3

This question has repeatedly appeared to me whilst studying certain linear differential equations with time-dependent coefficients.

Let $\phi(t) \in C^2(\Bbb R, \Bbb C)$; that is, $\phi(t)$ is a twice continuously differentialble complex valued function on the real line $\Bbb R$. Assume that

$\lim_{t \to \infty} \vert \phi(t) \vert = 0, \tag{1}$

and

$\lim_{t \to \infty} \vert \ddot \phi(t) \vert \to 0 \tag{2}$

as well. Then must we also have

$\lim_{t \to \infty} \vert \dot \phi(t) \vert \to 0? \tag{3}$

An answer to the above question as such would be most appreciated. There are, however, two more restricted questions the answers to which would suffice for my purposes:

I.) Suppose instead of (2) we assume the existence of a globally bounded, non-negative real function $b: \Bbb R \to \Bbb R_{\ge 0}$ such that

$\vert \ddot \phi(t) \vert \le b(t) \vert \phi(t) \vert \tag{4}$

for sufficiently large $t$. Is this hypothesis, in concert with (1), sufficient to force (3)? Readers should feel free to add various hypotheses on $b(t)$ if they desire, such as $b(t) \in C^k(\Bbb R, \Bbb R_{\ge 0})$ for some $k\ge 0$, or that $b(t)$ exhibits some specific functional behavior, e.g. $b(t) = e^{-t}$ for sufficiently large $t$.

I am particularly interested in the case $b(t) = B > 0$ a constant, so that

$\vert \ddot \phi(t) \vert \le B \vert \phi(t) \vert. \tag{5}$

2.) Suppose $c(t) \in C^k(\Bbb R, \Bbb C)$, $\Vert c(t) \Vert_k < \infty$, and

$\ddot \phi(t) + c(t)\phi(t) = 0; \tag{6}$

then the hypothesis $\lim_{t \to \infty} \vert \phi(t) \vert \to 0$ clearly implies (2); can we now show $\lim_{t \to \infty} \vert \dot \phi(t) \vert \to 0$?

Of particular interest to me is the case $k = 0$, that is, $c(t):\Bbb R \to \Bbb C$ is a bounded, continuous function.

It is clear that these questions are, more or less, in order of decreasing generality: (2) is a case of (1), itself a case of most widely scoped question stated at the beginning.

My own efforts on this problem focused primarily on case (2.) and equation (6). I looked a several things; for instance, (6) implies

$\dot \phi \ddot \phi(t) + c(t)\phi(t) \dot \phi(t) = 0, \tag{7}$

or

$\dfrac{1}{2} \dfrac{d(\dot \phi(t))^2}{dt} + c(t) \dfrac{1}{2} \dfrac{d (\phi(t))^2}{dt} = 0 \tag{8}$

or

$\dfrac{d(\dot \phi(t))^2}{dt} + c(t)\dfrac{d (\phi(t))^2}{dt} = 0, \tag{9}$

which leads to an integral relationship

$( (\dot \phi(t))^2 - (\dot \phi(t_0))^2 + \displaystyle \int_{t_0}^t c(s)\dfrac{d (\phi(s))^2}{ds} = E, \tag{10}$

a constant. I think perhaps (10) might be used to show $\dot \phi(t)$ becomes small for large $t$, since $\phi(t)$ does; but I haven't found a conclusive argument along these lines as of this writing.

I also tried looking at (1) and (2) directly, hoping to show that if $\ddot \phi(t)$ became very small, so that $\dot \phi(t)$ couldn't change too much, it ($\dot \phi$) would have to remain relatively small, lest $\phi(t)$ itself grow in a way not permitted by (1); but these are at the present more intuitive speculations rather than rigorous results.

I'm hoping someone can help me fill in the gaps . . . any insights offered will be seriously considered and appreciated, whether or not they provide a complete solution.

