Picard's Existence Theorem/Proof 2

Proof
Let us define the following series of functions:

Denote this sequence by $\sequence {y_k}_{k \mathop \in \N_0}$.

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

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:


 * $\displaystyle y_{n + 1} = y_0 + \sum_{k = 0}^n \paren {y_{k + 1} - y_k}$

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 $\displaystyle h = \frac A 2$.

For $a \le x \le h$ we have:

By taking supremum norm of both sides, we get:


 * $\displaystyle \norm {y_{n + 1} - y_n}_\infty \le \frac 1 2 \norm {y_n - y_{n - 1} }_\infty$

By induction, the inequality can be extended:


 * $(1): \quad \displaystyle \norm {y_{n + 1} - y_n} \le \frac 1 {2^n} \norm {y_1 - y_0}$

Therefore:

Same argument applies to $-h \le x \le a$.

Hence, $\sequence {y_k}_{k \mathop \in \N_0}$ converges in $\struct {\map {C^1} {\size{x - a} \le h}, \norm {\cdot}_\infty}$ to $y \in \map {C^1} {\size {x - a} \le h}$ absolutely.

Therefore, the sequence is convergent:


 * $\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_{n + 1} } x = x_0 + \lim_{n \mathop \to \infty} \int_a^x \map f {x, \map {y_n} x} \rd x$

To find the limit, consider the following sequence:


 * $\displaystyle \map {g_n} x = \map f {x, \map {y_n} x}$

The sequence $\sequence {g_n}_{n \mathop \in \N_0}$ is a sequence of partial sums $\displaystyle g_0 + \sum_{k \mathop = 0}^n \paren {g_{k + 1} - g_k}$.

It follows that:

So $\sequence {g_n}_{n \mathop \in \N_0}$ converges to some $g$ in $\struct {\map C {\size {x - a} \le h}, \norm {\, \cdot \,}_\infty}$ absolutely.

It follows that:

On the other hand, Riemann integral is a continuous maping.

From Continuous Map Preserves Convergent Sequences:

We conclude that:


 * $\displaystyle \map y x = y_0 + \int_a^x \map f {t, \map y t} \rd t$

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
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:


 * $\displaystyle \map {y_1} x - \map {y_1} {x_*} = \int_{x_*}^x \map {y_1'} t \rd t = \int_{x_*}^x \map {f_1} {t, \map {y_1} t} \rd t$


 * $\displaystyle \map {y_2} x - \map {y_2} {x_*} = \int_{x_*}^x \map {y_2'} t \rd t = \int_{x_*}^x \map {f_2} {t, \map {y_2} t} \rd t$

After taking the difference:


 * $\displaystyle \map {y_1} x - \map {y_2} x = \int_{x_*}^x \paren {\map {f_1} {t, \map {y_1} t} - \map {f_2} {t, \map {y_2} t}} \rd t$

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:


 * $\displaystyle B = \max_{t \mathop \in \closedint {x_*} {x_* \mathop + \frac 1 {A N} } } \size {\map {x_2} t - \map {x_1} t} \le 2 k$

Then $\forall x \in \closedint {x_*} {x_* + \dfrac 1 {AN}}$ we have:

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.