# Diverging solution to radial equation

I want to solve a seemingly simple eigenvalue problem. I have a fixed set of boundary conditions given and want to change the complex parameter omega in to minimize exponentially falling solutions for some large r value. However, for the regime of omega, in which I expect to get exponentially falling solutions, I only obtain exponentially growing solutions. Therefore, I can't even attempt to minimize the result at some finite but large r.

rstart = 2 + 0.001;
f[r_] := 1 - 1/r
eq = f[r] D[f[r] D[R[r], r],r] - (w^2 + f[r] ((l (l + 1))/r^2 + m^2))R[r] == 0 /. {l -> 1,m -> 4/10, w -> wr + I wi}
sol = ParametricNDSolve[{eq, R[rstart] == 1, R'[rstart] == 1},R, {r, rstart, 100}, {wr, wr}]
test = Table[Log[Abs[R[wr, wi][r]]] /. sol, {wr, 0.88, 1, 0.01}, {wi, 10^-4,10^-1, 10^-2}];
Plot[test, {r, rstart, 100}]

I have also looked at other people's posts, though, there the minimization was done with respect to the initial conditions, which are fixed in my case. I have also tried the same with an increased MaxStepSize, but also that was unsuccessful.

• Probably you have an implicit BC of the solution being zero at $r = \infty$, right? Just force it to be zero at the end of your range. – b3m2a1 Mar 18 '19 at 17:16
• But how would I force it to be zero, if I have the given initial conditions and want to determine w, by setting the solution to zero at infinity? – Nils Mar 18 '19 at 17:24
• "However, for the regime of omega, in which I expect to get exponentially falling solutions, I only obtain exponentially growing solutions. " Then are you sure the underlying model itself is correct? – xzczd Mar 22 '19 at 4:25
• @xzczd: it's possible that the OP's DE does have a growing and a decaying solution, but the condition of the problem is bad enough that integrating forward yields a linear combination of the growing and decaying solution, which is of course itself growing. – J. M.'s technical difficulties Mar 22 '19 at 12:09
• The ODE has the asymptotic limit R''[r] - (w^2 + m^2} R[r] == 0, which will have exponential or oscillatory solutions, depending on the value of (w^2 + m^2}. – bbgodfrey Mar 22 '19 at 12:17