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Consider this nonhomogenous wave equation: $$ \begin{cases}u_{tt}-u_{xx}=\sin(x)\\u(x,0)=\sin(x) \\ u_{t}(x,0)=0\end{cases}$$

Applying d'Alembert's formula and Duhamel's principle, one can easy show that $u(x,t)=\sin(x)$. My problem with this equation is that I must solve it using Fourier transform (as defined here), because applying FT on equation yields $$\hat u_{tt}(k,t)+4\pi^{2}k^2\hat u(k,t)= (\delta_{1/2\pi}-\delta_{-1/2\pi})/2$$ How to solve last equation? I've got problem finding particular solution.

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  • $\begingroup$ Hint set $u(t, x) = f(t)g(x).$ Because $\sin x$ is not a good function you cant' therefore its Fourier Transform in the usual sense except in the distributional sense. $\endgroup$ – Guy Fsone Aug 25 '17 at 12:47
  • $\begingroup$ @guyfaone What's wrong with using the FT of tempered distributions (i.e., on the space of Schwartz functions)? $\endgroup$ – Mark Viola Aug 25 '17 at 17:20
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$$\underbrace{\int_{-\infty}^\infty (u_{tt}(x,t)-u_{xx}(x,t))\,e^{i2\pi kx}\,dx}_{U_{tt}(k,t)+(2\pi k)^2U(k,t)}=\underbrace{\int_{-\infty}^\infty \sin(x)\,e^{i2\pi kx}\,dx}_{\frac1{2i}\,\frac1{2\pi}(\delta(2\pi k+1)-\delta(2\pi k-1))}$$

The general solution to the ODE

$$U_{tt}(k,t)+(2\pi k)^2U(k,t)=\frac1{2i}(\delta( k+1/2\pi)-\delta( k-1/2\pi))$$

is given by

$$U(k,t)=Ae^{2\pi k t}+Be^{-2\pi k t}+\frac{1}{(2\pi k)^2}\,\left(\frac1{2i}(\delta( k+1/2\pi)-\delta( k-1/2\pi) \right)$$

Given the initial conditions $U(k,0)=\frac1{2i}(\delta( k+1/2\pi)-\delta( k-1/2\pi))$ and $U_t(k,0)=0$, we find that $A=B=0$.

Taking the inverse Fourier transform yields

$$\begin{align} u(x,t)&=\int_{-\infty}^\infty U(k,t)e^{-i2\pi kx}\,dk\\\\ &=\frac{1}{2i(2\pi)^2}\int_{-\infty}^\infty \frac{\delta( k+1/2\pi)-\delta( k-1/2\pi)}{k^2} \,e^{-i2\pi kx}\,dk\\\\ &=\frac{1}{2i(2\pi)^2}\left((2\pi)^2(e^{ix}-e^{-ix}) \right)\\\\ &=\sin(x) \end{align}$$

as expected!

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