# $f(a)=f(b), f'(a)=f'(b)$, Existence of zeros of $f''(x)-\lambda (f'(x))^2 =0$ in $(a, b)$

Question:

$$f:[a, b]\to\mathbb R$$ is a function, which is contiunous and twice differentiable. If $$f(a)=f(b)$$ and $$f'(a)=f'(b)$$, for $$\forall\lambda\in\mathbb R$$, show that there exists at least one zero of the equation $$f''(x)-\lambda (f'(x))^2 =0$$ in the interval (a, b).

I first wanted to use the fact that $$\exists c_1\space\space s.t\space\space \frac{f(a)-f(b)}{a-b}=f'(c_1)=0$$ $$\exists c_2\space\space s.t\space\space \frac{f'(a)-f'(b)}{a-b}=f''(c_2)=0$$

However, I still do not know how these facts can be applied to the differential equation. Also, I multiplied LHS and RHS with $$e^{\lambda x}$$ but nothing happened. Could you please give me some key ideas about this problem? Thanks for answering.

Let $$F(x)=f'(x)e^{-\lambda(f(x)-f(a))},x\in[a,b].$$ And then you can check that: $$F(a)=F(b),F'(x)=e^{-\lambda(f(x)-f(a))}(f''(x)-\lambda(f'(x))^2)$$ Using Rolle's Theorem you can get the answer.
If you multiply for $$λ\ne 0$$ the equation with $$-λe^{-λf(x)}$$, then you will find that the condition can be reduced to $$(e^{-λf(x)})''=0$$ (See WKB approximation calculations and substitutions for Riccati equations for the inspiration.)
Now for $$g(x)=e^{-λf(x)}$$, $$g'(x)=-λf'(x)g(x)$$ you can again check that $$g(a)=g(b)$$ and $$g'(a)=g'(b)$$, and applying Rolle's theorem to the latter one you get $$g''(c)=0$$ for some $$c\in(a,b)$$.
The case $$λ=0$$, while separate, goes the same way starting from $$f'(a)=f'(b)$$.