Proving existence and uniqueness of a system Problem: Suppose $F: \mathbb{R} \to \mathbb{R}$ is strictly convex and $\mathcal{C}^1$. Suppose that $F(0) < F(x)$ for all $x \in \mathbb{R}$ and $F(0)<0$. Prove that the following has an unique solution:
$$
\begin{cases}
u'(x)=F(u(x)) \mbox{ for } x \in (0,1) \\
\int_0^1 u(x) dx =0
\end{cases}$$
Attempt: I tried to use contraction theorem defining the operator $Tu(x):=\int_0^xF(u(x))$ but I cannot continue. I noted that also $T$ is convex.
 A: I will develop the solution line suggested by Lutz Lehmann in the comments. Consider the (unique) maximal solution $u$ of the Cauchy problem
$$\left\{ \begin{align}
&u'=F(u) \\
&u(0)=0
\end{align}
\right.
$$
which is well-defined since $F$ is $C^1$. Now, by the strict convexity of $F$ and by the fact that $0$ is a minimum point we obtain that $F'$ is growing, $F'(x) >0$ for every $x>0$ and $F'(x) <0$ for every $x<0$. Therefore $\displaystyle \lim_{x \rightarrow \pm \infty} F(x)=\pm \infty$ and so $F$ has exactly one positive and one negative root (since $F(0)<0$), let us say $u_+$ and $u_-$. These are constant solutions of the ODE and thus
$$u_- < u(x)< u_+$$
for every $x \in \mathbb{R}$ (being bounded, $u$ is also globally defined). It follows that $u'=F(u) < 0$ and so by monotonicity $\displaystyle \exists\lim_{x \rightarrow \pm \infty} u(x)= u_{\mp}$. Note that
$$G: \mathbb{R} \rightarrow \mathbb{R}, \hspace{5pt} G(a)=\int_0^1 u(x+a) \mbox{ } dx$$
is continuous, strictly decreasing (since $G'(a)=\int_0^1 u'(x+a) \mbox{ } dx <0$) and $\displaystyle \lim_{a \rightarrow \pm \infty} G(a)=u_{\mp}$. By the intermediate value theorem, there exists an unique value of $a$ such that $G(a)=0$. Since the ODE is autonomous, $v(x)=u(x+a)$ is still a solution and so we have completed the existence part of the proof.
To show the uniqueness just observe that if $v$ is a solution of the original system then $u_- < v(0)<u_+$ (otherwise $\int_0^1 v$ would be strictly positive or negative) and extending $v$ to the only maximal solution of the ODE we get $ \displaystyle\lim_{x \rightarrow \pm \infty} v(x)= u_{\mp}$ and $v'<0$ (reasoning as above). Therefore it exists a (unique) value of $a \in \mathbb{R}$ such that $v(-a)=0$ and so $v(x-a)=u(x)$, by the uniqueness of the solution to the Cauchy problem. We conclude that
$$v(x)=u(x+a)$$
