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For a weak perturbation of a planar Hamiltonian system, integrate the first-order orbital derivative of the Hamiltonian along an unperturbed homoclinic orbit. This measures the first-order separation of stable and unstable manifolds on a transverse section. A simple zero as a parameter varies permits a nearby homoclinic connection by the implicit function theorem. The resulting leading balance is not a claim that the unperturbed separatrix remains an exact orbit after perturbation.

With the stated time rescaling, $\dot{x}={\epsilon }^3{u}^{\prime }$ and $\ddot{x}={\epsilon }^4{u}^{\prime \prime }$, where primes refer to $\tilde{t}$. Dividing by ${\epsilon }^4$ givesAt $\epsilon =0$, put $v=u^{\prime }$ and define the HamiltonianIts orbital derivative is $H^{\prime }=v(u^2-\kappa )+(\kappa -u^2)v=0$, so this is a conserved quantity. The Hamiltonian system has $u^{\prime }=H_v=v$, $v^{\prime }=-H_u=u^2-\kappa$.

For $\kappa >0$, set $s=\sqrt{\kappa }$. The potential $U(u)=\kappa u-u^3/3$ has a minimum at $-s$ and a maximum at $s$. The corresponding center equilibrium and saddle equilibrium have energies $H_c=-2s^3/3$ and $H_s=2s^3/3$. Levels $H_c<H<H_s$ contain closed ovals around $(-s,0)$, as well as separate unbounded components on the right. At $H=H_s$, the center's ovals limit to a homoclinic orbit to the saddle, turning at $u=-2s$. Its equation isThe same energy also has external unbounded saddle separatrices for $u>s$. Levels above $H_s$ pass over the potential barrier and are open. Along the upper halves $u^{\prime }=v>0$, so the closed loops are traversed clockwise. If $\kappa =0$, the two critical points coalesce into a degenerate equilibrium; if $\kappa <0$, the potential is strictly decreasing and there is no center or homoclinic loop.

Write $z=x+y$. When ${\mu }_1={\mu }_2=0$ the system gives $\dot{z}=y-z^3$ and $\dot{y}=-z^3$. Choosingmakes $V$ positive definite and radially unbounded, while its orbital derivative isThe set $\{\dot{V}=0\}$ is $z=0$. A trajectory remaining there must also have $\dot{z}=y=0$, so its largest invariant subset is the origin. The LaSalle invariance principle therefore proves that the origin is globally asymptotically stable. In the phase plane, $\dot{x}=0$ on $y=0$ and $\dot{y}=0$ on $x+y=0$; every nonstationary trajectory crosses the nested level curves of $V$ inward and tends to the origin.

At an equilibrium point, $y=0$ andHence the origin always exists, while $x=\pm \sqrt{{\mu }_1}$ exist for ${\mu }_1>0$. At $(x_{\ast },0)$ the Jacobian matrix isA stationary bifurcation occurs when a real eigenvalue passes through zero, hence onA Hopf bifurcation requires zero trace and positive determinant. At the origin this givesand on either nonzero branch it givesThus the Hopf locus consists of the negative ${\mu }_1$-axis and the ray ${\mu }_2=3{\mu }_1$ in the first quadrant; the stationary locus is the ${\mu }_2$-axis.

Now set ${\mu }_2=-1$, write $\mu ={\mu }_1$, and append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the extended centre manifold for a parameter is tangent to the $(x,\mu )$-plane, so put $y=h(x,\mu )$. Its centre-manifold invariance equation isUnder the ordering $\mu =O(x^2)$, solving through cubic order givesThis is the pitchfork bifurcation normal form: $x=0$ is stable for $\mu <0$ and unstable for $\mu >0$, while the two stable branches $x=\pm \sqrt{\mu }$ emerge for $\mu >0$. The bifurcation is therefore a supercritical pitchfork bifurcation.

Keep the same function $V$. For arbitrary $a,b$ its orbital derivative isLet $\gamma$ be a nonconstant orbit of the unperturbed system with period $T$. Sinceintegration over one period givesConsequently,If a perturbed periodic orbit persisted and converged to $\gamma$ as $(a,b)\to (0,0)$, its net change in $V$ over one period would vanish. The averaged first-integral obstruction to persistence of a periodic orbit would therefore requireWhen $ab<0$, the two leading terms have the same nonzero sign and their magnitude is bounded below by $r\min (A,B)(|a|+|b|)$. Thus no unperturbed periodic orbit persists for sufficiently small $a,b$ with $ab<0$.

Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

Forthe orbital derivative isChoosing $g^{\prime }(y)=\sin 2y$ cancels the mixed term. Taking $g(0)=0$ givessoThis function has positive definiteness near the origin, andFor $|x|<1$ and $|y|<1$, both terms are nonpositive and equality holds only at $(0,0)$. Thus $V$ is a strict Lyapunov function near the origin, and the Second Lyapunov theorem proves that the origin is asymptotically stable.