Eigenvalue/Orthogonal Eigenfunctions

• Apr 3rd 2011, 04:13 PM
jegues
Eigenvalue/Orthogonal Eigenfunctions
See figure attached for the problem statement as well as my attempt.

I'm confused as to how to do part B.

Do they want me to choose two values such as n=1, and n=2, compute their respective eigenfunctions and see if they are orthogonal?

Or do I have to do it for case such as, n=a and n=b?

Also, how do I compute the integral to verify that it's 0? Isn't it just going to be sin(something)*sin(somethingdifferent)?

Any ideas?
• Apr 4th 2011, 02:02 AM
Ackbeet
Quote:

Do they want me to choose two values such as n=1, and n=2, compute their respective eigenfunctions and see if they are orthogonal?

Or do I have to do it for case such as, n=a and n=b?
You have to do it for arbitrarily different eigenvalues. So, do the second, which will give you that the eigenfunctions corresponding to any two distinct eigenvalues are orthogonal.

Quote:

Also, how do I compute the integral to verify that it's 0? Isn't it just going to be sin(something)*sin(somethingdifferent)?
Technically, the inner product will be

$\displaystyle\langle y_{m}|y_{n}\rangle=\int_{0}^{1}\sin\left(\frac{(2m-1)\pi x}{2}\right)\sin\left(\frac{(2n-1)\pi x}{2}\right)dx.$

Assume $m\not=n,$ and show that the integral is zero. Then you're done. Make sense?
• Apr 4th 2011, 04:32 PM
jegues
Quote:

Originally Posted by Ackbeet
You have to do it for arbitrarily different eigenvalues. So, do the second, which will give you that the eigenfunctions corresponding to any two distinct eigenvalues are orthogonal.

Technically, the inner product will be

$\displaystyle\langle y_{m}|y_{n}\rangle=\int_{0}^{1}\sin\left(\frac{(2m-1)\pi x}{2}\right)\sin\left(\frac{(2n-1)\pi x}{2}\right)dx.$

Assume $m\not=n,$ and show that the integral is zero. Then you're done. Make sense?

So does the part inside the integrand become,

$\frac{1}{2}\left( cos( ((2m-1) - (2n-1))x) - cos(((2m-1) + (2n-1))x) \right)$

?
• Apr 4th 2011, 05:41 PM
Ackbeet