TheoremBase

The map Φ(t,a)(ξ+∫_a^t Φ(a,r)g(r)dr) is differentiated by the product rule and the fundamental theorem of calculus; the composition law shows it solves the problem, uniqueness identifies it with the solution, and pulling the constant factor Φ(t,a) into the integral gives the formula.

Proof

By Basic Facts about Intervals of the Real Line (v), [a,b][a,b] is an interval containing the two distinct points aa and bb; it is the interval II of Rules for Derivatives of Vector- and Matrix-Valued Maps, Rules for the Riemann Integral of Vector- and Matrix-Valued Maps and The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps throughout, and derivatives are taken relative to it. By The State-Transition Matrix of a Linear System (ii), the map U=Φ(⋅,a)U=\Phi(\cdot,a) is differentiable with U′(t)=A(t)Φ(t,a)U'(t)=A(t)\Phi(t,a) for every t∈[a,b]t\in[a,b], and by Basic Properties of the State-Transition Matrix (v), with aa for tt, the map V:[a,b]→Mn(R)V:[a,b]\to\mathcal{M}_{n}(\mathbb{R}), r↦Φ(a,r)r\mapsto\Phi(a,r), is differentiable; by Rules for Derivatives of Vector- and Matrix-Valued Maps (i) and Continuous Map Between Metric Spaces (ii), VV is continuous on [a,b][a,b]. By Basic Properties of the State-Transition Matrix (ii) and The State-Transition Matrix of a Linear System (iii), Φ(t,a)Φ(a,t)=Φ(t,t)=In\Phi(t,a)\Phi(a,t)=\Phi(t,t)=I_{n} and Φ(t,a)Φ(a,r)=Φ(t,r)\Phi(t,a)\Phi(a,r)=\Phi(t,r) for all r,t∈[a,b]r,t\in[a,b].

Proof of (i). Exactly one solution. By Linear Systems with Continuous Coefficients: Continuity, Global Lipschitz Bound and Well-Posedness of the Initial Value Problem (iii), with F(t,x)=A(t)x+g(t)F(t,x)=A(t)x+g(t), the initial value problem has exactly one solution xx; by Solution of an Initial Value Problem on [a,c][a,c] (iii) and Solution of an Initial Value Problem on [a,c][a,c] (i), with bb for cc, every differentiable y:[a,b]→Rny:[a,b]\to\mathbb{R}^{n} with y′(t)=A(t)y(t)+g(t)y'(t)=A(t)y(t)+g(t) for every t∈[a,b]t\in[a,b] and y(a)=ξy(a)=\xi is a solution, and so equals xx.

A candidate. Let h:[a,b]→Rnh:[a,b]\to\mathbb{R}^{n}, r↦Φ(a,r)g(r)=V(r)g(r)r\mapsto\Phi(a,r)g(r)=V(r)g(r); it is continuous on [a,b][a,b] by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v), with (R,dR)(\mathbb{R},d_{\mathbb{R}}) for its (X,d)(X,d) and [a,b][a,b] for its SS. By The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (i), with nn for qq, hh for xx and aa for cc, ∫ath\int_{a}^{t}h is defined for every t∈[a,b]t\in[a,b], and the map H:[a,b]→RnH:[a,b]\to\mathbb{R}^{n}, t↦∫atht\mapsto\int_{a}^{t}h (Maps Given by Integrals of Vector- and Matrix-Valued Maps (i)), is differentiable with H′(t)=h(t)H'(t)=h(t). For every k∈[n]k\in[n], ∫aahk=0\int_{a}^{a}h_{k}=0 by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii), so H(a)=0H(a)=0 by Riemann Integral of a Vector-Valued or Matrix-Valued Map (ii). The constant map t↦ξt\mapsto\xi is differentiable with derivative 00 by Rules for Derivatives of Vector- and Matrix-Valued Maps (ix), so z:t↦ξ+H(t)z:t\mapsto\xi+H(t) is differentiable with z′(t)=0+h(t)=h(t)z'(t)=0+h(t)=h(t) by Rules for Derivatives of Vector- and Matrix-Valued Maps (ii) (and Additive Laws for Vectors in Rq\mathbb{R}^q and for Real p×qp\times q Matrices: Group Laws and Rules for Differences (i)). Let y:[a,b]→Rny:[a,b]\to\mathbb{R}^{n}, t↦U(t)z(t)t\mapsto U(t)z(t). By Rules for Derivatives of Vector- and Matrix-Valued Maps (vi), yy is differentiable at every t∈[a,b]t\in[a,b], hence differentiable, and, by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v),

y′(t)=(A(t)Φ(t,a))z(t)+Φ(t,a)(Φ(a,t)g(t))=A(t)y(t)+(Φ(t,a)Φ(a,t))g(t)=A(t)y(t)+g(t).y'(t)=\bigl(A(t)\Phi(t,a)\bigr)z(t)+\Phi(t,a)\bigl(\Phi(a,t)g(t)\bigr)=A(t)y(t)+\bigl(\Phi(t,a)\Phi(a,t)\bigr)g(t)=A(t)y(t)+g(t).

