Approximate Solution of BVPs for 4th-Order IDEs by Using RKHS Method
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1 Applied Mathematical Sciences, Vol. 6, 01, no. 50, Approximate Solution of BVPs for 4th-Order IDEs by Using RKHS Method Mohammed Al-Smadi Mathematics Department, Al-Qassim University, Saudi Arabia Omar Abu Arqub Department of Mathematics, Al-Balqa Applied University, Jordan Nabil Shawagfeh Mathematics Department, University of Jordan, Jordan Abstract In this paper, reproducing kernel Hilbert space method is applied to approximate the solution of two-point boundary value problems for fourth-order integro-differential equations. The analytical solution is represented in the form of series in the reproducing kernel space. The n-term approximation is obtained and is proved to converge to the analytical solution. Moreover, the proposed method has an advantage that it is possible to pick any point in the interval of integration and as well the approximate solutions and its all derivatives up to order four will be applicable. Numerical examples are given to demonstrate the computation efficiency of the presented method. Results obtained by the method indicate the method is simple and effective. Mathematics Subject Classification: 45J05; 47B3; 34K8 Keywords: Reproducing kernel Hilbert space; Integro-differential equations; Boundary value problems 1 Introduction Mathematical modeling of real-life, physics, and engineering problems usually results in functional equations, for example, partial differential equations,
2 454 M. Al-Smadi, O. Abu Arqub, and N. Shawagfeh integral and integro-differential equations (IDEs), stochastic differential equations and others. Many mathematical formulation of physical phenomena contain IDEs, these equations arise in fluid dynamics, biological models and chemical kinetics [15]. IDEs are usually difficult to solve analytically so it is required to obtain an efficient approximate or numerical solution. Therefore, they have been of great interest by several authors. In literature, there exist numerical techniques such as Wavelet-Galerkin method [], Lagrange interpolation method [3], and Tau method [14] and semi analytical-numerical techniques such as Adomian s decomposition method [13, 6], Taylor polynomials [1], variational iteration method [5], homotopy perturbation method [8], and homotopy analysis method [4]. The boundary value problems (BVPs) for higher-order IDEs have been investigated by Morchalo [] and Agarwal [1] among others. Agarwal [1] discussed the existence and uniqueness of the solutions for these problems. In [1], no numerical method was presented. In this work, we apply reproducing kernel Hilbert space (RKHS) technique to develop a novel numerical method in the space W 5 [a, b] for obtaining approximations to the solution and all its derivatives up to order four for fourth-order IDEs [6]: x u (iv) (x) =f(x)+γu(x)+ h(x, t)g(u (t))dt, x, t [a, b], (1) subject to the boundary conditions a u (a) =α 0, u (a) =α 1, u (b) =β 0, u (b) =β 1, () where a, b, γ and α i, β i, i =0, 1 are real finite constants, u W 5 [a, b] is an unknown function to be determined, f W 1 [a, b], h (x, t) is continuous function on [a, b] [a, b], G (y) is continuous term in W 1 [a, b] asy = y (x) W 5 [a, b], a x b, <y< and is depending on the problem discussed, and W 1 [a, b], W 5 [a, b] are reproducing kernel spaces. Reproducing kernel theory has important application in numerical analysis, differential equations, integral and IDEs, probability and statistics and so on [3, 5, 6]. Recently, using the RKHS method, the authors in [3-1, 16-0, 7] discussed singular linear two-point BVP, singular nonlinear two-point periodic BVP, nonlinear system of BVPs, singular intergal equations, IDEs, and nonlinear partial differential equations. The rest of the paper is organized as follows: several reproducing kernel spaces are described in section. In section 3, a linear operator, a complete normal orthogonal system, and some essential results are introduced. Also, a method for the existence of solutions for IDE (1) and () based on reproducing kernel space is described. The numerical examples are presented in section 4. This article ends in section 5 with some concluding remarks.
