5. 2. Undamped Forces 179
V(:|:) _ZI
>. ‘IT
-1|’
Figure 8. Pendulum: potential energy
U
V I
-1r
5/
Figure 9. Pendulum: phase plane
The level set E'1(2) in —1r 5 a: 5 rr is the union of two curves that come
together at (—1r,0) and (1r,0). The trajectory on the top cm~ve is asymptotic to
(1r,0) as t goes to oo and is asymptotic to (—1r,0) as t goes to —oo. The bottom
orbit is asymptotic to the opposite xed points as t goes to plus and minus in nity.
This picture repeats itself in every strip —1r + n21r < :0 < rr + n21r. Such orbits are
called heteroclimic orbits, because they tend to different xed points at plus and
minus in nity.
The potential function V(:1:) 5 2 for all a:. Therefore, for C2 > 2, there are
two values of y for each rt in E'1(C2). Thus, E" (Cg) is the union of two cm'ves,
one with y > 0 and one with y < 0. The top trajectory has a: increasing to
00 as t goes to oo and the bottom trajectory has :1: decreasing to —oo as t goes
to oo. If we consider :1: as a periodic angular variable, then each of these two
trajectories is periodic and each just goes around and around in a single direction;
these two different trajectories go around in opposite directions. They are called
rotary solutions. See Figure 9. As noted in Chapter 1, for the pendulum, there are
periodic orbits near the origin and nonperiodic rotary orbits farther away.
Also as noted in Chapter 1, the period of the orbits surrounding the origin
depends on the amplitude. The plots of the three solutions, which are time periodic,
are plotted in Figure 10. This is also true for Example 5.1.
180 5. Phase Portraits Using Scalar Flinctions
41::
2.
O. t
16
_2f
__4<l a
Figure 10. Time plots of solutions with different amplitudes for the pendulum
showing the variation of the period _
Notice that, for any system of the form
:2: = y,
if : F(x)!
the matrix of partial derivatives at a xed point (a:", O) is
(P&a5l=@v&m 9'
with a characteristic elquation /\2 + V" (:c') = 0 and eigenvalues /\ = zt,/—V”(:|:').
Thus, if the point 2:‘ is a local minimum of V(:c) with V” (:1:') > 0, then the xed
point has purely imaginary eigenvalues and is a center for the linearized system. ln
fact, the analysis given in this section shows that such a xed point is surrounded
by periodic orbits, and so is a nonlinear center. If the point :i:‘ is a local ma.ximum
of V(:r) with V”(a:') < 0, then the xed point has one positive and one negative
real eigenvalue and is a saddle for both the linearized system and the nonlinear
system. Compa.re with the preceding examples.
Example 5.4. As a nal example, consider the system
i = 1/.
'y=—:c—a:2.
The potential energy is V $2 Is
and the energy function is
@=?+?
Eaw=wo+%2-
The xed points, which are critical points of the potential function, have :1: = 0, —1.
The potential function has V(0) = 0 and V(—1) = 1/6. Since V(0) = 0 is a local
minimum of the potential function, it corresponds to a center for the differential
5.2. Undamped Forces 181
equation; since V(—1) = 1/6 is a local maximum, it corresponds to a saddle for the
differential equation. See the graph of the potential function in Figure 11.
We need to consider level sets of the energy function for C6 < 0 = V(O), for
C1 with V(0) = 0 < C1 < 1/6 = V(—1), and for C2 > 1/6.
V(=v)
C2 T\‘ _ V 1:3 $C1
1 $1 -1 2:2 1
60’
Co _
Figure 11. Potential energy function, V, for Example 5.4
First, consider C1 with 0 < C1 < 1/6. There are three points a:1, 2;, and $3
with 2:1 < -1 < 2:2 < 0 < 2:3, such that V(:z:) = C1. The set
' {:c:V(a:)5C1}
has two parts: an interval (—oo,:r1] from minus in nity up to a value :21 and the
closed interval [:z:2,:r:3]. The corresponding level set E'1(C1) has only the value
of y = 0 for a: = 131,$g,!I.‘3, and two values of y for zr < 2:1 and I2 < a: < :03.
The part with $2 5 :1: 5 2:3 closes up for E'1(C1) to give a single closed curve
surrounding (0,0). The part with :1: 5 av, gives a ctu've that opens up to the left
in the plane to give a tra.jectory which comes and goes to in nity. See Figure 12.
For C2 > 1/6 and C0 < 0, the sets
{a::V(cc)5C6} and {a::V(a:)5C2}
form intervals from minus in nity up to a value :c_.; = V“1(C_,-). The level set
E'“1(C_.;) for j = 1, 2 is a. single curve that crosses the axis y = 0 once at :1: = :r_,-
and then runs off to z = —oo as t goes to both ztoo. See Figure 12.
For the value of 1/6, the line of constant height 1/6 is tangent to the potential
function at the local maximum at a: = -1. The potential function equals 1/6 at
a: = -1 and 1/2:
O=V(:v)—E1 =E1(2:t3+3a:2—1)
= %(a: +1)2(2:1: — 1).
For as = 0.5,—1, the only y is y = 0 on E‘1(1/6). For a: < —1 or -1 < zr < 0.5,
there are two values of y on E‘1(1/6). Thus, this level curve E"l(1/6) has a
homoclinic orbit to the right of the xed point :1: = -1; to the left of the xed
182 5. Phase Portraits Using Scalar Functions
U
-1
3
-1
- -1
Figure 12. Phase portrait for'l3xa.mple 5.4
point, there is an unbounded orbit on W’(—1,0) coming in from the left and a.n
unbounded orbit on W“(—1,0) going off to the left. See Figure 12. Figure 16 in
the next section shows the level sets E'1(O) and E" (1/6).
Exercises 5.2
1. Consider the system of differential equations,
i=2/,
y=—:c+:v3.
a. Find the xed points.
b. Find the potential function V(:1:) and draw its graph. Classify the type of
each xed point.
c. Draw the phase portrait for the system of differential equations.
2. Consider the system of differential equations,
i=2/.
3)=—I-l-I2.
a. Find the xed points.
“b. Find the potential function V(:c) and draw its graph. Classify the type of
each xed point.
c. Draw the phase portrait for the system of differential equations.
3. Consider the system of di erential equations for a pendulum with constant
torque,
='==y,
y = —sin(a:) + L,
forO<L<1.
5.3. Lyapunov Functions 183
a. Fi.nd the xed points. Hint: You will not be able to give exact numerical
values of :1: at the xed points, but merely give an expression for them. In
which intervals of [O,1r/2], [1r/2,-rr], [1r,31r/2], or [31r/2,21r] do they lie?
b. Find the potential function V(;r) and draw its graph. Classify the type
of each xed point. Hint: The critical points (with zero derivative) con'e-
spond to the xed points of the differential equation.
c. Draw the phase portrait for the system of differential equations.
d. How do the answers change if L = 1 or L > 1? How about 0 > L > -1,
L=—l,orL<—l?
4. Consider the system of differential equations,
1'=y,
y=—a:(:z:—1)(a:+2)(a:2 -_0)=-=5-i‘+11w3+9$2-18¢.
a. Find the xed points.
b. Find the potential function V(:c) and draw its graph. Classify the type of
each xed point. Hint: I/(-2) > V(1).
c. Draw the phase portrait for the system of di erential equations.
5. Consider the system of differential equations,
1% = y,
y = 2:3 + 2:2 — 22:.
a. Find the potential ftmction V(:c) and draw its graph.
b. Find the xed points of the system of differential equations and classify
the type of each.
c. Draw the phase portrait for the system of differential equations. Pay
special attention to the location of the stable and unstable manifolds of
the saddle xed points.
6. Consider the system of differential equations,
1: = yr
Q : '_vl($)>
where the potential function is given in Figure 13. Draw the phase portrait for
the system of differential equations. Pay special attention to the location of
the stable and unstable manifolds of the saddle xed points.
7. Give the w-limit sets and 0:-limit sets for Example 5.1.
5.3. Lyapunov Functions for Damped Systems
The preceding section considered physical systems that preserve energy. By adding
a damping term (or friction), the energy is no longer preserved but decreases and
the equations are called dissipative.
A damping term is a force caused by the motion of the particle; thus, the
damping G(:c,y) depends on the velocity y and possibly on the position a;. Since
the damping slows the particle down, a positive value of y results in a negative
184 5. Phase Portraits Using Scalar Functions
Figure 13. For Exercise 6 in Section 5.2
value of damping G(a:,y). The simplest type of damping depends linearly on the
velocity and is independent of position, yielding the system of differential equations
(5.1) a': = y,
17 = F($) - bill-
We start by adding a linear damping term to the pendulum equations consid-
ered in the preceding section. We use an argument about the basin of attraction
of the origin W’ (0) based on consideration of energy. Then, we abstract the ideas
from this example to de ne Lyapunov functions.
Example 5.5 (Pendulum with Damping). For the pendulum with linear damp-
ing, we have the equations
T = 14/,
y = -sin(:r) — by,
where b > O. The term —by is the damping. (If b = 0, we a.re back in the situation
with no damping.) The xed points are again y = 0 and :1: = n1r for any integer n.
