A two-dimensional linear system of differential equations x = ax+by,
y = cx+dy,
wherea,b,c, anddare constants, can be written compactly using vectors and matrices. Denoting
x(t)=
x(t) y(t)
, A=
a b c d
, the system can be written
x(t) y(t)
=
a b c d
x(t) y(t)
, or
x(t) =
a b c d
x(t).
We often write this simply as
x =Ax, (5.8)
where we have suppressed the understood dependence of x on t. We briefly reiterate the ideas introduced in the introduction, Section 5.1. A solution to the system (5.8) on an interval is a vector function x(t) = (x(t), y(t))T, that satisfies the system on the required interval. We can graph x(t) and y(t) vs.
t, which gives the state space representation or time seriesplots of the solu- tion. Alternatively, a solution can be graphed as a parametric curve, or vector function, in thexy plane. We call thexy plane thephase plane, and we call a solution curve plotted in the xy plane an orbit. Observe that a solution is a vector functionx(t) with componentsx(t) andy(t). In the phase plane, the
orbit is represented in parametric form and is traced out as time proceeds.
Thus, time is not explicitly displayed in the phase plane representation, but it is a parameter along the orbit. An orbit is traced out in a specific direction as time increases, and we usually denote that direction by an arrow along the curve. Furthermore, time can always be shifted along a solution curve. That is, ifx(t) is a solution, thenx(t−c) is a solution for any real numbercand it represents the same solution curve.
Our main objective is to find the phase portrait, or a plot of key orbits of the given system. We are particularly interested in theequilibrium solutions of (5.8). These are the constant vector solutions x∗ for which Ax∗ = 0. An equilibrium solution is represented in the phase plane as a point. The vector field vanishes at an equilibrium point. The time series representation of an equilibrium solution is two constant functions. If detA = 0 then x∗ = 0 is the only equilibrium of (5.8), and it is represented by the origin, (0,0), in the phase plane. We say in this case that the origin is anisolated equilibrium.
If detA= 0, then there will be an entire line of equilibrium solutions through the origin; each point on the line represents an equilibrium solution, and the equilibria are not isolated. Equilibrium solutions are important because the interesting behavior of the orbits occurs near these solutions. (Equilibrium solutions are also called critical points by some authors.)
Example 5.13
Consider the system
x = −2x−y, y = 2x−5y, which we write as
x=
−2 −1 2 −5
x.
The coefficient determinant is nonzero, so the only equilibrium solution is rep- resented by the origin,x(t) = 0,y(t) = 0. By substitution, it is straightforward to check that
x1(t)=
x(t) y(t)
= e−3t
e−3t
= 1
1
e−3t is a solution. Also
x2(t) =
e−4t 2e−4t
= 1
2
e−4t
is a solution. Each of these solutions has the form of a constant vector times a scalar exponential function of time t. Why should we expect exponential solutions? The two equations involve both x and y and their derivatives; a solution must make everything cancel out, and so each term must basically have the same form. Exponential functions and their derivatives both have the same form, and therefore exponential functions for both x and y are likely candidates for solutions. We graph these two independent solutionsx1(t) and x2(t) in the phase plane. See figure 5.5. Each solution, or orbit, plots as a ray traced from infinity (as timetapproaches−∞) into the origin (astapproaches +∞). The slopes of these ray-like solutions are defined by the constant vectors preceding the scalar exponential factor, the latter of which has the effect of stretching or shrinking the vector. Note that these two orbits approach the origin as time gets large, but they never actually reach it. Another way to look
y
x x2(t)
–x2(t) –x1(t)
x1(t)
Figure 5.5 x1(t) and x2(t) are shown as linear orbits (rays) entering the origin in the first quadrant. The reflection of those rays in the third quadrant are the solutions−x1(t) and−x2(t). Note that all four of these linear orbits approach the origin ast→+∞because of the decaying exponential factor in the solution. Ast→ −∞ (backward in time) all four of these linear orbits go to infinity.
at it is this. If we eliminate the parameter t in the parametric representation x=e−4t, y= 2e−4tofx2(t),say, theny= 2x, which is a straight line in thexy plane. This orbit is on one ray of this straight line, lying in the first quadrant.
