boundary value problems for fractional order …
TRANSCRIPT
BOUNDARY VALUE PROBLEMS FOR FRACTIONAL ORDERDIFFERENTIAL EQUATIONS: EXISTENCE THEORY AND
NUMERICAL SIMULATIONS
by
HASIB KHAN
A Dissertation
Submitted To The University Of Malakand In Partial Fulfillment Of Requirements For The Degree Of
DOCTOR OF PHILOSOPHY
In Mathematics
Supervisor: Prof. Dr Rahmat Ali Khan
DEPARTMENT OF MATHEMATICS UNIVERSITY OF MALAKAND, CHAKDARA,
PAKISTAN
2015
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AUTHORIZATION TO SUBMIT DISSERTATION
This dissertation of Mr. Hasib Khan submitted for the Degree of Doctor of Philosophy in Mathematics
and titled “BOUNDARY VALUE PROBLEMS FOR FRACTIONAL ORDER DIFFERENTIAL
EQUATIONS: EXISTENCE THEORY AND NUMERICAL SIMULATIONS” has been reviewed
in final form. Permission, as indicated by the signatures given below, is now granted to submit the copies
to University of Malakand for the final evaluation process.
Research Supervisor
Prof. Dr Rahmat Ali Khan ...............................................................................
Committee members
....................................................................................................................................................
....................................................................................................................................................
Chairman Department of Mathematics
Prof. Dr Gul Zaman .............................................................................
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Acknowledgement
First and foremost, I am grateful to Almighty Allah for all His blessings and mercies in every walks
of my life. Then, I do hereby greatly acknowledge hereby the Holy Prophet Muhammad (PBUH), whose
life and principles serve as a symbol of perpetual peace and knowledge for the whole mankind.
I am thankful to the prime minister on the reimbursement scheme which have supported me in the
fee of my PhD studies at the Mathematics department, University of Malakand, Chakdara, Dir (L).
I am also thankful to the higher education commission of Pakistan (HEC) for providing the scholars
a very good research and higher studies environment in the universities of Pakistan, which have enabled
me for the PhD degree in my homeland.
I am thankful to my respected supervisor Professor Dr Rahmat Ali Khan, Dean of Sciences, Uni-
versity of Malakand, Chakdara, for his guidance throughout my PhD, during course and research work.
His very kind support helped me getting along the track and out of the hurdles in my studies. Thanks
are also due to Professor Gul Zaman, Chairman of Mathematics Department, for providing such a nice
environment for the studies.
My sincere thanks to the Ex-Vice Chancellor and founder of Shaheed Benazir Bhutto University
Sheringal, Dir Upper, Professor Dr Jahandar Shah for his support during my PhD enrollment and the
provision of No Objection Certificate, in 2012. I pray for his long and peaceful life.
Thanks to Professor Dr Gul Hassan, Shaheed Benazir Bhutto University, Sheringal, Dir Upper, for
his help and support in my travel to Iran during his Chairmanship which has resolved a lot of difficulties
in my research work.
Thanks to Professor Dr Hossein Jafari, Department of Mathematical Sciences, University of South
Africa, for his support in different aspects of my studies. My special thanks to him for the hospitality
during my short stay at the Department of Mathematics, University of Mazandaran, Iran, and for the
financial support in the publications of my research as article processing charges. I found him a very
good person and a great mathematician. I am also thankful to Assistant Professor Dr Mohsen Alipour
and Dr Haleh Tajadodi for their help in different problems for the numerical programing. May Allah
bless all of them. Ameen!
Thanks also go to Professor Dr Dumitru Baleanu, Department of Mathematics Computer Science,
Cankaya University, Ankara, Turkey, for his help and support. He is indeed a kind human being and a
good professional. Thank you Sir.
I am also indebted to Professor Dr Xiao-Jun Yang, Department of Mathematics and Mechanics,
China University of Mining and Technology, for his help in the LF differential equations. He provided
me a lot of research materials and guided me for the work.
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Thanks are also extended to Professor Dr Mohammad Mehdi Rashidi, Tongji University, China, for
sharing of his research expertise and with whom I am going to interact for my research in which I will
work in the nearest future.
My humble thanks to my parents, whose unending support and prayers got me every thing of my
life. May Allah bless them with every happiness in their lives and hereafter!
My thanks to my brothers Suliman Khan (School Teacher) and Aziz Khan (PhD Scholar) for their
love and help in my life and a special thanks to my sisters for their support and prayers. The thanks are
extended to my wife for her support, help and prayers.
My love for my nephews Shah Fahad Khan, Shah Zaib Khan, Mashaal Khan, Amsaal Khan, Ahmad
Khan, Khalid Khan, Tanveer Ahmad Khan and Tauseef Ahmad Khan. They have given me good time in
playing Chess and kept me joyful when we were at home. May Allah succeed all of them. Ameen!
HASIB KHAN
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DEDICATION
I dedicate my PhD thesis to my dear parents who bolstered me long lasting and pray for their happiest
life.
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List of publications
This thesis consist of the following published papers:
1. Dumitru Baleanu, Hasib Khan, Hossein Jafari, Rahmat Ali Khan and Mohsen Alipour, “On existence
results for solutions of a coupled system of hybrid boundary valueproblems with hybrid conditions’, Ad-
vances in Difference Equations (2015) 2015:318
2 D. Baleanu, R. P. Agarwal, Hasib Khan, R. A. Khan and H. Jafari, “On the existence of solution
for fractional differential equations of order 3 < δ1 ≤ 4”, Advances in difference Equations 2015:362
(2015).
3. H. Jafari, D. Baleanu, Hasib Khan, R. A. Khan, and A. Khan, “Existence criterion for the solutions
of fractional order p–Laplacian boundary value problems” Boundary Value Problems 2015:164, (2015).
4. D. Baleanu, Hasib Khan, H. Jafari, and R. A. Khan, “On the Exact Solution of Wave Equations on
Cantor Sets,” Entropy 17, 6229–6237 (2015).
5. D. Baleanu, H. Jafari, Hasib Khan, and S. J. Johnston, “Results for solution of fractional coupled
hybrid boundary value problems,” Open Mathematics, 13, 601–608 (2015).
6. Hasib Khan, H. Jafari, R. A. Khan, H. Tajadodi, and S. J. Johnston, “Numerical Solutions of the
Nonlinear Fractional-Order Brusselator System by Bernstein Polynomials,” The Scientific World Jour-
nal Vol. 2014, Article ID 257484, 7 pages (2014).
7. Hasib Khan, M. Alipour, R. A. Khan, H. Tajadodi, and A. Khan, “On Approximate Solution of
Fractional Order Logistic Equations by OMBPs,” Journal of Mathematics and Computer Science 14,
222–232 (2015).
8. Hasib Khan, M. Alipour, H. Jafari, R. A. Khan, “Approximate Analytical Solution of a Coupled
System of Fractional Partial Differential Equations by Bernstein Polynomials,” Int. J. Appl. Comput.
Math., (2015). DOI 10.1007/s40819-015-0052-8
9. Hasib Khan, R. A. Khan, and M. Alipour “On the existence of solution for fractional boundary value
problem,” Journal of Fractional Calculus 6(1), 83–93, (2015).
10. R. A. Khan, Hasib Khan, “Existence of solution for a three point boudary value problem of frac-
tional differential equation,” Journal of Fractional Calculus 5(1), 156–164, (2014).
11. Existence and uniqueness of results for coupled system of fractional q–difference equations with
boundary conditions, J. Comput. Complex. Appl., 1(2), 79–88 (2015).
Accepted papers are:
1. M. Alipour, R. A. Khan, Hasib Khan “Applications of local fractional series expansion method to
vi
fractal Vehicular traffic flow model, Sindh University Research Journal (Accepted)
2. D. Baleanu, H. Jafari, R. A. Khan and Hasib Khan “On approximations of local fractional Schrodinger
equations on Cantor sets, Sindh University Research Journal (Accepted)
3. M. Alipour, R. A. Khan and Hasib Khan “Computational method Based on Bernstein polynomials
for solving a fractional optimal control problem”, Punjab University Journal of Mathematics (Accepted)
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Abstract
The study of fractional calculus has been initiated in the Seventeenth century and has received much
attention in the last few decades. Because of the fractional order derivatives, scientists have developed
excellent approach for the description of memory and hereditary properties of different problems in
science and engineering. Therefore, we see the applications of fractional calculus in the fields such
as; signal processing, diffusion process, physics, fluid mechanics, bioscience, chemistry, economics,
polymer rheology and many others.
In this thesis, we are concerned with the existence and uniqueness of positive solutions for different
classes of boundary value problems for fractional differential equations (FDEs). We also study numerical
solutions of FDEs and for some classes exact analytical solutions of local FDEs.
Existence and uniqueness theory for positive solutions is developed for following classes of bound-
ary value problems (BVPs) for FDEs:
Class of two point BVPs for FDEs, class of three point BVPs for FDEs, Class of multi point BVPs for
FDEs, a general class of BVPs with p-Laplacian operator, BVPs for coupled systems of FDEs, BVPs for
coupled system of fractional order differential integral equations, BVPs for coupled system of fractional
order q–difference equations, and BVPs for coupled systems of hybrid FDEs.
For numerical solutions, Bernstein polynomials (BPs) are used and operational matrices (OM) for
fractional order integrations and differentiations are developed. Based on these OM, numerical schemes
for numerical solutions are developed for the following classes of of FDEs;
fractional partial differential equations, coupled systems of FDEs, optimal control problems. We also
use B-Spline functions and develop operational matrices of B-Spline functions for the numerical solution
of a coupled system of FDEs.
We also study exact solutions of some local FDEs, we use different mathematical methods for differ-
ent local fractional (LF) problems. In this work, we produce iterative techniques for the approximation
of solutions of different problems in LF calculus and the efficiency of the schemes are tested by many
examples.
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Contents
1 Preliminaries 6
1.1 Fractional calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.1.1 q–Fractional calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.2 Local fractional calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.1.3 Local fractional Laplace Transform (LFLT) . . . . . . . . . . . . . . . . . . . . 9
1.2 Numerical Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2.1 Bernstein polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2.2 Approximation of a function by Bernstein polynomials . . . . . . . . . . . . . . 11
1.2.3 Linear Cardinal B-Spline function . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.2.4 Function approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.3 Fixed point theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Existence and uniqueness of solutions for FDEs 17
2.1 Two point BVPs for FDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.1.1 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2 Three point BVPs for FDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2.1 Existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.3 Multi point BVPs for FDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.3.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . 30
2.4 p–Laplacian BVP for FDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.4.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . 39
3 Coupled systems of BVPs for fractional differential equations 43
3.1 Coupled systems of BVPs for FDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.1.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . 47
3.2 Coupled systems of BVPs for fractional order q–difference equations . . . . . . . . . . . 53
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3.2.1 Existence of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.2.2 Uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
3.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4 Coupled systems of hybrid FDEs 61
4.1 Coupled system of hybrid FDEs I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4.1.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . 63
4.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.2 Coupled system of hybrid FDEs II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.2.1 Existence and uniqueness of solutions . . . . . . . . . . . . . . . . . . . . . . . 73
4.2.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5 Local fractional differential equations 79
5.1 Iterative technique for solutions of LF wave equations . . . . . . . . . . . . . . . . . . . 80
5.1.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5.2 Iteration technique for solutions of LF Schrodinger equation . . . . . . . . . . . . . . . 83
5.2.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
5.3 Iteration technique for the solution of fractal Vehicular Traffic flow model . . . . . . . . 87
5.3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
6 Numerical solutions of fractional differential equations 90
6.1 Operational Matrix for fractional order derivatives using BPs . . . . . . . . . . . . . . . 90
6.2 Numerical scheme for solutions of Fractional Telegraph equations . . . . . . . . . . . . 92
6.2.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
6.3 Numerical scheme for solutions of Fractional partial differential equations (FPDEs) . . . 95
6.3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
7 Fractional integral and product matrices of BPs 100
7.1 Operational matrix for fractional order integration . . . . . . . . . . . . . . . . . . . . 100
7.2 Operational matrix for multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
7.3 Fractional order Logistic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
7.3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
7.4 Fractional order Brusselator system (FOBS) . . . . . . . . . . . . . . . . . . . . . . . . 107
7.4.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
7.5 Optimal controle problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
x
7.5.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
8 B-Spline functions 115
8.1 Operational matrix for fractional integration . . . . . . . . . . . . . . . . . . . . . . . . 115
8.2 Operational matrix of multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
8.3 Fractional order Brusselator system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
8.3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
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List of Figures
6.1 Graph for the approximation of u(x, t) when m = 10 and α = 1.5, 1.75, 2.0 and the exact
solution in6.2.1, by our numerical technique (6.2.6)– (6.2.7). . . . . . . . . . . . . . . . . . . 93
6.2 Plot for error in our approximation for u(x, t), by the proposed technique and our numerical
technique (6.2.6), (6.2.7) for α = 2 with m = 10 in 6.2.1. . . . . . . . . . . . . . . . . . . . 93
6.3 Graph for the approximate values of u(x, t) for α = 1.5, 1.75, 2.0, m = 10 and the exact
solution at t = 0.25. The graph for our approximation is the dashed for α = 2 and the green line
represents the exact solution in 6.2.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
6.4 Graph for the approximation of u(x, t) when m = 10 and α = 1.5, 1.75, 2.0 and the exact
solution in example 6.2.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
6.5 Plot for the error in our approximation for u(x, t), by the proposed technique and our numerical
technique (6.2.6)– (6.2.7) for α = 2 with m = 10 in example 6.2.2. . . . . . . . . . . . . . . . 95
6.6 Graph for the approximate values of u(x, t) for α = 1.5, 1.75, 2.0, m = 10 and the exact
solution at t = 0.25. The graph for our approximation is the dashed for α = 2 and the green line
represents the exact solution in example 6.2.2. . . . . . . . . . . . . . . . . . . . . . . . . . 95
6.7 Error in our approximation for u(x, t), for m = 5 and α, β = 1. . . . . . . . . . . . . . . . . 97
6.8 Error in our approximation for v(x, t), for m = 5 and α, β = 1. . . . . . . . . . . . . . . . . 97
6.9 Plot of the approximation solutions u(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1. . . . . . . . . 97
6.10 Plot of the approximation solutions v(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1. . . . . . . . . 98
6.11 Error in our approximation for u(x, t), for m = 3 and α, β = 1. . . . . . . . . . . . . . . . . 98
6.12 Error in our approximation for v(x, t), for m = 3 and α, β = 1. . . . . . . . . . . . . . . . . 99
6.13 Plot of the approximation solutions u(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1. . . . . . . . . 99
6.14 Plot of the approximation solutions v(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1. . . . . . . . . 99
7.1 Approximation of FOLE in example 7.3.1, where the black line represents the exact solution,
Red line is for α = 1, Green line is for α = .99, Doted line is for α = .98, and x ∈ [0, 2] . . . . . 104
7.2 Error in our approximation for the FOLE in example 7.3.1 for α = 1 when x ∈ [0, 1]. . . . . . . 105
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7.3 Approximation of FOLE in example 2 via BPs operational matrices. Here Black line=Exact
solution, Green line is for α = 1, Doted line is for α = .99, and the Red line is for α = .95, and
x ∈ [0, 2]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
7.4 Error in our approximation of the FOLE in example 7.3.2 for α = 1 when x ∈ [0, 1]. . . . . . . 105
7.5 In this figure, we present comparison of our approximation with the exact solution in example
7.3.3, for different values of α. Here the Dashed line represents exact solution at α = 1, where
our approximations are: Green line is for α = 1, Red line is for α = .99, Gray line is for
α = .98, Black line is for α = .95, and x ∈ [0, 2]. . . . . . . . . . . . . . . . . . . . . . . . 106
7.6 Error in our approximate solution of the FOLE in example 7.3.3, when α = 1, x ∈ [0, 1]. . . . . 106
7.7 The exact solution: (black line) and approximation solutions when α = 1, β = 1 and m = 12
(dotted), m = 8(dashed). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
7.8 The exact solution: (black line) and approximation solutions when m = 12 and α = .98, β = 1
(dotted), α = .95, β = .99(dashed), α = 0.9, β = .98(Long-dashed). . . . . . . . . . . . . . 109
7.9 The exact solution: (black line) and approximation solutions when α = 1, β = 1 and m = 6
(dotted), m = 4(dashed). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
7.10 Fig.1. The exact solution: (black line) and approximation solutions when m = 6 and α =
1, β = .98 (dotted), α = .98, β = .95(dashed). . . . . . . . . . . . . . . . . . . . . . . . 110
7.11 Plot of x(t) for m = 7 and α = .8, .9, 1 in example 7.5.1. . . . . . . . . . . . . . . . . . . . 112
7.12 Plot of u(t) for m = 7 and α = .8, .9, 1 in example 7.5.1. . . . . . . . . . . . . . . . . . . . 113
7.13 Plot of x(t) for m = 8 and α = .8, .9, 1 in example 7.5.2. . . . . . . . . . . . . . . . . . . . 114
7.14 Plot of u(t) for m = 8 and α = .8, .9, 1 in example 7.5.2. . . . . . . . . . . . . . . . . . . . 114
8.1 The exact solution (black line) and approximation solutions when α = 1, β = 1 and J = 5
(dotted) and J = 4 (dashed). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
8.2 The exact solution (black line) and approximation solutions when J = 5 and α = 0.98, β = 1
(dotted) and α = 0.95, β = 0.98 (dashed). . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
8.3 The exact solution: (black line) and approximation solutions when α = 1, β = 1 and J = 5
(dotted), J = 4(dashed). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
8.4 The exact solution (black line) and approximation solutions when J = 5 and α = 0.98, β = 0.99
(dotted) and α = 0.97, β = 0.98 (dashed). . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
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Introduction
Differential equations have been considered by a large number of scientists in many different scientific
fields. The scientists used different techniques for the exploration of different important features of
differential equations, including; existence of positive solutions, numerical solutions and exact solutions.
For these types of studies different mathematical tools have been used.
The study of differential equations has been initiated in the seventeenth century by Isaac Newton
(1642–1727) and Gottfried Leibniz (1646–1716). The concept of the differential operator was then
generalized by Leibniz and Marquis de L’Hospital. Leibniz introduced the notion for nth derivative of
a function y(x) for n ∈ N by Dnydxn . L’Hospital raised a question to Leibniz ”what will be the result if
n = 12?” Leibniz replied that ”this is an apparent paradox from which, one day, useful consequences will
be drawn”. This leads to the concept of fractional derivatives and S. F. de Lacroix (1819) introduced the
fractional derivative and then a large number of scientists have shown their contributions in the fractional
calculus. Some of the useful and important contributions in fractional calculus are due to I. Podlubny [1],
R. P. Agarwal [45], D. Baleanu [76, 88], R. A. Khan [19, 22, 24, 33, 58, 73], H. Jafari [80, 83, 97, 98], X.
J. Yang [122, 123], K. Maleknejad [82, 100], A. Akgul and A. Kilicman [72], K. B. Oldhalm and J.
Spainer [9], M. A. E. Herzallah [63], B. Ahmad [13, 14, 41, 42, 49], M. Rehman [21, 22, 59], R. A. C.
Ferreira [40], W. Yang [34], G. A. Anastassiou [56], M. Benchohra [26], J. Sabatier [10], A. A. Kilbas
et al. [2], R. Hilfer [6], A. A. Kilbas [7], B. C. Dhage [66, 70, 71, 104] and G. Wang [103].
There are many types of fractional order derivatives, including; the Riemann, the Marchaud, the
RiemannLiouville, the Weyl, the Caputo, the Hadamard, the Granwald-Letnikov, the Liouville, the
Riesz-Feller. In this dissertation, we are related to the Riemann-Liouville and the Caputo fractional
derivatives.
Scientists of different fields are exploring different features of fractional calculus. Among the differ-
ent features, existence of solutions for FDEs, fractional order integro-differential equations, FDEs with
p-Laplacian operator, hybrid FDEs, coupled systems of FDEs, coupled system of hybrid FDEs, q–FDEs,
and many others have been considered by a large number of scientists using different mathematical meth-
ods. For the existence of solutions, scientists are utilizing different mathematical methods, including;
1
Schauder fixed point theorem, Banach’s Contraction Principle, Arzela Ascoli theorem, Dhage’s theorem,
monotone iterative technique, etc., [17–29, 31, 32, 63].
The numerical solutions of FDEs is another important aspect of the fractional calculus. In this
area, scientists are using different mathematical techniques, including; wavelets, operational matrices for
fractional order integration of different polynomials, operational matrices for fractional order derivative
of different polynomials, operational matrices for fractional order integration of B–Spline functions
[73, 74, 76, 77, 81, 82, 129].
The fractal curves [105], are everywhere continuous but nowhere differentiable and therefore, the
classical calculus cannot be used to interpret the motions in Cantor time-space [107]. Local FC [110–
113], is considered as one of the useful ways to handle the fractals and other functions that are con-
tinuous but non-differentiable. Mandelbrot [105, 117], described that fractal geometry is a workable
geometric middle ground between the excessive geometric order of Euclid and the geometric case of
general mathematics, and extensively illustrated wide range of applications of fractals in many scientific
fields like in, physics, engineering, mathematics, geophysics. Christianto and Rahul [118], gave the
derivation of Proca equations on Cantor sets. Hao et al. [119], investigated the Helmholtz and Diffusion
equations in Cantorian and Cantor-Type Cylindrical Coordinates. Carpinteri and Sapora [120], studied
diffusion problems in fractal media in Cantor sets. Zhang et al. studied LF wave equations under fixed
entropy. Many techniques are utilized for handling the LF problems in both ordinary and partial deriva-
tives. For instance, The Local fractional Laplace Transform [110], the LFLVIM [113] and many others.
These methods are widely used in different scientific fields [112–116].
Organization of the thesis: In Chapter 2, we give details about the preliminary results, including
the basic definitions and results from fractional calculus and provide some related theorems, includ-
ing; Banach’s contraction principle, Arzela Ascoli theorem, Dhage’s theorem, Krasnoselskii’s theorem,
Leray–Schauder Alternative. The introduction of the q–Fractional calculus, basic results about the Bern-
stein polynomials are also provided in this chapter.
In Chapter 3, we study sufficient conditions for existence and uniqueness of solutions (EUS) to some
classes of boundary value problems for FDEs. The first class of problems of our interest consist of FDEs
of order α ∈ (3, 4] with two point boundary conditions of the type:
Dαx(t) = −H(t, x(t)),
Dα1x(1) = 0 = I3−αx(0) = I4−αx(0), x(1) =Γ(α− α1)
Γ(α)Iα−α1H(t, x(t))(1),
(0.0.1)
where t ∈ [0, 1], α1 ∈ (1, 2], the fractional derivatives Dα, Dα1 are in the sense of Riemann–Liouville
of ordersα, α1, H : C([0, 1]×R → R) and Iα−α1H(t, x(t))(1) = 1Γ(α−α1)
∫ 10 (1−s)
α−α1−1H(s, x(s))ds.
2
We study existence of positive solutions (EPS). The second class of problems of our interest is the class
of three-point BVPs of the type:
Dαx(t) + f(t, x(t), x′(t)) = 0, 1 < α ≤ 2, (0.0.2)
x(0) = 0, Dpx(1) = δDpx(η), 0 < p < 1,
where 0 < µ < p < 1, 0 < η ≤ 1, Dα is Caputo’s fractional derivative of order α ∈ (1, 2]. We develop
upper and lower solutions (EUS) method for the existence of solutions. The third class of problems is
the class of multi-point BVPs:
Dαx(t) = f(t, x(t),Dpx(t)) 2 < α < 3, 0 < p < 1, (0.0.3)
x(0) = 0, Dpx(1) =m−2∑i=1
ζiDpx(ηi), x′′(1) = 0,
where Dα, Dp are Caputo’s fractional derivative of order q, p respectively. The fourth class of problems
is the higher order p-laplacian class:
Dγ(ϕp(Dβx(t))) + a(t)f(x(t)) = 0, 3 < β, γ ≤ 4, t ∈ [0, 1],
x(0) = x′′′(0), ηDαx(t)|t=1 = x′(0), ξx′′(1)− x′′(0) = 0, 0 < α < 1,
ϕp(Dβx(t))|t=0 = 0 = (ϕp(Dβx(t)))′|t=0, (0.0.4)
(ϕp(Dβx(t)))′′|t=1 =1
2(ϕp(Dβx(t)))′′|t=0,
(ϕp(Dβx(t)))′′′|t=0 = 0,
where Dβ , Dγ and Dα are Caputo’s fractional derivative of fractional order β, γ, α, and β, γ ∈ (3, 4],
α ∈ (0, 1), ϕp(s) = |s|p−2s, p > 1, ϕ−1p = ϕq, 1
p + 1q = 1.
In Chapter 4, we study two classes of coupled systems for EUS. The first class is the coupled systems
of BVPs for fractional order integro-differential equations (FDIEs) of orders α, β ∈ (3, 4] of the form
Dαx(t) = −G1(t, Iαx(t), Iβy(t)),
Dβy(t) = −G2(t, Iαx(t), Iβy(t)),(0.0.5)
subject to the boundary conditions
Dδx(1) = 0 = I3−αx(0) = I4−αx(0), x(1) =Γ(α− δ)
Γ(α)Iα−δG1(t, Iαx(t), Iβy(t))(1),
Dνv(1) = 0 = I3−βv(0) = I4−βv(0), v(1) =Γ(β − ν)
Γ(β)Iβ−νG2(t, Iαx(t), Iβy(t))(1),
where t ∈ [0, 1], δ, ν ∈ (1, 2]. The functions G1, G2 : [0, 1] × R × R → R, satisfy the Caratheodory
conditions. The fractional derivatives Dα, Dβ, Dδ, Dν are in Riemann–Liouville sense and Iαx(t) =
3
∫ t0
(t−s)α−1x(s)Γ(α) ds, Iβy(t) =
∫ t0
(t−s)β−1y(s)Γ(β) ds. The second class we study the EUS for a coupled sys-
tems of fractional order q–difference equations for α, β ∈ (n− 1, n], k ≤ n− 1, of the type:
Dαq x(t) = −f(t,Dk
q y(t),Dp1q y(t)),
Dβq y(t) = −g(t,Dk
qx(t),Dp2q x(t)),
(0.0.6)
subject to the following initial and boundary conditions
x(1) = 0, Dqx(1) = 0, Diqx(0) = 0, i = 2, 3, ..., n− 1,
y(1) = 0, Dqy(1) = 0, Djqy(0) = 0, j = 2, 3, ..., n− 1,
for t ∈ [0, 1] and n ≥ 3. Dγq , Dβ
q , Dp1q , Dp2
q , are Caputo’s fractional q–difference operators of order
γ, β, p1, p2 where p1, p2 ∈ (0, 1) and Dkq , Di
q, Djq are q–difference operators of order k ≤ n−1, i =
2, 3, . . . , n− 1, j = 2, 3, . . . , n− 1, respectively and f, g ∈ C([0, 1]×R2, R).
In Chapter 5, we study two systems of problems for coupled hybrid systems of FDEs. The first
coupled hybrid system of problems is given below:
Dα(x(t)
H(t, x(t), y(t))) = −K1(t, x(t), y(t)), α ∈ (2, 3],
Dβ(y(t)
G(t, x(t), y(t))) = −K2(t, x(t), y(t)), β ∈ (2, 3],
x(t)
H(t, x(t), y(t))|t=1 = 0, Dµ(
x(t)
H(t, x(t), y(t)))|t=δ1 = 0, x(2)(0) = 0,
y(t)
G(t, x(t), y(t))|t=1 = 0, Dν(
y(t)
G(t, x(t), y(t)))|t=δ2 = 0, y(2)(0) = 0,
(0.0.7)
where δ1, δ2 ∈ (0, 1), α, β ∈ (2, 3], µ, ν ∈ (0, 1), t ∈ [0, 1] and Dα, Dβ , Dµ, Dν are Caputo’s frac-
tional derivative of orders α, β, µ, ν, respectively, K1,K2 ∈ C([0, 1] × R2,R), H,G ∈ C([0, 1] ×
R2,R− {0}). x, y ∈ C([0, 1],R). The second coupled system of problems is:
Dα
(x
H(t, x, y)
)= −T1(t, x, y), α ∈ (n− 1, n],
Dβ
(y
G(t, x, y)
)= −T2(t, x, y), β ∈ (n− 1, n], (0.0.8)(
x
H(t, x, y)
)(i)
|t=0 = 0 =
(x
H(t, x, y)
)(1)
|t=1,x
H(t, x, y)|z=δ1 = δ1Iα−1
1 T1(t, x, y),(y
G(t, x, y)
)(j)
|z=0 = 0 =
(y
G(t, x, y)
)(1)
|t=1,y
G(t, x, y)|t=δ2 = δ2Iβ−1
1 T2(t, x, y),
i, j = 2, 3, . . . , n− 1,
t ∈ [0, 1]. where Dα, Dβ are Caputo’s fractional derivative of orders α, β respectively, T1, T2 ∈
C([0, 1]× R2,R), H,G ∈ C([0, 1]× R2,R− {0}) and x, y ∈ C([0, 1],R).
4
In Chapter 5, we study three problems from LF calculus and obtain their exact analytical solutions.
We use some mathematical methods to solve the LF wave equation, LF Schrodinger’s equation and
Vehicular traffic flow model in Cantore sets.
In Chapter 6, we develop numerical schemes for solutions of the fractional Telegraph equation (FTE)
and fractional partial differential equations with the help of OMBPs and also provide lustrative examples.
In Chapter 7, we develop operational matrices for fractional order integration of Bernstein polyno-
mials and use them for the numerical solutions of fractional order Logistic equations, fractional order
Brusselator systems and optimal control problems. We develop numerical techniques and provide illus-
trative examples.
In Chapter 8, we use B–Spline functions for the numerical solutions of the fractional order Brus-
selator systems. We develop a numerical technique by the using operational matrix of fractional order
integration of B–Spline functions and provide illustrative examples.
5
Chapter 1
Preliminaries
For the study of existence and uniqueness of solutions (EUS) to boundary value problems for FDEs,
numerical solutions and exact solutions of local FDEs, we use different tools from calculus, functional
and numerical analysis. In the following, we provide some basic definitions and known results, which
are useful in our study.
1.1 Fractional calculus
In this section, we recall some basic definitions and fundamental results from fractional calculus which
we use in this thesis.
Definition 1.1.1. [63] If f(t) ∈ L1(a, b), the set of all integrable functions, and α > 0, then the
Riemann-Liouville fractional integral of order α is defined by
Iαf(t) =1
Γ(α)
∫ t
a(t− s)α−1f(s)ds (1.1.1)
provided the integral converges.
Definition 1.1.2. [63] For α > 0, the Caputo fractional derivative of order α is defined by
Dαf(t) =1
Γ(n− α)
∫ t
0(t− s)n−α−1Dnf(s)ds, (1.1.2)
where n ∈ N is such that n− 1 < α < n and D = dds .
From this definition, we have:
Dαt f(t) =
1
Γ(n−α)
∫ t0
Dnf(s)(t−s)1+α−nds, n− 1 < α < n, n ∈ IN,
⌈n
dtnf(t), α = n.
(1.1.3)
Lemma 1.1.3. [7] For α, β > 0, the following relations hold:
Dαtβ =Γ(1 + β)
Γ(1 + β − α)tβ−α−1, β > n and Dαtk = 0, k = 0, 1, ..., n− 1.
6
Lemma 1.1.4. [56] Assume that a, b ≥ 0 and H ∈ L1[p, q], then Ia0+I
b0+H(t) = Ia+b
0+H(t) =
Ib0+I
a0+H(t) and cDb
0+Ib0+H(t) = H(t), for all t ∈ [p, q].
The following properties of fractional derivatives and integrals are useful for our study and are
available in [1, 8, 9]:
(i) DαxC = 0, (C is a constant),
(ii) Dαxx
β =
0, β ∈ IN0, β < ⌈α⌉ ,
Γ(1+β)Γ(1+β−α)x
β−α, β ∈ IN0, β ≥ ⌈α⌉ or β /∈ IN0, β > ⌊α⌋ ,(1.1.4)
(iii) IαxDα
xf(x) = f(x)−n−1∑k=0
f (k)(0+)xk
k!, n− 1 < α ≤ n, (1.1.5)
where Iαx is the fractional Riemann-Liouville integral defined by
Iαx f(x) =
1
Γ(α)
∫ x
0
f(t)
(x− t)1−αdt, α > 0. (1.1.6)
Lemma 1.1.5. [7] Fort β ≥ α > 0 and H(t) ∈ L1[a, b], the following holds
DαIβa+H(t) = Iβ−α
a+ H(t)
on the interval [a, b].
Lemma 1.1.6. [7] For H(t) ∈ C(0, 1), and n−1 < α ≤ n, solution of homogenous FDEDα0+H(t) = 0
is given by
H(t) = k1 + k2t+ k3t2 + ...+ knt
n−1, ki ∈ R, i = 1, 2, 3, ..., n. (1.1.7)
In some of our results, we deal with coupled systems of fractional order q–difference equations. For
this purpose, we need some basic definitions and results from q–Fractional calculus.
1.1.1 q–Fractional calculus
The q-Gamma function for real number α, is given by
Γq(α) =(1− q)(α−1)
(1− q)α−1,
where q ∈ R and satisfy Γq(α+ 1) = [α]qΓq(α),
(α− β)(γ) = αγ∞∏n=0
α− βqn
α− βqγ+n,
where n ∈ N and γ ∈ R.
For a function T , the q-derivative is defined by
(DqT )(x) =T (x)− T (qx)
(1− q)y, (DqT )(0) = lim
x→0(DqT )(x), x = 0.
7
For higher order q-derivative, we have
Dnq T (x) = Dq(D
n−1q T (x)).
The q-integral of a function defined in [0,1], is given by
IqT (x) =
∫ x
0T (s)dqs = x(1− q)
∞∑n=0
T (xqn)qn, x ∈ [0, 1].
For a ∈ [0, b], the q-integral from a to b is define by∫ b
af(x)dqx =
∫ b
0f(x)dqx−
∫ a
0f(x)dqx.
The fractional q-derivative in Caputo’s sense is defined by
Dαq,af(x) = I⌈α⌉−α
q,a D⌈α⌉q f(x),
where ⌈α⌉ denotes the smallest integer greater or equal to α.
Lemma 1.1.7. [40] Let α > 0, such that α ∈ R+ \ N. Then
Iαq Dα
q x(t) = x(t)−m−1∑k=0
tkDkqx(0)
Γq(k + 1),
where m is the smallest integer greater than or equal to α.
In some of our work, we study exact analytical solutions of some problems in fractal space based
on the techniques of local fractional(LF) calculus. We provide below some of the basic definitions and
results from LF calculus.
1.1.2 Local fractional calculus
A function f(x) is said to be a LF continuous function if f(x) satisfies
|f(x)− f(x0)| < aγ , (1.1.8)
where γ ∈ (0, 1], |x− x0| < b, for a, b > 0 and a, b ∈ R.
The LF derivative of a function f(x) ∈ Cγ(a, b) of order γ is defined by
Dγf(x)
dxγ=
∆γ(f(x)− f(x0))
(x− x0)γ, (1.1.9)
where
∆γ(f(x)− f(x0)) = Γ(1 + γ)(f(x)− f(x0)), (1.1.10)
8
andDγ
dxγ
(xmγ
Γ(mγ + 1)
)=
x(m−1)γ
Γ((m− 1)γ + 1), m ∈ N. (1.1.11)
The local fractional integral of a function g(t) on [a, b] is defined by
I(γ)[a,b]f(x) =
1
Γ(γ + 1)
∫ b
ag(x)(dx)γ =
1
Γ(γ + 1)lim
∆x→0
i=M−1∑i=0
g(xi)(∆xi)γ , (1.1.12)
where [a, b] is divided into M − 1 sub-intervals (xi, xi+1) and ∆xi = xi+1 − xi, with a = x0, b = xM
and
I(γ)0,x
(xmγ
Γ(mγ + 1)
)=
x(m+1)γ
Γ((1 +m)γ + 1). (1.1.13)
The following results describing Mittage-Leffler function and its variants in fractal space are useful in
our study
Eγ(xγ) =
∞∑j=0
xjγ
Γ(jγ + 1),
sinhγ(xγ) =
∞∑j=0
x(2jγ+1)
Γ((2jγ + 1)γ + 1),
coshγ(xγ) =
∞∑j=0
x2jγ
Γ(2jγγ + 1),
Dγ
dxγsinhγ(x
γ) = coshγ(xγ),
Dγ
dxγcoshγ(x
γ) = − sinhγ(xγ).
1.1.3 Local fractional Laplace Transform (LFLT)
In this subsection, we provide definition and some basic results related to the Yang-Laplace transforms
(LFLT). For a function f(t) satisfying the following inequality:
1
Γ(γ + 1)
∫ ∞
0|f(x)|(dx)γ < m <∞, (1.1.14)
the LFLT is defined by
Lγ{f(x)} =1
Γ(γ + 1)
∫ ∞
0Eγ(−sγxγ)f(x)(dx)γ , (1.1.15)
where γ ∈ (0, 1] and sγ = βγ + iγαγ . Here, iγ is the fractal imaginary unit and Re(sγ) = βγ > 0. We
use the following properties of the LFLT in our study:
Lγ{ah(x) + bg(x)} = aLγ{h(x)}+ bLγ{g(x)}, (1.1.16)
Lγ{h(t, x)}(mγ) = smγLγ{h(t, x)} − s(m−1)γLγh(0, x)− s(m−2)γh(0, x)(γ)
− s(m−3)γh(0, x)(2γ) . . .− h(0, x)((m−1)γ), (1.1.17)
9
Lγ{xmγ} =Γ(mγ + 1)
s(m+1)γ, (1.1.18)
coshγ(xγ) =
∞∑i=0
t2jγ
Γ(2jγ + 1). (1.1.19)
1.2 Numerical Analysis
In this section, we introduce two types of polynomial approximations, namely, Bernstein polynomials
and Linear Cardinal B-Spline functions, for numerical solutions of fractional order differential equa-
tions.
