Some important theorems on ODE's

see Braun, for an introductory exposition, and Coddington & Levinson and Birkhoff & Rota for a more advanced one

Definition:

A function $f(x,y)$ satisfies a ``Lipschitz Condition'' in the variable $y$ on a set $S\in {\mathbb{R}}^2$ provided $\exists$ constant $L$ such that

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

whenever $(x,y_1),(x,y_2)\in S$. $L$ is the Lipschitz constant.

Definition: A set $S\in {\mathbb{R}}^2$ is convex, if whenever $(x_1, y_1) \& (x_2,y_2)$ belong to $S$, the point $[(1-\lambda)x_1+\lambda x_2,(1-\lambda)y_1+\lambda y_2]$ also belongs to $S$ for each $\lambda$ when $0\le\lambda\le 1$. See Figure 1.

Figure 1: CONVEX AND NON-CONVEX EXAMPLES

We study the IVP

$$\left\{\begin{array}{l}y'=f(x,y) y(x_0)=y_0\end{array}\right. \mbox { a system of $n$ ODE's}$$ (3)

$$S\equiv\{(x,y)\vert a\le x\le b,y\in \mathbb{R}^n\}$$

$$a,b, , a\le x_0, \le b.$$

It has exactly 1 solution provided $f$ satisfies the following conditions of existence and uniqueness.

Existence & Uniqueness

Theorem. $f$ defined and continuous on $S$, convex (see 0.1.1), and satisfies a Lipschitz condition 0.1.1 $\Rightarrow$ for every $x_0\in[a,b]$ & every $y_0\in \mathbb{R}^n$ there exists 1 function $y(x)$ such that

a)

$y$ is continuous and differentiable for $x\in [a,b]$

b)

$y'(x)=f(x,y(x))$ for $x\in [a,b]$

c)

$y(x_0)=y_0$

$\Box$

Theorem. Suppose $f$ defined and convex. If there exist a constant $L>0$ such that

$$\displaystyle \bigg\vert\frac{df_i}{dy_j}\bigg\vert\le L\quad \forall(x,y)\in S$$

$\Rightarrow f$ satisfies a Lipschitz condition on $S$

$\Box$

Remark. In applications one finds that $f$ is usually continuous in $S$ and also continuously differentiable there, but could have the derivates $$\frac{df_i}{dy_j}$$ unbounded on $S$.
$\Rightarrow$ While (3) is still solvable the solution may only be defined in some $U(x_0)$ neighborhood of the initial $x_0\in[a,b]$.

Example)

$$\left\{\begin{array}{l}y'=y^2\\ y{(0)}=1\end{array}\right.$$

has solution $y=\displaystyle \frac{1}{1-x}$ defined only for $x<1$.

Continuous Dependence:

Theorem. Let $f: S\rightarrow \mathbb{R}^n$ continuous on $S$ also satisfying the Lipschitz condition 0.1.1

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

$\forall (x,y)\in S, \quad i=1,2.$    Let $a\le x_0\le b$. For the solution $y(x;s)$ of the initial value problem

$$\left\{ \begin{array}{l} y'=f(x,y)\\ y(x_0;s)=s \end{array}\right.$$

Figure 2: Trajectories $y_1$ and $y_2$ emanating from initial data $s_1$ and $s_2$, respectively

there holds the estimate

$$\vert\vert y(x; s_1)-y (x; s_2)\vert\vert\le e^{L\vert x-x_0\vert\vert}\vert s_1-s_2\vert\vert$$

for $a\le x\le b.$

Proof. See Figure 0.1.1

$$y(x;s)= s+\int^{x}_{x_{0}} f(t,y(t,s))dt,$$

$$y(x;s_1)-y(x_i; s_2)=s_1-s_2 + \int^{x}_{x_{0}}[f(t,y(t,s_1)) -f(t,y(t_1,s_2))]dt$$

thus

$$\vert\vert y(x;s_1)-y(x; s_2)\vert\vert\le \vert\vert s_1-s_2\ver... ..._{x_{0}}\vert\vert y(t;s_1)-y(t;s_2)\vert\vert dt\vert}_{\displaystyle \Phi(x)}$$ (4)

then $\Phi'(x)=\vert\vert y(t,s_1)-y(t,s_2)\vert\vert$. Thus by (4), for $x\ge x_0\\ \alpha(x)\le \vert\vert s_1-s_2\vert\vert$ with $\alpha\equiv \Phi'(x)-L\Phi (x)$

