Green's function for Helmholtz equation for the plane with a hole That is find $G$ which satisfies 
\begin{align}
(\nabla^2+k^2)G(\mathbf{x}, \mathbf{y},\omega) = \delta(\mathbf{x}- \mathbf{y})
\end{align}
subject to $$\frac{\partial G}{\partial y_n} = 0 ~~~~~\mathrm{on} ~~~~~~S$$ where $S$ is the unit circle centered on the origin. The domain should be taken to mean the exterior of $S$. This means the normal derivative is zero on the boundary. I use the notation with $\omega$ because it is implied that the forcing is at frequency $\omega$. This follows from the wave equation. Note that I would ideally like the 2D version of the above, that is $\mathbf{x} = (x_1,x_2)$.
 A: The Green's Function $G(\vec \rho,\vec \rho')$ that satisfies the $2$-D Helmholtz equation 
$$\nabla^2G(\vec \rho,\vec \rho')+k^2G(\vec \rho,\vec \rho')=\frac{\delta(\rho-\rho')\delta(\phi-\phi')}{\rho'} $$
inside a circle of radius $1$ and boundary condition 
$$\left.\frac{\partial G}{\partial \rho}\right|_{\rho =1}=0 \tag 1$$  
can be decomposed into the sum of the free-boundary Green's Function and a homogeneous solution.  We can write, therefore, 
$$G(\vec \rho,\vec \rho')=-\frac{i}{4}H_0^{(1)}(k|\vec \rho-\vec \rho'|)+\frac{i}{4}\sum_{n=-\infty}^{\infty}A_n(k,\rho')J_n(k\rho)e^{in(\phi-\phi')} \tag 2$$
where the free-boundary Green's Function is given in terms of the Hankel Function of the first kind and zero order, $H_0^{(1)}$, and the homogeneous term is expressed in terms of the standing cylindrical wave expansion in which $A_n$ is an expansion coefficient that we will find by enforcing the boundary condition $(1)$.
We will make use of the cylindrical wave expansion for the free-boundary Green's function $-\frac{i}{4}H_0^{(1)}(k|\vec \rho-\vec \rho'|)$
$$-\frac{i}{4}H_0^{(1)}(k|\vec \rho-\vec \rho'|)=-\frac{i}{4}\sum_{n=-\infty}^{\infty} J_n(k\rho_{<})H_n^{(1)}(k\rho_{>})e^{in(\phi-\phi')} \tag 3$$
where $\rho_{<}=\min(\rho,\rho')$ and $\rho_{>}=\max(\rho,\rho')$.  Substitution of $(3)$ into $(2)$ yields
$$G(\vec \rho,\vec \rho')=\frac{i}{4}\sum_{n=-\infty}^{\infty}\left(A_nJ_n(k\rho)-J_n(k\rho_{<})H_n^{(1)}(k\rho_{>})\right)e^{in(\phi-\phi')} \tag 4$$
Now, enforcing the boundary condition $(1)$ on $(4)$, we find the coefficient $A_n$ can be expressed as
$$A_n=\frac{H_n^{(1)'}(k)J_n(k\rho')}{J_n'(k)}$$
Finally, we have the Green's Function
$$\bbox[5px,border:2px solid #C0A000]{G(\vec \rho,\vec \rho')=\frac{i}{4}\sum_{n=-\infty}^{\infty}\left(\frac{H_n^{(1)'}(k)}{J_n'(k)}J_n(k\rho')J_n(k\rho)-J_n(k\rho_{<})H_n^{(1)}(k\rho_{>})\right)e^{in(\phi-\phi')} }$$
expressed in terms of a cylindrical wave expansion.
We can see immediately that the solution to the exterior problem is then
$$\bbox[5px,border:2px solid #C0A000]{G(\vec \rho,\vec \rho')=\frac{i}{4}\sum_{n=-\infty}^{\infty}\left(\frac{J_n'(k)}{H_n^{(1)'}(k)}H_n^{(1)}(k\rho')H_n^{(1)}(k\rho)-J_n(k\rho_{<})H_n^{(1)}(k\rho_{>})\right)e^{in(\phi-\phi')} }$$
