Solution of LaPlace's equation

Linear pde's can often be solved using a "separation of variables" technique. This is done by assuming φ[r,θ]  = A[r] G[θ].  As for Re=0 flows, we can often find the form of the angle function from the boundary conditions, in this case, the free stream flow.   

The boundary conditions for the sphere at infinity are = U cos[θ], = -U sin[θ].  Recall the definition - ∇ φ = v,   thus:   =- vth = - this suggests that the boundary conditions on  φ[r,θ] are  
φ[r->∞,θ] = -U r Cos[θ],  
at r=R, we have = 0, φ'[r=R] = 0

This suggests a trial form of phi as

We need to start the solution by substituting the trial solution into
the laplace equation

laplacetemp = Simplify[laplaceeq /. 
   {Derivative[a1_, a2_][φ][r, θ] :> 
     D[phitrial, {r, a1}, {θ, a2}], φ[r, θ] -> phitrial}]

Things are going well, the θ dependence divides out!  Let's get rid of it and rearrange the equation.

This can be solved readily with a solution of the form A[r] = rα because it is an Euler equation.  

Here is the far away boundary condition, φ[r,θ] --> - r Cos[θ] U (or = U Cos[θ])  , @ r-->infinity

The boundary condition on the sphere must be [r=R] = 0

Now solve for the constants

We can compute the velocity components   =- vth = -

 = - D[phianswer,r]