Proof of Existence and Uniqueness for Ordinary Differential Equations with Measurable Time Dependence

Throughout, $F$ is Carathéodory–Lipschitz with constant $\Lambda$ (Carathéodory–Lipschitz Right-Hand Sides (iv)); for a function $z:[a,b]\to\mathbb{R}^{n}$, $F(\cdot,z)$ is the function $r\mapsto F(r,z(r))$ of Composition of a Carathéodory–Lipschitz Function with a Measurable Path; a Carathéodory solution with initial value $\xi$ is a Carathéodory solution on $[a,b]$ of the integral equation with right-hand side $F$ and initial value $\xi$; and $C=C([a,b],\mathbb{R}^{n})$ is the set of functions $[a,b]\to\mathbb{R}^{n}$ continuous on $[a,b]$ (The Space of Continuous Maps on a Closed Interval and Its Weighted Uniform Distance (i), with $(\mathbb{R}^{n},d_{E})$ for $(Y,d_{Y})$). The triangle inequality for $\lVert\cdot\rVert$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi)) and $d_{E}(u,v)=\lVert u-v\rVert$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii)) are used without further mention. Integrals over $[s,t]$ of functions in $\mathcal{L}^{1}([a,b])$ or $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ are defined (Indefinite Lebesgue Integrals on a Closed Interval (i)), the functions $t\mapsto\int_{[a,t]}g\,d\lambda$ exist (Functions and Function-Valued Operators Given by Integrals over Subintervals (ii)), and such integrals are monotone in the integrand (Indefinite Lebesgue Integrals on a Closed Interval (iv)); since such an integral is the integral of the restriction to $[s,t]$ (Integrals over Subintervals of a Closed Interval: Notation (iii)), linearity of the integral (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)) applies to it, componentwise for integrands with values in $\mathbb{R}^{n}$. A real function continuous on $[a,b]$ is bounded, measurable and in $\mathcal{L}^{1}([a,b])$ (Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval (iii)); sums, constant multiples and norms of continuous functions are continuous (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 (iii), Continuity of Vector- and Matrix-Valued Maps: the Componentwise Criterion, Norms and Algebraic Operations (v)).

Step 1: the difference estimate. Let $z_{1},z_{2}\in C$ and $\zeta_{1},\zeta_{2}\in\mathbb{R}^{n}$, and let $m:[a,b]\to\mathbb{R}$ be the continuous function $r\mapsto\lVert z_{1}(r)-z_{2}(r)\rVert$. By Composition of a Carathéodory–Lipschitz Function with a Measurable Path (iii), $F(\cdot,z_{1})$ and $F(\cdot,z_{2})$ lie in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$. We show that for every $t\in[a,b]$

$$\Bigl\lVert\Bigl(\zeta_{1}+\int_{[a,t]}F\bigl(r,z_{1}(r)\bigr)\,dr\Bigr)-\Bigl(\zeta_{2}+\int_{[a,t]}F\bigl(r,z_{2}(r)\bigr)\,dr\Bigr)\Bigr\rVert\le\lVert\zeta_{1}-\zeta_{2}\rVert+\Lambda\int_{[a,t]}m\,d\lambda .\qquad(1)$$

The function $g=F(\cdot,z_{1})-F(\cdot,z_{2})$ lies in $\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ (Rules for Integrals of Vector- and Matrix-Valued Functions (iii)), and by linearity the difference of the two integrals in (1) is $\int_{[a,t]}g\,d\lambda$. By Indefinite Lebesgue Integrals on a Closed Interval (vii), by Composition of a Carathéodory–Lipschitz Function with a Measurable Path (iv) and monotonicity, and by linearity,

$$\Bigl\lVert\int_{[a,t]}g\,d\lambda\Bigr\rVert\le\int_{[a,t]}\lVert g(r)\rVert\,dr\le\int_{[a,t]}\Lambda\,m(r)\,dr=\Lambda\int_{[a,t]}m\,d\lambda ,$$

and (1) follows by the triangle inequality.

Step 2: every solution is continuous. Let $\xi\in\mathbb{R}^{n}$ and let $x$ be a Carathéodory solution with initial value $\xi$. By Carathéodory Solutions of an Initial Value Problem (i), $F(\cdot,x)\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ and $x(t)=\xi+\int_{[a,t]}F(r,x(r))\,dr$ for every $t\in[a,b]$. The function $t\mapsto\int_{[a,t]}F(r,x(r))\,dr$ is continuous on $[a,b]$ by Indefinite Lebesgue Integrals on a Closed Interval (ix), and so is its sum $x$ with the constant $\xi$. Thus $x\in C$.

