Limit sets of a gradient field

Question

I am trying to solve this question on J. Sotomayor's book on ODEs.

Define $X=\nabla f$, $f$ being defined in an open subset $\Delta \subset \mathbb R^n$. Prove that $X$ has no periodic orbits. And, if $X$ have only isolated singular points, show that is $p\in \Delta$ then the limit set of $p$ is empty or is a singular point.

About the first statement: if $\gamma$ is a (non-constant) periodic orbit, then, for some $T>0$, $\gamma(0)=\gamma(T)$. Therefore:

$$0=f(\gamma(T))-f(\gamma(0))=\int_0^T\nabla f(\gamma(t)) \cdot\gamma'(t)dt=\int_0^T\nabla f(\gamma(t)) \cdot \nabla f(\gamma(t)) dt =$$

$$=\int_0^T\vert{\nabla f(\gamma(t))\vert^2dt>0 }$$

and this is an absurd.

But I am having some troubles in the second part. I have some ideas.

If the orbit $\gamma_p$ passing through $p$ is not periodic then it is constant or it is injective. If $y_p$ is constant, $p$ is a singular point and $\omega(p)=p$.

The trouble is when $\gamma_p$ is one-to-one. What I've been trying to do is to prove that in this case $q \in \omega(p)$ only if

$$\lim_{t \to \infty} \gamma_p(t)=q$$

and then using the fact that $$f(q)-f(\gamma(0))=\int_0^\infty \vert\nabla f(\gamma(t))\vert^2dt$$

But the integral on the right side converges only if $$\lim_{t \to \infty} \nabla f(\gamma_p(t))=\nabla f(q)=0$$

and therefore $q$ is a singular point.

Is this correct? If it is, any hints of how to complete the missing step? It seems pretty intuitive to me, but I can't formalize it.

Answer 1

The second part of your argument is incorrect because you lack the hypothesis of $\gamma$ being maximal, this is, defined on the greatest open interval it can be defined. You do not know "a priori" if $\gamma$ is defined on $ (0, \infty) $, so that your above reasoning works. The point here is that you can map diffeomorphically the interval where $\gamma $ is defined to $(0, \infty)$ if and only if the curve is maximal. Argue by contradiction, and let $(0, b_1) $, $(0,b_2)$ with $b_1 < b_2$ be intervals where the integral curve is defined, such that $\gamma (0,b_1) \subset \gamma (0,b_2)$. Now map diffemorphically those intervals to $(0, \infty)$, so you get integral curves $x_1,x_2$ defined on $(0, \infty)$,which, by uniqueness of integral curves, are the same and, since they are defined on the same interval, they have the same image. Then necessarily $\gamma (0,b_1) = \gamma (0,b_2)$, which is not true.

However, the result is more general than what you proved: one point $q$ being limit (when time goes to the extreme of the maximal interval of definition of the curve) of a maximal integral curve of any differentiable vector field $X$ does necessarily satisfy $X(q) = 0$. Let $\gamma(t) $ be an integral curve of $X$ such that $\gamma(t) \to q$ when $t$ goes to infinity. Suppose by contradiction that $X(q) \not = 0$. Then there exists a regular curve $y$ solution of $ y ' (t) = X( y(t)), y(0)= q$. Thus you can construct another integral curve $z$ defined as $z(t) = x(t), t< a$ (after applying a diffeomorphism from the interval $(-\epsilon, \infty)$, and $z(t) = y(t), t \geq a$ (after applying a diffeomorphism from the interval $(-\epsilon, \epsilon)$ of definition of $y$). Hence, by uniqueness of integral curves, it is $z = x $, but the image of $z$ strictly contains the image of $x$, which contradicts the fact of $x$ being maximal.

Answer 2

I have to excuse myself, the reasoning in my preceeding answer is wrong, I just leave it there so people can see what the mistake is (quite a silly one, I have to admit): if a diffeomorphism $\varphi$ is applied to an integral curve $\gamma$, the resulting curve $ \gamma \circ \varphi$ is not necessarily an integral curve of the same field! For example, take in our case of a gradient vector field $\nabla f$, an integral curve $ x(t) $ defined on $ \left( 0, \infty \right)$. The curve $ y(t) = x ( \frac{1}{t})$ has the same image, but it is not an integral curve of $\nabla f$ because $ \dot{y} (t) = - \frac{1}{t^2} \dot{x}( \frac{1}{t}) = - \frac{1}{t^2} \nabla f ( y(t))$.

I am going to explain now the answer in a more general context than the one you asked for: for any differentiable vector field $X$ defined on an open set $\Delta \subseteq \mathbb{R}^n$, if $q$ is in the limit set of $p$, this is, if $\gamma(t)$ is the maximal integral curve which passes through $p$, $q = \lim\limits_{t \to \infty} \gamma(t)$, then $X(q) = 0$.

Argue by contradiction and assume $X(q) \not = 0$. Then, if $X= (x_1,...,x_n)$, $x_k (q) \not = 0$ for some $k$. By continuity, $\lim\limits_{t \to \infty} X(\gamma(t)) = X(q)$, in particular, for the $k$ coordinate $\lim\limits_{t \to \infty} x_k(\gamma(t)) = x_k(q)$. Hence, for $t \geq t_0$, $x_k ( \gamma(t)) $ does not change of sign. Now, consider the projection $\pi_k: \mathbb{R}^n \to \mathbb{R}$ over the $k$ coordinate and, reasoning as you do above:

$$ \pi_k ( \gamma (t) ) - \pi_k ( \gamma (t_0)) = \int_{t_0}^t \pi_k ( \gamma (s))'ds = \int_{t_0}^t e_k \gamma(s)' ds= \int_{t_0}^t \gamma_k(s)' ds= \int_{t_0}^t x_k (\gamma (s))ds. $$

Making now time go to infinity: $\pi_k (q) - \pi_k (\gamma(t_0)) = \int_{t_0}^{\infty} x_k (\gamma (s))ds $. Since the first member of the equality exists, the integral is well defined, so it must be $ \lim\limits_{t \to \infty} x_k (\gamma (t)) = x_k (q) = 0$, which is a contradiction.

In a nutshell, it must be $X(q)= 0$, otherwise, we get the preceding absurd.