Orthogonality and Weighting Function of Sturm-Liouville Equation

[JordanGo]

Homework Statement

A set of eigenfunctions y_n(x) satisfies the following Sturm-Liouville equation:
\frac{d(f(x)*y'_{m})}{dx}+\lambda*\omega*y_{m}=0
with following boundary conditions:
\alpha_{1}y+\beta_{1}y'=0
at x=a
\alpha_{2}y+\beta_{2}y'=0
at x=b
Show that the derivatives u_n(x)=y'_n(x) are orthogonal functions.
Determine the weighting function for these functions.
What boundary conditions are required for orthogonality?

Homework Equations

Orthogonal functions:
\int(dx*\omega*y_{n}(x)*y_{m}(x)=0
Integrate from a to b.

The Attempt at a Solution

I'm not sure how to start this problem, can someone point me in the right direction?

 

[Mark44]

Problems that involve differential equations should be posted in the Calculus & Beyond section, not in the Precalculus Math section. I am moving this thread to that section.

 

[JordanGo]

Sorry about that...
Anyway, well I found a way to prove orthogonality and ended up with:
(\lambda_{m}-\lambda_{n})\int(w*y_{n}*y_{m}) =0
(integral from a to b)
Now how do I find the weighting function?

 

[LCKurtz]

Sorry about that...
Anyway, well I found a way to prove orthogonality and ended up with:
(\lambda_{m}-\lambda_{n})\int(w*y_{n}*y_{m}) =0
(integral from a to b)
Now how do I find the weighting function?

Of course you must mean the definite integral$$
(\lambda_{m}-\lambda_{n})\int_a^b(w*y_{n}*y_{m}) =0$$That is a standard result in S-L theory, and the weight function is the $w$ in the integrand. However, your original post asked you to show the derivatives of the $y_n$ were orthogonal. I wondered when I saw your OP whether that was a typo or whether it was true. In any case, it isn't what you found the proof for.

 

[JordanGo]

I don't quiet understand what it means then to show if the derivatives are orthogonal...

 

[LCKurtz]

Of course you must mean the definite integral$$
(\lambda_{m}-\lambda_{n})\int_a^b(w\cdot y_{n}\cdot y_{m})\, dx =0$$That is a standard result in S-L theory, and the weight function is the $w$ in the integrand. However, your original post asked you to show the derivatives of the $y_n$ were orthogonal. I wondered when I saw your OP whether that was a typo or whether it was true. In any case, it isn't what you found the proof for.

I don't quiet understand what it means then to show if the derivatives are orthogonal...

Above, if $\lambda_m\ne \lambda_n$ then $\int_a^b(w\cdot y_{n}\cdot y_{m})\, dx =0$, which is what it means for $y_m$ and $y_n$ to be orthogonal with respect to the weight function $w$. For the derivatives to be orthogonal with respect to some weight function $f(x)$would mean $\int_a^bf\cdot y'_{n}\cdot y'_{m}\, dx =0$ if $\lambda_m\ne \lambda_n$.

 

[JordanGo]

Ok, so looking at the equation:
(\lambda_{m}-\lambda_{n})\int(f(x))y'_{n}y'_{m}=0
the only possibility is f(x)=0 because no matter what I do, I can't get terms to separate and moved to the right hand side.

 

[LCKurtz]

Ok, so looking at the equation:
(\lambda_{m}-\lambda_{n})\int(f(x))y'_{n}y'_{m}=0
the only possibility is f(x)=0 because no matter what I do, I can't get terms to separate and moved to the right hand side.

Assuming your mathematics is at an advanced enough level to be studying S-L problems and orthogonality, that comment is just silly. Above you stated that you understand how to get the orthogonality of the eigenfunctions $y_n(x)$ so presumably you have some idea of what is involved to prove orthogonality.

I have shown you what you need to prove. I don't know offhand how to solve your problem and I'm not inclined to spend the evening figuring it out. It might even be easy; I don't know. But I'm sure you need to use the given DE and boundary conditions somehow and maybe even use the orthogonality of the $y_n(x)$ themselves. You don't start with the conclusion. Good luck with it.

[Edit, added]:You might try mimicking the standard proof but changing it where required because your boundary conditions are different.