---
title: Picard's Existence and Uniqueness Theorem
module: Sequences and Series of Functions
moduleNumber: 7
lessonNumber: 4
order: 704
summary: >
  The Banach fixed-point theorem says a contraction of a complete metric space
  has exactly one fixed point, found by iterating from any start. Applied to the
  space of continuous functions with the uniform norm, it proves Picard's
  theorem: a first-order differential equation with a Lipschitz right-hand side
  has a unique local solution. Picard iteration constructs that solution
  explicitly, and worked examples show the Lipschitz condition is not optional.
topics: [Sequences and Series of Functions]
sources:
  - book: Lebl
    ref: "Ch. 6 — Sequences of Functions; §6.3 Picard's theorem; §7.6 Fixed point theorem"
  - book: Shkoller
    ref: "§3.1 Space of continuous functions; §3.2–3.3 Contraction mapping; §3.4 Fundamental theorem of ODEs"
  - book: Rosenlicht
    ref: "Ch. 8 — Differential equations"
draft: false
---

A first-order
differential equation $y' = F(x, y)$ with an initial condition $y(x_0) = y_0$
asks for a function whose slope at each point is dictated by $F$. When $F$ depends
only on $x$, the [fundamental
theorem](/real-analysis/riemann-integration/fundamental-theorem) integrates it.
When $F$ depends on $y$ as well, the unknown appears on both sides, and existence
of a solution is no longer automatic. Picard's theorem gives a clean sufficient
condition, and its proof is a fixed-point argument run in the space of continuous
functions.[^lebl-piece]

The main tool is the contraction mapping principle, proved first in full
generality and then specialized.

## The contraction mapping principle

A map that shrinks all distances has a single fixed point, and iterating the
map from any start converges to it.

> **Definition (Contraction).** Let $(X, d)$ be a metric space. A map
> $\varphi : X \to X$ is a **contraction** if there is a constant $k < 1$ with
> $$
> d\big(\varphi(p), \varphi(q)\big) \le k\, d(p, q) \qquad \text{for all } p, q \in X.
> $$
> A point $x$ with $\varphi(x) = x$ is a **fixed point**.

> **Theorem (Banach fixed-point theorem).** Let $(X, d)$ be a nonempty
> [complete](/real-analysis/metric-spaces/convergence-completeness) metric space
> and $\varphi : X \to X$ a contraction. Then $\varphi$ has a unique fixed point.

> **Proof.** Pick $x_0 \in X$ and iterate: $x_{n+1} = \varphi(x_n)$. The
> contraction bound compounds, $d(x_{n+1}, x_n) \le k^n d(x_1, x_0)$, so for
> $m > n$ the triangle inequality and the geometric series give
> $$
> d(x_m, x_n) \le \sum_{i=n}^{m-1} d(x_{i+1}, x_i) \le k^n d(x_1, x_0) \sum_{i=0}^{\infty} k^i = \frac{k^n}{1-k}\, d(x_1, x_0).
> $$
> Since $k < 1$, the right side goes to $0$, so $\{x_n\}$ is Cauchy; completeness
> gives $x = \lim_n x_n$. The map $\varphi$ is Lipschitz, hence continuous, so
> $$
> \varphi(x) = \varphi\Big(\lim_n x_n\Big) = \lim_n \varphi(x_n) = \lim_n x_{n+1} = x,
> $$
> and $x$ is a fixed point. For uniqueness, if $\varphi(x) = x$ and
> $\varphi(y) = y$, then $d(x, y) = d(\varphi(x), \varphi(y)) \le k\, d(x, y)$, and
> $k < 1$ forces $d(x, y) = 0$. $\;\square$

Both hypotheses are needed. Completeness supplies the limit of the iterates,
and the strict inequality $k < 1$ is what collapses the distance between two
candidate fixed points. A merely $1$-Lipschitz map, such as $x \mapsto x + 1$ on
$\mathbb{R}$, can have no fixed point at all.

