Proof of Solutions of an Initial Value Problem Are the Continuous Solutions of the Integral Equation

By Basic Facts about Intervals of the Real Line (v) and Basic Facts about Intervals of the Real Line (ii), $[a,c]$ is an interval containing the two distinct points $a$ and $c$, and $[a,c]\subseteq[a,b]$. $(\mathbb{R},d_{\mathbb{R}})$ is the metric space of Single-Variable Calculus on an Interval (iii), $(\mathbb{R}^{n},d_{E})$ by Euclidean Space and Real Matrices: Standing Notation (ii) and $(\mathcal{M}_{p\times q}(\mathbb{R}),d_{p\times q})$ by Euclidean Space and Real Matrices: Standing Notation (vi).

Proof of (i). We give the vector case; the matrix case is the same with $\mathcal{M}_{p\times q}(\mathbb{R})$, $d_{p\times q}$ and $X$ in place of $\mathbb{R}^{n}$, $d_{E}$ and $x$. Let $d$ be the product metric on $\mathbb{R}\times\mathbb{R}^{n}$. For $s\in[a,c]$ we have $s\in[a,b]$, so $(s,x(s))\in[a,b]\times\mathbb{R}^{n}$; let $g:[a,c]\to\mathbb{R}^{n}$ be the map $s\mapsto F(s,x(s))$ (which exists by Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii), with the identity map of $[a,c]$ of Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iii), a map into $[a,b]$, and $x$ for its $f$ and $g$).

Let $s_{0}\in[a,c]$. The choices are made in this order. First a real $\varepsilon>0$ is given. Then, as $F$ is continuous on $[a,b]\times\mathbb{R}^{n}$, it is continuous at $(s_{0},x(s_{0}))$ relative to $[a,b]\times\mathbb{R}^{n}$ by Continuous Map Between Metric Spaces (ii), and Continuous Map Between Metric Spaces (i) gives a real $\eta>0$ such that every $(s,v)\in[a,b]\times\mathbb{R}^{n}$ with $d\bigl((s_{0},x(s_{0})),(s,v)\bigr)<\eta$ satisfies $d_{E}\bigl(F(s,v),F(s_{0},x(s_{0}))\bigr)<\varepsilon$. Next, as $x$ is continuous on $[a,c]$, it is continuous at $s_{0}$ relative to $[a,c]$, and Continuous Map Between Metric Spaces (i), applied to $\eta$, gives a real $\delta_{1}>0$ such that every $s\in[a,c]$ with $d_{\mathbb{R}}(s_{0},s)<\delta_{1}$ satisfies $d_{E}(x(s),x(s_{0}))<\eta$. Let $\delta>0$ be the smaller of $\delta_{1}$ and $\eta$. Now let $s\in[a,c]$ with $d_{\mathbb{R}}(s_{0},s)<\delta$. Then $d_{\mathbb{R}}(s_{0},s)<\eta$ and, by Metric Space (iv), $d_{E}(x(s_{0}),x(s))=d_{E}(x(s),x(s_{0}))<\eta$; so $d\bigl((s_{0},x(s_{0})),(s,x(s))\bigr)<\eta$ by The Product Metric is a Metric (iii), and therefore $d_{E}(g(s),g(s_{0}))=d_{E}\bigl(F(s,x(s)),F(s_{0},x(s_{0}))\bigr)<\varepsilon$. So $g$ is continuous at $s_{0}$ relative to $[a,c]$, and, $s_{0}$ being arbitrary, continuous on $[a,c]$ by Continuous Map Between Metric Spaces (ii). This proves (i).

Proof of (ii). Let $h:[a,c]\to\mathbb{R}^{n}$ be the map $s\mapsto F(s,x(s))$ (as $g$ above, Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (iii)); being a map on the set $[a,c]$, $h$ is a set (Images of Sets under Functions Are Sets, and a Function with a Set Domain Is a Set (ii)), and $\int_{a}^{t}F(s,x(s))\,ds=\int_{a}^{t}h$. Derivatives of maps on $[a,c]$ are taken relative to $[a,c]$, as in Solution of an Initial Value Problem on $[a,c]$ (i).

