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 0≤t≤R, then

∥Z⁢(t)-I∥≤C⁢(eC⁢t-1)

when 0≤t≤R.

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)∥
≤C⁢∥Z⁢(t)∥
=C⁢∥I+∫0t𝑑s⁢Z′⁢(s)∥
≤C⁢∥I∥+C⁢∫0t𝑑s⁢∥Z′⁢(s)∥
≤C+C⁢∫0t𝑑s⁢∥Z′⁢(s)∥

For convenience, let us define f⁢(t)=∫0t𝑑s⁢∥Z′⁢(s)∥. Then we have f′⁢(t)≤C+C⁢f⁢(t) according to the foregoing derivation. By the product ruleMathworldPlanetmath,

dd⁢t⁢(e-C⁢t⁢f⁢(t))=e-C⁢t⁢(f′⁢(t)-C⁢f⁢(t)).

Since f′⁢(t)-C⁢f⁢(t)≤C, we have

dd⁢t⁢(e-C⁢t⁢f⁢(t))≤C⁢e-C⁢t.

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

e-C⁢t⁢f⁢(t)≤C⁢(1-e-C⁢t).

Multiplying both sides by eC⁢t and recalling the definition of f, we conclude

∫0t𝑑s⁢∥Z′⁢(s)∥≤C⁢(eC⁢t-1).

Finally, by the triangle inequality,

∥Z⁢(t)-I∥=∥∫0t𝑑s⁢Z⁢(s)∥≤∫0t𝑑s⁢∥Z⁢(s)∥.

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

∥Z⁢(t)-I∥≤C⁢(eC⁢t-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