eigenvalues and eigenfunctions

**PvtBillPilgrim** · March 27th 2009, 12:37 PM

Can someone help me with this?

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

Thanks in advance.

**HallsofIvy** · March 27th 2009, 06:11 PM

Originally Posted by PvtBillPilgrim

Can someone help me with this?

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

Thanks in advance.

That equation is the same as $x^2y"+ xy'+ (3-u)y= 0$ , 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.

**Danny** · March 28th 2009, 05:34 AM

Originally Posted by HallsofIvy

That equation is the same as $x^2y"+ xy'+ (3-u)y= 0$ ,(**) 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 $y = x^m$ in HallsofIvy equation (**) and look at the separate cases for $m \;(> 0, = 0, < 0)$ .

**HallsofIvy** · March 28th 2009, 07:11 AM

Originally Posted by danny arrigo

You might also want to try $y = x^m$ in HallsofIvy equation (**) and look at the separate cases for $m \;(> 0, = 0, < 0)$ .

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 $e^{(a+bi)t}= e^{at}(cos(bt)+ i sin(bt))$ than with $x^{a+bi}$ .

**Danny** · March 28th 2009, 07:23 AM

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 $e^{(a+bi)t}= e^{at}(cos(bt)+ i sin(bt))$ than with $x^{a+bi}$ .

In my experience, I find students have more of a problem transforming the DE under $x = \ln t$ , than using $y = x^m$ . Once the basic solutions are derived $x^a \sin b \ln x$ and $x^a \cos b \ln x$ , I find the students find these easier to use.

**Squiggles** · March 29th 2009, 02:19 PM

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

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

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 $f(x) = \Sigma{(Ck)Yk}$ 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

**Danny** · March 29th 2009, 02:47 PM

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;

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

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:

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

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

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

Now use your BC's.

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

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

**Squiggles** · March 29th 2009, 03:07 PM

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

)

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