I'm stuck with this problem. Given a domain $\Omega \subset \mathbb{R}^3$ where the function $u$ satisfies: $u_{xx} +u_{yy}+u_{zz} + k^2 u = 0$, I am asked to find the spherical mean over the sphere $\{(x, y, z)\in \mathbb{R}^3; ||(x-x_0, y-y_0, z-z_0)||=R\} \subset \Omega$.
I obviously thought of trying to adapt the mean value property for harmonic functions but to no avail. I then sought to find a general form of the solution using separation of variables. This way I believe that, if using spherical coordinates, the azimuthal factor of the solution is proportional $\Psi(\psi)=e^{in\psi}$, which means that when integrating out the azimuthal angle in $\int u d\sigma=-1/k^2\int \nabla^2 u d\sigma $ I will get zero by periodicity. I feel this solution is probably wrong. Any help would be appreciated. Thanks!
Let $v(r)$ be the mean on the sphere of radius $r$. Then $v'(r)$ is the mean of the normal derivative. Hence, $4\pi r^2 v'(r)$ is the flux of $\nabla u$ out of the sphere. By the divergence theorem, this flux is equal to the integral of $\Delta u$ over the ball bounded by the sphere. The latter integral is $$\int_{B_r}\Delta u = -k^2\int_{B_r} u = -4\pi k^2\int_0^r s^2 v(s)\,ds$$ Thus, $$r^2 v'(r) = -k^2\int_0^r s^2 v(s)\,ds \tag{1}$$ Differentiate $v$ to get an ODE: $$ r^2 v''(r) +2r v'(r)+k^2 r^2 v(r) =0 \tag{2}$$ Looks like the Bessel equation, except for the factor of $2$, which makes it a spherical Bessel equation: unlike the ordinary one, (2) has a rather elementary solution (sinc).