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This is a problem I stuck in Berkeley Problems in Mathematics, fall 1983:

Let $x(t)=(x_{1}(t)...x_{n}(t))$ be a differentiable function from $\mathbb{R}$ to $\mathbb{R}^{n}$. It satisfies a differential equation of the form $$x'(t)=f(x(t))$$

where $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is a continuous function. Assuming that $f$ satisfies the condition $$\langle f(y),y\rangle\le |y|^{2}$$ derive an inequality showing that $|x(t)|$ grows at most exponentially.

I am thinking about decompoising $f(y)$ into $y^{T}$ and $y$ directions, but this does not allow me to integrate $x'(t)$ from $t_{0}$ to $t$. If I can show $|f(y)|\le K|y|$ then the inequality would be immediate; but this is false as $f(y)$ can be orthogonal to $y$ at every point $y=x(t)$ and have arbitrarily large norm. So I do not know how to solve this in an elegant way. I can solve it in the most radical case that $x'(t)\cdot x(t)=0\forall t$ and $x'(t)=K(t)x(t)$, but I do not know how to deal with arbitrarily curves.

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up vote 3 down vote accepted

Let $V(t) = \frac{1}{2} \|x(t)\|^2$. Then $\dot{V}(t) = \langle x(t), \dot{x}(t) \rangle = \langle x(t), f(x(t)) \rangle \leq \|x(t)\|^2 = 2 V(t)$.

Now consider $\eta(t) = e^{-2t}V(t)$; we have $\dot{\eta}(t) \leq 0$, from which we get $\eta(t) \leq \eta(0)$, which is equivalent to $V(t) \leq V(0) e^{2t}$. Multiplying by $2$ and taking square roots gives $\|x(t)\| \leq \|x(0)\| e^t$.

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Thanks, this is extremely clear. – Bombyx mori Jul 22 '12 at 6:04

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