The linearization is
( ° ‘ )—cos(a:) —b '
bi\/b2+4
A direct check shows that the eigenvalues at 2: = :t1r are /\ _= -%——, and
these xed points are still saddles. At :1: = 0 or even multiples of 1r, — cos(0) = -1,
and the characteristic equation is /\2 + b/\ + 1, which has eigenvalues
,\=_?—°2bb2d_ 4’ = —g:ti 1—(b2/4).
These eigenvalues a.re complex for 0 < b < 2, and real for b 2 2. In any case, the
xed point is attracting. The linearization does not tell much about the size of the
basin of attraction, but it does say that nearby orbits tend to the xed point.
To nd out more about the basin of attraction, consider the energy obtained
for the system with no damping,
yz
E(w- :1) = 3 + (1 — @<>S(@))-
5.3. Lyapunov Functions 185
The time derivative along a solution is given as follows:
. d2
E = E + (1 — cos(a:))] = yy + sin(:r):t
= yl— $in(w) — In/l + Sin(w):u
= —by2 < 0.
ll
2
Figure 14. A Trajectory of damped pendulum crossing the level sets of
E(:1:,y), Example 5.5
1/
0
I
Figure 15. Phase Plane for Example 5.5
The time derivative of E is strictly negative, except when y = 0, or off the
line Z = { (zc, O) Thus, the energy is strictly decreasing except when the velocity
y is zero. If a trajectory has y = 0 but is not at a xed point, then 3) 76 0 and
the trajectory moves off the line y = O and E starts decreasing again. Thus, a
trajectory that starts with E(a:0,y0) < 2 and —1r < at < 1r, will have decreasing
energy along the whole trajectory. Whenever it crosses y = 0, it will have E = 0 for
186 5. Phase Portraits Using Scalar Functions
an instant, but will then start decreasing again. The energy must decrease toward
the minimum at (0,0); thus, it will tend asymptotically toward this xed point.
See Figm'es 14 and 15. Therefore, all these initial conditions are in the basin of
attraction of (0, 0).
De nition 5.6. Assume x‘ is a xed point for the differential equation x = F(x).
A real-valued function L is called a weak Lyapunov function for the differential
equation provided there is a region U about x‘ on which L is de ned and (i)
L(x) > L(x‘) for all x in U but distinct from x‘, and (ii) L(x) 3 0 for all x in
U. Notice that we allow the time derivative to be zero. In particular, the energy
function for a conservative differential equation is a weak Lyapunov function near a
loeal minimum of the energy function. The function L is called a Lyapunov function
or strict Lyapunov function on, a region U about x-" provided it is a weak Lyapunov
function which satis es L(x) < 0 for all x in U but distinct from x‘.
To use a Lyapunov function to analyze the phase portrait of a system of equa-
tions, somehow we must discover or invent the function L. However, we do not
need to be able to calculate the solutions in order to calculate L. In particular,
we do not have explicit representations of the solutions of the damped pendulum
problem.
The following result gives the rst result about stability of the xed points with
a weak Lyapunov function or Lyapunov function.
Theorem 5.1. Let x‘ be a xed point of the di erential equation x = F(x).
a. Assume that L is a weak Lyapunov function in a neighborhood U of x‘.
Then, x‘ is L-stable.
b. Assume that L is'~o (strict) Lyapunov function in a. neighborhood U of x‘.
Then, x" is attracting. Further assume that Lg > L(x‘) is a value for which
ULO = {x G U : L(x) 5 L0}
is contained inside U, is bounded, and is bounded away from the boundary of U
(i.e., containedin the interior of U). Then, the set ULO is contained in the basin
of attraction of x‘, ULO C W’(x').
Proof. (a) Let x(t) = ¢(t; xo) be a trajectory. Because the time derivative is less
than or equal to zero,
L(X(¢)) — L(X0) = K L(>=(s>>ds 3 0.
provided x(s) is in U for O g s 3 t. Thus, the solution cannot cross the level curves
to a higher value of L. Fix an e > 0. By taking a suf ciently small value of L0, the
set
U1,“ = {x E U: L(x) 3 L0} C {x : ||x— x"|| < e},
and it must be positively invariant. On the other hand, there is a 6 > 0 such that
{x : ||x— x"|| < 5} C Ugo.
Therefore, any initial condition xo with ||x0 —x‘ ll < 6 must have xo in ULo, ¢(t; xo)
in ULO for t Z 0, and so ||¢(t;x0) — x"|| < e for all t Z 0. Since e is arbitrarily
small, the xed point is L-stable.
5.3. Lyapunov Functions _ 187
(b) Now assume that L < 0. Take xo in U50, but not equal to x‘. If ¢(t; xo) did
not converge to x‘, then L(¢(t;x0)) § —K < 0 for t Z O. But then L(¢(t;x0)) 5
L(x0) — Kt goes to —oo. This is impossible since it must stay greater than L(x‘).
This contradiction shows that ¢(t;x0) must converge to x‘; that is, x‘ must be
attracting. U
Example 5.7. Consider
:i: = —y — 23,
u=¢—f-
Let L(:r:, y) = %(a:2 + ya), half of the distance from the origin squared. Then
L- = gs-+g—L yt=¢¢+w= wt-y-w“) +1/ti--1/3) = —w“ -11‘.
which is negative except at (0,0). Thus, the square of the distance to the origin
is strictly decreasing as t increases, and the trajectory must eventually limit on
the xed point (0,0). Therefore, (0,0) is asymptotically stable, and the basin of
attraction is the whole plane.
Unfortunately, Theorem 5.1 shows only that the pendulum with damping is
L-stable and not that it is asymptotically stable. To understand the key aspects
that make the damped pendulum asymptotically stable, we need to consider more
carefully the points where L = 0.
Theorem 5.2. Let x‘, L, and U be as in the preceding theorem, where L is
assumed to be n. weak Lyapunov function. De ne
ZU={X€UZL(X)=0}.
Assume that U is su iciently small sudh that, for any x1 E ZU \ {x"}, the tra-
jectory ¢(t;x1) moves of ZU into U \ ZU for small positivet (i.e., {x"} is the
largest positively invariant set in ZU). Then, x‘ is asymptotically stable.
Further assume that L0 > L(x") is a value for which
U50 = {X G U I L(x) s Lg}
is contained inside U, is bounded, and is bounded away from the boundary of U.
Then, the set U1“, is contained in the basin of attraction of x‘, U14, C W"(x").
The proof of this theorem is much the same as the analysis for the pendulum
with damping and is given in Example 5.5. A general proof is given in Section 5.7.
Sometimes we can use this theorem to prove that a xed point is asymptotically
stable even though its linearized equation is a center (has eigenvalues with zero
real part). When we want to emphasize the fact that the linearized system at
a fixed point is a center, we call such an attracting xed point weakly attracting
rather than attracting. When we want to emphasize that the linearized system
has an attracting xed point as well as the nonlinear system, we call the xed
point strongly attracting. In particular, we use this terminology when discussing
the Andronov—Hopf bifurcation in Section 6.4.
If we reverse the signs of the time derivatives, then we can get a result about
the unstable manifold of the xed point.
188 5. Phase Portraits Using Scalar Fbnctions
Theorem 5.3. Let x‘, L, and U be as in the preceding theorem, but now we assume
that
L(x) Z 0
for all x in U. We still assume that L(x) > L(x‘) for x in U but distinct from x’.
De ne
ZU={x€U:L(x)=0}.
Assume that U is su iciently small such that, for any x1 G ZU\{x‘}, the trajectory
¢(t;x1) moves o ZU into U \ ZU for small negative t (i.e., {x"} is the largest
negatively invariant set in ZU). Then, the unstable manifold of x‘ contains a whole
neighborhood of x‘; that is, x‘ is weakly repelling.
" Further, assume that L0 > L(x") is a value for which
Uis = {X E U 1 L(x) S Lo}
is contained inside U, is bounded, and is bounded away from the boundary of U.
Then, the set ULO is contained in the unstable manifold of x*, ULO C W“(x').
Example 5.8. Consider the basin of attraction for (0, 0) for the system formed
by adding a damping term to Example 5.4:
i1 = 1/.
y = —(:r + :22) — 0.3y.
The natural energy is obtained by considering the system without the term —0.3y
in the y equation:
_ L (:v,y) =12y2 + 12:1:2 + 13:1:3_
Let V(:r) = $2/2 + 2:3/3 be the potential energy term. The xed points have
a:=0,—1 with y=0.
V(:c) I L-1 e) |
U
/
U
Ls
-1.5 0.5
(=1) (b)
Figure 16. Example 5.8: (a) Graph of V(:i:). (b) Level sets L(:|:,y) = 0,
L(z,y) = %, and U.
5.3. Lyapunov Functions 189
The value of the potential function V(0) = 0, so we must consider values of :1:
for which V(:r) 2 0, i.e., 1: 2 -1.5. See Figure 16(a).
As before, L = L,:i; + L,y = (=t- + 12);, + y(—:1: - $2 - 0.31,) = -0.3 y2 g 0. The
set where L is zero is
ZU = {(:i:,0) G U}.
To make the xed point at (0, 0) be the maximal invariant set in the set ZU, we need
to exclude the other xed point at (-1, 0) from U. Since L(—1,0) = V(—1) = 1/6,
we take
U={ (a:,y):O$L(:1:,y)< J7>—l
See Figure 16(b). If we take any L0 < 1/6, U 1,0 is contained in the basin of attraction
for 0. Since L0 < 1/6 is arbitrary, all of U is contained in the basin of attraction
of 0. The set U is equal to the set of points inside the saddle connection for the
saddle xed point for the equations without damping of Example 5.4. Thus, all
these points a.re in the basin of attraction. See Figure 17 for a plot of trajectories
and the basin of attraction.
