Register to reply

Even and odd eigenfunctons

by Tomath
Tags: eigenfunctons
Share this thread:
Tomath
#1
Jun10-12, 07:04 PM
P: 8
1. The problem statement, all variables and given/known data
We are given the following Sturm-Liouville eigenvalueproblem:
(p(x)y')' + r(x)y = [itex]\lambda[/itex]y
y(-a) = y(a) = 0

on a symmetrisch interval I = [-a, a]. About p(x) and r(x) we are given that p(-x) = p(x) < 0 and r(-x) = r(x) [itex]\forall[/itex]x [itex]\in[/itex] [-a, a]. Show that every eigenfunction is either even or odd.


2. Relevant equations
-


3. The attempt at a solution
I was thinking of using the fact that for two different eigenvalues with their corresponding eigenfunctions w(x), v(x) the following identity holds:

$$\int_{-a}^{a} v(x) w(x) dt = 0$$

which hopefully implies that v(x) w(x) is an odd function. However, this doesn't really seem to work because both v(x) and w(x) can be even and then the identity above still holds (even though v(x) w(x) is even).

My second thought is trying to get some expression like the following:
$$\int_{-a}^{a} p'(x) w(x) dt = 0$$ or $$\int_{-a}^{a} r(x) w(x) dt = 0$$ since that would imply that w(x) is either even or odd. However I cannot seem to get any expression like that.

What am I missing in this problem?
Phys.Org News Partner Science news on Phys.org
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds
vela
#2
Jun11-12, 06:18 AM
Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,683
Try changing variables from x to -x. Use that to show that if y(x) is a solution, then y(-x) is also a solution.


Register to reply