Solving the 3D Nonhomogeneous Wave Equation

Question

I am attempting to solve the following differential equation: $$c^2 \nabla^2 u(\vec{r}, t)-\ddot{u}(\vec{r}, t)=f(\vec{r}, t)$$ ... where the forcing function $f(\vec{r}, t)= R_f(\vec{r})T_f(t)$ is known. My attempts have led me towards the Helmholtz equation, but I'm not sure what to do with it (or if it even is actually applicable in this scenario) given the presence of forcing.

If the DE was homogeneous, the strategy would be to take the Fourier transform of both sides, which would result in the second time derivative being replaced like so: $$c^2 \nabla^2 U(\vec{r}, \omega)+{\omega}^2U(\vec{r}, \omega)=0$$ ... in which case we rearrange to obtain the Helmholtz equation with $k=\omega/c$: $$\nabla^2 U = -k^2U$$ This can then be solved using the methods detailed on https://en.wikipedia.org/wiki/Helmholtz_equation.

My next thought after this was to try the same procedure on the nonhomogeneous form, i.e. take the Fourier Transform of both sides, netting us: $$c^2 \nabla^2 U(\vec{r}, \omega)+{\omega}^2U(\vec{r}, \omega)=R_f(\vec{r})\:\mathcal{F}\{T_f(t)\}(\omega)$$ I would then try the solution detailed in the same article (here's the specific section to look at), but our RHS's continued dependence on $\omega$ throws a wrench in those plans. Intuition says this is still the right direction to go in, but I'm not sure how to move forward from here, and I'd greatly appreciate any help towards the goal of solving this. Thank you in advance!

Answer 1

The Green (or Green's) function, $G(t,\vec r|t',\vec r')$, for the wave equation $\nabla^2 G-\frac1{c^2}\frac{\partial^2}{\partial t^2}G=\delta(t-t')\delta(\vec r-\vec r')$ is given by

$$G(t,\vec r|t',\vec r')=\frac{\delta(t-t'-|\vec r-\vec r'|/c)}{4\pi |\vec r-\vec r'|}$$

Therefore, the solution $\psi(\vec r,t)$ to the inhomogeneous wave equation with forcing function $R_f(\vec r)T_f(t)$ is

$$\begin{align} \psi(\vec r,t)&=\int_{-\infty}^\infty \int_{\mathbb{R}} R_f(\vec r')T_f(t')G(t,\vec r|t',\vec r')\,dt'\,d^3\vec r'\\\\ &=\int_{-\infty}^\infty \int_{\mathbb{R}} R_f(\vec r')T_f(t')\frac{\delta(t-t'-|\vec r-\vec r'|/c)}{4\pi |\vec r-\vec r'|}\,dt'\,d^3\vec r'\\\\ &=\int_{\mathbb{R}} R_f(\vec r')T_f(t-|\vec r-\vec r'|/c)\frac{1}{4\pi |\vec r-\vec r'|}\,d^3\vec r' \end{align}$$

Finally, $u=c^2\psi$.

Answer 2

If you instead take the Fourier transform in space, you get the second-order ODE

$$-4\pi c^2|\xi|^2\hat{u}(\xi,t)^2-\frac{d^2}{dt^2}\hat{u}(\xi,t) = \hat{f}(\xi,t)$$ Which you can solve using your preferred method for solving non-homogeneous second-order ODEs. If $u_0 = u(x,0), u_1 = \partial_t u(x,0)$, in the end you should get (setting all constants to $1$ for simplicity)

$$\hat{u}(\xi,t) = \cos(|\xi|t)\hat{u}_0 + \frac{\sin(|\xi|t)}{|\xi|}\hat{u}_1 - \int_0^t\frac{\sin(|\xi|(t-s))}{|\xi|}\hat{f}(s,\xi)ds$$

Taking the inverse Fourier transform of the distributions $\cos(|\xi|t)$ and $\sin(|\xi|t)|/|\xi|$ is generally nontrivial. But once you do this, you effectively recover a convolution formula for the solution, since $\mathcal{F}^{-1}(\hat{u}\hat{v}) = u*v$. In $\mathbb{R}^3$, this is called the Kirchoff formula:

$$u(x,t)=\frac{d}{dt}\left(\frac1t\int_{|x-y|=t}u_0(y)dS(y)\right) + \frac1t\int_{|x-y|= t}u_1(y)dS(y) -\frac1t\int_0^t\int_{|x-y|=t-s}f(s,y)dS(y)ds$$

Where $dS$ is the surface measure on the sphere (and again setting constants to 1).