If the initial point $(x_0,y_0)$ is not a critical point, then the corresponding trajectory is a curve in the xy-plane along which the point (x(t), y(t)) moves as t increases. It turns out that any trajectory not consisting of a single point is a nondegenerate curve with no self-intersections (Problem 30). We can exhibit qualitatively the behavior of solutions of the autonomous system in (1) by constructing a picture that shows its critical points together with a collection of typical solution curves or trajectories in the xy-plane. Such a picture is called a phase portrait (or phase plane picture) because it illustrates “phases” or xy-states of the system, and indicates how they change with time.

Another way of visualizing the system is to construct a slope field in the xy-phase plane by drawing typical line segments having slope

or a direction field by drawing typical vectors pointing the same direction at each point (x, y) as does the vector (F(x, y), G(x, y)). Such a vector field then indicates which direction along a trajectory to travel in order to “go with the flow” described by the system.

Figure 9.1.1 shows a direction field and phase portrait for the rabbit-squirrel system of Example 1. The direction field arrows indicate the direction of motion of the point (x(t), y(t)). We see that, given any positive initial numbers $x_0\ne 4$ and $y_0\ne 6$ of rabbits and squirrels, this point moves along a trajectory approaching the critical point (4, 6) as t increases.

Example 2

For the system

$$\begin{array}{rcl}{x}^{\text{′}}& =& x-y,\\ y^{\text{′}}& =& 1-x^2\end{array}$$ (7)

we see from the first equation that $x=y$ and from the second that $x=\pm 1$ at each critical point. Thus this system has the two critical points $(-1,-1)$ and (1, 1). The direction field in Fig. 9.1.2 suggests that trajectories somehow “circulate” counterclockwise around the critical point $(-1,-1),$ whereas it appears that some trajectories may approach, while others recede from, the critical point (1, 1). These observations are corroborated by the phase portrait in Fig. 9.1.3 for the system in (7).

Remark

One could carelessly write the critical points in Example 2 as $(\pm 1,\pm 1)$ and then jump to the erroneous conclusion that the system in (7) has four rather than just two critical points. When feasible, a sure-fire way to determine the number of critical points of an autonomous system is to plot the curves $F(x,y)=0$ and $G(x,y)=0$ and then note their intersections, each of which represents a critical point of the system. For instance, Fig. 9.1.4 shows the curve (line) $F(x,y)=x-y=0$ and the pair of lines $x=+1$ and $x=-1$ that constitute the “curve” $G(x,y)=1-x^2=0.$ The (only) two points of intersection $(-1,-1)$ and $(+1,+1)$ are then apparent.

The behavior of the trajectories near an isolated critical point of an autonomous system is of particular interest. Figure 9.1.5 is a close-up view of Fig. 9.1.1 near the critical point (4, 6), and similarly Figs. 9.1.6 and 9.1.7 are close-ups of Fig. 9.1.3 near the critical points $(-1,-1)$ and (1, 1), respectively. We notice immediately that although the systems underlying these phase portraits are nonlinear, each of these three magnifications bears a striking resemblance to one of the cases in our “gallery” Fig. 7.4.16 of phase plane portraits for linear constant-coefficient systems. Indeed, the three figures strongly resemble a nodal sink, a spiral source, and a saddle point, respectively.

These similarities are not a coincidence. Indeed, as we will explore in detail in the next section, the behavior of a nonlinear system near a critical point is generally similar to that of a corresponding linear constant-coefficient system near the origin. For this reason it is useful to extend the language of nodes, sinks, etc., introduced in Section 7.4 for linear constant-coefficient systems to the broader context of critical points of the two-dimensional system (1).

In general, the critical point $(x_{\star },y_{\star })$ of the autonomous system in (1) is called a node provided that

• Either every trajectory approaches $(x_{\star },y_{\star })$ as $t\to +\infty$ or every trajectory recedes from $(x_{\star },y_{\star })$ as $t\to +\infty ,$ and

• Every trajectory is tangent at $(x_{\star },y_{\star })$ to some straight line through the critical point.

As with linear constant-coefficient systems, a node is said to be proper provided that no two different pairs of “opposite” trajectories are tangent to the same straight line through the critical point. On the other hand, the critical point (4, 6) of the system in Eq. (5), shown in Figs. 9.1.1 and 9.1.5, is an improper node; as those figures suggest, virtually all of the trajectories approaching this critical point share a common tangent line at that point.

Likewise, a node is further called a sink if all trajectories approach the critical point, a source if all trajectories recede (or emanate) from it. Thus the critical point (4, 6) in Figs. 9.1.1 and 9.1.5 is a nodal sink, whereas the critical point $(-1,-1)$ in Figs. 9.1.3 and 9.1.6 is a source (more specifically, a spiral source). The critical point (1, 1) in Figs. 9.1.3 and 9.1.7, on the other hand, is a saddle point.

