
+ where
𝑝𝑝𝑝𝑝 +𝜗𝜗 −1
� �𝑢𝑢(𝑡𝑡𝑟𝑟𝑟𝑟 +𝜗𝜗 −𝑠𝑠 ) − 𝑢𝑢(𝑡𝑡𝑟𝑟𝑟𝑟 +𝜗𝜗 −(𝑠𝑠+1) )� 𝑏𝑏𝑠𝑠𝛼𝛼 � 𝑠𝑠=1
𝑏𝑏𝑠𝑠𝛼𝛼 = (𝑠𝑠 + 1)1−𝛼𝛼 − 𝑠𝑠1−𝛼𝛼 , ℎ� =
… (2.2)
𝑏𝑏 − 𝑎𝑎 , 𝑝𝑝 ∈ ℤ+ 𝑝𝑝𝑝𝑝
By the same way as in proof of proposition (2.1) in [61], we can complete the above proof. 2.3 Basic Quadrature Formulas: [25, 55,67] In this section we explain the most important rules in quadrature formulas for our 1
work which are Simpson’s ℎ and Adaptive Simpsons rules. In Simpson’s rule, the 3
integrand function on any interval, [𝑡𝑡𝑟𝑟−1 , 𝑡𝑡𝑟𝑟+1 ] , is approximate by a parabola curve that passes through three points (𝑡𝑡𝑟𝑟−1 , 𝑓𝑓𝑟𝑟−1 ), (𝑡𝑡𝑟𝑟 , 𝑓𝑓𝑟𝑟 ) and (𝑡𝑡𝑟𝑟+1 , 𝑓𝑓𝑟𝑟+1 ) where 𝑡𝑡𝑟𝑟 =
𝑡𝑡 𝑟𝑟−1 +𝑡𝑡 𝑟𝑟+1 2
, the result is:
31
Chapter Two
Block By Block Method
𝑡𝑡 𝑟𝑟+1
ℎ ℎ5 (4) � 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 = [𝑓𝑓(𝑡𝑡𝑟𝑟−1 ) + 4𝑓𝑓(𝑡𝑡𝑟𝑟 ) + 𝑓𝑓(𝑡𝑡𝑟𝑟+1 )] − 𝑓𝑓 (𝜉𝜉) ; 𝜉𝜉 ∈ (𝑡𝑡𝑟𝑟−1 , 𝑡𝑡𝑟𝑟+1 ) 90 3
𝑡𝑡 𝑟𝑟−1
Repeated this process 𝑁𝑁 (even)-subintervals of size ℎ on closed bounded interval [𝑎𝑎, 𝑏𝑏] , ℎ = (𝑏𝑏 − 𝑎𝑎)⁄𝑁𝑁 ; 𝑁𝑁 ≥ 2 with the set of grid points 𝑡𝑡𝑠𝑠 = 𝑎𝑎 + 𝑠𝑠ℎ, (𝑠𝑠 = ����� 0: 𝑁𝑁) ; Thus 𝑏𝑏
𝑁𝑁
ℎ � 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 = � 𝑊𝑊𝑠𝑠 𝑓𝑓(𝑡𝑡𝑠𝑠 ) + 𝐸𝐸𝑁𝑁 [𝑓𝑓] 3 𝑠𝑠=0
𝑎𝑎
… (2.3)
Where 𝑤𝑤0 = 𝑤𝑤𝑁𝑁 = 1 and 𝑤𝑤𝑠𝑠 = 3 − (−1)𝑠𝑠 for 𝑠𝑠 = 1,2, … , 𝑁𝑁 − 1 with global error: (𝑏𝑏 −𝑎𝑎) 5 (4) 𝐸𝐸𝑁𝑁 [𝑓𝑓] = − ℎ 𝑓𝑓 (𝜉𝜉) ; 𝑎𝑎 < 𝜉𝜉 < 𝑏𝑏 . 180
In this adaptive process, we divide the interval [𝑎𝑎, 𝑏𝑏] into two subintervals and decide whether each of them is to be divided into more subintervals. The first step in the procedure is to apply Simpson’s rule with step-size ℎ = (𝑏𝑏 − 𝑎𝑎)⁄2 . the result is: 𝑏𝑏
𝐼𝐼(𝑓𝑓) [𝑎𝑎, 𝑏𝑏] ≡ � 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 = 𝑆𝑆1 (𝑓𝑓; 𝑎𝑎, 𝑏𝑏) + 𝐸𝐸1 (𝑓𝑓; ℎ, 𝜉𝜉) 𝑆𝑆1 (𝑓𝑓; 𝑎𝑎, 𝑏𝑏) =
where:
𝑎𝑎
𝑏𝑏 − 𝑎𝑎 𝑎𝑎 + 𝑏𝑏 �𝑓𝑓(𝑎𝑎) + 4 𝑓𝑓 � � + 𝑓𝑓(𝑏𝑏)� 6 2
1 𝑏𝑏 − 𝑎𝑎 5 (4) 𝐸𝐸1 (𝑓𝑓; ℎ, 𝜉𝜉) = − � � 𝑓𝑓 (𝜉𝜉) for some 𝜉𝜉 𝑖𝑖𝑖𝑖 (𝑎𝑎, 𝑏𝑏) 90 2 ℎ
�⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯� ⋮ 𝑎𝑎___________________. ____________________𝑏𝑏 ⋮ 𝑎𝑎+𝑏𝑏 𝑐𝑐 = ℎ/2
2
ℎ/2
⋮�⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯� ⋮ �⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯�⋮
At the second level, divide [𝑎𝑎, 𝑏𝑏] up in the middle, let 𝑐𝑐 = (𝑎𝑎 + 𝑏𝑏)⁄2 step Size
𝑏𝑏−𝑎𝑎 4
=
ℎ
2
, thus:
𝐼𝐼(𝑓𝑓)[𝑎𝑎, 𝑏𝑏] = 𝐼𝐼(𝑓𝑓)[𝑎𝑎, 𝑐𝑐] + 𝐼𝐼(𝑓𝑓)[𝑐𝑐, 𝑏𝑏] ℎ ℎ = 𝑆𝑆1 (𝑓𝑓; 𝑎𝑎, 𝑐𝑐) + 𝐸𝐸1 �𝑓𝑓; , 𝜉𝜉1 � + 𝑆𝑆1 (𝑓𝑓; 𝑐𝑐, 𝑏𝑏) + 𝐸𝐸1 �𝑓𝑓; , 𝜉𝜉2 � 2 2 ℎ = 𝑆𝑆2 (𝑓𝑓; 𝑎𝑎, 𝑏𝑏) + 𝐸𝐸2 �𝑓𝑓; , 𝜉𝜉� 2 32
Chapter Two
Block By Block Method
where 𝑆𝑆2 (𝑓𝑓; 𝑎𝑎, 𝑏𝑏) = 𝑆𝑆1 (𝑓𝑓; 𝑎𝑎, 𝑐𝑐) + 𝑆𝑆1 (𝑓𝑓; 𝑐𝑐, 𝑏𝑏) 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 + 𝑐𝑐 𝑐𝑐 + 𝑏𝑏 = �𝑓𝑓 (𝑎𝑎) + 4𝑓𝑓 � � + 2𝑓𝑓(𝑐𝑐) + 4𝑓𝑓 � � + 𝑓𝑓(𝑏𝑏)� 12 2 2
ℎ ℎ ℎ −1 ℎ 5 (4) 𝐸𝐸2 �𝑓𝑓; , 𝜉𝜉� = 𝐸𝐸1 �𝑓𝑓; , 𝜉𝜉1 � + 𝐸𝐸1 �𝑓𝑓; , 𝜉𝜉2 � = � � �𝑓𝑓 (𝜉𝜉1 ) + 𝑓𝑓 (4) (𝜉𝜉2 )� 2 2 2 90 2 5 −1 ℎ 1 = � 𝑓𝑓 (4) (𝜉𝜉)� = 4 𝐸𝐸1 (𝑓𝑓; ℎ, 𝜉𝜉) 16 90 2
Moreover, the following formula is derived by using adaptive Simpson's rule: 𝑡𝑡 𝑟𝑟+1
� 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 ≅
𝑡𝑡 𝑟𝑟
ℎ �𝑓𝑓(𝑡𝑡𝑟𝑟 ) + 4𝑓𝑓(𝑡𝑡𝑟𝑟+1/2 ) + 𝑓𝑓(𝑡𝑡𝑟𝑟+1 )� 6
… (2.4)
where 𝑓𝑓(𝑡𝑡𝑟𝑟+1/2 ) can be found from the Newton-Gregory Forward-Difference formula:
where
𝑠𝑠 𝑠𝑠 𝑠𝑠 𝑠𝑠 𝑃𝑃𝑛𝑛 (𝑡𝑡) = 𝑓𝑓𝑟𝑟 + � � ∆𝑓𝑓𝑟𝑟 + � � ∆2 𝑓𝑓𝑟𝑟 + � � ∆3 𝑓𝑓𝑟𝑟 + ⋯ + � � ∆𝑛𝑛 𝑓𝑓𝑟𝑟 1 2 3 𝑛𝑛
𝑓𝑓𝑟𝑟 = 𝑓𝑓(𝑡𝑡𝑟𝑟 ) , 𝑡𝑡 𝑠𝑠 = 𝑡𝑡 𝑟𝑟 + 𝑠𝑠ℎ ; ∆𝑛𝑛 𝑓𝑓𝑟𝑟 = ∆𝑛𝑛−1 𝑓𝑓𝑟𝑟+1 − ∆𝑛𝑛−1 𝑓𝑓𝑟𝑟 ; 𝑛𝑛 ≥ 1
Hence putting 𝑠𝑠 =
1 2
and 𝑛𝑛 = 2 so to interpolate 𝑓𝑓�𝑡𝑡𝑟𝑟+1/2 � by 𝑃𝑃2 (𝑡𝑡 = 𝑡𝑡𝑠𝑠 ), thus:
1 𝑠𝑠 𝑠𝑠 𝑓𝑓 �𝑡𝑡𝑟𝑟+1 � ≅ 𝑃𝑃2 �𝑡𝑡𝑟𝑟 + ℎ� = 𝑓𝑓𝑟𝑟 + � � ∆𝑓𝑓𝑟𝑟 + � � ∆2 𝑓𝑓𝑟𝑟 2 1 2 2 3 1 3 = 𝑓𝑓𝑟𝑟 + 𝑓𝑓𝑟𝑟+1 − 𝑓𝑓𝑟𝑟+2 4 8 8
If take 𝑛𝑛 = 3 and also interpolate 𝑓𝑓�𝑡𝑡𝑟𝑟+1/2 � by 𝑃𝑃3 (𝑡𝑡 = 𝑡𝑡𝑠𝑠 ) we obtain:
1 1 5 𝑓𝑓�𝑡𝑡𝑟𝑟+1/2 � ≅ 𝑃𝑃3 �𝑡𝑡𝑟𝑟 + ℎ� = �5𝑓𝑓𝑟𝑟 + 15𝑓𝑓𝑟𝑟+1 − 𝑓𝑓𝑟𝑟+2 + 𝑓𝑓𝑟𝑟+3 � 2 16 16
Moreover, if putting 𝑛𝑛 = 4 and doing same steps we get: 1 𝑓𝑓 �𝑡𝑡𝑟𝑟+1 � ≅ 𝑃𝑃4 �𝑡𝑡𝑟𝑟 + ℎ� 2 2
1 = [35𝑓𝑓𝑟𝑟 + 35𝑓𝑓𝑟𝑟+1 − 70𝑓𝑓𝑟𝑟+2 + 28𝑓𝑓𝑟𝑟+3 − 5𝑓𝑓𝑟𝑟+4 ] 4 33
… (2.5) … (2.6)
… (2.7)
Chapter Two
Block By Block Method
2.4 A Parameter Estimation Procedure: Throughout this section we will use the notation [𝑡𝑡]ℎ = ⟦𝑡𝑡⁄ℎ⟧ℎ where ⟦∗⟧ is the greatest integer part function. From the following graph of the function [𝑡𝑡]ℎ , it is easy to see that [21, 58, 71]:
𝑡𝑡 − ℎ < [𝑡𝑡]ℎ ≤ 𝑡𝑡, For all 𝑡𝑡 ∈ 𝑅𝑅 and ℎ > 0 therefore [𝑡𝑡]ℎ → 𝑡𝑡 as ℎ → 0+ uniformly in 𝑡𝑡.
Graph of [𝑡𝑡]ℎ
Recall the historical function in our problem (1.32) for which all arbitrary orders are less than or equal to one, so we have only one historical function that is: 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡), For all 𝑡𝑡 such that 𝑎𝑎� ≤ 𝑡𝑡 ≤ 𝑎𝑎 where 𝜑𝜑: [𝑎𝑎�, 𝑎𝑎] → ℝ is given 1: 𝑚𝑚� with given continuous function on [𝑎𝑎�, 𝑎𝑎] and 𝑎𝑎� = 𝑎𝑎 − max�𝜏𝜏, 𝜏𝜏𝑗𝑗 : 𝑗𝑗 = ������ positive constants time delay 𝜏𝜏, 𝜏𝜏𝑗𝑗 ( ∀𝑗𝑗 = ������ 1: 𝑚𝑚). Consider a uniform grid {𝑡𝑡𝑠𝑠 = 𝑎𝑎 + 𝑠𝑠ℎ ∶ 𝑠𝑠 = 0,1,2, … , 𝑁𝑁, ℎ = (𝑏𝑏 − 𝑎𝑎)⁄𝑝𝑝𝑝𝑝 , (𝑝𝑝, 𝑁𝑁 ∈ ℤ+)} so to calculate 𝑢𝑢(𝑡𝑡 − 𝜏𝜏ℓ ) at 𝑡𝑡 = 𝑡𝑡𝑠𝑠 for all ℓ = 0,1,2, … (number of constant time delays {(𝜏𝜏 = 𝜏𝜏0 ), 𝜏𝜏1 , 𝜏𝜏2 , 𝜏𝜏3 , … }). Thus for any ( ℓ ∈ ℤ+ ∪ {0} ) we have: 𝑡𝑡𝑠𝑠 𝜏𝜏ℓ 𝑢𝑢(𝑡𝑡 − 𝜏𝜏ℓ )|𝑡𝑡=𝑡𝑡 𝑠𝑠 = 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏ℓ ) = 𝑢𝑢([𝑡𝑡𝑠𝑠 − 𝜏𝜏ℓ ]ℎ ) = 𝑢𝑢 �� − � ℎ� ℎ ℎ 𝜏𝜏ℓ 𝑡𝑡𝑠𝑠 1 𝜏𝜏ℓ 𝑡𝑡𝑠𝑠 𝑡𝑡𝑠𝑠 [𝜏𝜏ℓ ]ℎ = 𝑢𝑢 �� − � ℎ� = 𝑢𝑢 �� − � � ℎ� ℎ� = 𝑢𝑢 �� − � �� ℎ� ℎ ℎ ℎ ℎ ℎ ℎ ℎ 𝜏𝜏
Putting 𝑟𝑟̅ℓ = � ℓ� , for all ℓ = 0,1,2, … so on the other hand, we obtain: ℎ
𝑡𝑡𝑠𝑠−𝑟𝑟̅ℓ 𝑎𝑎 + (𝑠𝑠 − 𝑟𝑟̅ℓ )ℎ 𝑡𝑡𝑠𝑠 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏ℓ ) = 𝑢𝑢 �� − 𝑟𝑟̅ℓ � ℎ� = 𝑢𝑢 �� � ℎ� = 𝑢𝑢 �� � ℎ� ℎ ℎ ℎ = 𝑢𝑢�[𝑡𝑡𝑠𝑠−𝑟𝑟̅ℓ ]ℎ � = 𝑢𝑢�𝑡𝑡𝑠𝑠−𝑟𝑟̅ℓ �
Thus: 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏ℓ ) = 𝑢𝑢�𝑡𝑡𝑠𝑠−𝑟𝑟̅ℓ � , ℓ = 0,1,2, …; and note that:
𝑢𝑢(𝑡𝑡𝑘𝑘 ) = 𝜑𝜑(𝑡𝑡𝑘𝑘 ) for 𝑘𝑘 = −𝑟𝑟̅ℓ , −𝑟𝑟̅ℓ + 1, −𝑟𝑟̅ℓ + 2, … , −1,0
The above expression can be explained by giving the following example: 34
… (2.8)
Chapter Two
Block By Block Method
Take the closed bounded interval [𝑎𝑎, 𝑏𝑏] = [0,1] , 𝑁𝑁 = 10 with constant delay terms { 𝜏𝜏 = 𝜏𝜏0 = 0.1 , 𝜏𝜏1 = 0.34 , 𝜏𝜏2 = 0.25 } and 𝑝𝑝 = 2 (For Two Block) since by the equation (1.32) we have the initial condition 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 (is given) and ����� 2� = 0.34 . Thus 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) is also defined for all 𝜏𝜏𝑚𝑚𝑚𝑚𝑚𝑚 = max�𝜏𝜏, 𝜏𝜏𝑗𝑗 : 𝑗𝑗 = 1: 𝑡𝑡 ∈ [−0.34,0] .
So the step sizes for two blocks (𝑝𝑝 = 2) and 𝑁𝑁 = 10, [𝑎𝑎, 𝑏𝑏] = [0,1] have the ℎ = (𝑏𝑏 − 𝑎𝑎)/2𝑁𝑁 = 1/20 = 0.05 .Thus 𝜏𝜏
0.1
For ℓ = 0 (𝜏𝜏 = 𝜏𝜏0 = 0.1) , 𝑟𝑟̅0 = � 0 � = � � = 2, that is 𝑢𝑢(𝑡𝑡𝑘𝑘 ) = 𝜑𝜑(𝑡𝑡𝑘𝑘 ) for all 0.05 ℎ 𝑘𝑘 = −2, −1,0 . and 𝑢𝑢(𝑡𝑡−2 ) = 𝜑𝜑(𝑡𝑡−2 ) = 𝜑𝜑(𝑎𝑎 − 2ℎ) = 𝜑𝜑(−0.1) , is known from historical form 𝜑𝜑 because −0.1 lies in the interval [−0.34,0] which defines 𝜑𝜑 on it. Also, 𝑢𝑢(𝑡𝑡−1 ) = 𝜑𝜑(𝑡𝑡−1 ) = 𝜑𝜑(𝑎𝑎 − ℎ) = 𝜑𝜑(−0.05) , is known from historical form 𝜑𝜑. 𝑢𝑢(𝑡𝑡0 ) = 𝜑𝜑(𝑡𝑡0 ) = 𝑢𝑢0 , is given from initial conditions. So the computation of 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏0 ) for all 𝑠𝑠 = 0,1,2, … 𝑁𝑁 can be given as follows: 𝑠𝑠 = 0 , 𝑢𝑢(𝑡𝑡0 − 𝜏𝜏0 ) = 𝑢𝑢(𝑡𝑡−2 ) � … (𝑖𝑖) ⎧ 𝑠𝑠 = 1, 𝑢𝑢(𝑡𝑡1 − 𝜏𝜏0 ) = 𝑢𝑢(𝑡𝑡−1 ) ⎪ 𝑠𝑠 = 2, 𝑢𝑢(𝑡𝑡2 − 𝜏𝜏0 ) = 𝑢𝑢(𝑡𝑡0 )} … (𝑖𝑖𝑖𝑖) 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏0 ) = ⎨ 𝑠𝑠 = 3, 𝑢𝑢(𝑡𝑡3 − 𝜏𝜏0 ) = 𝑢𝑢(𝑡𝑡1 ) � … (𝑖𝑖𝑖𝑖𝑖𝑖) ⎪ ⋮ ⋮ ⎩𝑠𝑠 = 10, 𝑢𝑢(𝑡𝑡10 − 𝜏𝜏0 ) = 𝑢𝑢(𝑡𝑡8 )
Part 𝑖𝑖 is known from historical function, part 𝑖𝑖𝑖𝑖 is given from initial condition and part 𝑖𝑖𝑖𝑖𝑖𝑖 must be found by any calculation techniques. 𝜏𝜏
0.34
For ℓ = 1 (𝜏𝜏 = 𝜏𝜏1 = 0.34) , 𝑟𝑟̅1 = � 1 � = � � = 6 . Thus 𝑢𝑢(𝑡𝑡𝑘𝑘 ) = 𝜑𝜑(𝑡𝑡𝑘𝑘 ) for all 0.05 ℎ 𝑘𝑘 = −6, −5, −4, −3, −2, −1,0. that is 𝑢𝑢(𝑡𝑡−6 ) = 𝜑𝜑(𝑡𝑡−6 ) = 𝜑𝜑(𝑎𝑎 − 6ℎ) = 𝜑𝜑(−0.3), is known from historical form 𝜑𝜑 because −0.3 lies in [−0.34,0] which defines 𝜑𝜑 on it. Also for all 𝑢𝑢(𝑡𝑡−5 ), 𝑢𝑢(𝑡𝑡−4 ), 𝑢𝑢(𝑡𝑡−3 ), 𝑢𝑢(𝑡𝑡−2 ) and 𝑢𝑢(𝑡𝑡−1 ), and, 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 , can be put for all 𝑠𝑠 = 0,1,2, … , 𝑁𝑁: 𝑠𝑠 = 0,1,2, … 5; 𝑢𝑢(𝑡𝑡𝑠𝑠−6 ) = 𝜑𝜑(𝑎𝑎 − (𝑠𝑠 − 6)ℎ) 𝑢𝑢(𝑡𝑡6 − 𝜏𝜏1 ) = 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏1 ) = � 𝑠𝑠 = 6, 𝑠𝑠 > 6, 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏1 ) = 𝑢𝑢(𝑡𝑡𝑠𝑠−6 ) Part 𝑖𝑖 is known from historical function, part 𝑖𝑖𝑖𝑖 is given from initial part 𝑖𝑖𝑖𝑖𝑖𝑖 must be found by any calculation techniques. 𝜏𝜏
For ℓ = 2 (𝜏𝜏2 = 0.25) , 𝑟𝑟̅2 = � 2 � = � ℎ
0.25 0.05
� = 5. Thus:
condition and
𝑢𝑢(𝑡𝑡𝑘𝑘 ) = 𝜑𝜑(𝑡𝑡𝑘𝑘 ) for all
𝑘𝑘 = −5, −4, −3, −2, −1,0; that is by same techniques before; we get: 35
… (𝑖𝑖) … (𝑖𝑖𝑖𝑖) … (𝑖𝑖𝑖𝑖𝑖𝑖)
Chapter Two
Block By Block Method
𝑠𝑠 = 0,1,2, … 4; 𝑢𝑢(𝑡𝑡𝑠𝑠−5 ) = 𝜑𝜑(𝑎𝑎 − (𝑠𝑠 − 5)ℎ) … (𝑖𝑖) 𝑢𝑢(𝑡𝑡5 − 𝜏𝜏2 ) = 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏2 ) = � 𝑠𝑠 = 5, … (𝑖𝑖𝑖𝑖) … (𝑖𝑖𝑖𝑖𝑖𝑖) 𝑠𝑠 > 5, 𝑢𝑢(𝑡𝑡𝑠𝑠 − 𝜏𝜏2 ) = 𝑢𝑢(𝑡𝑡𝑠𝑠−5 )
Part 𝑖𝑖 is known from historical function, part 𝑖𝑖𝑖𝑖 is given from initial condition and part 𝑖𝑖𝑖𝑖𝑖𝑖 must be found by any calculation techniques; for all 𝑠𝑠 = ����� 0: 𝑁𝑁.
2.5 Solution Method for Higher-Fractional Order Linear VIDEs of Constant Multi-Time Retarded Delay: In this section, a new technique for solving linear Volterra integro-fractional differential equations in Caputo sense with constant multi-time Retarded delay applying Block-by-Block methods of difference block orders (Two, Three and Four) with the aid of finite difference approximation are presented. Recall (1.32) ����� for all 0 < 𝛼𝛼𝑖𝑖 (𝑖𝑖 = 1: 𝑛𝑛) ≤ 1 : 𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝑢𝑢(𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡)𝑢𝑢(𝑡𝑡 − 𝜏𝜏) 𝑖𝑖=1
𝑚𝑚
𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝑢𝑢(𝑥𝑥 − 𝜏𝜏𝑗𝑗 )𝑑𝑑𝑑𝑑
… (2.9)
𝑗𝑗 =1 𝑎𝑎
Under the following initial condition with given continuity historical function: 𝑢𝑢(𝑡𝑡0 = 𝑎𝑎) = 𝑢𝑢0 and 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) 1: 𝑚𝑚� 𝜑𝜑 ∈ 𝐶𝐶([𝑎𝑎�, 𝑎𝑎]; ℝ) for 𝑎𝑎� = 𝑎𝑎 − max �𝜏𝜏, 𝜏𝜏𝑗𝑗 : 𝑗𝑗 = ������
Now, consider the solution of (2.9) in the range 𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏 with 𝑏𝑏 − 𝑎𝑎 = 𝑁𝑁𝑁𝑁ℎ that is, we divide the interval [𝑎𝑎, 𝑏𝑏] in to 𝑁𝑁 equal intervals, each of which is then divided in to 𝑝𝑝-subinterval on length ℎ. For 𝑝𝑝𝑝𝑝 < 𝑘𝑘 ≤ 𝑝𝑝(𝑟𝑟 + 1) ; 𝑝𝑝 is some positive integer; 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 and set 𝑡𝑡 = 𝑡𝑡𝑘𝑘 we may write: 𝛼𝛼 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝑢𝑢(𝑡𝑡)�𝑡𝑡=𝑡𝑡 𝑘𝑘
𝑛𝑛 −1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡𝑘𝑘 ) � 𝑎𝑎𝑐𝑐 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝑢𝑢(𝑡𝑡)�𝑡𝑡=𝑡𝑡 + 𝑃𝑃0 (𝑡𝑡𝑘𝑘 ) 𝑢𝑢(𝑡𝑡𝑘𝑘 − 𝜏𝜏) 𝑖𝑖=1
𝑡𝑡 𝑝𝑝𝑝𝑝 𝑚𝑚
= 𝑓𝑓(𝑡𝑡𝑘𝑘 ) + 𝜆𝜆 � 𝑡𝑡 𝑘𝑘 𝑚𝑚
𝑎𝑎
𝑘𝑘
� 𝒦𝒦𝑗𝑗 (𝑡𝑡𝑘𝑘 , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
+ 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡𝑘𝑘 , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑡𝑡 𝑝𝑝𝑝𝑝 𝑗𝑗 =1
36
… (2.10)
Chapter Two
Block By Block Method
Now, by applying the block-stage idea at each sub-interval, we must compute the equation (2.10) at each point 𝑡𝑡𝑘𝑘 = 𝑡𝑡𝑝𝑝𝑝𝑝 +1 , 𝑡𝑡𝑝𝑝𝑝𝑝 +2 , 𝑡𝑡𝑝𝑝𝑝𝑝 +3 , … , 𝑡𝑡𝑝𝑝𝑝𝑝 +𝑝𝑝 = 𝑡𝑡𝑝𝑝(𝑟𝑟+1) for 𝑝𝑝-blocks order at each subinterval �𝑡𝑡𝑝𝑝𝑝𝑝 , 𝑡𝑡𝑝𝑝(𝑟𝑟+1) � . Moreover, by applying the proposition (2.2) for all fractional parts in the left hand sides of our equation at all points 𝑡𝑡 = 𝑡𝑡𝑝𝑝𝑝𝑝 +𝜗𝜗 for 𝜗𝜗 = 1,2,3, … , 𝑝𝑝 and all 𝑟𝑟 = 0,1,2,3, … , 𝑁𝑁 − 1, we obtain: 𝑝𝑝𝑝𝑝 +𝜗𝜗 −1
ℎ−𝛼𝛼 𝑛𝑛 𝛼𝛼 ��𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 − 𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −1 � + � �𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −𝑠𝑠 − 𝑢𝑢𝑝𝑝𝑝𝑝 −𝑠𝑠 �𝑏𝑏𝑠𝑠 𝑛𝑛 � Γ(2 − 𝛼𝛼𝑛𝑛 ) 𝑠𝑠=1
𝑛𝑛 −1
ℎ−𝛼𝛼 𝑛𝑛 −𝑖𝑖 + � 𝑃𝑃𝑖𝑖 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ) ��𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 − 𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −1 � Γ(2 − 𝛼𝛼𝑛𝑛 −𝑖𝑖 ) 𝑖𝑖=1 𝑝𝑝𝑝𝑝 +𝜗𝜗 −1
𝛼𝛼
+ � �𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −𝑠𝑠 − 𝑢𝑢𝑝𝑝𝑝𝑝 −𝑠𝑠 �𝑏𝑏𝑠𝑠 𝑛𝑛 −𝑖𝑖 � 𝑠𝑠=1
+ 𝑃𝑃0 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ) 𝑢𝑢�𝑡𝑡𝑝𝑝𝑝𝑝 +𝜗𝜗 − 𝜏𝜏� 𝑡𝑡 𝑝𝑝𝑝𝑝
𝑚𝑚
= 𝑓𝑓(𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 �𝑡𝑡𝑝𝑝𝑝𝑝 +𝜗𝜗 , 𝑥𝑥� 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑡𝑡 𝑝𝑝𝑝𝑝 +𝜗𝜗
𝑚𝑚
𝑎𝑎
𝑗𝑗 =1
+ 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡𝑝𝑝𝑝𝑝 +𝜗𝜗 , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑡𝑡 𝑝𝑝𝑝𝑝
𝑗𝑗 =1
… (2.11)
Now if we use the following quadrature rules to approximate the integral terms in equation (2.11) and applying the point notes about the delays term as a section (2.4), we obtain: 𝑎𝑎+𝑝𝑝𝑝𝑝 ℎ
�
𝑎𝑎
and
𝑝𝑝𝑝𝑝
𝒦𝒦𝑗𝑗 �𝑡𝑡𝑝𝑝𝑝𝑝 +𝜗𝜗 , 𝑥𝑥� 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 = ℎ � 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ, 𝑥𝑥ℓ ) 𝑢𝑢�𝑥𝑥ℓ − 𝜏𝜏𝑗𝑗 � ℓ=0
𝑎𝑎+(𝑝𝑝𝑝𝑝 +𝜗𝜗 )ℎ
�
𝑎𝑎 +𝑝𝑝𝑝𝑝 ℎ
𝑝𝑝𝑝𝑝
= ℎ � 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ, 𝑎𝑎 + ℓℎ)𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 ℓ=0
𝑝𝑝𝑝𝑝 +𝜗𝜗
�ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ , 𝑎𝑎 + ℓℎ)𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝒦𝒦𝑗𝑗 �𝑡𝑡𝑝𝑝𝑝𝑝 +𝜗𝜗 , 𝑥𝑥� 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 = ℎ � 𝑤𝑤 ℓ=𝑝𝑝𝑝𝑝
37
Chapter Two
Block By Block Method 𝜏𝜏 𝑗𝑗
Where 𝑟𝑟̅𝑗𝑗 = � � and 𝑤𝑤ℓ with 𝑤𝑤 � ℓ depend on the quadrature rule used, here we use Simpson’s
1 3
ℎ
ℎ with adative Simpson rules as in section (2.3), respectively. Noting
that the first integral vanishes at 𝑟𝑟 = 0, only, while the second integral is estimated by knowing the values of integrand at 𝑎𝑎 + 𝑝𝑝𝑝𝑝ℎ, 𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 1)ℎ, … , 𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ . that is the equation (2.11) becomes: 𝑝𝑝𝑝𝑝 +𝜗𝜗 −1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟𝜗𝜗 𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 − 𝐻𝐻𝑟𝑟𝜗𝜗 𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −1 + � ��𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −𝑠𝑠 − 𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −(𝑠𝑠+1) � �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,𝜗𝜗 ��
= 𝑓𝑓(𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ) − 𝑃𝑃0 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ) 𝑢𝑢𝑝𝑝𝑝𝑝 +𝜗𝜗 −𝑟𝑟̅0 𝑝𝑝𝑝𝑝
𝑚𝑚
ℓ=0
𝑗𝑗 =1
+ 𝜆𝜆ℎ � 𝑤𝑤ℓ � 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ , 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝑝𝑝𝑝𝑝 +𝜗𝜗
𝑚𝑚
+ 𝜆𝜆ℎ � 𝑤𝑤 � ℓ � 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗
For all
ℓ=𝑝𝑝𝑝𝑝
𝑗𝑗 =1
𝜗𝜗 = 1,2, … , 𝑝𝑝, 𝑝𝑝 is a order of the Blocks, 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 𝜏𝜏
(𝑁𝑁 ∈ ℤ+) and 𝑟𝑟̅𝑗𝑗 = � 𝑗𝑗 � for ℎ = ℎ
𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗
… (2.12)
=
−𝛼𝛼 𝑛𝑛 ⎧ ℎ ⎪ Γ(2 − 𝛼𝛼𝑛𝑛 )
𝑏𝑏−𝑎𝑎
∀ 𝑗𝑗 = 0,1,2, … , 𝑚𝑚 . Where
𝑁𝑁𝑁𝑁
if
𝑖𝑖 = 𝑛𝑛
−𝛼𝛼 ⎨𝑃𝑃 (𝑎𝑎 + (𝑝𝑝𝑝𝑝 + 𝜗𝜗)ℎ) ℎ 𝑛𝑛 −𝑖𝑖 ���������� otherwise 𝑖𝑖 = 1: 𝑛𝑛 − 1 ⎪ 𝑖𝑖 ) Γ(2 − 𝛼𝛼 𝑛𝑛−𝑖𝑖 ⎩
for all 𝑠𝑠 = 1,2, … , 𝑝𝑝𝑝𝑝 + 𝜗𝜗 − 1, and 𝑖𝑖 = 1,2, … , 𝑛𝑛: 𝐵𝐵𝑠𝑠𝑖𝑖
𝛼𝛼
𝑏𝑏𝑠𝑠 𝑛𝑛 = � 𝛼𝛼 𝑛𝑛 −𝑖𝑖 𝑏𝑏𝑠𝑠
if 𝑖𝑖 = 𝑛𝑛 otherwise
𝑛𝑛
𝑖𝑖 𝐻𝐻𝑟𝑟𝜗𝜗 = � 𝐶𝐶𝑟𝑟,𝜗𝜗 𝑖𝑖=1
for all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 and 𝜗𝜗 = 1,2, … , 𝑝𝑝.
… (2.13)
… (2.14) … (3.15)
For the choice to the 𝑝𝑝-order Block-by-Block method in equation (2.12), the following particular methods are given: 38
Chapter Two
Block By Block Method
2.5.1 Two Block Method: The general procedure for contracting two block method 𝑝𝑝 = 2, and for all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1, the integration over [𝑎𝑎, 𝑡𝑡2𝑟𝑟 ] can be accomplished by Simpson’s rule and the integral over each interval [𝑡𝑡2𝑟𝑟 , 𝑡𝑡2𝑟𝑟+1 ] approximate the integrand by quadratic interpolation polynomial at points 𝑡𝑡2𝑟𝑟 , 𝑡𝑡2𝑟𝑟+1 , 𝑡𝑡2𝑟𝑟+2 while the integral over �𝑡𝑡2𝑟𝑟 , 𝑡𝑡2(𝑟𝑟+1) � using Simpson’s rule also. Then through the use of equation (2.11) with equations (2.13-15), and motivated by our earlier derivation as in section (2.3), the useful past values of 𝑢𝑢 (which are needed) can be evaluated from the two equations (2.8) and The equation (2.9) can be written as:
First block
2𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢2𝑟𝑟+1 − 𝐻𝐻𝑟𝑟1 𝑢𝑢2𝑟𝑟 + � �[𝑢𝑢2𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢2𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
+ 𝑃𝑃0 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ) 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (2𝑟𝑟 + 1)ℎ) 𝑎𝑎+2𝑟𝑟ℎ 𝑚𝑚
+𝜆𝜆 � + 𝜆𝜆
Second block
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑
𝑗𝑗 =1 𝑎𝑎 𝑎𝑎+(2𝑟𝑟+1)ℎ 𝑚𝑚
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 … (2.16)
�
𝑗𝑗 =1
𝑎𝑎+2𝑟𝑟ℎ
2𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢2𝑟𝑟+2 − 𝐻𝐻𝑟𝑟2 𝑢𝑢2𝑟𝑟+1 + � �[𝑢𝑢2𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢2𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
+ 𝑃𝑃0 (𝑎𝑎 + (2𝑟𝑟 + 2)ℎ) 𝑢𝑢2𝑟𝑟+2−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (2𝑟𝑟 + 2)ℎ) 𝑎𝑎+2𝑟𝑟ℎ 𝑚𝑚
+𝜆𝜆 � + 𝜆𝜆
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 2)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑
𝑗𝑗 =1 𝑎𝑎 𝑎𝑎+(2𝑟𝑟+2)ℎ 𝑚𝑚
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 2)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 … (2.17)
�
𝑗𝑗 =1
𝑎𝑎+2𝑟𝑟ℎ
It will be noticed that the first integral in (2.16) with all integrals in (2.17) can be computed by Simpson’s
1 3
ℎ rule as in equation (2.3) and the remaining integral in 39
Chapter Two
Block By Block Method
(2.16) can be solved by the formula derived by using adaptive Simpson’s rule, that is, combine the equation (2.4) and (2.5), for all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 and 𝑗𝑗 = 1,2, … , 𝑚𝑚 : 𝑎𝑎+(2𝑟𝑟+1)ℎ
�
𝑎𝑎+2𝑟𝑟ℎ
𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑
ℎ �𝒦𝒦 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥2𝑟𝑟 ) 𝑢𝑢2𝑟𝑟−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 + 4𝒦𝒦𝑗𝑗 �𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥2𝑟𝑟+1/2 � 𝑢𝑢2𝑟𝑟+1/2−𝑟𝑟̅𝑗𝑗
≅
+ 𝒦𝒦𝑗𝑗 �𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥2𝑟𝑟+1 ) 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 ��
Now 𝑢𝑢2𝑟𝑟+1/2−𝑟𝑟̅𝑗𝑗 , for each 𝑗𝑗 and 𝑟𝑟, is not known, but it may be estimated by
quadratic interpolation as in equation (2.5), thus: 𝑢𝑢2𝑟𝑟+1/2−𝑟𝑟̅𝑗𝑗 ≅
3 3 1 𝑢𝑢2𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢2𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 8 4 8
For the First Block at (𝟐𝟐𝟐𝟐 + 𝟏𝟏) for 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 , becomes: 2𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢2𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢2𝑟𝑟 − � �[𝑢𝑢2𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢2𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
− 𝑃𝑃0 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ)𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅0 + 𝑓𝑓2𝑟𝑟+1 𝑚𝑚
2𝑟𝑟
ℎ + � � 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥ℓ ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0 𝑚𝑚
ℎ + � �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥2𝑟𝑟 )𝑢𝑢2𝑟𝑟−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 =1
3 1 3 +4𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥2𝑟𝑟+1/2 ) ∗ � 𝑢𝑢2𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢2𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 � 4 8 8
+𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑥𝑥2𝑟𝑟+1 ) 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 �
40
… (2.18)
Chapter Two
Block By Block Method
For the Second Block at (𝟐𝟐𝟐𝟐 + 𝟐𝟐) for 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 , becomes: 2𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢2𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢2𝑟𝑟+1 − � �[𝑢𝑢2𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢2𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
− 𝑃𝑃0 (𝑎𝑎 + (2𝑟𝑟 + 2)ℎ)𝑢𝑢2𝑟𝑟+2−𝑟𝑟̅0 + 𝑓𝑓2𝑟𝑟+2 𝑚𝑚 2𝑟𝑟+2
ℎ � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 2)ℎ , 𝑥𝑥ℓ ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 + � � 𝑤𝑤 3 𝑗𝑗 =1 ℓ=0
where for all 𝜗𝜗 = 1 and 2 with 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 : 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗
=
−𝛼𝛼 𝑛𝑛 ⎧ ℎ ⎪ Γ(2 − 𝛼𝛼𝑛𝑛 )
if
𝑖𝑖 = 𝑛𝑛
−𝛼𝛼 ⎨𝑃𝑃 (𝑎𝑎 + (2𝑟𝑟 + 𝜗𝜗)ℎ) ℎ 𝑛𝑛 −𝑖𝑖 otherwise 𝑖𝑖 = ���������� 1: 𝑛𝑛 − 1 ⎪ 𝑖𝑖 ) Γ(2 − 𝛼𝛼 𝑛𝑛−𝑖𝑖 ⎩ 𝑛𝑛
𝑖𝑖 𝐻𝐻𝑟𝑟𝜗𝜗 = � 𝐶𝐶𝑟𝑟,𝜗𝜗 𝑖𝑖=1
for all 𝑠𝑠 = 1,2,3, … ,2𝑟𝑟 + 𝜗𝜗 − 1 and 𝑖𝑖 = 1,2, … , 𝑛𝑛: 𝐵𝐵𝑠𝑠𝑖𝑖
and
𝛼𝛼
𝑖𝑖 = 𝑛𝑛 otherwise
𝑏𝑏𝑠𝑠 𝑛𝑛 = � 𝛼𝛼 𝑛𝑛 −𝑖𝑖 𝑏𝑏𝑠𝑠
𝑤𝑤ℓ = 3 − (−1)ℓ 𝑤𝑤 � ℓ = 3 − (−1)ℓ
where where
… (2.19)
… (2.20) … (2.21)
… (2.22)
ℓ = 1,2,3, … ,2𝑟𝑟 − 1 ℓ = 1,2,3, … ,2𝑟𝑟 + 1
Algorithm of Two Block Method (A2BM): The approximate solution for linear Volterra integro-fractional differential equations with constant multi-time Retarded delays (2.9) by applying two block procedure with the aid of finite difference approximation can be summarized by the following steps: Step1: i. ii. iii.
Set 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 [given from (2.9)] and input the historical function 𝜑𝜑(𝑡𝑡). Let ℎ = (𝑏𝑏 − 𝑎𝑎)/2𝑁𝑁 , 𝑁𝑁 ∈ ℤ+. 𝜏𝜏 𝑗𝑗 Evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2, … 𝑚𝑚 where 𝜏𝜏0 = 𝜏𝜏. ℎ
41
Chapter Two
Block By Block Method
Step2: for all 𝜗𝜗 = 1 and 2 with all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 : i.
ii.
𝑖𝑖 ����� Calculate 𝐶𝐶𝑟𝑟,𝜗𝜗 , 𝑖𝑖 = 1: 𝑛𝑛 , and 𝐻𝐻𝑟𝑟𝜗𝜗 from the equations (2.20) and (2.21). ����� Compute 𝐵𝐵𝑠𝑠𝑖𝑖 , 𝑖𝑖 = 1: 𝑛𝑛 , for all 𝑠𝑠 = 1,2, … ,2𝑟𝑟 + 𝜗𝜗 − 1 from the equation
(2.22). iii. Evaluate 𝐹𝐹𝑟𝑟𝜗𝜗 from: 𝐹𝐹𝑟𝑟𝜗𝜗 = 𝑓𝑓(𝑎𝑎 + (2𝑟𝑟 + 𝜗𝜗)ℎ) − 𝑃𝑃0 (𝑎𝑎 + (2𝑟𝑟 + 𝜗𝜗)ℎ)𝑢𝑢2𝑟𝑟+𝜗𝜗 −𝑟𝑟̅0 For all 𝒓𝒓 = 𝟎𝟎, 𝟏𝟏, 𝟐𝟐, … , 𝑵𝑵 − 𝟏𝟏 , doing steps 3-6: Step3: for the first block all 𝜗𝜗 = 1 calculate: i.
ii. iii.
𝑛𝑛 1 𝑖𝑖 𝑖𝑖 𝐴𝐴1 [𝑟𝑟] = − ∑2𝑟𝑟 𝑠𝑠=1�[𝑢𝑢2𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢2𝑟𝑟−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,1 �� , where 𝐴𝐴 [0] = 0
𝑄𝑄11 [𝑟𝑟] =
ℎ
2𝑟𝑟 ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 where
3 1 [0] =0. 𝑄𝑄1 ℎ 𝑄𝑄21 [𝑟𝑟] = �∑𝑚𝑚 𝑗𝑗 =1 𝒦𝒦𝑗𝑗 (𝑎𝑎 6
+ (2𝑟𝑟 + 1)ℎ , 𝑎𝑎 + 2𝑟𝑟ℎ)𝑢𝑢2𝑟𝑟−𝑟𝑟̅𝑗𝑗 1
3
3
1
+4 ∑𝑚𝑚 𝑗𝑗 =1 𝒦𝒦𝑗𝑗 �𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑎𝑎 + �2𝑟𝑟 + � ℎ� � 𝑢𝑢2𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢2𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 � 2
8
4
8
+ ∑𝑚𝑚 𝑗𝑗 =1 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 1)ℎ , 𝑎𝑎 + (2𝑟𝑟 + 1)ℎ) 𝑢𝑢2𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 �
Step4: Compute 𝑢𝑢2𝑟𝑟+1 = 𝑢𝑢(𝑡𝑡2𝑟𝑟+1 ), from:
𝐻𝐻𝑟𝑟1 𝑢𝑢2𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢2𝑟𝑟 + 𝐴𝐴1 [𝑟𝑟] + 𝑄𝑄11 [𝑟𝑟] + 𝑄𝑄21 [𝑟𝑟] + 𝐹𝐹𝑟𝑟1
Step5: for the second block 𝜗𝜗 = 2 calculate: i.
ii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴2 [𝑟𝑟] = − ∑2𝑟𝑟+1 𝑠𝑠=1 �[𝑢𝑢2𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢2𝑟𝑟+1−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,2 ��
𝑄𝑄12 [𝑟𝑟] =
ℎ
3
2𝑟𝑟+2 ∑𝑚𝑚 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (2𝑟𝑟 + 2)ℎ , 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤
Step6: Compute 𝑢𝑢2𝑟𝑟+2 = 𝑢𝑢(𝑡𝑡2𝑟𝑟+2 ) , from: With putting:
𝐻𝐻𝑟𝑟2 𝑢𝑢2𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢2𝑟𝑟+1 + 𝐴𝐴2 [𝑟𝑟] + 𝑄𝑄12 [𝑟𝑟] + 𝑄𝑄22 [𝑟𝑟] + 𝐹𝐹𝑟𝑟2
a. 𝑏𝑏𝑠𝑠𝛼𝛼 = (𝑠𝑠 + 1)1−𝛼𝛼 − 𝑠𝑠1−𝛼𝛼 ; 𝑠𝑠 = 0,1,2, … b. 𝑤𝑤ℓ = 3 − (−1)ℓ where ℓ = 1,2,3, … ,2𝑟𝑟 − 1 where ℓ = 1,2,3, … ,2𝑟𝑟 + 1 𝑤𝑤 � ℓ = 3 − (−1)ℓ
42
Chapter Two
Block By Block Method
2.5.2 Three Block Method: Here we take the order of block-by-block method 𝑝𝑝 = 3. then through the use of equation (2.10), the integration becomes over the sub-intervals [𝑎𝑎, 𝑡𝑡3𝑟𝑟 ] and [𝑡𝑡3𝑟𝑟 , 𝑡𝑡𝑘𝑘 ] where 𝑘𝑘 = 3𝑟𝑟 + 1, 3𝑟𝑟 + 2, 3𝑟𝑟 + 3 for all 𝑟𝑟 = 0,1, … , 𝑁𝑁 − 1. now equation (2.11), with using the notification of equations (2.13-15), can be written as: First block 3𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟+1 − 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟 + � �[𝑢𝑢3𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢3𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
+𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ) 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 1)ℎ) 𝑎𝑎+3𝑟𝑟ℎ 𝑚𝑚
+𝜆𝜆 � 𝑎𝑎
+𝜆𝜆
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
𝑎𝑎+(3𝑟𝑟+1)ℎ 𝑚𝑚
�
𝑎𝑎 +3𝑟𝑟ℎ
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
… (2.23)
Second block 3𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+2 − 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+1 + � �[𝑢𝑢3𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
+𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ) 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 2)ℎ) 𝑎𝑎+3𝑟𝑟ℎ 𝑚𝑚
+𝜆𝜆 � 𝑎𝑎
+𝜆𝜆
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
𝑎𝑎+(3𝑟𝑟+2)ℎ 𝑚𝑚
�
𝑎𝑎+3𝑟𝑟ℎ
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
43
… (2.24)
Chapter Two
Block By Block Method
Third block 3𝑟𝑟+2
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+3 − 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+2 + � �[𝑢𝑢3𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+2−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,3 ��
+𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ) 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 3)ℎ) 𝑎𝑎+3𝑟𝑟ℎ 𝑚𝑚
+𝜆𝜆 � 𝑎𝑎
+𝜆𝜆
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
𝑎𝑎+(3𝑟𝑟 +3)ℎ 𝑚𝑚
�
𝑎𝑎+3𝑟𝑟ℎ
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
… (2.25)
After some simple manipulation as in the section (2.5.1) for two blocks method, the 1 solution of integrations depends on quadrature formula: Simpson’s ℎ rule (2.3), 3 adaptive Simpson’s rule (2.4) and quadratic interpolation formula (2.6) using point’s 𝑡𝑡3𝑟𝑟 , 𝑡𝑡3𝑟𝑟+1 , 𝑡𝑡3𝑟𝑟+2 , 𝑡𝑡3𝑟𝑟 +3 . Motivated by our earlier works as in section (2.3), the useful past values of 𝑢𝑢 can be evaluated from the set of equations in (2.8). In practice, the value of 𝑟𝑟 divides each block into two parts. Therefore the resulting equations are: • For 𝒓𝒓 is even : For the First block at (𝟑𝟑𝟑𝟑 + 𝟏𝟏) becomes: 3𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟 − � �[𝑢𝑢3𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢3𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
− 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ)𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 1)ℎ) 𝑚𝑚
3𝑟𝑟
ℎ + � � 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0
𝑚𝑚
ℎ + � �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + 3𝑟𝑟ℎ)𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 =1
1 +4𝒦𝒦𝑗𝑗 �𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + �3𝑟𝑟 + � ℎ� 2
15 5 1 5 ∗ � 𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 16 16 16 16 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 1)ℎ)𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 � 44
… (2.26)
Chapter Two
Block By Block Method
For the Second block at (𝟑𝟑𝟑𝟑 + 𝟐𝟐) becomes: 3𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+1 − � �[𝑢𝑢3𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
− 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ)𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 2)ℎ) 𝑚𝑚 3𝑟𝑟+2
ℎ + � � 𝑤𝑤 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3
… (2.27)
𝑗𝑗 =1 ℓ=0
For the Third block at (𝟑𝟑𝟑𝟑 + 𝟑𝟑) becomes: 3𝑟𝑟+2
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+3 = 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+2 − � �[𝑢𝑢3𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+2−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,3 ��
− 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ)𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 3)ℎ) 𝑚𝑚 3𝑟𝑟+2
ℎ � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 + � � 𝑤𝑤 3 𝑗𝑗 =1 ℓ=0 𝑚𝑚
ℎ + � �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 2)ℎ)𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 =1
+ 4𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 5/2)ℎ) 15 5 1 5 ∗ � 𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 16 16 16 16 + 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 3)ℎ)𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 �
… (2.28)
• For 𝒓𝒓 is odd: For the First block at (𝟑𝟑𝟑𝟑 + 𝟏𝟏) becomes: 3𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟 − � �[𝑢𝑢3𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢3𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
− 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ)𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 1)ℎ) 𝑚𝑚 3𝑟𝑟+1
ℎ + � � 𝑍𝑍ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0
45
… (2.29)
Chapter Two
Block By Block Method
For the Second block at (𝟑𝟑𝟑𝟑 + 𝟐𝟐) becomes: 3𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+1 − � �[𝑢𝑢3𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
– 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ) 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 2)ℎ) 𝑚𝑚 3𝑟𝑟+1
ℎ + � � 𝑍𝑍ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0 𝑚𝑚
ℎ + � �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 1)ℎ)𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 =1
3 + 4𝒦𝒦𝑗𝑗 �𝑎𝑎 + (3𝑟𝑟 + 2)ℎ , 𝑎𝑎 + �3𝑟𝑟 + � ℎ � 2 15 5 1 5 ∗ � 𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 16 16 16 16
+𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 2)ℎ)𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 � For the Third block at (𝟑𝟑𝟑𝟑 + 𝟑𝟑) became: 3𝑟𝑟+2
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
… (2.30)
𝑖𝑖 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+3 = 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+2 − � �[𝑢𝑢3𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+2−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,3 ��
− 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ)𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 3)ℎ) 𝑚𝑚 3𝑟𝑟+2
ℎ + � � 𝑍𝑍ℓ̅ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0
Where for all 𝜗𝜗 = 1 , 2 and 3 with 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 : 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗
=
−𝛼𝛼 𝑛𝑛 ⎧ ℎ ⎪ Γ(2 − 𝛼𝛼𝑛𝑛 )
if
𝑖𝑖 = 𝑛𝑛
−𝛼𝛼 ⎨𝑃𝑃 (𝑎𝑎 + (2𝑟𝑟 + 𝜗𝜗)ℎ) ℎ 𝑛𝑛 −𝑖𝑖 ���������� otherwise 𝑖𝑖 = 1: 𝑛𝑛 − 1 ⎪ 𝑖𝑖 ) Γ(2 − 𝛼𝛼 𝑛𝑛 −𝑖𝑖 ⎩ 𝑛𝑛
𝑖𝑖 𝐻𝐻𝑟𝑟𝜗𝜗 = � 𝐶𝐶𝑟𝑟,𝜗𝜗 𝑖𝑖=1
46
… (2.31)
… (2.32) … (2.33)
Chapter Two
Block By Block Method
for all 𝑠𝑠 = 1,2,3, … ,3𝑟𝑟 + 𝜗𝜗 − 1 and 𝑖𝑖 = 1,2, … , 𝑛𝑛: 𝛼𝛼 𝑖𝑖 = 𝑛𝑛 𝑏𝑏𝑠𝑠 𝑛𝑛 𝑖𝑖 𝐵𝐵𝑠𝑠 = � 𝛼𝛼 𝑛𝑛 −𝑖𝑖 𝑏𝑏𝑠𝑠 otherwise
; 𝑤𝑤 �0 = 𝑤𝑤 � 3𝑟𝑟+2 = 1 𝑤𝑤0 = 𝑤𝑤3𝑟𝑟 = 1 �𝑤𝑤ℓ = 3 − (−1)ℓ where ℓ = 1,2,3, … ,3𝑟𝑟 − 1 𝑤𝑤 �ℓ = 3 − (−1)ℓ where ℓ = 1,2,3, … ,3𝑟𝑟 + 1 ̅ ; 𝑍𝑍0̅ = 𝑍𝑍3𝑟𝑟+3 =1 𝑍𝑍0 = 𝑍𝑍3𝑟𝑟+1 = 1 � 𝑍𝑍ℓ = 3 − (−1)ℓ where ℓ = 1,2,3, … ,3𝑟𝑟 ℓ 𝑍𝑍ℓ̅ = 3 − (−1) where ℓ = 1,2,3, … ,3𝑟𝑟 + 2
… (2.34)
Algorithm of Three Block Method (A3BM): The approximate solution for Linear Volterra integro-fractional differential equations with constant multi-time Retarded delay (2.9) by applying three Block method with the aid of finite difference approximation can be summarized by the following steps: Step1: i. ii. iii.
