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

Throughout we use the coordinate bound $|x^i|\le|x|$ from the elementary properties of the Euclidean norm, and the following consequence of linearity of the integral: if $0\le r\le t\le T$ and $g$ is bounded and measurable on $[0,T]$, then $\int_{[0,t]}g=\int_{[0,r]}g+\int_{[r,t]}g$, because $\mathbf{1}_{[0,t]}=\mathbf{1}_{[0,r]}+\mathbf{1}_{[r,t]}-\mathbf{1}_{\{r\}}$ and the integral of $g\,\mathbf{1}_{\{r\}}$ vanishes, the single point $r$ having Lebesgue measure zero. Here $\mathbf{1}_D$ is the function equal to $1$ on $D$ and $0$ elsewhere.

Part (a). Let $y:[0,T]\to\mathbb{R}^n$ be continuous and fix $i$. For a natural number $q$ put $t_k=kT/q$ for $k\in\{0,\dots,q\}$ and define $y^{(q)}:[0,T]\to\mathbb{R}^n$ by $y^{(q)}_s=y_{t_k}$ for $s\in[t_k,t_{k+1})$ and $k\in\{0,\dots,q-1\}$, and $y^{(q)}_T=y_T$.

For every real number $c$,

a finite union of intersections of intervals with sets that belong to the trace Borel $\sigma$-algebra by hypothesis 1. Hence $s\mapsto f^i(s,y^{(q)}_s)$ is measurable, and $|f^i(s,y^{(q)}_s)|\le|f(s,y^{(q)}_s)|\le K$ by hypothesis 2.

The map $y$ is continuous on the compact interval $[0,T]$, hence uniformly continuous. Every $s\in[0,T]$ satisfies $|s-u|\le T/q$ for the point $u\in\{t_0,\dots,t_q\}$ with $y^{(q)}_s=y_u$, so $|y^{(q)}_s-y_s|\le\omega(T/q)$ where $\omega$ is a modulus of uniform continuity of $y$, and therefore $|y^{(q)}_s-y_s|\to0$ as $q\to\infty$, uniformly in $s$. By hypothesis 3 and the coordinate bound,

so $f^i(s,y^{(q)}_s)$ converges to $f^i(s,y_s)$ for every $s\in[0,T]$. All these functions are bounded in absolute value by $K$, so by measurability of bounded pointwise limits, $s\mapsto f^i(s,y_s)$ is measurable; it is bounded by $K$, hence integrable over $[0,t]$ for every $t$.

Part (b), existence. Let $C$ denote the set of continuous maps $[0,T]\to\mathbb{R}^n$ equipped with the supremum metric $d_\infty(y,z)=\sup_{t\in[0,T]}|y_t-z_t|$, which is a complete metric space by completeness of the space of continuous vector-valued functions. Define $x^{(0)}_t=x_0$ for all $t$, and recursively

Assuming inductively that $x^{(j)}\in C$, part (a) shows that all these integrals exist. For $0\le r\le t\le T$ the $i$-th component of $x^{(j+1)}_t-x^{(j+1)}_r$ is $\int_{[r,t]}f^i(s,x^{(j)}_s)\,ds$, so by the norm bound for vector-valued integrals and hypothesis 2,

Hence $x^{(j+1)}\in C$, and by induction every iterate lies in $C$.

We show by induction on $j$ that

For $j=0$ this is the display above with $r=0$. Assuming it for $j$, the norm bound, hypothesis 3 and monotonicity give

the last integral being computed as a Riemann integral of a continuous integrand, which agrees with the Lebesgue integral by the compact-interval toolkit, and evaluated by the fundamental theorem of calculus.

Consequently $d_\infty(x^{(j+1)},x^{(j)})\le K\Lambda^jT^{j+1}/(j+1)!$. The series $\sum_{j\ge0}K\Lambda^jT^{j+1}/(j+1)!$ converges, its partial sums being bounded by $KT\,e^{\Lambda T}$, so for $j<j'$ the triangle inequality gives

a tail of a convergent series, which tends to $0$ as $j\to\infty$. Thus $(x^{(j)})$ is a Cauchy sequence in $C$ and converges to some $x\in C$; in particular $x$ is continuous.

Fix $t\in[0,T]$ and $i$. By hypothesis 3, the coordinate bound and monotonicity,

Since also $x^{(j+1),i}_t\to x^i_t$, letting $j\to\infty$ in the recursion gives

At $t=0$ all integrals vanish, so the value of $x$ at $0$ is $x_0$.

Part (b), uniqueness. Let $x$ and $\tilde{x}$ be continuous maps satisfying the integral equation with the same $x_0$. The map $u(t)=|x_t-\tilde{x}_t|$ is continuous, and $u(0)=0$. Subtracting the two equations componentwise and applying the norm bound and hypothesis 3,

The integrand is continuous, so the Lebesgue and Riemann integrals agree, and Gronwall's lemma in integral form, applied with $a=0$ and $b=\Lambda$, gives $u(t)\le0$ for every $t$. Since $u\ge0$, we get $u\equiv0$, that is $x=\tilde{x}$.

Part (c). For $0\le r\le t\le T$ the same computation as for the iterates gives $|x_t-x_r|\le\int_{[r,t]}|f(s,x_s)|\,ds\le K(t-r)$, and the case $t\le r$ follows by symmetry. $\blacksquare$