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Linear matrix ODEs have unique global solutions on a fixed interval

Statement

Let I=[a,b]R be a compact interval, let C:IMn(R) be continuous, let t0I, and let Y0Mn(R). Then the linear matrix initial value problem

Y(t)=C(t)Y(t),Y(t0)=Y0,

has a unique solution on all of I.

Facts & Assumptions

Given: The compact interval I=[a,b], the continuous matrix field C:IMn(R), the time t0I, and the initial matrix Y0.

[F1]

Picard-Lindelof gives a unique local solution for a continuous state-Lipschitz first-order system on an open domain (Picard-Lindelöf local existence and uniqueness for first-order systems).

[F2]

A solution of a first-order system is equivalent to a solution of the associated Volterra integral equation (A first-order initial value problem is equivalent to its Volterra integral equation).

[L1]

A solution whose graph approaches a compact interior region at a finite endpoint extends past that endpoint (A solution whose graph approaches a compact interior region at a finite endpoint extends past that endpoint).

[L2]

A continuous real-valued function on a nonempty compact metric space attains its maximum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L4]

Gronwall's integral inequality converts u(t)A+t0tm(s)u(s)ds into an exponential bound (Gronwall's integral inequality with variable and constant coefficients).

Proof

technique · direct
1.1

Extend C continuously to the open interval (a1,b+1) by setting C~(t):=C(a) for t<a, C~(t):=C(t) for tI, and C~(t):=C(b) for t>b. Regard Mn(R) as Rn2. Then the right-hand side F~(t,Y):=C~(t)Y is continuous on the open domain (a1,b+1)×Rn2 and is globally Lipschitz in Y on every compact time interval, because C~(t)(YZ)2C~(t)YZ2. Hence [F1] gives a unique local solution through (t0,Y0).

F1givenconstruct
2.1

Let J=(α,β) be the maximal interval of existence of that solution in (a1,b+1). Since C is continuous on compact I, [L2] supplies a bound C(t)M on all of I. For tJ[t0,b], the Volterra equation from [F2] and the norm estimate [L3] give the forward inequality below.

F2L2L3L4step 1.1

Y(t)2Y02+Mt0tY(s)2ds.

Therefore [L4] yields Y(t)2Y02eM(tt0) on J[t0,b].

2.2

For tJ[a,t0], the oriented Volterra equation gives Y(t)=Y0tt0C(s)Y(s)ds, so [L3] gives the reflected inequality below.

F2L2L3L4step 1.1

Y(t)2Y02+Mtt0Y(s)2ds.

The time-reflected part of [L4] therefore yields Y(t)2Y02eM(t0t) on J[a,t0].

3.1

Put R:=Y02eM(ba). If βb, then every sequence tjβ in J has graph points in the compact set [a,b]×B(0,R)(a1,b+1)×Mn(R) by steps 2.1 and 2.2. So [L1] extends the solution past β, contradicting maximality. The same argument at the left endpoint rules out αa. Therefore IJ, and the restriction of the maximal solution to I is defined on all of I.

L1step 2.1step 2.2
4.1

Step 3.1 gives existence on all of I, and uniqueness is the local uniqueness already supplied by [F1].

F1step 1.1step 3.1

Depends on

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