solving PDE equation like Helmholtz equation in 2D

In my project I need to solve following equation analytically could anyone help me ? As I read the other questions, my equation seems like Helmholtz equation $$\triangledown^2 u(x,y) - k_cu(x,y) =0 \quad x,y \in \Omega \\ u(x,L) = u(x,0), u(0,y) = 1 \quad x \in [0,L] , y \in [0,L]$$

here is my code in Mathematica

Dc = 1460
Kc = 9.41/(10^2)
LaplacEquation =  Dc*D[u[x, y], {x, 2}] + Dc* D[u[x, y], {y, 2}] - Kc*u[x, y] == 0
DSolve[LaplacEquation, {u[x, y], u[x, y]}, {x, y}]
• Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! – Michael E2 Jul 20 '16 at 16:23
• Just to clarify, you wish to solve this with the computer algebra software Mathematica? Have you tried DSolve[]? -- OTOH, if it is just a mathematics question, the site math.stackexchange.com is more appropriate. – Michael E2 Jul 20 '16 at 16:25
• @MichaelE2 thank you for your answer. Yes I want to solve it using Mathematica. I also tried DSolve[] but because non principal part $k_cu(x,y)$ DSolve[] could not find the answer – user1099913 Jul 20 '16 at 16:37
• Could you post the code for what you tried? – Feyre Jul 20 '16 at 16:43
• @MichaelE2 I have edited my post and added my code – user1099913 Jul 20 '16 at 17:04

This is a solution to the problem with exactly the boundary conditions as stated:

Dc = 1460
Kc = 9.41/(10^2)
LaplacEquation =
Dc*D[u[x, y], {x, 2}] + Dc*D[u[x, y], {y, 2}] - Kc*u[x, y] == 0

v[x] /.
First[DSolve[{LaplacEquation /. u -> (v[#] &), v == 1}, v[x], x]]

(* ==> E^(-0.0080282 x) (1. - 1. C + E^(0.0160564 x) C) *)

All I did here is observe that the boundary conditions can be satisfied if the solution has no dependence on y whatsoever. I then inserted that assumption into the Laplacian in the form of a replacement u -> (v[#]&) which states that the new function v depends only on x.

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}] 