Uniqueness of the solution of an ODE with zero integral mean and convex $f$ 
Consider the problem on $(0,1)$
  $$u(x)'=f(u(x))$$
  $$\int_0^1u(x)dx=0$$
  where $f$ is $C^1$ on the whole real line, strictly convex, has its minimum in $0$, and $f(0)<0$.
  Prove that the solution is unique.

I tried two ways: to reduce the problem to a Cauchy problem, showing that the condition on the mean, together with the properties of $f$, implies some 'initial-value' condition; or to apply the contraction principle (i.e. showing that the Fredholm operator associated to $f$ is a contraction). But I have not been able to follow any of these ways.
Can someone give a help, also explaining, if possible, useful strategies to apply in similar cases (i.e. 'modified' Cauchy problems)?
Thank you in advance.

Let $\phi_x(u_0)$ be the solution of the ode for $u(0)=u_0$.
Then $\phi_x(v_0)> \phi_x(u_0)$ for $v_0>u_0$ (by uniqueness of solutions to ode) for $x$ values for which both exist. So their integrals are also distinct. This shows uniqueness. 
To get existence note that if $f(u_0)<0$ then $\phi_x(u_0)$ exists and is  a decreasing function for all  $x\geq 0$. For $u_0=0$ the integral is therefore negative. If $f\geq -M$ then $\phi_x(u_0)\geq u_0-Mx$ so for $u_0$ large enough the integral becomes positive. By the IVT there is a unique $u_0$ for which the integral vanishes. Convexity is not of importance. Neither is the fact that the minimum of $f(u)$ is for $u=0$.
