Eigenstuff of Second Derivative

1. Oct 21, 2014

Hertz

Hi,

I'm trying to find the eigenvalues and eigenvectors of the operator $\hat{O}=\frac{d^2}{d\phi^2}$
Where $\phi$ is the angular coordinate in polar coordinates.

Since we are dealing with polar coordinates, we also have the condition (on the eigenfunctions) that $f(\phi)=f(\phi+2\pi)$.

This problem leads us to the differential equation $\hat{O}f=qf$ where q is the eigenvalue corresponding to the eigenfunction $f$.

The solution that I am looking at then concludes that the eigenfunctions take the form:$$f_q(x)=Ae^{\pm \sqrt{q} \phi}$$

But I don't see why this supposedly the solution instead of:$$f_q(x)=Ae^{\sqrt{q} \phi}+Be^{-\sqrt{q} \phi}$$

The second solution is the one I came up with, but I was unable to solve for the "spectrum of the operator" or the allowed values of q. That's why I tried to find an alternative solution.

I hope you can see my confusion. I understand how to find the spectrum given the first result, but I'm still not convinced that the first result is correct. After all, the general solution to the differential equation is what I got for my result. Furthermore, I can use the periodicity condition to solve for $B$ in terms of $A$ but for some reason my end result is not periodic. Can someone please provide some insight into the problem? Thank you

2. Oct 21, 2014

ShayanJ

The two solutions you mentioned are equivalent. You have two independent solutions which is the property of a second order ODE.
About your other problem. You need periodic solutions? OK, make the ones you have periodic. You should just require q to be negative!

3. Oct 21, 2014

Hertz

But the first solution has a single unknown whereas the second solution has two unknowns. Also, the first solution only has one term. It's a plus OR minus. It's as if they assume that either A or B has to be 0.

Good thinking on negative q's. I will look more into that. I'm still confused why my solution isn't periodic though. I set $f(\phi)=f(\phi+2\pi)$ and solved for B in terms of A and substituted it back into the original. I would think that the end product would be periodic, but it's not :(

4. Oct 21, 2014

ShayanJ

The first one is only a bad way of writing the second one!

By imposing the condition $q<0$, you're exponentials become imaginary which are linear combinations of sines and cosines! So q can't be any number, it has to be a negative real number and then the solutions become periodic.

5. Oct 21, 2014

Hertz

I don't think they are the same my friend.

-e
Oh or is it that they are just splitting the problem up into two separate eigenfunctions?! I.e. "For a given eigenvalue q, there are two eigenfunctions $f_+$ and $f_-$ which satisfy the differential equation." Then each eigenfunction has a single unknown and the periodicity condition can be used to find information about q. (Is there a less technical way to say "periodicity condition"?) Plus, the two eigenfunctions can be treated at the same time using the $\pm$ notation.

Can you do this?!?! This seems like a really hand-wavy argument to me!

Why do I need to impose this condition? Shouldn't I be able to solve for the appropriate values of $q$ just by assuming periodicity and then discovering the implications?

Here is my Mathematica code so you know I'm not making an algebra error:
Code (Text):

F[x_] := A E^{Sqrt[q] x} + B E^{-Sqrt[q] x}
eq1 := F[x] == F[x + 2 Pi];
f[x_] := F[x][[1]] /. Solve[eq1, B][[1]][[1]] // Simplify

In the first line I defined F as my general solution. In the second line I state the condition. And then in the third line I solve for B given the condition and substitute it into F, I define this new function as f.

So then, why is f not periodic? Is it because a sum of two exponentials simply can't be periodic? Then why don't I get an issue with solving the equation for B? (I did all this on paper too.)

Furthermore.... Even if I was able to solve for B and able to satisfy the periodicity condition, then what condition would I use to determine my allowed values of q??

Last edited: Oct 21, 2014