01 - Picard's Existence Theorem

Picard's Existence Theorem

Let $f : R \to R$ be a function defined on rectangle
$R := {(x, y) : ∣ x - a ∣ \leq h, ∣ y - b ∣ \leq k}$
Which satisfies

$P (i)$ - $(a)$ : $f$ is continuous in $R$ with $∣ f (x, y)∣ \leq M$ for all $(x, y) \in R$

$P (i)$ - $(b)$ : $M h \leq k$

$P (ii)$ : $f$ satisfies Lipschitz Condition in $R$

Then IVP
$y^{'} (x) = f (x, y (x)) with y (a) = b$
has a unique solution $y$ on interval $[a - h, a + h]$

Picard’s Theorem doesn’t give the best (largest) interval where existence/uniqueness holds and is a local result

Proof - Existence - Via Successive Approximation

Successive Approximation (Iteration)

Start of with constant function $y_{0} (x)$
$y_{0} (x) = b \forall x \in [a - h, a + h]$
Then for $n = 0, 1, \dots$ then
$y_{n + 1} (x) := b + \int_{a}^{x} f (t, y_{n} (t)) d t$
Equivalent to asking that $y_{n + 1}$ satisfies $y_{n + 1}^{'} (x) = f (x, y_{n} (x))$ and $y_{n + 1} (a) = b$

Link to original

Claim 1: Each $y_{n} (x)$ is well-defined and continuous on $[a - h, a + h]$ which satisfies
$∣ y_{n} (x) - b ∣ \leq k for all x \in ∣ a - h, a + h ∣$

Claim 1 - By Induction

True for $n = 0$ as it is constant
For $t \in [a - h, a + h]$ then $(t, y_{n} (t)) \in R$ so evaluate $f$ at this point

As $y_{n} : [a - h, a + h] \to [b - k, b + k]$ is continuous then
$t \mapsto (t, y_{n} (t)) is a continious function$
from $[a - h, a + h]$ to the rectangle $R$
As $f$ is continuous function from $R$ to $R$ then
$composition [a - h, a + h] ∋ t \mapsto f (t, y_{n} (t)) \in R is continious$
Hence it is integrable

So next iterate $y_{n + 1}$ is well-defined function from $[a - h, a + h] \to R$
So by properties of integration it is differentiable hence continuous

Hence we have
$∣ y_{n + 1} (x) - b ∣ \leq \int_{a}^{x} ∣ f (t, y_{n} (t))∣ d t \leq \int_{a}^{x} M d t = M ∣ x - a ∣ \leq M h \leq k$
for every $x \in [a - h, a + h]$
Note that we need modulus outside integrals for cases where $x \leq a$

Claim 2: Let $\hat{M}$ be so that $∣ f (x, b)∣ \leq \hat{M}$ for all $x \in [a - h, a + h]$
Let $L$ be so that the the Lipschitz Condition holds so

Define the differences between consecutive iterations as
$e_{n} (x) := y_{n} (x) - y_{n - 1} (x), for n = 1, 2, \dots$
which satisfy
$∣ e_{n} (x)∣ \leq \frac{L ^{n - 1} M ^}{n !} ∣ x - a ∣^{n}$
for every $x \in [a - h, a + h]$ and every $n \in N$
Note that we can choose $\hat{M} = M$ as the bound on $∣ f ∣$ (as $\hat{M} \leq M$ )

Claim 2 - By Induction

For $n = 1$ use $y_{0} (x) = b$ so
$e_{1} (x) = y_{1} (x) - y_{0} (x) = b + \int_{a}^{x} f (t, y_{0} (t)) d t - b = \int_{a}^{x} f (t, b) d t$
Hence
$∣ e_{1} (x)∣ \leq \int_{a}^{x} ∣ f (t, b)∣ d t \leq \int_{a}^{x} \hat{M} d t \leq \hat{M} ∣ x - a ∣$
Hence case is true for $n = 1$

Suppose claim holds for $n \in N$ then using the definition of Picard iterations
$e_{n + 1} (x) = y_{n + 1} (x) - y_{n} (x) = [b + \int_{a}^{x} f (t, y_{n} (t)) d t] - [b + \int_{a}^{x} f (t, y_{n - 1} (t)) d t] = \int_{a}^{x} f (t, y_{n}) (t) - f (t, y_{n - 1} (t)) d x$
Using Lipschitz Condition in $P (ii)$ and that $y_{n}, y_{n - 1}$ are contained in rectangle $R$
So for $∣ t - a ∣ \leq h$ then
$∣ f (t, y_{n} (t)) - f (t, y_{n - 1} (t))∣ \leq L ∣ y_{n} (t) - y_{n - 1} (t)∣ = L ∣ e_{n} (t)∣$
Combining with the induction assumption we get
$∣ e_{n + 1} (x)∣ \leq L \int_{a}^{x} ∣ e_{n} (t)∣ d t \leq L \int_{a}^{x} \frac{L ^{n - 1} M ^}{n !} ∣ t - a ∣^{n} d t = \frac{L ^{n} M ^}{( n + 1 )!} ∣ x - a ∣^{n + 1}$
So it is true by induction

Claim 3: The iterates $y_{n} (x) = y_{0} (x) + \sum_{j = 1}^{n} e_{j} (x)$ converge uniformly to
$continuous function y_{\infty} (x)$
on the interval $[a - h, a + h]$ and $y_{\infty} (x)$ is a solution of the recursive integral equation

Claim 3

Note that from Claim 2 then for every $n \in N$
$∣ e_{n} (x)∣ \leq \frac{L ^{n - 1} M ^}{n !} h^{n} =: M_{n} for all x \in [a - h, a + h]$
As $\sum_{n = 1}^{\infty} M_{n}$ converges by ratio test then by Weierstrass M-test then
$j = 1 \sum n e_{j} (x) converges uniformly on [a - h, a + h] as n \to \infty$
As $y_{0} (x)$ is just a fixed function (independent of $n$ ) then
$y_{n} (x) = y_{0} (x) + j = 1 \sum n e_{j} (x)$
also converges uniformly to limit function $y_{\infty} (x)$ on $[a - h, a + h]$

Uniform convergence combined with the continuity of the functions $y_{n}$ means
$y_{\infty} is also continuous$
Using Lipschitz-Condition and uniform convergence of $y_{n} (t)$ then
$t \in [a - h, a + h] sup ∣ f (t, y_{n} (t)) - f (t, y_{\infty} (t))∣ \leq L t \in [a - h, a + h] sup ∣ y_{n} (t) - y_{\infty} (t)∣ \to 0$
where the last step follows from $y_{n}$ converging uniformly

Since we can exchange limit and integrals for uniformly convergent sequences
$n \to \infty lim \int_{a}^{x} f (t, y_{n} (t)) d t = \int_{a}^{x} f (t, y_{n} (t)) d t = \int_{a}^{x} f (t, y_{\infty} (t)) d t$
So by passing the limit to the recursive equation to define $y_{n + 1}$ then
For all $x \in [a - h, a + h]$
$y_{\infty} (x) = n \to \infty lim y_{n + 1} (x) = b + n \to \infty lim \int_{a}^{x} f (t, y_{n} (t)) d t = b + \int_{a}^{x} f (t, y_{\infty} (x)) d t$
Hence $y_{\infty}$ is indeed a solution of the integral equation
So as the integral equation is equivalent to the IVP then the existence part of Picard’s Theorem is proven

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01 - Picard's Existence Theorem

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