03 - Picard's Theorem

Picard's Theorem for Non-Symmetric Rectangles

Replace condition $Mh\le k$ with

$$M\max (h_1,h_2)\le \min (k_1,k_2)$$

Boundedness Lemma for IVPs lemma

Let IVP refer to $y^{\prime }(x)=f(x,y(x))$ with $y(a)=b$

Let $f:R\to \mathbb{R}$ be so that $P(i)$ holds
Let $y(x)$ be any solution of the IVP where $y(x)$ is defined on interval $[a-h_1,a{+}_{h_2}]$ for some $0<h_{1,2}\le h$ then

$$|y(x)-b|\le k\text{ for all }x\in [a-h_1,a+h_2]$$

and the graph doesn’t leave $R$

Note that this doesn’t require the Lipschitz Condition

Proof - By Contradiction

Suppose the claim is wrong so there exists a solution $y:[a-h_1,a+h_2]\to \mathbb{R}$ of IVP
such that there exists some $x_1\in [a-h_{1}a+h_2]$ with $|y(x_1)-b|>k$

By symmetry WLOG that $x_1>a$
As the graph starts out in rectangle $R=[a-h,a+h]\times [b-k,b+k]$ but contains

$$(x_1,y(x_1))$$

with $x_1\in [a-h,a+h]$ which is outside rectangle then there must be a first point

$$(x_0,y(x_0))$$

where the graph intersects the boundary of the rectangle

Hence there is some $x_0\in (a,x_1)$ such that

$$|y(x_0)-b|=k\text{ while }|y(x)-b|<k\text{ for all }x\in [a,x_0)$$

Since for $t\in [a,x_0]$ the points $(t,y(t))$ are in rectangle where $|f|$ is bounded by $M$
So since $y$ satisfies the IVP and hence integral equation then we get

$$\begin{array}{rl}|y(x_0)-b|& =\left|{\int }_a^{x_0}f(t,y(t))dt\right|\le \left|{\int }_a^{x_0}|f(t,y(t))|dt\right|\le M|x_0-a|\\ & <M(x_1-a)\le M{h}_1\le Mh\le k\end{array}$$

which is a contradiction

Uniqueness of the Picard's Theorem

Let $y(x)$ and $\tilde{y}(x)$ be two solutions of the ODE to the same initial data then

$$y(x)=\tilde{y}(x)$$

Proof

Suppose $f$ is a continuous function which satisfies Lipschitz condition on rectangle

$$R=[c,d]\times [b-k,b+k]$$

Let $a\in [c,d]$ and $y(x),\tilde{y}(x)$b be solutions of same ODE

$$y^{\prime }(x)=f(x,y(x))\text{ and }y(a)=b,\tilde{y}(a)=\tilde{b}$$

where $b$ may not necessarily be the same as $\tilde{b}$ whose graph is in $R$

Consider the difference function

$$v(x)=|\tilde{y}(x)-y(x)|\ge 0$$

And we have that $y(x)$ and $\tilde{y}(x)$ are solutions of corresponding integral equations

$$y(x)=b+{\int }_a^{x}f(t,y(t))dt\text{ and }\tilde{y}(x)=\tilde{b}+{\int }_a^{x}f(t,\tilde{y}(t))dt$$

We also have Lipschitz condition

$$|f(t,y(t))-f(t,\tilde{y}(t))|\le L|y(t)-\tilde{y}(t)|\le Lv(t)$$

Hence we get

$$\begin{array}{rl}v(x)& =\left|b-\tilde{b}+{\int }_a^{x}f(t,y(t))-f(t,\tilde{y}(t))dt\right|\\ & \le |b-\tilde{b}|+\left|{\int }_a^x|f(t,y(t))-f(t,\tilde{y}(t))|dt\right|\\ & \le |b-\tilde{b}|+L\left|{\int }_a^{x}v(t)dt\right|\end{array}$$

So by using Gronwall’s Inequality then

$$|y(x)-\tilde{y}(x)|=v(x)\le |b-\tilde{b}|e^{L|x-a|}\text{ for all }x\in [c,d]$$

When $b=\tilde{b}$ then we get uniqueness of solutions on the whole interval

In addition to proving uniqueness part of Picard’s Theorem then
For close $b,\tilde{b}$ then

$$||y-\tilde{y}|{|}_{\sup }:=\underset{x\in [c,d]}{\sup }|y(x)-\tilde{y}(x)|$$

If $y(x),\tilde{y}(x)$ satisfy ODE with initial values $b,\tilde{b}$ with $|b-\tilde{b}|<\delta$ then

$$|y(x)-\tilde{y}(x)|\le |b-\tilde{b}|e^{L|x-a|}\le \delta e^{Lh}=ϵ$$

for any $x\in [c,d]$ with $h:=\max \{a-c,d-a\}$

Upper Existence Bound

Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be a function then define

$$T_{+}:=\sup \{t>a:\text{ there exists a solution of IVP on }[a,t)\}$$

Note that $T_{+}:=+\infty$ if for every $t>a$ then there is a solution on $[a,t)$

Lower Existence Bound

Let $f:{\mathbb{R}}^2\to \mathbb{R}$ be a function then define

$$T_{-}:=\inf \{t<a:\text{ there exists a solution of IVP on }(t,a]\}$$

Note that $T_{-}:=-\infty$ if for every $t<a$ then there is a solution on $(t,a]$