Applied Mathematics and Computation 342 (2019) 280–294
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Computational algorithm for solving fredholm time-fractional partial integrodifferential equations of dirichlet functions type with error estimates Mohammed Al-Smadi a, Omar Abu Arqub b,∗ a b
Department of Applied Science, Ajloun College, Al-Balqa Applied University, Ajloun 26816, Jordan Department of Mathematics, Faculty of Science, The University of Jordan, Amman 11942, Jordan
a r t i c l e
i n f o
Keywords: Computational algorithm Fractional calculus theory Partial integrodifferential equation Fredholm operator Caputo fractional derivative
a b s t r a c t Numerical modeling of partial integrodifferential equations of fractional order shows interesting properties in various aspects of science, which means increased attention to fractional calculus. This paper is concerned with a feasible and accurate technique for obtaining numerical solutions for a class of partial integrodifferential equations of fractional order in Hilbert space within appropriate kernel functions. The algorithm relies on the reproducing kernel Hilbert space method that provides the solutions in rapidly convergent series representations for the reproducing kernel based upon the Fourier coefficients of orthogonalization process. The Caputo fractional derivatives are introduced to address these issues. Moreover, the error estimate of the generated solutions is established as well as the convergence of the iterative method is investigated under some theoretical assumptions. The superiority and applicability of the present technique is illustrated by handling linear and nonlinear numerical examples. The outcomes obtained are compared with exact solutions and existing methods to confirm the effectiveness of the reproducing kernel method. © 2018 Elsevier Inc. All rights reserved.
1. Introduction The study of partial integrodifferential equation (PIDEs) with fractional parameters is an influential part of the theory and applications of fractional calculus which have become indispensable mathematical tools to describe and analyze various real world problems in a wide range of natural science, technology and engineering [1-5]. Mathematical modeling by fractional derivatives and integral operators is more appropriate at different scales of real systems than traditional derivatives and integration, which provide better description of structural and genetic characteristics of many dynamical and physical processes [6-14]. However, numerical solvability of Fredholm PIDEs of fractional-order is the focus of researchers’ attention through developing a suitable mathematical tool that classifies, simplifies, and provides specific information of the qualitative properties of the equations. Thereby utilizing accurate computational algorithms to deal with the complexity of fractional derivatives involved in these equations optimally. Indeed, it is not easy in many applications to obtain a closed form solution, especially in nonlinear situations, therefore an approximation of physical reality is required. On the other hand, for instance, reproducing kernel Hilbert space (RKHS), homotopy perturbation, differential transform, adomian decomposition,
∗
Corresponding author. E-mail address:
[email protected] (O.A. Arqub).
https://doi.org/10.1016/j.amc.2018.09.020 0 096-30 03/© 2018 Elsevier Inc. All rights reserved.
M. Al-Smadi, O.A. Arqub / Applied Mathematics and Computation 342 (2019) 280–294
281
variational iteration, finite difference, and collocation methods are the most commonly techniques used for handling the fractional PDEs generated by the modeling [15-25]. This study investigates the numerical solution and error estimate of the reproducing kernel method (RKM) for Fredholm PIDEs of fractional order in following form:
Dατ u(x, τ ) + μ1 ∂x u(x, τ ) + μ2 ∂x22 u(x, τ ) +
1 0
k(x, τ , ξ )∂x22 u(x, ξ )dξ = F (x, τ , u ),
(1)
with the nonhomogeneous initial and Dirichlet boundary conditions
as
u(x, 0 ) = ν (x ), u(0, τ ) = ϕ1 (τ ), u(1, τ ) = ϕ2 (τ ).
(2)
Here, (x,t) ∈ [0,1] × [0,1], the operator Dα τ u is the Caputo fractional derivative of order α with respect to τ and is defined
Dατ u(x, τ ) =
1 (1 − α )
τ 0
(τ − ξ )−α ∂ξ u(x, ξ )dξ ,
0 < ξ < τ,
(3)
0 < α ≤ 1, μ1 , μ2 ∈ R, ∂ x = ∂ /∂ x, ∂x22 = ∂ 2 /∂ x2 , F (· ) is continuos real-valued function such that Eqs. (1) and (2) satisfies the existence and uniqueness of the solutions, k(x, τ , ξ ) is arbitrary kernel function over the cube [0,1]3 , ν (x), ϕ 1 (τ ) and ϕ 2 (τ ) are continuos real-valued functions in the domain of interest, and u is an unknown function to be determined. The RK theory is a rich construct for several decades in computing numerical solutions with reasonable accuracy for various fields such that differential equations, integral equations, functional analysis, statistical learning theory, stochastic optimization, and so on [26-30]. The advantages of this method lie in: it requires much less space and time levels to obtain a smooth solution, has excellent properties in error estimation reflects high accuracy and reliability, provides uniform convergence rates to the exact solution, and offers a straightforward assumption to deal with various boundary conditions [31-53]. The basic motivation for the current paper is to investigate the error analysis of the iterative RKM for treating a class of PIDEs in Hilbert space. This work is arranged as follows. In Section 2, the RK spaces and associated kernels are provided to formulate the solution. In Section 3, basic assumptions, model formulation and implementation formulas in obtaining numerical solutions are discussed. In Section 4, the error analysis and convergence theorems for the RKM are proved. In Section 5, numerical comparisons between the obtained results and those of other methods are given to demonstrate the validity of the results in this analysis. The outcomes obtained confirm the applicability and robustness of the proposed method. The last section of this paper ends with a brief conclusion. 2. A brief excursion into reproducing kernel spaces Let H be a Hilbert space of real-valued functions defined on a nonempty set X . A function Q : X × X → R is a RK of H if the following are satisfied • Q(·, s ) ∈ H, for each s ∈ X , • ψ (· ), Q(·, s )H = ψ (s ), for each ψ ∈ H and each s ∈ X . A Hilbert space H of functions on a set X is called a RKHS if there exists a RK function Q : X × X → R. The RK function Q of H completely determines the Hilbert space H. Henceforth, for each ζ (∗ ) ∈ W, ||ζ ||2W is defined as ||ζ ||2W = ζ (∗ ), ζ (∗ )W , ˆ 1 , W 2 , W 3 }. Also, D = [0, 1] [0, 1], ∂ i+ j = (∂ i /∂ xi )(∂ j /∂ t j ), whenever i, j = 1,2, and for each where ∗ ∈ [0, 1] and W ∈ {W 1 , W 2
2
2
2
xi t j
u(∗, ◦ ) ∈ γ , ||u||2Y is also defined as ||u||2Y = u(∗, ◦ ), u(∗, ◦ )Y , where (∗, ◦ ) ∈ D, and ϒ ∈ {H,W}. Definition 1 [31]. Suppose that ζ lies in L2 [0,1]. The space W21 [0, 1] is defined as W21 [0, 1] = {ζ = ζ (s ) : ζ is absolutely continuous function on [0,1]}. Whilst, its metric system structure lies in
ζ1 (s ), ζ2 (s )W21 = ζ1 (0 )ζ2 (0 ) +
0
1
ζ1 (ξ )ζ2 (ξ )dξ .
