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$\textbf{Eigenvalue problem on a unit square $\Omega = [0,1]^2$ :}$

Consider the eigenvalue problem with the Dirichlet boundary condition that is, $$-Lu = \lambda u$$ where $$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$.

The boundary condition is that $u=0$ on $\partial \Omega$.

I am computing the $\textbf{eigenvalues}$ and $\textbf{eigenfunctions}$ of Laplacian numerically in Mathematica. Specially, I am interested about the $\textbf{multiplicity}$ of a specific eigenvalue. I already know that Tally or Count can certainly count the occurrances of eigenvalues in the list called vals.

{ℒ, ℬ} = {-Laplacian[u[x, y], {x, y}],
    DirichletCondition[u[x, y] == 0, True]};

{vals, funs} = 
  DEigensystem[{ℒ, ℬ}, 
   u[x, y], {x, 0, 1}, {y, 0, 1}, _];

Tally[vals]

In the eigenvalue problem there are infinite numbers of eigenvalues. So, We can not list all of them. I want to find out $\textbf{multiplicity}$ of a specific eigenvalue that counts the total number of occurrances. For example, the multiplicity of the eigenvalue $5 \pi^2$ is $2$. Similarly, I want to find out the multiplicity of the eigenvalue $50 \pi^2$ is $3$.

Thanking in advanced.

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With the exact solutions

u[{nx_, ny_}, {x_, y_}] = Sin[nx π x] Sin[ny π y];

and $\{n_x,n_y\}\in\mathbb{N}^2$, the eigenvalues are

λ[nx_, ny_] = (nx^2 + ny^2) π^2;

Test:

Assuming[Element[nx | ny, PositiveIntegers],
  {u[{nx, ny}, {x, 0}], u[{nx, ny}, {x, 1}], 
   u[{nx, ny}, {0, y}], u[{nx, ny}, {1, y}]} // FullSimplify]
(*    {0, 0, 0, 0}    *)

-D[u[{nx, ny}, {x, y}], {x, 2}] - D[u[{nx, ny}, {x, y}], {y, 2}] ==
λ[nx, ny] * u[{nx, ny}, {x, y}] // FullSimplify
(*    True    *)

The number of pairs $\{n_x,n_y\}$ equal to $\lambda/\pi^2$ is therefore given by OEIS A063725. You can calculate the multiplicity of a given value of $i=\lambda/\pi^2$ directly from the generating function,

g[q_] = (EllipticTheta[3, q] - 1)^2/4;
m[i_Integer?NonNegative] := SeriesCoefficient[g[q], {q, 0, i}]

For example, the multiplicity of $\lambda=50\pi^2$ is 3, as you know:

m[50]
(*    3    *)

There is no eigenvalue at $326\pi^2$, on the other hand:

m[326]
(*    0    *)

More generally, the $d$-dimensional generating function

g[d_, q_] = ((EllipticTheta[3, q] - 1)/2)^d;

allows us to find the multiplicity of the eigenvalue $\lambda=i\pi^d$ directly:

m[d_Integer?NonNegative, i_Integer?NonNegative] :=
  SeriesCoefficient[g[d, q], {q, 0, i}]

For example, the $d=17$-dimensional hypercube has $1\,416\,786\,753\,216$ eigenvalues $\lambda=250\pi^{17}$:

m[17, 250]
(*    1416786753216    *)

If you want the actual solutions, not just the number of solutions, PowersRepresentations is a good starting point (but it gives solutions containing zeros, which we need to filter out):

s[d_Integer?NonNegative, i_Integer?NonNegative] :=
  Select[PowersRepresentations[i, d, 2], Min[#] > 0 &]

For example, the eigenvalue $\lambda=50\pi^2$ in $d=2$ dimensions can be composed in two ordered ways, as you know:

s[2, 50]
(*    {{1, 7}, {5, 5}}    *)

Enumerating all permutations of these ordered combinations gives the full multiplicity of 3:

t[d_Integer?NonNegative, i_Integer?NonNegative] :=
  Join @@ Permutations /@ s[d, i]

t[2, 50]
(*    {{1, 7}, {7, 1}, {5, 5}}    *)

In $d=17$ dimensions the eigenvalue $\lambda=250\pi^{17}$ has 999 ordered solutions,

s[17, 250]
(*    {{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 6, 14},
       {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 10, 10},
       {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 15},
       ...
       {3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5},
       {3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6}}    *)

which then become the aforementioned $1\,416\,786\,753\,216$ unordered solutions from t[17, 250].

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  • $\begingroup$ We should work with the generating function explicitly because it gives a very efficient recipe, especially in the high-dimensional case. Just look at the given example for the 17-dimensional hypercube: manually summing and counting is going to be very tedious and difficult, whereas the call to m[17, 1000] takes less than half a second. The same is true, to some extent, for your case $d=2$ when you work with large values of $\lambda$. $\endgroup$
    – Roman
    Feb 16 '21 at 8:32
  • $\begingroup$ Sorry, this is not the right place for an introduction to generating functions or Jacobi theta functions. $\endgroup$
    – Roman
    Feb 16 '21 at 9:31

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