1.2.1 Bernstein polynomials
The Bernstein polynomials (BPs) [77] of degree m on the interval [0, 1] are defined by
Hi,m =
(m
i
)xi(1− x)m−i, 0 ≤ i ≤ m. (1.2.1)
These polynomials satisfy the following recursive relation:
Hi,m(x) = (1− x)Hi,m−1(x) + xHi−1,m−1(x), i = 0, 1 . . . ,m. (1.2.2)
Using the binomial expansion of (1− x)m−i, the BPs can be written in terms of a linear combination of
the basis functions as follows
Hi,m(x) =
(m
i
)xi(1− x)m−i =
(m
i
)xi
(m−i∑k=0
(−1)k(m− i
k
)xk
)
=
m−i∑k=0
(−1)k(m
i
)(m− i
k
)xi+k, i = 0, 1, . . . ,m. (1.2.3)
Let us denote ϕ(x) = [H0,m(x),H1,m(x), . . . ,Hm,m(x)], then we have ϕ(x) = ATm(x), where
Tm(x) = [1, x, x2, . . . , xm]T and A is (m+ 1)× (m+ 1) matrix defined by
A =
(−1)0(m0
)(−1)1
(m0
)(m−01−0
). . . (−1)m−0
(m0
)(m−0m−0
). . .
......
...
0 (−1)i(mi
). . . (−1)m−i
(mi
)(m−im−i
)...
. . . . . ....
0 . . . 0 (−1)m(mm
)
,
since det(A) = |A| = Πmi=0
(mi
)= 0, it means that A is an invertible matrix.
10
1.2.2 Approximation of a function by Bernstein polynomials
Any squared integrable function F(x) on the interval (0, 1) be approximated in terms of the basis func-
tions {H0,m(x),H1,m(x), . . . ,Hm,m(x)} of BPs (cf. [82]) as follows
F(x) ≈m∑i=0
ciHi,m(x) = CTϕ(x), (1.2.4)
where CT = [c0, c1, . . . , cm], χT (x) = [H0,m(x),H1,m(x), . . . ,Hm,m(x)]. The vector CT can be
obtained from (1.2.4) by using the following relation
CT ⟨ϕ(x), ϕ(x)⟩ = ⟨F(x), ϕ(x)⟩, (1.2.5)
where
⟨F(x), ϕ(x)⟩ =∫ 1
0F(x)ϕ(x)Tdx = [⟨F(x),H0,m⟩, ⟨F(x),H1,m(x)⟩, . . . , ⟨F(x),Hm,m(x)⟩], (1.2.6)
and ⟨ϕ(x), ϕ(x)⟩ = Q is known by dual matrix of ϕ(x) and is given by the following relation
Q = ⟨ϕ(x), ϕ(x)⟩ =∫ 1
0ϕ(x)ϕ(x)Tdx. (1.2.7)
The matrix Q is an invertible symmetric matrix of order (m+ 1)× (m+ 1) whose entries are given by
the relation
qi+1,j+1 =
∫ 1
0Hi,m(x)Hj,m(x)dx
=
(m
i
)(m
j
)∫ 1
0(1− x)2m−(i+j)xi+jdx
=
(mi
)(mj
)(2m+ 1)
(2mi+j
) i, j = 0, 1, . . . ,m.
From (1.2.5) and (1.2.7), it follows that
CT =
(∫ 1
0f(x)ϕ(x)Tdx
)Q−1. (1.2.8)
Similarly, we can approximate a function F(x, t) ∈ L2([0, 1]× [0, 1]) by BPs as follow
F(x, t) ≈m∑j=0
m∑i=0
CijBi,m(x)Hj,m(t)
=m∑j=0
m∑i=0
Hi,m(x)CijHj,m(t) = ϕT (x)C�ϕ(t),
(1.2.9)
where C� is a square matrix of (m+ 1)× (m+ 1), given by
11
C� =
c00 c01 . . . c0m
c1,0 c11 . . . c1m...
. . . . . ....
cm0 cm1 . . . cmm
,
where C� can be obtained from (1.2.9) by the relation
C� = Q−1
(∫ 1
0
∫ 1
0F(x, t)ϕ(x)ϕT (t)dxdt
)Q−1. (1.2.10)
1.2.3 Linear Cardinal B-Spline function
For a given scaling function in L2(R) the scaling function can be utilized to expand it. We know that the
scaling functions are defined R, therefore they can not be inside of the domain of the investigated issue.
To bypass this aspect, in this desertations we consider the B-spline scaling functions on [0, 1] [99, 129]:
Nm(x) =1
(m− 1)!
m∑k=0
(m
k
)(−1)k(x− k)m−1
+ , (1.2.11)
denotes the cardinal B-spline function of orderm ≥ 2 (degree m-1) for the knot sequence {. . . ,−1, 0, 1, . . .}
and sup[Nm(x)] = [0,m]. Also let N1(x) = χ[0,1](x).
Boor et al. [99] define explicit expression of N2(x) in the following form:
N2(x) =
2∑k=0
(2
k
)(−1)k(x− k)+ =
x, x ∈ [0, 1),
2− x, x ∈ [1, 2),
0, otherwise.
Suppose Nj,k(x) = N2(2jx − k), j, k ∈ Z and Bj,k = sup[Nj,k] = close {x : Nj,k = 0}. By
inspection we have that
Bj,k = [2−jk, 2−j(2 + k)], j, k ∈ Z. (1.2.12)
The set of indices is:
Cj = {k : Bj,k ∩ (0, 1) = 0} , j ∈ Z. (1.2.13)
In view of (1.2.13), we have min {Cj} = −1 and max {Cj} = 2j − 1, j ∈ Z. The support of Nj,k(x)
can be outside of [0, 1],therefore we have to define Nj,k(x) on [0, 1]. Thus, we conclude that
ϕj,k = Nj,k(x)χ[0,1](x), j ∈ Z, k ∈ Cj . (1.2.14)
12
As a result we have
ϕj,k(x) =2∑
i=0
(2
i
)(−1)i(2jx− (k + i))+ =
2jx− k, k
2j≤ x < k+1
2j,
2− (2jx− k), k+12j
≤ x < k+22j,
0, otherwise.
1.2.4 Function approximations
A function f(x) ∈ [0, 1] can be approximated by linear B-spline scaling functions for a fixed j = J , as
fJ(x) ≈2j−1∑k=−1
ckϕJ,k(x) = CTϕJ(x), (1.2.15)
where C and ϕJ(x) are (2j + 1) column vectors given by
C = [c−1, c0, . . . , c2j−1]T , (1.2.16)
ϕJ(x) = [ϕJ,−1(x), ϕJ,0(x), . . . , ϕJ,2j−1(x)]T . (1.2.17)
Assume that eJ(x) is the error of the approximation of f(x) by linear B-spline scaling functions then
|eJ(x)| = ∥f(x)− fJ(x)∥ = O(2−2J) (1.2.18)
For further details, See [100]. Then, ck can be obtained by
ck =
∫ 1
0f(x)ϕJ,k(x)dx, k = −1, . . . , 2j − 1, (1.2.19)
where ϕJ,k are the vector of dual basis of ϕJ . By using the linear combinations of ϕJ,k, the ϕJ,k are
obtained as:
ϕJ,k = P−1ϕJ , (1.2.20)
where
P =
∫ 1
0ϕJ(x)ϕ
TJ (x)dt =
1
2J−2
112
124
124
16
124
. . . . . . . . .
124
16
124
124
112
,
where P is a symmetric (2J + 1)× (2J + 1) matrix. Replacing (1.2.20) in (1.2.19) we get
cTk =
(∫ 1
0f(x)ϕTJ (x)dx
)P−1. (1.2.21)
13
In most of the results in this thesis, we investigate sufficient conditions for existence and uniqueness of
solutions to boundary value problems for fractional order differential equations. We use the results from
fixed point theory.
1.3 Fixed point theory
In this section, we provide some definitions and results from fixed point theory which we use in this
thesis.
Theorem 1.3.1. [31](The Arzela Ascoli Theorem) Let F be an equicontinuous, uniformly bounded fam-
ily of real valued functions f on an interval I (finite or infinite).Then F contains a uniformly convergent
sequence of function in f , converging to a function f∈C(I) where C(I) denotes the space of all con-
tinuous bounded functions on I . Thus, any sequence in F contains a uniformly bounded Convergent
subsequence on I and consequently F have a compact closure in C(I).
Definition 1.3.2. Let X be a metric space, and f : X → X , f is said to be a contraction if there exist
some constant 0 < k < 1, such that
d(f(x), f(y)) ≤ kd(x, y), ∀x, y ∈ X.
The infimum of such k’s is called the contraction coefficient.
Theorem 1.3.3. (Banach fixed point theorem) LetX be a complete metric space, and f be a contraction
on X . Then, there exists a unique fixed point of f , that is, f(x∗) = x∗.
Proof. For the existence of x∗, such that f(x∗) = x∗, assume some x0 ∈ X and define a recursive
sequence as xn+1 = f(xn), then the seqence {xn}∞n=1, is Cauchy. For this, assume that k be the
constructive coefficient of f , then:
d(x2, x1) = d(f(x1), f(x0)) ≤ kd(x1, x0),
d(x3, x2) = d(f(x2), f(x1)) ≤ kd(x2, x1) ≤ k2d(x1, x0),
......
d(xn+1, xn) ≤ knd(x1, x0),
Thus, for any p, q > p+ 1, we have:
d(xq, xp) ≤ d(xq, xq−1) + d(xq−1, xq−2) + . . .+ d(xp+1, xp) (1.3.1)
≤ (kq−1 + kq−2 + . . .+ xp)d(x1, x0) = kp(1− kq−1)
1− kd(x1, x0), (1.3.2)
as q → ∞, we have Kq−1 → 0. And we get d(xq, xp) ≤ kp
1−kd(x1, x0), this implies that the sequence is
a Cauchy sequence. Since, the metric spaceX is complete therefore, the sequence {xn}∞n will converge
14
to a point say z∗ in X . Thus, for any ϵ > 0, there is N sucn that d(z∗, xn) < ϵ2 ∀n > N . Hence,
d(f(z∗), z∗) ≤ d(f(z∗), (xn+1)) + d(z∗, xn+1) = (f(z∗), f(xn)) + d(z∗, xn+1) (1.3.3)
≤ kd(z∗, xn) + d(z∗, xn+1) <ϵ
2+ϵ
2= ϵ.
And this is valid for any ϵ > 0, that is, it must be d(f(z∗), z∗) = 0, this implies that f(z∗) = z∗. Thus,
z∗ is a fixed point of f .
Now to prove the uniqueness of the fixed point, assume that z is another fixed point of the f , then
we can have d(f(z), f(z∗)) = d(z, z∗), which is a contradiction, since d(f(z), f(z∗)) < kd(z, z∗) for
k < 1.
Lemma 1.3.4. [62] (Leray–Schauder Alternative) Let F : E → E be a completely continuous operator
(i.e., a map that is restricted to any bounded set in B is compact ). Let Y = {x ∈ E : x = µFx for some
0 < µ < 1}. Then either the set Y is unbounded or F has at least one fixed point.
Theorem 1.3.5. [31](Shauder-Tychnoff Fixed Point Theorem) Let X be a locally convex, topological
vector space. Let B be a compact, convex subset of X and T a continuous map of B into itself. Then Thas a fixed point x ∈ B.
Theorem 1.3.6. (Krasnoselskii Theorem) [32, 68, 69] Let E be a Banach space and B ⊂ E be a cone.
Assume that Q1, Q2 are open sets contained in E such that 0 ∈ Q1 and Q1 ⊂ Q2. Assume, further, that
F : B ∩ (Q2 \ Q1) → B is a completely continuous operator. If either
• (A1) ∥Fv∥ ≤ ∥v∥ for v ∈ B ∩ ∂Q1 and ∥Fv∥ ≥ ∥v∥ for v ∈ B ∩ ∂Q2, or
• (A2) ∥Fv∥ ≥ ∥v∥ for v ∈ B ∩ ∂Q1 and ∥Fv∥ ≤ ∥v∥ for v ∈ B ∩ ∂Q2,
then F has at least one fixed point in B ∩ (Q2 \ Q1).
Theorem 1.3.7. [63, 104] ( Dhage’s Theorem) Let Xa(0) and Xa(0) be open and closed balls in a
Banach algebra X centered at the origin 0 of radius a, for some real number a, and let A, B : Xa(0) →X be two operators satisfying the following:
• (1) A is Lipschitz with Lipschitz constant λ;
• (2) B is continuous and compact;
• (3) λM < 1, where M = ∥B(Xa(0))∥ = sup{∥B(x)∥ : x ∈ Xa(0)}.
Then, either
• (a) The equation A(x)B(x) = x has a solution in Xa(0), or
• (b) there is an element x ∈ X such that ∥x∥ = a satisfying µAxBx = x, for some 0 < µ < 1.
15
Definition 1.3.8. [63] Let X be a Banach space. A mapping F : X → X is called D– Lipschitzian if
there exist a continuous and nondecreasing function ψ : R→ R such that ∥Fx−Fy∥ ≤ ψ∥x− y∥ for
all x, y ∈ X , where ψ(0) = 0. If ψ is not necessarily nondecreasing and satisfies ψ(a) < a, for a > 0,
the mapping F is called a nonlinear contraction with a contraction function ψ.
Definition 1.3.9. [126] Let B be a solid cone in a real Banach space E , T : B0 → B0 be an operator and
0 < α < 1. Then T is called θ-concave operator if T (ku) ≥ kαT (x) for any 0 < k < 1 and x ∈ B0.
Lemma 1.3.10. [126] Assume that B is a normal solid cone in a real Banach space X , 0 < α < 1 and
T : P 0 → B0 is a α–concave Increasing operator. Then T has only one fixed point in B0.
16
Chapter 2
Existence and uniqueness of solutions forFDEs
Fractional calculus (FC) has recently attracted the attentions of researchers working in various scientific
fields. It is due to the applications of FC in mathematical modeling of various phenomena in many
scientific fields such as chemistry, physics, biophysics, biology, control theory, aerodynamics, physics,
blood flow problems, the fluid-dynamics, polymer rheology, traffic model, variation in thermodynamics,
economics, etc. [1, 2]. Fractional differential equations (FDEs) are capable of describing memory and
hereditary properties of certain important materials and processes. This area of research contributed in
the rapid popularity and development of the subject in the last few years. Theoretical aspects of the
FDEs such as existence, uniqueness, multiplicity, asymptotic behavior and periodicity of solutions have
recently been studied by many researchers, we refer to [11–13, 15, 16, 30] and the references therein.
In this chapter, we discuss existence and uniqueness of solutions to some important classes of
boundary value problems for FDEs. Most of the results of this chapter are published in international
journals [130, 132, 137, 138].
In section 2.1, we study existence of solutions to a class of two point BVPs for higher order FDEs
of order α ∈ (3, 4]. In section 2.2, we study a a general class of three point BVPs for FDEs where the
nonlinearity explicitly depends on the derivative and develop conditions for the EUS. In section 2.3, we
study a class of multi point BVPs for FDEs and develop sufficient conditions for the EUS. In section
2.4, we generalize the results to more general class p–Laplacian for FDEs and develop conditions for
the EUS.
17
2.1 Two point BVPs for FDEs
In this section, we study existence of positive solutions to a class of two point boundary value problems
for FDEs of order α ∈ (3, 4] of the form
Dαx(t) = −H(t, x(t)), t ∈ J = [0, 1]
Dα1x(1) = 0 = I3−αx(0) = I4−αx(0), x(1) =Γ(α− α1)
Γ(α)Iα−α1H(t, x(t))(1),
(2.1.1)
where α1 ∈ (1, 2], the fractional derivatives Dα, Dα1 are in Riemann–Liouville sense of orders α, α1,
respectively. The nonlinear function is assumed to be continuous, that is, H : C([0, 1]×R → R) and the
operator in the boundary conditions is given by Iα−α1H(t, x(t))(1) = 1Γ(α−α1)
∫ 10 (1−s)
α−α1−1H(s, x(s))ds.
Lemma 2.1.1. For s, t ∈ [0, 1], solutions of (2.1.1) are solutions of the following integral equation
x(t) =
∫ 1
0G(t, s)H(s, x(s))ds, (2.1.2)
where G(t, s) is Green’s function and is given by
G(t, s) =
−(t−s)α−1
Γν1+ tα−1(1−s)α−α1−1
Γ(α) + tα−2(1−s)α−1
Γ(α) , t ≥ s,
tα−1(1−s)α−α1−1
Γ(α) + tα−2(1−s)α−1
Γ(α) , t ≤ s.(2.1.3)
Proof. Applying the operator Iα0 on (2.1.1) and using lemma 1.1.6, we get the problem (2.1.1) in the
following equivalent integral form
x(t) = −IαH(t, x(t)) + c1tα−1 + c2t
α−2 + c3tα−3 + c4t
α−4. (2.1.4)
The boundary conditions I3−αx(0) = I4−αx(0) = 0 yield c3 = c4 = 0 and the boundary conditions
Dα1x(1) = 0 and x(1) = Γ(α−α1)Γ(α) Iα−α1H(t, x(t))(1) yield
c1 =
∫ 1
0
(1− s)α−α1−1H(s, x(s))ds
Γ(α)
c2 =
∫ 10 (1− s)α−1H(s, x(s))ds
Γ(α).
(2.1.5)
Hence, it follows that
x(t) =−∫ t
0
(t− s)α−1
Γ(α)H(s, x(s))ds+ tα−1
∫ 1
0
(1− s)α−α1−1
Γ(α)H(s, x(s))ds
+ tα−2
∫ 1
0
(1− s)α−1
Γ(α)H(s, x(s))ds =
∫ 1
0K(t, s)H(s, x(s))ds,
(2.1.6)
where G(t, s) is the Green’s functions which is given by (2.1.3).
Lemma 2.1.2. The Green’s function G(t, s) given by (2.1.3) satisfies the following properties
(A1) G(t, s) is continuous and G(t, s) ≥ 0 for each t, s ∈ J ;
(A2) maxt∈J G(t, s) = G(1, s) for each s ∈ J ;
(A3) mint∈[ 13,1]G(t, s) ≥ λ0G(1, s) for some λ0 ∈ (0, 1).
18
Proof. The continuity of the Green’s function G(t, s) is obvious from the definition in (2.1.3). For
t, s ∈ J such that t ≥ s, since s ≤ st implies that −(1 − s) ≤ −(1 − s
t ) and α − α1 − 1 < α − 1
implies that (1− s)α−α1−1 > (1− s)α−1. Hence, it follows that
G(t, s) =−(t− s)α−1
Γ(α)+tα−1(1− s)α−α1−1
Γ(α)+tα−2(1− s)α−1
Γ(α)
= −(1− s
t )α−1
Γ(α)tα−1 +
(1− s)α−α1−1
Γ(α)tα−1 +
(1− s)α−1
Γ(α)tα−2
≥ −(1− s)α−1
Γ(α)tα−1 +
(1− s)α−α1−1
Γ(α)tα−1 +
(1− s)α−1
Γ(α)tα−2
= ((1− s)α−α1−1 − (1− s)α−1)tα−1
Γ(α)+tα−2(1− s)α−1
Γ(α)≥ 0.
(2.1.7)
From (2.1.3), for t ≤ s it is obvious that G(t, s) ≥ 0. For (A2), we consider s, t ∈ J , such that t ≥ s.
For α ∈ (3, 4], α1 ∈ (1, 2], we have α−α1 − 1 ≤ α− 2, this implies that (1− s)α−α1−1 ≥ (1− s)α−2
and
∂
∂tG(t, s) =
−(α− 1)(t− s)α−2
Γ(α)+
(α− 1)tα−2(1− s)α−α1−1
Γ(α)+
(α− 2)tα−3(1− s)α−1
Γ(α)
= (α− 1)
[−(t− s)α−2 + tα−2(1− s)α−α1−1
Γ(α)
]+
(α− 2)tα−3(1− s)α−1
Γ(α)
= (α− 1)
[−(1− st )
α−2 + (1− s)α−α1−1
Γ(α)
]tα−2 +
(α− 2)(1− s)α−1
Γ(α)tα−3
≥ (α− 1)
[−(1− s)α−2 + (1− s)α−α1−1
Γ(α)
]tα−2 +
(α− 2)(1− s)α−1
Γ(α)tα−3 ≥ 0.
(2.1.8)
For the proof of (A3), utilizing (A1), (A2) in the following calculations:
mint∈[ 1
3,1]G(t, s) =
G(13 , s), s ∈ (0, 13 ],
G(13 , s), s ∈ [13 , 1],
=
−( 1
3−s)α−1+( 1
3)α−1(1−s)α−α1−1+( 1
3)α−2(1−s)α−1
Γ(α) , t ≥ s,
( 13)α−1(1−s)α−α1−1+( 1
3)α−2(1−s)α−1
Γ(α) , t ≤ s.(2.1.9)
For t ≥ s, we have
−(13 − s)α−1 + (13)α−1(1− s)α−α1−1 + (13)
α−2(1− s)α−1
−(1− s)α−1 + (1− s)α−α1−1 + (1− s)α−1
=−(13 − s)α−1 + (13)
α−2(1− s)α−α1−1[13 + (1− s)α
](1− s)α−α1−1
≥−(13 − 1
3)α−1 + (13)
α−2(1− 13)
α−α1−1[13 + (1− 1
3)α1]
(1− 13)
α−α1−1
= (1
3)α−2
[1
3+ (
2
3)α1
].
(2.1.10)
19
For t ≤ s, we deduce that
(13)α−1(1− s)α−α1−1 + (13)
α−2(1− s)α−1
(1− s)α−α1−1 + (1− s)α−1
=(13)
α−2(1− s)α−α1−1[13 + (1− s)α1
](1− s)α−α1−1 [1 + (1− s)α1 ]
≥(13)
α−2[13 + (1− 1
3)α1][
1 + (1− 13)
α1] =
(13)α−2
[13 + (23)
α1][
1 + (23)α1] .
(2.1.11)
Choose
λ0 = min{(1
3)α−2
[13+ (
2
3)α1],(13)
α−2[13 + (23)
α1][
1 + (23)α1] }
. (2.1.12)
Therefore, from (2.1.10), (2.1.11) and (2.1.12), we have λ0 ∈ (0, 1) such that
mint∈[ 1
3, 1]G(t, s) ≥ λ0max
t∈JG(t, s) = λ0G(1, s). (2.1.13)
2.1.1 Existence of solutions
In this section, we study existence of positive solutions to the BVPs (2.1.1) by using Krasnosel’ski1
fixed point theorem.
Consider the Banach space E = {x(t) : x(t) ∈ C(J )}, endowed with the norm ∥x(t)∥ =
maxt∈J |x(t)|. Define an operator T : E → E by:
T x(t) =−∫ t
0
(t− s)α−1
Γ(α)H(s, x(s))ds+ tα−1
∫ 1
0
(1− s)α−α1−1
Γ(α)H(s, x(s))ds
+ tα−2
∫ 1
0
(1− s)α−1
Γ(α)H(s, x(s))ds =
∫ 1
0K(t, s)H(s, x(s))ds.
(2.1.14)
In view of Lemma (2.1.1), solutions of the BVPs (2.1.1) are fixed points of T . Assume that the following
conditions hold:
(A1) There exist a real number k1 > 0 such that H(t, s) ≤ ξk1, when ever 0 ≤ x ≤ k1;
(A2) There exist a real number k2 > 0 such that H(t, s) ≥ νk2, when ever λ0k2 ≤ x ≤ k2, where λ0 is
constant defined by (2.1.12).
Use the notations ξ =[∫ 1
0 G(1, s)ds]−1
and ν =[∫ 1
13G(23 , s)ds
]−1.
Theorem 2.1.3. Assume that H(t, s) ≥ 0 and is continuous, then under the conditions (A1) and (A2),
the BVP (2.1.1) has at least one positive solution.
Proof. The continuity of the the operator T follows from the continuity of the Green’s function G(t, s)
and and that of H(t, x(t)) ∈ C(J × R → R). Define S = {x(t) ∈ E : ∥x∥ ≤ ∆} where ∆ =
20
maxt∈J H(t, s) + 1, then for any x(t) ∈ S, we have
|T x(t)| = | −∫ t
0
(t− s)α−1
Γ(α)H(s, x(s))ds+ tα−1
∫ 1
0
(1− s)α−α1−1
Γ(α)H(s, x(s))ds
+ tα−2
∫ 1
0
(1− s)α−1
Γ(α)H(s, x(s))ds|
≤ ∆
[(t− s)α
Γ(1 + α)|0t + tα−1 (1− s)α−α1
(α− α1)Γ(α)|01 + tα−2 (1− s)α
Γ(1 + α)|01]
= ∆
[tα
Γ(1 + α)+
tα−1
(α− α1)Γ(α)+
tα−2
Γ(1 + α)
]≤ ∆
[2
Γ(1 + α)+
1
(α− α1)Γ(α)
]<∞,
(2.1.15)
which implies that the operator T is bounded. Next, for x(t) ∈ S, t1, t2 ∈ J such that t2 > t1, we
have
|T x(t2)− T x(t1)|
= | −∫ t2
0
(t2 − s)α−1
Γ(ν1)H(s, x(s))ds+
∫ t1
0
(t1 − s)α−1
Γ(ν1)H(s, x(s))ds
+ (tα−12 − tα−1
1 )
∫ 1
0
(1− s)α−α1−1
Γ(α)H(s, x(s))ds
+ (tα−22 − tα−2
1 )
∫ 1
0
(1− s)α−1
Γ(α)H(s, x(s))ds|
≤ ∆
[tα2 − tα1Γ(1 + α)
+tα−12 − tα−1
1
(α− α1)Γ(α)+tα−22 − tα−2
1
Γ(1 + α)
].
(2.1.16)
From the definition of the norm and (2.1.16), we have: ∥T x(t2)− T x(t1)∥ → 0 as t1 → t2, which im-
plies that T is equicontinuous and by Arzela–Ascoli theorem, the operator T is completely continuous.
Consider a cone B = {x(t) ∈ E : x(t) ≥ 0 and mint∈[ 13,1] x(t) ≥ λ0∥x∥} in E , then for any x ∈ B,
we have that
mint∈[ 1
3,1](T x)(t) ≥ λ0
∫ 1
0G(1, s)H(s, x(s))ds = λ0max
t∈J
∫ 1
0G(t, s)H(s, x(s))ds = λ0∥T x∥,
(2.1.17)
this implies that T x(t) ∈ B. Let Q∞ = {x ∈ B : ∥x∥ > k1}, then for any x ∈ ∂Q1 implies that
∥x∥ = k1, so (A1) is satisfied for all x ∈ ∂Q1. For x ∈ B ∩ ∂Q1, we obtain
∥T x∥ = maxt∈J
∫ 1
0G(t, s)H(s, x(s))ds ≤ ξk1
∫ 1
0G(t, s)ds = k1, (2.1.18)
by (2.1.18), we get that ∥T x∥ ≤ ∥x∥ for x ∈ B ∩ ∂Q1. Assume Q2 = {x ∈ B : ∥x∥ < k2}, for
x ∈ ∂Q2, we have ∥x∥ = k2, this implies that the condition (C2) is satisfied for x ∈ B ∩ ∂Q2, further,
we have that
(T x)(23) =
∫ 1
0K(
2
3, s)H(s, x(s))ds ≥
∫ 1
13
G(2
3, s)H(s, x(s))ds ≥ νk2
∫ 1
13
G(2
3, s)ds = k2.
(2.1.19)
21
Thus, (2.1.19) implies ∥T x∥ ≥ ∥x∥ for x ∈ B ∩ ∂Q2. Therefore, by the help of lemma 1.3.6, the
operator T has a fixed point say x0 such that k1 ≤ ∥x0∥ ≤ k2.
Example 2.1.4. Consider the following BVP
Dαx(t) =
(∫ 1
13
G(2
3, s)ds
)−1(∫ 1
0G(1, s)ds
)−1 1 + 2maxx∈(0,1] |x(t)|2
, (2.1.20)
with the boundary conditions defined in (2.1.1).
For x ∈ Q2, we have
H(t, x(t)) ≥
(∫ 1
13
G(2
3, s)ds
)−1(∫ 1
0G(1, s)ds
)−1
(1 + 2x(t)
2) ≥ νx(t),
where ν =(∫ 1
13G(23 , s)ds
)−1and for x ∈ B ∩ ∂Q1, we obtain
H(t, x(t)) =
(∫ 1
13
G(2
3, s)ds
)−1(∫ 1
0G(1, s)ds
)−1
(1 + 2maxx∈(0,1] |x(t)|
2) ≤ ξk1,
for k1 =(∫ 1
13G(23 , s)ds
)−1(1+3maxt∈(0,1] |x(t)|
2 ). Therefore, the assumptions (A1) and (A2) are satis-
fied and hence by Theorem 2.1.3, the problem (2.1.20) has a solution.
Now in the following sections, we generalize the results to a class of three point boundary value
problems with a more general nonlinearity in the sense that the nonlinearity explicitly depends of the
first derivative.
2.2 Three point BVPs for FDEs
This section is devoted to the study of existence and uniqueness of solutions for three point BVPs for
FDEs of the type:
Dαx(t) + f(t, x(t), x′(t)) = 0, 1 < α ≤ 2,
x(0) = 0, Dpx(1) = δDpx(η), 0 < p < 1,(2.2.1)
where 0 < µ < p < 1, 0 < η ≤ 1, and Dα is Caputo’s fractional derivative of order α. Our results
are based on some classical results from fixed point theory. We impose some growth and continuity
conditions on f(t, x, x′). Let us use the following notations:
∆ =1− δη1−p
Γ(2− p), G1(t, s) =
−1
Γ(α)(t− s)α−1,
G2(t, s) =−δt
∆Γ(α− p)(η − s)α−p−1, G3(t, s) =
t
∆Γ(α− p)(1− s)α−p−1.
22
Lemma 2.2.1. For h(t) ∈ C[0, 1], 0 < δ < p < 1, and 0 < η ≤ 1, the BVP for fractional differential
equation
Dαx(t) + h(t) = 0, 1 < α ≤ 2,
x(0) = 0, Dpx(1) = δDpx(η), 0 < p < 1,(2.2.2)
has a solution of the form
x(t) =
∫ 1
0G(t, s)h(s)ds, (2.2.3)
where
G(t, s) =
G1(t, s) +G2(t, s) +G3(t, s) 0 ≤ s ≤ t ≤ η ≤ 1 ,
G2(t, s) +G3(t, s) 0 ≤ t ≤ s ≤ η ≤ 1 ,
G3(t, s) 0 ≤ t ≤ η ≤ s ≤ 1,
G1(t, s) +G2(t, s) +G3(t, s) 0 ≤ s ≤ η ≤ t ≤ 1 ,
G1(t, s) +G3(t, s) 0 ≤ η ≤ s ≤ t ≤ 1 ,
G3(t, s) 0 ≤ η ≤ t ≤ s ≤ 1 .
Proof. Applying the operator Iα on (2.2.2) and using lemma 1.1.6, we obtain
x(t) = −Iαh(t) + c1 + c2t. (2.2.4)
The boundary condition x(0) = 0 implies c1 = 0, thus we have
x(t) = −Iαh(t) + c2t, which implies Dpx(t) = −Iα−ph(t) + c2t1−p
Γ(2− p).
The boundary condition Dpx(1) = δDpx(η) yields c2 = 1∆(−δIα−ph(η) + Iα−ph(1)). Hence, (2.2.4)
takes the form
x(t) = −Iαh(t) +t
∆(−δIα−ph(η) + Iα−ph(1)), (2.2.5)
which can be rewritten as
x(t) =−1
Γ(α)
∫ t
0(t− s)α−1h(s)ds+
t
∆Γ(α− p)(−δ
∫ η
0(η − s)α−p−1h(s)ds
+
∫ 1
0(1− s)α−p−1h(s)ds) =
∫ 1
0G(t, s)h(s)ds.
Now, we study existence and uniqueness of solutions of the BVP (2.2.1)
23
2.2.1 Existence and uniqueness
Consider the space E = {x(t) ∈ C[0, 1] : x′(t) ∈ C[0, 1]} with the norm defined by ∥x∥ = maxt∈[0,1] |x(t)|+
maxt∈[0,1] |x′(t)|, E is a Banach space [23]. Define an operator T : E → E by:
T x(t) = −∫ t
0
(t− s)α−1
Γ(α)f(s, x(s), x′(s))ds
+t
∆Γ(α− p)(−δ
∫ η
0(η − s)α−p−1f(s, x(s), x′(s))ds+
∫ 1
0(1− s)α−p−1f(s, x(s), x′(s))ds)
(2.2.6)
In view of Lemma 2.2.1, solutions of the problem (2.2.1) are fixed points of the operator T . By conti-
nuity of f : [0, 1] × R × R → R and G on [0, 1] × [0, 1], the operator T is continuous and in view of
(2.2.6), we have
(T u)′(t) = −∫ t
0
(t− s)α−2
Γ(α− 1)f(s, x(s), x′(s))ds
+1
∆Γ(α− p)(−δ
∫ η
0(η − s)α−p−1f(s, x(s), x′(s))ds+
∫ 1
0(1− s)α−p−1f(s, x(s), x′(s))ds).
(2.2.7)
Assume that the following growth conditions hold:
(A1) There exists a nonnegative function m(t) ∈ L1([0, 1]) such that
|f(t, x(t), x′(t))| ≤ m(t) + a1|x(t)|λ1 + a2|x′(t)|λ2 ,
where a1, a2 ∈ R are nonnegative constants and 0 < λ1, λ1 < 1.
(A2) There exists a nonnegative function m(t) ∈ L1([0, 1]) such that
|f(t, x(t), x′(t)))| ≤ m(t) + a1|x(t)|λ1 + a2|x′(t)|λ2 ,
where a1, a2 ∈ R are nonnegative constants and λ1, λ1 > 1.
(A3) There exist a constant k > 0 such that
|f(t, x(t), x′(t)))− f(t, y(t), y′(t)))| ≤ k(|x(t)− y(t)|+ |x′(t)− y′(t)|),
for each t ∈ [0, 1] and all x, y, x′, y′ ∈ R .
24
For convenience in calculations, the following notations are introduced:
h1(t) =
∫ 1
0G(t, s)m(s)ds+
1
Γ(α− 1)
∫ t
0(t− s)α−2m(s)ds+
1
∆(
δ
Γ(α− p)
∫ η
0(η − s)α−p−1m(s)ds+
1
Γ(α− p)
∫ 1
0(1− s)α−p−1m(s)ds),
h2(t) =tα
Γ(1 + α)+
(t+ 1)(δηα−p + 1)
∆Γ(α− p+ 1)+
1
Γ(α)tα−1,
ϖ1(t) =1
Γα+ 1tα +
t
∆Γ(α− p+ 1)(δηα−p−1 + 1), ϖ2(t) =
tα−1
Γ(α)+
δηα−p + 1
∆Γ(α− p+ 1),
(2.2.8)
and ϖ = ϖ1 +ϖ2. Choose ρ ≥ max{3h1, (3k1h2)1
1−λ1 , (3k2h2)1
1−λ2 }, where k1 = maxt∈[0,1] a1(t),
k2 = maxt∈[0,1] a2(t). Consider a closed bounded subset B = {x(t) ∈ E : ∥x∥ ≤ ρ, t ∈ [0, 1]} of E .
Theorem 2.2.2. Under the assumption (A1) the BVP (2.2.1), has at least one solution.
Proof. For x ∈ B using the assumption (A1), we have:
|T x(t)| ≤ |∫ 1
0|G(t, s)|m(s)ds+ (a1(t)|x|λ1 + a2(t)|x′|λ2)
∫ 1
0|G(t, s)|ds
≤∫ 1
0|G(t, s)|m(s)ds+ (a1(t)|x|λ1 + a2(t)|x′|λ2)(
∫ t
0(|G1(t, s)) +G2(t, s)
+G3(t, s)|)ds+∫ η
t|G2(t, s) +G3(t, s)|ds+
∫ 1
η|G3(t, s)|ds)
=
∫ 1
0G(t, s)m(s)ds+ (a1(t)|x|λ1 + a2(t)|x′|λ2)(
tα
Γ(1 + α)+
δtηα−p
∆Γ(α− p+ 1)
+t
∆Γ(α− p+ 1))
≤∫ 1
0G(t, s)m(s)ds+ (k1|x|λ1 + k2|x′|λ2)(
tα
Γ(1 + α)
+δtηα−p
∆Γ(α− p+ 1)+
t
∆Γ(α− p+ 1)).
