We now introduce the concept of the phase plane, a useful way of visualizing solutions of planar systems (two dependent variables) of ODEs. We begin by considering the
system ·
x01 x02
¸
=
· 0 −1
1 0
¸ · x1 x2
¸ .
Observe that the coefficient matrix A is already in real canonical form, and etA =
· cos t − sin t sin t cos t
¸ .
If we were interested in the solution satisfying initial conditions x1(0) = 1 and x2(0) = 0, we would multiply etA by the vector [1 0] to obtain x1(t) = cos t and x2(t) = sin t. The most natural way to visualize these solutions is to simply graph both x1 and x2 as a function of t as in Figure 2.2. An alternate way to visualize the solution is to plot x2(t) versus x1(t) as a curve parametrized by time t. Such a plot is given in Figure 2.3, which illustrates the parametrized curve (cos t, sin t) for t ∈ [0, 2π]. In this case, the parametrized curve is a circle of radius 1 because x1(t) and x2(t) satisfy the relationship
x21+ x22 = cos2(t) + sin2(t) = 1
for all t. Notice that as t increases, the curve is traversed in the counterclockwise direction.
Other initial conditions would give rise to other parametric curves, giving a visual representation of the “flow” of the system. A parametric plot of x2(t) versus x1(t) for
x2
x1
Figure 2.3: Parametric plot of x2(t) = sin t versus x1(t) = cos t for t ∈ [0, 2π].
various choices of initial conditions is called a phase portrait or phase plane diagram for the system.
We now discuss three canonical examples of phase portraits for planar systems, and then explain how these special cases can be used to sketch phase portraits for more general planar systems. From the previous subsections, we know that every 2 × 2 matrix A can be written in the form A = P MP−1, where M has one of three forms:
• If A is diagonalizable, then
M =
· α 0 0 β
¸ .
• If A is non-diagonalizable and has a real, repeated eigenvalue α, then M =
· α 1 0 α
¸ .
• If A has complex-conjugate eigenvalues α ± βi, then M =
· α −β
β α
¸ .
CASE 1: Suppose x0 = Mx where M = diag{α, β}. The general solution is
x(t) = etMx(0); i.e.,
· x1(t) x2(t)
¸
=
· eαt 0 0 eβt
¸ · c1 c2
¸
=
· c1eαt c2eβt
¸ .
The exponential solutions x1 = c1eαt and x2 = c2eβt either grow or decay depending upon the signs of α and β. Suppose that both α and β are positive. Then both eαt and eβt will increase without bound as t → ∞. We can sketch the phase portrait by eliminating the parameter t. Notice that xβ1 = cβ1eαβt and xα2 = cα2eαβt. Assuming that c1 6= 0, taking the ratio gives
xα2
xβ1 = cα2 cβ1 = c,
where c is a constant. By algebra, we have x2 = cxβ/α1 , and graphing such power functions is easy. Figure 2.4 shows the possible phase portraits in cases where both α and β are positive. Notice that all trajectories point outward from the origin. In particular, if β = α (and thus β/α = 1), the trajectories in the phase plane are straight lines directed outward from the origin.
If both α and β are negative, then both x1and x2 will decay to zero exponentially fast as t → ∞. As indicated in Figure 2.5, the phase portraits are similar to the previous ones, except that all trajectories are directed inward towards the origin.
A more interesting situation occurs when α and β have different signs. In this case, we have exponential growth for one of our variables and exponential decay for the other variable. The possible phase portraits are shown in Figure 2.6.
The various phase portraits in Figures 2.4–2.6 all have one thing in common: the special solution corresponding to c1 = c2 = 0 is a constant solution in which we stay
“stuck” at the origin in the phase plane for all time t. Constant solutions of ODEs will be of particular interest to us, and we give them a special name.
Definition 2.2.1. An equilibrium of a system of ODEs is a solution that is constant.
That is, all dependent variables are constant for all time t.
Observation: For linear, constant-coefficient systems x0 = Ax of ODEs, x = 0 is always an equilibrium solution. In terms of the phase portrait, this means that the origin always corresponds to an equilibrium solution.
One way of further classifying equilibria is provided by the phase portraits dis-cussed above. Namely,
Definition 2.2.2. In Figure 2.4, the origin is called an unstable node—all trajec-tories point outward from the origin. In Figure 2.5, the origin is called a stable
x1
x2 x2
x1
x2
x1
Figure 2.4: Phase portraits corresponding to CASE 1 in which both α and β are positive. Left panel: 0 < α < β. Middle panel: 0 < α = β. Right panel: 0 < β < α.
x1
x2 x2
x1
x2
x1
Figure 2.5: Phase portraits corresponding to CASE 1 in which both α and β are negative. Left panel: β < α < 0. Middle panel: α = β < 0. Right panel: α < β < 0.
x1 x2
x1 x2
Figure 2.6: Phase portraits corresponding to CASE 1 in which both α and β have different signs. Left panel: β < 0 < α. Right panel: α < 0 < β.
node—all trajectories point inward to the origin. In Figure 2.6, the origin is called a saddle.
