I have the following question:

Determine if the function $F(x,y) = xy^{1/3}$ satisfy a Lipschitz condition on the rectangle $\{ (x,y) : |x| \leq h, |y| \leq k \}$ where $h > 0$ and $k > 0$?

If $b>0$ determine the region $|x| < h, |y|<k$ which has the largest value of $h$ in which Picard's theorem can be used to show that the initial-value problem

$y' = xy^{1/3}$, $y(0) = b$

has a unique solution (you may assume Picard's theorem, but should prove that the assumptions are satisfied) and find the solution explicitly.

If $b=0$ show that for any $c > 0$ there is a solution $y$ which is identically zero on $[-c,c]$ and positive when $|x| > c$. Find these solutions explicitly and show that the resulting solutions satisfy the ODE for all values of $x$.

In trying to determine whether or not this satisfies the Lipschitz condition or not I've done the following:

$|F(x,u) - F(x,v)| = |xu^{1/3} - xv^{1/3}| = |x||u^{1/3} - v^{1/3}| \leq h|u^{1/3} - v^{1/3}|$

But get stuck at this point as $h|u^{1/3} - v^{1/3}| \leq h|u-v|$ if and only if $u,v \geq 1$ - so I'm thinking this doesn't satisfy the Lipschitz condition but can't really formulate it.

For the two conditions of Picard's theorem to be satisfied, we must have:

$F(x,y)$ is continuous in $R$, where $R$ is the rectangle: $|x|<h$, $|y-b|<k$ and that $F$ is bounded by $M$ so $|F(x,y)| \leq M$ and $Mh \leq k$

The second condition being that $F$ satisfies a Lipschitz condition in $R$.

Beyond this I'm unsure what to do, thanks.

EDIT: Sorry, to include the version of Picard's theorem I'm using:

The ODE $y' = f(x,y)$ with $y(a) = b$ has a solution in the rectangle $R: |x-a| \leq h, |y-b| \leq k$ provided:

(i) $f$ is continuous in $R$, bounded by $M$ (so $|f(x,y)| \leq M$) and $Mh \leq k$

(ii) $f$ satisfies a Lipschitz condition in $R$.

Furthermore, this solution is unique.

  • $\begingroup$ Could you post the version of Picard's Theorem you're considering? $\endgroup$ – jkn Oct 22 '13 at 18:22
  • $\begingroup$ @jkn Sorry about that. I've edited the main post with the version I have. Thanks. $\endgroup$ – Noble. Oct 22 '13 at 18:27

The function $f(x, y)=xy^{1/3}$ is not Lipschitz on any domain containing $y=0$, for if we assume it is Lipshitz then it must satisfies $|xy^{1/3}|=|f(x, y)-f(x, 0)|\leq L|y|$ for some $L>0$. But $\displaystyle\lim_{y\rightarrow 0^{+}}\frac{xy^{1/3}}{y}=\displaystyle\lim_{y\rightarrow 0^{+}}\frac{x}{y^{2/3}}=\infty$ for $x>0$.


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.