#### Example 8.2.1.

Consider \(x'=-y-x^2\text{,}\) \(y'=-x+y^2\text{.}\) See Figure 8.3 for the phase diagram. Let us find the critical points. These are the points where \(-y-x^2 = 0\) and \(-x+y^2=0\text{.}\) The first equation means \(y = -x^2\text{,}\) and so \(y^2 = x^4\text{.}\) Plugging into the second equation we obtain \(-x+x^4 = 0\text{.}\) Factoring we obtain \(x(1-x^3)=0\text{.}\) Since we are looking only for real solutions we get either \(x=0\) or \(x=1\text{.}\) Solving for the corresponding \(y\) using \(y = -x^2\text{,}\) we get two critical points, one being \((0,0)\) and the other being \((1,-1)\text{.}\) Clearly the critical points are isolated.

Let us compute the Jacobian matrix:

\begin{equation*}
\begin{bmatrix}
-2x & -1 \\
-1 & 2y
\end{bmatrix} .
\end{equation*}

At the point \((0,0)\) we get the matrix \(\left[ \begin{smallmatrix} 0 & -1 \\ -1 & 0 \end{smallmatrix} \right]\) and so the two eigenvalues are \(1\) and \(-1\text{.}\) As the matrix is invertible, the system is almost linear at \((0,0)\text{.}\) As the eigenvalues are real and of opposite signs, we get a saddle point, which is an unstable equilibrium point.

At the point \((1,-1)\) we get the matrix \(\left[ \begin{smallmatrix} -2 & -1 \\ -1 & -2 \end{smallmatrix} \right]\) and computing the eigenvalues we get \(-1\text{,}\) \(-3\text{.}\) The matrix is invertible, and so the system is almost linear at \((1,-1)\text{.}\) As we have real eigenvalues and both negative, the critical point is a sink, and therefore an asymptotically stable equilibrium point. That is, if we start with any point \((x_i,y_i)\) close to \((1,-1)\) as an initial condition and plot a trajectory, it approaches \((1,-1)\text{.}\) In other words,

\begin{equation*}
\lim_{t \to \infty} \bigl( x(t), y(t) \bigr) = (1,-1) .
\end{equation*}

As you can see from the diagram, this behavior is true even for some initial points quite far from \((1,-1)\text{,}\) but it is definitely not true for all initial points.