/ / Article

Research Article | Open Access

Volume 2019 |Article ID 5602565 | https://doi.org/10.1155/2019/5602565

Ali Demir, Mine Aylin Bayrak, Ebru Ozbilge, "A New Approach for the Approximate Analytical Solution of Space-Time Fractional Differential Equations by the Homotopy Analysis Method", Advances in Mathematical Physics, vol. 2019, Article ID 5602565, 12 pages, 2019. https://doi.org/10.1155/2019/5602565

# A New Approach for the Approximate Analytical Solution of Space-Time Fractional Differential Equations by the Homotopy Analysis Method

Revised26 Mar 2019
Accepted14 Apr 2019
Published06 May 2019

#### Abstract

The motivation of this study is to construct the truncated solution of space-time fractional differential equations by the homotopy analysis method (HAM). The first space-time fractional differential equation is transformed into a space fractional differential equation or a time fractional differential equation before the HAM. Then the power series solution is constructed by the HAM. Finally, taking the illustrative examples into consideration we reach the conclusion that the HAM is applicable and powerful technique to construct the solution of space-time fractional differential equations.

#### 1. Introduction

The fractional calculus is a very attractive concept for scientists in various fields since it has a wide application in mathematics, physics, biology, and so on [1â€“4]. Various fractional derivatives are defined to get the solution of fractional differential equations [5, 6]. In order to get the solution, different methods are introduced and applied [7â€“11]. These methods include the finite difference method [12], the homotopy perturbation method (HPM) [13, 14], and the generalized differential transform method [15].

Since the mathematical models include space and time variables in the modelling of mathematical and physical problems, fractional derivatives are required. The HAM provides as simple way to construct truncated solution of various differential equations. The advantage of the HAM is that the accuracy of the HAM does not depend on small parameter in the considered equation [16â€“18]. Therefore, compare the other technique HAM has many advantage in the application [19â€“21].

The motivation of this paper is to determine the truncated solution of the following space-time FPDE by implementing the HAM with a new algorithm by which space-time fractional differential equation is reduced to either a space fractional differential equation or a time fractional differential equation making the problem easier to deal with by the HAM:where and the source function F depends on either or .

#### 2. Preliminaries

This section is about basic definitions and fundamental properties of the fractional calculus.

Riemann-Liouville fractional integral operator of order is introduced asSome of the properties of the operator , which we will need in our work, are as follows [1, 2]:If , thenThe Caputo fractional derivative of order is defined aswhere is the classical differential operator of order .

The definitions of the Mittag-Leffler function and the generalized Mittag-Leffler function are given, respectively, as follows:

#### 3. Basic Ideas of the Homotopy Analysis Method (HAM)

Let us consider the differential equationwhere is a nonlinear operator, and are independent variables, and is an unknown function. Liao [17] introduced the so-called zero-order deformation equation by the homotopy method as follows:with the initial conditionswhere , are parameters, is nonzero auxiliary function, and is an auxiliary linear operator with the following feature:Obviously, for and , we havewhere is an initial guess of and is an unknown function. As goes to , the solution changes from to . By Taylor series of based on is written aswhereIf the auxiliary linear operator, the initial guess, the auxiliary operator , and the auxiliary function are properly chosen, the series (20) converges at , then we haveDefine the vectorDifferentiating (16) -times with respect to the embedding parameter and then setting and finally dividing them by , we obtain the th-order deformation equation, with assumption subject to the following initial conditions:whereandWe can apply the operator to both sides of (24) to obtainUsing the Caputo derivative and the initial conditions, we obtainFinally, we will approximate the HAM solution (22) by the truncated seriesThe th-order deformation (24) is linear and thus can be easily solved, especially by means of symbolic computation software such as Matlab and Maple.

#### 4. Space-Time FPDE by the HAM

To construct the solution of problem (1)-(4) by the HAM, it is transformed into a space fractional differential problem or time fractional differential problem. The transformation we apply in this paper is motivated by the properties (7)-(9). Therefore let us take the problems with and without source functions into consideration:

(A) Transforming into Time-Fractional Differential Equation

(A1) No Source Function. In order to make the boundary condition (4) equal to zero, the transformation is applied to (1)-(2):In order to make the order of derivative on the right-hand side, integer, the transformation is applied to (31)-(32):Substituting (33) into (26), can be given byThe solution of th-order deformation (26) for becomesBy using (36) with the initial condition given by (34), we now successively obtainand so on. Hence, by using first four terms of the approximate solution can be obtained in the form of the series below:By the transformations , the solution of problem (1)-(2) is obtained.

(A2) Including the Source Function. In order to make the boundary condition (4) equal to zero, the transformation is applied to (1)-(2):In order to make the order of derivative on the right-hand side, integer, the transformation is applied to (42)-(43):Substituting (26) into (44), can be given byThe solution of th-order deformation (46) for becomesBy using (47) with the initial condition given by (45), we now successively obtainand so on. Hence, the truncated solutions of can be constructed in the series form by using the first four terms as follows:By the transformations , the solution of problem (1)-(2) is determined.

