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Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Let $G:[0,1{]}^2\to \mathbb{R}$ and $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ be continuous. For a fixed $R>0$, put $K=\max |G|$ and $M={\max }_{y\in [0,1],|s|\le R}|f(y,s)|$. The integral equationhas a continuous solution with $\Vert u{\Vert }_{\infty }\le R$ whenever $|\mu |KM\le R$. The integral kernel is $G$; it is not the null space of a linear map.

Define the right side as $Tu$. Its supremum norm is at most $|\mu |KM$, so it maps the closed radius-$R$ ball in $C([0,1])$ into itself. Uniform continuity of $f$ on its compact box gives continuity of $T$. Uniform continuity of $G$ makes the image equicontinuous as well as bounded. The Arzelà-Ascoli theorem makes the image relatively compact, and the Schauder fixed-point theorem supplies a fixed point.

No uniqueness follows. For $G=1$, $f(y,s)=1+s^2$, and $0<\mu <1/2$, every solution is constant and its value satisfies $c=\mu (1+c^2)$, which has two real roots. A suitable ball captures the small root while the other lies outside; the existence proof does not rule it out.

For the initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, continuity of $f$ and local Lipschitz continuity in $y$ ensure a unique local solution. The Lipschitz condition compares different state values at a common time: $\Vert f(t,u)-f(t,v)\Vert \le L\Vert u-v\Vert$ on a suitable neighborhood. Subtracting two integral equations and applying Gronwall's inequality proves uniqueness.

Continuity alone does not suffice. The scalar equation $y^{\prime }=2\sqrt{|y|}$ with $y(0)=0$ has the zero solution and, for every $a\ge 0$, the solution $y(t)=0$ up to $a$ and $y(t)=(t-a{)}^2$ thereafter. The non-Lipschitz behavior at zero permits an arbitrary waiting time. This illustrates why the hypotheses of the Picard-Lindelöf theorem carry real information.