[proofplan] We first use the Picard-Lindelof local existence and uniqueness theorem to obtain at least one local solution near the initial time. We then form the union of all relatively open solution intervals carrying a solution with the prescribed initial value. Local uniqueness forces any two such solutions to agree on overlaps, so the pointwise union defines a single solution on the union interval. Finally, the definition of this union gives maximality, and maximality gives uniqueness of the maximal solution. [/proofplan] [step:Use local Picard-Lindelof theory to start the construction] We use the Picard-Lindelof local existence and uniqueness theorem in its relative-interval form for continuous vector fields that are locally Lipschitz in the state variable, locally uniformly in time. Its hypotheses are exactly the continuity of $f: I \times E \to \mathbb{R}^n$ and the locally uniform state-Lipschitz condition, as now explicitly stated, applied at the point $(t_0,x_0) \in I \times E$; its conclusion gives a solution on a relatively open subinterval of $I$, with derivatives interpreted in the relative-domain sense described in the statement. Therefore there exists a relatively open interval $U_0 \subset I$ containing $t_0$ and a solution \begin{align*} u_0: U_0 \to E \end{align*} such that $u_0(t_0)=x_0$ and $\dot{u}_0(t)=f(t,u_0(t))$ at every point where the derivative is interpreted in the relative-domain sense. Let $\mathcal{S}$ denote the collection of all pairs $(U,u)$ such that $U \subset I$ is a relatively open interval containing $t_0$ and \begin{align*} u: U \to E \end{align*} is a solution of the [initial value problem](/page/Initial%20Value%20Problem) with $u(t_0)=x_0$. The pair $(U_0,u_0)$ belongs to $\mathcal{S}$, so $\mathcal{S}$ is nonempty. [/step] [step:Prove that any two local solutions in the family agree on their common domain] Let $(U,u)$ and $(V,v)$ be elements of $\mathcal{S}$. Define \begin{align*} W:=\{t \in U \cap V : u(t)=v(t)\}. \end{align*} The set $U \cap V$ is an interval in the relative topology of $I$ and contains $t_0$. Since $u(t_0)=x_0=v(t_0)$, the set $W$ is nonempty. We claim that $W=U \cap V$. Since $u$ and $v$ are continuous maps into $\mathbb{R}^n$, the map \begin{align*} h: U \cap V \to \mathbb{R}^n, \qquad h(t)=u(t)-v(t) \end{align*} is continuous, and hence $W=h^{-1}(\{0\})$ is closed in $U \cap V$. It remains to show that $W$ is open in $U \cap V$. Let $\tau \in W$. Then $u(\tau)=v(\tau)$. Applying the local uniqueness part of the Picard-Lindelof theorem at the initial point $(\tau,u(\tau))$ gives a relatively open interval $N \subset U \cap V$ containing $\tau$ on which the two solutions agree. Hence $N \subset W$, so $W$ is open in $U \cap V$. Because $U \cap V$ is an interval and therefore connected, the only nonempty subset of $U \cap V$ that is both open and closed in the relative topology is all of $U \cap V$. Thus $W=U \cap V$, and $u(t)=v(t)$ for every $t \in U \cap V$. [guided] The point of this step is to upgrade local uniqueness to uniqueness on the whole overlap. Local Picard-Lindelof uniqueness only says that two solutions with the same value at one time agree near that time. We must prove that this local agreement cannot stop somewhere inside the common interval. Take two solution pairs $(U,u)$ and $(V,v)$ from $\mathcal{S}$. Thus $U$ and $V$ are relatively open intervals in $I$, both contain $t_0$, and \begin{align*} u: U \to E \end{align*} and \begin{align*} v: V \to E \end{align*} solve the same differential equation with $u(t_0)=x_0=v(t_0)$. Define the agreement set \begin{align*} W:=\{t \in U \cap V : u(t)=v(t)\}. \end{align*} This set is nonempty because $t_0 \in W$. First, $W$ is closed in $U \cap V$. Indeed, define \begin{align*} h: U \cap V \to \mathbb{R}^n, \qquad h(t)=u(t)-v(t). \end{align*} Both $u$ and $v$ are continuous because they are solutions of an ODE with continuous right-hand side, so $h$ is continuous. Since $\{0\}$ is closed in $\mathbb{R}^n$, the