bound on matrix differential equation
Suppose that A and Z are two square matrices dependent
on a parameter which satisfy the differential equation
Z′(t)=A(t)Z(t) |
withh initial condition Z(0)=I.
Letting ∥⋅∥ denote the matrix operator norm
, we
will show that, if ∥A(t)∥≤C for some constant
C when 0≤t≤R, then
∥Z(t)-I∥≤C(eCt-1) |
when 0≤t≤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:
∥Z′(t)∥ | ≤∥A(t)∥∥Z(t)∥ | ||
≤C∥Z(t)∥ | |||
=C∥I+∫t0𝑑sZ′(s)∥ | |||
≤C∥I∥+C∫t0𝑑s∥Z′(s)∥ | |||
≤C+C∫t0𝑑s∥Z′(s)∥ |
For convenience, let us define f(t)=∫t0𝑑s∥Z′(s)∥.
Then we have f′(t)≤C+Cf(t) according to the
foregoing derivation. By the product rule,
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
∫t0𝑑s∥Z′(s)∥≤C(eCt-1). |
Finally, by the triangle inequality,
∥Z(t)-I∥=∥∫t0𝑑sZ(s)∥≤∫t0𝑑s∥Z(s)∥. |
Combining this with the inequality derived in the last paragraph produces the answer:
∥Z(t)-I∥≤C(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 |