A critical point $(x_{\star },y_{\star })$ of the autonomous system in (1) is said to be stable provided that if the initial point $(x_0,y_0)$ is sufficiently close to $(x_{\star },y_{\star }),$ then (x(t), y(t)) remains close to $(x_{\star },y_{\star })$ for all $t>0.$ In vector notation, with $\mathbf{\text{x}}(t)=(x(t),y(t)),$ the distance between the initial point ${\mathbf{\text{x}}}_0=(x_0,y_0)$ and the critical point ${\mathbf{\text{x}}}_{\star }=(x_{\star },y_{\star })$ is

Thus the critical point ${\mathbf{\text{x}}}_{\star }$ is stable provided that, for each $ϵ>0,$ there exists $\delta >0$ such that

for all $t>0.$ Note that the condition in (8) certainly holds if $\mathbf{\text{x}}(t)\to {\mathbf{\text{x}}}_{\star }$ as $t\to +\infty ,$ as in the case of a nodal sink such as the critical point (4, 6) in Figs. 9.1.1 and 9.1.5. Thus this nodal sink can also be described as a stable node.

The critical point $(x_{\star },y_{\star })$ is called unstable if it is not stable. The two critical points at $(-1,-1)$ and (1, 1) shown in Figs. 9.1.3, 9.1.6, and 9.1.7 are both unstable, because, loosely speaking, in neither of these cases can we guarantee that a trajectory will remain near the critical point simply by requiring the trajectory to begin near the critical point.

If $(x_{\star },y_{\star })$ is a critical point, then the equilibrium solution $x(t)\equiv x_{\star },y(t)\equiv y_{\star }$ is called stable or unstable depending on the nature of the critical point. In applications the stability of an equilibrium solution is often a crucial matter. For instance, suppose in Example 1 that x(t) and y(t) denote the rabbit and squirrel populations, respectively, in hundreds. We will see in Section 9.3 that the critical point (4, 6) in Fig. 9.1.1 is stable. It follows that if we begin with close to 400 rabbits and 600 squirrels—rather than exactly these equilibrium values—then for all future time there will remain close to 400 rabbits and close to 600 squirrels. Thus the practical consequence of stability is that slight changes (perhaps due to random births and deaths) in the equilibrium populations will not so upset the equilibrium as to result in large deviations from the equilibrium solutions.

It is possible for trajectories to remain near a stable critical point without approaching it, as Example 3 shows.

Example 3

Undamped mass-spring system Consider a mass m that oscillates without damping on a spring with Hooke’s constant k, so that its position function x(t) satisfies the differential equation $x″+{\omega }^{2}x=0$ (where ${\omega }^2=k/m$). If we introduce the velocity $y=dx/dt$ of the mass, we get the system

$$\begin{array}{l}{\displaystyle \frac{dx}{dt}}=y,\\ {\displaystyle \frac{dy}{dt}}=-{\omega }^{2}x\end{array}$$ (9)

with general solution

$$\begin{array}{rcrcr}x(t)& =& A\cos \omega t& +& B\sin \omega t,\\ y(t)& =& -A\omega \sin \omega t& +& B\omega \cos \omega t.\end{array}$$ (10a)
(10b)

With $C=\sqrt{A^2+B^2},A=C\cos \alpha ,$ and $B=C\sin \alpha ,$ we can rewrite the solution in (10) in the form

$$\begin{array}{rcr}x(t)& =& C\cos (\omega t-\alpha ),\\ y(t)& =& -\omega C\sin (\omega t-\alpha ),\end{array}$$ (11a)
(11b)

so it becomes clear that each trajectory other than the critical point (0, 0) is an ellipse with equation of the form

$${\displaystyle \frac{x^2}{C^2}}+{\displaystyle \frac{y^2}{{\omega }^2{C}^2}}=1.$$ (12)

As illustrated by the phase portrait in Fig. 9.1.8 (where $\omega =\frac{1}{2}$), each point $(x_0,y_0)$ other than the origin in the xy-plane lies on exactly one of these ellipses, and each solution (x(t), y(t)) traverses the ellipse through its initial point $(x_0,y_0)$ in the clockwise direction with period $P=2\pi /\omega .$ (It is clear from (11) that $x(t+P)=x(t)$ and $y(t+P)=y(t)$ for all t.) Thus each nontrivial solution of the system in (9) is periodic and its trajectory is a simple closed curve enclosing the critical point at the origin.

Figure 9.1.9 shows a typical elliptical trajectory in Example 3, with its minor semiaxis denoted by $\delta$ and its major semiaxis by $ϵ.$ We see that if the initial point $(x_0,y_0)$ lies within distance $\delta$ of the origin—so that its elliptical trajectory lies inside the one shown—then the point $(x(t),y(t))$ always remains within distance $ϵ$ of the origin. Hence the origin (0, 0) is a stable critical point of the system $x\text{′}=y,y\text{′}=-{\omega }^{2}x,$ despite the fact that no single trajectory approaches the point (0, 0). A stable critical point surrounded by simple closed trajectories representing periodic solutions is called a (stable) center.