Later, we will actually re-state this Definition in a more precise way. We remark that saddle equilibria are different from stable/unstable nodes in that only certain, special trajectories actually touch the equilibrium point. In Figure 2.6, these are the trajectories which lie along the x1 and x2 axes. The four trajectories which approach the origin as t → ±∞ are called separatrices. (Again, we will make this definition more precise later.)
CASE 2: Suppose x0 = Mx where M is the non-diagonalizable matrix M =
· α 1 0 α
¸
with α as a repeated, real eigenvalue. The phase portrait is more difficult to plot without computer assistance. To solve the system of ODEs, we decompose M as the sum of a diagonal matrix and a nilpotent matrix:
M = D + N =
Clearly D and N commute since D is a scalar multiple of the identity matrix, and you can also verify that N is nilpotent of order 2. It follows that
etM = etDetN = etD[I + tN] = The general solution of the system is therefore x(t) = etMx(0), or
· x1(t)
These solutions give a parametrization for the solution curves in the phase portrait, which appears in Figure 2.7. If α < 0, the trajectories approach the origin as t → ∞, and if α > 0, the trajectories are oriented outward from the origin. Notice that, unlike the case in which the origin is a saddle equilibrium, there is only one separatrix in this case (the x1-axis). In the left panel of Figure 2.7, the origin is a stable node, and in the right panel the origin is an unstable node.
We remark that the characteristic equation for M is (λ − α)2 = 0, which is a
“critical” case in the sense that we are just on the verge of having complex conju-gate roots. Indeed, quadratic equations with negative discriminants have complex conjugate solutions, and repeated real roots occur when the discriminant is 0.
x1
x2 x2
x1
Figure 2.7: Phase portraits corresponding to CASE 2, in which M has a repeated eigenvalue. Left panel: α = −1. Right panel: α = 1.
CASE 3: Suppose x0 = Mx where M is in real canonical form M =
· α −β
β α
¸ . We know that
etM = eαt
· cos βt − sin βt sin βt cos βt
¸ , which means that the general solution of the system is
· x1(t) x2(t)
¸
= etM
· c1 c2
¸
=
· c1eαtcos βt − c2eαtsin βt c1eαtsin βt + c2eαtcos βt
¸ .
Solutions involve combinations of exponential functions eαt with periodic functions sin βt and cos βt. If α < 0, the exponential factors will decay as t → ∞, while the other factors simply oscillate. If we graph some solution curves in the phase portrait, we will see that trajectories spiral inward to the origin. Likewise, if α > 0, trajectories will spiral outward from the origin. Finally, if α = 0, the exponential factor is constant and the solutions are purely oscillatory. In this case, the phase portrait consists of concentric circular trajectories.
For matrices M in real canonical form1, the orientation of the trajectories can be determined from the sign of β: clockwise if β < 0 and counter-clockwise if β > 0.
These various possibilities are summarized in Figure 2.8. The equilibrium at the origin is called a stable focus if trajectories spiral inward (α < 0), an unstable focus if trajectories spiral outward (α > 0), and a center if trajectories form concentric, closed curves (α = 0).
1If a matrix with complex conjugate eigenvalues is not in real canonical form, one simple way to determine the orientation of the trajectories is to plot a few “slope field vectors” as illustrated in the next example.
x2
x1
x2
x1
x2
x1
x2
x1
x2 x2
x1 x1
(a) (b)
(c) (d)
(e) (f)
Figure 2.8: Phase portraits corresponding to CASE 3, in which M has a complex conjugate eigenvalues. (a) Origin is a stable focus (α < 0) with trajectories oriented counter-clockwise (β > 0). (b) Origin is an unstable focus (α > 0) with trajectories oriented clockwise (β < 0). (c) Origin is a stable focus (α < 0) with trajectories oriented clockwise (β < 0). (d) Origin is an unstable focus (α > 0) with trajectories oriented counter-clockwise (β > 0). (e) Origin is a center (α = 0) with trajectories oriented counter-clockwise (β > 0). (f) Origin is a center (α = 0) with trajectories oriented clockwise (β < 0).
Remark: Above, we did not consider the possibility that the matrix M in the equation x0 = Mx has λ = 0 as an eigenvalue. In such cases, the origin is called a degenerate equilibrium. Notice that if λ = 0 is an eigenvalue, then det M = 0, which means that M is a singular matrix. It follows that the solutions of the equation Mx = 0 form a subspace of dimension at least 1, and any vector x in this subspace would be an equilibrium for our system of ODEs. In the remainder of the course, we will typically work only with systems which have isolated equilibrium points (defined later), as opposed to systems with infinitely many equilibrium points.
It is also worth noting that for the autonomous systems of ODEs that we have considered, solution trajectories in the phase portrait cannot intersect each other.
