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

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

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


I think Assuming[a>0 above is not even used by DSolve. Also with a being numerical, it does not solve it:


Maple 2017 solves it, and gives this

bc:=u(x,0)=0, u(x,a)=0, u(0,y)=sin(y), u(infinity,y)=0;
sol:=pdsolve({ode, bc}, u(x,y)) assuming a>0; 

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

Screen shot

Mathematica graphics

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

  • $\begingroup$ Perhaps, Mathematica is having difficulties, because the initial and boundary conditions are inconsistent. It is, of course, possible to Fourier-decompose the PDE in y, but the resulting solution does not converge uniformly. $\endgroup$
    – bbgodfrey
    Dec 28, 2017 at 5:14
  • $\begingroup$ @bbgodfrey oh, that was just me changing the PDE while playing around with it. The original one had Sin[y] there. But M still can't solve it. I should correct this now. Thanks. $\endgroup$
    – Nasser
    Dec 28, 2017 at 5:25

1 Answer 1


Using the symbolic boundary condition

u[t, y] == 0

instead of the infinite one, and then taking the limit $$ t \to \infty,$$ inside the sum seems to do the trick for me.

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] == Sin[y], u[t, y] == 0};
sol = Assuming[a > 0, DSolve[{ode, bc}, u[x, y], {x, y}]];
expr = (u[x, y] /. sol)[[1]][[1]];
limexpr = 
  Assuming[ a > 0 && K[1] > 0, Limit[expr, t -> \[Infinity]]];
Inactivate[Sum[limexpr, {K[1], 1, \[Infinity]}]]

This gives $$\underset{K[1]=1}{\overset{\infty }{\sum }}-\frac{2 \pi \sin (a) (-1)^{K[1]} K[1] e^{-\frac{\pi x K[1]}{a}} \sin \left(\frac{\pi y K[1]}{a}\right)}{\pi ^2 K[1]^2-a^2}$$

  • $\begingroup$ Thanks. btw, the original PDE had sin not cos. This does not affect your method. Just wanted to point I made change to the PDE now. $\endgroup$
    – Nasser
    Dec 28, 2017 at 5:29
  • $\begingroup$ I've updated the solution to reflect this :) $\endgroup$ Dec 30, 2017 at 3:27

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