2
$\begingroup$

Given an $n\times n$ matrix $A$ has all eigenvalues with negative real parts. Prove that if the system $\dot{x} = Ax + f(t)$ has a bounded solution, then all the solutions are bounded.

My attempt: I'm currently thinking of two ways to tackle this difficult problem.

1. Use Variation of Constants. Let $\phi(t)$ be the bounded solution with initial condition $x(0) = x_0$. Then $\phi(t) = H(t)H^{-1}(t_0)x_0 + H(t)\int_{t_0}^{t} H^{-1}(s)f(s)ds$. As $H(t) = e^{At} = $ finite sum of all the terms of the form $p(t)e^{at}cos(\beta t)$ or $p(t)e^{at}sin(\beta t)$ where $p(t) =$ polynomial of at most $n-1$ degree, $a+\beta\ i$ are eigenvalues of matrix A, so $a < 0$ in this case. From this info, I deduce that $H(t)$ is bounded, as well as $H^{-1}(t_0)$. Thus, the term $\int_{t_0}^{t} H^{-1}(s)f(s)ds$ must be bounded. Now, I think we can show that either this integral, by replacing the lower bound $t_0$ by any other number, would be bounded, or $f(t)$ is bounded and continuous. Are these really true, since I haven't been able to show either of these thoughts?

2. Use Uniqueness Theorem of Homogeneous ODE. To do that, assume there exists one solution $y(t)$ that is unbounded. So we have: $\phi(t) - y(t)$ is a solution to $\dot{x} = Ax$ with a suitable initial condition. But by uniqueness theorem, $\phi(t) - y(t) = k(t)$ where $k(t)$ is a bounded solution to the same initial value problem. So $y(t) = \phi(t)- k(t)$, which is bounded (contradiction). My biggest concern is whether such $k(t)$ exists and is bounded.

Can anyone please help review my two approaches above, and help complete the last step in either of them? I would really appreciate any help.

$\endgroup$
  • 1
    $\begingroup$ The general solution of $x'=Ax+f$ is $x=x_0+x_1$, where you can take as $x_1$ the bounded solution to the inhomogeneous problem, and $x_0$ is the general solution of $x'=Ax$, which is bounded by the assumption on $A$. $\endgroup$ – user138530 Oct 18 '15 at 7:17
  • $\begingroup$ @ChristianRemling: how do you know we can take $x_1$ as a bounded solution? Is it because of the 2nd assumption? Regards to the boundedness of homogeneous equation: I don't think we need that assumption on $A$ to have the boundedness property. Can you please help check if my 2nd approach above is correct? I think there must exist $k(t)$, as $k(t) = x_0e^{At}$ with $x(0) = x_0$ is the solution to the IVP. By uniqueness theorem, we would have: $\phi(t) - y(t) = x_0e^{At}$, so $y(t)$ should be bounded. This contradicts to the assumption. $\endgroup$ – user177196 Oct 18 '15 at 7:30
  • $\begingroup$ @ChristianRemling: can you please try out this problem as well? I really don't understand where I made the mistake:P math.stackexchange.com/questions/1484979/… $\endgroup$ – user177196 Oct 18 '15 at 7:31
  • 2
    $\begingroup$ What is the range on which the solutions should be bounded? I assume it's for $t\geq 0$, if it's for all $t\in R$ then the statement is false. $\endgroup$ – Shahar Even-Dar Mandel Oct 18 '15 at 7:40
  • $\begingroup$ Since the problem didn't really specify, the range is unfortunately, for all $t\in R$. Can you give a counterexample in that case? I wonder what made it true only when $t\geq 0$, as my 2nd approach above, which seems to work out, doesn't require any conditions on $t$. Did I miss something? $\endgroup$ – user177196 Oct 18 '15 at 7:50
0
$\begingroup$

I believe you can use two of the approaches at once. Note that $\phi(t)$ is the unique solution for $x(0) = x_0$ and it is bounded by assumption. Now, we need to show that any other solution, which will be unique for some arbitrary initial condition, is also bounded. So let's say $y(t)$ is the unique solution for an arbitrary $x(0) = x_1$. Now using the fact that $H(t)\int_{t_0}^{t} H^{-1}(s)f(s)ds$ and $H(t)$ are bounded for all $t \in [t_0, \infty)$, we can conclude the result.

Note: The vector $\int_{t_0}^{t} H^{-1}(s)f(s)ds$ may not be bounded. We only know that the vector $H(t)\int_{t_0}^{t} H^{-1}(s)f(s)ds$ is bounded.

$\endgroup$
  • $\begingroup$ Thank you for your great help, obareey! You're definitely right on the integral (my bad:P). Thanks for your correction! Did you see Christian Remling's approach above? I think it is the easiest, and slickest, one:) Btw, can you please give me some help on this problem: math.stackexchange.com/questions/1484979/… $\endgroup$ – user177196 Oct 18 '15 at 19:43
  • $\begingroup$ I think it is the same approach as mine, but with less words and calculations. I added the formulas because you already worked them out. Anyway, I agree with your comment. $\endgroup$ – obareey Oct 18 '15 at 19:48
  • $\begingroup$ Yeah, you're right. But that would help avoid going through all sorts of boundedness of the integral and the fundamental matrix solution:) Anyway, can you help with the problem in the link? $\endgroup$ – user177196 Oct 18 '15 at 19:51
  • $\begingroup$ did you consider your solutions only when $t\geq 0$? Now, I also think when $t\leq 0$, the question is false, as $H(t)$ is not bounded when $t< 0$. $\endgroup$ – user177196 Oct 19 '15 at 2:21
  • 1
    $\begingroup$ Yes, I made the causality assumption, which is generally made. It states that $x(t) = 0$ for $t < 0$. $\endgroup$ – obareey Oct 19 '15 at 4:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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