Set 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 [given from (2.9)] and input the historical function 𝜑𝜑(𝑡𝑡). Let ℎ = (𝑏𝑏 − 𝑎𝑎)/3𝑁𝑁 , 𝑁𝑁 ∈ ℤ+. 𝜏𝜏 𝑗𝑗
Evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2, … 𝑚𝑚 where 𝜏𝜏0 = 𝜏𝜏 ℎ
Step2: For all 𝜗𝜗 = 1 , 2 and 3 with all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 : i.
ii. iii.
𝑖𝑖 ����� Calculate 𝐶𝐶𝑟𝑟,𝜗𝜗 , (𝑖𝑖 = 1: 𝑛𝑛) and 𝐻𝐻𝑟𝑟𝜗𝜗 from the equations (2.32) and (2.33) respectively. ����� Compute 𝐵𝐵𝑠𝑠𝑖𝑖 , (𝑖𝑖 = 1: 𝑛𝑛) for all 𝑠𝑠 = 1,2, … ,3𝑟𝑟 + 𝜗𝜗 − 1 from the equation (2.34). Evaluate 𝐹𝐹𝑟𝑟𝜗𝜗 from: 𝐹𝐹𝑟𝑟𝜗𝜗 = 𝑓𝑓(𝑎𝑎 + (3𝑟𝑟 + 𝜗𝜗)ℎ) − 𝑃𝑃0 (𝑎𝑎 + (3𝑟𝑟 + 𝜗𝜗)ℎ)𝑢𝑢3𝑟𝑟+𝜗𝜗 −𝑟𝑟̅0 For all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 if 𝑟𝑟 is even doing the steps (3-8) else doing steps (9-14) doing steps 3-6:
47
Chapter Two
Block By Block Method
Step3: for the first block all 𝜗𝜗 = 1, 𝑟𝑟 − even calculate: i.
ii. iii.
𝑛𝑛 1 𝑖𝑖 𝑖𝑖 𝐴𝐴1 [𝑟𝑟] = − ∑3𝑟𝑟 𝑠𝑠=1�[𝑢𝑢3𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢3𝑟𝑟−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,1 �� , where 𝐴𝐴 [0] = 0 ℎ
3𝑟𝑟 ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤ℓ 3 where 𝑄𝑄11 [0] = 0 . ℎ 𝑄𝑄21 [𝑟𝑟] = ∑𝑚𝑚 �𝒦𝒦𝑗𝑗 (𝑎𝑎 + 6 𝑗𝑗 =1
𝑄𝑄11 [𝑟𝑟] =
𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ , 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗
(3𝑟𝑟 + 1)ℎ , 𝑎𝑎 + 3𝑟𝑟ℎ)𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗
+4𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + (3𝑟𝑟 + 1/2)ℎ) 15 5 1 5 ∗ � 𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 16 16 16 16 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ , 𝑎𝑎 + (3𝑟𝑟 + 1)ℎ) 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 �
Step4: Compute 𝑢𝑢3𝑟𝑟+1 = 𝑢𝑢(𝑡𝑡3𝑟𝑟+1 ) , from:
𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟 + 𝐴𝐴1 [𝑟𝑟] + 𝑄𝑄11 [𝑟𝑟] + 𝑄𝑄21 [𝑟𝑟] + 𝐹𝐹𝑟𝑟1
Step5: For the second block 𝜗𝜗 = 2 calculate: i.
ii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴2 [𝑟𝑟] = − ∑3𝑟𝑟+1 𝑠𝑠=1 �[𝑢𝑢3𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+1−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,2 ��
𝑄𝑄12 [𝑟𝑟] =
ℎ
3
3𝑟𝑟+2 ∑𝑚𝑚 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤
Step6: Compute 𝑢𝑢3𝑟𝑟+2 = 𝑢𝑢(𝑡𝑡3𝑟𝑟+2 ) , from:
𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+1 + 𝐴𝐴2 [𝑟𝑟] + 𝑄𝑄12 [𝑟𝑟] + 𝐹𝐹𝑟𝑟2
Step7: for the third block 𝜗𝜗 = 3 calculate: i.
ii. iii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴3 [𝑟𝑟] = − ∑3𝑟𝑟+2 𝑠𝑠=1 �[𝑢𝑢3𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+2−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,3 ��
𝑄𝑄13 [𝑟𝑟] = 𝑄𝑄23 [𝑟𝑟]
ℎ
3 ℎ
3𝑟𝑟+2 ∑𝑚𝑚 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤
= ∑𝑚𝑚 𝑗𝑗 =1�𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ , 𝑎𝑎 + (3𝑟𝑟 + 2)ℎ)𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 6
5 +4𝒦𝒦𝑗𝑗 �𝑎𝑎 + (3𝑟𝑟 + 3)ℎ , 𝑎𝑎 + �3𝑟𝑟 + � ℎ � 2 15 5 1 5 ∗ � 𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 16 16 16 16 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ , 𝑎𝑎 + (3𝑟𝑟 + 3)ℎ) 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 �
Step8: Compute 𝑢𝑢3𝑟𝑟+3 = 𝑢𝑢(𝑡𝑡3𝑟𝑟+3 ) , from:
𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+3 = 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+2 + 𝐴𝐴3 [𝑟𝑟] + 𝑄𝑄13 [𝑟𝑟] + 𝑄𝑄23 [𝑟𝑟] + 𝐹𝐹𝑟𝑟3 48
Chapter Two
Block By Block Method
Step9: for the first block all 𝜗𝜗 = 1 𝑟𝑟 − odd calculate: i.
ii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴1 [𝑟𝑟] = − ∑3𝑟𝑟 𝑠𝑠=1�[𝑢𝑢3𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢3𝑟𝑟−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,1 ��
𝑄𝑄11 [𝑟𝑟] =
ℎ
3
3𝑟𝑟+1 ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑍𝑍ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗
Step10: Compute 𝑢𝑢3𝑟𝑟+1 = 𝑢𝑢(𝑡𝑡3𝑟𝑟+1 ) , from:
𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢3𝑟𝑟 + 𝐴𝐴1 [𝑟𝑟] + 𝑄𝑄11 [𝑟𝑟] + 𝐹𝐹𝑟𝑟1
Step11: For the second block 𝜗𝜗 = 2 calculate: i.
ii. iii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴2 [𝑟𝑟] = − ∑3𝑟𝑟+1 𝑠𝑠=1 �[𝑢𝑢3𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+1−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,2 ��
𝑄𝑄22 [𝑟𝑟] =
ℎ
3 ℎ
3𝑟𝑟+1 ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑍𝑍ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗
𝑄𝑄22 [𝑟𝑟] = ∑𝑚𝑚 𝑗𝑗 =1 �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ , 𝑎𝑎 + (3𝑟𝑟 + 1)ℎ)𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 6
3 +4𝒦𝒦𝑗𝑗 �𝑎𝑎 + (3𝑟𝑟 + 2)ℎ , 𝑎𝑎 + �3𝑟𝑟 + � ℎ � 2 15 5 1 5 ∗ � 𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 16 16 16 16 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 2)ℎ , 𝑎𝑎 + (3𝑟𝑟 + 2)ℎ) 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅𝑗𝑗
Step12: Compute 𝑢𝑢3𝑟𝑟+2 = 𝑢𝑢(𝑡𝑡3𝑟𝑟+2 ) , from:
𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢3𝑟𝑟+1 + 𝐴𝐴2 [𝑟𝑟] + 𝑄𝑄12 [𝑟𝑟] + 𝑄𝑄22 [𝑟𝑟] + 𝐹𝐹𝑟𝑟2
Step13: For the third block 𝜗𝜗 = 3 calculate: i.
ii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴3 [𝑟𝑟] = − ∑3𝑟𝑟+2 𝑠𝑠=1 �[𝑢𝑢3𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢3𝑟𝑟+2−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,3 �� ℎ 3𝑟𝑟+3 ̅ 𝑄𝑄13 [𝑟𝑟] = ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑍𝑍ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 3)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3
Step14: Compute 𝑢𝑢3𝑟𝑟+3 = 𝑢𝑢(𝑡𝑡3𝑟𝑟+3 ) , from: with putting:
𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+3 = 𝐻𝐻𝑟𝑟3 𝑢𝑢3𝑟𝑟+2 + 𝐴𝐴3 [𝑟𝑟] + 𝑄𝑄13 [𝑟𝑟] + 𝐹𝐹𝑟𝑟3
a. 𝑏𝑏𝑠𝑠𝛼𝛼 = (𝑠𝑠 + 1)1−𝛼𝛼 − 𝑠𝑠1−𝛼𝛼 ; 𝑠𝑠 = 0,1,2, … b. 𝑤𝑤0 = 𝑤𝑤3𝑟𝑟 = 1 ; 𝑤𝑤 �0 = 𝑤𝑤 � 3𝑟𝑟+2 = 1 ℓ where ℓ = 1,2,3, … ,2𝑟𝑟 − 1 𝑤𝑤ℓ = 3 − (−1) ℓ where ℓ = 1,2,3, … ,2𝑟𝑟 + 1 𝑤𝑤 � ℓ = 3 − (−1) ̅ c. 𝑍𝑍0 = 𝑍𝑍3𝑟𝑟+1 = 1 ; 𝑍𝑍0̅ = 𝑍𝑍3𝑟𝑟+3 =1 ℓ where ℓ = 1,2,3, … ,3𝑟𝑟 𝑍𝑍ℓ = 3 − (−1) ℓ ̅ where ℓ = 1,2,3, … ,3𝑟𝑟 + 2 𝑍𝑍ℓ = 3 − (−1) 49
Chapter Two
Block By Block Method
2.5.3 Four Block Method: Here, we take the order of block-by-block method 𝑝𝑝 = 4. Then, the integrations in equation (2.10) became over the sub-interval [𝑎𝑎, 𝑡𝑡4𝑟𝑟 ] and [𝑡𝑡4𝑟𝑟 , 𝑡𝑡𝑘𝑘 ] where 𝑘𝑘 = 4𝑟𝑟 + 1, 4𝑟𝑟 + 2, 4𝑟𝑟 + 3, 4𝑟𝑟 + 4 for all 𝑟𝑟 = 0,1, … , 𝑁𝑁 − 1. Now equation (2.11), with using the notification of equations (2.13-15), can be written as: First block 4𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢4𝑟𝑟+1 − 𝐻𝐻𝑟𝑟1 𝑢𝑢4𝑟𝑟 + � �[𝑢𝑢4𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢4𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
+𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ)𝑢𝑢4𝑟𝑟+1−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 1)ℎ) 𝑎𝑎+4𝑟𝑟ℎ 𝑚𝑚
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑
+𝜆𝜆 �
𝑗𝑗 =1
𝑎𝑎
Second block
+𝜆𝜆
𝑎𝑎+(4𝑟𝑟+1)ℎ 𝑚𝑚
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ , 𝑥𝑥)𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑
�
𝑗𝑗 =1
𝑎𝑎 +3𝑟𝑟ℎ 4𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
… (2.35)
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢4𝑟𝑟+2 − 𝐻𝐻𝑟𝑟2 𝑢𝑢4𝑟𝑟+1 + � �[𝑢𝑢4𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
+𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 2)ℎ) 𝑢𝑢3𝑟𝑟+2−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 2)ℎ) 𝑎𝑎+4𝑟𝑟ℎ 𝑚𝑚
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 2)ℎ , 𝑥𝑥)𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑
+𝜆𝜆 �
𝑗𝑗 =1
𝑎𝑎
+𝜆𝜆
𝑎𝑎+(4𝑟𝑟+2)ℎ 𝑚𝑚
�
𝑎𝑎+4𝑟𝑟ℎ
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 2)ℎ , 𝑥𝑥)𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 𝑗𝑗 =1
50
… (2.36)
Chapter Two
Block By Block Method
Third block 4𝑟𝑟+2
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟3 𝑢𝑢4𝑟𝑟+3 − 𝐻𝐻𝑟𝑟3 𝑢𝑢4𝑟𝑟+2 + � �[𝑢𝑢4𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+2−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,3 ��
+𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ) 𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 3)ℎ) 𝑎𝑎+4𝑟𝑟ℎ 𝑚𝑚
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑥𝑥)𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑
+𝜆𝜆 �
𝑗𝑗 =1
𝑎𝑎
Fourth block 𝐻𝐻𝑟𝑟4 𝑢𝑢4𝑟𝑟+4
−
+𝜆𝜆
𝑎𝑎+(4𝑟𝑟+3)ℎ 𝑚𝑚
�
𝑎𝑎 +4𝑟𝑟ℎ
𝐻𝐻𝑟𝑟4 𝑢𝑢4𝑟𝑟+3
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑥𝑥)𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 𝑗𝑗 =1
4𝑟𝑟+3
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
… (2.37)
𝑖𝑖 + � �[𝑢𝑢4𝑟𝑟+4−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+3−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,4 ��
+𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 4)ℎ) 𝑢𝑢4𝑟𝑟+4−𝑟𝑟̅0 = 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 4)ℎ) 𝑎𝑎+4𝑟𝑟ℎ 𝑚𝑚
+𝜆𝜆 � 𝑎𝑎
+𝜆𝜆
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 4)ℎ , 𝑥𝑥) 𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �𝑑𝑑𝑑𝑑 𝑗𝑗 =1
𝑎𝑎+(4𝑟𝑟+4)ℎ 𝑚𝑚
�
𝑎𝑎 +4𝑟𝑟ℎ
� 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 4)ℎ, 𝑥𝑥)𝑢𝑢�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 𝑗𝑗 =1
… (2.38)
After some basic steps as in the section (2.5.1) for two blocks method, the treatment of integrations above depends on quadrature formula: Simpson’s
1 3
ℎ rule
(2.3), adaptive Simpson’s rule (2.4) and quadratic interpolation formula (2.7) using point’s 𝑡𝑡4𝑟𝑟 , 𝑡𝑡4𝑟𝑟+1 , 𝑡𝑡4𝑟𝑟+2 , 𝑡𝑡4𝑟𝑟 +3 , 𝑡𝑡4𝑟𝑟 +4 .
Motivated by our earlier works as in section (2.3), the useful past values u can be evaluated from the set of equations in (2.8), the resulting equations then are: 51
Chapter Two
Block By Block Method
For the First block at (𝟒𝟒𝟒𝟒 + 𝟏𝟏), for 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1, became: 4𝑟𝑟
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟1 𝑢𝑢4𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢4𝑟𝑟 − � �[𝑢𝑢4𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢4𝑟𝑟−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,1 ��
− 𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ)𝑢𝑢4𝑟𝑟+1−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 1)ℎ) 𝑚𝑚
4𝑟𝑟
ℎ + � � 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0 𝑚𝑚
ℎ + � �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ, 𝑎𝑎 + 4𝑟𝑟ℎ)𝑢𝑢4𝑟𝑟−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 =1
1 + 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 1/2)ℎ ) 8
70 5 35 ∗ � 𝑢𝑢4𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 35𝑢𝑢4𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 7 𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+4−𝑟𝑟̅𝑗𝑗 � 4 4 4 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 1)ℎ)𝑢𝑢4𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 �
For the Second block at (𝟒𝟒𝟒𝟒 + 𝟐𝟐) became: 4𝑟𝑟+1
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
… (2.39)
𝑖𝑖 𝐻𝐻𝑟𝑟2 𝑢𝑢4𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢4𝑟𝑟+1 − � �[𝑢𝑢4𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+1−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,2 ��
−𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 2)ℎ)𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 2)ℎ) 𝑚𝑚 4𝑟𝑟+2
ℎ + � � 𝑤𝑤 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 2)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 3 𝑗𝑗 =1 ℓ=0
52
… (2.40)
Chapter Two
Block By Block Method
For the Third block at (𝟒𝟒𝟒𝟒 + 𝟑𝟑) became: 4𝑟𝑟+2
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
𝑖𝑖 𝐻𝐻𝑟𝑟3 𝑢𝑢4𝑟𝑟+3 = 𝐻𝐻𝑟𝑟3 𝑢𝑢4𝑟𝑟+2 − � �[𝑢𝑢4𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+2−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,3 ��
−𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ)𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 3)ℎ) 𝑚𝑚 4𝑟𝑟+2
ℎ � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 + � � 𝑤𝑤 3 𝑗𝑗 =1 ℓ=0
𝑚𝑚
ℎ + � �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ , 𝑎𝑎 + (4𝑟𝑟 + 2)ℎ)𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 6 𝑗𝑗 =1
1 + 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 5/2)ℎ) 8
70 5 35 ∗ � 𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 35𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+4−𝑟𝑟̅𝑗𝑗 + 7𝑢𝑢4𝑟𝑟+5−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+6−𝑟𝑟̅𝑗𝑗 � 4 4 4 + 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 3)ℎ)𝑢𝑢3𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 �
For the Fourth block at (𝟒𝟒𝟒𝟒 + 𝟒𝟒) becomes: 4𝑟𝑟+3
𝑛𝑛
𝑠𝑠=1
𝑖𝑖=1
… (2.41)
𝑖𝑖 𝐻𝐻𝑟𝑟4 𝑢𝑢4𝑟𝑟+4 = 𝐻𝐻𝑟𝑟4 𝑢𝑢4𝑟𝑟+3 − � �[𝑢𝑢4𝑟𝑟+4−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+3−𝑠𝑠 ] �� 𝐵𝐵𝑠𝑠𝑖𝑖 𝐶𝐶𝑟𝑟,4 ��
− 𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 4)ℎ)𝑢𝑢4𝑟𝑟+4−𝑟𝑟̅0 + 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 4)ℎ) 𝑚𝑚 4𝑟𝑟+4
ℎ � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 4)ℎ , 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅ + � � 𝑤𝑤 𝑗𝑗 3 𝑗𝑗 =1 ℓ=0
Where for all 𝜗𝜗 = 1,2,3 and 4 with 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 : 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗
=
−𝛼𝛼 𝑛𝑛 ⎧ ℎ ⎪ Γ(2 − 𝛼𝛼𝑛𝑛 )
if
𝑖𝑖 = 𝑛𝑛
… (2.42)
−𝛼𝛼 ⎨𝑃𝑃 (𝑎𝑎 + (4𝑟𝑟 + 𝜗𝜗)ℎ) ℎ 𝑛𝑛 −𝑖𝑖 ���������� otherwise 𝑖𝑖 = 1: 𝑛𝑛 − 1 ⎪ 𝑖𝑖 ) Γ(2 − 𝛼𝛼 𝑛𝑛−𝑖𝑖 ⎩
53
… (2.43)
Chapter Two
Block By Block Method 𝑛𝑛
𝑖𝑖 𝐻𝐻𝑟𝑟𝜗𝜗 = � 𝐶𝐶𝑟𝑟,𝜗𝜗 𝑖𝑖=1
for all 𝑠𝑠 = 1,2,3, … ,4𝑟𝑟 + 𝜗𝜗 − 1 and 𝑖𝑖 = 1,2, … , 𝑛𝑛: 𝛼𝛼 𝑖𝑖 = 𝑛𝑛 𝑏𝑏𝑠𝑠 𝑛𝑛 𝑖𝑖 𝐵𝐵𝑠𝑠 = � 𝛼𝛼 𝑛𝑛 −𝑖𝑖 𝑏𝑏𝑠𝑠 otherwise
� 0 = 𝑤𝑤 �4𝑟𝑟+4 = 1 � 0 = 𝑤𝑤 �4𝑟𝑟+2 = 1 , 𝑤𝑤 𝑤𝑤0 = 𝑤𝑤4𝑟𝑟 = 1 ; 𝑤𝑤 𝑤𝑤ℓ = 3 − (−1)ℓ where ℓ = 1,2,3, … ,4𝑟𝑟 − 1 ℓ where ℓ = 1,2,3, … ,4𝑟𝑟 + 1 𝑤𝑤 �ℓ = 3 − (−1) ℓ 𝑤𝑤 �ℓ = 3 − (−1) where ℓ = 1,2,3, … ,4𝑟𝑟 + 3
… (2.44) … (2.45)
Algorithm of Four Block Method (A4BM): The approximate solution for Linear VIFDEs with constant multi-time Retarded delay (2.9) by using four block method with aid of finite difference approximation can be summarized by the following steps: Step1: i. ii. iii.
Set 𝑢𝑢(𝑡𝑡0 ) = 𝑢𝑢0 [given from (2.9)] and input the historical function 𝜑𝜑(𝑡𝑡) Let ℎ = (𝑏𝑏 − 𝑎𝑎)/4𝑁𝑁 , 𝑁𝑁 ∈ ℤ+. 𝜏𝜏 𝑗𝑗
Evaluate 𝑟𝑟̅𝑗𝑗 = � � for all 𝑗𝑗 = 0,1,2, … 𝑚𝑚 where 𝜏𝜏0 = 𝜏𝜏 ℎ
Step2: For all 𝜗𝜗 = 1 , 2 ,3 and 4 with all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 : i.
ii. iii.
𝑖𝑖 ����� Calculate 𝐶𝐶𝑟𝑟,𝜗𝜗 , 𝑖𝑖 = 1: 𝑛𝑛, and 𝐻𝐻𝑟𝑟𝜗𝜗 from the equations (2.43) and (2.44) respectively. ����� Compute 𝐵𝐵𝑠𝑠𝑖𝑖 , 𝑖𝑖 = 1: 𝑛𝑛, for all 𝑠𝑠 = 1,2, … ,4𝑟𝑟 + 𝜗𝜗 − 1 from the equation (2.45). Evaluate 𝐹𝐹𝑟𝑟𝜗𝜗 from: 𝐹𝐹𝑟𝑟𝜗𝜗 = 𝑓𝑓(𝑎𝑎 + (4𝑟𝑟 + 𝜗𝜗)ℎ) − 𝑃𝑃0 (𝑎𝑎 + (4𝑟𝑟 + 𝜗𝜗)ℎ)𝑢𝑢4𝑟𝑟+𝜗𝜗 −𝑟𝑟̅0
For all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1 , the steps (3-10) are done:
54
Chapter Two
Block By Block Method
Step3: For the first block all 𝜗𝜗 = 1 calculate: i.
ii. iii.
𝑛𝑛 1 𝑖𝑖 𝑖𝑖 𝐴𝐴1 [𝑟𝑟] = − ∑4𝑟𝑟 𝑠𝑠=1�[𝑢𝑢4𝑟𝑟+1−𝑠𝑠 − 𝑢𝑢4𝑟𝑟−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,1 �� , where 𝐴𝐴 [0] = 0
𝑄𝑄11 [𝑟𝑟] =
ℎ
3
4𝑟𝑟 ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗
where 𝑄𝑄11 [0] = 0 . ℎ
𝑄𝑄21 [𝑟𝑟] = ∑𝑚𝑚 𝑗𝑗 =1 �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (3𝑟𝑟 + 1)ℎ , 𝑎𝑎 + 3𝑟𝑟ℎ)𝑢𝑢3𝑟𝑟−𝑟𝑟̅𝑗𝑗 6
1 + 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 1/2)ℎ ) 8 70 5 35 ∗ � 𝑢𝑢4𝑟𝑟−𝑟𝑟̅𝑗𝑗 + 35𝑢𝑢4𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 7 𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+4−𝑟𝑟̅𝑗𝑗 � 4 4 4 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 1)ℎ , 𝑎𝑎 + (4𝑟𝑟 + 1)ℎ) 𝑢𝑢4𝑟𝑟+1−𝑟𝑟̅𝑗𝑗 �
Step4: Compute 𝑢𝑢4𝑟𝑟+1 = 𝑢𝑢(𝑡𝑡4𝑟𝑟+1 ) , from:
𝐻𝐻𝑟𝑟1 𝑢𝑢4𝑟𝑟+1 = 𝐻𝐻𝑟𝑟1 𝑢𝑢4𝑟𝑟 + 𝐴𝐴1 [𝑟𝑟] + 𝑄𝑄11 [𝑟𝑟] + 𝑄𝑄21 [𝑟𝑟] + 𝐹𝐹𝑟𝑟1
Step5: For the second block 𝜗𝜗 = 2 calculate: i.
ii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴2 [𝑟𝑟] = − ∑4𝑟𝑟+1 𝑠𝑠=1 �[𝑢𝑢4𝑟𝑟+2−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+1−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,2 ��
𝑄𝑄12 [𝑟𝑟] =
ℎ
3
4𝑟𝑟+2 ∑𝑚𝑚 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 2)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤
Step6: Compute 𝑢𝑢4𝑟𝑟+2 = 𝑢𝑢(𝑡𝑡4𝑟𝑟+2 ) , from:
𝐻𝐻𝑟𝑟2 𝑢𝑢4𝑟𝑟+2 = 𝐻𝐻𝑟𝑟2 𝑢𝑢4𝑟𝑟+1 + 𝐴𝐴2 [𝑟𝑟] + 𝑄𝑄12 [𝑟𝑟] + 𝐹𝐹𝑟𝑟2
Step7: for the third block 𝜗𝜗 = 3 calculate: i.
ii. iii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴3 [𝑟𝑟] = − ∑4𝑟𝑟+2 𝑠𝑠=1 �[𝑢𝑢4𝑟𝑟+3−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+2−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,3 ��
𝑄𝑄13 [𝑟𝑟] =
ℎ
3 ℎ
4𝑟𝑟+2 ∑𝑚𝑚 � ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤
𝑄𝑄23 [𝑟𝑟] = ∑𝑚𝑚 𝑗𝑗 =1 �𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 2)ℎ)𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 6
1 + 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ, 𝑎𝑎 + (4𝑟𝑟 + 5/2)ℎ ) 8 70 5 35 ∗ � 𝑢𝑢4𝑟𝑟+2−𝑟𝑟̅𝑗𝑗 + 35𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+4−𝑟𝑟̅𝑗𝑗 + 7 𝑢𝑢4𝑟𝑟+5−𝑟𝑟̅𝑗𝑗 − 𝑢𝑢4𝑟𝑟+6−𝑟𝑟̅𝑗𝑗 � 4 4 4 +𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 3)ℎ , 𝑎𝑎 + (4𝑟𝑟 + 3)ℎ) 𝑢𝑢4𝑟𝑟+3−𝑟𝑟̅𝑗𝑗 � 55
Chapter Two
Block By Block Method
Step8: Compute 𝑢𝑢4𝑟𝑟+3 = 𝑢𝑢(𝑡𝑡4𝑟𝑟+3 ) , from:
𝐻𝐻𝑟𝑟3 𝑢𝑢4𝑟𝑟+3 = 𝐻𝐻𝑟𝑟3 𝑢𝑢4𝑟𝑟+2 + 𝐴𝐴3 [𝑟𝑟] + 𝑄𝑄13 [𝑟𝑟] + 𝑄𝑄23 [𝑟𝑟] + 𝐹𝐹𝑟𝑟3
Step9: For the fourth block 𝜗𝜗 = 4 calculate: i.
ii.
𝑛𝑛 𝑖𝑖 𝑖𝑖 𝐴𝐴4 [𝑟𝑟] = − ∑4𝑟𝑟+3 𝑠𝑠=1 �[𝑢𝑢4𝑟𝑟+4−𝑠𝑠 − 𝑢𝑢4𝑟𝑟+3−𝑠𝑠 ] �∑𝑖𝑖=1 𝐵𝐵𝑠𝑠 𝐶𝐶𝑟𝑟,4 ��
𝑄𝑄14 [𝑟𝑟] =
ℎ
3
4𝑟𝑟+4 � ∑𝑚𝑚 𝑗𝑗 =1 ∑ℓ=0 𝑤𝑤ℓ 𝒦𝒦𝑗𝑗 (𝑎𝑎 + (4𝑟𝑟 + 4)ℎ, 𝑎𝑎 + ℓℎ) 𝑢𝑢ℓ−𝑟𝑟̅𝑗𝑗
Step10: Compute 𝑢𝑢4𝑟𝑟+4 = 𝑢𝑢(𝑡𝑡4𝑟𝑟+4 ) , from: with putting:
𝐻𝐻𝑟𝑟4 𝑢𝑢4𝑟𝑟+4 = 𝐻𝐻𝑟𝑟4 𝑢𝑢4𝑟𝑟+3 + 𝐴𝐴4 [𝑟𝑟] + 𝑄𝑄14 [𝑟𝑟] + 𝐹𝐹𝑟𝑟4
a. 𝑏𝑏𝑠𝑠𝛼𝛼 = (𝑠𝑠 + 1)1−𝛼𝛼 − 𝑠𝑠1−𝛼𝛼 ; 𝑠𝑠 = 0,1,2, … � 0 = 𝑤𝑤 �4𝑟𝑟+4 = 1 b. 𝑤𝑤0 = 𝑤𝑤4𝑟𝑟 = 1 ; 𝑤𝑤 � 0 = 𝑤𝑤 �4𝑟𝑟+2 = 1 ; 𝑤𝑤 𝑤𝑤ℓ = 3 − (−1)ℓ
where
ℓ = 1,2,3, … ,4𝑟𝑟 − 1
�ℓ = 3 − (−1)ℓ 𝑤𝑤
where
ℓ = 1,2,3, … ,4𝑟𝑟 + 3
𝑤𝑤 �ℓ = 3 − (−1)ℓ
where
ℓ = 1,2,3, … ,4𝑟𝑟 + 1
2.6 Implementation of the method: In this section, we applied the method presented in this chapter for solving linear VIFDEs with constant multi-time Retarded delays (2.9). We selected some examples in which the exact solution has already existed to show the effectiveness of the proposed algorithms (A2BM, A3BM and A4BM), Computer programming in MatLab (V .8) was written for all of them. The least square error with running time of programs was used for our comparison between the algorithms. Example (2.1): Recall test example (1), Take 𝑁𝑁 = 10 , ℎ = 1/10 and 𝑡𝑡𝑟𝑟 = 𝑎𝑎 + 𝑟𝑟ℎ ; 𝑟𝑟 = 0,1, … , 𝑁𝑁 − 1 . Let us take 𝛼𝛼 = 0.2 and run the Main programs (Block2, Block3 and Block4) respectively, as in the appendix. So:
56
Chapter Two
Block By Block Method
From equation (2.20) we evaluate 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗 by P2BM and Table (2.1) contains all the value of 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗 for the two Block method 𝜗𝜗 = 1,2. Table (2.1)
𝒓𝒓 0 1 2 3 4 5 6 7 8 9
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝝑𝝑 (𝒊𝒊 = 𝟏𝟏, 𝟐𝟐), for Two Block Method 𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟏𝟏
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟐𝟐
−1.952244637
−1.944922192
−1.932738454
−1.915723876
−1.893920987
−1.867384281
−1.836180088
−1.8003864
3.709453923 3.709453923 3.709453923 3.709453923
3.709453923 3.709453923 3.709453923 3.709453923
−1.760092684
−1.715399653
−1.666419016
−1.613273199
−1.55609504
−1.495027453
−1.430223077
−1.361843887
−1.290060797
−1.215053227
−1.137008657
−1.056122157
3.709453923 3.709453923 3.709453923 3.709453923 3.709453923 3.709453923
57
3.709453923 3.709453923 3.709453923 3.709453923 3.709453923 3.709453923
Chapter Two
Block By Block Method
From the equation (2.32) we evaluate 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗 by P3BM and Table (2.2) contains all value of 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗 for the three block method where 𝜗𝜗 = 1,2,3. Table (2.2)
𝒓𝒓 0 1 2 3 4 5 6 7 8 9
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝝑𝝑 ( 𝒊𝒊 = 𝟏𝟏, 𝟐𝟐) for Three Block Method
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟏𝟏
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟐𝟐
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟑𝟑
−2.118625842
−2.115094472
−2.109213214
−2.100988603
−2.090429777
−2.077548465
−2.062358981
−2.044878198
−2.025125539
−2.003122949
−1.978894873
−1.952468228
−1.923872375
−1.893139084
−1.8603025
−1.825399104
−1.788467675
−1.749549244
−1.708687049
−1.665926489
−1.62131507
−1.574902357
−1.526739915
−1.476881252
−1.425381762
−1.372298661
−1.317690925
−1.261619224
−1.204145853
−1.145334666
4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943
4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943
58
4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943 4.362610943
Chapter Two
Block By Block Method
From equation (2.43) we evaluate 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗 by P4BM and Table (2.3) contains all the value of 𝐶𝐶𝑟𝑟𝑖𝑖 ,𝜗𝜗 for the Four block method where 𝜗𝜗 = 1,2,3,4. Table (2.3)
𝒓𝒓 0 1 2 3 4 5 6 7 8 9
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟏𝟏
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝝑𝝑
(𝒊𝒊 = 𝟏𝟏, 𝟐𝟐) for Four Block Method 𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟐𝟐
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟑𝟑
𝑪𝑪𝒊𝒊𝒓𝒓 ,𝟒𝟒
−2.244644667
−2.242540203
−2.239034224
−2.234128922
−2.227827362
−2.220133483
−2.211052092
−2.200588866
−2.188750343
−2.175543922
−2.160977858
−2.145061252
−2.127804053
−2.109217046
−2.089311847
−2.068100896
−2.045597449
−2.021815571
−1.996770123
−1.970476759
−1.942951912
−1.914212782
−1.884277332
−1.85316427
−1.820893041
−1.787483813
−1.752957465
−1.717335576
−1.680640409
−1.642894895
−1.604122626
−1.564347833
−1.523595373
−1.481890715
−1.439259925
−1.395729643
−1.351327076
−1.306079973
−1.260016613
−1.213165784
4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796
4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796
59
4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796
4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796 4.894653796
Chapter Two
Block By Block Method
P2BM: Table (2.4) contains all the value of 𝐻𝐻𝑟𝑟𝜗𝜗 for two Block method, ����� calculating it from equation (2.21) where 𝑛𝑛 = 2 and for all 𝑟𝑟 = 0: 9. Table (2.4) 𝒓𝒓
0 1 2 3 4 5 6 7 8 9
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐), for Two Block Method 𝑯𝑯𝟏𝟏𝒓𝒓 𝑯𝑯𝟐𝟐𝒓𝒓 1.757209286 1.764531731 1.776715469
1.793730046
1.873273835
1.909067523
1.815532936
1.842069641
1.949361239
1.99405427
2.043034907
2.096180723
2.279230846
2.347610036
2.153358883
2.214426469
2.419393126
2.494400696
2.572445266
2.653331766
P3BM: Table (2.5) contains all value of 𝐻𝐻𝑟𝑟𝜗𝜗 for three Block method, ����� calculate it from equation (2.33) where 𝑛𝑛 = 2 and for all 𝑟𝑟 = 0: 9. Table (2.5) 0
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑) , for Three Block Method 𝑯𝑯𝟏𝟏𝒓𝒓 𝑯𝑯𝟐𝟐𝒓𝒓 𝑯𝑯𝟑𝟑𝒓𝒓 2.243985 2.247516472 2.253398
2
2.300252
𝒓𝒓
1 3 4 5 6 7 8 9
2.261622
2.272181167
2.285062
2.359488
2.383716071
2.410143
2.438739 2.537212 2.653924 2.787709 2.937229 3.100992
2.317732745 2.469471859 2.574143268 2.696684455 2.835871028 2.990312282 3.158465091
60
2.337485 2.502308 2.613062 2.741296 2.88573 3.04492
3.217276
Chapter Two
Block By Block Method
P4BM: Table (2.6) contains all value of 𝐻𝐻𝑟𝑟𝜗𝜗 for the four Block method, ����� calculating it from equation (2.44) where 𝑛𝑛 = 2 and for all 𝑟𝑟 = 0: 9. Table (2.6)
0
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏, (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒) , for Four Block Method 𝑯𝑯𝟏𝟏𝒓𝒓 𝑯𝑯𝟐𝟐𝒓𝒓 𝑯𝑯𝟑𝟑𝒓𝒓 𝑯𝑯𝟒𝟒𝒓𝒓 2.650009129 2.652113593 2.655619571 2.660524873
2
2.705903453
𝒓𝒓
1 3 4 5 6 7 8 9
2.666826433
2.674520313
2.683601704
2.766849742
2.785436749
2.805341948
2.849056346 2.951701884 3.073760755 3.214013387 3.371058423 3.54332672 𝜏𝜏 𝑗𝑗
2.719109873 2.872838225 2.980441013 3.107169983
2.733675938
2.749592543
2.897883672
2.924177036
3.010376463 3.14169633
3.2517589
3.290531169
3.588573822
3.634637182
3.41276308
2.69406493
3.455393871
2.826552899 3.041489525 3.177318219 3.330305963 3.498924152 3.681488012
Evaluate 𝑟𝑟̅𝑗𝑗 = � � for each 𝑗𝑗 = 0,1,2, … 𝑚𝑚, and ℎ = (𝑏𝑏 − 𝑎𝑎)/𝑝𝑝𝑝𝑝 to all constant ℎ
delay 𝜏𝜏𝑗𝑗 , where 𝜏𝜏0 = 𝜏𝜏, for 𝑝𝑝-(Two, Three and Four)-blocks. So: 𝜏𝜏 𝑗𝑗
1
thus:
1
thus:
1
thus:
For 2BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2,3, ℎ =
20
For 3BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2,3, ℎ =
30
For 4BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2,3, ℎ =
40
𝑟𝑟̅𝑗𝑗 = {16 , 4 , 8 , 12}
𝑟𝑟̅𝑗𝑗 = {24 , 6 , 12 , 18} 𝑟𝑟̅𝑗𝑗 = {32 , 8 , 16 , 24}
ℎ
𝜏𝜏 𝑗𝑗 ℎ
𝜏𝜏 𝑗𝑗 ℎ
𝜏𝜏 𝑗𝑗
The results in tables (2.7), (2.8) and (2.9) show the evaluation of 𝑟𝑟̅𝑗𝑗 = � � , and ℎ
𝑢𝑢 �𝑡𝑡𝑁𝑁𝑁𝑁 −𝑟𝑟̅𝑗𝑗 � for all 𝑗𝑗 = 0,1,2,3 and 𝑁𝑁 = 10 for 𝑝𝑝, (Two, Three and Four), Blockby-Block method, respectively, as explained in section 2.4.
