[Sturm-Liouville eigenvalues and eigenfunctions problem]

[GGGGc]

Homework Statement

This question is about the differential operator acting on functions of x in the range x ∈ [0, ∞). This is a generalization of the case covered in the notes where the range of x is finite. Here, one end of the range of the variable x is infinite.
How can I know these coefficients a_k? and get the corresponding eigenfunctions?

Relevant Equations

Consider the inhomogeneous eigenfunction equation: L tilde y = lambda y
we may define an operator in self-adjoint form L = wL tilde by means of a suitable weight function w(x) and the eigenfunction equation above becomes: Ly = lambda wy
We assume that we have boundary conditions on our functions that make L self-adjoint.

I have found that w(x) should be e^-x to make L self-adjoint.
and insert back get xL''+(x+1)L' +lambda L = 0
now it needs to assume a monic polynomial function, so I assume Ln = x^n+ sum from k=0 to n-1 (a_k*x^k)
get the 1st and 2nd order differential and insert back
I get lambda_n = (-nx^(n-1)*(n+x)-sum from k=0 to n-1 (a_k*k*x^(k-1)*(k+x))/(x^n + sum from k=0 to n-1 (a_k*x^k))
for n=0,1,2, i get lambda = 1, -(x+1)/(x+a_0), (-2x(x+2)-a_1(x+1))/(x^2+a_0+a_1x)
how can i know these coefficients a_k? and how can i get the corresponding eigenfunctions?

 

[pasmith]

First calculate \tilde{\mathcal{L}}(x^k) for k = 0, 1, 2.

Write <br /> L_n(x) = \sum_{k=0}^n a_kx^k so that by linearity <br /> \tilde{\mathcal{L}}(L_n) = \sum_{k=0}^n a_k \tilde{\mathcal{L}}(x^k) = \sum_{k=0}^n a_k \lambda_n x^k and compare cofficients of powers of x on each side. This will give you n+1 equations in the n+1 unknowns \lambda_n, a_{n-1}, a_{n-2},\dots, a_0 since a_n = 1 is known.

 

[GGGGc]

First calculate \tilde{\mathcal{L}}(x^k) for k = 0, 1, 2.

Write <br /> L_n(x) = \sum_{k=0}^n a_kx^k so that by linearity <br /> \tilde{\mathcal{L}}(L_n) = \sum_{k=0}^n a_k \tilde{\mathcal{L}}(x^k) = \sum_{k=0}^n a_k \lambda_n x^k and compare cofficients of powers of x on each side. This will give you n+1 equations in the n+1 unknowns \lambda_n, a_{n-1}, a_{n-2},\dots, a_0 since a_n = 1 is known.

Thanks for answering, but then what's the purpose to put in self-adjoint form? (ie. in (a), need to find w(x)) and is my lambda correct? I'm not really sure about your method, can you be specific?

 

[pasmith]

The point of obtaining the self-adjoint form is to give you practice in finding the self-adjoint form, and also to find the weight function w so you can verify as requested in the final part of the question that <br /> \langle L_1, L_0 \rangle_w = 0. Your result \lambda_0 = 1 is incorrect. You know that the monic polynomial of degree zero is the constant function 1, so you must have <br /> \tilde{\mathcal{L}}(1) = \lambda_0. This is the method I have suggested applied to the case n = 0. Does \lambda_0 = 1 work here?

Your other results in terms of rational functions of x are incomplete. Possibly for the right coefficients a_k they are correct, but the point is to find those coefficients as well as the eigenvalue; the method I have suggested will do this.

Giving any further hints would amount to doing the work for you, which is neither in your best interest nor permitted by forum rules.

 

[GGGGc]

thanks, I'll recheck that

 

[GGGGc]

The point of obtaining the self-adjoint form is to give you practice in finding the self-adjoint form, and also to find the weight function w so you can verify as requested in the final part of the question that <br /> \langle L_1, L_0 \rangle_w = 0. Your result \lambda_0 = 1 is incorrect. You know that the monic polynomial of degree zero is the constant function 1, so you must have <br /> \tilde{\mathcal{L}}(1) = \lambda_0. This is the method I have suggested applied to the case n = 0. Does \lambda_0 = 1 work here?

Your other results in terms of rational functions of x are incomplete. Possibly for the right coefficients a_k they are correct, but the point is to find those coefficients as well as the eigenvalue; the method I have suggested will do this.

Giving any further hints would amount to doing the work for you, which is neither in your best interest nor permitted by forum rules.

I've tried your method and I got lambda_n = sum from k=0 to n (k-k^2*x^-1), which seems correct as lambda_0 = 0, lambda_1 = 1-1/x, lambda_2 = 3-5/x, for L_0 =1, L_1 = x, but after I try question c, use integral from 0 to infinite (L_0*L_1*e^-x) I got 1, instead of 0. Can you provide some hints about which part I'm missing?
Here's how I get L_1: I insert back L_1 = x+a_0 and lambda_1 to the equation (1), L tilde, then I find a_0(1-1/x) = 0, so a_0 = 0, and L_1 = x.
Thanks in advance!