Classification of fixed points

Here, we concentrate on sketching the vector solutions, say $\boldsymbol{x}=(x(t) \quad y(t))^{\top}$ in the $x-y$ plane or the phase plane. The solution vector given by $\boldsymbol{x}(t)$ describes a motion in the phase plane as the independent variable (here, this is $t$ ) varies. Our objective is, given the solution to a linear system of differential equations, to draw a collection of trajectories in the $x-y$ plane such that we have enough to help us understand the behaviour of the system solution as time varies (both in backward, i.e. $t \rightarrow-\infty$ and forward time, i.e. $t \rightarrow \infty)$. A collection of enough trajectories is known as the phase portrait.

We look for fixed points at

or equivalently,

Notes:

• Fixed points are key to the qualitative behaviour of systems solutions.
• We examine what happens near fixed points (i.e. do the trajectories move away or towards the fixed point(s)?)
• 2D linear systems with constant coefficients always have a fixed point at the origin, $(0,0)$.
• We have the following sketching principle: trajectories in the phase plane never touch, merge or cross due to the existence and uniqueness of solutions. We assume that $f$ and $g$ are sufficiently smooth so that solutions exist. We therefore demand that $\boldsymbol{f}=(f, g)$ be continuously differentiable such that solutions $\boldsymbol{x}$ exist and are unique for any initial condition. This is not the strongest result for existence and uniqueness but it is sufficient for the purposes of this course. Now, given uniqueness, we know that no two trajectories can cross. This is because at a particular point $(x, y)$, the system gives a unique tangent vector given by $\boldsymbol{f}(\boldsymbol{x})$ which would be incompatible with having two solution curves go through a point. For the same reason, we cannot have two trajectories merging. That said, we can have more than one trajectory approach the same fixed point.

The phase portrait is determined by the eigenvalues and the eigenvectors of the determinant of $A$ in Eq. (14.18). For the 2D system, we solve Eq. (14.33) for the eigenvalues. We classify the various types of fixed points according to the relationship between the trace, $\tau$ and the determinant, $\Delta$ of the square matrix $A$ and we plot these on a trace-determinant plane (see Fig. 14.16).

We start off by identifying the different classes of fixed points. We have:

1. Attracting/repelling points

(a) Nodes

(b) Spirals

2. Saddles
3. Centres
4. Stars/degenerate nodes
5. Nonisolated fixed points

Next, we consider each one of the above cases separately and sketch the corresponding typical phase portrait. Firstly, let us write down an equation for $\lambda_{1,2}$. From Eq. (14.33) we have,

Attracting/repelling points

Nodes

These correspond to $\Delta>0$ and $\tau^{2}-4 \Delta>0$ such that Eq. (14.77) gives two real and distinct eigenvalues of the same sign. Further, if $\tau>0$ (recall that $\tau=\lambda_{1}+\lambda_{2}$ ) then $\lambda_{1}>\lambda_{2}>0$ and if $\tau<0$ then $\lambda_{1}<\lambda_{2}<0$. In the case of $\tau>0$ the fixed point is said to be an unstable node (or nodal source) and if $\tau<0$, the fixed point is a stable node (or nodal sink). We now look at the case of $\tau<0$ in more detail and use the sketching principles outlined for the rest of this section. For this case, both eigenvalues are negative and hence the exponentials in the general solution given by (14.45) represent decaying functions. Hence, as $t$ increases, the vector solution, $\boldsymbol{x}$ approaches the origin.

Example:

Consider the following linear system,

The eigenvalues are $\lambda_{1}=-3$ and $\lambda_{2}=-1$ and the general solution is,

If we write down Eq. (14.79) as two equations, one for $x$ and another for $y$ [recall that $\left.\boldsymbol{x}=\left(\begin{array}{ll}x & y\end{array}\right)^{\top}\right]$, we have:

What Eqs. (14.80) and (14.81) represent is a set of parametric equations (parametrised by $t$ ) and our objective is to plot these in the $x-y$ plane. The way we approach plotting the system given by (14.79) [or the equivalent equations given by Eqs. (14.80) and (14.81)] is described next.

