A bound on a solution of an ODE, given some bounds on endpoints

Question

Let $V : [a,b] \to \mathbb{R}$ be smooth, strictly increasing and $V(a) = 0$. Suppose that $f : [a,b] \to \mathbb{R}$ is smooth and satisfies $f^{\prime \prime} (x) + V(x) f(x) = 0$ on $[a,b]$. Can we then bound $\sup_{x \in [a,b]} |f(x)|$ in terms of $f(a) , f(b) , f^{\prime} (b), V(b)$? I intentionally don't put $f^{\prime} (a)$ into the list.

Answer 1

It will be more convenient (at least to me) to turn the problem right to left, restating it as follows:

Let $V\colon[a,b]\to\mathbb{R}$ be smooth, strictly decreasing and $V(b) = 0$. Suppose that $f\colon[a,b]\to\mathbb{R}$ is smooth and satisfies $f''(x)+V(x) f(x)=0$ on $[a,b]$. Can we then bound $\sup_{x \in [a,b]} |f(x)|$ in terms of $f(a),f'(a),f(b),V(a)$?

Such a bound does not exist. Indeed, take any natural $n$. Let $x_0:=0$. For all $k=0,\dots,n-1$, let $x_{k+1}:=x_k+k$. Let $a:=x_0=0$ and $b:=x_n$. For all $k=0,\dots,n-1$ and all $x\in[x_k,x_{k+1})$, let
$$f(x):=(-1)^k\,k\,\sin\frac{\pi(x-x_k)}k,$$ $$V(x):=\frac{\pi^2}{k^2},$$ with $f(x_n):=0$ and $V(x_n):=0$.

Then $f\in C^1[a,b]$, $V$ is nonincreasing, $V(b)=0$, $f''+Vf=0$ on $[a,b]\setminus\{x_1,\dots,x_n\}$, $f(a)=0$, $f'(a)=\pi$, $f(b)=0$, $V(a)=\pi^2$. However, $|f((x_{n-1}+x_n)/2)|=n-1$, which is unbounded as $n\to\infty$, even though the values of $f(a),f'(a),f(b),V(a)$ are fixed.

It remains to smooth $V$ and $f$ near the points $x_k$ appropriately, also modifying $V$ to make it strictly decreasing. (It should be clear that such qualitative conditions as smoothness and strictness cannot affect the presence or absence of a quantitative bound.)