3 Approximate solution of BVPs for 4th-order IDEs 455 Several reproducing kernel spaces In this section, several reproducing kernels needed are constructed in order to solve IDE (1) and () using RKHS method. Before the construction, we utilize the reproducing kernel concept. Definition 1. [9] Let E be a nonempty abstract set. A function K : E E C is a reproducing kernel of the Hilbert space H if 1. for each t E, K (,t) H.. for each t E and ϕ H, ϕ, K (,t) = ϕ (t). The last condition is called the reproducing property : the value of the function ϕ at the point t is reproducing by the inner product of ϕ with K (,t). A Hilbert space which possesses a reproducing kernel is called a RKHS [9]. Next, we first construct the space W 5 [a, b] in which every function satisfies the boundary conditions () and then utilize the space W 1 [a, b]. Definition. [8] W 5 [a, b] ={u (x) u(i),i =0, 1,, 3, 4 are absolutely continuous real-valued functions on [a, b], u (5) L [a, b], and u (a) =u (a) = u (b) =u (b) =0}. The inner product and norm in W 5 [a, b] are given by u, v W 5 = u (i) (a) v (i) (a)+ 1 b u (i) (b) v (i) (b)+ u (5) (y)v (5) (y)dy (3) and u W 5 = u, u W 5, respectively, where u, v W 5 [a, b]. Definition 3. [17] W 1 [a, b] ={u (x) u is absolutely continuous real-valued functions on [a, b] and u L [a, b]}. The inner product and norm in W 1 [a, b] are given by u, v W 1 = b a (u v + uv) dy and u W 1 = u, u W 1 respectively, where u, v W 1 [a, b]. a In [17], the authors have proved that the space W 1 reproducing kernel is [a, b] is a RKHS and its R x (y) = 1 [cosh (x + y b a) + cosh ( x + y b + a)]. sinh (b a) The space W 5 [a, b] is called a RKHS if for each fixed x [a, b] and any u (y) W 5 [a, b], there exist K (x, y) W 5 [a, b] (simply K x (y)) and y [a, b] such that u (y),k x (y) W 5 = u (x). Next theorem formulate the reproducing kernel K x (y).
4 456 M. Al-Smadi, O. Abu Arqub, and N. Shawagfeh Theorem 1. The space W 5 [a, b] is a RKHS and its reproducing kernel K x (y) is given by: 9 p i (x)y i, y x, K x (y) = 9 (4) q i (x)y i, y > x. Proof. Let K x (y) be the reproducing kernel function. By Eq. (3), we have u (y),k x (y) W 5 = u (i) (a) K (i) x (a)+ 1 u (i) (b) K (i) x (b)+ 4 ( 1) 4 i u (i) (y) K (9 i) x K (i) x (y) b a b a (a) = K (i) u (y) K(10) x (y) dy. Since K x (y) W 5 [a, b], it follows that x (b) = 0, i = 0, 1. Again, since u W 5 [a, b], one obtains (a) =0,K x (i) (a) =0,i =5, 6, and K x (i) (b) =0,i =5, 6, 7, then u (y),k x (y) W 5 = b u a (y)( K(10) x (y))dy. Now, for each x [a, b], if K x (y) also satisfies K x (10) (y) =δ (x y), where δ is the dirac-delta function, then u (y),k x (y) W 5 = u (x). Obviously, K x (y) u (i) (a) =u (i) (b) =0,i =0, 1. If K x () (a) K x (7) is the reproducing kernel of the space W 5 [a, b]. The characteristic equation of K x (10) (y) =δ (y x) isλ 10 = 0, and their characteristic values are λ = 0 with 10 multiple roots. So, let 9 p i (x)y i, y x, K x (y) = 9 q i (x)y i, y > x. On the other hand, let K x (y) satisfies K x (m) (x +0) = K x (m) (x 0), m = 0, 1,..., 8. Integrating K x (10) (y) =δ (x y) from x ε to x + ε with respect to y and let ε 0, we have the jump degree of K x (9) (y) aty = x given by K x (9) (x 0) K x (9) (x + 0) = 1. Through the last descriptions the unknown coefficient of Eq. (4) can be obtained. This completes the proof. Without loss of generality, the two rules K 1x (y) = 9 p i (x) y i and K x (y) = 9 q i (x) y i of the reproducing kernel K x (y) in Eq. (4) are obtained at a =0 and b = 1 in IDE (1) and () and are given as: 1 K 1x (y) = K y (x) = ( 1+x) y [5x 3 (8 15y)y + x 6 (19 1y)y + y 7 +xy 6 ( 9+y)+3x 5 y( y)+x 7 y( 4+3y)+5x 4 y(1 + 6y) +3x ( y y +1y 5 6y 6 + y 7) ]. We mention here that the kernel K x (y) is symmetric, that is K 1x (y) =K y (x) for each x and y.
