# Explaining Sturm-Liouville Problem?

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• Oct 19th 2009, 01:27 PM
cloakinghalk
Explaining Sturm-Liouville Problem?
I am having trouble figuring out how to solve sturm-liouville problems. An example of one type of these problems is find the eigen values and eigen vectors of y'' + λy=0 when y'(0)=0, y(1)+y'(1)=0. Another one would be y'' +y' +5λy=0 when y(0)+y'(0)=0 and y(1)+y'(1)=0. Now the way my teacher explained to solve these are you say y=e^(ax) and then sub this into your equation. Then solve for "a". The way the book explains how to do this is that λ=+or - a^2. Can anyone explain to me whats going on and then what to do after finding λ? thanks
• May 23rd 2010, 11:21 AM
GeoC
Sturm-Liouville is a special type of boundary value problem.

First consider $\displaystyle d2y/dx2 + \lambda*y$ with $\displaystyle \lambda$ not equal zero:

The auxiliary equation $\displaystyle r^2 + \lambda = 0$ has roots $\displaystyle r=\pm \sqrt{-\lambda}$.

Then the solution is of the form $\displaystyle y=c_{1}*e^{\sqrt{-\lambda}*x}+c_{2}*e^{-\sqrt{-\lambda}*x}$

Now use the boundary conditions to solve for the constants $\displaystyle c_{1}$ and $\displaystyle c_{2}$.

From the boundary conditions y'(0) = 0 and y(1) + y'(1) = 0 you can establish the following system of linear equations:

let $\displaystyle \alpha = \sqrt{-\lambda}$

1. $\displaystyle c_{1}\alpha - c_{2}\alpha = 0$;

2. $\displaystyle c_{1}e^\alpha (1 +\alpha) + c_{2}e^{-\alpha}(1 - \alpha) = 0$

3. Make sure you use the non-trivial solution to $\displaystyle c_{1}$... I will leave it to you to convince yourself that non-trivial solutions only exist for $\displaystyle \lambda > 0$.

The solutions will be in the form of trigonometric functions because the characteristic equation has complex roots.