ODELC1

From Exampleproblems

Show that $\begin{bmatrix} 1+x_1 \\ x_2^2 \\ \end{bmatrix}$ satisfies a Lipschitz condition when $x$ lies in any bounded domain $D$ (i.e. $| x | < M$ where $M$ is constant), but cannot satisfy a Lipschitz condition for all $x$ .

First find $f(\vec x, t)-f(\vec y,t) = \begin{bmatrix} x_1-y_1 \\ x_2^2-y_2^2 \\ \end{bmatrix}$ and $\vec x-\vec y=\begin{bmatrix} x_1-y_1 \\ x_2-y_2 \\ \end{bmatrix}$

$\frac{\partial f}{\partial x} = \begin{bmatrix} 1 & 0 \\ 0 & 2x_2 \\ \end{bmatrix}, \left|\frac{\partial f}{\partial x}\right|=1+2x_2\,$

If $x$ is in a bounded domain $| x | < M$ then $f$ satisfies a Lipschitz condition with constant $L=\max_{x_2\isin D} (1+2x_2) = 1+2M\,$ . If the domain is not bounded then the max will not be bounded and so $f$ will not be Lipschitz.

Main Page : Ordinary Differential Equations : Lipshitz Conditions

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