1. eigenvalues and eigenfunctions

Can someone help me with this?

How do I find the eigenvalues and eigenfunctions for the Sturm-Liouville problem
$\displaystyle x^2y''+xy'+3y=uy$;
$\displaystyle y(1)=0, y(2)=0$

2. Originally Posted by PvtBillPilgrim
Can someone help me with this?

How do I find the eigenvalues and eigenfunctions for the Sturm-Liouville problem
$\displaystyle x^2y''+xy'+3y=uy$;
$\displaystyle y(1)=0, y(2)=0$

That equation is the same as $\displaystyle x^2y"+ xy'+ (3-u)y= 0$, an "Euler type" or "equipotential" equation. The change of variable t= ln(x) changes it to an equation with constant coefficients. You should be able to find the general solution. Obviously, y(x)= 0, for all x, satifies the equation as well as y(1)= y(2)= 0. The eigenvalues are the values of u for which there are other, non-trivial, solutions. And the eigenfunctions are the non-trivial solutions.

3. Originally Posted by HallsofIvy
That equation is the same as $\displaystyle x^2y"+ xy'+ (3-u)y= 0$,(**) an "Euler type" or "equipotential" equation. The change of variable t= ln(x) changes it to an equation with constant coefficients. You should be able to find the general solution. Obviously, y(x)= 0, for all x, satifies the equation as well as y(1)= y(2)= 0. The eigenvalues are the values of u for which there are other, non-trivial, solutions. And the eigenfunctions are the non-trivial solutions.
You might also want to try $\displaystyle y = x^m$ in HallsofIvy equation (**) and look at the separate cases for $\displaystyle m \;(> 0, = 0, < 0)$.

4. Originally Posted by danny arrigo
You might also want to try $\displaystyle y = x^m$ in HallsofIvy equation (**) and look at the separate cases for $\displaystyle m \;(> 0, = 0, < 0)$.
The problem with that is that in order that the boundary conditions to be satisfied your m must be complex! I think most people, who have studied d.e.s with constant coefficients will be more comfortable with $\displaystyle e^{(a+bi)t}= e^{at}(cos(bt)+ i sin(bt))$ than with $\displaystyle x^{a+bi}$.

5. Originally Posted by HallsofIvy
The problem with that is that in order that the boundary conditions to be satisfied your m must be complex! I think most people, who have studied d.e.s with constant coefficients will be more comfortable with $\displaystyle e^{(a+bi)t}= e^{at}(cos(bt)+ i sin(bt))$ than with $\displaystyle x^{a+bi}$.
In my experience, I find students have more of a problem transforming the DE under $\displaystyle x = \ln t$, than using $\displaystyle y = x^m$. Once the basic solutions are derived $\displaystyle x^a \sin b \ln x$ and $\displaystyle x^a \cos b \ln x$, I find the students find these easier to use.

6. Hey I'm trying to answer the exact same problem for homework and am really confused..I get as far as;

$\displaystyle y = Ax^{ \sqrt{(3 - u)}} + Bx^{-\sqrt{(3 - u)}}$

But then I'm not sure what to do - I can see I'm going to get different results if u >3, <3, =3, but I don't know how to put that into an overall answer... at what point should I be using the boundary conditions?

Also, the 2nd part of the question asks me to expand a hypothetical f(x) in terms of the eigenfunctions - I assume in the format $\displaystyle f(x) = \Sigma{(Ck)Yk}$ where Ck are the sturm-louiville coefficients. But how can I do this if the format of Yk changes depending on whether u is > or < 3? Confused... :-s

7. Originally Posted by Squiggles
Hey I'm trying to answer the exact same problem for homework and am really confused..I get as far as;

$\displaystyle y = Ax^{ \sqrt{(3 - u)}} + Bx^{-\sqrt{(3 - u)}}$

But then I'm not sure what to do - I can see I'm going to get different results if u >3, <3, =3, but I don't know how to put that into an overall answer... at what point should I be using the boundary conditions?
The best thing to do is consider the following three cases:

$\displaystyle u = 3 + \omega^2,\;\; u = 3,\;\; u = 3 - \omega^2$. That corresponds to your three cases.

Cases 1: $\displaystyle u = 3 + \omega^2$ then

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

$\displaystyle y(1) = A 1^{\omega} + B 1^{-\omega} = 0$
$\displaystyle y(2) = A 2^{\omega} + B 2^{-\omega} = 0$

Two equations for the two unknowns $\displaystyle A\;\; \text{and}\;\;B$. The first says $\displaystyle B = -A$ whereas the second
$\displaystyle A 2^{\omega} -A 2^{-\omega} = 0$ so $\displaystyle A \left(2^{\omega} - 2^{-\omega}\right) = 0$ which gives A=0 leading to the trival solution $\displaystyle y = 0$. Similarly for the second case. It's the third case that's interesting. You try now.

8. Ah thank you, that makes sense! (I was trying to put all the possible solutions together before using the boundary conditions, d'oh! )