(2.2.9)
In view of (2.2.7), we have
|T ′x(t)| ≤ 1
Γ(α− 1)(
∫ t
0(t− s)α−2m(s)ds+ (k1|x|λ1 + k2|x′|λ2)
∫ t
0(t− s)α−2ds)
+1
∆(
δ
Γ(α− p)(
∫ η
0(η − s)α−p−1m(s)ds+ (k1|x|λ1 + k2|x′|λ2)
∫ η
0(η − s)α−p−1ds)
+1
Γ(α− p)(
∫ 1
0(1− s)α−p−1m(s)ds+ (k1|x|λ1 + k2|x′|λ2)
∫ 1
0(1− s)α−p−1ds))
=( 1
Γ(α− 1)
∫ t
0(t− s)α−2 +
1
∆(
δ
Γ(α− p)
∫ η
0(η − s)α−p−1
+1
Γ(α− p)
∫ 1
0(1− s)α−p−1)
)m(s)ds+ (
1
Γ(α)tα−1 +
1
∆
δηα−p + 1
Γ(α− p+ 1))(k1|x|λ1
+ k2|x′|λ2).
(2.2.10)
25
From the definition of the norm and (2.2.9), (2.2.10), we have:
∥T x(t)∥ ≤∫ 1
0G(t, s)m(s)ds+ (k1|x|λ1 + k2|x′|λ2)(
tα
Γ(1 + α)+
δtηα−p
∆Γ(α− p+ 1)
+t
∆Γ(α− p+ 1)) + (
1
Γ(α− 1)
∫ t
0(t− s)α−2 +
1
∆(
δ
Γ(α− p)
∫ η
0(η − s)α−p−1
+1
Γ(α− p)
∫ 1
0(1− s)α−p−1))m(s)ds+ (
1
Γ(α)tα−1 +
1
∆
δηα−p + 1
Γ(α− p+ 1))(k1|x|λ1
+ k2|x′|λ2)
≤ (
∫ 1
0G(t, s) +
1
Γ(α− 1)
∫ t
0(t− s)α−2 +
1
∆(
δ
Γ(α− p)
∫ η
0(η − s)α−p−1
+1
Γ(α− p)
∫ 1
0(1− s)α−p−1))m(s)ds+ (
tα
Γ(1 + α)+
(t+ 1)(δηα−p + 1)
∆Γ(α− p+ 1)
+1
Γ(α)tα−1)(k1|ρ|λ1 + k2|ρ|λ2).
(2.2.11)
Using (2.2.8) and (2.2.11), we have
∥T x(t)∥ ≤ h1 + (k1|ρ|λ1 + k2|ρ|λ2)h2 ≤ρ
3+ρ
3+ρ
3= ρ,
for h1 = supt∈[0,1] |h1(t)|, this implies that T B ⊂ B.
Now, we show that T is completely continuous operator. Let M = max{|f(t, x(t), x′(t))| : t ∈[0, 1], x ∈ B}, then
|T x(t)− T x(τ)| = |∫ 1
0G(t, s)f(s, x(s), x′(s))ds−
∫ 1
0G(τ, s)f(s, x(s), x′(s))ds|
≤M |∫ 1
0(G(t, s)−G(τ, s))ds|
=M |∫ t
0(t− s)α−1ds−
∫ τ
0(τ − s)α−1ds+
t− τ
∆Γ(α− p)(−δ
∫ η
0(η − s)α−p−1ds
+
∫ 1
0(1− s)α−p−1ds)| ≤M(
1
Γ(1 + α)(tα − τα) +
t− τ
∆Γ(α− p+ 1)(δηα−p + 1)),
(2.2.12)
and
|T ′x(t)− T ′x(τ)| = | − Iα−1(f(t, x(t), x′(t))− f(τ, x(τ), x′(τ))|
≤ M
Γ(α− 1)(
∫ t
0(t− s)α−2ds−
∫ τ
0(τ − s)α−2ds) =
M
Γ(α)(tα−1 − τα−1).
(2.2.13)
From the definition of the norm, (2.2.12) and (2.2.13), we have:
∥T x(t)− T x(τ)∥ ≤M(1
Γ(1 + α)(tα − τα) +
t− τ
∆Γ(α− p+ 1)(δηα−p + 1))
+M
Γ(α)(tα−1 − τα−1).
(2.2.14)
Since the functions tα−1, τα−1, tα, τα, are uniformly continuous on the interval [0,1], therefore, from
(2.2.14), it follows that T is equicontinuous and by Arzela-Ascoli theorem T is completely continuous.
Hence, by Schauder fixed point theorem, T has a fixed point x ∈ B.
26
Lemma 2.2.3. Under the assumption (A2), the BVP (2.2.1) has a solution.
Proof. The proof is similar to that of Theorem 2.2.2, so we exclude the proof.
Theorem 2.2.4. Assume that (A3) hold. If kϖ < 1, then the problem (2.2.1), has a unique solution.
Proof. In view of supposition (A3), we have the following estimates
|T x(t)− T y(t)| = 1
Γ(α)
∫ t
0(t− s)α−1|f(s, x(s), x′(s))− f(s, y(s), y′(s))|ds
+t
∆Γ(α− p)(δ
∫ η
0(η − s)α−p−1|f(s, x(s), x′(s))− f(s, y(s), y′(s))|ds
+
∫ 1
0(1− s)α−p−1|f(s, x(s), x′(s))− f(s, y(s), y′(s))|ds)
≤ 1
Γ(α+ 1)tα(k{|x− y|+ |x′ − y′|})) + t
∆Γ(α− p+ 1)(δηα−p−1(k(|x− y|
+ |x′ − y′|)) + (k{|x− y|+ |x′ − y′|}))
≤ k{|x− y|+ |x′ − y′|}( 1
Γ(α+ 1)tα +
t
∆Γ(α− p+ 1)(δηα−p−1 + 1))
= kϖ1∥x− y∥.
(2.2.15)
By the help of (2.2.7) we have the following estimates
|T ′x(t)− T ′y(t)| = | 1
Γ(α− 1)
∫ t
0(t− s)α−2|f(s, x(s), x′(s))− f(s, y(s), y′(s))|ds
+1
∆Γ(α− p)(−δ
∫ η
0(η − s)α−p−1|f(s, x(s), x′(s))− f(s, y(s), y′(s))|ds
+
∫ 1
0(1− s)α−p−1|f(s, x(s), x′(s))− f(s, y(s), y′(s))|)ds|
≤k{|x− y|+ |x′ − y′|}( tα−1
Γ(α)+
δηα−p + 1
∆Γ(α− p+ 1)) = kϖ2∥x− y∥.
(2.2.16)
Using the definition of the norm, (2.2.15) and (2.2.16), we have:
∥T x(t)− T y(t)∥ ≤ kϖ1∥x− y∥+ kϖ2∥x− y∥
= k(ϖ1 +ϖ2)∥x− y∥ = kϖ∥x− y∥.(2.2.17)
Thus, by contraction mapping principle, the BVP (2.2.1) has a unique solution.
Example 2.2.5. Consider the following FDE as a special case of the problem (2.2.1):
D32t x(t) =
(5+7|x(t)|+5|x′(t)|) cos(t)35 ,
x(0) = 0, D13x(1) = 1
10D13u(13).
(2.2.18)
For the unique solution of problem (2.2.18), we apply theorem (2.2.4) with f(t, x(t), x′(t))
= (5+7|x(t)|+5|x′(t)|) cos(t)35 , t ∈ [0, 1], α = 3
2 , p = 13 , δ = 1
10 , η = 13 . For x, x′, y, y′ ∈ R, we have
|f(t, x(t), x′(t))− f(t, y(t), y′(t))| ≤ 15{|x− y|+ |x′(t)− y′(t)|} which is condition (A4). By simple
calculations, we get ϖ = 2.4389 and kϖ = .4878 < 1. Thus by theorem (2.2.4) the problem (2.2.18)
has a unique solution.
27
In next section, we study further generalization of the problem (2.2.1).
2.3 Multi point BVPs for FDEs
In this section, we are concerned with a class of multi point boundary value problems (BVP) for FDE of
the type:
Dαx(t) = f(t, x(t),Dpx(t)), 2 < α < 3, 0 < p < 1,
x(0) = 0, Dpx(1) =
m−2∑i=1
ζiDpx(ηi), x′′(1) = 0, 0 < ηi, ζi ≤ 1, i = 0, 1, . . . ,m(2.3.1)
where Dα, Dp are Caputo’s fractional derivative of order α, p respectively.
Lemma 2.3.1. The BVP for fractional differential equation (2.3.1) is equivalent to the integral equation
x(t) =
∫ 1
0G(t, s)f(t, x(t),Dpx(t))dt, (2.3.2)
where
G(t, s) =
(t−s)α−1
Γ(α) + t∆(− 1
Γ(α−p)(1− s)α−p−1 + 1Γ(3−p)Γ(α−2)(1− s)α−3
+∑m−2
i=1ζi
Γ(α−p)(ηi − s)α−p−1 −∑m−2
i=1 ζi(ηi)2−p
Γ(3−p)Γ(α−p) (1− s)α−p−1)
− t2
2Γ(2−p)(1− s)α−3, s ≤ t,
t∆(− 1
Γ(α−p)(1− s)α−p−1 + 1Γ(3−p)Γ(α−2)(1− s)α−3
+∑m−2
i=1ζi(ηi−s)α−p−1
Γ(α−p) −∑m−2
i=1 ζi(η)2−p
Γ(3−p)Γ(α−p) (1− s)α−p−1)
− t2
2Γ(2−p)(1− s)α−3, t ≤ s.
(2.3.3)
Proof. Applying fractional integral of order α on the fractional differential equation (2.3.1), we have:
x(t) = Iαf(t, x(t),Dpx(t)) + c0 + c1t+ c2t2. (2.3.4)
Using the boundary conditions in (2.3.1), we obtain c0 = 0, c1 = 1∆(−Iα−pf(1) + 1
Γ(3−p)Iα−2f(1) +∑m−2
i=1 ζiIα−pf(ηi) −∑m−2
i=1ζi(ηi)
2−p
Γ(3−p) Iα−2f(1)), c2 = −Iα−2f(1)2 , where ∆ =
1−∑m−2
i=1 ζi(ηi)1−p
Γ(2−p) .
Hence (2.3.4) takes the form
x(t) =1
Γ(α)
∫ t
0(t− s)α−1f(s, x(s),Dpx(s))ds
+t
∆(
−1
Γ(α− p)
∫ 1
0(1− s)α−p−1f(s, x(s),Dpx(s))ds
+1
Γ(3− p)
1
Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds
(2.3.5)
28
+1
Γ(α− p)
m−2∑i=1
ζi
∫ ηi
0(ηi − s)α−p−1f(s, x(s),Dpx(s))ds
− 1
Γ(3− p)Γ(α− 2)
m−2∑i=1
ζi(ηi)2−p
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds)
− t2
2Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds.
For 0 < t < ηi, we have the following estimates
x(t) =
∫ t
0[
1
Γ(α)(t− s)α−1 +
t
∆(
−1
Γ(α− p)(1− s)α−p−1 +
(1− s)α−3
Γ(3− p)Γ(α− 2)
+
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)− 1
Γ(3− p)Γ(α− 2)
m−2∑i=1
ζi(ηi)2−p(1− s)α−3)
− t2(1− s)α−3
2Γ(α− 2)]f(s, x(s),Dpx(s))ds+
∫ η1
t{ t∆(
−1
Γ(α− p)(1− s)α−p−1
+1
Γ(3− p)Γ(α− 2)(1− s)α−3 +
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)
+1
Γ(3− p)Γ(α− 2)(1− s)α−3 +
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)
−∑m−2
i=1 ζi(ηi)2−p
Γ(3− p)Γ(α− 2)(1− s)α−3)− t2(1− s)α−3
2Γ(α− 2)}f(s, x(s),Dpx(s))ds
+m−2∑i=2
∫ ηi
ηi−1
{ t∆(
−1
Γ(α− p)(1− s)α−p−1 +
1
Γ(3− p)Γ(α− 2)(1− s)α−3
+1
Γ(α− p)
m−2∑i=1
ζi(ηi − s)α−p−1 −∑m−2
i=1 ζi(ηi)2−p(1− s)α−3
Γ(3− p)Γ(α− p))
− t2(1− s)α−3
2Γ(α− 2)}f(s, x(s),Dpx(s))ds+
∫ 1
ηm−2
{ t∆(
−1
Γ(α− 2)(1− s)α−p−1
+(1− s)α−3
Γ(3− p)Γ(α− 2)−∑m−2
i=1 ζi(ηi)2−p(1− s)α−3)
Γ(3− p)Γ(α− 2)
− t2
2Γ(α− 2)(1− s)α−3}f(s, x(s),Dpx(s))ds =
∫ 1
0G(t, s)f(s, x(s),Dpx(s))ds.
(2.3.6)
For ηi−1 < t < ηi, η0 = 0, ηm−1 = 1 where i = 1, 2, ...,m− 1, we have
x(t) =
i−1∑k=1
∫ ηk
ηk−1
[1
Γ(α)(t− s)α−1 +
t
∆(
−1
Γ(α− p)(1− s)α−p−1
+(1− s)α−3
Γ(3− p)Γ(α− 2)+
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)−∑m−2
i=1 ζi(ηi)2−p(1− s)α−3
Γ(3− p)Γ(α− 2))
− t2
2Γ(α− 2)(1− s)α−3]f(s, x(s),Dpx(s))ds+
∫ t
ηi−1
{∫ t
0[
1
Γ(α)(t− s)α−1
+t
∆(−(1− s)α−p−1
Γ(α− p)+
(1− s)α−3
Γ(3− p)Γ(α− 2)+
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)
29
−∑m−2
i=1 ζi(ηi)2−p(1− s)α−3
Γ(3− p)Γ(α− 2))− t2
2Γ(α− 2)(1− s)α−3]}f(s, x(s),Dpx(s))ds
+
∫ ηi
t{ t∆(−(1− s)α−p−1
Γ(α− p)+
(1− s)α−3
Γ(3− p)Γ(α− 2)+
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)
−∑m−2
i=1 ζi(ηi)2−p(1− s)α−3
Γ(3− p)Γ(α− 2))− t2
(1− s)α−3
2Γ(α− 2)}f(s, x(s),Dpx(s))ds
+
m−1∑k=i+1
∫ ηk
ηk−1
{ t∆(−(1− s)α−p−1
Γ(α− p)+
(1− s)α−3
Γ(3− p)Γ(α− 2)+
∑m−2i=1 ζi(ηi − s)α−p−1
Γ(α− p)
−∑m−2
i=1 ζi(ηi)2−p(1− s)α−3
Γ(3− p)Γ(α− 2))− t2
2Γ(α− 2)(1− s)α−3}f(s, x(s),Dpx(s))ds
=
∫ 1
0G(t, s)f(s, x(s),Dpx(s))ds.
2.3.1 Existence and uniqueness of solutions
Let I = [0, 1], and C(I) be the space of continuous functions defined on I . The space E = {x(t) ∈
C(I,R) : Dpx(t)∈C(I,R)} with the norm ∥x∥ = maxt∈I |x(t)|+maxt∈I |Dpx(t)|, is a Banach space
[23]. We assume the spermium value:
G∗ = supt∈I
∫ 1
0|G(t, s)|ds.
Lemma 2.3.2. Assume that the following hold:
(A1) f∈C([0, 1]×R×R,R),
(A2) there exist a constant k > 0 such that for each t ∈ I = [0, 1] and for all x, y ∈ R
|f(t, x,Dpx)− f(t, y,Dpy)|≤k{|x− y|+ |Dpx−Dpy|}, (2.3.7)
and
max{2kG∗, 2k{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− p+ 1)+
1
Γ(3− p)Γ(α− 1)
+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1)) +
1
Γ(3− p)Γ(α− 1)}} = r < 1,
then, the BVP (2.3.1) has a unique solution.
Proof. Define an operator T : E → E , by:
T x(t) =∫ 1
0G(t, s)f(s, x(s),Dpx(s))ds. (2.3.8)
30
We show that, T is a contraction mapping. For this, we have:
|T x− T y| =∫ 1
0|G(t, s)||f(s, x,Dpx)− f(s, y,Dpy)|ds
≤ supt∈I
G(t, s)
∫ 1
0|f(s, x,Dpx)− f(s, y,Dpy)|ds
≤ G∗k{|x− y|+ |Dpx−Dpy|}
≤G∗k(maxt∈I
|x− y|+maxt∈I
|Dpx−Dpy|)=G∗k∥x− y∥.
(2.3.9)
From (2.3.5), we have
DpT x(t) =∫ t
0
(t− s)α−p−1
Γ(α− p)f(s, x(s),Dpx(s))ds
+t1−p
∆Γ(2− p)(
∫ 1
0
−(1− s)α−p−1
Γ(α− p)f(s, x(s),Dpx(s))ds
+
∫ 1
0
(1− s)α−3
Γ(3− p)Γ(α− 2)f(s, x(s),Dpx(s))ds
+
∑m−2i=1 ζi
Γ(α− p)
∫ ηi
0(ηi − s)α−p−1f(s, x(s),Dpx(s))ds
−∑m−2
i=1 ζi(ηi)2−p
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds)
− t2−p
∫ 1
0
(1− s)α−3
Γ(3− p)Γ(α− 2)f(s, x(s),Dpx(s))ds,
(2.3.10)
which implies that
|DpT x−DpT y|
≤ 1
Γ(α− p)
∫ t
0(t− s)α−p−1|f(s, x(s),Dpx(s))− f(s, y(s),Dpy(s))|ds
+t1−p
∆Γ(2− p)(
1
Γ(α− p)
∫ 1
0(1− s)α−p−1|f(s, x(s),Dpx(s))− f(s, y(s),Dpy(s))|ds
+1
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3|f(s, x(s),Dpx(s))− f(s, y(s),Dpy(s))|ds
+
∑m−2i=1 ζi
Γ(α− p)
∫ ηi
0(ηi − s)α−p−1|f(s, x(s),Dpx(s))− f(s, y(s),Dpy(s))|ds
+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3|f(s, x(s),Dpx(s))− f(s, y(s),Dpy(s))|ds)
+t2−p
Γ(3− p)
∫ 1
0(1− s)α−3|f(s, x(s),Dpx(s))− f(s, y(s),Dpy(s))|ds
≤ tα−p
Γ(α− p+ 1)k{|x− y|+ |Dpx(s)−Dpy(s)|}+ t1−p
∆Γ(2− p)(
1
Γ(α− p+ 1){|x− y|
+ |Dpx(s)−Dpy(s)|}ds+ 1
Γ(3− p)Γ(α− 1){|x− y|+ |Dpx(s)−Dpy(s)|}
(2.3.11)
31
+
∑m−2i=1 (ηi)
α−p
Γ(α− p+ 1){|x− y|+ |Dpx(s)−Dpy(s)|}+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1){|x− y|
+ |Dpx(s)−Dpy(s)|}) + t2−p
Γ(3− p)
1
Γ(α− 1){|x− y|+ |Dpx(s)−Dpy(s)|}
≤ k{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− p+ 1)+
1
Γ(3− p)Γ(α− 1)
+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1)) +
1
Γ(3− p)Γ(α− 1)}∥x− y∥.
From (2.3.9) and (2.3.11), we have the following estimates
∥T x− T y∥ = maxt∈I
|T x− T y|+maxt∈I
|DpT x−DpT y|
≤ {kG∗∥x− y∥+ k{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− p+ 1)
+1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1))
(2.3.12)
+1
Γ(3− p)Γ(α− 1)}∥x− y∥
≤ max{2kG∗, 2k{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− p+ 1)
+1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1))
+1
Γ(3− p)Γ(α− 1)}}∥x− y∥ = r∥x− y∥.
Thus, T is contraction mapping and hence by Theorem 1.3.5 the BVP (2.3.1), has a unique fixed point
x.
Theorem 2.3.3. Assume that the following hold
(A1). f∈C(I × R× R);(A2). There exist h = h(t)∈C(I,R) and a continuous non-decreasing function ϕ : [0,∞] → (0,∞)
such that
|f(t, x,Dpx)|≤hϕ|(Dpx)| ∀ t ∈ I, x, y ∈ R; (2.3.13)
(A3). There exists constant r > 0, such that
max{2G∗h∗ϕ|Dpx(t)|, 2h∗ϕ|Dpx(t)|{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− p+ 1)
+1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1)) +
1
Γ(3− p)Γ(α− 1)}}
= r < 1,
where h∗ = sup{h(t), t ∈ I}, then (2.3.1), has a solution x(t) such that |x| < r for t ∈ [0, 1].
Proof. Step 1: Let {xn(t)} be a convergent sequence such that {xn(t) → x(t)} as n → ∞, then it is
easy to show that T (t, xn,Dpxn) → T (t, x,Dpx). Hence, it follows that
|T (t, xn,Dpxn)− T (t, x,Dpx)| → 0 or T (t, xn,Dpxn) → T (t, x,Dpx) as n→ ∞.
32
Step 2: let B = {x(t) ∈ E : ∥x∥ ≤ r} be a bounded subset of E . We show that T (B) is also bounded.
For x(t) ∈ B, we have
|T x| ≤ |∫ 10 G(t, s)f(s, x(s),D
px(s))ds| ≤ G∗ ∫ 10 |f(s, x(s),Dpx(s))|ds ≤ G∗h∗ϕ|Dpx(t)|.
(2.3.14)
From equation (2.3.10), we have
|DpT (x)| = | 1
Γ(α− p)
∫ t
0(t− s)α−p−1f(s, x(s),Dpx(s))ds
+t1−p
∆Γ(2− p)(
−1
Γ(α− p)
∫ 1
0(1− s)α−p−1f(s, x(s),Dpx(s))ds
+1
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds
+1
Γ(α− p)
m−2∑i=1
ζi
∫ ηi
0(ηi − s)α−p−1f(s, x(s),Dpx(s))ds
−∑m−2
i=1 ζi(ηi)2−p
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds)
− t2−p
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds|
≤h∗ϕ|Dpx(t)|{ 1
Γ(α− p+ 1)tα−pds+
t1−p
∆Γ(2− p)(
1
Γ(α− p+ 1)
+1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 (ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− p+ 1))
+t2−p
Γ(3− p)
1
Γ(α− 1)}
≤h∗ϕ|Dpx(t)|{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− 1)+
1
Γ(3− p)Γ(α− 1)
+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1)) +
1
Γ(3− p)Γ(α− 1)}.
(2.3.15)
Using the definition of the norm and (2.3.10) and (2.3.15), we obtain
∥T (x)∥ = maxt∈I
|T (x)|+maxt∈I
|DpT (x)|
≤G∗h∗ϕ|Dpx(t)|+ h∗ϕ|Dpx(t)|{ 1
Γ(α− p+ 1)
+1
∆Γ(2− p)(
1
Γ(α− p+ 1)+
1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)
+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1)) +
1
Γ(3− p)Γ(α− 1)}
≤max{2G∗h∗ϕ|Dpx(t)|, 2h∗ϕ|Dpx(t)|{ 1
Γ(α− p+ 1)
+1
∆Γ(2− p)(
1
Γ(α− p+ 1)+
1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 ζi(ηi)
α−p
Γ(α− p+ 1)
+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1)) +
1
Γ(3− p)Γ(α− 1)}} = r < 1,
33
which implies that T x ∈ B ∀ x ∈ B. Hence, T (B)⊆B.
Step 3: For equicontinuity, choose t1, t2 ∈ I such that t1 < t2 and x ∈ B, we have
|T x(t2)− T x(t1)| ≤∫ 1
0|G2(t, s)−G1(t, s)||f(s, x(s),Dpx(s))|ds
≤ h∗ϕ|Dpx(s)|∫ 1
0|G(t2, s)−G(t1, s)|ds.
This implies that
|T x(t2)− T x(t1)| → 0 as t1 → t2 . (2.3.16)
By (2.3.10) and the conditions (A1)− (A3), we have:
|DpT (x)(t2)−DpT (x)(t1)|
≤ | 1
Γ(α− p)(
∫ t2
0(t2 − s)α−p−1f(s)ds−
∫ t1
0(t1 − s)α−p−1f(s, x(s),Dpx(s))ds)
+t1−p2 − t1−p
1
∆Γ(2− p)(
−1
Γ(α− p)
∫ 1
0(1− s)α−p−1f(s, x(s),Dpx(s))ds
+1
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds
+1
Γ(α− p)
m−2∑i=1
ζi
∫ ηi
0(ηi − s)α−p−1f(s, x(s),Dpx(s))ds
−∑m−2
i=1 ζi(ηi)2−p
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds)
− t2−p2 − t2−p
1
Γ(3− p)Γ(α− 2)
∫ 1
0(1− s)α−3f(s, x(s),Dpx(s))ds|
≤h∗ϕ|Dpx(t)|{ 1
Γ(α− p+ 1)(tα−p
2 − tα−p1 ) +
t1−p2 − t1−p
1
∆Γ(2− p)(
1
Γ(α− p+ 1)
+1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 (ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1))
+t2−p2 − t2−p
1
Γ(3− p)
1
Γ(α− 1)}.
This implies that
|DpT (x)(t2)−DpT (x)(t1)| → 0 as t1 → t2, (2.3.17)
Thus, by (2.3.16) and (2.3.17), we have
∥T x(t2)− T x(t1)∥ = maxt∈I
|T x(t2)− T x(t1)|+maxt∈I
|DpT (x)(t2)−DpT (x)(t1)|
≤ h∗ϕ|Dpx(s)|∫ 1
0|G(t2, s)−G(t1, s)|ds
+ h∗ϕ|Dpx(t)|{ tα−p2 − tα−p
1
Γ(α− p+ 1)+t1−p2 − t1−p
1
∆Γ(2− p)(
1
Γ(α− p+ 1)
+1
Γ(3− p)Γ(α− 1)+
∑m−2i=1 (ηi)
α−p
Γ(α− p+ 1)+
∑m−2i=1 ζi(ηi)
2−p
Γ(3− p)Γ(α− 1))
+(t2−p
2 − t2−p1 )
Γ(3− p)Γ(α− 1)}.
34
Therefore, ∥T x(t2)−T x(t1)∥ → 0 as t1 → t2. By Arzela Ascoli theorem, the operator T is completely
continuous and hence by Shauder fixed point theorem, T has a fixed point x ∈ B, which is a solution of
the BVP (2.3.1).
Example 2.3.4. Consider a special case of the FDE (2.3.1):
D5/2x(t) = f(t, x(t),D0.5x(t))
x(0) = 0, D0.5x(1) =
m−2∑i=1
ζiD0.5x(ηi) = 1/5, x′′(1) = 0.(2.3.18)
where, ζi = 1/2, ηi = 1/2, m = 3 and f(t, x(t),D0.5x(t)) = 140et(1+3|x(t)|+5|D0.5x(t)|) . It follows that
|f(t, x,Dpx)− f(t, y,Dpy)| ≤ 1
40et(
3|x| − 3|up|+ 5|y| − 5|Dpy|(1 + 3|x|+ 5|up|)(1 + 3|y|+ 5|Dpy|)
)
≤ 1
40(5|x−Dpx|+ 5|y −Dpy|) = 1
8(|x−Dpx|+ |y −Dpy|).
(2.3.19)
From (2.3.19), we have k = 18 , and by simple calculations, we obtain G∗ = 2.55, 2kG∗ = 0.64 < 1, and
2k{ 1
Γ(α− p+ 1)+
1
∆Γ(2− p)(
1
Γ(α− p+ 1)+
1
Γ(3− p)Γ(α− 1)
+1
Γ(α− p+ 1)
m−2∑i=1
ζi(ηi)α−p +
1
Γ(3− p)Γ(α− 1)
m−2∑i=1
ζi(ηi)2−p)
+1
Γ(3− p)Γ(α− 1)} < 0.4752 < 1.
(2.3.20)
Therefore, the conditions (A1) and (A2) are satisfied and hence by Lemma (2.3.2), the BVP (2.3.18),
has a unique solution.
Now we study more general class of BVP for FDEs
2.4 p–Laplacian BVP for FDEs
In this section, we study existence and uniqueness of solutions of the following p-Laplacian BVP for
FDEs
Dγ(ϕp(Dβx(t))) + a(t)f(x(t)) = 0, 3 < β, γ ≤ 4, t ∈ [0, 1],
x(0) = x′′′(0), ηDαx(t)|t=1 = x′(0), ξx′′(1)− x′′(0) = 0, 0 < α < 1,
ϕp(Dβx(t))|t=0 = 0 = (ϕp(Dβx(t)))′|t=0, (2.4.1)
(ϕp(Dβx(t)))′′|t=1 =1
2(ϕp(Dβx(t)))′′|t=0,
(ϕp(Dβx(t)))′′′|t=0 = 0,
where Dβ , Dγ and Dα are Caputo’s fractional derivative of fractional order β, γ, α, β, γ ∈ (3, 4],
α ∈ (0, 1) and ϕp(s) = |s|p−2s, p > 1. The p–Laplacian operator satisfies ϕ−1p = ϕq, 1
p + 1q = 1.
35
Lemma 2.4.1. Let f(t) ∈ C[0, 1], the BVP
Dβx(t) = f(t), 3 < β ≤ 4, (2.4.2)
x(0) = x′′′(0), ηDαx(1) = x′(0), ξx′′(1)− x′′(0) = 0, 0 < α < 1, (2.4.3)
has a solution of the form
x(t) =
∫ 1
0G(t, s)f(t)ds, (2.4.4)
where
G(t, s) =
tη1− η
Γ(2−α)
[(1−s)β−α−1
Γ(β−α) + ξ(1−ξ)Γ(3−α)
(1−s)β−3
Γ(β−2)
]+ t2ξ
2(1−ξ)(1−s)β−3
Γ(β−2) + 1Γ(β)(t− s)β−1, 0 < s ≤ t < 1,
tη1− η
Γ(2−α)
[(1−s)β−α−1
Γ(β−α) + ξ(1−ξ)Γ(3−α)
(1−s)β−3
Γ(β−2)
]+ t2ξ
2(1−ξ)(1−s)β−3
Γ(β−2) , 0 < t ≤ s < 1.
(2.4.5)
Proof. Applying the operator Iα on the differential equation in (2.4.2), we obtain
x(t) = Iβf(t) + c1 + c2t+ c3t2 + c4t
3. (2.4.6)
The boundary conditions x(0) = 0 = x′′′(0) yield to c1 = 0 = c4, and by using the boundary condition
ξx′′(1) − x′′(0) = 0, we have c3 = ξ2(1−ξ)I
β−2h(1). Using ηDαx(1) = x′(0), we obtain c2 =
η1− η
Γ(2−α)
[Iβ−αf(1) + ξ
(1−ξ)Γ(3−α)Iβ−2f(1)
]. Substituting the values of c1, c2, c3, c4, in (2.4.6), we
get
x(t) = Iβf(t) +tη
1− ηΓ(2−α)
[Iβ−αh(1) +
ξ
(1− ξ)Γ(3− α)Iβ−2f(1)
](2.4.7)
+t2ξ
2(1− ξ)Iβ−2f(1).
The integral form is given as
x(t) =
∫ t
0
(t− s)β−1f(s)ds
Γ(β)+
t2ξ
2(1− ξ)
∫ 1
0
(1− s)β−3h(s)ds
Γ(β − 2)
+tη
1− ηΓ(2−α)
[
∫ 1
0
(1− s)β−α−1f(s)ds
Γ(β − α)(2.4.8)
+ξ
(1− ξ)Γ(3− α)
∫ 1
0
(1− s)β−3f(s)ds
Γ(β − 2)] =
∫ 1
0G(t, s)f(s)ds.
Lemma 2.4.2. For 3 < β, γ ≤ 4, assume that a, x : (0, 1) → [0,+∞) and f : [0,+∞) → [0,+∞) are
continuous functions, the BVP with p–Laplacian operator
Dγ(ϕp(Dβx(t))) + a(t)f(x(t)) = 0, 3 < β, γ ≤ 4, 0 < α < 1 t ∈ [0, 1],
x(0) = x′′′(0), ηDαy(t)|t=1 = x′(0), ξx′′(1)− x′′(0) = 0,
ϕp(Dβx(t))|t=0 = 0 = (ϕp(Dβx(t)))′|t=0, (2.4.9)
(ϕp(Dβx(t)))′′|t=1 =1
2(ϕp(Dβx(t)))′′|t=0,
(ϕp(Dβx(t)))′′′|t=0 = 0,
36
has a solution of the form
x(t) =
∫ 1
0G(t, s)ϕq(
∫ 1
0H(s, z)a(z)f(x(z))dz)ds, (2.4.10)
where
H(t, s) =
− (t−x)γ−1
Γγ + t2
Γ(γ−2)(1− s)γ−3, 0 < x ≤ t < 1,
t2
Γ(γ−2)(1− s)γ−3, 0 < t ≤ x < 1,(2.4.11)
and G(t, s) is given by (2.4.5).
Proof. Applying integral Iγ on the differential equation in (2.4.9) and using lemma 9, we obtain
ϕp(Dβx(t)) = −Iγa(t)f(x(t)) + c1 + c2t+ c3t2 + c4t
3. (2.4.12)
The boundary conditions ϕp(Dβx(0)) = (ϕp(Dβx(0)))′ = (ϕp(Dαx(0)))′′′ = 0 lead to c1 = c2 =
c4 = 0. From (2.4.12) and c1 = c2 = c4 = 0, we deduce
(ϕp(Dβx(t)))′′ = −Iγ−2a(t)f(x(t)) + 2c3. (2.4.13)
The boundary condition (ϕp(Dβx(1)))′′ = 12(ϕp(D
βx(0)))′′, yields c3 = 1Γ(γ−2)
∫ 10 (1−s)
γ−3a(z)f(x(z))dz.
Consequently, (2.4.12) takes the form
ϕp(Dβx(t)) = − 1
Γ(γ)
∫ 1
0(t− s)γ−1a(s)f(x(s))ds
+t2
Γ(γ − 2)
∫ 1
0(1− s)γ−3a(s)f(x(s))ds (2.4.14)
=
∫ 1
0H(t, s)a(s)f(x(s))ds.
The BVP (2.4.9) reduces to the following problem
Dβx(t) = ϕq(
∫ 1
0H(t, s)a(s)f(x(s))ds) (2.4.15)
x(0) = x′′′(0), ηDαx(1) = x′(0), ξx′′(1)− x′′(0) = 0, 0 < α < 1,
which in view of Lemma 2.4.1, yields the required result, as under:
x(t) =
∫ 1
0G(t, s)ϕq(
∫ 1
0H(s, z)a(z)f(x(z))dz)ds. (2.4.16)
Lemma 2.4.3. Let 3 < γ ≤ 4. The Green’s function H(t, s) is a continuous function and satisfies
(A ) H(t, s) ≥ 0, H(t, s) ≤ H(1, s), for t, s ∈ (0, 1];
(B) H(t, s) ≥ tγ−1H(1, s) for t, s ∈ (0, 1].
37
Proof. The continuity of H(t, s) is ensured by its definition in (2.4.11). For (A), consider the case when
0 < s ≤ t ≤ 1, we have
H(t, s) = −(t− s)γ−1
Γ(γ)+
t2
Γ(γ − 2)(1− s)γ−3
= −tγ−1(1− s
t )γ−1
Γ(γ)+
t2
Γ(γ − 2)(1− s)γ−3 (2.4.17)
≥ − tγ−1(1− s)γ−1
Γ(γ)+
tγ−1
Γ(γ − 2)(1− s)γ−3
≥ tγ−1(1− s)γ−1
Γ(γ)Γ(γ − 2)[Γ(γ)− Γ(γ − 2)] > 0.
In case, when 0 < t ≤ s ≤ 1, the H(t, s) > 0 is obvious. Now, for H(t, s) ≤ H(1, s), for t, s ∈ (0, 1],
it follow that
∂
∂tH(t, s) =
− (t−s)γ−2
Γ(γ−1) + 2 tΓ(γ−2)(1− s)γ−3, 0 < s ≤ t ≤ 1,
2tΓ(γ−2)(1− s)γ−3, 0 < t ≤ s ≤ 1.
For x, t ∈ (0, 1], such that 0 < s ≤ t ≤ 1. From (2.4.18), we deduce
∂
∂tH(t, s) = −(t− s)γ−2
Γ(γ − 1)+
2t
Γ(γ − 2)(1− s)γ−3
= −tγ−2(1− s
t )γ−2
Γ(γ − 1)+
2t
Γ(γ − 2)(1− s)γ−3 (2.4.18)
≥ − tγ−2(1− s)γ−2
Γ(γ − 1)+
2tγ−2
Γ(γ − 2)(1− s)γ−3
≥ tγ−2(1− s)γ−2
Γ(γ − 1)Γ(γ − 2)[2Γ(γ − 1)− Γ(γ − 2)] ≥ 0,
from (2.4.18), we have ∂∂tH(t, s) ≥ 0. In case, when 0 < t ≤ s ≤ 1, (2.4.18), implies that ∂
∂tH(t, s) ≥ 0
is obvious which implies that H(t, s) is an increasing function w.r.t t. Hence H(t, s) ≤ H(1, s).
For part (B), consider the case, when 0 < x ≤ t ≤ 1, we proceed:
H(t, s)
H(1, s)=
−(t−x)γ−1
Γ(γ) + t2
Γ(γ−2)(1− s)γ−3
−(1−s)γ−1
Γ(γ) + 1Γ(γ−2)(1− s)γ−3
=
−tγ−1(1−xt)γ−1
Γ(γ) + t2
Γ(γ−2)(1− s)γ−3
−(1−s)γ−1
Γ(γ) + 1Γ(γ−2)(1− s)γ−3
(2.4.19)
≥−tγ−1(1−s)γ−1
Γ(γ) + tγ−1
Γ(γ−2)(1− s)γ−3
−(1−s)γ−1
Γ(γ) + 1Γ(γ−2)(1− s)γ−3
= tγ−1
−(1−s)γ−1
Γ(γ) + 1Γ(γ−2)(1− s)γ−3
−(1−s)γ−1
Γ(γ) + 1Γ(γ−2)(1− s)γ−3
= tγ−1.
