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Solve the following 3-D wave equation using Fourier transform $$PDE: u_{tt}=C^2[u_{xx}+u_{yy}+u_{zz}],\qquad-\infty<x,y,z<\infty,\qquad t>0$$ $$BC: u(x,y,z,t)\rightarrow 0\qquad as \qquad r^2=x^2+y^2+z^2\rightarrow \infty \qquad $$ $$IC: u(x,y,z,0)=f(r),\qquad r=\sqrt{x^2+y^2+z^2} , \qquad u_t(x,y,z,0)=0 $$

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Let me rewrite your problem as this \begin{align*} PDE &:\frac{\partial^2}{\partial t^2}u=c^2\nabla^2u,\qquad\nabla^2\equiv\frac{\partial^2}{\partial x_1^2}+\frac{}{}\frac{\partial^2}{\partial x_2^2}+\frac{\partial^2}{\partial x_3^2}\\ BC &: u\rightarrow 0~\text{as}~ x^2+y^2+z^2\rightarrow \infty\\ IC &: u(x,t)=u_0(r)\\ u_t(x,t)=0 \end{align*} Solution:

We apply Fourier transform to each space dimension by letting \begin{align*} U(\lambda,t)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(x,t)e^{i(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3)}dx_1dx_2dx_3\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(x,t)e^{i\lambda x}dx \end{align*} It is understood that $$ dx=dx_1dx_2dx_3, x=(x_1,x_2,x_3), \lambda=(\lambda_1,\lambda_2,\lambda_3)$$

If we take the 3-D transform of the PDE we will get $$\frac{\partial^2}{\partial t^2}U=-c^2\lambda^2U,$$ where $\lambda^2=\lambda_1^2 +\lambda_2^2 +\lambda_3^2$. The solution to the ODE is $$ U(\lambda,t)=A(\lambda)\cos{c\lambda t}+B(\lambda)\sin{c\lambda t} $$ Applying IC, we find $B(\lambda)=0$ and $A(\lambda)=U(\lambda,0)$, where \begin{align*} U(\lambda,0)&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u_0(r)e^{i\lambda x}dx\\ &=\int_{0}^{\infty}dr\int_{0}^{\pi}r^2\sin{\theta}d\theta\int_{0}^{2\pi}d\phi u_0(r)e^{i\lambda r\cos{\theta}} \end{align*} in spherical coordinates. [We have oriented the coordinate system so that $\theta$ is the angle the vector $x$ makes relative to a fixed vector $\lambda$.]

\begin{align*} U(\lambda,0)&=2\pi\int_{0}^{\infty}dr r^2u_0(r)\int_{0}^{\pi}d(-\cos{\theta})e^{i\lambda r\cos{\theta}}\\ &=2\pi\int_{0}^{\infty}dr r^2u_0(r)\biggl(e^{i\lambda r\cos{\theta}}/(-i\lambda r)\biggl)\biggl|_{0}^{\pi}\\ &=4\pi \int_{0}^{\infty}u_0(r)\frac{\sin{\lambda r}}{\lambda}dr \equiv U_0(\lambda), \end{align*} which is a function of $\lambda$ only. Thus $$U(\lambda,t)=U_0(\lambda)\cos{c\lambda t}. $$ The inverse Fourier transform is \begin{align*} u(x,t)&=\frac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}U_0(\lambda)\cos{c\lambda t}e^{-i\lambda x}d\lambda\\ &=\frac{1}{(2\pi)^3}\int_{0}^{\infty}d\lambda\int_{0}^{2\pi}2\pi \sin{\theta}\lambda^2 U_0(\lambda)\cos{c\lambda t}e^{-i\lambda r\cos{theta}}d\theta\\ &=\frac{2}{(2\pi)^2}\int_{0}^{\infty}\lambda d\lambda U_0(\lambda)\cos{c\lambda t}\sin{\lambda r}/r, \end{align*} Since $\sin{\lambda r}\cos{c\lambda t}=\frac{1}{2}[ \sin{\lambda(r-ct)}+\sin{\lambda(r+ct)}]$ and

\begin{align*} u(x,0)&=u_0(r)=\frac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}U_0(\lambda)e^{-i\lambda x}d\lambda\\ &=\frac{2}{(2\pi)^2}\int_{0}^{\infty}\lambda d\lambda U_0(\lambda)\sin{\lambda r}/r, \end{align*} we have \begin{align*} ru(x,t)&=\frac{2}{(2\pi)^2}\int_{0}^{\infty}\lambda d\lambda U_0(\lambda)\sin{\lambda (r-ct)} +\frac{2}{(2\pi)^2}\int_{0}^{\infty}\lambda d\lambda U_0(\lambda)\sin{\lambda (r+ct)}\\ &=\frac{1}{2}(r-ct)u_0(r-ct)+\frac{1}{2}(r+ct)u_0(r+ct). \end{align*}

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Just take 3 dimensional Fourier transform on the equation, and then it will be solved. Let $\hat{u}$ be the 3-D Fourier transform of $u$ and the variables $x,y,z$ will transform to $s_1,s_2,s_3$. Take Fourier on the original equation, we have $$ \hat{u}_{tt}=-C^2(s_1^2+s_2^2+s_3^2)\hat{u}\\ \hat{u}(s_1,s_2,s_3,0)=\hat{f}\\ \hat{u}_t(s_1,s_2,s_3,0)=0 $$ This is a 2nd order ODE. It can be solved by classical methods (without lose of generality we assume $C\ge0$): $$ \hat{u}=\hat{f}\cos(Ct\sqrt{s_1^2+s_2^2+s_3^2}) $$ Then, take inverse Fourier transform on it, you can obtain the solution of original PDE. I omit it here because it is trivial.

PS: Note that when you take inverse Fourier transform, the zero boundary condition will be used.

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