Proof of Variation of Constants with Bounded Measurable Forcing and the Two-Parameter Fundamental Solution

Throughout, continuous and measurable for real functions on a compact interval are as in the statement; a continuous real function on $[0,T]$ is measurable for the trace Borel $\sigma$-algebra by claim 4 of Measurability of Countable Suprema, Bounded Pointwise Limits, Monotone Functions, and Continuous Functions, and bounded by Continuous Real-Valued Functions on a Compact Interval are Bounded; sums, scalar multiples and products of continuous real functions on $[0,T]$ are continuous by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, so finite linear combinations and finite sums of products of continuous functions are continuous. Sums, scalar multiples and products of bounded measurable real functions are bounded measurable by claims 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Integrals of continuous functions over compact subintervals are the same whether read as Riemann integrals (as in Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations) or as Lebesgue integrals, by claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval; we read them as Lebesgue integrals below. Two conventions on subintervals: if $f$ is measurable for the trace Borel $\sigma$-algebra on $[0,T]$ then its restriction to $[0,t]$ is measurable for the trace Borel $\sigma$-algebra on $[0,t]$ (a preimage $S\cap[0,T]$ with $S$ Borel restricts to $S\cap[0,t]$), and the Lebesgue integral of a bounded measurable function over $[0,r]\subseteq[0,t]$ is the same whether $[0,r]$ is regarded as a subinterval of $[0,t]$ or of $[0,T]$, both being the integral of the zero extension over the real line by claim 2 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval. A bounded measurable real function on a compact interval is Lebesgue integrable there (its positive and negative parts are bounded measurable, and the integral of a nonnegative function bounded by $C$ over an interval of length $L$ is at most $CL$ by monotonicity). A constant matrix passes through a componentwise integral by linearity of the integral, since each component of $Mh(u)$ is a finite linear combination of the components of $h(u)$; likewise a constant vector passes through a matrix-valued componentwise integral, each component of $N(u)\xi$ being a finite linear combination of the entries of $N(u)$. Bounded entries bound the norm: if every entry of a $k\times k$ matrix $M$ has absolute value at most $C$, each absolute row sum is at most $kC$, so $\lVert M\rVert_{\mathrm{rs}}^{2}\le k(kC)^{2}$ and $\lVert M\rVert_{\mathrm{rs}}\le k^{3/2}C$; this is why the bounds $\bar{A}$ and $\bar{\Phi}$ of the statement exist. The term Riemann integral below refers to that definition.

Submultiplicativity of $\lVert\cdot\rVert_{\mathrm{rs}}$. For $k\times k$ matrices $M$, $M'$ and each row index $i$, $\sum_{j}|(MM')_{ij}|\le\sum_{j}\sum_{p}|M_{ip}||M'_{pj}|=\sum_{p}|M_{ip}|\sum_{j}|M'_{pj}|\le\lVert M'\rVert_{\mathrm{rs}}\sum_{p}|M_{ip}|$, because each row sum $\sum_j|M'_{pj}|$ is at most the Euclidean norm of the vector of row sums of $M'$. Squaring and summing over $i$ gives $\lVert MM'\rVert_{\mathrm{rs}}\le\lVert M\rVert_{\mathrm{rs}}\,\lVert M'\rVert_{\mathrm{rs}}$.

Claim 1. $\Phi(t,t)=\Phi(t)\Phi(t)^{-1}=I_k$ by the definition of the inverse. By submultiplicativity, $\lVert\Phi(t,u)\rVert_{\mathrm{rs}}\le\lVert\Phi(t)\rVert_{\mathrm{rs}}\lVert\Psi(u)\rVert_{\mathrm{rs}}\le\bar{\Phi}^{2}$. For fixed $t$, each entry of $u\mapsto\Phi(t)\Psi(u)$ is a finite linear combination of the entries of $\Psi(u)$, which are continuous on $[0,T]$ by claim 2 of Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations, so it is continuous by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. For the integral identity, fix $0\le\tau\le t\le T$ and $v\in\mathbb{R}^k$. By claim 2 of Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations, $\Psi(t)=I_k-\int_{[0,t]}\Psi(u)A(u)\,du$ and $\Psi(\tau)=I_k-\int_{[0,\tau]}\Psi(u)A(u)\,du$, so, reading the entrywise integrals of the continuous integrand $\Psi A$ as Riemann integrals (claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval) and using additivity of the Riemann integral over the adjacent intervals $[0,\tau]$ and $[\tau,t]$,

Multiplying on the left by the constant matrix $\Phi(t)$, applying to $v$, and using $\Phi(t)\Psi(t)=I_k$ and claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product, we get $\Phi(t,\tau)v-v=\int_{[\tau,t]}\Phi(t)\Psi(u)A(u)v\,du=\int_{[\tau,t]}\Phi(t,u)A(u)v\,du$.