Step 3: comparison of two solutions. Let $\xi,\eta\in\mathbb{R}^{n}$, and let $x$ and $y$ be Carathéodory solutions with initial values $\xi$ and $\eta$. By Step 2, $x,y\in C$, so the function $u:t\mapsto\lVert x(t)-y(t)\rVert$ is continuous on $[a,b]$, hence bounded and measurable. By the integral equations for $x$ and $y$ and by (1) with $z_{1}=x$, $z_{2}=y$, $\zeta_{1}=\xi$ and $\zeta_{2}=\eta$,

$$u(t)\le\lVert\xi-\eta\rVert+\Lambda\int_{[a,t]}u\,d\lambda\qquad\bigl(t\in[a,b]\bigr).$$

Gronwall's lemma (Gronwall's Lemma for Bounded Measurable Functions (i), with $\alpha=\lVert\xi-\eta\rVert$ and $\beta=\Lambda\ge0$) gives

$$\lVert x(t)-y(t)\rVert\le\lVert\xi-\eta\rVert\exp\bigl(\Lambda(t-a)\bigr)\qquad\bigl(t\in[a,b]\bigr).\qquad(2)$$

Step 4: an exponential weight. Let $\varphi:[a,b]\to\mathbb{R}$ be the function $t\mapsto\exp(2\Lambda(t-a))$. By the chain rule and the derivatives of $\exp$ and of affine functions (Chain Rule for Real Functions on Intervals (i), Basic Properties of the Exponential Function (vii), Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (ii), Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (i), Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (v), Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (iv)), $\varphi$ is differentiable on $[a,b]$ with $\varphi'(t)=2\Lambda\,\varphi(t)$; hence it is continuous on $[a,b]$ (The Increment Form of the Derivative, and Differentiability Implies Continuity (ii)). Moreover $\varphi(a)=1$ and $\varphi(t)>0$ for every $t\in[a,b]$ (Basic Properties of the Exponential Function (i), Basic Properties of the Exponential Function (iii)). The function $\psi=\tfrac12\varphi$ is differentiable on $[a,b]$ with derivative $\Lambda\varphi$ (Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (v)), a continuous function. Let $t\in[a,b]$. By linearity, by Indefinite Lebesgue Integrals on a Closed Interval (xi) applied to the continuous function $\Lambda\varphi$, and by the fundamental theorem of calculus (Rules for Oriented Riemann Integrals of Real-Valued Maps: Sums, Scalar Multiples, Constants, Continuous Maps, Additivity and Evaluation by an Antiderivative (iv), Rules for Oriented Riemann Integrals of Real-Valued Maps: Sums, Scalar Multiples, Constants, Continuous Maps, Additivity and Evaluation by an Antiderivative (vi)),

$$\Lambda\int_{[a,t]}\varphi\,d\lambda=\int_{a}^{t}\Lambda\varphi=\int_{a}^{t}\psi'=\psi(t)-\psi(a)=\tfrac12\bigl(\varphi(t)-1\bigr)\le\tfrac12\,\varphi(t).\qquad(3)$$

Let $w=1/\varphi$; it is continuous on $[a,b]$ (Continuity of Identity Maps, Compositions and Reciprocals (iii)), $w(t)>0$ and $w(t)\varphi(t)=1$ for every $t\in[a,b]$. Let $d_{w}$ be the weighted uniform distance on $C$ with weight $w$. Since $(\mathbb{R}^{n},d_{E})$ is complete ($\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (iv)), $(C,d_{w})$ is a nonempty complete metric space (The Weighted Uniform Distance Is a Metric, and Is Complete When the Target Is Complete (i), The Weighted Uniform Distance Is a Metric, and Is Complete When the Target Is Complete (ii), The Weighted Uniform Distance Is a Metric, and Is Complete When the Target Is Complete (iii)).

Step 5: existence by a contraction. Let $\xi\in\mathbb{R}^{n}$. For $z\in C$, let $T(z):[a,b]\to\mathbb{R}^{n}$ be the function

$$T(z)(t)=\xi+\int_{[a,t]}F\bigl(r,z(r)\bigr)\,dr\qquad\bigl(t\in[a,b]\bigr),$$

defined because $F(\cdot,z)\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ (Composition of a Carathéodory–Lipschitz Function with a Measurable Path (iii)), and continuous on $[a,b]$ by Indefinite Lebesgue Integrals on a Closed Interval (ix). So $T(z)\in C$, and $T:C\to C$ is a function (formed from the functions of Functions and Function-Valued Operators Given by Integrals over Subintervals (iii) and Functions and Function-Valued Operators Given by Integrals over Subintervals (iv)). If $T(z)=z$, then $z$ is a Carathéodory solution with initial value $\xi$, by Carathéodory Solutions of an Initial Value Problem (i).

