Proof of Variation of Constants with Integrable Forcing

Throughout, $\Phi=\Phi_{A}$. For $a\le u\le v\le b$, the integral over $[u,v]$ of an element of $\mathcal{L}^{1}([a,b])$ or of $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ is the integral over $[a,b]$ of the integrand times $\mathbf{1}_{[u,v]}$, componentwise for vector-valued integrands (Indefinite Lebesgue Integrals on a Closed Interval (i)); so linearity and constant matrix factors (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (viii), Rules for Integrals of Vector- and Matrix-Valued Functions (iii), Rules for Integrals of Vector- and Matrix-Valued Functions (v)) hold for integrals over $[u,v]$ as well. A continuous function from $[a,b]$ into $\mathbb{R}^{q}$ or $\mathcal{M}_{p\times q}(\mathbb{R})$ has continuous components or entries (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (i), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (ii), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)), which are measurable, bounded and integrable (Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii)); so the function is measurable (Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (i), Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (ii)), it lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{q})$ if it is vector-valued (Integrable Vector- and Matrix-Valued Functions and Their Integrals (i)), and its norm is continuous (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iii)), hence bounded. Finally, $[a,b]\times[a,b]$ carries the product $\sigma$-algebra $\mathcal{B}_{[a,b]}\otimes\mathcal{B}_{[a,b]}$ and the product measure $\lambda_{[a,b]}\otimes\lambda_{[a,b]}$ of Product Measures, Lebesgue Measure on ℝⁿ and Image Measures: Notation (ii), which is defined because $\lambda_{[a,b]}$ is finite, hence $\sigma$-finite (Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (i), Measure, Measure Space, and Probability Measure (iv)).

Step 1: two companions of $\Phi$. Put

$$P(s)=A(s)\Phi(s,a),\qquad Q(s)=\Phi(a,s)\qquad(s\in[a,b]).$$

The function $s\mapsto\Phi(s,a)$ is differentiable with derivative $P(s)$ at every $s$ (The State-Transition Matrix of a Linear System (ii)), and $Q$ is differentiable (Basic Properties of the State-Transition Matrix (v)); both are therefore continuous (Rules for Derivatives of Vector- and Matrix-Valued Maps (i)), and so is $P$ (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)). By the composition rule (Basic Properties of the State-Transition Matrix (ii)) and $\Phi(s,s)=I_{n}$ (The State-Transition Matrix of a Linear System (iii)),

$$\Phi(t,s)=\Phi(t,a)\,Q(s),\qquad\Phi(s,a)\,Q(s)=I_{n}\qquad(s,t\in[a,b]).$$

We call these two identities the factorization. By the preliminaries, and since operator norms are nonnegative (Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (i)), there are constants $K_{A},K_{P},K_{Q}\ge0$ with $\lVert A(s)\rVert\le K_{A}$, $\lVert P(s)\rVert\le K_{P}$ and $\lVert Q(s)\rVert\le K_{Q}$ for every $s\in[a,b]$.

Step 2: integrability ((i)). Put $h(s)=Q(s)g(s)$ for $s\in[a,b]$. As $g$ is measurable (an element of $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$) and $Q$ is continuous, hence measurable, $h$ and $\lVert h\rVert$ are measurable (Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (ii), Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (iii)). By Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii),

$$\lVert h(s)\rVert\le\lVert Q(s)\rVert\,\lVert g(s)\rVert\le K_{Q}\lVert g(s)\rVert\qquad(s\in[a,b]),$$

and $\lVert g\rVert\in\mathcal{L}^{1}([a,b])$ (Rules for Integrals of Vector- and Matrix-Valued Functions (viii)); so $\lVert h\rVert\in\mathcal{L}^{1}([a,b])$ by domination (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (v)), and $h\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ by the integrability criterion (Rules for Integrals of Vector- and Matrix-Valued Functions (i)). Now fix $t\in[a,b]$. By the factorization, $\Phi(t,s)g(s)=\Phi(t,a)h(s)$ for every $s$, so $s\mapsto\Phi(t,s)g(s)$ lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ (Rules for Integrals of Vector- and Matrix-Valued Functions (v)). This proves (i). Moreover the constant factor $\Phi(t,a)$ passes through the integral, so that, with $y:[a,b]\to\mathbb{R}^{n}$ the function (Functions and Function-Valued Operators Given by Integrals over Subintervals (ii))

$$y(t)=\int_{[a,t]}h(s)\,ds\qquad(t\in[a,b]),$$

we have

$$x(t)=\Phi(t,a)\bigl(\xi+y(t)\bigr)\qquad(t\in[a,b]).$$

Step 3: continuity ((iii)). As $h\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$, the function $y$ is continuous on $[a,b]$ by Indefinite Lebesgue Integrals on a Closed Interval (ix), and so is $x=\Phi(\cdot,a)(\xi+y)$ (Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (iv)).