Plotting method: We first look at the behaviour of the solution when each one of the constants i.e. $c_{1}$ or $c_{2}$ (not both) are set to zero. These are called the half-line, straight, or exponential solutions. Let us start by setting $c_{2}=0$ and assigning $c_{1}$ a number, e.g. let $c_{1}=1$. Then, the solution is,

This looks more manageable to plot now. We choose a value for $t$ (usually $t=0$ is the best choice). This value will give us a point in the $x$ - $y$ plane. For example, by setting $t=0$, $e^{-t}=1$ and we have the point $(1,-1)$. This isn't a trajectory yet, it's just a point. We obtain a trajectory by moving forward and backward in time. Let us move forward first since this makes more sense (physically). Note that $c_{2}$ is still equal to 0 and $c_{1}=1$. As $t \rightarrow \infty, e^{-t}$ becomes smaller and smaller. At a later time, say $t \approx 0.70, e^{-t} \approx 0.5$; then, since $e^{-t}$ multiplies the vector $\boldsymbol{v}_{\mathbf{1}}=\left(\begin{array}{ll}1 & -1\end{array}\right)^{\top}$, the point now becomes approximately equal to $0.5\left(\begin{array}{ll}1 & -1\end{array}\right)^{\top}=\left(\begin{array}{ll}0.5 & -0.5\end{array}\right)^{\top}$ given by $(0.5,-0.5)$ in the $x-y$ plane. This behaviour is shown in Fig. 14.1(a): the red point is what Eq. (14.82) gives if $t=0$ and the blue point is what Eq. (14.82) gives for $t \approx 0.70$. These points lie on the same line (this is given by the vector $\boldsymbol{v}_{\mathbf{1}}$ ). This tells us that, as $t$ increases, the points move along the same straight line towards the origin. We can therefore obtain a lot of information from the eigenvalues and eigenvectors of the matrix $A$ :

(a)

(c)

(b)

(d)

Figure 14.1: (a) Behaviour of individual points as $t \rightarrow \infty$ corresponding to $c_{1}=1$ and $c_{2}=0$ in Eq. (14.79) as $t$ increases. (b) Trajectory corresponding to $c_{1}=1$ and $c_{2}=0$ as $t$ increases. (c) Trajectories corresponding to $c_{1}= \pm 1$ and $c_{2}=0$ as $t$ increases. (d) Trajectories corresponding to $c_{1}= \pm 1$ and $c_{2}=0$ as $t$ increases (brown lines) and trajectories corresponding to $c_{2}= \pm 1$ and $c_{1}=0$ as $t$ increases (purple lines)

• Each eigenvalue gives the direction of motion. If the eigenvalue is negative, the motion is inwards towards the origin. Conversely, if the eigenvalue is positive, the motion is outwards away from the origin.
• The corresponding eigenvector gives the line on which the points travel. For the $c_{2}=0$ and $c_{1}=1$ case, the trajectory is represented by a straight line going through the blue and red points. This is shown in Fig. 14.1(b).

We now apply the same ideas using $c_{1}=-1$ and $c_{2}=0$ in Eq. (14.79). The result is a mirror image of the trajectory we plotted in Fig. 14.1(b); this is shown in Fig. 14.1(c). Now, for the case of real and distinct eigenvalues we can get an additional two half lines. These correspond to setting $c_{1}=0$ and $c_{2}= \pm 1$. Doing this and following the same method as above, we find another two straight line solutions that correspond to

these solutions approach the origin as well (this is because the second eigenvalue is also negative, $\left.\lambda_{2}=-1\right)$. Their trajectories are shown in Fig. 14.1(d).

What about the rest of the solutions i.e. when $c_{1}$ and $c_{2}$ are both nonzero? They are represented by a combination of the two solutions, $\boldsymbol{x}_{\mathbf{1}}$ and $\boldsymbol{x}_{\mathbf{2}}$. These trajectories are shown in blue in Fig. 14.2 (the direction field is also depicted). The plot in Fig. 14.2 was generated by numerically solving the ODEs. However, usually when plotting the phase

Figure 14.2: Phase portrait for $\lambda_{1}<\lambda_{2}<0$ : asymptotically stable sink node.

portrait, we simply want a sketch of the phase portrait without the need of numerical simulations. To sketch the solutions, first plot the exponential solutions as accurately as possible. To plot the rest of the solutions we proceed as follows. Say we start at a point far away from the origin and closer to the brown line rather than the purple. Then, the trajectory will look almost parallel to the brown line. As $t$ increases, the blue trajectory follows the direction of the straight line solution (i.e. towards the origin) and, as it gets closer and closer to the purple line, it turns around to follow the direction of the purple line. Following the same idea for a few other points far away from the origin, it is possible to obtain the phase portrait as shown in Fig. 14.2. For the case where both eigenvalues are negative, all trajectories in this case approach the fixed point.

