Solving the Sturm-Liouville System: Finding Eigenvalues and Eigenfunctions

[sassie]

Homework Statement

Evaluate and find the eigenvalues and eigenfunctions for:

$$-\frac{d}{dx}[(1+x^{2})\frac{dy}{dx}]=\lambda\frac{1}{1+x^{2}}y}$$

Homework Equations

The Attempt at a Solution

Okay, what I did was to expand the given equation to get:

$$((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0$$ <---- edit typo

But then I get on how to get the needed general solution. I have tried substitution by letting $$t=1+x^{2}$$ and then using the chain rule to get the first and second derivatives etc. I tried that to get a nice equation, but it ended up with a messier equation. Now I am really stuck and don't know what to do. Your help is very much appreciated.

 

[Mark44]

Homework Statement

Evaluate and find the eigenvalues and eigenfunctions for:

$$-\frac{d}{dx}[(1+x^{2})\frac{dy}{dx}]=\lambda\frac{1}{1+x^{2}}y}$$

Homework Equations

The Attempt at a Solution

Okay, what I did was to expand the given equation to get:

$$((1+x^{2})^{2})y''+2x(1+x^{2}) + \lambda y =0$$

But then I get on how to get the needed general solution. I have tried substitution by letting $$t=1+x^{2}$$ and then using the chain rule to get the first and second derivatives etc. I tried that to get a nice equation, but it ended up with a messier equation. Now I am really stuck and don't know what to do. Your help is very much appreciated.

It looks like you made an error when you differentiated. I get
$$-(1 + x^2)y'' -2x \frac{dy}{dx} = \lambda \frac{1}{1 + x^2} y' - \frac{2x}{(1 + x^2)^2} y'$$

After multiplying both sides by (1 + x²)² I don't get what you show.

 

[sassie]

It looks like you made an error when you differentiated. I get
$$-(1 + x^2)y'' -2x \frac{dy}{dx} = \lambda \frac{1}{1 + x^2} y' - \frac{2x}{(1 + x^2)^2} y'$$

After multiplying both sides by (1 + x²)² I don't get what you show.

Okay, what I did was (sorry I did skip a few steps):

LHS = $$-(1+x^2)y''-2xy'-(\lambda \frac{1}{1 + x^2}) y=0$$

I then multiplied through by $$(1+x^2)$$

$$((1+x^{2})^{2})y''+2x(1+x^{2}) + \lambda y =0$$

Now, I understand how you got the left hand side of the equation you gave, I am a little confused about how you got the right hand side.

 

[sassie]

But can anyone help me with this question? I've been working on it for 5 hours straight and still I can't a solution! This question is just so much harder than ones I have handled previously!

 

[lanedance]

ok so the first step is to solve the general ode

then find the eigenvalues and functions from the general solution, using the boundary conditions, where are you up to?

EDIT - re-read, so you're trying to find the general solutions...

 

[sassie]

I haven't even got the general ODE and that's what I have been having problems with. I have absolutely no idea how to obtain it :'(

I know it should be easy but for the life of me, I can't do it

 

[sassie]

ok so the first step is to solve the general ode

then find the eigenvalues and functions from the general solution, using the boundary conditions, where are you up to?

EDIT - re-read, so you're trying to find the general solutions...

Re: EDIT

Yes, so I am trying to get the general solutions to the given equation but I do not know how to.

 

[Mark44]

Okay, what I did was (sorry I did skip a few steps):

LHS = $$-(1+x^2)y''-2xy'-(\lambda \frac{1}{1 + x^2}) y=0$$

I then multiplied through by $$(1+x^2)$$

$$((1+x^{2})^{2})y''+2x(1+x^{2}) + \lambda y =0$$

Now, I understand how you got the left hand side of the equation you gave, I am a little confused about how you got the right hand side.

I should not have differentiated the right side. I erred by thinking incorrectly that we were differentiating both sides. Sorry for the bad advice - it was late at night and I misread your equation.

Your equation still has an error, though. It is missing a y' factor in the second term.
$$((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0$$

 

[sassie]

I should not have differentiated the right side. I erred by thinking incorrectly that we were differentiating both sides. Sorry for the bad advice - it was late at night and I misread your equation. Your equation was correct as it was.