,, 2 -S y
.-. r
iii’ 7*: »
i-
_g
_
-2
Figure 17. Basin of attraction for Example 5.8
Remark 5.9. Only special types of equations have Lyapunov functions. There is
no general way of nding such a function. We have given a method of nding one
for a system from an undamped force with “damping” added. If the equation under
consideration is formed by adding terms to an equation that is known to have a
Lyapunov function L, then sometimes a slight variation of L will be a Lyapunov
function for the new equation. The book by LaSalle and Lefschetz [LaS61] has
many such examples.
190 5. Phase Portraits Using Scalar Phnctions
Exercises 5.3
1. Consider the system
i = 2/.
y = —.1: + :23 — y.
a. Use a Lyapunov function to nd a (reasonably large) region contained in
the basin of attraction for 0.
b. Discuss why solutions go to the xed point.
.. c. Draw the phase portrait by hand.
d. Use a computer program to draw the phase portrait for representative
initial conditions. Be sure to include enough initial conditions to reveal
the important features of the system, including the stable and unstable
manifolds of any saddle xed points.
2. Consider the system
-7-‘=1/.
y=:z:—a:3—y.
a. Use a Lyapunov function to nd a (reasonably large) region contained in
the basin of attraction for the xed point sinks.
b. Discuss why solutions go to the xed point.
c. Draw the phase portrait by hand.
d. Use a computer program to d.raw the phase portrait for representative
initial conditions‘-.' Be sure to include enough initial conditions to reveal
the important features of the system, including the stable and unstable
manifolds of any saddle xed points.
3. Consider the system of differential equations
:i: = -13 + zcyz,
17 = —2w2y — 1/3-
Let L(:r, y) = $2 + yz.
a. Show that is a Lyapunov function.
b. What does L tell about the solutions of the system?
4. Consider the Lotka—Volterra equation for a food chain of three species given by
ii = 375l(7'1_ a1m— aizrvz) = Iiwi.
552 = I2(T2 + 02111 — 1122112 - 0-2a$3l = $21112,
is = Is(Ta + 432912 - assf sl = -Tsws,
where all the r, > 0 and a,,- > 0. The quantities w,- are de ned to be the
quantities in the parentheses on the same line. (Notice that, (i) $1 and :22
have a negative effect on the growth rate of 2:1; (ii) 2:1 has a positive effect
and 22 and :23 have a negative effect on the growth rate of 2:2; and (iii) ax;
has a positive effect and $3 have a negative effect on the growth rate of 2:3.
5.4. Bounding Functions 191
That is why it is called a food chain.) Assume that there is an equilibrium
(r1.1=2.Ia) = (P1.P2.Ps) with all the P1 > 0. so
1'1 = <111P1 + a12P2,
1'2 = —¢121P1 + 1122112 + 112:-11731
1'3 = —¢1a2P2 + ¢lsaPs-
a. Show that the quantities w, can be rewritten as
1111 = -a11($1" P1) - ¢l12($2 -P2),
11/2 = 21(I1 - P1) - ‘122($2 — P2) - ¢12a($a - Pa)»
we = ¢1s2(12 — P2) - '1as($a - Pa)-
b. De ne the function
3
I/(I1, 912.1:-\)'= Z ¢e($¢ - Pi 1I1(11)l.
-i=1
where the c.- > 0 are chosen so that
2 _ m and 2 = an
61 021 62 032'
Show that 3
L = — Z¢a0»=‘i(1F1' — Pi)2-
i=1
c. Show that the basin of attraction of the xed point (p1,p2,p3) includes
the whole rst octant
{(1!1,$3,223)2$1> 0, 222 > 0,173 > 0}.
5. Assume that L is a weak Lyapunov’ mction for the system x = F(x) in IR".
Assume that the forward orbit from X0 is bounded. Prove that L(y) = 0 at all
points y in w(x0).
5.4. Bounding Functions
In this section, for a given differential equation x = F(x), we try to pick a real-
valued function L(x) to determine the positively invariant sets and limit sets. The
time derivative L is usually both positive and negative in different parts of the
phase space so it is not a Lyapunov function, but we are still able to use it to nd
regions that are positively invariant. We call any such real-valued function a test
function. We call L a bounding function if some value C ca.n be used to locate a
positively invariant set surrounded by L_1(C).
De nition 5.10. A test function L from R" to IR is called a bounding function
for x = F(x) and a subset U of IR" provided that there is is a connected piece
B of L‘1(C) for some value C’ such that U is ‘the set of all points inside B with
B the boundary of U and they satisfy the following conditions.
(i) The sets U and B are bounded (do not go oil‘ to in nity).
(ii) The gradient VL(x) 96 0 at all points x of B.
(iii) If x is a point of B, then ¢(x; t) is in U \ B for all small enough t > 0.
192 5. Phase Portraits Using Scalar Functions
If conditions (i)—(iii) are satis ed, then the set UUB is then positively invariant
and the level set B bounds the forward trajectories starting inside. Note that
condition (ii) implies that B is a smooth level curve in dimension two and a smooth
level surface in dimension three.
If the time derivative L is nonzero (and so of one sign) on the level set B,
then the trajectories cross this level set in one direction. So, (iil’) if L(x) < 0 and
VL(x) points to the outside of U, then condition (iii) is satis ed.
Example 5.11. This example has a limit set that contains both a xed point and
orbits that have both their w-limit and oz-limit at the xed point. Consider
i = 2/.
11=.$—2==3+v(1=’ —I“-—y’)-
If the term containing y were not there in the 1) equation, the energy function would
be ‘
L($.11) = i_ 22i+ $4 + ~y2z.
The time derivative of this L for the whole original equations is given by
L = (—a: + 2a:3)y + y(a: — 2:23 + y(:c2 — $4 — 312))
Z 0 when L < 0,
= —2y2L = 0 when L = 0,
3 0 when L > 0.
Because its derivative is both positive and negative, it is not a Lyapunov function.
If L(zQ,y0) = 0,-then the orbit stays on the level set L'1(0), which contains
the xed point at the origin. This level set is bounded and one-dimensional, so all
the trajectories must converge to the xed point at the origin as t goes to plus or
minus in nity; that is, for these points, w(:|:Q,y0) = a(a:0, yo) = 0 and
(z:o,yQ) e W-"(0) n W“(o).
Let
1" = {(M1) = I/(m/) = 0}-
Since the stable and unstable manifolds of the origin are curves and contain F, they
must equal I‘,
F = W‘(0) = W“(0).
Any C > 0 and B = L'1(C) satis es conditions (i)—(iii) and so L is a
bounding function for the region inside L_1(C). (Note that L(x) is only 3 O, but
it still satis es (iii).)
If C = L(:c0,1/0) > 0, then the trajectory has to stay inside L‘1(C) and so is
bounded. The value L(¢(a:0, yo)) decreases along the orbit with its time derivative
equals to zero only when the orbit crosses the a:-axis (where y = 0). This orbit
¢(:z:q,y0) continues to cross the level set L'1(C’) for 0 < C’ < C’, going in toward
the level set L'1(0) = F. Thus, these orbits spiral down and accumulate on I‘;
for any point q E I‘, there is a sequence of times tn going to in nity such that
4b(t,,; (:z:0,1/0)) is close to q, so q G w(a:0,yQ). Since this is true for each such point
5.4. Bounding Functions 193
q 6 F, w(:1:o, yo) = 1". See Figure 18. For negative time, the orbit goes off to in nity
and a(a:0,y0) = 0.
L/
Figure 18. Phase portrait for Example 5.11
For a point (a:1,y1) with '1/3 < L(J:1,y1) < 0 and :21 > 0, the orbit spirals
out and accumulates on the right half of I‘,
OJ(I]_,y]) = F O {($,y) ZZ 2
See Figure 18. Going backward in time, the orbit approaches the xed point
(1/\/5, 0), so the a-limit is a(:r1,y1) = {(1/\/§,0)}.
Similarly, for a point (:c2,'y2) with ‘1/3 < L(a:;,y;) < 0 and .122 < 0, the orbit
spirals out and accumulates on the left half of F,
w(a:;,y2) = F r'1{(m,y) :1: 5 0}.
Also a(:c2,'yg) = {(-1/\/§,0)}.
For any of these orbits, whenever the limit set is nonempty, it contains one of
the xed points.
The next example illustrates that it is not necessary for both sides of the
unstable manifold to return, as in the preceding example.
Example 5. 12. Consider
rv = 1/,
1) = :1: — 1:2 + y(3a:2 — 21:3 — 33/2),
and L(a:,y) = —%i + 3; + In this case, only the right side of the unstable
manifold of the saddle point at the origin is in the stable manifold:
I‘+ = W“(0) H W"(0) = {(:1:,y) : L(a:,y) = 0, :1: Z 0}.
If -1/6 < L(x0,y0) < 0 and :50 > O, then w(:r0,y0) = l"+.
In each of the preceding two examples, the divergence of the vector eld at the
saddle point (origin) is zero, which is not the usual way this type of behavior occurs.