For the case of two negative, real and distinct eigenvalues, i.e. $\lambda_{1}<\lambda_{2}<0$, the fixed point at the origin is known as an asymptotically stable sink node.

$\lambda_{1}>\lambda_{2}>0$

For two positive, real, and distinct eigenvalues as $t \rightarrow \infty$, the vector solution approaches $\infty$ as well. So, now, all the trajectories need to be drawn moving away from the origin. The phase portrait of the general solution therefore, looks like the one given in Fig. 14.2 with the exception that the arrows are reversed and pointing away from $(0,0)$. Of course the eigenvectors corresponding to the eigenvalues of matrix $A$ dictate the line which the exponential solutions (half lines) follow but the general qualitative behaviour is similar. A phase portrait of this behaviour is shown in Fig. 14.3; the matrix used to generate the plot is given by $A=\left(\begin{array}{cc}5 & -1 \\ 3 & 1\end{array}\right)$.

Figure 14.3: Phase portrait for $\lambda_{1}>\lambda_{2}>0$ : unstable source node.

For the case of two positive, real, and distinct eigenvalues, i.e. $\lambda_{1}>\lambda_{2}>0$, the fixed point at the origin is known as an unstable source node.

Note that the eigenvalue which has the largest absolute value represents the fast direction while the other one is the slow direction. The trajectories far away from the fixed point are almost parallel to the fast direction and those that are close to the fixed point are tangent to the slow direction.

Spirals

Complex eigenvalues take the following form:

The different sub-cases we consider here depend on the real part of the complex eigenvalue, namely, the value of $\mu$. The value of $\mu$ may be negative or positive $(\mu=0$ is considered in a separate class of fixed points).

The general solution vector to the linear system whose matrix $A$ has complex eigenvalues is given by the product of $e^{\mu t}$ [where $\mu$ is the real part of the eigenvalues in (14.84)] and a sinusoidal function (sines and cosines) with argument $\omega t$. The sinusoidal function comes from having complex exponents (for which, we use Euler's formula). Since, the sinusoidal function alone simply oscillates for all time, $t$, then, the only reason a vector solution will approach the fixed point at the origin is if the real exponential (associated with $\mu$, i.e. $e^{\mu t}$ ) decays as time increases. If the real exponential grows as time increases, the trajectory of the vector solution will move further and further away from the origin. $\boldsymbol{\mu}<\mathbf{0}$

We now consider a linear system with complex eigenvalues with the real part of the eigenvalue being negative. This represents $\mu<0$ in (14.84).

Example:

Consider the following linear system,

The eigenvalues are $\lambda=-1 \pm 2 i$. The real part gives us a decaying function, $e^{-t}$ and the imaginary part gives a sine/cosine function. The latter are periodic and bounded (with period $2 \pi$ ). The decaying function makes a trajectory starting at any point away from the origin to spiral steadily inwards and towards the origin. A phase portrait exhibiting this behaviour is shown in Fig. 14.4. Note that the spiral can either approach the fixed

Figure 14.4: Phase portrait for $\lambda=\mu \pm \omega i$ with $\mu<0$ : asymptotically stable spiral sink.

point clockwise or anti-clockwise. To determine the direction of rotation, we compute the vector field $\dot{\boldsymbol{x}}$ at a particular point in time. This will indicate the direction at that phase point on the phase plane. A shortcut way of determining the direction of rotation is by calculating the difference of the off-diagonal of the matrix $A$. The off-diagonal entries in matrix $A$ are shown in red below,

Hence, the difference of the off-diagonal is given by $b-c$.

• If $b-c>0$, the direction of rotation is clockwise; - If $b-c<0$, the direction of rotation is anti-clockwise.

In this case, the difference of the off-diagonal is -6 which gives a anti-clockwise rotation as shown in Fig. 14.4.

For the case of complex eigenvalues i.e. $\lambda=\mu \pm \omega i$ with $\mu<0$, the behaviour is an asymptotically stable spiral sink, which spirals steadily towards the origin.