$$
% caption: Iterating a contraction from any start walks the values in toward the
% single fixed point where the graph meets the diagonal; slope below one keeps
% the steps shrinking.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-0.15,0) -- (4,0) node[anchor=north east] {$x$};
\draw[black, ->] (0,-0.15) -- (0,4) node[anchor=north east] {};
\draw[black, thick] (0,0) -- (3.7,3.7);
\node[black, anchor=south east, font=\scriptsize] at (3.7,3.6) {$y=x$};
% contraction map: slope 0.45, intercept 1.4 -> fixed point at x=1.4/0.55=2.545
\draw[acc, very thick] (0,1.4) -- (3.7,3.065);
\node[acc, anchor=north west, font=\scriptsize] at (3.1,3.0) {map};
% cobweb from x0=0.2
% x0=0.2 -> y=1.49
\draw[black] (0.2,0.2) -- (0.2,1.49) -- (1.49,1.49) -- (1.49,2.07) -- (2.07,2.07) -- (2.07,2.33) -- (2.33,2.33) -- (2.33,2.45) -- (2.45,2.45);
\fill[black] (2.545,2.545) circle (2pt);
\node[black, anchor=north west, font=\scriptsize] at (2.6,2.5) {fixed point};
\node[black, anchor=north, font=\scriptsize] at (0.2,0) {start};
\end{tikzpicture}
$$

The proof is constructive: it names the fixed point as a limit of iterates and
bounds the distance to it after $n$ steps by
$d(x, x_n) \le \dfrac{k^n}{1-k}\, d(x_1, x_0)$. This is the basis of Newton's
method and many numerical solvers.

## The space of continuous functions

Running the fixed-point theorem on functions needs a complete metric space made
of them. The [uniform
norm](/real-analysis/function-sequences/pointwise-uniform-convergence) supplies
exactly one.

> **Definition (The space $C([a,b])$).** Let $C([a,b])$ be the continuous
> functions $f : [a, b] \to \mathbb{R}$ with the **uniform metric**
> $$
> d(f, g) = \|f - g\|_{[a,b]} = \sup_{x \in [a,b]} |f(x) - g(x)|.
> $$
> Convergence in this metric is uniform convergence.

> **Theorem.** $C([a,b])$ with the uniform metric is a complete metric space.

> **Proof.** A Cauchy sequence in this metric is [uniformly
> Cauchy](/real-analysis/function-sequences/pointwise-uniform-convergence), so it
> converges uniformly to some bounded function $f$, and the uniform limit of
> continuous functions is
> [continuous](/real-analysis/function-sequences/interchange-of-limits). Hence
> $f \in C([a,b])$ and the sequence converges to it in the uniform metric.
> $\;\square$

Completeness of $C([a,b])$ is what the Banach theorem requires when the metric
space is a space of functions; the uniform norm, the uniform Cauchy criterion,
and continuity of the uniform
limit are just what its hypotheses demand. A closed subset of $C([a,b])$ is again complete in the
subspace metric, and that is where the solution will be sought.

## Picard's theorem

Rewrite the differential equation as an integral equation. If $f$ solves
$f'(x) = F(x, f(x))$ with $f(x_0) = y_0$, then integrating both sides gives

$$
f(x) = y_0 + \int_{x_0}^{x} F\big(t, f(t)\big)\,\d t,
$$

and conversely any continuous $f$ satisfying this equation is differentiable with
the right derivative and initial value. The right-hand side defines an operator
$T$ on functions,

$$
T(f)(x) = y_0 + \int_{x_0}^{x} F\big(t, f(t)\big)\,\d t,
$$

and a solution is precisely a fixed point of $T$. Solving the differential
equation becomes finding $f$ with $T(f) = f$.

$$
% caption: The integral operator $T$ sends a candidate $f$ to a new function; a
% solution of the differential equation is exactly a fixed point of this loop.
\begin{tikzpicture}[scale=1.0, font=\footnotesize,
  box/.style={draw, minimum width=30mm, minimum height=13mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, draw=acc] (f) at (0,0) {candidate $f$};
\node[box] (tf) at (5.6,0) {$y_0 + \int F(t,f)\,dt$};
\draw[->, acc, thick] (f) to[bend left=22] node[midway, above, font=\scriptsize] {apply $T$} (tf);
\draw[->, acc, thick] (tf) to[bend left=22] node[midway, below, font=\scriptsize] {feed back} (f);
\node[black, anchor=north, font=\scriptsize] at (2.8,-1.15) {solution: the $f$ this loop leaves unchanged};
\end{tikzpicture}
$$

> **Theorem (Picard's existence and uniqueness).** Let $I, J \subset \mathbb{R}$
> be closed bounded intervals with interiors $I^\circ, J^\circ$, and let
> $(x_0, y_0) \in I^\circ \times J^\circ$. Suppose $F : I \times J \to \mathbb{R}$
> is continuous and **Lipschitz in the second variable**: there is $L$ with
> $$
> |F(x, y) - F(x, z)| \le L\,|y - z| \qquad \text{for all } y, z \in J,\ x \in I.
> $$
> Then there is $h > 0$ with $[x_0 - h, x_0 + h] \subset I$ and a unique
> differentiable $f : [x_0 - h, x_0 + h] \to J$ with $f'(x) = F(x, f(x))$ and
> $f(x_0) = y_0$.

