i) Determine if the function $F(x,y) = xy^{(1/3)}$ satisfies a Lipschitz condition on the rectangle ${(x,y): mod(x) \leq h , mod(y) \leq k}$ where $h>0, k>0$.
ii) If $b>0$ determine the region $\mod(x) < h, \mod(y) < k$ 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).
iii) Find the solution explicitly and sketch this solution for various values of $b$.
iv) If $b=0$ show that for any $c>0$ there is a solution y which is identically zero on $[-c,c]$ and positive when $\mod(x) > c$. Find these solutions explicitly and show that the resulting solutions satisfy the ODE for all values of $x$.
I've done the first part and shown that the function isn't Lipschitz but then now that I've got that I don't understand how to do the rest of the question.
Best Answer
An answer from a previous post: The function $f(x, y)=xy^{1/3}$ is not Lipschitz on any rectangle 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 $x>0$.