Can you explain why?
Suppose a planar system of ODEs has a coefficient matrix A which is not in one of the three canonical forms we discussed above. To sketch the phase portrait, we need to determine which canonical form A is similar to, and this is accomplished by finding the eigenvalues and eigenvectors. If A is any 2 × 2 matrix with non-zero eigenvalues, then the associated phase portrait is always a “skewed” version of one of the portraits in the above figures. We illustrate this via two examples.
Example. Sketch the phase portrait for x0 = Ax, where A =
· 0 −4
1 0
¸ .
Solution: The characteristic equation is λ2+ 4 = 0, from which we infer that the eigenvalues are complex conjugate: λ = ±2i. We will put A in real canonical form by finding an eigenvector for λ = 2i, the eigenvalue with positive imaginary part.
Row-reducing the matrix (A − λI) gives A − λI = free variable. Next, express the eigenvector as a sum of its real and imaginary parts:
· w1
which contains the imaginary part of w in its first column and the real part of w in and the real canonical form for A is given by
M = P−1AP =
· 0 −2 2 0
¸ . Routine calculation yields the matrix exponential
etA = P etMP−1 = P from which it follows that the general solution of our system is
· x1(t) The parameter t can be eliminated by algebra, leading to the relationship
x21+ 4x22 = c21 + 4c22,
where the right hand side is an arbitrary non-negative constant. You should recognize this as the equation of an ellipse. The phase portrait is sketched in Figure 2.9, and we see that the origin is a center. One way to determine the orientation of the trajectories is to pick a few convenient points in the phase plane and sketch the associated “slope field” vectors. For example, suppose we start from the point (x, y) = (1, 0) on the x-axis in the phase plane. To determine the direction of motion from that point, we multiply the coefficient matrix A by (1, 0), obtaining (0, 1). The vector (0, 1), which points straight upward, is tangent to the solution trajectory passing through (1, 0).
This indicates that trajectories are oriented counter-clockwise, as shown in Figure 2.9.
Notice that our phase portrait is simply a “skewed” version of our canonical example of a center equilibrium (which had circular trajectories).
Example. Sketch the phase portrait for x0 = Ax, where A =
· 2 −3 1 −2
¸ .
The characteristic equation is λ2 − 1 = 0, so we have real eigenvalues λ = ±1 with different sign. We immediately conclude that the origin in our phase portrait will be a saddle. You should show that the eigenvalues λ = 1 and λ = −1 give rise to
x1 x2
Figure 2.9: Phase portrait showing the elliptic trajectories x21+ 4x22 = constant.
eigenvectors ·
3 1
¸
and
· 1 1
¸ , respectively. By Proposition 2.1.13, the general solution is
· x1 x2
¸
= c1et
· 3 1
¸
+ c2e−t
· 1 1
¸ . Consequence: If we set c1 = 0 and plot the trajectories
c2e−t
· 1 1
¸
in the phase plane, we find that this is a parametrization of a line of slope 1 as t varies from −∞ to ∞. The value of c2 merely selects our “starting point” as we traverse the line. Similarly, if we had set c2 = 0, we would have obtained the parametrization of a line of slope 13. In other words the lines spanned by the two eigenvectors of A form the separatrices in our phase portrait. This tells us precisely how to “skew” our canonical example of a saddle equilibrium to obtain a sketch of the phase portrait, which is shown in Figure 2.10. Notice that the motion along the separatrix corresponding to the negative eigenvalue is directed inward to the origin, while the motion along the separatrix corresponding to the positive eigenvalue is directed outward from the origin.
Before continuing our investigation of the “geometry” associated with planar, constant-coefficient systems, we follow up our earlier remark that the origin is always an equilibrium of the linear, homogeneous constant-coefficient system x0 = Ax. This admittedly narrow class of systems is not quite as exclusive as it may appear. Having an equilibrium at the origin is not a severe restriction at all, because equilibria can always be relocated via a change of variables. For example, consider the
constant-x2
x1
Figure 2.10: Phase portrait showing the skewed saddle The separatrices are the lines spanned by the eigenvectors [1, 1] and [3, 1].
coefficient system x0 = Ax − b, where b is a constant vector. Assuming that A is an invertible matrix (i.e., λ = 0 is not an eigenvalue), we may solve for equilibrium solutions by setting x0 = 0. Equivalently, any vector x satisfying Ax = b is an equilibrium. Let x∗ = A−1b denote the equilibrium solution. Then our original system of ODEs can be written as x0 = Ax − Ax∗ = A(x − x∗). Making the substitution y = x − x∗, we obtain a new system y0 = Ay. The new constant-coefficient system has its equilibrium at the origin, and we can solve it using the techniques discussed in previous sections. We may then recover the solution of our original system by writing x = y + x∗.
The system x0 = Ax − b is actually a special case of the inhomogeneous systems we will learn how to solve soon.