61
Chapter Two
Block By Block Method
Table (2.7)
𝒓𝒓
𝑵𝑵𝑵𝑵
𝒕𝒕
0
0
0
1 2 3 4 5 6 7 8 9 10
Two Block Method for Constant Time Delay 𝑵𝑵𝑵𝑵 = 𝟐𝟐𝟐𝟐 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎. 𝟓𝟓 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝝉𝝉𝟎𝟎 = 𝝉𝝉 = 𝟎𝟎. 𝟖𝟖 𝒓𝒓�𝟎𝟎 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟒𝟒
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟒𝟒 𝒓𝒓�𝟐𝟐 = 𝟖𝟖
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏
0.125
−0.955
−0.755
−0.395
0.28
−0.92
−0.68
−0.28
1
0.05
3
0.15
−0.155
−0.995
−0.875
−0.595
5
0.25
−0.395
𝑢𝑢11
−0.955
−0.755
7
0.35
−0.595
𝑢𝑢12
−0.995
−0.875
9
0.45
−0.755
𝑢𝑢13
𝑢𝑢11
−0.955
11
0.55
−0.875
𝑢𝑢12
−0.995
13
0.65
−0.955
𝑢𝑢13
𝑢𝑢11
15
0.75
−0.995
17
0.85
𝑢𝑢11
19
0.95
2 4 6 8 10 12 14 16 18 20
0.1 0.2 0.3
−0.02
−0.98
−0.28
𝒖𝒖𝟎𝟎
−0.82 −0.92
−0.68
−0.5
𝑢𝑢12
−0.68
𝑢𝑢22
−0.82
𝑢𝑢32
𝑢𝑢12
−0.92
𝑢𝑢42
𝑢𝑢22
−0.98
𝑢𝑢52
𝑢𝑢32
𝑢𝑢12
𝒖𝒖𝟎𝟎
𝑢𝑢62
𝑢𝑢42
𝑢𝑢22
0.9
𝑢𝑢12
𝑢𝑢72
𝑢𝑢52
𝑢𝑢32
1
𝑢𝑢22
𝑢𝑢82
𝑢𝑢62
𝑢𝑢42
0.4 0.5 0.6 0.7 0.8
𝑢𝑢14 𝑢𝑢15 𝑢𝑢16 𝑢𝑢17 𝑢𝑢18
𝑢𝑢12
62
−0.98
−0.5
𝒖𝒖𝟎𝟎
𝑢𝑢14 𝑢𝑢15 𝑢𝑢16
−0.82 −0.92 −0.98 𝒖𝒖𝟎𝟎
𝑢𝑢12 𝑢𝑢13 𝑢𝑢14
Chapter Two
Block By Block Method
Table (2.8)
𝒓𝒓
𝑵𝑵𝑵𝑵
𝒕𝒕
0
0
0
1
2
3
4
5
6
7
8
Three Block Method for Constant Time Delay 𝑵𝑵𝑵𝑵 = 𝟑𝟑𝟑𝟑 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝝉𝝉𝟎𝟎 = 𝝉𝝉 = 𝟎𝟎. 𝟖𝟖 𝒓𝒓�𝟎𝟎 = 𝟐𝟐𝟐𝟐
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟔𝟔
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟒𝟒 𝒓𝒓�𝟐𝟐 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏
0.28
−0.92
−0.68
−0.28
1
0.033333333
0.175556
−0.94444
−0.73111
−0.35778
3
0.1
−0.02
−0.98
−0.82
−0.5
2
0.066666667
0.075556
−0.96444
−0.77778
−0.43111
4
0.133333333
−0.11111
−0.99111
−0.85778
−0.56444
6
0.2
−0.28
𝒖𝒖𝟎𝟎
−0.92
−0.68
5
0.166666667
−0.19778
7
0.233333333
−0.35778
9
0.3
−0.5
8
0.266666667
−0.43111
10
0.333333333
−0.56444
12
0.4
−0.68
11
0.366666667 0.433333333
−0.73111
15
0.5
−0.82
0.466666667
16
0.533333333
−0.85778
18
0.6
−0.92
17
0.566666667
19
0.633333333
−0.94444
21
0.7
−0.98
20
0.666666667 0.733333333
−0.99111
24
0.8
𝒖𝒖𝟎𝟎
23
0.766666667
𝑢𝑢13
−0.98
−0.82
−0.85778
𝑢𝑢23
𝒖𝒖𝟎𝟎
−0.92
−0.99778
−0.89111
𝑢𝑢11
−0.94444
𝑢𝑢13
−0.98
−0.96444
𝑢𝑢32
𝑢𝑢12
𝑢𝑢14
𝑢𝑢12
−0.99111
𝑢𝑢23
𝒖𝒖𝟎𝟎
𝑢𝑢42
𝑢𝑢22
𝑢𝑢15
𝑢𝑢13
−0.99778 𝑢𝑢11
𝑢𝑢52
𝑢𝑢32
𝑢𝑢12
𝑢𝑢16
𝑢𝑢14
𝑢𝑢12
𝑢𝑢53 𝑢𝑢62 𝑢𝑢63
63
−0.77778
−0.99111
𝑢𝑢43
−0.99778
−0.96444
𝑢𝑢12
𝑢𝑢33
−0.96444
22
−0.73111
𝑢𝑢13
−0.89111
−0.62444
−0.94444
𝑢𝑢22
−0.77778
−0.89111
𝑢𝑢11
𝑢𝑢12
−0.62444
13 14
−0.99778
𝑢𝑢33 𝑢𝑢42 𝑢𝑢43
𝑢𝑢13 𝑢𝑢22 𝑢𝑢23
Chapter Two
𝒓𝒓
Block By Block Method
𝑵𝑵𝑵𝑵
𝒕𝒕
25
0.833333333
27
0.9
𝝉𝝉𝟎𝟎 = 𝝉𝝉 = 𝟎𝟎. 𝟖𝟖 𝒓𝒓�𝟎𝟎 = 𝟐𝟐𝟐𝟐
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟔𝟔
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟒𝟒 𝒓𝒓�𝟐𝟐 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏
0.866666667
𝑢𝑢12
𝑢𝑢72
𝑢𝑢52
𝑢𝑢32
28
0.933333333
𝑢𝑢12
𝑢𝑢18
𝑢𝑢16
𝑢𝑢14
30
1
26
9
0.966666667
29
10
Three Block Method for Constant Time Delay 𝑵𝑵𝑵𝑵 = 𝟑𝟑𝟑𝟑 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 � 𝑢𝑢11
𝑢𝑢13
𝑢𝑢17
𝑢𝑢73
𝑢𝑢22
𝑢𝑢82 𝑢𝑢83
𝑢𝑢23
Table (2.9)
𝑢𝑢15
𝑢𝑢13
𝑢𝑢53
𝑢𝑢33
𝑢𝑢62
𝑢𝑢42
𝑢𝑢63
𝑢𝑢43
Four Block method for Constant Time delay 𝒓𝒓 0 1
𝑵𝑵𝑵𝑵
𝒕𝒕
0
0
1 2 3 4 5
2
6 7 8 9
3
10 11 12
4
13 14
0.025
𝑵𝑵𝑵𝑵 = 𝟒𝟒𝟒𝟒 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝝉𝝉𝟎𝟎 = 𝝉𝝉 = 𝟎𝟎. 𝟖𝟖 𝒓𝒓�𝟎𝟎 = 𝟑𝟑𝟑𝟑
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟖𝟖
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟒𝟒 𝒓𝒓�𝟐𝟐 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟐𝟐𝟐𝟐
−0.155
−0.995
−0.875
−0.395
−0.08875
−0.98875
−0.84875
−0.33875
0.05
−0.21875
−0.99875
−0.89875
−0.44875
0.1
−0.33875
−0.155
−0.93875
−0.54875
0.15
−0.44875
−0.28
−0.96875
−0.63875
0.2
−0.54875
𝒖𝒖𝟎𝟎
−0.98875
−0.71875
0.25
−0.63875
𝑢𝑢12
−0.99875
−0.78875
0.3
−0.71875
𝑢𝑢14
−0.155
−0.84875
0.35
−0.78875
−0.28
−0.89875
0.075 0.125 0.175 0.225 0.275 0.325
−0.28
−0.08875
−0.395
−0.21875
−0.5
−0.33875
−0.595
𝑢𝑢11
−0.68
𝑢𝑢13
−0.755
𝑢𝑢12
𝑢𝑢22
64
−0.92
−0.955 −0.98
−0.995
−0.08875 −0.21875
−0.5
−0.595 −0.68
−0.755 −0.82
−0.875
Chapter Two
Block By Block Method Four Block method for Constant Time delay
𝒓𝒓
𝑵𝑵𝑵𝑵 15 16 17
5
18 19 20 21
6
22 23 24 25
7
26 27 28
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟖𝟖
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟒𝟒 𝒓𝒓�𝟐𝟐 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟐𝟐𝟐𝟐
0.4
−0.84875
𝑢𝑢24
𝒖𝒖𝟎𝟎
−0.93875
0.45
−0.89875
𝑢𝑢32
𝑢𝑢12
−0.96875
0.5
−0.93875
𝑢𝑢34
𝑢𝑢14
−0.98875
0.55
−0.96875
𝑢𝑢42
𝑢𝑢22
−0.99875
0.6
−0.98875
𝑢𝑢44
𝑢𝑢24
𝒖𝒖𝟎𝟎
0.65
−0.99875
𝑢𝑢52
𝑢𝑢32
𝑢𝑢12
0.7
−0.155
𝑢𝑢54
𝑢𝑢34
𝑢𝑢14
0.375 0.425 0.475 0.525 0.575 0.625 0.675
−0.82
38 39 40
𝑢𝑢15
−0.08875
𝑢𝑢11
37
𝑢𝑢43
−0.995
0.825
35
𝑢𝑢14
−0.98
33 34
𝑢𝑢33
−0.955
−0.33875
0.8
𝑢𝑢13
−0.92
0.775
0.75
𝑢𝑢23
−0.875
31
30
36
10
𝝉𝝉𝟎𝟎 = 𝝉𝝉 = 𝟎𝟎. 𝟖𝟖 𝒓𝒓�𝟎𝟎 = 𝟑𝟑𝟑𝟑
−0.21875
32
9
𝑵𝑵𝑵𝑵 = 𝟒𝟒𝟒𝟒 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
0.725
29 8
𝒕𝒕
𝑢𝑢53 𝑢𝑢16
−0.33875 𝑢𝑢11
𝑢𝑢13 𝑢𝑢12
𝑢𝑢23 𝑢𝑢13
𝑢𝑢33 𝑢𝑢14
−0.92
−0.955 −0.98
−0.995
−0.08875 𝑢𝑢11
𝑢𝑢13 𝑢𝑢12
−0.28
𝑢𝑢62
𝑢𝑢42
𝑢𝑢22
𝒖𝒖𝟎𝟎
𝑢𝑢64
𝑢𝑢44
𝑢𝑢24
𝑢𝑢63 𝑢𝑢17
𝑢𝑢43 𝑢𝑢15
𝑢𝑢23 𝑢𝑢13
0.85
𝑢𝑢12
𝑢𝑢72
𝑢𝑢52
𝑢𝑢32
0.9
𝑢𝑢14
𝑢𝑢74
𝑢𝑢54
𝑢𝑢34
0.875 0.925
𝑢𝑢13
𝑢𝑢73
𝑢𝑢12
𝑢𝑢18
𝑢𝑢53 𝑢𝑢16
𝑢𝑢33 𝑢𝑢14
0.95
𝑢𝑢22
𝑢𝑢82
𝑢𝑢62
𝑢𝑢42
1
𝑢𝑢24
𝑢𝑢84
𝑢𝑢64
𝑢𝑢44
0.975
𝑢𝑢83
𝑢𝑢23
65
𝑢𝑢63
𝑢𝑢43
Chapter Two
Block By Block Method
Table (2.10) presents a comparison between the exact solution and numerical solution by Block-by-Block method of Blocks (Two, Three and Four) it also includes the least square error and running time. Table (2.10) Exact Solution
𝒕𝒕
0
−1
Numerical Solutions with 𝜶𝜶 = 𝟎𝟎. 𝟐𝟐 𝑵𝑵 = 𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑
𝟒𝟒𝟒𝟒𝟒𝟒
−1
−1
−1
0.1
−0.98
−0.97635
−0.977912726
−0.978607804095
0.3
−0.82
−0.81077
−0.814863838
−0.816629886249
−0.492488862
−0.495092445539
0.2 0.4 0.5
−0.92 −0.68 −0.5
0.6
−0.28
0.8
0.28
0.7 0.9 1
−0.02 0.62 1
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
−0.91344 −0.66841 −0.48641 −0.26476 −0.00344
−0.916319921 −0.673577007 −0.271593146
0.297728
−0.274513088242
−0.0140506765528
0.62961315
0.626256491039
1.015982
1.008762548
1.579232
2.181983
1.8433975 𝑒𝑒 − 003
−0.675796271816
−0.01087725 0.289756224
0.637499
−0.91757409781
5.6023045 𝑒𝑒 − 004
0.286359335126 1.00569475002 2.3845195 𝑒𝑒 − 004 2.365222
The results in Table (2.11) show the least square errors and running times for Block-by-Block method of Blocks (Two, Three and Four) with different values of step size ℎ and 𝑁𝑁.
66
Chapter Two
Block By Block Method
Table (2.11) Numerical Solutions with 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎
𝒉𝒉 𝑩𝑩𝑩𝑩𝑩𝑩 𝟐𝟐𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
(𝑵𝑵 = 𝟏𝟏𝟏𝟏)
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝜶𝜶 = 𝟎𝟎. 𝟐𝟐
(𝑵𝑵 = 𝟏𝟏𝟏𝟏𝟏𝟏)
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 (𝑵𝑵 = 𝟓𝟓𝟓𝟓𝟓𝟓)
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
1.8433975 1.710247 5.8260113 1.579232 7.673566 36.301438 𝑒𝑒 − 003 𝑒𝑒 − 005 𝑒𝑒 − 007 5.6023045 4.9098366 1.6429057 2.181983 11.346510 54.017770 𝑒𝑒 − 004 𝑒𝑒 − 006 𝑒𝑒 − 007 2.3845195 2.0185362 6.6789993 2.365222 14.684924 75.732070 𝑒𝑒 − 004 𝑒𝑒 − 006 𝑒𝑒 − 008
Example (2.2): Recall test example (2), which is linear VIFDE of constant multi0: 𝑁𝑁 − 1. time Retarded delay, take 𝑁𝑁 = 10, ℎ = 0.1, 𝑡𝑡𝑟𝑟 = 𝑎𝑎 + 𝑟𝑟ℎ, 𝑟𝑟 = ����������� P2BM: Table (2.12) contains all the value of 𝐻𝐻𝑟𝑟𝜗𝜗 for the two Block method, calculating it from equation (2.21) where number of fractional parts 𝑛𝑛 = 3 and for all 𝑟𝑟 = ����������� 0: 𝑁𝑁 − 1. Table (2.12)
𝒓𝒓
0
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐) for Two Block Method 𝑯𝑯𝟏𝟏𝒓𝒓
14.544931
14.64266537
14.93586849
15.03360287
1
14.74039974
3
15.13133724
2 4 5 6 7 8 9
𝑯𝑯𝟐𝟐𝒓𝒓
14.83813412 15.22907162
15.32680599
15.42454037
15.52227474
15.62000911
15.71774349
15.81547786
15.91321224
16.01094661
16.10868099
16.20641536
16.30414973
16.40188411
67
Chapter Two
Block By Block Method
P3BM: Table (2.13) contains all value of 𝐻𝐻𝑟𝑟𝜗𝜗 for three Block method ����������� calculate it from equation (2.33) where 𝑛𝑛 = 3 and for all 𝑟𝑟 = 0: 𝑁𝑁 − 1. Table (2.13) 𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑) for Three Block Method 𝑯𝑯𝟏𝟏𝒓𝒓 𝑯𝑯𝟐𝟐𝒓𝒓 𝑯𝑯𝟑𝟑𝒓𝒓
𝒓𝒓 0
21.06734799
2
21.49130867
1 3 4
7 9
21.63262889
21.70328901
21.77394912 21.98592946 22.1979098
22.33923003
22.40989014
22.76319071
22.83385082
22.55121037
8
21.56196878
21.34998844
22.12724969
6
21.20866821
21.27932833
21.91526935
5
21.1380081
22.97517105
22.62187048 23.04583116
21.42064855 21.84460923 22.05658957 22.26856991 22.48055025 22.69253059 22.90451093 23.11649127
P4BM: Table (2.14) contains all value of 𝐻𝐻𝑟𝑟𝜗𝜗 for four Block method ����������� calculate it from equation (2.44) where 𝑛𝑛 = 3 and for all 𝑟𝑟 = 0: 𝑁𝑁 − 1. Table (2.14)
𝒓𝒓
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏, (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒) for Four Block Method 𝑯𝑯𝟒𝟒𝒓𝒓 𝑯𝑯𝟏𝟏𝒓𝒓 𝑯𝑯𝟐𝟐𝒓𝒓 𝑯𝑯𝟑𝟑𝒓𝒓
0
27.41266998
27.46880364
2
27.86173924
27.9178729
1 3 4 5 6 7 8 9
27.63720461
27.5249373
27.58107096
27.97400656
28.03014022
27.69333827
27.74947193
28.08627387
28.14240753
28.19854119
28.53534314
28.59147679
28.31080851 28.75987777 28.9844124
29.20894703 29.43348166
28.36694216 28.81601142 29.04054605 29.26508068 29.48961531
68
28.42307582 28.64761045 28.87214508 29.09667971 29.32121434 29.54574897
27.80560559 28.25467485 28.47920948 28.70374411 28.92827874 29.15281337 29.377348
29.60188263
Chapter Two
Block By Block Method 𝜏𝜏
𝑗𝑗 Evaluate 𝑟𝑟̅𝑗𝑗 = � � for each 𝑗𝑗 = 0,1,2, … 𝑚𝑚, and ℎ = (𝑏𝑏 − 𝑎𝑎)/𝑝𝑝𝑝𝑝 to all constant ℎ delay 𝜏𝜏𝑗𝑗 , where 𝜏𝜏0 = 𝜏𝜏, for 𝑝𝑝 (Two, Three and Four)-blocks. So:
𝜏𝜏 𝑗𝑗
1
thus:
1
thus:
1
thus:
For 2BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2,3, ℎ =
20
For 3BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2,3, ℎ =
30
For 4BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � for all j = 0,1,2,3, ℎ =
40
𝑟𝑟̅𝑗𝑗 = {6,4,14,12} 𝑟𝑟̅𝑗𝑗 = {9,6,21,18}
𝑟𝑟̅𝑗𝑗 = {13,8,28,24}
ℎ
𝜏𝜏 𝑗𝑗 ℎ
𝜏𝜏 𝑗𝑗 ℎ
𝜏𝜏 𝑗𝑗
The results in tables (2.15), (2.16) and (2.17) show the evaluation of 𝑟𝑟̅𝑗𝑗 = � � , and ℎ
𝑢𝑢 �𝑡𝑡𝑁𝑁𝑁𝑁 −𝑟𝑟̅𝑗𝑗 � for all 𝑗𝑗 = 0,1,2,3 and 𝑁𝑁 = 10 for 𝑝𝑝, (Two, Three and Four), Blockby-Block method, respectively, as explained in section 2.4. Table (2.15)
𝒓𝒓
𝑵𝑵𝑵𝑵
𝒕𝒕
0
0
0
1 2 3 4 5 6
Two Block method for Constant Time delay 𝑵𝑵𝑵𝑵 = 𝟐𝟐𝟐𝟐 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝝉𝝉 = 𝝉𝝉𝟎𝟎 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝒓𝒓�𝟎𝟎 = 𝟔𝟔 0.427
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟒𝟒 0.608
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒓𝒓�𝟐𝟐 = 𝟏𝟏𝟏𝟏 −0.057
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏 0.016
1
0.05
0.515625
0.703375 −0.025375 0.066375
3
0.15
0.703375
0.900125
0.066375
0.191125
5
0.25
0.900125
𝑢𝑢11
0.191125
0.342875
7
0.35
𝑢𝑢11
𝑢𝑢12
0.342875
0.515625
9
0.45
𝑢𝑢13
0.515625
0.703375
11
0.55
𝑢𝑢14
0.703375
0.900125
2 4 6 8 10
0.1 0.2 0.3
0.608 0.801
0.801 𝒖𝒖𝟎𝟎
𝒖𝒖𝟎𝟎
𝑢𝑢12
0.4
𝑢𝑢12
𝑢𝑢22
0.5
𝑢𝑢22
𝑢𝑢32
𝑢𝑢12 𝑢𝑢13
69
0.016 0.125 0.264 0.427 0.608
0.125 0.264 0.427 0.608 0.801
Chapter Two
Block By Block Method
𝒓𝒓
7 8 9 10
𝑵𝑵𝑵𝑵
𝒕𝒕
12
0.6
13
0.65
15
0.75
17
0.85
19
0.95
14 16 18 20
Two Block method for Constant Time delay 𝑵𝑵𝑵𝑵 = 𝟐𝟐𝟐𝟐 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝝉𝝉 = 𝝉𝝉𝟎𝟎 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝒓𝒓�𝟎𝟎 = 𝟔𝟔
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟒𝟒
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒓𝒓�𝟐𝟐 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏
𝑢𝑢14
𝑢𝑢15
0.900125
𝑢𝑢11
𝑢𝑢16
𝑢𝑢11
𝑢𝑢32
𝑢𝑢42
0.7
𝑢𝑢42
𝑢𝑢52
0.8
𝑢𝑢52
0.9 1
0.801
𝒖𝒖𝟎𝟎
𝒖𝒖𝟎𝟎
𝑢𝑢12
𝑢𝑢62
𝑢𝑢12
𝑢𝑢22
𝑢𝑢62
𝑢𝑢72
𝑢𝑢22
𝑢𝑢32
𝑢𝑢72
𝑢𝑢82
𝑢𝑢32
𝑢𝑢42
𝑢𝑢15 𝑢𝑢16
𝑢𝑢17 𝑢𝑢18
𝑢𝑢17
𝑢𝑢12 𝑢𝑢13
𝑢𝑢12
𝑢𝑢14
𝑢𝑢13
Table (2.16) Three Block Method for Constant Time delay 𝒓𝒓
𝑵𝑵𝑵𝑵
𝒕𝒕
0
0
1
2
2
3
𝑵𝑵𝑵𝑵 = 𝟑𝟑𝟑𝟑 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
0
𝝉𝝉 = 𝝉𝝉𝟎𝟎 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝒓𝒓�𝟎𝟎 = 𝟗𝟗 0.427
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟔𝟔 0.546037037
0.066666667
0.546037037
0.671296296 −0.01262963 0.085037037
4
0.133333333
0.671296296
6
0.2
0.801
1
0.033333333
3
0.1
5
0.166666667
0.48562963 0.608
0.735703704
7
0.233333333
0.866962963
9
0.3
𝒖𝒖𝟎𝟎
8
0.266666667
0.608
0.735703704
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒓𝒓�𝟐𝟐 = 𝟐𝟐𝟐𝟐 −0.057
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏 0.016
−0.037037037 0.04862963 0.016
0.125
0.801
0.04862963
0.168296296
𝒖𝒖𝟎𝟎
0.125
0.264
0.214703704
0.37037037
0.866962963 0.085037037 0.214703704 𝑢𝑢11
0.93337037
𝑢𝑢12
70
𝑢𝑢13
0.168296296 0.315962963 0.264
0.427
Chapter Two
Block By Block Method Three Block Method for Constant Time delay
𝒓𝒓
4
5
6
7
8
9
10
𝑵𝑵𝑵𝑵
𝒕𝒕
10
0.333333333
12
0.4
𝑵𝑵𝑵𝑵 = 𝟑𝟑𝟑𝟑 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝝉𝝉 = 𝝉𝝉𝟎𝟎 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝒓𝒓�𝟎𝟎 = 𝟗𝟗
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟔𝟔
0.366666667
𝑢𝑢12
𝑢𝑢22
13
0.433333333
𝑢𝑢21
𝑢𝑢31
15
0.5
11
𝑢𝑢11
𝑢𝑢13
𝑢𝑢21
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒓𝒓�𝟐𝟐 = 𝟐𝟐𝟐𝟐 0.315962963
𝑢𝑢23
0.427
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟏𝟏𝟏𝟏 0.48562963
0.37037037
0.546037037
0.48562963
0.671296296
0.608
0.801
0.735703704
0.93337037 𝑢𝑢11
0.608
0.466666667
𝑢𝑢22
𝑢𝑢32
0.546037037 0.735703704
16
0.533333333
𝑢𝑢31
𝑢𝑢41
0.671296296 0.866962963
18
0.6
14
𝑢𝑢33
𝑢𝑢23
0.566666667
𝑢𝑢32
𝑢𝑢42
19
0.633333333
𝑢𝑢41
𝑢𝑢51
0.866962963
21
0.7
𝑢𝑢53
𝒖𝒖𝟎𝟎
17
𝑢𝑢33
𝑢𝑢43
0.666666667
𝑢𝑢42
𝑢𝑢52
22
0.733333333
𝑢𝑢51
𝑢𝑢61
24
0.8
20
𝑢𝑢43
0.801
𝒖𝒖𝟎𝟎
0.93337037
𝑢𝑢12
𝑢𝑢11
𝑢𝑢21
𝑢𝑢13
0.766666667
𝑢𝑢52
𝑢𝑢62
𝑢𝑢12
𝑢𝑢22
25
0.833333333
𝑢𝑢61
𝑢𝑢71
𝑢𝑢21
𝑢𝑢31
27
0.9
23
𝑢𝑢63
𝑢𝑢53
𝑢𝑢13
𝑢𝑢23
0.866666667
𝑢𝑢62
𝑢𝑢72
𝑢𝑢22
𝑢𝑢32
28
0.933333333
𝑢𝑢71
𝑢𝑢81
𝑢𝑢31
𝑢𝑢41
30
1
26
29
0.966666667
𝑢𝑢63
𝑢𝑢73 𝑢𝑢82
𝑢𝑢72
𝑢𝑢83
𝑢𝑢73
71
𝑢𝑢23 𝑢𝑢32 𝑢𝑢33
𝑢𝑢33 𝑢𝑢42 𝑢𝑢43
Chapter Two
Block By Block Method
Table (2.17) Four Block Method for Constant Time delay 𝒓𝒓 0 1
𝑵𝑵𝑵𝑵
𝒕𝒕
0
0
3
0.075
5
0.125
7
0.175
9
0.225
11
0.275
13
0.325
15
0.375
𝑢𝑢12
17
0.425
𝑢𝑢14
19
0.475
𝑢𝑢22
21
0.525
𝑢𝑢24
2
6
10 12
4
14 16
5
18 20
6
0.384328125
0.025
8 3
𝝉𝝉 = 𝝉𝝉𝟎𝟎 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝒓𝒓�𝟎𝟎 = 𝟏𝟏𝟏𝟏
1
4 2
𝑵𝑵𝑵𝑵 = 𝟒𝟒𝟒𝟒 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
22
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟖𝟖 0.608
0.515625
0.751953125 −0.005859375 0.094703125
0.1
0.561390625
0.15
0.655359375
0.2
0.751953125
0.25
0.850421875
𝑢𝑢12
0.3
0.950015625
𝑢𝑢14
0.35
𝑢𝑢11
𝑢𝑢22
𝑢𝑢13
𝑢𝑢24
0.5
0.55
0.016
0.655359375 −0.042453125 0.040109375
0.470796875
0.45
−0.057
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟐𝟐𝟐𝟐
0.427
0.05
0.4
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒓𝒓�𝟐𝟐 = 𝟐𝟐𝟐𝟐
0.703375
−0.025375
0.066375
0.801
0.016
0.125
0.608
0.850421875
0.040109375
0.157171875
0.703375
0.950015625
0.094703125
0.226765625
0.801
𝑢𝑢11
0.157171875
0.302734375
𝑢𝑢13
0.226765625
0.384328125
𝑢𝑢12
0.302734375
0.470796875
𝑢𝑢23
0.384328125
0.561390625
𝑢𝑢13
0.470796875
0.655359375
𝑢𝑢33
0.561390625
0.751953125
𝑢𝑢14
0.655359375
0.850421875
0.900125 𝒖𝒖𝟎𝟎
0.900125 𝒖𝒖𝟎𝟎
𝑢𝑢12
𝑢𝑢32
𝑢𝑢23
𝑢𝑢34
𝑢𝑢13
72
𝑢𝑢42
0.066375 0.125
0.191125 0.264
0.342875 0.427
0.515625 0.608
0.703375
0.191125 0.264
0.342875 0.427
0.515625 0.608
0.703375 0.801
0.900125
Chapter Two
Block By Block Method Four Block Method for Constant Time delay
𝒓𝒓
𝒕𝒕
23
0.575
25
0.625
27
0.675
𝑢𝑢42
29
0.725
𝑢𝑢44
31
0.775
𝑢𝑢52
33
0.825
𝑢𝑢54
35
0.875
𝑢𝑢62
37
0.925
𝑢𝑢64
39
0.975
24 7
26 28
8
30 32
9
34 36
10
𝑵𝑵𝑵𝑵 = 𝟒𝟒𝟒𝟒 , 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 , 𝒖𝒖 �𝒕𝒕𝑵𝑵𝑵𝑵−𝒓𝒓�𝒋𝒋 �
𝑵𝑵𝑵𝑵
38 40
0.6
0.65 0.7
0.75 0.8
0.85 0.9
0.95 1
𝝉𝝉 = 𝝉𝝉𝟎𝟎 = 𝟎𝟎. 𝟑𝟑𝟑𝟑 𝒓𝒓�𝟎𝟎 = 𝟏𝟏𝟏𝟏
𝝉𝝉𝟏𝟏 = 𝟎𝟎. 𝟐𝟐 𝒓𝒓�𝟏𝟏 = 𝟖𝟖
𝑢𝑢33
𝑢𝑢44
𝑢𝑢32 𝑢𝑢34
𝝉𝝉𝟐𝟐 = 𝟎𝟎. 𝟕𝟕𝟕𝟕 𝒓𝒓�𝟐𝟐 = 𝟐𝟐𝟐𝟐
𝝉𝝉𝟑𝟑 = 𝟎𝟎. 𝟔𝟔 𝒓𝒓�𝟑𝟑 = 𝟐𝟐𝟐𝟐
𝑢𝑢43
0.751953125
0.950015625
𝑢𝑢15
0.850421875
𝑢𝑢11
𝑢𝑢53
0.950015625
𝑢𝑢16
𝑢𝑢11
𝑢𝑢14
𝑢𝑢52
𝑢𝑢43
𝑢𝑢54
0.801
𝒖𝒖𝟎𝟎
0.900125
𝑢𝑢12
𝒖𝒖𝟎𝟎
𝑢𝑢14
𝑢𝑢13 𝑢𝑢12
𝑢𝑢15
𝑢𝑢62
𝑢𝑢12
𝑢𝑢22
𝑢𝑢53
𝑢𝑢64
𝑢𝑢14
𝑢𝑢24
𝑢𝑢63 𝑢𝑢17
𝑢𝑢13 𝑢𝑢12
𝑢𝑢23 𝑢𝑢13
𝑢𝑢16
𝑢𝑢72
𝑢𝑢22
𝑢𝑢32
𝑢𝑢63
𝑢𝑢74
𝑢𝑢24
𝑢𝑢34
𝑢𝑢73 𝑢𝑢18
𝑢𝑢23 𝑢𝑢13
𝑢𝑢33 𝑢𝑢14
𝑢𝑢17
𝑢𝑢82
𝑢𝑢32
𝑢𝑢42
𝑢𝑢73
𝑢𝑢84
𝑢𝑢34
𝑢𝑢44
𝑢𝑢83
𝑢𝑢72
𝑢𝑢33
𝑢𝑢43
Table (2.18) presents a comparison between the exact solution and numerical solution by Block-by-Block method of Blocks (Two, Three and Four) it also includes the least square error and running time.
73
Chapter Two
Block By Block Method
Table (2.18) 𝒕𝒕
0
Numerical Solutions
Exact Solution 1
(𝑵𝑵 = 𝟏𝟏𝟏𝟏 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏)
𝟐𝟐𝟐𝟐𝟐𝟐
𝟑𝟑𝟑𝟑𝟑𝟑
𝟒𝟒𝟒𝟒𝟒𝟒
1
1
1
0.1
1.199
1.19103385754
1.191248472
1.197467045
0.3
1.573
1.54748128853
1.549278739
1.567171535
0.2 0.4 0.5 0.6 0.7 0.8 0.9 1
1.392 1.736 1.875
1.37539421274 1.70245092284 1.8347189331
1.984
1.93821587645
2.088
2.03320943124
2.057 2.071 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
2
2.00658016006 2.01130346114 1.93393037225 1.9307746 𝑒𝑒 − 002
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
1.376237284 1.705411901 1.839064511 1.944132657 2.014266533 2.042906831 2.023337062 1.948760566
2.363461
1.3444422 𝑒𝑒 − 002 2.684010
1.388478563 1.727797711 1.864655312 1.971895252 2.043596223 2.073747755 2.056252432 1.984947853 1.1963821 𝑒𝑒 − 003 3.799347
The result in Table (2.19) shows the least square errors and running times for Block-by-Block method of Blocks (Two, Three and Four), with different values of step size ℎ and 𝑁𝑁. Table (2.19)
𝒉𝒉
𝑩𝑩𝑩𝑩𝑩𝑩
𝟎𝟎. 𝟏𝟏 (𝑵𝑵 = 𝟏𝟏𝟏𝟏)
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝟎𝟎. 𝟎𝟎𝟎𝟎 (𝑵𝑵 = 𝟏𝟏𝟏𝟏𝟏𝟏)
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 (𝑵𝑵 = 𝟓𝟓𝟓𝟓𝟓𝟓)
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝟐𝟐𝟐𝟐𝟐𝟐
1.9307746 𝑒𝑒 − 002
2.363461
2.7668159 11.371778 𝑒𝑒 − 004
4.0061146 𝑒𝑒 − 005
53.681539
𝟒𝟒𝟒𝟒𝟒𝟒
1.1963821 𝑒𝑒 − 003
3.799347
6.0712919 21.062310 𝑒𝑒 − 005
8.7613087 𝑒𝑒 − 006
111.617415
𝟑𝟑𝟑𝟑𝟑𝟑
1.3444422 𝑒𝑒 − 002
2.684010
1.1396414 16.375843 𝑒𝑒 − 004
74
1.6466935 𝑒𝑒 − 005
78.268611
Chapter Two
Block By Block Method
Example (2.3): Consider test example (3), Take 𝑁𝑁 = 10, ℎ = 0.1; 𝑡𝑡𝑟𝑟 = 𝑎𝑎 + 𝑟𝑟ℎ, 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1. From equations (2.21, 2.33 and 2.44) in which the number of fractional parts is 𝑛𝑛 = 3. Tables (2.20, 2.21 and 2.22) contains all value of 𝐻𝐻𝑟𝑟𝜗𝜗 for Two, Three and Four Blocks method, respectively, by running the Main programs (2Block, 3Block and 4Block) as in the appendix Table (2.20) 𝒓𝒓 0
Two Block method 𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐) 𝑯𝑯𝟏𝟏𝒓𝒓 𝑯𝑯𝟐𝟐𝒓𝒓 16.04203836
16.42440086
16.58669211
16.15224525
1 2
16.27964955
16.76676013
3
16.96488617
17.18139666
4
17.41666397
17.67110741
5
17.94519421
18.23944074
6
18.55441375
18.89073185
7
19.249067
19.63014624
8
20.03475345
20.46373137
9
𝒓𝒓
15.94892232
Table (2.21)
20.91798365
Three Block method 𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑)
𝑯𝑯𝟏𝟏𝒓𝒓
𝑯𝑯𝟐𝟐𝒓𝒓
𝑯𝑯𝟑𝟑𝒓𝒓
0
20.91945114
20.84444217
20.91945114
2
21.5746246
21.44071357
21.5746246
22.59369976
22.39755457
1
21.20276064
3
22.03723692
5
23.24805453
4 6 7 8 9
24.00532241 24.87155455 25.85389265 26.96064045
21.0985674
21.20276064
21.87277716
22.03723692
23.01877011
23.24805453
23.74111328 24.57028569 25.51305815 26.57733839
75
22.59369976 24.00532241 24.87155455 25.85389265 26.96064045
Chapter Two
Block By Block Method
Table (2.22) 𝒓𝒓
Four Block method 𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏, (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒)
𝑯𝑯𝟏𝟏𝒓𝒓
𝑯𝑯𝟐𝟐𝒓𝒓
0
25.09046353
25.14173415
2
25.68210452
25.7854923
1 3 4 5 6 7 8 9
25.33429649
𝑯𝑯𝟑𝟑𝒓𝒓
𝑯𝑯𝟒𝟒𝒓𝒓
25.1994558
25.26363854
25.89552543
26.0122467
25.4114478
25.49511471
26.13570305
26.26594564
26.40302987
26.54701542
27.37285601
27.55974724
27.75408075
27.95595201
26.69796629 28.16546117
26.85595081 28.3827132
29.08204889
29.33141965
31.31851984
31.63870852
30.13012712
30.41369601
27.02104174 28.60781787 29.58913201 30.70617894 31.96847291
25.58532348
27.19331624 28.84088987 29.85532104 31.00773273 32.3079931
As the two examples before, we can calculate all values of 𝑟𝑟̅𝑗𝑗 = �𝜏𝜏𝑗𝑗 /ℎ� for
all j = 0,1,2, … 𝑚𝑚 , ℎ = (𝑏𝑏 − 𝑎𝑎)/𝑝𝑝𝑝𝑝 and 𝑢𝑢 �𝑡𝑡𝑁𝑁𝑁𝑁 −𝑟𝑟̅𝑗𝑗 � for Blocks-method (Two,
Three and Four) by running the Main (2Block, 3Block and 4Block) programs, respectively.
Table (2.26) presents a comparison between the exact solution and numerical solution by Block-by-Block method of Blocks (Two, Three and Four), also includes the least square error and running time from running the (2Block, 3Block and 4Block) programs, respectively. Also, the result in Table (2.27) shows the least square errors and running times for Block-by-Block of (Two, Three and Four) with different values of 𝑁𝑁 and ℎ ) by running the Main (2Block, 3Block and 4Block) programs, respectively.
76
Chapter Two
Block By Block Method
Table (2.26) Exact Solution
𝒕𝒕
0
0
Numerical Solutions (𝑵𝑵 = 𝟏𝟏𝟏𝟏) 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒 0
0
0
0.1
−0.1999
−0.199844869
−0.199866436
−0.199876447
0.3
−0.5919
−0.590767381
−0.591219573
−0.591427537
−0.3984
0.2
−0.7744
0.4
−0.9375
0.5
−1.0704
0.6
−1.1599
0.7
−1.1904
0.8
−1.1439
0.9
−1
1
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
−0.398027016 −0.771922109 −0.932969524 −1.063004512
−0.398174621 −0.772917937 −0.934799726 −1.06600491
−0.398242842 −0.773374003 −0.93563499 −1.06737016
−1.148735456
−1.153281573
−1.155344819
−1.122174858
−1.131075395
−1.135097271
−1.174482277 −0.971351299 1.7535342 𝑒𝑒 − 003
−1.180984735 −0.983119497 6.1119235 𝑒𝑒 − 004
2.159592
2.649447
−1.183929093 −0.988427076 2.8800225 𝑒𝑒 − 004 3.574060
Table (2.27) Numerical Solutions 𝒉𝒉
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 (𝑵𝑵 = 𝟏𝟏𝟏𝟏)
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎(𝑵𝑵 = 𝟏𝟏𝟏𝟏𝟏𝟏) 𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 , (𝑵𝑵 = 𝟓𝟓𝟓𝟓𝟓𝟓)
𝟐𝟐𝟐𝟐𝟐𝟐
1.7535342 𝑒𝑒 − 003
2.159592
3.1618076 10.765615 𝑒𝑒 − 005
2.1242876 𝑒𝑒 − 006
48.674457
𝟒𝟒𝟒𝟒𝟒𝟒
2.8800225 𝑒𝑒 − 004
3.574060
5.0009049 21.140608 𝑒𝑒 − 006
3.3489826 𝑒𝑒 − 007
99.062459
𝑩𝑩𝑩𝑩𝑩𝑩 𝟑𝟑𝟑𝟑𝟑𝟑
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
6.1119235 𝑒𝑒 − 004
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
2.649447
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
1.0755989 14.047329 𝑒𝑒 − 005
77
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
7.2090685 𝑒𝑒 − 007
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
69.558011
Chapter Two
Block By Block Method
Example (2.4): Consider test example (4), which is a linear higher fractional VIDE of constant multi-time Retarded delay with variable coefficients. Take 𝑁𝑁 = 10, ℎ = 0.1; 𝑡𝑡𝑟𝑟 = 𝑎𝑎 + 𝑟𝑟ℎ 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 − 1. From equations (2.21, 2.33 and 2.44) which the number of fractional parts is 𝑛𝑛 = 3. Tables (2.28, 2.29 and 2.30) contains all the value of 𝐻𝐻𝑟𝑟𝜗𝜗 for Two, Three and Four Blocks method, respectively, by running the programs Main (2Block, 3Block and 4Block), see the appendix. Table (2.28) 𝒓𝒓
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐), for Two Block method 𝑯𝑯𝟏𝟏𝒓𝒓
𝑯𝑯𝟐𝟐𝒓𝒓
0
8.450271294
8.150111651
2
7.283165451
7.006332206
1 3 4 5
7.855353793 6.736096716
5.968447122
5.72816332
5.273445123
8
4.480372059
9
6.472797254
6.21678944
6 7
7.566274674
5.496351203 5.059901703
4.856200976
4.662847588 4.309332107
4.150314049
4.003934259
78
Chapter Two
Block By Block Method
Table (2.29)
𝒓𝒓
𝑯𝑯𝟏𝟏𝒓𝒓
𝑯𝑯𝟐𝟐𝒓𝒓
𝑯𝑯𝟑𝟑𝒓𝒓
0
10.56035275
10.32932624
10.10081768
2
9.213713776
8.999134475
8.787630598
1 3 4 5 6
9.874912405
9.651698662
8.579306389
5.51585045
7.589282964
7.219596956
6.86652112
9
8.172631725
7.780017267
7.402432315 6.370577761
9.431267678
8.374269622
7.974507903
7 8
𝒓𝒓
𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑), for Three Block method
7.040912987
6.696566835
6.531200545
6.214859264
5.918805693
6.064211286
5.778820175
5.644438447
5.393252563
5.276847822
Table (2.30) 𝑯𝑯𝝑𝝑𝒓𝒓 , 𝒉𝒉 = 𝟎𝟎. 𝟏𝟏 , (𝝑𝝑 = 𝟏𝟏, 𝟐𝟐, 𝟑𝟑, 𝟒𝟒) , for Four Block method 𝑯𝑯𝟏𝟏𝒓𝒓
𝑯𝑯𝟐𝟐𝒓𝒓
𝑯𝑯𝟑𝟑𝒓𝒓
𝑯𝑯𝟒𝟒𝒓𝒓
0
12.36077174
12.16942693
11.97955749
11.79120077
2
10.87348269
10.69504127
10.51840187
10.3436101
1 3 4 5 6 7 8 9
11.60439506 10.17071275 9.499044996 8.861750471 8.262444244 7.705121576 7.194197918 6.734553097
11.41917963 9.999757768 9.336363806 8.708213238 8.119012645 7.572858431 7.074277823 6.628274182
79
11.23559473
11.05368165
9.830794324
9.663872835
8.557108476
8.408497766
7.443566314
7.317320438
9.175883607 7.978269343 6.957641223 6.52562409
9.01766012
7.840282395 6.844371238 6.426694684
Chapter Two
Block By Block Method
As the first two examples before, we can calculate all the values of 𝑟𝑟̅𝑗𝑗 = �𝜏𝜏𝑗𝑗 /ℎ� for
all j = 0,1,2, … 𝑚𝑚 , ℎ = (𝑏𝑏 − 𝑎𝑎)/𝑝𝑝𝑝𝑝 and 𝑢𝑢 �𝑡𝑡𝑁𝑁𝑁𝑁 −𝑟𝑟̅𝑗𝑗 � for Blocks-method (Two,
Three and Four) by running the Main (2Block, 3Block and 4Block) programs, respectively. While here the constant multi-time delays are: 𝜏𝜏 = 𝜏𝜏0 = 0.45 , 𝜏𝜏1 = 0.7 and 𝜏𝜏2 = 0.45 so for all j = 0,1,2 : 𝜏𝜏 𝑗𝑗
For 2BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � , ℎ =
1
ℎ 𝜏𝜏 𝑗𝑗
20 1
ℎ
40
For 3BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � , ℎ = ℎ 𝜏𝜏 𝑗𝑗
For 4BM: evaluate 𝑟𝑟̅𝑗𝑗 = � � , ℎ =
30 1
thus: 𝑟𝑟̅𝑗𝑗 = {32,10,12} thus:
𝑟𝑟̅𝑗𝑗 = {48,15,18}
thus: 𝑟𝑟̅𝑗𝑗 = {64,20,25}
Table (2.34) presents a comparison between the exact solution and the numerical solution by Block-by-Block method of Blocks (Two, Three and Four) it also includes the least square error and running time from running the (2Block, 3Block and 4Block) programs, respectively. Table (2.34)
0
Exact Solution 1
0.2
1.016
𝒕𝒕
0.1 0.3 0.4 0.5
1.002
1.003480325
1.003233883
1.002385233
1.054
1.061587523
1.060055214
1.056147656
1.128 1.25
0.6
1.432
0.8
2.024
0.7 0.9 1
Numerical Solution 𝐰𝐰𝐰𝐰𝐰𝐰𝐰𝐰 𝑵𝑵 = 𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒 1 1 1
1.686 2.458 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
3
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
1.020153004
1.019384629
1.139658593
1.137125233
1.266362373
1.262579583
1.453755065
1.448459477
1.713880244
1.706804177
2.058704614
2.049605131
2.499969561
2.488690187
3.048984903
3.035559992
7.0966253 𝑒𝑒 − 003
3.8569111 𝑒𝑒 − 003
1.723161
2.514574
80
1.017138205 1.131394146 1.254888697 1.438656653 1.694716024 2.035059673 2.471616619 0.016192241 7.3167336 𝑒𝑒 − 004 4.402013
Chapter Two
Block By Block Method
Also, the result in Table (2.35) shows the least square errors and running times for Block-by-Block of (Two, Three and Four) with different values of 𝑁𝑁 and ℎ by running the Main (2Block, 3Block and 4Block) programs, respectively. Table (2.35)
Numerical Solutions 𝒉𝒉 𝑩𝑩𝑩𝑩𝑩𝑩 𝟐𝟐𝟐𝟐𝟐𝟐 𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎 (𝑵𝑵 = 𝟏𝟏𝟏𝟏) 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 (𝑵𝑵 = 𝟏𝟏𝟏𝟏𝟏𝟏) 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
𝒉𝒉 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 , (𝑵𝑵 = 𝟓𝟓𝟓𝟓𝟓𝟓) 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻 /𝑺𝑺𝑺𝑺𝑺𝑺
7.0966253 𝑒𝑒 − 003
1.723161
5.1607132 𝑒𝑒 − 005
12.889662
3.1309167 𝑒𝑒 − 006
54.529503
7.3167336 𝑒𝑒 − 004
4.402013
7.8364894 𝑒𝑒 − 006
22.591391
4.6567582 𝑒𝑒 − 007
115.079308
3.8569111 𝑒𝑒 − 003
2.514574
1.7158103 𝑒𝑒 − 005
16.714084
1.0278157 𝑒𝑒 − 006
79.947837
2.7 Discussion: Finding the exact solution for Linear Volterra Integro-Fractional Differential equations of constant multi-time Retarded delay with variable coefficients is difficult or mostly impossible and we need more mathematical computation. In this chapter we used Block-by-Block Method to solve (VIFDEs) with multi-time delay, and it contained Two, Three and Four. The order of Caputo derivatives in the range of (0,1) for each Block, and the computer program MatLab was written and tested for several examples, moreover the comparison of computing accuracy and speed the least square error and running time of the associative program were given in tabular forms, from which the following points are concluded: 1. The numerical experiments show that the Fourth Block method is the most popular method that gives the best approximation to the exact solution amongst the other Blocks Methods. 2. We get a good accuracy if we choose 𝑁𝑁 sufficiently large (small step size ℎ) see tables (2.11, 19, 27 and 35). 3. Two Block method is faster, and lower in accuracy than the other Blocks Methods.
81
CHAPTER THREE
Least Square Orthogonal Method
Chapter Three
Least Square Orthogonal Method
3.1 Introduction: Orthogonal polynomials and their properties are extremely important in approximation theory; also they have many important applications in such areas as mathematical physics and many others. Although, they play a central role in modern numerical analysis in particular with least-squareiu techniques. As well as they are defined in such a way that the interpolation gives the best fit over the entire region. One of the main properties of the orthogonal series is its fast rate of convergence [32, 38, 43]. In mathematics, the classical orthogonal sets are most commonly occurring in applications such as The Chebyshev, Legendre, Laguerre and Hermite Polynomials. We can see the importance of orthogonal polynomials, particularly in the numerical treatments for integral and integro-differential equations with the aid of the least squares data fitting. In fact, these ideas can be generalized from integral to fractional equations with multi-time delays. AL-Rawi [8] and Qassim [65] used Laguerre polynomial to approximate the solution of the first kind integral equations depending on the principle of the leastsquares data fitting. On the other hand Al-Ani [5], Kalwi [7] applied this technique to treat numerically FIE's of second kind using Laguerre, Hermit and Legendre and Chebyshev polynomials, respectively. Also, in [62] Chebyshev and Legendre orthogonal were used with the aid of least square data to solve LVIFDE’s. Moreover, Abbas Saadatmandi with Mehdi Dehgan [1] and Memudu W.W with Taiwo O.A [44] used orthogonal polynomials to solve numerically the higher-order linear Fredholm integro-differential difference equations. In this chapter, we develop a good new framework to obtain the numerical solution of higher-fractional order VIDE’s of multi-time delay with variable coefficients by using orthogonal polynomials including Chebyshev and Legendre polynomials with the aid of least square techniques. Moreover, the involved integral operators in this method were evaluated numerically by applying Clenshaw-Curtis formula. Finally, LU-factorization method has been used to determine the values of coefficients from the resulting normal equations. In order to express these solutions, algorithms with computer programs in MatLab are written and many numerical examples are given for illustration.