Step 4: integrals of $P$. Let $a\le u\le t\le b$ and $v\in\mathbb{R}^{n}$. We show

$$\int_{[u,t]}P(s)v\,ds=\bigl(\Phi(t,a)-\Phi(u,a)\bigr)v .$$

The function $s\mapsto\Phi(s,a)$ is differentiable with the continuous derivative $P$, so by the fundamental theorem of calculus (The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (iii)) the Riemann integral $\int_{u}^{t}P$ is defined and equals $\Phi(t,a)-\Phi(u,a)$. Its entries are the Riemann integrals $\int_{u}^{t}P_{ik}$ (Riemann Integral of a Vector-Valued or Matrix-Valued Map (iv)), and these equal $\int_{[u,t]}P_{ik}\,d\lambda$ because $P_{ik}$ is continuous (Indefinite Lebesgue Integrals on a Closed Interval (xi)). Since the $i$-th component of $P(s)v$ is $\sum_{k=1}^{n}P_{ik}(s)v_{k}$, integrating term by term gives the claim. With $u=a$, $v=\xi$ and $\Phi(a,a)=I_{n}$ it reads

$$\int_{[a,t]}P(s)\xi\,ds=\Phi(t,a)\xi-\xi .$$

Step 5: exchanging the order of integration. Fix $t\in[a,b]$. We show

$$\int_{[a,t]}P(s)y(s)\,ds=\Phi(t,a)\,y(t)-\int_{[a,t]}g(s)\,ds ,$$

the left integrand being continuous by Steps 1 and 3. Fix $i\in[n]$, let

$$D=\{(s,u)\in[a,b]\times[a,b]:u\le s\le t\},$$

and let $H:[a,b]\times[a,b]\to\mathbb{R}$ be the function

$$H(s,u)=\mathbf{1}_{D}(s,u)\sum_{k=1}^{n}P_{ik}(s)\,h_{k}(u)=\mathbf{1}_{D}(s,u)\bigl(P(s)h(u)\bigr)_{i}.$$

It is measurable, since $D=\{s-u\ge0\}\cap\{s\le t\}$ and $H$ are built by arithmetic from the projections and their compositions with the measurable functions $P_{ik}$ and $h_{k}$ (The Product σ-Algebra: Rectangles, Generators, Sections, Projections, Pairings and the Swap (vi), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii), Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (i), Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i), Arithmetic of Measurable Real- and [0,∞]-Valued Maps (ii), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v), Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)).

For all $s,u\in[a,b]$, by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iv) and Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii),

$$|H(s,u)|\le\lVert P(s)h(u)\rVert\le K_{P}\lVert h(u)\rVert .$$

The function $G(s,u)=K_{P}\lVert h(u)\rVert$ is nonnegative and measurable, as above. Its section at $u$ is the constant $K_{P}\lVert h(u)\rVert$, with integral $(b-a)K_{P}\lVert h(u)\rVert$ (Basic Properties of the Integral of Nonnegative Measurable Maps (ii), Basic Properties of the Integral of Nonnegative Measurable Maps (iv), Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (i)), so Tonelli's theorem (Tonelli's and Fubini's Theorems (i)) gives

$$\int G\,d(\lambda_{[a,b]}\otimes\lambda_{[a,b]})=(b-a)K_{P}\int_{[a,b]}\lVert h(u)\rVert\,du<\infty ,$$

using $\lVert h\rVert\in\mathcal{L}^{1}([a,b])$ from Step 2 (Basic Properties of the Integral of Nonnegative Measurable Maps (iv), Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)). Thus $G\in\mathcal{L}^{1}(\lambda_{[a,b]}\otimes\lambda_{[a,b]})$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)), and $H\in\mathcal{L}^{1}(\lambda_{[a,b]}\otimes\lambda_{[a,b]})$ by domination (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (v)).

For $s,u\in[a,b]$ the sections of $H$ are

$$H_{s}(u)=\mathbf{1}_{[a,s]}(u)\bigl(P(s)h(u)\bigr)_{i}\quad(s\le t),\qquad H^{u}(s)=\mathbf{1}_{[u,t]}(s)\bigl(P(s)h(u)\bigr)_{i}\quad(u\le t),$$

and zero for $s>t$, respectively $u>t$. Every section is integrable, with

$$\int_{[a,b]}H_{s}\,d\lambda=\Bigl(P(s)\int_{[a,s]}h(u)\,du\Bigr)_{i}=\bigl(P(s)y(s)\bigr)_{i},\qquad\int_{[a,b]}H^{u}\,d\lambda=\Bigl(\bigl(\Phi(t,a)-\Phi(u,a)\bigr)h(u)\Bigr)_{i}$$

for $s\le t$ and $u\le t$ (by the rules for integrals over $[u,v]$ above and Step 4 with $v=h(u)$). So no section is exceptional, and Fubini's theorem in both orders (Tonelli's and Fubini's Theorems (ii), Tonelli's and Fubini's Theorems (iii)) equates the integrals over $[a,b]$ of the two section integrals, both being $\int H\,d(\lambda_{[a,b]}\otimes\lambda_{[a,b]})$. Rewriting integrals of functions vanishing outside $[a,t]$ as integrals over $[a,t]$,

$$\int_{[a,t]}\bigl(P(s)y(s)\bigr)_{i}\,ds=\int_{[a,t]}\Bigl(\bigl(\Phi(t,a)-\Phi(u,a)\bigr)h(u)\Bigr)_{i}\,du .$$

By the factorization, $\Phi(u,a)h(u)=\Phi(u,a)Q(u)g(u)=g(u)$, so the right integrand is the $i$-th component of $\Phi(t,a)h(u)-g(u)$. As $i$ was arbitrary, linearity and the constant factor $\Phi(t,a)$ give the claimed identity.

