Picard's Existence Theorem

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:

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:

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

We also have that:
 * $\left|{f \left({t, y_{n-1} \left({t}\right)}\right) dt - f \left({t, y_{n-2} \left({t}\right)}\right)}\right| \le A \left|{y_{n-1} \left({t}\right) - y_{n-2} \left({t}\right)}\right|$

by the Lipschitz condition.

By the induction hypothesis, it follows that:

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

So:

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


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

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:

From Power Series over Factorial, it follows that $\displaystyle b + M h + \cdots + \frac {M A^{n-1} h^n} {n!} + \cdots$ 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 eqn $y' = f \left({x, y}\right)$.

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 $0$ and so $Y \left({x}\right) = y \left({x}\right)$ in $\left({a - h \, . \, . \, a + h}\right)$.

It is also known as the Picard-Lindelöf Theorem or the Cauchy-Lipschitz Theorem, after, and.

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