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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
asked Dec 11, 2012 in Mathematics by Vijay (15 points) [ no revision ]
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

1 Answer

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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 doraemonpaul (20 points) [ no revision ]

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