82
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3.2 Orthogonal polynomials: [51, 52, 65] Orthogonal polynomials were classes of polynomials {𝐺𝐺𝑖𝑖 ; 𝑖𝑖 = 0,1,2, … } defined over a range [𝑎𝑎, 𝑏𝑏] that an orthogonality relation 𝑏𝑏
� 𝑤𝑤 (𝑡𝑡 )𝐺𝐺𝑖𝑖 (𝑡𝑡)𝐺𝐺𝑗𝑗 (𝑡𝑡 )𝑑𝑑𝑑𝑑 = 𝛿𝛿𝑖𝑖𝑖𝑖 𝐶𝐶𝑗𝑗
𝑎𝑎
Where 𝑤𝑤(𝑡𝑡) is a weighting function and 𝛿𝛿𝑖𝑖𝑖𝑖 is the kronecker delta (equal to 1 if 𝑖𝑖 = 𝑗𝑗 and to 0 otherwise) if these relationships hold for all 𝑖𝑖, the family of polynomials {𝐺𝐺𝑖𝑖 (𝑡𝑡)} constitute a set of orthogonal polynomials. Among these orthogonal polynomials there are two important ones, distinguished by their particular simplicity, these are: Chebyshev and Legendre polynomials. 3.2.1 Shifted Chebyshev Polynomials: [36, 52, 54] The well-known Chebyshev polynomials of order (𝑟𝑟 ∈ ℤ+) are defined on the interval 𝑥𝑥 ∈ [−1,1] and can be determined with the aid of the following simple recurrence formula: 𝒯𝒯𝑟𝑟+1 (𝑥𝑥) = 2𝑥𝑥 𝒯𝒯𝑟𝑟 (𝑥𝑥) − 𝒯𝒯𝑟𝑟−1 (𝑥𝑥),
𝑟𝑟 = 1,2,3, …
Where 𝒯𝒯0 (𝑥𝑥) = 1 and 𝒯𝒯1 (𝑥𝑥) = 𝓍𝓍 . While the Chebyshev polynomials on the interval 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] , it is necessary to shift defining domain by means of the following linear transformation: 𝓍𝓍 = 2 �
𝑡𝑡 =
𝑡𝑡 − 𝑎𝑎 �−1 𝑏𝑏 − 𝑎𝑎
𝑏𝑏 − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 𝑥𝑥 + 2 2
,
,
𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏
− 1 ≤ 𝑥𝑥 ≤ 1
… (3.1)
The shifted Chebyshev polynomials in 𝑡𝑡 are obtained as follows:
… (3.2)
𝒯𝒯0 (𝑡𝑡) = 1
,
𝒯𝒯1 (𝑥𝑥) =
2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 𝑏𝑏 − 𝑎𝑎
2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 𝒯𝒯𝑟𝑟+1 (𝑡𝑡) = 2 � � 𝒯𝒯𝑟𝑟 (𝑡𝑡) − 𝒯𝒯𝑟𝑟−1 (𝑡𝑡), 𝑏𝑏 − 𝑎𝑎
𝑟𝑟 = 0,1,2, …
� … (3.3)
The family of shifted Chebyshev polynomials is orthogonal with Chebyshev weighting function 1/�(𝑡𝑡 − 𝑎𝑎)(𝑏𝑏 − 𝑡𝑡) , on the interval [𝑎𝑎, 𝑏𝑏], that is: 83
Chapter Three
Least Square Orthogonal Method 𝑏𝑏
1 � = � 2 𝜋𝜋𝛿𝛿𝑟𝑟ℓ �(𝑡𝑡 − 𝑎𝑎)(𝑏𝑏 − 𝑡𝑡) 𝜋𝜋 𝑎𝑎 𝒯𝒯𝑟𝑟 (𝑡𝑡)𝒯𝒯ℓ (𝑡𝑡)𝑑𝑑𝑑𝑑
for 𝑟𝑟 ≠ 0, ℓ ≠ 0 for
𝑟𝑟 = ℓ = 0
Where 𝛿𝛿𝑟𝑟ℓ is the kronecker delta. On the other hand, the shifted Chebyshev polynomials can also be defined in terms of the sums: 𝑟𝑟 ⌊ ⌋ 2
𝑟𝑟−2𝑘𝑘 𝑟𝑟 (−1)𝑘𝑘 2𝑟𝑟−2𝑘𝑘 𝑟𝑟 − 𝑘𝑘 𝑡𝑡 − 𝑎𝑎 𝒯𝒯𝑟𝑟 (𝑡𝑡) = � , � � �2 � � − 1� 𝑘𝑘 2 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 − 𝑘𝑘 𝑘𝑘=0
𝑟𝑟 ≥ 1
Also, the 𝑛𝑛-derivative of 𝑟𝑟- Chebyshev polynomials on any bounded interval [𝑎𝑎, 𝑏𝑏] is formulated as: 𝑑𝑑 𝑛𝑛 𝒯𝒯𝑟𝑟 (𝑡𝑡 ) = 𝑑𝑑𝑡𝑡 𝑛𝑛 𝑟𝑟−𝑛𝑛 ⌊ ⌋ 2
𝑟𝑟−2𝑘𝑘−𝑛𝑛 ⎧ 𝑟𝑟 (𝑟𝑟 − 𝑘𝑘 − 1)! 2 𝑛𝑛 𝑡𝑡 − 𝑎𝑎 𝑘𝑘 𝑟𝑟−2𝑘𝑘 ⎪ � � �2 � � − 1� 𝑖𝑖𝑖𝑖 𝑟𝑟 > 𝑛𝑛 ⎪ 2 � (−1) 2 𝑏𝑏 − 𝑎𝑎 𝑘𝑘! (𝑟𝑟 − 2𝑘𝑘 − 𝑛𝑛)! 𝑏𝑏 − 𝑎𝑎 𝑘𝑘=0
⎨ 𝑟𝑟−1 2 𝑟𝑟 � 2 𝑟𝑟! � ⎪ ⎪ 𝑏𝑏 − 𝑎𝑎 ⎩0
𝑖𝑖𝑖𝑖 𝑟𝑟 = 𝑛𝑛
3.2.2 Shifted Legendre Polynomials: [36, 52, 54]
𝑖𝑖𝑖𝑖 𝑟𝑟 < 𝑛𝑛
… (3.4)
The set of Legendre polynomials over the interval 𝑥𝑥 ∈ [−1,1] , a simple calculation formula for it of 𝑟𝑟-order is a recursive formulation: 𝒫𝒫𝑟𝑟+1 (𝑥𝑥) =
2𝑟𝑟 + 1 𝑟𝑟 𝑥𝑥𝒫𝒫𝑟𝑟 (𝑥𝑥) − 𝒫𝒫 (𝑥𝑥) 𝑟𝑟 + 1 𝑟𝑟 + 1 𝑟𝑟−1
, 𝑟𝑟 = 1,2, …
Where 𝒫𝒫0 (𝑥𝑥) = 1 and 𝒫𝒫1 (𝑥𝑥) = 𝑥𝑥 . While the Legendre polynomials on the interval 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] , it is necessary to shift the defining domain by using (3.1) and (3.2) to obtain the shifted Legendre polynomials in 𝑡𝑡 as the following: 2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 𝒫𝒫0 (𝑡𝑡) = 1 , 𝒫𝒫1 (𝑡𝑡) = ⎫ 𝑏𝑏 − 𝑎𝑎 … (3.5) (2𝑟𝑟 + 1)(2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏) 𝑟𝑟 ⎬ 𝒫𝒫𝑟𝑟 (𝑡𝑡) − 𝒫𝒫 (𝑡𝑡) , 𝑟𝑟 = 1,2, … 𝒫𝒫𝑟𝑟+1 (𝑡𝑡) = (𝑟𝑟 + 1)(𝑏𝑏 − 𝑎𝑎) ⎭ 𝑟𝑟 + 1 𝑟𝑟−1 The family of shifted Legendre polynomials is orthogonal with Legendre weighting function 1 , on the interval [𝑎𝑎, 𝑏𝑏] that is: 84
Chapter Three
Least Square Orthogonal Method 𝑏𝑏
� 𝒫𝒫𝑟𝑟 (𝑡𝑡)𝒫𝒫ℓ (𝑡𝑡)𝑑𝑑𝑑𝑑 =
𝑎𝑎
𝑏𝑏 − 𝑎𝑎 𝛿𝛿 2𝑟𝑟 + 1 𝑟𝑟ℓ
Where 𝛿𝛿𝑟𝑟ℓ is the kronecker delta. The shifted Legendre polynomial can also be defined in terms of the sums 𝑟𝑟 ⌊ ⌋ 2
𝑟𝑟−2𝑘𝑘 1 𝑡𝑡 − 𝑎𝑎 𝑟𝑟 2𝑟𝑟 − 2𝑘𝑘 𝑘𝑘 𝒫𝒫𝑟𝑟 (𝑡𝑡) = 𝑟𝑟 �(−1) � � � 𝑟𝑟 ≥ 1 � �2 � � − 1� 𝑘𝑘 𝑟𝑟 2 𝑏𝑏 − 𝑎𝑎 𝑘𝑘=0
The 𝑛𝑛-derivative of the 𝑟𝑟-Legendre polynomials on any closed bounded interval [𝑎𝑎, 𝑏𝑏] are formulated as: 𝑑𝑑 𝑛𝑛 𝒫𝒫𝑟𝑟 (𝑡𝑡) 𝑑𝑑𝑡𝑡 𝑛𝑛 𝑟𝑟−𝑛𝑛 ⌊
=
⌋
⎧1 2 𝑟𝑟−2𝑘𝑘−𝑛𝑛 (2𝑟𝑟 − 2𝑘𝑘 )! (−1)𝑘𝑘 2 𝑛𝑛 𝑡𝑡 − 𝑎𝑎 ⎪ 𝑖𝑖𝑖𝑖 𝑟𝑟 > 𝑛𝑛 ⎪ 2𝑟𝑟 � 𝑘𝑘! (𝑟𝑟 − 𝑘𝑘 )! (𝑟𝑟 − 2𝑘𝑘 − 𝑛𝑛)! �𝑏𝑏 − 𝑎𝑎� �2 �𝑏𝑏 − 𝑎𝑎� − 1� 𝑘𝑘=0
⎨ (2𝑟𝑟)! 2 𝑟𝑟 � � ⎪ ⎪ 2𝑟𝑟 𝑟𝑟! 𝑏𝑏 − 𝑎𝑎 ⎩0
𝑖𝑖𝑖𝑖 𝑟𝑟 = 𝑛𝑛
𝑖𝑖𝑖𝑖 𝑟𝑟 < 𝑛𝑛
… (3.6)
4.3 Gauss and Clenshaw-Curtis Formulas: [16, 20, 24, 36, 52] 𝑁𝑁-Gaussian quadrature rules are given for numerically evaluating integrals on any closed bounded interval. Let 𝑤𝑤 be a given weight function over an interval [𝑎𝑎, 𝑏𝑏] and let 𝐺𝐺𝑤𝑤(𝑁𝑁) be the corresponding 𝑁𝑁-open Gauss-quadrature formula
For the integral
𝑁𝑁
(𝑁𝑁) 𝐺𝐺𝑤𝑤
= � 𝑤𝑤𝑟𝑟 𝑓𝑓(𝑡𝑡𝑟𝑟 )
𝑏𝑏
𝑏𝑏
𝑟𝑟=0
(𝑁𝑁)
(𝑁𝑁)
… (3.8)
𝐼𝐼𝐼𝐼 = � 𝑔𝑔(𝑡𝑡)𝑑𝑑𝑑𝑑 = � 𝑤𝑤(𝑡𝑡) 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝑎𝑎
For various 𝑁𝑁, where
(𝑁𝑁) 𝑡𝑡𝑟𝑟
𝑎𝑎
… (3.7)
are called nodes, here, associated with zeros of (𝑁𝑁)
orthogonal polynomials are integration points and 𝑤𝑤𝑟𝑟 is called weights of the quadrature formula related the orthogonal polynomials. In practice, we try to (𝑁𝑁)
minimize the error 𝐼𝐼𝐼𝐼 − 𝑤𝑤𝑟𝑟 𝑔𝑔 by taking various values of 𝑁𝑁 .
85
Chapter Three
Least Square Orthogonal Method
3.3.1 Gauss-Chebyshev Quadrature Formula: Let the weight function be 𝑤𝑤(𝑡𝑡) = 1/�(𝑡𝑡 − 𝑎𝑎)(𝑏𝑏 − 𝑡𝑡) on the interval [𝑎𝑎, 𝑏𝑏]. Then, 𝑁𝑁-Gaussian quadrature formula for the integral 𝑏𝑏
𝑏𝑏
𝐼𝐼𝐼𝐼 = � 𝑔𝑔(𝑡𝑡 )𝑑𝑑𝑑𝑑 = � 𝑎𝑎
𝑎𝑎
(𝑁𝑁) 𝑡𝑡𝑟𝑟
𝑓𝑓 (𝑡𝑡 )
�(𝑡𝑡 − 𝑎𝑎)(𝑏𝑏 − 𝑡𝑡 )
𝑑𝑑𝑑𝑑
����� Classified by the zeros (simply 𝑡𝑡𝑟𝑟 (𝑟𝑟 = 0: 𝑁𝑁) ) of the Chebyshev polynomial: The 𝑵𝑵 −Open Gauss-Chebyshev quadrature rule for 𝐼𝐼𝐼𝐼 has the form: (𝑁𝑁)
𝐺𝐺𝑤𝑤
𝑁𝑁−1
𝜋𝜋 = � 𝑓𝑓(𝑡𝑡𝑟𝑟 ) 𝑁𝑁
⎫ ⎪
… (3.9) ⎬ 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 2𝑟𝑟 + 1 𝜋𝜋 where 𝑡𝑡𝑟𝑟 = � � 𝑧𝑧𝑟𝑟 + � � and 𝑧𝑧𝑟𝑟 = cos � �⎪ 2 2 𝑁𝑁 2 ⎭ The 𝑵𝑵-Closed Gauss-Chebyshev quadrature rule for 𝐼𝐼𝐼𝐼 has the form: 𝑟𝑟=0
𝑁𝑁−1
𝜋𝜋 = � 𝑁𝑁
(𝑁𝑁) 𝐺𝐺𝑤𝑤
′′
⎫ ⎪
𝑓𝑓(𝑡𝑡𝑟𝑟 )
… (3.10) ⎬ 𝜋𝜋 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 where 𝑡𝑡𝑟𝑟 = � � 𝑧𝑧𝑟𝑟 + � � and 𝑧𝑧𝑟𝑟 = cos �𝑟𝑟 � ⎪ 2 ⎭ 2 2 Where the double prime (") on the summation sign implies that the first and end terms are halved . 3.3.2 Gauss-Legendre Quadrature Formula: Let the weight function is 𝑤𝑤(𝑡𝑡) = 1 on the interval [𝑎𝑎, 𝑏𝑏].Then, 𝑏𝑏 𝑁𝑁-quadrature formula for the integral 𝐼𝐼𝐼𝐼 = ∫𝑎𝑎 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 has the form: (𝑁𝑁)
Where
𝑏𝑏−𝑎𝑎
𝑡𝑡𝑟𝑟 = �
2
𝐺𝐺𝑤𝑤
𝑏𝑏+𝑎𝑎
� 𝑧𝑧𝑟𝑟 + �
2
𝑟𝑟=0
𝑁𝑁−1
𝑏𝑏 − 𝑎𝑎 (𝑁𝑁) = � 𝑤𝑤𝑟𝑟 𝑓𝑓(𝑡𝑡𝑟𝑟 ) 2 𝑟𝑟=0
� and 𝑧𝑧𝑟𝑟 are (𝑁𝑁)
polynomials 𝐿𝐿𝑁𝑁 (𝑡𝑡). the coefficients 𝑤𝑤𝑟𝑟 formula:
the 𝑟𝑟-th
zeros
𝑁𝑁-Legendre
can be calculated using the following
2 (1 − 𝑧𝑧𝑟𝑟2 )[𝐿𝐿′𝑁𝑁 (𝑧𝑧𝑟𝑟 )]2 3.3.3 Cleanshaw-Curtis Integration Formula: (𝑁𝑁)
𝑤𝑤𝑟𝑟
of
… (3.11)
=
… (3.12)
The Clenshaw-Curtis quadrature formula is the formula (3.7) based on the extreme Chebyshev zeros (as formed in 3.10) and an expansion of the integrand in terms of Chebyshev polynomials. The general form for 𝐼𝐼𝐼𝐼 in equation (3.8) can be formed as follows: 𝑁𝑁
′′ 𝑏𝑏 − 𝑎𝑎 (𝑁𝑁) 𝐺𝐺𝑤𝑤 𝑔𝑔 = � 2
𝑟𝑟=0
86
(𝑁𝑁)
(𝑁𝑁)
𝑤𝑤𝑟𝑟 𝑔𝑔(𝑡𝑡𝑟𝑟 )
… (3.13)
Chapter Three
Least Square Orthogonal Method
Where the double prime on the summation symbol is that the first and end term are halved and (𝑁𝑁)
𝑡𝑡𝑟𝑟
=�
𝑏𝑏 − 𝑎𝑎 (𝑁𝑁) 𝑏𝑏 + 𝑎𝑎 � 𝑧𝑧𝑟𝑟 + � � 2 2
(𝑁𝑁)
𝑤𝑤𝑟𝑟
𝑀𝑀
′′ 4 = � 𝑁𝑁
ℓ=0
,
𝑣𝑣ℓ cos �ℓ 𝑟𝑟
1 𝑣𝑣ℓ = � 1 − ℓ2 0
(𝑁𝑁)
𝑧𝑧𝑟𝑟
= cos �𝑟𝑟
𝜋𝜋 � ; 𝑀𝑀 ∈ ℤ+ 𝑀𝑀
𝜋𝜋 � 2
; if ℓ is evn
; if ℓ is odd
3.4 Solution Technique for Linear VIFDEs of Constant multi-Time Retarded Delay: In the present section, a new technique for solving VIFDEs with multi time Delay and variable coefficients applying least-square data fitting with the aid of orthogonal polynomials, here Chebyshev and Legendre polynomials here been presented. We need the following four basic lemmas before starting the solution of general form of our problem:
Lemma (3.1): [56]
The Caputo fractional derivative for order 𝛼𝛼 where 𝑚𝑚 − 1 < 𝛼𝛼 < 𝑚𝑚, 𝑚𝑚 = ⌈𝛼𝛼⌉ of 𝑡𝑡−𝑎𝑎 shifed Chebyshev polynomial of degree 𝑟𝑟 (≥ 1) 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) = 𝒯𝒯𝑟𝑟 �2 � � − 1�, on 𝑏𝑏 −𝑎𝑎 interval [a, b] can be formulated: ⌊𝑟𝑟 ⁄2⌋
𝑟𝑟 2 𝑚𝑚 Γ(𝑟𝑟 − 𝑘𝑘) 𝑟𝑟−2𝑘𝑘 𝐶𝐶 𝛼𝛼 ∗ (𝓍𝓍) = � 2 𝑀𝑀(𝑡𝑡; 𝑟𝑟, 𝑘𝑘, 𝑚𝑚) � (𝑡𝑡 − 𝑎𝑎)𝑚𝑚−𝛼𝛼 � (−1)𝑘𝑘 𝑎𝑎 𝐷𝐷𝑡𝑡 𝒯𝒯𝑟𝑟 2 𝑏𝑏 − 𝑎𝑎 Γ(𝑘𝑘 + 1)
where
𝑘𝑘 =0
… (3.14)
𝑀𝑀(𝑡𝑡; 𝑟𝑟, 𝑘𝑘, 𝑚𝑚) 0 , 𝑚𝑚 > 𝑟𝑟 − 2𝑘𝑘 ⎧ 1 ⎪ , 𝑚𝑚 = 𝑟𝑟 − 2𝑘𝑘 Γ(𝑚𝑚 − 𝛼𝛼 + 1) = 𝑟𝑟−2𝑘𝑘−𝑚𝑚 ⎨ 𝑡𝑡 − 𝑎𝑎 ℓ (−1)ℓ+𝑟𝑟−2𝑘𝑘−𝑚𝑚 ⎪ � �2 � �� , 𝑚𝑚 < 𝑟𝑟 − 2𝑘𝑘 ( ) 𝑏𝑏 − 𝑎𝑎 Γ ℓ + 𝑚𝑚 − α + 1 Γ(𝑟𝑟 − 2𝑘𝑘 − 𝑚𝑚 − ℓ + 1) ⎩ ℓ=0
We can see the complete proof in details.
87
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Least Square Orthogonal Method
Lemma (3.2): [62] The Caputo fractional derivative for order 𝛼𝛼 where 𝑚𝑚 − 1 < 𝛼𝛼 < 𝑚𝑚 ,𝑚𝑚 = ⌈𝛼𝛼⌉,
of shifed Legendre polynomial of degree 𝑟𝑟 (≥ 1), 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) = 𝒫𝒫𝑟𝑟 �2 �
𝑏𝑏−𝑎𝑎
interval [a, b] can be formulated: 𝐶𝐶 𝛼𝛼 𝑎𝑎 𝐷𝐷𝑡𝑡
𝒫𝒫𝑟𝑟∗ (𝑡𝑡) =
Where
𝑡𝑡−𝑎𝑎
� − 1�, on
⌊𝑟𝑟 ⁄2⌋
1 2 𝑚𝑚 Γ(2𝑟𝑟 − 2𝑘𝑘 + 1) 𝑚𝑚−𝛼𝛼 𝑘𝑘 (𝑡𝑡 (−1) − 𝑎𝑎) � 𝑀𝑀(𝑡𝑡; 𝑟𝑟, 𝑘𝑘, 𝑚𝑚) � � 2𝑟𝑟 𝑏𝑏 − 𝑎𝑎 Γ(𝑘𝑘 + 1)Γ(𝑟𝑟 − 𝑘𝑘 + 1) 𝑘𝑘=0
… (3.15)
𝑀𝑀(𝑡𝑡; 𝑟𝑟, 𝑘𝑘, 𝑚𝑚) 0 , 𝑚𝑚 > 𝑟𝑟 − 2𝑘𝑘 ⎧ 1 ⎪ , 𝑚𝑚 = 𝑟𝑟 − 2𝑘𝑘 Γ(𝑚𝑚 − 𝛼𝛼 + 1) = 𝑟𝑟−2𝑘𝑘−𝑚𝑚 ⎨ (−1)ℓ+𝑟𝑟−2𝑘𝑘−𝑚𝑚 𝑡𝑡 − 𝑎𝑎 ℓ ⎪ � �� , 𝑚𝑚 < 𝑟𝑟 − 2𝑘𝑘 �2 � Γ(ℓ + 𝑚𝑚 − α + 1)Γ(𝑟𝑟 − 2𝑘𝑘 − 𝑚𝑚 − ℓ + 1) 𝑏𝑏 − 𝑎𝑎 ⎩ ℓ=0
We can see the complete proof in details.
Lemma (3.3): For a fixed constant delay 𝜏𝜏 > 0 , a shifted 𝑚𝑚-Legendre polynomials on a closed bounded interval [𝑎𝑎, 𝑏𝑏] as formed in (3.5). The 𝜏𝜏-delay 𝑚𝑚- Legendre polynomial has the following representation: 𝓟𝓟(𝑡𝑡 − 𝜏𝜏) = 𝑯𝑯𝜏𝜏 𝓟𝓟(𝑡𝑡),
𝑡𝑡 − 𝜏𝜏 > 𝑎𝑎
… (3.16)
Where 𝓟𝓟(𝜂𝜂) is the column vector of dimension (𝑚𝑚 + 1 × 1) express and as [𝓟𝓟𝟎𝟎 (𝜂𝜂), 𝓟𝓟𝟏𝟏 (𝜂𝜂), … , 𝓟𝓟𝒎𝒎 (𝜂𝜂) ]𝑇𝑇 for 𝜂𝜂 = 𝑡𝑡 or 𝑡𝑡 − 𝜏𝜏 , aid, 𝑯𝑯𝜏𝜏 is a constant lower triangular matrix for any fixed 𝜏𝜏 defined as: 1 ℎ ⎛ 1,0 𝑯𝑯𝜏𝜏 = ⎜ ℎ2,0 ⋮ ⎝ℎ𝑚𝑚 ,0
0 1 ℎ2,1 ⋮ ℎ𝑚𝑚 ,1
0 0 1 ⋮
ℎ𝑚𝑚 ,2
… … …
0 0 0 ⋮
… … ℎ 𝑚𝑚 ,𝑚𝑚 −1
88
0 0⎞ 0⎟ ⋮ 1⎠ 𝑚𝑚 +1×𝑚𝑚 +1
Chapter Three
Least Square Orthogonal Method
While each element in constant matrix 𝑯𝑯𝜏𝜏 can be found by the following steps as in proof:
Recall the equation (3.5) and replacing 𝑡𝑡 with 𝑡𝑡 − 𝜏𝜏 , 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] and 𝜏𝜏 > 0 ,we get expanding and rearranging: 𝑟𝑟
𝒫𝒫𝑟𝑟 (𝑡𝑡 − 𝜏𝜏) = � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡)
… (3.17)
𝑘𝑘=0
where 𝑡𝑡 > 𝑎𝑎 + 𝜏𝜏 and 𝑟𝑟 ≥ 0. we assume that ℎ𝑟𝑟,𝑘𝑘 = 0 for all 𝑟𝑟 < 𝑘𝑘 , from the first few polynomials:
For 𝒓𝒓 = 𝟎𝟎 : from equation (3.5) we have 𝒫𝒫0 (𝑡𝑡 − 𝜏𝜏) = 1 and from the other hand in (3.17) we obtain, 0
� ℎ0,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡) = ℎ0,0 𝒫𝒫0 (𝑡𝑡) = ℎ0,0
𝑘𝑘=0
For 𝒓𝒓 = 𝟏𝟏 : from equation (3.17) we have 𝒫𝒫1 (𝑡𝑡 − 𝜏𝜏) = ℎ1,0 𝒫𝒫0 (𝑡𝑡) + ℎ1,1 𝒫𝒫1 (𝑡𝑡)
By equation (4.5), 𝒫𝒫0 (𝑡𝑡) = 1 and 𝒫𝒫1 (𝑡𝑡) = So
𝒫𝒫1 (𝑡𝑡 − 𝜏𝜏) =
2𝑡𝑡−𝑎𝑎−𝑏𝑏 𝑏𝑏−𝑎𝑎
thus ℎ0,0 = 1
… (3.18)
and
2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 2𝜏𝜏 2𝜏𝜏 − = 𝒫𝒫1 (𝑡𝑡) − 𝒫𝒫 (𝑡𝑡) 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 0 𝑏𝑏 − 𝑎𝑎
ℎ1,0 = −
2𝜏𝜏 𝑏𝑏 − 𝑎𝑎
and
For 𝒓𝒓 ≥ 𝟐𝟐 : Rewriting equation (3.5) as follows:
ℎ1,1 = 1
… (3.19)
𝑟𝑟 + 1 𝑟𝑟 2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 𝒫𝒫𝑟𝑟 (𝑡𝑡) = 𝒫𝒫𝑟𝑟+1 (𝑡𝑡) + 𝒫𝒫 (𝑡𝑡), 𝑟𝑟 ≥ 1 … (3.20) 2𝑟𝑟 + 1 2𝑟𝑟 + 1 𝑟𝑟−1 𝑏𝑏 − 𝑎𝑎
Since, from equation (3.17), using equation (3.5) with (3.20) and replacing 𝑡𝑡 by 𝑡𝑡 − 𝜏𝜏 , we have
89
Chapter Three
Least Square Orthogonal Method
𝑟𝑟+1
� ℎ𝑟𝑟+1,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡) = 𝒫𝒫𝑟𝑟+1 (𝑡𝑡 − 𝜏𝜏)
𝑘𝑘=0
=
2𝑟𝑟 + 1 2𝜏𝜏 𝑟𝑟 2𝑟𝑟 + 1 2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 𝒫𝒫𝑟𝑟 (𝑡𝑡 − 𝜏𝜏) − 𝒫𝒫𝑟𝑟 (𝑡𝑡 − 𝜏𝜏) − 𝒫𝒫 (𝑡𝑡 − 𝜏𝜏) 𝑟𝑟 + 1 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 + 1 𝑟𝑟−1 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 + 1 𝑟𝑟
𝑟𝑟
𝑘𝑘=0
𝑘𝑘=0
2𝑟𝑟 + 1 2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 2𝜏𝜏 = � � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡)� �ℎ𝑟𝑟,0 + � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡)� − 𝑟𝑟 + 1 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎
𝑟𝑟−1
𝑟𝑟 − � ℎ𝑟𝑟−1,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡) 𝑟𝑟 + 1
In more details, we can write the above equation as follows:
𝑘𝑘=0
𝑟𝑟+1
� ℎ𝑟𝑟+1,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡)
𝑘𝑘=0
𝑟𝑟
𝑘𝑘 2𝑟𝑟 + 1 𝑘𝑘 + 1 = 𝒫𝒫𝑘𝑘+1 (𝑡𝑡) + 𝒫𝒫 (𝑡𝑡)� �ℎ𝑟𝑟,0 𝒫𝒫1 (𝑡𝑡) + � ℎ𝑟𝑟,𝑘𝑘 � 2𝑘𝑘 + 1 𝑘𝑘−1 𝑟𝑟 + 1 2𝑘𝑘 + 1 𝑟𝑟
𝑘𝑘=1
𝑟𝑟−1
2𝜏𝜏 𝑟𝑟 − � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡)� − � ℎ𝑟𝑟−1,𝑘𝑘 𝒫𝒫𝑘𝑘 (𝑡𝑡) 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 + 1 𝑘𝑘=0
𝑘𝑘=0
… (3.21)
By equating the coefficient of 𝒫𝒫𝑘𝑘 (𝑡𝑡), 𝑘𝑘 = 0,1,2, … , 𝑟𝑟, 𝑟𝑟 + 1 on both sides of equation (3.21) for all 𝑟𝑟 = 1,2, … , 𝑚𝑚 while from the first we find ℎ0,0 , ℎ1,0 and ℎ1,1 and assum that ℎ𝑟𝑟,𝑘𝑘 = 0 ∀ 𝑟𝑟 < 𝑘𝑘 . so: For 𝒌𝒌 = 𝟎𝟎 : the coefficients on both sides in equation (3.20) which make 𝒫𝒫0 (𝑡𝑡) is: ℎ𝑟𝑟+1,0 =
2𝑟𝑟 + 1 2𝑟𝑟 + 1 2𝜏𝜏 𝑟𝑟 ℎ𝑟𝑟,1 − ℎ � � ℎ𝑟𝑟,0 − 3(𝑟𝑟 + 1) 𝑟𝑟 + 1 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 + 1 𝑟𝑟−1,0
For 𝑟𝑟 = 1,2, … , 𝑟𝑟 − 1
… (3.22)
For 𝒓𝒓 ≥ 𝟐𝟐 : the coefficients of 𝒫𝒫𝑘𝑘 (𝑡𝑡) in equation (3.21), after using simple induction, we get: ℎ𝑟𝑟+1,𝑘𝑘 =
2𝑟𝑟 + 1 𝑘𝑘 + 1 𝑘𝑘 2𝜏𝜏 ℎ𝑟𝑟,𝑘𝑘+1 + ℎ𝑟𝑟,𝑘𝑘−1 − ℎ � � 𝑟𝑟 + 1 2𝑘𝑘 + 3 2𝑘𝑘 − 1 𝑏𝑏 − 𝑎𝑎 𝑟𝑟,𝑘𝑘 𝑟𝑟 − ℎ … (3.23) 𝑟𝑟 + 1 𝑟𝑟−1,𝑘𝑘 90
Chapter Three
Least Square Orthogonal Method
For 𝒌𝒌 = 𝒓𝒓 : The coefficient 𝒫𝒫𝑘𝑘 (𝑡𝑡) in equation (3.21) we obtain: ℎ𝑟𝑟+1,𝑟𝑟 =
2𝑟𝑟 + 1 2𝜏𝜏 𝑟𝑟 ℎ𝑟𝑟,𝑟𝑟−1 − � � 𝑟𝑟 + 1 2𝑟𝑟 − 1 𝑏𝑏 − 𝑎𝑎
For 𝒌𝒌 = 𝒓𝒓 + 𝟏𝟏 : we have it from equation (3.21) and using (3.18): ℎ𝑟𝑟+1,𝑟𝑟+1 = ℎ𝑟𝑟,𝑟𝑟 = ℎ𝑟𝑟−1,𝑟𝑟−1 = ⋯ = ℎ1,1 = ℎ0,0 = 1
… (3.24) … (3.25)
Clearly we have ℎ𝑟𝑟,𝑟𝑟 = 1for all 𝑟𝑟 = 0,1,2, … , 𝑚𝑚 thus equation (3.18, 3.19, 3.22-25) defines all elements in constant matrix 𝑯𝑯𝜏𝜏 in equation (3.16) which is the lower triangular matrix of dimension m + 1 × m + 1 .
Lemma (3.4): For a fixed constant delay 𝜏𝜏 > 0 , a shifted 𝑚𝑚-Chebyshev polynomials on a closed bounded interval [𝑎𝑎, 𝑏𝑏] as formed in (3.3). The 𝜏𝜏-delay 𝑚𝑚-Chebyshev polynomials have the following representation: 𝓣𝓣(𝑡𝑡 − 𝜏𝜏) = 𝑯𝑯𝜏𝜏 𝓣𝓣(𝑡𝑡),
𝑡𝑡 − 𝜏𝜏 > 𝑎𝑎
… (3.26)
Where 𝓣𝓣(𝜂𝜂) is the column vector of dimension (𝑚𝑚 + 1 × 1) express as [𝓣𝓣𝟎𝟎 (𝜂𝜂), 𝓣𝓣𝟏𝟏 (𝜂𝜂), … , 𝓣𝓣𝒎𝒎 (𝜂𝜂) ]𝑇𝑇 for 𝜂𝜂 = 𝑡𝑡 or 𝑡𝑡 − 𝜏𝜏 , aid, 𝑯𝑯𝜏𝜏 is a constant lower triangular matrix for any fixed 𝜏𝜏 which is defined as: 1 ℎ ⎛ 1,0 𝑯𝑯𝜏𝜏 = ⎜ ℎ2,0 ⋮ ⎝ℎ𝑚𝑚 ,0
0 1 ℎ2,1 ⋮ ℎ𝑚𝑚 ,1
0 0 1 ⋮
ℎ𝑚𝑚 ,2
… … …
0 0 0 ⋮
… … ℎ 𝑚𝑚 ,𝑚𝑚 −1
0 0⎞ 0⎟ ⋮ 1⎠ 𝑚𝑚 +1×𝑚𝑚 +1
While each element in constant matrix 𝑯𝑯𝜏𝜏 can be found by the following steps as in proof: Recall the equation (3.3) and replacing 𝑡𝑡 with 𝑡𝑡 − 𝜏𝜏 init, 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏]and 𝜏𝜏 > 0 ,we get expanding and rearranging: 𝑟𝑟
𝒯𝒯𝑟𝑟 (𝑡𝑡 − 𝜏𝜏) = � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) 𝑘𝑘=0
… (3.27)
Where 𝑡𝑡 > 𝑎𝑎 + 𝜏𝜏 and 𝑟𝑟 ≥ 0. We assume that ℎ𝑟𝑟,𝑘𝑘 = 0 for all 𝑟𝑟 < 𝑘𝑘 , from the first few polynomials: 91
Chapter Three
Least Square Orthogonal Method
For 𝒓𝒓 = 𝟎𝟎 : from equation (3.3) we have 𝒯𝒯0 (𝑡𝑡 − 𝜏𝜏) = 1 and from the other hand in (3.27) we obtain, 0
� ℎ0,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) = ℎ0,0 𝒯𝒯0 (𝑡𝑡) = ℎ0,0
𝑘𝑘=0
For 𝒓𝒓 = 𝟏𝟏 : from equation (3.27) we have
thus
ℎ0,0 = 1
… (3.28)
𝒯𝒯1 (𝑡𝑡 − 𝜏𝜏) = ℎ1,0 𝒯𝒯0 (𝑡𝑡) + ℎ1,1 𝒯𝒯1 (𝑡𝑡)
By equation (4.3), 𝒯𝒯0 (𝑡𝑡) = 1 and 𝒯𝒯1 (𝑡𝑡) = 𝒯𝒯1 (𝑡𝑡 − 𝜏𝜏) =
2𝑡𝑡−𝑎𝑎−𝑏𝑏 𝑏𝑏 −𝑎𝑎
and
2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 2𝜏𝜏 2𝜏𝜏 − = 𝒯𝒯1 (𝑡𝑡) − 𝒯𝒯 (𝑡𝑡) 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 0 𝑏𝑏 − 𝑎𝑎
so by equating the above equation, we get ℎ1,0 = −
2𝜏𝜏 𝑏𝑏 − 𝑎𝑎
and
For 𝒓𝒓 ≥ 𝟐𝟐 : Rewriting equation (3.3) as follows: 2�
ℎ1,1 = 1
2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 � 𝒯𝒯𝑟𝑟 (𝑡𝑡) = 𝒯𝒯𝑟𝑟+1 (𝑡𝑡) + 𝒯𝒯𝑟𝑟−1 (𝑥𝑥), 𝑏𝑏 − 𝑎𝑎
… (3.29) 𝑟𝑟 ≥ 1
… (3.30)
Since, from equation (3.27), using equation (3.3) with (3.30) and replacing 𝑡𝑡 by 𝑡𝑡 − 𝜏𝜏 , we get: 𝑟𝑟+1
� ℎ𝑟𝑟+1,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) = 𝒯𝒯𝑟𝑟+1 (𝑡𝑡 − 𝜏𝜏)
𝑘𝑘=0
2𝜏𝜏 2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 = 2� 𝒯𝒯 (𝑡𝑡 − 𝜏𝜏) − 𝒯𝒯𝑟𝑟−1 (𝑡𝑡 − 𝜏𝜏) � 𝒯𝒯𝑟𝑟 (𝑡𝑡 − 𝜏𝜏) − 2 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 𝑏𝑏 − 𝑎𝑎 𝑟𝑟 𝑟𝑟 2𝜏𝜏 2𝑡𝑡 − 𝑎𝑎 − 𝑏𝑏 = 2� � �ℎ𝑟𝑟,0 + � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡)� − 2 � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑟𝑟−1
− � ℎ𝑟𝑟−1,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡)
𝑘𝑘=0
𝑘𝑘=0
In more details, we can write the above equation as follows:
92
𝑘𝑘=0
Chapter Three
Least Square Orthogonal Method
𝑟𝑟+1
𝑟𝑟
𝑘𝑘=0
𝑘𝑘=0
� ℎ𝑟𝑟+1,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) = 2ℎ𝑟𝑟,0 𝒯𝒯1 (𝑡𝑡) + � ℎ𝑟𝑟,𝑘𝑘 [𝒯𝒯𝑘𝑘+1 (𝑡𝑡) + 𝒯𝒯𝑘𝑘−1 (𝑡𝑡)] 𝑟𝑟
𝑟𝑟−1
𝑘𝑘=0
𝑘𝑘=0
4𝜏𝜏 − � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) − � ℎ𝑟𝑟−1,𝑘𝑘 𝒯𝒯𝑘𝑘 (𝑡𝑡) 𝑏𝑏 − 𝑎𝑎
… (3.31)
By equating the coefficient of 𝒯𝒯𝑘𝑘 (𝑡𝑡), 𝑘𝑘 = 0,1,2, … , 𝑟𝑟, 𝑟𝑟 + 1 on both sides of equation (3.31) for all 𝑟𝑟 = 1,2, … , 𝑚𝑚 while from the first we find ℎ0,0 , ℎ1,0 and ℎ1,1 and assum that ℎ𝑟𝑟,𝑘𝑘 = 0 ∀ 𝑟𝑟 < 𝑘𝑘 . so: For 𝒌𝒌 = 𝟎𝟎 : the coefficient on both sides in equation (3.31) which make 𝒯𝒯0 (𝑡𝑡) is: ℎ𝑟𝑟+1,0 = ℎ𝑟𝑟,1 −
4𝜏𝜏 ℎ − ℎ𝑟𝑟−1,0 𝑏𝑏 − 𝑎𝑎 𝑟𝑟,0
… (3.32)
For 𝒌𝒌 = 𝟏𝟏 : the coefficient on both sides in equation (3.31) which make 𝒯𝒯1 (𝑡𝑡) is: ℎ𝑟𝑟+1,1 = 2ℎ𝑟𝑟,0 + ℎ𝑟𝑟,2 −
4𝜏𝜏 ℎ − ℎ𝑟𝑟−1,1 𝑏𝑏 − 𝑎𝑎 𝑟𝑟,1
… (3.33)
For 𝒌𝒌 = 𝟐𝟐, 𝟑𝟑, … , 𝒓𝒓 − 𝟏𝟏 and 𝒓𝒓 ≥ 𝟑𝟑: The coefficient 𝒯𝒯𝑘𝑘 (𝑡𝑡) in equation (3.31), after using simple induction, we obtain: ℎ𝑟𝑟+1,𝑘𝑘 = ℎ𝑟𝑟,𝑘𝑘−1 + ℎ𝑟𝑟,𝑘𝑘+1 −
4𝜏𝜏 ℎ − ℎ𝑟𝑟−1,𝑘𝑘 𝑏𝑏 − 𝑎𝑎 𝑟𝑟,𝑘𝑘
For 𝒌𝒌 = 𝒓𝒓: The coefficient 𝒯𝒯𝑟𝑟 (𝑡𝑡) in equation (3.31) gives: ℎ𝑟𝑟+1,𝑟𝑟 = ℎ𝑟𝑟,𝑟𝑟−1 −
4𝜏𝜏 𝑏𝑏 − 𝑎𝑎
For 𝒌𝒌 = 𝒓𝒓 + 𝟏𝟏 : we have it from equation (3.31) and using (3.28): ℎ𝑟𝑟+1,𝑟𝑟+1 = ℎ𝑟𝑟,𝑟𝑟 = ℎ𝑟𝑟−1,𝑟𝑟−1 = ⋯ = ℎ1,1 = ℎ0,0 = 1
… (3.34) … (3.35) … (3.36)
Clearly we have ℎ𝑟𝑟,𝑟𝑟 = 1for all 𝑟𝑟 = 0,1,2, … , 𝑚𝑚 thus equation (3.28, 3.29, 3.32-35) defined all elements in constant matrix 𝑯𝑯𝜏𝜏 in equation (3.26) which is the lower triangular matrix of dimension m + 1 × m + 1 .
93
Chapter Three
Least Square Orthogonal Method
The Solution Technique of higher Fractional Order Linear VIDEs of multi-
time Retarded delay with variable coefficients: Recall equation (1.32): 𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝑢𝑢(𝑡𝑡)
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝑢𝑢(𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡)𝑢𝑢(𝑡𝑡 − 𝜏𝜏) 𝑖𝑖=1
𝑚𝑚
𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝑢𝑢(𝑥𝑥 − 𝜏𝜏𝑗𝑗 )𝑑𝑑𝑑𝑑 , 𝑗𝑗 =1 𝑎𝑎
𝑡𝑡 ∈ 𝐼𝐼 = [𝑎𝑎, 𝑏𝑏]
With property that 𝛼𝛼𝑛𝑛 > 𝛼𝛼𝑛𝑛 −1 > 𝛼𝛼𝑛𝑛−2 > 𝛼𝛼𝑛𝑛 −3 > ⋯ > 𝛼𝛼1 ≥ 𝛼𝛼0 = 0 , initial
conditions: 𝑢𝑢(𝑘𝑘) (𝑡𝑡) = 𝑢𝑢𝑘𝑘 ; 𝑘𝑘 = 0,1,2, … , 𝜇𝜇 − 1 and 𝜇𝜇 = ⌈𝛼𝛼𝑛𝑛 ⌉ and 𝜇𝜇-time
continuous differentiable historical function 𝑢𝑢(𝑡𝑡) = 𝜑𝜑(𝑡𝑡) ; 𝑡𝑡 ∈ [𝑎𝑎�, 𝑎𝑎] where
𝑎𝑎� = 𝑎𝑎 − max�𝜏𝜏, 𝜏𝜏𝑗𝑗 : 𝑗𝑗 = 1: 𝑚𝑚� . The epitome of this method is to approximate the
solution of equation (1.32) by the form
𝑁𝑁
𝑢𝑢(𝑡𝑡) ≅ 𝑢𝑢𝑁𝑁 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝐺𝐺𝑟𝑟 (𝑡𝑡)
… (3.37)
𝑟𝑟=0
The coordinate functions 𝐺𝐺𝑟𝑟 (𝑡𝑡) are usually chosen as orthogonal polynomials
(Chebyshev or Legendre) and 𝐶𝐶𝑟𝑟 's are undetermined constant coefficients for all 𝑟𝑟.
The delay points can be defined using equation (3.37) and historical functions with basic idea in the delays of orthogonality lemmas (3.3 and 3.4) 𝜑𝜑(𝑡𝑡 − 𝜏𝜏∗ )
𝑁𝑁
𝑟𝑟
𝑟𝑟=0
𝑘𝑘=0
if
𝑢𝑢(𝑡𝑡 − 𝜏𝜏∗ ) ≅ 𝑢𝑢𝑁𝑁 (𝑡𝑡 − 𝜏𝜏∗ ) = � 𝜏𝜏 ∗ 𝐺𝐺𝑘𝑘 (𝑡𝑡) if � 𝐶𝐶𝑟𝑟 � ℎ𝑟𝑟,𝑘𝑘
𝑡𝑡 − 𝜏𝜏∗ ≤ 𝑎𝑎
𝑡𝑡 − 𝜏𝜏∗ > 𝑎𝑎
… (3.38)
By substituting 𝑢𝑢𝑁𝑁 as formed (3.37 and 3.38) in equation (1.32) with 𝑢𝑢 ,we obtain:
94
Chapter Three 𝑁𝑁
Least Square Orthogonal Method 𝑛𝑛 −1
𝛼𝛼
𝛼𝛼
� 𝐶𝐶𝑟𝑟 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡)� 𝑟𝑟=0
𝜑𝜑(𝑡𝑡 − 𝜏𝜏) ; 𝑁𝑁
𝑟𝑟
𝑟𝑟=0
𝑘𝑘=0
𝑖𝑖=1
if 𝑡𝑡 − 𝜏𝜏 ≤ 𝑎𝑎
+𝑃𝑃0 (𝑡𝑡) ∗ � � 𝜏𝜏 𝐺𝐺𝑘𝑘 (𝑡𝑡) if 𝑡𝑡 − 𝜏𝜏 > 𝑎𝑎 � 𝐶𝐶𝑟𝑟 � ℎ𝑟𝑟,𝑘𝑘 𝑡𝑡
𝑚𝑚
= 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) ∗ 𝑗𝑗 =1 𝑎𝑎
+R 𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ )
⎧
𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 �
𝑁𝑁
𝑟𝑟
𝜏𝜏
; if 𝑥𝑥 − 𝜏𝜏𝑗𝑗 ≤ 𝑎𝑎
⎫
𝑗𝑗 ⎨� 𝐶𝐶𝑟𝑟 � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) ; if 𝑥𝑥 − 𝜏𝜏𝑗𝑗 > 𝑎𝑎 ⎬ ⎩𝑟𝑟=0 𝑘𝑘=0 ⎭
𝑑𝑑𝑑𝑑 … (3.39)
Where R 𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ ) is the error function involved which depends on 𝑡𝑡 and the way that the constant coefficients 𝐶𝐶̅ = [𝐶𝐶0 , 𝐶𝐶1 , … , 𝐶𝐶𝑁𝑁 ] are chosen. Now the equation (3.39) can be written as: 𝑁𝑁
R 𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ ) = � 𝐶𝐶𝑟𝑟 𝜓𝜓𝑟𝑟 (𝑡𝑡) − ℱ(𝑡𝑡) 𝑟𝑟=0
… (3.40)
Where 𝜓𝜓𝑟𝑟 (𝑡𝑡) and ℱ(𝑡𝑡) are defined for fixed any point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], here we have 𝑚𝑚 + 1 constant delays 𝜏𝜏 = 𝜏𝜏0 , 𝜏𝜏1 , … , 𝜏𝜏𝑚𝑚 . Thus, we have 𝑚𝑚 + 2 basic parts which construct 𝜓𝜓𝑟𝑟 (𝑡𝑡) and ℱ(𝑡𝑡) as follows: 𝜓𝜓𝑟𝑟 (𝑡𝑡 ) 𝐼𝐼0 (𝑡𝑡 ) 𝜓𝜓 𝑟𝑟 ⎧ 𝐼𝐼1 𝜓𝜓 (𝑡𝑡 ) ⎪ ⎪ 𝑟𝑟𝐼𝐼 2 = 𝜓𝜓𝑟𝑟 (𝑡𝑡 ) ⋮ ⎨ 𝐼𝐼𝑚𝑚 ⎪ ⎪ 𝜓𝜓𝑟𝑟 (𝑡𝑡 )
𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) ≤ 𝑎𝑎
𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ 𝑎𝑎
𝑖𝑖𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝜏𝜏 ′𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ 𝑎𝑎 ⋮ 𝑖𝑖𝑖𝑖 𝑚𝑚 − 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 𝑎𝑎𝑛𝑛𝑛𝑛 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ 𝑎𝑎 𝐼𝐼𝑚𝑚 +1 ⎩𝜓𝜓𝑟𝑟 (𝑡𝑡 ) 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 … (3.41)
And
ℱ𝐼𝐼0 (𝑡𝑡 ) ⎧ 𝐼𝐼1 ℱ (𝑡𝑡 ) ⎪ ⎪ 𝐼𝐼 2 ℱ (𝑡𝑡 ) = ℱ (𝑡𝑡 ) ⋮ ⎨ 𝐼𝐼𝑚𝑚 ( ) ⎪ ⎪ ℱ 𝑡𝑡 ⎩ℱ𝐼𝐼𝑚𝑚 +1 (𝑡𝑡 )
𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) ≤ 𝑎𝑎 𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ 𝑎𝑎 𝑖𝑖𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡 𝑜𝑜𝑜𝑜 𝜏𝜏 ′𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ 𝑎𝑎 ⋮ 𝑖𝑖𝑖𝑖 𝑚𝑚 − 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ≤ 𝑎𝑎 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝜏𝜏 ′𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 … (3.42)
95
Chapter Three
Least Square Orthogonal Method
Each basic part in (3.41) and (3.42) contain cases. Let 𝛿𝛿 be the number of 𝜏𝜏's which satisfies the condition (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 for any fixed 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], so the number of cases +1 here is equal to �𝑚𝑚𝑚𝑚+1−𝛿𝛿 �. For finding each case inside each basic part, let ℓ be any nonnegative integer number which takes the values {0,1,2, … , 𝑚𝑚 − 1}while ℓ + 1takes {1,2, … , 𝑚𝑚} and {𝜐𝜐0 , 𝜐𝜐1 , 𝜐𝜐2 … , 𝜐𝜐ℓ } also takes the values from the positive integer set {0,1,2, … , 𝑚𝑚} providing that 𝜐𝜐0 < 𝜐𝜐1 < 𝜐𝜐2 … < 𝜐𝜐ℓ . Thus: 𝑛𝑛−1
𝛼𝛼
𝐼𝐼
𝛼𝛼
𝜓𝜓𝑟𝑟0 (𝑡𝑡) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) 𝑚𝑚
𝑖𝑖=1 𝑡𝑡
… (3.43)
ℱ𝐼𝐼0 (𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − 𝑃𝑃0 (𝑡𝑡) 𝜑𝜑(𝑡𝑡 − 𝜏𝜏0 ) … (3.44) 𝑗𝑗 =1 𝑎𝑎
�𝜐𝜐 ,𝜐𝜐 …𝜐𝜐 � 𝐼𝐼 0 1 ℓ (𝑡𝑡) 𝜓𝜓𝑟𝑟ℓ+1
+
ℱ
=
𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐺𝐺𝑟𝑟 (𝑡𝑡 )
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡 ) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡 ) 𝑖𝑖=1
if ℓ = 0 ; 𝑟𝑟 ⎧ 𝑡𝑡 𝜏𝜏 0 ⎪if 𝜐𝜐0 = 0; 𝑃𝑃0 (𝑡𝑡 ) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑡𝑡) − � ⎪ o. w. ; 𝜆𝜆 � 𝑘𝑘=0
0
𝑟𝑟
�
𝑎𝑎 𝑗𝑗 =𝜐𝜐 1 ,𝜐𝜐 2 ,…,𝜐𝜐 ℓ
⎨ 𝑡𝑡 𝑟𝑟 𝜏𝜏 𝑗𝑗 ⎪ � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑 ⎪ if 𝜐𝜐0 ≠ 0; −𝜆𝜆 � ⎩ 𝑘𝑘=0 𝑎𝑎 𝑗𝑗 =𝜐𝜐 0 ,𝜐𝜐 1 ,…,𝜐𝜐 ℓ
�𝜐𝜐 0 ,𝜐𝜐 1 …𝜐𝜐 ℓ � 𝐼𝐼ℓ+1
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡 ) +
𝐼𝐼
𝛼𝛼
𝑡𝑡
𝑚𝑚
⎧ if 𝜐𝜐 = 0 ; 𝜆𝜆 � ⎪ 0 ⎪ 𝑎𝑎
�
𝑗𝑗 =1 𝑗𝑗 ≠𝜐𝜐 1 ,𝜐𝜐 2 …,𝜐𝜐 ℓ 𝑡𝑡 𝑚𝑚
ℱ
= 𝑓𝑓(𝑡𝑡)
𝑘𝑘=0
… (3.45)
𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑
⎨ � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − 𝑃𝑃0 (𝑡𝑡 )𝜑𝜑(𝑡𝑡 − 𝜏𝜏0 ) ⎪ if 𝜐𝜐0 ≠ 0 ; 𝜆𝜆 � ⎪ 𝑗𝑗 =1 𝑎𝑎 ⎩ 𝑗𝑗 ≠𝜐𝜐 0 ,𝜐𝜐 1 ,…,𝜐𝜐 ℓ 𝑛𝑛−1
𝑟𝑟
𝛼𝛼
𝜏𝜏
0 𝜓𝜓𝑟𝑟𝑚𝑚 +1 (𝑡𝑡) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡 ) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡 ) + 𝑃𝑃0 (𝑡𝑡 ) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑡𝑡 )
𝐼𝐼𝑚𝑚 +1 (𝑡𝑡)
𝜏𝜏 𝑗𝑗
𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑖𝑖=1
𝑡𝑡 𝑚𝑚
𝑟𝑟
𝑎𝑎 𝑗𝑗 =1
𝑘𝑘=0
𝜏𝜏
𝑘𝑘=0
𝑗𝑗 −𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑
96
… (3.46)
… (3.47)
… (3.48)
Chapter Three
Least Square Orthogonal Method
For more illustration of the above expression we can explain it by the following testing constant 𝝉𝝉 −delays examples: Suppose we have three different constant delays, say 𝜏𝜏 = 𝜏𝜏0 , 𝜏𝜏1 and 𝜏𝜏2 ,which are all constant positive real number. Fixing the point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], here we have 𝑚𝑚 + 2 = 4 basic parts for: 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝜓𝜓𝑟𝑟 (𝑡𝑡): 𝜓𝜓𝑟𝑟0 (𝑡𝑡), 𝜓𝜓𝑟𝑟1 (𝑡𝑡), 𝜓𝜓𝑟𝑟2 (𝑡𝑡) and ℱ𝑟𝑟 (𝑡𝑡): ℱ𝑟𝑟 0 (𝑡𝑡), ℱ𝑟𝑟 1 (𝑡𝑡), ℱ𝑟𝑟 2 (𝑡𝑡)
+1 While each part contains 𝛾𝛾-cases, 𝛾𝛾 = �𝑚𝑚𝑚𝑚+1−𝛿𝛿 �, where 𝛿𝛿 is number of 𝜏𝜏′s which satisfies (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎. Thus, for any fixed collocation part we do the following: i. If (𝑡𝑡 − 𝜏𝜏0 ) and (𝑡𝑡 − 𝜏𝜏1 ) and (𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎 . Here 𝛿𝛿 = 0 and 𝑚𝑚 = 2 that is 3 𝛾𝛾 = �3−0 � = 1 , so we have only one case and use historical function for all 𝑢𝑢(𝑡𝑡 − 𝜏𝜏𝑘𝑘 ) = 𝜑𝜑(𝑡𝑡 − 𝜏𝜏𝑘𝑘 ) for 𝑘𝑘 = 0,1,2 . And using equations (3.43 and 3.44) to 𝐼𝐼 𝐼𝐼 find 𝜓𝜓𝑟𝑟0 (𝑡𝑡) and ℱ𝑟𝑟 0 (𝑡𝑡) . So, for that fixed collocation point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], equation (3.39) becomes: 𝑁𝑁
𝛼𝛼 � 𝐶𝐶𝑟𝑟 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) 𝑟𝑟=0
ii.
