A uniqueness result for a BVP over a semi-infinite interval

Question

Suppose $f:[0,\infty) \to \mathbb{R}$ is continuous, and consider the ODE $$y''(x)=f(x)y(x) $$ with the boundary conditions $y(0)=y(\infty)=0$. Under what conditions on $f$ is $y \equiv 0$ the only solution? I know the result is true for $f(x) \in \{1,x,x^2\}$, since I have some understanding of exponentials, Airy functions and parabolic cylinder functions. But I'm looking for a stronger result. Thank you!

Answer 1

One sufficient condition is $f(x)\ge0$. Multiply the equation by $y(x)$ and integrate between $0$ and $t>0$: $$ \int_0^ty''(x)\,y(x)\,dx=\int_0^tf(x)\,(y(x))^2\,dx\ge0. $$ Integrating by parts the left hand side we get $$ y'(t)\,y(t)-\int_0^t(y'(x))^2\,dx\ge0\implies (y(x)^2)'\ge0. $$ It follows that $y^2$ is increasing and does not converge to $0$ at $\infty$ unless it is identically $0$.

The condition is far from necesary, as $f(x)=-1$ shows