Solutions of (5.8) the formx(t) =veλt, whereλis a real constant andvis
a constant, real vector, are calledlinear orbits because they plot as rays in thexy-phase plane.
We are ready to make some observations about the structure of the solution set to the two-dimensional linear system (5.8). All of these properties can be extended to three, or evenn, dimensional systems.
1. (Superposition)Ifx1(t) andx2(t) are any solutions andc1andc2are any constants, then the linear combinationc1x1(t) +c2x2(t) is a solution.
2. (General Solution) If x1(t) and x2(t) are two linear independent solu- tions (i.e., one is not a multiple of the other), then all solutions are given byx(t) =c1x1(t) +c2x2(t), wherec1 andc2 are arbitrary constants. This combination is called the general solutionof (5.8).
3. (Existence-Uniqueness)The initial value problem x=Ax, x(t0)=x0,
wherex0is a fixed vector, has a unique solution valid for all−∞< t <+∞. The existence-uniqueness property actually guarantees that there are two independent solutions to a two-dimensional system. Let x1 be the unique so- lution to the initial value problemx1 =Ax1, x1(0)= (1,0)T and x2 be the unique solution to the initial value problemx2 =Ax2, x(0)=(0,1)T. These must be independent. Otherwise they would be proportional and we would have
x1(t) =kx2(t),
for allt, wherekis a nonzero constant. But if we take t= 0, we would have (1,0)T =k(0,1)T,
which is a contradiction.
The question is how to determine two independent solutions so that we can obtain the general solution. This is a central issue we address in the sequel.
One method to solve a two-dimensional linear system is to eliminate one of the variables and reduce the problem to a single second-order equation.
Example 5.14
(Method of Elimination) Consider
x = 4x−3y, y = 6x−7y.
Differentiate the first and then use the second to get
x = 4x−3y= 4(4x−3y)−3(6x−7y)
= −2x+ 9y=−2x+ 9(−1 3x+4
3x)
= −3x+ 10x,
which is a second-order equation. The characteristic equation isλ2+3λ−10 = 0 with rootsλ=−5,2.Thus
x(t) =c1e−5t+c2e2t. Then
y(t) =−1 3x+4
3x= 3c1e−5t+2 3c2e2t. We can write the solution in vector form as
x(t) = x(t)
y(t)
=c1 e−5t
3e−5t
+c2 e2t
23e2t
. In this form we can see that two independent vector solutions are
x1(t) = e−5t
3e−5t
, x2(t) = e2t
23e2t
,
and the general solution is a linear combination of these, x(t) = c1x1(t) + c2x2(t). However simple this strategy appears in two dimensions, it does not work as easily in higher dimensions, nor does it expose methods that are easily adaptable to higher-dimensional systems. Therefore we do not often use the elimination method.
But we point out features of the phase plane. Notice that x1 graphs as a linear orbit in the first quadrant of thexy phase plane, along the ray defined by of the vector (1,3)T. It enters the origin ast→ ∞ because of the decaying exponential factor. The other solution,x2, also represents a linear orbit along the direction defined by the vector (1,2/3)T. This solution, because of the increasing exponential factore2t, tends to infinity ast→+∞. Figure 5.6 shows the linear orbits. Figure 5.7 shows several orbits on the phase diagram obtained by taking different values of the arbitrary constants in the general solution.
The structure of the orbital system near the origin, where curves veer away and approach the linear orbits as time goes forward and backward, is called asaddle pointstructure. The linear orbits are sometimes calledseparatricesbecause they separate different types of orbits. All orbits approach the separatrices as time gets large, either negatively or positively.
y
x x2(t)
–x2(t)
–x1(t)
x1(t)
Figure 5.6 Linear orbits in Example 5.14 representing the solutions corre- sponding to x1(t) and x2(t), and the companion orbits −x1(t) and −x2(t).
These linear orbits are called separatrices.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1.5
−1
−0.5 0 0.5 1 1.5 2
x
y
Figure 5.7 Phase portrait for the system showing a saddle point at the origin.