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡)� = 𝑓𝑓(𝑡𝑡) 𝑖𝑖=1 𝑡𝑡
2
+𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − 𝑃𝑃0 (𝑡𝑡)𝜑𝜑(𝑡𝑡 − 𝜏𝜏0 ) + 𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝑐𝑐̅) 𝑎𝑎 𝑗𝑗 =1
If one constant delay (𝜏𝜏0 , 𝜏𝜏1 , 𝜏𝜏2 ) satisfies (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 and the other is not. 3 That is (𝛿𝛿 = 1, 𝑚𝑚 = 2) and 𝛾𝛾 = �3−1 � = 3 that we have three cases 𝐼𝐼1 𝐼𝐼1 in 𝜓𝜓𝑟𝑟 (𝑡𝑡) and ℱ𝑟𝑟 (𝑡𝑡) , here ℓ = 0 so we have only [𝜐𝜐0 ] which takes the value from the positive integer set {0,1,2} that is: 𝐼𝐼
[0]
𝐼𝐼
[1]
𝐼𝐼
[2]
𝐼𝐼
[0]
𝐼𝐼
[1]
𝐼𝐼
[2]
𝐼𝐼 𝐼𝐼 𝜓𝜓𝑟𝑟1 (𝑡𝑡) ∶ 𝜓𝜓𝑟𝑟1 (𝑡𝑡) , 𝜓𝜓𝑟𝑟1 (𝑡𝑡) , 𝜓𝜓𝑟𝑟1 (𝑡𝑡) and ℱ𝑟𝑟 1 (𝑡𝑡) ∶ ℱ𝑟𝑟 1 (𝑡𝑡) , ℱ𝑟𝑟 1 (𝑡𝑡) , ℱ𝑟𝑟 1 (𝑡𝑡) Thus:
a. If (𝑡𝑡 − 𝜏𝜏0 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏1 , 𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎. From equations (3.45, 3.46) we find [0]
𝐼𝐼 𝜓𝜓𝑟𝑟1
(𝑡𝑡) and
[0]
𝐼𝐼 𝜓𝜓𝑟𝑟1
(𝑡𝑡) =
[0]
𝐼𝐼 ℱ𝑟𝑟 1
[0]
𝐼𝐼 ℱ𝑟𝑟 1
(𝑡𝑡) respectively.
𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐺𝐺𝑟𝑟 (𝑡𝑡) + 𝑡𝑡
𝑛𝑛 −1
𝛼𝛼 � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) + 𝑖𝑖=1
2
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥)𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 𝑎𝑎 𝑗𝑗 =1
97
𝑟𝑟
𝜏𝜏
0 𝑃𝑃0 (𝑡𝑡) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑡𝑡)
𝑘𝑘=0
Chapter Three
Least Square Orthogonal Method
b. If (𝑡𝑡 − 𝜏𝜏1 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎. from equations (3.45 and 3.46) we 𝐼𝐼
[1]
𝐼𝐼
[1]
can find 𝜓𝜓𝑟𝑟1 (𝑡𝑡) and ℱ𝑟𝑟 1 (𝑡𝑡), respectively. [1]
𝐼𝐼 𝜓𝜓𝑟𝑟1
(𝑡𝑡) =
[1]
𝐼𝐼 ℱ𝑟𝑟 1
𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐺𝐺𝑟𝑟 (𝑡𝑡)
+
𝑡𝑡
𝑛𝑛−1
𝛼𝛼 � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) 𝑖𝑖=1
𝑡𝑡
𝑟𝑟
𝜏𝜏
1 − 𝜆𝜆 � 𝒦𝒦1 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑘𝑘=0
𝑎𝑎
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥)𝜑𝜑(𝑥𝑥 − 𝜏𝜏2 )𝑑𝑑𝑑𝑑 − 𝑃𝑃0 (𝑡𝑡) 𝜑𝜑(𝑡𝑡 − 𝜏𝜏0 ) 𝑎𝑎
c. If (𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 ) ≤ 𝑎𝑎. from equations (3.45, 3.46) we can find [2] 𝐼𝐼1
[2]
𝐼𝐼 𝜓𝜓𝑟𝑟1
(𝑡𝑡) and
𝜓𝜓𝑟𝑟 (𝑡𝑡) = [2] 𝐼𝐼1
ℱ𝑟𝑟
[2]
𝐼𝐼 ℱ𝑟𝑟 1
𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐺𝐺𝑟𝑟 (𝑡𝑡)
(𝑡𝑡), respectively.
+ 𝑡𝑡
𝑛𝑛−1
𝛼𝛼 � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) 𝑖𝑖=1
𝑡𝑡
𝑟𝑟
𝜏𝜏
2 − 𝜆𝜆 � 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑎𝑎
𝑘𝑘=0
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � 𝒦𝒦1 (𝑡𝑡, 𝑥𝑥)𝜑𝜑(𝑥𝑥 − 𝜏𝜏1 )𝑑𝑑𝑑𝑑 − 𝑃𝑃0 (𝑡𝑡) 𝜑𝜑(𝑡𝑡 − 𝜏𝜏0 ) 𝑎𝑎
So for that fixed collocation point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] ; R 𝑁𝑁 (𝑡𝑡; 𝑐𝑐̅) can be constract from equation (3.40): 𝐼𝐼
iii.
[0]
𝐼𝐼
[0]
⎧𝜓𝜓𝑟𝑟1 (𝑡𝑡)⎫ ⎧ℱ𝑟𝑟 1 (𝑡𝑡) 𝑁𝑁 ⎪ [1] ⎪ ⎪ [1] R 𝑁𝑁 (𝑡𝑡; 𝑐𝑐̅) = � 𝐶𝐶𝑟𝑟 𝜓𝜓 𝐼𝐼1 (𝑡𝑡) − ℱ 𝐼𝐼1 (𝑡𝑡) ⎨ 𝑟𝑟 [2] ⎬ ⎨ 𝑟𝑟 [2] 𝑟𝑟=0 ⎪ 𝐼𝐼1 ⎪ ⎪ 𝐼𝐼 ⎩𝜓𝜓𝑟𝑟 (𝑡𝑡)⎭ ⎩ ℱ𝑟𝑟 1 (𝑡𝑡)
if (𝑡𝑡 − 𝜏𝜏0 ) > 𝑎𝑎 ⎫ ⎪ if (𝑡𝑡 − 𝜏𝜏1 ) > 𝑎𝑎 ⎬ ⎪ if (𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 ⎭
If two constant delays (𝜏𝜏0 , 𝜏𝜏1 , 𝜏𝜏2 ) satisfy (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 and the other is not. 3 That is (𝛿𝛿 = 2, 𝑚𝑚 = 2) and 𝛾𝛾 = �3−2 � = 3 that we have also three cases 𝐼𝐼 𝐼𝐼 in 𝜓𝜓𝑟𝑟2 (𝑡𝑡) and ℱ𝑟𝑟 2 (𝑡𝑡), here ℓ = 1 so we have only [𝜐𝜐0 𝜐𝜐1 ], 𝜐𝜐0 < 𝜐𝜐1 and 𝜐𝜐0 , 𝜐𝜐1 ∈ {0,1,2} that is: 𝐼𝐼
[01]
𝐼𝐼
[02]
𝐼𝐼
[12]
𝐼𝐼 𝜓𝜓𝑟𝑟2 (𝑡𝑡): 𝜓𝜓𝑟𝑟2 (𝑡𝑡), 𝜓𝜓𝑟𝑟2 (𝑡𝑡), 𝜓𝜓𝑟𝑟2 (𝑡𝑡) 𝐼𝐼
[01]
𝐼𝐼
[02]
𝐼𝐼
[12]
𝐼𝐼 ℱ𝑟𝑟 2 (𝑡𝑡): ℱ𝑟𝑟 2 (𝑡𝑡), ℱ𝑟𝑟 2 (𝑡𝑡), ℱ𝑟𝑟 2 (𝑡𝑡) Thus:
98
and
Chapter Three
Least Square Orthogonal Method
a. If (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎 . from equations (3.45, 3.46) we can 𝐼𝐼
[01]
𝐼𝐼
[01]
find 𝜓𝜓𝑟𝑟2 (𝑡𝑡) and ℱ𝑟𝑟 2 (𝑡𝑡) respectively. [01]
𝐼𝐼 𝜓𝜓𝑟𝑟2
(𝑡𝑡) =
− [01]
𝐼𝐼 ℱ𝑟𝑟 2
𝑛𝑛−1
𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐺𝐺𝑟𝑟 (𝑡𝑡)
𝛼𝛼 + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) + 𝑖𝑖=1 𝑡𝑡 𝑟𝑟 𝜏𝜏 1 𝜆𝜆 � 𝒦𝒦1 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑎𝑎
𝑟𝑟
𝜏𝜏
0 𝑃𝑃0 (𝑡𝑡) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑡𝑡)
𝑘𝑘=0
𝑘𝑘=0
𝑡𝑡
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥)𝜑𝜑(𝑥𝑥 − 𝜏𝜏2 )𝑑𝑑𝑑𝑑 𝑎𝑎
b. If (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏1 ) ≤ 𝑎𝑎. from equations (3.45 and 3.46) we can find [02] 𝐼𝐼2
𝜓𝜓𝑟𝑟
[02]
𝐼𝐼 𝜓𝜓𝑟𝑟2
(𝑡𝑡) and
[02]
𝐼𝐼 ℱ𝑟𝑟 2
(𝑡𝑡), respectively.
𝑛𝑛−1
𝑟𝑟
𝜏𝜏 0 (𝑡𝑡) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡𝛼𝛼 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑡𝑡) 𝑖𝑖=1
𝑡𝑡
𝑟𝑟
𝑘𝑘=0
𝜏𝜏
2 − 𝜆𝜆 � 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑑𝑑𝑑𝑑 [02] 𝐼𝐼2
ℱ𝑟𝑟
𝑎𝑎
𝑘𝑘=0
𝑡𝑡
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 � 𝒦𝒦1 (𝑡𝑡, 𝑥𝑥)𝜑𝜑(𝑥𝑥 − 𝜏𝜏1 )𝑑𝑑𝑑𝑑 𝑎𝑎
c. If (𝑡𝑡 − 𝜏𝜏1 , 𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏0 ) ≤ 𝑎𝑎. from equations (3.45, 3.46) we can 𝐼𝐼
[12]
𝐼𝐼
[12]
find 𝜓𝜓𝑟𝑟2 (𝑡𝑡) and ℱ𝑟𝑟 2 (𝑡𝑡), respectively. [12]
𝐼𝐼 𝜓𝜓𝑟𝑟2
(𝑡𝑡) =
[12]
𝐼𝐼 ℱ𝑟𝑟 2
𝐶𝐶 𝛼𝛼 𝑛𝑛 𝑎𝑎 𝐷𝐷𝑡𝑡 𝐺𝐺𝑟𝑟 (𝑡𝑡) +
−
𝑡𝑡
𝑛𝑛 −1
𝛼𝛼
� 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) 𝑖𝑖=1
𝑟𝑟
𝜏𝜏 1 𝜆𝜆 � �𝒦𝒦1 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥) 𝑘𝑘=0 𝑎𝑎
(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) − 𝑃𝑃0 (𝑡𝑡) 𝜑𝜑(𝑡𝑡 − 𝜏𝜏0 )
99
𝑟𝑟
𝜏𝜏
2 + 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥)� 𝑑𝑑𝑑𝑑
𝑘𝑘=0
Chapter Three
Least Square Orthogonal Method
So for that fixed collocation point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏] ; 𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝑐𝑐̅) can be constracted from equation (3.40). [01]
[01]
𝐼𝐼 ⎧𝜓𝜓𝑟𝑟2
𝐼𝐼 ⎧ℱ𝑟𝑟 2
(𝑡𝑡)⎫ (𝑡𝑡) ⎪ [02] ⎪ ⎪ [02] R 𝑁𝑁 (𝑡𝑡; 𝑐𝑐̅) = � 𝐶𝐶𝑟𝑟 𝜓𝜓 𝐼𝐼2 (𝑡𝑡) − ℱ 𝐼𝐼2 (𝑡𝑡) ⎨ 𝑟𝑟 [12] ⎬ ⎨ 𝑟𝑟 [12] 𝑟𝑟=0 ⎪ 𝐼𝐼2 ⎪ ⎪ 𝐼𝐼 ⎩𝜓𝜓𝑟𝑟 (𝑡𝑡)⎭ ⎩ ℱ𝑟𝑟 2 (𝑡𝑡) 𝑁𝑁
if (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 ) > 𝑎𝑎 ⎫ ⎪ if (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 ⎬ ⎪ if (𝑡𝑡 − 𝜏𝜏1 , 𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 ⎭
If all constant delays satisfy (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎. That is (𝛿𝛿 = 3, 𝑚𝑚 = 2) and 3 (3.47, 3.48) 𝛾𝛾 = �3−3 � = 1 , we have only one case. Then using equations
iv.
𝐼𝐼
𝐼𝐼
𝐼𝐼
to find 𝜓𝜓𝑟𝑟3 (𝑡𝑡) and ℱ𝑟𝑟 3 (𝑡𝑡) , respectively. 𝑛𝑛 −1
𝛼𝛼
𝑟𝑟
𝛼𝛼
𝜏𝜏
0 𝜓𝜓𝑟𝑟3 (𝑡𝑡) = 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑡𝑡)
𝑖𝑖=1
𝑡𝑡
𝑘𝑘=0
𝑟𝑟
𝜏𝜏 1 𝐺𝐺𝑘𝑘 (𝑥𝑥) −𝜆𝜆 � �𝒦𝒦1 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝑘𝑘=0 𝑎𝑎
𝐼𝐼
ℱ𝑟𝑟 3 (𝑡𝑡) = 𝑓𝑓(𝑡𝑡)
𝑟𝑟
𝜏𝜏
2 + 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝐺𝐺𝑘𝑘 (𝑥𝑥)� 𝑑𝑑𝑑𝑑
𝑘𝑘=0
In the example before, we discuss the types as all constant delay is different. If the equality in constant delays appears, we must discuss the following types: Type (1): If 𝜏𝜏 = 𝜏𝜏0 is equal to one of 𝜏𝜏1 or 𝜏𝜏2 , say 𝜏𝜏0 = 𝜏𝜏1 = 𝜏𝜏∗ (the other is the same). For any fixed point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], here we have (𝑚𝑚 + 2) − 𝜂𝜂 basic parts 𝜂𝜂 is number of equality (here we have one equality so 𝜂𝜂 = 1) thus we have 4 − 1 = 3 basic parts. So: 𝐼𝐼
𝐼𝐼
𝐼𝐼
𝐼𝐼
𝐼𝐼
𝐼𝐼
𝜓𝜓𝑟𝑟 (𝑡𝑡): 𝜓𝜓𝑟𝑟0 (𝑡𝑡), 𝜓𝜓𝑟𝑟1 (𝑡𝑡), 𝜓𝜓𝑟𝑟2 (𝑡𝑡) and ℱ𝑟𝑟 (𝑡𝑡): ℱ𝑟𝑟 0 (𝑡𝑡), ℱ𝑟𝑟 1 (𝑡𝑡), ℱ𝑟𝑟 2 (𝑡𝑡)
+1−𝜂𝜂 3+1−1 2 Each part containing 𝛾𝛾-cases, 𝛾𝛾 = �𝑚𝑚𝑚𝑚+1−𝜂𝜂 � = �3+1−1−𝛿𝛿 � = �2−𝛿𝛿 � where 𝛿𝛿 is −𝛿𝛿
number of constant delays which satisfy (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 . Thus: i.
If (𝑡𝑡 − 𝜏𝜏∗ ) and (𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎 . that is, here 𝛿𝛿 = 0, so 𝛾𝛾 = �22� = 1 we have 𝐼𝐼
𝐼𝐼
one case and use equation (3.43 and 3.44) to find 𝜓𝜓𝑟𝑟0 (𝑡𝑡) and ℱ𝑟𝑟 0 (𝑡𝑡) . So for that fixed point 𝑡𝑡 ∈ [𝑎𝑎, 𝑏𝑏], equation (3.39) becomes: 100
Chapter Three
Least Square Orthogonal Method
⎡ ⎤ 𝑛𝑛 −1 ⎢ 𝛼𝛼 ⎥ 𝛼𝛼 � 𝐶𝐶𝑟𝑟 ⎢ 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝐺𝐺𝑟𝑟 (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝐺𝐺𝑟𝑟 (𝑡𝑡)⎥ 𝑟𝑟=0 𝑖𝑖=1 ⎢����������������������� ⎥ 𝐼𝐼 𝜓𝜓 𝑟𝑟 0 (𝑡𝑡) ⎣ ⎦ 𝑁𝑁
𝑡𝑡
= 𝑓𝑓(𝑡𝑡) + 𝜆𝜆 �[𝒦𝒦1 (𝑡𝑡, 𝑥𝑥)𝜑𝜑(𝑥𝑥 − 𝜏𝜏∗ ) + 𝒦𝒦2 (𝑡𝑡, 𝑥𝑥)𝜑𝜑(𝑥𝑥 − 𝜏𝜏2 )]𝑑𝑑𝑑𝑑 − 𝑃𝑃0 (𝑡𝑡)𝜑𝜑(𝑡𝑡 − 𝜏𝜏∗ ) ����������������������������������������������� 𝑎𝑎 𝐼𝐼
ℱ𝑟𝑟 0 (𝑡𝑡)
+ 𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝑐𝑐̅)
ii.
If one of them (𝑡𝑡 − 𝜏𝜏∗ ) or (𝑡𝑡 − 𝜏𝜏2 ) satisfy > 𝑎𝑎 and the other not. So, 𝛿𝛿 = 1 𝐼𝐼 𝐼𝐼 2 thus 𝛾𝛾 = �2−1 � = 2 cases in 𝜓𝜓𝑟𝑟1 (𝑡𝑡) and ℱ𝑟𝑟 1 (𝑡𝑡), here ℓ = 0 so we have only [𝜐𝜐0 ] which takes the value from integer set {0,1} that is: 𝐼𝐼
[0]
𝐼𝐼
[1]
𝐼𝐼
[0]
𝐼𝐼
[1]
𝐼𝐼 𝐼𝐼 𝜓𝜓𝑟𝑟1 (𝑡𝑡) ∶ 𝜓𝜓𝑟𝑟1 (𝑡𝑡) , 𝜓𝜓𝑟𝑟1 (𝑡𝑡) and ℱ𝑟𝑟 1 (𝑡𝑡) ∶ ℱ𝑟𝑟 1 (𝑡𝑡) , ℱ𝑟𝑟 1 (𝑡𝑡) Thus:
a. If (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎 , that means (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎 which is the same as in distance parts (iii, a). b. If (𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏∗ ) ≤ 𝑎𝑎, that is meaning (𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 and (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 ) ≤ 𝑎𝑎 which is the same as in distance parts (i, c). iii. If all satisfy (𝑡𝑡 − 𝜏𝜏∗ and 𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎. so here 𝛿𝛿 = 2 thus 𝛾𝛾 = �20� = 1 cases 𝐼𝐼 𝐼𝐼 in 𝜓𝜓𝑟𝑟2 (𝑡𝑡) and ℱ𝑟𝑟 2 (𝑡𝑡), which means all 𝑡𝑡 − 𝜏𝜏0 > 𝑎𝑎, 𝑡𝑡 − 𝜏𝜏1 > 𝑎𝑎 and 𝑡𝑡 − 𝜏𝜏2 > 𝑎𝑎 which is the same as in distance parts (iv).
Type (2): If all are equal 𝜏𝜏 = 𝜏𝜏0 = 𝜏𝜏1 = 𝜏𝜏2 (say 𝜏𝜏∗ ) so we have (𝑚𝑚 + 2) − 𝜂𝜂 = (2 + 2) − 2 = 2
That is two basic parts thus: 𝐼𝐼
𝐼𝐼
𝐼𝐼
𝐼𝐼
𝜓𝜓𝑟𝑟 (𝑡𝑡): 𝜓𝜓𝑟𝑟0 (𝑡𝑡), 𝜓𝜓𝑟𝑟1 (𝑡𝑡) and ℱ𝑟𝑟 (𝑡𝑡): ℱ𝑟𝑟 0 (𝑡𝑡), ℱ𝑟𝑟 1 (𝑡𝑡)
All (𝑡𝑡 − 𝜏𝜏∗ ) ≤ 𝑎𝑎 that means: (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 , 𝑡𝑡 − 𝜏𝜏2 ) ≤ 𝑎𝑎 , This is the same as in distance parts (i). ii. All (𝑡𝑡 − 𝜏𝜏∗ ) > 𝑎𝑎 that means: (𝑡𝑡 − 𝜏𝜏0 , 𝑡𝑡 − 𝜏𝜏1 , 𝑡𝑡 − 𝜏𝜏2 ) > 𝑎𝑎 , which is the same as in distance parts (iv). The main points here are how to find the coefficients 𝐶𝐶𝑟𝑟 's (𝑟𝑟 = 0,1, … , 𝑁𝑁) of 𝑢𝑢𝑁𝑁 (𝑡𝑡) in equation (3.37) such that 𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ ) is minimized. The general least squares techniques insist on minimizing the norm of the error i.
101
Chapter Three
Least Square Orthogonal Method
functions 𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ ) , where 𝐶𝐶̅ = (𝐶𝐶0 , 𝐶𝐶1 , … , 𝐶𝐶𝑁𝑁 ) by introducing a weight function 𝑤𝑤(𝑡𝑡) on interval [𝑎𝑎, 𝑏𝑏], we now wish to minimize 𝐼𝐼(𝐶𝐶̅ ): 𝑏𝑏
𝐼𝐼(𝐶𝐶̅ ) = � 𝑤𝑤(𝑡𝑡)|𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ )|2 𝑑𝑑𝑑𝑑 𝑎𝑎
The necessary condition for 𝐼𝐼(𝐶𝐶̅ ) to be minimum, are: 𝑏𝑏
𝜕𝜕𝜕𝜕(𝐶𝐶̅ ) 𝜕𝜕𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝐶𝐶̅ ) ̅ = 2 � 𝑤𝑤(𝑡𝑡)𝑅𝑅𝑁𝑁 (𝑡𝑡; 𝐶𝐶 ) 𝑑𝑑𝑑𝑑 = 0 ; 𝑠𝑠 = 0,1,2, … , 𝑁𝑁 𝜕𝜕𝑐𝑐𝑠𝑠 𝜕𝜕𝐶𝐶𝑠𝑠 𝑎𝑎
Putting (3.39) into equation (3.49), we obtain: 𝑏𝑏
… (3.49)
𝑁𝑁
� 𝑤𝑤(𝑡𝑡)𝜓𝜓𝑠𝑠 (𝑡𝑡) �� 𝐶𝐶𝑟𝑟 𝜓𝜓𝑟𝑟 (𝑡𝑡) − ℱ(𝑡𝑡)� 𝑑𝑑𝑑𝑑 = 0
𝑎𝑎
𝑟𝑟=0
After some simple manipulation, the following linear system is included: 𝑁𝑁
� 𝐶𝐶𝑟𝑟 𝑎𝑎𝑠𝑠𝑠𝑠 = 𝑏𝑏𝑠𝑠 ,
𝑟𝑟=0
Where
𝑠𝑠 = 0,1,2, … , 𝑁𝑁
… (3.50)
𝑏𝑏
⎫ 𝔞𝔞𝑠𝑠𝑠𝑠 = � 𝑤𝑤(𝑡𝑡)𝜓𝜓𝑠𝑠 (𝑡𝑡)𝜓𝜓𝑟𝑟 (𝑡𝑡)𝑑𝑑𝑑𝑑⎪ ⎪ 𝑎𝑎 𝑏𝑏
… (3.51)
⎬ 𝔟𝔟𝑠𝑠 = � 𝑤𝑤(𝑡𝑡)𝜓𝜓𝑠𝑠 (𝑡𝑡)ℱ(𝑡𝑡)𝑑𝑑𝑑𝑑 ⎪ ⎪ ⎭ 𝑎𝑎
Rewrite equation (3.50) in matrix form as:
where 𝔞𝔞00 𝔞𝔞10 𝐴𝐴 = � ⋮ 𝔞𝔞𝑁𝑁0
𝔞𝔞01 𝔞𝔞11 ⋮ 𝔞𝔞𝑁𝑁1
… 𝔞𝔞0𝑁𝑁 … 𝔞𝔞1𝑁𝑁 �, ⋮ ⋮ … 𝔞𝔞𝑁𝑁𝑁𝑁
𝐴𝐴𝐴𝐴 = 𝐵𝐵
𝐶𝐶0 𝐶𝐶 𝐶𝐶 = � 1 � , ⋮ 𝐶𝐶𝑁𝑁
𝔟𝔟0 𝔟𝔟 𝐵𝐵 = � 1 � ⋮ 𝔟𝔟𝑁𝑁
… (3.52)
In this technique the initial conditions of equation (1.32) are added as new rows in the system (3.50), these rows can be formed as: 𝑁𝑁
(𝑘𝑘)
� 𝐶𝐶𝑟𝑟 �𝐺𝐺𝑟𝑟 (𝑡𝑡)�
𝑟𝑟=0
In matrix form, this gives
𝑡𝑡=𝑎𝑎
= 𝑢𝑢𝑘𝑘 , 102
𝑘𝑘 = 0,1,2, … , 𝜇𝜇 − 1
Chapter Three
Least Square Orthogonal Method
𝐺𝐺𝐺𝐺 = 𝑈𝑈
Where
… (3.53)
… 𝐺𝐺𝑁𝑁 (𝑎𝑎) 𝐺𝐺 (𝑎𝑎) 𝐺𝐺1 (𝑎𝑎) 𝑢𝑢0 𝐶𝐶0 ⎡ 0′ ⎤ ′ ′ 𝑢𝑢1 … 𝐺𝐺𝑁𝑁 (𝑎𝑎) 𝐺𝐺1 (𝑎𝑎) 𝐺𝐺 (𝑎𝑎) 𝐶𝐶1 ⎥, , 𝑈𝑈 = 𝐺𝐺 = ⎢ 0 ⋮ 𝐶𝐶 = � � � ⋮ � ⋮ ⋮ ⋮ ⋮ ⎢ (𝜇𝜇 ⎥ (𝜇𝜇 −1) (𝜇𝜇 −1) −1) 𝑢𝑢𝜇𝜇 −1 𝐶𝐶𝑁𝑁 (𝑎𝑎) 𝐺𝐺1 (𝑎𝑎) … 𝐺𝐺𝑁𝑁 (𝑎𝑎)⎦ ⎣𝐺𝐺0 Obtaining a new matrix by adding (3.53) to(3.52), it yields 𝐷𝐷𝐷𝐷 = 𝐸𝐸 … (3.54) where 𝐵𝐵 𝐴𝐴 and 𝐸𝐸 = �…� 𝐷𝐷 = �…� 𝑈𝑈 (𝑁𝑁+𝜇𝜇 +1)×1 𝐺𝐺 (𝑁𝑁+𝜇𝜇 +1)×(𝑁𝑁+1) To determine the constant coefficients 𝐶𝐶𝑟𝑟 's in equation (3.54), store the matrix 𝐷𝐷 and compute 𝐷𝐷𝑇𝑇 𝐷𝐷 and 𝐷𝐷𝑇𝑇 𝐸𝐸 then use LU-factorization procedure to solve: [𝐷𝐷𝑇𝑇 𝐷𝐷]𝐶𝐶 = [𝐷𝐷𝑇𝑇 𝐸𝐸] .
Then substitute the values 𝐶𝐶𝑟𝑟 's in equation (3.37), the approximate solution is obtained for higher order fractional linear VIDEs of constant multi-time Retarded Delay with variable coefficients (1.32).
3.4.1 Using Chebyshev Polynomials: In this section, we can take, instead of trial function 𝐺𝐺𝑟𝑟 (𝑡𝑡), the shifted Chebyshev
polynomial 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) = 𝒯𝒯𝑟𝑟 �2 �
𝑡𝑡−𝑎𝑎
𝑏𝑏−𝑎𝑎
� − 1� to approximate the solution 𝑢𝑢(𝑡𝑡) of higher
fractional order VIDEs of constant multi-time Retarded delay with variable coefficients equation (1.32) by forming: 𝑁𝑁
𝑢𝑢𝑁𝑁 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒯𝒯𝑟𝑟 �2 � 𝑟𝑟=0
𝑡𝑡 − 𝑎𝑎 � − 1� , 𝑏𝑏 − 𝑎𝑎
𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏
By substituting 𝑢𝑢𝑁𝑁 (𝑡𝑡) with equation (1.32), and applying the same stages described in section (3.4), we conclude the system in (3.50 and 3.51), with: • using ℳ −Open Gauss Chebyshev Formula (3.9) , we have: 𝔞𝔞𝑠𝑠𝑠𝑠
ℳ−1
𝜋𝜋 ⎫ = � 𝜓𝜓𝑠𝑠 �𝑡𝑡𝜌𝜌 �𝜓𝜓𝑟𝑟 �𝑡𝑡𝜌𝜌 � ⎪ ℳ 𝜌𝜌 =0 ℳ−1
⎬ 𝜋𝜋 𝔟𝔟𝑠𝑠 = � 𝜓𝜓𝑠𝑠 �𝑡𝑡𝜌𝜌 � ℱ(𝑡𝑡𝜌𝜌 ) ⎪ ℳ ⎭ 𝜌𝜌=0 103
… (3.55)
Chapter Three
Least Square Orthogonal Method
Where, ℳ is the number of Chebyshev zeros we can take it and 𝑡𝑡𝜌𝜌 = �
𝑏𝑏 − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 � 𝑧𝑧𝜌𝜌 + � � ; 2 2
𝑧𝑧𝜌𝜌 = cos �
2𝜌𝜌 + 1 𝜋𝜋 � ℳ 2
… (3.56)
• using ℳ −Closed Gauss-Chebyshev Formula (3.10), we have: 𝔞𝔞𝑠𝑠𝑠𝑠
ℳ
″ 𝜋𝜋 = � ℳ
𝔟𝔟𝑠𝑠 =
𝜌𝜌=0 ℳ
″ 𝜋𝜋 � ℳ 𝜌𝜌=0
𝜓𝜓𝑠𝑠 �𝑡𝑡𝜌𝜌 �𝜓𝜓𝑟𝑟 �𝑡𝑡𝜌𝜌 �
𝜓𝜓𝑠𝑠 �𝑡𝑡𝜌𝜌 �ℱ(𝑡𝑡𝜌𝜌 )
⎫ ⎪
… (3.57)
⎬ ⎪ ⎭
Where, ℳ is the number of Chebyshev extrema zeros we can take it and 𝑡𝑡𝜌𝜌 = �
𝑏𝑏 − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 � 𝑧𝑧𝜌𝜌 + � � 2 2
;
𝑧𝑧𝜌𝜌 = 𝑐𝑐𝑐𝑐𝑐𝑐 �
𝜌𝜌𝜌𝜌 � ℳ
… (3.58)
• To evaluate all cases 𝜓𝜓𝑟𝑟 (𝑡𝑡) in equations (3.43, 3.45 and 3.47) at 𝑡𝑡 = 𝑡𝑡𝜌𝜌 for all 𝑟𝑟 and 𝜌𝜌, for open and closed Gauss-Chebyshev formulas, that is: 𝐼𝐼
𝑛𝑛 −1
𝛼𝛼
𝛼𝛼
�𝜓𝜓𝑟𝑟0 (𝑡𝑡)�𝑡𝑡=𝑡𝑡 = � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) � 𝜌𝜌
𝑖𝑖=1
at 𝑡𝑡=𝑡𝑡 𝜌𝜌
Using lemma (3.1) to evaluate the fractional differentiation of shifted Chebyshev polynomials 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) at all points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 and all different 𝑟𝑟 ≥ 1 . �𝜐𝜐 0 ,𝜐𝜐 1 ,…,𝜐𝜐 ℓ � 𝐼𝐼ℓ+1 (𝑡𝑡 )� �𝜓𝜓𝑟𝑟 𝑡𝑡=𝑡𝑡 𝜌𝜌
+
=
𝛼𝛼 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡 ) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 � 𝒯𝒯𝑟𝑟∗ (𝑡𝑡 ) ��������������������� 𝑖𝑖=1 𝑖𝑖
if ℓ = 0 ; 0 𝑟𝑟 ⎧ 𝑡𝑡 𝑟𝑟 𝜏𝜏 0 𝜏𝜏 𝑗𝑗 ⎪if 𝜐𝜐0 = 0 ; 𝑃𝑃0 (𝑡𝑡 ) � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘∗ (𝑡𝑡 ) − � o. w; 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘∗ (𝑥𝑥) 𝑑𝑑𝑑𝑑 ⎪ ������������� 𝑘𝑘=0 ⎪ 𝑘𝑘=0 ������������������������������� 𝑎𝑎 𝑗𝑗 =𝜐𝜐 1 ,…,𝜐𝜐 ℓ 𝐴𝐴
𝐵𝐵
⎨ 𝑡𝑡 𝑟𝑟 𝜏𝜏 𝑗𝑗 ⎪ � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑘𝑘∗ (𝑥𝑥) 𝑑𝑑𝑑𝑑 ⎪ if 𝜐𝜐0 ≠ 0 ; − 𝜆𝜆 � ⎪ 𝑘𝑘=0 ��������������������������� 𝑎𝑎 𝑗𝑗 =𝜐𝜐 0 ,𝜐𝜐 1 ,…,𝜐𝜐 ℓ ⎩ ����������������������������������������������������� 𝐶𝐶 𝑖𝑖𝑖𝑖
104
Chapter Three
Least Square Orthogonal Method
The following stages can be applied for the above equation: i.
As before, using lemma (3.1) to evaluate the fractional differentiation of shifted Chebyshev polynomials at all points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 and all different 𝑟𝑟 ≥ 1 . Second part of the equation can be evaluated as: 𝜏𝜏 0 for all (𝑟𝑟 ≥ 1 and If 𝜐𝜐0 = 0 : Part 𝑨𝑨 , using lemma (3.4) to find ℎ𝑟𝑟,𝑘𝑘
ii.
𝑘𝑘 = 0,1,2, … , 𝑟𝑟) and we solve part 𝑩𝑩 by Clenshaw-Curtis rule for integration as in (3.3.3) after using lemma (3.4). If 𝜐𝜐0 ≠ 0 : Part𝑪𝑪, by Clenshaw-Curtis rule for integration as in (3.3.3) and 𝜏𝜏
𝑗𝑗 using lemma (3.4) to find ℎ𝑟𝑟,𝑘𝑘 for all (𝑟𝑟 ≥ 1 and 𝑘𝑘 = 0,1,2, … , 𝑟𝑟) for 1: 𝑚𝑚) , then we get the solution for part 𝑪𝑪 . each 𝜏𝜏𝑗𝑗 ′𝑠𝑠 (𝑗𝑗 = ������
𝑛𝑛−1
𝑟𝑟
𝛼𝛼 𝛼𝛼 𝜏𝜏 0 𝐼𝐼 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) �𝜓𝜓𝑟𝑟𝑚𝑚 +1 (𝑡𝑡)�𝑡𝑡=𝑡𝑡 = � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 � 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡) � ℎ𝑟𝑟,𝑘𝑘 𝜌𝜌 ��������������������� ����� �������� 𝑘𝑘=0 𝑖𝑖=1 𝑖𝑖
𝑡𝑡 𝑚𝑚
𝑖𝑖𝑖𝑖
𝑟𝑟
𝜏𝜏
𝑗𝑗 − 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝒯𝒯𝑟𝑟∗ (𝑡𝑡) 𝑑𝑑𝑑𝑑 𝑘𝑘=0 ��������������������� 𝑎𝑎 𝑗𝑗 =1
𝑖𝑖𝑖𝑖𝑖𝑖
The following stages can be applied for the above equation:
i. ii. iii.
As before, using lemma (3.1) to evaluate the fractional differentiation of shifted Chebyshev polynomials at all 𝑡𝑡 = 𝑡𝑡𝜌𝜌 and all different 𝑟𝑟 ≥ 1 . 𝜏𝜏
0 Using lemma (3.4) to find ℎ𝑟𝑟,𝑘𝑘 for all (𝑟𝑟 ≥ 1 and 𝑘𝑘 = 0,1,2, … , 𝑟𝑟) for
delay 𝜏𝜏0 . Solve this part by Clenshaw-Curtis rule for integration as in (3.3.3) after 𝜏𝜏
𝑗𝑗 using lemma (3.4) for ℎ𝑟𝑟,𝑘𝑘 .
• To evaluate all the cases ℱ(𝑡𝑡) in equations (3.44, 3.46 and 3.48) at 𝑡𝑡 = 𝑡𝑡𝜌𝜌 for all 𝑟𝑟 and 𝜌𝜌 , for open and closed Gauss-Chebyshev formulas, that is: 𝑡𝑡 𝑚𝑚
[ℱ𝐼𝐼0 (𝑡𝑡)]𝑡𝑡=𝑡𝑡 𝜌𝜌 = 𝑓𝑓(𝑡𝑡) 𝜑𝜑(𝑡𝑡 � −��� 𝜏𝜏0 ) 𝑃𝑃0 (𝑡𝑡) � + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − �� ����� 𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖 ������������������� 𝑎𝑎 𝑗𝑗 =1 𝑖𝑖𝑖𝑖
105
Chapter Three
Least Square Orthogonal Method
The following stages can be applied for the above equation:
i. ii.
We calculate 𝑓𝑓(𝑡𝑡) at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 .
iii.
after using historical function, 𝜑𝜑(𝑡𝑡).
We solve this part by Clenshaw-Curtis rule for integration as in (3.3.3)
This part evaluates at all 𝑡𝑡 = 𝑡𝑡𝜌𝜌 after using historical function.
�𝜐𝜐 0 ,𝜐𝜐 1 ,…,𝜐𝜐 ℓ �
�ℱ𝐼𝐼ℓ+1
(𝑡𝑡 )�
𝑡𝑡=𝑡𝑡 𝜌𝜌
(𝑡𝑡 ) = 𝑓𝑓 � 𝑖𝑖
+
𝑡𝑡
𝑚𝑚
⎧ if 𝜐𝜐0 = 0 ; 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 ⎪ 𝑗𝑗 =1 ⎪ 𝑎𝑎 ⎪ 𝑗𝑗 ≠𝜐𝜐 1 ,…,𝜐𝜐 ℓ ����������������������� ⎪ 𝑖𝑖𝑖𝑖 𝑡𝑡
𝑚𝑚
⎨ )��� if 𝜐𝜐0 ≠ 0 ; 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − 𝑃𝑃 𝑡𝑡� 𝜑𝜑(𝑡𝑡 � −��� 𝜏𝜏0 ) �� 0 (� ⎪ 𝐵𝐵 𝑗𝑗 =1 ⎪ 𝑎𝑎 𝑗𝑗 ≠𝜐𝜐 1 ,…,𝜐𝜐 ℓ ⎪ ����������������������� ⎪ ����������������������������������� 𝐴𝐴 ⎩ 𝑖𝑖𝑖𝑖𝑖𝑖
The following stages can be applied for the above equation:
i. ii.
We calculate 𝑓𝑓(𝑡𝑡) at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 .
iii.
after using historical function, 𝜑𝜑(𝑡𝑡).
If 𝜐𝜐0 = 0 ; We solve part (ii) by Clenshaw-Curtis rule for integration (3.3.3) If 𝜐𝜐0 ≠ 0 so we have:
A. we solve part A, by Clenshaw-Curtis rule for integration (3.3.3) after using historical function, 𝜑𝜑(𝑡𝑡).
B. Evaluate part B at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 after using historical function, 𝜑𝜑(𝑡𝑡).
While [ℱ𝐼𝐼𝑚𝑚 +1 (𝑡𝑡)]𝑡𝑡=𝑡𝑡 𝜌𝜌 = 𝑓𝑓(𝑡𝑡) , this equation can be calculated 𝑓𝑓(𝑡𝑡) at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 .
106
Chapter Three
Least Square Orthogonal Method
The Algorithm (ADCH): The approximate solution of higher fractional order of linear VIDEs with constant multi-time Retarded delay with variable coefficients (1.32) with the use of Chebyshev polynomials in least-square orthogonal technique can be summarized by the following steps: Step 1: Fixing the zeroth-Chebyshev orthogonal points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 in [𝑎𝑎, 𝑏𝑏], which are defined by equations (3.56) for open formed and (3.58) for closed formed. Step 2:
For each constant delay {𝜏𝜏 = 𝜏𝜏0 , 𝜏𝜏1 , 𝜏𝜏2 , 𝜏𝜏3 … , 𝜏𝜏𝑚𝑚 } find 𝐻𝐻 (𝑗𝑗 = ������ 0: 𝑚𝑚) by using the equations (3.29, 3.32-36).
the
matrix
𝜏𝜏 𝑗𝑗
Step 3:
Using step 2 and for each 𝑡𝑡 = 𝑡𝑡𝜌𝜌 :
a. Evaluate 𝜓𝜓𝑟𝑟 (𝑡𝑡) for shifted Chebyshev polynomial at fixed point by discussing all basic parts as in equation (3.41) and then for cases using equations (3.43, 3.45 and 3.47) for all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 . b. Compute ℱ(𝑡𝑡) by discussing all basic parts as in equation (3.42) and then for cases using equations (3.44, 3.46 and 3.48).
Step 4:
By equations (3.55) and (3.57) and computing the elements of matrices 𝐴𝐴 = [𝑎𝑎𝑠𝑠𝑠𝑠 ]𝑁𝑁+1×𝑁𝑁+1 and 𝐵𝐵 = [𝑏𝑏𝑠𝑠𝑠𝑠 ]𝑁𝑁+1×1 for open and closed Gauss- Chebyshev formula, respectively. Step 5:
Constructing the initial condition matrices 𝐺𝐺 and 𝑈𝑈 using equation (3.53).
Step 6:
Using steps (4 and 5) to construct the matrix 𝐷𝐷 and 𝐸𝐸 which represent in linear system (3.54).
Step 7:
Applying LU-factorization technique to construct system in step 6 for finding ����� 𝑁𝑁) , after multiplying both sides by 𝐷𝐷𝑇𝑇 , as defined constant coefficients 𝐶𝐶𝑟𝑟 (𝑟𝑟 = 0: in equation (3.54). Step 8:
To obtain the approximate solution 𝑢𝑢𝑁𝑁 (𝑡𝑡) of 𝑢𝑢(𝑡𝑡), substitute 𝐶𝐶𝑟𝑟 's in equation (3.37) where 𝐺𝐺𝑟𝑟 is 𝒯𝒯𝑟𝑟∗ . 107
Chapter Three
Least Square Orthogonal Method
3.4.2 Using Legendre Polynomials: In this part, we can take, instead of trial function 𝐺𝐺𝑟𝑟 (𝑡𝑡), the shifted Legendre
polynomial, 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) = 𝒫𝒫𝑟𝑟 �2 �
equation (1.32) by forming:
𝑡𝑡−𝑎𝑎
𝑏𝑏−𝑎𝑎
𝑁𝑁
� − 1�, to approximate the solution
𝑢𝑢𝑁𝑁 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒫𝒫𝑟𝑟 �2 � 𝑟𝑟=0
𝑡𝑡 − 𝑎𝑎 � − 1� , 𝑏𝑏 − 𝑎𝑎
𝑢𝑢(𝑡𝑡) of
𝑎𝑎 ≤ 𝑡𝑡 ≤ 𝑏𝑏
By applying the same stages as in Chebyshev polynomials technique section (3.4.1) and using Gauss-Legendre formula (3.11), we get:
𝔞𝔞𝑠𝑠𝑠𝑠
ℳ−1
𝑏𝑏 − 𝑎𝑎 (ℳ) ⎫ = � 𝑤𝑤𝜌𝜌 𝜓𝜓𝑠𝑠 �𝑡𝑡𝜌𝜌 �𝜓𝜓𝑟𝑟 �𝑡𝑡𝜌𝜌 � ⎪ 2 𝜌𝜌=0 ℳ−1
… (3.59)
⎬ 𝑏𝑏 − 𝑎𝑎 (ℳ) 𝔟𝔟𝑠𝑠 = � 𝑤𝑤𝜌𝜌 𝜓𝜓𝑠𝑠 �𝑡𝑡𝜌𝜌 � ℱ(𝑡𝑡𝜌𝜌 ) ⎪ 2 ⎭ 𝜌𝜌=0
where, ℳ is the number of Legendre zeros we can take it and 𝑡𝑡𝜌𝜌 = �
𝑏𝑏 − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 � 𝑧𝑧𝜌𝜌 + � � 2 2 (ℳ)
Here 𝑧𝑧𝜌𝜌 is the ℳ-Th zeros of 𝑃𝑃ℳ (𝑡𝑡) and 𝑤𝑤𝜌𝜌 (ℳ)
𝑤𝑤𝜌𝜌
=
�1 −
2
2 ′ 𝑧𝑧𝜌𝜌2 ��𝑃𝑃ℳ (𝑧𝑧𝜌𝜌 )�
… (3.60)
are defined as in equation below:
, 𝜌𝜌 = 0,1,2, … , ℳ − 1
To evaluate all cases 𝜓𝜓𝑟𝑟 (𝑡𝑡) in equations (3.43, 3.45 and 3.47) at 𝑡𝑡 = 𝑡𝑡𝜌𝜌 for all 𝑟𝑟 and 𝜌𝜌, for Legendre formula, that is: 𝐼𝐼 �𝜓𝜓𝑟𝑟0 (𝑡𝑡)�𝑡𝑡=𝑡𝑡
𝜌𝜌
=
𝛼𝛼 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 𝒫𝒫𝑟𝑟∗ (𝑡𝑡)
𝑛𝑛−1
𝛼𝛼
+ � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) � 𝑖𝑖=1
𝑡𝑡=𝑡𝑡 𝜌𝜌
Apply lemma (3.2) to evaluate the fractional differentiation of shifted Legendre polynomial at all points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 and all different 𝑟𝑟 ≥ 1 .
