# Picard's Existence Theorem/Proof

## Theorem

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

Let $\exists M \in \R: \forall x, y \in D: \left|{f \left({x, y}\right)}\right| < M$.

Let $f \left({x, y}\right)$ satisfy in $D$ the Lipschitz condition in $y$:

$\left|{f \left({x, y_1}\right) - f \left({x, y_2}\right)}\right| \le A \left|{y_1 - y_2}\right|$

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

Let the rectangle $R$ be defined as $\left\{{\left({x, y}\right) \in \R^2: \left|{x - a}\right| \le h, \left|{y - b}\right| \le k}\right\}$ such that $M h \le k$.

Let $R \subseteq D$.

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

$y' = f \left({x, y}\right)$

has one and only one solution $y = y \left({x}\right)$ for which $b = y \left({a}\right)$.

## Proof

Let us define the following series of functions:

 $\displaystyle y_0 \left({x}\right)$ $=$ $\displaystyle b$ $\displaystyle y_1 \left({x}\right)$ $=$ $\displaystyle b + \int_a^x f \left({t, y_0 \left({t}\right)}\right) \, \mathrm d t$ $\displaystyle y_2 \left({x}\right)$ $=$ $\displaystyle b + \int_a^x f \left({t, y_1 \left({t}\right)}\right) \, \mathrm d t$ $\displaystyle$ $\ldots$ $\displaystyle$ $\displaystyle y_n \left({x}\right)$ $=$ $\displaystyle b + \int_a^x f \left({t, y_{n-1} \left({t}\right)}\right) \, \mathrm d t$

What we are going to do is prove that $\displaystyle y \left({x}\right) = \lim_{n \to \infty} y_n \left({x}\right)$ 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 $y = y_n \left({x}\right)$ lies in the rectangle $R$.

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

Suppose $y = y_{n-1} \left({x}\right)$ lies in $R$.

Then:

 $\displaystyle \left\vert {y_n \left({x}\right) - b}\right\vert$ $=$ $\displaystyle \left\vert{\int_a^x f \left({t, y_{n-1} \left({t}\right)}\right) \, \mathrm d t}\right\vert$ $\displaystyle$ $\le$ $\displaystyle M \left\vert{x - a}\right\vert$ $\displaystyle$ $\le$ $\displaystyle M h$ $\displaystyle$ $<$ $\displaystyle k$

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

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

### Bounded Nature of Adjacent Differences

We will show that:

$\displaystyle \left|{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right| \le \frac {M A^{n-1}} {n!}\left|{x - a}\right|^n$

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:

$\displaystyle y_n \left({x}\right) - y_{n-1} \left({x}\right) = \int_a^x \left({f \left({t, y_{n-1} \left({t}\right)}\right) - f \left({t, y_{n-2} \left({t}\right)}\right)}\right) \, \mathrm d t$

We also have that:

$\left|{f \left({t, y_{n-1} \left({t}\right)}\right) - f \left({t, y_{n-2} \left({t}\right)}\right)}\right| \le A \left|{y_{n-1} \left({t}\right) - y_{n-2} \left({t}\right)}\right|$

by the Lipschitz condition.

By the induction hypothesis, it follows that:

$\displaystyle \left|{f \left({t, y_{n-1} \left({t}\right)}\right) - f \left({t, y_{n-2} \left({t}\right)}\right)}\right| \le \frac {M A^{n-1} \left|{t - a}\right|^{n-1}} {\left({n - 1}\right)!}$

So:

 $\displaystyle \left\vert{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right\vert$ $\le$ $\displaystyle \frac {M A^{n-1} } {\left({n - 1}\right)!} \left\vert{\int_a^x \left\vert{t - a}\right\vert^{n-1} \, \mathrm d t}\right\vert$ $\displaystyle$ $=$ $\displaystyle \frac {M A^{n-1} } {n!} \left\vert{x - a}\right\vert^n$

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

$\displaystyle \left|{y_1 \left({x}\right) - b}\right| \le \left|{\int_a^x f \left({t, b}\right) \, \mathrm d t}\right| \le M \left|{x - a}\right|$

Thus by induction:

$\displaystyle \left|{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right| \le \frac {M A^{n-1} } {n!}\left|{x - a}\right|^n$

for all $n$.

