Int. Journal of Math. Analysis, Vol. 8, 24, no. 22, 93-97 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/.2988/ijma.24.442 The Representation of Energy Equation by Laplace Transform Taehee Lee and Hwajoon Kim Kyungdong University Dept. of Cyber Security/ School of IT Engineering Goseong 29-75, Gangwon, Korea Corresponding author Copyright c 24 Taehee Lee and 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 Integral transform methods can be used for solving partial differential equations, and the vibrating/semi-infinite string is the typical model of wave equation. In this article, we have checked the representation of energy equation by Laplace transform. Mathematics Subject Classification: 35L5, 35L35, 44A Keywords: energy equation, Laplace transform Introduction It is a well-known fact that every sufficiently smooth solution of the wave equation satisfies the energy equation. Normally, d L L dt 2 [ρ u 2 t + T u 2 x ]dx =[T u x u t ] L + ρ qu t dx () is called the energy equation for the wave equation in the form ρ u tt = T u xx + ρ q(x, t). (2) The representation 2 [ρ u 2 t + T u 2 x ]dx
94 Taehee Lee and Hwajoon Kim is called the energy integral of the function u(x, t) on the interval [,L], and it is denoted by E(t). Since V (t) = 2 T u 2 x dx and K(t) = 2 ρ u 2 t dx, we can rewrite the energy integral E(t) as E(t) =K(t)+V (t), where V (t)/k(t) is the potential/kinetic energy respectively. This is the famous energy integral, and the energy equation () means that the rate of change of the total energy of the string is equal to the rate at which external forces do work on the string. On the other hand, the small plane transverse vibrations of a perfectly flexible string are expressed by u tt = c 2 u xx, <x<l, t>, (3) where the constant c is given by c = T /ρ, ρ is the linear density of the string, and the T is initial magnitude of a force. The displacement function satisfies the boundary conditions u(, t)=, u(l, t) =. To find the initial conditions, we consider the motion of the center of mass of a segment. To specify this motion, we must specify its initial position and initial velocity as u(x, ) = f(x), u t (x, ) = g(x). If distributed external forces act on the string, the equation (3) changes to (2) because of the total external force. Several researches have been pursued for integral transforms related to differential equations[, 3-4, 6-7, 9-2, 4-5], and for energy equation[2, 5, 8, 3, 6]. [8] has dealt with Lie symmetry analysis of pipe flow energy equation, [3] has proposed mathematical modeling of wind turbine in a wind energy conversion system, and [6] is checking the energy decay for a wave equation with nonlinear boundary memory and damping source. Among these researches, [2] and [5] are noticeable in the researches for the energy equation. In this article, we have checked the representation of the energy equation by Laplace transform.
The representation of the energy equation by Laplace transform 95 2 The representation of the energy equation by Laplace transform We have pursued the representation of differential equations by Laplace transform or Elzaki s[4,, 4]. In the following theorem, we would like to check the representation of the energy equation by Laplace transform. Theorem 2. The solution u(x, t) of the energy equation d L L dt 2 [ρ u 2 t + T u 2 x]dx =[T u x u t ] L + ρ qu t dx can be represented by [u(x, t)] = 2 A(s)sinh( s c x) cρ e s c x e s c x (sf(x)+g(x)+q) dx s e c x + cρ e s c x (sf(x)+g(x)+q) dx, (4) where Q(x, s) = [q(x, t)], h is a hyperbolic function, and a constant A. Proof. Clearly u x u xt = t ( 2 T u 2 x) and by integration of parts, T u xx u t dx =[T u x u t ] L T u x u tx dx. Hence the energy equation () can be rewritten as ρ u t u tt dx = T u xx u t dx + We divide the equation (5) by u t, obtaining ρ u tt dx = T u xx dx + Differentiating with respect to x, we obtain ρ u tt = T u xx + ρ q(x, t). ρ qu t dx. (5) ρ q dx. Let us write U(x, s) = [u(x, t)]. Then the above equation can be expressed by ρ [s 2 U sf(x) g(x)] = T 2 U x 2 + ρ Q(x, s)
96 Taehee Lee and Hwajoon Kim for Q[(x, s) = [q(x, t)]. Organizing the equality, we have ρ s 2 2 U U T x = ρ sf(x)+ρ 2 g(x)+ρ Q. (6) Since c 2 = T /ρ, a general solution U h (x, s) of the homogeneous ODE of (6) is U h (x, s) =A(s)e s c x + B(s)e s c x for constants A and B. Next, let us find the form of any particular solution U p (x, s) of (6). To begin with, let us find the Wronskian W of e s c x and e s c x. Clearly W =/c and so, U p (x, s) = e s c x c e s c x ρ (sf(x)+g(x)+q) dx +e s c x c e s c x ρ (sf(x)+g(x)+q) dx for Q[(x, s) = [q(x, t)]. From u(, t) =, U(, s) = [u(, t)] = () =. Hence U(,s)=A(s)+B(s) c ρ (sf() + g() + Q(,s)) dx + c ρ (sf() + g() + Q(,s)) dx = for constants A and B. This implies B(s) = A(s). Thus we have [u(x, t)] = A(s)e s c x A(s)e s c x cρ e s c x e s c x (sf(x)+g(x)+q) dx s e c x + cρ e s c x (sf(x)+g(x)+q) dx for Q(x, s) = [q(x, t)] and for a constant A. This is the same result as the equation (4). References [] H. A. Agwa, F. M. Ali and A. Kilicman, A new integral transform on time scales and its applications, Adv. Differ. eqs., 6 (22), -4. [2] Bradshaw, Po, D. H. Ferriss, and N. P. Atwell, Calculation of boundarylayer development using the turbulent energy equation, J. of Fluid Mech. 28 (967), 593-66. [3] Ig. Cho and Hj. Kim, The solution of Bessel s equation by using integral transforms, Appl. Math. Sci., 7 (23), 669-675.
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