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?


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.