Is there a trick to make Mathematica solve

$ $ {\frac {\partial^{2} u}{\partial {x}^{2}}} +{\frac {\partial ^{2} u}{\partial {y}^{2}}} =0$ $ with one boundary condition at $ \infty$ ?

Boundary conditions are

$ $ \begin{align*} u \left( 0,y \right) &=\cos \left( y \right) \ u \left( x,0 \right) &=0 \ u \left( x,a \right) &=0\ u \left( \infty ,y \right) &=0 \end{align*} $ $

Maple can solve this analytically, but one must tell it to assume $ a>0$ . But I can’t get Mathematica 11.2 to solve it. Even when $ a$ is given a numerical value. This is last problem listed here

Mathematica can solve it only by removing the $ \infty$ boundary condition and making it finite.

Here is the code

`ClearAll[x,y,a] ode=D[u[x,y],{x,2}]+D[u[x,y],{y,2}]==0; bc={u[x,0]==0,u[x,a]==0,u[0,y]==Cos[y],u[Infinity,y]==0}; Assuming[a>0,DSolve[{ode,bc},u[x,y],{x,y}]] `

I think `Assuming[a>0`

above is not even used by `DSolve`

. Also with `a`

being numerical, it does not solve it:

`ClearAll[x,y,a] a=2; ode=D[u[x,y],{x,2}]+D[u[x,y],{y,2}]==0; bc={u[x,0]==0,u[x,a]==0,u[0,y]==Cos[y],u[Infinity,y]==0}; DSolve[{ode,bc},u[x,y],{x,y}] `

Maple 2017 solves it, and gives this

`ode:=diff(u(x,y),x$ 2)+diff(u(x,y),y$ 2)=0; bc:=u(x,0)=0, u(x,a)=0, u(0,y)=cos(y), u(infinity,y)=0; sol:=pdsolve({ode, bc}, u(x,y)) assuming a>0; `

$ $ u \left( x,y \right) =\sum _{n=1}^{\infty }-2\,{\frac {\pi\, \left( \left( -1 \right) ^{n}\cos \left( a \right) -1 \right) n}{{\pi}^{2}{n }^{2}-{a}^{2}}\sin \left( {\frac {\pi\,yn}{a}} \right) {{\rm e}^{-{ \frac {\pi\,xn}{a}}}}} $ $

Screen shot

Any one knows of a trick to make Mathematica solve this symbolically?