04 - Setup for Picard's Theorem (Paired System)

Pair System of First Order ODEs

Consider a pair first so for functions $y_1,y_2$ then

$$\begin{array}{rl}{y}_1^{\prime }(x)& =f_1(x,y_1(x),y_2(x))\\ y_2^{\prime }(x)& =f_2(x,y_1(x),y_2(x))\end{array}$$

with initial data

$$y_1(a)=b_1,y_2(a)=b_2$$

Then we can use vector notation

$$\underline{y}=\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right),\underline{f}=\left(\begin{array}{c}{f}_1\\ f_2\end{array}\right),\underline{b}=\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)$$

So we can write the system of equations as

$$\begin{array}{rl}{\underline{y}}^{\prime }(x)& =\underline{f}(x,\underline{y}(x))\\ \underline{y}(a)& =\underline{b}\end{array}$$

Ball of Radius $k$

$$B_k(\underline{b})=\{\underline{y}\in {\mathbb{R}}^2:||\underline{y}-\underline{b}|{|}_1\le k\}=\{(y_1,y_2)\in {\mathbb{R}}^2:|y_1-b_1|+|y_2-b_2|\le k\}$$

Note that we are using the $l^1$ norm so that $||\underline{y}||=|y_1|+|y_2|$

Region Neighbourhood for ODEs

Instead of rectangle $R$ we use

$$S=[a-h,a+h]\times B_k(\underline{b})\subset {\mathbb{R}}^3$$

Solving Second Order ODEs using Picard

Suppose we have IVP for second-order ODE

$$y^{\prime \prime }=F(x,y(x),y^{\prime }(x))\text{ with }y(a)=b,y^{\prime }(a)=c$$

Note that this can also be extended to $n$-th order linear ODEs

Solution

Reduce it to a first-order paired system by

$$y_1(x)=y(x)\text{ and }{y}_2(x)=y^{\prime }(x)$$

Note that $y(s)$ solves above IVP if and only if $\underline{y}(x)=(y_1(x),y_2(x))$ satisfies

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

where $\underline{f}$ is defined as

$$\underline{f}(x,y_1,y_2):=(y_2,F(x,y_1,y_2){)}^T$$

So $\underline{f}$ is continuous if and only if $F(x,y_1,y_2)$ is continuous
Similarly if $F$ satisfies the Lipschitz condition then $\underline{f}$ also does with $L=L_2+1$

Hence we can apply Picard’s Theorem