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Consider the cubic feedback systemAt the origin its eigenvalues are $0,-2,-1$. With $v=(x+y)/2$ and $w=(x-y)/2$, the center subspace is the $v$ axis and the two remaining directions are strictly stable. The equations becomeThe symmetry $(v,w,z)\mapsto (-v,-w,z)$ allows a center manifold graph with $w(v)$ odd and $z(v)$ even. Write $w=c_3{v}^3+O(v^5)$ and $z=d_2{v}^2+d_4{v}^4+O(v^6)$. The invariance equation for a graph requires $w^{\prime }(v)\dot{v}=\dot{w}$ and $z^{\prime }(v)\dot{v}=\dot{z}$, rather than setting the stable-coordinate derivatives to zero.

The quadratic $z$ equation gives $d_2=1$; the cubic $w$ equation gives $-2c_3+(a+1)/2=0$. At fourth order the $z$ invariance equation is $(a-1)v^4=-d_4{v}^4$, soSubstitution into the $v$ equation yieldsIn particular, the quintic coefficient is $[(3a+1)c_3-d_4]/2$, which records both the cubic stable graph and the quartic lag in $z$. For $a<1$, the leading odd term points toward zero on both sides; for $a>1$ it points outward. At $a=1$ the cubic vanishes but the quintic coefficient is one and still points outward. Since the transverse directions are strictly stable, the centre manifold theorem gives local asymptotic stability exactly when $a<1$.

The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. The Jacobian matrix isAt the origin its eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle equilibrium for every allowed $\mu$. At either nonzero fixed point the characteristic polynomial is ${\lambda }^2+\lambda +2{\mu }^2$, givingThus both are stable nodes for $0<|\mu |<1/(2\sqrt{2})$ and stable focuses for $|\mu |>1/(2\sqrt{2})$. At equality the repeated eigenvalue is $-1/2$ and the matrix is not scalar, so there is only one independent eigenvector: the fixed point is a stable improper node.

The stable manifold of the origin is $x=0$, exactly: there $\dot{y}=-y$. Conversely, a trajectory converging to the origin with $x\ne 0$ would eventually have $\dot{x}/x={\mu }^2-y>{\mu }^2/2$, which precludes convergence of $x$ to zero.

For the local unstable manifold, write $y=h(x)=a{x}^2+b{x}^4+O(x^6)$. Reflection symmetry makes $h$ even, and the invariance equation for a graph requiresComparing quadratic and quartic terms gives $2a{\mu }^2=1-a$ and $4b{\mu }^2-2a^2=-b$. Therefore