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:


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?


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 $y\to0$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ – Julián Aguirre Feb 22 '15 at 15:04
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.