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Consider the simple linear inhomogeneous differential equation $\dot{x}(t) = u(t)$, with $x(0) \in \mathbb{R}$. Assume that there exists a unique $x_{\infty} \in \mathbb{R}$ such that $\lim\limits_{t\to\infty} x(t) = x_{\infty}$ for each initial value $x(0)$.

My question is: What do we know about $u(t)$ in this case? The obvious hypothesis is that asymptotically $u(t) \to x_{\infty}-x(t)$, but I can't seem to prove it. I would appreciate any help or a counter example. I feel the key to the proof (if such exists) must be in the observation that convergence to $x_{\infty}$ occurs regardless of the initial value $x(0)$.

For what it's worth, this question is related to feedback control system design, but I am beginning to investigate the necessary structure of controllers achieving set point control.

Aero
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3 Answers3

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Sorry but the question is most unclear. If indeed $\dot x(t)=u(t)$ then $$x(t)=x(0)+\int_0^tu(s)\mathrm ds.$$ If, as the notation strongly suggests, $u(t)$ does not depend on $x(t)$ but only on $t$, then, whatever the function $u$ can be, one cannot have that $x(t)\to x_\infty$ for every $x(0)$, for some limit $x_\infty$ independent of $x(0)$.

Or, one assumes that $u(t)$ actually depends on $x(t)$, then, first, one should write the evolution equation as $$\dot x(t)=u(x(t)),$$ and, second, the asymptotics of such a dynamical system are well known. For example, if $u(\xi)$ has the sign of $\xi_*-\xi$, then $x(t)\to\xi_*$ when $t\to+\infty$, for every $x(0)$, that is, the situation described in the post occurs, with $x_\infty=\xi_*$.

Did
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  • I admit it is perhaps too vague to merit a precise answer. My point was this: What do we know about $u$ if all we know is asymptotic behavior of $x$, independently of the initial state? Of course, intuitively there is bound to be a feedback of the regulated variable, but what kind? I was not willing to explicitly specify it as an evolution equation at the outset. – Aero Sep 28 '14 at 12:29
  • Did you actually read the answer? Did you notice that it proves an impossibility result? You seem again to be talking to yourself as if my answer contained only some vague opinions... – Did Sep 28 '14 at 12:31
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You're basically saying that given a differentiable function with finite limit at infinity, then the derivative tends to zero.

There are a lot of counterexamples to this.

For example take as $x(t)$ the constant function zero, and add some frequent "bumps" with decreasing height, but high derivative. This tends to zero, but the derivate doesn't converge.

gebruiker
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Exodd
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  • Thanks for the response, but actually that is not ENTIRELY what I am saying. For every initial condition, the same function $u$ should yield $x_{\infty}$ as a final value. I am aware of the counter examples you referred to (such as http://math.stackexchange.com/questions/162078/if-a-function-has-a-finite-limit-at-infinity-does-that-imply-its-derivative-goe?lq=1), but in this case we assume more. – Aero Sep 27 '14 at 11:17
  • Sorry, I misread it. – Exodd Sep 27 '14 at 11:20
  • That's ok, any ideas for solving this? – Aero Sep 27 '14 at 11:24
  • Now I have a problem:

    Let's say $y(t)$ is a solution with initial condition $y(0)=0$, and limit $\lim_{t\to\infty}y(t)=x_{\infty}$.

    Then $y(t)+c$ is a solution with initial condition $y(0)=c$, but the limit is $x_{\infty}+c$..

    Doesn't this mean that there's no solution with that constraint?

     – Exodd Sep 27 '14 at 11:31
  • Actually no, $u(t) = x_{\infty} - x(t)$ satisfies the conditions for each $x(0)$. Like I said, this is a typical situation in set point feedback control problems, and I am wondering if this stated structure for $u$ is actually necessary. – Aero Sep 27 '14 at 11:58
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Contrary to my initial intuition, the answer to this question is not related to the initial conditions, but the smoothness of $u(\cdot)$.

If one assumes that $u$ is feebly oscillating at infinity, then the result follows from certain well-known Tauberian theorems: Clearly $u(t) \to 0$ in the sense of Cesàro by our assumption. A Tauberian theorem then ensures that $u(t) \to 0$ in the ordinary sense, which was the claim. See e.g. Theorem 4.2.5 in W. Arendt et al. "Vector-valued Laplace Transforms and Cauchy Problems", Birkhäuser 2010, p. 249.