Oh, okay! I was pondering why your RHS was the way it was. But can you help me with getting the equation into a form where I can the general solution? I REALLY need a hint to get me on the right track. I am sure I can solve for the eigenvalues and eigenfunctions myself, it's just been getting the general solution which has been causing me problems.

EDIT re: error sorry, that was a typo

 

[lanedance]

yeah that's tricky, mucked around with a few things, series was pretty nasty, even if you solve for the lambda = 0 case you get an inverse tangent...

one thing i did notice however is that if you use the chain rule with some substitution t(x), then using chain rule, with dashes for x derivative & dots for t derivative
$$y' = \frac{d}{dx}y(t(x)) = \frac{dy(t)}{dt}\frac{dt(x)}{dx} = \dot{y} t'$$

so then it would be nice to cancel the x term in the brackets, so i started with
$$t'(x) = \frac{dt(x)}{dx} = 1+x^2$$

and things evolved nicely form there...

 

[lanedance]

the nice part is that after cancelling the term, you're left with:
$$\frac{d}{dx}(\dot{y}) = \ddot{y} t'(x) = \ddot{y}(1+x^2)$$

and unless i missed something, everything simplifies a heap... ;)

 

[sassie]

the nice part is that after cancelling the term, you're left with:
$$\frac{d}{dx}(\dot{y}) = \ddot{y} t'(x) = \ddot{y}(1+x^2)$$

and unless i missed something, everything simplifies a heap... ;)

Okay, thanks. I'll try that out :)

 

[Mark44]

sassie, see the end of my post #8. Your equation in the first post did have an error in it.

 

[TheFirstOrder]

sassie, see the end of my post #8. Your equation in the first post did have an error in it.

Thanks, I did notice that mistake :)

the nice part is that after cancelling the term, you're left with:
$$\frac{d}{dx}(\dot{y}) = \ddot{y} t'(x) = \ddot{y}(1+x^2)$$

and unless i missed something, everything simplifies a heap... ;)

Hey lanedance, is it possible for you to put up more of your working? I seem to be getting some thing different

 

[lanedance]

apologies - think i made a mistake writing down the ODE, so it doesn't cancel like i thought;(

 

[sassie]

apologies - think i made a mistake writing down the ODE, so it doesn't cancel like i thought;(

sorry, that was my fault. i wrote down the wrong ode, but mark44 pointed out the mistake. look's like I'm back at square 1.

 

[Mark44]

I spent a little time reviewing S-L equations this morning. In most of the examples I saw there were boundary conditions. Are there boundary conditions in your problem?

 

[sassie]

I spent a little time reviewing S-L equations this morning. In most of the examples I saw there were boundary conditions. Are there boundary conditions in your problem?

yes,

y(0)=0 and y(1)=0

 

[lanedance]

so the original ODE is:

$$-\frac{d}{dx}[(1+x^{2})\frac{dy}{dx}]=\lambda\frac{1}{1+x^{2}}y}$$
$$((1+x^{2})^{2})y''+2x(1+x^{2})y' + \lambda y =0$$ <---- edit typo

ok think I've may have made some headway on the ODE, the lambda x=0 case is easy to solve & gives an inverse tangent function, which may be a hint of what direction to go...

now assume we are going to simplify by making a substitution t(x), then
$$y' = \dot{y}t'$$
$$y'' = \ddot{y} (t')^2 + \dot{y} t''$$

so going back to the DE and substituting in gives
$$((1+x^{2})^{2})(\ddot{y}(t')^2 + \dot{y}t'') +2x(1+x^{2})\dot{y}t' + \lambda y =0$$

collecting terms
$$((t')^2(1+x^{2})^{2})\ddot{y} + (t''(1+x^2)^2+t'2x(1+x^2))\dot{y} + \lambda y =0$$

so then it looks worth trying
$$t' = \frac{1}{1+x^2}$$

then differentiating again gives
$$t'' = \frac{-2x}{(1+x^2)^2}$$

then see how that goes subbing in, hopefully will simplify down for a better DE for t...

if this works, you'll need to be careful with your boundary conditions, on determining the eigenvalues & eigenfunctions...

 

[lanedance]

so did you get to $\ddot{y}+\lambda y = 0$?