= Peano existence theorem
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{title2=$y'=f(t,y),\quad y(t_0)=y_0$}
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Let $U\subset\mathbb R\times\mathbb R^d$ be an <open set>, $d<\infty$, and $f:U\to\mathbb R^d$ a <continuous function>. Through every $(t_0,y_0)\in U$ there is a local <solution of a differential equation>: a <continuously differentiable function> $y:I\to\mathbb R^d$ on an open <interval> containing $t_0$, satisfying the displayed <initial value problem> pointwise. No <Lipschitz condition> is assumed.<Ordinary Differential Equations and Dynamical Systems by Gerald Teschl>[Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, §2.7, Theorem 2.19, printed p. 56: forward existence and the analogous backward result]{ref}
<Continuity> of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and <initial condition> actually be followed? Prediction also needs uniqueness. The <delayed solutions of a square-root differential equation> below show how the same rule and <initial condition> can leave a departure time undetermined. The <Picard-Lindelöf theorem> supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.
For the <initial value problem> $y'=2\sqrt{|y|}$, $y(0)=0$, a forward <solution of a differential equation> can stay zero until any $c\geq0$, then follow $y=(t-c)^2$. The <delayed solutions of a square-root differential equation> verify this freedom exactly.
\Image[/peano-delayed-solution-fan.png]
{title=One initial condition admits an arbitrary forward waiting time}
{description=Colored curves leave zero at the marked waits $c=0,1,2,3$; the dashed curve stays zero forever. Every curve satisfies the same original differential equation for $t\geq0$.}
Local existence gives neither a global <solution of a differential equation> nor a bound on all future values. The <continuous function> $f(t,y)=y^2$ already permits <finite-time blowup>; the <finite-dimensional continuation criterion> explains the additional issue of extending a trajectory within its domain.
Proof outline: choose a <compact set> $K=[t_0-a,t_0+a]\times\overline B_r(y_0)\subset U$. By the <extreme value theorem>, $M=\max_K\|f\|<\infty$. Take $0<h\leq a$ with $Mh\leq r$. Forward <Euler method> polygons remain in the <closed ball> and satisfy $\|y_n(t)-y_n(s)\|\leq M|t-s|$. The <Arzelà-Ascoli theorem> yields a <convergent subsequence> with <uniform convergence>. <Uniform continuity> of $f$ on $K$ makes their <integral equation> defect tend uniformly to zero, so the limit satisfies
$$
y(t)=y_0+\int_{t_0}^t f(s,y(s))\,ds.
$$
The <fundamental theorem of calculus> gives a <continuously differentiable function>. Repeat backward and join at $t_0$, where both <derivatives> equal $f(t_0,y_0)$.<Ordinary Differential Equations and Dynamical Systems by Gerald Teschl>[Teschl, §2.7, printed pp. 54–56, equations (2.69), (2.71)–(2.76): polygon bounds, compactness, and vanishing integral defect]{ref}
The <compact set> here uses a <finite-dimensional vector space>; bounded closed balls in an arbitrary <Banach space> need not be compact. When $M=0$, the constant <solution of a differential equation> $y=y_0$ works on the entire time <interval> in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.
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