Robert Lewis
  • 72,871
  • I'm confused, can't we just take $\int_0^\infty \phi'(t),dt = \phi(\infty)-\phi(0)=-\phi(0)$ and therefore since the integral converges $\phi' \to 0$? –  Jun 28 '17 at 23:44
  • @james.nixon: what if $\phi$ oscillates? or is complex and loops around zero. If you've got a rigorious proof, pleas post it! – Robert Lewis Jun 28 '17 at 23:47
  • @james.nixon: I guess you're right if $\phi(t) \ge 0$. Maybe. – Robert Lewis Jun 28 '17 at 23:49
  • If it loops around zero, it doesn't matter. We're still constraining $\phi$ to tend to zero. This implies the above integral converges, which is just the Fundamental theorem of Calculus. Then a necessary condition for an integral at infinity to converge is that it tends to zero, that's just elementary. –  Jun 28 '17 at 23:49
  • 1
    Are you familiar with the result (which appears as exercises in Rudin and Spivak's Calculus) that if $|f|\le M_0$ and $|f''|\le M_2$ on $(0,\infty)$, then $|f'|\le 2\sqrt{M_0M_2}$? (So far as I know, this result applies only to real-valued functions.) – Ted Shifrin Jun 28 '17 at 23:56
  • 4
    @james.nixon No, that's not elementary, it's false. – zhw. Jun 28 '17 at 23:57
  • @zhw Supposing $f$ is a continuous function. If $\int_0^\infty f(t),dt$ converges, then $f(t)\to 0$ as $t \to \infty$. What am I missing here? It's the continuous parallel of $\sum a_n < \infty \Rightarrow a_n \to 0$. –  Jun 28 '17 at 23:59
  • @TedShifrin: no I am not but I would certainly like to learn more about it. Have you an online reference? Thanks a lot, this is the kind of thing I'm looking for--in the complex case. – Robert Lewis Jun 29 '17 at 00:01
  • @james.nixon, maybe absolutle convergence needs to be looked at here? – Robert Lewis Jun 29 '17 at 00:03
  • @zhw.: see my last comment to james.nixon. – Robert Lewis Jun 29 '17 at 00:03
  • @RobertLewis Really? We'd need absolute convergence? What's an example of a conditionally convergent integral $\int_0^\infty f(t),dt$ where $f$ is continuous and $f(t) \not \to 0$? –  Jun 29 '17 at 00:05
  • @RobertLewis Absolute convergence won't help. Pay close attention to Ted Shifrin. – zhw. Jun 29 '17 at 00:05
  • 2
    It's proved by using Taylor's Theorem with remainder, considering $f(a+h) = f(a)+f'(a)h+f''(c)h^2/2$ for positive $a,h$. One solves for $f'(a)$, estimates using the triangle inequality, and minimizes the resulting expression. I don't see any complex analogue at the moment. – Ted Shifrin Jun 29 '17 at 00:07
  • 1
    @james.nixon Consider triangular spikes of base $1/n^3$ and height $n$ marching off to $\infty.$ – zhw. Jun 29 '17 at 00:07
  • @zhw. believe me I am paying attention! Any leads you've got, preferably web-accessable, would be most welcome! – Robert Lewis Jun 29 '17 at 00:07
  • @james.nixon: very helpful example! – Robert Lewis Jun 29 '17 at 00:08
  • 1
    @TedShifrin Just apply what you posted to the real and imaginary parts. – zhw. Jun 29 '17 at 00:08
  • @zhw. I'll think about what you just said to Ted! – Robert Lewis Jun 29 '17 at 00:10
  • 1
    @zhw Ahhh that makes perfect sense. Now that I think about it, we can even construct an analytic example $\cos(x^2)$ comes to mind. Hmmm, wonder why I thought that... Thinking too quickly is always a mistake... –  Jun 29 '17 at 00:13
  • @james.nixon: well, after the comments I sure thought I was thinking too quickly! Scary! Cheers! And thanks for the input! – Robert Lewis Jun 29 '17 at 00:15
  • 1
    Ah, duh, thanks @zhw :) I'm still jetlagged from a 24-hour travel day yesterday ... That's my excuse, anyhow. – Ted Shifrin Jun 29 '17 at 00:17

1 Answers1

3

Here is a well known result:

Theorem: If $f:(a, \infty) \to\mathbb{R} $ is a function such that $f(x) \to L$ as $x\to\infty$ and $|f''(x)|\leq K$ for all $x>a$ then $f'(x) \to 0$ as $x\to\infty$.

Based on your question it is easy to see that both real and imaginary parts of $\phi$ satisfy the hypotheses of the above theorem and therefore derivatives of the real and imaginary parts tend to $0$ as the argument $t\to\infty$. It follows that $|\phi'(t) |\to 0$ as $t\to\infty$.

J. E. Littlewood gave a wonderful proof of the theorem mentioned in the beginning.

Paramanand Singh
  • 92,128
  • thank you so much! Exactly what I need! Can you shoot me a reference (online if possible) to a proof? – Robert Lewis Jun 29 '17 at 03:18
  • @RobertLewis: I edited to give a link to a proof. There are many proofs available. See this question (https://math.stackexchange.com/q/730411/72031) and list of linked questions to it – Paramanand Singh Jun 29 '17 at 03:20