Given the 2D wave equation in polar coordinates:
$$u\_{\rho\rho}+\dfrac{1}{\rho}u\_{\rho}+\dfrac{1}{\rho^2}u\_{\theta\theta}=\dfrac{1}{a^2}u\_{tt}$$
with $u=u(\rho,\theta,t),(\rho,\theta,t)\in [0,c]\times(-\pi,+\pi]\times [0,+\infty)$
and boundary conditions:
$$u(0,\theta,t)=\nu(t),\forall \theta\in(-\pi,+\pi], t\in [0,+\infty)$$
$$u(\rho,\theta,0)=0,\forall \rho\in[0,c).\forall t\in[0,+\infty)$$
$$u\_t(\rho,\theta,0)=0,\forall\rho\in[0,c],\forall\theta\in(-\pi,+\pi]$$
where $\nu(t)$ is a gaussian random noise
$\nu(t)\sim \mathcal{N}(\mu,\,\sigma^{2})\,.$
If the noise is in the center of the disk,the term $u\_{\theta\theta}$ vanishes. Is it possible, and how, to calculate $u(\rho,t)$ ?



2D wave equation with gaussian boundary condition
You have then
$
\rho^2u\_{\rho\rho}+\rho u\_{\rho}=\rho^2 a^{-2} u\_{tt}.
$
Looking for a solution of the form $\phi(t) w(\rho)$, we obtain
$$
\bigl(\rho^2w''(\rho)+\rho w'(\rho)\bigr)\phi(t)=\rho^2 a^{-2} \phi''(t)w(\rho).
$$
If $\phi''(t)=-\lambda^2\phi(t)$, i.e. $\phi(t)= \alpha e^{i\lambda t}+\beta e^{-i\lambda t}$,
we get
$$
\rho^2w''(\rho)+\rho w'(\rho)+\rho^2a^{-2}\lambda^{2} w(\rho)=0.
$$
By rescaling or assuming for simplicity that $a=\lambda$, we recognize a Bessel equation with parameter 0: the Bessel equation with parameter $\nu$ is 
$$
\rho^2w''(\rho)+\rho w'(\rho)+(\rho^2-\nu^2) w(\rho)=0,
$$
is an ODE for which $0$ is a regular singular point. We have two independent solutions of the Bessel equation with parameter $\nu$, the Bessel function of the first and second kind, classical special functions. It is possible to do the same analysis even if the term $u\_{\theta\theta}$ does not vanish by Fourier expansion and this explains the importance of Bessel functions for the wave equation. As pointed out by @AHusain, the above answer is useless without implementing the boundary condition for all $t\ge 0$ at $\rho=0$:The Bessel function $J\_0$ is an entire function such that $J\_0(0)=1$ (whereas $Y\_0$ has a singularity at $0$) and writing the Fourier transform of $\nu$ we get
$$
\nu(t)=\int\_{\mathbb R} e^{2πi\lambda t}\hat\nu(\lambda) d\lambda,\quad\text{where $\hat \nu$ is also a Gaussian function.}
$$
Since the equation is linear, we may superpose solutions and we get to solve for each $\lambda$
$$
\rho^2w\_\lambda''(\rho)+\rho w\_\lambda'(\rho)+\rho^2a^{-2}4π^2\lambda^{2} w\_\lambda(\rho)=0,\quad w\_\lambda(0)=1,
\tag{$\ast$}
$$
and then to obtain a solution given formally by 
$$
u(\rho, t)=\int\_{\mathbb R} e^{2πi\lambda t}w\_\lambda(\rho)\hat\nu(\lambda) d\lambda.
$$
Equation $(\ast)$ can be solved by rescaling: setting $w\_\lambda(\rho)=v\_\lambda(\alpha \rho)$,
we find
$$
\rho^2\alpha^2v\_\lambda''(\alpha\rho)+\rho \alpha v\_\lambda'(\alpha\rho)+\rho^2a^{-2}4π^2\lambda^{2} v\_\lambda(\alpha\rho)=0,\quad v\_\lambda(0)=1,
$$
so that with $r=\alpha \rho$,
$$
r^2v\_\lambda''(r)+rv\_\lambda'(r)+r^2\alpha^{-2}a^{-2}4π^2\lambda^{2} v\_\lambda(r)=0,\quad v\_\lambda(0)=1.
$$
Choosing $\alpha= a^{-1}2π \lambda$ such that $1=\alpha^{-2}a^{-2}4π^2\lambda^{2}, $
we get $$
v\_\lambda (r)= J\_0(r).
$$


