If you accept that the general solution can be constructed as an (infinite) Fourier series,
$$\sum_{m=-\infty}^{\infty}\phi_m(r) \exp(i m \theta),$$
then you can obtain the expression for $\phi_m(r)$ as follows:
eqn =
0 == Simplify@
Laplacian[ Laplacian[ ϕ[r] Exp[I m θ], {r, θ}, "Polar"], {r, θ}, "Polar"];
DSolve[eqn, ϕ, r]
(*
==> {{ϕ ->
Function[{r},
r^-m C[1] + r^(2 - m) C[2] + r^(2 + m) C[3] + r^m C[4]]}}
*)
The integration constants in this result can then be chosen differently for each $m$. This solves the homogeneous problem (if the constants are chosen to satisfy the given boundary conditions).
You can do the same for the inhomogeneous problem with specific $i, j$, e.g., $i=2$, $j=3$:
eqn1 =
r^2 Exp[I 3 θ] ==
Simplify@
Laplacian[Laplacian[ ϕ[r] Exp[I 3 θ], {r, θ}, "Polar"], {r, θ}, "Polar"];
DSolve[eqn1, ϕ, r]
(*
==> {{ϕ ->
Function[{r}, r^6/189 + C[1]/r^3 + C[2]/r + r^3 C[3] + r^5 C[4]]}}
*)
eqn2 =
r^2 Exp[-I 3 θ] ==
Simplify@
Laplacian[ Laplacian[ ϕ[r] Exp[-I 3 θ], {r, θ}, "Polar"], {r, θ}, "Polar"];
DSolve[eqn2, ϕ, r]
(*
==> {{ϕ ->
Function[{r}, r^6/189 + C[1]/r^3 + C[2]/r + r^3 C[3] + r^5 C[4]]}}
*)
Here, I broke the $\cos(3\theta)$ up into two parts containing $\exp(\pm 3 i\theta)$ and solved the two separately. Then you have to add both solutions and divide by two to get the solution to with $r^2 \cos(3 \theta)$ as the inhomogeneous term.
Finally, using the same superposition principle as above for some special choices of $i,j$, I tried the general solution for arbitrary $i,j$ in the following form:
eqn3 =
r^i Cos[j θ] ==
Simplify@
Laplacian[ Laplacian[ ϕ[r] Cos[j θ], {r, θ}, "Polar"], {r, θ}, "Polar"];
DSolve[eqn3, ϕ, r]
(*
==> {{ϕ ->
Function[{r},
r^(4 + i)/((-4 - i + j) (-2 - i + j) (2 + i + j) (4 + i + j)) +
r^-j C[1] + r^(2 - j) C[2] + r^(2 + j) C[3] + r^j C[4]]}}
*)
So for the inhomogeneous equation, all we had to do is choose the same $\theta$ dependence for the unknown solution as appears in the given right-hand side.
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