A solution solves the integral equation. Let $x$ be a solution on $[a,c]$: by Solution of an Initial Value Problem on $[a,c]$ (i) and Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (i), $x$ is differentiable, $x'(t)=F(t,x(t))=h(t)$ for every $t\in[a,c]$, and $x(a)=\xi$. By Rules for Derivatives of Vector- and Matrix-Valued Maps (i), with $[a,c]$ in place of $I$, $x$ is continuous at every $t\in[a,c]$ relative to $[a,c]$, hence continuous on $[a,c]$ by Continuous Map Between Metric Spaces (ii). By (i), $h$ is continuous on $[a,c]$. The derivative $x':[a,c]\to\mathbb{R}^{n}$ of Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (iv) has the same values as $h$, so $x'=h$ and $x'$ is continuous on $[a,c]$. Let $t\in[a,c]$. By The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (iii), with $[a,c]$ in place of $I$ and $a$ and $t$ in place of $u$ and $v$, the integral $\int_{a}^{t}h=\int_{a}^{t}x'$ is defined and equals $x(t)-x(a)=x(t)-\xi$. Hence $x(t)=\xi+\int_{a}^{t}h$.

A continuous solution of the integral equation is a solution. Let $x$ be continuous on $[a,c]$, and suppose that for every $t\in[a,c]$ the integral $\int_{a}^{t}h$ is defined and $x(t)=\xi+\int_{a}^{t}h$. By (i), $h$ is continuous on $[a,c]$. By The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (i), with $a$ and $c$ in place of $a$ and $b$, $a$ as the base point and $h$ in place of $x$, $\int_{a}^{t}h$ is defined for every $t\in[a,c]$, and the map $Y:[a,c]\to\mathbb{R}^{n}$, $t\mapsto\int_{a}^{t}h$ (which exists by Maps Given by Integrals of Vector- and Matrix-Valued Maps (i), with $n$ for its $q$, $[a,c]$ for its $E$ and $S$, $a$ for its $c$ and $h$ for its $x$), is differentiable, with $Y'(t)=h(t)$ for every $t\in[a,c]$. By Rules for Derivatives of Vector- and Matrix-Valued Maps (ix), the constant map $K:[a,c]\to\mathbb{R}^{n}$, $t\mapsto\xi$, which exists by Maps Formed Pointwise: Constant Maps, Pairs of Values, Maps Applied to Values, and Families of Maps (i), is differentiable at every $t\in[a,c]$ with derivative $0$. Since $x(t)=K(t)+Y(t)$ for every $t\in[a,c]$, $x$ is the pointwise sum $K+Y$, and by Rules for Derivatives of Vector- and Matrix-Valued Maps (ii) it is differentiable at every $t\in[a,c]$, with

$$x'(t)=0+h(t)=F\bigl(t,x(t)\bigr).$$

For every $k\in[n]$, the integral $\int_{a}^{a}h_{k}$ of the component map $h_{k}$ is defined and equals $0$, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (viii) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (vii); so, by Riemann Integral of a Vector-Valued or Matrix-Valued Map (ii), $\int_{a}^{a}h$ is the point of $\mathbb{R}^{n}$ all of whose components are $0$, that is, $\int_{a}^{a}h=0$, and $x(a)=\xi+0=\xi$. Thus $x$ is differentiable on $[a,c]$ (Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (iii)) and is a solution on $[a,c]$ by Solution of an Initial Value Problem on $[a,c]$ (i). This proves (ii).

Proof of (iii). The argument for (ii) applies word for word, with $X$, $\Xi$, $\mathcal{M}_{p\times q}(\mathbb{R})$, the zero matrix $0_{p\times q}$, Solution of an Initial Value Problem on $[a,c]$ (ii), Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (ii) and the matrix case of (i) in place of $x$, $\xi$, $\mathbb{R}^{n}$, $0$, Solution of an Initial Value Problem on $[a,c]$ (i), Derivative of a Vector-Valued or Matrix-Valued Map on an Interval (i) and its vector case, using the matrix halves of Rules for Derivatives of Vector- and Matrix-Valued Maps (i), Rules for Derivatives of Vector- and Matrix-Valued Maps (ix), Rules for Derivatives of Vector- and Matrix-Valued Maps (ii), The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (iii) and The Fundamental Theorem of Calculus and the Mean Value Inequality for Vector- and Matrix-Valued Maps (i), with Maps Given by Integrals of Vector- and Matrix-Valued Maps (ii) in place of Maps Given by Integrals of Vector- and Matrix-Valued Maps (i), and with the entry maps $h_{ij}$ ($i\in[p]$, $j\in[q]$) and Riemann Integral of a Vector-Valued or Matrix-Valued Map (iv) in place of the component maps $h_{k}$ ($k\in[n]$) and Riemann Integral of a Vector-Valued or Matrix-Valued Map (ii).