Nonlinear Analysis and Differential Equations, Vol. 4, 2016, no. 2, 95-107 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/nade.2016.51144 Adaptation of Taylor s Formula for Solving System of Differential Equations Nourah Nizar AlRasheed c/o Saudi Aramco, Dhahran 31311, Saudi Arabia Copyright c 2015 Nourah Nizar AlRasheed. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper, the residual power series (RPS) method is presented for solving a class of differential equations system based on Taylor s series formula. The analytical solution is provided in the form of convergent series with easily computable components. The RPS method obtains the series expansion of the solution and reproduces the exact solution when the solution is polynomial. Several examples are given to demonstrate the computation efficiency of the proposed technique. Results obtained by the method show that the RPS method is a powerful tool that provides exact or very close approximate solution for a wide range of phenomena arising in natural sciences, modelled in terms of differential equations system. Mathematics Subject Classification: 35F05, 35F31, 74H15, 41A58 Keywords: Series expansions, System of differential equations, Power series method, Approximate solutions 1 Introduction System of differential equations (SDEs) are extensively used in modeling of complex phenomena arising in applied science and engineering such as elasticity, electromagnetic and fluid dynamics [11-13, 15, 17-19,]. Generally, SDEs do not have an explicit solution so it is required to obtain an efficient approximate method to deal with the solutions. However, there have been lots efforts in
96 Nourah Nizar AlRasheed giving exact as well as approximate solution relating different kinds of SDEs. Therefore, the numerical solutions of such differential equations systems have been highly studied by many authors. The Adomian decomposition method (ADM) [9], the homotopy perturbation method (HPM) [1], the reproducing kernel method (RKM), [25-30], the homotopy analysis method (HAM) [20], the spectral method (SM) [5-7], the residual power series method (RPSM) [2,8], the generalized differential transform method (GDTM) [3, 16, 32], and other methods [4, 10, 21-24, 33] have been used to obtain analytical approximation to linear and nonlinear problems. The purpose of this work is to extend the application of the residual power series method for obtaining analytical solution for the system of first-order differential equations in the following form subject to the initial conditions y 1 (x) = F 1 (x, y (x)), a x b, y 2 (x) = F 2 (x, y (x)), a x b,. (1) y n (x) = F n (x, y (x)), a x b, y i (x 0 ) = α i,0, i = 1, 2,..., n, (2) where a, b and α i,0 are real-finite constants, F i (x, y) are linear continuousanalytiacl functions on the given interval, and y i are unknown functions of independent variable x to be determined. Also, we assume that F i satisfies all the necessary requirements for the existence of a unique solution of Eqs. (1) and (2) on [a, b]. The RPS method has been developed as an efficient numerical as well as analytical method to determine the coefficients of power series solutions for different types of differential equations [14, 31]. The RPS method is an alternative procedure for obtaining analytical Taylor series solution for SDEs. It is a powerful tool to construct power series solutions for strongly linear and nonlinear equations without linearization, perturbation or discretization that computes the coefficients of the power series solutions by chain of linear equations of one variable. However, the method provides the solution in terms of a rapidly convergent series with easily computable components. This approach is simple and needs less effort to achieve the results. It does not require any convertibilities switching from first to higher order in which the solutions and all its derivatives are applicable for each arbitrary point in the given interval. Thus, the RPS method can be applied directly to the given problems by choosing an appropriate value for the initial guess approximation. The reminder of the paper is organized as follows. In Section 2, basic idea of the residual power series (RPS) method is presented as well as extended to
