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:

 $\|Z(t)-I\|\leq C\left(e^{Ct}-1\right).$
Title bound on matrix differential equation BoundOnMatrixDifferentialEquation 2013-03-22 18:59:00 2013-03-22 18:59:00 rspuzio (6075) rspuzio (6075) 4 rspuzio (6075) Theorem msc 34A30