# Prove that $\dot{x} = Ax + f(t)$ has all bounded solutions

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.

• 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 $A$. – user138530 Oct 18 '15 at 7:17
• @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. – user177196 Oct 18 '15 at 7:30
• @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/… – user177196 Oct 18 '15 at 7:31
• 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. – Shahar Even-Dar Mandel Oct 18 '15 at 7:40
• 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? – user177196 Oct 18 '15 at 7:50

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.
• 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 $t< 0$. – user177196 Oct 19 '15 at 2:21
• Yes, I made the causality assumption, which is generally made. It states that $x(t) = 0$ for $t < 0$. – obareey Oct 19 '15 at 4:16