[proofplan] We build a compact tube around the reference trajectory and choose parameter values near $p_0$ so that $F$ has one common state-Lipschitz constant on that tube. [Uniform continuity](/page/Uniform%20Continuity) of $F$ on the compact tube makes the parameter error uniformly small. For any nearby solution, the integral form of the equation gives a Gronwall-type estimate both forward and backward from $t_0$. Choosing the initial and parameter neighbourhoods small enough keeps the nearby solution inside the tube, and the continuation criterion prevents finite-time escape before reaching the endpoints $a$ and $b$. [/proofplan]

[step:Choose a compact tube around the reference trajectory inside $U$]Define \begin{align*} d_0:=\operatorname{dist}(C,\mathbb{R}^n\setminus U). \end{align*} By hypothesis, $d_0>0$. Choose a radius $r>0$ such that $0<r<d_0/2$, and define the closed $r$-neighbourhood of $C$ by \begin{align*} K:=\{z\in \mathbb{R}^n:\operatorname{dist}(z,C)\leq r\}. \end{align*} Since $C$ is compact, $K$ is closed and bounded in $\mathbb{R}^n$, hence compact. If $z\in K$, then there exists $c\in C$ with $|z-c|\leq r$ after replacing $r$ by any slightly larger radius and then passing to the limit; therefore \begin{align*} \operatorname{dist}(z,\mathbb{R}^n\setminus U)\geq d_0-r>0. \end{align*} Thus $K\subset U$. Also $x([a,b])\subset C\subset K$, and the smaller tube \begin{align*} K_{r/2}:=\{z\in \mathbb{R}^n:\operatorname{dist}(z,C)\leq r/2\} \end{align*} is a compact subset of the interior of $K$.[/step]

[guided]The purpose of this step is to create a compact region in state space where all later estimates will take place. The reference solution stays in $C$, and $C$ sits a positive distance from the complement of $U$. Define \begin{align*} d_0:=\operatorname{dist}(C,\mathbb{R}^n\setminus U). \end{align*} The hypothesis says exactly that $d_0>0$. Choose $r$ with $0<r<d_0/2$, and set \begin{align*} K:=\{z\in \mathbb{R}^n:\operatorname{dist}(z,C)\leq r\}. \end{align*} Because $C$ is compact, this closed neighbourhood is bounded and closed in $\mathbb{R}^n$, so it is compact by the [Heine-Borel theorem](/theorems/309). We also need $K$ to remain inside $U$. If $z\in K$, then $z$ lies within distance at most $r$ of $C$. Since every point of $C$ is at distance at least $d_0$ from $\mathbb{R}^n\setminus U$, the triangle inequality gives \begin{align*} \operatorname{dist}(z,\mathbb{R}^n\setminus U)\geq d_0-r>0. \end{align*} Hence $z\notin \mathbb{R}^n\setminus U$, so $z\in U$. Thus $K\subset U$. Finally, since $x([a,b])\subset C$, the reference solution lies in $K$, and the smaller compact tube \begin{align*} K_{r/2}:=\{z\in \mathbb{R}^n:\operatorname{dist}(z,C)\leq r/2\} \end{align*} leaves a margin of size $r/2$ before the boundary of $K$.[/guided]