(4)
Theorem 1 [31]. The Hilbert space W21 [0, 1] is a complete RK with structure kernel function {1 }
Rs (τ ) = 1 + min (s, τ ).
(5)
ˆ 1 [0, 1] can be defined as ζ1 (x ), ζ2 (x ) ˆ 1 = ζ1 (0 )ζ2 (0 ) + Similarly, the space W 2 W
min(x, ς ), where [0,1] is the domain space in x direction.
2
1
0 ζ1 ( ξ ) ζ 2 ( ξ ) d ξ
{1} and Rˆς (x ) = 1 +
Definition 2 [32]. Suppose that ζ lies in L2 [0,1]. The space W22 [0, 1] is defined as W22 [0, 1] = {ζ = ζ (s ) : ζ , ζ are absolutely continuous functions on [0,1] and ζ (0) = 0}. Whilst, its metric system structure lies in
ζ1 (s ), ζ2 (s )W22 =
1 ζ1(i) (0 )ζ2(i) (0 ) + i=0
1 0
ζ1 (ξ )ζ2 (ξ )dξ .
(6)
282
M. Al-Smadi, O.A. Arqub / Applied Mathematics and Computation 342 (2019) 280–294
Theorem 2 [32]. The Hilbert space W22 [0, 1] is a complete RK with structure kernel function {2 }
Rs ( τ ) =
⎧ sτ 2 τ3 ⎪ ⎨sτ + − ,
τ ≤ s,
2
6 3 ⎪ s s τ ⎩sτ + − , 2 6
(7)
2
τ > s.
Definition 3 [33]. Suppose that ζ lies in L2 [0,1]. The space W23 [0, 1] is defined as W23 [0, 1] = {ζ = ζ (x ) : ζ , ζ , ζ are absolutely continuous functions on [0,1] and ζ (0) = ζ (1) = 0}. Whilst, its metric system structure lies in
1 = ζ1(i) (0 )ζ2(i) (0 ) + ζ1 (1 )ζ2 (1 ) +
1
ζ1 (s ), ζ2 (s )W23
i=0
ζ1 (ξ )ζ2 (ξ )dξ .
(8)
0
Theorem 3 [33]. The Hilbert space W23 [0, 1] is a complete RK with structure kernel function
{3 }
Rs
s τ 2 s 126 − τ 3 − s3 + s s3 − 10τ 3 + 5τ 24 − s3 + 5τ s τ 3 − 24 + τ s3 − 24 , 1 (τ ) =
120 τ τ s2 126 − τ 3 − s3 + τ τ 3 − 10s3 + 5s 24 − τ 3 + 5s τ y3 − 24 + s τ 3 − 24 ,
τ ≤ s, τ > s. (9)
∂x33 ∂τ33 u
Definition 4. Suppose that lies in The Hilbert space W (D ) can be defined as W (D ) = {u = u(x, τ ) : is complete continuous function in D and u(x, 0) = u(0, τ ) = u(1, τ ) = 0}. Whilst, its metric system structure lies in
u1 (x, τ ), u2 (x, τ )W =
1 j=0
1
1
0
+
∂x22 ∂τ22 u
∂τj j u1 (x, 0 ), ∂τj j u2 (x, 0 )W23
+
L2 ( D ).
0
1
∂τ 2 ∂ 2
j u xj 1
(0, τ )∂τ 2 ∂ 2
j u xj 2
(0, τ ) + ∂τ 2 u1 (1, τ )∂τ 2 u2 (1, τ ) dτ 2
2
j=0 1
0
∂x33 ∂τ22 u1 (x, τ )∂x33 ∂τ22 u2 (x, τ )dxdτ .
(10)
Theorem 4. The Hilbert space W (D ) is a complete RK with structure kernel function {3 }
{2 }
R(y,s) (x, τ ) = Ry (x )Rs (τ ),
(11)
such that for any u(x, τ ) ∈ W (D ), we have u(x, τ ), R(y,s) (x, τ ) are the kernel functions of W23 [0, 1] and W22 [0, 1], respectively.
W
{3}
{2}
= u(y,s) and R(y,s) (x, τ ) = R(x,τ ) (y,s), where Ry (x ) and Rs (τ )
Proof. By using properties of the inner product spaces W32 [0, 1] and W22 [0, 1] with respect to the differentials dx and dτ , respectively, one can find
u(x, τ ), R{y3} (x )R{s2} (τ )W =
1 j=0
∂τj j u(x, 0 ), ∂τj j R{y3} (x )R{s2} (0 )W23
+
1
1
1
+
1
j=0
1 0
(0, τ )∂τ22 ∂
{2 }
{3 }
{2 }
(0 )Rs (τ ) + ∂τ22 u(1, τ )∂τ22 Ry (1 )Rs (τ ) dτ
∂x33 ∂τ22 u(x, τ )∂x33 ∂τ22 R{y3} (x )R{s2} (τ )dxdτ
1
0
1
∂τ22 ∂
j {3 } R xj y
∂τj j u(x, 0 ), R{y3} (x )∂τj j R{s2} (0 )W23
+
j u xj
j=0
0
j=0
=
1
0
+ =
1 0
∂τ22 ∂
j u xj
{2 }
(0, τ )∂τ22 Rs (τ )∂
j {3 } R xj y
0
∂x33 ∂τ22 u(x, τ )∂x33 R{y3} (x )∂τ22 R{s2} (τ )dxdτ
∂τj j u(x, 0 ), R{y3} (x )W23 ∂τj j R{s2} (0 )
{2 }
(0 ) + ∂τ22 u(1, τ )Ry (1 )∂τ22 Rs (τ ) dτ
j=0 1
{3 }
M. Al-Smadi, O.A. Arqub / Applied Mathematics and Computation 342 (2019) 280–294
+
=
0
1 j=0
=
1 j=0
1
∂τ22 R{s2} (τ )∂τ22
1
∂xjj u(0, τ )∂xjj R{y3} (0 ) + u(1, τ )R{y3} (1 ) +
j=0
∂τj j u(y, 0 )∂τj j R{s2} (0 ) + ∂τj j u(y, 0 )∂τj j R{s2} (0 ) +
1
0
1
0
1 0
283
∂x33 u(x, τ )∂x33 R{y3} (x )dx dτ
∂τ22 R{s2} (τ )∂τ22 u(x, τ ), R{y3} (x )W23 dτ ∂τ22 R{s2} (τ )∂τ22 u(y, τ )dτ
{2 }
= u(y, τ ), Rs (τ )W 2 = u(y, s ). 2
(12)
Thus, u(x, τ ), R(y,s) (x, τ )W = u(y,s). Whilst on the other hand, R(y,s) (x, τ ) = R(y,s) (ξ , ζ ), R(x,τ ) (ξ , ζ )W = R(x,τ ) (ξ , ζ ), R(y,s) (ξ , ζ )W = R(x,τ ) (y,s). Definition 5. Suppose that ∂ x ∂ τ u lies in L2 (D ). The Hilbert space H (D ) can be defined as H (D ) = {u = u(x, τ ) : u is complete continuous function in D}. Whilst, its metric system structure lies in
u1 (x, τ ), u2 (x, τ )H = u1 (x, 0 ), u2 (x, 0 )Wˆ 1 +
2
1
0
∂τ u1 (0, τ )∂τ u2 (0, τ )dτ +
1
0
0
1
∂x2τ u1 (x, τ )∂x2τ u2 (x, τ )dxdτ . (13)
Theorem 5. The Hilbert space H (D ) is a complete RK with structure kernel function {1 }
{1 }
r(y,s) (x, τ ) = Rˆy (x )Rs (τ ),
(14) ˆ {1}
{1}
such that for any u(x, τ ) ∈ H (D ), we have u(x, τ ), r(y,s) (x, τ )H = u(y,s) and r(y,s) (x, τ ) = r(x,τ ) (y,s), where Ry (x ) and Rs (τ ) ˆ 1 [0, 1] and W 1 [0, 1], respectively. are the kernel functions of W 2
2
3. Statement of the problem To implement the proposed method, the nonhomogeneous constraints conditions should be homogenized by suitable transformation to lie the mentioned conditions in the space W(D). However, for the convenience of the reader, u(x,t) is still denote to solution of the converted equation. So, let
Dατ u(x, τ ) + μ1 ∂x u(x, τ ) + μ2 ∂x22 u(x, τ ) +
1 0
k(x, τ , ξ )∂x22 u(x, ξ )dξ ≡ p(x, τ , u(x, τ ) ),
(15)
with the homogeneous initial and Dirichlet conditions:
u(x, 0 ) = 0, u ( 0, τ ) = u ( 1, τ ) = 0.