> **Proof sketch.** Assume $x_0 = 0$. Since $I \times J$ is compact and $F$ is
> continuous, $F$ is bounded, say $|F| \le M$. Pick $\alpha > 0$ with
> $[-\alpha, \alpha] \subset I$ and $[y_0 - \alpha, y_0 + \alpha] \subset J$, and
> set
> $$
> h = \min\left\{ \alpha, \frac{\alpha}{M + L\alpha} \right\}.
> $$
> Let $Y = \{\, f \in C([-h, h]) : f([-h,h]) \subset J \,\}$, a closed subset of
> the complete space $C([-h,h])$, hence itself complete. The bound $|F| \le M$
> makes $T$ map $Y$ into $Y$: for $|x| \le h$,
> $$
> |T(f)(x) - y_0| = \left| \int_0^x F(t, f(t))\,\d t \right| \le M h \le M\cdot\frac{\alpha}{M+L\alpha} \le \alpha.
> $$
> The Lipschitz condition makes $T$ a contraction on $Y$: for $f, g \in Y$,
> $$
> |T(f)(x) - T(g)(x)| \le \int_0^x |F(t,f(t)) - F(t,g(t))|\,\d t \le L h\, d(f,g) \le \frac{L\alpha}{M + L\alpha}\, d(f, g),
> $$
> and $C := \dfrac{L\alpha}{M + L\alpha} < 1$; taking the supremum over $x$ gives
> $d(T(f), T(g)) \le C\, d(f, g)$. The Banach theorem supplies a unique
> $f \in Y$ with $T(f) = f$, and the
> [fundamental theorem](/real-analysis/riemann-integration/fundamental-theorem)
> makes $f$ differentiable with $f' = F(x, f(x))$ and $f(0) = y_0$. $\;\square$

The theorem is local. The interval half-width $h$ shrinks as the bound $M$ or the
Lipschitz constant $L$ grows, so a fast-changing field yields only a short
guarantee. The constant $C < 1$ is engineered by the choice of $h$: making the
interval short enough forces the operator to contract.

## Picard iteration

The proof of the Banach theorem is constructive, so the same iteration builds the
solution. Starting from the constant $f_0(x) = y_0$, the **Picard iterates** are

$$
f_{k}(x) = y_0 + \int_{x_0}^{x} F\big(t, f_{k-1}(t)\big)\,\d t,
$$

and they converge uniformly to the solution.

> **Worked example (Picard iterates for the exponential).** For $f'(x) = f(x)$,
> $f(0) = 1$, take $F(x, y) = y$ and $L = 1$. Starting from $f_0(x) = 1$,
> $$
> f_1(x) = 1 + \int_0^x 1\,\d t = 1 + x, \qquad f_2(x) = 1 + \int_0^x (1 + t)\,\d t = 1 + x + \frac{x^2}{2},
> $$
> $$
> f_3(x) = 1 + \int_0^x \left(1 + t + \frac{t^2}{2}\right)\d t = 1 + x + \frac{x^2}{2} + \frac{x^3}{6},
> $$
> $$
> f_4(x) = 1 + \int_0^x \left(1 + t + \frac{t^2}{2} + \frac{t^3}{6}\right)\d t = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}.
> $$
> The iterates are the partial sums of the [Taylor
> series](/real-analysis/differentiation/taylors-theorem) for $e^x$, and their
> uniform limit is the exponential. Picard's theorem thus reproves the existence
> and uniqueness of $e^x$ as the solution of $y' = y$.

