An [ordinary differential equation](/page/Ordinary%20Differential%20Equation) describes a rule for instantaneous motion, but a rule alone does not choose a path. The equation $\dot{x}=f(x)$ says that a moving point should have velocity $f(x)$ when it is at position $x$; it does not say where the point starts. An initial value problem adds that missing datum. The central question is whether the velocity rule and the starting point determine a solution, whether that solution is unique, and how long it survives.

The first warning is that a plausible-looking velocity field may fail to determine the future. Consider a particle on the real line whose speed is proportional to the square root of its distance from the origin. If it starts at the origin, does it stay there, or can it begin moving later? The differential equation allows both behaviours, so the initial condition alone is not enough unless the vector field has the right regularity.

[example: Non-Unique Departure from Rest] Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=2\sqrt{|x|}$, and impose $x(0)=0$. The constant curve $x(t)=0$ is a solution because $\dot{x}(t)=0$ for every $t$, while \begin{align*} f(x(t))=f(0)=2\sqrt{|0|}=0. \end{align*} Now fix $a\ge 0$ and define $x_a(t)=0$ for $t\le a$, while $x_a(t)=(t-a)^2$ for $t>a$. Since $0\le a$, we have $x_a(0)=0$. The curve is continuous at $t=a$ because the left value is $0$ and \begin{align*} \lim_{t\downarrow a}(t-a)^2=0. \end{align*} It is also differentiable at $t=a$: for $h<0$, \begin{align*} \frac{x_a(a+h)-x_a(a)}{h}=\frac{0-0}{h}=0, \end{align*} and for $h>0$, \begin{align*} \frac{x_a(a+h)-x_a(a)}{h}=\frac{h^2-0}{h}=h, \end{align*} so both one-sided difference quotients tend to $0$. For $t<a$, $\dot{x}_a(t)=0$ and $f(x_a(t))=f(0)=0$. At $t=a$, the calculation above gives $\dot{x}_a(a)=0$, and again $f(x_a(a))=0$. For $t>a$, differentiating $x_a(t)=(t-a)^2$ gives \begin{align*} \dot{x}_a(t)=2(t-a). \end{align*} Since $t>a$, we have $t-a>0$, so \begin{align*} f(x_a(t))=2\sqrt{|(t-a)^2|}=2\sqrt{(t-a)^2}=2(t-a). \end{align*} Thus every $x_a$ solves $\dot{x}=2\sqrt{|x|}$ on $\mathbb{R}$ with $x_a(0)=0$, and the parameter $a\ge 0$ gives infinitely many distinct solutions. The vector field itself is continuous, but it is not Lipschitz on any interval containing $0$. Indeed, for $h>0$, \begin{align*} \frac{|f(h)-f(0)|}{|h-0|}=\frac{|2\sqrt{h}-0|}{h}=\frac{2}{\sqrt{h}}, \end{align*} and this quantity becomes arbitrarily large as $h\downarrow 0$. The initial condition is therefore not the source of non-uniqueness; the failure is the missing Lipschitz control at the origin. [/example]

This example sets the agenda for the subject. Initial value problems are not merely a notation for differential equations with starting data. They are the place where analysis decides whether deterministic laws really determine motion, how regularity assumptions enter, and what kinds of singular behaviour can appear even in finite-dimensional systems.

## Definition

To define the problem precisely, we must specify the time interval, the state space, the vector field, the initial time, and the initial state. The state space is usually an open subset of Euclidean space, because solutions should be allowed to move slightly in every coordinate direction while remaining inside the domain where the vector field is defined.

[definition: Initial Value Problem] Let $E \subset \mathbb{R}^n$ be open, let $I \subset \mathbb{R}$ be an interval, let $t_0 \in I$, let $x_0 \in E$, and let $f: I \times E \to \mathbb{R}^n$ be a vector field. The initial value problem with vector field $f$ and initial condition $(t_0,x_0)$ is the problem of finding an interval $J \subset I$ with $t_0 \in J$ and a continuous map $x: J \to E$ such that $x$ is differentiable on the interior $J^\circ$, satisfies $\dot{x}(t)=f(t,x(t))$ for all $t \in J^\circ$, and satisfies $x(t_0)=x_0$. [/definition]

This page uses the non-autonomous formulation as the parent notion: the velocity may depend on both time and state. The autonomous equation $\dot{x}=V(x)$ is recovered by taking $f(t,x)=V(x)$, so the geometric phase-space picture is a special case rather than a competing definition.

