= 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 give
$$
T_{\rm loc}(x)=\frac1k\left[\arctan\frac hk-\arctan\frac xk\right],\qquad
Y(x)=h\exp\left\{-\frac\lambda k\left[\arctan\frac hk-\arctan\frac xk\right]\right\},
$$
and 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: $Y'(x)=\lambda Y(x)/(x^2+k^2)$. 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 $\xi^2+k^2\le5d^2$, so $T_{\rm loc}\ge1/(5d)$. Consequently
$$
|P'(x)|\le C\frac{e^{-\lambda/(5d)}}{d^2}\longrightarrow0\qquad(d\to0).
$$
For 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>. Thus
$$
\boxed{\text{a unique nearby attracting periodic orbit exists for }\mu<0\text{, for either sign of }\nu.}
$$
Here “all $\nu$” means all return parameters in the neighborhood where the stated local/global construction applies, not arbitrarily distant global return geometries.
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