$\mu>0$

Now, if the real part of the eigenvalue is positive, i.e. $\mu>0$ in Eq. (14.84), we have a similar behaviour to the phase portrait shown in Fig. 14.4 but, since the exponential function is growing with increasing time, we expect the trajectories to move away from the origin and approach $\infty$ as $t \rightarrow \infty$. A phase portrait showing this behaviour is shown in Fig. 14.5. The trajectory is unstable as it spirals steadily away from the origin.

Figure 14.5: Phase portrait for $\lambda=\mu \pm \omega i$ with $\mu>0$ : unstable spiral source.

For the case of complex eigenvalues i.e. $\lambda=\mu \pm \omega i$ with $\mu>0$, the behaviour is an unstable spiral source, which spirals steadily away from the origin.

Saddles

Next we discuss the case where $\Delta<0$; Eq. (14.77) gives $\lambda_{1}>0$ and $\lambda_{2}<0$ (both real, opposite sign). Note that nodes occur when the eigenvalues are real but they are either both positive or both negative. In such a case, as $t \rightarrow \infty$ the solution associated with the negative eigenvalue approaches the origin while the solution associated with the positive eigenvalue moves away from the origin.

Example:

Consider the following linear system,

The eigenvalues are $\lambda_{1}=3$ and $\lambda_{2}=-1$ and the general solution is,

Just as above, by setting $c_{1}=0$ and keeping $c_{2}$ nonzero, we obtain the solution that is associated only with $\left(\begin{array}{ll}1 & 1\end{array}\right)^{\top} e^{-t}$. This tells us that the straight line solutions associated with this solution approach the origin because the eigenvalue is negative (this is shown in Fig. 14.6 with blue arrows). Similarly, by setting $c_{2}=0$ and $\overline{\text { keeping }} c_{1}$ nonzero, we

Figure 14.6: Exponential (half line) solutions corresponding to $c_{1}=0$ (blue arrows) and $c_{2}=0$ (red arrows).

obtain the solution associated only with $\left(\begin{array}{ll}1 & 5\end{array}\right)^{\top} e^{3 t}$. Since the eigenvalue is positive, we expect the trajectories to move away from the origin with increasing time (this is shown in Fig. 14.6 with red arrows). Note that the straight lines associated with $c_{1}$ nonzero (and $\left.c_{2}=0\right)$ have a steeper slope because of the eigenvector $\left[\left(\begin{array}{ll}1 & 5\end{array}\right)^{\top}\right.$ is steeper than $\left.\left(\begin{array}{ll}1 & 1\end{array}\right)^{\top}\right]$. What we have therefore is a stable trajectory and an unstable trajectory.

To complete the phase portrait, we sketch the rest of the solutions by starting at a point which is closer to the stable half line solutions [e.g. start at the point $(-10,-9.5)]$. As time increases that point follows the trajectory that is closest to it (i.e. the stable one) and hence moves towards the origin. As it approaches the origin however, it also approaches the unstable half line which directs it away from the origin. This is shown in Fig. 14.7 with a red curve. Again, following the same ideas, we can sketch several more trajectories thus

Figure 14.7: Exponential (half line) solutions corresponding to $c_{1}=0$ (blue arrows) and $c_{2}=0$ (red arrows) and a trajectory (red curve) passing through the point $(-10,-9.5)$.

getting a complete picture of the qualitative behaviour of the solution given by (14.88). This is shown in Fig. 14.8.

For the case of one positive and one negative, real eigenvalue, i.e. $\lambda_{1}<0<\lambda_{2}$, the fixed point at the origin is known as a saddle.

Centres

We have already seen that for $\tau^{2}-4 \Delta<0$ (with $\Delta>0$ ), the eigenvalues are complex $(\lambda=\mu \pm \omega i)$ and the trajectories are represented by stable or unstable spirals (spiral sinks and sources, respectively). We show here what the trajectories look like if the real part, i.e. $\mu$ is 0 . If $\mu=0$ then, points on the $x-y$ plane rotate within paths which surround the origin but the points never approach or move away from $\mathbf{0}$, for all time $t$.

