$\dot u(t) = - \nabla V(u(t)) $ unique solution if $V$ is convex

Question

I found this statement in a book I am reading:

If $V: \mathbb{R}^n \rightarrow \mathbb{R}$ is differentiable and convex, then the differential equation $$\dot u(t) = - \nabla V(u(t)) $$ has a unique solution $u: [0, + \infty) \rightarrow \mathbb{R}$ for every initial value $u_0$.

How can I prove it?

Answer 1

I think the standard Gronwall's argument will work:

Let $u$ and $v$ be two solutions of the same initial value problem, and let $w = u - v$. Then $w(0) = 0$ and $w$ satisfies

$$ w' = u' - v' = -\nabla V u + \nabla V v. $$

Now, consider $f(t) = \left<w(t),w(t)\right>$, we want to show this is zero. We know that $f(0) = 0$. Calculating $f'$ gives $$ f'(t) = 2\left<w'(t),w(t)\right> = -2\left<\nabla V u - \nabla V v,u - v\right> \leq - 2 C \left<u-v,u-v\right> = -2C f(t) $$ by the definition of convexity.

Now, just use Gronwall's inequality to complete the proof.