Eigenvalues for a non self adjoint operator

Click For Summary
SUMMARY

The discussion focuses on obtaining eigenvalues for a non-self-adjoint second-order linear differential operator L acting on polynomial functions y(n). The method involves equating coefficients of equal powers of x to derive the nth eigenvalue without the need for a weight function to convert L into self-adjoint form. Participants confirmed that the eigenvalues can be calculated directly from the polynomial expansion, and orthogonality of the resulting polynomial terms can be achieved using the Gram-Schmidt algorithm if necessary, although it was concluded that orthogonalization was not required for this specific method.

PREREQUISITES
  • Understanding of second-order linear differential operators
  • Familiarity with polynomial expansions and eigenvalue problems
  • Knowledge of the Gram-Schmidt orthogonalization process
  • Basic concepts of self-adjoint operators in linear algebra
NEXT STEPS
  • Study the properties of non-self-adjoint operators in differential equations
  • Learn about polynomial eigenfunction expansions in linear differential equations
  • Explore the Gram-Schmidt algorithm for orthogonalization of polynomial bases
  • Investigate weight functions and their role in transforming operators to self-adjoint form
USEFUL FOR

Mathematicians, physicists, and engineers working with differential equations, particularly those interested in eigenvalue problems and polynomial expansions in non-self-adjoint contexts.

qtm912
Messages
37
Reaction score
1
TL;DR
Trying to obtain eigenvalues for a non self adjoint linear second order differential operator , but unsure about how (if) to use the weight function
Hi all- I am trying to obtain eigenvalues for an equation that has a very simple second order linear differential operator L acting on function y - so it looks like :
L[y(n)] = Lambda (n) * y(n)

Where y(n) can be written as a sum of terms in powers of x up to x^n

but I find L is non self adjoint. As indicated the question involves assuming that the form of y is a polynomial say in x and use "the equating of coefficients of equal powers method" to work out the coefficients of the x powers and the nth eigenvalue for the expansion. That should be manageable but unsure if it is necessary first to use a weight function w (to convert L to self adjoint form) and if so at what stage. Initial thought was that the eigenvalues are as given by this calculation but that the eigenfunction expansion would have to be calculated as above and then dividing it by the weight function - not sure really if any of this is on the right track.

Guidance appreciated.
 
Physics news on Phys.org
Do what you said: write ##y(n)## as polynomial, apply ##L## and compare the terms. No weight functions or self adjoint needed. The equation ##Ly=\lambda y## is the same for any linear function ##L##.
 
Thanks fresh_42 - thinking about it some more it lines up with what I had in mind - for the eigenvalues anyway. But we have not applied any weight function; now the operator by assumption was not self adjoint so does this not mean that the polynomial terms I later derive in the y expansion that is expressed as a sum of polynomial eigenfunctions in x, will no longer be orthogonal. To make them orthogonal the weight function would play a role would it? If not how would one proceed to obtain an expansion of polynomials in terms of an orthogonal basis.
 
Noted - and after thinking about it further I realized that orthogonalisation in this case was not necessary using this method of deriving the coefficients. Your comments were very helpful in clearing some doubts thanks - the topic has been fully addressed and can be closed..
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K