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The local Lipschitz criterion is the following:

Let $c>0$ and $f \in C([a,b] \times [y_0-c, y_0+c])$.

If $f$ satisfies in $[a,b] \times [y_0-c,y_0+c]$ the Lipschitz criterion as for $y$, uniformly as for $t$,

$$\exists L \geq 0: \forall t \in [a,b] \ \forall y_1, y_2 \in [y_0-c,y_0+c]:$$

$$|f(t,y_1)-f(t,y_2)| \leq L|y_1-y_2|$$

then the ODE $(1) \left\{\begin{matrix} y'(t)=f(t,y(t))\\ y(a)=y_0 \end{matrix}\right.$ is solved uniquely, at least at the interval $[a,b']$

where $A=\max_{a \leq t \leq b , y_0-c \leq y \leq y_0+c} |f(t,y)| \ $ and

$b'=\min \{ b, a+ \frac{c}{A}\}$.

Remark: The continuity of $f, f \in C([a,b] \times \mathbb{R})$ suffices to ensure the existence of a solution of the ODE $(1)$ at an interval $[a,c], c>a$. But, it doesn't ensure us the uniqueness.

For example, $f(y)=\sqrt{|y|}$ doesn't satisfy the local criterion of Lipschitz as for $y$ at none interval that contains $0$.

I tried to show the latter as follows:

$$\frac{|f(t,y_1)-f(t,y_2)|}{|y_1-y_2|}=\frac{|\sqrt{|y_1|}-\sqrt{|y_2|}|}{|y_1-y_2|}=\frac{|\sqrt{|y_1|}-\sqrt{|y_2|}|}{|\sqrt{|y_1|}-\sqrt{|y_2|}||\sqrt{|y_1|}+\sqrt{|y_2|}|}=\frac{1}{|\sqrt{|y_1|}+\sqrt{|y_2|}|}=\frac{1}{\sqrt{|y_1|}+\sqrt{|y_2|}}$$

Is it right so far? And how do we justify that $f$ doesn't satisfy the local condition of Lipschitz as for $y$ at none of the intervals that contains $0$.

Does it hold because of the fact that it can be that $y_1=y_2=0$?

Also how can we find the intervals at which the local Lipschitz condition is satified?

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First question

Since $y_1$ and $y_2$ are as close to $0$ as you want, it is impossible to have $1/(\sqrt{|y_1|}+\sqrt{|y_2|})$ bounded by a constant.

Second question

Most of the times, the easiest way to show that $f(x,y)$ satisfies a Lipschitz condition is to show that $\partial f/\partial y$ is bounded. In this case (for $y>0$) $f'(y)=1/(2\,\sqrt{y})$, which is not bounded as $y\to0$.

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  • $\begingroup$ For the first part: It could also be that the interval we are looking at, that contains also $0$, is big... right? In this case, do we consider $y_1, y_2$ that are close to $0$? $$$$ So could we say that $f'(y)=\frac{1}{2 \sqrt{y}} \to +\infty$ as $y \to 0$ and so the local condition of Lipschitz isn't satified for an interval, if it contains $0$? $$$$ Also in order to find intervals for which the condition is satified, do we have to find the $y$ fo which there is a $M \in \mathbb{R}$ such that $\frac{1}{2 \sqrt{y}} \leq M$ ? $\endgroup$ – evinda Feb 20 '15 at 18:35
  • $\begingroup$ $\sqrt{|y|}$ is Lipschitz on any interval $[a,\infty)$ with $a>0$. Problems arise around $y=0$ because its derivative converges to $\infty$ as $y\to0$. $\endgroup$ – Julián Aguirre Feb 20 '15 at 18:40
  • $\begingroup$ So could we say that since $f'(y)=\frac{1}{2 \sqrt{|y|}} \to +\infty$ as $y \to +\infty$, the function $\sqrt{|y|}$ isn't Lipschitz on any interval that contains $0$? $$$$ Also since $y$ is in an absolute value, couldn't we say that $\sqrt{|y|}$ is also Lipschitz on any interval $(-\infty,-a]$ with $a>0$? Or am I wrong? $\endgroup$ – evinda Feb 20 '15 at 18:45
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    $\begingroup$ Yes, that is correct. $\endgroup$ – Julián Aguirre Feb 22 '15 at 15:04
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    $\begingroup$ That is irrelevant. Peano's theorem guarantees existence of solution of $y'=f(x,y)$ if $f$ is continuous. $\endgroup$ – Julián Aguirre Feb 22 '15 at 21:27

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