Functional analysis/integration problem?

If you assume y is twice differentiable, I think it is necessary and sufficient for y to satisfy the following two conditions:

(1) y'' + cy = 0 for all x.

(2) y'(a) = y'(-a) = 0.

Here is an argument: Integrating y'g' by parts gives:

int_{-a}^a y'g' = [y'(a)g(a) - y'(-a)g(-a)] - int_{-a}^{a} y''g

So:

0 = [y'(a)g(a)-y'(-a)g(-a)] - int_{-a}^{a}[g(y'' + cy)]dx

So we see the above equation is satisfied for all g whenever the above two conditions hold. It remains only to show that these conditions are _necessary_. Fix any interval [d,e] in (-a,a), and define h(x) as the function that is 1 on the interval [d,e] and zero everywhere else. Then there is a differentiable function g(x) that closely approximates h(x), and so we need:

0 = [y'(a)h(a)-y'(-a)h(-a)] - int_{-a}^a [h(y'' + cy)]dx = 0 + int_{d}^{e} (y'' + cy) dx

So for any interval [d,e] in [-a,a] we need:

0 = int_{d}^{e} (y'' + cy) dx.

Differentiating the above with respect to e gives the first condition y'' + cy = 0 for all x. The second condition then follows because we need y'(a)g(a)-y'(-a)g(-a)=0 for all differentiable functions g.

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Hmm, I think the only non-zero functions y that satisfy the above 2 conditions are when c>0, we get:

y = B*cos(sqrt(c)t)

where B is any real number, and c must be a positive number that satisfies a*sqrt(c) = pi/2 + k*pi for some integer k.

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Actually, I think there is one more solution when c<0 and a satisfies e^(sqrt(-c)a)=1, and y has the form y = A e^(sqrt(-c)t) - Ae^(-sqrt(-c)t). But these seem to be the only ones.

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