# Picard's Existence Theorem

## Theorem

Let $\map f {x, y} : \R^2 \to \R$ be continuous in a region $D \subseteq \R^2$.

Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.

Let $\map f {x, y}$ satisfy in $D$ the Lipschitz condition in $y$:

$\size{\map f {x, y_1} - \map f {x, y_2} } \le A \size {y_1 - y_2}$

where $A$ is independent of $x, y_1, y_2$.

Let the rectangle $R$ be defined as $\set {\tuple {x, y} \in \R^2: \size {x - a} \le h, \size {y - b} \le k}$ such that $M h \le k$.

Let $R \subseteq D$.

Then $\forall x \in \R: \size {x - a} \le h$, the first order ordinary differential equation:

$y' = \map f {x, y}$

has one and only one solution $y = \map y x$ for which $b = \map y a$.

## Proof 1

Let us define the following series of functions:

 $\ds \map {y_0} x$ $=$ $\ds b$ $\ds \map {y_1} x$ $=$ $\ds b + \int_a^x \map f {t, \map {y_0} t} \rd t$ $\ds \map {y_2} x$ $=$ $\ds b + \int_a^x \map f {t, \map {y_1} t} \rd t$ $\ds$ $\ldots$ $\ds$ $\ds \map {y_n} x$ $=$ $\ds b + \int_a^x \map f {t, \map {y_{n - 1} } t} \rd t$

What we are going to do is prove that $\ds \map y x = \lim_{n \mathop \to \infty} \map {y_n} t$ is the required solution.

There are five main steps, as follows:

### The curve lies in the rectangle

We will show that for $a - h \le x \le a + h$, the curve $y = \map {y_n} x$ lies in the rectangle $R$.

That is, that $b - k < y < b + k$.

Suppose $y = \map {y_{n - 1} } x$ lies in $R$.

Then:

 $\ds \size {\map {y_n} x - b}$ $=$ $\ds \size {\int_a^x \map f {t, \map {y_{n - 1} } t} \rd t}$ $\ds$ $\le$ $\ds M \size {x - a}$ $\ds$ $\le$ $\ds M h$ $\ds$ $<$ $\ds k$

Clearly $y_0$ lies in $R$, and the argument holds for $y_1$.

So by induction, $y = \map {y_n} x$ lies in $R$ for all $n \in \N$.

### Bounded Nature of Adjacent Differences

We will show that:

$\ds \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$

This is also to be proved by induction.

Suppose that this holds for $n-1$ in place of $n$.

Let this be the induction hypothesis.

We have:

$\ds \map {y_n} x - \map {y_{n - 1} } x = \int_a^x \paren {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \rd t$

We also have that:

$\size {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \le A \size {\map {y_{n - 1} } t - \map {y_{n - 2} } t}$

by the Lipschitz condition.

By the induction hypothesis, it follows that:

$\ds \size {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \le \frac {M A^{n - 1} \size {t - a}^{n - 1} } {\paren {n - 1}!}$

So:

 $\ds \size {\map {y_n} x - \map {y_{n - 1} } x}$ $\le$ $\ds \frac {M A^{n - 1} } {\paren {n - 1}!} \size {\int_a^x \size {t - a}^{n - 1} \rd t}$ $\ds$ $=$ $\ds \frac {M A^{n - 1} } {n!} \size {x - a}^n$

For the base case, we use $n = 1$:

$\ds \size {\map {y_1} x - b} \le \size {\int_a^x \map f {t, b} \rd t} \le M \size {x - a}$

Thus by induction:

$\ds \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$

for all $n$.

### Uniform Convergence of Sequence

Next we show that the sequence $\sequence {\map {y_n} x}$ converges uniformly to a limit for $a - h \le x \le a + h$.

From Bounded Nature of Adjacent Differences above, we have:

 $\ds$  $\ds b + \paren {\map {y_1} x - b} + \cdots + \paren {\map {y_n} x - \map {y_{n - 1} } x} + \cdots$ $\ds$ $\le$ $\ds b + M h + \cdots + \frac {M A^{n - 1} h^n} {n!} + \cdots$

From Radius of Convergence of Power Series over Factorial, it follows that $b + M h + \cdots + \dfrac {M A^{n - 1} h^n} {n!} + \cdots$ is absolutely convergent for all $h$.

