How can numerically compute eigenvalues of an ordinary differential equation in MATLAB?

Hello,

I need to compute (numerically) the eigenvalues (L) of this singular ODE,

, subject to

Is it possible to use the Matlab function bvp4c? Or another?

Best regards,

Lemuel

[Torsten]

https://math.stackexchange.com/questions/2507694/what-numerical-techniques-are-used-to-find-eigenfunctions-and-eigenvalues-of-a-d

[Lemuel Carlos Ramos Arzola]

The approach discussed in the link is quite weak and inefficient. I need a robust approach.

Answer 1

[Lemuel Carlos Ramos Arzola]

Hello,

Thanks for the answers.

I found the same problem here:

https://www.mathworks.com/matlabcentral/answers/405977-singular-jacobian-with-bvp4c-used-to-solve-eigenvalue-problem

and the answer completely satisfies me

[Torsten]

I'm surprised that you are satisfied with the answer since you don't know the third boundary condition corresponding to your eigenvalue, I guess.

[Lemuel Carlos Ramos Arzola]

Hello Torsten,

One of the results of the algorithm is the "true" value of the eigenvalue. This is the objective. Then, when I change the initial guess (the third boundary condition), the algorithm compute another eigenvalue. Is a robust approach, similar to a roots calculation method (eg. Newton-Raphson, Secant, etc).

[Torsten]

But as far as I see, you won't get an eigenvalue for an arbitrary choice of the third boundary condition.

E.g. if you have the ODE

y''+L*y = 0

y(0)=y(2*pi)=0,

the eigenvalues and eigenfunctions are L_n = (n/2)^2 and y_n(x) = sin(n*x/2) (n=1,2,3,...).

So if you choose y'(0)=1 as third boundary condition at x=0, e.g., every function y(x)=a*sin(sqrt(L)*x) with a*sqrt(L)=1 is a solution of the ODE, not only those for which a=2/n and L=(n/2)^2 (n=1,2,3.,,,).

[Lemuel Carlos Ramos Arzola]

Dear Torsten,

Thanks for all your comments.

The singular ODE (the original in this question) arises in the context of heat convection in tubes. This problem is know as Graetz's problem.

Recently I "found" that a third boundary condition exist, namely, y(0)=1. With this "new" condition, the MATLAB function bvp4c can also find unknown parameters in the ODE, in our case, L.

I only need to change the initial guess for the unknown parameter L, not the third boundary condition as I said wrongly in the previous comment .

Following the algorithm shown in this link, and changing the initial guess for L, I found the eigenvalues that I needed.

https://www.mathworks.com/matlabcentral/answers/405977-singular-jacobian-with-bvp4c-used-to-solve-eigenvalue-problem

Although I had some difficulties with the singularity at x = 0.

Answer 2

[Bjorn Gustavsson]

Have a look at what you can do with chebfun. It seem to cover eigenvalue/eigenfunctions of ODEs in some detail:

Chebfun ODE-eigenvalue examples

HTH

[Lemuel Carlos Ramos Arzola]

Thanks for your answer but the examples does not cover my problem.

Answer 3

[Torsten]

https://www.wolframalpha.com/input/?i=y''(x)%2B1%2Fx*y'(x)%2Ba*(1-x%5E2)*y(x)%3D0,+y'(0)%3D0

So you are left with the problem to find "a" such that

L_(0.25*(sqrt(a)-2)) (x) = 0 for x=sqrt(a).

[Lemuel Carlos Ramos Arzola]

Hi Torsten,

The link does not work.

Best regards,

Lemuel

[Torsten]

For me, the link works.

Don't just click, but copy.

Best wishes

Torsten.