@mmcdara Yes, with *u*(*x*,0)=0 it's possible to solve the problem in Maple by Fourier series.

Here is the outline of the idea.

We reflect the domain about the line *y*=0 and thus obtain a rectangle. We solve Laplace's equation on the rectangle by taking for boundary values on the new edges the negatives of the values of the opposite pages. Then the solution will be zero automatically along the line *y*=0 due to symmetry.

A detail to be taken care of is that the rectangle's edges are not parallel to the coordinate axes. To apply the Fourier series, we rotate the rectangle by 45 degrees to bring its edges into alignment with the coordinate axes. The PDE does not change since the Laplacian operator is rotationally invariant.

To avoid the rotation step, it would be *much easier to rotate the original domain*. That is, instead of the triangle shown in the OP's drawing, we take a triangle formed by the coordinate axes and the line x+y=1 (or ax+by=c for some constants a, b, c). Then the reflection of the triangle across the hypotenuse will produce a rectangle in which we may apply the standard Fourier series in the straightforward way.

Even then, there is an extra snag that needs to be taken care of. The Fourier series method requires homogeneous boundary conditions, that is, *u*=0, along the four edges of the rectangle. But our boundary data is not zero along the edges. To take care of that, we introduce a function *φ(x,y)* defined over the rectangle so that *φ* matches the given boundary conditions along the rectangle's edges. If we let *w*(x,y) = *u*(x,y) - *φ(x,y),* then *w* will be zero over the rectangle's edges, and therefore Fourier series may be applied to calculate it. We note, however, since *u* satisfies the Laplace's equation, *w *satisfies the Poisson equation *w*_{xx} + *w*_{yy} = −*φ*_{xx} −*φ*_{yy}. which may be solved by Fourier series just as easily.

Finally, there remains the question of how one constructs the function *φ(x,y).* That is the subject of a topic that I posted here a few years ago.

There is some work involved in implementing the steps outlined above. I haven't bothered to do the work considering that this approach applies to the very limited case when the domain is a *right triangle*, the boundary condition on the triangle's hypotenuse is zero, and also the fact that the OP has specified no such boundary condition. In general such problems are best solved through the method of *finite elements* which is not available in Maple.