(B) Transforming into Space-Fractional Differential Equation

(B1) No Source Function. By the transformation to (1), (3), and (4) are transformed into the problem below:Substituting (26) into (53), can be given byThe solution of th-order deformation (56) for becomesBy using (54)-(55), we now successively obtainand so on. Hence, the truncated solutions of can be constructed in the series form by using the first four terms below:By the transformation , the solution of problem (1), (3), and (4) is determined.

(B2) Including Source Function. Let us consider the problem below:Substituting (26) into (63), can be given byThe solution of th-order deformation (66) for becomesBy using (64)-(65), we now successively obtainand so on. Hence, the truncated solutions of can be constructed in the series form by using the first four terms below:By the transformation , the solution of problem (1), (3), and (4) is determined.

#### 5. Examples

Example 1. Let us take the homogeneous problem.
(A1) Including No Source FunctionTransforming into time-fractional differential equation, (73)-(74) are reduced as follows:Using the HAM and by the transformations , we obtain the following solution of problem (73)-(76):In a particular case for and , the HAM solution of (73)-(76) coincides with the exact solution .
(B1) Including No Source FunctionTransforming into space-fractional differential equation, (80), (82), and (83) are reduced to the problem below:Using the HAM and applying the transformation , we obtain the following solution of problem (80), (82), and (83):which coincides with the exact solution.

In Tables 1 and 2, the absolute errors have been worked out for various values of and . As can be seen from the tables, for the fixed values of and as increases, the absolute error decreases, while as increases, the absolute error also increases. Moreover as increases to 1, increases to 2 and the absolute error decreases. In addition, the absolute error takes the minimum values at the auxiliary values , for case A and case B, respectively.

 â€‰ = 0.5 = 0.75 = 1 = 0.5 = 0.75 = 1 = 1.5 = 1.75 = 2 = 1.5 = 1.75 = 2 0.3 0.2 0.68841 0.03769 0.00004 0.81871 0.04127 0.00031 â€‰ 0.4 0.27514 0.01312 0.00004 0.37117 0.00200 0.00029 â€‰ 0.6 0.08483 0.04706 0.00003 0.16617 0.03354 0.00026 â€‰ 0.8 0.03374 0.07006 0.00003 0.03721 0.05572 0.00022 0.6 0.2 1.02204 0.10261 0.00044 1.16631 0.08697 0.00472 â€‰ 0.4 0.55654 0.06402 0.00041 0.67960 0.07240 0.00443 â€‰ 0.6 0.33977 0.02800 0.00037 0.45762 0.04762 0.00397 â€‰ 0.8 0.19646 0.00200 0.00031 0.31112 0.02465 0.00335 0.9 0.2 1.21882 0.11304 0.00636 1.34963 0.05854 0.02261 â€‰ 0.4 0.76521 0.12078 0.00598 0.89617 0.11277 0.02125 â€‰ 0.6 0.55660 0.10408 0.00536 0.69660 0.12060 0.01904 â€‰ 0.8 0.41428 0.08328 0.00452 0.56229 0.11704 0.01607
 â€‰ â€‰ â€‰ â€‰ â€‰ = 0.5 = 0.75 = 1 = 0.5 = 0.75 = 1 = 1.5 = 1.75 = 2 = 1.5 = 1.75 = 2 0.3 0.2 0.07111 0.02540 0.00000 0.07111 0.02540 0.00000 â€‰ 0.4 0.19034 0.07464 0.00000 0.19032 0.07464 0.00000 â€‰ 0.6 0.34423 0.14093 0.00000 0.34412 0.14092 0.00000 â€‰ 0.8 0.53422 0.22399 0.00000 0.53372 0.22390 0.00001 0.6 0.2 0.09598 0.03428 0.00000 0.09598 0.03428 0.00000 â€‰ 0.4 0.25693 0.10075 0.00000 0.25691 0.10076 0.00000 â€‰ 0.6 0.46467 0.19024 0.00000 0.46452 0.19023 0.00000 â€‰ 0.8 0.72113 0.30235 0.00000 0.72045 0.30223 0.00001 0.9 0.2 0.12957 0.04627 0.00000 0.12957 0.04627 0.00000 â€‰ 0.4 0.34681 0.13600 0.00000 0.34679 0.13601 0.00000 â€‰ 0.6 0.62724 0.25680 0.00000 0.62703 0.25678 0.00000 â€‰ 0.8 0.97342 0.40813 0.00000 0.97250 0.40797 0.00001

Example 2. Let us take the nonhomogeneous problem.
(A2) Including Source Function Reduction to time-fractional differential equation, (88)-(89) transform to the following:Using the HAM and by the transformations , we obtain the following solution of problem (88)-(91):For and , the HAM solution of (88)-(91) coincides with the exact solution .
(B2) Including Source Function Transforming (95), (97), and (98) into the following problem, we haveUsing the HAM and the transformation , we obtain the following solution of problem (95), (97), and (98):