So in three dimensions, we can find an explicit solution to the homogeneous wave equation $\partial_t^2 u - \Delta u=0$ that is a function of $r$, $t$ if our data $f(x)=u(x,0)$, $g(x)=\partial_tu(x,0)$ is radially symmetric. If we write $U(r,t)$ as the solution to the 1-dimensional wave equation \begin{equation} \partial_t^2 U-\partial_r^2U=0,\quad rf(r)=U(r,0),\quad rg(r)=\partial_tU(r,0), \end{equation} where we extend $f(-r)=f(r)$, $g(-r)=g(r)$, then we see that $u(r,t)=\frac{1}{r}U(r,t)$ is the solution to our 3D wave equation. $U$ can be written explicitly as a function of $r$, $t$, so therefore, $u$ also has an explicit representation.
I am wondering if we can do something similar with the 2D wave equation with radially symmetric initial conditions. Is there a similar trick I can use to write the solution as an explicit function of $r$, $t$?
The answer to this is a simple modificaition of the 3D case. If we write $U(r,t)$ as the solution to the 1-dimensional wave equation \begin{equation} \partial_t^2 U-\partial_r^2U=0,\quad rf(2r)=U(r,0),\quad 2rg(2r)=\partial_tU(r,0), \end{equation} where we extend $f(-r)=f(r)$, $g(-r)=g(r)$, then we see that $U(r,t)=ru(2r,2t)$. Therefore, $u(r,t)=2\frac{U(r/2,t/2)}{r}$.