108
Chapter Three
Least Square Orthogonal Method
�𝜐𝜐 0 ,𝜐𝜐 1 …𝜐𝜐 ℓ � 𝐼𝐼ℓ+1 (𝑡𝑡)� �𝜓𝜓𝑟𝑟
+
𝑡𝑡=𝑡𝑡 𝜌𝜌
𝛼𝛼
𝑛𝑛−1
𝛼𝛼
= � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 � 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) ��������������������� 𝑖𝑖=1 𝑖𝑖
if ℓ = 0 ; 0 𝑟𝑟 ⎧ 𝑡𝑡 𝑟𝑟 𝜏𝜏 0 ∗ 𝜏𝜏 𝑗𝑗 (𝑡𝑡) (𝑡𝑡) ⎪if 𝜐𝜐0 ; 𝑃𝑃0 � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑟𝑟 −� 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) 𝑑𝑑𝑑𝑑 o. w; 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 ⎪ ������������� 𝑘𝑘=0 ⎪ 𝑘𝑘=0 ������������������������������� 𝑎𝑎 𝑗𝑗 =𝜐𝜐 1 ,…,𝜐𝜐 ℓ 𝐴𝐴
𝐵𝐵
⎨ 𝑡𝑡 𝑟𝑟 𝜏𝜏 𝑗𝑗 ⎪ ∗ ⎪ if 𝜐𝜐0 ≠ 0 ; − 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑟𝑟 (𝑡𝑡) 𝑑𝑑𝑑𝑑 ⎪ 𝑘𝑘=0 ��������������������������� 𝑎𝑎 𝑗𝑗 =𝜐𝜐 0 ,𝜐𝜐 1 …𝜐𝜐 ℓ ⎩ ��������������������������������������������������� 𝐶𝐶 𝑖𝑖𝑖𝑖
The following stages can be applied for the above equation: i.
First, using lemma (3.2) to evaluate the fractional differentiation of shifted Legendre polynomial at all points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 and all different 𝑟𝑟 ≥ 1. The second part of the equation can be evaluated as the following: 𝜏𝜏 0 for all (𝑟𝑟 ≥ 1 and If 𝜐𝜐0 = 0 : Part 𝑨𝑨 , using lemma (3.3) to find ℎ𝑟𝑟,𝑘𝑘 𝑘𝑘 = 0,1,2, … , 𝑟𝑟). And we solve part 𝑩𝑩 by Clenshaw-Curtis rule for integration as in (3.3.3) after using lemma (3.3). If 𝜐𝜐0 ≠ 0 : Part 𝑪𝑪 , by Clenshaw-Curtis rule for integration as in (3.3.3) and 𝜏𝜏 𝑗𝑗 using lemma (3.3) to find ℎ𝑟𝑟,𝑘𝑘 for all (𝑟𝑟 ≥ 1 and 𝑘𝑘 = 0,1,2, … , 𝑟𝑟) and for each 𝜏𝜏𝑗𝑗 .
ii.
𝐼𝐼 �𝜓𝜓𝑟𝑟𝑚𝑚 +1 (𝑡𝑡)�𝑡𝑡=𝑡𝑡
𝜌𝜌
=
𝑛𝑛−1
𝑟𝑟 𝛼𝛼 𝜏𝜏 0 + � 𝑃𝑃𝑖𝑖 (𝑡𝑡) 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛 −𝑖𝑖 � 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) + 𝑃𝑃0 (𝑡𝑡) � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) ��������������������� �������� ����� 𝑘𝑘=0 𝑖𝑖=1 𝑖𝑖 𝑖𝑖𝑖𝑖 𝑡𝑡 𝑚𝑚 𝑟𝑟 𝜏𝜏 𝑗𝑗 − 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) � ℎ𝑟𝑟,𝑘𝑘 𝒫𝒫𝑟𝑟∗ (𝑡𝑡) 𝑑𝑑𝑑𝑑 𝑘𝑘=0 ��������������������� 𝑎𝑎 𝑗𝑗 =1
𝛼𝛼 � 𝐶𝐶𝑎𝑎 𝐷𝐷𝑡𝑡 𝑛𝑛
The following stages can be applied for the above equation:
i. ii. iii.
𝑖𝑖𝑖𝑖𝑖𝑖
As before, using lemma (3.2) to evaluate the fractional differentiation of shifted Legendre polynomial at all 𝑡𝑡 = 𝑡𝑡𝜌𝜌 and all different 𝑟𝑟 ≥ 1 . 𝜏𝜏 0 Using lemma (3.3) to find ℎ𝑟𝑟,𝑘𝑘 for (𝑟𝑟 ≥ 1 and 𝑘𝑘 = 0,1,2, … , 𝑟𝑟) and constant delay 𝜏𝜏0 . Solving this part by Clenshaw-Curtis rule for integration as (3.3.3) after 𝜏𝜏 𝑗𝑗 using lemma (3.3) for ℎ𝑟𝑟,𝑘𝑘 . 109
Chapter Three
Least Square Orthogonal Method
To evaluate all cases ℱ(𝑡𝑡) in equations (3.44, 3.46 and 3.48) at 𝑡𝑡 = 𝑡𝑡𝜌𝜌 for all 𝑟𝑟 and 𝜌𝜌 , for Legendre polynomial formulas, that is: 𝑡𝑡 𝑚𝑚
[ℱ𝐼𝐼0 (𝑡𝑡)]𝑡𝑡=𝑡𝑡 𝜌𝜌 = 𝑓𝑓(𝑡𝑡) 𝜑𝜑(𝑡𝑡 � −��� 𝜏𝜏0 ) 𝑃𝑃0 (𝑡𝑡) � + 𝜆𝜆 � � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − �� ����� 𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖 ������������������� 𝑎𝑎 𝑗𝑗 =1 𝑖𝑖𝑖𝑖
The following stages can be applied for the above equation:
i. ii. iii.
We calculate 𝑓𝑓(𝑡𝑡) at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 .
We solve this part by Clenshaw-Curtis rule for integration as in (3.3.3) after using historical function, 𝜑𝜑(𝑡𝑡). This part evaluates at all 𝑡𝑡 = 𝑡𝑡𝜌𝜌 after using historical function, 𝜑𝜑(𝑡𝑡).
�𝜐𝜐 0 ,𝜐𝜐 1 …𝜐𝜐 ℓ �
�ℱ𝐼𝐼ℓ+1
(𝑡𝑡)�
𝑡𝑡=𝑡𝑡 𝜌𝜌
= 𝑓𝑓(𝑡𝑡) � 𝑖𝑖
+
⎧ if 𝜐𝜐0 = 0 ; ⎪ ⎪ ⎪ ⎪
𝑡𝑡
𝜆𝜆 � 𝑎𝑎
𝑚𝑚
� 𝑗𝑗 =1
𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑
𝑗𝑗 ≠𝜐𝜐 1 ,…,𝜐𝜐 ℓ ����������������������� 𝑡𝑡
𝑚𝑚
𝑖𝑖𝑖𝑖
⎨ 𝜑𝜑(𝑡𝑡 � −��� 𝜏𝜏0 ) � 𝒦𝒦𝑗𝑗 (𝑡𝑡, 𝑥𝑥) 𝜑𝜑�𝑥𝑥 − 𝜏𝜏𝑗𝑗 � 𝑑𝑑𝑑𝑑 − �� 𝑃𝑃0 (𝑡𝑡) if 𝜐𝜐0 ≠ 0 ; 𝜆𝜆 � ����� ⎪ 𝐵𝐵 𝑗𝑗 =1 ⎪ 𝑎𝑎 𝑗𝑗 ≠𝜐𝜐 0 ,𝜐𝜐 1 ,…,𝜐𝜐 ℓ ⎪ ����������������������� ⎪ ������������������������������������� 𝐴𝐴 ⎩ 𝑖𝑖𝑖𝑖𝑖𝑖
The following stages can be applied for the above equation:
i. ii. iii.
We calculate 𝑓𝑓(𝑡𝑡) at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 .
If 𝜐𝜐0 = 0 : We solve part (ii) it by Clenshaw-Curtis rule for integration (3.3.3) after using historical function, 𝜑𝜑(𝑡𝑡). If 𝜐𝜐0 ≠ 0 so we have: A. we solve part A by Clenshaw-Curtis rule for integration (3.3.3) after using historical function, 𝜑𝜑(𝑡𝑡). B. Evaluate part B at all collocation points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 after using historical function, 𝜑𝜑(𝑡𝑡).
[ℱ𝐼𝐼𝑚𝑚 +1 (𝑡𝑡)]𝑡𝑡=𝑡𝑡 𝜌𝜌 = 𝑓𝑓 (𝑡𝑡 ) , This equation can be calculated 𝑓𝑓(𝑡𝑡)
points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 .
110
at all collocation
Chapter Three
Least Square Orthogonal Method
The Algorithm (ADLP): The approximate solution of higher fractional order of linear VIDEs of constant multi-time Retarded delay with variable coefficients (1.32) with the use of Legendre polynomial in least-square orthogonal technique can be summarized by the following steps: Step 1: a. Fixing the zeroth-Legendre orthogonal points 𝑡𝑡 = 𝑡𝑡𝜌𝜌 in [𝑎𝑎, 𝑏𝑏], which are defined by equation (3.11). (ℳ)
b. Calculate the weighted Legendre constant 𝑤𝑤𝜌𝜌 by 2 (ℳ) 𝑤𝑤𝜌𝜌 = 𝜌𝜌 = 0,1,2, … , ℳ − 1 2 , ′ 2 �1 − 𝑧𝑧𝜌𝜌 ��𝐿𝐿ℳ (𝑧𝑧𝜌𝜌 )� where ℳ is the number of zeroth of 𝐿𝐿ℳ (𝑡𝑡). Step 2: For each constant delay {𝜏𝜏 = 𝜏𝜏0 , 𝜏𝜏1 , 𝜏𝜏2 , 𝜏𝜏3 … , 𝜏𝜏𝑚𝑚 } find the matrix 𝐻𝐻𝜏𝜏 𝑗𝑗 (𝑗𝑗 = ������ 0: 𝑚𝑚) by using the equations (3.19, 3.22-25). Step 3: Using step 2 and for each 𝑡𝑡 = 𝑡𝑡𝜌𝜌 : a. Evaluate 𝜓𝜓𝑟𝑟 (𝑡𝑡) for shifted Legendre polynomial at fixed point by discussing all basic parts as in equation (3.41) and then for cases using equations (3.43, 3.45 and 3.47) for all 𝑟𝑟 = 0,1,2, … , 𝑁𝑁 . b. Compute ℱ(𝑡𝑡) by discussing all basic parts as in equation (3.42) and then for cases using equations (3.44, 3.46 and 3.48). Step 4: By equation (3.59) and computing the elements of matrices 𝐴𝐴 = [𝑎𝑎𝑠𝑠𝑠𝑠 ]𝑁𝑁+1×𝑁𝑁+1 and 𝐵𝐵 = [𝑏𝑏𝑠𝑠𝑠𝑠 ]𝑁𝑁+1×1 Gauss-Legendre formula, respectively. Step 5: Constructing the initial condition matrices 𝐺𝐺 and 𝑈𝑈 using equation (3.53). Step 6: Using steps (4 and 5) to construct the matrix 𝐷𝐷 and 𝐸𝐸 which represent in linear system (3.54). Step 7: Applying LU-factorization technique to construct system in step 6 for finding ����� 𝑁𝑁) , after multiplying both sides by 𝐷𝐷𝑇𝑇 , as defined constant coefficients 𝐶𝐶𝑟𝑟 (𝑟𝑟 = 0: in equation (3.54). Step 8: To obtain the approximate solution 𝑢𝑢𝑁𝑁 (𝑡𝑡) of 𝑢𝑢(𝑡𝑡) , substitute 𝐶𝐶𝑟𝑟 's in equation (3.37) with 𝐺𝐺𝑟𝑟 is 𝒫𝒫𝑟𝑟∗ 111
Chapter Three
Least Square Orthogonal Method
3.5 Implementation of the method: Here, some numerical examples are presented for a linear Volterra IntegroFractional Differential Equation with multi-time Delay. Their results are obtained by running programs Main Delay OpenCheb, Main Delay ClosedCheb and Main Delay Legendre, respectively. Example (3.1): Recall test example (1) which is the linear fractional VIDE of constant multitime Retarded delay. Take 𝑁𝑁 = 2, ℳ = 5 and 𝐼𝐼𝐼𝐼 = 18 (numerical approximation number in method, general Clenshaw-Curtis formula) assume the approximate solution in the form: 2
2
𝑢𝑢2 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒯𝒯𝑟𝑟 (2𝑡𝑡 − 1) and
𝑢𝑢2 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒫𝒫𝑟𝑟 (2𝑡𝑡 − 1)
𝑟𝑟=0
𝑟𝑟=0
By applying lemmas (3.3 and 3.4) for this example, we determine the different matrices 𝐻𝐻 with respect to each different constant 𝜏𝜏-delay for Legendre polynomials, and Chebyshev polynomial, using m.file Matlab programs: OrthogonalTauCheb and OrthogonalTauLeg. Here:𝜏𝜏 = 𝜏𝜏0 = 0.8 , 𝜏𝜏1 = 0.6 , 𝜏𝜏2 = 0.4 , 𝜏𝜏3 = 0.2 : For Legendre orthogonal polynomials:
𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏
1 = �−1.6 3.84
1 𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏 2 = �−0.8 0.96
0 1 −4.8 0 1 −2.4
0 0� 1 0 0� 1
,
,
𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏 1
1 = �−1.2 2.16
1 𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏 3 = �−0.4 0.24
0 1 −1.2
0 0� 1
1 = �−1.2 2.88
0 1 −4.8
0 0� 1
For Chebyshev orthogonal polynomials:
𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 2
1 = �−1.6 5.12 1 = �−0.8 1.28
0 0 1 0� −6.4 1 0 0 1 0� −3.2 1
0 0 1 0� −3.6 1
,
𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 1 ,
𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 3
1 = �−0.4 0.32
0 0 1 0� −1.6 1
Here two different values of fractional order 𝛼𝛼 are taken for testing: 112
Chapter Three
i.
Least Square Orthogonal Method
For 𝛼𝛼 = 0.2, apply the algorithms [ADCH and ADLP] to find the approximate solution for our problem, so run the programs Main Delay OpenCheb, Main Delay ClosedCheb and Main Delay Legendre to find the constant coefficients 𝐶𝐶𝑟𝑟 's, which is present in the table (3.1): Table (3.1)
𝑪𝑪𝒓𝒓
LSOM DCH DOH DLE
𝑪𝑪𝟎𝟎
𝑪𝑪𝟏𝟏
𝑪𝑪𝟐𝟐
−0.25006037
0.99991022
0.24997042
−0.33337826
0.9999205
0.33329872
−0.25005997
0.99991211
0.24997144
Thus, from table (3.1) and the approximation expression the following approximate formulas become:
ii.
𝑢𝑢2𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑡𝑡 ) = 1.99976336𝑡𝑡 2 + 0.00005708 𝑡𝑡 – 1.00000017 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑢𝑢2 (𝑡𝑡 ) = 1.99977152 𝑡𝑡 2 + 0.0000527 𝑡𝑡 – 1.00000064 𝐿𝐿𝐿𝐿𝐿𝐿 𝑢𝑢2 (𝑡𝑡) = 1.99979232𝑡𝑡 2 + 0.00004868 𝑡𝑡 − 1.00000004
For 𝛼𝛼 = 0.85, Assume the approximation solution the same as before, by applying the algorithms [ADCP and ADLP] and runing programs of Main Delay OpenCheb, Main Delay ClosedCheb and Main Delay Legendre, the result of least square orthogonal methods are obtained and table (3.2) presents the values 𝐶𝐶𝑟𝑟 's, respectively. Table (3.2)
𝑪𝑪𝒓𝒓
LSOM DCH DOH DLE
𝑪𝑪𝟎𝟎
𝑪𝑪𝟏𝟏
𝑪𝑪𝟐𝟐
−0.25000418
0.99997939
0.24999607
−0.33333778
0.99999044
0.33333026
−0.24999348
0.99998073
0.24999671
Thus, from table (3.2) and the approximation expression the following approximate formulas become:
113
Chapter Three
Least Square Orthogonal Method
𝑢𝑢2𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑡𝑡 ) = 1.99996856 𝑡𝑡 2 − 0.00000978𝑡𝑡 − 0.9999875 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑢𝑢2 (𝑡𝑡 ) = 1.99997368 𝑡𝑡 2 − 0.00001222 𝑡𝑡 − 0.9999775 𝐿𝐿𝐿𝐿𝐿𝐿 𝑢𝑢2 (𝑡𝑡) = 1.99998156 𝑡𝑡 2 − 0.00000068 𝑡𝑡 − 0.99999796
Table (3.3) shows a comparison between the exact solution 𝑢𝑢(𝑡𝑡) and approximate solutions 𝑢𝑢2 (𝑡𝑡) for all closed and open Chebyshev and Legendre polynomials, respectively, for fractional order 𝛼𝛼 = 0.2 . Table (3.3)
𝒕𝒕
0
Exact Solution −1
Least-Square Orthogonal Method for 𝜶𝜶 = 𝟎𝟎. 𝟐𝟐 DCH
DOH
DLE
−1.00000017
−1.00000064
−1.00000004
0.1
−0.98
−0.9799968284
−0.9799976548
−0.9799972488
0.3
−0.82
−0.8200043436
−0.8200053932
−0.8200041272
0.2 0.4 0.5
−0.92 −0.68 −0.5
−0.9199982196 −0.6800152004 −0.50003079
−0.9199992392 −0.6800161168 −0.50003141
−0.9199986112 −0.6800137968 −0.50002762
0.6
−0.28
−0.2800511124
−0.2800512728
−0.2800455968
0.8
0.28
0.2798940444
0.2798952928
0.2799059888
0.7 0.9 1.0
−0.02 0.62 1
L.S.E
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖 R. Time/Sec
−0.0200761676 0.6198595236 0.99982027
−0.0200757052 0.6198617212 0.99982358
−0.0200677272 0.6198755512 0.99984096
7.28883 𝑒𝑒 − 08
7.08507 𝑒𝑒 − 08
5.72651 𝑒𝑒 − 08
6.234342
6.143791
6.248377
3.12473 𝑒𝑒 − 09
2.97454𝑒𝑒 − 09
114
2.721 𝑒𝑒 − 09
Chapter Three
Least Square Orthogonal Method
Table (3.4) shows a comparison between the exact solution 𝑢𝑢(𝑡𝑡) and approximate solutions 𝑢𝑢2 (𝑡𝑡) for all closed and open Chebyshev and Legendre polynomials, respectively, for fractional order 𝛼𝛼 = 0.85 . Table (3.4)
𝒕𝒕
Exact
0
−1
Least-Square Orthogonal Method for 𝜶𝜶 = 𝟎𝟎. 𝟖𝟖𝟖𝟖 DCH
DOH
DLE
−0.9999875
−0.9999775
−0.99999796
0.1
−0.98
−0.9799887924
−0.9799789852
−0.9799982124
0.3
−0.82
−0.8199932636
−0.8199835348
−0.8199998236
0.2 0.4 0.5
−0.92 −0.68 −0.5
−0.9199907136 −0.6799964424 −0.50000025
−0.9199809968 −0.6799865992 −0.49999019
−0.9199988336 −0.6800011824 −0.50000291
0.6
−0.28
−0.2800046864
−0.2799943072
−0.2800050064
0.8
0.28
0.2799845544
0.2799958792
0.2799896944
0.7 0.9 1.0
−0.02 0.62 L.S.E
1
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖 R. Time/Sec
−0.0200097516
−0.0199989508
0.6199782316 0.99997128
0.6199901828 0.99998396
2.08052 𝑒𝑒 − 09
2.26006 𝑒𝑒 − 09
6.247662
5.903073
1.5364 𝑒𝑒 − 08
1.46812 𝑒𝑒 − 08
−0.0200074716 0.6199864916 0.99998292
6.79912 𝑒𝑒 − 10
1.87959 𝑒𝑒 − 08 6.190474
The result in Table (3.5) shows the least square errors , running times and the values residual equations 𝑅𝑅2 (𝑡𝑡; 𝐶𝐶̅ ) are also included by applying the formula (3.40), for all open and closed Chebyshev and Legendre respectively for two different values of fractional order 𝛼𝛼 = 0.2 and 𝛼𝛼 = 0.85 and three different 𝐼𝐼𝐼𝐼. 115
Chapter Three
Least Square Orthogonal Method
Table (3.5) Least-Square Orthogonal Method 𝑰𝑰𝑰𝑰
𝜶𝜶 LSOM DCH
18
DOH DLE DCH
19
DOH DLE DCH
20
DOH DLE
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
7.28883 𝑒𝑒 − 08 7.08507 𝑒𝑒 − 08 5.72651 𝑒𝑒 − 08
𝜶𝜶 = 𝟎𝟎. 𝟐𝟐
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
6.234342
2.721 𝑒𝑒 − 09
6.248377
2.97454 𝑒𝑒 − 09
4.74955 𝑒𝑒 − 11
1.03847 𝑒𝑒 − 09
4.11228 𝑒𝑒 − 11
1.54 𝑒𝑒 − 15
2.0533 𝑒𝑒 − 16
1.54 𝑒𝑒 − 15
1.01332 𝑒𝑒 − 15
/𝑺𝑺𝑺𝑺𝑺𝑺
3.12473 𝑒𝑒 − 09
1.27629 𝑒𝑒 − 09 1.24605 𝑒𝑒 − 09
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻
4.52648 𝑒𝑒 − 11 2.0533 𝑒𝑒 − 16 1.075 𝑒𝑒 − 16
6.143791
6.650073 5.948829 6.211659 6.156947 5.842171 6.108334
116
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
2.08052 𝑒𝑒 − 09 2.26006 𝑒𝑒 − 09 6.79912 𝑒𝑒 − 10 3.73272 𝑒𝑒 − 11
4.08726 𝑒𝑒 − 11
1.27376 𝑒𝑒 − 11
𝜶𝜶 = 𝟎𝟎. 𝟖𝟖𝟖𝟖
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
/𝑺𝑺𝑺𝑺𝑺𝑺
1.5364 𝑒𝑒 − 08
6.247662
1.87959 𝑒𝑒 − 08
6.190474
1.46812 𝑒𝑒 − 08
2.61727 𝑒𝑒 − 10 2.4931 𝑒𝑒 − 10
3.20092 𝑒𝑒 − 10
1.2326 𝑒𝑒 − 32
6.45527 𝑒𝑒 − 16
1.3332 𝑒𝑒 − 16
1.02357 𝑒𝑒 − 14
1.2326 𝑒𝑒 − 32
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻
6.45527 𝑒𝑒 − 16
5.903073
7.049830 6.109064 6.247900 7.134330 6.359825 6.221093
Chapter Three
Least Square Orthogonal Method
Example (3.2): Recall example (2), take 𝑁𝑁 = 3 , ℳ = 5 and 𝐼𝐼𝐼𝐼 = 18 (numerical approximation number in method general Clenshaw-Curtis formula) and assume the approximate solution in the form: 3
𝑢𝑢3 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒯𝒯𝑟𝑟 (2𝑡𝑡 − 1) and 𝑟𝑟=0
3
𝑢𝑢3 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒫𝒫𝑟𝑟 (2𝑡𝑡 − 1) 𝑟𝑟=0
By applying lemmas (3.3 and 3.4) for this example we determine the different matrices 𝐻𝐻 with respect to each different constant 𝜏𝜏-delay for Legendre polynomials, and Chebyshev polynomial, using m.file Matlab programs: OrthogonalTauCheb and OrthogonalTauLeg. Here 𝜏𝜏 = 𝜏𝜏0 = 0.33 , 𝜏𝜏1 = 0.6 , 𝜏𝜏2 = 0.72 , 𝜏𝜏3 = 0.2 : For Legendre orthogonal polynomials: 0 0 1 −0.66 1 0 𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏 = � 0.6534 −1.98 1 −1.37874 3.267 −3.3 0 1 0 −1.44 1 0 𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏 2 = � 3.1104 −4.32 1 −8.90496 15.552 −7.2
0 0 � 0 1 0 0 � 0 1
For Chebyshev orthogonal polynomials: 1 −0.66 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 = � 0.8712 −3.12998 1 −1.44 =� 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 2 4.1472 −16.26394
0 0 1 0 −2.64 1 5.2272 −3.96 0 0 1 0 −5.76 1 24.8832 −8.64
0 0 � 0 1 0 0 � 0 1
𝐿𝐿𝐿𝐿𝐿𝐿
𝐻𝐻𝜏𝜏 1
𝐿𝐿𝐿𝐿𝐿𝐿
𝐻𝐻𝜏𝜏 3
1 −1.2 =� 2.16 −5.52 1 −0.4 =� 0.24 −0.56
0 1 −3.6 10.8 0 1 −1.2 1.2
0 0 1 −6 0 0 1 −2
0 1 0 −1.2 1 0 =� 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 1 2.88 −4.8 1 −10.512 17.28 −7.2 0 1 0 −0.4 1 0 =� 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 3 0.32 −1.6 1 −1.456 1.92 −2.4
0 0 � 0 1 0 0 � 0 1 0 0 � 0 1 0 0 � 0 1
Furthermore, after running the programs, table (3.6) and the approximation expression the following approximate formulas are obtained: 𝑢𝑢3𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑡𝑡) = −1.000358048 𝑡𝑡 4 + 0.000173232 𝑡𝑡 2 + 1.999963018 𝑡𝑡 + 1.000000019 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑢𝑢3 (𝑡𝑡) = −1.000353248 𝑡𝑡 4 + 0.000158272 𝑡𝑡 2 + 1.999978118 𝑡𝑡 + 1.000000479 𝐿𝐿𝐿𝐿𝐿𝐿 𝑢𝑢3 (𝑡𝑡) = −1.00037236 𝑡𝑡 4 + 0.00019482 𝑡𝑡 2 + 1.999961324 𝑡𝑡 + 1.000000088
117
Chapter Three
Least Square Orthogonal Method
Table (3.6) 𝑪𝑪𝒓𝒓
𝑪𝑪𝟎𝟎
LSOM DCH
DLE
𝑪𝑪𝟐𝟐
𝑪𝑪𝟑𝟑
1.6874346
0.53115029
−0.18754548 −0.031261189
1.7499526
0.54991051
−0.25006062 −0.050018618
1.6874385
DOH
𝑪𝑪𝟏𝟏
0.53115261
−0.18754645 −0.031261039
Table (3.7) 𝒕𝒕
0
Least-Square Orthogonal Method
Exact Solution
DCH
DOH
DLE
1
1.000000019
1.000000479
1.000000088
0.1
1.199
1.19899769507
1.19899952027
1.19899779624
0.3
1.573
1.57299484798
1.57299862118
1.57299596528
1.875
1.87498008
1.87498495
0.2 0.4 0.5 0.6 0.7 0.8 0.9 1.0
1.392 1.736
1.3919966875
1.73599002825
1.3919996075
1.73599444185
1.39199716672 1.73599195776 1.87498291
1.984
1.98396285495
1.98396802615
1.98396658784
2.088
2.0878979813
2.0879034045
2.08790318368
2.057 2.071
L.S.E
2
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖 R. Time/Sec
2.05693620482
2.05694155082
2.05694075712
2.07084603613
2.07085146773
8.92873 𝑒𝑒 − 08
8.29109 𝑒𝑒 − 08
8.31094 𝑒𝑒 − 08
9.885762
9.700046
9.948770
1.999778221
9.54403 𝑒𝑒 − 10
1.999783621
6.98591 𝑒𝑒 − 10
118
2.07085163336 1.999783872
7.54664 𝑒𝑒 − 10
Chapter Three
Least Square Orthogonal Method
Apply the algorithms [ADCH and ADLP] to find the approximate solution of our problem, and run the programs Main Delay OpenCheb, Main Delay ClosedCheb and Main Delay Legendre to find the constant coefficients Cr 's, which are present in Table (3.6). Also, Table (3.7) shows a comparison between the exact solution 𝑢𝑢(𝑡𝑡) and approximate solutions 𝑢𝑢3 (𝑡𝑡) for all closed and open Chebyshev and Legendre polynomials respectively.
Example (3.3):
Recall test example (5) which is the linear fractional VIDE of constant multitime Retarded delay with variable coefficients and the maximum arbitrary order is greater than 1. Take 𝑁𝑁 = 2, ℳ = 7 and 𝐼𝐼𝐼𝐼 = 15 (numerical approximation number in method general Clenshaw-Curtis formula) and assume the approximate solution in the form: 2
𝑢𝑢2 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒯𝒯𝑟𝑟 (2𝑡𝑡 − 1) and 𝑟𝑟=0
2
𝑢𝑢2 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒫𝒫𝑟𝑟 (2𝑡𝑡 − 1) 𝑟𝑟=0
By applying the lemma (3.3 and 3.4) for this example we determine the different matrices 𝐻𝐻 with respect to each different constant 𝜏𝜏-delay for Legendre polynomials, and Chebyshev polynomial, using m.file Matlab programs: OrthogonalTauCheb and OrthogonalTauLeg. Here: 𝜏𝜏 = 𝜏𝜏0 = 0.45 , 𝜏𝜏1 = 0.7 𝜏𝜏2 = 0.45 : For Legendre orthogonal polynomials:
𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏
1 = � −0.9 1.215
0 1 −2.7
0 1 𝐿𝐿𝐿𝐿𝐿𝐿 0� 𝐻𝐻𝜏𝜏 1 = �−1.4 1 2.94
0 0 1 0� −4.2 1
For Chebyshev orthogonal polynomials: 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒
1 = �−0.9 1.62
0 0 1 0� −3.6 1
𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 1
1 = �−1.4 3.92
0 1 −5.6
𝐿𝐿𝐿𝐿𝐿𝐿 𝐻𝐻𝜏𝜏 2
1 = � −0.9 1.215
0 1 𝐶𝐶ℎ𝑒𝑒 0� 𝐻𝐻𝜏𝜏 2 = �−0.9 1 1.62
0 0 1 0� −2.7 1
0 0 1 0� −3.6 1
apply the algorithms [ADCH and ADLP] to find the approximate solution of our problem, and run the programs Main Delay OpenCheb, Main Delay ClosedCheb and Main Delay Legendre to find the constant coefficients 𝐶𝐶𝑟𝑟 's, which are present in table (3.8)
119
Chapter Three
Least Square Orthogonal Method
Table (3.8) 𝑪𝑪𝒓𝒓
LSOM DCH DOH DLE
𝑪𝑪𝟎𝟎
1.1260698 1.1260698 1.0010648
𝑪𝑪𝟏𝟏
0.50124201 0.50112436 0.50131998
𝑪𝑪𝟐𝟐
0.37531972 0.37529134 0.50044674
Furthermore, after running the programs, table (3.8) and the approximation expression, the following approximate formulas are obtained: 𝑢𝑢2𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑡𝑡 ) = 3.00255776𝑡𝑡 2 − 2.00007374 𝑡𝑡 + 1.00014751 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑢𝑢2 (𝑡𝑡 ) = 3.00233072 𝑡𝑡 2 − 2.000082 𝑡𝑡 + 1.00023678 𝐿𝐿𝐿𝐿𝐿𝐿 𝑢𝑢2 (𝑡𝑡) = 3.00268044 𝑡𝑡 2 − 2.00004048 𝑡𝑡 + 1.00019156
Table (3.9)
𝒕𝒕
0
Least-Square Orthogonal Method
Exact Solution
DCH
DOH
DLE
1
1.00014751
1.00023678
1.00019156
0.1
0.83
0.8301657136
0.8302518872
0.8302143164
0.3
0.67
0.6703555864
0.6704219448
0.6704206556
0.2 0.4 0.5 0.6
0.72 0.68 0.75
0.7202350724
0.7203136088
0.6805272556
0.6805768952
0.75075008
0.75077846
0.7202906816 0.6806042384 0.75084143
0.88
0.8810240596
0.8810266392
0.8811322304
0.8
1.32
1.3217254844
1.3216628408
1.3218746576
1.0
2
0.7 0.9
1.0
1.63
L.S.E
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖 R. Time/Sec
1.0713491944
1.0713214328
1.6321529296
1.6320508632
2.00263153
2.0024855
1.84779 𝑒𝑒 − 06
1.72837 𝑒𝑒 − 06
7.151392
7.814615
5.19708 𝑒𝑒 − 07
5.94394 𝑒𝑒 − 07
1.0714766396 1.6323262844 2.00283152
2.1823 𝑒𝑒 − 06
5.31708 𝑒𝑒 − 07 7.528204
Table (3.9) shows a comparison between the exact solution 𝑢𝑢(𝑡𝑡) and approximate solutions 𝑢𝑢2 (𝑡𝑡) for all closed and open Chebyshev and Legendre polynomials respectively. 120
Chapter Three
Least Square Orthogonal Method
The result in table (3.10) shows the least square errors, running times and the values residual equations 𝑅𝑅2 (𝑡𝑡; 𝐶𝐶̅ ) which are also included by applying the formula (3.40), for all open and closed Chebyshev and Legendre respectively for three different values of ℳ-zeros. Table (3.10)
Least-Square Orthogonal Method 𝓜𝓜
LSOM
𝟕𝟕
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
𝟏𝟏𝟏𝟏
𝐃𝐃𝐃𝐃𝐃𝐃
DOH
DLE 2.1823 𝑒𝑒 − 06
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
1.84779 𝑒𝑒 − 06
1.72837 𝑒𝑒 − 06
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
7.151392
7.814615
5.19708 𝑒𝑒 − 07
5.94394 𝑒𝑒 − 07
5.31708 𝑒𝑒 − 07 7.528204
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
1.82216 𝑒𝑒 − 06
1.79489 𝑒𝑒 − 06
2.28374 𝑒𝑒 − 06
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
11.806297
13.670802
14.787157
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
514945 𝑒𝑒 − 07
5.46566 𝑒𝑒 − 07
5.321 𝑒𝑒 − 07
Table (3.11) Least-Square Orthogonal Method 𝑰𝑰𝑰𝑰
LSOM
DCH
DOH
DLE
DCH
DOH
DLE
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
1.84779 𝑒𝑒 − 05
1.72837 𝑒𝑒 − 05
2.1823 𝑒𝑒 − 05
8.62817 𝑒𝑒 − 32
8.62817 𝑒𝑒 − 32
8.62817 𝑒𝑒 − 32
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
7.993982
7.734929
7.596129
7.562767
7.855947
8.010992
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
15
5.19708 𝑒𝑒 − 06
5.94394 𝑒𝑒 − 06
20
5.31708 𝑒𝑒 − 06
121
2.26794 𝑒𝑒 − 29
2.26794 𝑒𝑒 − 29
1.72317 𝑒𝑒 − 29
Chapter Three
Least Square Orthogonal Method
Example (3.4): Recall example (3) which is the linear fractional VIDE of constant multi-time Retarded delay with variable coefficients and the maximum arbitrary order is less than 1, take 𝑁𝑁 = 4 , ℳ = 9 and 𝐼𝐼𝐼𝐼 = 15. and assume the approximate solution in the form: 4
𝑢𝑢4 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒯𝒯𝑟𝑟 (2𝑡𝑡 − 1) 𝑟𝑟=0
4
and 𝑢𝑢4 (𝑡𝑡) = � 𝐶𝐶𝑟𝑟 𝒫𝒫𝑟𝑟 (2𝑡𝑡 − 1) 𝑟𝑟=0
By applying the lemma (3.3 and 3.4) for this example we determine the different matrices 𝐻𝐻 with respect to each different constant 𝜏𝜏-delay for Legendre polynomials, and Chebyshev polynomial, using m.file Matlab programs: OrthogonalTauCheb and OrthogonalTauLeg. Here: 𝜏𝜏 = 𝜏𝜏0 = 0.1, 𝜏𝜏1 = 0.9, 𝜏𝜏2 = 0.3 . Thus:
For Legendre orthogonal polynomials
𝐿𝐿𝐿𝐿𝐿𝐿
𝐻𝐻𝜏𝜏
𝐿𝐿𝐿𝐿𝐿𝐿
𝐻𝐻𝜏𝜏 1
1 ⎡ −0.2 ⎢ = ⎢ 0.06 ⎢−0.22 ⎣ 0.207
0 0 0 1 0 0 −0.6 1 0 0.3 −1 1 −0.74 0.7 −1.4
0 1 0 0 ⎡ −1.8 1 0 0 ⎢ 4.86 1 0 −5.4 =⎢ −9 1 24.3 ⎢−16.38 ⎣ 62.127 −107.46 56.7 −12.6 𝐿𝐿𝐿𝐿𝐿𝐿
𝐻𝐻𝜏𝜏 2
1 ⎡ −0.6 ⎢ = ⎢ 0.54 ⎢−1.14 ⎣ 2.367
0 0 0 1 0 0 −1.8 1 0 2.7 −3 1 −5.58 6.3 −4.2
0 0⎤ 0⎥⎥ 0⎥ 1⎦ 0 0⎤ 0⎥⎥ 0⎥ 1⎦
0 0⎤ ⎥ 0⎥ 0⎥ 1⎦
For Chebyshev orthogonal polynomials
𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 1
1 0 ⎡ −0.2 1 ⎢ −0.8 = ⎢ 0.08 0.48 ⎢−0.632 ⎣ 0.6528 −1.856
1 ⎡ −1.8 ⎢ = ⎢ 6.48 ⎢ −28.728 ⎣135.8208 𝐻𝐻𝜏𝜏𝐶𝐶ℎ𝑒𝑒 2
0 1 −7.2 38.88 −201.024
1 ⎡ −0.6 ⎢ = ⎢ 0.72 ⎢−2.664 ⎣ 6.7968
0 0 0 0 1 0 −1.2 1 0.96 −1.6
0 0 0 0 1 0 −10.8 1 77.76 −14.4
0 1 −2.4 4.32 −11.712
0 0 0 0 1 0 −3.6 1 8.64 −4.8
0 0⎤ 0⎥⎥ 0⎥ 1⎦
0 0⎤ ⎥ 0⎥ 0⎥ 1⎦
0 0⎤ 0⎥⎥ 0⎥ 1⎦
apply the algorithms [ADCH and ADLP] to find the approximate solution of our problem, and run the programs Main Delay OpenCheb, Main Delay ClosedCheb and Main Delay Legendre to find the constant coefficients 𝐶𝐶𝑟𝑟 's, which are present in Table (3.11):
122
Chapter Three
Least Square Orthogonal Method
Table (3.12) LSOM DCH
DOH
DOH
𝑪𝑪𝟎𝟎
−0.72684647
−0.7268453
−0.80024568
𝑪𝑪𝟐𝟐
0.21864049
0.21864242
0.28556526
0.0078168609
0.014292785
𝑪𝑪𝒓𝒓
𝑪𝑪𝟏𝟏 𝑪𝑪𝟑𝟑 𝑪𝑪𝟒𝟒
−0.56289516 0.062505219
0.0078156442
−0.5628926 0.062507061
−0.600397 0.1000094
Furthermore, after running the programs, table (3.11) and the approximation expression, the following approximate formulas are obtained:
𝑢𝑢2𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (𝑡𝑡) = 1.0004024576𝑡𝑡 4 − 0.0006379072𝑡𝑡 3 − 0.00062352 𝑡𝑡 2 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂
𝑢𝑢2
− 1.9999209124 𝑡𝑡 − 0.0000003948
(𝑡𝑡) = 1.0005581952𝑡𝑡 4 − 0.0008904384𝑡𝑡 3 − 0.000501824𝑡𝑡 2 − 1.9999370108𝑡𝑡 − 0.0000004801
𝐿𝐿𝐿𝐿𝐿𝐿 𝑢𝑢2 (𝑡𝑡) = 1.00049494992𝑡𝑡 4 − 0.00080189984𝑡𝑡 3 − 0.00053979032 𝑡𝑡 2
− 1.99992845976 𝑡𝑡 − 0.00000003503
Table (3.12) shows a comparison between the exact solution 𝑢𝑢(𝑡𝑡) and approximate solutions 𝑢𝑢4 (𝑡𝑡) for all closed and open Chebyshev and Legendre polynomials respectively.
123
Chapter Three
Least Square Orthogonal Method
Table (3.13) Least-Square Orthogonal Method 𝒕𝒕
Exact
0
DCH
DOH
DLE
0
−3.94800000025𝑒𝑒 − 07
−4.80100000005 𝑒𝑒 − 07
−3.49999999816𝑒𝑒 − 08
−0.3984
−0.398413977
−0.398414186
−0.398412942
0.1
−0.1999
0.3
−0.5919
0.2 0.4 0.5 0.6 0.7 0.8 0.9 1.0
−0.7744 −0.9375 −1.0704 −1.1599
−0.199899319 −0.591946749
−0.774498275
−0.937671316
−0.937670859
−1.070663039
−1.070663336
−1.16027273
−1.190897939
−1
−1.000780277
1.50997 𝑒𝑒 − 07
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
18.115769
4.59485 𝑒𝑒 − 08
−0.199899031 −0.591944796 −0.774496436 −0.937668516 −1.0706605
−1.160273679
−1.160270668
−1.144533165
−1.144532727
1.4949 𝑒𝑒 − 07
1.49259 𝑒𝑒 − 07
21.246168
19.646624
−1.190898524
−1.144535249
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
−0.591946268
−0.774499046
−1.1904 −1.1439
−0.199900034
−1.000771558 2.96428 𝑒𝑒 − 08
−1.19089611
−1.000775235
4.11714 𝑒𝑒 − 08
The result in Table (3.13) shows the least square errors, running times and the values residual equations 𝑅𝑅4 (𝑡𝑡; 𝐶𝐶̅ ) which are also included by applying the formula (3.40), for all open and closed Chebyshev and Legendre respectively for two different values of 𝐼𝐼𝐼𝐼. 124
Chapter Three
Least Square Orthogonal Method
Table (3.14) Least-Square Orthogonal Method IN
LSOM 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
1.50997 𝑒𝑒 − 07
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
18.115769
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐
𝐃𝐃𝐃𝐃𝐃𝐃
DOH
DLE
1.4949 𝑒𝑒 − 07
1.49259 𝑒𝑒 − 07
21.246168
19.646624
4.59485 𝑒𝑒 − 08
2.96428 𝑒𝑒 − 08
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
1.85499 𝑒𝑒 − 15
3.17348 𝑒𝑒 − 15
3.38067 𝑒𝑒 − 15
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
18.480263
20.031036
20.068469
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
7.72289 𝑒𝑒 − 15
7.06176 𝑒𝑒 − 15
4.11714 𝑒𝑒 − 08
7.90073 𝑒𝑒 − 15
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
1.48682 𝑒𝑒 − 31
1.48682 𝑒𝑒 − 31
5.54973 𝑒𝑒 − 17
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
18.692873
19.919238
20.037919
𝑹𝑹𝟐𝟐 = 𝑳𝑳. 𝑺𝑺. 𝑬𝑬𝒖𝒖
3.6 Discussion:
5.41191 𝑒𝑒 − 30
5.41191 𝑒𝑒 − 30
3.35504 𝑒𝑒 − 15
In this chapter, three numerical algorithms have been applied to solve Volterra Integro-Fractional Differential Equations with multi times Delay. For each algorithm a computer program was written and several examples were included for illustration and good results were achieved. The least square error, least square error function (𝑢𝑢) and running time were all also given in tabular forms; thus the following points have been identified. 1. The good results depend on the number of approximate parts of integral 𝐼𝐼𝐼𝐼 (if we take 𝐼𝐼𝐼𝐼 a large number the exact solution of 𝑢𝑢𝑁𝑁 (𝑡𝑡)) and the number of the orthogonal polynomials 𝑁𝑁 are obtained with suitable number of polynomial zeros ℳ . 2. By running the programs of MainDelayOpen/ClosedCheb and MainDelayLegendre, the least square error function in Chebychev polynomials gives more accurate solutions than Legendre polynomial method, see table (3.5), (3.11) and (3.14), so it is better. 3. In general, the solution by Chebyshev polynomials is easier and faster than of Legendre polynomials. 4. From example (3.1 and 3.3) if we take 𝐼𝐼𝐼𝐼 = 25 we obtain the exact solution 𝑢𝑢2 (𝑡𝑡) which is obtained only for open and closed Chebyshev. But not in Legendre method. 125
CHAPTER FOUR
Conclusions and Recommendations
Chapter Four
Conclusions and Recommendations
4.1 Conclusions: The approximate numerical methods for general linear fractional Volterra integro-differential equations of constant multi-time Retarded delay type with variable coefficients were introduced by applying and discussing several techniques. In this work, for any arbitrary derivative order, numerical algorithms were built derived from two different types of methods two, three and four Block-by-Block methods and the least-square orthogonal (Legendre and Chebyshev) polynomials methods for computing the state problem of linear VIFDEs of constant multi-time Retarded delay type. Finally, programs were written in MatLab (V. 8) and examples were solved and good results were achieved. A comparison is also made between these numerical techniques depending on the least square error which was calculated from the numerical solution against the perfect solution, and the running time of the associated main programs: MainBlock2, MainBlock3, MainBlock4, MainDelayClosedCheb, MainDelayOpenCheb and MainDelay Legendre (see appendix). The comparison is made as follows: Table (4.1) and (4.2) show a comparison between the least-square error and the running time for solving test example (1): (for two different value 𝛼𝛼 = 0.2
and 𝛼𝛼 = 0.85 ), respectively using Block-by-Block (two, three and four) block
techniques and least-square Chebyshev and Legendre polynomial method. Moreover, Table (4.3) illustrates the comparison between the two methods: Four block-by-block method and least-square Chebyshev polynomial for solving test example (2); While, Table (4.4) shows the different values of least-square error and the running time for solving test example (2), by applying two block-by-block method and least-square Legendre polynomial. Finally, Table (4.5) lists a comparison between the least square error and the running time for solving test example (3), by least-square polynomial method different values of trial leastsquare Closed Chebyshev polynomial and increased the number of approximate parts of integral 𝐼𝐼𝐼𝐼 (if we take 𝐼𝐼𝐼𝐼 in a large number the exact solution of 𝑢𝑢𝑁𝑁 (𝑡𝑡))
and the number of the orthogonal polynomials 𝑁𝑁 are obtained with suitable number of polynomial zeros ℳ.
126
Chapter Four
Conclusions and Recommendations
Table (4.1) 𝜶𝜶 = 𝟎𝟎. 𝟐𝟐
Test example (1) Methods Block-by-Block 𝑵𝑵 = 𝟐𝟐𝟐𝟐 Least-square polynomial 𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝟐𝟐𝟐𝟐𝟐𝟐
4.6145612 𝑒𝑒 − 04
𝟒𝟒𝟒𝟒𝟒𝟒
5.749388 𝑒𝑒 − 05
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺 1.877271
1.3700142 𝑒𝑒 − 04
𝟑𝟑𝟑𝟑𝟑𝟑
1.54
DCH
2.823965 3.600834
𝑒𝑒 − 15
6.156947
1.54 𝑒𝑒 − 15
DOH
5.842171
1.01332 𝑒𝑒 − 15
DLE
6.108334
Table (4.2) Test example (1) Methods Block-by-Block 𝑵𝑵 = 𝟐𝟐𝟐𝟐 Least-square polynomial 𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐
𝟐𝟐𝟐𝟐𝟐𝟐
𝑳𝑳. 𝑺𝑺. 𝑬𝑬 ⋯⋯ ⋯⋯
𝟑𝟑𝟑𝟑𝟑𝟑 𝟒𝟒𝟒𝟒𝟒𝟒 DCH
DOH DLE
𝜶𝜶 = 𝟎𝟎. 𝟖𝟖𝟖𝟖
⋯⋯
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺 ⋯⋯ ⋯⋯ ⋯⋯
1.2326 𝑒𝑒 − 32
7.134330
1.3332 𝑒𝑒 − 16
6.221093
1.2326 𝑒𝑒 − 32
6.359825
Table (4.3) Methods Test example (2) 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
𝟒𝟒𝟒𝟒𝟒𝟒
𝑵𝑵 = 𝟐𝟐𝟐𝟐
1.2001039 𝑒𝑒 − 003 5.027485
127
𝑫𝑫𝑫𝑫𝑫𝑫
𝑰𝑰𝑰𝑰 = 𝟐𝟐𝟐𝟐
4.33052 𝑒𝑒 − 016 12.504279
Chapter Four
Conclusions and Recommendations
"Table (4.4) Methods
𝟐𝟐𝟐𝟐𝟐𝟐
𝑵𝑵 = 𝟏𝟏𝟏𝟏
Test example ( 2 ) 𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
DLE
1.9307746 𝑒𝑒 − 002 2.156559
𝑰𝑰𝑰𝑰 = 𝟏𝟏𝟏𝟏
9.51669 𝑒𝑒 − 005 13.127563
"Table (4.5) 𝑫𝑫𝑫𝑫𝑫𝑫
Method
(Test Example 3)
Values (𝑵𝑵, 𝓜𝓜, 𝑰𝑰𝑰𝑰)
𝑵𝑵 = 𝟒𝟒
𝓜𝓜 = 𝟗𝟗
𝓜𝓜 = 𝟏𝟏𝟏𝟏
𝓜𝓜 = 𝟗𝟗 𝑵𝑵 = 𝟓𝟓 𝓜𝓜 = 𝟏𝟏𝟏𝟏
𝑳𝑳. 𝑺𝑺. 𝑬𝑬
𝑹𝑹. 𝑻𝑻𝑻𝑻𝑻𝑻𝑻𝑻/𝑺𝑺𝑺𝑺𝑺𝑺
2.30379 𝑒𝑒 − 09
20.447211
𝐼𝐼𝐼𝐼 = 16
1.16485 𝑒𝑒 − 06
𝐼𝐼𝐼𝐼 = 20
1.85499 𝑒𝑒 − 15
𝐼𝐼𝐼𝐼 = 18 𝐼𝐼𝐼𝐼 = 16 𝐼𝐼𝐼𝐼 = 18 𝐼𝐼𝐼𝐼 = 20 𝐼𝐼𝐼𝐼 = 16
1.15213 𝑒𝑒 − 06 2.14049 𝑒𝑒 − 09 3.17348 𝑒𝑒 − 15 1.1601 𝑒𝑒 − 06
19.036918 20.548094 21.153596 21.526016 23.804280 24.513687
𝐼𝐼𝐼𝐼 = 18
2.24925 𝑒𝑒 − 09
𝐼𝐼𝐼𝐼 = 16
1.15896 𝑒𝑒 − 06
29.547065
3.16511 𝑒𝑒 − 15
33.410376
𝐼𝐼𝐼𝐼 = 20 𝐼𝐼𝐼𝐼 = 18 𝐼𝐼𝐼𝐼 = 20
3.11202 𝑒𝑒 − 15 2.23637 𝑒𝑒 − 09
128
24.641987 25.32769
31.763066
Chapter Four
Conclusions and Recommendations
The comparison between the numerical solutions obtained by the methods: Block-by-Block method and least-square orthogonal polynomials methods with the exact solution for some test examples have been illustrated in the following figures. Figures (4.1) and (4.2) show a comparison between the exact and numerical solutions of VIFDEs with constant multi time delay of arbitrary order 𝛼𝛼 = 0.2, which was presented in testing examples (1) and application example using blockby-block method (two, three and four) and least-square orthogonal (Legendre and Chebyshev) polynomials methods, respectively. 1.5 1 0.5 0 -0.5 0
0.2
0.4
0.6
0.8
1
1.2
-1 -1.5 exact
2BM
3BM
4BM
Figure (4.1)
1.5 1 0.5 0 0
0.2
0.4
0.6
0.8
1
-0.5 -1 -1.5 exact
DCHR
DOHR
Figure (4.2) 129
DLER
1.2
Chapter Four
Conclusions and Recommendations
Figure (4.3) show a comparison between the exact solutions of VIFDEs with constant multi time delay of order value 𝛼𝛼 = 0.85 which was presented in testing
examples (1) and using least-square orthogonal method but the Block-by-Block method can't applied to compute the solution at this fractional order, which is greater than 1. 1.5
1
0.5
0 0
0.2
0.4
0.6
0.8
1
1.2
-0.5
-1
-1.5 exact
DCHR
DOHR
DLER
Figure (4.3)
Figures (4.4) and (4.5) show a comparison between the exact solutions of Volterra integro-higher fractional differential equations with constant multi-time Delay of Retarded types which was presented in test examples (2), while the maximum arbitrary order was less than one, so we used block-by-block method and the leastsquare orthogonal polynomials method, respectively.
