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I want to solve the following wave equation: $$ \frac {\partial^2}{\partial t^2}u=c^{2}\frac {\partial^2}{\partial x^2}u,\quad -\infty<x<\infty,\text{ }0<t<\infty $$ subject to the initial condition $u(x,0)=0$ and $\partial u/\partial t=\delta(x)$ at $t=0$, and the boundary conditions $u(x,t)\to 0$ as $x\to \pm \infty$.

What I've done is:

Denote the Fourier transform with respect to $x$ by $\mathcal {F}_{x}$. The we get $$ \frac {\partial^2}{\partial t^2}U(k,t)+c^{2}k^{2}U(k,t)=0, $$ where $U(k,t)=\mathcal {F}_{x}\left[u(x,t)\right]$. The solution to the ODE for $U$ is given by $$ U(k,t)=A(k)\cos (ckt)+B(k)\sin (ckt). $$ The initial condition suggests that $A(k)=U(k,0)=0$ and $ckB(k)=U_{t}(k,0)=\mathcal {F}_{x}\left[\delta(x)\right]=1$. Therefore, we get $$ U(k,t)=\frac {1}{ck}\sin (ckt). $$ The inverse Fourier transform of $U(k,t)$ gives $$ u(x,t)=\frac {1}{2\pi}\int_{-\infty}^{\infty}\frac {1}{ck}\sin (ckt)e^{-ikx}\, dk=\frac {1}{2\pi}\int_{-\infty}^{\infty}\frac {1}{i2ck}e^{-ik(x-ct)}\, dk-\frac {1}{2\pi}\int_{-\infty}^{\infty}\frac {1}{i2ck}e^{-ik(x+ct)}dk $$ I wonder how to go from here. It seems like these two integrals depend on the sign of $x-ct$ and $x+ct$.

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1 Answer 1

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The integrals do depend on the signs of $x\pm ct$. We exploit the symmetry to write

$$\begin{align} u(x,t)&=\frac1{4\pi c}\int_{-\infty}^\infty \frac{\sin(k(x+ct))}{k}\,dk-\frac1{4\pi c}\int_{-\infty}^\infty \frac{\sin(k(x-ct))}{k}\,dk\\\\ &=\frac1{4 c}\left( \text{sgn}(x+ct)-\text{sgn}(x-ct)\right)\\\\ &=\begin{cases}\frac1{2c}&,|x|\le ct\\\\0&,\text{elsewhere}\end{cases} \end{align}$$

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  • $\begingroup$ Should it be $1/2c$ for $|x|\leq ct$? $\endgroup$ Commented Nov 12, 2022 at 19:32
  • $\begingroup$ Okay, I feel like when $-ct<x<ct$, $x-ct<0$, which means that sgn$(x-ct)=-1$ instead of $0$. $\endgroup$ Commented Nov 12, 2022 at 20:16
  • $\begingroup$ Indeed. Good catch. I've edited accordingly. $\endgroup$
    – Mark Viola
    Commented Nov 12, 2022 at 20:20

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