Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it

If $\delta >b-t_0$ replace $\delta$ by $b-t_0$; then the interval $(b-\delta ,b)\cap [t_0,b)$ is nonempty. Choose any $t$ in it; the hypothesis gives $(t,x(t))\in K$, so $K$ is nonempty. The norm function $z\mapsto \parallel F(z){\parallel }_2$ is continuous on the compact metric subspace $K$: given $z_0\in K$ and $\epsilon >0$, continuity of $F$ at $z_0$ [L7] gives ${\delta }_0>0$ with $\parallel F(z)-F(z_0){\parallel }_2<\epsilon$ whenever $d_2(z,z_0)<{\delta }_0$, and the reverse triangle inequality [L5] then gives $\mid \parallel F(z){\parallel }_2-\parallel F(z_0){\parallel }_2\mid \le \parallel F(z)-F(z_0){\parallel }_2<\epsilon$. Hence [L6] gives $M:=\max \{\parallel F(z){\parallel }_2:z\in K\}<+\mathrm{\infty }$.

For $s,t$ with $b-\delta <s<t<b$: since $x$ solves the initial value problem, [L3] gives $x(t)=x_0+{\int }_{t_0}^{t}F(u,x(u))du$ and $x(s)=x_0+{\int }_{t_0}^{s}F(u,x(u))du$, hence $x(t)-x(s)={\int }_s^{t}F(u,x(u))du$; the integrand is continuous in $u$ on $[s,t]$ because $u\mapsto F(u,x(u))$ is a composition of continuous maps (the components of $x$ are continuous by differentiability, [L7]) and $(u,x(u))\in K$ on this interval, so its components are integrable on $[s,t]$ by [L8]. By [L4], $\parallel x(t)-x(s){\parallel }_2\le {\int }_s^t\parallel F(u,x(u)){\parallel }_{2}du\le M(t-s)$.

The limit $x_b:={\lim }_{t\uparrow b}x(t)$ exists. Indeed, choose $k_0\ge 1$ with $1\mathrm{/}{k}_0<\min \{\delta ,b-t_0\}$ [L7] and put $y_k:=x(b-1\mathrm{/}k)$ for $k\ge k_0$; by step 2.1, $\parallel y_k-y_l{\parallel }_2\le M\mathrm{\mid }1\mathrm{/}k-1\mathrm{/}l\mathrm{\mid }$ for all $k,l\ge k_0$, so $(y_k)$ is a Cauchy sequence in ${\mathbb{R}}^n$ and converges, by [L5], to some $x_b\in {\mathbb{R}}^n$. Given $\eta >0$, choose $k\ge k_0$ with $1\mathrm{/}k<\eta \mathrm{/}(4(M+1))$ and $\parallel y_k-x_b{\parallel }_2<\eta \mathrm{/}2$; then for every $t\in (\max \{b-\delta ,b-\eta \mathrm{/}(4(M+1))\},b)$, step 2.1 gives $\parallel x(t)-x_b{\parallel }_2\le \parallel x(t)-y_k{\parallel }_2+\parallel y_k-x_b{\parallel }_2\le M(\mathrm{\mid }t-b\mathrm{\mid }+1\mathrm{/}k)+\eta \mathrm{/}2<M\cdot \eta \mathrm{/}(2(M+1))+\eta \mathrm{/}2<\eta$. Hence ${\lim }_{t\uparrow b}x(t)=x_b$.

The point $(b,x_b)$ lies in $K$, hence in $D$: for $t\in (b-\delta ,b)\cap [t_0,b)$ the points $(t,x(t))$ lie in $K$ and converge to $(b,x_b)$ as $t\uparrow b$ (in the product metric, the first coordinate trivially and the second by step 3.1); since $K$ is closed [L6], its complement is open, so a point outside $K$ would have a ball around it disjoint from $K$, which is impossible for $(b,x_b)$ because every such ball contains some $(t,x(t))$.

