Past exam of the differential equations 1 course of the University of Oxford / 2017 / 2 / b / ii / Solution

For example, choose $h=1/4$ and $k=1/2$ around $(a,b)=(0,0)$. On the rectangle, $1+y\ge 1/2$, so $f=-\sqrt{1+y}$ is continuous, $|f|\le \sqrt{3/2}$ and $|\partial f/\partial y|\le 1/\sqrt{2}$. The mean value theorem therefore supplies a Lipschitz constant $L=1/\sqrt{2}$. These choices even satisfy $Mh<k$ and $Lh<1$.

While $y>-1$, separation of variables gives $2\sqrt{1+y}=2-x$. Therefore the uniquely determined branch up to its first arrival at $-1$ isIt reaches $y=-1$ at $x=2$, with derivative zero. The restriction $x\le 2$ is essential: extending the same quadratic beyond two would make its derivative positive and violate the original differential equation. Beyond that time the waiting-time continuation at a non-Lipschitz equilibrium is nonunique.