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For $\nu =1,\kappa =0$, the first integral isIts minimum $-1/4$ occurs at the two centres. Levels between $-1/4$ and $0$ give a periodic orbit around each centre equilibrium; positive levels give larger periodic orbits surrounding both wells. The zero level is the figure-eight separatrix through the saddle, with two homoclinic loops.

On $E=0$, $y^2=x^2(1-x^2/2)$. The printed parametrization $x=\pm \sqrt{2}\mathrm{sech}z$, $y=\pm \sqrt{2}\mathrm{sech}z\tanh z$, taking matching signs, satisfies this identity because $1-{\mathrm{sech}}^{2}z={\tanh }^{2}z$. It also has $dx/dz=-y$ and $dy/dz=-(x-x^3)$, so its parameter is physical reversed time, $z=-t+\text{constant}$, rather than $z=t$. It is nonetheless exactly a parametrization of the requested trajectory. In physical forward time one can use $x(t)=\sqrt{2}\mathrm{sech}t$, $y(t)=-\sqrt{2}\mathrm{sech}t\tanh t$ on the right loop.

Approach to the origin requires $z\to \pm \infty$, hence infinite physical time. Equivalently $|y|\sim x$ near the right-loop endpoint, so the elapsed time contains the divergent integral ${\int }_0^{\delta }dx/x$. The point at $x=\sqrt{2}$ is reached in finite time from any interior point; it is the asymptotic saddle endpoints, not that turning point, that cause the infinite total traversal time.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

Let $F$ be a continuously differentiable planar vector field with an isolated equilibrium point $p$, and let the twice continuously differentiable first integral $H$ satisfy $F\cdot \nabla H=0$ and $\nabla H(p)=0$. If the Hessian matrix $D^{2}H(p)$ is positive or negative definite, then $p$ is a centre equilibrium.

For a positive-definite matrix, sufficiently short rays from $p$ have strictly increasing $H$ with radial derivative bounded below by a positive constant times distance. Every sufficiently small positive level difference therefore meets each ray exactly once, and the implicit function theorem makes these intersections a simple closed curve. These regular level sets shrink to $p$; use $-H$ in the negative-definite case. The vector field is tangent and nonzero on each such compact curve. Its speed is bounded away from zero, so a traversal takes finite time and each curve is a periodic orbit. An arbitrary strict local extremum without these regularity and nondegeneracy conditions does not by itself supply this argument.

Multiply the ordinary differential equation by $f^{\prime }$ and integrate. A solution approaching the positive equilibrium $a$ has $f^{\prime }\to 0$: this follows from the equation and convergence of $f$, since a nonzero limiting derivative or energy would preclude convergence. Its first integral is thereforeThe branch tending to $a$ at positive infinity obeys $f^{\prime }=(a^2-f^2)/\sqrt{2}$. Separation yieldsThe constant solution is $A=0$. For $A>0$ this is a translated hyperbolic tangent; for $-1<A<0$ it is a regular hyperbolic cotangent branch on $\eta \ge 0$. The condition $A>-1$ excludes both an interior pole and a divergent value at the finite endpoint. The opposite sign in the first integral moves away from the positive equilibrium as $\eta$ increases, except at the constant solution.

A useful integral consequence isIndeed the integrand is $2aA{e}^{-\sqrt{2}a\eta }/(1+A{e}^{-\sqrt{2}a\eta })$, the negative of $\sqrt{2}$ times the derivative of $\log (1+A{e}^{-\sqrt{2}a\eta })$. An integral deficit can therefore fix the translation parameter uniquely, even when the differential equation and far value leave it free.

In the dilute-vortex description of the two-dimensional XY model, $K={\rho }_s/(k_{B}T)$ is dimensionless stiffness and $y$ is phase vortex fugacity. A phase vortex has logarithmic energy $\pi {\rho }_s\log (L/a)$, while its placement entropy is $2k_B\log (L/a)$. Small phase vortex pairs screen the logarithmic interaction, lowering the stiffness. Near $K=2/\pi$, the rescaled variables $X=2-\pi K$, $Y=4\pi y$ obey $X^{\prime }=Y^2$, $Y^{\prime }=XY$ to leading order. The invariant $X^2-Y^2$ leads to an essential correlation-length singularity $\xi \sim a\exp (C/\sqrt{T-T_c})$, not a finite power-law exponent.

Here $\ell =\log (r/a)$ increases toward larger observation distances, and the vortex fugacity $y$ is a dimensionless weight per unit core, conventionally initialized as $y_0\simeq e^{-E_c/(k_{B}T)}$ with the microscopic measure fixed. Numerical coefficients depend on this normalization. In the displayed convention, differentiating $X=2-\pi K$ using the original flow givesOnly replacing the prefactor $(2-X{)}^2/4$ by one near $X=0$ gives the scaled leading model. Its first integral follows directly:Thus this first integral belongs to the stated scaled model, not exactly to the original $K,y$ equations. For $Y\ge 0$, the critical separatrix is $X=-Y$. The bound side has $X<-Y$ and flows to $Y=0$, $X<0$; the unbinding side ultimately runs to large vortex fugacity, where the dilute approximation ceases to be quantitative.

To derive the unusual correlation length scaling without fitting an exponent, write the invariant on a nearby unbinding trajectory as $X^2-Y^2=-c^2$. Then $X^{\prime }=X^2+c^2$ and the time to cross fixed small limits $-b$ to $b$ is $2\arctan (b/c)/c\sim \pi /c$. If a physical one-parameter path crosses the separatrix transversely, its $c^2$ is proportional to the positive reduced temperature. Since distance grows as $e^{\ell }$, this gives $\xi \sim a\exp (b_0/\sqrt{t})$, with a nonuniversal $b_0>0$. This is the dimensionless-temperature form of the singularity above.