Hence, by the Weierstrass M-Test:

$b + \paren {\map {y_1} x - b} + \cdots + \paren {\map {y_n} x - \map {y_{n - 1} } x} + \cdots$

converges uniformly for $a - h \le x \le a + h$.

Since its terms are continuous functions of $x$, its sum $\ds \lim_{n \mathop \to \infty} \map {y_n} x = \map y x$ is also continuous from Combination Theorem for Sequences.

### Solution Satisfies Differential Equation

We now show that $y = \map y x$ satisfies the differential equation $y' = \map f {x, y}$.

Since:

$\map {y_n} x$ converges uniformly to $\map y x$ in the open interval $\openint {a - h} {a + h}$ from Uniform Convergence of Sequence above
$\size {\map f {x, y} - \map f {x, y_n} } \le A \size {y - y_n}$ from the Lipschitz condition in $y$

it follows that $\map f {x, \map {y_n} x}$ tends uniformly to $\map f {x, \map y x}$.

Letting $n \to \infty$ in:

$\ds \map {y_n} x = b + \int_a^x \map f {t, \map {y_{n - 1} } t} \rd t$

we get:

$\ds \map y x = b + \int_a^x \map f {t, \map y t} \rd t$

The integrand $\map f {t, \map y t}$ is a continuous function of $t$.

Therefore the integral has the derivative $\map f {x, y}$.

Also, we have that $\map y a = b$.

### Uniqueness of Solution

We now show that the solution $y = \map y x$ that we have found is the only solution where $\map y a = b$.

Aiming for a contradiction, suppose there is another such solution, $y = \map Y x$, say.

Let $\size {\map Y x - \map y x} \le B$ when $\size {x - a} \le h$. (Certainly we could take $B = 2 k$.)

Then:

$\ds \map Y x - \map y x = \int_a^x \paren {\map f {t, \map Y t} - \map f {t, \map y t} } \rd t$

But:

$\size {\map f {t, \map Y t} - \map f {t, \map y t} } \le A \size {\map Y t - \map y t} \le A B$

So:

$\size {\map Y t - \map y t} \le A B \size {x - a}$

Repeating the argument, we can get successive estimates for the upper bound of $\size {\map Y x - \map y x}$ in $\openint {a - h} {a + h}$.

This gives:

$\ds \frac {A^2 B} {2!} \size {x - a}^2, \ldots, \frac {A^n B} {n!} \size {x - a}^n, \ldots$

But this sequence tends to $0$.

So $\map Y x = \map y x$ in $\openint {a - h} {a + h}$.

This contradicts the supposition that $\map Y x$ and $\map y x$ are different.

Hence by Proof by Contradiction it follows that $\map y x$ is unique.

$\blacksquare$

## Proof 2

Let us define the following series of functions:

 $\ds \map {y_0} x$ $=$ $\ds b$ $\ds \map {y_1} x$ $=$ $\ds b + \int_a^x \map f {t, \map {y_0} t} \rd t$ $\ds \map {y_2} x$ $=$ $\ds b + \int_a^x \map f {t, \map {y_1} t} \rd t$ $\ds$ $\ldots$ $\ds$ $\ds \map {y_n} x$ $=$ $\ds b + \int_a^x \map f {t, \map {y_{n-1} } t} \rd t$

Denote this sequence by $\sequence {y_k}_{k \mathop \in \N_0}$.

What we are going to do is prove that $\ds \map y x = \lim_{n \mathop \to \infty} \map {y_n} x$ is the required solution.

### The curve lies in the rectangle

We will show that for $a - h \le x \le a + h$, the curve $y = \map {y_n}x$ lies in the rectangle $R$.

That is, that $b - k < y < b + k$.

Suppose $y = \map {y_{n - 1} } x$ lies in $R$.

Then:

 $\ds \size {\map {y_n} x - b}$ $=$ $\ds \size {\int_a^x \map f {t, \map {y_{n - 1} } t} \rd t}$ $\ds$ $\le$ $\ds M \size {x - a}$ $\ds$ $\le$ $\ds M h$ $\ds$ $<$ $\ds k$

Clearly $y_0$ lies in $R$, and the argument holds for $y_1$.

So by induction, $y = \map {y_n} x$ lies in $R$ for all $n \in \N$.