The critical point $(x_{\star },y_{\star })$ is called asymptotically stable if it is stable and, moreover, every trajectory that begins sufficiently close to $(x_{\star },y_{\star })$ also approaches $(x_{\star },y_{\star })$ as $t\to +\infty .$ That is, there exists $\delta >0$ such that

where ${\mathbf{\text{x}}}_0=(x_0,y_0),{\mathbf{\text{x}}}_{\star }=(x_{\star },y_{\star }),$ and $\mathbf{\text{x}}(t)=(x(t),y(t))$ is a solution with $\mathbf{\text{x}}(0)={\mathbf{\text{x}}}_0$.

Now suppose that x(t) and y(t) denote coexisting populations for which $(x_{\star },y_{\star })$ is an asymptotically stable critical point. Then if the initial populations $x_0$ and $y_0$ are sufficiently close to $x_{\star }$ and $y_{\star },$ respectively, it follows that both

That is, x(t) and y(t) actually approach the equilibrium populations $x_{\star }$ and $y_{\star }$ as $t\to +\infty ,$ rather than merely remaining close to those values.

For a mechanical system as in Example 3, a critical point represents an equilibrium state of the system—if the velocity $y=x\text{′}$ and the acceleration $y\text{′}=x″$ vanish simultaneously, then the mass remains at rest with no net force acting on it. Stability of a critical point concerns the question whether, when the mass is displaced slightly from its equilibrium, it

1. Moves back toward the equilibrium point as $t\to +\infty$,

2. Merely remains near the equilibrium point without approaching it, or

3. Moves farther away from equilibrium.

In Case 1 the critical [equilibrium] point is asymptotically stable; in Case 2 it is stable but not asymptotically so; in Case 3 it is an unstable critical point. A marble balanced on the top of a soccer ball is an example of an unstable critical point. A mass on a spring with damping illustrates the case of asymptotic stability of a mechanical system. The mass-and-spring without damping in Example 3 is an example of a system that is stable but not asymptotically stable.

Example 4

Damped mass-spring system Suppose that $m=1$ and $k=2$ for the mass and spring of Example 3 and that the mass is attached also to a dashpot with damping constant $c=2.$ Then its displacement function x(t) satisfies the second-order equation

$$x″(t)+2x\text{′}(t)+2x(t)=0.$$ (15)

With $y=x\text{′}$ we obtain the equivalent first-order system

$$\begin{array}{l}{\displaystyle \frac{dx}{dt}}=y,\\ {\displaystyle \frac{dy}{dt}}=-2x-2y\end{array}$$ (16)

with critical point (0, 0). The characteristic equation $r^2+2r+2=0$ of Eq. (15) has roots $-1+i$ and $-1-i,$ so the general solution of the system in (16) is given by

$$\begin{array}{rcl}x(t)& =& e^{-t}(A\cos t+B\sin t)=C{e}^{-t}\cos (t-\alpha ),\\ y(t)& =& e^{-t}[(B-A)\cos t-(A+B)\sin t]\\ & =& -C\sqrt{2}{e}^{-t}\sin \left(t-\alpha +\frac{1}{4}\pi \right),\end{array}$$ (17a)
(17b)

where $C=\sqrt{A^2+B^2}$ and $\alpha ={\tan }^{-1}(B/A).$ We see that x(t) and y(t) oscillate between positive and negative values and that both approach zero as $t\to +\infty .$ Thus a typical trajectory spirals inward toward the origin, as illustrated by the spiral in Fig. 9.1.10.

It is clear from (17) that the point (x(t), y(t)) approaches the origin as $t\to +\infty ,$ so it follows that (0, 0) is an asymptotically stable critical point for the system $x\text{′}=y,y\text{′}=-2x-2y$ of Example 4. Such an asymptotically stable critical point—around which the trajectories spiral as they approach it—is called a stable spiral point (or a spiral sink). In the case of a mass–spring—dashpot system, a spiral sink is the manifestation in the phase plane of the damped oscillations that occur because of resistance.

If the arrows in Fig. 9.1.10 were reversed, we would see a trajectory spiraling outward from the origin. An unstable critical point—around which the trajectories spiral as they emanate and recede from it—is called an unstable spiral point (or a spiral source). Example 5 shows that it also is possible for a trajectory to spiral into a closed trajectory—a simple closed solution curve that represents a periodic solution (like the elliptical trajectories in Fig. 9.1.8).

