My issue with using eigenfunction expansion for ODEs

1. Jul 13, 2012

ericm1234

I have previously taken PDE's and ODE's. I understand obtaining the equation y''+lambda*y=0 (lambda then giving the eigenvalues). But I've encountered now the use of eigenfunction expansion for an ODE; and what I don't understand, is that in solving it they're making some assumption that y''+y=0 (the homogeneous problem) is actually y''+lambda*y=0, hence where the 'eigenfunctions for the homogeneous problem' are coming from.

2. Jul 13, 2012

tiny-tim

hi ericm1234!

any polynomial differential equation can be factored into linear factors

eg, y'' + Ay' + By = 0 can written as (D2 + AD + B)y = 0, which can then be factored into (D - λ1)(D - λ2)y = 0, the two independent solutions of which are (D - λ1)y = 0 and (D - λ2)y = 0

3. Jul 13, 2012

HallsofIvy

Staff Emeritus
What, exactly is your problem? Certainly "y''+ y= 0" is of the form "y''+ lambda y= 0" with lambda= 1.

4. Jul 13, 2012

ericm1234

Tiny Tim:
I mean..I sort of see that. Not sure how it relates though to my problem..
HallsofIvy: My problem is understanding where the "eigenfunctions of the corresponding homogeneous problem" are coming from. I agree that you could say lambda=1 and attain y''+y=0. In this problem, the eigenfunctions of the corresponding homogeneous problem are cos(nx) ( the boundary conditions were y(0)=y(pie)=0 ).
It SEEMS TO ME that this is really ONLY implied by the equation y''+lambda y=0 , certainly not by lambda being allowed to be =1. I say this because, it is in utilizing the B.C.'s that we attain sin(lambda x)= 0, in turn Sqrt(lambda) pie=n pie, etc.
I guess I am missing something obvious in understanding the 'difference' between my memorization of the process from my PDE class and your assigning of lambda=1 simply.

5. Jul 13, 2012

AlephZero

We aren't too good at guessing games here.

If you tell us what problem you are trying to solve and the way you are solving it, step by step, somebody can probably help. Without that information, you know what you remembered (or mis-remembered) from your PDE class, but we dont!

6. Jul 13, 2012

ericm1234

y''+y=ax+cosx, y(0)=y(pie)=0 using eigenfunction expansion.
As I said, my issue is not solving this problem; it's understanding why eigenfunction expansion is a valid method, when in doing the eigenfunction expansion, you need "eigenfunctions of the homogeneous problem". Ok, so y''+y=0 is the homogeneous problem.
But the "eigenfunctions of the homogeneous problem", being cos(nx), come from the problem y''+lambda y=0, then using the B.C.'s, finding the appropriate values for lambda.
So, my question is essentially, how does y''+y=0 imply "homogeneous eigenfunctions" being cos(nx)?

7. Jul 14, 2012

tiny-tim

y'' + λ2y = 0

(D2 + λ2)y = 0

(D + iλ)(D - iλ)y = 0

so the independent solutions are (D + iλ)y = 0 and (D - iλ)y = 0

ie dy/y = iλ and dy/y = -iλ

ie y = eiλt and y = e-iλt

so the general solution is y = Aiλt + Be-iλt

which we can also write as y = Acosλt + Bsinλt (different A and B, of course)

8. Jul 14, 2012

ericm1234

Yes Tim I see that..but in y''+y=0, there is no lambda.

9. Jul 14, 2012

tiny-tim

λ = 1

10. Jul 14, 2012

HallsofIvy

Staff Emeritus
Yes, there is. $\lambda= 1$ as I pointed out and you agreed. By tiny-tim's "eigenvalue" analysis, the general solution is $y= Ae^{ix}+ Be^{-ix}$ or, equivalently, $y(x)= C cos(x)+ B sin(x)$.

I still don't understand what your difficulty is.

11. Jul 14, 2012

AlephZero

I think you (and possibly Tiny Tim as well) are getting confused by the terminology here.

The way I see it, you have TWO eigenfunctions, $e^{ix}$ and $e^{-ix}$, or $\cos x$ and $\sin x$ if you prefer.

So the general solution is a linear combination of the two eigenfunctions, $A \cos x + B \sin x$, and the boundary conditions fix the values of A and B.

The $\lambda$ comes in if you start from one step further back (which is hardly necessary for this problem). Assume the equation $y'' + y = 0$ has a solution of general form $y = Ae^{\lambda x}$.
Substituting in the differential equation, you get $A(\lambda^2 + 1) = 0$.
So either $A = 0$ (which is not a very interesting solution) or $\lambda^2 + 1 = 0$. That is why $\lambda = \pm i$ are called "eigenvalues". Eigenvalue is just the German word for "specail value" - i.e. the values when you get "interesting" solutions. Another translation is "characteristic values", and $\lambda^2 + 1 = 0$ is sometimes called the "characteristic equation".

12. Jul 14, 2012

tiny-tim

i don't think i used any terminology (certainly not "eigenfunction") …

i just let the equations speak for themselves

13. Jul 14, 2012

Thanks guys