(16)
For the conduct of proceedings, we define the fractional differential linear operator : W (D ) → H (D ) such that
u(x, τ ) := Dατ u(x, τ ) + μ1 ∂x u(x, τ ) + μ2 ∂x22 u(x, τ ) +
1 0
k(x, τ , ξ )∂x22 u(x, ξ )dξ .
(17)
Consequently, we have the following equivalent form of the fractional Fredholm PIDE:
u(x, τ ) = p(x, τ , u(x, τ )).
(18)
Based on the Schwarz inequality, it is easy to see that : W (D ) → H (D ) is bounded linear operator, whereas 2 ≤ Mu 2 for a positive integer M. Now, to build an orthogonal function systems of the space W (D ), u(x, τ ) W 1 W 2
we choose a countable dense subset {(xi , τi )}∞ in D, and define ϕi (x, τ ) = r(xi ,τi ) (x, τ ) and ψ i (x,τ ) = ∗ ϕ i (x,τ ), where i=1
∗ : H (D ) → W (D ) is the adjoint operator of and is uniquely determined. The normalized orthonormal function systems {ψ¯ i (x, τ )}∞ of the space W (D ) is constructed from the process of the Gram-Schmidt orthogonalization of {ψi (x, τ )}∞ as i=1 i=1 follows
ψ¯ i (x, τ ) =
i
μik ψk (x, τ ).
(19)
k=1
is a complete function system in W (D ) with Lemma 1. The sequence {ψi (x, τ )}∞ i=1
ψi (x, τ ) = (y,s) R(x, τ )(y,s)=(x ,τ ) . i
i
Proof. Here, (y,s) indicates that the operator applies to the function of (y,s). Indeed
ψi (x, τ ) = ∗ ϕi (x, τ ) = ∗ ϕi (y, s ), R(x,τ ) (y, s ) W
(20)
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M. Al-Smadi, O.A. Arqub / Applied Mathematics and Computation 342 (2019) 280–294
= ϕi (y, s ), (y,s) R(x,τ ) (y, s ) H = (y,s) R(x,τ ) (y, s ) (y,s )=(xi ,τi ) = (y,s) R(y,s) (x, τ ) ∈ W (D ). (y,s )=(x ,τ ) i
(21)
i
Now, for each fixed u ∈ W (D ), let u(x,τ ), ψ i (x,τ )W = 0, i = 1,2, …. Then, u(x, τ ), ψ i (x,τ )W = u(x, τ ),
∗ ϕ i (x,τ )W = u(x, τ ), ϕ i (τ )H = u(xi ,τ i ) = 0. Whilst, {(xi , τi )}∞ is dense on D, we must have u(x, τ ) = 0 from the i=1 existence of − 1 , it follows that u = 0. Lemma 2. The sequence {R(xi ,τi ) (x, τ )}∞ is a linearly independent in the space W (D ). i=1 Proof. It is adequate to show that {R(xi ,τi ) (x, τ )}m is a linearly independent for each m ≥ 1. In fact, if {ci }m satisfies i=1 i=1 m c R ( x, τ ) = 0 , taking h ( x, τ ) ∈ W ( D ) such that h ( x , τ ) = δ for each l = 1,2, , m . Then k l l l,k k i=1 i (xi ,τi )
0=
hk (x, τ ),
m
ci R(xi ,τi ) (x, τ )
i=1
=
m i=1
=
m
W
ci hk (x, τ ), R(xi ,τi ) (x, τ ) W c i h k ( x i , τi ) = c k .
(22)
i=1
Thus ck = 0 for k = 1,2, , m.
In the next theorem, μik are the orthogonalization coefficients of the orthonormal systems ψ¯ i (x, τ ), which are obtained directly from the Gram-Schmidt orthogonalization process of Eq. (19). Theorem 6. Suppose that Ai = ik=1 μik Pk . If u ∈ W (D ) is the solution of Eqs. (18) and (16), then
u(x, τ ) =
∞
Ai ψ¯ i (x, τ ).
(23)
i=1
Proof. Since, u(x, τ ), ϕ i (x,τ )W = u(xi ,τ i ) for each u ∈ W (D ), whilst, {ψ¯ i (x, τ )}∞ , then it is a convergent in the sense of · W . Thus, i=1
u(x, τ ) =
∞ i u(x, τ ), ψ¯ i (x, τ )
W
i=1 k=1
=
∞
u(x, τ ),
i=1
=
i
∞
i=1
Ai ψ¯ i (x, τ ) is the Fourier series expansion about
ψ¯ i (x, τ )
μik ψk (x, τ )W ψ¯ i (x, τ )
k=1
∞ i
μik u(x, τ ), ∗ ϕk (x, τ )W ψ¯ i (x, τ )
i=1 k=1
= =
∞ i
μik u(xk , τk )ψ¯ i (x, τ )
i=1 k=1 ∞
Ai ψ¯ i (x, τ ).
(24)
i=1
In other words,
∞
i=1
Ai ψ¯ i (x, τ ) is the exact solution of Eqs. (18) and (16).
The finite domain D can be divided into a p × q mesh point with the space step size x =
1 p
toward the x direction of
[0,1] and the time step size t = toward the t direction of [0,1], respectively, where p and q are positive integers. Anyhow, the grid points (xl ,τ m ) in the space-time domain D are defined simultaneously as 1 q
(xl , τm ) = (l x, mτ ), l = 0, 1, 2, · · · , p, m = 0, 1, 2, · · · , q.