5 Approximate solution of BVPs for 4th-order IDEs Numerical method and convergence analysis In this section, the existence of the solution and an iterative method of obtaining the solution of IDE (1) and () are developed in the RKHS W 5 [a, b]. To do this, we define a differential operator L : W 5 [a, b] W 1 [a, b] such that Lu (x) =u (4) (x) γu(x). After homogenization of the boundary conditions (), the IDE (1) and () can be converted into the equivalent form as follows: Lu (x) =F (x, T u(x)),x [a, b], u (a) =u (a) =u (b) =u (b) =0, where F (x, T u(x)) = f (x) +Tu(x), Tu(x) = x h(x, t)g(u (t))dt, u (x) a W 5 [a, b], and F (x, T u (x)) W 1 [a, b]. It is clear that L is a bounded linear operator. Now, we construct an orthogonal function system. Put ϕ i (x) =R xi (x) and ψ i (x) = L i ϕ (x), where L is the adjoint operator of L. In terms of the properties of reproducing kernel K x (y), one obtains u (x),ψ i (x) W 5 = Lu (x),ϕ i (x) W 1 = Lu(x i ), i =1,,... The orthonormal system { ψi (x) } of the space W 5 [a, b] can be derived from Gram-Schmidt orthogonalization process of {ψ i (x)} as follows: ψ i (x) = i β ik ψ k (x), where β ik are orthogonalization coefficients such that β ii > 0, i =1,,... Through the next theorem the subscript y by the operator L indicates that the operator L applies to the function of y. Theorem. If {x i } is dense on [a, b], then {ψ i (x)} is a complete function system of W 5 [a, b] and ψ i (x) =L y K x (y) y=xi. Proof. Clearly, ψ i (x) =L i ϕ (x) = L i ϕ (y),k x (y) = ϕ i (y),l y K x (y) = L y K x (y) y=xi. Now, for each fixed u (x) W 5 [a, b], let u (x),ψ i (x) =0, i =1,,..., that is u (x),l ϕ i (x) = Lu ( ),ϕ i ( ) = Lu (x i ) = 0. Note that {x i } is dense on [a, b], therefore Lu (x) = 0. It follows that u (x) = 0 from the existence of L 1. So, the proof of the Theorem is complete. The structure of the next two theorems are as follows: firstly, we will give the representation of the exact solution of IDE (1) and () in the space W 5 [a, b]. After that, the convergence of approximate solution u n (x) to the analytic solution will be proved. Theorem 3. If {x i } is dense on [a, b] and the solution of the IDE (1) and () is unique, then this solution satisfies the form: u (x) = i β ik F (x k,tu(x k )) ψ i (x). (5) k=1 k=1
6 458 M. Al-Smadi, O. Abu Arqub, and N. Shawagfeh Proof. From Theorem, it is easy to see that { ψi (x) } is the complete orthonormal basis of the space W 5 [a, b]. Thus, u (x) can be expanded in the Fourier series as u (x) = u (x), ψi (x) ψi (x). Since v (x),ϕ i (x) = v (x i ) for each v (x) W 1 [a, b]. Hence, we have u (x) = u (x), ψi (x) ψi (x) = k=1 i β ik u (x),ψ k (x) ψ i (x) = β ik Lu (x),ϕ k (x) ψ i (x) = k=1 k=1 i β ik u (x),l ϕ k (x) ψ i (x) = i β ik F (x, T u(x)),ϕ k (x) ψ i (x) = F (x k,tu(x k )) ψ i (x). The proof of the theorem is complete. i k=1 i β ik k=1 Remark 1. If IDE (1) and () is linear, then the analytical solution can be obtained directly from Eq. (5). In the nonlinear case, the approximation solution can be obtained using the following iterative method: according to Eq. (5), the representation of the solution of IDE (1) and () can be denoted by u (x) = B i ψi (x), where B i = i β ik F (x k,u k 1 (x k ),Tu k 1 (x k )). In fact, k=1 B i are unknown, we will approximate B i using known A i. For a numerical computations, we define initial function u 0 (x 1 ), put u 0 (x 1 )=u(x 1 ), and define the n-term approximation to u (x) by: u n (x) = n A i ψi (x), (6) where the coefficients A i of ψ i (x), i =1,,..., n are given as: A i = i k=1 β ik F (x k,tu k 1 (x k )). (7) In the iteration process of Eq. (6), we can guarantee that the approximation u n (x) satisfies the boundary conditions (). Next, we will proof u n (x) in the iterative formula (6) is converge to the exact solution u (x) of IDE (1) and (). The Lemma 1 through Lemma 5 are collected for future use. Lemma 1. If u (x) W 5 [a, b], then there exists M>0 such that u(x) C M u(x) W 5. Proof. For any x, y [a, b], we have u (i) (x) = u(y),k x (i) (y) W 5, i =0, 1,..., 4. By the expression of K x (y), clearly, K x (i) (y) W 5 M i. Thus, u (i) (x) = u(y),k x (i) (y) W 5 K x (i) (y) W 5 u(x) W 5 M i u(x) W 5, i =0, 1,..., 4. Hence, u(x) C M u(x) W 5, where M = M 0 + M M 4.