The generic properties of a planar differential equation that cause the homoclinic
194 5. Phase Portraits Using Scalar Functions
connection to attract nearby orbits are the following: (i) There is a saddle xed
point p with at least one branch I‘+ of the unstable manifold W“(p) contained in
the stable manifold W"(p), F"‘ C W“(p) OW‘ (p). (ii) Assume that the divergence
of the vector eld at p is negative. Then orbits starting at qo in the region bounded
by I‘+ have
w(qq) = {p} U I‘+.
Example 5.13. This is an example of a system of differential equations on R3
with the limit set equal to an ellipsoid together with a line segment. Consider the
system of differential equations
a’: = — y + 2 :1: z,
1') = 6 1: + 2 1/ Z) .I
2=l-—a:2—y2—z2+z(a:2+y2) (1; é (:z:2+y2) -22),
which in cylindrical coordinates is
é = 16»
1" = 2 1' z,
i=l—1'2—z2+r2z (l—§1'2—z2).
This example is motivated by an equation considered in [Guc83] used as a bifur-
cation from a xed point, with a pair of purely imagi.nary eigenvalues and zero as
the other eigenvalue.
We rst consider the system in the (1-,z)-space for r Z 0. The xed points are
(r, z) = (0,:l:1) and 4_(1,"0)|_. (The xed point (1,0) corresponds to a periodic orbit
in R3.) We use the test function '
L(r,z) = 'r (1 — £13 — zz).
Notice that 2 = l — 1'2 — 22 + 1'2 L. The time derivative of L is
5-,. =. ~1. I-1 ——r —z2)+'r(—§'r1"—2z2)
/X
=2zL——'rMQ 2-21-z(l—1-2—z2+rzL)
'-t'oAW
=2zL— Ix) N i (1—l'r2—z2)—2'/'2z2L
3
>0 when L<0,1'>0, and 2:750,
=—21‘2z2L= =0 when L=O,
<0 when L>0, r>0, and z9é0.
The level set
L_1(O)={(O,z)}U{('r,z):l=%1'2+z2,1'2 0},
is invariant, which is the z-axis together with a semi-ellipse.
Points inside the ellipse with 1» > 0 and Z ee 0 have L > 0 and L < 0. The
only point here that stays on the set {z = 0} is the xed point (1, 0). Therefore,
if we start at a point inside the ellipse pg = (T0, zq) 96 (1,0) and 1'0 > 0, then the
5.5. Gradient Systems 195
orbit d>(t; pg) converges to the set L“(0), the semi-ellipse together with the z-axis
between z = -1 and 1. Therefore,
w(pq)C{(O,z):—l§z§1}U{('r,z):l=%1-2+z2, 1'ZO}CL_1(O).
Turning to the ow on R3, the only xed points are (az, y, z) = (0, 0, :i:1). There
is a periodic orbit for 2 = 0, 1' = 1, and 0 arbitrary. The semi-ellipse in the plane
corresponds to an ellipsoid in R3,
L_l(0)={(0,0,Z)}U{(I,y,z) : l = -2 (:c2+y2) +22}-
If we start at a point qq inside the ellipsoid, but not on the z-axis and not on the
periodic orbit with z = 0 and 'r = 1, then by the analysis in R2,
w(q0)C{(O,O,z):—l§z$1}U{(:|:,y,z):l=-;(:c2+y2)+z2}.
Since the orbit goes past the xed points (0,0, il) slowly, it seems likely that the
w-limit set contains the whole ellipsoid together with the line segment in the z-axis.
We give further examples of boimding functions when considering periodic or-
bits in Section 6.2 and chaotic attractors in Chapter 7.
Exercises 5.4
l. Consider the system of differential equations
:i: = y, 0
3] = —(a:+rc2) — y(3y2 + 3:122 +2:1:3 — 1).
Give the w-limit sets and 0:-limit sets. Hint: Use the test function
L(a:,y) = E13,/2 + 51:22 + 512:3.
2. Suppose that A is a closed and l)OU11d6Cl set that is invariant by a ow ¢(t; x)
and there are no nonempty invariant closed proper subsets of A.
a. Prove that every trajectory is dense in A.
b. Prove that w(x0) = 0z(X9) = A for each xo in A.
5.5. Gradient Systems
In this section we consider a type of equations for which all the 01- and w-limit sets
are xed points and there can be no periodic orbits. Let G(x) be a real-valued
function on R", and let
VG(x) = 6£(x)
6131
5
a6TG..(")
196 5. Phase Portraits Using Scalar Flinctions
be its gradient of G at the point x (written as a column vector). The associated
system of differential equations
. 6G
$1 = —aT:l(Xl,
. 8G
fin — '-Tnnbi),
or in vector notation x = —VG(x),
--
is called a gradient system of di erential equations.
Any point x‘ where VG(x*) = 0 is called a critical point of the function and
is a xed point of the gradient system of differential equations.
Theorem 5.4. Consider a gradient system of di erential equations x = s—VG(x),
where G is a real-valued function. Then, G is a Lyapunov function (at least near
a local minimum) for the gradient system of di erential equations formed from G.
More precisely, G(x) 3 0 at all points and G(x) = 0 if and only if x is a xed
point. Away from the red points, G is strictly decreasing along trajectories.
Proof.
- “ ac. ac 2
Q. -M=
or in vector notation
G = VG(x) - >1 = VG(x) - (—VG(x))
= —||VG(><)||’ 5 0-
This time derivative is zero only at points where VG(x) = 0; that is, at xed
points of the differential equation. Away from the xed points, G < 0 and G is
strictly decreasing. El
Calculus books usually mention that the gradient is perpendicular to any vector
tangent to a level surface, so the vectors tangent to the level surface are just those
with VG(x) - v = 0.
Theorem 5.5. Let x = —VG(x) be a gradient system of differential equations.
Then, the vector eld VG(x) is perpendicular to the level sets of G at any point
that is not a. critical point (i.e., it is perpendicular to any vector tangent to the level
set).
Proof. Any tangent vector v that is tangent to the level set, can be given as the
tangent vector to a curve lying in the level set. Therefore, there is a cu.rve x(s)
5.5. Gradient Systems 197
d
with G(x(s)) E. G0, a constant. Letting v = $x(s)|s=0,
d
0 - EG0| s=0
d
: TsG(x(s))l3=Q
= VG(x(O)) — v.
Therefore, VG(x(0)) is perpendicular to v. Cl
Example 5.14. Consider the real-valued function associated to the two-welled
potential,
G($, y) = _2I 4I 2
It has critical points at (;l:1, 0) and (0,0). The points (il, O) are local rni.|1'una. and
(0,0) is a saddle. The gradient system is
:t = 1: — 2:3,
1? = -11-
The solutions ¢(t; (O, yD)) = (0, 'y0e“) all tend directly to the saddle point at the ori-
gin. Thus, the stable manifold of the origin is the y-axis. Any solution ¢(t; ($0,310))
with 2:0 > 0 cannot cross the stable manifold of the origin, and so must stay in
the right half-plane where z > 0. Therefore, all these points must converge to the
xed point (1, 0). Similarly, if J10 < 0, then ¢>(t; (a:Q,yQ)) converges to the attracting
xed point at (-1, O). See Figure 19 for the level sets of G and a few representative
trajectories.
n
U
21
'¢'€7»\
"Ilia-i.l9
-2
Figure 19. Level sets and trajectories for Example 5.14
198 5. Phase Portraits Using Scalar Fimctions
There is a close connection between the type of critical point of the function
G and the type of xed point for the associated gradient system of differential
equations.
Theorem 5.6. Let x‘ be a critical point of the real-valued function G and, there-
fore, a L'ced point of the associated gradient system of di ’erential equations. Then,
all the eigenvalues of the xed point of the system of di erential equations are real.
(i) If the critical point is a local minimum of G’, then it is an asymptotically stable
xed point (sink) for the associated gradient system of di erential equations. (ii) If
the critical point is a saddle of G, then it is a saddle xed point for the associated
gradient system of di erential equations, and so it is unstable. (iii) If the critical
point is a local maximum of G, then it is a source for the associated gradient system
of di erential equations, and so it is unstable. .
Proof. The vector eld giving the differential equation is
F(x) = -vG<=<> = (-;_6,;G;<’><)) .
Therefore, the linearization at xed points is
aic _
DF(x') ~ <"5;iEC;(X l).
which is the negative of the Hessian of G. This matrix is symmetric, so it has real
eigenvalues.
At a minimum of G, all the eigenvalues of the Hessian are positive, so all
the eigenvalues of DF-(,<) are negative; therefore, it is a sink and is asymptotically
stable. At a maximum of G, all the eigenvalues of the Hessian are negative, so all
the eigenvalues of DF,, are positive; therefore, it is a so1u"ce. At a saddle of G’,
there are some positive and some negative eigenvalues of the Hessian and also of
DFX, so it is a saddle xed point. E]
Theorem 5.7. Let x = —VG(x) be a gradient system of differential equations. Let
z be an a-limit point or an w-limit point. Then, z is a xed point for the gradient
system of di erential equations.
