Numerical analysis of the one-dimensional Wave equation subject to a boundary integral specification

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1 Numerical analysis of the one-dimensional Wave equation subject to a boundary integral specification B. Soltanalizadeh a, Reza Abazari b, S. Abbasbandy c, A. Alsaedi d a Department of Mathematics, University of Houston, 48 Calhoun Rd, Houston, TX 7724, USA b Department of Mathematics, University of Tabriz, Tabriz, Iran c Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran d Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 2589, Saudi Arabia Abstract In this research a numerical technique is developed for the one-dimensional Wave equation that combines classical and integral boundary conditions. New matrix formulation technique with arbitrary polynomial bases are proposed for the numerical/analytical solution of this kind of partial differential equation. Not only the exact solutions have been achieved by the known forms of the series solutions, but also for the finite terms of series, the corresponding numerical approximations have been computed. We give a simple and efficient algorithm based on an iterative process for numerical solution of the method. Some numerical examples are included to demonstrate the validity and applicability of the technique. 2 Mathematics subject classification: 35Lxx; 35Qxx; 65M2. Keywords : Wave equation; Nonlocal boundary conditions; Expansion methods; Matrix formulation. Introduction In 98, Ortiz and Samara [] proposed an operational technique for the numerical solution of nonlinear ordinary differential equations with some sup- Corresponding author: babak@math.uh.edu

2 2 plementary conditions based on the Tau method [2]. The stabiliaty and convergence of the Tau approximations have been proved in [28]. During recent years many authors has been used this method for solving various types of equations. For example in [3] this method has been used for linear ordinary differential eigenvalue problems and in [5, 6] has been used for partial differential equations and integral and integro-differential equations[7, 8]. Abbasbandi in [7] has been used this method for the system of nonlinear Volterra integrodifferential equations. In 963, nonlocal boundary equation have been presented by Cannon [8] and Batten [9] independently. Then, parabolic initial-boundary problems with nonlocal integral conditions for parabolic equations were investigated by Kamynin [2] and Ionkin [24]. There are many physical phenomena that formulated in nonlocal form of PDEs. The development of numerical techniques for the solution of the hyperbolic non-local boundary value problems has been an important research topic in many branches of science and engineering. Hyperbolic initial-boundary value problems in one dimension which involve non-local boundary conditions are studied by several authors. Beilin [25] investigated the non-local analogue to classical mixed problems, which involve initial, boundary integral conditions. Bouziani [26] studied the existence, uniqueness and continuous for a hyperbolic partial differential equation. The wave equation is an important second-order linear partial differential equation of waves, such as sound waves, light waves and water waves. It arises in fields such as acoustics, electromagnetics, and fluid dynamics. In this paper a new matrix formulation is presented for the problem of obtaining numerical/analytical approximations to u(x, t) which satisfies the wave equation: with the initial condition and boundary condition u tt αu xx = f(x, t), < x < L, < t T () L u(x, ) = r(x), x L, (2) u t (x, ) = s(x), x L, (3) u(, t) = p(t), < t T, (4) k(x)u(x, t)dx = m(t), < t T, (5) where the functions f(x, t), r(x), s(x), k(x), p(t) and q(t) and the constants α, L and T are known. The function f(x, t) is often called the source function because in practice it describes the effects of the sources of waves on the

3 3 medium carrying them. Physical examples of source functions include the force driving a wave on a string, or the charge or current density in the Lorenz gauge of electromagnetism. Beilin proved existence and uniqueness of classical solution of ()-(5) and found it s Fourier representation [25]. In [27], Dehghan at.el presented several finite differential schemes for numerical solution of ()-(5). These three level techniques are based on two second-order (one explicit and one weighted) schemes and a fourth-order technique(a weighted explicit). Also in [28], the shifted Legendre Tau method technique is developed for the solution of the studied model. In [29], variational iteration method is used for solving the studied model. In [3], A numerical technique based on an integro-differential equation and local interpolating functions have been presented. Authors of [32] presented a numerical scheme based on finite difference and spectral methods. W. Ang, presented an other numerical method[3]. A new matrix formulation technique with arbitrary polynomial bases has been proposed for the numerical/analytical solution of the Heat equation with nonlocal boundary condition []. Authors of [9] presented a matrix formulation method based on shifted standard and shifted Chebyshev bases for solving Wave equation with a boundary integral condition. Similar problems can be found in [, 2,?, 2, 22, 3, 4] 2 Description of Matrix formulation In Eqs. ()-(5), the functions f(x, t), p(x), k(x), g(t) and m(t) generally are not polynomials. We assume that these functions are polynomial or they can be approximated by polynomials to any degree of accuracy. For this purpose, one may use one or two variate Taylor or Chebyshev series or other suitable methods. So we can write: f(x, t) n i= n j= f ijx i t j = X T F T, r(x) n i= r ix i = X T R, i= k ix i = X T K, j= m jt j = MT, s(x) n i= s ix i = X T S, k(x) n g(t) n j= g jt j = GT, m(t) n where X = [, x, x 2,..., x n ] T, T = [, t, t 2,..., t n ] T, S = [s, s, s 2,..., s n ] T, R = [r, r, r 2,..., r n ] T, K = [k, k, k 2,..., k n ] T, G = [g, g, g 2,..., g n ] T, M = [m, m, m 2,..., m n ] T, F = [F, F, F 2,..., F n ], F i = [f i, f i, f 2i,..., f ni ] T, i =,, 2,..., n. (7) (6)