(25)
Remember here the following markers: P = P (x, τ , u(x, τ )), Pk = P (xk , τk , u(xk , τk )), and = P (xk , τk , un (xk , τk )), whenever k = 1,2,3, …, ∞. According the proposed methodology, the following algorithm is given to summarize the procedure in finding the approximate solutions: Pkn
M. Al-Smadi, O.A. Arqub / Applied Mathematics and Computation 342 (2019) 280–294
285
Algorithm 1 To approximate un (x,τ ) of Eq. (18) and (16) on D, the following steps should be taken. Step Step Step Step Step Step Step Step Step
1: 2: 3: 4: 5: 6: 7: 8: 9:
Choose n = pq collocation points in the domain D; Set ψi (xi , τi ) = (y,s) R(x, τ )|(y,s)=(xi ,τi ) ; Obtain the orthogonalization coefficients μik ; Set ψ¯ i (xi , τi ) = il=1 μik ψi (xi , τi ), i = 1, 2, ..., n; Choose an initial approximation u0 (x1 ,τ 1 ); Set i = 1; Set Bi = il=1 μik Pkk−1 ; Set ui (xi , τi ) = ik=1 nj=1 Bk ψ¯ k (xk , τk ); If i < η, then set i = i + 1 and go to step 7, else stop.
For numerical computations, let (x1 ,τ 1 ) = (0,0), then from the Dirichlet conditions of Eq. (16), the value of u(x1 ,τ 1 ) is known. Set u0 (x1 ,τ 1 ) = u(x1 ,τ 1 ) and define the n-term numerical solution of u(x, τ ) using the truncating version as
un (x, τ ) =
n
Ai ψ¯ i (x, τ ).
(26)
i=1
∞ ¯ Since W (D ) is a Hilbert space, then the series i=1 Ai ψi (x, τ ) < ∞. Thus, we can guarantee that the numerical solution un (x,τ ) satisfies the constraints conditions of Eq. (16). 4. Error analysis Suppose that un W is bounded, {(xi , τi )}∞ is dense on D, and the solution of Eqs. (18) and (16) is unique. Then the i=1 n-term numerical solution un (x,τ ) must converges to the exact solution u(x, τ ) as n → ∞. Lemma 3. The partial derivatives of the numerical solution ∂ i i ∂ j un (x, τ ) are converging uniformly to the partial derivatives of j
x
τ
the exact solution ∂ i i ∂ j u(x, τ ), whenever i = 0,1,2, j = 0,1 as n → ∞. j
x
τ
Proof. Since W (D ) is a Hilbert space, from Eqs. (23) and (26), it is follows that, u − un W → 0 as n → ∞. Again, since
i j ∂ i ∂ j u(x, τ ) − ∂ i i ∂ j j un (x, τ ) = u(y, s ) − un (y, s ), ∂ i i ∂ j j R(x,τ ) (y, s ) x τ x τ x τ W ≤ u − un W ∂ i i ∂ j j R(x,τ ) (y, s ) x
≤ Mi, j u − un W .
τ
W
(27)
Thus, |∂ i i ∂ j u(x, τ ) − ∂ i i ∂ j un (x, τ )| → 0 as n → ∞. j
x
τ
j
x
τ
For errors estimations, the following assumptions have been made in defining the proof of requested two next theorems: 1 1 hx = n−1 with xi = (i − 1)hx , hτ = n−1 with τ j = (j − 1)hτ , where i = 1,2, …, n and j = 1,2, …, n. In the case of two variables, we may use a norm that represents the maximum absolute value at each point, that is, u(x, τ ) ∞ = max(x,τ )∈D |u(x, τ )|. Theorem 7. Let u(x, τ ) and un (x,τ ) be the exact and the numerical solutions of Eqs. (18) and (16), respectively. If u(x, τ ) − un (x,τ ) is the nature error at (x, τ ) ∈ D in the space W (D ), then there is a positive constant C such that
u(x, τ ) − un (x, τ ) ∞ ≤ C h2x + hx hτ .
(28)
Proof. In each subdomain [xi , xi+1 ] × [τ j , τ j+1 ] ⊂ D, one can write
∂x u(x, τ ) − ∂x un (x, τ ) = ∂x u(x, τ ) − ∂x u xi , τ j + ∂x un xi , τ j − ∂x un (x, τ ) + ∂x u xi , τ j − ∂x un xi , τ j .
(29)
Expand ∂ x u(x,τ ) about the point (xi ,τ j ) using Taylor series of function of two variables, we get
∂x u(x, τ ) = ∂x u xi , τ j + (hx ∂x + hτ ∂τ )∂x u xi + ζ hx , τ j + ζ hτ + . . . , ζ ∈ [0, 1].
(30)
By means of the continuation of ∂ xt u and ∂x22 u on D, one can get
∂x u(x, τ ) − ∂x un xi , τ j = O(hx + hτ ). ∞
(31)
On the other hand, we know that
τ x 2
∂x un xi , τ j − ∂x un (x, τ ) ≤ ∂y2 un y, τ j dy + |∂xs un (x, s )|ds. xi
τj
(32)
In the means of ∞ th norm, it follows that
∂x un xi , τ j − ∂x un (x, τ ) = O(hx + hτ ). ∞
(33)
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Again, given any ε > 0, using Lemma 3, there is an n sufficiently large such that
∂ x u x i , τ j − ∂ x u n x i , τ j < ε . ∞
(34)
As ε is arbitrary and by combining Eqs. (29)–(34) for the chosen value of n, we must have
∂x u(x, τ ) − ∂x un (x, τ ) ∞ = O(hx + hτ ).
(35)
Using the integral property for differentiable functions, we have
u(x, τ ) − un (x, τ ) = u(xi , τ ) − un (xi , τ ) +
x xi
(∂y u(y, τ ) − ∂y un (y, τ ) )dy.
(36)
Using Eqs. (35) and (36) and applying Lemma 3, we can straightforwardly see that
u(x, τ ) − un (x, τ ) ∞ ≤ C h2x + hx hτ ,
(37)
where C > 0. Theorem 8. Let u(x, τ ) and un (x,τ ) be the exact and the numerical solutions of Eqs. (18) and (16), respectively. If u(x, τ ) − un (x,τ ) is nature error at (x, τ ) ∈ D in the space W (D ) with ∂ 33 ∂τ u(x, τ ), ∂ 22 ∂ 22 u(x, τ ) ∈ C (D ) such that ∂ 33 ∂τ u(x, τ ) ∞ x
x
and ∂ 22 ∂ 22 u(x, τ ) ∞ are bounded, then there is a positive constant M such that x
τ
x
τ
u(x, τ ) − un (x, τ ) ∞ ≤ M hτ h3x + h2τ h2x .
