# Sturm Liouville Eigenfunctions??

• Mar 14th 2009, 12:25 AM
DJDorianGray
Sturm Liouville Eigenfunctions??
Problem: find the eigenvectors and eigenfunctions of the following Sturm-Liouville problem:

$\displaystyle \displaystyle \frac{d}{dx} \left(x \frac{dy}{dx} \right) + \frac{\lambda y}{x} = 0, \qquad 1 \leq x \leq 2, \qquad y(1)=y(2) = 0$

What I did: since $\displaystyle x \neq 0$ I can multiply by $\displaystyle x$ and rewrite as:

$\displaystyle \displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \lambda y = 0$

I substitute $\displaystyle y = x^k$ and find $\displaystyle k = \pm \sqrt{-\lambda}$, so the general solution would be:

$\displaystyle \displaystyle y = A x^{\sqrt{-\lambda}} + B x^{-\sqrt{-\lambda}}$

The condition $\displaystyle y(1) = 0$ gives $\displaystyle A + B = 0$. The second condition gives:

$\displaystyle \displaystyle y(2) = A \left( 2^{\sqrt{-\lambda}} - 2^{- \sqrt{-\lambda}} \right) = 0$

So i get $\displaystyle 2^{\sqrt{\lambda}} = \pm 1$ and here I am stuck, because the Sturm Liouville eigenvalues are supposed to be real.... any suggestions on where I am wrong??

Thank you :)
• Mar 14th 2009, 07:08 AM
Jester
Quote:

Originally Posted by DJDorianGray
Problem: find the eigenvectors and eigenfunctions of the following Sturm-Liouville problem:

$\displaystyle \displaystyle \frac{d}{dx} \left(x \frac{dy}{dx} \right) + \frac{\lambda y}{x} = 0, \qquad 1 \leq x \leq 2, \qquad y(1)=y(2) = 0$

What I did: since $\displaystyle x \neq 0$ I can multiply by $\displaystyle x$ and rewrite as:

$\displaystyle \displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \lambda y = 0$

I substitute $\displaystyle y = x^k$ and find $\displaystyle k = \pm \sqrt{-\lambda}$, so the general solution would be:

$\displaystyle \displaystyle y = A x^{\sqrt{-\lambda}} + B x^{-\sqrt{-\lambda}}$

The condition $\displaystyle y(1) = 0$ gives $\displaystyle A + B = 0$. The second condition gives:

$\displaystyle \displaystyle y(2) = A \left( 2^{\sqrt{-\lambda}} - 2^{- \sqrt{-\lambda}} \right) = 0$

So i get $\displaystyle 2^{\sqrt{\lambda}} = \pm 1$ and here I am stuck, because the Sturm Liouville eigenvalues are supposed to be real.... any suggestions on where I am wrong??

Thank you :)

You should consider three cases for $\displaystyle \lambda$: $\displaystyle <0, =0, \text{and}\; >0$ or $\displaystyle \lambda = -\omega^2$, $\displaystyle \lambda = 0$ and $\displaystyle \lambda = \omega^2$ since the solutions of the ODE are very different for these cases.
• Mar 14th 2009, 07:23 AM
Opalg
Quote:

Originally Posted by DJDorianGray
Problem: find the eigenvectors and eigenfunctions of the following Sturm-Liouville problem:

$\displaystyle \displaystyle \frac{d}{dx} \left(x \frac{dy}{dx} \right) + \frac{\lambda y}{x} = 0, \qquad 1 \leq x \leq 2, \qquad y(1)=y(2) = 0$

What I did: since $\displaystyle x \neq 0$ I can multiply by $\displaystyle x$ and rewrite as:

$\displaystyle \displaystyle x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \lambda y = 0$

I substitute $\displaystyle y = x^k$ and find $\displaystyle k = \pm \sqrt{-\lambda}$, so the general solution would be:

$\displaystyle \displaystyle y = A x^{\sqrt{-\lambda}} + B x^{-\sqrt{-\lambda}}$

The condition $\displaystyle y(1) = 0$ gives $\displaystyle A + B = 0$. The second condition gives:

$\displaystyle \displaystyle y(2) = A \left( 2^{\sqrt{-\lambda}} - 2^{- \sqrt{-\lambda}} \right) = 0$

So i get $\displaystyle 2^{\sqrt{\lambda}} = \pm 1$ and here I am stuck, because the Sturm Liouville eigenvalues are supposed to be real.... any suggestions on where I am wrong??

Don't be put off by complex solutions. You can often combine them to form real solutions. In this case, $\displaystyle y=x^{\pm k}=e^{\pm k\ln x}$ solves the equation for every complex value of k. The solution $\displaystyle y=Ae^{k\ln x} + Be^{-k\ln x}$ satisfies the boundary conditions if $\displaystyle k=n\pi i/\ln 2$, and then $\displaystyle y = \sin(n\pi\ln x/\ln2)$. The eigenvalue for this eigenfunction is $\displaystyle \lambda = (n\pi/\ln 2)^2$.
• Mar 15th 2009, 02:18 AM
DJDorianGray
My mistake, I did not see that lambda values were actually real (Headbang) thank you guys