Question: Maple 2018 gives wrong solution to Laplace PDE outside DISK

This is using Maple 2018.

I noticed when solving Laplace PDE on disk, that no condition is needed to tell Maple if one is asking for solution inside the disk or outside. So how does Maple know which one it is?

It turned out Maple gives the same solution for the PDE outside as inside, which is wrong.

The solution to Laplace PDE inside a disk of radius 1 is

Which Maple gives correctly. But it also give the same above solution for outside the disk. The solution outside the disk should have r^(-n) and not r(n).  Like this

Basically Maple ignores assumption on `r`. This is what I tried

restart;
pde := (diff(r*(diff(u(r, theta), r)), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0:
bc := u(1, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi):
sol_inside:=pdsolve([pde, bc], u(r, theta), HINT = boundedseries) assuming r<1;

restart;
pde := (diff(r*(diff(u(r, theta), r)), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0:
bc := u(1, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi):
sol_outside:=pdsolve([pde, bc], u(r, theta)) assuming r>1;

Both the above gives the same answer.

How then does one solve the Laplace PDE outside the disk using Maple? And why is assumptions on "r" seems to be ignored in the above?