# Elementary properties of gradient systems

Consider $$x_0\in\mathbb{R}^n$$ and a $$C^{1,1}$$ function $$f:\mathbb{R}^n\rightarrow\mathbb{R}$$ (that is, a differentiable function whose gradient is Lipschitz function). Consider the system $$\begin{cases} \dot{x}(t)=-\nabla f(x(t)),\\ x(0)=x_0. \end{cases}$$ Let $$\gamma_{x_0}(t)$$ be a trajectory (orbit) of the above system which starts at the point $$x_0$$ is defined for all $$t$$ in the maximal interval $$[0,T_{x_0})$$ ($$T_{x_0}\in(0,+\infty]$$). I am interested in the following elementary properties of the trajectory (orbit) $$\gamma_{x_0}(t)$$:

1. Consider the mapping $$t\mapsto\rho(t):=f(\gamma_{x_0}(t))$$. Prove that if $$\nabla f(x_0)\ne 0$$ then $$\rho(t)$$ is strictly deacreasing.

2. The $$\omega-$$limit set of $$\gamma_{x_0}$$ is defined by $$\Omega(\gamma_{x_0}):=\{\gamma_\infty:\exists (t_n)\rightarrow\infty, \lim\gamma_{x_0}(t_n)=\gamma_\infty\}.$$ Prove that if $$\gamma_{x_0}$$ is bounded (that is, there exists a bounded set $$B$$ such that $$\gamma_{x_0}(t)\in B$$ for all $$t$$) then:

• $$\Omega(\gamma_{x_0})$$ is either singleton or infinite;

• $$\nabla f(\gamma_\infty)=0$$ for all $$\gamma_\infty\in\Omega(\gamma_{x_0})$$;

• $$f$$ is constant on $$\Omega(\gamma_{x_0})$$;

• $$\text{dist}(\gamma_{x_0}(t),\Omega(\gamma_{x_0}))\leq\text{dist}(\gamma_{x_0}(t), S)$$, where $$S:=\{x:\nabla f(x)=0\}.$$ Moreover, $$\text{dist}(\gamma_{x_0}(t), S)\rightarrow 0$$ as $$t\rightarrow\infty$$.

• If $$\gamma_{x_0}$$ has finite length, that is $$\text{length}(\gamma):=\int_{0}^{T_{x_0}}\|\dot{\gamma}(t)\| dt<+\infty$$ then $$\Omega(\gamma_{x_0})$$ is a singleton which is denoted by $$\gamma_\infty$$ and $$\lim_{t\rightarrow\infty}\gamma_{x_0}(t)=\gamma_\infty.$$

My attempt

The derivative of $$\rho$$ is given by $$\rho^{\prime}(t)=\langle\nabla f(\gamma_{x_0}(t)),\dot{\gamma}(t)\rangle=-\|\nabla f(\gamma_{x_0}(t)\|^2=-\|\dot{\gamma}(t)\|^2\leq 0.$$ Hence $$\rho(t)$$ is decreasing. But I cannot know how to prove $$\rho(t)$$ is strictly decreasing.

I have tried to prove these properties but it is uneasy to prove them. I would be grateful if someone could help me to make clear all the properties. Thank you for all kind help.

• I am grateful if someone coud give me the references for the above properties. Commented May 28, 2016 at 15:24
• Are you sure about the second last point, since $\Omega(\gamma_{x_0}) \subset S$, the inclusion must be the other way. Commented May 30, 2016 at 4:33
• Oops, replace 'inclusion' by 'order' in my previous comment. Commented May 30, 2016 at 4:46
• Do you agree with my edits? Commented Jul 18, 2023 at 8:39

Here are hints for parts of the questions:

1) Note that if $\nabla f(x_0) \neq 0$ then $\dot{\gamma}(t) \neq 0$ for all $t \ge 0$. To see this, if $\dot{\gamma}(t_0) = 0$ for some $t_0 >0$, then $t \mapsto \gamma(t_0)$ and $t \mapsto \gamma(t)$ would be two solutions passing through $\gamma(t_0)$ and by uniqueness they must be the same hence a contradiction. It follows that $\dot{\rho}(t) <0$ for all $t \ge 0$.