### Existence

The sequence $\sequence {y_k}_{k \mathop \in N_0}$ can be expressed as a telescoping series:

$\ds y_{n + 1} = y_0 + \sum_{k \mathop = 0}^n \paren {y_{k + 1} - y_k}$

The theorem contains more variables $\paren {\set {x, y_1, y_2} }$ and parameters $\paren {\set{h, k, M, A} }$ than inequality constraints.

Thus, more relations between them can be chosen without affecting the constraints.

Choose $h = \dfrac A 2$.

For $a \le x \le h$ we have:

 $\ds \size {\map {y_{n+1} } x - \map {y_n} x}$ $=$ $\ds \size {\int_a^x \map f {t, \map {y_n} t} - \map f {t, \map {y_{n - 1} } t} \rd t}$ $\ds$ $\le$ $\ds \int_a^x \size {\map f {t, \map {y_n} t} - \map f {t, \map {y_{n - 1} } t} } \rd t$ Absolute Value of Definite Integral $\ds$ $\le$ $\ds \int_a^x A \size {\map {y_n} t - \map {y_{n - 1} } t} \rd t$ Definition of Lipschitz Condition (Real Function) $\ds$ $\le$ $\ds \int_a^x A \norm {y_n - y_{n - 1} }_\infty \rd t$ Definition of Supremum Norm $\ds$ $=$ $\ds A \norm {y_n - y_{n-1} }_\infty \paren {x - a}$ $\ds$ $\le$ $\ds A \norm {y_n - y_{n-1} }_\infty h$ $\ds$ $=$ $\ds \frac 1 2 \norm {y_n - y_{n - 1} }_\infty$

By taking supremum norm of both sides, we get:

$\norm {y_{n + 1} - y_n}_\infty \le \dfrac 1 2 \norm {y_n - y_{n - 1} }_\infty$

By induction, the inequality can be extended:

$(1): \quad \norm {y_{n + 1} - y_n} \le \dfrac 1 {2^n} \norm {y_1 - y_0}$

Therefore:

 $\ds \sum_{n = 0}^\infty \norm {y_{n + 1} - y_n}$ $\le$ $\ds \norm {y_1 - y_0} \sum_{n \mathop = 0}^\infty \frac 1 {2^n}$ $\ds$ $<$ $\ds \infty$ Sum of Infinite Geometric Progression

Same argument applies to $-h \le x \le a$.

Hence, $\sequence {y_k}_{k \mathop \in \N_0}$ converges in $\struct {\map {C^1} {\size{x - a} \le h}, \norm {\cdot}_\infty}$ to $y \in \map {C^1} {\size {x - a} \le h}$ absolutely.

Therefore, the sequence is convergent:

$\ds \map y x = \lim_{n \mathop \to \infty} \map {y_{n + 1} } x = x_0 + \lim_{n \mathop \to \infty} \int_a^x \map f {x, \map {y_n} x} \rd x$

To find the limit, consider the following sequence:

$\map {g_n} x = \map f {x, \map {y_n} x}$

The sequence $\sequence {g_n}_{n \mathop \in \N_0}$ is a sequence of partial sums $\ds g_0 + \sum_{k \mathop = 0}^n \paren {g_{k + 1} - g_k}$.

It follows that:

 $\ds \norm {\map {g_{k + 1} } x - \map {g_k} x}$ $=$ $\ds \norm {\map f {x, \map {y_{k+1} } x} - \map f {x, \map {y_k} x} }$ $\ds$ $\le$ $\ds L \norm {\map {y_{k + 1} } x - \map {y_k} x}$ assumption in theorem $\ds$ $\le$ $\ds L \norm {y_{k + 1} - y_k}_\infty$ Definition of Supremum Norm $\ds$ $\le$ $\ds \frac 1 {2^k} \norm {y_1 - y_0}_\infty$ from $(1)$

So $\sequence {g_n}_{n \mathop \in \N_0}$ converges to some $g$ in $\struct {\map C {\size {x - a} \le h}, \norm {\, \cdot \,}_\infty}$ absolutely.

It follows that:

 $\ds \map g x$ $=$ $\ds \lim_{n \mathop \to \infty} \map {g_n} x$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map f {x, \map {y_n} x}$ $\ds$ $=$ $\ds \map f {x, \map y x}$

On the other hand, a Riemann integral is a continuous mapping.