Claim 2. Fix $t\in[0,T]$. Each component of $u\mapsto\Phi(t,u)g(u)$ is a finite sum of products of a continuous (hence bounded measurable) entry of $\Phi(t,u)$ with a bounded measurable component of $g$, hence bounded measurable, and $|\Phi(t,u)g(u)|\le\bar{\Phi}^{2}|g(u)|$ by claim 1 and the norm bound of the statement. Put $y(t)=\int_{[0,t]}\Psi(u)g(u)\,du\in\mathbb{R}^k$; the integrand is bounded measurable by the same argument, and $\int_{[0,t]}\Phi(t,u)g(u)\,du=\Phi(t)y(t)$ by writing $\Phi(t,u)g(u)=\Phi(t)(\Psi(u)g(u))$ (claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product) and passing the constant matrix $\Phi(t)$ through the integral. Hence $x(t)=\Phi(t)(\xi+y(t))$. By claim (i) of Integration by Parts for Indefinite Lebesgue Integrals on a Compact Interval (applied componentwise with $T$ there equal to $T$, $u_0=0$ and $f$ a component of $\Psi g$, which is Lebesgue integrable over $[0,T]$ as a bounded measurable function), each component of $y$ is continuous on $[0,T]$; each component of $x$ is then a finite sum of products of continuous functions, hence continuous, hence bounded measurable. The norm bound follows from $|\Phi(t)\xi|\le\bar{\Phi}|\xi|$, from $|\int_{[0,t]}\Phi(t,u)g(u)\,du|\le\int_{[0,t]}|\Phi(t,u)g(u)|\,du\le\bar{\Phi}^{2}\int_{[0,t]}|g(u)|\,du$ by Norm Bound for a Vector-Valued Lebesgue Integral over a Compact Interval and monotonicity, and from the triangle inequality.

It remains to verify the integral equation; for $t=0$ both sides equal $\xi$. Fix $t\in(0,T]$ and indices $i,j\in\{1,\dots,k\}$. On $[0,t]$, by claim 1 of Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations the entry $\Phi_{ij}$ satisfies $\Phi_{ij}(r)=\delta_{ij}+\int_{[0,r]}(A\Phi)_{ij}(u)\,du$ with $\delta_{ij}$ the entry of $I_k$, and $y_j(r)=0+\int_{[0,r]}(\Psi g)_j(u)\,du$, both integrands being bounded measurable, hence Lebesgue integrable over $[0,t]$. By claim (ii) of Integration by Parts for Indefinite Lebesgue Integrals on a Compact Interval on the interval $[0,t]$ (with the restrictions of the integrands to $[0,t]$, measurable and integrable there by the conventions above, and with $u_r=\Phi_{ij}(r)$, $v_r=y_j(r)$ for $r\in[0,t]$),

Summing over $j$ and using linearity of the integral, $\sum_j(A\Phi)_{ij}(u)y_j(u)=(A(u)\Phi(u)y(u))_i$ (claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product) and $\sum_j\Phi_{ij}(u)(\Psi(u)g(u))_j=(\Phi(u)\Psi(u)g(u))_i=g_i(u)$, we obtain

Moreover $\Phi(t)\xi=\xi+\int_{[0,t]}A(u)\Phi(u)\xi\,du$ by claim 1 of Fundamental Solution and Variation of Constants for Linear Ordinary Differential Equations applied to the constant vector $\xi$ (passing $\xi$ through the integral componentwise). Adding the two identities and using $x(u)=\Phi(u)(\xi+y(u))$ gives $x(t)=\xi+\int_{[0,t]}(A(u)x(u)+g(u))\,du$, the integrand $u\mapsto A(u)x(u)+g(u)$ being bounded measurable since $A$ has continuous entries and $x$ continuous components.

Claim 3. Let $z$ be as stated and put $w=z-x$, a bounded measurable map with $w(t)=\int_{[0,t]}A(u)w(u)\,du$ for all $t$, by subtracting the two integral equations (linearity of the integral). The function $\phi(t)=|w(t)|$ is bounded and measurable on $[0,T]$ (Norm Bound for a Vector-Valued Lebesgue Integral over a Compact Interval records that the norm of a bounded measurable map is bounded measurable), and by that lemma, the norm bound of the statement and monotonicity of the integral,

By Gronwall's Lemma for Bounded Measurable Functions with $a=0$ and $c=\bar{A}$, $\phi(t)\le0$ for all $t$, so $w=0$ and $z=x$.