Writing out the Modified Helmholtz equation in spherically symmetric co-ordinates

Note that $\nabla^2 \psi(r)\;$=$\;\frac{d^{2} \psi}{d r^{2}}+\frac{2}{r} \frac{d \psi}{d r}$=$\frac{1}{r} \frac{d^{2}}{d r^{2}}(r \psi)$ in the spherically symmetric case.

LapSphSym[F[r]] := (1/r)*D[r*F[r], {r, 2}]
eqmHrr1 = LapSphSym[F[r]] - m^2*F[r] == DiracDelta[r - r1]
Assuming[Element[{r, r1, F, m}, Reals], DSolve[eqmHrr1, F[r], r]]

MMa 12.0 output

(** {{F[r] -> C[1]/(E^(m*r)*r) + (E^(m*r)*C[2])/r - 
 (E^((-m)*r - m*r1)*(-E^(2*m*r) + E^(2*m*r1))*r1*HeavisideTheta[r - r1])/
  (2*m*r)}} **)


[1] Is this the correct approach, including DiracDelta[r-r1] , if "r" is a (spherically symmetric) vector?

[2] How to input the boundary conditions to get the Green's functions?

  • 1
    $\begingroup$ Somewhat related: mathematica.stackexchange.com/q/209662/1871 As shown there, GreenFunction knows how to handle the problem in Cartesian coordinates, but doesn't seem to know how to handle the spherical coordinates, at least now. $\endgroup$ – xzczd Jul 9 at 1:52

A partial attempt is documented below, but it needs further work to get at the solutions.

eqmHrr1 = -LapSphSym[F[r]] - m^2*F[r] == (1/r^2)*DiracDelta[r - r1];
sol = TrigExpand[Assuming[Element[{r, r1, F[r], m}, Reals], DSolve[eqmHrr1, F[r], r]]]
FullSimplify[Simplify[sol, Assumptions -> {0 < r < Infinity && r1 < r < Infinity}]]
(** {{F[r] -> (C[1]/E^(I*m*r) + E^(I*m*r)*C[2] - 
  Sin[m*(r - r1)]/(m*r1))/r}}**)

How does one enforce limits at infinity, not include the origin, so that C[1] and C[2] can be determined? The correct solution(s) is/are: $\displaystyle{\frac{e^{\pm \iota~ m~|r - r1|}}{4\pi |r -r1|}}$

| improve this answer | |
  • 1
    $\begingroup$ (1/r^2)*DiracDelta[r - r1] it is not the same as in 1D case. It describes singularity distributed on a sphere r=r1. It can be electric charge on a soap bubble for instance. $\endgroup$ – Alex Trounev Jul 13 at 9:54
  • $\begingroup$ Thanks. How should the DiracDelta term written correctly, i.e. is there a way to write DiracDlelta[$\vec{r}$ - $\vec{r1}$] using "r" as a scalar in spherically symmetric coordinates? Alternatively, what's the correct way to cast this equation so the solution can be validated in 3D within Mma? Just in case, It's not essential to use "r" as a scalar. $\endgroup$ – arny Jul 13 at 19:28
  • $\begingroup$ There is a good answer to your question on math.stackexchange.com/questions/2235615/… $\endgroup$ – Alex Trounev Jul 13 at 20:19
  • $\begingroup$ Also there is a good explanation about 3D Dirac delta on farside.ph.utexas.edu/teaching/jk1/Electromagnetism.pdf $\endgroup$ – Alex Trounev Jul 13 at 20:33

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