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Homebound on matrix differential equation

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# bound on matrix differential equation

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

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

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

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

when $0\leq t\leq R$.

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

$\displaystyle\|Z^{{\prime}}(t)\|$ | $\displaystyle\leq\|A(t)\|\|Z(t)\|$ | ||

$\displaystyle\leq C\|Z(t)\|$ | |||

$\displaystyle=C\left\|I+\int_{0}^{t}ds\,Z^{{\prime}}(s)\right\|$ | |||

$\displaystyle\leq C\|I\|+C\int_{0}^{t}ds\,\|Z^{{\prime}}(s)\|$ | |||

$\displaystyle\leq C+C\int_{0}^{t}ds\,\|Z^{{\prime}}(s)\|$ |

For convenience, let us define $f(t)=\int_{0}^{t}ds\,\|Z^{{\prime}}(s)\|$. Then we have $f^{{\prime}}(t)\leq C+Cf(t)$ according to the foregoing derivation. By the product rule,

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

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

${d\over dt}\left(e^{{-Ct}}f(t)\right)\leq Ce^{{-Ct}}.$ |

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

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

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

$\int_{0}^{t}ds\,\|Z^{{\prime}}(s)\|\leq C\left(e^{{Ct}}-1\right).$ |

Finally, by the triangle inequality,

$\|Z(t)-I\|=\left\|\int_{0}^{t}ds\,Z(s)\right\|\leq\int_{0}^{t}ds\,\|Z(s)\|.$ |

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

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

## Mathematics Subject Classification

34A30*no label found*

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