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I am trying to solve the following:

Assume that $E$ is a smooth function in $\mathbb{R}^n$, and let $\nabla E$ denote the gradient, show that

$$x' = \nabla E(x)$$

has no nonconstant periodic solution.

So I start by assuming that $x(t)$ is a nonconstant periodic solution with minimal period $T$. Then $$\frac{d}{dt}E(x(t)) = \nabla E(x(t))x'(t) = |\nabla E(x(t))|^2 \geq 0.$$

Since $x(t)$ is periodic, we have $x(0) = x(T)$ and thus $E(x(0)) = E(x(T))$. I'm told this implies that $|\nabla E(x(t))|^2 = 0$ for all $t \in [0,T]$, which, if it is, really helps me out because then I've reached a contradiction. However, I haven't been able to understand why $E(x(0)) = E(x(T)) \implies |\nabla E(x(t)) |^2 = 0$. Hoping that someone can help clarify this.

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You're almost done, you just need to integrate :

$$x(T)-x(0) = \int_0^T \frac{dE(x(s))}{ds} ds = \int_0^T |\nabla E (x(s))|^2 ds,$$

so if $x(T) = x(0)$, then the integral is zero; since the integrand is nonnegative, is must be also be zero.

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