Limit of solution of linear system of ODEs as $t\to \infty$ I am completely stuck on the following problem:
Consider the linear system: $x'(t)=A(t)x(t)$ where $A(t)$ is an $n$ by $n$ matrix. Assume that $\lim_{t\to \infty}A(t)=B$. Suppose that each eigenvalue of $B$ has strictly negative real part. Let $y(t)$ be a solution to the linear system. Does anyone how to show: $\lim_{t\to \infty}|y(t)|=0$
I know that if $y(t)$ is a solution to the linear system and all eigenvalues of $A$ have  strictly negative real parts, then $y(t)\to 0$ as $t\to \infty$. However, this is true when $A$ is independent of $t$. The issue for this problem is that $A$ is dependent on $t$, and it's the limiting matrix $B$ that has eigenvalues with strictly negative real parts. I don't know how to approach this problem. Thanks to anyone who help me out!
 A: Eigenvalues depend continuously on the matrix, so your condition that $\lim_{t \to \infty} A(t)$ exists and the real parts of the eigenvalues are all negative, this tells you there is some $t_0$ so that if $t>t_0$ then the matrix $A(t)$ has eigenvalues with all negative parts.  Moreover, there is some negative number $\kappa$ so that the real parts of the eigenvalues of $A(t)$ are less than $\kappa$ for all $t > t_0$. 
From here the proof is quite similar to the case where the matrix $A$ is constant, for example the argument from the Hubbard and West textbook (TAM 18, Theorem 7.6.1) generalizes immediately.
The key idea is to find a positive-definite Hermitian metric ($\cdot$) on $\mathbb C^n$ so that $Re(A(t)v \cdot v) \leq \kappa v \cdot v$ for all $t \geq t_0$ and all $v$, and $\kappa$ is some negative real number.  There are many ways to do this, one is it use the coordinate system that puts $A(\infty)$ in its Jordan form, then appropriately re-scale the higher-order eigenvectors to be sufficiently-short. You then use the standard Hermitian metric for that coordinate system. 
Anyhow, that does the job since $(v \cdot v)' = 2Re(A(t)\cdot v) \leq \kappa v \cdot v$ so the length of $v$ decays exponentially. 
edit: It's not clear to me what you would like more detail on.  The coordinate system perhaps?  Say $X^{-1} A(\infty) X = J$ is the Jordan form of the matrix $A(\infty)$.  This mean $J$ has eigenvalues down the diagonal and $1$'s (or $0$'s) immediately off the diagonal.  The matrix $X$ consists of eigenvectors and higher-order eigenvectors for $A(\infty)$.  If you scale-down those higher-order eigenvectors (doing column operations on $X$) you get a new matrix $Y$ and $Y^{-1} A(\infty) Y$ still has the eigenvalues running down the diagonal, but the non-zero off-diagonal terms are much smaller.  You can do this to make the off-diagonal terms as small as you like -- in particular, much smaller than the eigenvalues themselves.  That is a stable statement.  In any sufficiently-small neighbourhood of $A(\infty)$ in the space of matrices, your matrices will have relatively large numbers down the diagonal, and the off-diagonal terms will be relatively small.  That suffices to get the last inequality above. 
A: The answer is positive. You can consider your system as
$$
\dot x=Bx+C(t)x,\tag{1}
$$
where $\lim_{t\to\infty}\|C(t)\|=0$, use the integral representation of the solution and the fact that the fundamental matrix solution $\Phi(t)$ to $\dot x=Bx$ satisfies the estimate $\|\Phi(t)\|\leq Ce^{-\mu t}$ for some $\mu>0$.
A word of caution: If matrix $B$ is stable (i.e., $\mbox{Re}\,\lambda\leq 0$) then the condition $\lim_{t\to\infty}\|C(t)\|=0$ is not enough to guarantee the stability of the trivial solution.
Edit: Outline of a proof. Let
$$
\dot \Phi(t)=B\Phi(t). 
$$
In $(1)$ I make the substitution $x=\Phi(t)z$ and find that
$$
\dot z=\Phi^{-1}(t)C(t)\Phi(t)z.
$$
I integrate, return to the original variable and get that instead of $(1)$ I have
$$
x(t)=\Phi(t)x_0+\int_{t_0}^t\Phi(t-\tau+t_0)C(\tau)x(\tau)d\tau.
$$
Now you estimate the norm $\|x\|$ in the usual way, use a Grownwall's inequality and get the conclusion (it does require some work to fill in the gaps).