38
The proof of the result for the case of 0 < t ≤ s ≤ 1 is similar, therefore, we omit it.
Assume that the following conditions are satisfied:
(A1) 0 <∫ 10 H(1, s)a(s)ds < +∞;
(A2) There exist 0 < δ < 1 and ρ > 0 such that f(t) ≤ δLϕp(t), for 0 ≤ t ≤ ρ,
where 0 < L ≤ (ϕp(ϖ)δ∫ 10 H(1, s)a(s)ds)−1;
(A3) There exist b > 0, such that f(t) ≤Mϕp(t), for x > b,
0 < M < (ϕp(ϖ2α−1)
∫ 1
0H(1, s)a(s)ds)−1;
(A4) f(t) is non-decreasing with respect to t;
(A5) There exist 0 ≤ β < 1 such that f(kt) ≥ (ϕp(k))βf(t), for any 0 ≤ k < 1, and 0 < t < +∞.
2.4.1 Existence and uniqueness of solutions
Theorem 2.4.4. Under the assumption (A1) and (A2), the BVP with p-Laplacian operator (2.4.1) has at
least one positive solution.
Proof. Consider a closed convex set K1 = {x(t) ∈ C[0, 1] : 0 ≤ x(t) ≤ ρ on [0, 1]} [128]. Define an
operator T : K1 → C[0, 1] by
T x(t) =∫ 1
0G(t, s)ϕq(
∫ 1
0H(s, z)a(z)f(x(z))dz)ds. (2.4.20)
By (2.4.2), x(t) is a solution of the BVP with p-Laplacian operator (2.4.1) if and only if x(t) is a fixed
point of T defined by (2.4.20). The compactness of the operator T can be easily shown. Consider,∫ 1
0G(t, s)ds =
∫ t0 (t− s)β−1ds
Γ(β)+
tη
1− ηΓ(2−α)
[
∫ 10 (1− s)β−α−1ds
Γ(β − α)
+ξ∫ 10 (1− s)β−3ds
(1− ξ)Γ(3− α)Γ(β − 2)] +
t2ξ
2(1− ξ)
∫ 10 (1− s)β−3ds
Γ(β − 2)(2.4.21)
=(t− s)β|0tΓ(1 + β)
+tη
1− ηΓ(2−α)
[(1− s)β−α|01Γ(β − α+ 1)
+ξ
(1− ξ)Γ(3− α)
(1− s)β−2|01Γ(β − 1)
] +t2ξ
2(1− ξ)
(1− s)β−2|01Γ(β − 1)
=tβ
Γ(1 + β)+
tη
1− ηΓ(2−α)
[1
Γ(β − α+ 1)
+ξ
(1− ξ)Γ(3− α)
1
Γ(β − 1)] +
t2ξ
2(1− ξ)
1
Γ(β − 1)
39
≤ 1
Γ(1 + β)+
η
1− ηΓ(2−α)
[1
Γ(β − α+ 1)
+ξ
(1− ξ)Γ(3− α)
1
Γ(β − 1)] +
ξ
2(1− ξ)
1
Γ(β − 1)= ϖ,
where ϖ = 1Γ(1+β) +
η1− η
Γ(2−α)
[1
Γ(β−α+1) +ξ
(1−ξ)Γ(3−α)1
Γ(β−1)
]+ ξ
2(1−ξ)1
Γ(β−1) . For any x ∈ K1,
using (A2) and (2.4.3), we obtain
T (x(t)) =
∫ 1
0G(t, s)ϕq(
∫ 1
0H(s, z)a(z)f(x(z)dz))ds
≤∫ 1
0G(t, s)ϕq(
∫ 1
0H(1, z)a(z)Lδϕp(ρ)dz)ds
≤ ϖϕq(
∫ 1
0H(1, z)a(z)Lδϕp(ρ)dz) (2.4.22)
= ϖϕq(Lδ
∫ 1
0H(1, z)a(z)dz)ρ ≤ ρ,
this implies that T (K1) ⊆ K1. Therefore, by Schauder’s fixed point theorem, T has a fixed point
x(t) ∈ K1.
Theorem 2.4.5. Under the assumptions (A1), (A3) the BVP with p-Laplacian operator (2.4.1) has at
least one positive solution.
Proof. Let b > 0 be as given in (A3). Define F∗ = max0≤x≤b f(x), then F∗ ≥ f(x) for 0 ≤ x ≤ b. In
view of (J3), we have
ϖ12α−1ϕq(M)ϕq(
∫ 1
0H(1, s)a(x)dx) < 1. (2.4.23)
Choose b∗ > b, large enough such that
ϖ12α−1(ϕq(F∗) + ϕq(M)b∗)ϕq(
∫ 1
0H(1, z)a(z))dz) < b∗. (2.4.24)
Define K2 = {x(t) ∈ C[0, 1] : 0 ≤ x(t) ≤ b∗ on [0, 1]}, S1 = {t ∈ [0, 1] : 0 ≤ x(t) ≤ b},
S2 = {t ∈ [0, 1] : b < x(t) ≤ b∗}, then S1 ∪ S2 = [0, 1] and S1 ∩ S2 = φ. For t ∈ K2, (A3) implies
f(x(t)) ≤Mϕp(x(t)) ≤Mϕp(b∗). Further, for t ∈ [0, 1], we have:
0 ≤ T (x(t)) =
∫ 1
0G(t, s)ϕq(
∫ 1
0H(s, z)a(z)f(x(z))dz)ds (2.4.25)
≤ ϖϕq(
∫S1
H(1, z)a(z)f(x(z))dz +
∫S2
H(1, z)a(z)f(x(z))dz)
≤ ϖϕq(F∗∫S1
H(1, z)a(z))dz +Mϕp(b∗)
∫S2
H(1, z)a(z))dz)
≤ ϖϕq(F∗ +Mϕp(b∗))ϕq(
∫ 1
0H(1, z)a(z))dz).
From (2.4.24) and using the inequality (a+ b)r ≤ 2r(ar + br) for any a, b, r > 0, (2.4.25) gets the form
0 ≤ T y(t) ≤ ϖ12α−1(ϕq(F∗) + ϕq(M)b∗)ϕq(
∫ 1
0H(1, z)a(z))dz) ≤ b∗, (2.4.26)
40
thus T (K2) ⊆ K2. Hence by Schauder’s fixed point theorem T has a fixed point x ∈ K2, which is the
positive solution of FDE with BVP and p–Laplacian operator (2.4.1).
Theorem 2.4.6. Assume that (A1), (A4), (A5) hold. Then the BVP with p-Laplacian operator (2.4.1)
has a unique positive solution.
Proof. Consider the set B = {x(t) ∈ C[0, 1] : y(t) ≥ 0 on [0, 1]}, where B is a normal solid cone in
C[0, 1] with B0 = {x(t) ∈ C[0, 1] : x(t) > 0 on [0, 1]}. Let T : B0 → C[0, 1] be defined by (2.4.20),
we prove that T is θ–concave and increasing operator. For x1, x2 ∈ B0 with x1 ≥ x2 the assumption
(A4) implies that the operator T is an increasing operator, i.e., T (x1(t)) ≥ T (x2(t)) on t ∈ [0, 1]. By
the help of f(kx) ≥ ϕp(kθ)f(x), we proceed
T (ky(t)) ≥∫ 1
0G(t, s)ϕq(
∫ 1
0H(t, x)ϕp(k
θ)a(z)f(x(z))dz)ds (2.4.27)
= kθ∫ 1
0G(t, s)ϕq(
∫ 1
0H(t, x)a(z)f(x(z))dz)ds = kθT (y(t)),
which implies that T is θ–concave operator and by the help of Lemma 2.4.4, the operator T has a unique
fixed point in B0, which is the unique positive solution of the FDE with BVP and p-Laplacian operator
(2.4.1).
Example 2.4.7. Consider the following BVP as a special case of the FDE with BVP and p–Laplacian
operator (2.4.1):
D3.5(ϕp(D3.5x(t))) + tx(t) = 0,
x(0) = 0 = x′′′(0),1
2D0.5x(t)|t=1 − x′(0) = 0,
1
2x′′(1)− x′′(0) = 0,
ϕp(D3.5x(t))|t=0 = 0 = (ϕp(D3.5x(t)))′|t=0, (2.4.28)
(ϕp(D3.5x(t)))′′|t=1 =1
2(ϕp(D3.5x(t)))′′, (ϕp(D3.5x(t)))′′|t=0 = 0,
where, α = β = 3.5, α = 0.5, ξ = γ = 12 , a(t) = t, f(x(t)) = x(t). By simple computation, we
obtain 0 < L ≤ 1.9092, choose L = 1.5, δ = 1, p = 2, then the conditions (A1), (A2) are satisfied.
Hence, by Theorem 2.4.4, the BVP with p-Laplacian operator (2.4.28) has at least one positive solution.
Example 2.4.8. Consider the following BVP as a special case of the FDE with BVP and p–Laplacian
operator (2.4.1):
D3.5(ϕp(D3.5x(t))) + t 3√x(t) = 0,
x(0) = 0 = x′′′(0), 0.1D0.5x(t)|t=1 − x′(0) = 0, 0.1x′′(1)− x′′(0) = 0,
ϕp(D3.5x(t))|t=0 = 0 = (ϕp(D3.5x(t)))′|t=0, (2.4.29)
(ϕp(D3.5x(t)))′′|t=1 =1
2(ϕp(D3.5x(t)))′′, (ϕp(D3.5x(t)))′′|t=0 = 0,
where, α = β = 3.5, ξ = η = 0.1, a(t) = t, f(x(t)) = 3√x(t). By simple computations, we get
M < 4.4792, choosing M = 4.00, b = 1 and p = q = 2, we see that (2.4.29) satisfy (A1) and (A3).
41
Hence by Theorem 2.4.5, the FDE with BVP and p-Laplacian operator (2.4.29) has at least one positive
solution.
Example 2.4.9. Consider the following BVP as a special case of the p–Laplacian BVP for FDE (2.4.1)
D3.5(ϕp(D3.5x(t))) + tx(t) = 0,
x(0) = 0 = x′′′(0),1
3D0.5x(t)|t=1 − x′(0) = 0,
1
3x′′(1)− x′′(0) = 0, (2.4.30)
ϕp(D3.5x(t))|t=0 = 0 = (ϕp(D3.5x(t)))′|t=0,
(ϕp(D3.5x(t)))′′|t=1 =1
2(ϕp(D3.5x(t)))′′, (ϕp(D3.5x(t)))′′|t=0 = 0.
Here, β = γ = 3.5, α = 0.5, ξ = γ = 13 , a(t) = t, f(x(t)) = x(t). It is clear that (2.4.30) satisfy
conditions (A1) and (A4). Also considering θ = 12 , (A5) is satisfied. Thus, by Theorem 2.4.6, BVP
(2.4.30) has a unique solution.
42
Chapter 3
Coupled systems of BVPs for fractionaldifferential equations
Existence of solutions for coupled systems of BVPs for FDEs have attracted the attention of researcher
quite recently. For example, Rehman and Khan [59], studied EPS for a coupled system of FDEs in the
Caputo’s sense. Li and Liu [60], discussed EUS for coupled system of FDEs by utilizing by the help of
some classical results. Ntouyas and Obaid [61], discussed a coupled system FDEs with integral BVP in
Caputo’s sense. Xu and Liu [101], developed iterative techniques for the existence and approximation
solutions for a coupled systems of two point BVPs for FDEs. Wei et al. [102], used monotone itera-
tive techniques for Initial value problems for FDEs, where the fractional differential operators were in
Riemann–Liouville Sense. Wang et al. [103], discussed the existence of solutions for FDEs by using
Monotone iterative method.
In this chapter, we study two types of coupled systems for existence and uniqueness of solutions.
In section 3.1, we study existence and uniqueness (EU) of maximal and minimal solutions of a general
class of coupled systems of BVPs and also provide error estimates. In section 3.2, we study existence
and uniqueness of solutions to coupled systems of BVPs of fractional order q–difference equations. The
results of this section are published [140]. The basic concept of the fractional q-difference calculus
are available in the research works of Al- Salam [44] and Agarwal [45]. Recently, q–difference FDEs
have studied by many scientists, for example, Ferreira [40], studied non-trivial solutions for fractional
q-difference boundary value problems. Ahmad et al. [42], studied nonlinear fractional q-difference
equations with nonlocal and sub-strip type boundary conditions. Ahmad and Ntouyas [41], considered
fractional q-difference hybrid equations and inclusions with Dirichlet boundary conditions. Yang [34],
studied positive solutions for BVPs involving nonlinear fractional q-difference equations. Zhao et al.
[35], studied q–difference equations with nonlocal q–integral boundary conditions. The detail study
43
about q–difference equations can be found in [36, 38, 39, 45–51, 53–55, 67].
3.1 Coupled systems of BVPs for FDEs
In this section, we develop monotone iterative technique for existence and uniqueness of minimal and
maximal solutions of the following coupled systems of BVPs
Dαx(t) = −G1(t, Iαx(t), Iβy(t)), α, β ∈ (3, 4]
Dβy(t) = −G2(t, Iαx(t), Iβy(t)),(3.1.1)
with conditions
Dδx(1) = 0 = I3−αx(0) = I4−αx(0), x(1) =Γ(α− δ)
Γ(α)Iα−δG1(t, Iαx(t), Iβy(t))(1),
Dνy(1) = 0 = I3−βy(0) = I4−βv(0), y(1) =Γ(β − ν)
Γ(β)Iβ−νG2(t, Iαx(t), Iβy(t))(1),
where t ∈ [0, 1], δ, ν ∈ (1, 2]. The functions G1, G2 : [0, 1] × R × R → R, satisfy the Caratheodory
conditions. The fractional derivatives Dα, Dβ, Dδ, Dν are in Riemann–Liouville sense and Iαx(t) =∫ t0
(t−s)α−1x(s)Γ(α) ds, Iβy(t) =
∫ t0
(t−s)β−1y(s)Γ(β) ds.
Let E = C[0, 1], with the norm ∥x∥ = maxt∈[0,1] |x(t)|, E ×E with the norm ∥(x, y)∥ = ∥x∥+∥y∥,
then obviously, E , E × E are Banach spaces. Let B be a cone in E and define partially order in E
as; for u, v ∈ E , v ≥ u if and only if v − u ∈ B. Further, define partially order in E × E as, for
(u, v), (x, y) ∈ E ×E , (u, v) ≥ (x, y) if and only if u ≥ x, v ≥ y. Then E , E ×E are partially ordered
Banach spaces. The following conditions are assumed:
(A1) G1, G2 : [0, 1]× R2 → R satisfy Caratheodory conditions;
(A2) (x0, y0) ∈ E × E , (x∗, y∗) ∈ E × E are lower and upper solutions respectively of the problem
(6.3.1), with x0 ≤ x∗, y0 ≤ y∗;
(A3) For any x, y, x1, y1 ∈ S with x ≥ x1, y ≥ y1, there exist nonnegative constants L1, L2, M1, M2
such that
0 ≤ G1(t, x(t), y(t))− G1(t, x1(t), y1(t)) ≤ L1(x− x1) + L2(y − y1),
0 ≤ G2(t, x(t), y(t))− G2(t, x1(t), y1(t)) ≤ N1(x− x1) +N2(y − y1).
Lemma 3.1.1. For s, t ∈ [0, 1], solutions of the coupled system (3.1.1) are solutions of the following
coupled systems of integral equations
x(t) =
∫ 1
0G1(t, s)G1(s, Iαx(s), Iβy(s))ds, (3.1.2)
y(t) =
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds, (3.1.3)
44
where Iαx(t) =∫ t0
(t−s)α−1x(s)Γ(α) ds, Iβy(t) =
∫ t0
(t−s)β−1y(s)Γ(β) ds and Gi(t, s)(i = 1, 2) are Green’s
functions given by
G1(t, s) =
−(t−s)α−1
Γα + tα−1(1−s)α−δ−1
Γ(α) + tα−2(1−s)α−1
Γ(α) , t ≥ s,
tα−1(1−s)α−δ−1
Γ(α) + tα−2(1−s)α−1
Γ(α) , t ≤ s,(3.1.4)
and
G2(t, s) =
−(t−s)β−1
Γβ + tβ−1(1−s)β−ν−1
Γ(β) + tβ−2(1−s)β−1
Γ(β) , t ≥ s,
tα−1(1−s)β−ν−1
Γ(β) + tβ−2(1−s)β−1
Γ(β) , t ≤ s.(3.1.5)
Proof. Applying the operator Iα0 on (3.1.1), we get
x(t) = −IαG1(t, Iαx(t), Iβy(t)) + c1tα−1 + c2t
α−2 + c3tα−3 + c4t
α−4,
y(t) = −IβG1(t, Iαx(t), Iβy(t)) + d1tβ−1 + d2t
β−2 + d3tβ−3 + d4t
β−4.(3.1.6)
Initial conditions I3−αx(0) = I4−αx(0) = 0 = I3−βy(0) = I4−βy(0) in (3.1.1) implies that c3 =
c4 = d3 = d4 = 0. Applying conditions Dδx(1) = 0 = Dνy(1) on (3.1.6), we get
c1 =
∫ 1
0
(1− s)α−δ−1G1(s, Iαx(s), Iβy(s))ds
Γ(α), (3.1.7)
d1 =
∫ 1
0
(1− s)β−ν−1G2(s, Iαx(s), Iβy(s))ds
Γ(β). (3.1.8)
From conditions x(1) = Γ(α−δ)Γ(α) Iα−δG1(t, Iαx(t), Iβy(t))(t = 1), y(1) = Γ(β−ν)
Γ(β) Iβ−νG2 (t, Iαx(t),
Iβy(t))(t = 1), we deduce that
c2 =
∫ 10 (1− s)α−1G2(s, Iαx(s), Iβy(s))ds
Γ(α), (3.1.9)
d2 =
∫ 10 (1− s)β−1G2(s, Iαx(s), Iβy(s))ds
Γ(β). (3.1.10)
By substituting the values of c1, c2, c3, c4, d1, d2, d3, d4 in the coupled system (3.1.6), we have
x(t) = −∫ t
0
(t− s)α−1
Γ(α)G1(s, Iαx(s), Iβy(s))ds+ tα−1
∫ 1
0
(1− s)α−δ−1
Γ(α)G1(s, Iαx(s), Iβy(s))ds
+ tα−2
∫ 1
0
(1− s)α−1
Γ(α)G1(s, Iαx(s), Iβy(s))ds =
∫ 1
0G1(t, s)G1(s, Iαx(s), Iβy(s))ds,
(3.1.11)
and
y(t) =
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds, (3.1.12)
where Gi(i = 1, 2) are the Green’s functions defined by (3.1.4), (3.1.5) respectively.
45
Lemma 3.1.2. For t, s ∈ [0, 1], α, β ∈ (3, 4], ν, δ ∈ (1, 2]. The Green’s functions defined by (3.1.4), (3.1.5)
satisfy the following:
(A1) Gi(t, s) ≥ 0, for i = 1, 2;
(A2)∫ 10 G1(t, s) ≤ 2
(α−δ)Γ(α) and∫ 10 G2(t, s) ≤ 2
(β−ν)Γ(β) .
Proof. ConsiderG1(t, s), for t, s ∈ [0, 1] such that t ≥ s. Since s ≤ st implies that −(1−s) ≤ −(1− s
t )
and α− δ − 1 < α− 1 implies that (1− s)α−δ−1 > (1− s)α−1, and
G1(t, s) =−(t− s)α−1
Γα+tα−1(1− s)α−δ−1
Γ(α)+tα−2(1− s)α−1
Γ(α)
=−tα−1(1− s
t )α−1
Γα+tα−1(1− s)α−δ−1
Γ(α)+tα−2(1− s)α−1
Γ(α)
(3.1.13)
≥ −tα−1(1− s)α−1
Γα+tα−1(1− s)α−δ−1
Γ(α)+tα−2(1− s)α−1
Γ(α)
=tα−1
Γ(α)((1− s)α−δ−1 − (1− s)α−1) +
tα−2(1− s)α−1
Γ(α)≥ 0.
From (3.1.4), for t ≤ s it is obvious that G1(t, s) ≥ 0. Similarly, we can show that G2(t, s) ≥ 0. This
completes the proof of (A1). For (A2), we consider s, t ∈ [0, 1], such that t ≥ s, we have:∫ 1
0G1(t, s)ds =
∫ 1
0(−(t− s)α−1
Γα+tα−1(1− s)α−δ−1
Γ(α)+tα−2(1− s)α−1
Γ(α))ds
=−(t− s)α
αΓα|01 +
tα−1(1− s)α−δ
(α− δ)Γ(α)|01 +
tα−2(1− s)α
αΓ(α)|01
= − tα
αΓ(α)+
tα−1
(α− δ)Γ(α)+
tα−2
αΓ(α)=
tα−2
α(α− δ)Γ(α)((t− t2)α+ α+ t2δ − δ)
≤ tα−2
α(α− δ)Γ(α)(tα+ α+ t2δ − δ) ≤ 1
α(α− δ)Γ(α)(α+ α+ δ − δ) =
2
(α− δ)Γ(α).
(3.1.14)
For s ≥ t, we have∫ 1
0G1(t, s)ds =
∫ 1
0(tα−1(1− s)α−δ−1
Γ(α)+tα−2(1− s)α−1
Γ(α))ds =
tα−1(1− s)α−δ
(α− δ)Γ(α)|01 +
tα−2(1− s)α
αΓ(α)|01
=tα−1
(α− δ)Γ(α)+
tα−2
αΓ(α)=
tα−2
α(α− δ)Γ(α)(tα+ α− δ) ≤ 1
α(α− δ)Γ(α)(2α− δ).
(3.1.15)
Since 1α(α−δ)Γ(α)(2α − δ) < 2
(α−δ)Γ(α) . Therefore, we have∫ 10 G1(t, s)ds ≤ 2
(α−δ)Γ(α) . Similarly, we
can show that∫ 10 G2(t, s)ds ≤ 2
(β−ν)Γ(β) .
46
3.1.1 Existence and uniqueness of solutions
In view of Lemma (3.1.1), the coupled systems (3.1.6) is equivalent to the following system of integral
equations:
x(t) =
∫ 1
0G1(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx(s), Iβy(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds)ds,
y(t) =
∫ 1
0G2(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx(s), Iβy(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds)ds.
(3.1.16)
Define an operator T : B × B → E × E by
T (x(t), y(t)) = (T1(x(t), y(t)), T2(x(t), y(t))), (3.1.17)
where,
T1(x(t), y(t)) =∫ 1
0G1(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx(s), Iβy(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds)ds,
(3.1.18)
T2(x(t), y(t)) =∫ 1
0G2(t, s)G2(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx(s), Iβy(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds)ds.
(3.1.19)
It is clear that the fixed points of the operator T are solutions of the system (3.1.1).
Now, we define the lower and upper solutions for the operator equation
T (x, y) = (x, y). (3.1.20)
We say a lower solution to the pair (x0, y0) of the operator equation (3.1.20), if (I − T )(x0, y0) ≤ 0,
and (x∗, y∗) be the upper solution of the operator equation (3.1.20) if (I − T )(x∗, y∗) ≥ 0.
The following notations are defined for the simplification in calculations:
M1 =2
(α− δ)Γ(α)(
L1
Γ(1 + α)
2
(α− δ)Γ(α)+
L2
Γ(1 + β)
2
(β − ν)Γ(β))(
N1
Γ(1 + α)+
N2
Γ(1 + β)),
(3.1.21)
M2 =2
(β − ν)Γ(β)(
L1
Γ(1 + α)
2
(α− δ)Γ(α)+
L2
Γ(1 + β)
2
(β − ν)Γ(β))(
N1
Γ(1 + α)+
N2
Γ(1 + β)),
(3.1.22)
and M = M1 +M2.
47
Theorem 3.1.3. Assume that the conditions (A1) and (A2) hold and M < 1, then the system (3.1.1) has
a minimal solution (x∗, y∗) and maximal solution (x∗∗, y∗∗). Assume the initial iterations x0, x∗0 and
form the following iterative sequence for each n ∈ Z,
xn(t) =
∫ 1
0G1(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαxn−1(s), Iβyn−1(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαxn−1(s), Iβyn−1(s))ds),
(3.1.23)
x∗n(t) =
∫ 1
0G1(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx∗n−1(s), Iβy∗n−1(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx∗n−1(s), Iβy∗n−1(s))ds),
(3.1.24)
yn(t) =
∫ 1
0G2(t, s)G2(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαn−1(s), Iβyn−1(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαxn−1(s), Iβyn−1(s))ds),
(3.1.25)
y∗n(t) =
∫ 1
0G2(t, s)G2(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx∗n−1(s), Iβy∗n−1(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx∗n−1(s), Iβy∗n−1(s))ds),
(3.1.26)
and
x0 ≤ x1 ≤ . . . ≤ xn ≤ . . . ≤ x∗n ≤ . . . ≤ x∗1 ≤ x∗0, (3.1.27)
y0 ≤ y1 ≤ . . . ≤ yn ≤ . . . ≤ y∗n ≤ . . . ≤ y∗1 ≤ y∗0, (3.1.28)
where
x∗ = limn→∞
xn(t), y∗ = limn→∞
yn(t), y∗ =
∫ 1
0G2(t, s)G2(s, Iαx∗(s), Iβy∗(s))ds, (3.1.29)
x∗∗ = limn→∞
x∗n(t), y∗∗ = lim
n→∞y∗n(t), y
∗∗ =
∫ 1
0G2(t, s)G2(s, Iαx∗∗(s), Iβy∗∗(s))ds, (3.1.30)
and errors estimations for the minimal, maximal solutions are given by
∥x∗(t)− xn(t)∥+ ∥y∗(t)− yn(t)∥ ≤ Mn
1−M{∥x1(t)− x0(t)∥+ ∥y1(t)− y0(t)∥},
∥x∗(t)− x∗∗n (t)∥+ ∥y∗(t)− y∗∗n (t)∥ ≤ Mn
1−M{∥x∗0(t)− x∗1(t)∥+ ∥y∗0(t)− y∗1(t)∥}.
Proof. Assume (x0, y0) ∈ E × E as the lower solution of (3.1.1), then we have:
x0(t) ≤∫ 1
0G1(t, s)G1(s, Iαx(s), Iβy(s))ds, y0(t) ≤
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds,
(3.1.31)
this leads us to
Iαx0(t) ≤ Iα
∫ 1
0G1(t, s)G1(s, Iαx(s), Iβy(s))ds,
Iβy0(t) ≤ Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds.
(3.1.32)
48
Thus
x0(t) ≤∫ 1
0G1(t, s)G1(s, Iαx(s), Iβy(s))ds
≤∫ 1
0G1(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx(s), Iβy(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds)ds,
= T1(x, y),
(3.1.33)
and
y0(t) ≤∫ 1
0G2(t, s)G1(s, Iαx(s), Iβy(s))ds
≤∫ 1
0G2(t, s)G1(s, Iα
∫ 1
0G1(t, s)
G1(s, Iαx(s), Iβy(s))ds, Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβy(s))ds)ds,
= T2(x, y).
(3.1.34)
Therefore, we have x0 lower solution of T1. From (A3),we have that for any x, y ∈ B such that
x0 ≤ x, y0 ≤ y, we get T1(x0, y0) ≤ T1(x, y), i.e., T1 is an increasing operator. Also from (3.1.23), for
any n ∈ Z+, we have
xn(t) = T1(xn−1(t), yn−1(t), yn(t) = T2(xn−1(t), yn−1(t)). (3.1.35)
By the help of assumption A2, we have:
x0 ≤ x1 ≤ . . . ≤ xn, y0 ≤ y1 ≤ . . . ≤ yn. (3.1.36)
For any x, y, z, p ∈ B such that x ≤ y, z ≤ p, it is obvious Iαx ≤ Iαy, Iβz ≤ Iβp and using C3 we
get G1(t, Iαx, Iβz) ≤ G1(t, Iαy, Iβp), G2(t, Iαx, Iβz) ≤ G2(t, Iαy, Iβp),∫ 1
0G1(t, s)G1(s, Iαx(s), Iβz(s))ds ≤
∫ 1
0G1(t, s)G1(s, Iαy(s), Iβp(s))ds,
Iα
∫ 1
0G1(t, s)G1(s, Iαx(s), Iβz(s))ds ≤ Iα
∫ 1
0G1(t, s)G1(s, Iαy(s), Iβp(s))ds,
(3.1.37)
and ∫ 1
0G2(t, s)G2(s, Iαx(s), Iβz(s))ds ≤
∫ 1
0G2(t, s)G2(s, Iαy(s), Iβp(s))ds,
Iβ
∫ 1
0G2(t, s)G2(s, Iαx(s), Iβz(s))ds ≤ Iβ
∫ 1
0G2(t, s)G2(s, Iαy(s), Iβp(s))ds.
(3.1.38)
49
It follows from (A3), (A2), that
0 ≤ G1(t, Iα
∫ 1
0K1(s, ξ)G1(ξ, Iαy, Iβp)dξ, Iβ
∫ 1
0K2(s, ξ)G2(ξ, Iαy, Iβp)dξ)
− G1(t, Iα
∫ 1
0K1(s, ξ)G1(ξ, Iαx, Iβz)dξ, Iβ
∫ 1
0K2(s, ξ)G2(ξ, Iαx, Iβz)dξ)
≤ L1
Γ(α)
∫ t
0(t− s)α−1
∫ 1
0K1(s, ξ)(N1(Iαy − Iαx) +N2(Iβp− Iβz))dξds
+L2
Γ(β)
∫ t
0(t− s)β−1
∫ 1
0K1(s, ξ)(N1(Iαy − Iαx) +N2(Iβp− Iβz))dξds.
(3.1.39)
Consequently, we have
∥G1(t, Iα
∫ 1
0K1(s, ξ)G1(ξ, Iαy, Iβp)dξ, Iβ
∫ 1
0K2(s, ξ)G2(ξ, Iαy, Iβp)dξ)
− G1(t, Iα
∫ 1
0K1(s, ξ)G1(ξ, Iαx, Iβz)dξ, Iβ
∫ 1
0K2(s, ξ)G2(ξ, Iαx, Iβz)dξ)∥
≤ L1
Γ(1 + α)
2
(α− δ)Γ(α)(
N1
Γ(1 + α)∥y − x∥+ N2
Γ(1 + β)∥p− z∥)
+L2
Γ(1 + β)
2
(β − ν)Γ(β)(
N1
Γ(1 + α)∥y − x∥+ N2
Γ(1 + β)∥p− z∥).
≤ (L1
Γ(1 + α)
2
(α− δ)Γ(α)+
L2
Γ(1 + β)
2
(β − ν)Γ(β))(
N1
Γ(1 + α)
+N2
Γ(1 + β))(∥y − x∥+ ∥p− z∥).
(3.1.40)
Using (3.1.21), (3.1.40), we have
∥T1(y(t), p(t))− T1(x(t), y(t))∥ ≤ M1(∥y − x∥+ ∥p− z∥). (3.1.41)
Similarly, we can have
∥T2(y(t), p(t))− T2(x(t), y(t))∥ ≤ M2(∥y − x∥+ ∥p− z∥). (3.1.42)
By the help of (3.1.41), (3.1.42), we deduce
∥T (y(t), p(t))− T (x(t), y(t))∥ ≤ M(∥y − x∥+ ∥p− z∥). (3.1.43)
Using (3.1.23), (3.1.25) and (3.1.43), we can have that
∥x2 − x1∥+ ∥y2 − v1∥ = ∥T (x1, y1)− T (x0, x0)∥ ≤ M(∥x1 − x0∥+ ∥y1 − y0∥). (3.1.44)
50
Similarly, we can get
∥x3 − x2∥+ ∥y3 − y2∥ = ∥T (x2, y2)− T (x1, y1)∥ ≤ M(∥x2 − x1∥+ ∥y2 − y1∥)
≤ M2(∥x1 − x0∥+ ∥y1 − y0∥),
∥x4 − x3∥+ ∥y4 − y3∥ ≤ M3(∥x1 − x0∥+ ∥y1 − y0∥),
∥x5 − x4∥+ ∥y5 − y4∥ ≤ M4(∥x1 − x0∥+ ∥y1 − y0∥),
∥x6 − x5∥+ ∥y6 − y5∥ ≤ M5(∥x1 − x0∥+ ∥y1 − y0∥),...
......
∥xn+1 − xn∥+ ∥yn+1 − yn∥ ≤ Mn(∥x1 − x0∥+ ∥y1 − y0∥).
(3.1.45)
For n, p ∈ Z+, we have that
∥xn+p − xn∥+ ∥yn+p − yn∥
≤ ∥xn+p − xn+p−1∥+ ∥yn+p − vn+p−1∥+ ∥xn+p−1 − xn+p−2∥
+ ∥yn+p−1 − yn+p−2∥+ . . .+ ∥xn+1 − x0n∥+ ∥yn+1 − yn∥
≤ Mn+p−1(∥x1 − x0∥+ ∥y1 − y0∥) +Mn+p−2(∥x1 − x0∥+ ∥y1 − y0∥) + . . .
+Mn(∥x1 − x0∥+ ∥y1 − y0∥)
= Mn 1−Mp
1−M(∥x1 − x0∥+ ∥y1 − y0∥).
(3.1.46)
For M ∈ (0, 1), we have that ∥xn+p − xn∥ + ∥yn+p − yn∥ → 0 as n → +∞. This implies that
∥xn+p − xn∥ → 0, ∥yn+p − yn∥ → 0 as n → +∞. Thus an = {xn}, bn = {yn} are Cauchy
sequences in B. Let x∗(t) = limn→∞ xn(t), y∗(t) = limn→∞ xn(t) thus T1(x∗(t), y∗(t)) = x∗ and
T2(x∗(t), y∗(t)) = y∗. From (3.1.46) we have, p → +∞, implies that ∥x∗ − xn∥ + ∥y∗ − yn∥ ≤Mn
1−M(∥x1 − x0∥+ ∥y1 − y0∥). Since the operators T1, T2 are increasing operators, therefore, we have
the sequence x0 ≤ x1 ≤ . . . ≤ xn ≤ . . . ≤ x∗n ≤ . . . ≤ x∗2 ≤ x∗1 ≤ x∗0.
Thus, the system (3.1.1) has a pair of solution (x∗(t), y∗(t)) = T (x∗(t), y∗(t)). Similarly, we can
show for the upper solution of the system (3.1.1), i.e., x∗∗ = limn→+∞ x∗n, y∗∗(t) = limn→+∞ x∗n. That
is, p→ +∞ implies ∥x∗n(t)−x∗∗(t)∥+ ∥y∗n(t)− y∗∗(t)∥ ≤ Mn
1−M(∥x1 −x0∥+ ∥y1− y0∥). Therefore,
(x∗, y∗) is a minimal solution and (x∗∗, y∗∗) is the maximal solution of the CSFDIEs (3.1.1).
Lemma 3.1.4. Under the assumptions C1 − C3 and M < 1. The minimal and maximal solutions of the
system (3.1.1) are unique.
Proof. Let us assume that (x∗, y∗) , (x∗∗, y∗∗) be the minimal, maximal solutions respectively for the
operator equation (3.1.16). Let there exist another pair of minimal solution(µ0, λ0) ∈ B × B. Then
µ0 ≤ T1(µ0, λ0) = µ1 and λ0 ≤ T2(µ0, λ0) = λ1. Consequently, we are having the sequence µ0 ≤µ1 ≤ . . . ≤ µn and λ0 ≤ λ1 ≤ . . . ≤ λn such that limn→∞ µn = µ∗ and limn→∞ λn = λ∗. Where
∥µ∗−xn∥+∥λ∗−yn∥ ≤ Mn
1−M(∥µ0−x0∥+∥λ0−y0∥). As n→ +∞ then ∥µ∗−x∗∥+∥λ∗−y∗∥ → 0, i.e.,
∥µ∗ − x∗∥ → 0, ∥λ∗ − y∗∥ → 0 consequently, we have µ∗ → x∗ and λ∗ → y∗. Therefore the minimal
51
solution (x∗, y∗) of the operator equation (3.1.16) is unique. Similarly, we can show the uniqueness of
the maximal solution (x∗∗, y∗∗).
Example 3.1.5. Consider the following system of coupled FDIEs as a special case of the coupled system
(3.1.1), for the EU of minimal and maximal solutions:
D3.5x(t) = −sin2(t)I3.5x(t) + cos2(t)I3.5y(t)
50 + t,
D3.5y(t) = −sin2(t)I3.5x(t) + cos2(t)I3.5y(t)
70 + t,
D1.3x(1) = 0 = I−0.5x(0) = I2.2x(0), x(1) =Γ(2.2)
Γ(3.5)I2.2G1(t, I3.5x(t), I3.5y(t))(1),
D1.3y(1) = 0 = I−0.5y(0) = I2.2y(0), y(1) =Γ(2.2)
Γ(3.5)I2.2G2(t, I3.5x(t), I3.5y(t))(1).