Proof. This proof is essentially the same as the proof of Theorem 4.4. Assume
that z is in w(x0). Then, G(¢(t;x0)) is decreasing. There is a sequence of times t,-
going to in nity such that ¢(t,-;xO) approaches z. We can take the times so that
t1 <t2<"'<tj <bj.|.1<--',SO
G(¢(i1;Xo)) > G(¢(¢2;x0)) > > G(¢('5a';Xo)) > G(¢(ij+1;Xo)) > -
By continuity of G, G(¢(tj; x0)) approaches G(z). The value G(d>(t; x0)) can never
get below G'(z) because it is strictly decreasing. Therefore, G(¢(t;xq)) must go to
zero. Again, by continuity, C(¢(tJ-;x0)) converges to C(z) so this must be zero;
that is, z must be a xed point. Cl
The theorem says that the w-limit points must be xed points. In fact, most
points go to sinks, since the stable manifolds of saddles are lower dimensional. Since
sinks are local minima of G, most points have w-limits at minima of G. Thus, going
5.6. Applications 7 V 199
along the trajectories of the gradient system of differential equations is one way to
nd a local minimum of G.
Exercises 5.5
1. Consider the real-valued function
G($1,Ig) = -21% + 12%.
Plot the phase portrait for the gradient system >2 = —VG(x). Include in the
sketch some representative level sets of G.
2. Consider the real-valued function
G(a=1. rm‘) = as? — 2:‘.: + mg
Plot the phase portrait for the gradient system x = —VG(,,). Include in the
sketch some representative level sets of G.
3. Consider the real-valued fimction
G($l > $2) = a311:? —2:r§+3a:§ .
Plot the phase portrait for the gradient system x = -VG(,,). Include in the
sketch some representative level sets of G.
4. Show that any differential equation 1': = f(2) on the line is a gradient differential
equation. What is the function G(a:) such that f(:l:) = —G"(a:)?
5.6. Applications
5.6.1. Nonlinear Oscillators. In Section 5.2, we discussed undamped forces
and showed how they preserved an energy function. However, we did not discuss
the relationship between the different equations and physical oscillators. In this
section, we discuss such connections.
We consider an equation of the form
-'15 = f(w)
or the corresponding system
(5-2) i = u.
2? = f(1)-
The function _f(x) is the force acting on the particle. (We have taken the mass
equal to one or incorporated it into the function f for simplicity.) If this is some
type of spring or restoring force, then often the force for —a: is equal to —f (a:)
(i.e., f (—a:) = — f(as) and f is an odd function). The harmonic oscillator or linear
Oscillator has f(:z:) = —w2a:, which is a linear odd restoring force. The pendulum
Eiven in Example 5.3 has f (2) = — sin(a:), which is a nonlinear odd restoring force.
Example 5.1 has f = :1: — $3 which is an odd function, but is repelling for small
displacements and is only attracting for displacements with |:c| > 1. In Example
200 5. Phase Portraits Using Scalar Fhnctions
5.4, the force f(2) = —:c - 22 is not an odd function and the eifect of positive and
negative displacements from :1: = 0 is very di erent.
If x(t) = d>(t;x0) is periodic, then each of the coordinates :0,-(t) is periodic
and the motion is an oscillation. Any nonlinear system of the form of equation
(5.2) that has a periodic orbit for some initial condition can be called an undamped
nonlinear oscillator or just an oscillator. With this general de nition, Examples
5.1, 5.3, and 5.4 are undamped nonlinear oscillators. However, since the force for
Example 5.4 is not odd and the orbits do not surround the origin, this system is
not what is usually called an oscillator.
Du ing equation
The simplest odd function contains ,both linear and cubic terms,
f(a:) = aa: + bra‘.
The differential equation with this cubic force is called the Du ing equation. For
small values of |:z:|, the linear term is largest and dominates, while for large values
of |:r|, the cubic term is bigger. The potential function is given by
24
V(a:)= —a.%—b%.
Comparing with the examples in Section 5.2, notice that for a < 0, the origin
is a center. If both a < 0 and b < 0, then the restoring force becomes stronger for
larger values of |a:|. This is called a hard spring. In this case, the potential fimction
has a single critical point at :2 = 0, and all the orbits are periodic.
If a < 0 but b > =0, then the restoring force becomes weaker for larger values
of ]:c|. In fact for su iciently large values of |:c|, the force is a repelling force rather
than an attracting one. For this reason, it is called a soft spring. The potential
function has three critical points: :1: = 0 and :t\/ I“/bl. The point a: = 0 is a local
minimum for V, so it is a center for the system of differential equations. The points
:1: = :i:,/I“/bl are maxima. for V, so they are saddles for the system of differential
equations. In this system, the origin is surrounded by periodic orbits, but other
orbits go o ' to in nity. This system has some similarity to -the pendulum equation.
If a > 0, then the force is a repelling for small displacements. The potential
has a local maximum at a: = 0, and the origin is a. saddle. If b < 0, then for
larger values of |a:|, the force is a restoring force; it is what we called the two-well
potential. Thus, there are two center xed points at (:c,y) = (i\/%,0). See
Example 5.1. There are some periodic orbits that surround one of the xed points
(:t,/ |°/bl, 0), and other periodic orbits that surround all three xed points.
If both a > 0, and b > 0, then the origin is the only critical point, and the origin
is the only xed point and is a saddle. This system is repelling for all displacements.
Six-twelve potential
Interatornic interactions are often described by a model using what is called
the sia:-twelve potential, given by
"<”l= rma. ‘ b
5.6. Applications 201
for the force b
a.
f($) - ‘E + F,
where a > 0 and b > 0. The potential has a unique critical point at (“/b)1/6, which
is a minimum. As z goes to zero, V(a:) goes to plus in nity. As :0 goes to in nity,
V(a:) goes zero. Thus, there is a center for the system of differential equations that
is surrounded by periodic orbits. Points with E(:i:, y) = V(:c) +3/2/2 > 0 have motion
that is 11l’lbOl.111ClC(l. We leave the details of the phase portrait as a.n exercise.
5.6.2. Interacting Populations. We have considered competitive and predator-
prey models for two interacting populations. In this section, we consider interac-
tions with more populations, where the equations are still of the type called Lotka-
Volterra equations,
n.
(5.3) :i;,- = 12,- r; + Eaijmj for 1 5 i 3 n.
j=1
A general reference for more results on these systems is [Hof88]. The book by
Hirsch and Smale [I-Iir74] considers systems where the part inside the parentheses
can be a nonlinear function. In this section, we let R1 = {x : an > 0 for all i}.
We next show that if there is no fixed point in R1, then the w-limit set of a
point has to be contained in the set with at least one of the populations equal to
zero. This result is a general type of competitive exclusion.
Theorem 5.8 (Exclusion property). Assume that a Lotka— Volterra system (5.3)
has no xed point in R1. Then for any xo G R1, w(xo) R1 = 9) and or(x0) R1 =
6I
Proof. The sets {r + Ax : x E R1 } and {0} are each convex and disjoint since
there are no xed points. A standard theorem in the theory of convex sets says
that there is a cER"\ {0} such that c-(r+Ax)<c-0=0fora1l xe R1.
We consider the Lyapunov function
L(x) = 20,- ln(:r:,-).
The time derivative is as follows:
(r,-+(A_x),-)=c-(r+Ax) <0.
Since this is strictly negative on all of R1, w(xo) F1 R1 = 0 and o(x0) F1 R1 = (ll
for any xo E R1. U
The following theorem is a generalization of the fact that the time average of
a periodic orbit for a predator-prey system is equal to the xed point.
Theorem 5.9. Consider a Lotka— Volterra system (5.3) where there is an orbit
X(t) =. ¢(t;x0) with xo G R1 such that
0<aS_a:,~(t)5A<oo forl i n.
Denote the time average of the orbit by
z(T) = 1/Tx(¢)ai.
T0
Then there has to be a red point of (5.3) p G R1 in the w-limit of ii*~3
Fhrther, if there is a unique red point p 6 R1, then z(T) converges to p as '*l‘.-’
goes to in nity.
Proof. We integrate ~; from 0 to T.
(1, 1n(w.(T)> —1n(=».(0>) = 1 /T h(t) d,
T T 0‘ I," '
= 1/T r,~dt+Za,-i/T :1:-(t)dt
= T; + 204'," Zj(T).
.7'
Since a 3 :r:;(T) 3 A, the limit of the left side of (*) is zero. Si-nce a 3 z,-(T) 3 A,
there has to be a subsequence of times Tn such that z(T,,) converges to a point p
with a.3p,-3A and
0 = r + Ap.
So p is a xed point in R1. If the xed point is unique, then it is isolated. Thus
z cannot have any other w-limit points and the limit must exist. U
_ 5.6.3. Replicator equations. In this chapter, we have considered the popula-
tion models given by the Lotka—Volterra equations. They are examples of equations
of the form :i:; = xi f,-(x), called Volterra equations. For the Lotka—Volterra equa-
tions, the functions fi are linear. In this subsection, we consider replicator equations
where the functions _f,- contain quadratic terms.
Evolutionary game theory developed by applying some concepts from biology
to the situation in game theory where a population competes against itself. The
replicator differential equations adds continuous adjustment toward Nash equilibria
into evolutionary game theory. Thus, the replicator differential equations introduce
dynamics into game theory and a type of stability for some Nash equilibria. We
content ourselves by introducing the replicator differential equations, giving a few
basic results, and presenting some examples. The book [GinO0] by Gintis discusses
the connection of the replicator differential equations with evolutionary game the-
ory. The book [Hof88] by Hofbauer and Sigmund gives a much more complete
treatment of the replicator differential equations. In writing this section, we have
also consulted the notes by Schecter [Sch06] that are used for a course taught using
[Gin00].
