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