MHB Fourier transform to solve the wave equation

[Markov2]

I need to use the Fourier transform to solve the wave equation:

$\begin{aligned} & {{u}_{tt}}={{c}^{2}}{{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0, \\
& u(x,0)=f(x), \\
& {{u}_{t}}(x,0)=g(x).
\end{aligned}
$

So I have $\dfrac{{{\partial }^{2}}F(u)}{\partial {{t}^{2}}}=-{{c}^{2}}{{w}^{2}}F(u)$ which gives $F(u(x,w))(t)=c_1\cos(wct)+c_2\sin(wct)$ and $F(u(x,0))=F(f )$ (1) and $\dfrac{\partial F(u(x,0))}{\partial t}=g(x)$ (2). So by using (1) I get $F(u(x,0))=c_1=F( f)$ and $F_t(u(x,0))=c_2cw=g(x)$ so $F(u(x,w))=F( f)\cos(wct)+\dfrac{g(x)}{cw}\sin(wct).$

Well I want to know if I'm correct so far. After this, I'm having problems with the inverse! :(

 

[Sudharaka]

I need to use the Fourier transform to solve the wave equation:

$\begin{aligned} & {{u}_{tt}}={{c}^{2}}{{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0, \\
& u(x,0)=f(x), \\
& {{u}_{t}}(x,0)=g(x).
\end{aligned}
$

So I have $\dfrac{{{\partial }^{2}}F(u)}{\partial {{t}^{2}}}=-{{c}^{2}}{{w}^{2}}F(u)$ which gives $F(u(x,w))(t)=c_1\cos(wct)+c_2\sin(wct)$ and $F(u(x,0))=F(f )$ (1) and $\dfrac{\partial F(u(x,0))}{\partial t}=g(x)$ (2). So by using (1) I get $F(u(x,0))=c_1=F( f)$ and $F_t(u(x,0))=c_2cw=g(x)$ so $F(u(x,w))=F( f)\cos(wct)+\dfrac{g(x)}{cw}\sin(wct).$

Well I want to know if I'm correct so far. After this, I'm having problems with the inverse! :(

Hi Markov, :)

I think you can find the method of solving the wave equation using the Fourier transform if you Google something like, "wave equation and Fourier series". You may find the answer to your question http://www.iam.ubc.ca/%7Esospedra/05-separation.pdf(Scroll down, and at the end the Fourier series method is given).

Kind Regards,
Sudharaka.