ODEs: Lipschitz-continuity and an IVP

We have   $y'=f(x,y)$ 

with $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2, \left (\begin{array}{c} y_1 \\ y_2 \end{array} \right ) \mapsto \left (\begin{array}{c} g(y_1) \\ h(y_1) y_2 \end{array} \right ) $,

where $g: \mathbb{R} \rightarrow \mathbb{R}$ is Lipschitz-continuous and $h: \mathbb{R} \rightarrow \mathbb{R}$ is continuous.



Give an example for functions $g$ and $h$ such that $f$ is not Lipschitz-continuous (so the requirements for Picard-Lindelof are not met).



Prove that the IVP    $y(x_0)=y_0$    for    $x_0 \in \mathbb{R}$    and    $y_0 \in \mathbb{R}^2$    has a unique solution on an open interval $J$ with $x_0 \in J$.


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  • I don't think I understand the last step, how does exponentiating and using z^1 = y^1 yield z^2 = y^2 ?

  • Exponentiating yields that z2 = z2_0 * exp( int h(z1(x))). Since z2_0 = y2_0 and z1 = y1, we have z2 = y2_0 exp( int h(z2(x))) = y2.

  • Sorry, that should read z2 = y2_0 exp( int h(y1(x)) ) = y2.

The answer is accepted.
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