preimage $h^{-1}(\{0\})$ is closed in $U \cap V$. But $h^{-1}(\{0\})=W$. Second, $W$ is open in $U \cap V$. Let $\tau \in W$. Then $u(\tau)=v(\tau)$. At the point $(\tau,u(\tau))$, the hypotheses of the Picard-Lindelof uniqueness theorem are still valid: $f$ is continuous on $I \times E$, and the explicitly stated local uniform state-Lipschitz condition applies in a relatively open time neighbourhood of $\tau$ and an open state neighbourhood of $u(\tau)$. Therefore two solutions of the same equation that take the same value at time $\tau$ must agree on some relatively open interval $N \subset U \cap V$ containing $\tau$. Hence every point of $W$ has a neighbourhood inside $U \cap V$ contained in $W$, so $W$ is open. Now $U \cap V$ is an interval, hence connected. The set $W$ is nonempty, open, and closed in $U \cap V$. Connectedness forces $W=U \cap V$. Therefore the two solutions agree everywhere on their common domain. [/guided] [/step] [step:Define the maximal candidate by taking the union of all local solution intervals] Define \begin{align*} J:=\bigcup_{(U,u)\in \mathcal{S}} U. \end{align*} Since every $U$ appearing in the union is relatively open in $I$, the set $J$ is relatively open in $I$. Since every such $U$ is an interval containing $t_0$, their union is again an interval containing $t_0$: if $a,b \in J$ and $c \in I$ satisfies $a \le c \le b$, then $a \in U_a$ and $b \in U_b$ for some $(U_a,u_a),(U_b,u_b)\in\mathcal{S}$; both intervals contain $t_0$, so their union contains the interval between $a$ and $b$, and hence $c \in J$. Define \begin{align*} x: J \to E \end{align*} as follows. For $t \in J$, choose a pair $(U,u)\in\mathcal{S}$ with $t \in U$, and set $x(t):=u(t)$. This definition is independent of the chosen pair by the overlap agreement proved above. Thus $x$ is a well-defined map from $J$ to $E$. [/step] [step:Verify that the union map is a solution on $J$] Let $\tau \in J$. By definition of $J$, there exists $(U,u)\in\mathcal{S}$ such that $\tau \in U$. Since $U$ is relatively open in $I$, there is a relatively open neighbourhood $N \subset J$ of $\tau$ with $N \subset U$. On $N$, the definition of $x$ gives $x|_N=u|_N$. Because $u$ is a solution, the map $u$ has the required differentiability on $N$ and satisfies \begin{align*} \dot{u}(t)=f(t,u(t)) \end{align*} for all relevant $t \in N$. Since $x|_N=u|_N$, the map $x$ has the same differentiability at $\tau$ and satisfies \begin{align*} \dot{x}(\tau)=f(\tau,x(\tau)). \end{align*} Also $x(t_0)=x_0$, because every pair in $\mathcal{S}$ has initial value $x_0$. Therefore \begin{align*} x: J \to E \end{align*} is a solution of the initial value problem on the relatively open interval $J$. [/step] [step:Show that every relatively open solution is a restriction of the union solution] Let \begin{align*} y: K \to E \end{align*} be any solution of the same initial value problem, where $K \subset I$ is a relatively open interval containing $t_0$. Then $(K,y)\in\mathcal{S}$ by definition of $\mathcal{S}$. Hence $K \subset J$ by the definition of $J$ as the union of all domains in $\mathcal{S}$. For each $t \in K$, the definition of $x$ using the pair $(K,y)$ gives $x(t)=y(t)$. Equivalently, \begin{align*} y=x|_K. \end{align*} Thus every solution on a relatively open interval with the prescribed initial value is the restriction of $x$. [/step] [step:Deduce uniqueness of the maximal solution] Suppose \begin{align*} \tilde{x}: \tilde{J} \to E \end{align*} is another solution with the same maximal restriction property, where $\tilde{J} \subset I$ is a relatively open interval containing $t_0$. Applying the restriction property of $x$ to $\tilde{x}$ gives $\tilde{J}\subset J$ and $\tilde{x}=x|_{\tilde{J}}$. Applying the restriction property of $\tilde{x}$ to $x$ gives $J\subset \tilde{J}$ and $x=\tilde{x}|_J$. Hence $J=\tilde{J}$ and $x=\tilde{x}$ on this common interval. Therefore the solution constructed above is the unique maximal solution of the initial value problem. [/step]