Proof of A Priori Bounded Solutions of Locally Lipschitz Ordinary Differential Equations Exist Globally

We use the componentwise inequalities $|x^{i}|\le|x|\le\sum_l|x^{l}|$ and $d(x,y)=|x-y|$ of claim 1 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals, and the reverse triangle inequality $\bigl||x|-|y|\bigr|\le d(x,y)$, which follows from the triangle inequality for the Euclidean distance (Euclidean Distance is a Metric on $\mathbb{R}^n$) since $|x|=d(x,0)\le d(x,y)+d(y,0)$ and symmetrically.

The truncation. Define $\pi:\mathbb{R}^{k}\to\mathbb{R}^{k}$ by $\pi(x)=x$ if $|x|\le\rho+1$ and $\pi(x)=\frac{\rho+1}{|x|}\,x$ otherwise. Then $|\pi(x)|\le\rho+1$ for all $x$, and $\pi(x)=x$ exactly when $|x|\le\rho+1$.

$\pi$ is Lipschitz with constant $2$: $d(\pi(x),\pi(y))\le2\,d(x,y)$ for all $x,y$. If $|x|,|y|\le\rho+1$ this is clear. If $|x|,|y|>\rho+1$, then with $s=\rho+1$,

so, using $|y|\,\bigl|\frac1{|x|}-\frac1{|y|}\bigr|=\frac{\bigl||y|-|x|\bigr|}{|x|}\le\frac{d(x,y)}{|x|}$ and $|x|>s$,

If $|x|\le\rho+1<|y|$: $\pi(y)$ lies on the segment from $0$ to $y$ with $|\pi(y)|=\rho+1$, and $d(y,\pi(y))=|y|-(\rho+1)\le|y|-|x|\le d(x,y)$, so $d(\pi(x),\pi(y))=d(x,\pi(y))\le d(x,y)+d(y,\pi(y))\le2\,d(x,y)$; the case $|y|\le\rho+1<|x|$ is symmetric.

Composition: if $h:[a,b]\to\mathbb{R}^{k}$ has continuous components, so does $\pi\circ h$: given a point $t$ and $\varepsilon>0$, choose $\delta>0$ so that $|s-t|<\delta$ makes every $|h^{l}(s)-h^{l}(t)|<\varepsilon/(2k)$; then for each component $i$,

The truncated equation. Define $\widetilde{F}(t,x)=F(t,\pi(x))$. Composition continuity (hypothesis (i) of Global Existence and Uniqueness for Lipschitz Ordinary Differential Equations in Integral Form) holds for $\widetilde{F}$, since $\widetilde F(t,h(t))=F(t,(\pi\circ h)(t))$ with $\pi\circ h$ continuous. And $\widetilde{F}$ is globally Lipschitz: $|\pi(x)|,|\pi(y)|\le\rho+1$, so by hypothesis (ii),

By Global Existence and Uniqueness for Lipschitz Ordinary Differential Equations in Integral Form there is exactly one $\widetilde x:[a,b]\to\mathbb{R}^{k}$ with continuous components satisfying $\widetilde x(t)=\xi+\int_a^t\widetilde F(r,\widetilde x(r))\,dr$ on $[a,b]$.

The bound $|\widetilde x|\le\rho$. The function $t\mapsto|\widetilde x(t)|$ is continuous (reverse triangle inequality and componentwise continuity, as above). Let

$S$ contains $a$ because $|\widetilde x(a)|=|\xi|\le\rho$ (hypothesis (iii)); $S$ is an interval containing $[a,c]$ for each of its members $c$; and by continuity $|\widetilde x|\le\rho+1$ on $[a,c^{*}]$, so $c^{*}\in S$. On $[a,c^{*}]$ we have $\pi(\widetilde x(r))=\widetilde x(r)$, so if $c^{*}>a$ the restriction of $\widetilde x$ is a solution on $[a,c^{*}]$ in the sense of the statement, and hypothesis (iii) gives $|\widetilde x(t)|\le\rho$ on $[a,c^{*}]$; if $c^{*}=a$ this bound holds at $t=a$ anyway. If $c^{*}<b$, then since $|\widetilde x(c^{*})|\le\rho$ and $|\widetilde x|$ is continuous, there is $c'\in(c^{*},b]$ with $|\widetilde x|<\rho+1$ on $[c^{*},c']$, whence $c'\in S$, contradicting $c^{*}=\sup S$. Therefore $c^{*}=b$, and by the argument just given (now on $[a,b]$), $|\widetilde x(t)|\le\rho$ for all $t\in[a,b]$.

Conclusion. Since $|\widetilde x|\le\rho\le\rho+1$ everywhere, $\pi\circ\widetilde x=\widetilde x$, so $\widetilde F(r,\widetilde x(r))=F(r,\widetilde x(r))$ and $\widetilde x$ is a solution on $[a,b]$ of the original equation, with $|\widetilde x|\le\rho$. Conversely, if $h$ is any solution on $[a,b]$, hypothesis (iii) gives $|h|\le\rho$, hence $\pi\circ h=h$ and $h$ satisfies the truncated equation; by the uniqueness in Global Existence and Uniqueness for Lipschitz Ordinary Differential Equations in Integral Form, $h=\widetilde x$. This proves existence and uniqueness of the solution $x=\widetilde x$. $\blacksquare$