(3.1.47)
From (3.1.47), we have that β = α = 3.5, δ = ν = 1.3. G1(t, Iαx(t), Iβy(t)) = sin2(t)I3.5x(t)+cos2(t)I3.5y(t)50+t ,
G2(t, Iαx(t), Iβy(t)) = sin2(t)I3.5x(t)+cos2(t)I3.5y(t)70+t . For any (u, v), (x, y) ∈ E × E , such that (u, v) ≥
(x, y), we have
0 ≤ G1(t, Iαu(t), Iβv(t))− G1(t, Iαx(t), Iβy(t))
≤ 1
50(Iαu(t)− Iαx(t)) +
1
50(Iβv(t)− Iβy(t)),
0 ≤ G2(t, Iαu(t), Iβv(t))− G2(t, Iαx(t), Iβy(t))
≤ 1
70(Iαu(t)− Iαx(t)) +
1
70(Iβv(t)− Iβy(t)).
(3.1.48)
From (3.1.48), we have L1 = L2 = 150 , M1 = M2 = 1
70 . From these values, one can easily calculate
M = 1.26414 × 10−6 < 1. Therefore, by 3.1.4, there exist unique minimal and maximal solutions of
the problem (3.1.47).
Example 3.1.6. Consider the following system of FDIEs for error estimations:
D3.5x(t) = −sin2(t)I3.5x(t) + cos2(t)I3.5y(t)
50 + t,
D3.5y(t) = −sin2(t)I3.5x(t) + cos2(t)I3.5y(t)
70 + t,
D1.3x(1) = 0 = I−0.5x(0) = I2.2x(0), x(1) =Γ(2.2)
Γ(3.5)I2.2G1(t, I3.5x(t), I3.5y(t))(1),
D1.3y(1) = 0 = I−0.5y(0) = I2.2y(0), y(1) =Γ(2.2)
Γ(3.5)I2.2G2(t, I3.5x(t), I3.5y(t))(1).
(3.1.49)
From (3.1.49), we have that β = α = 3.5, δ = ν = 1.3, M = 1.26414× 10−6. We consider the lower
solution (−0.5,−0.5) and the upper solution (0.5, 0.5). For n = 1, we deduce that
∥x∗(t)− xn(t)∥+ ∥y∗(t)− yn(t)∥ ≤ Mn
1−M{∥x1(t)− x0(t)∥+ ∥y1(t)− y0(t)∥}
≤ 2.52828× 10−6,
∥x∗(t)− x∗∗n (t)∥+ ∥y∗(t)− y∗∗n (t)∥ ≤ Mn
1−M{∥x∗0(t)− x∗1(t)∥+ ∥y∗0(t)− y∗1(t)∥}
≤ 2.52828× 10−6.
52
The errors can further be decreased by increasing the iteration values.
3.2 Coupled systems of BVPs for fractional order q–difference equations
We study existence and uniqueness of solutions for a coupled systems of fractional order q-difference
equations of the form:
Dαq x(t) = −f(t,Dk
q y(t),Dp1q y(t)), γ ∈ (n− 1, n], k ≤ n− 1,
Dβq y(t) = −g(t,Dk
qx(t),Dp2q x(t)), β ∈ (n− 1, n], k ≤ n− 1,
(3.2.1)
with initial and boundary conditions
x(1) = 0, Dqx(1) = 0, Diqx(0) = 0, i = 2, 3, ..., n− 1,
y(1) = 0, Dqy(1) = 0, Djqy(0) = 0, j = 2, 3, ..., n− 1,
for t ∈ [0, 1] and n ≥ 3. Dγq , Dβ
q , Dp1q , Dp2
q , are Caputo’s fractional q–difference operators of order
α, β, p1, p2 where p1, p2 ∈ (0, 1) and Dkq , Di
q, Djq are q–difference operators of order k ≤ n−1, i =
2, 3, . . . , n− 1, j = 2, 3, . . . , n− 1, respectively; and f, g ∈ C([0, 1]× R2,R).
Lemma 3.2.1. For a given h(t) ∈ C([0, 1],R), the BVP
Dαq x(t) = −h(t), n− 1 < α ≤ n, n ≥ 3, t ∈ [0, 1] (3.2.2)
x(1) = 0, Dqx(1) = 0, Diqx(0) = 0, i = 2, 3, ..., n− 1, (3.2.3)
has a solution x(t) of the form
x(t) =−1
Γq(α)
∫ t
0(t− qs)α−1h(s)dqs+
1
Γq(α)
∫ 1
0(1− qs)α−1h(s)dqs
− 1
Γq(α− 1)
∫ 1
0(1− qs)α−2h(s)dqs+
t
Γq(α− 1)
∫ 1
0(1− qs)α−2h(s)dqs.
(3.2.4)
Proof. Applying fractional q integral Iαq on (3.2.2) and using lemma 1.1.7, we obtain
x(t) = −Iαq h(t) + a0 + a1t+ a2t
2 + ...+ an−1tn−1. (3.2.5)
Using the given conditions (3.2.3), we get
a0 = Iαq h(t)|t=1 − Iα−1
q h(t)|t=1, a1 = Iα−1q h(t)|t=1, a2 = a3 = ... = an−1 = 0.
Thus, (3.2.5) takes the form
x(t) = −Iαq h(t) + Iα
q h(t)|t=1 − Iα−1q h(t)|t=1 + tIα−1
q h(t)|t=1. (3.2.6)
53
The q-fractional integral form of (3.2.6) is
x(t) =−1
Γq(α)
∫ t
0(t− qs)α−1h(s)dqs+
1
Γq(α)
∫ 1
0(1− qs)α−1h(s)dqs
− 1
Γq(α− 1)
∫ 1
0(1− qs)α−2h(s)dqs+
t
Γq(α− 1)
∫ 1
0(1− qs)α−2h(s)dqs.
(3.2.7)
Conversely, applying fractional q–differential operator Dαq on (3.2.7) and using the basic property
Dαq Iα
q h(t)(t) = h(t), we have
Dαq x(t) = Dα
q
(− 1
Γq(α)
∫ t
0(t− qs)α−1h(s)dqs+
1
Γq(α)
∫ 1
0(1− qs)α−1h(s)dqs
− 1
Γq(α− 1)
∫ 1
0(1− qs)α−2h(s)dqs+
t
Γq(α− 1)
∫ 1
0(1− qs)α−2h(s)dqs)
= −Dαq Iα
q h(t) = −h(t),
and the boundary conditions (3.2.3) are also true, which can be easily verified from (3.2.7).
Consider E = {x(t) : Dpqx(t) ∈ C and x(t) ∈ C(n)}, with the norm, ∥x∥ = maxk≤n−1
maxt∈[0,1] |Dkqx(t)|+maxt∈[0,1] |D
pqx(t)|, where k ∈ N, p ∈ (0, 1); the space E × E equipped with the
norm, ∥(x, y)∥ = ∥x∥+ ∥y∥. Obviously, E , E ×E are Banach spaces. Let us define T : E ×E → E ×E
by
T (x, y)(t) = (T1(y), T2(x))(t),
where,
T1(y(t)) =−1
Γq(α)
∫ t
0(t− qs)α−1f(s,Dk
q y(s),Dp1q y(s))dqs
+1
Γq(α)
∫ 1
0(1− qs)α−1f(s,Dk
q y(s),Dp1q y(s))dqs
− 1
Γq(α− 1)
∫ 1
0(1− qs)α−2f(s,Dk
q y(s),Dp1q y(s))dqs
+t
Γq(α− 1)
∫ 1
0(1− qs)α−2f(s,Dk
q y(s),Dp1q y(s))dqs,
(3.2.8)
and
T2(x(t)) =−1
Γq(β)
∫ t
0(t− qs)β−1g(s,Dk
qx(s),Dp2q x(s))dqs
+1
Γq(β)
∫ 1
0(1− qs)β−1g(s,Dk
qx(s),Dp2q x(s))dqs
− 1
Γq(β − 1)
∫ 1
0(1− qs)β−2g(s,Dk
qx(s),Dp2q x(s))dqs
+t
Γq(β − 1)
∫ 1
0(1− qs)β−2g(s,Dk
qx(s),Dp2q x(s))dqs.
(3.2.9)
54
We define the following terms for our convenience:
ϖ1 =1
Γq(α− k + 1)+
1
Γq(α− p1)+
1
Γ(2− p1)Γq(α), (3.2.10)
ϖ2 =1
Γq(β − k + 1)+
1
Γq(β − p2)+
1
Γ(2− p2)Γq(β), (3.2.11)
and
M = min{1− (k1ϖ1 + λ1ϖ2), 1− (λ2ϖ2 +ϖ1k2)}. (3.2.12)
3.2.1 Existence of solutions
Theorem 3.2.2. For f, g ∈ C( [0, 1]×R2,R), let there exist real constantsm1, m2, k1, k2, λ1, λ2 such
that for all u1, v1 ∈ R, we have
|f(t,Dkq y(t),Dp1
q y(t))| ≤ m1 + k1|Dkq y(t)|+ k2|Dp1
q y(t)|,
|g(t,Dkqx(t),Dp2
q x(t))| ≤ m2 + λ1|Dkqx(t)|+ λ2|Dp2
q x(t)|.
In addition,
k1ϖ1 + λ1ϖ2 < 1, λ2ϖ2 +ϖ1k2 < 1,
where ϖ1, ϖ2 are defined by (3.2.10), (3.2.11), respectively. Then, the system (3.2.1) has a positive
solution.
Proof. It is enough to show that the operator T has a fixed point. We divide the proof in three steps.
Step 1: We prove that the operator T maps a bounded set into a bounded. For this, consider Sρ =
{(x, y)(t) : ∥(x, y)∥ ≤ ρ} ⊆ E × E . Then for any (x, y)(t) ∈ Sρ and t ∈ [0, 1], there exist constants
M1,M2 such that
|f(t,Dkq y(t),Dp1
q y(t))| ≤ M1, |g(t,Dkqx(t),Dp2
q x(t))| ≤ M2,
then, by the virtue of (3.2.3), we get
|DkqT1(y(t))| = | − 1
Γq(α− k)
∫ t
0(t− qs)α−k−1f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ M1
Γq(α− k + 1),
(3.2.13)
and
|Dp1q T1(y(t))| = | − 1
Γq(α− p1)
∫ t
0(t− qs)α−p1−1f(s,Dk
q y(s),Dp1q y(s))dqs
+t1−p1
Γ(2− p1)Γq(α− 1)
∫ 1
0(1− qs)α−2f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ M1{1
Γq(α− p1)+
1
Γ(2− p1)Γq(α)}.
(3.2.14)
55
Using (3.2.13), (3.2.14) and definition of the norm, we deduce that
∥T1(y)∥ ≤ M1{1
Γq(α− k + 1)+
1
Γq(α− p1)+
1
Γ(2− p1)Γq(α)}, (3.2.15)
∥T2(x)∥ ≤M2{1
Γq(β − k + 1)+
1
Γq(β − p2)+
1
Γ(2− p2)Γq(β)}. (3.2.16)
From (3.2.15), (3.2.16), we have:
∥T (x, y)∥ ≤ {M1{1
Γq(α− k + 1)+
1
Γq(α− p1)+
1
Γ(2− p1)Γq(α)}
+M2{1
Γq(β − k + 1)+
1
Γq(β − p2)+
1
Γ(2− p2)Γq(β)}}.
(3.2.17)
Thus ∥T (x, y)∥ ≤ ϖ1M1 +ϖ2M2 and hence T is uniformly bounded.
Step 2: In this step, we show that the operator T is an equicontinuous operator. For this, we assume
(x, y)(t) ∈ Sρ and t1, t2 ∈ [0, 1] with t1 < t2. Then, from (3.2.8), we have the following estimates:
|DkqT1(y(t2))−Dk
qT1(y(t1))|
= | − 1
Γq(α− k)
∫ t2
0(t2 − qs)α−k−1f(s,Dk
q y(s),Dp1q y(s))dqs
+1
Γq(α− k)
∫ t1
0(t1 − qs)α−k−1f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ M1tα−k2 − tα−k
1
Γq(α− k + 1),
(3.2.18)
and
|Dp1q T1(y(t2)) − Dp1
q T1(y(t1))|
= | − 1
Γq(α− p1)
∫ t2
0(t2 − qs)α−p1−1f(s,Dk
q y(s),Dp1q y(s))dqs
+1
Γq(α− p1)
∫ t1
0(t1 − qs)α−p1−1f(s,Dk
q y(s),Dp1q y(s))dqs (3.2.19)
+t1−p12 − t1−p1
1
Γ(2− p1)Γq(α− 1)
∫ 1
0(1− qs)α−2f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ M1tα−k2 − tα−k
1
Γq(α− k + 1)+
t1−p12 − t1−p1
1
Γ(2− p1)Γq(α).
From (3.2.18), (3.2.19), we get
∥T1(y(t2))− T1(y(t1))∥ = | − 1
Γq(α− k)
∫ t2
0(t2 − qs)α−k−1f(s,Dk
q y(s),Dp1q y(s))dqs
+1
Γq(α− k)
∫ t1
0(t1 − qs)α−k−1f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ M1{tα−k2 − tα−k
1
Γq(α− k + 1)+
tα−k2 − tα−k
1
Γq(α− k + 1)+
t1−p12 − t1−p1
1
Γ(2− p1)Γq(α)},
(3.2.20)
∥T2(x(t2))− T2(x(t1))∥ ≤ M2{tβ−k2 − tβ−k
1
Γq(β − k + 1)+
tβ−k2 − tβ−k
1
Γq(β − k + 1)+
t1−p22 − t1−p2
1
Γ(2− p1)Γq(β)}. (3.2.21)
56
From (3.2.20)–(3.2.21), we deduce that
∥T (x, y)(t2)− T (x, y)(t1)∥
≤ M1{tα−k2 − tα−k
1
Γq(α− k + 1)+
tα−k2 − tα−k
1
Γq(α− k + 1)+
t1−p12 − t1−p1
1
Γ(2− p1)Γq(α)}
+M2{tβ−k2 − tβ−k
1
Γq(β − k + 1)+
tβ−k2 − tβ−k
1
Γq(β − k + 1)+
t1−p22 − t1−p2
1
Γ(2− p1)Γq(β)},
(3.2.22)
which tends to zero independent of (x, y) as t1 → t2. So by Arzela Ascoli theorem, we have that T is
completely continuous operator.
Step 3: In this step, we show that the set:
S∗ = {(x, y)(t) ∈ E × E : µT (x, y)(t) = (x, y)(t), 0 < µ < 1},
is a bounded set. For this, we take (x, y) ∈ S∗. Then (x, y) = µT(x, y), implies that
x(t) = µT1(y(t)), y(t) = µT2(x(t)),
|Dkqx(t)| ≤
1
Γq(α− k + 1)(m1 + k1|Dk
q y(t)|+ k2|Dp1q y(t)|), (3.2.23)
and
|Dp1q x| ≤ { 1
Γq(α− p1)+
1
Γ(2− p1)Γq(α)}(m1 + k1|Dk
q y(t)|+ k2|Dp1q y(t)|). (3.2.24)
Therfore, by (3.2.10), (3.2.23), (3.2.24), we have
∥x∥ ≤ { 1
Γq(α− k + 1)+
1
Γq(α− p1)+
1
Γ(2− p1)Γq(α)}(m1 + k1|Dk
q y(t)|+ k2|Dp1q y(t)|)
= ϖ1(m1 + k1|Dkq y(t)|+ k2|Dp1
q y(t)|).(3.2.25)
Similarly, from (3.2.9) we have
∥y∥ ≤ ϖ2(m2 + λ1|Dkqx(t)|+ λ2|Dp2
q x(t)|). (3.2.26)
From (3.2.25), (3.2.26), it follows that
∥(x, y)∥ ≤ ϖ1m1 +ϖ2m2 + (k1ϖ1 + λ1ϖ2)∥x∥+ (λ2ϖ2 +ϖ1k2)∥y∥. (3.2.27)
Consequently, we get
∥(x, y)∥ ≤ ϖ1m1 +ϖ2m2
M, (3.2.28)
where M is given by (3.2.12). It shows that S∗ is a bounded set. Thus, Leray-Schauder alternative
affirms that the operator T , has a fixed point. Hence, the problem (3.2.1) has at least one positive
solution.
57
3.2.2 Uniqueness of solutions
In the following result, we discuss the uniqueness of solution for the problem (3.2.1) with the help of
Banach’s contraction principle.
Theorem 3.2.3. Let f, g ∈ C([0, 1],R2,R) and there exist real constants µ1, µ2, ν1, ν2 such that for all
t ∈ [0, 1] and x1, x2, y1, y2 ∈ R,
|f(t,Dkq y2,Dp1
q y2)−f(t,Dkq y1,Dp1
q y1)| ≤ µ1|Dkq y2 −Dk
q y1|+ µ2|Dp1q y2 −Dp1
q y1|,
and
|g(t,Dkqx2,Dp2
q x2)−g(t,Dkqx1,Dp2
q x1)| ≤ ν1|Dkqx2 −Dk
qx1|+ ν2|Dp2q x2 −Dp2
q x1|.
in addition, provided that
ϖ1(µ1 + µ2) +ϖ2(ν1 + ν2) < 1,
where ϖ1, ϖ2 are defined by (3.2.10)–(3.2.11). Then, the problem (3.2.1) has a unique solution.
Proof. We define the terms, supt∈[0,1] f(t, 0, 0) = N1 <∞, supt∈[0,1] g(t, 0, 0) = N2 <∞ and the set
Sν = {(x, y)(t) ∈ E × E : ∥(x, y)∥ ≤ ν}, where
ν >N1ϖ1 +N2ϖ2
1−ϖ1(µ1 + µ2)−ϖ2(ν1 + ν2).
Let (x, y) ∈ Sν . Then
|DkqT1(y(t))| = | − 1
Γq(α− k)
∫ t
0(t− qs)α−k−1f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ 1
Γq(α− k)
∫ t
0(t− qs)α−k−1{(|f(s,Dk
q y(s),Dp1q y(s))− f(s, 0, 0)|)
+ |f(s, 0, 0)|}dqs
≤ 1
Γq(α− k + 1){(|f(s,Dk
q y(s),Dp1q y(s))− f(s, 0, 0)|) + |f(s, 0, 0)|}
≤ 1
Γq(α− k + 1){µ1|Dk
q y|+ µ2|Dp1q y|+N1}
≤ { 1
Γq(α− k + 1)(µ1 + µ2)ν +N1},
(3.2.29)
and
|Dp1q T1(y(t))| = | − 1
Γq(α− p1)
∫ t
0(t− qs)α−p1−1f(s,Dk
q y(s),Dp1q y(s))dqs
+t1−p1
Γ(2− p1)Γq(α− 1)
∫ 1
0(1− qs)α−2f(s,Dk
q y(s),Dp1q y(s))dqs|
≤ 1
Γq(α− p1)
∫ t
0(t− qs)α−p1−1{|f(s,Dk
q y(s),Dp1q y(s))− f(s, 0, 0)|
(3.2.30)
58
+ |f(s, 0, 0)|}dqs+ t1−p1
∫ 1
0
(1− qs)α−2
Γ(2− p1)Γq(α− 1){|f(s,Dk
q y(s),Dp1q y(s))
− f(s, 0, 0)|+ |f(s, 0, 0)|}dqs|
≤ (1
Γq(α− p1 + 1)+
1
Γ(2− p1)Γq(α)){µ1|Dk
q y|+ µ2|Dp1q y|+N1},
taking the norm, we get
∥T1(y)∥ ≤ ϖ1{(µ1 + µ2)ν +N1}, (3.2.31)
similarly,
∥T2(x)∥ ≤ ϖ2{(ν1 + ν2)ν +N2}. (3.2.32)
Consequently, we have
∥T (x, y)∥ ≤ ν. (3.2.33)
Therefore, T Sν ⊂ Sν . Now, we show that T is contraction. For this, we have
|DkqT1(y2(t))−Dk
qT1(y1(t))|
= | − 1
Γq(α− k)
∫ t
0(t− qs)α−k−1f(s,Dk
q y2(s),Dp1q y2(s))− f(s,Dk
q y1(s),Dp1q y1(s))dqs|
≤ 1
Γq(α− k + 1){µ1|Dk
q y2(s)−Dkq y1(s)|+ µ2|Dp1
q y2(s)−Dp1q y1(s)|},
(3.2.34)
and
|Dp1q T1(y2(t))−Dp1
q T1(y1(t))|
= | 1
Γq(α− p1)
∫ t
0(t− qs)α−p1−1|f(s,Dk
q y2(s),Dp1q y2(s))− f(s,Dk
q y1(s),Dp1q y1(s))|dqs
+t1−p1
Γ(2− p1)Γq(α− 1)
∫ 1
0(1− qs)α−2|f(s,Dk
q y2(s),Dp1q y2(s))− f(s,Dk
q y1(s),Dp1q y(s))|dqs
≤ (1
Γq(α− p1 + 1)+
1
Γ(2− p1)Γq(α)){µ1|Dk
q y2(s)−Dkq y1(s)|+ µ2|Dp1
q y2(s)−Dp1q y1(s)|}.
(3.2.35)
It follows from (3.2.34)–(3.2.35), we get
∥T1(y2)− T1(y1)∥
≤ { 1
Γq(α− k + 1)+
1
Γq(α− p1 + 1)+
1
Γ(2− p1)Γq(α)}{µ1|Dk
q y2(s)−Dkq y1(s)|
+ µ2|Dp1q y2(s)−Dp1
q y1(s)|}
= ϖ1{µ1|Dkq y2(s)−Dk
q y1(s)|+ µ2|Dp1q y2(s)−Dp1
q y1(s)|}
≤ ϖ1(µ1 + µ2){|Dkq y2(s)−Dk
q y1(s)|+ |Dp1q y2(s)−Dp1
q y1(s)|}.
(3.2.36)
Similarly, we can obtain
∥T2(x2)− T2(x1)∥ ≤ ϖ2(ν1 + ν2){|Dkqx2(s)−Dk
qx1(s)|+ |Dp1q x2(s)−Dp1
q x1(s)|}. (3.2.37)
59
From (3.2.36) − (3.2.37), we deduce that
∥T (x2, y2)− T (x1, y1)∥ ≤ (ϖ1(µ1 + µ2) +ϖ2(ν1 + ν2)){∥x2 − x1∥+ ∥y2 − y1∥}. (3.2.38)
whereϖ1(µ1+µ2)+ϖ2(ν1+ν2) < 1. Thus, T is contraction and Banach’s fixed point theorem affirms
the fixed point of T . So the problem (3.2.1) has a unique positive solution.
3.2.3 Examples
In this section, we are concerned with two illustrative examples for the applications of our work, i.e.,
Theorems 3.2.2 and 3.2.3. In these examples, we consider n = 3.
Example 3.2.4. We investigate the existence of positive solution for the following problem of q-difference
equations
D2.5.5 x(t) = − t+D2
0.5y +D0.50.5y
70,
D2.5.5 y(t) = − t+D2
0.5x+D0.50.5x
60,
x(1) = 0, D0.5x(1) = 0, D(2)0.5x(0) = 0,
y(1) = 0, D0.5y(1) = 0, D(2)0.5y(0) = 0.
(3.2.39)
β = γ = 2.5, p1 = p2 = 0.5 = q, k = 2, k1 = k2 = 170 , λ1 = λ2 = 1
60 . From these values and
(3.2.11)–(3.2.12), we find thatϖ1 = ϖ2 = 27.2018 and k1ϖ1+λ1ϖ2 = λ2ϖ2+ϖ1k2 = 0.84196 < 1.
Therefore, by Theorem 3.2.2, the coupled system (3.2.39), has a solution.
Example 3.2.5. Consider the problem:
D2.5.5 x(t) = −
(t+
D20.5y
100+
D0.50.5y
110
),
D2.5.5 y(t) = −
(t+
D20.5x
150+
D0.50.5x
160
),
x(1) = 0, D0.5x(1) = 0, D20.5x(0) = 0,
y(1) = 0, D0.5y(1) = 0, D20.5y(0) = 0.
(3.2.40)
From (3.2.40), we have
|f(t,D20.5y2,D0.5
0.5y2)−f(t,D20.5y1,D0.5
0.5y1)| ≤1
100|D2
0.5y2−D20.5y1|+
1
110|D0.5
0.5y2−D0.50.5y1|, (3.2.41)
|g(t,D20.5y2,D0.5
0.5y2)−g(t,D20.5y1,D0.5
0.5y1)| ≤1
150|D2
0.5y2−D20.5y1|+
1
160|D0.5
0.5y2−D0.50.5y1|, (3.2.42)
and (3.2.41)–(3.2.42) implies that µ1 = 1100 , µ2 = 1
110 , ν1 = 1150 , ν2 = 1
160 . Using these values, we
find that ϖ1(µ1 + µ2) + ϖ2(ν1 + ν2) = 0.870663 < 1. Therefore, by Theorem 3.2.3, the coupled
system (3.2.40), has a unique solution.
60
Chapter 4
Coupled systems of hybrid FDEs
Recently, the theory on EUS and multiplicity of nonlinear BVPs for FDEs have attracted much attentions
and is fast growing area of research, we refer to [57, 58] and the references therein for some of valuable
and recent achievements in the study of existence theory for BVPs corresponding to FDEs. Existence
of positive solution (EPS) for coupled system of hybrid FDEs were considered by many scientists, for
example, Hedrih [65] and Dhage [66,71] explored some aspects of Hybrid differential equations, Ahmad
et al. [62], studied a coupled system of hybrid FDEs by classical approach, they considered the fractional
derivative in Caputo’s sense. Herzallah and Baleanu [63], studied existence of solutions for HFDEs in
Caputo’s sense. Ahmad and Ntouyas [64], discussed the existence of solutions for hybrid Hadamard
FDEs with initial value problems by a fixed point theorem due to Dhage.
In this chapter, we consider the EUS for two types of coupled systems of HFDEs. The results of this
chapter are published in international journal [131, 134]. In section 4.1, we consider the EUS for the
coupled system I. The results for EUS are obtained in subsection 4.1.1 and are applied to two coupled
systems of HFDEs in subsection 4.1.2. In section 4.2, the coupled system of HFDEs II is considered
for the EUS. The results are obtained in the subsection 4.2.1 and applied to two problems in subsection
4.2.2.
4.1 Coupled system of hybrid FDEs I
In this section, we study EUS for the following coupled systems of HFDEs
Dα(x(t)
H(t, x(t), y(t))) = −K1(t, x(t), y(t)), α ∈ (2, 3],
Dβ(y(t)
G(t, x(t), y(t))) = −K2(t, x(t), y(t)), β ∈ (2, 3], (4.1.1)
61
x(t)
H(t, x(t), y(t))|t=1 = 0, Dµ(
x(t)
H(t, x(t), y(t)))|t=δ1 = 0, x(2)(0) = 0,
y(t)
G(t, x(t), y(t))|t=1 = 0, Dν(
y(t)
G(t, x(t), y(t)))|t=δ2 = 0, y(2)(0) = 0,
where, t ∈ [0, 1]. δ1, δ2 ∈ (0, 1), α, β ∈ (2, 3], and Dα, Dβ , Dµ, Dν are Caputo’s fractional derivative
of orders α, β, µ, ν respectively, K1,K2 ∈ C([0, 1]×R2,R), H,G ∈ C([0, 1]×R2,R−{0}) and x, y ∈
C([0, 1],R). Here we remark that H(t, x, y) can not be of O(x) and G(t, x(t), y(t)) can not be of O(t).
Because if H(t, x, y) is of O(x) and G(t, x(t), y(t)) is of O(t), then the system (4.1.1) will reduce to a
coupled system of algebraic equations and will lose its hybrid nature. It is investigated that dynamics of
multi deformable bodies such as beams, plates, membranes etc coupled by standard light fractional order
discrete continuous layers is described by coupled fractional order hybrid system dynamics [65]. These
are generally perturbed FDEs and are mainly classified as perturbed differential equations of the first and
second type. The Perturbation itself is of two types, namely, linear and quadratics whose details can be
studied in [65, 66, 71]. Here, our system 4.1.1 is not an artificial one but in fact a quadratic perturbation
of the second type. These types of systems are generally tackled with the use of hybrid fixed point
theory. The significance of the hybrid system of differential equations lies in the fact that they include
several dynamical systems as special cases. For example, if we choose H(t, x, y) = 1, G(t, x, y) = 1,
as constant functions, then our system 4.1.1 reduce to a class of coupled system of boundary value
problems for FDEs of the type:
Dαx(t) = −K1(t, x(t), y(t)), α ∈ (2, 3],
Dβy(t) = −K2(t, x(t), y(t)), β ∈ (2, 3],
x(t)|t=1 = 0, Dµx(t)|t=δ1 = 0, x(2)(0) = 0,
y(t)|t=1 = 0, Dνy(t)|t=δ2 = 0, z(2)(0) = 0.
Lemma 4.1.1. For x ∈ AC3[0, 1] and any h(t) ∈ C[0, 1], the solution of the hybrid FDE:
Dα(x(t)
H(t, x, z)) = −h(t), α ∈ (2, 3], (4.1.2)
x(t)
H(t, x, y)|t=1 = 0, Dµ(
x(t)
H(t, x, y))|t=δ1 = 0, x(2)(0) = 0, µ ∈ (0, 1),
is given by
x(t) = H(t, x, y)[−∫ t
0
(t− s)α−1
Γ(α)h(s)ds+
∫ 1
0
(1− s)α−1
Γ(α)h(s)ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1h(s)ds]. (4.1.3)
Proof. Applying the operator Iα (4.1.2), we obtain
x(t)
H(t, x, y)= −Iαh(t) + c1 + c2t+ c3t
2, (4.1.4)
62
by the use of initial condition x(2)(0) = 0, we get c3 = 0, and hence, (4.1.4) takes the form
x(t)
H(t, x, y)= −Iαh(t) + c1 + c2t. (4.1.5)
Applying Caputo’s fractional derivative of order µ, on (4.1.5), we get
Dµ(x(t)
H(t, x, y)) = −Iα−µh(t) + c2
t1−µ
Γ(2− µ). (4.1.6)
The boundary condition Dµ( x(t)H(t,x,α))|t=δ1 = 0, yields c2 = Iα−µ
∆1Z(δ1), where ∆1 =
δ1−µ1
Γ(2−µ) and the
boundary condition xH(t,x,y) |t=1 = 0 yields c1 = IαZ(1) − Iα−µ
∆1Z(δ1). Substituting the values of
c1, c2, in (4.1.6), we get:
x(t)
H(t, x, y)= −Iαh(t) + IαZ(1) + (t− 1)
Iα−µ
∆1Z(δ1). (4.1.7)
The equivalent integral form of (4.1.7) is:
x(t) = H(t, x, y)[−∫ t
0
(t− s)α−1
Γ(α)h(s)ds+
∫ 1
0
(1− s)α−1
Γ(α)h(s)ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1h(s)ds
]. (4.1.8)
Thus, the proof is completed.
4.1.1 Existence and uniqueness of solutions
In this section, we are concerned with the EUS for the coupled system of HFDEs (4.1.1) by using
some classical results. For this, consider the Banach space E = {x(t) : x(t) ∈ C[0, 1]} with norm
∥x∥ = maxt∈[0,1] |x(t)| and (E × E , ∥(., .)∥) with norm ∥(x, y)∥ = ∥x∥ + ∥y∥. We define an operator
T : E × E → E × E by
T (x, y)(t) = (T1(x, y)(t), T2(x, y)(t)), (4.1.9)
where
T1(x, y)(t) = H(t, x, y)[−∫ t
0
(t− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, x(s), y(s))ds (4.1.10)
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds
],
and
T2(x, y)(t) = G(t, x, y)[−∫ t
0
(t− s)β−1
Γ(β)K2(s, x(s), y(s))ds
+
∫ 1
0
(1− s)β−1
Γ(β)K2(s, x(s), y(s))ds (4.1.11)
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1K2(s, x(s), y(s))ds
],
63
where ∆1 =δ1−µ1
Γ(2−µ) , ∆2 =δ1−ν2
Γ(2−ν) . In view of Lemma 4.1.1, the solutions of the coupled system of
HFDEs (4.1.1), is a fixed point of the operator T . Define the following terms:
N1 =2
Γ(1 + α)+
2δα−µ1
∆1Γ(α− µ+ 1), (4.1.12)
N2 =2
Γ(1 + β)+
2δβ−ν2
∆2Γ(β − ν + 1). (4.1.13)
Theorem 4.1.2. Assume the following:
• (A1) There exists real constants µi > 0 (for i = 1, 2, 3, 4) such that |H(t, x1, y1)−H(t, x2, y2)| ≤µ1|x1−x2|+µ2|y1−y2|, |G(t, x1, y1)−G(t, x2, y2)| ≤ µ3|x1−x2|+µ4|y1−y2|, for (t, xj , yj) ∈([0, 1]× R× R), j = 1, 2.
• (A2) There exist functions Mi(t) ∈ C[0, 1] (for i = 1, 2) such that |K1(t, x, y)| ≤ M1(t), |K2(t, x, y)| ≤M2(t) for (t, x, y) ∈ ([0, 1]× R× R).
• (A3) γ(N1∥M1∥+N2∥M2∥) < 1, for γ = µ1 + µ2 + µ3 + µ4.
• (A4)y
1−ρ
(H0N1∥M1∥1−λN1∥M1∥ + G0N2∥M2∥
1−λN2∥M2∥
)< r for H0 = maxt∈[0,1] |H(t, 0, 0)| and G0 = maxt∈[0,1] |G(t, 0, 0)|
and ρ = max{ λN1∥M1∥1−λN1∥M1∥ ,
λN2∥M2∥1−λN2∥M2∥}, where λ ∈ (0, 1).
then, the coupled system of HFDEs (4.1.1), has a solution.
Proof. We give the proof in the following four steps. The operators T1, T2 given in (4.1.10), (4.1.11)
are equivalent to:
T1(x(t), y(t)) = f1(x(t), y(t))g1(x(t), y(t)) = (x(t), y(t)) t ∈ [0, 1], (4.1.14)
T2(x(t), y(t)) = f2(x(t), y(t))g2(x(t), y(t)) = (x(t), y(t)) t ∈ [0, 1]. (4.1.15)
Thus, the operator T defined in (4.1.9) can be expressed in the form:
T (x, y)(t) = (T1(x, y)(t), T2(x, y)(t)) = A(x(t), y(t))B(x(t), y(t)), (4.1.16)
where F1 = (f1, f2), F2 = (g1, g2) and fi, gi : Br(0) → E for i = 1, 2 are defined by
f1(x(t), y(t)) = H(t, x, y), f2(x(t), y(t)) = G(t, x, y), (4.1.17)
g1(x(t), y(t)) = −∫ t
0
(t− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, x(s), y(s))ds (4.1.18)
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds,
64
g2(x(t), y(t)) = −∫ t
0
(t− s)β−1
Γ(β)K2(s, x(s), y(s))ds
+
∫ 1
0
(1− s)β−1
Γ(β)K2(s, x(s), y(s))ds (4.1.19)
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1K2(s, x(s), y(s))ds,
Step 1. The operators f1, f2 are Lipschitz on E . For this, let (x1, z1), (x2, z2) ∈ E × E and t ∈ [0, 1];
then by (A1) we get
|f1(x1, y1)(t)− f1(x2, y2)(t)| = |H(x1, y1)−H(x2, y2)|
≤ µ1|x1(t)− y1(t)|+ µ2|x2(t)− y2(t)| (4.1.20)
≤ (µ1 + µ2)∥(x1, y1)− (x2, y2)∥.
Similarly, we obtain
|f2(x1, y1)(t)− f2(x2, y2)(t)| = |G(x1, y1)− G(x2, y2)|
≤ µ3|x1(t)− y1(t)|+ µ2|x4(t)− y2(t)| (4.1.21)
≤ (µ3 + µ4)∥(x1, y1)− (x2, y2)∥.
From (4.1.20) and (4.1.21), we obtain
|A(x1, y1)(t) − A(x2, y2)(t)| = |(f1, f2)(x1, y1)− (f1, f2)(x2, y2)|
≤ (µ1 + µ2 + µ3 + µ4)∥(x1, y1)− (x2, y2)∥. (4.1.22)
Thus, (4.1.22) implies that F1 is Lipschitz on E × E with Lipschitz constant µ1 + µ2 + µ3 + µ4.
Step II In this step, we show that the operator F2 is continuous. For this, let us assume that {xn}, {yn}be convergent sequences such that xn → x, yn → y as n→ +∞. From the continuity of K1, it follows
that K1(t, xn(t), yn(t)) → K1(t, x(t), y(t)), K2(t, xn(t), yn(t)) → K2(t, x(t), y(t)). By Lebesgue
dominated theorem, we have
limn→∞
g1(xn, yn) = limn→∞
(−∫ t
0
(t− s)α−1
Γ(α)K1(s, xn(s), yn(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, xn(s), yn(s))ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, xn(s), yn(s))ds) (4.1.23)
= −∫ t
0
(t− s)α−1
Γ(α)limn→∞
K1(s, xn(s), yn(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)limn→∞
K1(s, xn(s), yn(s))ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1 lim
n→∞K1(s, xn(s), yn(s))ds
65
= −∫ t
0
(t− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds
= g1(x, y).
