Let's say we have a 2nd-order linear ODE $$\mathcal{L}\,y(x)=0,$$ whose differential operator $\mathcal{L}$ is parity even. We want to solve it for homogeneous Dirichlet boundary condition (b.c.) $y(-a)=y(a)=0$ for some general $a$.

Since $\mathcal{L}$ is linear and parity definite, we can always construct even or odd solutions, either of which can satisfy the above b.c.. So we formally have both even and odd solutions for the b.c. above. We don't care about nonexistence. Let's assume the uniqueness of solution. That means our solution is even/odd/both even and odd. So it's just zero for the latter. (I think even/odd cases are only possible for specific $a$?)

Is this argument correct? If so, is the assumed uniqueness the only possible loophole for a counterexample?

BTW, $y''+y=0$ can have $\sin{x},\cos{x}$ solutions only for special isolated $a$, not some general or continuum of $a$.


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