Singular Jacobian with BVP4c used to solve eigenvalue problem

I have been trying to solve an eigenvalue problem of the form:

y"+y/x+(Lambda^2)*(1-x^2)y=0

which is a Sturm-Liouville problem that occurs e.g. in the case of heat convection in tubes and channels. The task here is to find the eigenvalues Lambda in order to describe the temperature in the fluid.

As I understand one way to solve this problem is by intorudcing the parameter Lambda into the definition of the differential equation:

function dydx=evp1(x,y,lambda); 
    dydx=[y(2)
         -y(2)/x-lambda*(1-x^2)*y(1)];
end
function res=evp1bc1(ya,yb,lambda); 
    res=[ya(1)-1;yb(1);ya(2)];     
end
function v=evp1Guess1(x); 
    v=[cos(x);sin(x)]; 
end
clear all; close all; echo on; format long;
a=0; b=1; N=10; lambda=0.5;
solInit1=bvpinit(linspace(a,b,N),@evp1Guess1,lambda);
solN1=bvp4c(@evp1,@evp1bc1,solInit1);

This short and elegant code by Zaitsev (in the Book: Handbook of Ordinary DIfferential Equations) finds the eigenvalues for simpler eigenvalue problems (e.g. y"+Lambda.y=0). However, for my problem it creates an error message referring to a singular Jacobian which I assume is due to the presence of the y(1)/x term in the equation at x=0. Is there a way to get around this problem?

[Tanya Sharma]

Hi Saeid,

I am also solving an eigenvalue problem using bvp4c.

As we can pass the unknown eigenvalue as a parameter in bvp4c and give it some initial guess. But in turn we only get a single value of the parameter in return using

sol.parameters

But an ODE can have as many eigenvalues. But this algorithm restricts our eigenvalue to only one value.

How can I obtain all the possible eigenvalues for the ODE?

[Saeid]

You're right, the problem can indeed have an infinite number of eigenvalues. What I eventually came up with may not be the most efficient/professional solution but I had to define a vector of initial guesses, then ran a loop over this vector and set those guesses as the initital guess of Lambda for the ODE, then solved it for every initial guess. Many of the eigenvalues, e.g., from the guesses number say, 25, 26, and 27, will be the same, and you need to to eliminate the repeated ones and store them only once. As I said, not very elegant, but it did the job!

However, you can also use PDEPE which is a far more powerful tool, and after finding it I kind of lost interest in the eigenvalue game.

[Tanya Sharma]

Hello Saeid,

Actually finding the eigenvalues is very important in my analysis. Can you share the structure of the loop for your guess vector. It may help me as I am stuck with only one eigenvalue.

Thanks in advance.

Answer 1

[Torsten]

Use a=(a small value) instead of a=0.

Best wishes

Torsten.

[Merrill Yeung]

This simple example illustrates how to solve a BVP with an unkown parameter (here actually is an eigenvalue problem). Usually for a 2rd order ODE, only two boundary conditions are avalable. My question is how to use bvp4v to solve problems like that. (This example needs 3 boundary conditions.)