# Math Help - Eigenvalues and Green's Formula

1. ## Eigenvalues and Green's Formula

Consider the eigenvalue problem in with the DE is

$\varphi^{(4)}-\lambda\varphi=0, \ \ \ 0

and the boundary conditions are

$\varphi(0)=\varphi''(0)=\varphi(L)=\varphi''(L)=0$

Can Green's Formula be used to show all the eigenvalues are real?

$\displaystyle A=-\frac{d^2}{dx^2}$

$\displaystyle I=\int_0^L[(A^2\varphi)\bar{\varphi}-\varphi(A^2\bar{\varphi})]dx$

Not sure if this is correct and what to do next.

2. Your "Green's Formula" looks like the typical inner product used to show that a self-adjoint operator has real eigenvalues. You might want to double-check that formula, though, as you seem to be assuming that the adjoint of $A^{2}$ is $A^{2}.$ That's what you'd be wanting to prove! I would try to show that the operator $D^{4}$, acting on any function satisfying your boundary conditions, is a self-adjoint operator. Then you know that the eigenvalues are real.

Alternatively, you could go the long route: find the eigenvalues, and get a condition on them that must be true in order not to have trivial solutions (since eigenvectors aren't allowed, by definition, to be zero), and show that that condition implies the eigenvalues are real.

Those are my 3/4 baked thoughts.

3. Originally Posted by Ackbeet
Your "Green's Formula" looks like the typical inner product used to show that a self-adjoint operator has real eigenvalues. You might want to double-check that formula, though, as you seem to be assuming that the adjoint of $A^{2}$ is $A^{2}.$ That's what you'd be wanting to prove! I would try to show that the operator $D^{4}$, acting on any function satisfying your boundary conditions, is a self-adjoint operator. Then you know that the eigenvalues are real.
How do I do this?

4. Here's how I would do it, actually: take a look at this link. You can see that for your BVP, the function vanishes at the endpoints, so this formal definition of the adjoint works. Make sure to look at the example (Stürm-Liouville operator) to get a feel for how to do it. Your problem is considerably simpler.

5. Showing the eigenvalues are all real.

$\displaystyle I=\int_0^L[(A^2\varphi)\bar{\varphi}-\varphi(A^2\bar{\varphi})]dx$

$A^2\bar{\varphi}=\bar{\lambda}\bar{\varphi}$

$\displaystyle I=\int_0^L[(\lambda\varphi)\bar{\varphi}-\varphi(\bar{\lambda}\bar{\varphi})]dx$

From Complex Analysis, we (meaning me and others who know) know $\varphi\bar{\varphi}=\varphi^2$

$\displaystyle (\lambda-\bar{\lambda})\int_0^L\varphi^2 \ dx=0$

Phi is continuous and not 0 everywhere on the interval [0, L], so

$\displaystyle\int_0^L\varphi^2 \ dx>0$

$
(\lambda-\bar{\lambda})=0\Rightarrow \lambda=\bar{\lambda}$

Therefore, all eigenvalues are real.

6. Have you shown that $A$ is self-adjoint? If not, then how do you know in advance that $I=0?$

7. Originally Posted by Ackbeet
Have you shown that $A$ is self-adjoint? If not, then how do you know in advance that $I=0?$
On another piece of paper, I used Green's Formula to show the integral was 0.

8. I'd be curious to see this Green's Formula. Can you provide a reference for it? All the Green's formulae I know of are vector integral equations.

9. Originally Posted by Ackbeet
I'd be curious to see this Green's Formula. Can you provide a reference for it? All the Green's formulae I know of are vector integral equations.
$\displaystyle\int f^{(4)}g \ dx= f'''g-fg'''+f'g''-f''g'+\int fg^{(4)}dx$

$\displaystyle\Rightarrow \int f^{(4)}g-fg^{(4)}dx= f'''g-fg'''+f'g''-f''g'$

From the boundary conditions given, this integral is 0.

10. Oh, I see. Integration by parts a whole bunch of times. Hmm. Never heard that called Green's Formula. Which book calls it that?

Nice derivation, by the way.

11. Originally Posted by Ackbeet
Oh, I see. Integration by parts a whole bunch of times. Hmm. Never heard that called Green's Formula. Which book calls it that?

Nice derivation, by the way.
It is called that in Elementary PDE by Berg and McGregor. It has to do with Green's first or second equation. It came by through LaGrange but formalized by Green.

12. How do I show if the eigenfunctions of different eigenvalues are orthogonal using Green's Formula (they may not be orthogonal too)?

13. Well, you want to form the inner product

$\displaystyle\langle\phi|\psi\rangle=\int_{0}^{L}\ bar{\phi}\psi\,dx,$

where $\phi,\psi$ are eigenvectors corresponding to different eigenvalues, and you want to show that that inner product is zero. I think the usual procedure here is to look at the expression

$\langle\phi|A\psi\rangle-\langle\psi|A\phi\rangle=\lambda_{\psi}\langle\phi |\psi\rangle-\lambda_{\phi}\langle\psi|\phi\rangle.$

The expression on the left looks like your Green's formula, so I think you might be able to show that it's zero. And then, since the eigenvalues are different (and real), I think your result will follow. You might need the self-adjoint-ness of the operator, though.