I am trying to solve the following Helmholtz equation, with some specific conditions.
$$b(x,y,z) - \bigg(\frac{\partial^2 b}{\partial x^2} + \frac{\partial^2 b}{\partial y^2} +\frac{\partial^2 b}{\partial z^2}\bigg) = 0$$
Compared to the standard Helmholtz equation, the sign here is negative (which can be accomplished with setting a complex number $k = 1i$ in the standard Helmholtz equation).
The conditions I have are as follows:
\begin{align} b(x,y,0) &= 0\\ b(x,y,z) &= b(y,x,z) = b(x,-y,z) = b(-x,y,z) = b(-x,-y,z) \\ b(x,y,z) &= -b(x,y,-z)\\ b(\pm \infty,\pm \infty,\pm \infty) &= 0%\\ \end{align}
Basically a non-explosive solution. I expect also $\large\frac{\partial b}{\partial z}$ to be negative close to origin. The solution according to wolfram is as follows (F = b, k = 1i, l and m doesn't need to be integers in my case):
The solution I am trying to get for $b(0,0,z)$ is supposed to look like (according to my design):
I am taking only a few terms from the infite summation in order to fit to experimental data. Unfortunately it looks impossible to reach to a solution that starts from 0, growing then decaying, unless I make a two piece solution-one for the growing part, one for the decaying part (with some non-smoothness at second derivatives at a few points).
I am puzzled that I cannot generate an easy damping solution. What am I missing here?
There's no solution for the problem. Just consider the $z$ direction, for which separation of variables can give you an independent 1D boundary value problem. The $z$ part of the Helmholtz equation is
$$b_z(z)-b_z''(z)=0.$$
If you solve this equation, you'll get
$$b_z(z)=A\cosh(z)+B\sinh(z),$$
where the symmetry requested by $b(x,y,z) = -b(x,y,-z)$ implies $A=0$.
$\sinh$ is unbounded at infinities, which contradicts your boundary conditions, prescribing that the function vanishes at both infinities.