130
Chapter Four
Conclusions and Recommendations
2.5
2
1.5
1
0.5
0 0
0.2
0.4
0.6 exact
DCHR
0.8
1
1.2
4BM
Figure (4.4)
2.5
2
1.5
1
0.5
0 0
0.2
0.4
0.6 exact
DLER
Figure (4.5)
131
0.8 2BM
1
1.2
Chapter Four
Conclusions and Recommendations
From the present results (Tables and Figures) and the information in the analytical technique in chapter two and the numerical methods in chapters three and four, the following conclusions are drawn: 1. In general, the numerical methods which are used in chapter three and four have proved their effectiveness in solving higher order of linear VIFDEs with constant multi-time Retarded delay in Caputo sense and finding good results. 2. For some special types of our problem, the analytical methods which are used in chapter two provide good and the exact solutions. 3. Block-by-Block methods cannot be used to compute the solution of test example (1 with 𝛼𝛼 > 0.5 ), and test example (5) because this method is
derived from all the fractional order that lies between 0 and 1 ,while leastsquare orthogonal methods can solve any arbitrary order successfully, see
Tables (4.1-5). 4. Block-by-Block method is faster than the least-square orthogonal methods for small 𝑁𝑁 .
5. The result of Closed Chebyshev polynomials for solving our problem is better than the other numerical methods. 6. The least-square orthogonal methods give better results than the block-byblock methods when the type of the fractional order on state is greater than 1.
7. A disadvantage of least square orthogonal polynomial method is that if (𝑁𝑁, 𝓜𝓜) is large, then the solution of 𝑁𝑁-algebraic equation for the unknown
coefficients is complicated (Long running time).
132
Chapter Four
Conclusions and Recommendations
4.2 Recommendations: The following points are recommended for future works: 1. Extending the use of all the methods that are used in this thesis to include a non-linear (Volterra and Fredholm) fractional integro-differential Equations of constant multi-time Retarded delays with variable coefficients. 2. Using other types of basic function, such as Laguerre or Hermite polynomials in least-square orthogonal polynomials. 3. Using linear programming technique to compute the solution of linear VIFDEs of Retarded delay. 4. Use some analytical methods for solving special case of our problem such as successive approximation method and variational iteration methods. 5. Using different definitions of fractional derivative such as Miller-Ross (sequential fractional derivative) instead of Caputo derivative to compute our problem.
133
Appendix
Appendix
Programs
Main program of Two Block (Block2) clc clear all format long g syms t; w=0.2; % ab=input('Input the interval [a,b] = '); ab=[0,1] ; % N=input('Input the number of sub-interval N = '); N=20 ; U1=zeros(N,1) ; U2=zeros(N+1,1) ; % p=input('Input the number of Blocks p = '); p=2 ; % hO=input('The hestorical function hO(t) = '); hO=(2*(t^2))-1 ; % U2(1,1)=input('The initial condition u0(a) = '); U2(1,1)=-1 ; alfa=[2*w w 0] ; coffi=[1 -cos(t) (t.^3)]; Tau=[0.8 0.2 0.4 0.6]; tic Np=N*p ; h=(ab(2)-ab(1))/Np ; n=length(alfa)-1 ; m=length(Tau)-1 ; tr1=zeros(N,1);tr2=zeros(N+1,1); tr1(1:N,1)=ab(1)+(1:p:Np)*h ; tr2(1:N+1,1)=ab(1)+(0:p:Np)*h; [F1,F2]=Frs2(tr1,tr2,N); [tr1,F1],[tr2,F2] C=zeros(n,p,N) ; H=zeros(N,p) ; B=zeros(Np+p-1,n) ; for r=0:N-1 for l=1:p S=0.0; for i=1:n if i~=n al=alfa(i+1); R=h^(-al)/gamma(2-al); C(i,l,r+1)=(subs(coffi(i+1),t,(ab(1)+(p*r+l)*h)))*R; else al=alfa(n-i+1); C(i,l,r+1)=h^(-al)/gamma(2-al); end S=S+C(i,l,r+1); end H(r+1,l)=S;
134
Appendix
Programs
end end for i=1:n if i~=n al=alfa(i+1); af=1-al; B(1:Np+p-1,i)=(1+(1:Np+p-1)).^af-(1:Np+p-1).^af; else al=alfa(1); af=1-al; B(1:Np+p-1,n)=(1+(1:Np+p-1)).^af-(1:Np+p-1).^af; end end C , H , B Tg=zeros(1,m+1); for ii=1:m+1 Tg(1,ii)=subs((Tau(1,ii)-mod(Tau(1,ii),h))/h); end mTg=max(Tg) Z=zeros(mTg,1); for i=1:mTg Z(i,1)=subs(hO,t,ab(1)-i*h); end Tg , Z T=zeros(Np+1,1); for r=0:N-1 pr=p*r ; T(pr+1,1)=U2(r+1,1); ti=tr1(r+1,1); FF1=F1(r+1,1) ; S=0.0; if r~=0 for s=1:pr BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,1,r+1); end S=S+(T(pr-s+2,1)-T(pr-s+1,1))*BC; end end Tg1=pr+1-Tg(1,1); if Tg1<0 TR1=(subs(coffi(end),t,ti))*Z(-Tg1,1); else TR1=(subs(coffi(end),t,ti))*T(Tg1+1,1); end Sj1=0.0; if r~=0 w1=ones(1,pr+1);w1(2:pr)=3-(-1).^(1:pr-1); for j=1:m Ss=0.0; for s=0:pr ts=ab(1)+s*h; Tg2=s-Tg(1,j+1); if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end Ss=Ss+w1(1,s+1)*Krs(j,ti,ts)*TR2; end Sj1=Sj1+Ss; end Sj1=(h/3)*Sj1; end Sj2=0.0; for j=1:m S1=0.0;S2=0.0;S3=0.0; Tg2=pr-Tg(1,j+1) ;if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end ts=ab(1)+pr*h; S1=Krs(j,ti,ts)*TR2;
135
Appendix
Programs
T11=pr-Tg(1,j+1); if T11<0 T21=Z(-T11,1); else T12=pr+1-Tg(1,j+1);if T12<0 T22=Z(-T12,1); else T13=pr+2-Tg(1,j+1);if T13<0 T23=Z(-T13,1); else ts=ab(1)+(pr+1/2)*h; S2=4*Krs(j,ti,ts)*(3*T21/8+3*T22/4-T23/8); T31=pr+1-Tg(1,j+1);if T31<0 T33=Z(-T31,1); else ts=ab(1)+(pr+1)*h; S3=Krs(j,ti,ts)*T33; Sj2=Sj2+(S1+S2+S3); end Sj2=(h/6)*Sj2; U1(r+1,1)=U2(r+1,1)+(FF1-S-TR1+Sj1+Sj2)/H(r+1,1);
T21=T(T11+1,1); end T22=T(T12+1,1); end T23=T(T13+1,1); end
T33=T(T31+1,1); end
T(pr+2,1)=U1(r+1,1) ; ti=tr2(r+2,1); FF2=F2(r+2,1) ; SS=0.0; for s=1:pr+1 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,2,r+1); end SS=SS+(T(pr-s+3,1)-T(pr-s+2,1))*BC; end Tg2=pr+2-Tg(1,1); if Tg2<0 TR2=(subs(coffi(end),ti))*Z(-Tg2,1); else TR2=(subs(coffi(end),ti))*T(Tg2+1,1); end Sj2=0.0; w2=ones(1,pr+3);w2(2:pr+2)=3-(-1).^(1:pr+1); for j=1:m Ss=0.0; for s=0:pr+2 ts=ab(1)+s*h; Tg22=s-Tg(1,j+1); if Tg22<0 TR22=Z(-Tg22,1); else TR22=T(Tg22+1,1); end Ss=Ss+w2(1,s+1)*Krs(j,ti,ts)*TR22; end Sj2=Sj2+(h/3)*Ss; end U2(r+2,1)=U1(r+1,1)+(FF2-SS-TR2+Sj2)/H(r+1,2); end U1 , U2 toc disp(' THE TABLE ') disp('the points ti The Exact Solutions u(t)The Block Two Solutions uN(t)') exact=sym(zeros(1,1)); exact=(2*(t^2))-1 ; % exact=input('the exact functions ui(t) = ES=0;h=(ab(2)-ab(1))/N; TI=(ab(1):h:ab(2))' ; T1=subs(exact,t,TI); T2=subs(U2); ES=sum((T1-T2).^2); disp(subs(vpa([TI T1 T2 ],12))) LSE=subs(vpa([ES],8))
136
');
Appendix
Programs
Main program of Three Block (Block3) clc clear all format long g syms t w=0.2; % ab=input('Input the interval [a,b] = '); ab=[0,1] ; % N=input('Input the number of sub-interval N = '); N=20 ; U1=zeros(N,1) ; U2=zeros(N,1) ; U3=zeros(N+1,1) ; % p=input('Input the number of Blocks p = '); p=3 ; % hO=input('The hestorical function hO(t) = '); hO=(2*(t^2))-1 ; % U3(1,1)=input('The initial condition u0(a) = '); U3(1,1)=-1 ; alfa=[2*w w 0] ; coffi=[1 -cos(t) (t.^3)]; Tau=[0.8 0.2 0.4 0.6]; tic Np=N*p ; h=(ab(2)-ab(1))/Np ; n=length(alfa)-1 ; m=length(Tau)-1 ; tr1=zeros(N,1); tr2=zeros(N,1); tr3=zeros(N+1,1); tr1(1:N,1)=ab(1)+(1:p:Np)*h ; tr2(1:N,1)=ab(1)+(2:p:Np)*h; tr3(1:N+1,1)=ab(1)+(0:p:Np)*h; [F1,F2,F3]=Frs3(tr1,tr2,tr3,N); [tr1,F1],[tr2,F2],[tr3,F3] C=zeros(n,p,N) ; H=zeros(N,p) ; B=zeros(Np+p-1,n) ; for r=0:N-1 for l=1:p S=0.0; for i=1:n if i~=n al=alfa(i+1); R=h^(-al)/gamma(2-al); C(i,l,r+1)=(subs(coffi(i+1),t,(ab(1)+(p*r+l)*h)))*R; else al=alfa(n-i+1); C(i,l,r+1)=h^(-al)/gamma(2-al); end S=S+C(i,l,r+1); end H(r+1,l)=S; end end for i=1:n if i~=n al=alfa(i+1); af=1-al; B(1:Np+p-1,i)=(1+(1:Np+p-1)).^af-(1:Np+p-1).^af; else al=alfa(1); af=1-al; B(1:Np+p-1,n)=(1+(1:Np+p-1)).^af-(1:Np+p-1).^af; end end C , H , B Tg=zeros(1,m+1);
137
Appendix
Programs
for ii=1:m+1 Tg(1,ii)=subs((Tau(1,ii)-mod(Tau(1,ii),h))/h); end mTg=max(Tg) Z=zeros(mTg,1); for i=1:mTg Z(i,1)=subs(hO,t,ab(1)-i*h); end Tg , Z T=zeros(Np+1,1) ; for r=0:N-1 pr=p*r ; T(pr+1,1)=U3(r+1,1); if mod(r,2) == 0 ti=tr1(r+1,1); FF1=F1(r+1,1) ; S=0.0; if r~=0 for s=1:pr BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,1,r+1); end S=S+(T(pr-s+2,1)-T(pr-s+1,1))*BC; end end Tg1=pr+1-Tg(1,1); if Tg1<0 TR1=(subs(coffi(end),t,ti))*Z(-Tg1,1); else TR1=(subs(coffi(end),t,ti))*T(Tg1+1,1); end Sj1=0.0; if r~=0 w1=ones(1,pr+1);w1(2:pr)=3-(-1).^(1:pr-1); for j=1:m Ss=0.0; for s=0:pr ts=ab(1)+s*h; Tg2=s-Tg(1,j+1); if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end Ss=Ss+w1(1,s+1)*Krs(j,ti,ts)*TR2; end Sj1=Sj1+Ss; end Sj1=(h/3)*Sj1; end Sj2=0.0; for j=1:m S1=0.0;S2=0.0;S3=0.0; Tg2=pr-Tg(1,j+1); if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end ts=ab(1)+pr*h; S1=Krs(j,ti,ts)*TR2; T11=pr-Tg(1,j+1); if T11<0 T21=Z(-T11,1); else T21=T(T11+1,1); end T12=pr+1-Tg(1,j+1);if T12<0 T22=Z(-T12,1); else T22=T(T12+1,1); end T13=pr+2-Tg(1,j+1);if T13<0 T23=Z(-T13,1); else T23=T(T13+1,1); end T14=pr+3-Tg(1,j+1);if T14<0 T24=Z(-T14,1); else T24=T(T14+1,1); end ts=ab(1)+(pr+1/2)*h; S2=4*Krs(j,ti,ts)*(5*T21+15*T22-5*T23+T24)/16; T31=pr+1-Tg(1,j+1);if T31<0 T33=Z(-T31,1); else T33=T(T31+1,1); end ts=ab(1)+(pr+1)*h; S3=Krs(j,ti,ts)*T33; Sj2=Sj2+(S1+S2+S3);
138
Appendix
Programs
end Sj2=(h/6)*Sj2; U1(r+1,1)=U3(r+1,1)+(FF1-S-TR1+Sj1+Sj2)/H(r+1,1); T(pr+2,1)=U1(r+1,1) ; ti=tr2(r+1,1); FF2=F2(r+1,1) ; SS=0.0; for s=1:pr+1 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,2,r+1); end SS=SS+(T(pr-s+3,1)-T(pr-s+2,1))*BC; end Tg2=pr+2-Tg(1,1); if Tg2<0 TR2=(subs(coffi(end),t,ti))*Z(-Tg2,1); else TR2=(subs(coffi(end),t,ti))*T(Tg2+1,1); end Sj3=0.0; w2=ones(1,pr+3);w2(2:pr+2)=3-(-1).^(1:pr+1); for j=1:m Ss=0.0; for s=0:pr+2 ts=ab(1)+s*h; Tg22=s-Tg(1,j+1); if Tg22<0 TR22=Z(-Tg22,1); else TR22=T(Tg22+1,1); end Ss=Ss+w2(1,s+1)*Krs(j,ti,ts)*TR22; end Sj3=Sj3+Ss; end Sj3=(h/3)*Sj3; U2(r+1,1)=U1(r+1,1)+(FF2-SS-TR2+Sj3)/H(r+1,2); T(pr+3,1)=U2(r+1,1) ; ti=tr3(r+2,1); FF3=F3(r+2,1) ; SA=0.0 ; for s=1:pr+2 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,3,r+1); end SA=SA+(T(pr-s+4,1)-T(pr-s+3,1))*BC; end Tg3=pr+3-Tg(1,1); if Tg3<0 TR30=(subs(coffi(end),t,ti))*Z(-Tg3,1); else TR30=(subs(coffi(end),t,ti))*T(Tg3+1,1); end Sj3=0.0; w3=ones(1,pr+3);w3(2:pr+2)=3-(-1).^(1:pr+1); for j=1:m Ss=0.0; for s=0:pr+2 ts=ab(1)+s*h; Tg31=s-Tg(1,j+1); if Tg31<0 TR3=Z(-Tg31,1); else TR3=T(Tg31+1,1); end Ss=Ss+w3(1,s+1)*Krs(j,ti,ts)*TR3; end Sj3=Sj3+Ss; end Sj3=(h/3)*Sj3; Sj4=0.0; for j=1:m
139
Appendix
Programs
S1=0.0;S2=0.0;S3=0.0; Tg2=pr+2-Tg(1,j+1);if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); ts=ab(1)+(pr+2)*h; S1=Krs(j,ti,ts)*TR2; T11=pr+2-Tg(1,j+1);if T11<0 T21=Z(-T11,1); else T21=T(T11+1,1); T12=pr+3-Tg(1,j+1);if T12<0 T22=Z(-T12,1); else T22=T(T12+1,1); T13=pr+4-Tg(1,j+1);if T13<0 T23=Z(-T13,1); else T23=T(T13+1,1); T14=pr+5-Tg(1,j+1);if T14<0 T24=Z(-T14,1); else T24=T(T14+1,1); ts=ab(1)+(pr+5/2)*h; S2=4*Krs(j,ti,ts)*(5*T21+15*T22-5*T23+T24)/16; T31=pr+3-Tg(1,j+1);if T31<0 T33=Z(-T31,1); else T33=T(T31+1,1); ts=ab(1)+(pr+3)*h; S3=Krs(j,ti,ts)*T33; Sj4=Sj4+(S1+S2+S3); end Sj4=(h/6)*Sj4; U3(r+2,1)=U2(r+1,1)+(FF3-SA-TR30+Sj3+Sj4)/H(r+1,3); else ti=tr1(r+1,1); FF1=F1(r+1,1) ; S=0.0 ; for s=1:pr BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,1,r+1); end S=S+(T(pr-s+2,1)-T(pr-s+1,1))*BC; end Tg1=pr+1-Tg(1,1); if Tg1<0 TR1=(subs(coffi(end),t,ti))*Z(-Tg1,1); else TR1=(subs(coffi(end),t,ti))*T(Tg1+1,1); end Sj1=0.0; z1=ones(1,pr+2);z1(2:pr+1)=3-(-1).^(1:pr); for j=1:m Ss=0.0; for s=0:pr+1 ts=ab(1)+s*h; Tg2=s-Tg(1,j+1); if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end Ss=Ss+z1(1,s+1)*Krs(j,ti,ts)*TR2; end Sj1=Sj1+Ss; end Sj1=(h/3)*Sj1; U1(r+1,1)=U3(r+1,1)+(FF1-S-TR1+Sj1)/H(r+1,1); T(pr+2,1)=U1(r+1,1) ; ti=tr2(r+1,1); FF2=F2(r+1,1) ; SS=0.0; for s=1:pr+1 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,2,r+1); end SS=SS+(T(pr-s+3,1)-T(pr-s+2,1))*BC; end Tg2=pr+2-Tg(1,1); if Tg2<0 TR20=(subs(coffi(end),t,ti))*Z(-Tg2,1); else TR20=(subs(coffi(end),t,ti))*T(Tg2+1,1); end
140
end
end end end end
end
Appendix
Programs
Sj2=0.0; z2=ones(1,pr+2);z2(2:pr+1)=3-(-1).^(1:pr); for j=1:m Ss=0.0; for s=0:pr+1 ts=ab(1)+s*h; Tg22=s-Tg(1,j+1); if Tg22<0 TR22=Z(-Tg22,1); else TR22=T(Tg22+1,1); end Ss=Ss+z2(1,s+1)*Krs(j,ti,ts)*TR22; end Sj2=Sj2+Ss; end Sj2=(h/3)*Sj2; Sj5=0.0; for j=1:m S1=0.0;S2=0.0;S3=0.0; Tg2=pr+1-Tg(1,j+1);if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end ts=ab(1)+(pr+1)*h; S1=Krs(j,ti,ts)*TR2; T11=pr+1-Tg(1,j+1);if T11<0 T21=Z(-T11,1); else T21=T(T11+1,1); end T12=pr+2-Tg(1,j+1);if T12<0 T22=Z(-T12,1); else T22=T(T12+1,1); end T13=pr+3-Tg(1,j+1);if T13<0 T23=Z(-T13,1); else T23=T(T13+1,1); end T14=pr+4-Tg(1,j+1);if T14<0 T24=Z(-T14,1); else T24=T(T14+1,1); end ts=ab(1)+(pr+1.5)*h; S2=4*Krs(j,ti,ts)*(5*T21+15*T22-5*T23+T24)/16; T31=pr+2-Tg(1,j+1);if T31<0 T33=Z(-T31,1); else T33=T(T31+1,1); end ts=ab(1)+(pr+2)*h; S3=Krs(j,ti,ts)*T33; Sj5=Sj5+(S1+S2+S3); end Sj5=(h/6)*Sj5; U2(r+1,1)=U1(r+1,1)+(FF2-SS-TR20+Sj2+Sj5)/H(r+1,2); T(pr+3,1)=U2(r+1,1) ; ti=tr3(r+2,1); FF3=F3(r+2,1) ; S=0.0 ; for s=1:pr+2 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,3,r+1); end S=S+(T(pr-s+4,1)-T(pr-s+3,1))*BC; end Tg3=pr+3-Tg(1,1); if Tg3<0 TR30=(subs(coffi(end),t,ti))*Z(-Tg3,1); else TR30=(subs(coffi(end),t,ti))*T(Tg3+1,1); end Sj3=0.0; z3=ones(1,pr+4);z3(2:pr+3)=3-(-1).^(1:pr+2); for j=1:m Ss=0.0; for s=0:pr+3 ts=ab(1)+s*h; Tg31=s-Tg(1,j+1); if Tg31<0 TR3=Z(-Tg31,1); else TR3=T(Tg31+1,1); end Ss=Ss+z3(1,s+1)*Krs(j,ti,ts)*TR3; end Sj3=Sj3+Ss; end Sj3=(h/3)*Sj3; U3(r+2,1)=U2(r+1,1)+(FF3-S-TR30+Sj3)/H(r+1,3); end end
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U1 , U2 , U3 toc THE TABLE disp(' ') disp(' the points ti The Exact Solutions u(t) The Block Two Solutions uN(t)') exact=sym(zeros(1,1)); exact=(2*(t^2))-1 ; % exact=input('the exact functions ui(t) = '); ES=0;h=(ab(2)-ab(1))/N; TI=(ab(1):h:ab(2))' ; T1=subs(exact,t,TI); T2=subs(U3); ES=sum((T1-T2).^2); disp(subs(vpa([TI T1 T2 ],12))) LSE=subs(vpa([ES],8))
Main program Four Block (Block4) clc clear all format long g syms t w=0.2; % ab=input('Input the interval [a,b] = '); ab=[0,1] ; % N=input('Input the number of sub-interval N = '); N=20; U1=zeros(N,1) ; U2=zeros(N,1) ; U3=zeros(N,1) ; U4=zeros(N+1,1) ; % p=input('Input the number of Blocks p = '); p=4 ; % hO=input('The hestorical function hO(t) = '); hO=(2*(t^2))-1; % U4(1,1)=input('The initial condition u0(a) = '); U4(1,1)=-1 ; alfa=[2*w w 0] ; coffi=[1 -cos(t) (t.^3)]; Tau=[0.8 0.2 0.4 0.6]; tic Np=N*p ; h=(ab(2)-ab(1))/Np ; n=length(alfa)-1 ; m=length(Tau)-1 ; tr1=zeros(N,1); tr2=zeros(N,1); tr3=zeros(N,1); tr4=zeros(N+1,1); tr1(1:N,1)=ab(1)+(1:p:Np)*h ; tr2(1:N,1)=ab(1)+(2:p:Np)*h; tr3(1:N,1)=ab(1)+(3:p:Np)*h ; tr4(1:N+1,1)=ab(1)+(0:p:Np)*h; [F1,F2,F3,F4]=Frs4(tr1,tr2,tr3,tr4,N); [tr1,F1],[tr2,F2],[tr3,F3],[tr4,F4] C=zeros(n,p,N) ; H=zeros(N,p) ; B=zeros(Np+p-1,n) ; for r=0:N-1 for l=1:p S=0.0; for i=1:n if i~=n al=alfa(i+1); R=h^(-al)/gamma(2-al); C(i,l,r+1)=(subs(coffi(i+1),t,(ab(1)+(p*r+l)*h)))*R; else al=alfa(n-i+1); C(i,l,r+1)=h^(-al)/gamma(2-al); end S=S+C(i,l,r+1); end H(r+1,l)=S; end
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end for i=1:n if i~=n al=alfa(i+1); af=1-al; B(1:Np+p-1,i)=(1+(1:Np+p-1)).^af-(1:Np+p-1).^af; else al=alfa(1); af=1-al; B(1:Np+p-1,n)=(1+(1:Np+p-1)).^af-(1:Np+p-1).^af; end end C , H , B Tg=zeros(1,m+1); for ii=1:m+1 Tg(1,ii)=subs((Tau(1,ii)-mod(Tau(1,ii),h))/h); end mTg=max(Tg) Z=zeros(mTg,1); for i=1:mTg Z(i,1)=subs(hO,t,ab(1)-i*h); end Tg , Z T=zeros(Np+1,1); for r=0:N-1 pr=p*r ; T(pr+1,1)=U4(r+1,1); ti=tr1(r+1,1); FF1=F1(r+1,1) ; S=0.0; if r~=0 for s=1:pr BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,1,r+1); end S=S+(T(pr-s+2,1)-T(pr-s+1,1))*BC; end end Tg1=pr+1-Tg(1,1); if Tg1<0 TR1=(subs(coffi(end),t,ti))*Z(-Tg1,1); else TR1=(subs(coffi(end),t,ti))*T(Tg1+1,1); end Sj1=0.0; if r~=0 w1=ones(1,pr+1);w1(2:pr)=3-(-1).^(1:pr-1); for j=1:m Ss=0.0; for s=0:pr ts=ab(1)+s*h; Tg2=s-Tg(1,j+1); if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end Ss=Ss+w1(1,s+1)*Krs(j,ti,ts)*TR2; end Sj1=Sj1+Ss; end Sj1=(h/3)*Sj1; end Sj2=0.0; for j=1:m S1=0.0;S2=0.0;S3=0.0; Tg2=pr-Tg(1,j+1) ;if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end ts=ab(1)+pr*h; S1=Krs(j,ti,ts)*TR2; T11=pr-Tg(1,j+1) ;if T11<0 T21=Z(-T11,1); else T21=T(T11+1,1); end
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T12=pr+1-Tg(1,j+1);if T12<0 T22=Z(-T12,1); else T22=T(T12+1,1); end T13=pr+2-Tg(1,j+1);if T13<0 T23=Z(-T13,1); else T23=T(T13+1,1); end T14=pr+3-Tg(1,j+1);if T14<0 T24=Z(-T14,1); else T24=T(T14+1,1); end T15=pr+4-Tg(1,j+1);if T15<0 T25=Z(-T15,1); else T25=T(T15+1,1); end ts=ab(1)+(pr+1/2)*h; S2=(1/8)*Krs(j,ti,ts)*(35*T21/4+35*T22-70*T23/4+7*T24-5*T25/4); T31=pr+1-Tg(1,j+1);if T31<0 T33=Z(-T31,1); else T33=T(T31+1,1); end ts=ab(1)+(pr+1)*h; S3=Krs(j,ti,ts)*T33; Sj2=Sj2+(S1+S2+S3); end Sj2=(h/6)*Sj2; U1(r+1,1)=U4(r+1,1)+(FF1-S-TR1+Sj1+Sj2)/H(r+1,1); T(pr+2,1)=U1(r+1,1) ; ti=tr2(r+1,1); FF2=F2(r+1,1) ; SS=0.0; for s=1:pr+1 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,2,r+1); end SS=SS+(T(pr-s+3,1)-T(pr-s+2,1))*BC; end Tg2=pr+2-Tg(1,1); if Tg2<0 TR2=(subs(coffi(end),ti))*Z(-Tg2,1); else TR2=(subs(coffi(end),ti))*T(Tg2+1,1); end Sj3=0.0; w2=ones(1,pr+3);w2(2:pr+2)=3-(-1).^(1:pr+1); for j=1:m Ss=0.0; for s=0:pr+2 ts=ab(1)+s*h; Tg22=s-Tg(1,j+1); if Tg22<0 TR22=Z(-Tg22,1); else TR22=T(Tg22+1,1); end Ss=Ss+w2(1,s+1)*Krs(j,ti,ts)*TR22; end Sj3=Sj3+Ss; end Sj3=(h/3)*Sj3; U2(r+1,1)=U1(r+1,1)+(FF2-SS-TR2+Sj3)/H(r+1,2); T(pr+3,1)=U2(r+1,1); ti=tr3(r+1,1); FF3=F3(r+1,1) ; S=0.0; for s=1:pr+2 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,3,r+1); end S=S+(T(pr-s+4,1)-T(pr-s+3,1))*BC; end Tg1=pr+3-Tg(1,1); if Tg1<0 TR1=(subs(coffi(end),t,ti))*Z(-Tg1,1); else TR1=(subs(coffi(end),t,ti))*T(Tg1+1,1); end Sj4=0.0; w3=ones(1,pr+3);w3(2:pr+2)=3-(-1).^(1:pr+1); for j=1:m
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Ss=0.0; for s=0:pr+2 ts=ab(1)+s*h; Tg2=s-Tg(1,j+1); if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end Ss=Ss+w3(1,s+1)*Krs(j,ti,ts)*TR2; end Sj4=Sj4+Ss; end Sj4=(h/3)*Sj4; Sj5=0.0; for j=1:m S1=0.0;S2=0.0;S3=0.0; Tg2=pr+2-Tg(1,j+1);if Tg2<0 TR2=Z(-Tg2,1); else TR2=T(Tg2+1,1); end ts=ab(1)+(pr+2)*h; S1=Krs(j,ti,ts)*TR2; T11=pr+2-Tg(1,j+1);if T11<0 T21=Z(-T11,1); else T21=T(T11+1,1); end T12=pr+3-Tg(1,j+1);if T12<0 T22=Z(-T12,1); else T22=T(T12+1,1); end T13=pr+4-Tg(1,j+1);if T13<0 T23=Z(-T13,1); else T23=T(T13+1,1); end T14=pr+5-Tg(1,j+1);if T14<0 T24=Z(-T14,1); else T24=T(T14+1,1); end T15=pr+6-Tg(1,j+1);if T15<0 T25=Z(-T15,1); else T25=T(T15+1,1); end ts=ab(1)+(pr+5/2)*h; S2=(1/8)*Krs(j,ti,ts)*(35*T21/4+35*T22-70*T23/4+7*T24-5*T25/4); T31=pr+3-Tg(1,j+1);if T31<0 T33=Z(-T31,1); else T33=T(T31+1,1); end ts=ab(1)+(pr+3)*h; S3=Krs(j,ti,ts)*T33; Sj5=Sj5+(S1+S2+S3); end Sj5=(h/6)*Sj5; U3(r+1,1)=U2(r+1,1)+(FF3-S-TR1+Sj4+Sj5)/H(r+1,3); T(pr+4,1)=U3(r+1,1) ; ti=tr4(r+2,1); FF4=F4(r+2,1) ; SS=0.0; for s=1:pr+3 BC=0.0; for i=1:n BC=BC+B(s,i)*C(i,4,r+1); end SS=SS+(T(pr-s+5,1)-T(pr-s+4,1))*BC; end Tg2=pr+4-Tg(1,1); if Tg2<0 TR2=(subs(coffi(end),ti))*Z(-Tg2,1); else TR2=(subs(coffi(end),ti))*T(Tg2+1,1); end Sj6=0.0; w4=ones(1,pr+5);w4(2:pr+4)=3-(-1).^(1:pr+3); for j=1:m Ss=0.0; for s=0:pr+4 ts=ab(1)+s*h; Tg22=s-Tg(1,j+1); if Tg22<0 TR22=Z(-Tg22,1); else TR22=T(Tg22+1,1); end Ss=Ss+w4(1,s+1)*Krs(j,ti,ts)*TR22; end Sj6=Sj6+Ss; end Sj6=(h/3)*Sj6; U4(r+2,1)=U3(r+1,1)+(FF4-SS-TR2+Sj6)/H(r+1,4); end
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U1 , U2 ,U3 , U4 toc THE TABLE disp(' ') disp(' the points ti The Exact Solutions u(t) The Block Two Solutions uN(t)') exact=sym(zeros(1,1)); exact=(2*(t^2))-1 ; % exact=input('the exact functions ui(t) = '); ES=0;h=(ab(2)-ab(1))/N; TI=(ab(1):h:ab(2))' ; T1=subs(exact,t,TI); T2=subs(U4); ES=sum((T1-T2).^2); disp(subs(vpa([TI T1 T2 ],12))) LSE=subs(vpa([ES],8))
Subprograms of Block Methods Frs2 function [F01,F02]=Frs2(t1,t2,N1); syms z format long g w=0.2; fun=((4*(z.^(2*(1-w))))/(gamma(3-(2*w))))-((4*cos(z)*(z.^(2-w)))/(gamma(3w)))+(sin(z)*((0.5*(z^4))-((4/15)*(z^3))-((23/50)*(z^2))))+((123/25)*(exp(z)))-(123/25)+((28/5)*(z))-((43/25)*(z^2))+((118/75)*(z^3))((49/15)*(z^4))+((8/5)*(z^5)); F01=zeros(N1,1) ; F02=zeros(N1+1,1) ; tf1=t1(:,1) ; F01(:,1)=subs(fun,z,tf1) ; tf2=t2(:,1) ; F02(:,1)=subs(fun,z,tf2) ;
Frs3 function [F01,F02,F03]=Frs3(t1,t2,t3,N1); syms z format long g w=0.2; fun=((4*(z.^(2*(1-w))))/(gamma(3-(2*w))))-((4*cos(z)*(z.^(2-w)))/(gamma(3w)))+(sin(z)*((0.5*(z^4))-((4/15)*(z^3))-((23/50)*(z^2))))+((123/25)*(exp(z)))-(123/25)+((28/5)*(z))-((43/25)*(z^2))+((118/75)*(z^3))((49/15)*(z^4))+((8/5)*(z^5)); F01=zeros(N1,1) ; F02=zeros(N1,1) ; F03=zeros(N1+1,1) ; tf1=t1(:,1) ; F01(:,1)=subs(fun,z,tf1) ; tf2=t2(:,1) ; F02(:,1)=subs(fun,z,tf2) ; tf3=t3(:,1) ; F03(:,1)=subs(fun,z,tf3) ;
Frs4 function [F01,F02,F03,F04]=Frs4(t1,t2,t3,t4,N1); syms z format long g w=0.2; fun=((4*(z.^(2*(1-w))))/(gamma(3-(2*w))))-((4*cos(z)*(z.^(2-w)))/(gamma(3w)))+(sin(z)*((0.5*(z^4))-((4/15)*(z^3))-((23/50)*(z^2))))+((123/25)*(exp(-
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z)))-(123/25)+((28/5)*(z))-((43/25)*(z^2))+((118/75)*(z^3))((49/15)*(z^4))+((8/5)*(z^5)); F01=zeros(N1,1) ; F02=zeros(N1,1) ; F03=zeros(N1,1) ; F04=zeros(N1+1,1) tf1=t1(:,1) ; F01(:,1)=subs(fun,z,tf1) ; tf2=t2(:,1) ; F02(:,1)=subs(fun,z,tf2) ; tf3=t3(:,1) ; F03(:,1)=subs(fun,z,tf3) ; tf4=t4(:,1) ; F04(:,1)=subs(fun,z,tf4) ;
;
Krs function kernel=Krs(pp,tt,xx); tz=tt;xz=xx; kernels=[-xz*sin(tz),exp(xz-tz),(tz+(xz^2))]; kernel=kernels(1,pp);
Main program of Delay Closed Chebyshev Polynomial (MainDelay ClosedCheb) %
Closed-Chebyshev Program For Four Constant Delays
clc clear all format long g syms t x ab=[0,1] ; % ab=input('Input the interval [a,b] = '); ww=0.85; N=2 ; % N=input('Input the number of expantion terms N = '); hO=(2*(t^2))-1 ; % hO=input('The hestorical function hO(t) = '); alfa=[2*ww ww 0] ; % alfa=input('The alfa part = '); coffi=[1 -cos(t) t^3]; % from the first one we begining Tau=[0.8 0.6 0.4 0.2]; % The delay constant tau M=5 ; % number of points (Gauss-Legendre Zeroth) IN=20; % numbers of parts to solve the integrals numerically n=length(alfa)-1 ; % n is the number of Fractional term alfa part m=length(Tau)-1 ; % m is the number of Kernels term mu=max(ceil(alfa)) ; IC=zeros(mu,1); for i=1:mu IC(i,1)=input('input the initial condition ui ='); end tic Z=zeros(M+1,1); X=zeros(M+1,1); Z(1:M+1,1)=cos(pi*((0:M))/M); X(1:M+1,1)=((ab(2)-ab(1))*Z((0:M)+1,1)+(ab(2)+ab(1)))/2; lTau=length(Tau);H=zeros(N+1,N+1,lTau);d=ab(2)-ab(1); for ll=1:lTau H(:,:,ll)=OrthogonalTauCheb(Tau(ll),N,d); end
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TM=chebyshevM(ab(1),ab(2),N,x); KR=KERS(1:m,t,x); F=FUNC(t,X); D=zeros(M+1,N+1); for r=0:N for l=0:M tx=X(l+1,1); D(l+1,r+1)=sum(subs(coffi(1:end1),tx).*caputoCheb(ab(1),ab(2),r,tx,alfa(1:n))); end end DP=zeros(M+1,N+1); DF=zeros(M+1,1); for l=0:M tx=X(l+1,1); for r=0:N P1=0.0;P2=0.0;P3=0.0;P4=0.0;P5=0.0;P6=0.0;P7=0.0;P8=0.0; P9=0.0;P10=0.0;P11=0.0;P12=0.0;P13=0.0;P14=0.0;P15=0.0;P16=0.0; if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) P1=0.0; elseif ((tx-Tau(1)<=ab(1))&&((tx-Tau(2)&tx-Tau(3) & tx-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1);INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; end P2=-S1; elseif((tx-Tau(2)<=ab(1))&&((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P3=S12-S22; elseif ((tx-Tau(3)<=ab(1))&&((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P4=S12-S22; elseif ((tx-Tau(4)<=ab(1))&&((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P5=S12-S22; elseif(((tx-Tau(1)& tx-Tau(2))<=ab(1)) &&((tx-Tau(3)& tx-Tau(4))>ab(1))) S1=0.0; for k=0:r
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Appendix
Programs KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT;
end P6=-S1; elseif(((tx-Tau(1) & tx-Tau(3))<=ab(1))&&((tx-Tau(2) & tx-Tau(4))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P7=-S2; elseif(((tx-Tau(1)& tx-Tau(4))<=ab(1)) &&((tx-Tau(2) & tx-Tau(3))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P8=-S2; elseif(((tx-Tau(2)& tx-Tau(3))<=ab(1)) &&((tx-Tau(1) & tx-Tau(4))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P9=S0-S2; elseif(((tx-Tau(2)& tx-Tau(4))<=ab(1)) &&((tx-Tau(1) & tx-Tau(3))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P10=S0-S2; elseif(((tx-Tau(3)& tx-Tau(4))<=ab(1)) &&((tx-Tau(1) & tx-Tau(2))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P11=S0-S2; elseif(((tx-Tau(1) & tx-Tau(2)& tx-Tau(3))<=ab(1)) && (tx-Tau(4)>ab(1))) S0=0.0; for k=0:r KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P12=-S0; elseif (((tx-Tau(1) & tx-Tau(2)& tx-Tau(4))<=ab(1)) && (tx-Tau(3)>ab(1))) S0=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P13=-S0;
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elseif (((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))<=ab(1)) &&(tx-Tau(2)>ab(1))) S0=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P14=-S0; elseif (((tx-Tau(2) & tx-Tau(3)& tx-Tau(4))<=ab(1)) && (tx-Tau(1)>ab(1))) S0=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; end P15=S0; else % if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1)) S1=0.0;S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; GL=subs(coffi(end),t,x)*TM(k+1,1); FF=subs(GL,x,tx); S2=S2+H(r+1,k+1,1)*FF; end P16=S2-S1; end DP(l+1,r+1)=D(l+1,r+1)+P1+P2+P3+P4+P5+P6+P7+P8+P9+P10+P11+P12+P13+P14+P15+P16; end end for l=0:M tx=X(l+1,1); PF1=0.0;PF2=0.0;PF3=0.0;PF4=0.0;PF5=0.0;PF6=0.0;PF7=0.0;PF8=0.0; PF9=0.0;PF10=0.0;PF11=0.0;PF12=0.0;PF13=0.0;PF14=0.0;PF15=0.0;PF16=0.0; if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF1=INTh-gL; elseif ((tx-Tau(1)<=ab(1))&& ((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF2=-gL; elseif ((tx-Tau(2)<=ab(1))&& ((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF3=INTh; elseif ((tx-Tau(3)<=ab(1))&& ((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF4=INTh; elseif ((tx-Tau(4)<=ab(1))&& ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF5=INTh; elseif (((tx-Tau(1)& tx-Tau(2))<=ab(1))&&((tx-Tau(3) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF6=INTh-gL; elseif (((tx-Tau(1)& tx-Tau(3))<=ab(1)) &&((tx-Tau(2)& tx-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tx);
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Programs
INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF7=INTh-gL; elseif (((tx-Tau(1) &tx-Tau(4))<=ab(1)) &&((tx-Tau(2) &tx-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF8=INTh-gL; elseif (((tx-Tau(2) &tx-Tau(3))<=ab(1))&& ((tx-Tau(1) &tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF9=INTh; elseif (((tx-Tau(2)& tx-Tau(4))<=ab(1))&&((tx-Tau(1) & tx-Tau(3))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx);INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF10=INTh; elseif (((tx-Tau(3)& tx-Tau(4))<=ab(1)) &&((tx-Tau(1) &tx-Tau(2))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF11=INTh; elseif (((tx-Tau(1) &tx-Tau(2) & tx-Tau(3))<=ab(1)) && (tx-Tau(4)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF12=INTh-gL; elseif (((tx-Tau(1) &tx-Tau(2) & tx-Tau(4))<=ab(1)) &&(tx-Tau(3)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF13=INTh-gL; elseif (((tx-Tau(1) & tx-Tau(3)& tx-Tau(4))<=ab(1)) && (tx-Tau(2)>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF14=INTh-gL; elseif (((tx-Tau(2)& tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(1)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx);INT=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF15=INT; else % if (((tx-Tau(1) & tx-Tau(2) & tx-Tau(3)&tx-Tau(4))>ab(1))) PF16=0.0; end DF(l+1,1)=F(l+1,1)+PF1+PF2+PF3+PF4+PF5+PF6+PF7+PF8+PF9+PF10+PF11+PF12+PF13+PF1 4+PF15+PF16; end DD=DP.';FF=DF; A=zeros(N+1,N+1);B=zeros(N+1,1); for r=0:N for s=0:N SR=0.0; for j=0:M if j==0 | j==M SR=SR+(1/2)*(DD(r+1,j+1)*DD(s+1,j+1)); else SR=SR+DD(r+1,j+1)*DD(s+1,j+1); end end A(r+1,s+1)=(pi/M)*SR; end
151
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Programs
SF=0.0; for j=0:M if j==0 | j==M SF=SF+(1/2)*(DD(r+1,j+1)*DF(j+1,1)); else SF=SF+DD(r+1,j+1)*DF(j+1,1); end end B(r+1,1)=(pi/M)*SF; end ICC=initialconditionCheb(ab(1),ab(2),N,mu); CB=cell(2,1);CB{1}=B;CB{2}=IC;BB=cell2mat(CB); CA=cell(2,1);CA{1}=A;CA{2}=ICC;AA=cell2mat(CA); AAA=AA'*AA;BBB=AA'*BB; C=zeros(N+1,1); [L0,U0]=lu(AAA);C=vpa(U0\(L0\BBB),8) toc disp(' ') disp('The Numerical Solution is uN= ') TMt=chebyshevM(ab(1),ab(2),N,t); u=sym(zeros(1));u=sum(C.*TMt); pretty(simplify(vpa(u,10))) disp(' ') % The following steps using to find L.S.E. using exact with % approximation solutions. exact=input('the exact function u(t) = '); nt=10; ht=(ab(2)-ab(1))/nt;T0=[ab(1):ht:ab(2)].'; for i=0:nt tt=T0(i+1,1); ES(1,i+1)=(subs(exact,tt)-subs(u,tt)).^2; end LSE=sum(ES); % The following steps using to find L.S.E.f for Residuals Dn=zeros(nt+1,N+1); for r=0:N for l=0:nt tt=T0(l+1,1); Dn(l+1,r+1)=sum(subs(coffi(1:end1),tt).*caputoCheb(ab(1),ab(2),r,tt,alfa(1:n))); end end Df=zeros(nt+1,N+1); Ff=zeros(nt+1,1); Ft=FUNC(t,T0); for l=0:nt tt=T0(l+1,1); for r=0:N Pt1=0.0;Pt2=0.0;Pt3=0.0;Pt4=0.0;Pt5=0.0;Pt6=0.0;Pt7=0.0;Pt8=0.0; Pt9=0.0;Pt10=0.0;Pt11=0.0;Pt12=0.0;Pt13=0.0;Pt14=0.0;Pt15=0.0;Pt16=0.0; if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) Pt1=0.0; elseif ((tt-Tau(1)<=ab(1))&&((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; end
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Programs
Pt2=-S1; elseif ((tt-Tau(2)<=ab(1))&& ((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt3=S12-S22; elseif ((tt-Tau(3)<=ab(1)) &&((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt4=S12-S22; elseif ((tt-Tau(4)<=ab(1)) &&((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt5=S12-S22; elseif (((tt-Tau(1) &tt-Tau(2))<=ab(1))&&((tt-Tau(3) & tt-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; end Pt6=-S1; elseif (((tt-Tau(1) &tt-Tau(3))<=ab(1))&&((tt-Tau(2) & tt-Tau(4))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt7=-S2; elseif (((tt-Tau(1) &tt-Tau(4))<=ab(1))&&((tt-Tau(2) & tt-Tau(3))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt8=-S2; elseif (((tt-Tau(2) &tt-Tau(3))<=ab(1))&&((tt-Tau(1) & tt-Tau(4))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT;
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Programs
end Pt9=S0-S2; elseif (((tt-Tau(2) &tt-Tau(4))<=ab(1))&&((tt-Tau(1) & tt-Tau(3))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt10=S0-S2; elseif (((tt-Tau(3) &tt-Tau(4))<=ab(1))&&((tt-Tau(1) & tt-Tau(2))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt11=S0-S2; elseif (((tt-Tau(1) & tt-Tau(2)& tt-Tau(3))<=ab(1)) && (tt-Tau(4)>ab(1))) S0=0.0; for k=0:r KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt12=-S0; elseif (((tt-Tau(1)& tt-Tau(2) & tt-Tau(4))<=ab(1)) && (tt-Tau(3)>ab(1))) S0=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt13=-S0; elseif (((tt-Tau(1) & tt-Tau(3)& tt-Tau(4))<=ab(1)) && (tt-Tau(2)>ab(1))) S0=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt14=-S0; elseif (((tt-Tau(2) &tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(1)>ab(1))) S0=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; end Pt15=S0; else % if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1)) S1=0.0;S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; GL=subs(coffi(end),t,x)*TM(k+1,1); FF=subs(GL,x,tt); S2=S2+H(r+1,k+1,1)*FF; end Pt16=S2-S1; end
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Appendix
Programs
Df(l+1,r+1)=Dn(l+1,r+1)+Pt1+Pt2+Pt3+Pt4+Pt5+Pt6+Pt7+Pt8+Pt9+Pt10+Pt11+Pt12+Pt1 3+Pt14+Pt15+Pt16; end end for l=0:nt tt=T0(l+1,1); Pn1=0.0;Pn2=0.0;Pn3=0.0;Pn4=0.0;Pn5=0.0;Pn6=0.0;Pn7=0.0;Pn8=0.0; Pn9=0.0;Pn10=0.0;Pn11=0.0;Pn12=0.0;Pn13=0.0;Pn14=0.0;Pn15=0.0;Pn16=0.0; if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn1=INTh-gL; elseif ((tt-Tau(1)<=ab(1)) &&((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1))) gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn2=-gL; elseif ((tt-Tau(2)<=ab(1)) &&((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn3=INTh; elseif ((tt-Tau(3)<=ab(1)) &&((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn4=INTh; elseif ((tt-Tau(4)<=ab(1)) &&((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn5=INTh; elseif (((tt-Tau(1)& tt-Tau(2))<=ab(1))&&((tt-Tau(3) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn6=INTh-gL; elseif (((tt-Tau(1) &tt-Tau(3))<=ab(1))&&((tt-Tau(2) & tt-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn7=INTh-gL; elseif (((tt-Tau(1)& tt-Tau(4))<=ab(1))&&((tt-Tau(2) & tt-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn8=INTh-gL; elseif (((tt-Tau(2)&tt-Tau(3))<=ab(1)) &&((tt-Tau(1) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn9=INTh; elseif (((tt-Tau(2) &tt-Tau(4))<=ab(1))&&((tt-Tau(1) & tt-Tau(3))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt);INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn10=INTh; elseif (((tt-Tau(3) &tt-Tau(4))<=ab(1)) &&((tt-Tau(1) &tt-Tau(2))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn11=INTh; elseif (((tt-Tau(1)& tt-Tau(2) & tt-Tau(3))<=ab(1)) && (tt-Tau(4)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn12=INTh-gL;
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Programs
elseif (((tt-Tau(1) & tt-Tau(2) &tt-Tau(4))<=ab(1)) && (tt-Tau(3)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn13=INTh-gL; elseif (((tt-Tau(1)& tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(2)>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn14=INTh-gL; elseif (((tt-Tau(2)& tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(1)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt);INT=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn15=INT; else % if (((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) Pn16=0.0; end Ff(l+1,1)=Ft(l+1,1)+Pn1+Pn2+Pn3+Pn4+Pn5+Pn6+Pn7+Pn8+Pn9+Pn10+Pn11+Pn12+Pn13+Pn 14+Pn15+Pn16; end DDf=Df.';Ff; LSEF=subs(((C'*DDf).'-Ff).^2); LSEf=sum(LSEF); disp(' THE TABLE ') disp('the points ti the Exact Solutions u(t)The Numerical Solutions uN(t)') T1=subs(exact,T0);T2=subs(u,T0); disp(subs(vpa([T0 T1 T2 ],12))) LSE=vpa([LSE],6) , LSEf=vpa([LSEf],6)
Main Delay Open Chebyshev Polynomial (MainDelayOpenCheb) %
Open-Chebyshev Program For Four Constant Delays
clc clear all format long g syms t x ab=[0,1] ; % ab=input('Input the interval [a,b] = '); ww=0.85; N=2 ; % N=input('Input the number of expantion terms N = '); hO=(2*(t^2))-1 ; % hO=input('The hestorical function hO(t) = '); alfa=[2*ww ww 0] ; % alfa=input('The alfa part = '); coffi=[1 -cos(t) t^3]; % from the first one we begining Tau=[0.8 0.6 0.4 0.2]; % The delay constant tau M=5 ; % number of points (Gauss-Legendre Zeroth) IN=20; % numbers of parts to solve the integrals numerically n=length(alfa)-1 ; % n is the number of Fractional term alfa part m=length(Tau)-1 ; % m is the number of Kernels term mu=max(ceil(alfa)) ; IC=zeros(mu,1); for i=1:mu IC(i,1)=input('input the initial condition ui ='); end tic Z=zeros(M,1); X=zeros(M,1); Z(1:M,1)=cos(pi*(2*(0:M-1)+1)/(2*M));
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Programs