In the game theoretic context of the replicator equation, individuals within a
large population can choose from a nite number of choices of actions {s1, . . . , sn}.
A square n X n matrix A is speci ed that gives the payoff of playing one action
against another: a,-j is the payoff for the individual playing action s; against another
5. 6. Applications 203
playing sj. This is called a symmetric game because the playoff for the second player
playing sj against s,- is aji. The matrix A is not necessarily symmetric.
Let
S= {P= (P1.---,Pn)T 1m 20 for all @2311 =1}
be the simplex of probability vectors. An element p E S is called a state of the
population and corresponds to individuals within a large population playing the
action s,- with probability pj. The payoff of an individual who takes action si
against a population with state p is assumed to be
Z], 04-_-;Pj = (°i)TAP = 9* - AP,
where e‘ is the standard unit vector with a one in the 13"“-place. In the same way,
the payoff of a state q playing against a state p is assumed to be
2”. q.-at-J-P5 = QTAP = q - AP-
For the replicator system of di erential equations, the relative rate of growth
15¢/P, is set equal to the difference between the payoff of playing action si and the
payoif of playing p,
(5.4) pi = p,- (ei - Ap — p - Ap) for 1 5 i 5 n.
Thus, if ei - Ap > _p - Ap, then the proportion of the population playing action .9,-
increases, and if e‘ - Ap < p - Ap, then the proportion of the population playing
action s; decreases.
In the analysis of the dynamics, we use the “faces” of S where certain p,- are
zero. For a nonempty subset I of {1, . . . ,n}, let
S1={p€S:p;>OifiEI,pj=Oifj¢I},
§;={p€S:p,->0f0ra.lli€I} and
Notethat I(P)={i=p1~>0}-
S=U{S1:IC{1,...,n}}
and
cl(S])={p€S:pj=Oifj¢I}.
We have the following straightforward result.
Theorem 5.10. The replicator system of equations (5.4) has the following prop-
erties.
a. S is invariant.
b. For any nonempty subset I of{1,...,n}, cl(S;), S; and $1 are invariant.
Proof. (a)
%(ZiPt) = 2iPiei -AP — gimp - AP
= P-AP — (Zip.-) (P-AP)
= (1 — Zia) (P'AP)-
204 5. Phase Portraits Using Scalar Functions
It follows that the set of the vectors S’ = {p 6 IR" : 3,1», = 1 } is invariant. The
derivative pi = 0 on the hyper-plane {pi = 0}, so it is invariant and a solution
starting with Pi 2 0 cannot cross into the region with pi < O. Thus, a. solution
starting in S must stay in both S’ and the region {p : pi Z O for 1 3 i 3 n},
i.e., S is invariant.
(b) A solution starting in cl(S1) starts and remains in the subset of S with
p,- = 0 for i ¢ I, so in cl(S;), i.e., cl(S;) is invariant.
Similarly, the set cl(S1) \ S; = {p G cl(S;) : p,- = O for somei 6 I} is
invariant. It follows that S; = cl(S1) \ (cl(S;) \S;) is invariant.
Since S; = Um, S,-, S; is also invariant. El
Theorem 5.11. A point 13 is a zed point of (5-'4) -if and only if ei - A13 = 1') - A15
whenever 15, > 0. So, e‘ - A13 has the same value for all i with pi > 0.
The proof follows directly from the replicator system of differential equations
(5.4).
Theorem 5.12. If the replicator system of di erential equations (5.4) does not have
any red points in the interior int(S) = S(1_____,,}, then w(p) C 6(S) = S \ int(S)
and a(p) C c'9(S).
Remark 5.15. Theorem 5.8 gives a similar result for Lotka—Volterra equations in
arbitrary dimensions. If there is no fixed point in R1 = {x : a:,~ > O for all i},
then w(p) H R1 = (D for any p.
Proof. We consider the case for the w-limit set, and the proof for the a-limit set is
similar. Also, because the boundary is invariant, if p G 6(S), then w(p) C 6((S).
Therefore, we can assume that p E int(S). We next construct a strictly decreasing
Lyapunov function on int(S).
For the proof, we use the following two convex sets: A = {y G IR" : yl =
-- - = y,,} and_ W = A(int(S)). The set A is a subspace and so is convex; W is
the linear image of a convex set, so is convex. It follows from Theorem 5.11 that
p G int(S) is a xed point if and only if Ap G A. Because there are no xed points
in the interior, W F1 A = 0. Since these two convex sets are disjoint, a standard
theorem in the theory of convex sets says there is a c G IR" such that c - z < c - y
for all zé W and all y G A. But c-y=Z,-c,y,- =y1 For c-z <y1 2,6;
to hold for both large positive and negative y1, we need 2,0, = O. Since each
z 6 W equals Ax for x 6 int(S), c - Ax < 0 for all x 6 int(S).
Using this c, de ne V(x) = E,e;ln(z;) on int(S). Then
l./(x) = Zc;(A.x),- — x-Axzq = c-Ax < 0.
fly the argument of Theorem 5.1(b), if there were an x G w(p) O int(S), then
V(x) = 0. Thus, w(p) n int(S) = ill and w(p) C 6(5). Cl
Example 5.16. When A is 2 x 2, the differential equation is determined by a
scalar equation by setting a: = pl and P2 = 1 — a;. The differential equation for
5. 6. Applications 205
_01 21 is
a':=:c[—a:+2(1—:r:)+:v2 —2:z:(1—:v) —(1—:r:)2]
= 2 [—a:(1 — :1:)+ 2(1 — :c)2 — (1 — a:)2]
=:v(1—:c)[—:c+1—a:]
= :c(l — :c)(1— 21:).
The xed points are 2 = 0, 1/2, 1. The two end points 0 and 1 are repelling, and
1/g is asymptotically stable.
Example 5.17. The differential equation for is
:i:=:c[:c,—:r:2—2(1—:c)2]
= :z:(1 — a:)(3a: — 2).
The xed points are a: = 0, 2/3, 1. The two end points 0 and 1 are asymptotically
stable and 2/3 is repelling.
0 -1 has aj,- = —a.,J- so p-Ap =
Example 5.18. The 3 X 3 matrix ( 1 0 —
I-5 0-I C)»-l>—* L-Z
—p - Ap = 0, and the differential equations are p,- = pi (e1 - Ap). Therefore,
P1 = P1 (—P2 +Pa).
P2 = P2 (P1 — Pa).
I33 = Pa ("P1 + P2)-
Letting P3 = 1 — pl — pg, we get a system with two variables,
151 = P1(1 — P1— 2P2),
I32 = P2(2P1 +212 — 1)~
The xed points are (0,0), (1, 0), (0,1), and (1/3,1/3). A direct check shows that
(0,0), (1,0), and (0,1), are saddles and (1/3, 1/3) is a center.
For the original equations, E‘/E (—p1pgP3) = 0, so L = —p1pgp3 is constant along
solutions so is an integral of motion. For equations with two variables, the integral
becomes L(P1,P2) = —p1pg(1 —p1 — pg), which has a minimu.rn at (1/3,1/3) and
is a weak Lyap1mov function for this xed point. Therefore (1/3, 1/3) is Lyapunov
stable but not asymptotically stable.
06—
Example 5.19. (Hofbauer and Sigmund) The 3 x 3 matrix ( —3 0 ) has
—1 3 601%
the system of differential equations
I51 = P1 (5112 - 4P3 — 3P1P2 + 5PrPs - 3112173),
P2 = P2 (—3P1 + 5P3 — 3P1P2 + 5P1Pa — 8P2P_a).
Pa = Pa (—P1 + 3102 — 3171172 + 5P1P3 — 3P2P3)-
206 5. Phase Portraits Using Scalar Flmctions
Setting Pa = 1 - P1 - P2.
P1 = P1 (-4 + 9P1 + 2P2 — 5Pi + 39%).
P2=P2(5-3P1-13P2-5Pi+3P§)-
This system has xed points at (0,0), (1,0), (0,1), (0»5/8)‘ (4/5,0), and (1/3,1/3).
The matrix of partial derivatives is
-4 + 18p] + 2pg — 15p? + 8p§ 2p; + l6p1pg
DF(Pl»P2) = < —3pg — 10P1P2 5 — 3171 — 26pg — 5p? + 244%) '
The matrix at the various xed points and their stability type are as follows:
ob | -. 3- saddle,
DF(0'o) = l
(1,0) is asymptotically stable,
O
< 1, 0) is repelling,
\/\ |_n _23> ,
DF(1|o) = l (§,0) is a saddle,
O
3)»DF(o.1> = C005
_ a
DF(%‘°) = {KIends 52) ,
O
"Fro 2) = KZ _0E) , (0, g) I.S a saddle,
a\_ ii
E (%, is asymptotically stable.
”Fes> = _-1m0u"‘? jg) ,
Z‘ ~w' e 9
The characteristic equation at (1/3, 1/3) is /\2 + 3%/\ + 3% = 0 with eigenvalues — % :i:
365-1. Thus, (1/3,1/3) is asymptotically stable. See Figure 20 for the phase portrait.
