Question: solving Laplace inside disk problem. pdsolve

I am trying to see if Maple can solve Laplace PDE inside the disk in polar coordinates. Standard textbook problem. Radius of disk is `a`. The boundary conditions on the disk is `f(theta)`. One of the conditions needed also is that the solution is finite in the center of the disk.

I do not know how to tell Maple that the solution should be finite in the center of the disk. If I do not give this conditions, Maple gives me strange looking solution, which does not look like anything close to the standard series solution one gets from hand solution. There is not even a series solution.

This is what I tried

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta);
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0

Now, how to tell it that `u(0,theta)` is bounded? So that the `ln(r)` solution do not show up? Adding `u(0,theta)<infinity` to the boundary conditions, gives error

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta),u(0,theta)<infinity;
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0

The standard solution to this PDE is

Where `c0` and `cn` and `kn` above are found from boundary conditions at $u(a,\theta)$.

How can one get Maple to give the above solution? How to tell it that $u$ is bounded at $r=0$?