Question is that to prove if $$u'' + e^u = −x \text{ for } 0< x <1$$, then u cannot attain a minimum in $(0,1)$

Step I tried:

First, $u'' + e^u <0.$ I assume exists a $c \in (a,b)$ s.t. $u(c)=m,$ where $m$ is the minimum. Then, we have $u'(c)=0$ and $u''(c)>0$.

I want to show $$u(c)'' + e^{u(c)} >0$$ to get contradiction. Can I just say $u(c)'' + e^{u(c)} >0$ since we get that assume and done the proof? Or some thing need to be explain or prove?

  • $\begingroup$ $u'' < 0$ for all $x$ in the interval. If $u$ is continuous and differentiable, and it obtains its minimum in an open interval, we require $u'=0$ and $u'' > 0$ $\endgroup$
    – user317176
    Sep 28 at 10:43
  • $\begingroup$ That can't be determined. Nothing says that this function achieves a maximum either. It could be monotonic. $\endgroup$
    – user317176
    Sep 28 at 11:01
  • $\begingroup$ Try a simple contradiction: If $u$ has a minimum $x_0\in (0,1)$, then $u''(x_0) \ge 0$. But as mentioned in the first comment, $u'' < 0$ everywhere, so there cannot be any minimum in $(0,1)$. $\endgroup$
    – DominikS
    Sep 28 at 11:18

1 Answer 1


It seems that the point of the exercise is to show that there is no minimum on the open interval $(0,1)$.

As already pointed out in the comments, $u''<0$ on $(0,1)$, therefore the function is concave. If it has a minimum on $[0,1]$, this will have to be on the boundary $x\in\{0, 1\}$, but not on the open interval $x\in (0,1)$, which is a general property of concave functions on $\mathbb R$, see Note 2.

Note 1: due to a lack of initial or boundary conditions, you cannot rule out a minimum on the boundary.

Note 2: A function with $u'' < 0$ can have at most one maximum or minimum on the interior $(0,1)$, because $u'$ is strictly monotonic and can thus only be zero at most once. This point where $u'=0$ will have $u'' < 0$, hence any extremum in $(0,1)$ will be a strict maximum. This leaves the boundary points $\{0, 1\}$ as the only remaining candidates for minima, but they are not in the open interval $(0,1)$.

  • $\begingroup$ Thank you! I think I be more clearly about it. $\endgroup$
    – Apple
    Sep 29 at 3:40

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