# Picard's Existence Theorem/Proof 2

## 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

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 $\struct {\map {\CC^1} {\size{x - a} \le h}, \norm {\cdot}_\infty}$ to $y \in \map {\CC^1} {\size {x - a} \le h}$ 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 $\struct {\map \CC {\size {x - a} \le h}, \norm {\, \cdot \,}_\infty}$ 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.

$\blacksquare$