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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?

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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.

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  • $\begingroup$ 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. $\endgroup$
    – mlk
    Commented Jan 17, 2022 at 15:54
  • $\begingroup$ @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? $\endgroup$ Commented Jan 17, 2022 at 15:58
  • $\begingroup$ 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. $\endgroup$
    – mlk
    Commented Jan 17, 2022 at 16:06
  • $\begingroup$ I actually meant 'speed' while using the term velocity, and have edited the question. Sorry for the abuse in terms. $\endgroup$
    – MMH..
    Commented Jan 17, 2022 at 16:20
  • $\begingroup$ @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. $\endgroup$ Commented Jan 17, 2022 at 16:45

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