Application of Darboux’s theorem:Prove that:there exists $\xi \in (0,1)$, s.t. $f”(\xi ) = f(\xi )$

Question

In my textbook, there is an interesting problem in the section of Darboux's theorem:

• $f(x)$ is second-order differentiable on $[0,1]$

• $f(0)=2, f '(0)=0, f(1)=e+e^{-1}$

How to prove that:there exists $\xi \in (0,1)$, s.t. $$f''(\xi ) = f(\xi )$$

I noticed that it's required to construct a function and apply Darboux's theorem, just like the exercises of mean value theorem , and I found that maybe the solutions to $f''=f$:
$$y = {c_1}{e^x} + {c_2}{e^{ – x}}$$ is related to the problem since $f(1)=e+e^{-1}$.But I just fail to construct a nice function to do so.Can you give me some hint? And are there some special techniques to find such functions? (Maybe solving some related differential equations is one of them) Thank you very much.

Answer 1

Yes, you go over the solutions of this ODE in using them to straighten the problem. Consider $h(x)=f(x)\cosh(1-x)+f'(x)\sinh(1-x)$, where $\cosh(1-x)$ and $\sinh(1-x)$ are basis solution for the ODE $f''(x)=f(x)$.

Then one can compute

1. $h(0)=f(0)\cosh(1)+f'(0)\sinh(1)=2\cosh(1)$
2. $h(1)=f(1)\cosh(0)+f'(1)\sinh(0)=2\cosh(1)=e+e^{-1}$

It follows from the theorem of Rolle that there is a point $ξ\in(0,1)$ with \begin{align} 0=h'(ξ)&=[f'(ξ)\cosh(1-ξ)-f(ξ)\sinh(1-ξ)]+[f''(ξ)\sinh(1-ξ)-f'(ξ)\cosh(1-ξ)] \\ &=(f''(ξ)-f(ξ))\sinh(1-ξ). \end{align}