(38)
Proof. In each subdomain [xi , xi+1 ] × [τ j , τ j+1 ] ⊂ D, one can write
∂x22 ∂τ u(x, τ ) − ∂x22 ∂τ un (x, τ )
= ∂x22 ∂τ u(x, τ ) − ∂x22 ∂τ u xi , τ j + ∂x22 ∂τ un xi , τ j − ∂x22 ∂τ un (x, τ ) + ∂x22 ∂τ u xi , τ j − ∂x22 ∂τ un xi , τ j .
(39)
Expand ∂ 22 ∂τ u(x, τ ) about the point (xi ,τ j ) using Taylor series of function of two variables, we get x
∂x22 ∂τ u(x, τ ) = ∂x22 ∂τ u xi , τ j + (hx ∂x + hτ ∂τ )∂x22 ∂τ u xi + ζ hx , τ j + ζ hτ + . . . , ζ ∈ [0, 1].
(40)
By means of the continuation of ∂ 33 ∂τ u and ∂x22 ∂τ22 u on D, one can show that x
2
∂x2 ∂τ u(x, τ ) − ∂x22 ∂τ un xi , τ j = O(hx + hτ ). ∞
(41)
On the other aspect as well, one can write
t 2 x 3
∂x2 ∂τ un xi , τ j − ∂x22 ∂τ un (x, τ ) ≤ ∂x3 ∂τ un y, τ j dy + ∂x22 ∂s22 un (x, s )ds. xi
(42)
tj
In the means of ∞ th norm, it follows that
2
∂x2 ∂τ un xi , τ j − ∂x22 ∂τ un (x, τ ) = O(hx + hτ ). ∞
(43)
Again, given any ε > 0, using Lemma 3, there is an n sufficiently large such that
2
∂x2 ∂τ u xi , τ j − ∂x22 ∂τ un xi , τ j < ε . ∞
(44)
As ε is arbitrary and by combining Eqs. (39)–(44) for the chosen value of n, we must have
2 ∂x2 ∂τ u(x, τ ) − ∂x22 ∂τ un (x, τ ) = O(hx + hτ ). ∞
(45)
Using the integral property for differentiable functions, we have
∂x ∂τ u(x, τ ) − ∂x ∂τ un (x, τ ) = ∂x ∂τ u(xi , τ ) − ∂x ∂τ un (xi , τ ) + ∂τ u(x, τ ) − ∂τ un (x, τ ) = ∂τ u(xi , τ ) − ∂τ un (xi , τ ) + u(x, τ ) − un (x, τ ) = u(x, τi ) − un (x, τi ) +
τ τi
x xi
x xi
∂y22 ∂τ u(y, τ ) − ∂y22 ∂τ un (y, τ ) dy.
(∂y ∂τ u(y, τ ) − ∂y ∂τ un (y, τ ) )dy.
(∂s u(x, s ) − ∂s un (x, s ) )ds.
(46) (47) (48)
Using Eqs. (45) and (48) and applying Lemma 3, we can straightforwardly see that
u(x, τ ) − un (x, τ ) ∞ ≤ M hτ h3x + h2τ h2x ,
(49)
where M > 0. Theorem 9. Let u(x, τ ) and un (x,τ ) be the exact and the numerical solutions of Eqs. (18) and (16), respectively. If u(x, τ ) − un (x,τ ) is the nature error at (x, τ ) ∈ D in the space W (D ), then {u(x, τ ) − un (x, τ )}∞ is decreasing in the sense of the norm of W (D ) n=1 and u(x, τ ) − un (x,τ ) W → 0 as n → ∞.
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Proof. Using prevoius results described in this paper, one can write
u(x, τ ) − un (x, τ ) = 2 W
∞ i=n+1
i
2
μik Pk
.
(50)
k=1
Consequently {u(x, τ ) − un (x, τ )}∞ is decreasing in the sense of the norm of · W . In addition, since n=1 2 → 0 as n → ∞. convergent. Thus, u(x, τ ) − un (x, τ ) W
∞
i=1
Ai ψ¯ i (x, τ ) is
5. Numerical experiments In this section, we implement the RKM to solve linear and nonlinear examples of fractional Fredholm PIDEs. Also, the numerical results obtained are compared with exact solutions and other solutions found in the literature, which reveal the simplicity and applicability of the present method for various physical issues in fractional analysis. In the process of computation, all the symbolic and numerical computations are performed by using Mathematics 10 software package. Example 1. Consider the following linear fractional Fredholm PIDE:
Dατ u(x, τ ) − ∂x u(x, τ ) − ∂x22 u(x, τ ) +
1 0
cos(x + τ )∂x22 u(x, s )ds = f (x, τ ),
(51)
with the homogeneous initial and Dirichlet boundary conditions
u(x, 0 ) = 0, u ( 0, τ ) = 0, u ( 1, τ ) = 0,
(52)
where 0 ≤ x, τ ≤ 1, 1 < α ≤ 2, and f(x, τ ) is chosen so that the exact solution is given by u(x, τ ) = (1 − x)2 ln(x + 1)τ 3α . Example 2. Consider the following nonlinear fractional Fredholm PIDE:
Dατ u(x, τ ) − u(x, τ )(3 + u(x, τ ) )2 + ∂x u(x, τ ) − ∂x22 u(x, τ ) +
1
0
(xτ s )∂x22 u(x, s )ds = f (x, τ ),
(53)
with the nonhomogeneous initial and Dirichlet boundary conditions
u(x, 0 ) = 0, u(0, τ ) = 0, u(1, τ ) = sin (2(π − 3 ) )τ 1+α ,
(54)
where 0 ≤ x, τ ≤ 1, 1 < α ≤ 2, and f(x, τ ) is chosen so that the exact solution is given by u(x, τ ) = sin (2x2 (π − 3x3 ))τ 1 + α . Example 3. Consider the following nonlinear fractional Fredholm PIDE:
Dατ u(x, τ ) + cos (u(x, τ ) ) − 2∂x u(x, τ ) + ∂x22 u(x, τ ) −
0
1
sα sinh (x + τ )∂x22 u(x, s )ds = f (x, τ ),
(55)
with the nonhomogeneous initial and Dirichlet boundary conditions
u(x, 0 ) = xex−1 , u(0, τ ) = τ α +2 , u(1, τ ) = 1 + τ α +2 ,
(56)
where 0 ≤ x, τ ≤ 1, 1 < α ≤ 2, and f(x, τ ) is chosen so that the exact solution is given by u(x, τ ) = xex − 1 + τ α + 2 . The presented algorithm in this work is used successfully to solve the equations mentioned in the previous examples numerically. The related parameters for the algorithm are listed as follows: Parameters
Description
D = [0, 1] [0, 1] p = 20 q = 20 x = 1/20 t = 1/20 (xl ,τ m ) = (lx,mτ ) α ∈ (0,1] R(y,s) (x, τ ) r(y,s) (x, τ )
The The The The The The The The The
direct product between space and time direction domains number of mishpoint in the x direction domain of [0,1] number of mishpoint in the τ direction domain of [0,1] space step size in the x direction domain of [0,1] space step size in the x direction domain of [0,1] two-dimensional partition of the domain D order of the fractional derivative in the Caputo sense structure of the RK function of the Hilbert space W (D ) structure of the RK function of the Hilbert space H (D )
For more detail and clarification, through this section, we will do the following main analyzes: • The absolute errors of approximating u(x, τ ) are tabulated for Examples 1, 2, and 3. • The absolute errors of approximating ∂ τ u(x,τ ), ∂ x u(x,τ ), and ∂ 22 u(x, τ ) are tabulated for Example 2. x • The numerical comparisons between the current and existing results are tabulated for Examples 1, 2 and 3.
