If $g$ is 2 times differentiable in $[a,b]$ and $g''+g'\,g=g$ and $g(a)=g(b)=0$, prove that $g=0$.

Question

Let $g:[a,b]\rightarrow \mathbb{R}$ two times differentiable such that $$g''(x)+g'(x)\,g(x)=g(x),~x\in [a,b]$$ and $g(a)=g(b)=0$. Prove that $g(x)=0$ for all $x\in [a,b].$

Attempt. It seemed like one of those exercises where multiplying be a suitable factor we get a derivative. I started by multiplying with $g$, after with $e^g$ but I dind't get what I expected. Am I on the wrong path?

Thanks in advance for the help!

Answer 1

Hint: Consider what the equation says when $x$ is a global maximum or global minimum of $g$.

A full answer is hidden below.

Suppose $x\neq a,b$ is a global maximum of $g$. Then $g'(x)=0$ and $g''(x)\leq 0$. But the given equation then says that $g''(x)=g(x)$ so $g(x)\leq 0$. Since $g(a)=g(b)=0$, this means the maximum value of $g$ can only be $0$. Similarly, the minimum value of $g$ is $0$, so $g(x)=0$ for all $x$.

Answer 2

We have

$$ g'' +(g'-1)g=0\Rightarrow-\frac{g''}{1-g'} = g $$

then

$$ \ln(1-g')=\int g(\tau)d\tau $$

and then

$$ g' = 1 - e^{\int g(\tau) d\tau} $$

Now if $g$ is twice continuous $g'$ should be null for at least one $a < \eta < b$ or

$g'(\eta) = 0\Rightarrow e^{\int_0^{\eta}g(\tau)d\tau } = 1\Rightarrow g(\tau) = 0$ for $\tau \in (a,b)$ because otherwise $ e^{\int_0^{\eta}g(\tau)d\tau } \ne 0$ and then there is not possible the existence of such $\eta$ which is an absurd due to Rolle's theorem.