[step:Obtain uniform Lipschitz and parameter-continuity bounds on the tube]Choose an open neighbourhood $Q_0\subset P$ of $p_0$ such that $\overline{Q_0}$ is compact and $\overline{Q_0}\subset P$. Put \begin{align*} S:=[a,b]\times K\times \overline{Q_0}. \end{align*} The set $S$ is compact. For each $(\tau,z,\pi)\in S$, the local uniform Lipschitz hypothesis gives open neighbourhoods $A_{\tau,z,\pi}\subset I$ of $\tau$, $V_{\tau,z,\pi}\subset U$ of $z$, and $R_{\tau,z,\pi}\subset P$ of $\pi$, together with a constant $L_{\tau,z,\pi}\geq 0$, such that \begin{align*} |F(t,u,p)-F(t,v,p)|\leq L_{\tau,z,\pi}|u-v| \end{align*} whenever $t\in A_{\tau,z,\pi}$, $u,v\in V_{\tau,z,\pi}$, and $p\in R_{\tau,z,\pi}$. A finite subcover of $S$ by these product neighbourhoods gives constants $L_1,\ldots,L_N$. Let $L_{\mathrm{loc}}:=\max\{L_1,\ldots,L_N\}$. Let $\rho>0$ be a Lebesgue number for this finite cover with respect to the product Euclidean metric on $S$. If $t\in [a,b]$, $p\in \overline{Q_0}$, and $u,v\in K$ satisfy $|u-v|<\rho$, then $(t,u,p)$ and $(t,v,p)$ lie in one member of the finite cover, and hence \begin{align*} |F(t,u,p)-F(t,v,p)|\leq L_{\mathrm{loc}}|u-v|. \end{align*} Since $F$ is continuous on the compact set $S$, define \begin{align*} M:=\sup\{|F(t,z,p)|:(t,z,p)\in S\}<\infty. \end{align*} For pairs with $|u-v|\geq \rho$, the triangle inequality gives \begin{align*} |F(t,u,p)-F(t,v,p)|\leq 2M\leq \frac{2M}{\rho}|u-v|. \end{align*} Therefore, with \begin{align*} L:=\max\{L_{\mathrm{loc}},2M/\rho\}, \end{align*} one has \begin{align*} |F(t,u,p)-F(t,v,p)|\leq L|u-v| \end{align*} for all $t\in [a,b]$, all $u,v\in K$, and all $p\in \overline{Q_0}$. Define the parameter modulus \begin{align*} \omega:Q_0\to [0,\infty) \end{align*} by \begin{align*} \omega(p):=\sup\{|F(t,z,p)-F(t,z,p_0)|:t\in [a,b],\, z\in K\}. \end{align*} The set $S$ is compact, and $F$ is continuous on $S$. Hence $F$ is uniformly continuous there, so \begin{align*} \lim_{p\to p_0}\omega(p)=0. \end{align*}[/step]

[guided]The local Lipschitz hypothesis is not yet enough for the later Gronwall estimate, because Gronwall needs one constant that works for every state pair in the whole tube $K$. We first choose an open neighbourhood $Q_0\subset P$ of $p_0$ whose closure is compact and contained in $P$, and define \begin{align*} S:=[a,b]\times K\times \overline{Q_0}. \end{align*} This set is compact because $[a,b]$, $K$, and $\overline{Q_0}$ are compact. For each $(\tau,z,\pi)\in S$, local uniform Lipschitz continuity gives neighbourhoods $A_{\tau,z,\pi}\subset I$, $V_{\tau,z,\pi}\subset U$, and $R_{\tau,z,\pi}\subset P$, and a constant $L_{\tau,z,\pi}\geq 0$, such that \begin{align*} |F(t,u,p)-F(t,v,p)|\leq L_{\tau,z,\pi}|u-v| \end{align*} whenever $t\in A_{\tau,z,\pi}$, $u,v\in V_{\tau,z,\pi}$, and $p\in R_{\tau,z,\pi}$. Compactness of $S$ gives finitely many such product neighbourhoods covering $S$. Denote their Lipschitz constants by $L_1,\ldots,L_N$, and set \begin{align*} L_{\mathrm{loc}}:=\max\{L_1,\ldots,L_N\}. \end{align*} There is one subtle point: a finite cover only controls pairs $u,v$ lying in the same state neighbourhood. To handle this, let $\rho>0$ be a Lebesgue number for the finite cover in the product Euclidean metric on $S$. If $t\in [a,b]$, $p\in \overline{Q_0}$, $u,v\in K$, and $|u-v|<\rho$, then the two points $(t,u,p)$ and $(t,v,p)$ lie in one member of the finite cover. Hence \begin{align*} |F(t,u,p)-F(t,v,p)|\leq L_{\mathrm{loc}}|u-v|. \end{align*} For pairs with $|u-v|\geq \rho$, we use boundedness instead of local Lipschitz continuity. Since $F$ is continuous