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Table 1 Numerical results of Example 1 over the domain D. x 0.2
0.4
0.6
0.8
τ 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1
α = 0.25
α = 0.5 −7
2.4517378172 × 10 1.0068974805 × 10 − 6 3.6568085576 × 10 − 6 5.2265248627 × 10 − 6 3.8433517302 × 10 − 7 1.5881829145 × 10 − 6 4.1812198902 × 10 − 6 6.6754172819 × 10 − 6 4.8295446630 × 10 − 7 1.9852286431 × 10 − 6 4.5710106969 × 10 − 6 8.3442716024 × 10 − 6 7.6096123482 × 10 − 7 3.1372436507 × 10 − 6 6.2821933886 × 10 − 6 7.2744873015 × 10 − 6
α = 0.75 −7
α=1 −7
1.5261361141 × 10 6.2733225125 × 10 − 7 1.4444393802 × 10 − 6 2.6367898263 × 10 − 6 1.5454542912 × 10 − 7 6.3527316584 × 10 − 7 2.3011731108 × 10 − 6 4.2523196829 × 10 − 6 2.4046375024 × 10 − 7 9.9136899367 × 10 − 7 2.4627234230 × 10 − 6 5.6701669127 × 10 − 6 2.4350759519 × 10 − 7 1.0039179683 × 10 − 6 3.3303018844 × 10 − 6 6.3061465143 × 10 − 6
1.4488633981 × 10 5.7556859299 × 10 − 7 1.3313032090 × 10 − 6 2.4032814807 × 10 − 6 1.4702270429 × 10 − 7 5.9468676381 × 10 − 7 1.3747351234 × 10 − 6 2.5365273079 × 10 − 6 2.2220068063 × 10 − 7 8.4117309523 × 10 − 7 2.1264004694 × 10 − 6 4.0370123571 × 10 − 6 2.2828837042 × 10 − 7 9.1607514607 × 10 − 7 3.1846580166 × 10 − 6 5.9293586943 × 10 − 6
3.8636357335 × 10 − 8 1.6675920599 × 10 − 7 3.8396489848 × 10 − 7 1.1203634600 × 10 − 6 6.3920743698 × 10 − 8 2.6352846466 × 10 − 7 6.1170424459 × 10 − 7 1.1384510721 × 10 − 6 1.1590907189 × 10 − 7 4.7645487433 × 10 − 7 1.0970425672 × 10 − 6 2.0026251845 × 10 − 6 1.8263069634 × 10 − 7 7.5293847612 × 10 − 7 1.7477264132 × 10 − 6 3.2296098857 × 10 − 6
Table 2 Numerical results of Example 2 over the domain D. x
τ
α = 0.25
α = 0.5
α = 0.75
α=1
0.2
0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1
1.1348209378 × 10 − 7 3.4536525256 × 10 − 7 6.0349072754 × 10 − 7 7.8266513655 × 10 − 7 2.9873164479 × 10 − 7 5.8746328959 × 10 − 7 2.9381383903 × 10 − 6 6.7650958324 × 10 − 6 4.3355174501 × 10 − 7 7.1212725217 × 10 − 7 3.2509031934 × 10 − 6 8.3215603017 × 10 − 6 1.3622602173 × 10 − 7 4.0833356582 × 10 − 7 7.1597049187 × 10 − 7 8.7811593217 × 10 − 7
6.2165745462 × 10 − 8 2.2272324531 × 10 − 7 5.1058638422 × 10 − 7 7.6192883366 × 10 − 7 7.3192883006 × 10 − 8 4.6025220061 × 10 − 7 7.1477361442 × 10 − 7 8.2349521973 × 10 − 7 8.7491254401 × 10 − 8 5.8210871242 × 10 − 7 7.0777646213 × 10 − 7 8.5048904873 × 10 − 7 6.5018063868 × 10 − 8 2.8308561819 × 10 − 7 6.3752220231 × 10 − 7 7.8701223904 × 10 − 7
3.5681807104 × 10 − 8 1.6999109573 × 10 − 7 2.8301122686 × 10 − 7 4.5262233501 × 10 − 7 4.9037830092 × 10 − 8 2.4714026441 × 10 − 7 5.1396291017 × 10 − 7 6.1333256827 × 10 − 7 4.9915927629 × 10 − 8 3.7583865735 × 10 − 7 6.6431206035 × 10 − 7 7.8659381716 × 10 − 7 4.1850974755 × 10 − 8 2.0254884508 × 10 − 7 3.9662203028 × 10 − 7 5.7166369921 × 10 − 7
1.9318178612 × 10 − 8 7.9409145481 × 10 − 8 1.8284042793 × 10 − 7 3.3377086405 × 10 − 7 3.0438449538 × 10 − 8 8.2548974575 × 10 − 8 1.9128773570 × 10 − 7 5.3826831436 × 10 − 7 7.7862248476 × 10 − 8 9.3151184247 × 10 − 8 4.2175379815 × 10 − 7 6.7344857852 × 10 − 7 2.9419182381 × 10 − 8 5.4969535761 × 10 − 8 3.4300686974 × 10 − 7 5.6095066127 × 10 − 7
α = 0.5
α = 0.75
α=1
0.4
0.6
0.8
Table 3 Numerical results of Example 3 over the domain D. x 0.2
0.4
0.6
0.8
τ 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1
α = 0.25 −7
2.1481720019 × 10 3.7618395348 × 10 − 7 7.8837016381 × 10 − 7 9.2102295605 × 10 − 7 2.3060667798 × 10 − 7 4.6185275342 × 10 − 7 8.3059900688 × 10 − 7 9.2770 0 08437 × 10 − 7 2.4228618283 × 10 − 7 5.7487348215 × 10 − 7 8.3465992732 × 10 − 7 9.6970870809 × 10 − 7 3.8441028138 × 10 − 7 7.5561297496 × 10 − 7 9.4068380233 × 10 − 7 9.7281126025 × 10 − 7
−7
1.7316726658 × 10 3.2897109742 × 10 − 7 3.7624231769 × 10 − 7 6.5192825016 × 10 − 7 1.8570705294 × 10 − 7 3.5384139235 × 10 − 7 4.7487348215 × 10 − 7 7.0289546992 × 10 − 7 2.1523048063 × 10 − 7 4.0340084096 × 10 − 7 6.6351371325 × 10 − 7 7.2248646541 × 10 − 7 2.2459373499 × 10 − 7 4.8018477255 × 10 − 7 7.7543305937 × 10 − 7 7.9805772261 × 10 − 7
−7
1.0106011983 × 10 1.6388196750 × 10 − 7 2.3905796608 × 10 − 7 5.6454397540 × 10 − 7 1.8216430611 × 10 − 7 2.6030658914 × 10 − 7 2.9418508190 × 10 − 7 6.6086147825 × 10 − 7 1.9960884796 × 10 − 7 3.6230799268 × 10 − 7 5.7804554793 × 10 − 7 6.7307997707 × 10 − 7 2.1082921586 × 10 − 7 3.8440626803 × 10 − 7 6.8396660828 × 10 − 7 7.7692915114 × 10 − 7
1.1098062724 × 10 − 8 1.3475516667 × 10 − 8 2.3987084829 × 10 − 8 4.6951033335 × 10 − 8 1.5559383669 × 10 − 8 3.2416693935 × 10 − 8 1.4942838745 × 10 − 7 6.2012285851 × 10 − 8 2.2169823188 × 10 − 8 1.3613113948 × 10 − 7 2.1304279661 × 10 − 7 7.6931983681 × 10 − 8 2.6399003035 × 10 − 8 1.4999640176 × 10 − 7 2.8080854547 × 10 − 7 7.7035008652 × 10 − 8
• The approximate solutions are sketched for the Examples 1 and 2. • The absolute value of ε n (x,τ ) are plotted for Example 3. • The effect of number of nodes on the values of u(x, τ )∞ are tabulated for Example 3. With a view to demonstrate the agreement between the exact and the RK approximate solutions, Tables 1, 2, and 3 show absolute errors to the approximate solutions of Examples 1, 2, and 3, respectively, at different values of (x, τ ) in D when α ∈
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Table 4 The partial derivatives absolute errors of Example 2 over the domain D.