$T$ is a contraction with constant $\tfrac12$. Let $z_{1},z_{2}\in C$, let $\delta=d_{w}(z_{1},z_{2})$, which is $\ge0$ as $d_{w}$ is a metric, and let $m(r)=\lVert z_{1}(r)-z_{2}(r)\rVert$. By The Space of Continuous Maps on a Closed Interval and Its Weighted Uniform Distance (iii), we have $w(r)\,m(r)\le\delta$, that is, $m(r)\le\delta\,\varphi(r)$, for every $r\in[a,b]$. For $t\in[a,b]$, (1) with $\zeta_{1}=\zeta_{2}=\xi$, monotonicity and linearity, and (3) give

$$\lVert T(z_{1})(t)-T(z_{2})(t)\rVert\le\Lambda\int_{[a,t]}m\,d\lambda\le\delta\,\Lambda\int_{[a,t]}\varphi\,d\lambda\le\tfrac{\delta}{2}\,\varphi(t),$$

so $w(t)\,d_{E}\bigl(T(z_{1})(t),T(z_{2})(t)\bigr)\le\tfrac{\delta}{2}$. As $d_{w}(T(z_{1}),T(z_{2}))$ is the value of the left-hand side at some $t_{0}\in[a,b]$ (The Space of Continuous Maps on a Closed Interval and Its Weighted Uniform Distance (iii)),

$$d_{w}\bigl(T(z_{1}),T(z_{2})\bigr)\le\tfrac12\,d_{w}(z_{1},z_{2}),$$

and $T$ is a contraction. By the contraction mapping theorem (Contraction Mapping Theorem on a Nonempty Complete Metric Space (i)), $T$ has a fixed point, which is a Carathéodory solution with initial value $\xi$.

Step 6: proof of the clauses. (i). By Carathéodory Solutions of an Initial Value Problem (ii), the elements of $\mathrm{Sol}(F,\xi)$ are the Carathéodory solutions with initial value $\xi$. Step 5 gives one. If $x$ and $y$ are two, then (2) with $\eta=\xi$ gives $\lVert x(t)-y(t)\rVert\le0$, so $x(t)=y(t)$ for every $t\in[a,b]$ (Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (iii)), and $x=y$.

(ii). This is Step 2 for $x=x_{\xi}$.

(iv). This is (2) for $x=x_{\xi}$ and $y=x_{\eta}$.

(iii). Let $x=x_{\xi}$, let $h:r\mapsto\lVert F(r,0)\rVert$, a nonnegative function in $\mathcal{L}^{1}([a,b])$, and let $u:t\mapsto\lVert x(t)\rVert$, continuous on $[a,b]$ by (ii), hence bounded and measurable. For $r\in[a,b]$, Composition of a Carathéodory–Lipschitz Function with a Measurable Path (iv) (with the constant function $0$ for $y$) and the triangle inequality give $\lVert F(r,x(r))\rVert\le h(r)+\Lambda u(r)$. For $t\in[a,b]$, the integral equation, the triangle inequality, Indefinite Lebesgue Integrals on a Closed Interval (vii), monotonicity and linearity give

$$u(t)\le\lVert\xi\rVert+\int_{[a,t]}\bigl\lVert F\bigl(r,x(r)\bigr)\bigr\rVert\,dr\le\lVert\xi\rVert+\int_{[a,t]}h\,d\lambda+\Lambda\int_{[a,t]}u\,d\lambda\le\lVert\xi\rVert+\int_{[a,b]}h\,d\lambda+\Lambda\int_{[a,t]}u\,d\lambda ,$$

the last step because $\int_{[a,b]}h\,d\lambda=\int_{[a,t]}h\,d\lambda+\int_{[t,b]}h\,d\lambda$ (Indefinite Lebesgue Integrals on a Closed Interval (iii)) and $\int_{[t,b]}h\,d\lambda\ge0$ by monotonicity, as $h\ge0$. Gronwall's lemma (Gronwall's Lemma for Bounded Measurable Functions (i), with $\alpha=\lVert\xi\rVert+\int_{[a,b]}h\,d\lambda$ and $\beta=\Lambda$) gives (iii).

(v). Let $x=x_{\xi}$ and $s,t\in[a,b]$. The function $F(\cdot,x)\in\mathcal{L}^{1}([a,b];\mathbb{R}^{n})$ satisfies $\lVert F(r,x(r))\rVert\le K$ for every $r\in[a,b]$, and by the integral equation $x(t)-x(s)=\int_{[a,t]}F(r,x(r))\,dr-\int_{[a,s]}F(r,x(r))\,dr$; so Indefinite Lebesgue Integrals on a Closed Interval (x) gives $\lVert x(t)-x(s)\rVert\le K\,|t-s|$. $\blacksquare$