Adaptation of Taylor s formula... 97 provide the approximate series solutions for system (1) and (2). In Section 3, some examples are given to illustrate the capability of the RPS method. Results reveal that only few terms are required to deduce the approximate solutions which are found to be accurate and efficient. Finally, the conclusions of this work are introduced in Section 4. 2 Residual Power Series Method In this section, we present a brief description of the standard RPS method in order to obtain the series solution for system (1) and (2), which will be used in the remainder of this work. The RPS method consists in expressing the solutions as a power series expansion about the initial point x = x 0. Thus, we suppose that the solutions of system (1) and (2) have the following form y i (x) = c ij (x x 0 ) j, i = 1, 2,..., n, (3) j=0 where y i,j (x) = c ij (x x 0 ) j, i = 1, 2,..., n, j = 0, 1, 2,..., are terms of approximations. From the initial conditions (2), we have the initial guesses as y i,0 (x 0 ) = α i,0. Hence, the approximate solutions is given by y i (x) = α i,0 + j=1 c ij(x x 0 ) j, i = 1, 2,..., n. The kth-truncated series y k i (x) can be calculated by y k i (x) = α i,0 + c ij (x x 0 ) j, i = 1, 2,..., n. (4) j=1 Regarding applying the RPS method, the kth-residual function is given by Res k i (x) = d dx yk i (x) F i (x, y k i (x)), i = 1, 2,..., n, (5) and the th residual function is given by Res i (x) = lim k Res k i (x) = y i (x) F i (x, y i (x)), i = 1, 2,..., n. (6) It should be pointed out that Res i (x), i = 1, 2,..., n, are infinitely differentiable functions at x = x 0. That is, Res i (x) = 0 for each x (0, 1). On the d other hand, s Res dx s i (x 0 ) = ds Res k dx s i (x 0 ) = 0, for s = 0, 1,..., k. This relation is a fundamental rule in the RPS method and its applications. Now, by substituting the kth-truncated series yi k (x) into Eq. (5) yields that Res k i (x) = jc ij (x x 0 ) j 1 F i (x, α i,0 + c ij (x x 0 ) j ), i = 1, 2,..., n. (7) j=1 j=1
98 Nourah Nizar AlRasheed Consequently, if we set k = 1 and then use the fact Res 1 i (x 0 ) = 0, then we have c i,1 = F i (x 0, α i,0 ), i = 1, 2,..., n. (8) Thus, the first approximate solutions is given by y 1 i (x) = α i,0 + F i (x 0, α i,0 )(x x 0 ), i = 1, 2,..., n. (9) Now, if we differentiate both sides of Eq. (7) with respect to x and set k = 2, then we have d 2 dx Res2 i (x) = j(j 1)c ij (x x 0 ) j 2 d 2 dx F i(x, α i,0 + c ij (x x 0 ) j ), j=2 = 2c i,2 F i (x, j=1 2 jc ij (x x 0 ) j 1 ), i = 1, 2,..., n. j=1 and if we use the fact d dx Res2 i (x 0 ) = 0, then we have c i,2 = 1 2 F i(x 0, c i,1 ), i = 1, 2,..., n, (10) where c i,1 are given in Eq. (8) for i = 1, 2,..., n. Thus, the second approximate solutions is given by y 2 i (x) = α i,0 + F i (x 0, α i,0 )(x x 0 ) + 1 2 F i(x 0, c i,1 )(x x 0 ) 2, i = 1, 2,..., n. (11) For the third approximate solutions, if we differentiate both sides of Eq. (7) twice with respect to x, set k = 3 and use the fact d2 dx 2 Res 3 i (x 0 ) = 0, then we have c i,3 = 1 3! F i(x 0, c i,2 ), i = 1, 2,..., n, (12) where c i,2 are given in Eq. (10) for i = 1, 2,..., n. By the same technique, the process can be repeated to generate a sequence of approximate solutions for system (1) and (2). That is, for k = m, we have c i,m = 1 m! F i(x 0, c i,(m 1) ), i = 1, 2,..., n. (13) Moreover, higher accuracy can be achieved by evaluating more components of the solution. It will be convenient to have a notation for the error in the approximation y i (x) y k i (x). The kth remainder solutions is given by Rem k i (x) = y i (x) yi k (x) = 1 m! F i(x 0, c i,(m 1) )(x x 0 ) m, i = 1, 2,..., n. m=k+1