Example 5

Consider the system

$$\begin{array}{l}{\displaystyle \frac{dx}{dt}}=-ky+x(1-x^2-y^2),\\ {\displaystyle \frac{dy}{dt}}=kx+y(1-x^2-y^2).\end{array}$$ (18)

In Problem 21 we ask you to show that (0, 0) is its only critical point. This system can be solved explicitly by introducing polar coordinates $x=r\cos \theta ,y=r\sin \theta ,$ as follows. First note that

$${\displaystyle \frac{d\theta }{dt}}=\frac{d}{dt}\left(\arctan \frac{y}{x}\right)={\displaystyle \frac{x{y}^{\text{′}}-x^{\text{′}}y}{x^2+y^2}}.$$

Then substitute the expressions given in (18) for $x\text{′}$ and $y\text{′}$ to obtain

$${\displaystyle \frac{d\theta }{dt}}={\displaystyle \frac{k(x^2+y^2)}{x^2+y^2}}=k.$$

It follows that

$$\theta (t)=kt+{\theta }_0,\text{where }{\theta }_0=\theta (0).$$ (19)

Then differentiation of $r^2=x^2+y^2$ yields

$$\begin{array}{rcl}2r{\displaystyle \frac{dr}{dt}}& =& 2x{\displaystyle \frac{dx}{dt}}+2y{\displaystyle \frac{dy}{dt}}\\ & =& 2(x^2+y^2)(1-x^2-y^2)=2r^2(1-r^2),\end{array}$$

so $r=r(t)$ satisfies the differential equation

$${\displaystyle \frac{dr}{dt}}=r(1-r^2).$$ (20)

In Problem 22 we ask you to derive the solution

$$r(t)={\displaystyle \frac{r_0}{\sqrt{r_0^2+(1-r_0^2)e^{-2t}}}},$$ (21)

where $r_0=r(0).$ Thus the typical solution of Eq. (18) may be expressed in the form

$$\begin{array}{rcl}x(t)& =& r(t)\cos (kt+{\theta }_0),\\ y(t)& =& r(t)\sin (kt+{\theta }_0).\end{array}$$ (22)

If $r_0=1,$ then Eq. (21) gives $r(t)\equiv 1$ (the unit circle). Otherwise, if $r_0>0,$ then Eq. (21) implies that $r(t)\to 1$ as $t\to +\infty .$ Hence the trajectory defined in (22) spirals in toward the unit circle if $r_0>1$ and spirals out toward this closed trajectory if $0<r_0<1.$ Figure 9.1.11 shows a trajectory spiraling outward from the origin and four trajectories spiraling inward, all approaching the closed trajectory $r(t)\equiv 1$.

Under rather general hypotheses it can be shown that there are four possibilities for a nondegenerate trajectory of the autonomous system

The four possibilities are these:

1. (x(t), y(t)) approaches a critical point as $t\to +\infty$.

2. (x(t), y(t)) is unbounded with increasing t.

3. (x(t), y(t)) is a periodic solution with a closed trajectory.

4. (x(t), y(t)) spirals toward a closed trajectory as $t\to +\infty$.

As a consequence, the qualitative nature of the phase plane picture of the trajectories of an autonomous system is determined largely by the locations of its critical points and by the behavior of its trajectories near its critical points. We will see in Section 9.2 that, subject to mild restrictions on the functions F and G, each isolated critical point of the system $x\text{′}=F(x,y),y\text{′}=G(x,y)$ resembles qualitatively one of the examples of this section—it is either a node (proper or improper), a saddle point, a center, or a spiral point.

Consider a first-order differential equation of the form

which may be difficult or impossible to solve explicitly. Its solution curves can nevertheless be plotted as trajectories of the corresponding autonomous two-dimen-sional system

Most ODE plotters can routinely generate phase portraits for autonomous systems. Those appearing in this chapter were plotted using programs that are free for educational use. For instance, the Matlab program pplane illustrated in Fig. 9.1.20 can be found at math.rice.edu/~dfield. Another freely available and user-friendly Matlab-based ODE package with similar graphical capabilities is Iode (www.math.uiuc.edu/iode).

For example, to plot solution curves for the differential equation

we plot trajectories of the system

The result is shown in Fig. 9.1.21.

Plot similarly some solution curves for the following differential equations.

1. ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{4x-5y}{2x+3y}}$

2. ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{4x-5y}{2x-3y}}$

3. ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{4x-3y}{2x-5y}}$

4. ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{2xy}{x^2-y^2}}$

5. ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x^2+2xy}{y^2+2xy}}$

Now construct some examples of your own. Homogeneous functions like those in Problems 1 through 5—rational functions with numerator and denominator of the same degree in x and y—work well. The differential equation

of this form generalizes Example 5 in this section but would be inconvenient to solve explicitly. Its phase portrait (Fig. 9.1.22) shows two periodic closed trajectories—the circles $r=1$ and $r=2.$ Anyone want to try for three circles?