For example, consider the following linear system,

The eigenvalues are $\lambda= \pm i$. We can find two linearly independent solutions corresponding to one eigenvalue, say $\lambda_{1}=i$, by separating the 'trial' solution into real and imaginary

Figure 14.8: Phase portrait for $\lambda_{1}<0<\lambda_{2}$ : saddle.

parts (see Subsec. 14.2.1 for details). We find that the general solution to (14.89) is,

Suppose that for given $c_{1}$ and $c_{2}$ and at $t=0$, the solution (14.90) gives us the point $(1,2)$ on the $x$ - $y$ plane (shown by a black, circle marker in Fig. 14.9). Since the period of the functions is $2 \pi$, at $t=2 \pi$ the solution will give the same value, i.e. the point $(1,2)$. This means that the marker moves in some given trajectory and at every $2 \pi$ intervals, it returns to its original position. This motion can only be described by a rotating path. This is shown in Fig. 14.9: the trajectory passing through the point $(1,2)$ is described by an ellipse. The ellipse (note the circle is a special case of the ellipse) is the trajectory we are looking for when the eigenvalues are purely imaginary. We can of course draw several ellipses around and/or inside the one in Fig. 14.9 to show more trajectories corresponding to arbitrary points in the $x-y$ plane. The only thing left to determine however is whether the direction of rotation will be clockwise or anti-clockwise. For our particular example here given by the system in Eq. (14.89), the off-diagonal difference is 3 hence the direction of rotation is clockwise. A phase portrait and the corresponding direction field is shown in Fig. 14.10. As far as the stability is concerned, this behaviour is (neutrally) stable. Note that we do not refer to the fixed point as asymptotically stable because the points that move within the ellipses never approach the origin: the trajectory stays within the ellipse for all time $t$.

For the case of purely imaginary eigenvalues i.e. $\lambda= \pm \omega i$, the behaviour is a stable centre, centred at the origin.

Figure 14.9: Trajectory passing through the point $(1,2)$ corresponding to the solution given by (14.90).

Figure 14.10: Phase portrait for $\lambda= \pm \omega i$ (i.e. purely imaginary eigenvalues): neutrally stable centre.

Star/degenerate cases

For the linear system (14.18) whose matrix $A$ gives the same eigenvalue twice, in Subsec. 14.2.4, we considered two cases:

(i) complete eigenvalues (when $A=\lambda I$ );

(ii) defective eigenvalues (when $A \neq \lambda I$ ). For each one of the two cases above we will consider $\lambda<0$ and $\lambda>0$. Recall that we only have one eigenvalue in these cases and it is real.

Complete eigenvalues

Let us think back to how we got the general solution for complete eigenvalues. We were able to identify that any vector is an eigenvector and, hence, easily find two linearly independent eigenvectors, namely,

The general solution in such a case therefore takes the following form,

Since any vector is an eigenvector and each eigenvector is associated with the same eigenvalue, we can sketch the phase portrait of the solution (14.92) by plotting various straight lines in the phase plane. The stability of the behaviour depends on the nature of $\lambda$.

For $\lambda<0$, all the trajectories decay to the origin. This is shown in Fig. 14.11: all the trajectories are straight line solutions and the equilibrium point is asymptotically stable.

For the case of repeated, complete eigenvalues with $\lambda<0$, the behaviour is an asymptotically stable star node, where all trajectories decay to the origin.

Figure 14.11: Phase portrait for repeated, complete eigenvalues with $\lambda<0$ : asymptotically stable star.

For $\lambda>0$, the phase portrait looks very similar to the one shown in Fig. 14.11 but all trajectories grow away from the origin hence the arrows point outward. This behaviour is shown in Fig. 14.12. For the case of repeated, complete eigenvalues with $\lambda>0$, the behaviour is an unstable star node, where all trajectories grow away from the origin.

Figure 14.12: Phase portrait for repeated, complete eigenvalues with $\lambda>0$ : unstable star.

Defective eigenvalues

For the defective case, we can only find one eigenvector corresponding to the eigenvalue and therefore we cannot express the general solution as in Eq. (14.92). Instead, we had to come up with a different expression for the second solution, namely

where we had to use $\boldsymbol{v}_{\mathbf{1}}$ to find $\boldsymbol{v}_{\mathbf{2}}$ (see Subsec. 14.2.4 for details). When sketching the phase portrait for such a case, we consider the same ideas as in real and distinct eigenvalues, i.e. we start off by plotting the 'easy' solutions represented by straight, half lines. Contrary to the real + distinct case however, where we had two independent exponential solutions, in this case, we only have one which is associated with $\boldsymbol{v}_{\mathbf{1}}$. We proceed by looking at a specific example for $\lambda<0$.

