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

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!

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 $x$.

• Very nice proof. Is it true that it would be enough to assume that $g(a)=0$ (and not assume that $g(b)=0$ also)? Commented May 30, 2018 at 22:11
• No, since then there could be a maximum or minimum at $b$. For instance, $g(x)=x$ is a counterexample for $[a,b]=[0,1]$. Commented May 30, 2018 at 22:15
• More generally, the theory of ODEs gives a solution on $[a,a+\epsilon]$ for some $\epsilon>0$ given any initial values of $g(a)$ and $g'(a)$. If you pick $g(a)=0$ and $g'(a)\neq 0$, then for sufficiently small $\epsilon$, $g(a+\epsilon)$ will have the same sign as $g'(a)$. Commented May 30, 2018 at 22:18
• Regarding your first argumeny, that $g$ has a total maximum at $x\in (a,b)$: couldn't $x$ be a for example? Commented May 30, 2018 at 22:54

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.

• I would to know what are the mistakes of my argument. The simple unknown negation is not a good teacher. Commented May 31, 2018 at 7:48
• How does $\int_0^\eta g(\tau)d\tau=0$ mean that $g(\tau)=0$ for all $\tau$? That doesn't follow at all... Commented May 31, 2018 at 15:44
• Thanks for the argument. Note that $g'(\eta) = 0$ then $g(\eta) \ne 0$ (local maximum/minimum) otherwise $g = 0$ Commented May 31, 2018 at 16:19
• It should exists at least one $\eta$ which verifies $g'(\eta) = 0, g(\eta)\ne 0$ Commented May 31, 2018 at 16:22