# Stability of certain second order ODE

I am having a hard time determining the motion of $$X$$ in the ODE $$X''+\nabla f(X)=0$$ with initial conditions arbitrary $$X(0)$$ and zero velocity, i.e. $$X'(0) = 0$$. $$X$$ is in $$\mathbb{R}^{n}$$, and $$f$$ must be strongly convex with Lipschitz gradient on $$\mathbb{R}^{n}$$. Specifically, I am curious if $$X$$ converges to a point as $$t \rightarrow \infty$$ if the speed of $$X$$ $$(|\dot{X}|)$$ strictly increases. I have tried some energy-function approach, but it's not helping at all. Is there a well-known theory or countercase that can tackle this problem?

Suppose that $$n=1$$ and $$f(x)=(x+1)^2$$. Then $$f$$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $$X(t)$$ does not converge as $$t\to\infty$$.

On the other hand, still for $$n=1$$, suppose that the speed $$|X'|$$ is strictly increasing.

Then either $$X'>0$$ or $$X'<0$$ on $$(0,\infty)$$. (Otherwise, the continuous function $$X'$$ will take the value $$0$$ at some point of $$(0,\infty)$$, which will contradict the conditions that $$X'(0)=0$$ and $$|X'|$$ is strictly increasing.)

So, the continuous function $$X$$ is either strictly increasing or strictly decreasing on $$(0,\infty)$$. So, the function $$X\colon[0,\infty)\to X([0,\infty))$$ has an inverse $$X^{-1}$$. Letting now $$v:=X'\circ X^{-1}$$, we have $$X'(t)=v(X(t))$$ for all $$t\in[0,\infty)$$, whence $$X''(t)=v'(X(t))X'(t)=v'(X(t))v(X(t))$$. Now the ODE can be rewritten as $$v'(x)v(x)+f'(x)=0$$ for $$x$$ in the interval $$X([0,\infty))$$, or as $$\begin{equation*} \dfrac d{dx}\,(v(x)^2)=-2f'(x), \end{equation*}$$ which implies $$v(x)^2=2(f(0)-f(x))$$, because $$v(0)=v(X(0))=X'(0)=0$$. So, $$\begin{equation*} X'(t)^2=v(X(t))^2=2(f(0)-f(X(t))) \tag{1} \end{equation*}$$ for $$t\in[0,\infty)$$.

Since $$f$$ is strongly convex, we have $$f(x)\to\infty$$ as $$|x|\to\infty$$. Noting that the left-hand side of (1) is $$\ge0$$, we conclude that the function $$X$$ must be bounded. Since $$|X'|$$ is strictly increasing, and either $$X'>0$$ or $$X'<0$$ on $$(0,\infty)$$, it follows that $$X'$$ is either increasing or decreasing (from $$X'(0)=0$$), and hence $$X$$ is either (i) convex and increasing or (ii) concave and decreasing, on $$(0,\infty)$$. So, the bounded function function $$X$$ must be constant. In view of the condition $$X(0)=0$$, we conclude that $$X=0$$ identically.

Thus, any solution $$X$$ of the ODE with strictly increasing speed must be the constant $$0$$. On the other hand, if the constant $$0$$ is indeed a solution of the ODE, then we must have $$\nabla f(0)=f'(0)=0$$.

Now, for any dimension $$n$$, if $$\nabla f(0)=0$$, then the constant $$0$$ is the only solution of the ODE.

• I was just going to write a similar counterexample. Then I noticed that technically the answer might be yes, since I don't think that any example exists in which the velocity is monotonically increasing.
– mlk
Commented Jan 17, 2022 at 15:54
• @mlk : Right, that would be hard, if at all possible, to do. However, the velocity is a vector (if $n\ge2$). So, how would you then interpret the condition that the velocity is increasing? Commented Jan 17, 2022 at 15:58
• I simply assumed that velocity meant the same thing as speed in the question, because I also could find no other interpretation. And the people who always make a strict distinction between the two are usually physicists or engineers and those would never miss the harmonic oscillator as the very first example of such an ODE.
– mlk
Commented Jan 17, 2022 at 16:06
• I actually meant 'speed' while using the term velocity, and have edited the question. Sorry for the abuse in terms. Commented Jan 17, 2022 at 16:20
• @MMH.. : It is now shown that, at least for $n=1$, the only solution to the ODE with nondecreasing speed is the constant $0$. This suggests that (i) the same holds for any $n$ and (ii) the increasing-speed requirement is not natural. Commented Jan 17, 2022 at 16:45