Similarly, we obtain
limn→0
g2(xn, yn) = limn→∞
(−∫ t
0
(t− s)β−1
Γ(β)K2(s, xn(s), yn(s))ds
+
∫ 1
0
(1− s)β−1
Γ(β)K2(s, xn(s), yn(s))ds
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1K2(s, xn(s), yn(s))ds)),
= −∫ t
0
(t− s)β−1
Γ(β)limn→∞
K2(s, xn(s), yn(s))ds
+
∫ 1
0
(1− s)β−1
Γ(β)limn→∞
K2(s, xn(s), yn(s))ds
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1 lim
n→∞K2(s, xn(s), yn(s))ds))
= −∫ t
0
(t− s)β−1
Γ(β)K2(s, x(s), y(s))ds
+
∫ 1
0
(1− s)β−1
Γ(β)K2(s, x(s), y(s))ds
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1K2(s, x(s), y(s))ds
= g2(x, y).
By the help of (4.1.23), (4.1.24) we proved that the operator F2 = (g1, g2) is continuous for all t ∈ [0, 1].
Step III In this step, we show that the operator F2 = (g1, g2) is compact on Br(0).
|g1(x, y)(t)| = | −∫ t
0
(t− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds| (4.1.24)
≤∫ t
0
(t− s)α−1
Γ(α)|M1(s)|ds+
∫ 1
0
(1− s)α−1
Γ(α)|M1(s)|ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1|M1(s)|ds
≤( 1
Γ(1 + α)+
1
Γ(1 + α)+
2δα−µ1
∆1Γ(α− µ+ 1)
)∥M1∥
= N1∥M1∥.
66
Taking supremum over t ∈ [0, 1], we have
∥g1(x, y)∥ ≤ N1∥M1∥ ∀ (x, y) ∈ Br(0)× Br(0). (4.1.25)
Similarly, we obtain
|g2(x(t), y(t))| = | −∫ t
0
(t− s)β−1
Γ(β)K2(s, x(s), y(s))ds
+
∫ 1
0
(1− s)β−1
Γ(β)K2(s, x(s), y(s))ds
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1K2(s, x(s), y(s))ds|
≤∫ t
0
(t− s)β−1
Γ(β)|M2(s)|ds+
∫ 1
0
(1− s)β−1
Γ(β)|M2(s)|ds (4.1.26)
+(t− 1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1|M2(s)|ds
≤( 1
Γ(1 + β)+
1
Γ(1 + β)+
2δβ−ν2
∆1Γ(β − ν + 1)
)∥M2∥
= N2∥M2∥.
Taking the supremum of (4.1.26), we get
∥g2(x, y)∥ ≤ N2∥M2∥ ∀ (x, y) ∈ Br(0)× Br(0).
From (4.1.25) and (4.1.26),it follows that
∥F(x, y)∥ ≤ N1∥M1∥+N2∥M2∥ ∀ (x, y) ∈ Br(0)× Br(0), (4.1.27)
which implies that F2(Br(0)× Br(0)) is a uniformly bounded set in E × E .Now, we prove that the operator F2 is an equicontinuous. For this, let us assume 0 ≤ t1 ≤ t2 ≤ 1, then
we have
|g1(x, y)(t2) − g1(x, y)(t1)| = |∫ t2
0
(t2 − s)α−1
Γ(α)K1(s, x(s), y(s))ds
−∫ t1
0
(t1 − s)α−1
Γ(α)K1(s, x(s), y(s))ds
+(t2 − t1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds|) (4.1.28)
≤∫ t2
0
(t2 − s)α−1
Γ(α)M1(s)ds−
∫ t1
0
(t1 − s)α−1
Γ(α)M1(s)ds
+(t2 − t1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1M1(s)ds)
≤( tα2 − tα1Γ(1 + α)
+(t2 − t1)δ
α−ν1
∆1Γ(α− µ+ 1)
)∥M1∥.
67
Similarly, we have
|g2(x, y)(t1) − g2(x, y)(t2)| = |∫ t2
0
(t2 − s)β−1
Γ(β)K2(s, x(s), y(s))ds
−∫ t1
0
(t1 − s)β−1
Γ(β)K2(s, x(s), y(s))ds
+(t2 − t1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1K2(s, x(s), y(s))ds|
≤∫ t2
0
(t2 − s)β−1
Γ(β)M2(s)ds−
∫ t1
0
(t1 − s)β−1
Γ(β)M2(s)ds (4.1.29)
+(t2 − t1)
∆2Γ(β − ν)
∫ δ2
0(δ2 − s)β−ν−1M2(s)ds
≤( tβ2 − tβ1Γ(1 + β)
+(t2 − t1)δ
β−ν2
∆2Γ(β − ν + 1)
)∥M2∥.
From (4.1.28) and (4.1.29), it follows that
|F(x, y)(t1)−F(x, y)(t2)| ≤[tα2 − tα1Γ(1 + α)
+(t2 − t1)δ
α−ν1
∆1Γ(α− µ+ 1)
]∥M1∥ (4.1.30)
+
[tβ2 − tβ1Γ(1 + β)
+(t2 − t1)δ
β−ν2
∆2Γ(β − ν + 1)
]∥M2∥
which implies that F(x, y)(t1) − F(x, y)(t2) → 0 as t1 → t2. This shows that the operator B is an
equicontinuous and by Arzela Ascoli theorem, the operator B is a compact.
Step IV We have γF2,max < 1, where F2,max = ∥F2(Br(0) × Br(0))∥ = sup{∥F2(x, y)∥ : x, y ∈Br(0) × Br(0)}, and gi,max = ∥gi(Br(0))∥ = sup{∥gi(x) : x ∈ Br(0)∥} for i = 1, 2. Using (4.1.27)
and hypothesis (A3), we have
F2,max ≤ γ(N1∥M1∥+N2∥M2∥) < 1, ∀ (x, y) ∈ Br(0)× Br(0)). (4.1.31)
Now, we prove that the operator T maps bounded subset of E×E into a bounded set. For this, assume that
(x, y) ∈ Br(0)×Br(0) and t ∈ [0, 1], λ ∈ (0, 1) such that ∥(x, y)∥ = r and (x, y) = λA(x, y)F(x, y),
using (A2), (A4), we proceed
|x(t)| = λ|H(t, x, y)||(−∫ t
0
(t− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds)| (4.1.32)
≤ λ|H(t, x, y)−H(t, 0, 0) +H(t, 0, 0)||(−∫ t
0
(t− s)α−1
Γ(α)M1(s)ds
+
∫ 1
0
(1− s)α−1
Γ(α)M1(s)ds+ (t− 1)
∫ δ1
0
(δ1 − s)α−µ−1M1(s)ds)
∆1Γ(α− µ)|
≤ λ(µ1|x(t)|+ µ2|y(t)|+H0)(2
Γ(1 + α)+
2δα−µ1
∆1Γ(α− µ+ 1))∥M1∥
= λ1(µ1|x(t)|+ µ2|y(t)|+H0)N1∥M1∥.
68
Taking supremum over t ∈ [0, 1] and using (A4), we obtain
∥x∥ ≤ λ1(µ1∥x∥+ µ2∥y∥+H0)N1∥M1∥. (4.1.33)
Which implies that
∥x∥ ≤ λ(µ2∥y∥+H0)
1− λµ1N1∥M1∥N1∥M1∥. (4.1.34)
Similarly, we have
∥y∥ ≤ λ(µ1∥x∥+ G0)
1− λµ2N2∥M2∥N2∥M2∥. (4.1.35)
By the help of (4.1.34) and (4.1.35), we get
∥(x, y)∥ = ∥x∥+ ∥y∥ ≤ λ(µ2∥y∥+H0)
1− λµ1N1∥M1∥N1∥M1∥
+ λ(µ1∥x∥+ G0)
1− λµ2N2∥M2∥N2∥M2∥ (4.1.36)
= λµ2∥y∥
1− λN1∥M1∥N1∥M1∥+ λ
H0
1− λµ1N1∥M1∥N1∥M1∥
+ λµ1∥x∥
1− λµ2N2∥M2∥N2∥M2∥+ λ
G0
1− λµ2N2∥M2∥N2∥M2∥.
From (4.1.36), we obtain
∥(x, y)∥ ≤ 1
1− ρ
(λH0N1∥M1∥
1− λµ1N1∥M1∥+
λG0N2∥M2∥1− µ2λN2∥M2∥
)< r, (4.1.37)
which contradicts our supposition ∥(x, y)∥ = r. Thus, the operator (x, y) = F1(x, y)F2(x, y) has a
solution in Br(0) × Br(0) which is a solution of the system (4.1.1). This completes the proof of the
theorem.
Theorem 4.1.3. Assume that (A1) and (A2) hold. If µ(N1+N2)(∥M1∥+∥M2∥) < 1, then the coupled
system of HFDEs (4.1.1), has a unique solution.
Proof. Define H0 = supt∈[0,1] |H(t, 0, 0)|, G0 = supt∈[0,1] |G(t, 0, 0)| and µ = µ1 + µ2 + µ3 + µ4.
Choose ρ ≥ µ(H0+G0)(N1+N2)(∥M1∥+∥M2∥)1−µ(N1+N2)(∥M1∥+∥M2∥) where N1, N2 are defined in (4.1.12), (4.1.13) respectively.
Define a set Sρ = {(x, y)(t) ∈ E × E : ∥(x, y)∥ ≤ ρ}. Then, for (x, y) ∈ Sρ, we have
|T1(x(t), y(t))| = |H(t, x, y)||(−∫ t
0
(t− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+
∫ 1
0
(1− s)α−1
Γ(α)K1(s, x(s), y(s))ds
+(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1K1(s, x(s), y(s))ds)|
≤ |H(t, x, y)−H(t, 0, 0) +H(t, 0, 0)|(∫ t
0
(t− s)α−1
Γ(α)∥M1(s)∥ds (4.1.38)
+
∫ 1
0
(1− s)α−1
Γ(α)∥M1(s)∥ds+
(t− 1)
∆1Γ(α− µ)
∫ δ1
0(δ1 − s)α−µ−1∥M1(s)∥ds)
69
≤ (µ1|x(t)|+ µ2|y(t)|+H0)(2
Γ(1 + α)+
2δα−µ1
∆1Γ(α− µ+ 1))∥M1∥
= (|µ1x(t)|+ µ2|y(t)|+H0)N1∥M1∥.
Taking supremum over t ∈ [0, 1], we obtain
∥T1(x, y)∥ ≤ (µ1 + µ2)(∥x∥+ ∥y∥+H0)N1∥M1∥ (4.1.39)
≤ (µ1 + µ2)(β +H0)N1∥M1∥.
Similarly, we have
∥T2(x, y)∥ ≤ (µ3 + µ4)(∥x∥+ ∥y∥+ G0)N2∥M2∥ (4.1.40)
≤ (µ3 + µ4)(β + G0)N2∥M2∥.
By the help of (4.1.39), (4.1.40), we get
∥T (x, y)∥ ≤ (µ1 + µ2)(β +H0)N1∥M1∥
+ (µ3 + µ4)(β + G0)N2∥M2∥ (4.1.41)
≤ (µ1 + µ2 + µ3 + µ4)(β +H0 + G0)(N1
+ N2)(∥M1∥+ ∥M2∥) ≤ β.
Therefore, (4.1.40) implies that the operator T is a bounded operator. Now, for (x1, y1)(t),
(x2, y2)(t) ∈ E × E and any t ∈ [0, 1], we have
|T1(x2, y2)(t)− T1(x1, y1)(t)|
≤ |H(t, x2, y2)−H(t, x1, y1)|{∫ t
0
(t− s)α−1
Γ(α)|M1(s)|ds (4.1.42)
+
∫ 1
0
(1− s)α−1
Γ(α)|M1(s)|ds+ |t− 1|
∫ δ1
0
(δ1 − s)α−µ−1|M1(s)|ds∆1Γ(α− µ)
}
≤ (µ1|x2 − x1|+ µ2|y2 − y1|)( 2
Γ(1 + α)+
2δα−µ1
∆1Γ(α− µ+ 1)
)|M(t)|
≤ (µ1 + µ2)N1|M1(t)|(|x2 − x1|+ y2 − y1|).
Taking the supremum over t ∈ [0, 1], we obtain
∥T1(x2, y2) − T1(x1, y1)∥ (4.1.43)
≤ (µ1 + µ2)N1∥M1∥(∥x2 − x1∥+ ∥y2 − y1∥).
Similarly, we obtain
∥T2(x2, y2) − T2(x1, y1)∥ (4.1.44)
≤ (µ3 + µ4)N2∥M2∥(∥x2 − x1∥+ ∥y2 − y1∥).
70
From (4.1.43) and (4.1.44), it follows that
∥T (x2, y2) − T (x1, y1)∥ (4.1.45)
≤ µ(N1 +N2)(∥M1∥+ ∥M2∥)(∥x2 − x1∥+ ∥y2 − y1∥),
which implies that T is a contraction and by the help of Banach’s contraction principle the operator Thas a unique fixed point which is the unique solution of the coupled system of HFDEs (4.1.1).
4.1.2 Examples
Example 4.1.4. Consider the following coupled system of HFDEs as a special case of the problem
(4.1.1):
D2.5( x(t)
100 + (|x(t)|+|y(t)|) sin(t)50
)=
1 + x(t) sin(t)
1 + |x(t)|+ |y(t)|, (4.1.46)
D2.5( y(t)
100 + (|x(t)|+|y(t)|) cos(t)50
)=
1 + x(t) cos(t)
1 + |x(t)|+ |y(t)|,
with initial and boundary conditions as defined in (4.1.1), for δ1 = δ2 = µ = ν = 0.5 and t ∈ [0, 1].
From (4.1.46), we have H(t, x, y) = 100 + (|x(t)|+|y(t)|) sin(t)50 , G(t, x, y) = 100 + (|x(t)|+|y(t)|) cos(t)
50 .
|H(t, x2, y1) − H(t, x1, y1)| ≤ 150(∥x2 − x1∥ + ∥y2 − y1∥), |G(t, x2, y1) − G(t, x1, y1)| ≤ 1
50(∥x2 −x1∥+∥y2−y1∥) which implies µi = 1
50 for i = 1, 2, 3, 4.By using these values, we deduce ρ ≥ 67.7022,
we choose ρ = 70. Thus, by the Theorem 4.1.2, the coupled system of HFDEs (4.1.46) has a solution
in B70(0)× B70(0).
Example 4.1.5. Consider the following coupled system of HFDEs as a special case of the problem
(4.1.1):
D2.5( x(t)
10 + (|x(t)|+|y(t)|) sin(t)50
)=
1 + x(t) sin(t)
2 + |x(t)|+ |y(t)|, (4.1.47)
D2.5( y(t)
10 + (|x(t)|+|y(t)|) cos(t)50
)=
1 + x(t) cos(t)
2 + |x(t)|+ |y(t)|,
with initial and boundary conditions as defined in (4.1.1), for δ1 = δ2 = µ = ν = 0.5. From
system (4.1.47), we have H(t, x, y) = 10 + (|x(t)|+|y(t)|) sin(t)50 , G(t, x, y) = 10 + (|x(t)|+|y(t)|) cos(t)
50 .
|H(t, x2, y1)−H(t, x1, z1)| ≤ 150(∥x2−x1∥+∥y2−y1∥), |G(t, x2, y1)−G(t, x1, z1)| ≤ 1
50(∥x2−x1∥+∥y2 − y1∥) which implies µi = 1
50 for i = 1, 2, 3, 4. By using these values, we deduce ρ ≥ 6.77022, we
choose ρ = 7. µ(N1 +N2)(∥M1∥+ ∥M2∥) ≤ 0.252917 < 1. Therefore, the system (4.1.47) satisfies
the assumptions of Theorem 4.1.3. Thus, the system (4.1.47), has a unique solution in B7(0)× B7(0).
71
4.2 Coupled system of hybrid FDEs II
In this section, we are interested in the unique solution of the coupled system of HFDEs of the type:
Dα
(x
H(t, x, y)
)= −K1(t, x, y), α ∈ (n− 1, n],
Dβ
(y
G(t, x, y)
)= −K2(t, x, y), β ∈ (n− 1, n],(
x
H(t, x, y)
)(i)
|t=0 = 0 =
(x
H(t, x, y)
)(1)
|t=1,x
H(t, x, y)|t=δ1 = δ1Iα−1
1 T1(t, x, y),(y
G(t, x, y)
)(j)
|t=0 = 0 =
(x
G(t, x, y)
)(1)
|t=1,y
G(t, x, y)|t=δ2 = δ2Iβ−1
1 T2(t, x, y),
i, j = 2, 3, . . . , n− 1,
(4.2.1)
where Dα, Dβ are Caputo’s fractional derivative of orders α, β respectively, K1,K2 ∈ C([0, 1]×R2,R),
H,G ∈ C([0, 1]× R2,R− {0}) and x, y ∈ C([0, 1],R).
Lemma 4.2.1. Let h(t) be a continuous function then the mild solution of
Dα
(x
H(t, x, y)
)= −h(t), α ∈ (n− 1, n], (4.2.2)
with (x
H(t, x, y)
)(i)
|t=0 = 0 =
(x(t)
H(t, x, y)
)(1)
|t=1,x
H(t, x, y)|z=δ1 = δ1Iα−1h(1), (4.2.3)
for i = 2, . . . , n− 1, is given by
x(t) = H(t, x, y)[−∫ t
0
(t− s)α−1
Γ(α)h(s)ds+
∫ δ
0
(δ − s)α−1
Γ(α)h(s)ds
+ t
∫ 1
0
(1− s)α−2
Γ(α− 1)h(s)ds].
(4.2.4)
Proof. Applying the operator Iα0 on (4.2.2) and using Lemma 1.1.6, we obtain
x(t)
H(t, x, y)= −Iαh(t) + c1 + c2t+ c3t
2 + . . .+ cntn−1. (4.2.5)
Initial conditions(
xH(t,x,y) |z=0
)(i)= 0, for i = 2, . . . , n − 1 implies c3 = . . . = cn = 0, and (4.2.5)
gets the formx(t)
H(t, x, y)= −Iαh(t) + c1 + c2t. (4.2.6)
Applying the boundary condition(
x(t)H(t,x,y) |t=1
)(1)= 0, on (4.2.6), we have c2 = Iα−1h(1) and (4.2.6)
gets the formx(t)
H(t, x, y)= −Iαh(t) + c1 + tIα−1h(1). (4.2.7)
72
By the help of condition xH(t,x,y) |t=δ1 = δ1Iα−1h(1), and (4.2.7), we get c0 = Iαh(δ1). Putting the
values of c0, c1, . . . , cn in (4.2.7), we get
x(t)
H(t, x, y)= −Iαh(t) + Iαh(δ) + tIα−1h(1), (4.2.8)
and the integral form of the solution x(t), is given by
x(t) = H(t, x, y)[−∫ t
0
(t− s)α−1
Γ(α)h(s)ds+
∫ δ
0
(δ − s)α−1
Γ(α)h(s)ds
+ t
∫ 1
0
(1− s)α−2
Γ(α− 1)h(s)ds].
(4.2.9)
4.2.1 Existence and uniqueness of solutions
In this section, our interest is in two lemmas for the EMS and unique solution of the coupled system
(4.2.1). For these two lemmas, we divide this section in two subsections. Consider the Banach space
E = {x(t) : x(t) ∈ C[0, 1]} with norm ∥x∥ = max{|x(t)| for t ∈ [0, 1]} and (E × E , ∥(., .)∥) with
norm ∥(x, y)∥ = ∥x∥+ ∥y∥. We define an operator T : E × E → E × E by
T (x, y)(t) = (T1(x, y)(t), T2(x, y)(t)), (4.2.10)
where
T1(x, y)(t) = H(t, x, y)[−∫ s
0
(t− s)α−1
Γ(α)K1(s, x, y)ds+
∫ δ1
0
(δ1 − s)α−1
Γ(α)K1(s, x, y)ds
+ t
∫ 1
0
(1− s)α−2
Γ(α− 1)T1(s, x, y)ds],
(4.2.11)
and
T2(x, y)(t) = G(t, x, y)[−∫ z
0
(z − y)β−1
Γ(β)K2(s, x, y)ds+
∫ δ2
0
(δ2 − y)β−1
Γ(β)K2(s, x, y)ds
+ t
∫ 1
0
(1− y)β−2
Γ(β − 1)K2(s, x, y)ds],
(4.2.12)
We define the following terms for simplifying our calculations
N1 =1
Γ(1 + α)+
δα1Γ(1 + α)
+1
Γ(α), (4.2.13)
N2 =1
Γ(1 + β)+
δβ2Γ(1 + β)
+1
Γ(β). (4.2.14)
Assume the following two conditions:
(A1) The functions H(t, x, y),G(t, x, y) are continuous and bounded such that |H(t, x, y)| ≤ µ1,
73
|G(t, x, y)| ≤ µ2 for all (t, x, y) ∈ ([0, 1]× R× R) and µ1, µ2 ∈ R;
(A2) There exist real constants λj , ηj , ξj > 0 for j = 1, 2 such that |K1(t, x, y)| ≤ γ1 + η1|x| + ξ1|y|,
|K2(t, x, y)| ≤ γ2 + η2|x| + ξ2|y| for all x, y ∈ R and t ∈ [0, 1] also µ1N1(γ1 + η1|x| + ξ1|y|) +
µ2N2(γ2 + η2|x|+ ξ2|y|) < +∞.
Theorem 4.2.2. Assume that (A1), (A2) are satisfied and there exist real constants M1,M2 ∈ R, such
that
|K1(t, x, y)| ≤ M1, |K2(t, x, y)| ≤ M2,
for all t ∈ [0, 1], then the system (4.2.1), has a solution.
Proof. The proof is given in the following three steps.
Step 1: In this step, we prove that the operator T maps bounded subset of E × E into a bounded. For
this, we consider (x, y) ∈ E × E for t ∈ [0, 1], then (4.2.11) deduce the following:
|T1(x, y)(t)| = |H(t, x, y)|[| −∫ t
0
(t− s)α−1
Γ(α)K1(s, x, y)ds
+
∫ δ1
0
(δ1 − s)α−1
Γ(α)K1(s, x, y)ds+ z
∫ 1
0
(1− y)α−2
Γ(α− 1)K1(s, x, y)ds|]
≤ µ1
[1
Γ(1 + α)+
δα1Γ(1 + α)
+1
Γ(α)
]M1 = µ1N1M1,
(4.2.15)
similarly, (4.2.12) implies that:
|T2(x, y)(t)| ≤ µ2
[1
Γ(1 + β)+
δβ2Γ(1 + β)
+1
Γ(β)
]M2 = µ2N2M2. (4.2.16)
Using the definition of the norm and (4.2.15), (4.2.16), we have:
∥T (x, y)(t)∥ ≤ µ1N1M1 + µ2N2M2 <∞, (4.2.17)
thus, T is uniformly bounded.
Step 2: Now, to show that T is an equicontinuous operator. For this, let (x, y)(t) ∈ E × E and t1, t2 ∈[0, 1], such that t1 < t2. Then, we have
|T1(x, y)(t2)− T1(x, y)(t1)| = µ1[|∫ t2
0
(t2 − s)α−1
Γ(α)K1(s, x, y)ds
−∫ t1
0
(t1 − y)α−1
Γ(α)K1(s, x, y)ds+ (t2 − t1)
∫ 1
0
(1− y)α−2
Γ(α− 1)K1(s, x, y)ds|]
≤ µ1
[tα2 − tα1Γ(1 + α)
+t2 − t1Γ(α)
]M1,
(4.2.18)
similarly, we can get
|T2(x,y)(t2)− T2(x, y)(t1)| ≤ µ2
[tβ2 − tβ1Γ(1 + β)
+t2 − t1Γ(β)
]M2, (4.2.19)
74
by the help of (4.2.18) and (4.2.19), we have
|T (x, y)(t2)− T (x, y)(t1)| ≤ µ1
[tα2 − tα1Γ(1 + α)
+t2 − t1Γ(α)
]M1
+ µ2
[tβ2 − tβ1Γ(1 + β)
+t2 − t1Γ(β)
]M2,
(4.2.20)
∥T (x, y)(t2)− T (x, y)(t1)∥ → 0 as t2 → t1. By step 1, step 2 and Arzela Ascoli theorem, we have Tis completely continuous operator.
Step 3: Here, we show that the set
S∗ = {(x, y)(t) ∈ E × E : µT (x, y)(t) = (x, y)(t)},
where 0 < µ < 1, is bounded. From the set S∗, we have µT1(x, y)(t) = x(t) and µT2(x, y)(t) = x(t),
consequently,
|x(t)| ≤ µ|T1(x, y)(t)|
≤ |H(t, x, y)|[| −∫ t
0
(t− s)α−1
Γ(α)K1(s, x, y)ds
+
∫ δ1
0
(δ1 − y)α−1
Γ(α)K1(s, x, y)ds+ t
∫ 1
0
(1− s)α−2
Γ(α− 1)K1(s, x, y)ds|]
≤ µ1
[1
Γ(1 + α)+
δα1Γ(1 + α)
+1
Γ(α)
](γ1 + η1|x|+ ξ1|y|)
= µ1N1(γ1 + η1|x|+ ξ1|y|),
(4.2.21)
and
|y(t)| ≤ µ|T2(x, y)(t)|
≤ |G(t, x, y)|[| −∫ t
0
(t− s)β−1
Γ(β)K2(s, x, y)ds+
∫ δ2
0
(δ2 − s)β−1
Γ(β)K2(s, x, y)ds
+ t
∫ 1
0
(1− y)β−2
Γ(β − 1)K2(s, x, y)ds|]
≤ µ2
(1
Γ(1 + β)+
δβ2Γ(1 + β)
+1
Γ(β)
)(γ2 + η2|x|+ ξ2|y|)
≤ µ2N2(γ2 + η2|x|+ ξ2|y|).
(4.2.22)
From (4.2.21) and (4.2.22), we have
∥x∥ ≤ µ1N1(γ1 + η1|x|+ ξ1|y|), (4.2.23)
∥y∥ ≤ µ2N2(γ2 + η2|x|+ ξ2|y|), (4.2.24)
consequently,
∥(x, y)∥ ≤ µ1N1(γ1 + η1|x|+ ξ1|y|) + µ2N2(γ2 + η2|x|+ ξ2|y|) < +∞, (4.2.25)
75
this implies that S∗ is bounded. Thus, by the help of Step 1, 2, 3 we have that T is completely continuous
and by the help of lemma 1.3.4, T has fixed the point (x, y)(t) and is the solution of the coupled system
of HFDEs (4.2.1).
Theorem 4.2.3. Assume that K1,K2 ∈ C([0, 1]×R2,R) and there exist positive real constants γ1, γ2, γ3, γ4,
such that for all t ∈ [0, 1] and x1, x2, y1, y2 ∈ R,
|K1(t, x, y)−K1(t, x1, y1)| ≤ γ1|x− y1|+ γ2|y − y1|,
and
|K2(t, x, y)−K2(t, x1, y1)| ≤ γ3|x− y1|+ γ4|y − y1|,
with the condition that
µ1N1(γ1 + γ2) + µ2N2(γ3 + γ4) < 1,
where N1,N2 are defined by (4.2.13), (4.2.14). Then the problem (4.2.1), has a unique solution.
Proof. Assume that supt∈[0,1]K1(t, 0, 0) = ζ1 < ∞, supt∈[0,1]K2(t, 0, 0) = ζ2 < ∞ and the set
Sν = {(x, y)(t) ∈ E × E : ∥(x, y)∥ ≤ ν}, such that
ν ≥ µ1N1ζ1 + µ2N2ζ21− µ1N1(γ1 + γ2)− µ2N2(γ3 + γ4)
. (4.2.26)
From (4.2.11), we have:
|T1(x, y)(t)| ≤ |H(t, x, y)|[| −∫ z
0
(z − y)α−1
Γ(α)K1(s, x, y)ds
+
∫ δ1
0
(δ1 − s)α−1
Γ(α)K1(s, x, y)ds+ t
∫ 1
0
(1− y)α−2
Γ(α− 1)K1(s, x, y)ds|]
≤ µ1[
∫ t
0
(t− s)α−1
Γ(α)(|K1(s, x, y)−K1(s, 0, 0)|+ |K1(s, 0, 0)|)dy
+
∫ δ1
0
(δ1 − s)α−1
Γ(α)(|K1(s, x, y)−K1(s, 0, 0)|+ |K1(s, 0, 0)|)ds
+ t
∫ 1
0
(1− s)α−2
Γ(α− 1)(|K1(s, x, y)−K1(s, 0, 0)|+ |K1(s, 0, 0)|)ds]
≤ µ1
(1
Γ(1 + α)+
δα1Γ(1 + α)
+1
Γ(α)
)(γ1|x|+ γ2|y|+ ζ1)
≤ µ1N1((γ1 + γ2)ν + ζ1).
(4.2.27)
Similarly, (4.2.12)
|T2(x, y)(t)| ≤ µ2N2((γ3 + γ4)ν + ζ2). (4.2.28)
From (4.2.26), (4.2.27) and (4.2.28), we have
∥T (x, y)∥ ≤ ν. (4.2.29)
76
Let, (x1, y1)(t), (x2, y2)(t) ∈ E × E and for any t ∈ [0, 1], we have
|T1(x2, y2)(t)− T1(x1, y1)(t)|
≤ |H(t, x, y)|[∫ z
0
(z − y)α−1
Γ(α)(|K1(s, x2, y2)−K1(s, x1, y1)|)ds
+
∫ δ1
0
(δ1 − s)α−1
Γ(α)(|K1(s, x2, y2)−K1(s, x1, y1(y))|)dy
+ z
∫ 1
0
(1− s)α−2
Γ(α− 1)(|K1(s, x2, y2)−K1(s, x1, y1(y))|)ds]
≤ µ1
(1
Γ(1 + α)+
δα1Γ(1 + α)
+1
Γ(α)
)(γ1|x2 − x1|+ γ2|y2 − y1|)
≤ µ1N1(γ1 + γ2)(|x2 − x1|+ |y2 − y1|).
(4.2.30)
Similarly, from (4.2.12), we have:
|T2(x2, y2)(t)− T2(x1, y1)(t)| ≤ µ2N2(γ3 + γ4)(|x2 − x1|+ |y2 − y1|), (4.2.31)
Therefore, from (4.2.30), (4.2.32), we have:
∥T (x2, y2)− T (x1, y1)∥ ≤ (µ1N1(γ1 + γ2) + µ2N2(γ3 + γ4))(|x2 − x1|+ |y2 − y1|). (4.2.32)
Therefore, the function T is a contraction mapping and consequently, T fixes a point which is the unique
solution of (4.2.1).
4.2.2 Examples
Example 4.2.4. Consider the following coupled system of HFDEs as a special case of (4.2.1):
D3.5
(x
1 + | sinx|+ | cos y|
)= − t+ | sinx|+ | cos y|
40(1 + 2t),
D3.5
(y
1 + | sinx|+ | cos y|
)= − t+ | sinx|+ | cos y|
40(1 + 3t),(
x
H(t, x, y)
)(i)
|t=0 = 0 =
(x
H(t, x, y)
)(1)
|t=1,x
H(t, x, y)|t=0.5 = 0.5I2.5K1(t, x, y)|t=1,(
y
G(t, x, y)
)(i)
|t=0 = 0 =
(y
G(t, x, y)
)(1)
|t=1,y
G(t, x, y)|t=0.5 = 0.5I2.5K2(t, x, y)|t=1.
(4.2.33)
In this system, β = α = 3.5, δ1 = δ2 = 0.5, H = G = 1 + | sinx| + | cos y| ≤ 3 = µ1 = µ2,
H,G ∈ R− {0}. |K1| ≤ 140 + |x|
40 + |y|40 and |K2| ≤ 1
40 + |x|40 + |y|
40 , λi = ηi = ξi =140 for i = 1, 2 and
µ1N1(t1 + η1|x| + ξ1|y|) + µ2N2(t2 + η2|x| + ξ2|y|) < +∞ for any x, y ∈ C([0, 1],R). Therefore,
conditions (A1), (A2), are satisfied and by Theorem 4.2.2, the problem (4.2.33) has a solution.
77
Example 4.2.5. In this example, Theorem 4.2.3, is utilized for the unique solution of the following
coupled system of HFDEs:
D3.5
(x
1 + | sinx|+ | cos y|
)= − t+ | sinx|+ | cos y|
35(1 + 2t),
D3.5
(y
1 + | sinx|+ | cos y|
)= − t+ | sinx|+ | cos y|
35(1 + 3t),(
x
H(t, x, y)
)(i)
|t=0 = 0 =
(x
H(t, x, y)
)(1)
|t=1,x
H(t, x, y)|t=0.5 = 0.5I2.5K1(t, x, y)|t=1,(
y
G(t, x, y)
)(i)
|t=0 = 0 =
(y
G(t, x, y)
)(1)
|t=1,y
G(t, x, y)|t=0.5 = 0.5I2.5K2(t, x, y)|t=1.
(4.2.34)
In this coupled system of HFDEs, we have α = β = 3.5, δ1 = δ2 = 0.5, K1(t, 0, 0) ≤ 135 ,K2(t, 0, 0) ≤
135 and
|K1(t, x, y)−K1(t, x1, y1)| ≤1
35(|x− y1|+ |y − y1|),
|K2(t, x, y)−K2(t, x1, y1)| ≤1
35(|x− y1|+ |y − y1|),
(4.2.35)
from (4.2.35), we have γ1 = γ2 = γ3 = γ4 = 135 , N1 = N2 = 0.394472, ζ1 = ζ2 = 1, ultimately, we
get ν ≥ 0.0782001, and µ1N1(γ1 + γ2) + µ2N2(γ3 + γ4) = 0.135248 < 0.2. Therefore, by Theorem
4.2.3, the uniqueness of solution of the coupled system of HFDEs (4.2.34), in S0.1(0), is varified.
78
Chapter 5
Local fractional differential equations
Local fractional derivative was defined by Kolwankar and Gangal [108] and got the attention of scientists
in the fields like physics and engineering based on the fractals. A recent definition of derivatives on
fractals have been introduced by Golmankhanreh and Baleanu [109]. Where the fractal curves [105],
are everywhere continuous but nowhere differentiable and therefore, the classical calculus cannot be
used to interpret the motions in Cantor time-space [107]. LFC [110–113], started to be considered as
one of the useful way to handle the fractals and other functions that are continuous but non-differentiable.
Mandelbrot [105, 117], described that fractal geometry is a workable geometric middle ground between
the excessive geometric order of Euclid and the geometric chaos of general mathematics and extensively
illustrated wide range of applications fractals in many scientific fields like in, physics, engineering,
mathematics, geophysics. Christianto and Rahul [118], gave the derivation of Proca equations on Cantor
sets. Hao et al. [119], investigated the Helmholtz and Diffusion equations in Cantorian and Cantor-Type
Cylindrical Coordinates. Carpinteri and Sapora [120], studied diffusion problems in fractal media in
Cantor sets. Zhang et al. studied LF wave equations under fixed entropy. Many techniques are utilized
for handling the LF problems in both ordinary and partial derivatives. For instance, The Local fractional
Laplace Transform [110], the local fractional variational iteration method (LFLVIM) [113] and many
others. These methods are widely used in different scientific fields [106, 112–116, 120–122, 124]. This
area of research is much popular in the community of scientists and we are continuously observing
recent developments in it.
In this chapter, we are concerned with the analytical solutions of three problems in LF calculus. The
problems are; LF wave equations (LFWE), LF Schrodinger equation (LFSE) and a Fractal Vehicular
Traffic flow model (FVTFM). The results of this chapter are [134, 141, 142]. This chapter has three
sections. In section 5.1, an iterative technique for the solutions of the wave equations is developed by
the help of local fractional variational iteration method (LFLVIM) and the scheme is applied on two ex-
79
amples of LFWEs. In section 5.2, solutions of LFSE are obtained by the help of Yang Laplace transform
(LFLT) and Series expansion method (SEM). The iterative technique is applied on two examples. In sec-
tion 5.3, an iterative technique for solutions of FVTFM is developed which is applied for the solutions
of two illustrative examples in subsection 5.3.1.
5.1 Iterative technique for solutions of LF wave equations
In this section, we develop an iterative scheme for analytic solutions of the following LFWEs
vmγt (t, x)− p(x)vnγx (t, x) = 0, (5.1.1)
where m, n are orders of LF partial derivatives with respect to t and x respectively.
Applying the LF variation iteration method for the correction LF operator for (5.1.1), we have
vm+1(t, x) = vm(t, x) + I(γ)0,t
(X (x)γ
Γ(γ + 1)
)(vmγ
m,t(t, x)− p(x)vnγm,x(t, x)), (5.1.2)
where X (x)γ
Γ(γ+1) , is the Lagrange multiplier and (5.1.2) leads to
vm+1(t, x) = vm(t, x) + I(γ)0,t
(X (x− t)γ
Γ(γ + 1)
)(vmγ
m,t(t, x)− p(x)vnγm,x(t, x)). (5.1.3)
Applying LFLT on (5.1.3), we have
Lγ{vm+1(t, x)} = Lγ{vm(t, x)}+ Lγ{X (x)γ
Γ(γ + 1)}Lγ{(vmγ
m,t(t, x)− p(x)vnγm,x(t, x)}
= Lγ{vm(t, x)}+ Lγ{X (x)γ
Γ(γ + 1)}{smγLγ{vm(t, x)} (5.1.4)
− s(m−1)γvm(0, x)− s(m−2)γvm(0, x)(γ)
− s(m−3)γvm(0, x)(2γ) . . .− vm(0, x)((m−1)γ) − p(x)Lγ{vnγm,x(t, x)}}.
