[proofplan] We convert the initial value problem into an equivalent integral equation via Picard's formulation, then construct approximate solutions using the Euler polygonal method (Cauchy polygons). The Arzelà–Ascoli theorem extracts a uniformly convergent subsequence, and continuity of $f$ ensures the limit satisfies the integral equation. The key ingredients are: (1) the uniform bound $|f| < M$ on $\mathcal{D}$ gives equicontinuity, (2) the Arzelà–Ascoli theorem provides a convergent subsequence, and (3) passing the limit through the integral uses only continuity of $f$, not Lipschitz continuity — which is why uniqueness is not obtained. [/proofplan] [step:Reformulate the IVP as an integral equation] A function $x: I \to \mathbb{R}^n$ defined on an interval $I$ containing $t_0$ is a solution of $\dot{x} = f(x, t)$, $x(t_0) = x_0$ if and only if $x$ is continuous and satisfies the integral equation \begin{align*} x(t) = x_0 + \int_{t_0}^{t} f(x(s), s) \, d\mathcal{L}^1(s) \end{align*} for all $t \in I$. The forward direction follows from the Fundamental Theorem of Calculus applied to the continuously differentiable function $x$; the reverse direction follows because the right-hand side is differentiable (since $s \mapsto f(x(s), s)$ is continuous) with derivative $f(x(t), t)$. [guided] Why reformulate the problem as an integral equation? Because the integral formulation works with continuous functions rather than differentiable ones, and the space $C(I, \mathbb{R}^n)$ of continuous functions has good compactness properties (via Arzelà–Ascoli) that the space of differentiable functions does not. A function $x: I \to \mathbb{R}^n$ defined on an interval $I$ containing $t_0$ satisfies $\dot{x} = f(x, t)$, $x(t_0) = x_0$ if and only if it satisfies \begin{align*} x(t) = x_0 + \int_{t_0}^{t} f(x(s), s) \, d\mathcal{L}^1(s) \end{align*} for all $t \in I$. For the forward direction: if $x$ is differentiable with $\dot{x}(t) = f(x(t), t)$, integrating both sides from $t_0$ to $t$ and using $x(t_0) = x_0$ yields the integral equation. For the reverse: if $x$ satisfies the integral equation, then the right-hand side is differentiable in $t$ (since $f(x(\cdot), \cdot)$ is continuous, being a composition of continuous functions), and differentiating gives $\dot{x}(t) = f(x(t), t)$. Setting $t = t_0$ gives $x(t_0) = x_0$. [/guided] [/step] [step:Define the time interval and construct Euler polygon approximations] Set $T := \min(\alpha, \beta / M)$ and let $I_T := [t_0 - T, t_0 + T]$. For each integer $k \geq 1$, define the $k$-th Euler polygon approximation as follows. Partition $[t_0, t_0 + T]$ into $k$ equal subintervals of width $h_k := T/k$, with grid points $t_j^{(k)} := t_0 + jh_k$ for $j = 0, 1, \ldots, k$. Define \begin{align*} x_k: [t_0, t_0 + T] &\to \mathbb{R}^n \end{align*} by $x_k(t_0) := x_0$ and, for $t \in (t_j^{(k)}, t_{j+1}^{(k)}]$, \begin{align*} x_k(t) := x_k(t_j^{(k)}) + f(x_k(t_j^{(k)}), t_j^{(k)}) \cdot (t - t_j^{(k)}). \end{align*} That is, $x_k$ is piecewise linear, following the slope $f(x_k(t_j^{(k)}), t_j^{(k)})$ on each subinterval. The construction for $[t_0 - T, t_0]$ is analogous, working backwards from $t_0$. [guided] The Euler polygon method is the simplest numerical approximation to an ODE: on each small subinterval, we freeze the vector field at the left endpoint and follow a straight line with that slope. The resulting function $x_k$ is piecewise linear and therefore continuous. The idea is that as $k \to \infty$ (and the step size $h_k \to 0$), these approximations should converge to a true solution — provided we can extract a convergent subsequence. For each $k \geq 1$, set $h_k := T/k$ and define grid points $t_j^{(k)} := t_0 + jh_k$ for $j = 0, 1, \ldots, k$. The Euler polygon $x_k: [t_0, t_0 + T] \to \mathbb{R}^n$ is defined inductively by $x_k(t_0) := x_0$ and \begin{align*} x_k(t) := x_k(t_j^{(k)}) + f(x_k(t_j^{(k)}), t_j^{(k)}) \cdot (t - t_j^{(k)}) \end{align*} for $t \in (t_j^{(k)}, t_{j+1}^{(k)}]$. This is well-defined provided each evaluation point $(x_k(t_j^{(k)}), t_j^{(k)})$ lies in $\mathcal{D}$, which we verify in the next step. [/guided] [/step] [step:Verify that each Euler polygon stays inside $\mathcal{D}$] We show by induction on $j$ that $|x_k(t) - x_0| < \beta$ for all $t \in I_T$, so that $(x_k(t), t) \in \mathcal{D}$ and the construction is valid. At $t = t_0$, $x_k(t_0) = x_0$. On the subinterval $(t_j^{(k)}, t_{j+1}^{(k)}]$, assuming $(x_k(t_j^{(k)}), t_j^{(k)}) \in \mathcal{D}$, we have $|f(x_k(t_j^{(k)}), t_j^{(k)})| < M$. Therefore \begin{align*} |x_k(t) - x_0| &\leq \sum_{i=0}^{j} |f(x_k(t_i^{(k)}), t_i^{(k)})| \cdot h_k + |f(x_k(t_j^{(k)}), t_j^{(k)})| \cdot (t - t_j^{(k)}) \\ &< M \cdot (j \cdot h_k + (t - t_j^{(k)})) \\ &= M \cdot (t - t_0) \\ &\leq M \cdot T \leq M \cdot \frac{\beta}{M} = \beta. \end{align*} Hence each polygon remains in $\mathcal{D}$. [/step] [step:Establish uniform boundedness and equicontinuity of the sequence $(x_k)$] From the previous step, $|x_k(t) - x_0| < \beta$ for all $t \in I_T$ and all $k$, so the family $\{x_k\}_{k \geq 1}$ is uniformly bounded in $C(I_T, \mathbb{R}^n)$. For equicontinuity: let $t, t' \in I_T$ with $t < t'$. Since $x_k$ is piecewise linear with slopes bounded by $M$, \begin{align*} |x_k(t') - x_k(t)| \leq M |t' - t| \end{align*} for all $k \geq 1$. This Lipschitz bound is uniform in $k$, so the family is equicontinuous. [/step] [step:Extract a uniformly convergent subsequence via Arzelà–Ascoli] The family $\{x_k\}_{k \geq 1} \subset C(I_T, \mathbb{R}^n)$ is uniformly bounded and equicontinuous. The interval $I_T$ is compact, and $\mathbb{R}^n$ is complete. By the [Arzelà–Ascoli Theorem](/theorems/???), there exists a subsequence $(x_{k_m})_{m \geq 1}$ and a continuous function $x: I_T \to \mathbb{R}^n$ such that $x_{k_m} \to x$ uniformly on $I_T$. Since $|x_k(t) - x_0| < \beta$ for all $k$ and $t$, the limit satisfies $|x(t) - x_0| \leq \beta$, so $(x(t), t) \in \overline{\mathcal{D}}$ for all $t \in I_T$. For $|t - t_0| < T$ (strict inequality), $(x(t), t) \in \mathcal{D}$. [guided] This is the step where continuity of $f$ (rather than Lipschitz continuity) suffices. The [Arzelà–Ascoli Theorem](/theorems/???) requires two conditions on a family of functions in $C(K, \mathbb{R}^n)$ (where $K$ is compact): **uniform boundedness** and **equicontinuity**. We verify both. **Uniform boundedness.** From the previous step, $|x_k(t) - x_0| < \beta$ for all $k \geq 1$ and all $t \in I_T$. Hence every $x_k$ takes values in the closed ball $\overline{B}(x_0, \beta)$, which gives $\sup_{k, t} |x_k(t)| \leq |x_0| + \beta < \infty$. **Equicontinuity.** Each Euler polygon $x_k$ is piecewise linear with slopes bounded by $M$ (the bound on $|f|$ in $\mathcal{D}$). Therefore $|x_k(t') - x_k(t)| \leq M|t' - t|$ for all $t, t' \in I_T$ and all $k$. This Lipschitz bound is uniform in $k$, so the family $\{x_k\}$ is equicontinuous. Since $I_T$ is a compact interval and $\mathbb{R}^n$ is complete, the [Arzelà–Ascoli Theorem](/theorems/???) guarantees the existence of a subsequence $(x_{k_m})_{m \geq 1}$ and a continuous function $x: I_T \to \mathbb{R}^n$ such that $x_{k_m} \to x$ uniformly on $I_T$. The uniform limit inherits the bound: since $|x_{k_m}(t) - x_0| < \beta$ for all $m$ and $t$, taking $m \to \infty$ gives $|x(t) - x_0| \leq \beta$. For $|t - t_0| < T$ (strict inequality), $(x(t), t)$ lies in the interior