### Uniform Convergence of Sequence

Next we show that the sequence $\left \langle {y_n \left({x}\right)} \right \rangle$ converges uniformly to a limit for $a - h \le x \le a + h$.

From Bounded Nature of Adjacent Differences above, we have:

 $\displaystyle$  $\displaystyle b + \left({y_1 \left({x}\right) - b}\right) + \cdots + \left({y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right) + \cdots$ $\displaystyle$ $\le$ $\displaystyle b + M h + \cdots + \frac {M A^{n-1} h^n} {n!} + \cdots$

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:

$b + \left({y_1 \left({x}\right) - b}\right) + \cdots + \left({y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right) + \cdots$

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

Since its terms are continuous functions of $x$, its sum $\displaystyle \lim_{n \to \infty} y_n \left({x}\right) = y \left({x}\right)$ is also continuous from Combination Theorem for Sequences.

### Solution Satisfies Differential Equation

We now show that $y = y \left({x}\right)$ satisfies the differential equationn $y' = f \left({x, y}\right)$.

Since:

$y_n \left({x}\right)$ converges uniformly to $y \left({x}\right)$ in the open interval $\left({a - h \,.\,.\, a + h}\right)$ from Uniform Convergence of Sequence above
$\left|{f \left({x, y}\right) - f \left({x, y_n}\right)}\right| \le A \left|{y - y_n}\right|$ from the Lipschitz condition in $y$

it follows that $f \left({x, y_n \left({x}\right)}\right)$ tends uniformly to $f \left({x, y \left({x}\right)}\right)$.

Letting $n \to \infty$ in:

$\displaystyle y_n \left({x}\right) = b + \int_a^x f \left({t, y_{n-1} \left({t}\right)}\right) \, \mathrm d t$

we get:

$\displaystyle y \left({x}\right) = b + \int_a^x f \left({t, y \left({t}\right)}\right) \, \mathrm d t$

The integrand $f \left({t, y \left({t}\right)}\right)$ is a continuous function of $t$.

Therefore the integral has the derivative $f \left({x, y}\right)$.

Also, we have that $y \left({a}\right) = b$.

### Uniqueness of Solution

We now show that the solution $y = y \left({x}\right)$ that we have found is the only solution where $y \left({a}\right) = b$.

Suppose there is another such solution, $y = Y \left({x}\right)$, say.

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

Then:

$\displaystyle Y \left({x}\right) - y \left({x}\right) = \int_a^x \left({f \left({t, Y \left({t}\right)}\right) - f \left({t, y \left({t}\right)}\right)}\right) \, \mathrm d t$

But:

$\left|{f \left({t, Y \left({t}\right)}\right) - f \left({t, y \left({t}\right)}\right)}\right| \le A \left|{Y \left({t}\right) - y \left({t}\right)}\right| \le AB$

So:

$\left|{Y \left({t}\right) - y \left({t}\right)}\right| \le AB \left|{x - a}\right|$

Repeating the argument, we can get successive estimates for the upper bound of $\left|{Y \left({x}\right) - y \left({x}\right)}\right|$ in $\left({a - h \,.\,.\, a + h}\right)$.

This gives:

$\displaystyle \frac {A^2 B}{2!} \left|{x - a}\right|^2, \ldots, \frac {A^n B}{n!} \left|{x - a}\right|^n, \ldots$

But this sequence tends to $0$.

So $Y \left({x}\right) = y \left({x}\right)$ in $\left({a - h \,.\,.\, a + h}\right)$.

$\blacksquare$

## Source of Name

This entry was named for Charles Émile Picard.

It is also known as the Picard-Lindelöf Theorem or the Cauchy-Lipschitz Theorem, after Ernst Leonard Lindelöf, Augustin Louis Cauchy and Rudolf Lipschitz.

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