Consider the following linear system,

The system (14.94) gives a repeated eigenvalue of $\lambda=-3$. The general solution is given by:

The exponential solution follows the direction of $\left(\begin{array}{ll}1 & 1\end{array}\right)^{\top}$ and, since the associated eigenvalue is negative, the trajectories decay to the origin. This is the only easy solution and is shown in Fig. 14.13. We now explore what happens if both $c_{1}$ and $c_{2}$ are nonzero

Figure 14.13: Exponential solution associated with $\boldsymbol{v}_{\mathbf{1}}$ : since $\lambda<0$, the trajectories decay to the origin.

both in forward and backward time. Trajectories starting far away from the origin are almost parallel to the exponential solution (shown in Fig. 14.13) and follow its direction (in this case all trajectories move towards the origin). So, this case is very similar to real and distinct eigenvalues with the exception that there is only one independent eigenvector which trajectories tend to follow. Since we only have one exponential solution, before we can proceed to sketch the phase portrait, we need to know whether the trajectories move clockwise or anti-clockwise. Knowledge of the direction of rotation and the fact that trajectories cannot cross or touch allows us to sketch the phase portrait correctly. Again, we calculate the direction of rotation using the difference of the off-diagonal of the matrix defining the linear system. From Eq. (14.94), this is calculated as -8 hence the direction of rotation is anti-clockwise. There is only one way to sketch the rest of the trajectories then and this is shown in Fig. 14.14 along with the direction field.

Note that the nature of this type of solutions has a characteristic S-shape where we observe a large degree of rotation in the trajectories. This is not very surprising: the discriminant of the characteristic equation corresponding to repeated eigenvalues is exactly at the transition point between nodes, as observed in the case of real and distinct eigenvalues, and spirals, as observed for complex eigenvalues. The phase portrait for defective eigenvalues is represented by nodes with a large degree of rotation resembling spirals.

For the case of repeated, defective eigenvalues with $\lambda<0$, the behaviour is described by an asymptotically stable improper (or defective) node, where all trajectories decay to the origin.

Figure 14.14: Phase portrait for repeated, defective eigenvalues with $\lambda<0$ : asymptotically stable improper (or defective) node.

For the case where $\lambda>0$, the same ideas apply with the exception that the behaviour is unstable and all trajectories grow away from the origin. An example of this behaviour is shown in Fig. 14.15.

Figure 14.15: Phase portrait for repeated, defective eigenvalues with $\lambda>0$ : unstable improper (or defective) node. For the case of repeated, defective eigenvalues with $\lambda>0$, the behaviour is described by an unstable improper (or defective) node, where all trajectories move away from the origin.

Nonisolated fixed points

In Eq. (14.77), when $\tau \neq 0$ and $\Delta=0$, we have two cases: (i) $\lambda_{1}=0$ and $\lambda_{2}>0$ or (ii) $\lambda_{1}<0$ and $\lambda_{2}=0$. For such a system, there exists a line of fixed points (i.e. the origin is not the only fixed point anymore). For case (i) the nonzero eigenvalue is positive, so we have a line of unstable fixed points while for case (ii) since the nonzero eigenvalue is negative, we obtain a line of (neutrally) stable fixed points. Finally, when both eigenvalues are zero $(\tau=0, \Delta=0)$ the result is a plane of fixed points (think various scattered points in the phase plane) in the case where the matrix is complete (this implies that $A$ is the zero matrix) and a line of fixed point where the matrix is defective.

Trace-determinant plane

The fixed points for the 2D linear system are easily classified according to the trace and determinant of the matrix $A$ defining the system as we have seen earlier. For this reason, a plot of the trace against the determinant (commonly referred to as the trace-determinant plane) is known as the classification diagram (see Fig. 14.16).

We note that in the trace-determinant plane, some classes of fixed points are represented inside whole regions. In particular these are the saddles, unstable or stable nodes and spirals. We also have the so-called borderline cases, which exist along a certain border in the classification diagram. These borders are coloured clearly in Fig. 14.16:

• $\tau-4 \Delta=0$ : stable/unstable stars and degenerate nodes (corresponding to defective $\lambda$ ) exist there;
• $\tau=0$ for $\Delta>0$ : neutrally stable centres occur;
• $\Delta=0$ : line $(\tau>0, \tau<0)$ or plane $(\tau=0)$ of nonisolated fixed points.

These ideas, especially the borderline cases become important in nonlinear systems and the stability of the fixed points in the presence of bifurcations ${ }^{27}$ where we observe qualitative changes in the behaviour of the system.