# Wave equation

1. Jul 29, 2008

### pk415

We are supposed to work this using Laplace transforms

$$U_{tt}=9U_{xx}; -infty<x<infty$$

$$U(x,0)=sinx$$

$$U_t(x,0)=0$$

The attempt at a solution

$$Let L=\hat{U}$$

$$L[U_{tt}]=s^2\hat{U}-s(sinx)$$

$$L[9U_{xx}]=9\hat{U}_{xx}$$

$$s^2\hat{U}-s(sinx)=9\hat{U}_{xx}$$

$$\hat{U}_{xx}-\frac{s^2}{9}\hat{U}=-\frac{s}{9}sinx$$

This is a second order nonhomogeneous equation where the homogeneous solution is

$$\hat{U}_h=A(s)e^{\frac{s}{3}x}+B(s)e^{-\frac{s}{3}x}$$

and the particular solution is

$$\hat{U}_p=\frac{s}{s^2+9}sinx$$

Then we have

$$\hat{U}=A(s)e^{\frac{s}{3}x}+B(s)e^{-\frac{s}{3}x}+\frac{s}{s^2+9}sinx$$

We know that

$$\lim_{x\rightarrow\infty}U=0$$

$$\lim_{x\rightarrow{-\infty}}U=0$$

That second one should say limit as x approaches negative infinity, cant figure out how to get latex to do that

$$\lim_{x\rightarrow\infty}\hat{U}=0$$

$$\lim_{x\rightarrow\infty}\hat{U}=0$$

Second one negative infinity again

If we plug these in we get

$$A(s)=\frac{\lim_{x\rightarrow\infty}\frac{s}{s^2+9}sinx}{\lim_{x\rightarrow\infty}e^{-\frac{s}{3}x}}}$$

$$A(s)=0$$

I'm not sure if that last step is legal, but if it is you can use the same reasoning for B(s)
Any ideas?