02 - Setup for Picard's Theorem

Rectangular Neighbourhood for the Initial Value Problem

We seek a solution $y(x)$ which is defined on interval $[a-h,a+h]$ for suitable $h>0$
We also require the graph to be contained in rectangle $R$ around point $(x,y)=(a,b)$

$$R=\{(x,y):|x-a|\le h,|y-b|\le k\}$$

Lipschitz Condition (with constant $L$)

Function $f:[a-h,a+h]\times [b-k,b+k]$ satisfies Lipschitz Condition

If there exists $L>0$ such that

$$|f(x,y_1)-f(x,y_2)|\le L|y_1-y_2|$$

for all $x\in [a-h,a+h]$ and for all $y_1,y_2\in [b-k,b+k]$

Stronger than being continuous in the second variable
Weaker then continuously differentiable

Way to show a function $f$ satisfies the condition

Suppose on $R$ that $f$ is differentiable with respect to $y$, with

$$|f_y(x,y)|\le K\text{ for some }K\text{ and all }(x,y)\in R$$

For each $x\in [a-h,a+h]$, we can apply MVT to function $[b-k,b+k]\ni y\mapsto f(x,y)$
So that for any $y,\tilde{y}\in [b-k,b+k]$, there exists $\xi$ between $y$ and $\tilde{y}$ such that

$$|f(x,\tilde{y})-f(x,y)|=|f_y(x,\xi )(y-\tilde{y})|\le K|y-\tilde{y}|$$

So $f$ satisfies the Lipschitz Condition on rectangle $R$ with $L=K$

Assumptions and Properties of function $f:R\to \mathbb{R}$

Properties

1. $f$ is a function such that $(x,y)\mapsto f(x,y)$

Assumptions

1. $f$ is continuous in $R$ ($R$ is bounded so this guarantees that $f$ is bounded on $R$) with

$$M:=\underset{R}{\sup }|f|=\underset{R}{\max }|f|$$

Don't get confused with $x\mapsto f(x,y(x))$

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Rewriting the IVP into an equivalent Integral Equation

As $y(x)$ is differentiable and satisfies IVP on interval $[a-h,a+h]$ then

$$y(x)\text{ is certainly continuous so }f\text{ is continuous}$$

As $x\mapsto f(x,y(x))$ is a continuous function on $[a-h,a+h]$ it is integrable

Hence

$$y(x)-y(a)=[y(t){]}_a^x={\int }_a^{x}f(t,y(t))dt\text{ for any }x\in [a-h,a+h]$$

By rearranging then we get any solution of IVP satisfies

$$y(x)=b+{\int }_a^{x}f(t,y(t))dt\text{ for any }x\in [a-h,a+h]$$

Converse Argument $y:[a-h,a+h]\to [b-k,b+k]$ is a continuous function satisfying integral equation Then

If

$$y(a)=b$$

Since the integrand $t\mapsto f(t,y(t))$ is continuous then by
Fundamental Theorem of Calculus that $y$ is differentiable in every $x\in [a-h,a+h]$ with

$$y^{\prime }(x)=f(x,y(x))$$

So $y$ is a solution of the IVP

Successive Approximation (Iteration)

Start of with constant function $y_0(x)$

$$y_0(x)=b\forall x\in [a-h,a+h]$$

Then for $n=0,1,\cdots$ then

$$y_{n+1}(x):=b+{\int }_a^{x}f(t,y_n(t))dt$$