# Solving the wave-equation using a Fourier-transformation

I am trying to solve the following partial differential equation using a Fourier-transformation.

\begin{align} u_{tt}&=u_{xx} &t>0, x>0, \\ u_x(0,t) &= 0 &t \geq 0, \\ u(x,0) &= 0 &x>0,\\ u_t(x,0) &= g(x) &x>0.\end{align}

By using $U(\omega,t) = \mathscr{F}\{ u(x,t)\}$ and $G(\omega) = \mathscr{F}\{g(x)\}.$ We can transform the PDE into the following system of Fourier Transformations: \begin{align} U_{tt}(\omega,t)+\omega^2U(\omega,t)&=0, \\ -i\omega \cdot U(\omega) &= 0, \\ U(\omega,0) &= 0, \\ U_t(\omega,0) &= G(\omega). \end{align}

Then, we can see from the first equation that $U(\omega,t) = A(\omega)\sin(\omega t)+B(\omega)\cos(\omega t).$ This is because the first equation is actually an ODE. Then I want to use initial conditions to find the functies of $A$ and $B$. Using the Dirichlet condition: \begin{align} U(\omega,0) = A(\omega)\cdot0+B(\omega)\cdot1=0 \Rightarrow B(\omega) = 0. \end{align} Then by using the Neumann condition: \begin{align} U_t(\omega,0) = \omega \cdot A(\omega)\cos(\omega \cdot 0) =\omega \cdot A(\omega) = G(\omega). \end{align} This is where I'm stuck. Ideally, I'd like to divide both sides by $\omega$ as to get an expression for $A$. However, then I have to take into account that $\omega$ can be equal to $0$. I don't know what to do with that. Can anyone give me a nudge in the right direction?

Don't worry about the $0$. You have already found that $$U(\omega,t) = G(\omega)\frac{\sin(\omega t)}{\omega}=G(\omega)\cdot t\operatorname{sinc}(\omega t)$$ (noting that $t>0$).
Now use the convolution theorem and the fact that $\tau \operatorname{sinc}(\frac{\omega \tau}{2})$ is the Fourier transform of $\operatorname{rect}(\frac{x}{\tau})$ (the rectangular function) using $\tau=2t$.
• Why don't I have to worry about the $0$? – K.Kamal Jul 1 '18 at 16:30
• @K.Kamal Because $\operatorname{sinc}(\omega t)$ is defined for all $\omega\neq 0$ and has a well-defined inverse Fourier transform. Meaning that you can just ignore the $0$, as you don't need that to retrieve its inverse Fourier transform. – The Phenotype Jul 1 '18 at 17:10