Taking the γ order local fractional variation of (5.1.4) w.r.t T , and assuming that the term p(x)vnγx (t, x)
be invariant, we have
δ(γ)Lγ{vm+1(t, x)} = δ(γ)Lγ{vm(t, x)}+ Lγ{X (x)γ
Γ(γ + 1)}δ(γ){smγLγ{vm(t, x)}
− s(m−1)γvm(0, x)− s(m−2)γvm(0, x)(γ) (5.1.5)
− s(m−3)γvm(0, x)(2γ) . . .− vm(0, x)((m−1)γ)}.
From (5.1.5), we obtain the Lagrange-Multiplier as follows
X (x)γ
Γ(γ + 1)=
−1
smγ, (5.1.6)
80
and by the help of (5.1.4) and (5.1.6), we have the following relation
Lγ{vm+1(t, x)} = Lγ{vm(t, x)} − 1
smγ{smγLγ{vm(t, x)}
− s(m−1)γvm(0, x)− s(m−2)γvm(0, x)(γ) (5.1.7)
− s(m−3)γvm(0, x)(2γ) . . .− vm(0, x)((m−1)γ) − p(x)vnγm,x(s, x)}.
Consequently, we have solution of (5.1.1) as given below:
v(t, x) = limm→∞
L−1γ {Lγ{vm(s, x)}}. (5.1.8)
5.1.1 Applications
As an application of the iterative technique (5.1.8), we consider the following examples.
Example 5.1.1. Consider the following wave equation on Cantor sets:
∂2γ
∂t2γv(t, x)− C
∂2γ
∂x2γv(t, x) = 0, (5.1.9)
subject to the following initial-boundary conditions
∂γ
∂tγv(0, x) = 0, v(0, x) = Eγ(x
γ). (5.1.10)
Using (5.1.7), we obtain
Lγ{vm+1(t, x)} = Lγ{vm(t, x)} − 1
s2γ{s2γLγ{vm(t, x)} (5.1.11)
− s(γ)vm(0, x)− vm(0, x)(γ) − C∂2γ
∂x2γvm(s, x)}.
Using the initial condition (5.1.10), we get
{v0(s, x)} = Lγ{v0(0, x)} =Eγ(x
γ)
sγ. (5.1.12)
From (5.1.10) to (5.1.12), we have the first iteration
Lγ{v1(t, x)} = Lγ{v0(t, x)} −1
s2γ{s2γLγ{v0(t, x)}
− s(γ)v0(0, x)− v0(0, x)(γ) − C
∂2γ
∂x2γv0(s, x)}
=1
s2γ
[sγEγ(x
γ) + CEγ(x
γ)
sγ
](5.1.13)
=Eγ(x
γ)
sγ+ C
Eγ(xγ)
s3γ= v1(s, x).
81
For the second iteration, we utilize (5.1.10) and (5.1.13) and obtain
Lγ{v2(t, x)} = Lγ{v1(t, x)} −1
s2γ{s2γLγ{v1(t, x)}
− s(γ)v1(0, x)− v1(0, x)(γ) − C
∂2γ
∂x2γv1(s, x)} (5.1.14)
=1
s2γ
[sγEγ(x
γ) + C
(Eγ(x
γ)
sγ+ C
Eγ(xγ)
s3γ
)]=
Eγ(xγ)
sγ+ C
Eγ(xγ)
s3γ+ C2Eγ(x
γ)
s5γ= v2(s, x).
For the third iteration, using (5.1.10), (5.1.12) and (5.1.14), we have
Lγ{v3(t, x)} = Lγ{v2(t, x)} −1
s2γ{s2γLγ{v2(t, x)}
− s(γ)v2(0, x)− v2(0, x)(γ) − C
∂2γ
∂x2γv2(s, x)} (5.1.15)
=1
s2γ
[sγEγ(x
γ) + C
(Eγ(x
γ)
sγ+ C
Eγ(xγ)
s3γ+ C2Eγ(x
γ)
s5γ
)].
=Eγ(x
γ)
sγ+ C
Eγ(xγ)
s3γ+ C2Eγ(x
γ)
s5γ+ C3Eγ(x
γ)
s7γ= v3(s, x).
Continuing this process upto nth iteration, we obtain
Lγ{vn(t, x)} =Eγ(x
γ)
sγ+ C
Eγ(xγ)
s3γ+ C2Eγ(x
γ)
s5γ+ C3Eγ(x
γ)
s7γ+ . . . (5.1.16)
+ Cn Eγ(xγ)
s(2n+1)γ.
Applying L−1γ on (5.1.16), we get
vn(t, x) = Eγ(xγ) + C
t2γEγ(xγ)
Γ(2γ + 1)+ C2 t4γEγ
Γ(4γ + 1)+ C3Eγ(x
γ)
s7γ+ . . .+ Cn Eγ(x
γ)
s(2n+1)γ
= Eγ(xγ) lim
n→∞
n∑j=0
Cj t2jγ
Γ(2jγ + 1)= Eγ(x
γ) coshγ(Ctγ). (5.1.17)
Example 5.1.2. Consider the following wave equation on Cantor sets:
∂2γv(t, x)
∂t2γ− xγ
Γ(γ + 1)
∂2γv(t, x)
∂x2γ= 0, (5.1.18)
the initial-boundary condition are as under
∂γv(0, x)
∂tγ= 0, v(0, x) =
x2γ
Γ(2γ + 1). (5.1.19)
From (5.1.19), we get
v(s, x) = Lγ{v0(0, x)} =x2γ
sγΓ(2γ + 1). (5.1.20)
82
From (5.1.7), (5.1.18) and (5.1.20), we get the first iteration
Lγ{v1(t, x)} = Lγ{v0(t, x)} −1
s2γ{s2γLγ{v0(t, x)} (5.1.21)
− s(γ)v0(0, x)− v0(0, x)(γ) − xγ
Γ(γ + 1)
∂2γ
∂x2γv0(s, x)}
=x2γ
sγΓ(2γ + 1)− xγ
s3γΓ(γ + 1)= v1(s, x).
For the second iteration, using (5.1.19) and (5.1.21), we obtain
Lγ{v2(t, x)} = Lγ{v1(t, x)} −1
s2γ{s2γLγ{v0(t, x)}
− s(γ)v1(0, x)− v1(0, x)(γ) − xγ
Γ(γ + 1)
∂2γ
∂x2γv1(s, x)} (5.1.22)
=x2γ
sγΓ(2γ + 1)− xγ
s3γΓ(γ + 1)= v2(s, x).
For the third iteration, using (5.1.19) and (5.1.22), to get
Lγ{v2(t, x)} = Lγ{v1(t, x)} −1
s2γ{s2γLγ{v0(t, x)}
− sγv1(0, x)− v1(0, x)(γ) − xγ
Γ(γ + 1)
∂2γ
∂x2γv1(s, x)} (5.1.23)
=x2γ
sγΓ(2γ + 1)− xγ
s3γΓ(γ + 1)= v2(s, x).
Continuing this process upto nth iteration, we get
Lγ{vn+1(t, x)} =x2γ
sγΓ(2γ + 1)− xγ
s3γΓ(γ + 1). (5.1.24)
Applying L−1γ on (5.1.24), we deduce the following result as the solution of (5.1.18), (5.1.19), by the
help of proposed method LFLVIM and our scheme in (5.1.8):
v(t, x) =x2γ
Γ(2γ + 1)− t2γxγ
Γ(2γ + 1)Γ(γ + 1). (5.1.25)
5.2 Iteration technique for solutions of LF Schrodinger equation
In this section, we develop an iterative technique for the analytical solutions of LFSE by the help of two
methods, namely; LFSEM and LFLT. For this, consider the following LFDE:
v(nγ)t (t, x)− λγv(t, x) = 0, (5.2.1)
where v(nγ)t (t, x) = ∂nγ
∂tnγ v(t, x) and λγ is a linear operator. The function v(t, x), can be considered as
multi-term separated function of the variables t and x as under:
v(t, x) =∞∑j=0
Tj(t)Xj(x), (5.2.2)
83
where Tj(t) and Xj(x) are LF continuous function. Expressing Tj(t) = tjγ
Γ(1+jγ) , then (5.2.2) takes the
form
v(t, x) =∞∑j=0
tjγ
Γ(1 + jγ)Xj(x). (5.2.3)
Taking LFLT of (5.2.4), we obtain
v(t, s) =∞∑j=0
tjγ
Γ(1 + jγ)Xj(s). (5.2.4)
By the help of (5.2.5), we produce
Lγ{v(nγ)t (t, x)} =∞∑j=0
tjγ
Γ(1 + jγ)Xj+n(s), (5.2.5)
Lγ{λγv(t, x)} =∞∑j=0
tjγ
Γ(1 + jγ)(λγXj)(s). (5.2.6)
By the use of (5.2.5), (5.2.6), in (5.2.1) yields
∞∑j=0
tjγ
Γ(1 + jγ)Xj+n(s) =
∞∑j=0
tjγ
Γ(1 + jγ)λγXj(s). (5.2.7)
From (5.2.7), we have the following recursion relation
Xj+n(s) = (λγXj)(s). (5.2.8)
From (5.2.8), we have
Xj+1(s) = (λγXj)(s), (5.2.9)
and
Xj+2(s) = (λγXj)(s). (5.2.10)
Consequently, we get the solution of (5.2.1), as under:
v(x, s) =
∞∑j=0
tjγ
Γ(jγ + 1)Xj(s). (5.2.11)
Where the convergent relation is
limj→∞
(tjγ
Γ(1 + jγ))Xj(s) = 0. (5.2.12)
And the solution of (5.2.1) is obtained by the use of (5.2.9), as below
v(t, x) =∞∑j=0
tjγ
Γ(1 + jγ)L−1γ {Xj(s)}. (5.2.13)
84
5.2.1 Applications
In this subsection we utilize the iteration scheme (5.2.9) for the solution of LF Schrodinger equations.
Example 5.2.1. Consider the LFSE
∂γ
∂tγv(t, x) +
hγ2iγm
∂2γ
∂x2γv(t, x) =
1
iγhγv(t, x), (5.2.14)
where γ ∈ (0, 1], with the initial condition
v0(0, x) =x2γ
Γ(1 + 2γ)= X0(x). (5.2.15)
In this example, the operator λγ is given by
λγ =1
iγhγ− hγ
2iγm
∂2γ
∂x2γ. (5.2.16)
Applying the LFLT on (5.2.15), we deduce
X0(s) =x2γ
sγΓ(1 + 2γ). (5.2.17)
Applying the operator λγ on (5.2.17) and using the iteration formula (5.2.9), we get:
X1(s) =
(1
iγhγ− hγ
2iγm
∂2γ
∂x2γ
)x2γ
sγΓ(1 + 2γ)=
1
iγhγ
x2γ
sγΓ(1 + 2γ)− hγ
2iγmsγ. (5.2.18)
For the second iteration, applying the operator λγ on (5.2.18) and the recursion relation (5.2.9), we
proceed
X2(s) =
(1
iγhγ− hγ
2iγm
∂2γ
∂x2γ
)(
1
iγhγ
x2γ
sγΓ(1 + 2γ)− hγ
2iγmsγ) (5.2.19)
=1
i2γh2γ
x2γ
sγΓ(1 + 2γ)− 2
2i2γmsγ.
For the third iteration, applying the operator λγ on (5.2.19) and using the recursion relation (5.2.9), we
deduce
X3(s) =
(1
iγhγ− hγ
2iγm
∂2γ
∂x2γ
)(
1
i2γh2γ
x2γ
sγΓ(1 + 2γ)− 2
2i2γmsγ) (5.2.20)
=1
i3γh3γ
x2γ
sγΓ(1 + 2γ)− 3
2i3γhγmsγ.
Continuing this process, we get the following approximations:
X4(s) =1
i4γh4γ
x2γ
sγΓ(1 + 2γ)− 4
2i4γh2γmsγ,
X5(s) =1
i5γh5γ
x2γ
sγΓ(1 + 2γ)− 5
2i5γh3γmsγ,
X6(s) =1
i6γh6γ
x2γ
sγΓ(1 + 2γ)− 6
2i6γh4γmsγ, (5.2.21)
......
...
Xj(s) =1
ijγhjγ
x2γ
sγΓ(1 + 2γ)− j
2ijγhj−2γ msγ
, (5.2.22)
85
therefore, the solution of the LFSE (5.2.14), (5.2.15) is approximated by the help of (5.2.13) and (5.2.22),
as below
v(t, x) =
∞∑j=0
tjγ
Γ(jγ + 1)L−1γ { 1
ijγhjγ
x2γ
sγΓ(1 + 2γ)− j
2ijγhj−2γ msγ
}
=∞∑j=0
tjγ
Γ(jγ + 1){ 1
ijγhjγ
x2γ
Γ(1 + 2γ)− j
2ijγhj−2γ m
}.
Example 5.2.2.
∂γ
∂tγv(t, x) + i
∂2γ
∂x2γv(t, x) = 0, (5.2.23)
with initial condition
v(0, x) = Eγ(xγ) = X0(x). (5.2.24)
Applying LFLT on (5.2.24), we have
X0(s) =Eγ(x
γ)
sγ. (5.2.25)
From (5.2.23), we have the operator λγ = −i ∂2γ
∂x2γ . Applying the operator λγ on (5.2.25), we deduce
X1(s) = −i ∂2γ
∂x2γEγ(x
γ)
sγ= −iEγ(x
γ)
sγ. (5.2.26)
For the second approximation, we apply the operator λγ on (5.2.26) and using (5.2.9), we have
X2(s) = −i ∂2γ
∂x2γ
((−1)i
Eγ(xγ)
sγ
)= (−1)2i2
Eγ(xγ)
sγ. (5.2.27)
Continuing this process, we have
X3(s) = −i ∂2γ
∂x2γ
((−1)2i2
Eγ(xγ)
sγ
)= (−1)3i3
Eγ(xγ)
sγ, (5.2.28)
X4(s) = −i ∂2γ
∂x2γ
((−1)3i3
Eγ(xγ)
sγ
)= (−1)4i4
Eγ(xγ)
sγ, (5.2.29)
X5(s) = −i ∂2γ
∂x2γ
((−1)4i4
Eγ(xγ)
sγ
)= (−1)5i5
Eγ(xγ)
sγ, (5.2.30)
......
...
Xj(s) = (−1)jijEγ(x
γ)
sγ. (5.2.31)
From (5.2.13) and (5.2.31), we have the approximation of solution of the probelem (5.2.23), (5.2.24) as
under
v(t, x) =
∞∑j=0
tjγ
Γ(1 + jγ)L−1γ {(−1)jij
Eγ(xγ)
sγ}
= Eγ(xγ)Eγ(−it)γ = Eγ(x− it)γ . (5.2.32)
86
5.3 Iteration technique for the solution of fractal Vehicular Traffic flowmodel
In this section, we develop iterative scheme for analytic solution of FVTFM which is also known by
Lighthill-Richards model [123]. The model equation is given by
∂nγ
∂tnγv(t, x) + ρ
∂mγ
∂xmγv(t, x) = 0, (5.3.1)
where γ ∈ (0, 1), with the conditions
v(0, x) = f(x), v(t, 0) = g(t). (5.3.2)
For this, consider the following equation
vnγt (t, x) = ρλγv(t, x), (5.3.3)
where vnγt (t, x) = Dnγ
dtnγ v(t, x) and λγ is a linear operator. Expressing v(t, x) as a multi-term separated
functions of independent variables Ti and Xi as below:
v(t, x) =∞∑j=0
Tj(t)Xj(x), (5.3.4)
where Tj(t) and Xj(x) are local fractional continuous functions. Expressing the function Tj(t) =
tjγ
Γ(1+jγ) , then (5.3.4) takes the form:
v(t, x) =
∞∑j=0
cjtjγ
Γ(1 + jγ)Xj(x), (5.3.5)
assuming cj = 1, in (5.3.5), we have
v(t, x) =
∞∑j=0
tjγ
Γ(1 + jγ)Xj(x). (5.3.6)
From (5.3.6), we have the following equalities:
vnγt (t, x) =
∞∑j=0
tjγ
Γ(1 + jγ)Xj+n(x), (5.3.7)
λγv(t, x) =
∞∑j=0
tjγ
Γ(1 + jγ)λγXj(x). (5.3.8)
Using (5.3.7) and (5.3.8) in (5.3.3), we obtain
∞∑j=0
tjγ
Γ(1 + jγ)Xj+n(x) =
∞∑j=0
tjγ
Γ(1 + jγ)λγXj(x). (5.3.9)
87
From (5.3.9),we have the following recursion
Xj+n(s) = λγXj(s). (5.3.10)
By the recursion formula (5.3.10), we have
Xj+1(s) = λγXj(s), (5.3.11)
and
Xj+2(s) = λγXj(s). (5.3.12)
Finally, we get the solutions of (5.3.3) as below
v(x, s) =∞∑j=0
tjγ
Γ(1 + jγ)Xj(s). (5.3.13)
Where the convergent condition is
limj→∞
tjγ
Γ(1 + jγ)Xj(s) = 0. (5.3.14)
Therefore, the solution of (5.3.3), is obtained by the help of (5.3.11), as
v(t, x) =
∞∑j=0
tjγ
Γ(1 + jγ)L−1γ {Xj(s)}. (5.3.15)
5.3.1 Applications
This subsection is reserved for the application of the recursion scheme (5.3.11).
Example 5.3.1. Consider the FVTFM:
∂γ
∂tγv(t, x) + ρ
∂γ
∂xγv(t, x) = 0, (5.3.16)
where γ ∈ (0, 1], with the initial condition
v(0, x) = Eγ(xγ) = X0(x). (5.3.17)
In this model, the operator λγ is
λγ = −ρ ∂γ
∂xγ. (5.3.18)
By the help of (5.3.11), we deduce
X1(x) = (−1)ρ∂γ
∂xγEγ(x
γ) = (−1)Eγ(xγ). (5.3.19)
Using (5.3.11) and (5.3.19), we deduce the second iteration
X2(x) = (−1)ρ∂γ
∂xγ(−1)Eγ(x
γ) = (−1)2Eγ(xγ). (5.3.20)
88
Continuing the process of approximating terms, we deduce
X3(x) = (−1)ρ∂γ
∂xγ(−1)2Eγ(x
γ) = (−1)3Eγ(xγ), (5.3.21)
X4(x) = (−1)ρ∂γ
∂xγ(−1)3Eγ(x
γ) = (−1)4Eγ(xγ), (5.3.22)
......
Xj(x) = (−1)jEγ(xγ). (5.3.23)
The solution of FVTFM (5.3.16), (5.3.17), is approximated by the help of (5.3.12) and (5.3.23), as:
v(t, x) =
∞∑j=0
tjγ
Γ(1 + jγ)(−1)jEγ(x
γ)
= Eγ(xγ)
∞∑j=0
t2jγ
Γ(1 + 2jγ)ρ2j −
∞∑j=0
ρ2j+1t(2j+1)γ
Γ(1 + (2j + 1)γ)
(5.3.24)
= Eγ(xγ)(cosγ(ρt
γ)− sinγ(ρtγ)).
Example 5.3.2. Consider the following FVTFM:∂γ
∂tγv(t, x) + ρ
∂γ
∂xγv(t, x) = 0, (5.3.25)
where γ ∈ (0, 1], with the initial condition
v(0, x) = sinhγ(xγ) = X0(x). (5.3.26)
From (5.3.25), (5.3.26) we have ρ = 1 and the operator λγ is
λγ = − ∂γ
∂xγ. (5.3.27)
And
X0(x) = sinhγ(xγ). (5.3.28)
For the first approximation of the model (5.3.25), (5.3.26), we use (5.3.11), (5.3.27) and (5.3.28) as
under
X1(x) = − ∂γ
∂xγ(sinhγ(x
γ)) = − coshγ(xγ). (5.3.29)
For the second approximation of the model (5.3.25), (5.3.26), using (5.3.11), (5.3.27) and (5.3.29), and
deduce
X2(x) = − ∂γ
∂xγ(− coshγ(x
γ)) = − sinhγ(xγ). (5.3.30)
Consequently, we get the following approximation
v(t, x) = limm→∞
(sinhγ(xγ)
m∑j=0
(−1)jt2jγ
Γ(1 + 2jγ)
− coshγ(xγ)
m∑j=0
(−1)jt(2j+1)γ
Γ(1 + (2j + 1)γ)) (5.3.31)
= sinhγ(xγ) cosγ(t
γ)− coshγ(xγ) sinγ(t
γ).
89
Chapter 6
Numerical solutions of fractionaldifferential equations
Numerical solutions of FDEs have studied by many researchers such as; Yousefi and Behroozifar [84],
developed operational matrices using Bernstein polynomials (BPs), Rostamy et al. [83], studied multi
term FDEs by the using operational matrices of BPs, Jafari et al. [81], studied Abel differential equation
of fractional order using Homotopy analysis method, Baleanu et al. [88], studied fractional quadratic
Riccati differential equations via operational matrices using Bernstein polynomials(OMBPs), Alipour
and Rosatamy [89], used OMBPs for solving Abel’s integral equation, Alipour and Rostamy [74], ob-
tained numerical solutions of nonlinear FDE via OMBPs.
In literature, some good and interesting results on the numerical solutions of telegraph equations
(TE) via different numerical tools are available, for example, Inc et al. [72], studied numerical solutions
of one dimensional Telegraph equation (TE) of second kind by reproducing kernel in Hilbert space,
Srivastava et al. [85], studied the solutions of TE by reduced differential transform method, Garg et
al. [86], studied space time FTE by generalized differential transform method.
In this chapter, we develop numerical scheme for solutions of FTE and a coupled system of farctional
partial differential equation (FPDE) by using OMBPs for fractional derivatives. Some of the results of
this chapter are published [136]. This chapter has three sections, in section 6.1, we develop OMBPs for
fractional derivatives. In section 6.2 use the OMBPs and develop s numerical technique for solutions of
FTEs. In section 6.3, a numerical technique for the solutions of a coupled system of FPDEs is developed
using OMBPs.
6.1 Operational Matrix for fractional order derivatives using BPs
90
The operational matrix of fractional order derivative is followed by (1.1.3) as
0Dαϕ(t) =1
Γ(n− α)
∫ t
0(t− x)n−α−1χ(n)
m (x)dx =1
Γ(n− α)tn−α−1 ∗ χ(n)
m (t), (6.1.1)
where the operator ∗ denotes the convolution product. Thus, we have
0Dαϕ(t) = A1
Γ(n− α)
(tn−α−1 ∗ T (n)
m (t))= ADαTm(x) = A [Dα1,Dαt, . . . ,Dαtm]T . (6.1.2)
where Dαti is as defined in (1.1.4). Therefore we can write,
DαTm(t) = HT (6.1.3)
where H is a square diagonal matrix of order (m+1)×(m+1) and T is a row matrix of order (m+1)×1.
The entries of these matrices are given by
Hij =
Γ(j+1)
Γ(j+1−α) i, j = ⌈α⌉ , . . . ,m and i = j,
0 otherwise,(6.1.4)
and
T ij =
0 i = 0, . . . , ⌈α− 1⌉ ,
ti−α i = ⌈α⌉ , . . . ,m.(6.1.5)
Now we approximating ti−α for i = ⌈α⌉ , . . . ,m using BPs as in (1.2.4), we have
ti−α ≈MTi ϕm(t) (6.1.6)
where Mi is a row matrix of order (m+ 1)× 1 and can be obtained by the relation
Mi = Q−1
(∫ 1
0ti−αϕm(t)dt
)= Q−1Mi (6.1.7)
where M i =[M i,0,M i,1, . . . ,M i,m
]T and
Mi,j =
∫ 1
0ti−αHj,m(t)dt =
m!Γ(j + i− α+ 1)
j!Γ(i+m− α+ 2)(6.1.8)
where i = ⌈α⌉ , . . . ,m and j = 0, 1, . . . ,m. Now we assume a square matrix V of order (m+1)×(m+
1) having zero vectors at first ⌈α⌉ columns andWi in its (⌈α⌉+1)th columns, for i = (⌈α⌉+1), . . . ,m.
Consequently we have
Dαϕm(t) ≈ Dαϕm(t) ≈ AHV Tϕm(t) (6.1.9)
where
Dα ≈ AHV T (6.1.10)
is the operational matrix of fractional derivative of order α using BPs.
91
6.2 Numerical scheme for solutions of Fractional Telegraph equations
In this section, we develope a numerical technique for solutions of fractional Telegraph equations (FTE)
via operational matrices of fractional order derivatives and operational matrix of multiplication based on
BPs. Consider the FT equation
∂αu(x, t)
∂tα− C2∂
2u(x, t)
∂x2+ a
∂u(x, t)
∂x+ bu(x, t) = f(x, t), α ∈ (1, 2] (6.2.1)
with initial conditions
u(x, 0) = f1(x), ut(x, 0) = f2(x),
and boundary conditions u(0, t) = g1(t), u(1, t) = g2(t), where x, t ∈ [0, 1], and a, b ∈ R. Using
(1.2.9), we have
u(x, t) ≈ ϕT (x)C�1ϕ(t), f(x, t) ≈ ϕT (x)C�2ϕ(t) (6.2.2)
and by (6.1.9) and (6.1.10), we have
∂2
∂x2u(x, t) ≈ (
∂2
∂x2ϕ(x))TC�1ϕ(t) ≈ (D2ϕ(x))
TC�1ϕ(t) = ϕ(x)TDT2 C�1ϕ(t),
∂
∂xu(x, t) ≈ (
∂
∂xϕ(x))TC�1ϕ(t) = (Dϕ(x))TC�1ϕ(t) = ϕ(x)TDTC�1ϕ(t)
∂α
∂tαu(x, t) ≈ ϕ(x)TC�1Dαϕ(t),
u(x, 0) = θ1(x) ≈ F T1 ϕ(x), ut(x, 0) = θ2(x) ≈ F T
2 ϕ(x).
(6.2.3)
Using (6.2.3) in (6.3.1), we have
ϕ(x)TC�1Dαϕ(t)− C2ϕ(x)TDT2 C�1ϕ(t) + aϕ(x)TDTC�1ϕ(t) + bϕ(x)TC�1ϕ(t)
≈ ϕ(x)TC�2ϕ(t)(6.2.4)
with initial conditions
ϕ(x)TC�1ϕ(0) ≈ F T1 ϕ(x), ϕ(x)TC�1D1ϕ(0) ≈ F T
2 ϕ(x)
ϕ(0)TC�1ϕ(t) ≈ GT1 ϕ(t), ϕ(1)TC�1ϕ(t) ≈ GT
2 ϕ(t).(6.2.5)
We use Tau method for solutions of the system, so we get from (6.2.4) and (6.2.5), the following system∫ 1
0
∫ 1
0Hi(x)ϕ(x)
T (C�1Dα − C2DT2 C�1 + aDTC�1 + bC�1 − C�2)ϕ(t)Hj(t)dxdt = 0,
i = 0, 1, . . . ,m− 4
(6.2.6)
with
C�1ϕ(0)−F1 = 0, C�1D1ϕ(0)−F2 = 0, ϕ(0)TC�1 −GT1 = 0, ϕ(1)TC�1 −GT
2 = 0. (6.2.7)
Solving this system of algebraic equations for the unknown C�1 , C�2 and putting in (6.3.2), give us our
required approximations.
92
6.2.1 Applications
In this section, we provide some applications of the technique (6.2.6), (6.2.7) developed in the previous
section.
Example 6.2.1. Choose f1(x) = x, f2(x) = 0, f(x, t) = 1 − x − t2 and the constants C2 = 1, a =
−1, b = −1. For α = 2, the exact solution is given by u(x, t) = x+ t2. For the purpose of comparison,
we compare the exact solution with the solution obtained using the technique (6.2.6), (6.2.7). In Figure
6.1, the comparison of our scheme for α = 1.5, 1.75, 2 with the exact solution is given. In Figure 6.2, we
have given error in our scheme for by plotting the absolute value of the difference of our approximated
value and that of exact solution. We have considered the value of the approximating terms of BPs m =
10. In Figure 6.3, we have given the sketch of our approximation for different values of α = 1.5, 1.75, 2
and the exact solution for t = 0.25.
0.00.5
1.0
0.0
0.5
1.0
0
1
2
Figure 6.1: Graph for the approximation of u(x, t) when m = 10 and α = 1.5, 1.75, 2.0 and the exact solution
in6.2.1, by our numerical technique (6.2.6)– (6.2.7).
0.0
0.5
1.0 0.0
0.5
1.00
5.´10-8
1.´10-7
Figure 6.2: Plot for error in our approximation for u(x, t), by the proposed technique and our numerical technique
(6.2.6), (6.2.7) for α = 2 with m = 10 in 6.2.1.
93
0.2 0.4 0.6 0.8 1.0
1.0
1.5
2.0
Figure 6.3: Graph for the approximate values of u(x, t) for α = 1.5, 1.75, 2.0, m = 10 and the exact solution at
t = 0.25. The graph for our approximation is the dashed for α = 2 and the green line represents the exact solution
in 6.2.1.
Example 6.2.2. We assume a FTE with initial conditions θ1(x) = 0, θ2(x) = 0 and the constants
C2 = 2, a = 1, b = 1, with f(x, t) = 12t2x2 − 4t4 + 2xt4 + t4x2. For α = 2 the exact solution is
u(x, t) = t4x2, for α = 2. and comparison with the exact solution is given for the illustration of our
technique (6.2.6), (6.2.7). In Figure 6.4, the comparison of our scheme for α = 1.5, 1.75, 2 with the
exact solution is given. In Figure 6.5, we have given error in our scheme for by plotting the absolute
value of the difference of our approximated value and that of exact solution. We have considered the
value of the approximating terms of BPs m = 10. In Figure 6.6, we have given the sketch of our
approximation for different values of α = 1.5, 1.75, 2 and the exact solution for t = 0.25.
0.0
0.5
1.00.0
0.5
1.0
0.0
0.2
0.4
Figure 6.4: Graph for the approximation of u(x, t) when m = 10 and α = 1.5, 1.75, 2.0 and the exact solution
in example 6.2.2.
94
0.0 0.5 1.0
0.0
0.5
1.0
0
5.´10-6
0.00001
Figure 6.5: Plot for the error in our approximation for u(x, t), by the proposed technique and our numerical
technique (6.2.6)– (6.2.7) for α = 2 with m = 10 in example 6.2.2.
0.2 0.4 0.6 0.8 1.0
0.02
0.04
0.06
0.08
0.10
0.12
Figure 6.6: Graph for the approximate values of u(x, t) for α = 1.5, 1.75, 2.0, m = 10 and the exact solution at
t = 0.25. The graph for our approximation is the dashed for α = 2 and the green line represents the exact solution
in example 6.2.2.
6.3 Numerical scheme for solutions of Fractional partial differential equa-tions (FPDEs)
The numerical solutions of FPDEs is one of the important and hot research area which have got a good
attention of the scientists round the world [76–79, 81, 83, 84]. Motivated by the recent contributions of
the scientists [73,83,84], we develop a scheme for numerical solutions of the following coupled system
of FPDEs using OMBPs K1
∂αu(x,t)∂xα +K2
∂βv(x,t)∂tβ
= 0,
K3∂βv(x,t)
∂xβ +K4∂αu(x,t)
∂tα = 0,
(6.3.1)
with initial conditions
u(0, t) = f(t), v(0, t) = g(t).
where x, t ∈ [0, 1], α, β ∈ (0, 1] and K1,K2,K3,K4, are constants. Using (1.2.9), we have
u(x, t) = ϕT (x)C1ϕ(t), v(x, t) = ϕT (x)C2ϕ(t). (6.3.2)
95
By the use of (6.1.10) and (6.3.2), we obtain
∂α
∂xαu(x, t) ≈ (
∂α
∂xαϕ(x))TC1ϕ(t) ≈ (Dαϕ(x))
TC1ϕ(t) = ϕ(x)TDTαC1ϕ(t),
∂β
∂xβv(x, t) ≈ (
∂β
∂xβϕ(x))TC2ϕ(t) ≈ (Dβϕ(x))
TC2ϕ(t) = ϕ(x)TDTβC2ϕ(t),
∂α
∂tαu(x, t) ≈ ϕ(x)TC1Dαϕ(t),
∂β
∂tβv(x, t) ≈ ϕ(x)TC2Dβϕ(t),
u(0, t) = f(t) ≈ F Tϕ(t), v(0, t) = g(t) ≈ GTϕ(t).
(6.3.3)
Using (6.3.3) in (6.3.1), we have
K1ϕ(x)TDT
αC1ϕ(t) +K2ϕ(x)TC2Dβϕ(t) = 0,
K3ϕ(x)TDT
βC2ϕ(t) +K4ϕ(x)TC1Dαϕ(t) = 0,
(6.3.4)
with initial conditions
ϕ(0)TC1ϕ(t) = F Tϕ(t), ϕ(0)TC2ϕ(t) = GTϕ(t). (6.3.5)
By using Tau method [38], we obtain the following system of algebraic equations∫ 1
0
∫ 1
0Bi(x)ϕ(x)
T(K1DT
αC1 +K2C2Dβ
)ϕ(t)Bj(t)dxdt = 0,∫ 1
0
∫ 1
0Bi(x)ϕ(x)
T(K3DT
βC2 +K4C1Dα
)ϕ(t)Bj(t)dxdt = 0,(
ϕ(0)TC1 − F T)
Q = 0,(ϕ(0)TC2 −GT
)Q = 0,
(6.3.6)
where i = 0, 1, . . . ,m − 1 and j = 0, 1, 2, . . . ,m. Solving this system of algebraic equations for the
unknown C1, C2 and putting in (6.3.2), we get the desired result.
6.3.1 Applications
Example 6.3.1. Here we consider our coupled system with initial conditions f(t) = sin(t), g(t) =
cos(t) and the constants K1 = 1,K2 = 1,K3 = 1,K4 = −1. Since with α = β = 1 our system has
the exact solutions for u(x, t) = ex sin(t) and v(x, t) = ex cos(t). In Figures 6.7 and 6.8, we show the
errors of obtained results for m = 5 and α = β = 1. So, we can see the approximate solutions are good
agreement with analytical solutions.In [73], this example has been solved and errors in their approximate
solutions for x and v have been recorded by 10−3 with m = 5. While in our results which are based on
the OMBPs, are better as compared to [73]. In present work, the errors in the approximate solutions for
x, v are rounded by 10−4 for the same m. Also, figures 6.9 and 6.10 show the approximate solutions for
α = 1, β = 0.8, 0.9, 1 and m = 3. These figures show that for fixed α = 1, as β approaches close to 1,
the approximate solutions approach to the solutions for β = 1 as expected.
96
Figure 6.7: Error in our approximation for u(x, t), for m = 5 and α, β = 1.
Figure 6.8: Error in our approximation for v(x, t), for m = 5 and α, β = 1.
Figure 6.9: Plot of the approximation solutions u(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1.
97
Figure 6.10: Plot of the approximation solutions v(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1.
Example 6.3.2. Here we consider our coupled system with initial conditions f(t) = −t2, g(t) = 0 and
the constants K1 = 1,K2 = −1,K3 = 1,K4 = 1. Since with α = β = 1 our system has the exact
solutions for u(x, t) = x2 − t2 and v(x, t) = 2xt. The errors of approximate solutions with m = 3
and α = β = 1 are shown in figures 6.11 and 6.12. The results show the approximate solutions have
high accuracy. In [73], this example has been solved and errors in their approximate solutions for x and
V have been approximated by 10−3. While in our results which are based on OMBPs, are too better
as compared to [73]. In present work, the errors in the approximate solutions for x, V are rounded
by 10−15. Also, figures 6.13 and 6.14 show the approximate solutions for α = 1, β = 0.8, 0.9, 1 and
m = 3. Similar to previous example, These figures show that for α = 1, as β approaches close to 1, the
approximate solutions approach to the solutions for β = 1 as expected.
Figure 6.11: Error in our approximation for u(x, t), for m = 3 and α, β = 1.
98
Figure 6.12: Error in our approximation for v(x, t), for m = 3 and α, β = 1.
Figure 6.13: Plot of the approximation solutions u(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1.
Figure 6.14: Plot of the approximation solutions v(x, t), for m = 3, α = 1 and β = 0.8, 0.9, 1.
99
Chapter 7
Fractional integral and product matricesof BPs
In this chapter, we propose Bernstein polynomials (BPs) to achieve the numerical solutions of nonlinear
fractional-order chaotic system known by fractional order Brusselator system (FOBS).
By the help of OMBPs for fractional integration and multiplications, a numerical techniques is estab-
lished for the solution of fractional order Logistic equation (FOLE), fractional order Brusselator system
(FOBS) and for an optimal controle problem (OCP). This is joint work with Professor Dumitru Baleanu,
Professor Hossein Jafari, Assistant Professor Mohsen Alipour and Haleh Tajadodi [135, 139, 143]. In
section 7.1, we discuss OMBPs for fractional integration. In section 7.2, detail about OMBPs for mul-
tiplication is given. In section 7.3, a numerical technique for the solution of fractional order Logistic
equation is produced which is applied on two examples in subsection 7.3.1. In section 7.4, a numerical
technique for the solution of fractional order Brusselator system(FOBS) is developed which is applied
for the solution of two problems of FOBS in subsection 7.4.1. In section 7.5, a numerical technique for
the solution of a fractional order optimal controle problem (OCP) is obtained. The technique for OCP is
applied for the solution of two problems of fractional OCPs in subsection 7.5.1.
7.1 Operational matrix for fractional order integration
Lemma 7.1.1. For ϕ(t) = [B0,m(t), B1,m(t), . . . , Bm,m(t)]T , the OMBPs for fractional integral Iα for
α an arbitrary real valued number, is given by:
Iαt ϕ(t) ≈ Iαϕ(t), (7.1.1)
where Iα = ADET , is a square matrix of (m+ 1)× (m+ 1).
