{title2=$u(t_0)=u_0$}
{wiki}

An <initial value problem> combines an <ordinary differential equation> with an <initial condition>, for example $y'=f(t,y)$ 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> $u'=A(t)u$, 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> $y'=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution is
$$
y(t)=T_a+(y_0-T_a)e^{-k(t-t_0)}.
$$
Take $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log6\approx17.92$ 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 $|f(t,u)-f(t,v)|\leq L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y'=y^2$, $y(0)=1$, 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>:
$$
y'=\sqrt{|y|},\qquad y(0)=0.
$$
Besides $y(t)\equiv0$, each delay $a\geq0$ gives a solution for $t\geq0$:
$$
y_a(t)=\begin{cases}0,&0\leq t\leq a,\\(t-a)^2/4,&t>a.\end{cases}
$$
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, $y_a'=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative <derivative> there would violate the equation. Since $y'\geq0$, a forward solution starting from zero stays nonnegative.

\Image[/ode-nonunique-solutions.png]
{title=One initial value permits infinitely many delayed-start solutions without a Lipschitz condition}

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 https://ocw.mit.edu/courses/18-034-honors-differential-equations-spring-2009/a8c15198f81e85870da85da9792565b9_MIT18_034s09_lec02.pdf[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.


 Initial value problem (source code)