 Solution

Put $y=\dot{x}$. The planar autonomous differential equation isIts equilibria have $y=0$, $x^2=p$. There are none for $p<0$, one at $p=0$, and two $x_{\pm }=\pm \sqrt{p}$ for $p>0$. At an equilibrium $x_{\ast }$ the stability matrix isWriting $s=\sqrt{p}>0$, $x_{+}=s$ is always a saddle equilibrium. At $x_{-}=-s$, the determinant is $2s>0$ and the trace is $s-q$: it is attracting for $q>s$ and repelling for $q<s$.

At $p=0$ and $q\ne 0$, the eigenvalues are $0,-q$; the nonzero quadratic term and parameter unfolding give a saddle-node bifurcation. At $p>0$, the trace crosses zero at $q=s$, with eigenvalues $\pm i\sqrt{2s}$, giving a Hopf bifurcation. It is nondegenerate: around $x=-s$ the equation is $\ddot{\xi }+(q-s)\dot{\xi }+2s\xi -{\xi }^2+\xi \dot{\xi }=0$. At $q=s$, a small oscillatory expansion of leading amplitude $r$ has $⟨\xi {\dot{\xi }}^2⟩=r^4/8+O(r^5)$, so the first nonlinear damping is positive and nonzero. For example, its second-order oscillatory expansion is $\xi =r\cos (\Omega t)+r^2/(2{\Omega }^2)-r^2\cos (2\Omega t)/(6{\Omega }^2)-r^2\sin (2\Omega t)/(6\Omega )$ with ${\Omega }^2=2s$. Thus the small attracting limit cycle is on the $q<s$ side. At $(p,q)=(0,0)$ the matrix has a double zero eigenvalue and a nontrivial Jordan block, the Bogdanov–Takens bifurcation organizing the two curves. The local bifurcation curves areA change between node and focus away from these curves is not itself a local bifurcation.

For the requested global exclusion, suppose $x(t)$ were periodic with period $T$. Integrate the original equation over a period. The integrals of $\ddot{x}$, $q\dot{x}$, and $x\dot{x}=\frac{1}{2}(d/dt)x^2$ vanish. ThereforeThe left side is nonnegative and the right side is negative when $p<0$. This contradiction proves there are no periodic orbits for $p<0$, independently of $q$.