4 4 Therefore we consider approximate solution of the form U n (x, t) = n n u ij x i t j = X T UT, (8) i= j= where U = [U, U, U 2,..., U n ], with U i = [u i, u i, u 2i,..., u ni ] T. The matrix U is an (n + ) (n + ) matrix which contains (n + ) 2 unknown coefficients of U n (x, t). To find these unknowns, we proceed as follows. We first consider the initial condition u(x, ) = r(x), x L. then du attention to Eq. (6) and Eq.(8), we obtain which implies X T U = X T R, U = R. (9) since X is a basis vector. From Eq. (9), we can find the first column of U. Now consider u(, t) = p(t), < t T. () Substituting Eq. (6) and (8) in Eq. (), we get X T UT = P T. () where X = [,,,..., ]. Since T is a basis vector, we have Now we consider the nonlocal boundary condition L X T U = P. (2) k(x)u(x, t)dx = m(t). We substitute from Eqs.(6) and (8) and obtain L ( n h= k hx h )( n i= n j= u ijx i t j )dx = n j= m jt j, hence n j= m jt j = ( n = n h= i= ( n h= n i= n j= k hu ij t j L xh+i dx n j= k hu ij t j v h+i+ ), < t T. ) (3)

5 5 or equivalently where with and d i = DUT = MT. (4) D = [d, d, d 2,..., d n ], n k h v h+i+, i =,, 2,..., n, h= v h+i+ = The equivalent from (4) is Lh+i+, h, i =,, 2,..., n. h + i + DU = M, (5) Since T is a basis vector. Now, we recall the following lemma from [4, 8], to write Eq. () in the matrix form to determine remainder equations. This lemma is proved by induction. Lemma 2.. The effect of r repeated differentiation on coefficients vector a = [a, a, a 2,..., a n ] of a polynomial y n (x) = ax is the same as that of postmultiplication of a by the matrix η r : d r dx r y n(x) = aη r X, where η is the (n + ) (n + ) operational matrix of derivation : η = n (n+)(n+) Corollary 2.. By using lemma 2., we have u xx = X T (η T ) 2 UT, u t = X T UηT, u tt = X T Uη 2 T. (6) Now, consider the u t (x, ) = s(x), x L, (7)

6 6 Applying Eqs. (6), (8) and (6), we obtain where T () = [,,,..., ] T. Finally, consider the UηT () = S, x L, (8) u tt αu xx = f(x, t), < x < L, < t T. (9) Therefore substituting from Eqs. (6), (8) and (6) in Eq. (6), leads to or Since X and T are basis vectors. X T Uη 2 T αx T (η T ) 2 UT = X T F T, Uη 2 α(η T ) 2 U = F. (2) 3 Determining final linear equations In this section, we arrange the linear equations obtained in the previous section, to have a system of (n + )(n + ) equations for the (n + )(n + ) unknowns. We find (n + ) equations from Eq. (8). Note that, since the Eq. () and boundary conditions (4) and (5) are not defined for x [, ] and t [, T ], we chose n equations from Eq. (2), n equations from Eq. (3), n equations from Eq. (8) and n(n 2) equations from Eq. (2). By Eq. (9), we have and from Eq. (8), we have and from Eq. (2), we get from Eq. (5), we get U(i, ) = r i, i =,, 2,..., n, (2) U(, j) = p j, j =, 2,..., n, (22) U(i, ) = s i, i =, 2,..., n, (23) n d i U(i, j) = m j, j =, 2,..., n. (24) i= Finally, for j =, 2,..., n By Eq. (2), we have ( ) U(i, j) = f(i, j 2) + α(i + 2)(i + )U(i + 2, j 2), (j )(j) i =,..., n 2, (25) U(n, j) = (f(n, j)). (j )(j)

