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.
Best Answer
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.