of $\mathcal{D}$. Why does uniqueness fail here? Because the Arzelà–Ascoli theorem guarantees only the existence of *a* convergent subsequence, not that *every* subsequence converges to the same limit. Different subsequences of Euler polygons may converge to different solutions of the integral equation. Ruling out this possibility requires the Lipschitz condition of the Picard–Lindelöf theorem, which makes the Picard iteration operator a contraction with a unique fixed point. [/guided] [/step] [step:Show the limit $x$ satisfies the integral equation] Each Euler polygon satisfies \begin{align*} x_k(t) = x_0 + \int_{t_0}^{t} f(x_k(\tau_k(s)), \tau_k(s)) \, d\mathcal{L}^1(s), \end{align*} where $\tau_k(s) := t_j^{(k)}$ for $s \in (t_j^{(k)}, t_{j+1}^{(k)}]$ is the left endpoint of the subinterval containing $s$. This holds because on each subinterval the slope of $x_k$ is exactly $f(x_k(t_j^{(k)}), t_j^{(k)})$. We must show that as $m \to \infty$, \begin{align*} \int_{t_0}^{t} f(x_{k_m}(\tau_{k_m}(s)), \tau_{k_m}(s)) \, d\mathcal{L}^1(s) \to \int_{t_0}^{t} f(x(s), s) \, d\mathcal{L}^1(s). \end{align*} Since $f$ is continuous on the compact set $\overline{\mathcal{D}}$, it is uniformly continuous. We have $|\tau_{k_m}(s) - s| \leq h_{k_m} \to 0$ and $|x_{k_m}(\tau_{k_m}(s)) - x(s)| \leq |x_{k_m}(\tau_{k_m}(s)) - x_{k_m}(s)| + |x_{k_m}(s) - x(s)| \leq M h_{k_m} + \|x_{k_m} - x\|_\infty \to 0$ uniformly in $s$. By uniform continuity of $f$, the integrand $f(x_{k_m}(\tau_{k_m}(s)), \tau_{k_m}(s)) \to f(x(s), s)$ uniformly in $s \in I_T$. Passing the limit through the integral (justified since the convergence is uniform and the domain of integration is bounded), \begin{align*} x(t) = \lim_{m \to \infty} x_{k_m}(t) = x_0 + \int_{t_0}^{t} f(x(s), s) \, d\mathcal{L}^1(s). \end{align*} [guided] The key technical point is passing the limit through the integral. We need $f(x_{k_m}(\tau_{k_m}(s)), \tau_{k_m}(s)) \to f(x(s), s)$ uniformly in $s$, not just pointwise. Here is why this holds: 1. $\tau_{k_m}(s) \to s$ uniformly because $|\tau_{k_m}(s) - s| \leq h_{k_m} = T/k_m \to 0$. 2. $x_{k_m}(\tau_{k_m}(s)) \to x(s)$ uniformly because \begin{align*} |x_{k_m}(\tau_{k_m}(s)) - x(s)| &\leq |x_{k_m}(\tau_{k_m}(s)) - x_{k_m}(s)| + |x_{k_m}(s) - x(s)| \\ &\leq M \cdot |\tau_{k_m}(s) - s| + \|x_{k_m} - x\|_\infty \\ &\leq M h_{k_m} + \|x_{k_m} - x\|_\infty \to 0. \end{align*} 3. Since $f$ is continuous on the compact set $\overline{\mathcal{D}}$, it is uniformly continuous. The uniform convergence of the arguments therefore implies uniform convergence of $f$ applied to them. With uniform convergence of the integrand on the bounded interval $[t_0, t]$, we may interchange limit and integral: \begin{align*} x(t) = x_0 + \lim_{m \to \infty} \int_{t_0}^{t} f(x_{k_m}(\tau_{k_m}(s)), \tau_{k_m}(s)) \, d\mathcal{L}^1(s) = x_0 + \int_{t_0}^{t} f(x(s), s) \, d\mathcal{L}^1(s). \end{align*} Why does uniqueness fail? Because the Arzelà–Ascoli theorem guarantees only the existence of *a* convergent subsequence, not that *every* subsequence converges to the same limit. Different subsequences of Euler polygons may converge to different solutions. Ruling out this possibility requires the Lipschitz condition of the Picard–Lindelöf theorem, which makes the integral operator a contraction with a unique fixed point. [/guided] [/step] [step:Conclude existence on the interval $|t - t_0| < T$] By the equivalence established in the first step, the continuous function $x: I_T \to \mathbb{R}^n$ satisfying the integral equation is a solution of the initial value problem $\dot{x} = f(x, t)$, $x(t_0) = x_0$ on the open interval $|t - t_0| < T = \min(\alpha, \beta/M)$. This completes the proof. [/step]