Extend $x$ to the continuous curve $\bar{x}:[t_0,b]\to {\mathbb{R}}^n$ by $\bar{x}(t):=x(t)$ for $t<b$ and $\bar{x}(b):=x_b$; it is continuous by step 3.1. Let $0<h_0<\delta$. For $s\in (b-h_0,b)$ and $t\in (s,b)$, step 2.1 and additivity of the integral [L3] give $(x(t)-x(s))-{\int }_s^{b}F(u,\bar{x}(u))du=-{\int }_t^{b}F(u,\bar{x}(u))du$; since $(u,\bar{x}(u))\in K$ for every $u\in [s,b]$ (for $u<b$ by hypothesis, for $u=b$ by step 4.1), one has $\parallel F(u,\bar{x}(u)){\parallel }_2\le M$ there, and the norm inequality [L4] gives $\parallel (x(t)-x(s))-{\int }_s^{b}F(u,\bar{x}(u))du{\parallel }_2\le {\int }_t^b\parallel F(u,\bar{x}(u)){\parallel }_{2}du\le M(b-t)\to 0$ as $t\uparrow b$. With $x(t)\to x_b$ (step 3.1) this yields $x_b-x(s)={\int }_s^{b}F(u,\bar{x}(u))du$. Hence the left difference quotient satisfies $\parallel \frac{\bar{x}(b)-\bar{x}(s)}{b-s}-F(b,x_b){\parallel }_2\le {\sup }_{s\le u\le b}\parallel F(u,\bar{x}(u))-F(b,x_b){\parallel }_2⟶0(s\uparrow b)$ because for every $\eta >0$ continuity of $\bar{x}$ at $b$ and of $F$ at $(b,x_b)$ makes the displayed integrand smaller than $\eta$ for all $u\in [s,b]$ once $s$ is sufficiently close to $b$; [L6] ensures each supremum is finite. So $\bar{x}$ is differentiable at every point of $[b-h_0,b]$ as a function on that interval, with ${\bar{x}}^{\mathrm{\prime }}(t)=F(t,\bar{x}(t))$ for $t<b$ and left derivative $F(b,x_b)$ at $t=b$; it is therefore, for every such $h_0$, a solution of the initial value problem $z^{\mathrm{\prime }}=F(t,z)$, $z(b)=x_b$ on $[b-h_0,b]$ in the sense of [L1] and [L3].

Apply the local existence conclusion of Picard-Lindelöf [L1] at $(b,x_b)\in D$, whose hypotheses on $F$ are the assumed continuity and local Lipschitz continuity in the state variable [L2]. It gives a solution through $(b,x_b)$ on some interval around $b$; choose $h>0$ so $[b-h,b+h]$ is contained in that interval and denote the restricted solution by $y$. Then $y$ is a solution on $[b-h,b+h]$ through $(b,x_b)$.

Let $h^{\mathrm{\prime }}:=\min \{h,\delta \mathrm{/}2\}>0$, where $h$ is the Picard-Lindelöf half-length of step 5.2; by step 5.1 the curve $\bar{x}$ is a solution on $[b-h^{\mathrm{\prime }},b]$ through $(b,x_b)$, and the restriction of $y$ is a solution on $[b-h,b]\supseteq [b-h^{\mathrm{\prime }},b]$ through the same point; by the uniqueness clause of [L1] they agree on $[b-h^{\mathrm{\prime }},b]$. Define $\tilde{x}:[t_0,b+h)\to {\mathbb{R}}^n$ by $\tilde{x}(t):=\bar{x}(t)$ for $t\le b$ and $\tilde{x}(t):=y(t)$ for $b<t<b+h$. Then $\tilde{x}$ is continuous, and it is differentiable with ${\tilde{x}}^{\mathrm{\prime }}(t)=F(t,\tilde{x}(t))$ at every $t\in [t_0,b+h)$: for $t<b$ and for $t>b$ this is the respective solution property, while at $t=b$ the left difference quotient tends to $F(b,x_b)$ by step 5.1 and the right difference quotient is the two-sided derivative $y^{\mathrm{\prime }}(b)=F(b,x_b)$ of the solution $y$; the two agree, so the derivative of $\tilde{x}$ at $b$ exists and equals $F(b,x_b)=F(b,\tilde{x}(b))$. With $\epsilon :=h$ this is an extension as asserted in clause 1.

Under the hypotheses of clause 2, the argument of steps 1.1-6.1 applies with $K$ replaced by the given compact $K_0\subseteq U$: the bound of step 1.1 is $M:=\max \{\parallel G(z){\parallel }_2:z\in K_0\}<+\mathrm{\infty }$ by [L6], the Lipschitz estimate of step 2.1 becomes $\parallel x(t)-x(s){\parallel }_2\le M(t-s)$ because $F(u,x(u))=G(x(u))$ and $x(u)\in K_0$ for $u\in (b-\delta ,b)\cap [t_0,b)$, the limit point of step 4.1 lies in $K_0$ since $K_0$ is compact hence closed [L6], and $(b,x_b)\in \mathbb{R}\times U=D$ because $x_b\in K_0\subseteq U$; the local solution of step 5.2 and the gluing of step 6.1 are then carried out for the same field $F(t,z)=G(z)$ at $(b,x_b)$. This proves clause 2 with the same $\epsilon =h$.