DSolve
has limited abilities with PDEs. I can get part way there, maybe, but something is fishy about the result Mathematica returns.
First, since your equation is linear and homogeneous and has a simple relationship among the coefficients, we can rewrite it as a BVP over a unit square:
$$\nabla^2u + a\, u=0, u(x,0)=u(x,1),\ u(0,y)=1$$
It is apparently still too general for DSolve
to handle, but if we impose further boundary constraints, namely,
$$u(x,0)=u(x,1)=u(0,y)=u(1,y)=1\,,$$
we can get a solution for various values of a
, similar to the Helmholtz example in the docs for DSolve
. Luckily, the dependence on a
is "obvious," and we can get an expression in terms of a
:
pde = a * u[x, y] + Laplacian[u[x, y], {x, y}] == 0;
bcs = {u[x, 0] == 1, u[x, 1] == 1, u[0, y] == 1, u[1, y] == 1};
Block[{a = 5},
{dsol0} = DSolve[{pde, bcs}, u, {x, y}]
];
dsol = dsol0 /. {5 -> a, -5 -> -a};
u[x, y] /. dsol // TraditionalForm
If you doubt it, you can try other values of a
:
Block[{a = 167},
DSolve[{pde, bcs}, u[x, y], {x, y}]
] // TraditionalForm
There are a couple of problems. Numerically this is hard to evaluate, so checking it is difficult. And it does not look symmetric in x
and y
, even though the problem is. (Maybe someone will recognize this as a common sum, but for me, it would take more time than I can give at the moment. It's not a sum Mathematica can do.)
This is the best I can do.
My reformulation has more restrictive boundary conditions than the OP's, and in my mind there is some doubt whether the solution is correct. But maybe it will help the OP.
DSolve[]
? -- OTOH, if it is just a mathematics question, the site math.stackexchange.com is more appropriate. $\endgroup$