 $\ds \lim_{n \mathop \to \infty} \int_a^x \map f {t, \map {y_n} t} \rd t$ $=$ $\ds \lim_{n \mathop \to \infty} \int_a^x \paren {\map {g_0} t + \sum_{k \mathop = 0}^{n - 1} \paren {\map {g_{k + 1} } t - \map {g_k} t} } \rd t$ $\ds$ $=$ $\ds \int_a^x \map g t \rd t$ $\ds$ $=$ $\ds \int_a^x \map f {t, \map y t} \rd t$

We conclude that:

$\ds \map y x = y_0 + \int_a^x \map f {t, \map y t} \rd t$

where:

$\map y a = y_0 + 0 = b$

and, by Fundamental Theorem of Calculus:

$\map {y'} x = 0 + \map f {x, \map y x}$

for all $x \in \R : \size {x - a} \le h$.

### Uniqueness

Aiming for a contradiction, suppose that the solution to IVP is not unique.

Then, for the same initial conditions there exists a non-empty subset of $R$ where solutions differ.

Let $y_1, y_2$ be solutions to IVP for $x \in \R : \size {x - a} \le h$.

Let $x_* := \max \set {x \in \R : \size {x - a} \le h : \map {y_1} t = \map {y_2} t, \forall t \le x }$

Then:

$\ds \map {y_1} x - \map {y_1} {x_*} = \int_{x_*}^x \map {y_1'} t \rd t = \int_{x_*}^x \map {f_1} {t, \map {y_1} t} \rd t$
$\ds \map {y_2} x - \map {y_2} {x_*} = \int_{x_*}^x \map {y_2'} t \rd t = \int_{x_*}^x \map {f_2} {t, \map {y_2} t} \rd t$

After taking the difference:

$\ds \map {y_1} x - \map {y_2} x = \int_{x_*}^x \paren {\map {f_1} {t, \map {y_1} t} - \map {f_2} {t, \map {y_2} t}} \rd t$

Let $N \in \R$ be such that:

$N > \max \set {1, \dfrac 1 A, \dfrac 1 {A \paren {a \mathop + h \mathop - x_*} } }$

For all cases it holds that:

$x_* + \dfrac 1 {A N} < a + h$

Let:

$\ds B = \max_{t \mathop \in \closedint {x_*} {x_* \mathop + \frac 1 {A N} } } \size {\map {x_2} t - \map {x_1} t} \le 2 k$

Then $\forall x \in \closedint {x_*} {x_* + \dfrac 1 {AN} }$ we have:

 $\ds \size {\map {x_2} x - \map {x_1} x}$ $=$ $\ds \size {\int_{x_*}^x \paren {\map f {\map {x_2} t, t} - \map f {\map {x_1} t, t} } \rd t}$ $\ds$ $\le$ $\ds \int_{x_*}^x \size {\map f {\map {x_2} t, t} - \map f {\map {x_1} t, t} } \rd t$ Absolute Value of Definite Integral $\ds$ $\le$ $\ds \int_{x_*}^x A \size {\map {x_2} t - \map {x_1} t} \rd t$ Definition of Lipschitz Condition (Real Function) $\ds$ $\le$ $\ds \int_{x_*}^x A B \rd t$ $\ds$ $=$ $\ds A B \paren {x - x_*}$ $\ds$ $\le$ $\ds A B \paren {x_* + \frac 1 {A N} - x_*}$ $\ds$ $=$ $\ds \frac {A B} {A N}$ $\ds$ $=$ $\ds \frac B N$

Thus:

$\forall t \in \closedint {x_*} {x_* + \dfrac 1 {A N} } : \size {\map {x_1} t - \map {x_2} t} \le \dfrac B N$

and $B \le \dfrac B N$ or $N \le 1$.

This brings us to a contradiction.

Hence our assumption that the solution to IVP is not unique was false.

Hence the result, by Proof by Contradiction.

$\blacksquare$

## Also known as

Picard's Existence Theorem is also known as:

the Picard-Lindelöf Theorem, after Charles Émile Picard and Ernst Leonard Lindelöf
the Cauchy-Lipschitz Theorem, after Augustin Louis Cauchy and Rudolf Otto Sigismund Lipschitz.

Some sources give this as Picard's Theorem but there are other theorems with this appellation so it is better to disambiguate.

## Source of Name

This entry was named for Charles Émile Picard.