Using NDEigensystem to solve coupled eigenvalue problem

Question

I want to find the Eigenvalues and Eigenfunctions for the following Eigenvalue problem

I tried to solve this numerically

pars=Rationalize[{L-> 6.9,A->0.0133,a0->410,
c1a->0.003,c2a->0.002,c1p->-2.5 10^-6,c2p->4.3 10^-6}];

I defined to Operator like above

op={-(a0^2/A)D[\[Phi]2[xi],xi], -A D[\[Phi]1[xi],xi]}/.pars

I am not shure how to use the

DirichletCondition

but I tired this

bc={DirichletCondition[\[Phi]2[xi]==c1p \[Phi]1[xi],xi==0],
DirichletCondition[\[Phi]2[xi]==c2p \[Phi]1[xi],xi==L]}/.pars

Unfortunately Mathematica could not understand my input

NDEigensystem[Flatten[{op,bc}],{\[Phi]1[xi],\[Phi]2[xi]},{xi,0,L}/.pars,4]

Did I make a mistake?

Thanks!

Answer 1

I know this question is old and has an explicit analytic solution given, but here is a numerical method to confirm the roots, which works for cross-coupling of the boundary conditions as seen here.

I have a package for numerically calculating solutions of eigenvalue problems using the Evans function via the method of compound matrices, which is hosted on github. See my answers to other questions or the github for some more details.

First we install the package (only need to do this the first time):

Needs["PacletManager`"]
PacletInstall["CompoundMatrixMethod", 
    "Site" -> "http://raw.githubusercontent.com/paclets/Repository/master"]

Then we first need to turn the ODEs into a matrix form $\mathbf{y}'=\mathbf{A} \cdot \mathbf{y}$, using my function ToMatrixSystem:

Needs["CompoundMatrixMethod`"]

pars = Rationalize[{L -> 6.9, A -> 0.0133, a0 -> 410, c1a -> 0.003, 
    c2a -> 0.002, c1p -> -2.5 10^-6, c2p -> 4.3 10^-6}];
eqns = {-(a0^2/A) D[ϕ2[xi], xi] == λ ϕ1[xi], -A D[ϕ1[xi], xi] == λ ϕ2[xi]}
bcs = {ϕ2[0] == c1p ϕ1[0], ϕ2[L] == c2p ϕ1[L]}
sys = ToMatrixSystem[eqns, bcs, {ϕ1, ϕ2}, {xi, 0, L}, λ] /. pars

The object sys contains the matrix $\mathbf{A}$, as well as similar matrices for the boundary conditions and the range of integration.

Now the function Evans will calculate the Evans function (also known as the Miss-Distance function) for any given value of $\lambda$; this is an analytic function whose roots coincide with eigenvalues of the original equation.

FindRoot will then find solutions for a given start point:

FindRoot[Evans[λ, sys], {λ, 1}]
 (* {λ -> -12.5116} *)

And this gives the same values as for the analytic solution given in the other answer.

λ/.FindRoot[Evans[λ, sys], {λ, -12 + # I}] & /@ Range[-1860, 1860, 186] // Quiet
(* {-12.5116 - 1866.74 I, -12.5116 - 1680.07 I, -12.5116 - 1493.39 I, -12.5116 - 1306.72 I, 
  -12.5116 - 1120.05 I, -12.5116 - 933.372 I, -12.5116 - 746.697 I, -12.5116 - 560.023 I, 
  -12.5116 - 373.349 I, -12.5116 - 186.674 I, -12.5116, -12.5116 + 186.674 I, 
  -12.5116 + 373.349 I, -12.5116 + 560.023 I, -12.5116 + 746.697 I, -12.5116 + 933.372 I, 
  -12.5116 + 1120.05 I, -12.5116 + 1306.72 I, -12.5116 + 1493.39 I, -12.5116 + 1680.07 I, 
  -12.5116 + 1866.74 I} *)

Answer 2

I have computed the analytic solution of the Eigenvalue problem above

With the Eigenvalues

\[Beta] -> Log[(Sqrt[A + a0 c1p] Sqrt[A - a0 c2p])/Sqrt[(A - a0 c1p) (A + a0 c2p)]]

ev[k_] := (a0 (I k \[Pi] + \[Beta]))/L

with some constant beta and the Eigenfunctions

ef[xi_,k_] := {{Cosh[(xi ev[k])/a0]/c1p-(a0 Sinh[(xi ev[k])/a0])/A},
              {Cosh[(xi ev[k])/a0]-(A Sinh[(xi ev[k])/a0])/(a0 c1p)}}