Example 5.2 Consider the system \[ \dot{x} = xy,\quad \dot{y} = -y - x^2. \] The linear normal form (based on the linearisation at the origin) has the constant matrix \[ A = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \mbox{diag} \; (0, -1). \] Then the eigenpairs are \[ \lambda_1 = 0, e_1 = \begin{pmatrix} 1 \cr 0 \end{pmatrix},\qquad \lambda_2 = -1, e_2 = \begin{pmatrix} 0 \cr 1 \end{pmatrix}. \] Since the matrix \(A\) is already in normal form, no coordinate transformation is needed. Now the centre manifold takes the form \begin{equation}\label{eqwc} y = h(x)=a x^2 + b x^3 + c x^4 + O(x^5), \end{equation} There is no constant term, because the centre manifold passes through the origin; there is no linear term, because this manifold should be tangent to \(e_1\) (or equivalently \(E^c\), the centre manifold of the linearised system). We can determine the coefficients by comparing two ways for calculating \(\dot y\). Directly from the \(\dot y \) equation of the system \begin{equation}\label{eqyc} \dot{y} = -y - x^2 =-(a x^2 + b x^3 + c x^4) -x^2+O(x^5). \end{equation} On the other hand, differentiating (\ref{eqwc}) w.r.t \(t\) gives \(\dot{y} = d{h(x)}/{dt} = \dot{x} h'(x)\), i.e., \begin{equation}\label{eqhc} x(ax^2+bx^3+cx^4+\cdots)(2ax+3bx^2+4cx^3+\cdots) = 2ax^4 + \cdots. \end{equation} Equating coefficients of \(x^2\), \(x^3\) and \(x^4\) in (\ref{eqyc}) and (\ref{eqhc}) gives \[ -a-1=0, \quad -b=0, \quad -c=2a^2, \] i.e. \(a=-1\), \(b=0\) and \(c=-2\). Thus the centre manifold is \(y=-x^2-2x^4+O(x^5)\) and the dynamics on the centre manifold is \[ \dot x = xh(x)=-x^3-2x^5+O(x^7) . \] Thus \(\dot x <0\) if \(x>0\) and \(\dot x >0\) if \(x<0\). So the origin is stable and the solutions look like a stable node, but the motion onto the centre manifold in the \(y\)-direction is much faster than the motion on the centre manifold, leading to a phase portrait as shown in Figure 5.4.

Example 5.3 Consider the system \[ \dot{x} = y-x+xy,\qquad \dot{y} = x-y-x^2. \] First we have to convert the linearised system \[ \begin{pmatrix} \dot{x} \cr \dot{y} \end{pmatrix} = A \begin{pmatrix} x \cr y\end{pmatrix},\qquad A = \begin{pmatrix} -1 & 1 \cr 1 & -1 \end{pmatrix} \] into normal form. It is easy to calculate the eigenpairs of \(A\), \[ \lambda_1 = 0,\quad e_1=\begin{pmatrix} 1 \cr 1 \end{pmatrix},\quad \lambda_2 = -2, \quad e_2 = \begin{pmatrix} -1 \cr 1\end{pmatrix}. % U= [e_1, e_2]^{-1} = \begin{pmatrix} % 1 & - 1\cr 1 & 1 \end{pmatrix}^{-1} % =\frac{1}{2}\begin{pmatrix} 1 & 1 \cr -1 & 1\end{pmatrix}. \] Let \[ U=[e_1, e_2]^{-1} = \begin{pmatrix} 1 & - 1\cr 1 & 1 \end{pmatrix}^{-1} =\frac{1}{2}\begin{pmatrix} 1 & 1 \cr -1 & 1\end{pmatrix}, \] the change of variable from \((x,y)\) to \((u,v)\) (in normal form) is \[ \begin{pmatrix} u \cr v \end{pmatrix}=U \begin{pmatrix} x \cr y \end{pmatrix} = \begin{pmatrix} (x+y)/2 \cr (y-x)/2 \end{pmatrix},\qquad \mbox{or} \qquad \begin{pmatrix} x \cr y \end{pmatrix} = U^{-1} \begin{pmatrix} u \cr v \end{pmatrix} = [e_1,e_2]\begin{pmatrix} u \cr v \end{pmatrix}= \begin{pmatrix} u-v \cr u+v \end{pmatrix}. \] The new system in \((u,v)\) is \[ \dot{u} = uv-v^2, \dot{v} = -2v+uv-u^2. \] The centre manifold is parametrised by \(v=h(u) = au^2+bu^3+cu^4+\cdots\), then \begin{align*} \dot{v} &= (2au+3bu^2+4cu^3+\cdots)\dot{u} \cr & =(2au+3bu^2+4cu^3+\cdots)\big( u(au^2+bu^3+cu^4+\cdots)-(au^2+bu^3+cu^4+\cdots)^2\big) \cr & = 2a^2u^4+\cdots \end{align*} and on the other hand \begin{align*} \dot{v} &= -2v+uv-u^2\cr &= -2(au^2+bu^3+cu^4+\cdots)+u(au^2+bu^3+cu^4+\cdots)-u^2\cr &=(2a+1)u^2+(2b-a)u^3+(3c-b)u^4+\cdots . \end{align*} Comparing the coefficients of \(u^2, u^3\) and \(u^4\) of the two expressions of \(\dot{v}\), we get \[ a = -\frac{1}{2},\quad b=-\frac{1}{4},\quad c=-\frac{3}{8},\quad \mbox{ or }\quad v = -\frac{1}{2}u^2-\frac{1}{4}u^3-\frac{3}{8}u^4+\cdots. \] The dynamics on the centre manifold is \[ \dot{u} = uv-v^2 = -\frac{1}{2}u^3-\frac{1}{4}u^4-\frac{3}{8}u^5+\cdots, \] which is stable if \(u\) is small. Going back to the original coordinates, the centre manifold is approximately \[ y-x = -\frac{1}{4}(x+y)^2-\frac{1}{16}(x+y)^3-\frac{3}{64}(x+y)^4+\cdots \]