Applied Mathematical Sciences, Vol. 8, 2014, no. 151, 7529-7534 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2014.49784 The Shifted Data Problems by Using Transform of Derivatives Hwajoon Kim* Kyungdong University School of IT Engineering Yangju 482-010, Gyeonggi, Korea * Corresponding author Copyright c 2014 Hwajoon Kim. This is an open access article 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 We have checked the shifted data problems by using transform of derivatives. In particular, we have put emphasis on the representation of (y ) and (y ) to f(a) and f (a) for any number a. Mathematics Subject Classification: 35A22, 44A10 Keywords: shifted data problems, Laplace transform, Elzaki transform 1 Introduction In literature, the existing transforms are hardly dealing with the shifted date problems. Hence we would like to check the shifted data problems in this article. The Laplace transforms of derivatives of the given function f(t) are represented by (f ) = s (f) f(0) and (f ) = s 2 (f) sf(0) f (0), and almost all initial value problems are dealing with ones at zero. In order to overcome this restriction, we would like to check the shifted data problems with relation to integral transforms[1-14]. Additionally, we have checked the method by using Elzaki transform[2, 4, 6, 8, 10, 14] as well. Of course, the proposed method can be applied to another integral transforms. In theorem 2.1, we have checked that the solution of the shifted data y + ay + by = r(t), y(k) = c 0, y (k) = c 1 has the form of y(t) = 1 [Y 1 (t k)]
7530 Hwajoon Kim where k > 0, (y 1 ) = Y 1 and y 1 (t 1 ) = y(t). Similarly, we have obtained the solution y(t) = T 1 [Y 1 (t k)] by Elzaki transform. examples as well. With relation to the theorem, we have checked some 2 The shifted data problems by using transform of derivatives We have checked the shifted data problems by using transform of derivatives. To begin with, let us see the definition of Elzaki transform and related results. Definition 2.1 The Elzaki transform of the functions belonging to a class A, where A = {f(t) M, k 1, k 2 > 0 such that f(t) < Me t /k j, if t ( 1) j [0, )} where f(t) is denoted by E[f(t)] = T (u) and defined as or equivalently, T (u) = u 2 f(ut)e t dt, k 1, k 2 > 0, 0 T (u) = u f(t)e t/u dt, u (k 1, k 2 ). 0 Let us put E(f(t)) = T (u). Then the following results can be obtained from the definition and simple calculations[5, 8, 10]. 1) E[f (t)] = T (u)/u uf(0) 2) E[f (t)] = T (u)/u 2 f(0) uf (0) 3) E[tf (t)] = u 2 d [T (u)/u uf(0)] u[t (u)/u uf(0)] du 4) E[t 2 f (t)] = u 4 d2 [T (u)/u uf(0)] du2 5) E[tf (t)] = u 2 d du [T (u)/u2 f(0) uf (0)] u[t (u)/u 2 f(0) uf (0)] 6) E[t 2 f (t)] = u 4 d2 du 2 [T (u)/u2 f(0) uf (0)] Theorem 2.2 Let t = t 1 +k for k > 0. Then the solution of the shifted data y +ay +by = r(t), y(k) = c 0, y (k) = c 1 has the form of y(t) = 1 [Y 1 (t k)] where (y 1 ) = Y 1 and y 1 (t 1 ) = y(t). Similarly, we can obtain the solution by Elzaki transform. y(t) = T 1 [Y 1 (t k)]
The shifted data problems by using transform of derivatives 7531 Proof. Let us set t = t 1 + k. Then the shifted data can be rewritten as y + ay + by = r(t), y(k) = c 0, y (k) = c 1 (1) y 1 + ay 1 + by 1 = r(t 1 + k), y 1 (0) = c 0, y 1(0) = c 1 for y 1 (t 1 ) = y(t). Taking Laplace transform on both sides, we have s 2 Y 1 sc 0 c 1 + asy 1 c 0 = R(t 1 + k) for (r(t)) = R(s). Organizing the equality, we have Thus (s 2 + as)y 1 = R(s + k) + (s + 1)c 0 + c 1. Y 1 = 1 s(s + a) [R(s + k) + (s + 1)c 0 + c 1 ]. (2) Taking the inverse Laplace transform, we obtain y(t 1 ) = 1 (Y 1 ). Since t 1 = t k, the solution of the equation (1) has a form of for (y 1 ) = Y 1 and for y 1 (t 1 ) = y(t). y(t) = 1 [Y 1 (t k)] (3) If we change the Laplace transform (y) to Elzaki transform T (u), then by [3], the equation (2) is changed to Y 1 = u2 [R(u + k) + (1 + au)c 0 + uc 1 ] 1 + au + bu 2 for Y 1 = T (u) and for T (u) = E[f(t)]. Similarly, t 1 = t k gives the solution for y 1 (t 1 ) = y(t). y(t) = T 1 [Y 1 (t k)] (4) Example 2.3 Solve y y = t, y(3) = 1, y (3) = 1. Solution. (By Laplace transform) Let us set t = t 1 + 3. Then the given equation can be expressed by y 1 y 1 = t 1 + 3, y 1 (0) = 1, y 1(0) = 1 (5) for y 1 (t 1 ) = y(t). Taking Laplace transform, we have s 2 Y 1 s 1 Y 1 = 1 s 2 + 2 s
