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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.

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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.

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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 $f'(a)=f'(b)$.

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