If $f(a) = f(b)$ and $f'(a) = f'(b)$, prove that for every real number $λ$ the equation $f''(x) − λ(f'(x))^2 = 0$ has at least one solution in $(a,b)$

Question

Let $f : [a, b] → R$ be a function, continuous on $[a, b]$ and twice differentiable on $(a, b)$. If $f(a) = f(b)$ and $f'(a) = f'(b)$, prove that for every real number $λ$ the equation $f''(x) − λ(f'(x))^2 = 0$ has at least one solution in the interval $(a, b)$

Quite clearly, Rolle's theorem could be useful because of the given conditions. Do you have other hints?

Answer 1

Let $g(x) = f'(x) e^{-\lambda f(x)}$ for all $x\in (a,b)$. Then, $g$ is $C^1$ on $(a,b)$, $C^0$ on $[a,b]$ and : $$\forall x\in(a,b), g'(x) =( f''(x) - \lambda f'(x)^2)e^{-\lambda f(x)}$$

so $f''(x) - \lambda f'(x)^2 = 0$ iff $g'(x) = 0$. From $f(a) = f(b)$ and $f'(a) = f'(b)$, we have $g(a) = g(b)$, so Rolle's theorem gives us what we want.

Remark : Note that you need $f$ to be $C^1$ on $[a,b]$ and twice differentiable on $(a,b)$ for Rolle's theorem to apply.