bound on matrix differential equation

Suppose that $A$ and $Z$ are two square matrices dependenton a parameter which satisfy the differential equation

$$Z^{\prime }(t)=A(t)Z(t)$$

withh initial condition $Z(0)=I$.Letting $\parallel \cdot \parallel$ denote the matrix operator norm, wewill show that, if $\parallel A(t)\parallel \le C$ for some constant$C$ when $0\le t\le R$, then

$$\parallel Z(t)-I\parallel \le C\left(e^{Ct}-1\right)$$

when $0\le t\le R$.

We begin by applying the product inequality for the norm,then employing the triangle inequality (both in the sum andintegral forms) after expressing$Z$ as the integral of its derivative:

$\parallel Z^{\prime }(t)\parallel$$\le \parallel A(t)\parallel \parallel Z(t)\parallel$

$\le C\parallel Z(t)\parallel$

$=C\parallel I+{\displaystyle {\int }_0^t}𝑑s{Z}^{\prime }(s)\parallel$

$\le C\parallel I\parallel +C{\displaystyle {\int }_0^t}𝑑s\parallel Z^{\prime }(s)\parallel$

$\le C+C{\displaystyle {\int }_0^t}𝑑s\parallel Z^{\prime }(s)\parallel$

For convenience, let us define $f(t)={\int }_0^t𝑑s\parallel Z^{\prime }(s)\parallel$.Then we have $f^{\prime }(t)\le C+Cf(t)$ according to theforegoing derivation. By the product rule,

$$\frac{d}{dt}\left(e^{-Ct}f(t)\right)=e^{-Ct}(f^{\prime }(t)-Cf(t)).$$

Since $f^{\prime }(t)-Cf(t)\le C$, we have

$$\frac{d}{dt}\left(e^{-Ct}f(t)\right)\le C{e}^{-Ct}.$$

Taking the integral from $0$ to $t$ of both sides andnoting that $f(0)=0$, we have

$$e^{-Ct}f(t)\le C\left(1-e^{-Ct}\right).$$

Multiplying both sides by $e^{Ct}$ and recalling thedefinition of $f$, we conclude

$${\int }_0^t𝑑s\parallel Z^{\prime }(s)\parallel \le C\left(e^{Ct}-1\right).$$

Finally, by the triangle inequality,

$$\parallel Z(t)-I\parallel =\parallel {\int }_0^t𝑑sZ(s)\parallel \le {\int }_0^t𝑑s\parallel Z(s)\parallel .$$

Combining this with the inequality derived in thelast paragraph produces the answer:

$$\parallel Z(t)-I\parallel \le C\left(e^{Ct}-1\right).$$