Keywords: Burger's equation, viscous equation, Laplace transform. 1 Introduction ... tion y + p(x)y + q(x)y = 0 on I, and W is the Wronskian of y1, y2. Theorem 2.1 ...
Int. Journal of Math. Analysis, Vol. 8, 2014, no. 31, 1543 - 1548 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ijma.2014.46185
The Representation on Solutions of Burger’s Equation by Laplace Transform Hwajoon Kim Kyungdong University School of IT Engineering Goseong 219-705, Gangwon, Korea Tarig M. Elzaki King Abdulaziz University Department of Mathematics Jeddah 21589, Saudi Arabia c 2014 Hwajoon Kim and Tarig M. Elzaki. This is an open access article Copyright 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 Navier-Stokes equation, dropping the pressure term, we obtain Burger’s equation. The equation frequently occurs in modeling of gas dynamics and traffic flow, and it divides inviscid form and viscous one. In this article, we have checked the representation of Burger’s equation by Laplace transform.
Mathematics Subject Classification: 35L70, 35L20, 44A10 Keywords: Burger’s equation, viscous equation, Laplace transform
1
Introduction
Burger’s equation is an important partial differential equation from fluid mechanics, and it has a form of ut + u · ux = ν · uxx
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Hwajoon Kim and Tarig M. Elzaki
for a given velocity u and viscosity coefficient ν. We call it viscous Burger’s equation, and in case of ν = 0, we call it inviscid Burger’s equation. Normally, the solution of Burger’s equation can be obtained by Hopf-Cole transformation which transforms Burgers equation into a linear parabolic equation[4], and the numerical solution has been still pursued because of comparing it with the exact solution. In this article, we would like to approach the topic by using Laplace transform. Although the proposed method is somewhat restrictive, we have tried to come at it with different angle. Hence, this try has a meaning in the way that integral transform method is applied solving nonlinear differential equation. With relation to this topic, several researches have been pursued for integral transforms related to differential equations[3, 5-7, 10-16], and for nonlinear equation[2, 8-9]. Benton researched exact solutions of the one-dimensional Burger’s equation[1], Elzaki has proposed Elzaki homotopy perturbation method to find the solution of Burger’s equation[9], and Chen applied the increment linerarization technique(ILT) and Laplace transform to solve the equation[4]. Chen employed Laplace transform to control linear terms, and ILT to control nonlinear term. However, since this method is some complicated, we would like to propose the representation of Burger’s equation by Laplace transform only.
2
The representation on solutions of the Burger’s equation by Laplace transform
Would let us check the representation on solutions of the burger’s equation by Laplace transform. The tool used is a Lagrange’s method, and it gives a particular solution yp of y + p(x)y + q(x)y = r(x) on open interval I in the form y2 r y1 r dx + y2 dx (1) yp (x) = −y1 W W where y1 , y2 form a basis of solutions of the corresponding homogeneous equation y + p(x)y + q(x)y = 0 on I, and W is the Wronskian of y1 , y2 . Theorem 2.1 The solution u(x, t) of the Burger’s equation ut + u · ux = ν · uxx with u(x, 0) = u0 can be represented by s U(x, s) = A(s)eαx + B(s)eβx + u0 ν
(2)
The representation of the Burger’s equation by Laplace transform
1545
where u0 is a constant, u is a given velocity, and ν is viscosity coefficient. If u(x, 0) = f (x), the solution can be expressed by = eαx
1 β−α
e−αx f (x) dx − eβx
1 β−α
e−βx f (x) dx
where α and β are roots of the characteristic equation νλ2 − uλ − s = 0. Proof. Taking Laplace transform with respect to t on both sides and writing U(x, s) = £[u(x, t)], we have sU − u(x, 0) + u ·
∂2U ∂U =ν 2 ∂x ∂x
for u(x, t) considered as a function of x. Organizing this equality, we have ν
∂2U ∂U − sU = −u(x, 0). −u· 2 ∂x ∂x
(3)
Putting λ is the solutions of the characteristic equation νλ2 − uλ − s = 0, we have
u±
(4)
√
u2 + 4sν . 2ν Let us put the values of λ as α and β. Since the equation (3) contains only a derivative with respect to x, a general solution is λ=
U(x, s) = A(s)eαx + B(s)eβx + Up
(5)
for Up is a particular solution of the equation (3). Calculating the Wronskian W of eαx and eβx , we have W = (β − α)e(α+β)x . By Lagrange’s method (1), we have αx
Up = e
−eβx
eβx u0 dx (β − α)e(α+β)x
eαx u0 dx (β − α)e(α+β)x
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Hwajoon Kim and Tarig M. Elzaki
= eαx
u0 −αx 1 u0 −βx 1 e (− ) − eβx e (− ) β−α α β−α β u0 −u0 + = α(β − α) β(β − α)
= u0 [
1 −β + α s ] = −u0 = u0 αβ(β − α) αβ ν
because of the equation (4). Hence we have s U(x, s) = A(s)eαx + B(s)eβx + u0 . ν
(6)
In case of u(x, 0) = f (x), we have Up = eαx βx
−e = eαx
1 β−α
eβx f (x) dx (β − α)e(α+β)x
eαx f (x) dx (β − α)e(α+β)x
e−αx f (x) dx − eβx
1 β−α
e−βx f (x) dx
where α and β are roots of the equation (4). Since the equation (6) can be rewritten as Up =
u0 (α − β)ν 3 √ , (αν − u)(βν − u) u2 + 4sν
it is clear that Laplace transform of the inviscid Burger’s equation has a form of U(x, s) = A(s)eαx + B(s)eβx . In the above equality, we note that the Wronskian of eαx and eβx can be rewritten as √ u2 + 4sν u x eν . − ν Concomitantly, we can obtain the relations of £(u · ux ). Corollary 2.2 Let us u be considered as a function of x and t, and let ux is a function of x only. Then
£(u · ux ) = −u(x, 0)( ux dt)]t=0 − £(ut · 1 (b) £(u · ux ) = [£(u + ux )2 − £(u2 + ux 2 )] 2
(a)
ux dt) + £(us ·
ux dt)
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The representation of the Burger’s equation by Laplace transform
(c)
s£(u · ux ) = £(u t ux ) + £(uuxt ) + u(x, 0)ux (x, 0)
(d) £(u · ux ) = (e) £(u · ux ) =
e−st e−st ((ux )2 + uuxx ) dxdt e−st (ut ux + uuxt) dtdt.
Proof. (a)
∞
£(u · ux ) = = ue
−st
(
ux dt)]∞ 0
−
0 ∞
0
ue−st · ux dt −st
(ut − us )e
= −u(x, 0)( ux dt)]t=0 − £(ut · (b) Clear. (c) £(u · ux ) =
∞ 0
( ux dt) dt
ux dt) + £(us ·
ux dt).
uux · e−st dt,
and by the integration of parts and simple calculation, we obtain s£(u · ux ) = £(ut ux ) + £(uuxt) + u(x, 0)ux (x, 0).
(d) Since uux = [(ux )2 + uuxx ] dx, £(u · ux ) =
=
e−st
((ux )2 + uuxx ) dxdt
e−st ((ux )2 + uuxx) dxdt.
(e) In the way similar to (d), we have the result.
In corollary 2.2, note that ux dt = u = ux dx. For example, if u = xt2 , then ux = t2 , ux dt = (1/3)t3 but ux dx = t2 x.
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