Adaptation of Taylor s formula... 99 In fact, it often happens that the remainders become smaller and smaller, approaching zero as k gets large. The concept of accuracy refers to how closely a computed or measured value agrees with the truth value. To show the accuracy of the present method, two types of errors are reported in this work. The first one is called the exact error Ext k i (x) while the second is called the relative error Reli k (x), which are defined, respectively, by Ext k i (x) = y i (x) y k i (x), Rel k i (x) = Ext k i (x)/ y i (x), i = 1, 2,..., n. 3 Numerical Examples In this section, some numerical examples are considered to validate the reliability, accuracy and efficiency of the RPS method that provides an analytical approximate solution in terms of an infinite power series. The examples reflect the behavior of the solution with different non-homogeneous terms. Results obtained are compared with the exact solution of each example and are found to be in good agreement with each other. In the process of computation, all computations performed by Mathematica software package. Example 3.1 Consider the following first-order differential equation subject to the initial condition y (x) y(x) = e 2x, 0 x 1, (14) y (0) = 2. (15) To apply the RPS method, select the initial guess of the approximation such that y 0 (0) = 2. Then, the kth-truncated series solution of Eq. (14) is given by y k (x) = 2 + c 1s x s = 2 + c 11 x + c 12 x 2 + c 13 x 3 +..., s=1 whereas the kth-truncated series is given by Res k (x) = sc 1s x s 1 (2 + s=1 c 1s x s ) e 2x. (16) If we set k = 1 and then use the fact Res 1 (0) = 0, then we have c 11 = 3. That is, y 1 (x) = 2 + 3x. Now, differentiating both sides of Eq. (16) with respect to x, and setting k = 2 yields d 2 2 dx Res2 (x) = s(s 1)c 1s x s 2 sc 1s x s 1 2e 2x. s=2 s=1 s=1
100 Nourah Nizar AlRasheed By using the fact d dx Res2 (0) = 0, we have 2c 12 c 11 = 2, that is, c 12 = 5 2. Thus, y 2 (x) = 2 + 3x + 5 2 x2. Again, differentiating both sides of Eq. (16) twice with respect to x, and setting k = 3 yields d 2 dx 2 Res3 (x) = 3 s(s 1)(s 2)c 1s x s 3 s=3 3 s(s 1)c 1s x s 2 4e 2x. As well as according to the fact d2 Res 3 (0) = 0, we have 6c dx 2 13 2c 12 = 4, and then the coefficient c 13 = 3. Therefore, the 3rd-truncated series solution 2 y 3 (x) = 2 + 3x + 5 2 x2 + 3 2 x3. Now, differentiating both sides of Eq. (16) again with respect to x, and setting k = 4 yields d 3 dx 3 Res4 (x) = 4 s!c 1s x s 4 s=4 s=2 4 s(s 1)(s 2)c 1s x s 3 8e 2x. s=3 d Using the fact 3 Res 4 (0) = 0, we have 24c dx 3 14 6c 13 = 8, and then the coefficient c 14 = 17. Therefore, the 4th-truncated series solution 24 y4 (x) = 2+3x+ 5 2 x2 + 3 2 x3 + 17 24 x4. By continuing with similar fashion, the 6th-truncated series solution is given by y 6 (x) = 2 + 3x + 5 2 x2 + 3 2 x3 + 17 24 x4 + 11 40 x5 + 13 144 x6 = 2 + 3x + 5 2 x2 + 9 6 x3 + 17 24 x4 + 33 120 x5 + 65 720 x6 = (1 + 1) + (x + 2x) + 1 2 (x2 + 4x 2 ) + 1 3! (x3 + 8x 3 ) = + 1 4! (x4 + 16x 4 ) + 1 5! (x5 + 32x 5 ) + 1 6! (x6 + 64x 6 ) (1 + x + 12 x2 + 13! x3 + 14! x4 + 15! x5 + 16! ) x6 + (1 + 2x + 12 (2x)2 + 13! (2x)3 + 14! (2x)4 + 15! (2x)5 + 16! ) (2x)6 = 6 n=0 1 n! xn + 6 n=0 1 n! (2x)n. Correspondingly, the general form of the kth-truncated series solution is given by y k 1 1 (x) = n! xn + n! (2x)n, n=0 n=0 which coincide with the exact solution y(x) = e x + e 2x.
