 Solution

A persistent saddle loop must return with the same energy. The Melnikov energy-balance method therefore requires, at first order,For the normalized orbit of part (c), put $z=\tanh (\tau /\sqrt{2})$. Since $u_0=-2+3z^2$, $v_0^2=18(1-z^2{)}^2{z}^2$ and $d\tau =\sqrt{2}dz/(1-z^2)$,For $s=\sqrt{\alpha }$ the general orbit has $I_0(s)=s^{5/2}{I}_0$ and $I_1(s)=s^{7/2}{I}_1$. The energy mismatch is therefore proportional to $I_0(s)[\beta -(5/7)s]$, whose derivative with respect to $\beta$ is nonzero. This simple zero selects the saddle-loop curve:The remainder can be sharpened. Reversal of $\tau$ and $v$ interchanges $\epsilon$ and $-\epsilon$ without changing the saddle-loop condition. Smoothness of the homoclinic splitting thus makes the selected $\beta$ an even function of $\epsilon$, givingThe leading coefficient alone is enough to separate it from the Hopf curve ${\mu }_H=\sqrt{\lambda }$. The integral hint in the PDF equals $2/(k+1)$ for nonnegative even integers $k$; for odd $k$ its integrand is odd and the integral is zero. Only even powers enter the calculation above.

Near the double-zero point, and with the stipulated absence of further bifurcations, the distinct open regions are:

On the homoclinic curve the cycle becomes a saddle loop; its positive saddle trace $\mu +\sqrt{\lambda }$ makes the loop repelling on the cycle side. At the Hopf curve the unstable cycle shrinks into the negative equilibrium point. The node–focus discriminant curves merely change the spiral appearance and are not additional bifurcations. The impossibility of a periodic orbit for $\lambda <0$ can also be seen without a picture: at a local minimum of periodic $x(t)$ one would have $y=0$ and $\ddot{x}=-\lambda +x^2>0$, but at its maximum the same strictly positive expression would be required to be nonpositive.