# Sturm-Liouville problem

• May 9th 2009, 11:23 AM
math_lete
Sturm-Liouville problem
Hi i'm having a bit of trouble working out how to do this question. As i've not done Sturm-Liouville theory before and finding it hard to work it out.
The question is
y''-2y'+(1+lamda)y=0 with homogeneous bc on x=0,1
Write in S-L form
Write the orthagonality condition.

Now i'm having trouble working out how this is done could anybody be kind enough to go through the steps needed to work this out?
Thanks very much
• May 9th 2009, 11:46 AM
HallsofIvy
Quote:

Originally Posted by math_lete
Hi i'm having a bit of trouble working out how to do this question. As i've not done Sturm-Liouville theory before and finding it hard to work it out.
The question is
y''-2y'+(1+lamda)y=0 with homogeneous bc on x=0,1
Write in S-L form
Write the orthagonality condition.

Now i'm having trouble working out how this is done could anybody be kind enough to go through the steps needed to work this out?
Thanks very much

The "Sturm-Liouville form" is $\frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)+ (q(x)+ \lambda)y= 0$.

You need to find p(x) so that $\frac{d}{dx}\left(p(x)\frac{dy}{dx}\right)= p(x)\frac{d^2y}{dx^2}+ \frac{dp}{dx}\frac{dy}{dx}= p(x)\frac{d^2y}{dx^2}- 2p(x)\frac{dy}{dx}$.

In other words, you need to find p(x) so that $\frac{dp}{dx}= -2p$, then multiply the equation by that p. That's essentially the same as finding an "integrating factor".
• May 9th 2009, 12:03 PM
TheEmptySet
Quote:

Originally Posted by math_lete
Hi i'm having a bit of trouble working out how to do this question. As i've not done Sturm-Liouville theory before and finding it hard to work it out.
The question is
y''-2y'+(1+lamda)y=0 with homogeneous bc on x=0,1
Write in S-L form
Write the orthagonality condition.

Now i'm having trouble working out how this is done could anybody be kind enough to go through the steps needed to work this out?
Thanks very much

So we solve the associated equation

$r^2-2r+(1+\lambda)=0$

Now if we complete the square we get

$(r-1)^2+\lambda \iff (r-1)^2=-\lambda \iff r=1 \pm\sqrt{-\lambda}$

Now we have 3 cases: $\lambda > 0$ $\lambda = 0$ $\lambda < 0$

If $\lambda = 0$

we get $y=c_1+c_2x$ Now if you impose the boundry conditions you get the trivial solution $y=0$

If you have $\lambda < 0$ we get

$y=c_1e^{-(1+\sqrt{-\lambda})x}+c_2e^{-(1-\sqrt{\lambda})x}$

now if x=0 we get

$0=c_1+c_2 \iff c_1=-c_2$

Now if x=1 we get

$0=c_1 e^{-(1+\sqrt{-\lambda})}+c_2 e^{-(1-\sqrt{-\lambda}) }$

$0=-c_2 e^{-(1+\sqrt{-\lambda})}+c_2 e^{-(1-\sqrt{-\lambda}) }$

$0=c_2 (-e^{-(1+\sqrt{-\lambda})}+e^{-(1-\sqrt{-\lambda})} )$

Since the exponential is a 1-1 they are not equal so $c_2=0$

and we only get the trivial solution

So finally we get to the last case

$\lambda > 0$ we get

$y=c_1e^{-x}\cos(x\sqrt{-\lambda})+c_2e^{-x}\sin(x\sqrt{-\lambda})$

So when x=0 we get

$0=c_1$ so the equation reduces to

$y=c_2e^{-x}\sin(x\sqrt{-\lambda})$

When x=1 we get

$0=c_2e^{-1}\sin(\sqrt{-\lambda}) \iff 0=\sin(\sqrt{-\lambda})$

Now we take the inverse sine of both sides to get

$\pi \cdot k =\sqrt{-\lambda} \mbox{ for }k \in \mathbb{Z}$

This gives

$\pi^2 k^2=-\lambda \mbox{ for } k \in \mathbb{Z}^+$

$y_k=c_k\sin(\pi k x)$