Picard's Existence Theorem

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

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

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 $$\left\{{\left({x, y}\right) \in \reals^2: \left|{x - a}\right| \le h, \left|{y - b}\right| \le k}\right\}$$ such that $$M h \le k$$.

Let $$R \subseteq D$$.

Then $$\forall x \in \reals: \left|{x - a}\right| \le h$$, the 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 $$y \left({x}\right) = \lim_{n \to \infty} y_n \left({x}\right)$$ 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 $$\left|{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right| \le \frac {M A^{n-1}} {n!}\left|{x - a}\right|^n$$.

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

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:

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

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, $$\left|{y_n \left({x}\right) - y_{n-1} \left({x}\right)}\right| \le \frac {M A^{n-1}} {n!}\left|{x - a}\right|^n$$ 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 $$b + M h + \cdots + \frac {M A^{n-1} h^n} {n!} + \cdots$$ 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 $$\lim_{n \to \infty} y_n \left({x}\right) = y \left({x}\right)$$ is also continuous from Combination Theorem for Sequences.

Solution Satisfies Differential Equation
We now show that $$y = y \left({x}\right)$$ satisfies the differential equation $$y' = f \left({x, y}\right)$$.

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

we get $$y \left({x}\right) b + \int_a^x f \left({t, y \left({t}\right)}\right) dt$$.

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 $$Y \left({x}\right) - y \left({x}\right) = \int_a^x \left({f \left({t, Y \left({t}\right)}\right) - f \left({t, y \left({t}\right)}\right)}\right) dt$$.

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

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