P2
1.0
0.8
0.6
0.4
0.2
0-1 0.2 0.4 ' 0.6 o.a 1.0 P1
Figure 20. Example 5.19
5.6. Applications 207
5.6.4. Monetary policy. A model of N. Obst for in ation and monetary policy
is given in [Chi84]. The amount of the national product is Q, the “price” of the
national product is P, and the value of the national product is PQ. The money
supply and demand are given by M, and Md, and the variable p. = M4/M, is the
ratio. It is assumed that Md = a.PQ where a > 0 is a constant, so it = °PQ/M,
The following rates of change are given:
%% = p is the rate of in ation,
5% = q is the rate of growth of national product, and
iM, at -" m is the rate of monetarY exPansion '
We assume q is a constant, and p and m are variables. The rate of change of
in ation is assumed to be given by
ddpt -_h( M8Ms _ Md _-hu ll),
where h > 0 is a constant.
Obst 3Ig§6d that the policy that sets m should not be a function of p but a
function of — Ii, so of ,u. For simplicity of discussion, we assume that m = m1p.+m-Q
with mi > 0. Differentiating ln(p) with respect to t, we get the following:
ln(/1) = 1I1(@) +lI1(P) + h1(Q) — ln(Ms),
1dp_ 1dP 1dQ 1dM,_
it at _0+ P at + Q at M, at "H" (”“”+m°)'
Combining, we have the system of nonlinear equations
—ddzt) = h (1 — /1) ,
£d =(P+q—m1#-mollh
At the equilibrium, /,1‘ = 1 and p‘ = m1 +1110 — q. To avoid de ation, the monetary
policy should be made with ml + mo 2 q, i.e., the monetary growth needs to be
large enough to sustain the growth of the national product. The system can be
linearized at (_p‘, pf) by forming the matrix of partial derivatives,
0 —h _ 0 —h
,U- (p+q—m1I1—'"10l-mill (pan-) 1 “ml D
The determinant is h > 0, and the trace is —m1 < 0. Therefore, the eigenvalues at
the equilibrium, ‘ml/2:h\/ml/4 — h, have negative real parts. For 4h > mf, they are
complex, “ml/2 :t i \/ h, — mi/4. Since the real parts of the eigenvalues are nonzero,
the linear terms dominate the behavior of solutions near the equilibrium (just like
for critical points of a real-valued function) and solutions near the equilibrium spiral
in toward the equilibrium. See Figure 21.
208 5. Phase Portraits Using Scalar Emotions
.‘I1
'1
. P‘ ,
Figure 21. Monetary policy
By itself, the linearization does not tell what happens to solutions far from the
equilibrium, but another method shows that all solutions with no > 0 do converge
to the equilibrium. We form a real-valued Lyapunov function (by judicious guessing
and integrating), '’
Lu». 1») = ‘Q2 + (q- m1—mo)P+h# - Mum)-
This function is de ned on the upper half-plane, /1 > 0, and has a minimmn at
(p",/,1‘). The time derivative of L along solution of the differential equation is
given as follows:
_adn_L = p_ddpt ’ _ '-“ _ m°)_ddpt +hd_d_pt _ _ph d_do_t
+(‘1 m‘
= (P+q—m1 -molh-(1"#)+h(#-1)(P+q-m1#+m1-mo)
= hf!‘ - 1)(m1 - mi/-1)
= -h;m1(P- '" llz S 9-
This function decreases along solutions, and it can be shown that it must go to the
minimum value so the solution goes to the equilibrium.
5.6.5. Neural Networks. A neural network is a model for the human brain,
in which there are many neurons that separately decay to some steady state but
are coupled together. Hop eld [Hop82] discussed some models and constructed a
Lyapunov function for his model. Let N be the number of neurons, with the state
of the 1"’ neuron given by the real variable :0,-. The constants q > 0 measure the
decay of the 2"" neuron without any stimulation. There is a matrix T = (t,_,-),
called the connection matr-z':c, which determines whether the jm neuron affects the
i‘l‘ neuron or not, and in what manner. If t,~_,- > 0, then the action is excitatory,
if t,~_,- < 0, then it is inhibitory, and if t,-_,- = O, then there is no direct interaction.
Typically, there are many neurons which do not interact with each other, so there
are many 0’s in the connection matrix. A function fj ($1), called the neuron response
function, measures the effective action of the j‘h neuron on other neurons, with the
net effect of the _'i"‘ neuron on the -it“ neuron given by ti“,-f,~(:1:,-). We assume that
5. 6. Applications 209
each of the functions fj is differentiable, _f_,; (an) > 0, and bounded above and below.
Finally, there are constant inputs bi. Combining these effects, we assume that the
di erential equation for the i‘h neuron is given by
N
it = -6: $1" + 2%,; fj($j) + bi.
i=1
or the system of differential equations
(5.5) i'c=—Cx+Tf(x)+b,
where C is the diagonal matrix with entries c,- along the diagonal, the function f
has the form, f(x) = (f1(:r:1), . . .,f,,(a:,,))T, and b = (b1,...,b,,)T. For simplicity
in the following, we assume the T is invertible and B = T'1b, so
J'c= —Cx+Tf(x)+TB.
For the case when T is symmetric, Hop eld [Hop82] was able to construct
a Lyapunov function which forces the trajectories to tend to xed points. (It is
possible to have more than one xed point with just these assumptions.) The
assumption that T is symmetric is not very realistic, and recently A. Williams
[VVil02] was able to generalize to the case for which T is not symmetric with other
appropriate assumptions. Let
1~"<=<>=Zc. /Of4(=4) r.-*<s)d_~»
Then, the Lyapunov function is given by
(5.6) V(x) = F(x) - 2 [f(x) + B]TT [f(x) + B].
We give this result in the next theorem.
Theorem 5.13. Consider the system of differential equations given by equation
(5.5) with C a diagonal matrias. Assume that each of the functions f, is di er-
entiable, bounded above and below, and that f{(w,) > 0. Further assume that the
connection matria: T is invertible and symmetric. Then, the function V(x) given
by equation (5.6) is a Lyapunov function, with l’/(x) < 0 except at the fired points.
Thus, all solutions tend to a flrceal points as time goes to in nity.
Proof. Taking the time derivative of V, we get
V- (X) = Z ‘ii fi_ 1(fi(=¢i)) f1(==t)I'=. i - Zlflxl + BlTT%8f(X) 1.1"
1' i ‘
= [c x]TDf(x) st - [T f(x) + T 131* 011,, st
= [Cx - Tf(x) - TB]T 0r(,,, X
= —J.(-r Df(x)
Since Df(x) is a diagonal matrix with positive entries f,’(:c,), ll/(x) 3 0 and l-/(x) is
zero if and only if x is zero. This is what we need to prove. El
Remark 5.20. Notice that, if T is not symmetric, then the derivative of the second
term gives the symmetric part of T and the substitution for at cannot be made.
210 5. Phase Portraits Using Scalar Functions
The neural network gives an output for any input b. The next theorem says
that distinct inputs yield distinct outputs.
Theorem 5.14. Consider the system of differential equations given by equation
(5.5) with C, T, and f xed. Assume that each of the coordinate functions f, is
di erentiable, bounded above and below, and that f,((:n,) > 0. Assume that bl and
b2 are two inputs with bl 74 b2. Then, any steady state for b‘ is distinct from
any steady state for b2
Proof. Assume that 21' is a. steady state for bi for j = 1, 2. Then,
0 = -cs‘ +Tr(s1) +b1 and
0 = _c>2’ + "rr(>2’) + if,
so '
op-<1 - 221+ T [—f(i1) + r(>-8)] = bl T b2.
Notice that, if ii‘ = >12, then the left side of the last equation is zero, so bl = b2.
Thus, if bl 76 b2, then it‘ 76 5:2. El
Exercises 5.6
1. Consider the system of differential equations with the six-twelve potential
:3 = ya
. , ”_ =""";§a +?b-
where a. > 0 and b >l 0.
a. Find the potential function V(a:) and draw its graph. Classify the type of
the xed point.
b. Draw the phase portrait for the system of differential equations.
2. Consider the replicator equations for the matrix A =
a. Determine the scalar replicator equations.
b. Determine the xed points and their stability.
5.7. Theory and Proofs
Lyapunov functions
Restatement of Theorem 5.2. Let x‘ be o xed point, and let L be a weak
Lyapunov function in a neighborhood U of x‘. De ne
ZU={x€U:I'/(x)=0}.
Assume that U is su iciently small such that, for any x1 E ZU\{x‘}, the trajectory
¢(t; x1) moves o ZU into U \ Zu for small positive t (i.e., {x"} is the largest
positively invariant set in ZU). Then, x‘ is asymptotically stable.
Further assume that L0 > L(x') is a value for which
ULO = {KG UIL(X) s L0}
5. 7. Theory and Proofs 211
is contained inside U, is bounded, and is bounded away from the boundary of U.
Then, the set ULO is contained in the basin of attraction of x‘, ULO C W’(x").
Proof. Let L‘ = L(x'), and L0 > L‘ be such that
ULO = {x E U: L(x)5 L0}
is contained inside U, is bounded, and is bounded away from the boimdary of U.
We show that ULO is contained in W“’(x").
Let xo be a point in UL“ \ {x'}. The set ULU is positively invariant, so the
whole forward orbit ¢(t;x0) must be contained in ULO. Since L 3 0, the value
L(¢>(t;x0)) is decreasing as t increases and is bounded below by L‘. Therefore, it
must have a limit value L0,, as t goes to in nity. Since L(¢(t;x0)) is decreasing,
L(¢(t;x<>)) 2 Loo for an r 2 0.
We complete the proof using two different ways of expressing the argument.
