Picard's Existence Theorem/Proof 1

Proof
Let us define the following series of functions:

What we are going to do is prove that $\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_n} t$ 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:

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

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

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:

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

So:

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


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

Thus by induction:
 * $\displaystyle \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$

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:

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 $\displaystyle \lim_{n \mathop \to \infty} \map {y_n} x = \map y x$ 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:
 * $\displaystyle \map {y_n} x = b + \int_a^x \map f {t, \map {y_{n - 1} } t} \rd t$

we get:
 * $\displaystyle \map y x = b + \int_a^x \map f {t, \map y t} \rd t$

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 $\map y a = b$.

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

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

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.