Aero
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  • This seems completely offtopic, which is odd since you are the one who asked the question... – Did Sep 28 '14 at 11:52
  • Actually $u(t) \to 0$ is sufficient for the purposes of the original question. Indeed, trivially then $u(t) \to x_{\infty}-x(t)$ as $t \to \infty$. – Aero Sep 28 '14 at 12:27
  • No, as shown in my answer, "trivially". Are you talking to yourself? – Did Sep 28 '14 at 12:28
  • No need for hostility. If we know that $x(t) \to x_{\infty} \in \mathbb{R}$, then by the differential equation specification, $u$ is Cesaro convergent towards zero. If we assume sufficient regularity, then $u$ is convergent towards zero in the normal sense, which really just (trivially) means that $u$ and $x-x_{\infty}$ are equal at infinity. – Aero Sep 28 '14 at 12:40
  • Sorry to insist but did you actually read my answer? It explains why the properties you consider as being trivially true are in fact impossible in the model you describe in the question, and it also suggests an alternative model where they might become true. I am asking because, until now, your reaction to these mathematical explanations have been nonexistent (an attitude which, in mathematical circles, is considered as one of the most deliberately hostile one can imagine). – Did Sep 28 '14 at 13:02
  • Yes I did - actually already at the outset - but I guess I still fail to grasp your point. I acknowledge that $u$ can't be just "anything", but isn't that evident from the differential equation specification? Is there something wrong with this answer of mine that we are commenting on? It seems to me that this is just about as general as it gets, without explicitly specifying a functional form for $u$. Would you agree? – Aero Sep 28 '14 at 13:12
  • And yet the point is so simple to grasp... Quote A: "If ... u(t) does not depend on x(t) but only on t, then..." Quote B: "Or, one assumes that u(t) actually depends on x(t), then..." Until you tell us that you consider situation A or that you consider situation B, we are stalled. – Did Sep 28 '14 at 13:16
  • Oh. But does it have to be A or B, a priori? I was looking for the necessary structure for $u$, given the differential equation and its asymptotic behavior. Refer to the answer - $u(t)$ tends asymptotically to the feedback law $x_{\infty}-x(t)$; no structural assumptions on $u$ beyond smoothness are necessary. – Aero Sep 28 '14 at 13:26
  • "no structural assumptions on u beyond smoothness are necessary" Really? Which part of "If ... $u(t)$ does not depend on $x(t)$ but only on $t$, then, whatever the function $u$ can be, one cannot have that $x(t)\to x_\infty$ for every $x(0)$, for some limit $x_\infty$ independent of $x(0)$" do you fail to understand? (And yes "it has to be A or B, a priori" (and also a posteriori) since A reads "u depends on t only" and B reads "u does not depend on t only".) – Did Sep 28 '14 at 13:53
  • This is getting pointless. Which part of this answer are you refuting? – Aero Sep 28 '14 at 14:14
  • Rrrrright... Since I now asked essentially the same (trivial) question several times and since, each time, you (carefully) avoided to answer, this can only mean that you are more interested in sophistic rhetorics than in mathematics. Your call (but be aware that this is not the topic of the site). – Did Sep 28 '14 at 14:20
  • Well, now. Did YOU answer my questions (posed twice) in this discussion? No. Did I answer your question? No. But I gave you the reason for not doing so: I do not want to choose a specific structure for $u$. In fact, I wanted it to be part of the solution, as I explained. This is not mere rhetorics. As far as I can see, the present answer is just about as close as it gets in terms of that structure. Your opinion has been noted - thank you - now please move on. – Aero Sep 28 '14 at 16:00
  • You asked no question apart from "does it have to be A or B, a priori?", which is absurd, as already mentioned. But you now admit that you did not care about specifying case A or case B although I explained patiently why this was absolutely necessary (is it so difficult to explain whether $u(t)$ is really $u(t)$, or $u(x(t))$? apparently it is). So: pure rhetorics and proud ignorance. Nice combo on your part. (Maybe you should answer your own questions instead of posting them on the site, making everybody lose their time?) – Did Sep 28 '14 at 16:08
  • "Which part of this answer are you refuting?" "Is there something wrong with this answer of mine that we are commenting on?" Are you blind? – Aero Sep 28 '14 at 16:12
  • "Which part of this answer are you refuting?" All of it, in the sense that you posted a collection of words with no direct link to the question asked. It is so easy to exhibit u "feebly oscillating at infinity" such that the limit of x at infinity does depend on x(0)... say, u=0 everywhere, that your answer seems randomly composed by a computer. "Is there something wrong with this answer of mine that we are commenting on?" See above. "Are you blind?" No thanks, but a bit puzzled by your willing deafness, I must admit. – Did Sep 28 '14 at 17:13