TheoremBase

Linear Systems with Continuous Coefficients: Continuity, Global Lipschitz Bound and Well-Posedness of the Initial Value Problem

For continuous coefficients on [a,b], the linear right-hand sides A(t)x+g(t) and A(t)X+XB(t)+G(t) are continuous, globally Lipschitz in the state with constant a bound of the coefficient norms, and their initial value problems have exactly one solution.

Statement

In the settings of Single-Variable Calculus on an Interval and Euclidean Space and Real Matrices: Standing Notation, let a,b∈Ra,b\in\mathbb{R} with a<ba<b, and let n,q∈Nn,q\in\mathbb{N}. R×Rn\mathbb{R}\times\mathbb{R}^{n} carries the product metric of dRd_{\mathbb{R}} and dEd_{E}, and R×Mn×q(R)\mathbb{R}\times\mathcal{M}_{n\times q}(\mathbb{R}) the product metric of dRd_{\mathbb{R}} and dn×qd_{n\times q}, metrics by The Product Metric is a Metric (i). A map defined on [a,b][a,b], on [a,b]×Rn[a,b]\times\mathbb{R}^{n} or on [a,b]×Mn×q(R)[a,b]\times\mathcal{M}_{n\times q}(\mathbb{R}) is continuous on that set for these metrics, for dRd_{\mathbb{R}} on R\mathbb{R}, dEd_{E} on Rn\mathbb{R}^{n} and dp×rd_{p\times r} on Mp×r(R)\mathcal{M}_{p\times r}(\mathbb{R}) (p,r∈Np,r\in\mathbb{N}). ∥⋅∥\lVert\cdot\rVert is the Euclidean norm on vectors and the operator norm on matrices, as in Euclidean Space and Real Matrices: Standing Notation (vi).

Let A:[a,b]→Mn(R)A:[a,b]\to\mathcal{M}_{n}(\mathbb{R}) and g:[a,b]→Rng:[a,b]\to\mathbb{R}^{n} be maps. There is exactly one map F:[a,b]×Rn→RnF:[a,b]\times\mathbb{R}^{n}\to\mathbb{R}^{n} with F(t,x)=A(t)x+g(t)F(t,x)=A(t)x+g(t) for all t∈[a,b]t\in[a,b] and x∈Rnx\in\mathbb{R}^{n}. Likewise, let A:[a,b]→Mn(R)A:[a,b]\to\mathcal{M}_{n}(\mathbb{R}), B:[a,b]→Mq(R)B:[a,b]\to\mathcal{M}_{q}(\mathbb{R}) and G:[a,b]→Mn×q(R)G:[a,b]\to\mathcal{M}_{n\times q}(\mathbb{R}) be maps. There is exactly one map F:[a,b]×Mn×q(R)→Mn×q(R)F:[a,b]\times\mathcal{M}_{n\times q}(\mathbb{R})\to\mathcal{M}_{n\times q}(\mathbb{R}) with F(t,X)=A(t)X+XB(t)+G(t)F(t,X)=A(t)X+XB(t)+G(t) for all t∈[a,b]t\in[a,b] and X∈Mn×q(R)X\in\mathcal{M}_{n\times q}(\mathbb{R}).

Let A:[a,b]→Mn(R)A:[a,b]\to\mathcal{M}_{n}(\mathbb{R}) and g:[a,b]→Rng:[a,b]\to\mathbb{R}^{n} be continuous on [a,b][a,b], and let F:[a,b]×Rn→RnF:[a,b]\times\mathbb{R}^{n}\to\mathbb{R}^{n} be the map of (i) for AA and gg, F(t,x)=A(t)x+g(t)F(t,x)=A(t)x+g(t). Then FF is continuous on [a,b]×Rn[a,b]\times\mathbb{R}^{n}; there is K∈RK\in\mathbb{R} with K≥0K\ge0 and ∥A(t)∥≤K\lVert A(t)\rVert\le K for every t∈[a,b]t\in[a,b]; for every such KK,

∥F(t,x)−F(t,y)∥≤K ∥x−y∥(t∈[a,b], x,y∈Rn).\lVert F(t,x)-F(t,y)\rVert\le K\,\lVert x-y\rVert\qquad\bigl(t\in[a,b],\ x,y\in\mathbb{R}^{n}\bigr).

Let AA, gg and FF be as in (ii). For every ξ∈Rn\xi\in\mathbb{R}^{n} the initial value problem of The Initial Value Problem for a First-Order System of Ordinary Differential Equations (i) with right-hand side FF and initial value ξ\xi has exactly one solution.

Let A:[a,b]→Mn(R)A:[a,b]\to\mathcal{M}_{n}(\mathbb{R}), B:[a,b]→Mq(R)B:[a,b]\to\mathcal{M}_{q}(\mathbb{R}) and G:[a,b]→Mn×q(R)G:[a,b]\to\mathcal{M}_{n\times q}(\mathbb{R}) be continuous on [a,b][a,b], and let F:[a,b]×Mn×q(R)→Mn×q(R)F:[a,b]\times\mathcal{M}_{n\times q}(\mathbb{R})\to\mathcal{M}_{n\times q}(\mathbb{R}) be the map of (i) for AA, BB and GG, F(t,X)=A(t)X+XB(t)+G(t)F(t,X)=A(t)X+XB(t)+G(t). Then FF is continuous on [a,b]×Mn×q(R)[a,b]\times\mathcal{M}_{n\times q}(\mathbb{R}); there are KA,KB∈RK_{A},K_{B}\in\mathbb{R} with KA,KB≥0K_{A},K_{B}\ge0, ∥A(t)∥≤KA\lVert A(t)\rVert\le K_{A} and ∥B(t)∥≤KB\lVert B(t)\rVert\le K_{B} for every t∈[a,b]t\in[a,b]; for all such KAK_{A} and KBK_{B},

∥F(t,X)−F(t,Y)∥≤(KA+KB) ∥X−Y∥(t∈[a,b], X,Y∈Mn×q(R)).\lVert F(t,X)-F(t,Y)\rVert\le(K_{A}+K_{B})\,\lVert X-Y\rVert\qquad\bigl(t\in[a,b],\ X,Y\in\mathcal{M}_{n\times q}(\mathbb{R})\bigr).

Let AA, BB, GG and FF be as in (iv). For every Ξ∈Mn×q(R)\Xi\in\mathcal{M}_{n\times q}(\mathbb{R}) the initial value problem of The Initial Value Problem for a First-Order System of Ordinary Differential Equations (ii) with right-hand side FF and initial value Ξ\Xi has exactly one solution.

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