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

Let $T>0$ be a real number and $k\ge1$ a natural number. Let $A$ assign to each $t\in[0,T]$ a real $k\times k$ matrix $A(t)$ (real matrix) all of whose entries are continuous functions of $t$ on $[0,T]$, the interval being regarded as a subset of the real line with the absolute value metric and $\mathbb{R}$ carrying the same metric. Let $\Phi$ be the fundamental solution of $A$ on $[0,T]$ and $\Psi(t)=\Phi(t)^{-1}$ its inverse (claims 1 and 2 of that theorem), and define the two-parameter fundamental solution by

with the matrix product. For a real matrix $M$ with $k$ rows and $k$ columns put $\lVert M\rVert_{\mathrm{rs}}=\bigl(\sum_{i=1}^{k}(\sum_{j=1}^{k}|M_{ij}|)^{2}\bigr)^{1/2}$ (the Euclidean norm of the vector of absolute row sums, an ad hoc norm used only in this lemma and called the row-sum vector norm here), so that $|Mv|\le\lVert M\rVert_{\mathrm{rs}}\,|v|$ for every $v\in\mathbb{R}^{k}$ by claim 3 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product, where $|\cdot|$ is the Euclidean norm on Euclidean space $\mathbb{R}^{k}$ and $Mv$ the matrix-vector product. Let $\bar{A}\ge0$ and $\bar{\Phi}\ge0$ be real numbers with $\lVert A(t)\rVert_{\mathrm{rs}}\le\bar{A}$, $\lVert\Phi(t)\rVert_{\mathrm{rs}}\le\bar{\Phi}$ and $\lVert\Psi(t)\rVert_{\mathrm{rs}}\le\bar{\Phi}$ for all $t\in[0,T]$; such numbers exist because every entry of $A$, $\Phi$ and $\Psi$ is continuous on $[0,T]$, hence bounded by Continuous Real-Valued Functions on a Compact Interval are Bounded, and a matrix whose entries are bounded by $C$ has row-sum vector norm at most $k^{3/2}C$ (each absolute row sum being at most $kC$). A map $[0,T]\to\mathbb{R}^{k}$ is called bounded measurable when its components are bounded and measurable with respect to the trace Borel $\sigma$-algebra on $[0,T]$ and the Borel $\sigma$-algebra of the real line; for such maps, $\int_{[a,b]}\cdot\,du$ denotes the componentwise Lebesgue integral over the compact interval $[a,b]\subseteq[0,T]$, equal to $0$ when $a=b$. Let $I_{k}$ be the identity matrix.

1. (Two-parameter fundamental solution.) $\Phi(t,t)=I_{k}$ and $\lVert\Phi(t,u)\rVert_{\mathrm{rs}}\le \bar{\Phi}^{2}$ for all $t,u\in[0,T]$; for fixed $t$ every entry of $u\mapsto\Phi(t,u)$ is continuous on $[0,T]$; and for all $0\le\tau\le t\le T$ and every $v\in\mathbb{R}^{k}$,

2. (Existence.) Let $\xi\in\mathbb{R}^{k}$ and let $g:[0,T]\to\mathbb{R}^{k}$ be bounded measurable. Then $u\mapsto\Phi(t,u)g(u)$ is bounded measurable for every $t$, the map

has continuous, hence bounded measurable, components, $|x(t)|\le\bar{\Phi}|\xi|+\bar{\Phi}^{2}\int_{[0,t]}|g(u)|\,du$ for every $t$, and

3. (Uniqueness.) If $z:[0,T]\to\mathbb{R}^{k}$ is bounded measurable and $z(t)=\xi+\int_{[0,t]}(A(u)z(u)+g(u))\,du$ for every $t\in[0,T]$, then $z=x$ on $[0,T]$.