7 7 The Eq. (2)-(25) generate a system of (n + )(n + ) equations. By solving this system of equations, the unknown coefficients of U can be calculated. For solving this system, we introduce a simple interesting process. we know that the first column of U by Eq. (2) and the first row of U by Eq. (22) and the second column by Eq. (23) were obtained. In every step, we obtain one column of U by an interesting process. for example for finding the second column of U, we know that the value of U(, ) was calculated. thus from Eq. (25) the values of U(2, ), U(3, ),..., U(n, ) can be found. now, if we utilize Eq. (24), then the value of U(n, ) will be found. Here the 3th column of U have been obtained. By the similar process, we obtain the remainder columns of U. Therefore the corresponding algorithm can be introduced to under case: Algorithm step : Choose n N as the degree of approximate solution. step 2: Determine the vectors K, P, G, D and M and the matrix F. step 3: Set X = [, x, x 2, x 3,..., x n ] T, T = [, t, t 2, t 3,..., t n ] T. step 4: For i =,, 2,..., n, set U(i, ) = r i. step 5: For j =, 2,..., n, set U(, j) = p j. step 4: For i =, 2,..., n, set U(i, ) = s i. step 6: For j = 2, 3,..., n U(n, j) = f(n, j ). (j )(j) i=,2,3,...,n-2 do U(i, j) = (j )(j) End do U(n, j) = d n (m j n i= d iu(i, j)). step 7: Set U n (x, t) = X T UT. step 8: End ( ) f(i, j )+α(i+2)(i+)u(i+2, j ). 4 Numerical Examples In this section, we applied the process presented in this paper and solved five examples. These examples are chosen such that their exact solutions are known. The numerical computations have been done by the software Matlab. Let e n (x, t) = u n (x, t) u(x, t), we calculate the following norms of the error for different values of n. E = e n (x, t) = max{ e n (x, t), x L, t T }. (26) Example 4.. ([33])Consider the heat equation presented in Eqs. ()-(5) with f(x, t) = (2x 5 + 2x 3 2x 2 ) (2x 3 + 6x 2)(t 2 t), r(x) =, s(x) = x 2 x 3 x 5, k(x) =,

8 8 t(t ) g(t) =, m(t) = 2 α =, L =, T =. Authors [33] solved this example by two forms of MOL and obtained the results of Table. Table. The norm u exact u MOL xi, in [33] x MOL MOL Selecting n = 2 and using Eqs. (8) and (9), we get U(, ) = 2, U(, ) =, U(2, ) = 4, U(, ) =, U(, 2) = 8. Now from Eq. (2) and then Eq. (2), we obtain and also then and by using Eq. (7), we have U(, ) =, U(2, ) = 4, U(, 2) =, U(2, 2) = 32, U = , U(x, t) = 2 + t + 8t 2 + 8x 2 + 4x 2 t + 32x 2 t 2, which it is the exact solution of the problem.

9 9 Example 4.2. Consider the Eqs. ()-(5) with f(x, t) = (x 2 2)e t, r(x) = s(x) = x 2, k(x) =, g(t) =, Applying Eqs. (8)-(9), we obtain and then m(t) = et, α = L =, T =.5. 3 U(, j) =, j =, 2,..., n, U(, ) =, U(, ) =, U(2, ) =, U(3, ) =... = U(n, ) =, U(, ) =, U(2, ) =, U(3, ) =... = U(n 2, ) =, By continuing this process, we get U(n, ) = (f(n, )) =, U(n, ) = (n + )( 3 3 ) =. u(x, t) = x 2 ( + t + t2 tn ! n! ), which is the Taylor expansion of the u(x, t) = x 2 e t, which is the exact solution of this example. The numerical results of this example reported in Tables 2.. and 2.2. and plot of corresponding error function is shown in Fig.. Table 2.. Absolute errors of the presented method for n=5 x t =. t =.3 t =

10 Table 2.2. Maximum errors for x = :.5 :., t = :.25 :.5 n = 5 n = n = 3 n = 5 E x e(x,t) t x Fig.: Plot of error function from Example 4.2. for n=5. Example 4.3. Consider the Eqs. ()-(5) with f(x, t) = e (x+t), r(x) = e x, s(x) = e x, k(x) =, p(t) = e t, m(t) = (e )e t, α = 2, L = T =, using Eqs. (8) and (9), we get U(i, ) =, i =,, 2,..., n, i! U(, j) =, j =, 2,..., n, j! and from Eq. (2), we have U(i, ) =, i =, 2,..., n, i! U(i, 2) =, i =, 2,..., n 2, 2!i!