## Solution Intervals and Basic Forms

The interval $I$ is part of the data when asking for a solution on a prescribed time domain. In many applications, however, the interval is not known in advance; the question is to find an interval around $t_0$ on which a solution exists. To make that question meaningful, we need a separate definition of a solution on whatever interval is under consideration.

[definition: Solution of an Initial Value Problem] Let $E \subset \mathbb{R}^n$ be open, let $I \subset \mathbb{R}$ be an interval, let $t_0 \in I$, let $x_0 \in E$, and let $f: I \times E \to \mathbb{R}^n$. A solution of the initial value problem on an interval $J \subset I$ with $t_0 \in J$ is a continuous map $x: J \to E$ such that $x$ is differentiable on $J^\circ$, satisfies $\dot{x}(t)=f(t,x(t))$ for all $t \in J^\circ$, and satisfies $x(t_0)=x_0$. [/definition]

When $f$ is continuous, any solution has continuous derivative on the interior of its solution interval, because $t \mapsto f(t,x(t))$ is continuous there. Thus solutions are automatically $C^1$ away from endpoints in the standard existence theory.

A basic organizing question is whether the velocity rule changes with the clock itself or only with the current state. If the same state always determines the same velocity, then the initial time can be normalized to $0$ and the problem is governed by a vector field on state space alone.

In that situation the unknown interval is not just bookkeeping: it records how long the solution curve can be followed while it remains inside the state space $E$. The formal problem therefore asks for both a curve and a time interval containing the normalized initial time, with the curve's tangent at each interior time prescribed by the vector field at its current position.

[definition: Autonomous Initial Value Problem] Let $E \subset \mathbb{R}^n$ be open, let $V: E \to \mathbb{R}^n$ be a vector field, and let $x_0 \in E$. The autonomous initial value problem with initial state $x_0$ is the problem of finding an interval $I \subset \mathbb{R}$ with $0 \in I$ and a continuous map $x: I \to E$ such that $x$ is differentiable on $I^\circ$, satisfies $\dot{x}(t)=V(x(t))$ for all $t \in I^\circ$, and satisfies $x(0)=x_0$. [/definition]

The autonomous case matches the geometric picture of a vector field placing an arrow at each point of the state space. A solution is a curve whose tangent vector agrees with the arrow it meets. Time-dependent forcing, seasonal effects, and control inputs break this picture because the arrow at a fixed state may vary with time, so we need a separate formulation that keeps time as an explicit input.

[definition: Non-Autonomous Initial Value Problem] Let $E \subset \mathbb{R}^n$ be open, let $I \subset \mathbb{R}$ be an interval, let $f: I \times E \to \mathbb{R}^n$ be a vector field, and let $(t_0,x_0) \in I \times E$. The non-autonomous initial value problem is the problem $\dot{x}(t)=f(t,x(t))$ on $J^\circ$ and $x(t_0)=x_0$ for an unknown continuous map $x: J \to E$ that is differentiable on $J^\circ$, where $J \subset I$ is an interval containing $t_0$. [/definition]

A non-autonomous problem can be converted into an autonomous problem in one higher dimension by adjoining time as a state variable. This device is useful, but it does not erase the distinction: the original state variable and the clock play different roles in applications. Once solutions exist, we also need language for the geometric curve seen in state space.

[definition: Trajectory] Let $E \subset \mathbb{R}^n$ be open, let $I \subset \mathbb{R}$ be an interval, and let $x: I \to E$ be a solution of an initial value problem. The trajectory of $x$ is the image set $x(I)=\{x(t):t\in I\}\subset E$. [/definition]

The trajectory forgets the speed at which the curve is traversed. For autonomous equations, this is often natural: the phase portrait is a picture of possible paths in state space, while the parameter $t$ records how the motion moves along those paths. The simplest solvable example already shows the three questions of existence, uniqueness, and dependence on initial state.