ODELC2

From Exampleproblems

Show that $\begin{bmatrix} \sqrt{|x_1|} \\ x_2 \\ \end{bmatrix}$ is continuous for all $x$ , but does not satisfy a Lipschitz condition in any domain $D$ which contains $x = 0$ .

$f(\vec x,t) = \begin{bmatrix} \sqrt{|x_1|} \\ x_2 \\ \end{bmatrix}$

If $x_{1,2}\ne 0$ then $\frac{d}{dx_1} \sqrt{|x_1|} = \frac{1}{2}x_1^{-1/2}\,$ and $\frac{d}{dx_2} x_2 = 1\,$ are defined. If $x_{1,2}=0\,$ then $\frac{d}{dx_1}0=\frac{d}{dx_2}0=0\,$ are defined, so $f$ is continuous.

But $\frac{\partial f}{\partial x} = \begin{bmatrix} \frac{1}{2} x_1^{-1/2} & 0 \\ 0 & 1 \\ \end{bmatrix}, \left|\frac{\partial f}{\partial x} \right| = \frac{1}{2} x_1^{-1/2}+1\,$

is only defined if $x_1\ne 0$ , so $f$ is Lipschitz only if $x = 0$ is not in the domain.

Main Page : Ordinary Differential Equations : Lipshitz Conditions

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