Even and Odd Eigenfunctions in Sturm-Liouville Problems

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SUMMARY

The discussion centers on the Sturm-Liouville eigenvalue problem defined by the equation (p(x)y')' + r(x)y = λy with boundary conditions y(-a) = y(a) = 0. It is established that the functions p(x) and r(x) are even functions, satisfying p(-x) = p(x) < 0 and r(-x) = r(x) for all x in the interval [-a, a]. The conclusion drawn is that every eigenfunction corresponding to this problem is either an even or odd function, as demonstrated by the symmetry in the problem and the properties of the eigenfunctions.

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Homework Statement


We are given the following Sturm-Liouville eigenvalueproblem:
(p(x)y')' + r(x)y = \lambday
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) \forallx \in [-a, a]. Show that every eigenfunction is either even or odd.


Homework Equations


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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?
 
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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.
 

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