τ
x 0
∂ τ u(x,τ ) 3.4006446540 × 10 4.3060653250 × 10 − 7 6.0785691623 × 10 − 8 5.6920478680 × 10 − 8 5.2187639123 × 10 − 8 6.6494346169 × 10 − 7 3.7259673166 × 10 − 7 4.5299457463 × 10 − 7 3.7720236477 × 10 − 7 3.7360933885 × 10 − 8 6.7282243440 × 10 − 7 2.2348130575 × 10 − 8 9.1631132728 × 10 − 7 1.8893095404 × 10 − 8 5.7003720356 × 10 − 7 2.6685123655 × 10 − 7 7.7553065926 × 10 − 7 6.4909024121 × 10 − 8 3.1491619194 × 10 − 8 9.4767703466 × 10 − 7 6.8750834270 × 10 − 7 1.3926593070 × 10 − 8 9.6044556560 × 10 − 7 5.0906667320 × 10 − 8
0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1 0.25 0.5 0.75 1
0.2
0.4
0.6
0.8
1
∂ x u(x,τ ) −7
∂x22 u(x, τ ) −7
2.0134365401 × 10 − 5 7.4576790550 × 10 − 6 7.4665700330 × 10 − 6 1.1623060665 × 10 − 5 2.6052786611 × 10 − 6 9.1328168700 × 10 − 6 3.8549834779 × 10 − 6 7.3642966800 × 10 − 5 7.6817990891 × 10 − 5 2.5266067672 × 10 − 6 5.1828709520 × 10 − 6 6.5988424311 × 10 − 5 6.9638437374 × 10 − 6 8.6092383568 × 10 − 5 2.6931004989 × 10 − 6 6.4629659556 × 10 − 5 1.1830748677 × 10 − 6 5.9533688984 × 10 − 5 2.1185282320 × 10 − 5 2.5411119825 × 10 − 6 2.7777673850 × 10 − 6 1.2664295100 × 10 − 5 9.0488643036 × 10 − 6 8.9514790339 × 10 − 6
6.6051913607 × 10 5.9948623291 × 10 − 7 5.2127840790 × 10 − 6 9.0055558374 × 10 − 7 5.8757251174 × 10 − 7 3.3943841370 × 10 − 7 4.7409025799 × 10 − 7 5.0241629160 × 10 − 6 9.0838141806 × 10 − 6 8.7844623014 × 10 − 6 2.8648976716 × 10 − 7 3.7588328935 × 10 − 7 9.0133190529 × 10 − 7 9.0 040 088888 × 10 − 7 6.7882561523 × 10 − 6 2.2486422896 × 10 − 7 2.5899027012 × 10 − 7 8.2560423829 × 10 − 7 2.1614034400 × 10 − 6 7.2799216575 × 10 − 6 9.5722096937 × 10 − 6 3.5889430024 × 10 − 7 8.2763853748 × 10 − 7 3.8069936491 × 10 − 6
Table 5 Numerical comparison of absolute errors of Example 1 at τ = 0.4. Node
IRKM
RDTM
RKM
x
α = 0.25
α = 0.75
α = 0.25
α = 0.75
α = 0.25
0.2 0.4 0.6 0.8
2.2562 × 10 − 7 3.7214 × 10 − 5 1.0061 × 10 − 3 7.8266 × 10 − 3
6.0770 × 10 − 7 3.3865 × 10 − 5 7.4723 × 10 − 3 2.1932 × 10 − 3
3.2508 × 10 − 6 5.9441 × 10 − 6 1.1167 × 10 − 6 1.5041 × 10 − 6
2.3936 × 10 − 6 4.4392 × 10 − 6 5.3378 × 10 − 6 5.0517 × 10 − 6
1.0148 1.4347 1.8902 2.7163
× 10 − 6 × 10 − 6 × 10 − 6 × 10 − 6
α = 0.75 3.6304 × 10 − 7 1.2913 × 10 − 7 2.7791 × 10 − 7 8.2229 × 10 − 8
{0.25,0.5,0.75,1}. From these tables, the error estimate shows that the accuracy of the numerical solutions is closely related to the fill time τ m = mτ , m = 0,1, ..., q, where τ = 1q , and fill distance xl = lx, l = 0,1, ..., p, where x = 1p . Thereby, more accurate numerical solutions can be obtained using more mesh points. The main advantage of this algorithm lies in the fact that it is possible to pick any point in the domain D, and as well, the n-term approximate partial derivatives
⎧ n ⎪ ⎪ ∂ un (x, τ ) = Ai ψ¯ i (x, τ ), τ ⎪ ⎨ i=1
n ⎪ ⎪ ⎪ Ai ∂xωω ψ¯ i (x, τ ), ω = 1, 2, ⎩∂xωω un (x, τ ) =
(57)
i=1
are applicable. However, in Table 4, numerical results of approximating derivatives for Example 2 at various values of (x, τ ) ∈ D have been tabulated. Indeed, such approximate partial derivatives are converge uniformly to exact partial derivatives and completely agree well with Lemma 3. To further show the advantage of the RKM, we now present comparison experiments for Examples 1, 2, and 3 at various 0 ≤ x ≤ 1 and various 1 < α ≤ 2 when τ = 0.4. The numerical methods that are used for comparison include the implicit Runge-Kutta method (IRKM) and the reduced differential transform method (RDTM). Anyhow, Tables 5–7 show comparisons between the absolute errors of our method together with the other aforementioned methods. It is clear from the tables that, the RKM in comparison is much better with a view to accuracy and applicability followed by the RDTM and the IRKM, respectively. The geometric behaviors of the memory and hereditary properties of the RK solutions and their level characteristics are also investigated here. Anyhow, the comparisons of between the computational values of the RK solutions for different values of α ∈ {0.25,0.5,0.75,1} for Example 1 have been depicted on the domain D as shown in Fig. 1, whilst, Fig. 2 show the comparisons of for Example 2 on the same domain and at the same fractional level. It is clear from the Figs. 1 and 2 that each of the graphs are nearly coinciding and similar in their behaviors with good agreement with the RK solutions when the ordinary derivatives are considered. Whilst, one can note that the RK solutions
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Table 6 Numerical comparison of absolute errors of Example 2 at τ = 0.4. Node
IRKM
RDTM
RKM
x
α = 0.25
α = 0.75
α = 0.25
α = 0.75
0.2 0.4 0.6 0.8
3.4605 × 10 − 6 0.6061 × 10 − 5 6.5129 × 10 − 3 1.9501 × 10 − 3
3.9888 × 10 − 6 2.9740 × 10 − 5 3.1367 × 10 − 3 5.7110 × 10 − 3
9.8075 × 10 − 7 7.0301 × 10 − 6 9.9534 × 10 − 6 6.3494 × 10 − 6
