Helmholtz Equation Exterior of a Sphere with Dirichlet Boundary Conditions I tried to solve the Helmoltz Equation $\Delta \psi=\lambda\psi$ on the exterior of a sphere of radius $R$ in 3 dimensions with Dirichlet b.c.. I know that in the $\lambda=0$ case the problem is solved by
$$\psi(x, y)=\frac{1}{4\pi|y-x|}-\frac{1}{4\pi\frac{|x|}{R}|y-x^{*}|},$$
where $x^{*}=\frac{R^{2}x}{|x|^{2}}$ is the Kelvin transform of the point $x$ ($x$, $y\in\mathbb{R}^3$).
Does the same procedure holds to solve the general $\lambda$ case? And so is
$$\psi(x, y)=\frac{e^{-\sqrt{\lambda}|y-x|}}{4\pi|y-x|}-\frac{e^{-\sqrt{\lambda}\frac{|x|}{R}|y-x^{*}|}}{4\pi\frac{|x|}{R}|y-x^{*}|}$$
the solution for $\lambda\in\mathbb{R}$?
 A: So basically, we would like to solve the BVP
$$(\nabla^2+k^2)u=0\\u(\boldsymbol r)=0~~\text{when}~~|\boldsymbol r|=R$$
The $k^2$ is a matter of convenience. If $\lambda$ in your equation is actually a positive real number, we can take $k$ to be purely imaginary. Keep in mind that the equation is not fully determined, since this is a second order equation for which we have only supplied one boundary condition.
I've taken the position of the charged particle to be at the origin. We can just use a coordinate shift later to be able to put the charged object at any position we want. Based on how I'm interpreting your question, I think we can go ahead and assume that $u$ has some nice symmetry, i.e, the angular dependence vanishes, $\partial_\theta u=\partial_\phi u=0$. In that case, the Laplacian takes the form
$$\nabla^2u=\frac{1}{r^2}\left(r^2\frac{\partial u}{\partial r}\right)=\frac{2}{r}\partial_ru+\partial_r^2u$$
So now instead of a PDE, we actually have an ODE
$$u''(r)+\frac{2}{r}u'(r)+k^2u(r)=0$$
So... how to solve this? Well, out of the blue, we can try an ansatz
$$u(r)=v(r)/r$$
If you plug this in for $u$ and go through the calculations, you will find that it simplifies to
$$\frac{v''(r)+k^2v(r)}{r}=0$$
Assuming $r\neq 0$ we can multiply both sides and solve for $v$ to get
$$v(r)=c_1\sin(kr)+c_2\cos(kr)$$
And thus
$$u(r)=c_1\frac{\sin(kr)}{r}+c_2\frac{\cos(kr)}{r}$$
Our boundary condition reads
$$c_1\sin(kR)+c_2\cos(kR)=0\implies c_2=-c_1\tan(kR)$$
Hence
$$u(r)=c_1\left(\frac{\sin(kr)}{r}-\tan(kR)\frac{\cos(kr)}{r}\right)$$
We can now shift this equation by some vector $\boldsymbol{r}_0$, to get
$$u(\boldsymbol{r})=c_1\left(\frac{\sin(k|\boldsymbol{r}-\boldsymbol{r}_0|)}{|\boldsymbol{r}-\boldsymbol{r}_0|}-\tan(kR)\frac{\cos(k|\boldsymbol{r}-\boldsymbol{r}_0|)}{|\boldsymbol{r}-\boldsymbol{r}_0|}\right)$$
Maybe this helps?