100
Proof. The operational matrices of fractional Integration of the vector ϕ(t) can be approximated as [83]:
0Iαt ϕ(t) ≃ Iαϕ(t), (7.1.2)
where Iα is the (m + 1) × (m + 1) Riemann-Liouville fractional operational matrix of integration for
BPs.
0Iαt ϕ(t) =
1
Γ(α)
∫ t
0(t− τ)α−1ϕm(τ)dτ =
1
Γ(α)tα−1 ∗ ϕ(t), (7.1.3)
where ∗ denotes the convolution product and by substituting ϕ(t) = ATm(t) and from Eq. (1.1.4) we
get:
1
Γ(α)tα−1 ∗ ϕ(t) =
1
Γ(α)tα−1 ∗ (ATm(t)) =
1
Γ(α)A(tα−1 ∗ Tm(t))
=A
Γ(α)
[tα−1 ∗ 1, tα−1 ∗ t, . . . , tα−1 ∗ t
]T= A [Iα1, Iαt, . . . , Iαtm]T
= A
[0!
Γ(1 + α)tα,
1!
Γ(α+ 2)tα+1, . . . ,
m!
Γ(α+m+ 1)tα+m
]T= ADTm,
where D and Tm are given by
D =
0!
Γ(1+α) 0 . . . 0
0 1!Γ(α+2) . . . 0
......
. . ....
0 0 . . . m!Γ(α+m+1)
and Tm =
tα
tα+1
...
tα+m
.
Now we approximate tk+α by m+ 1 terms of the Bernstein basis
tα+i ≃ ETi ϕm(t), (7.1.4)
where,
Ei = Q−1
(∫ 1
0tα+iϕ(t)dt
)(7.1.5)
= Q−1
[∫ 1
0tα+iB0,m(t)dt,
∫ 1
0tα+iB1,m(t)dt,. . . ,
∫ 1
0tα+iBm,m(t)dt
]T=Q−1Ei,
Ei = [Ei,0, Ei,1, . . . , Ei,m] and
Ei,j =
∫ 1
0tα+iBi,j(t)dt =
m!Γ(i+ j + α+ 1)
j!Γ(i+m+ α+ 2), i, j = 0, 1, . . . ,m, (7.1.6)
where E is an (m+1)× (m+1) matrix that has vector Q−1Ei for ith columns. Therefore, we can write
Iαϕ(t) = AD[ET0 ϕ(t), E
T1 ϕ(t), . . . , E
Tmϕ(t)]
T = ADETϕ(t). (7.1.7)
Finally, we obtain
0Iαt ϕ(t) ≃ Iαϕ(t) (7.1.8)
where
Iα = ADET , (7.1.9)
is operational matrix for fractional integration.
101
7.2 Operational matrix for multiplication
Lemma 7.2.1. Suppose y(t) ≈ cTϕ(t), x(t) ≈ DTϕ(t) and C(m+1)×(m+1) be the operational matrix of
product using BPs for vector c. Then the approximation of product x(t)y(t) can be expressed by
x(t)y(t) ≈ ϕ(t)Cd. (7.2.1)
Proof. The operational matrices for the product C is given by
CTϕ(t)ϕ(t)T ≃ ϕ(t)T C, (7.2.2)
where C is an (m+ 1)× (m+ 1) matrix. So we have
CTϕ(t)ϕ(t)T = CTϕ(t)(Tm(t)TAT
)= [CTϕ(t), t(CTϕm(t)), . . . , tm(CTϕm(t))]AT
=
[n∑
i=0
ciBi,m(t),
n∑i=0
citBi,m(t), . . . ,
n∑i=0
citmBi,m(t)
]. (7.2.3)
Now, we approximate all functions tkBi,n(t) in terms of {Bi,m}mi=0 for i, k = 0, 1, . . . ,m. By (1.2.4),
we have
tkBi,m(t) ≃ eTk,iϕm(t). (7.2.4)
where ek,i = [e0k,i, e1k,i, . . . , e
mk,i]
T , then we obtain the components of the vector of ek,i
ejk,i = Q−1
(∫ 1
0tkBi,m(t)ϕ(t)dt
)= Q−1
[∫ 1
0tkBi,m(t)B0,m(t)dt,
∫ 1
0tkBi,m(t)B1,m(t)dt,. . . , tkBi,m(t)Bm,m(t)dt
]T=
Q−1
2m+ k + 1
[ (m0
)(2m+ki+k
) , (m1
)(2m+ki+k+1
) , . . . , (mm
)(2m+ki+k+m
)]T i, k = 0, 1, . . . ,m. (7.2.5)
Finally, we obtain
n∑i=0
citkBi,m(t) =
n∑i=0
ci
n∑j=0
ejk,iBj,m(t)
=
n∑j=0
Bj,m(t)
(n∑
i=0
ciejk,i
)
= ϕm(t)T
[n∑
i=0
cie0k,i,
n∑i=0
cie1k,i, . . . ,
n∑i=0
ciemk,i
]T= ϕm(t)T [ek,0, ek,1, . . . , ek,m]C = ϕm(t)TVk+1C, (7.2.6)
where Vk+1(k = 0, 1 . . . ,m) is an (m+ 1)× (m+ 1) matrix that has vectors ek,i(i = 0, 1, . . . ,m) for
each column. If we choose an (m+ 1)× (m+ 1) matrix
C = [V1c, V1c, . . . , Vm+1c], then from (7.2.3) and (7.2.6), we write
CTϕ(t)ϕ(t)T ≃ ϕ(t)T CAT , (7.2.7)
where C = CAT is the operation matrix of product.
102
Corollary 7.2.1. Suppose y(t) ≈ cTϕ(t) and C(m+1)×(m+1) be the operational matrix of product using
BPs for vector c, then the approximation of function yk(t), (k ∈ N) is given by
yk(t) ≈ ϕ(t)Ck. (7.2.8)
where Ck ≈ Ck−1c.
7.3 Fractional order Logistic equation
In this section, we are interested to develop a numerical technique for solutions of fractional order
Logistic equations (FOLE) using operational matrices of fractional order integration and multiplication
based on BPs
Dαx(t) = βx(t)(1− x(t)), t > 0, β > 0, α ∈ (0, 1] (7.3.1)
with initial condition
x(0) = b (7.3.2)
We expand fractional derivative in (6.3.1) by the Bernstein basis ϕ as follows:
Dαx(t) = HTϕ(t) (7.3.3)
where
HT = [c0, c1, . . . , cm], ϕT = [Υ0,m,Υ1,m, . . . ,Υm,m], (7.3.4)
are unknowns. Using initial condition (7.3.2), (1.1.5) and (7.3.1), we have
x(t) = 0Iαt Dα
t x(t) + x(0) ≈ (HTIα +DT )ϕ(t) = GTϕ(t), (7.3.5)
where (HTIα+DT ) = GTα and Iα is the fractional operational matrix for integration of order α defined
by (7.1.9) and DT = [b, b, . . . , b]. By substituting the approximations (7.4.3), (7.4.5) in (7.3.1), we have
HTϕ(t) ≈ βGTϕ(t)(Id −GTϕ(t)) = β(GTϕ(t)−GTϕ(t)ϕT (t)G)
= β(GTϕ(t)− ϕT (t)GG) = β(GT −GT GT )ϕ(t)(7.3.6)
where Id is the identity matrix. Thus from (7.3.6), we obtain the following system
HT − β(GT −GT GT ) = 0 (7.3.7)
from this system we getH and hence we approximate the solution x(t) of FOLE (7.3.1) by using (7.3.5).
103
7.3.1 Applications
Here we consider three examples of FOLE for the numerical solutions in order to check the accuracy of
our scheme. We consider the solutions for different values of α and compare with the exact solutions in
these examples.
Example 7.3.1. We consider the following FOLE [96]:
Dαx(t) =1
2x(t)(1− x(t)), t > 0, 0 < α ≤ 1,
x(0) =1
2,
(7.3.8)
for the numerical solutions for different values of α by our prescribed scheme. The solution is plotted in
comparison with the exact solution
x(t) =Exp( t2)
(1 + Exp( t2)).
Our approximations are too close to the exact solution in interval [0, 1] and beyond this interval gradual
deviation is observed. We provide the solutions in [0, 2], by 7.1 and the error plot is given in interval
[0, 1], by 7.2.
0.5 1.0 1.5 2.0x
0.55
0.60
0.65
0.70
yHxL
Figure 7.1: Approximation of FOLE in example 7.3.1, where the black line represents the exact solution, Red
line is for α = 1, Green line is for α = .99, Doted line is for α = .98, and x ∈ [0, 2] .
Example 7.3.2. We consider a FOLE [96]:
Dαx(t) =1
4x(t)(1− x(t)), t > 0, 0 < α ≤ 1,
x(0) =1
3,
(7.3.9)
for the numerical solutions for different values of α by our prescribed scheme. The approximate solu-
tions are plotted in comparison with the exact solution
x(t) =Exp( t4)
(2 + Exp( t4))
104
0.2 0.4 0.6 0.8 1.0x
0.0002
0.0004
0.0006
0.0008
yHxL
Figure 7.2: Error in our approximation for the FOLE in example 7.3.1 for α = 1 when x ∈ [0, 1].
which shows too better results in interval [0, 1], and beyond this interval little deviation is observed. We
provide the solutions in interval [0, 2], by 7.3.3, and the error plot is given in interval [0, 1], by 7.4.
0.5 1.0 1.5 2.0x
0.35
0.40
0.45
yHxL
Figure 7.3: Approximation of FOLE in example 2 via BPs operational matrices. Here Black line=Exact solution,
Green line is for α = 1, Doted line is for α = .99, and the Red line is for α = .95, and x ∈ [0, 2].
0.2 0.4 0.6 0.8 1.0x
0.00005
0.00010
0.00015
yHxL
Figure 7.4: Error in our approximation of the FOLE in example 7.3.2 for α = 1 when x ∈ [0, 1].
105
Example 7.3.3. We consider a FOLE:
Dαx(t) =1
3x(t)(1− x(t)), t > 0, 0 < α ≤ 1,
x(0) =1
3,
(7.3.10)
for the numerical solutions for different values of α by our prescribed scheme. The solution is plotted in
comparison with the exact solution
x(t) =Exp( t3)
(2 + Exp( t3)).
The solutions obtained are tending close to the exact solution in interval [0, 1] when α approaches 1 and
beyond this interval deviation is observed. We provide the solutions in interval [0, 2], by 7.5 and the
error plot is given in the interval [0, 1], by 7.6.
0.5 1.0 1.5 2.0x
0.35
0.40
0.45
0.50
yHxL
Figure 7.5: In this figure, we present comparison of our approximation with the exact solution in example 7.3.3,
for different values of α. Here the Dashed line represents exact solution at α = 1, where our approximations are:
Green line is for α = 1, Red line is for α = .99, Gray line is for α = .98, Black line is for α = .95, and x ∈ [0, 2].
0.2 0.4 0.6 0.8 1.0x
0.00005
0.00010
0.00015
0.00020
0.00025
0.00030
0.00035
yHxL
Figure 7.6: Error in our approximate solution of the FOLE in example 7.3.3, when α = 1, x ∈ [0, 1].
106
7.4 Fractional order Brusselator system (FOBS)
Chaos theory is considered to be an important tool for viewing and understanding our universe and
different ways are utilized in order to reduce problems produced by the unusual behaviors of chaotic
system including chaos control [92,93]. In literature, several authors have considered the chaotic system
FOBS recently. Such as: Gafiychuk and Datsko [95], investigated stability of FOBS. Wang and Li [94],
proved that the solution of FOBS has a limit cycle using numerical method. Jafari et al. [81], used the
variational iteration method for the investigation of the approximate solutions of a FOBS.
Here we are interested in getting the numerical solutions of the nonlinear FOBS:
Dαt x(t) = a− (µ+ 1)x(t) + x2(t)y(t), (7.4.1)
Dβt y(t) = µx(t)− x2(t)y(t),
with initial conditions
x(0) = c1, y(0) = c2 (7.4.2)
via operational matrices of fractional order integration and multiplication of BPs. Here, a > 0, µ > 0,
α, β ∈ (0, 1], and c1, c2 are constants and Dα,Dβ represent Caputo’s derivatives of order α, β respec-
tively. Expanding Dαy(t) in the basis of Bernstein polynomials, we have
Dαy(t) = CTϕ(t), (7.4.3)
where
CT = [c0, c1, . . . , cm], ϕT = [B0,m, B1,m, . . . , Bm,m], (7.4.4)
are unknowns. Using initial condition (7.4.2), (1.1.5) and (7.1.9), we approximate x(t) by
x(t) = 0Iαt Dα
t x(t) + x(0) ≈ (CT1 Iα +DT )ϕ(t) = GTϕ(t), (7.4.5)
where (CT1 Iα +DT ) = GT
α and Iα is the operational matrix for fractional order integration of order α
defined by (7.1.9) and
DT = [x0, x0, . . . , x0]. (7.4.6)
Similarly, we approximate y(t) from (7.4.1) by BPs as
y(t) = 0Iαt D
βt x(t) + y(0) ≈ (CT
2 Iβ + dT2 )ϕ(t) = HTϕ(t), (7.4.7)
where (CT2 Iβ +DT ) = HT
β and Iβ is the fractional operational matrix of integration of order β defined
by (7.1.9) and
dT2 = [y0, y0, . . . , y0]. (7.4.8)
107
Substituting (7.4.3)–(7.4.5) and (7.4.7) in (7.4.1), we get
CT1 ϕ(t) = ATϕ(t)− (µ+ 1)GTϕ(t) +HTϕ(t)Gϕ(t)ϕT (t)GT
1 ,
CT2 ϕ(t) = µGTϕ(t)−HTϕ(t)GTϕ(t)ϕT (t)G.
(7.4.9)
Now using matrix of multiplication (7.2.7) in (7.4.9), we have:
CT1 ϕ(t) = ATϕ(t)− (µ+ 1)GTϕ(t) + ϕT (t)HGG, (7.4.10)
CT2 ϕ(t) = µGTϕ(t)− ϕT (t)HGG.
From (7.4.10) we have the following system(CT1 −AT + (µ+ 1)GT −GT GT HT
)ϕ(t) = 0, (7.4.11)(
CT2 − µGT +GT GT HT
)ϕ(t) = 0. (7.4.12)
Using the independent property of BPs, we get
CT1 −AT + (µ+ 1)GT −GT GT HT = 0, (7.4.13)
CT2 − µGT +GT GT HT = 0, (7.4.14)
solving this system for the vectors C1, C2, we can approximate x(t) and y(t) from (7.4.5), (7.4.7)
respectively.
7.4.1 Applications
Below we use the presented approach in order to solve two examples.
Example 7.4.1. We consider FOBS given by [81]:
Dαt x(t) = −2x(t) + x(t)2y(t), (7.4.15)
Dβt y(t) = x(t)− x(t)2y(t),
with initial conditions x(0) = 1, y(0) = 1.
Fig. 7.7 presents comparison between exact and approximate solution obtained by the help of BPs
for x(t), y(t) at α = 1, β = 1 when m = 8, 12 and Fig. 7.8 presents comparison between exact solution
and our approximate solution by BPs for (t), y(t) at m = 12 and different values of α and β.
Example 7.4.2. We demonstrate accuracy of the presented numerical technique by considering the FOBS
given by [81]:
Dαx(t) = 0.5− 1.1x(t) + x(t)2y(t), (7.4.16)
Dβy(t) = 0.1x(t)− x(t)2y(t),
with initial conditions x(0) = 0.4, y(0) = 1.5.
108
0.2 0.4 0.6 0.8 1t
0.3
0.4
0.5
0.6
0.7
0.8
0.9
xHtL
0.2 0.4 0.6 0.8 1t
1.025
1.05
1.075
1.1
1.125
1.15
1.175
yHtL
Figure 7.7: The exact solution: (black line) and approximation solutions when α = 1, β = 1 and m = 12
(dotted), m = 8(dashed).
0.2 0.4 0.6 0.8 1t
0.3
0.4
0.5
0.6
0.7
0.8
0.9
xHtL
0.2 0.4 0.6 0.8 1t
1.025
1.05
1.075
1.1
1.125
1.15
1.175
yHtL
Figure 7.8: The exact solution: (black line) and approximation solutions when m = 12 and α = .98, β = 1
(dotted), α = .95, β = .99(dashed), α = 0.9, β = .98(Long-dashed).
Fig. 7.9 demonstrates the exact solution together with the approximate solutions x(t), y(t) for
α, β = 1 and different values of m = 4, 6. Definitely, by increasing the value of m of Bernstein
basis, the approximate values of x(t), y(t) converge to the exact solutions. The approximate solutions
x(t), y(t) together with the exact solution for m = 6 and different values of α, β plotted in Fig. 7.10 we
see that as α approaches 1, the numerical solution converges to exact solution.
0.2 0.4 0.6 0.8 1t
0.45
0.5
0.55
0.6
0.65
0.7
xHtL
0.2 0.4 0.6 0.8 1t
1.15
1.25
1.3
1.35
1.4
1.45
1.5
yHtL
Figure 7.9: The exact solution: (black line) and approximation solutions when α = 1, β = 1 andm = 6 (dotted),
m = 4(dashed).
109
0.2 0.4 0.6 0.8 1t
0.45
0.5
0.55
0.6
0.65
0.7
xHtL
0.2 0.4 0.6 0.8 1t
1.15
1.25
1.3
1.35
1.4
1.45
1.5
yHtL
Figure 7.10: Fig.1. The exact solution: (black line) and approximation solutions when m = 6 and α = 1, β =
.98 (dotted), α = .98, β = .95(dashed).
7.5 Optimal controle problem
In this section, we are interested in the solution of the fractional order OCP:
Minimize:
J(x, u, b) =
∫ b
0L(t, x(t), u(t))dt, (7.5.1)
subjected to the dynamical system:
M1x(t) +M2Dαx(t) = f(t, x(t), u(t)), α ∈ (0, 1], (7.5.2)
x(0) = x0, x(b) = xb, (7.5.3)
where M1,M2 are constants and f, L are multi-variable polynomials of t, x(t), u(t). The scheme is
produced by the help of operational matrices for fractional order integration of Bernstein polynomi-
als(BPs), see for detail [90, 127]. Expending x(t) by Bernstein basis, as follows:
x(t) ≈ CT1 ϕ(t), (7.5.4)
where CT1 = [µ0, µ1, . . . , µm]. Integrating (7.5.4) and using initial condition in (7.5.2) we have
x(t) ≈ CT1 I1ϕ(t) + x0 = CT
1 I1ϕ(t) +XT0 ϕ(t) = HT
1 ϕ(t), (7.5.5)
were XT0 = [x0, x0, . . . , x0] and HT
1 = CT1 I1 +XT
0 . Similary, by the use of (1.1.5) and (7.5.4) we can
have the approximation
Dαx(t) = I1−αx(t) ≈ I1−αCT1 ϕ(t) = HT
2 ϕ(t), (7.5.6)
where HT2 = ΛT
1 I1−α. Expension of u(t) by Bernstein basis yield:
u(t) ≈ CT2 ϕ(t), (7.5.7)
110
where CT2 = [ν0, ν1, . . . , νm]. Thus
J(x, u, T ) ≈∫ b
0L(t,HT
1 ϕ(t), CT2 ϕ(t))dt. (7.5.8)
Using the corollaries 7.2.1, 7.2.1, we can approximate L(t,HT1 ϕ(t), C
T2 ϕ(t)) by:
L(t,HT1 ϕ(t), C
T2 ϕ(t)) ≈ L(H1, C2)ϕ(t), (7.5.9)
where L : R(m+1)×1 × R(m+1)×1 → R1×(m+1). Thus,
J∗(H1, C2) =
∫ b
0L(H1, C2)ϕ(t)dt, (7.5.10)
by the help of (7.5.4)–(7.5.7), the dynamical system (7.5.2), becomes
M1CT1 ϕ(t) +M2HT
2 ϕ(t) = f(t,HT1 ϕ(t), C
T2 ϕ(t)). (7.5.11)
By the help of Lemma 7.2.1, 7.2.1, we can get the approximation
f(t,HT1 ϕ(t), C
T2 ϕ(t)) ≈ F(H1, C2)ϕ(t), (7.5.12)
where F : R(m+1)×1×R(m+1)×1 → R1×(m+1). By the help of (7.5.11), (7.5.12), we have the following
equation:
M1CT1 ϕ(t) +M2HT
2 ϕ(t)−F(H1, C2)ϕ(t) = 0, (7.5.13)
and thus the dynamical system (7.5.2) gets the form
M1CT1 +M2HT
2 −F(HT1 , C
T2 ) = 0. (7.5.14)
Defining, the Lagrangian function as:
L∗[H1, C2, λ] = J∗[H1, C2] + (M1CT1 +M2HT
2 −F(H1, C2))λ+ λ(HT1 ϕ(b)− xb), (7.5.15)
where λ = [C0, C1, . . . , Cm]T , λ are the unknown Lagrange multipliers. For the minimization of the
optimal control problem (7.5.1), (7.5.2) we have the following necessary conditions
∂
∂C2L∗[H1, C2, λ] = 0,
∂
∂H1L∗[H1, C2, λ] = 0,
∂
∂λL∗[H1, C2, λ] = 0, HT
1 ϕ(b)− xb = 0.
(7.5.16)
From here, we can get C2, H1 and ultimately we can approximate x(t), u(t) from (7.5.5), (7.5.7),
respectively.
111
7.5.1 Applications
Example 7.5.1. We consider the OCP
Minimize J =
∫ 1
0(tu(t)− (α+ 2)x(t))2dt, (7.5.17)
subjected to the dynamical system
x′(t) +Dαx(t) = u(t) + t2, (7.5.18)
with initial and boundary conditions
x(0) = 0, x(1) =2
Γ(α+ 3), (7.5.19)
the exact solution is given by x(t) = 2tα+2Γ(α+3) , u(t) = 2tα+1
Γ(α+2) .
For m = 7, the approximate solutions of the states functions and the control functions are plotted in
the figures 7.11, fign1.2opc. These figures show that as α→ 1 the approximate solutions get close to the
exact solution. Also, the error in the norm L2(∥f(t)∥L2[a,b] =√∫ b
a |f(t)|2dt) for approximate solutions
obtained are given in tables 1, 2 for x(t), u(t) respectively for different values of approximating terms
m.
Figure 7.11: Plot of x(t) for m = 7 and α = .8, .9, 1 in example 7.5.1.
Table 1. ∥x(t)− xm(t)∥ for different values of α and m in example 7.5.1.
m α = 1 α = 0.9 α = 0.8
7 2.89933 ∗ 10−17 7.5999 ∗ 10−7 1.89168 ∗ 10−6
9 2.41324 ∗ 10−17 1.65651 ∗ 10−7 4.27512 ∗ 10−7
12 1.03578 ∗ 10−17 2.89815 ∗ 10−8 7.7282 ∗ 10−8
112
Figure 7.12: Plot of u(t) for m = 7 and α = .8, .9, 1 in example 7.5.1.
Table 2. ∥u(t)− um(t)∥ for different values of α and m in example 7.5.1.
m α = 1 α = 0.9 α = 0.8
7 1.93584 ∗ 10−16 0.0000120377 0.0000314255
9 3.53488 ∗ 10−16 4.09026 ∗ 10−6 0.0000112193
12 8.26356 ∗ 10−17 1.16833 ∗ 10−6 3.38702 ∗ 10−6
Example 7.5.2. We consider the OCP
Minimize J =
∫ 1
0(u(t)− x(t))2dt, (7.5.20)
subjected to the dynamical system
x′(t) +Dαx(t) = u(t)− x(t) +6tα + 2
Γ(α+ 3), (7.5.21)
with initial and boundary conditions
x(0) = 0, x(1) =6
Γ(α+ 4), (7.5.22)
the exact solution is given by x(t) = u(t) = 6tα+3Γ(α+4)
In figures 7.13 and 7.14, we can observe plots of the approximate solutions of the states function and
the control function for m = 8 and α = 0.8, 0.9, 1.0 . Moreover, in tables 3 and 4, we can see the errors
in the norm L2 for the obtained solutions.
Table 3. ∥x(t)− xm(t)∥ for different values of α and m in example 7.5.2.
m α = 1 α = 0.9 α = 0.8
4 1.11714 ∗ 10−18 0.0000288778 0.0000575561
8 7.53703 ∗ 10−19 7.17619 ∗ 10−8 1.80442 ∗ 10−7
12 1.06883 ∗ 10−20 3.14699 ∗ 10−9 8.29973 ∗ 10−9
113
Figure 7.13: Plot of x(t) for m = 8 and α = .8, .9, 1 in example 7.5.2.
Figure 7.14: Plot of u(t) for m = 8 and α = .8, .9, 1 in example 7.5.2.
Table 4. ∥u(t)− um(t)∥ for different values of α and m in example 7.5.2.
m α = 1 α = 0.9 α = 0.8
4 5.87904 ∗ 10−18 0.000112289 0.000221961
8 9.3817 ∗ 10−19 2.92602 ∗ 10−7 7.30618 ∗ 10−7
12 7.79996 ∗ 10−20 1.31311 ∗ 10−8 3.42575 ∗ 10−8
By these examples, we see that our numerical technique is highly efficient.
114
Chapter 8
B-Spline functions
In literature, several authors have considered the Fractional order Brusselator system recently. Like as;
Gafiychuk and Datsko [95], investigated stability of the fractional Brusselator system. Wang and Li [94],
proved that the solution of FOBS has a limit cycle using numerical method. Jafari et al. [81], used the
variational iteration method to investigate the approximate solutions of this system.
In this chapter, we discuss the numerical solution of FOBS by the OM of B-Spline functions for the
fractional integration and multiplcation matrix. In section 8.1, we present OM for fractional integration.
In section 8.2, a product matrix is presented. In section 8.3, a numerical technique for the solution of
FOBS is developed which is applied in subsection 8.3.1.
8.1 Operational matrix for fractional integration
The operational matrix of fractional integration of the vector ϕJ can be approximated as:
0Iαx ϕJ(x) ≃ IαϕJ(x), (8.1.1)
where Iα is the (2J + 1) × (2J + 1) operational matrix of fractional integration for linear B-spline
function. We obtain the matrix Iα as follows:
Iα =
∫ 1
0(0Iα
x ϕJ(x))ϕTJ (x)dx =
(∫ 1
0(0Iα
x ϕJ(x))ϕTJ (x)dx
)P−1, (8.1.2)
where
E =
∫ 1
0(0Iα
x ϕJ(x))ϕTJ (x)dx. (8.1.3)
115
In (8.1.3), E is a (2J + 1)× (2J + 1) matrix given by
E =
∫ 10 (0I
αx ϕj,−1(x))ϕ
Tj,−1(x)dx . . .
∫ 10 (0I
αx ϕj,−1(x))ϕ
Tj,2j−1
(x)dx...
. . ....∫ 1
0 (0Iαx ϕj,2j−1(x))ϕ
Tj,−1(x)dx . . .
∫ 10 (0I
αx ϕj,2j−1(x))ϕ
Tj,2j−1
(x)dx
, (8.1.4)
And we have 0Iαx ϕj,k(x) as follows:
0Iαx ϕj,k(x) = 0Iα
x
(2∑
i=0
(2
i
)(−1)i(2Jx− (k + i))+
)
=
0, x ≤ k2J,
2−Jα(2Jx−k)α+1
Γ(α+2) , k2J
≤ x < k+12J,
2−Jα((2Jx−k)α+1−2(2Jx−(k+1))α+1)Γ(α+2) , k+1
2J≤ x < k+2
2J,
2−Jα((2Jx−k)α+1−2(2Jx−(k+1))α+1+(2Jx−(k+2))α+1)Γ(α+2) , x ≥ k+2
2J.
8.2 Operational matrix of multiplication
It is always necessary to assess the product of ϕ(t) and ϕ(t)T , which is called the product matrix for
linear B-spline function. The operational matrices for the product C is given by
CTϕJ(t)ϕJ(t)T ≃ ϕJ(t)
T C, (8.2.1)
where C is an (m+ 1)× (m+ 1) matrix. So we have
CTϕJ(x)ϕJ(x)T = cTϕJ(x)[ϕj,−1(x), ϕj,0(x), . . . , ϕj,2j−1(x)] (8.2.2)
=
2j−1∑i=−1
ciϕj,i(x)ϕj,−1(x),2j−1∑i=−1
ciϕj,iϕj,0(x), . . . ,2j−1∑i=−1
ciϕj,iϕj,2j−1(x)
,Now, we approximate the product of ϕj,i(x)ϕj,k(x) in terms of ϕj,i for i, k = −1, . . . , 2j − 1, as
ϕj,i(x)ϕj,k(x) ≈ eTk,iϕJ , (8.2.3)
where ek,i = [e0k,i, e1k,i, . . . , e
mk,i]
T , then we obtain the components of the vector of ek,i
ejk,i = P−1
∫ 1
0ϕj,i(x)ϕj,k(x)ϕJ(x)dx (8.2.4)
= P−1
∫ 10 ϕj,i(x)ϕj,k(x)ϕj,−1(x)dx∫ 10 ϕj,i(x)ϕj,k(x)ϕj,0(x)dx
...∫ 10 ϕj,i(x)ϕj,k(x)ϕj,2j−1(x)dx
, (8.2.5)
116
Thus we obtain finally
2j−1∑i=−1
ciϕj,i(x)ϕj,k(x) ≈2j−1∑i=−1
ci
2j−1∑n=−1
ek,in ϕj,n
=2j−1∑n=−1
ϕj,n
2j−1∑i=−1
ciek,in
= ϕTJ
∑2j−1i=−1 cie
k,i−1∑2j−1
i=−1 ciek,i0
...∑2j−1i=−1 cie
k,i2j−1
= ϕTJ [ek,−1, ek,0, . . . , ek,2j−1]C
= ϕTJ Ck+2C, (8.2.6)
where Ck+2(k = −1, 0, . . . , 2j − 1) is an (2J + 1) × (2J + 1) matrix that has vectors ek,i(i =
−1, 0, . . . , 2j − 1) for each column. Therfore we obtain the operational matrix of product C = Ck+2C.
Corollary 8.2.1. If y(t) = CTϕJ(t), consequently we can get the approximate function for yk(t), using
BPs as follows:
yk(t) = ϕJ(t)T Ck, (8.2.7)
where Ck = Ck−1C and C is (2j+1)×(2j+1) operational matrix of product using B–spline functions.
8.3 Fractional order Brusselator system
Here, we are interested in getting the numerical solution of the nonlinear fractional Brusselator system:
Dαt x(t) = a− (µ+ 1)x(t) + x2(t)y(t), (8.3.1)
Dβt y(t) = µx(t)− x2(t)y(t),
with initial conditions
x(0) = c1, y(0) = c2, (8.3.2)
provided that a > 0, µ > 0, α, β ∈ (0, 1], and c1, c2 are constants. Moreover, Dα,Dβ represent Caputo’s
derivative of order α, β respectively. For this, we give a numerical technique for the solutions of FOBS,
based on the operational matrices (OMs) of functions approximations, OM for multiplications and OM
for fractional order integration. By the proposed method the FOBS system (8.3.1), will be replaced in a
system of algebraic equations. For this purpose, we take a start from functions approximation (1.2.15),
as under:
Dαx(t) ≃ CT1 ϕJ(t), (8.3.3)
117
CT1 = [c−1, c0, . . . , c2j−1]
T and ϕJ as defined in (1.2.17). Using initial condition (8.3.4) and (1.1.4), we
approximate x(t) by
x(t) = 0Iαt Dα
t x(t) + x(0) ≈ (CT1 Iα + dT1 )ϕJ(t) = CTϕJ(t), (8.3.4)
where dT1 = [x0, x0, . . . , x0] and C = CT1 Iα + dT1 . Similarly, approximating
Dαy(t) ≈ CT2 ϕJ(t), (8.3.5)
where CT2 = [g−1, g0, . . . , g2j−1]
T and ϕJ as defined in (1.2.17). Using initial condition (8.3.2) and
(1.1.4), we approximate y(t) by
y(t) = 0Iαt D
βt y(t) + y(0) ≈ (CT
2 Iβ + dT2 )ϕJ(t) = GTϕJ(t), (8.3.6)
where dT2 = [y0, y0, . . . , y0] and G = CT2 Iβ + dT2 . Using (8.2.7), (8.3.6), we are approximating the
terms x2(t), x2(t)y(t):
x2(t) = x(t)x(t) ≈ CTϕJ(t)ϕTJ (t)C ≈ ϕTJ (t)CC = CT CTϕJ(t), (8.3.7)
and
y(t)x2(t) ≃ GTϕJ(t)ϕTJ (t)CC ≃ ϕTJ (t)GCC = CT CT GTϕJ(t). (8.3.8)
By the help of (8.3.3), (8.3.7), (8.3.8), the FOBS (8.3.1), (8.3.2), gets the form:
CT1 ϕJ(t) = ATϕJ(t)− (µ+ 1)CTϕJ(t) + CT CT GTϕJ(t), (8.3.9)
CT2 ϕJ(t) = µGTϕJ(t)− CT CT GTϕJ(t). (8.3.10)
Finally, we get the following algebraic system:
CT1 −AT + (µ+ 1)CT − CT CT GT = 0, (8.3.11)
CT2 − µGT + CT CT GT = 0, (8.3.12)
solving this system for C1, C2 we can approximate the solutions x(t) and y(t) from (8.3.4), (8.3.6)
respectively.
8.3.1 Applications
Below we use the presented approach in order to solve fractional fractional order Brusselator system.
Example 8.3.1. We consider fractional order Brusselator system given by [81]:
Dαt x(t) = −2x(t) + x(t)2y(t), (8.3.13)
Dβt y(t) = x(t)− x(t)2y(t),
with initial conditions x(0) = 1, y(0) = 1.
118
0.2 0.4 0.6 0.8 1t
0.4
0.6
0.8
xHtL
0.2 0.4 0.6 0.8 1t
1.025
1.05
1.075
1.1
1.125
1.15
1.175
yHtL
Figure 8.1: The exact solution (black line) and approximation solutions when α = 1, β = 1 and J = 5 (dotted)
and J = 4 (dashed).
0.2 0.4 0.6 0.8 1t
0.3
0.4
0.5
0.6
0.7
0.8
0.9
xHtL
0.2 0.4 0.6 0.8 1t
1.025
1.05
1.075
1.1
1.125
1.15
1.175
yHtL
Figure 8.2: The exact solution (black line) and approximation solutions when J = 5 and α = 0.98, β = 1
(dotted) and α = 0.95, β = 0.98 (dashed).
In Figure 8.1 presents comparison between exact and approximate solution obtained by the help of
the linear B-spline function for x(t), y(t) at α = 1, β = 1 when J = 4, 5 and Figure 8.2 presents
comparison between exact solution and our approximate solution for x(t), y(t) at J = 5 and different
values of α and β.
Example 8.3.2. We demonstrate accuracy of the presented numerical technique by considering the FOBS
given by [81]:
Dαx(t) = 0.5− 1.1x(t) + x(t)2y(t), (8.3.14)
Dβy(t) = 0.1x(t)− x(t)2y(t),
with initial conditions x(0) = 0.4, y(0) = 1.5.
Figure 8.3 demonstrates the exact solution together with the approximate solutions x(t), y(t) for
α, β = 1 and different values of J = 4, 5. Definitely, by increasing the value of J of B-spline basis, the
approximate values of x(t), y(t) converge to the exact solutions. The approximate solutions x(t), y(t)
together with the exact solution for J = 5 and different values of α, β plotted in Figure 8.4 we see that
as α approaches 1, the numerical solution converges to exact solution.
119
0.2 0.4 0.6 0.8 1t
0.45
0.5
0.55
0.6
0.65
0.7
xHtL
0.2 0.4 0.6 0.8 1t
1.15
1.25
1.3
1.35
1.4
1.45
1.5
yHtL
Figure 8.3: The exact solution: (black line) and approximation solutions when α = 1, β = 1 and J = 5 (dotted),
J = 4(dashed).
0.2 0.4 0.6 0.8 1t
0.45
0.5
0.55
0.6
0.65
0.7
xHtL
0.2 0.4 0.6 0.8 1t
1.15
1.25
1.3
1.35
1.4
1.45
1.5
yHtL
Figure 8.4: The exact solution (black line) and approximation solutions when J = 5 and α = 0.98, β = 0.99
(dotted) and α = 0.97, β = 0.98 (dashed).
Nomenclature
In this theses we have used the following abbreviations:
FDEs for fractional differential equations, ES for existence of solutions, EUS for existence and unique-
ness of solutions, BVP for boundary value problem, HFDEs for hybrid fractional differential equations,
CSFDIEs for coupled system of fractional differential integral equations, FTE for fractional Telegraph
equation, FPDE for fractional partial differential equation, OM for operational matrix, OMBPs opera-
tional matrices for Bernstein polynomials, OCP for optimal controle problem, FOBS for fractional order
Brusselator system, LF for local fractional, LFSE for local fractional Schrodinger equation, SEM for
series expension method, LFLT for Yang laplace transform, LFLVIM for local fractional laplace vari-
ational iteration method, FVTFM for Fractal Vehicular Traffic flow model and LFWEs for the local
fractional wave equations. Other abbreviations have explained in the relevant work.
We declare that the thesis is our original work and before the submission of the final copy, Professor
Dr Rahmat Ali Khan has done the proof reading.
120
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132