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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.

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  • $\begingroup$ 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
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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 $x>0$.

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