X(1:M,1)=((ab(2)-ab(1))*Z((0:M-1)+1,1)+(ab(2)+ab(1)))/2; lTau=length(Tau);H=zeros(N+1,N+1,lTau);d=ab(2)-ab(1); for ll=1:lTau H(:,:,ll)=OrthogonalTauCheb(Tau(ll),N,d); end TM=chebyshevM(ab(1),ab(2),N,x); KR=KERS(1:m,t,x);F=FUNC(t,X); D=zeros(M,N+1); for r=0:N for l=0:M-1 tx=X(l+1,1); D(l+1,r+1)=sum(subs(coffi(1:end1),tx).*caputoCheb(ab(1),ab(2),r,tx,alfa(1:n))); end end DP=zeros(M,N+1); DF=zeros(M,1); for l=0:M-1 tx=X(l+1,1); for r=0:N P1=0.0;P2=0.0;P3=0.0;P4=0.0;P5=0.0;P6=0.0;P7=0.0;P8=0.0; P9=0.0;P10=0.0;P11=0.0;P12=0.0;P13=0.0;P14=0.0;P15=0.0;P16=0.0; if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) P1=0.0; elseif ((tx-Tau(1)<=ab(1))&&((tx-Tau(2)&tx-Tau(3)&tx-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; end P2=-S1; elseif((tx-Tau(2)<=ab(1)) && ((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P3=S12-S22; elseif((tx-Tau(3)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P4=S12-S22; elseif((tx-Tau(4)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT;
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Programs
end P5=S12-S22; elseif(((tx-Tau(1)&tx-Tau(2))<=ab(1)) && ((tx-Tau(3) & tx-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; end P6=-S1; elseif(((tx-Tau(1)&tx-Tau(3))<=ab(1)) && ((tx-Tau(2) & tx-Tau(4))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P7=-S2; elseif(((tx-Tau(1)&tx-Tau(4))<=ab(1)) && ((tx-Tau(2) & tx-Tau(3))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P8=-S2; elseif(((tx-Tau(2)&tx-Tau(3))<=ab(1)) && ((tx-Tau(1) & tx-Tau(4))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P9=S0-S2; elseif(((tx-Tau(2)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(3))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P10=S0-S2; elseif(((tx-Tau(3)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(2))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P11=S0-S2; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))<=ab(1)) && (tx-Tau(4)>ab(1))) S0=0.0; for k=0:r KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P12=-S0; elseif(((tx-Tau(1)&tx-Tau(2) & tx-Tau(4))<=ab(1)) && (tx-Tau(3)>ab(1))) S0=0.0;
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Programs
for k=0:r KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P13=-S0; elseif(((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(2)>ab(1))) S0=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P14=-S0; elseif(((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(1)>ab(1))) S0=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; end P15=S0; else % if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1)) S1=0.0;S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; GL=subs(coffi(end),t,x)*TM(k+1,1); FF=subs(GL,x,tx); S2=S2+H(r+1,k+1,1)*FF; end P16=S2-S1; end DP(l+1,r+1)=D(l+1,r+1)+P1+P2+P3+P4+P5+P6+P7+P8+P9+P10+P11+P12+P13+P14+P15+P16; end end for l=0:M-1 tx=X(l+1,1); PF1=0.0;PF2=0.0;PF3=0.0;PF4=0.0;PF5=0.0;PF6=0.0;PF7=0.0;PF8=0.0; PF9=0.0;PF10=0.0;PF11=0.0;PF12=0.0;PF13=0.0;PF14=0.0;PF15=0.0;PF16=0.0; if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF1=INTh-gL; elseif((tx-Tau(1)<=ab(1)) && ((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF2=-gL; elseif((tx-Tau(2)<=ab(1)) && ((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF3=INTh; elseif((tx-Tau(3)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF4=INTh; elseif((tx-Tau(4)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF5=INTh; elseif(((tx-Tau(1)&tx-Tau(2))<=ab(1)) && ((tx-Tau(3) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tx);
159
Appendix
Programs
INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF6=INTh-gL; elseif(((tx-Tau(1)&tx-Tau(3))<=ab(1)) && ((tx-Tau(2) & tx-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF7=INTh-gL; elseif(((tx-Tau(1)&tx-Tau(4))<=ab(1)) && ((tx-Tau(2) & tx-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF8=INTh-gL; elseif(((tx-Tau(2)&tx-Tau(3))<=ab(1)) && ((tx-Tau(1) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF9=INTh; elseif(((tx-Tau(2)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(3))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx);INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF10=INTh; elseif(((tx-Tau(3)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(2))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF11=INTh; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))<=ab(1)) && (tx-Tau(4)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF12=INTh-gL; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))<=ab(1)) && (tx-Tau(3)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF13=INTh-gL; elseif(((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(2)>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF14=INTh-gL; elseif(((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(1)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx);INT=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF15=INT; else % if (((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) PF16=0.0; end DF(l+1,1)=F(l+1,1)+PF1+PF2+PF3+PF4+PF5+PF6+PF7+PF8+PF9+PF10+PF11+PF12+PF13+PF1 4+PF15+PF16; end DD=DP.'; FF=DF; A=zeros(N+1,N+1);B=zeros(N+1,1); for r=0:N for s=0:N SR=0.0; for j=0:M-1 SR=SR+DD(r+1,j+1)*DD(s+1,j+1); end A(r+1,s+1)=(pi/M)*SR;
160
Appendix
Programs
end SF=0.0; for j=0:M-1 SF=SF+DD(r+1,j+1)*FF(j+1,1); end B(r+1,1)=(pi/M)*SF; end ICC=initialconditionCheb(ab(1),ab(2),N,mu); CB=cell(2,1);CB{1}=B;CB{2}=IC;BB=cell2mat(CB); CA=cell(2,1);CA{1}=A;CA{2}=ICC;AA=cell2mat(CA); AAA=AA'*AA;BBB=AA'*BB; C=zeros(N+1,1); [L0,U0]=lu(AAA);C=vpa(U0\(L0\BBB),8) toc disp(' ') disp('The Numerical Solution is uN= ') TMt=chebyshevM(ab(1),ab(2),N,t); u=sym(zeros(1));u=sum(C.*TMt); pretty(simplify(vpa(u,10))) disp(' ') % The following steps using to find L.S.E. using exact with % approximation solutions. exact=input('the exact function u(t) = '); nt=10; ht=(ab(2)-ab(1))/nt;T0=[ab(1):ht:ab(2)].'; for i=0:nt tt=T0(i+1,1); ES(1,i+1)=(subs(exact,tt)-subs(u,tt)).^2; end LSE=sum(ES); % The following steps using to find L.S.E.f for Residuals Dn=zeros(nt+1,N+1); for r=0:N for l=0:nt tt=T0(l+1,1); Dn(l+1,r+1)=sum(subs(coffi(1:end1),tt).*caputoCheb(ab(1),ab(2),r,tt,alfa(1:n))); end end Ft=FUNC(t,T0); Df=zeros(nt+1,N+1); Ff=zeros(nt+1,1); for l=0:nt tt=T0(l+1,1); for r=0:N Pt1=0.0;Pt2=0.0;Pt3=0.0;Pt4=0.0;Pt5=0.0;Pt6=0.0;Pt7=0.0;Pt8=0.0; Pt9=0.0;Pt10=0.0;Pt11=0.0;Pt12=0.0;Pt13=0.0;Pt14=0.0;Pt15=0.0;Pt16=0.0; if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) Pt1=0.0; elseif((tt-Tau(1)<=ab(1)) && ((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; end Pt2=-S1; elseif((tt-Tau(2)<=ab(1)) && ((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))>ab(1))) S12=0.0;S22=0.0;
161
Appendix
Programs
for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt3=S12-S22; elseif((tt-Tau(3)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt4=S12-S22; elseif((tt-Tau(4)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt5=S12-S22; elseif(((tt-Tau(1)&tt-Tau(2))<=ab(1)) && ((tt-Tau(3) & tt-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; end Pt6=-S1; elseif(((tt-Tau(1)&tt-Tau(3))<=ab(1)) && ((tt-Tau(2) & tt-Tau(4))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt7=-S2; elseif(((tt-Tau(1)&tt-Tau(4))<=ab(1)) && ((tt-Tau(2) & tt-Tau(3))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt8=-S2; elseif(((tt-Tau(2)&tt-Tau(3))<=ab(1)) && ((tt-Tau(1) & tt-Tau(4))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt9=S0-S2; elseif(((tt-Tau(2)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(3))>ab(1)))
162
Appendix
Programs
S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt10=S0-S2; elseif(((tt-Tau(3)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(2))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt11=S0-S2; elseif(((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))<=ab(1)) && (tt-Tau(4)>ab(1))) S0=0.0; for k=0:r KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt12=-S0; elseif(((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))<=ab(1)) && (tt-Tau(3)>ab(1))) S0=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt13=-S0; elseif(((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(2)>ab(1))) S0=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt14=-S0; elseif(((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(1)>ab(1))) S0=0.0; for k=0:r GL=subs(coffi(end),t,x)*TM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; end Pt15=S0; else % if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1)) S1=0.0;S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*TM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; GL=subs(coffi(end),t,x)*TM(k+1,1); FF=subs(GL,x,tt); S2=S2+H(r+1,k+1,1)*FF; end Pt16=S2-S1; end Df(l+1,r+1)=Dn(l+1,r+1)+Pt1+Pt2+Pt3+Pt4+Pt5+Pt6+Pt7+Pt8+Pt9+Pt10+Pt11+Pt12+Pt1 3+Pt14+Pt15+Pt16; end end
163
Appendix
Programs
for l=0:nt tt=T0(l+1,1); Pn1=0.0;Pn2=0.0;Pn3=0.0;Pn4=0.0;Pn5=0.0;Pn6=0.0;Pn7=0.0;Pn8=0.0; Pn9=0.0;Pn10=0.0;Pn11=0.0;Pn12=0.0;Pn13=0.0;Pn14=0.0;Pn15=0.0;Pn16=0.0; if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn1=INTh-gL; elseif((tt-Tau(1)<=ab(1)) && ((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1))) gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn2=-gL; elseif((tt-Tau(2)<=ab(1)) && ((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn3=INTh; elseif((tt-Tau(3)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn4=INTh; elseif((tt-Tau(4)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn5=INTh; elseif(((tt-Tau(1)&tt-Tau(2))<=ab(1)) && ((tt-Tau(3) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn6=INTh-gL; elseif(((tt-Tau(1)&tt-Tau(3))<=ab(1)) && ((tt-Tau(2) & tt-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn7=INTh-gL; elseif(((tt-Tau(1)&tt-Tau(4))<=ab(1)) && ((tt-Tau(2) & tt-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn8=INTh-gL; elseif(((tt-Tau(2)&tt-Tau(3))<=ab(1)) && ((tt-Tau(1) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn9=INTh; elseif(((tt-Tau(2)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(3))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt);INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn10=INTh; elseif(((tt-Tau(3)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(2))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn11=INTh; elseif(((tt-Tau(1)&tt-Tau(2) & tt-Tau(3))<=ab(1)) && (tt-Tau(4)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn12=INTh-gL; elseif(((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))<=ab(1)) && (tt-Tau(3)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn13=INTh-gL;
164
Appendix
Programs
elseif(((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(2)>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn14=INTh-gL; elseif(((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(1)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt);INT=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn15=INT; else % if (((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) Pn16=0.0; end Ff(l+1,1)=Ft(l+1,1)+Pn1+Pn2+Pn3+Pn4+Pn5+Pn6+Pn7+Pn8+Pn9+Pn10+Pn11+Pn12+Pn13+Pn 14+Pn15+Pn16; end DDf=Df.'; Ff; LSEF=subs(((C'*DDf).'-Ff).^2); LSEf=sum(LSEF); disp(' ') disp(' the points ti Numerical Solutions uN(t)') T1=subs(exact,T0);T2=subs(u,T0); disp(subs(vpa([T0 T1 T2 ],12))) LSE=vpa([LSE],6) , LSEf=vpa([LSEf],6)
THE TABLE the Exact Solutions u(t)
The
Main Delay Legender Polynomial (MainDelayLeg) %
Legendre Program For Four Constant Delays
clc clear all format long g syms t x ab=[0,1] ; % ab=input('Input the interval [a,b] = '); ww=0.85; N=2 ; % N=input('Input the number of expantion terms N = '); hO=(2*(t^2))-1 ; % hO=input('The hestorical function hO(t) = '); alfa=[2*ww ww 0] ; % alfa=input('The alfa part = '); coffi=[1 -cos(t) t^3]; % from the first one we begining Tau=[0.8 0.6 0.4 0.2]; % The delay constant tau M=5 ; % number of points (Gauss-Legendre Zeroth) IN=20; % numbers of parts to solve the integrals numerically n=length(alfa)-1 ; % n is the number of Fractional term alfa part m=length(Tau)-1 ; % m is the number of Kernels term mu=max(ceil(alfa)) ; IC=zeros(mu,1); for i=1:mu IC(i,1)=input('input the initial condition ui ='); end tic Z=zeros(M,1); X=zeros(M,1); W=zeros(M,1); [Z(1:M,1),W(1:M,1)]=ZerosLeg(M); X(1:M,1)=((ab(2)-ab(1))*Z((0:M-1)+1,1)+(ab(2)+ab(1)))/2;
165
Appendix
Programs
lTau=length(Tau);H=zeros(N+1,N+1,lTau);d=ab(2)-ab(1); for ll=1:lTau H(:,:,ll)=OrthogonalTauLeg(Tau(ll),N,d); end LM=legendreM(ab(1),ab(2),N,x); KR=KERS(1:m,t,x); F=FUNC(t,X); D=zeros(M,N+1); for r=0:N for l=0:M-1 tx=X(l+1,1); D(l+1,r+1)=sum(subs(coffi(1:end1),tx).*caputoLeg(ab(1),ab(2),r,tx,alfa(1:n))); end end DP=zeros(M,N+1); DF=zeros(M,1); for l=0:M-1 tx=X(l+1,1); for r=0:N P1=0.0;P2=0.0;P3=0.0;P4=0.0;P5=0.0;P6=0.0;P7=0.0;P8=0.0; P9=0.0;P10=0.0;P11=0.0;P12=0.0;P13=0.0;P14=0.0;P15=0.0;P16=0.0; if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) P1=0.0; elseif((tx-Tau(1)<=ab(1)) && ((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; end P2=-S1; elseif((tx-Tau(2)<=ab(1)) && ((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P3=S12-S22; elseif((tx-Tau(3)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT; end P4=S12-S22; elseif((tx-Tau(4)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S22=S22+INT;
166
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Programs
end P5=S12-S22; elseif(((tx-Tau(1)&tx-Tau(2))<=ab(1)) && ((tx-Tau(3) & tx-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; end P6=-S1; elseif(((tx-Tau(1)&tx-Tau(3))<=ab(1)) && ((tx-Tau(2) & tx-Tau(4))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P7=-S2; elseif(((tx-Tau(1)&tx-Tau(4))<=ab(1)) && ((tx-Tau(2) & tx-Tau(3))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P8=-S2; elseif(((tx-Tau(2)&tx-Tau(3))<=ab(1)) && ((tx-Tau(1) & tx-Tau(4))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P9=S0-S2; elseif(((tx-Tau(2)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(3))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P10=S0-S2; elseif(((tx-Tau(3)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(2))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S2=S2+INT; end P11=S0-S2; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))<=ab(1)) && (tx-Tau(4)>ab(1))) S0=0.0; for k=0:r KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P12=-S0; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))<=ab(1)) && (tx-Tau(3)>ab(1))) S0=0.0;
167
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Programs
for k=0:r KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P13=-S0; elseif(((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(2)>ab(1))) S0=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S0=S0+INT; end P14=-S0; elseif(((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(1)>ab(1))) S0=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tx); S0=S0+H(r+1,k+1,1)*FL; end P15=S0; else % if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1)) S1=0.0;S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tx))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tx); S1=S1+INT; GL=subs(coffi(end),t,x)*LM(k+1,1); FF=subs(GL,x,tx); S2=S2+H(r+1,k+1,1)*FF; end P16=S2-S1; end DP(l+1,r+1)=D(l+1,r+1)+P1+P2+P3+P4+P5+P6+P7+P8+P9+P10+P11+P12+P13+P14+P15+P16; end end for l=0:M-1 tx=X(l+1,1); PF1=0.0;PF2=0.0;PF3=0.0;PF4=0.0;PF5=0.0;PF6=0.0;PF7=0.0;PF8=0.0; PF9=0.0;PF10=0.0;PF11=0.0;PF12=0.0;PF13=0.0;PF14=0.0;PF15=0.0;PF16=0.0; if ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF1=INTh-gL; elseif((tx-Tau(1)<=ab(1)) && ((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF2=-gL; elseif((tx-Tau(2)<=ab(1)) && ((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF3=INTh; elseif((tx-Tau(3)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF4=INTh; elseif((tx-Tau(4)<=ab(1)) && ((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF5=INTh; elseif(((tx-Tau(1)&tx-Tau(2))<=ab(1)) && ((tx-Tau(3) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tx);
168
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Programs
INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF6=INTh-gL; elseif(((tx-Tau(1)&tx-Tau(3))<=ab(1)) && ((tx-Tau(2) & tx-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF7=INTh-gL; elseif(((tx-Tau(1)&tx-Tau(4))<=ab(1)) && ((tx-Tau(2) & tx-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF8=INTh-gL; elseif(((tx-Tau(2)&tx-Tau(3))<=ab(1)) && ((tx-Tau(1) & tx-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF9=INTh; elseif(((tx-Tau(2)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(3))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx);INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF10=INTh; elseif(((tx-Tau(3)&tx-Tau(4))<=ab(1)) && ((tx-Tau(1) & tx-Tau(2))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF11=INTh; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(3))<=ab(1)) && (tx-Tau(4)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF12=INTh-gL; elseif(((tx-Tau(1) & tx-Tau(2) & tx-Tau(4))<=ab(1)) && (tx-Tau(3)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF13=INTh-gL; elseif(((tx-Tau(1) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(2)>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tx); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tx); PF14=INTh-gL; elseif(((tx-Tau(2) & tx-Tau(3) & tx-Tau(4))<=ab(1)) && (tx-Tau(1)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tx);INT=ClenshawCurtis(Gg1,IN,35,ab(1),tx); PF15=INT; else % if (((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) PF16=0.0; end DF(l+1,1)=F(l+1,1)+PF1+PF2+PF3+PF4+PF5+PF6+PF7+PF8+PF9+PF10+PF11+PF12+PF13+PF1 4+PF15+PF16; end DD=DP.'; FF=DF; A=zeros(N+1,N+1);B=zeros(N+1,1); for r=0:N for s=0:N SR=0.0; for j=0:M-1 SR=SR+W(j+1,1)*DD(r+1,j+1)*DD(s+1,j+1); end A(r+1,s+1)=((ab(2)-ab(1))/2)*SR;
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Programs
end SF=0.0; for j=0:M-1 SF=SF+W(j+1,1)*DD(r+1,j+1)*FF(j+1,1); end B(r+1,1)=((ab(2)-ab(1))/2)*SF; end ICC=initialconditionLeg(ab(1),ab(2),N,mu); CB=cell(2,1);CB{1}=B;CB{2}=IC;BB=cell2mat(CB); CA=cell(2,1);CA{1}=A;CA{2}=ICC;AA=cell2mat(CA); AAA=AA'*AA;BBB=AA'*BB; C=zeros(N+1,1); [L0,U0]=lu(AAA);C=vpa(U0\(L0\BBB),8) toc disp(' ') disp('The Numerical Solution is uN= ') LMt=legendreM(ab(1),ab(2),N,t); u=sym(zeros(1));u=sum(C.*LMt); pretty(simplify(vpa(u,10))) disp(' ') % The following steps using to find L.S.E. using exact with % approximation solutions. exact=input('the exact function u(t) = '); nt=10; ht=(ab(2)-ab(1))/nt;T0=[ab(1):ht:ab(2)].'; for i=0:nt tt=T0(i+1,1); ES(1,i+1)=(subs(exact,tt)-subs(u,tt)).^2; end LSE=sum(ES); % The following steps using to find L.S.E.f for Residuals Dn=zeros(nt+1,N+1); for r=0:N for l=0:nt tt=T0(l+1,1); Dn(l+1,r+1)=sum(subs(coffi(1:end1),tt).*caputoLeg(ab(1),ab(2),r,tt,alfa(1:n))); end end Ft=FUNC(t,T0); Df=zeros(nt+1,N+1); Ff=zeros(nt+1,1); for l=0:nt tt=T0(l+1,1); for r=0:N Pt1=0.0;Pt2=0.0;Pt3=0.0;Pt4=0.0;Pt5=0.0;Pt6=0.0;Pt7=0.0;Pt8=0.0; Pt9=0.0;Pt10=0.0;Pt11=0.0;Pt12=0.0;Pt13=0.0;Pt14=0.0;Pt15=0.0;Pt16=0.0; if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) Pt1=0.0; elseif((tt-Tau(1)<=ab(1)) && ((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; end Pt2=-S1; elseif((tt-Tau(2)<=ab(1)) && ((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))>ab(1))) S12=0.0;S22=0.0;
170
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Programs
for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt3=S12-S22; elseif((tt-Tau(3)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt4=S12-S22; elseif((tt-Tau(4)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))>ab(1))) S12=0.0;S22=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S12=S12+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S22=S22+INT; end Pt5=S12-S22; elseif(((tt-Tau(1)&tt-Tau(2))<=ab(1)) && ((tt-Tau(3) & tt-Tau(4))>ab(1))) S1=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; end Pt6=-S1; elseif(((tt-Tau(1)&tt-Tau(3))<=ab(1)) && ((tt-Tau(2) & tt-Tau(4))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt7=-S2; elseif(((tt-Tau(1)&tt-Tau(4))<=ab(1)) && ((tt-Tau(2) & tt-Tau(3))>ab(1))) S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt8=-S2; elseif(((tt-Tau(2)&tt-Tau(3))<=ab(1)) && ((tt-Tau(1) & tt-Tau(4))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt9=S0-S2; elseif(((tt-Tau(2)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(3))>ab(1)))
171
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Programs
S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt10=S0-S2; elseif(((tt-Tau(3)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(2))>ab(1))) S0=0.0;S2=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S2=S2+INT; end Pt11=S0-S2; elseif(((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))<=ab(1)) && (tt-Tau(4)>ab(1))) S0=0.0; for k=0:r KL=KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt12=-S0; elseif(((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))<=ab(1)) && (tt-Tau(3)>ab(1))) S0=0.0; for k=0:r KL=KR(1,2)*H(r+1,k+1,3); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt13=-S0; elseif(((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(2)>ab(1))) S0=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S0=S0+INT; end Pt14=-S0; elseif(((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(1)>ab(1))) S0=0.0; for k=0:r GL=subs(coffi(end),t,x)*LM(k+1,1); FL=subs(GL,x,tt); S0=S0+H(r+1,k+1,1)*FL; end Pt15=S0; else % if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1)) S1=0.0;S2=0.0; for k=0:r KL=KR(1,1)*H(r+1,k+1,2)+KR(1,2)*H(r+1,k+1,3)+KR(1,3)*H(r+1,k+1,4); gg=(subs(KL,t,tt))*LM(k+1,1); INT=ClenshawCurtis(gg,IN,35,ab(1),tt); S1=S1+INT; GL=subs(coffi(end),t,x)*LM(k+1,1); FF=subs(GL,x,tt); S2=S2+H(r+1,k+1,1)*FF; end Pt16=S2-S1; end Df(l+1,r+1)=Dn(l+1,r+1)+Pt1+Pt2+Pt3+Pt4+Pt5+Pt6+Pt7+Pt8+Pt9+Pt10+Pt11+Pt12+Pt1 3+Pt14+Pt15+Pt16; end end
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for l=0:nt tt=T0(l+1,1); Pn1=0.0;Pn2=0.0;Pn3=0.0;Pn4=0.0;Pn5=0.0;Pn6=0.0;Pn7=0.0;Pn8=0.0; Pn9=0.0;Pn10=0.0;Pn11=0.0;Pn12=0.0;Pn13=0.0;Pn14=0.0;Pn15=0.0;Pn16=0.0; if ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn1=INTh-gL; elseif((tt-Tau(1)<=ab(1)) && ((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))>ab(1))) gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn2=-gL; elseif((tt-Tau(2)<=ab(1)) && ((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn3=INTh; elseif((tt-Tau(3)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn4=INTh; elseif((tt-Tau(4)<=ab(1)) && ((tt-Tau(1) & tt-Tau(2) & tt-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn5=INTh; elseif(((tt-Tau(1)&tt-Tau(2))<=ab(1)) && ((tt-Tau(3) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn6=INTh-gL; elseif (((tt-Tau(1) & tt-Tau(3))<=ab(1)) && ((tt-Tau(2) & ttTau(4))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn7=INTh-gL; elseif(((tt-Tau(1)&tt-Tau(4))<=ab(1)) && ((tt-Tau(2) & tt-Tau(3))>ab(1))) Gg=KR(1,3)*subs(hO,t,x-Tau(4));Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn8=INTh-gL; elseif(((tt-Tau(2)&tt-Tau(3))<=ab(1)) && ((tt-Tau(1) & tt-Tau(4))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn9=INTh; elseif(((tt-Tau(2)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(3))>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt);INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn10=INTh; elseif(((tt-Tau(3)&tt-Tau(4))<=ab(1)) && ((tt-Tau(1) & tt-Tau(2))>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn11=INTh; elseif(((tt-Tau(1)&tt-Tau(2) & tt-Tau(3))<=ab(1)) && (tt-Tau(4)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,x-Tau(3)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn12=INTh-gL; elseif(((tt-Tau(1) & tt-Tau(2) & tt-Tau(4))<=ab(1)) && (tt-Tau(3)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt);
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Pn13=INTh-gL; elseif(((tt-Tau(1) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(2)>ab(1))) Gg=KR(1,2)*subs(hO,t,x-Tau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt); INTh=ClenshawCurtis(Gg1,IN,35,ab(1),tt); gl=coffi(end)*subs(hO,t,t-Tau(1)); gL=subs(gl,t,tt); Pn14=INTh-gL; elseif(((tt-Tau(2) & tt-Tau(3) & tt-Tau(4))<=ab(1)) && (tt-Tau(1)>ab(1))) Gg=KR(1,1)*subs(hO,t,x-Tau(2))+KR(1,2)*subs(hO,t,xTau(3))+KR(1,3)*subs(hO,t,x-Tau(4)); Gg1=subs(Gg,t,tt);INT=ClenshawCurtis(Gg1,IN,35,ab(1),tt); Pn15=INT; else % if (((tx-Tau(1) & tx-Tau(2) & tx-Tau(3) & tx-Tau(4))>ab(1))) Pn16=0.0; end Ff(l+1,1)=Ft(l+1,1)+Pn1+Pn2+Pn3+Pn4+Pn5+Pn6+Pn7+Pn8+Pn9+Pn10+Pn11+Pn12+Pn13+Pn 14+Pn15+Pn16; end DDf=Df.'; Ff; LSEF=subs(((C'*DDf).'-Ff).^2); LSEf=sum(LSEF); disp(' ') disp(' the points ti Numerical Solutions uN(t)') T1=subs(exact,T0);T2=subs(u,T0); disp(subs(vpa([T0 T1 T2 ],12))) LSE=vpa([LSE],6) , LSEf=vpa([LSEf],6)
THE TABLE the Exact Solutions u(t)
Subprograms of (ClosedCheb, openCheb and Leg) Subprograms of (ClosedCheb) function DD=caputoCheb(a1,b1,j1,t1,alfa1); la=length(alfa1);DD=sym(zeros(1,la)); for ia=1:la if j1==0 & alfa1(ia)==0 DD(1,ia)=1; elseif j1>=1 m=ceil(alfa1(ia)); F=floor(j1/2); sum=0;mul=1; mull=j1*((2/(b1-a1))^m)*((t1-a1)^(m-alfa1(ia)))/2; for r=0:F mul1=1;mu=1; jr=j1-2*r; mu=(-1)^r*(2^jr)*gamma(j1-r)/gamma(r+1); if m>jr MM=0; elseif m==jr MM=1/gamma(m-alfa1(ia)+1); else sum1=0; for k=0:jr-m sum1=sum1+((-1)^(k+jr-m)*(2*(t1-a1)/(b1a1))^k)/(gamma(k+m-alfa1(ia)+1)*gamma(jr-m-k+1)); end MM=sum1;
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Programs end mul1=mu*MM; sum=sum+mul1; end mul=mull*sum; DD(1,ia)=mul;
else DD(1,ia)=0; end end function C=chebyshev(a1,b1,m1,t1); if m1==0 C=1; else %F1=subs((m1-mod(m1,2))/2); F1=floor(m1/2); sumc=0; for r=0:F1 mm=m1-2*r; sumc=sumc+(-1)^r*(2^mm)*((2.*(t1-a1)./(b1-a1)-1).^mm)*(factorial(m1-r1)/(factorial(r)*factorial(mm))); end C=sumc*m1/2; end function CM=chebyshevM(a1,b1,N1,t1); CM=sym(zeros(N1+1,1)); for m1=0:N1 if m1==0 CM(m1+1,1)=1; else %F1=subs((m1-mod(m1,2))/2); F1=floor(m1/2); sum=0; for r=0:F1 mm=m1-2*r; sum=sum+(-1)^r*(2^mm)*((2.*(t1-a1)./(b1-a1)-1).^mm)*(factorial(m1r-1)/(factorial(r)*factorial(mm))); end CM(m1+1,1)=sum*m1/2; end end function I=ClenshawCurtis(gg,N,N1,a,b); tk=cos((0:N).*pi./N); tk0=((b-a)/2)*tk+((b+a)/2); gk=zeros(N+1,1); gk(:,1)=subs(gg,tk0(1,:)); v=zeros(1,N1+1); for s=0:N1 if mod(s,2) == 0 v(1,s+1)=1/(1-s.^2); else v(1,s+1)=0.0; end end C=zeros(N+1,N1+1); for k=0:N for s=0:N1 C(k+1,s+1)=cos((s*k*pi)/N); end end w=zeros(1,N+1);
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for k=0:N S1=0.0; for s=0:N1 S=v(1,s+1)*C(k+1,s+1); if s==0 | s==N1 S=S/2; else S=S; end S1=S1+S; end w(1,k+1)=(4/N)*S1; end for k=0:N if k==0 | k==N gk(k+1,1)=gk(k+1,1)/2; else gk(k+1,1)=gk(k+1,1); end end I=((b-a)/2)*w*gk; function DT=diffchebyshev(a1,b1,n1,m1,t1); if m1
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HC(i+1,1)=-Tab; else HC(i+1,1)=HC(i,2)-2*Tab*HC(i,1)-HC(i-1,1); end if i>=2 HC(i+1,2)=2*HC(i,1)+HC(i,3)-2*Tab*HC(i,2)-HC(i-1,2); end if i>3 for k=3:i-1 HC(i+1,k)=HC(i,k-1)+HC(i,k+1)-2*Tab*HC(i,k)-HC(i-1,k); end end if i>2 HC(i+1,i)=HC(i,i-1)-2*Tab; end end HCC=subs(HC);
Subprograms of (OpenCheb) function HCC=OrthogonalTauCheb(tau,im,dab); Tab=2*tau/dab;HC=sym(eye(im+1)); for i=1:im if i==1 HC(i+1,1)=-Tab; else HC(i+1,1)=HC(i,2)-2*Tab*HC(i,1)-HC(i-1,1); end if i>=2 HC(i+1,2)=2*HC(i,1)+HC(i,3)-2*Tab*HC(i,2)-HC(i-1,2); end if i>3 for k=3:i-1 HC(i+1,k)=HC(i,k-1)+HC(i,k+1)-2*Tab*HC(i,k)-HC(i-1,k); end end if i>2 HC(i+1,i)=HC(i,i-1)-2*Tab; end end HCC=subs(HC); function DD=caputoCheb(a1,b1,j1,t1,alfa1); la=length(alfa1);DD=sym(zeros(1,la)); for ia=1:la if j1==0 & alfa1(ia)==0 DD(1,ia)=1; elseif j1>=1 m=ceil(alfa1(ia)); F=subs((j1-mod(j1,2))/2); % i.e. F=floor(j1/2); sum=0;mul=1; mull=j1*((2/(b1-a1))^m)*((t1-a1)^(m-alfa1(ia)))/2; for r=0:F mul1=1;mu=1; jr=j1-2*r; mu=(-1)^r*(2^jr)*gamma(j1-r)/gamma(r+1); if m>jr MM=0; elseif m==jr MM=1/gamma(m-alfa1(ia)+1); else
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sum1=0; for k=0:jr-m sum1=sum1+((-1)^(k+jr-m)*(2*(t1-a1)/(b1a1))^k)/(gamma(k+m-alfa1(ia)+1)*gamma(jr-m-k+1)); end MM=sum1; end mul1=mu*MM; sum=sum+mul1; end mul=mull*sum; DD(1,ia)=mul; else DD(1,ia)=0; end end function C=chebyshev(a1,b1,m1,t1); if m1==0 C=1; else F1=subs((m1-mod(m1,2))/2); % i.e. F1=floor(m1/2); sumc=0; for r=0:F1 mm=m1-2*r; sumc=sumc+(-1)^r*(2^mm)*((2.*(t1-a1)./(b1-a1)-1).^mm)*(factorial(m1-r1)/(factorial(r)*factorial(mm))); end C=sumc*m1/2; end function CM=chebyshevM(a1,b1,N1,t1); CM=sym(zeros(N1+1,1)); for m1=0:N1 if m1==0 CM(m1+1,1)=1; else F1=subs((m1-mod(m1,2))/2); % i.e. F1=floor(m1/2); sum=0; for r=0:F1 mm=m1-2*r; sum=sum+(-1)^r*(2^mm)*((2.*(t1-a1)./(b1-a1)-1).^mm)*(factorial(m1r-1)/(factorial(r)*factorial(mm))); end CM(m1+1,1)=sum*m1/2; end end function I=ClenshawCurtis(gg,N,N1,a,b); tk=cos((0:N).*pi./N); tk0=((b-a)/2)*tk+((b+a)/2); gk=zeros(N+1,1); gk(:,1)=subs(gg,tk0(1,:)); v=zeros(1,N1+1); for s=0:N1 if mod(s,2) == 0 v(1,s+1)=1/(1-s.^2); else v(1,s+1)=0.0; end end C=zeros(N+1,N1+1); for k=0:N for s=0:N1 C(k+1,s+1)=cos((s*k*pi)/N);
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end end w=zeros(1,N+1); for k=0:N S1=0.0; for s=0:N1 S=v(1,s+1)*C(k+1,s+1); if s==0 | s==N1 S=S/2; else S=S; end S1=S1+S; end w(1,k+1)=(4/N)*S1; end for k=0:N if k==0 | k==N gk(k+1,1)=gk(k+1,1)/2; else gk(k+1,1)=gk(k+1,1); end end I=((b-a)/2)*w*gk; function DT=diffchebyshev(a1,b1,n1,m1,t1); if m1
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Subprograms of (Legendre) function HLL=OrthogonalTauLeg(tau,im,dab); Tab=2*tau/dab; HL=sym(eye(im+1)); for i=1:im if i==1 HL(i+1,1)=-Tab; else l=i-1; HL(i+1,1)=((2*l+1)/(3*(l+1)))*HL(i,2)((2*l+1)/(l+1))*Tab*HL(i,1)-(l/(l+1))*HL(i-1,1); end if i>2 li=i-1; for k=1:li-1 HL(i+1,k+1)=((2*li+1)/(li+1))*(((k+1)/(2*k+3))*HL(i,k+2)+(k/(2*k1))*HL(i,k)-Tab*HL(i,k+1))-(li/(li+1))*HL(i-1,k+1); end end if i~=1 l=i-1; HL(i+1,i)=((2*l+1)/(l+1))*((l/(2*l-1))*HL(i,i-1)-Tab); end end HLL=subs(HL); function [Zz,Ww]=ZerosLeg(mm); Zz=zeros(mm,1);Ww=zeros(mm,1); switch mm case 1 P=[1 0]; case 2 P=[3/2 0 -1/2]; case 3 P=[5/2 0 -3/2 0]; case 4 P=[35/8 0 -15/4 0 3/8]; case 5 P=[63/8 0 -35/4 0 15/8 0]; case 6 P=[231/16 0 -315/16 0 105/16 0 -5/16]; case 7 P=[429/16 0 -693/16 0 315/16 0 -35/16 0]; case 8 P=[6435/128 0 -3003/32 0 3465/64 0 -315/32 0 35/128]; case 9 P=[12155/128 0 -6435/32 0 9009/64 0 -1155/32 0 315/128 0]; case 10 P=[46189/256 0 -109395/256 0 45045/128 0 -15015/128 0 3465/256 0 63/256]; case 11 P=[88179/256 0 -230945/256 0 109395/128 0 -45045/128 0 15015/256 0 693/256 0]; case 12 P=[676039/1024 0 -969969/512 0 2078505/1024 0 -255255/256 0 225225/1024 0 -9009/512 0 231/1024]; case 13 P=[1300075/1024 0 -2028117/512 0 4849845/1024 0 -692835/256 0 765765/1024 0 -45045/512 0 3003/1024 0]; case 14 P=[5014575/2048 0 -16900975/2048 0 22309287/2048 0 -14549535/2048 0 4849845/2048 0 -765765/2048 0 45045/2048 0 -429/2048]; case 15
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P=[5204880276848639/1099511627776 0 -35102025/2048 0 50702925/2048 0 37182145/2048 0 14549535/2048 0 -2909907/2048 0 255255/2048 0 -6435/2048 0]; case 16 P=[300540195/32768 0 -4879575259545599/137438953472 0 7655885321011199/137438953472 0 -185910725/4096 0 334639305/16384 0 20369349/4096 0 4849845/8192 0 -109395/4096 0 6435/32768]; case 17 P=[583401555/32768 0 -300540195/4096 0 8539256704204799/68719476736 0 -7655885321011199/68719476736 0 929553625/16384 0 -66927861/4096 0 20369349/8192 0 -692835/4096 0 109395/32768 0]; case 18 P=[2268783825/65536 0 -5199797385953279/34359738368 0 4508102925/16384 0 -4411154475/16384 0 5019589575/32768 0 -1673196525/32768 0 156165009/16384 0 -14549535/16384 0 2078505/65536 0 -12155/65536]; case 19 P=[4418157975/65536 0 -20419054425/65536 0 5199797385953279/8589934592 0 -5514936621465599/8589934592 0 6938146072166399/17179869184 0 5019589575/32768 0 557732175/16384 0 -66927861/16384 0 14549535/65536 0 230945/65536 0]; case 20 P=[34461632205/262144 0 -5501419619942399/8589934592 0 5687278390886399/4294967296 0 -49589132175/32768 0 136745788725/131072 0 7631960679383039/17179869184 0 7895131737292799/68719476736 0 -557732175/32768 0 334639305/262144 0 -4849845/131072 0 46189/262144]; otherwise Zz='Unknown Zeros input more zeros'; Ww='input more zeros to find Ww'; end Zz(:,1)=roots(P); syms x L=legendre(-1,1,mm,x); DL=diff(L,'x',1);J(:,1)=subs(DL,Zz(:,1)); Ww(:,1)=2./(((1-Zz.^2)).*(J.^2)); function DP=caputoLeg(a1,b1,j1,t1,alfa1); la=length(alfa1);DP=sym(zeros(1,la)); for ia=1:la if j1==0 & alfa1(ia)==0 DP(1,ia)=1; elseif j1>=1 m=ceil(alfa1(ia)); F=subs((j1-mod(j1,2))/2); % i.e. F=floor(j1/2); sum=0;mul=1; mull=(((2/(b1-a1))^m)*((t1-a1)^(m-alfa1(ia))))/(2^j1); for r=0:F mul1=1;mu=1;jr=j1-2*r; mu=(-1)^r*gamma(2*j1-2*r+1)/(gamma(r+1)*gamma(j1-r+1)); if m>jr MM=0; elseif m==jr MM=1/gamma(m-alfa1(ia)+1); else sum1=0; for k=0:jr-m sum1=sum1+((-1)^(k+jr-m)*(2*(t1-a1)/(b1a1))^k)/(gamma(k+m-alfa1(ia)+1)*gamma(jr-m-k+1)); end MM=sum1; end mul1=mu*MM; sum=sum+mul1; end mul=mull*sum;
181
Appendix
Programs DP(1,ia)=mul;
else DP(1,ia)=0; end end function I=ClenshawCurtis(gg,N,N1,a,b); tk=cos((0:N).*pi./N); tk0=((b-a)/2)*tk+((b+a)/2); gk=zeros(N+1,1); gk(:,1)=subs(gg,tk0(1,:)); v=zeros(1,N1+1); for s=0:N1 if mod(s,2) == 0 v(1,s+1)=1/(1-s.^2); else v(1,s+1)=0.0; end end C=zeros(N+1,N1+1); for k=0:N for s=0:N1 C(k+1,s+1)=cos((s*k*pi)/N); end end w=zeros(1,N+1); for k=0:N S1=0.0; for s=0:N1 S=v(1,s+1)*C(k+1,s+1); if s==0 | s==N1 S=S/2; else S=S; end S1=S1+S; end w(1,k+1)=(4/N)*S1; end for k=0:N if k==0 | k==N gk(k+1,1)=gk(k+1,1)/2; else gk(k+1,1)=gk(k+1,1); end end I=((b-a)/2)*w*gk; function DT=difflegendre(a1,b1,n1,m1,t1); if m1
182
Appendix
Programs
end function LM1=legendreM(a1,b1,N1,x1); LM1=sym(zeros(N1+1,1)); for n1=0:N1 if n1==0 LM1(n1+1,1)=1; else F1=subs((n1-mod(n1,2))/2); % i.e. F1=floor(n1/2); suml=0; for r=0:F1 nn=n1-2*r; suml=suml+((-1)^r)*((2.*(x1-a1)./(b1-a1)-1).^nn)*(factorial(2*n12*r)/(factorial(r)*factorial(n1-r)*factorial(nn))); end LM1(n1+1,1)=suml*(1/(2.^n1)); end end function F1=FUNC(t,X1); % syms t ww=0.85; fun=((4*(t.^(2*(1-ww))))/(gamma(3-(2*ww))))-((4*cos(t)*(t.^(2-ww)))/(gamma(3ww)))+(sin(t)*((0.5*(t^4))-((4/15)*(t^3))-((23/50)*(t^2))))+((123/25)*(exp(t)))-(123/25)+((28/5)*(t))-((43/25)*(t^2))+((118/75)*(t^3))((49/15)*(t^4))+((8/5)*(t^5)); F1=subs(fun,X1); function JJ=initialconditionLeg(a1,b1,N1,mi); J1=zeros(mi,N1+1); for k1=0:mi-1 for r1=0:N1 J1(k1+1,r1+1)=difflegendre(a1,b1,k1,r1,a1); end end JJ=subs(J1); function Kernels=KERS(p,t,x); % syms x t Kers=[t+(x^2) , exp(x-t) , -x*sin(t)]; Kernels=Kers(1,p);
183
References
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LVIFDEs LVIFDEs least-square orthogonal (Block-by-Block method method LVIFDEs V8.1
اﻟﻤﻌﺎﻟﺠﺎت اﻟﻌﺪدﯾﺔ ﻟﺤﻞ اﻟﻤﻌﺎدﻻت ﻓﻮﻟﺘﯿﺮا اﻟﺘﻜﺎﻣﻠﯿﺔ- اﻟﺘﻔﺎﺿﻠﯿﺔ اﻟﺨﻄﯿﺔ ذات رﺗﺐ ﻛﺴﺮﯾﺔ ﻣﻊ زﻣﻦ -اﻟﺘﺒﺎطﺆﯾﺔ ﻣﻦ ﻧﻮع اﻟﺮﯾﺘﺎردﯾﺔ
رﺳﺎﻟﺔ ﻣﻘﺪﻣﺔ اﻟﻰ ﻣﺠﻠﺲ ﻓﺎﻛﻠﺘﻲ اﻟﻌﻠﻮم وﺗﺮﺑﯿﺔ اﻟﻌﻠﻮم ﺳﻜﻮل اﻟﻌﻠﻮم ﻓﻰ ﺟﺎﻣﻌﺔ اﻟﺴﻠﯿﻤﺎﻧﯿﺔ ﻛﺠﺰء ﻣﻦ ﻣﺘﻄﻠﺒﺎت ﻧﯿﻞ ﺷﮭﺎدة ﻣﺎﺟﺴﺘﯿﺮ ﻓﻲ ﻋﻠﻮم اﻟﺮﯾﺎﺿﯿﺎت
ﻣﻦ ﻗﺒﻞ
ﻣﯿﺮان ﺑﯿﺎن ﻣﺤﻤﺪ اﻣﯿﻦ ﺑﻜﺎﻟﻮرﯾﻮس ﻓﻲ اﻟﺮﯾﺎﺿﯿﺎت-ﺟﺎﻣﻌﺔ اﻟﺴﻠﯿﻤﺎﻧﯿﺔ2008-
ﺑﺎﺷﺮاف
د .ﺷﺎزاد ﺷﻮﻗﻲ اﺣﻤﺪ اﻻﺳﺘﺎذ اﻟﻤﺴﺎﻋﺪ
ﺷﻌﺒﺎن ١٤٣٧
ﺣﻮزﯾﺮان ٢٠١٦
Block-by-Block method V8
2008-