11 U(n, 2) =, i =, 2,..., n 2, 2!(n )! then by using Eq. (2) and above numerical result, we get U(n, 2) = e 2!2! 2!3! 2!4!... 2!(n )! 2!n!. By using this recursive scheme, the remainder values of U(i, j) were obtained. Now by using Eq. (7), we get u(x, t) + t + 2! t n! tn +x + xt + 2! xt n! xtn + + 2! x2 + 2! x2 t + 2!2! x2 t n!n! xn t n +..., which is the Taylor expansion of the u(x, t) = e x+t, which is the exact solution of this example. The numerical results of the absolute errors with n = 5,, 5 obtained by using our described method are given in Tables 3.. and 3.2. and plot of corresponding error function is shown in Fig.2. Table 3.. Absolute errors of the presented method n=2 x t =. t =.3 t = Table 3.2. Maximum errors for x = :.5 :., t = :.25 :.5 n = 5 n = n = 5 n = 2 E

12 2 x e(x,t) t x Fig.2: Plot of error function from Example 4.3. for n=2. Example 4.4. ([5, 6])Consider the Eqs. ()-(5) with f(x, t) =, r(x) =, s(x) = πcos(πx), p(t) =, m(t) = sin(πt), α =, L =, T =.5. For this example, we solved the problem, by applying the technique described in preceding section we obtain the closed form of the solution as follows u n (x, t) x(t 2 t 3 + t4 2! t5 3! +...) x2 (t 2 t 3 + t4 2! t5 3! +...) which is the Taylor expansion of the u(x, t) = cos(πx)sin(πt), which is the exact solution of this example. The numerical results of the absolute errors with n = 5, 5, 8, 2 obtained by using our described method are given in Table 3. and the plot of corresponding error function is shown in Fig.3. Table 4.. Absolute errors of Tau and Finite difference methods for t=.5

13 3 x The optimal Explicit [6] minimum errors of[5] Table 4.2. Absolute errors of the presented method for n=3 x t =. t =.3 t = Table 4.3. Maximum errors for x = :.5 :., t = :.25 :.5 n = n = 2 n = 25 n = 26 E

14 4 x e(x,t) t x Fig.3: Plot of error function from Example 4.4. for n=26. Example 4.5. ([5, 6])Consider the Eqs. ()-(5) with f(x, t) =, r(x) = cos(πx), s(x) =, p(t) = cos(πt), m(t) = α =, L =, T =.25. For this example, we solved the problem, by applying the technique described in preceding section we obtain the closed form of the solution as follows u n (x, t) = ( 2 π2 t 2 + 4! π4 t 4 6! π6 t ) 2 π2 x 2 ( 2 π2 t 2 + 4! π4 t 4 6! π6 t ) + 24 π4 x 4 ( 2 π2 t 2 + 4! π4 t 4 6! π6 t ), which is the Taylor expansion of the u(x, t) = (cos(π(x+t)))+(cos(π(x t))), 2 which is the exact solution of this example. The numerical results of the absolute errors with n = 5, 5, 8, 2 obtained by using our described method are given in Table 3. and the plot of corresponding error function is shown in Fig.4. Table 5. Maximum errors for x = :.5 :., t = :.25 :.5 n = n = 2 n = 25 n = 26 E

15 5 x e(x,t) y x Fig.4: Plot of error function from Example 4.5. for n=26. 5 Conclusions In this paper, we focus on the Wave equation with nonlocal boundary condition and present an operational matrix formulation to solving this equation. By using this method, numerical/analytical results obtained by a simple iterative process. This method reduce the computational difficulties of the other methods and all the calculations can be made with simple iterative process. Also we can increase the accuracy of the series solution by increasing the number of terms in the series solution. Consequently, it is seen that this method can be an alternative way for the solution of partial differential equations that have no analytic solutions. References [] E.L. Ortiz, H. Samara, An operational approach to the Tau method for the numerical solution of nonlinear differential equations, Computing 27 (98), [2] E.L. Ortiz, The Tau method, SIAM. J. Numer. Anal., 6 (969),

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