$$
% caption: The Picard iterates for $y'=y$, $y(0)=1$ are the exponential's Taylor
% partial sums; each iterate matches the solution to one more order and hugs it
% over a wider range.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-2.2,0) -- (2.2,0) node[anchor=north east] {$x$};
\draw[black, ->] (0,-0.3) -- (0,4) node[anchor=north east] {$y$};
\draw[black!75, very thick] plot[smooth, domain=-1.9:1.35, samples=80] (\x, {exp(\x)});
\node[black!75, anchor=south east, font=\scriptsize] at (1.3,3.75) {$e^x$};
\draw[acc!45, thick, dashed] plot[domain=-1.9:1.75, samples=2] (\x, {1+\x});
\node[acc!70, anchor=west, font=\scriptsize] at (1.85,1.5) {$f_1$};
\draw[acc, thick, dashed] plot[smooth, domain=-1.9:1.55, samples=60] (\x, {1+\x+\x*\x/2});
\node[acc, anchor=south west, font=\scriptsize] at (-1.75,1.15) {$f_2$};
\draw[acc, very thick, dashed] plot[smooth, domain=-1.9:1.45, samples=60] (\x, {1+\x+\x*\x/2+\x*\x*\x/6});
\node[acc, anchor=north west, font=\scriptsize] at (0.85,1.25) {$f_3$};
\fill[black] (0,1) circle (1.8pt);
\end{tikzpicture}
$$

## Necessity of the hypotheses

The Lipschitz condition and continuity of $F$ each rule out a specific failure.

- **Without Lipschitz, uniqueness fails.** For $f'(x) = 2\sqrt{|f(x)|}$,
  $f(0) = 0$, the field $F(x, y) = 2\sqrt{|y|}$ is continuous but not Lipschitz in
  $y$ near $0$. Both $f(x) = 0$ and $f(x) = x^2$ (for $x \ge 0$) solve it, so the
  initial value does not determine the solution. The infinite slope of
  $\sqrt{|y|}$ at $y = 0$ is what prevents a contraction estimate.
- **Without continuity, existence fails.** For $y' = \varphi(x)$ with
  $\varphi = 0$ on the rationals and $\varphi = 1$ elsewhere, no solution exists:
  a solution's derivative would be $\varphi$, but $\varphi$ has no [intermediate
  value property](/real-analysis/differentiation/mean-value-theorem), which every
  derivative must have.
- **Lipschitz holds but only locally, so $h$ is finite.** For $f'(x) = f(x)^2$,
  $f(0) = 1$, the solution $f(x) = \dfrac{1}{1-x}$ exists only on $(-\infty, 1)$
  and blows up at $x = 1$. The map $y \mapsto y^2$ is Lipschitz on any bounded
  $J$ but not on all of $\mathbb{R}$, and the interval of existence is
  necessarily short near the blow-up.

$$
% caption: Dropping the Lipschitz condition costs uniqueness: from the same start
% at the origin, both the zero solution and a parabola satisfy the same equation.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-0.3,0) -- (3.6,0) node[anchor=north east] {$x$};
\draw[black, ->] (0,-0.3) -- (0,3) node[anchor=north east] {$y$};
\draw[acc, very thick] plot[domain=0:3.1, samples=60] (\x, {\x*\x/3.2});
\node[acc, anchor=south east, font=\scriptsize] at (3.0,2.55) {$x^2$};
\draw[black!70, very thick] (0,0) -- (3.4,0);
\node[black!70, anchor=south west, font=\scriptsize] at (2.4,0.05) {$f=0$};
\fill[black] (0,0) circle (2pt);
\node[black, anchor=south east, font=\scriptsize] at (0,0.1) {same start};
\end{tikzpicture}
$$

## Summary

| Ingredient | Supplied by | Role in Picard's proof |
| --- | --- | --- |
| Uniform norm | [pointwise/uniform lesson](/real-analysis/function-sequences/pointwise-uniform-convergence) | the metric on $C([a,b])$ |
| Uniform Cauchy $\Rightarrow$ converges | same | Cauchy iterates have a limit |
| Uniform limit is continuous | [interchange lesson](/real-analysis/function-sequences/interchange-of-limits) | the limit is a genuine function in $C([a,b])$ |
| Fundamental theorem | [integration module](/real-analysis/riemann-integration/fundamental-theorem) | integral equation $\Leftrightarrow$ differential equation |
| Banach fixed point | the contraction principle above | existence and uniqueness of the solution |

The differential equation is recast as a fixed-point problem, the space of
continuous functions is complete because uniform limits stay
continuous, and the Lipschitz condition turns integration into a contraction. The
solution is not only shown to exist and to be unique; the iteration that proves it
also computes it.

[^lebl-piece]: Lebl, §6.3 and §7.6. The two sections prove the same theorem twice — once by hand with the uniform norm and once through the abstract fixed-point theorem — and the treatment here follows the second, cleaner route while keeping the explicit iterates and examples of the first. Rosenlicht, Ch. 8, places the existence-uniqueness result at the head of the theory of differential equations.