Adaptation of Taylor s formula... 101 Example 3.2 Consider the following system of first-order differential equations { y 1 (x) + 2xy 1 (x) + y 2 (x) = 2x cos(x), 0 x 1, (17) y 2(x) + y 2 (x) 3y 1 (x) = sin(x) 2 cos(x), 0 x 1, subject to the initial conditions y 1 (0) = 1, y 2 (0) = 0. (18) To apply the RPS method, if we select the initial guesses of the approximation such that y 0 1 (0) = 1 and y 0 2 (0) = 0. Then, the kth-truncated series solutions of Eq. (17) are given by y k 1 (x) = 1 + y k 2 (x) = c 1s x s = 1 + c 11 x + c 12 x 2 + c 13 x 3 +..., s=1 c 2s x s = c 21 x + c 22 x 2 + c 23 x 3 + c 24 x 4 +..., s=1 whereas the kth-truncated series solutions are given by Res k 1 (x) = k s=1 sc 1sx s 1 + k s=0 2c 1sx s+1 + k s=1 c 2sx s 2x cos(x), Res k 2 (x) = k s=1 sc 2sx s 1 + k s=1 c 2sx s 3(1 + k s=1 c 1sx s ) (19) sin(x) + 2 cos(x). If we set k = 1 and use the facts Res 1 1 (0) = 0, Res 1 2 (0) = 0, then we have c 11 = 0 and c 21 = 1. Therefore, the 1st-truncated series solutions are y 1 1 (x) = 1 and y 1 2 (x) = x. Now, by differentiating both sides of Eq. (19) with respect to x, and setting k = 2 yields d dx Res2 1 (x) = 2 s=2 s(s 1)c 1sx s 2 + 2 s=0 2(s + 1)c 1sx s + 2 s=1 sc 2sx s 1 + 2(x sin(x) cos(x)), d dx Res2 2 (x) = 2 s=2 s(s 1)c 2sx s 2 + 2 s=1 sc 2sx s 1 2 s=1 3sc 1sx s 1 cos(x) 2 sin(x); By using the fact d dx Res2 i (0) = 0, i = 1, 2, we have that 2c 12 + c 21 = 0 and 2c 22 + c 21 3c 11 = 1, then the coefficients c 12 = 1 2 and c 22 = 0. Therefore,
102 Nourah Nizar AlRasheed the 2nd-truncated series solutions are y 2 1 (x) = 1 1 2 x2 and y 2 2 (x) = x. Again, by differentiating both sides of Eq. (19) twice with respect to x, and setting k = 3 yields d 2 Res 3 dx 2 1 (x) = 3 s=3 s(s 1)(s 2)c 1sx s 3 + 3 s=1 2s(s + 1)c 1sx s 1 + 3 s=2 s(s 1)c 2sx s 2 + 2(x cos(x) + 2 sin(x)), d 2 Res 3 dx 2 2 (x) = 3 s=3 s(s 1)(s 2)c 2sx s 3 + 3 s=2 s(s 1)c 2sx s 2 3 s=2 3s(s 1)c 1sx s 2 + sin(x) 2 cos(x); As well as according to the fact d2 Res 3 dx 2 i (0) = 0, i = 1, 2, we have 6c 13 + 4c 11 + 2c 22 = 0 and 6c 23 + 2c 22 6c 12 = 2, then the coefficients c 13 = 0 and c 23 = 1 6. Therefore, the 3rd-truncated series solution y3 1(x) = 1 1 2 x2 and y 3 2(x) = x 1 6 x3. Now, by differentiating both sides of Eq. (19) again with respect to x, and setting k = 4 yields d 3 Res 4 dx 3 1 (x) = 4 s=4 s!c 1sx s 4 + 4 s=2 2s(s + 1)(s 1)c 1sx s 2 + 4 s=3 s(s 1)(s 2)c 2sx s 3 + 2(x sin(x) + 3 cos(x)), d 3 Res 4 dx 3 2 (x) = 4 s=4 s!c 2sx s 4 + 4 s=3 s(s 1)(s 2)c 2sx s 3 4 s=3 3s(s 1)(s 2)c 1sx s 3 + cos(x) + 2 sin(x). Using the fact d3 dx 3 Res 4 i (0) = 0, i = 1, 2, we have 4!c 14 + 12c 12 + 6c 23 + 6 = 0 and 4!c 24 + 6c 23 186c 13 + 1 = 0, then the coefficients c 14 = 1 4! and c 14 = 0. Therefore, the 4th-truncated series solutions are y 4 1(x) = 1 1 2 x2 + 1 4! x4 and y 4 2(x) = x 1 6 x3. By continuing with similar fashion, the 7th-truncated series solutions are given by y 7 1(x) = 1 1 2 x2 + 1 4! x4 1 6! x6 = y 7 2(x) = x 1 6 x3 + 1 5! x5 1 7! x7 = 3 ( 1) s x 2s (2s)!, s=0 3 s=0 ( 1) s+1 x 2s+1 (2s + 1)!. Correspondingly, the general form of the nth-truncated series solutions are
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