7 Approximate solution of BVPs for 4th-order IDEs 459 Lemma. If u n (x) W 5 u (x), x n y as n, and F (x, w) is continuous in [a, b] with respect to x, w for x [a, b] and w (, ), then F (x n,tu n 1 (x n )) F (y, Tu (y)) as n. Proof. Firstly, we will prove that u n (x n ) u (y) in the sense of W 5. Since u n (x n ) u (y) = u n (x n ) u n (y)+u n (y) u (x) u n (x n ) u n (y) + u n (y) u (y). By reproducing property of K x (y), we have u n (x),k xn (x) K y (x) W 5 u n (x) W 5 K xn (x) K y (x) W 5. From the symmetry of K x (y), it follows that K xn (x) K y (x) W 5 0asn. Hence, u n (x n ) u n (y) 0as soon as x n y. On the other hand, by Lemma 1, we know that u n (x) is convergent uniformly to u (x). Thus, for any x, y [a, b], it holds that u n (y) u (y) C 0 as soon as u n (y) u (y) W 5 0asn. Therefore, u n (x n ) u (y) in the sense of W 5 as x n y and n. Thus, by means of the continuation of Tu( ), it is obtained that Tu n (x n ) Tu(y) asn. Hence, by the continuity of F, we have F (x n,tu n 1 (x n )) F (y, Tu (y)) as n. Lemma 3. {u n } n=1 norm of W 5 [a, b]. in Eq. (6) is monotone increasing in the sense of the Proof. By Theorem, { } ψ i is the complete orthonormal system in the space W 5 [a, b]. Hence, we have u n W = n A 5 i ψi (x), n A i ψi (x) W 5 = n (A i ). Therefore, u n W 5 is monotone increasing. Lemma 4. Lu n (x j )=F (x j,tu j 1 (x j )), j n. Proof. The proof will be obtained by induction. If j n, then Lu n (x j )= n A i L ψ i (x j )= n A i L ψi (x),ϕ j (x) = n A W 1 i ψi (x),ψ j (x). The orthogonality of { ψi (x) } gives j W 5 β jl Lu n (x l )= n A i ψ j i (x), β jl ψ l (x) W 5 l=1 l=1 = n A i ψi (x), ψ j (x) = A W 5 j = j β jl F (x l,tu l 1 (x l )). Now, if j = 1, then l=1 Lu n (x 1 )=F (x 1,Tu 0 (x 1 )). Again, if j =, then β 1 Lu n (x 1 )+β Lu n (x )= β 1 F (x 1,Tu 0 (x 1 )) + β F (x,tu 1 (x )). So, Lu n (x )=F(x,Tu 1 (x )). By induction, we have Lu n (x j )=F (x j,tu j 1 (x j )). Lemma 5. Lu n (x j )=Lu (x j ), j n. Proof. From Lemma 1, u n (x) converge uniformly to u (x). Let u n (x) = P n u (x), where P n is an orthogonal projector from the space W 5 [a, b] to Span
8 460 M. Al-Smadi, O. Abu Arqub, and N. Shawagfeh {ψ 1,ψ,..., ψ n }. Thus, Lu n (x j )= Lu n (x),ϕ j (x) = u n (x),l j ϕ (x) = Pn u (x),ψ j (x) = u (x),p n ψ j (x) = u (x),ψ j (x) = Lu (x),ϕ j (x) = Lu (x j ). Theorem 4. If {x i } is dense on [a, b] and u n W 5 is bounded, then u n (x) in the iterative formula (6) convergent to the exact solution u (x) of IDE (1) and () in the space W 5 [a, b] and u (x) = A i ψi (x), where A i is given by Eq. (7). Proof. First of all, we will prove the convergence of u n (x). By Eq. (6), we have u n+1 (x) =u n (x)+a n+1 ψn+1 (x). From the orthogonality of { ψi (x) }, it follows that u n+1 W = u 5 n W +(A 5 n+1 ) = u n 1 W +(A 5 n ) +(A n+1 ) =... = u 0 W + n+1 (A 5 i ). From Lemma 3, the sequence u n W 5 is monotone increasing. Due to the condition that u n W 5 is bounded, u n W 5 is convergent as n. Then, there exists a constant c such that (A i ) = c. It implies that A i = i k=1 β ik F (x k,tu k 1 (x k )) l,,,... Since (u m u m 1 ) (u m 1 u m )... (u n+1 u n ) it follows that if m>n, then u m (x) u n (x) W = 5 u m (x) u m 1 (x) u n+1 (x) u n (x) W = u 5 m (x) u m 1 (x) W u n+1 (x) u n (x) W = m (A 5 i ). Consequently, as n, m, we have i=n+1 m u m (x) u n (x) W 0as (A 5 i ) 0. Considering