Fourier transform to solve the wave equation

Markov

Member
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

Well-known member
MHB Math Helper
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 here(Scroll down, and at the end the Fourier series method is given).

Kind Regards,
Sudharaka.