How to prove the orthogonality of eigenfunctions in Sturm-Louisville problems?

In summary, the conversation discusses how eigenfunctions corresponding to different eigenvalues are orthogonal, as shown by Sturm-Louisville eigenvalue theory. The conversation also considers the specific case of Sturm-Louisville form and the boundary conditions needed to show that the overlap integral of two eigenfunctions is zero. The conversation ultimately concludes that the second boundary condition can be used to simplify the overlap integral and show that it is indeed zero.
  • #1
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[SOLVED] orthogonal eigenfunctions

From Sturm-Louisville eigenvalue theory we know that eigenfunctions corresponding to different eigenvalues are orthogonal. For example,

[tex]\Phi_{xx} + \lambda \Phi = 0[/tex]

would be of Sturm-Louisville form (note: [itex]\Phi_{xx}[/itex] represents the second derivative of Phi with respect to x)

[tex]\frac{d}{dx}[p(x) \frac{d \Phi}{dx}] + (q(x) + \sigma(x) \lambda)\Phi = 0[/tex]

where q = 0, p = 1, and [itex]\sigma[/itex] = 1. The boundary conditions are

BC1: [tex]\Phi(0) = 0[/tex]

and

BC2: [tex]\Phi_x(L) + h \Phi(L,t) = 0 [/tex]

which means that there can be one eigenfunction corresponding to a negative eigenvalue and the rest are sines relating to the positive eigenvalues, specifically that the eigenfunctions would be

[tex]\Phi_n(x) = \left\{ {\begin{array}{*{20}c}
{sinh \sqrt{s_1}x} & {n=1} \\
{sin\sqrt{\lambda_n}x} & {n > 1} \\
\end{array}} \right.[/tex]

Basically all I need to do is show that

[tex]\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = 0[/tex]

with integration by parts I have gotten to the point

[tex]\int_0^L sin \sqrt{\lambda_n}x sinh \sqrt{s_1} x dx = \frac{\sqrt{s_e}cosh(\sqrt{s_e}L)sin(L\sqrt{\lambda}) - \sqrt{\lambda_n}cos(\sqrt{\lambda_n}L)sinh(L\sqrt{s_e})}{s_e+\lambda}[/tex]

and mathematica gives the same thing. I don't see this being zero any time soon. Any suggestions?
 
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  • #2
I think it is zero. Use the information on lambda and s that you get by imposing the second BC.
 
  • #3
The second BC gives transcendental equations though.

[tex] tan(\sqrt{\lambda}x) = -\sqrt{\lambda}/h[/tex]

and

[tex]tanh(\sqrt{s}L) = -\sqrt{s}/h[/tex]

which require numerical methods. I'm pretty sure this can be done analytically.
 
  • #4
You don't have to solve them, you have to USE them to show that your overlap integral is zero.
 
  • #5
I'm probably going to feel stupid when I finally figure it out, but I still don't really see what to do. The second BC will tell us that the eigenvalues are different, which is why Sturm-Louisville would say that the integral must be zero. Beyond that though, I can't see how it directly influences the integration.
 
  • #6
I can write your overlap integral as
[tex]\frac{\sinh(\sqrt{s_e}L)\sin(\sqrt{\lambda_n}L)}{s_e+\lambda_n}\left[
\sqrt{s_e}\coth(\sqrt{s_e}L) - \sqrt{\lambda_n}\cot(\sqrt{\lambda_n}L)}\right][/tex]
Now look at the equations you get from the boundary conditions, and use them to simplify each term in the square brackets ...
 
Last edited:
  • #7
Shoot dang Avodyne, nice algebra trick and nice foresight. Thanks for the help.
 
  • #8
You're welcome!
 

What are orthogonal eigenfunctions?

Orthogonal eigenfunctions are a set of functions that are both orthogonal (perpendicular) and eigenfunctions (satisfy a specific equation) of a given linear operator. In other words, these functions are both mutually perpendicular and satisfy a particular mathematical relationship.

What is the significance of orthogonal eigenfunctions in mathematics?

Orthogonal eigenfunctions have many applications in mathematics, particularly in the fields of differential equations and functional analysis. They are also useful in physics, engineering, and other sciences as they represent important physical quantities and can be used to solve complex problems.

How are orthogonal eigenfunctions calculated?

The process of finding orthogonal eigenfunctions involves solving a differential or integral equation using techniques such as separation of variables, Fourier series, or Gram-Schmidt orthogonalization. These methods are used to find a set of functions that satisfy the given equation and are mutually perpendicular to each other.

Can orthogonal eigenfunctions exist in any system?

No, orthogonal eigenfunctions can only exist in certain systems that satisfy specific mathematical requirements. These systems must have a well-defined inner product (a way to measure the "angle" between two functions) and a set of functions that form a complete basis (can represent any other function within the system).

What is the difference between orthogonal eigenfunctions and normal eigenfunctions?

The main difference between orthogonal and normal eigenfunctions is that orthogonal eigenfunctions are mutually perpendicular, while normal eigenfunctions are not. This means that orthogonal eigenfunctions have a specific mathematical relationship with each other, while normal eigenfunctions do not necessarily have this property.

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