0.9531 2.8355 4.7208 6.3084
× 10 − 7 × 10 − 7 × 10 − 6 × 10 − 6
α = 0.25
α = 0.75
8.9131 × 10 − 7 4.7827 × 10 − 7 5.8929 × 10 − 7 4.5327 × 10 − 7
9.2235 × 10 − 8 3.4485 × 10 − 7 9.1382 × 10 − 8 6.1761 × 10 − 7
Table 7 Numerical comparison of absolute errors of Example 3 at τ = 0.4. Node
IRKM
x
α = 0.25
α = 0.75
α = 0.25
RDTM
α = 0.75
RKM
0.2 0.4 0.6 0.8
8.4107 × 10 − 6 2.3303 × 10 − 5 4.6189 × 10 − 3 7.7245 × 10 − 3
5.3267 × 10 − 6 2.8486 × 10 − 5 0.5955 × 10 − 3 1.7392 × 10 − 3
2.6943 × 10 − 7 5.1746 × 10 − 7 3.0592 × 10 − 6 6.6047 × 10 − 6
5.8620 6.9282 4.2018 2.8756
× 10 − 7 × 10 − 7 × 10 − 6 × 10 − 6
α = 0.25
α = 0.75
3.3518 × 10 − 7 4.4204 × 10 − 7 4.9765 × 10 − 7 6.7801 × 10 − 7
7.2724 × 10 − 7 2.7659 × 10 − 7 3.5893 × 10 − 7 1.0102 × 10 − 7
Fig. 1. Comparisons of between the computational values of the RK solutions at various values of (x, τ ) ∈ D when α ∈ {0.25, 0.5, 0.75, 1} for Example 1: black α = 1; blue: α = 0.75; green: α = 0.5; red: α = 0.25. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)
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Fig. 2. Comparisons of between the computational values of the RK solutions at various values of (x, τ ) ∈ D when α ∈ {0.25,0.5,0.75,1} for Example 2: black α = 1; blue: α = 0.75; green: α = 0.5; red: α = 0.25. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)
are continuously depends on the time-fractional derivatives assumed. As a fast scan, the process described by the RKM is slightly more skewed to the down and drag to the right than that modeled by the standard equations (when ordinary derivatives are considered). In the meantime, from the graphs, it can be seen that the RK solutions are stable and convergent. The graph planning of the absolute error function ε n (x,τ ) = |u(x, τ ) − un (x,τ )| and its level characteristics is discussed geometrically at the middle space x direction. To specific, the n-term of ε n (x,t) is computed as follows:
20 εn (0.5, τ ) = 0.5e−0.5 + τ α+2 − Ai ψ¯ i (0.5, τ ). i=1
(58)
Fig. 3 gives the relevant data of the RK results at various value of (0.5, τ ) ∈ D when α ∈ {0.25, 0.5, 0.75, 1} for Example 3. Regarding the convergence speed, it is obvious that the difference between the exact and the RK nodal values decreases initially till a maximum level fractional derivatives values are reached. The last target is to show the effect of number of nodes on behavior of u(x, τ )∞ in order to check numerically the order of convergence of the RKM. Anyhow, Table 8 gives the relevant data for Example 3 at n ∈ {25,10 0,225,40 0,625}. It is observed that the reduction in the step size, n = pq = 1x 1τ , of D results in a reduction in the error and correspondingly an improvement in the accuracy of the obtained solution. This goes in agreement with the known fact that the error is monotone decreasing where more accurate solutions are achieved using a reduction in the step size, whilst, the cost to be paid while going in this direction is the rapid increase in the number of iterations required for convergence.
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Fig. 3. The absolute value of the error function ε n (0.5,τ ) of the RK solutions at various values of τ ∈ [0,1] when α ∈ {0.25,0.5,0.75,1} for Example 3: black α = 1; blue: α = 0.75; green: α = 0.5; red: α = 0.25. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.) Table 8 The effect of number of nodes on the values of u(x, τ )∞ in Example 3 over the domain D.
x
τ
α = 0.25
α = 0.5
α = 0.75
α=1
1/5 1/10 1/15 1/20 1/25
1/5 1/10 1/15 1/20 1/25
4.61176 × 10 − 4 7.42581 × 10 − 5 1.35079 × 10 − 6 2.14002 × 10 − 7 9.51853 × 10 − 8
3.56984 × 10 − 4 6.54091 × 10 − 5 9.60787 × 10 − 6 4.70130 × 10 − 7 1.89679 × 10 − 8
3.77140 × 10 − 4 5.42023 × 10 − 5 1.50232 × 10 − 6 6.75828 × 10 − 7 2.76580 × 10 − 8
6.27042 × 10 − 5 2.47893 × 10 − 6 7.37730 × 10 − 7 5.63577 × 10 − 8 4.99624 × 10 − 9
6. Conclusions In this paper, the RKM is successfully used in solving a class of fractional PIDEs with initial and Dirichlet boundary conditions generated by an actual model. Error analysis of the RKM has also been discussed in details. The numerical solution converges uniformly to the exact solution. Furthermore, the RKM can be easily applied to a wide class of fractional PIDEs. Numerical examples have been compared with exact solution and other methods to validate the agreement between theoretical statements and numerical solutions in term of convergence. The obtained results confirm the reliability and versatility of the proposed method. Acknowledgment The authors would like to express their gratitude to the unknown referees for carefully reading the paper and their helpful comments. References [1] F. Mainardi, Fractional Calculus and Waves in Linear Viscoelasticity, Imperial College Press, London, UK, 2010.
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