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This initial value problem uses a square root and absolute value to define a continuous function on all of ${\mathbb{R}}^2$. For every finite waiting time $c\ge 0$, a forward solution of a differential equation on $t\ge 0$ isThe constant solution of a differential equation $y_{\infty }(t)=0$ is another choice. For example, at $t=3$, waits $c=0,1,2,3$ give values $9,4,1,0$. These are dimensionless mathematical trajectories; no measured physical waiting time is being supplied.

Verification uses the original ordinary differential equation, including the join. Before departure, both $y_c^{\prime }$ and $2\sqrt{|y_c|}$ vanish. For $t>c$, their common value is $2(t-c)$. For $c>0$, the difference quotient at the join isThus the derivative exists at $c$, equals the rate law there, and is continuous across the join. For $c=0$ the same computation gives the right derivative at the starting endpoint. For $c>0$, the left and right second derivatives at departure are $0$ and $2$: these forward solutions of a differential equation are continuously differentiable functions, without being twice differentiable at departure. The local two-sided statement of the Peano existence theorem is separate from this explicitly forward family.

The rate law tends to zero as $y\to 0$, but its secant slopes from zero are unbounded:No finite local Lipschitz constant exists at zero. At $\epsilon =10^{-2},10^{-4},10^{-6}$ the quotient is $20,200,2000$, even though the rates are shrinking to zero. Continuity controls the rate itself; Lipschitz continuity would control its change relative to the state displacement.

There is also a numerical lesson. The forward Euler method starting exactly at zero gives $Y_{k+1}=Y_k+2\Delta t\sqrt{|Y_k|}=0$ at every step. It reproduces the constant solution of a differential equation for every positive step size, while the moving trajectories remain equally valid. A computation producing one trajectory cannot establish uniqueness. Likewise, the separable differential equation calculation that divides by $\sqrt{|y|}$ only describes the moving pieces: checking the original equation recovers the zero pieces and their joins.

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

Literally, the printed hypothesis with positive $C$ is impossible for pairs at distance less than one, even when $U$ is constant. Use the corrected log-Lipschitz modulus $\mu$ from part (e). Local existence for the continuous finite-dimensional vector field follows from the Peano existence theorem. Boundedness gives $|X(t)-X(s)|\le \Vert U{\Vert }_{\infty }|t-s|$. Thus a trajectory cannot escape to infinity in finite time; at a finite endpoint it has a limit, and local existence there extends it. Solutions exist globally.

For uniqueness, let $X,Y$ have the same initial point and set $\delta (t)=|X(t)-Y(t)|$. This absolutely continuous function satisfies, almost everywhere,This is the Osgood uniqueness criterion, rather than the usual linear Gronwall inequality, because the velocity need not be Lipschitz continuous. To see the argument directly, put $w_{\epsilon }=\delta +\epsilon$. As long as $w_{\epsilon }\le 1$, monotonicity of $\mu$ givesFor $z_{\epsilon }=1-\log w_{\epsilon }$, this becomes $z_{\epsilon }^{\prime }\ge -C{z}_{\epsilon }$. Integration givesFor any fixed finite interval, this upper bound tends uniformly to zero as $\epsilon \downarrow 0$. Taking $\epsilon$ small prevents a first exit through $w_{\epsilon }=1$, so the bound remains valid throughout that interval. Thus $\delta =0$. Backward time has the same proof. Every initial point has exactly one global trajectory.

The decisive fact is ${\int }_0^{1}dr/\mu (r)=\infty$. For different nearby initial points, the same comparison givesuntil the separation reaches one. This quantitative continuous dependence makes the flow a homeomorphism, although it need not be a smooth diffeomorphism.

The preceding estimate makes $(x_k)$ equicontinuous, andmakes it uniformly bounded. The Arzelà-Ascoli theorem therefore supplies a subsequence converging uniformly to a continuous function $x:[0,1]\to {\mathbb{R}}^n$. On every mesh interval,including an endpoint by using the adjacent interval. Consequently $y_k\to x$ uniformly along the same subsequence.

All $x_k(t)$ and $y_k(t)$ lie in one compact ball. The restriction of the continuous vector field $F$ to that ball is uniformly continuous, so $F\circ y_k\to F\circ x$ uniformly. Passing to the limit in the defining equation givesThe fundamental theorem of calculus now makes $x$ differentiable and yields $x^{\prime }(t)=F(x(t))$. This proves the required special case of the Peano existence theorem; uniqueness need not hold because $F$ was assumed continuous but not locally Lipschitz continuous.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set and $f:U\to {\mathbb{R}}^d$ a continuous function. Suppose $f$ is locally Lipschitz in the state $y$, with a Lipschitz constant uniform on a sufficiently small time-state neighborhood of each point. Then each $(t_0,y_0)\in U$ determines a unique local solution of a differential equation to the displayed initial value problem, on an interval around $t_0$. This is the local theorem in Ordinary Differential Equations and Dynamical Systems by Gerald Teschl.

For example, $y^{\prime }=y$, $y(0)=1$, has the solution of a differential equation $y(t)=e^t$: its derivative is itself and its starting value is $1$. The Lipschitz constant is $L=1$, since $|u-v|=|f(t,u)-f(t,v)|$. Thus uniqueness follows even before computing $y(1)=e\approx 2.71828183$. Picard iteration below constructs this exponential function one polynomial at a time.

The Lipschitz condition has real content. For the continuous function involving a square root,each waiting time $c\ge 0$ gives a different forward solution of a differential equation:The constant solution of a differential equation $y(t)=0$ is another choice. At $t=2$, delays $c=0,1,2$ give $y=4,1,0$ respectively. Before the join, both sides of the ordinary differential equation are zero; afterwards, $y_c^{\prime }=2(t-c)=2\sqrt{y_c}$. Both one-sided derivatives at $c$ are zero, so the join is valid. All these trajectories can also be extended by zero to $t<0$. The failed Lipschitz condition is visible in the quotientFor $\epsilon =10^{-2},10^{-4},10^{-6}$ this quotient is $20,200,2000$. Continuity still allows local existence by the Peano existence theorem; it cannot choose a waiting time. MIT's 18.034 Lecture 2, Example 2.3 develops the same waiting-time mechanism for a different ordinary differential equation, $y^{\prime }=3y^{2/3}$.

The useful promise is that an initial condition and a sufficiently regular local rule select one trajectory. It justifies comparing models and following trajectories without an arbitrary choice at each instant. A locally Lipschitz continuous rule need not keep that trajectory bounded: $y^{\prime }=1+y^2$ meets the theorem's hypotheses everywhere but has finite-time blowup. Longer existence is governed by the finite-dimensional continuation criterion.