7532 Hwajoon Kim for (y 1 ) = Y 1. Simplification gives By s-shifting theorem, we have Y 1 = 1 s 2 (s 2 1) + 3 s(s 2 1) = 1 s 2 1 1 s + 3s 2 s 2 1 3 s. y 1 (t) = 1 (Y 1 ) = e t + sin ht 1 t 1 + 3 cos ht 1 3 for h is hyperbolic function. Since t 1 = t 3, we can easily find the solution y(t) as y(t) = e t 3 + sin h(t 3) + 3 cos h(t 3) t. (6) To verify this solution, differentiating y in order to get y y, we would obtain the result y y = t. (By Elzaki transform) Taking Elzaki transform on the equation (5), we have T (u) = u 3 3u 2 + 2u3 1 u + 4u2 2 1 u 2 for E(y 1 (t)) = T (u). Taking the inverse Elzaki transform on both sides, we have y 1 (t 1 ) = t 1 3 + 2 sin ht 1 + 4 cos ht 1 for h is the hyperbolic function. Since y 1 (t 1 ) = y(t), by t 1 = t 3, we obtain the solution y(t) = t + 2e t 3 + 2 cos h(t 3). Surely, this is the same result with the equation (6). Example 2.4 Solve the initial value problem y + 3y 4y = 6e 2t 2, y(1) = 4, y (1) = 5. Solution. Setting t = t 1 + 1, we have y 1 + 3y 1 4y 1 = 6e 2t 1, y 1 (0) = 4, y 1(0) = 5. Thus, Laplace transform of the equation is (s 1)(s + 4)Y 1 = 4s + 17 + 6 s 2
The shifted data problems by using transform of derivatives 7533 for y 1 (t 1 ) = y(t). Simplification and partial fraction expansion gives Y 1 = 3 s 2. Taking the inverse transform, we obtain Since t 1 = t 1, we have the answer y 1 = 3e t 1 + e 2t 1. y = 3e t 1 + e 2(t 1). In the above solution, we note that 4s + 17 + 6 s 2 (s 1)(s + 4) = 21/5 s 1 1/5 s + 4 + 1 s 2 6/5 /5 s + 4 = 3 s 2. References [1] HC. Chae and HJ. Kim, The Validity Checking on the Exchange of Integral and Limit in the Solving Process of PDEs, Int. J. of Math. Anal, 8 (2014), 1089-1092. [2] Ig. Cho and Hj. Kim, The solution of Bessel s equation by using integral transforms, Appl. Math. Sci., 7 (2013), 6069-6075. [3] Ig. Cho and Hj. Kim, The Laplace transform of derivative expressed by Heviside function, Appl. Math. Sci., Vol. 90 (2013), 4455-4460. [4] T. M. Elzaki and Hj. Kim, The Solution of Burger s Equation by Elzaki Homotopy Perturbation Method, Appl. Math. Sci., 8 (2014), 2931-2940. [5] T. M. Elzaki, S. M. Ezaki and E. M. A. Hilal, ELzaki and Sumudu Transform for Solving some Differential Equations, Glob. J. of Pure & Appl. Math., 8 (2012), 167-173. [6] Kh. Jung and Hj. Kim, The practical formulas for differentiation of integral transforms, Int. J. of Math. Anal., 8 (2014), 471-480.
7534 Hwajoon Kim [7] Hj. Kim and Tarig M. Elzaki, The Representation on Solutions of Burger s Equation by Laplace Transform, Int. J. of Math. Anal., 8 (2014), 1543-1548. [8] Hj. Kim, The time shifting theorem and the convolution for Elzaki transform, Int. J. of Pure & Appl. Math. 87 (2013), 261-271. [9] Hj. Kim, The solution of Euler-Cauchy equation expressed by differential operator using Laplace transform, Int. J. of Pure & Appl. Math., 84 (2013). [10] Hj. Kim, A note on the shifting theorems for the Elzaki transform, Int. J. of Math. Anal., 8 (2014), 481-488. [11] Th. Lee and Hj. Kim, The representation of energy equation by Laplace transform, Appl. math. Sci 8 (2014), 1093-1097. [12] Sb. Nam and Hj. Kim, The Representation on Solutions of the Sine- Gordon and Klein-Gordon Equations by Laplace Transform, Appl. Math. Sci. 8 (2014), 4433-4440. [13] Yc. Song and Hj. Kim, Legendre s equation expressed by the initial value by using integral transforms, Appl. Math. Sci., 8 (2014), 531-540. [14] Yc. Song and Hj. Kim, The solution of Volterra integral equation of the second kind by using the Elzaki transform, Appl. Math. Sci., 8 (2014), 525-530. Received: September 15, 2014; October 28, 2014