on the compact set $S$, the number \begin{align*} M:=\sup\{|F(t,z,p)|:(t,z,p)\in S\} \end{align*} is finite. The triangle inequality gives \begin{align*} |F(t,u,p)-F(t,v,p)|\leq |F(t,u,p)|+|F(t,v,p)|\leq 2M\leq \frac{2M}{\rho}|u-v|. \end{align*} Thus the constant \begin{align*} L:=\max\{L_{\mathrm{loc}},2M/\rho\} \end{align*} works for every $t\in [a,b]$, every $u,v\in K$, and every $p\in \overline{Q_0}$. Finally, define \begin{align*} \omega:Q_0\to [0,\infty) \end{align*} by \begin{align*} \omega(p):=\sup\{|F(t,z,p)-F(t,z,p_0)|:t\in [a,b],\, z\in K\}. \end{align*} Uniform continuity of $F$ on the compact set $S$ implies that this supremum tends to $0$ as $p\to p_0$. This is the parameter error that will appear in the Gronwall estimate.[/guided]

[step:Derive the two-sided integral estimate while the nearby solution remains in the tube]Fix $y_0\in U$ and $p\in Q_0$. By the existence and uniqueness theorem for maximal solutions, applied to the vector field \begin{align*} F_p:I\times U\to \mathbb{R}^n,\qquad (t,z)\mapsto F(t,z,p), \end{align*} there is a unique maximal solution \begin{align*} y:J\to U \end{align*} with $t_0\in J$, $y(t_0)=y_0$, and $\dot y(t)=F(t,y(t),p)$ on its interval of definition. Let $J_K\subset J\cap [a,b]$ be any interval containing $t_0$ such that $y(t)\in K$ for all $t\in J_K$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. By the [integral form of an initial value problem](/theorems/8311) [citetheorem:8311], for $t\in J_K$ with $t\geq t_0$, \begin{align*} y(t)-x(t)=y_0-x_0+\int_{t_0}^{t}\bigl(F(s,y(s),p)-F(s,x(s),p_0)\bigr)\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms, applying the triangle inequality, adding and subtracting $F(s,x(s),p)$, using the Lipschitz bound on $K$, and using the definition of $\omega(p)$ give \begin{align*} |y(t)-x(t)|\leq |y_0-x_0|+\int_{t_0}^{t}\bigl(L|y(s)-x(s)|+\omega(p)\bigr)\,d\mathcal{L}^1(s). \end{align*} Thus \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+(t-t_0)\omega(p)\bigr)e^{L(t-t_0)} \end{align*} for $t\in J_K\cap [t_0,b]$. Similarly, for $t\in J_K$ with $t\leq t_0$, the integral equation gives \begin{align*} y(t)-x(t)=y_0-x_0-\int_{t}^{t_0}\bigl(F(s,y(s),p)-F(s,x(s),p_0)\bigr)\,d\mathcal{L}^1(s), \end{align*} Taking Euclidean norms, applying the triangle inequality, splitting the integrand into the state error and parameter error, using the Lipschitz bound on $K$, using the definition of $\omega(p)$, and applying the elementary integral Gronwall estimate on $[t,t_0]$ yield \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+(t_0-t)\omega(p)\bigr)e^{L(t_0-t)}. \end{align*} Consequently, with \begin{align*} T:=\max\{b-t_0,t_0-a\}, \end{align*} every such $t\in J_K$ satisfies \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+T\omega(p)\bigr)e^{LT}. \end{align*}[/step]