Step 6: the integral equation ((ii)). By Step 2 and the definition of $P$,

$$A(s)x(s)=P(s)\xi+P(s)y(s)\qquad(s\in[a,b]),$$

a continuous function by Steps 1 and 3; so $r\mapsto A(r)x(r)$ lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$, and so does $r\mapsto A(r)x(r)+g(r)$ (Rules for Integrals of Vector- and Matrix-Valued Functions (iii)). For $t\in[a,b]$, Steps 4 and 5 give

$$\int_{[a,t]}\bigl(A(s)x(s)+g(s)\bigr)\,ds=\bigl(\Phi(t,a)\xi-\xi\bigr)+\Bigl(\Phi(t,a)y(t)-\int_{[a,t]}g(s)\,ds\Bigr)+\int_{[a,t]}g(s)\,ds=x(t)-\xi ,$$

the last equality by Step 2. Since $F(r,x(r))=A(r)x(r)+g(r)$ for $r\in[a,b]$, $x$ is a Carathéodory solution on $[a,b]$ of the integral equation with right-hand side $F$ and initial value $\xi$, that is, $x\in\mathrm{Sol}(F,\xi)$ (Carathéodory Solutions of an Initial Value Problem (ii)). This is (ii).

Step 7: uniqueness ((iv)). Let $z\in\mathrm{Sol}(F,\xi)$. By Carathéodory Solutions of an Initial Value Problem (ii) and the definition of $F$, the function $r\mapsto A(r)z(r)+g(r)$ lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ and

$$z(t)=\xi+\int_{[a,t]}\bigl(A(s)z(s)+g(s)\bigr)\,ds\qquad(t\in[a,b]);$$

by Step 6 the same holds for $x$. Let $w=z-x$. Then $r\mapsto A(r)w(r)$ is the difference of $r\mapsto A(r)z(r)+g(r)$ and $r\mapsto A(r)x(r)+g(r)$, so it lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$, and subtracting the two integral equations gives

$$w(t)=\int_{[a,t]}A(s)w(s)\,ds\qquad(t\in[a,b]).$$

So $w$ is continuous by Indefinite Lebesgue Integrals on a Closed Interval (ix), and $\varphi(t)=\lVert w(t)\rVert$ is continuous, hence measurable, bounded and in $\mathcal{L}^{1}([a,b])$ (Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii)). Since $\lVert A(s)w(s)\rVert\le K_{A}\varphi(s)$ for $s\in[a,b]$ (Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii)), the norm bound Indefinite Lebesgue Integrals on a Closed Interval (vii) and monotonicity Indefinite Lebesgue Integrals on a Closed Interval (iv) give

$$\varphi(t)\le\int_{[a,t]}\lVert A(s)w(s)\rVert\,ds\le K_{A}\int_{[a,t]}\varphi(s)\,ds\qquad(t\in[a,b]).$$

Gronwall's lemma (Gronwall's Lemma for Bounded Measurable Functions (i), with constant term $0$ and coefficient $K_{A}$) gives $\varphi(t)\le0$, so $\lVert w(t)\rVert=0$ and $w(t)=0$ for every $t$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii)); that is, $z=x$. Hence $\mathrm{Sol}(F,\xi)\subseteq\{x\}$, and with $x\in\mathrm{Sol}(F,\xi)$ from (ii), $\mathrm{Sol}(F,\xi)=\{x\}$.

Step 8: the bound ((v)). Let $C$ be as there and $t\in[a,b]$. By Rules for the Operator Norm: Bound, Least Constant, Definiteness, Homogeneity, Triangle Inequality and Products (ii) and the hypothesis on $C$,

$$\lVert\Phi(t,a)\xi\rVert\le\lVert\Phi(t,a)\rVert\,\lVert\xi\rVert\le C\lVert\xi\rVert,\qquad\lVert\Phi(t,s)g(s)\rVert\le C\lVert g(s)\rVert\quad(s\in[a,b]).$$

The function $s\mapsto\Phi(t,s)g(s)$ lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ by (i), so its norm lies in $\mathcal{L}^{1}([a,b])$ (Rules for Integrals of Vector- and Matrix-Valued Functions (viii)). By the definition of $x$, Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi), the norm bound Indefinite Lebesgue Integrals on a Closed Interval (vii) and monotonicity Indefinite Lebesgue Integrals on a Closed Interval (iv),

$$\lVert x(t)\rVert\le\lVert\Phi(t,a)\xi\rVert+\int_{[a,t]}\lVert\Phi(t,s)g(s)\rVert\,ds\le C\lVert\xi\rVert+C\int_{[a,t]}\lVert g(s)\rVert\,ds=C\Bigl(\lVert\xi\rVert+\int_{[a,t]}\lVert g(s)\rVert\,ds\Bigr).$$