bound on matrix differential equation


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

Z(t)=A(t)Z(t)

withh initial conditionMathworldPlanetmath Z(0)=I. Letting denote the matrix operator normMathworldPlanetmath, we will show that, if A(t)C for some constant C when 0tR, then

Z(t)-IC(eCt-1)

when 0tR.

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

Z(t) A(t)Z(t)
CZ(t)
=CI+0t𝑑sZ(s)
CI+C0t𝑑sZ(s)
C+C0t𝑑sZ(s)

For convenience, let us define f(t)=0t𝑑sZ(s). Then we have f(t)C+Cf(t) according to the foregoing derivation. By the product ruleMathworldPlanetmath,

ddt(e-Ctf(t))=e-Ct(f(t)-Cf(t)).

Since f(t)-Cf(t)C, we have

ddt(e-Ctf(t))Ce-Ct.

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

e-Ctf(t)C(1-e-Ct).

Multiplying both sides by eCt and recalling the definition of f, we conclude

0t𝑑sZ(s)C(eCt-1).

Finally, by the triangle inequality,

Z(t)-I=0t𝑑sZ(s)0t𝑑sZ(s).

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

Z(t)-IC(eCt-1).
Title bound on matrix differential equation
Canonical name BoundOnMatrixDifferentialEquation
Date of creation 2013-03-22 18:59:00
Last modified on 2013-03-22 18:59:00
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 4
Author rspuzio (6075)
Entry type Theorem
Classification msc 34A30