[guided]We now compare the nearby solution $y$ with the reference solution $x$ only on time intervals where $y$ is already known to stay in $K$. This avoids a circular argument: the estimate is first conditional, and the next step will use it to prove that $y$ cannot leave the tube. Fix $y_0\in U$ and $p\in Q_0$. For this fixed parameter, define \begin{align*} F_p:I\times U\to \mathbb{R}^n,\qquad (t,z)\mapsto F(t,z,p). \end{align*} The map $F_p$ is continuous and locally Lipschitz in the state variable because these properties are inherited from $F$. Hence the existence and uniqueness theorem for maximal ODE solutions gives a unique maximal solution \begin{align*} y:J\to U \end{align*} with $t_0\in J$ and $y(t_0)=y_0$. Let $J_K\subset J\cap [a,b]$ be an interval containing $t_0$ such that $y(t)\in K$ for every $t\in J_K$. For $t\in J_K$ with $t\geq t_0$, the integral formulation of the initial value problem gives \begin{align*} y(t)=y_0+\int_{t_0}^{t}F(s,y(s),p)\,d\mathcal{L}^1(s) \end{align*} and \begin{align*} x(t)=x_0+\int_{t_0}^{t}F(s,x(s),p_0)\,d\mathcal{L}^1(s). \end{align*} Subtracting these identities gives \begin{align*} y(t)-x(t)=y_0-x_0+\int_{t_0}^{t}\bigl(F(s,y(s),p)-F(s,x(s),p_0)\bigr)\,d\mathcal{L}^1(s). \end{align*} We split the integrand into a state error and a parameter error: \begin{align*} F(s,y(s),p)-F(s,x(s),p_0)=F(s,y(s),p)-F(s,x(s),p)+F(s,x(s),p)-F(s,x(s),p_0). \end{align*} Because $y(s)\in K$, $x(s)\in C\subset K$, and $p\in Q_0$, the common Lipschitz estimate gives \begin{align*} |F(s,y(s),p)-F(s,x(s),p)|\leq L|y(s)-x(s)|. \end{align*} The definition of $\omega(p)$ gives \begin{align*} |F(s,x(s),p)-F(s,x(s),p_0)|\leq \omega(p). \end{align*} Therefore \begin{align*} |y(t)-x(t)|\leq |y_0-x_0|+\int_{t_0}^{t}\bigl(L|y(s)-x(s)|+\omega(p)\bigr)\,d\mathcal{L}^1(s). \end{align*} The elementary integral Gronwall estimate applied to the nonnegative function \begin{align*} g:[t_0,t]\to [0,\infty),\qquad s\mapsto |y(s)-x(s)| \end{align*} gives \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+(t-t_0)\omega(p)\bigr)e^{L(t-t_0)}. \end{align*} The same comparison must also be made backward in time, because $t_0$ need not equal $a$. If $t\leq t_0$, then the integral identities on $[t,t_0]$ give \begin{align*} y(t)-x(t)=y_0-x_0-\int_{t}^{t_0}\bigl(F(s,y(s),p)-F(s,x(s),p_0)\bigr)\,d\mathcal{L}^1(s). \end{align*} On $[t,t_0]$, taking Euclidean norms, applying the triangle inequality, splitting the integrand into the state error and parameter error, using the common Lipschitz estimate, using the definition of $\omega(p)$, and applying the elementary integral Gronwall estimate yield \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+(t_0-t)\omega(p)\bigr)e^{L(t_0-t)}. \end{align*} If \begin{align*} T:=\max\{b-t_0,t_0-a\}, \end{align*} then both the forward and backward estimates are summarized by \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+T\omega(p)\bigr)e^{LT} \end{align*} for every $t\in J_K$.[/guided]

[step:Choose neighbourhoods forcing the nearby solutions to stay in the compact tube]Choose neighbourhoods $N\subset U$ of $x_0$ and $Q\subset Q_0$ of $p_0$ such that \begin{align*} \bigl(|y_0-x_0|+T\omega(p)\bigr)e^{LT}<r/2 \end{align*} for every $(y_0,p)\in N\times Q$. This is possible because $\omega(p)\to 0$ as $p\to p_0$. Fix $(y_0,p)\in N\times Q$, and let \begin{align*} y:J\to U \end{align*} be the corresponding maximal solution. Define \begin{align*} J_*:=\{t\in J\cap [a,b]: y(s)\in K \text{ for every } s \text{ between } t_0 \text{ and } t\}. \end{align*} The set $J_*$ contains $t_0$ because $y(t_0)=y_0$ and, after shrinking $N$ if necessary, $N\subset K_{r/2}\subset K$. On $J_*$ the estimate from the previous step applies, so \begin{align*} |y(t)-x(t)|<r/2 \end{align*} for all $t\in J_*$. Since $x(t)\in C$, this implies \begin{align*} \operatorname{dist}(y(t),C)<r/2 \end{align*} for all $t\in J_*$. Hence $y(t)\in K_{r/2}$ for all $t\in J_*$. We now make the no-exit argument explicit. Let \begin{align*} \beta:=\sup\{t\in J\cap [t_0,b]: y(s)\in K \text{ for every } s\in [t_0,t]\}. \end{align*} The preceding estimate shows that $y(t)\in K_{r/2}$ for every $t\in J\cap [t_0,\beta)$ for which the interval $[t_0,t]$ lies in $J$. Since $K_{r/2}$ has positive distance from $\mathbb{R}^n\setminus K$ and $y$ is continuous, $y$ cannot first leave $K$ at time $\beta$ while $\beta\in J$ and $\beta<b$. Hence the obstruction to reaching $b$ cannot be exit from $K$. The fixed-parameter field $F_p:I\times U\to\mathbb{R}^n$ is continuous and locally Lipschitz in the state variable, so the [continuation criterion for ordinary differential equations](/theorems/8314) [citetheorem:8314] applies to the maximal solution $y:J\to U$. Let $\gamma:=\sup(J\cap [t_0,b])$. If $\gamma<b$, then $\gamma\in [a,b]\subset I$ is not the right endpoint of the ambient time interval $I$. The no-exit estimate above gives $y(t)\in K_{r/2}$ for all $t\in J\cap [t_0,\gamma)$, so the image of $y$ approaching $\gamma$ remains in the compact set $K_{r/2}\subset U$. This contradicts the continuation criterion. Applying the analogous left-endpoint argument to \begin{align*} \alpha:=\inf\{t\in J\cap [a,t_0]: y(s)\in K \text{ for every } s\in [t,t_0]\} \end{align*} excludes failure before $a$. Therefore $[a,b]\subset J$, so $y:[a,b]\to U$ exists on all of $[a,b]$.[/step]

[guided]The estimate from the previous step was conditional: it was valid only while $y$ stayed in $K$. We now turn that conditional estimate into an actual no-exit statement. Define the right-hand reachable endpoint \begin{align*} \beta:=\sup\{t\in J\cap [t_0,b]: y(s)\in K \text{ for every } s\in [t_0,t]\}. \end{align*} On every interval $[t_0,t]$ with $t<\beta$, the two-sided estimate applies and gives \begin{align*} |y(t)-x(t)|<r/2. \end{align*} Because $x(t)\in C$, this implies $y(t)\in K_{r/2}$. The set $K_{r/2}$ lies strictly inside $K$: its distance from $\mathbb{R}^n\setminus K$ is at least $r/2$. Therefore continuity of $y$ prevents a first exit from $K$ while $y$ is still defined and while the endpoint is smaller than $b$. It remains to rule out the other possible obstruction, namely that the maximal solution stops existing before $b$ even though it has not exited the compact tube. For the fixed parameter $p$, the vector field \begin{align*} F_p:I\times U\to \mathbb{R}^n,\qquad (t,z)\mapsto F(t,z,p) \end{align*} is continuous and locally Lipschitz in the state variable. Hence the continuation criterion for ordinary differential equations [citetheorem:8314] applies. Let $\gamma:=\sup(J\cap [t_0,b])$. If $\gamma<b$, then $\gamma\in [a,b]\subset I$, so $\gamma$ is not the right endpoint of the ambient time interval $I$. As $\gamma$ is approached from within $J$, the no-exit estimate keeps the image of $y$ inside the compact set $K_{r/2}\subset U$. This contradicts the continuation criterion. The backward-time argument is analogous but uses the left endpoint. Define \begin{align*} \alpha:=\inf\{t\in J\cap [a,t_0]: y(s)\in K \text{ for every } s\in [t,t_0]\}. \end{align*} The backward estimate keeps $y$ in $K_{r/2}$ before any possible exit from $K$, and the continuation criterion excludes a left endpoint of $J\cap [a,b]$ larger than $a$. Thus $[a,b]\subset J$, and the solution exists on the whole interval.[/guided]

[step:Pass the estimate to the whole interval and prove uniqueness and uniform convergence]For every $(y_0,p)\in N\times Q$, the preceding step gives a maximal solution whose interval contains $[a,b]$, hence a solution \begin{align*} y:[a,b]\to U \end{align*} defined on the whole interval. If $\tilde y:[a,b]\to U$ is any other solution of the same initial value problem, then the uniqueness part of the maximal-solution theorem applied to $F_p$ implies that $\tilde y$ agrees with the maximal solution on $[a,b]$. Thus the solution on $[a,b]$ is unique. The estimate from the two-sided comparison applies to every $t\in [a,b]$. Hence \begin{align*} \sup_{t\in [a,b]}|y(t)-x(t)|\leq \bigl(|y_0-x_0|+T\omega(p)\bigr)e^{LT}. \end{align*} Since $|y_0-x_0|\to 0$ and $\omega(p)\to 0$ as $(y_0,p)\to (x_0,p_0)$, we obtain \begin{align*} \lim_{(y_0,p)\to (x_0,p_0)}\sup_{t\in [a,b]}|y(t)-x(t)|=0. \end{align*} This is precisely [uniform convergence](/page/Uniform%20Convergence) of the nearby solutions to the reference solution on $[a,b]$.[/step]

[guided]Fix $(y_0,p)\in N\times Q$. The previous step proved that the maximal solution for the fixed vector field \begin{align*} F_p:I\times U\to\mathbb{R}^n, \qquad (t,z)\mapsto F(t,z,p), \end{align*} has an interval of definition containing $[a,b]$. Therefore it restricts to a solution \begin{align*} y:[a,b]\to U \end{align*} of the initial value problem with $y(t_0)=y_0$. We also need uniqueness on the finite interval $[a,b]$, not just existence of one maximal solution. Let \begin{align*} \tilde y:[a,b]\to U \end{align*} be another solution with the same initial condition. The uniqueness part of the maximal-solution theorem applied to $F_p$ says that any solution with initial data $(t_0,y_0)$ agrees with the unique maximal solution on the common part of their domains. Since $[a,b]$ is contained in the maximal interval, $\tilde y(t)=y(t)$ for every $t\in [a,b]$. Hence the solution on $[a,b]$ is unique. It remains to prove the stated continuity of the solution with respect to $(y_0,p)$. Since $y$ exists on all of $[a,b]$ and the no-exit argument kept it inside the tube, the two-sided estimate applies for every $t\in [a,b]$: \begin{align*} |y(t)-x(t)|\leq \bigl(|y_0-x_0|+T\omega(p)\bigr)e^{LT}. \end{align*} Taking the supremum over $t\in [a,b]$ gives \begin{align*} \sup_{t\in [a,b]}|y(t)-x(t)|\leq \bigl(|y_0-x_0|+T\omega(p)\bigr)e^{LT}. \end{align*} As $(y_0,p)\to (x_0,p_0)$, the first term $|y_0-x_0|$ tends to $0$, and the parameter modulus satisfies $\omega(p)\to 0$. Therefore \begin{align*} \lim_{(y_0,p)\to (x_0,p_0)}\sup_{t\in [a,b]}|y(t)-x(t)|=0. \end{align*} This is exactly uniform convergence on $[a,b]$, and the theorem is proved.[/guided]