 Solution

For $\mu <0$, put $k=\sqrt{-\mu }$. There are no local equilibrium points and $\dot{x}=x^2+k^2>0$, so every incoming point passes through the local region. The same sections giveand again $P(x)=\nu +cY(x)+O(Y^2)$. The global return supplied by the problem closes the passage into a periodic orbit if this map has a fixed point.

Its derivative is strongly contracting: $Y^{\prime }(x)=\lambda Y(x)/(x^2+k^2)$. To see the smallness uniformly near the codimension-two point, put $d=\sqrt{x^2+k^2}$ and assume $2d<h$. The passage includes $[d,2d]$, on which ${\xi }^2+k^2\le 5d^2$, so $T_{\mathrm{loc}}\ge 1/(5d)$. ConsequentlyFor a sufficiently small entry interval and all sufficiently small $\nu$ of either sign, $P$ maps the interval into itself and has derivative bounded strictly below one. The mean value theorem bounds differences by a factor strictly below one, while the endpoint signs of $P(x)-x$ give existence of a fixed point. ThusHere “all $\nu$” means all return parameters in the neighborhood where the stated local/global construction applies, not arbitrarily distant global return geometries.