the completeness of i=n+1 the space W 5 [a, b], there exists a u (x) W 5 [a, b] such that u n (x) u(x) as n in the sense of W 5. Secondly, we will prove that u (x) is the solutions of IDE (1) and (). From Lemmas 4 and 5, since {x i } is dense on [a, b], for any x [a, b], there exists subsequence { } x nj, such that xnj x as j. It is clear that Lu ( ) ( x nj = F xnj,tu nj 1 (x k ) ). Hence, let j, by lemma and the continuity of F, we have Lu (x) =F (x, T u (x)). That is, u (x) satisfies IDE (1). Since ψ i (x) W 5 [a, b], u (x) satisfies the boundary conditions (). In other words, u (x) is the solution of IDE (1) and (), where u (x) = A i ψi (x) and A i is given by Eq. (7). The approximate solution u N n (x) can be obtained by taking finitely many terms in the series representation of u n (x) and u N n (x) = N i β ik F (x k,tu n 1 (x k )) ψ i (x). k=1
9 Approximate solution of BVPs for 4th-order IDEs Numerical Examples In order to have a clear overview of our method, two examples with known exact solutions are studied to demonstrate the accuracy of the present method. Results obtained by the method are compared with the analytical solution of each example and are found to be in good agreement with each other. Through this paper the numerical computation performed by using Mathematica 7.0 software package. Example 1. Consider the linear Volterra IDE [6]: u (iv) (x) =x(1 + e x )+3e x + u(x) x u(t)dt, 0 x, t 1, 0 subject to the boundary conditions u (0) = 1, u (0) = 1, u(1) = 1 + e, u (1) = e. The exact solution is u (x) =1+xe x. Using RKHS method, we choose 11 points on [0, 1]. The numerical results at some selected gird points are given in Table 1. Table 1. Numerical results of u(x) for Example 1: x Ex. solution App. solution Absolute error Relative error Example. Consider the nonlinear Volterra IDE [6]: u (iv) (x) =1+ x e t u (t)dt, 0 x, t 1, subject to the boundary conditions 0 u (0) = 1, u (0) = 1, u(1) = e, u (1) = e. The exact solution is u (x) =e x. Using RKHS method, we choose 11 points on [0, 1]. The numerical results at some selected gird points are given in Table.
10 46 M. Al-Smadi, O. Abu Arqub, and N. Shawagfeh Table. Numerical results of u(x) for Example : x Ex. solution App. solution Absolute error Relative error As we mention, it is possible to pick any point in [0, 1] and as well the approximate solutions and its all derivative up to order four will be applicable using the same previous partition. Next, the absolute error of u (m) (x), m = 0, 1,..., 4 for Example at some selected gird points are given in Table 3. Table 3. Absolute error of u (m) (x) for Example : m x =0.16 x =0.48 x =0.64 x = Conclusions In this paper, the RKHS method was employed to solve the IDE (1) and (). The numerical results show that the present method is an accurate and reliable analytical technique for two-point BVPs of fourth-order IDEs. References [1] R.P. Agarwal, Boundary value problems for higher order integrodifferential equations Nonlinear Analysis, Theory, Meth. Appl. 7 (3) (1983), [] A. Avudainayagam and C. Vani, Wavelet-Galerkin method for integrodifferential equations, Applied Numerical Mathematics, 3 (000), [3] A. Berlinet and C. Thomas-Agnan, Reproducing Kernel Hilbert Space in Probability and Statistics, Kluwer Academic Publishers, 004.
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