Helmholtz-Dirichlet Green's Function for unit ball - Explicit formula? The formula for the Green's function for Laplace's equation on the unit ball with Dirichlet boundary conditions is well-known:
$$ 
G(x,y) = \frac{1}{4\pi}\left(\frac{1}{|x-y|} - \frac{1}{\Big|x|y|-\dfrac{y}{|y|}\Big|}\right)\label{1}\tag{1}
$$
I was wondering whether an analogous explicit form is known for the operator $-\Delta-z^2$ for $z^2\in \mathbb C\setminus\mathbb R_+$.
The free fundamental solution for $z\neq 0$ (in 3d) is explicitly known to be $\frac{1}{4\pi}\frac{e^{-z|x-y|}}{|x-y|}$, but incorporating the boundary conditions seems nontrivial...
 A: I was also interested in this question, but I think that for the Helmholtz equation there is no such simple formula for the Green function of the unit ball.
Suppose that, for each $x_0\in B_1(0)$, one looks for $G(x,x_0)$ such that
$$
\Delta G(\cdot,x_0)+k^2 G(\cdot,x_0)=\delta_{x_0},\quad x\in B_1(0),
\tag{1}
$$
and
$$
G(\cdot,x_0)|_{\partial B_1(0)}\equiv0.\tag{2}
$$
If one argues by the method of symmetries then one would look for a candidate of the form
$$ 
-4\pi G(x,x_0)=\frac{e^{ik|x-x_0|}}{|x-x_0|} - A\frac{e^{ik|x-x^*_0|}}{|x-x^*_0|},\quad |x|\leq 1.\tag{$\star$}
$$
The quantities $A\in\mathbb{C}$ and $x_0^*\in \mathbb{R}^3\setminus\overline{B}_1(0)$ must be chosen so that
$$
G(x,x_0)=0,\quad \mbox{if $|x|=1$},
$$
since (1) is always satisfied.
Thus we need to impose
$$
|x-x_0^*|=|A||x-x_0|,\quad \forall\;|x|=1.
$$
At this point the argument coincides with that for the Laplace equation, so one is led to pick $x_0^*=x_0/|x_0|$, the usual symmetric point of $x_0\in B_1(0)$, and for this choice one has
$$
|x-x_0^*|=|x_0|^{-1}\,|x-x_0|,\quad |x|=1,
$$
and thus $|A|=|x_0|^{-1}$. But going back to $(\star)$ this means that we have to choose
$$
A=\frac{e^{ik(1-\frac1{|x_0|})|x-x_0|}}{|x_0|},
$$
which is not a constant number unless $k=0$.
Curiously, one can find a reference in the web which asserts that a simple explicit formula is given by
$$
-4\pi G(x,x_0)=\frac{e^{ik|x-x_0|}}{|x-x_0|} - \frac{e^{ik|x_0|\,|x-x^*_0|}}{|x_0|\,|x-x^*_0|}, \quad |x|\leq1.
$$
See p. 127 of the PhD Thesis of P. Teeravarapaug  (Oklahoma State Univ, 1977), available at
https://core.ac.uk/download/pdf/215229677.pdf
However, this assertion seems not to be right, since that function will not satisfty the equation (1). In fact, the summand $u(x)=\frac{e^{ik|x_0|\,|x-x^*_0|}}{|x_0|\,|x-x^*_0|}$ is only a solution of
$$
\Delta u+k^2|x_0|^2 u =0.
$$