Take the initial value problem

$$\left\{ \begin{array}{l} \Phi'(x)=\alpha(x)+L\Phi(x)\\ \Phi(x_0)=0 \end{array}\right.$$

for $x\ge x_0\Rightarrow\Phi(x)=e^{L(x-x_0)}\int^x_{x_0}\alpha(t) e^{-L(t-x_0)}dt.$

Since $\alpha\le \vert\vert s_1-s_2\vert\vert$, then

$$0\le \vert\vert\Phi\vert\vert\le e^{L(x-x_0)}\vert\vert s_1-s_2\vert\vert\int^x_{x_{0}}e^{-L(t-x_0)} dt$$

$$=\frac{1}{L}\vert\vert s_1-s_2\vert\vert[e^{L(x-x_0)}-1]\quad x\ge x_0$$

Since $\alpha=\Phi'-L\Phi\Rightarrow \vert\vert y(x,s_1)-y(x,s_2)\vert\vert=\Phi'(x)=\alpha(x)+ L\Phi(x)\le \vert\vert s_1-s_2\vert\vert e^{L\vert x-x_0\vert}$ $\Box$

The above theorem can be sharpened: under extra continuity, the solution of the IVP actually depends on initial value in a continuously differentiable manner:

Theorem. If, in addition to assumptions in previous theorem and if the Jacobian $D_y f(x,y)\equiv [\partial f_i/\partial y_j]$ exists on $S$, and is continuous and bounded,

$$\vert\vert D_yf(x,y)\vert\vert\le L$$    for$$(x,y)\in S,$$

$\Rightarrow$ the solution $y(x,s)$ of $y'=f(x,y), y(x_0, s)=s$ is continuously differentiable for all $x\in [x_0,b]$ and all $s\in \mathbb{R}^n$

$\Box$

Example)

$$y'=1+\sin(xy)=f(x,y)$$

$$S=\{(x,y)\vert\le x\le 1, -\infty<y<\infty\}$$

$$\frac{\partial f}{\partial y}=x\cos(xy)\Rightarrow L=1$$

$\therefore$ for any $(x_0, y_0)$ with $0<x_0<1\exists$ a $Y(x)$ and associated IVP on some interval $[x_0-\alpha, x_0+\alpha]\in [0,1]$.

Example) $y'=\displaystyle \frac{2x}{a^2}y^2\qquad y(0)=1 \qquad a>0$ const.

$Y(x)=\displaystyle \frac{a^2}{a^2-x^2}\quad -a<x<a$. Solution depends on size of $a$.

To determine $L$ take $\frac{\partial f}{\partial y}(x,y)=\frac{4xy}{a^2}$ $\therefore$ to have $L$ finite on $S$, $S$ must be bounded in $x$ and $y$, say $-c\le x\le c, -b\le y\le b$.

Theorem. (Existence)

$$\mbox {For } \left\{\begin{array}{l} y'=f(x,y)\\ y(x_0)=y_0 \end{array}\right.$$ (5)

if $f$ is continuous in a rectangle $S$ with center at $(x_0, y_0)$, say,

$$S=\left\{(x,y):\vert x-x_0\vert\le \alpha, \vert y-y_0\vert\le \beta\right\}$$

then the initial value problem (5) has a solution $y(x)$ for $\vert x-x_0\vert\le \min (\alpha, \beta/M)$, where $M$ is the maximum of $\vert f(x,y)\vert$ in $S$.

Example)

$$y'=(x+\sin y)^2$$

$$y(0)=3$$

$$-1\le x\le 1$$

$f(x)=(x+\sin y)^2$ and $(x_0, y_0)=(0,3)$

$$S=\{(x,y):\vert x\vert\le \alpha, \vert y-3\vert\le \beta\}$$

If $(x,y)\vert\le (\alpha + 1)^2\equiv M$. Want $\min (\alpha, \beta/M)\ge 1$ so set $\alpha=1$. Then $M=4$ and everything is consistent if $\beta\ge 4$. So theorem asserts that a solution exists on $\vert x\vert\le\min (\alpha, \beta/M)=1$.

$\Box$

A useful theorem: Consider

$$\left\{\begin{array}{l} Y'=T(x)Y\\ Y(a)= I \end{array}\right.$$ (6)

where $T(x)$ is an $n\times n$    matrix

Theorem. If $T(x)$ is continuous on $[a,b]$ and $k(x)\equiv \vert\vert T(x)\vert\vert\Rightarrow$ solution $Y(x)$ of (6) satisfies

$$\vert\vert Y(x)-I\vert\vert\le e^{\int^{x}_{a}k(t) dt}-1\quad x\ge a$$

Proof: exercise. $\Box$