# Laplace transform in solving 2d wave equation

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I have the following wave equation $\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=\dfrac{1}{c^2}\dfrac{\partial^2 u}{\partial t^2}$ with boundary conditions at $x=0,\ u=\sin(\omega t )\sin(n\pi y)$ at $x=l,\ u=0$ and at $y=0,\ u=0;\ y=1, u=0$. I derived results taking Laplace transform in time $\left(\dfrac{\partial^2 U}{\partial x^2}+\dfrac{\partial^2 U}{\partial y^2}=\dfrac{s^2 U}{c^2}\right)$, and it has the solution $\dfrac{\sin(k(l-x))}{\sin(kl)}\sin(\omega t)\sin(n\pi y)+\mbox{transients}$. The result doesn't seem to give traveling wave solution but a standing wave solution in $x$ and $y$.

I'm wondering if I should take a double Laplace transform in $x$ and $t$. Can someone please let me know if that approach will help? Also, will I be able to capture traveling waves, say before it reaches the boundary at $x=l$?

PS: $U$ is Laplace transform of $u$ in $t$. The initial conditions are 0.

This post imported from StackExchange Mathematics at 2014-06-09 19:14 (UCT), posted by SE-user vijay
If you prescribe initial conditions then there is only one solution, you cannot expect to just arbitrarily get "travelling waves". If you don't prescribe initial conditions, then observe that if $u$ is a solution to the boundary value problem and $v$ solves $v_{xx} + v_{yy} = v_{tt}/c^2$ with zero boundary conditions, then $u+v$ also solves the boundary value problem. So you can play with the frequencies of $v$ to get a "traveling wave".

This post imported from StackExchange Mathematics at 2014-06-09 19:14 (UCT), posted by SE-user Willie Wong

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The best method should be using separation of variables rather than using Laplace transform.

The general solution is of the form $u(x,y,t)=\int_0^\infty\int_0^\infty C_1(r,s)\sin xr\sin ys\sin(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_2(r,s)\sin xr\sin ys\cos(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_3(r,s)\sin xr\cos ys\sin(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_4(r,s)\sin xr\cos ys\cos(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_5(r,s)\cos xr\sin ys\sin(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_6(r,s)\cos xr\sin ys\cos(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_7(r,s)\cos xr\cos ys\sin(ct\sqrt{r^2+s^2})~dr~ds+\int_0^\infty\int_0^\infty C_8(r,s)\cos xr\cos ys\cos(ct\sqrt{r^2+s^2})~dr~ds$

Now substitute the conditions $u(0,y,t)=\sin n\pi y\sin\omega t$ , $u(l,y,t)=0$ , $u(x,0,t)=0$ and $u(x,1,t)=0$ for eliminating some of the arbitrary functions.

This post imported from StackExchange Mathematics at 2014-06-09 19:14 (UCT), posted by SE-user doraemonpaul
answered Dec 22, 2012 by (20 points)

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