Moreover y(a)=Φ(a,a)(ξ+0)=Inξ=ξy(a)=\Phi(a,a)(\xi+0)=I_{n}\xi=\xi by The State-Transition Matrix of a Linear System (iii), Additive Laws for Vectors in Rq\mathbb{R}^q and for Real p×qp\times q Matrices: Group Laws and Rules for Differences (i) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v). Hence y=xy=x.

The formula. Let t∈[a,b]t\in[a,b]. The map r↦Φ(t,r)g(r)r\mapsto\Phi(t,r)g(r) (Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii)) has the values (Φ(t,a)Φ(a,r))g(r)=Φ(t,a)h(r)\bigl(\Phi(t,a)\Phi(a,r)\bigr)g(r)=\Phi(t,a)h(r), by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v). As ∫ath\int_{a}^{t}h is defined, Rules for the Riemann Integral of Vector- and Matrix-Valued Maps (vi), with Φ(t,a)\Phi(t,a) for LL, aa and tt for uu and vv, and hh for xx, shows that ∫atΦ(t,r)g(r) dr\int_{a}^{t}\Phi(t,r)g(r)\,dr is defined and equals Φ(t,a)H(t)\Phi(t,a)H(t). By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v),

x(t)=y(t)=Φ(t,a)ξ+Φ(t,a)H(t)=Φ(t,a)ξ+∫atΦ(t,r)g(r) dr.x(t)=y(t)=\Phi(t,a)\xi+\Phi(t,a)H(t)=\Phi(t,a)\xi+\int_{a}^{t}\Phi(t,r)g(r)\,dr .

This proves (i).

Proof of (ii). Let O:[a,b]→Mq(R)O:[a,b]\to\mathcal{M}_{q}(\mathbb{R}) be the constant map with value 0q0_{q}, continuous on [a,b][a,b] by Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv) and Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v). For Y∈Mn×q(R)Y\in\mathcal{M}_{n\times q}(\mathbb{R}), Y0q=0n×qY0_{q}=0_{n\times q} by Products of Real Matrices with Zero Matrices, Negatives and Differences (i), so the right-hand side of Linear Systems with Continuous Coefficients: Continuity, Global Lipschitz Bound and Well-Posedness of the Initial Value Problem (v), with AA, OO and GG for AA, BB and GG, is (t,Y)↦A(t)Y+G(t)(t,Y)\mapsto A(t)Y+G(t) (Additive Laws for Vectors in Rq\mathbb{R}^q and for Real p×qp\times q Matrices: Group Laws and Rules for Differences (i)), and the initial value problem has exactly one solution XX; as in the proof of (i), every differentiable Y:[a,b]→Mn×q(R)Y:[a,b]\to\mathcal{M}_{n\times q}(\mathbb{R}) with Y′(t)=A(t)Y(t)+G(t)Y'(t)=A(t)Y(t)+G(t) for every t∈[a,b]t\in[a,b] and Y(a)=ΞY(a)=\Xi equals XX, by Solution of an Initial Value Problem on [a,c][a,c] (iii) and Solution of an Initial Value Problem on [a,c][a,c] (ii). The rest of the proof of (i) applies word for word, with GG, Ξ\Xi, Mn×q(R)\mathcal{M}_{n\times q}(\mathbb{R}), the zero matrix 0n×q0_{n\times q}, the map h:r↦Φ(a,r)G(r)h:r\mapsto\Phi(a,r)G(r) into Mn×q(R)\mathcal{M}_{n\times q}(\mathbb{R}) and its entry maps hijh_{ij} (i∈[n]i\in[n], j∈[q]j\in[q]) in place of gg, ξ\xi, Rn\mathbb{R}^{n}, 00, the map hh into Rn\mathbb{R}^{n} and its component maps hkh_{k} (k∈[n]k\in[n]); using the matrix halves of The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (i), Rules for Derivatives of Vector- and Matrix-Valued Maps (ix) and Rules for Derivatives of Vector- and Matrix-Valued Maps (ii), Maps Given by Integrals of Vector- and Matrix-Valued Maps (ii) and Riemann Integral of a Vector-Valued or Matrix-Valued Map (iv) in place of Maps Given by Integrals of Vector- and Matrix-Valued Maps (i) and Riemann Integral of a Vector-Valued or Matrix-Valued Map (ii), Rules for Derivatives of Vector- and Matrix-Valued Maps (v) in place of Rules for Derivatives of Vector- and Matrix-Valued Maps (vi), the rules Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i), Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (ii) and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv) in place of Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v), and the identity ∫atMA(r) dr=M∫atA\int_{a}^{t}MA(r)\,dr=M\int_{a}^{t}A of Rules for the Riemann Integral of Vector- and Matrix-Valued Maps (vi), with Φ(t,a)\Phi(t,a) for MM and hh for the map AA of that lemma. This proves (ii).

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