2) Since $\gamma_{x_0}$ is bounded, it is clear that $\Omega(\gamma_{x_0})$ is non empty (the sequence $\gamma_{x_0}(n)$ must have an accumulation point). In particular, there is at least one $\omega$-limit point.

It is not too difficult to show that the $\omega$-limit is closed, hence compact.

Suppose $\gamma_\infty^k$ are $\omega$-limit points for $k=1,...,m$, for $m>1$. Choose $\epsilon>0$ such that the open sets $U_k=B(\gamma_\infty^k, \epsilon)$ are disjoint. Since $K=\Omega(\gamma_{x_0}) \setminus ( U_1 \cup \cdots \cup U_m)$ is non empty (since $\gamma_{x_0}$ is continuous) and compact, it follows that the trajectory must intersect $K$ infinitely often, hence $K$ contain another distinct $\omega$-limit point. It follows that there are an infinite number of $\omega$-limit points.

Suppose $\gamma_\infty \in \Omega(\gamma_{x_0})$ and $\nabla f(\gamma_\infty) \neq 0$. Let $K$ be a compact set containing the trajectory $\gamma_{x_0}$. It follows that there is some $M$ such that $\|\dot{\gamma_{x_0}}(t)\| \le M$, hence $t \mapsto {\gamma_{x_0}}(t)$ is uniformly continuous. Let $\epsilon = { 1\over 2} \| \nabla f(\gamma_\infty) \|$ and choose $\delta>0$ such that if $x \in B(\gamma_\infty, \delta)$ then $\| \nabla f(x) \| > \epsilon$.

There is a sequence $t_n \to \infty$ such that $\gamma_{x_0}(t_n) \in B(\gamma_\infty, {1 \over 2}\delta)$, and since $\gamma_{x_0}$ is uniformly continuous, there is some $T>0$ such that $\gamma_{x_0}(t) \in B(\gamma_\infty, \delta)$ for all $t \in [t_n, t_n+T]$. Note that $\rho(t_n+T) -\rho(t_n) \le - T\epsilon^2$. Since $t \mapsto \rho(t)$ is non increasing, we see that $\rho(t_n) \downarrow -\infty$, which is a contradiction and so $\nabla f(\gamma_\infty) = 0$.

Hence $\Omega(\gamma_{x_0}) \subset S$.

Since $\rho$ is non increasing, it follows that $f$ is constant on the $\omega$-limit set.

It is straightforward to show that if $d(\gamma_{x_0}(t), S) \not\to 0$, then there is an $\omega$-limit point $p$ that is not in $S$. This is a contradiction.

For the last point, suppose $\Omega(\gamma_{x_0})$ contains two distinct points $\gamma_1, \gamma_2$. Let $U_k = B(\gamma_k, {1 \over 3} \|\gamma_1-\gamma_2\| )$ (a diagram might help here). Let $t_n \to \infty$ be a sequence of points such that $\gamma_{x_0}(t_n) \in U_1$, and let $t_n' = \inf \{ t \ge t_n | \gamma_{x_0}(t) \in U_2 \}$.

By taking a subsequence if necessary we can assume that $t_{n+1} > t_n'$.

Then ${1 \over 3} \|\gamma_1-\gamma_2\| \le \|\gamma_{x_0}(t_n)- \gamma_{x_0}(t_n') \| \le \int_{t_n}^{t_n'} \| \dot{\gamma_{x_0}}(\tau)\| d \tau$.

Since the intervals $[t_n,t_n']$ are disjoint we see that $\int_0^t \| \dot{\gamma_{x_0}}(\tau)\| d \tau$ is unbounded.

It follows from this that if the length of the path is bounded then the $\omega$-limit set is a singleton and a compactness argument shows that the path converges to this limit.

• Dear copper.hat. Thank you for your kind work. Commented May 30, 2016 at 22:48
• Thanks. It is what I mean. Thanks a lot. Commented May 31, 2016 at 16:14
• Dear copper.hat. Could you make clear why the continuity of $\gamma_{x_0}$ implies $K:=\Omega(\gamma_{x_0})\setminus(U_1\cup U_2...\cup U_m)\ne\emptyset$. Thanks Commented May 31, 2016 at 16:17
• If $K$ was empty then after some time the trajectory must remain in exactly one of the $U_k$, which would contradict the $\gamma_\infty^k$ being $\omega$-limit points. Commented May 31, 2016 at 16:21