The first is closer to the argument given for the damped pendulum and the second
uses the properties of the w-limit set.
First Argument: If Loo = L‘, then we are done, since
L"(L') nU = {x" }.
Assume that Loo > L‘. In order for L(¢(t;x0)) to have L0,, as a limit as t
goes to in nity, L(¢(t;x0)) must go to zero. The orbit ¢(t;xo) is bounded, so it
must keep coming near some point z. (This actually uses the fact that UL, is
compact.) Therefore, there is a sequence of times tn going to in nity such that
¢(t,,;x0) approaches z as n goes to in nity. Therefore, L(¢(t,,;x0)) limits on L(z)
a.nd also Loo, so L(z) = Loo. Because L(¢(t; xo)) goes to zero, L(z) = 0 and z must
be a point in -
zu n L'1(L,,°) <3 ULO.
Since z 96 x‘, for small t > 0, ¢(t;z) must leave ZU and L(¢(f;z)) must be a
value less than Leo. Then, ¢>(t+ t,,;x0) = ¢(t;¢(t,,;x0)) converges to ¢(t;z) and
L(¢(t + t,,;x0)) converges to L(¢(t; z)) which is less than Loo. This contradicts the
fact that L(¢(t;xo)) Z Loo for all t Z 0. Therefore, L0,, = L‘, and the trajectory
converges to x‘.
Second Argument: Because the trajectory is contained inside U1,0, the w-limit
set w(x0) is nonempty. For any point z in w(x0), there is a sequence of times t,, going
to in nity such that ¢(t,,;x0) approaches z. Therefore, the limit of L(¢(t,,; x0)) is
L(z) and also Loo. Thus, L(z) = Loo for any point in w(x0).
Take any point z in u.)(Xq). The limit set is invariant, so ¢(t; z) must remain in
w(x0) for all t Z 0,
L(¢(t; z)) -E Loo,
§L<¢<t Z» E 0,
and ¢(t;z) must stay in ZU for all t Z 0. Since x‘ is the only point in ZU whose
whole forward orbit stays inside ZU, the only point in w(x0) must be x‘. This
shows that w(x0) = {x‘ }, and the trajectory ¢(t; xo) must converge to x‘. El
—
Chapter 6
Periodic Orbits
Although several systems already considered have contained periodic orbits, they
have not been the focus of the study until this chapter. In the rst section, we give
the basic de nitions and some simple examples that have periodic orbits. These
examples are ones that are simple in polar equations, so we can use one-dimensional
analysis of the radial component to show that there is a periodic orbit. Section
6.2 presents a general result about differential equations in the plane, called the
Poi.ncaré—Bendixson theorem. This theorem states that an orbit, which stays away
from xed points and does not go to in nity, must approach a periodic orbit.
Although, this result holds only in two dimensions, it is an important way to show
that certain systems must have periodic orbits. Most of the applications given of
this theorem have a repelling xed point and use a test function to show that there
is a positively invariant bounded region:.There must be at least one periodic orbit
in this bounded region. Section 6.3 considers a certain class of nonlinear self-excited
oscillators that have unique attracting periodic orbits. These equations again have
a repelling xed point at the origin, but have an attracting force only outside a
vertical strip. Orbits with large amplitudes must cross the strip where energr is
added to the system, but spend more time in the region where energy is taken out of
the system. A calculation is needed to show that the net effect on large amplitude
orbits is to lose energy when a complete trip around the xed point at the origin is
made.
The next two sections consider mechanisms of creating periodic orbits by vary-
ing parameters. The rst "bifurcation" is the creation of a periodic orbit coming
out of a xed point, called the Andronov—Hopf bifurcation. There are many sys-
tems arising from applications for which the A.ndronov—Hopf bifurcation applies, of
which we give a couple in this chapter; in Chapter 7, we note that it also occurs for
the Lorenz system. The second “bifurcation” is the creation of a periodic orbit from
an orbit with both a- and w- limit sets at a xed point; this is called a homoclinic
bifurcation. This bifurcation occurs less often than the Andronov—Hopf bifurcation,
213
214 6. Periodic Orbits
but it is another mechanism for the creation of periodic orbits; it, too, arises for
the Lorenz system considered in Chapter 7.
Section 6.6 shows the relationship between the divergence of the vector eld of
a di erential equation and the way the ow changes the area or volume of a region
in the phase space. This result is used to show that certain systems of differential
equations in the plane do not have periodic orbits contained entirely in a region of
the phase plane. In Chapter 7, these results are used in the discussion of “chaotic
attractors.” In Section 6.7, we return to the discussion of stable periodic orbits and
the connection with a. function obtained by going around once near the periodic
orbit. This function is what is called the Poincaré map. It is introduced in Section
6.1 in terms of simple examples, and it plays the central role in the analysis of the
Lienard equations in Section 6.3. _
6.1. Introduction to Periodic Orbits
Let ¢(t;xq) be the ow of the differential equation >2 = F(x). In Section 3.1, we
de ned a periodic po-int of period T to be a point xo such that ¢(T;xo) = xo, but
¢(t;x0) 79 x9 for 0 < t < T. If X0 is periodic, with period T, then the set of all the
points {¢(t;x0) : 0 3 t 3 T} is called a periodic orbit or closed orbit It is called a
periodic orbit because the ow is periodic in time, d>(t + T; xo) = ¢>(t;xo) for all t.
It is called a closed orbit because the orbit “closes” up on itself after time T and
the set of points on the whole orbit {¢(t; xo) : —oo < t < oo} is a closed set. (The
de nition of a closed set is given in Appendix A.2.)
De nition 6.1. Just as a xed point can be L-stable or asymptotically stable, so
a periodic orbit can have different types of stability.
A periodic orbit 'y é {¢(t; xo) : 0 3 t 3 T} is called orbitally L-stable provided
that the following condition holds: Given any e > 0, there is a 6 > 0 such that, if
xo is an initial condition within a. distance 6 of 'y, then ¢(t; xo) is within a distance
e of 7 for all t Z 0. Thus, an orbit is orbitally L-stable if all solutions which start
near the orbit stay near it for all positive time.
The periodic orbit is called orbitally w-attracting provided there is a 61 > 0
such that any initial condition xo within 61 of 'y has the distance between ¢(t;xq)
and '7 go to zero as t goes to in nity (i.e., w(x0) = '7).
The periodic orbit is called orbitally asymptotically stable provided it is orbitally
L-stable and orbitally w-attracting. Notice, that the de nition does not require that
the orbit approach any particular solution in the closed orbit, but merely the set of
all points in the orbit. This last fact is the reason the adverb “orbitally” is used in
the term. A.n orbitally asymptotically stable periodic orbit is also called a. periodic
sink or an attracting periodic orbit.
An orbit is called orbitally unstable provided it is not orbitally L-stable; in
other words, there is a distance so > O, initial conditions xj arbitrarily close to '1,
and times tj > 0, such that the distance from ¢(t,-;x,-) to 'y is greater than co > 0.
Finally, an orbit is called repelling if all orbits move away: A periodic orbit
is called repelling provided that, as t goes to minus in nity, it satis es conditions
similar to those for orbitally L-stable and orbitally w attracting, that is, given e > 0,
there is a 6 > 0 such that, if xo is within 6 of 1, then ¢(t;x0) is within c of 'y for
6.1. Introduction to Periodic Orbits V 215
all t 3 0, and the distance between ¢(t;x0) and 7 goes to zero as t goes to minus
in nity (i.e., Oz(Xq) = '7). A repelling periodic orbit is also called a periodic source.
Notice that the periodic orbits for the pendulum are orbitally L-stable but not
orbitally asymptotically stable (because they occur in a whole band of periodic
orbits of varying periods).
Periodic orbits in the plane can either be contained in a band of periodic orbits,
like the pendulum equation, or they can be isolated in the sense that nearby orbits
are not periodic. The latter case is called a limit cycle: A lim-it cycle is an isolated
periodic orbit for a system of differential equations in the plane. On each side of
a limit cycle, the other trajectories can be either spiraling in toward the periodic
orbit or spiraling away (i.e., on each side, either the cv- or w-limit set equals the
periodic orbit). If trajectories on both sides are spiraling in, then the periodic orbit
is attracting or orbitally asymptotically stable in the sense previously de ned. If
trajectories on both sides are spiraling away, then the periodic orbit is repelling
in the sense previously de ned. Finally, the case of attracting from one side and
repelling from the other side is orbitally unstable and is called orbitally semistable.
The periodic orbit for the system of equations given earlier in Example 4.2
is the simplest example of an attracting limit cycle. It is analyzed using polar
coordinates.
In the rest of this section, we consider one example in the plane, which is easy
to analyze using polar coordinates, and we mention the example of the Van der Pol
system of differential equations, which we consider in Section 6.3. In the rest of the
chapter, we consider more complicated situations which have periodic orbits.
Example 6.2. As in Example 4.2, the following system of equations can be an-
alyzed using polar coordinates. In rectangular coordinates, the example is given
W
¢=y+r0—r’—f)M—w’—f%
1= -1-+1<1~1” ~1 >(4-=2-11*).
which, in polar coordinates, has the form
1" = r(1 — r2)(4 — r2),
6=-L
We do not nd explicit solutions, but just consider the signs of 1":
1" > 0 if 0 < 1- < 1,
1* < 0 if 1 < 1- < 2,
1* > 0 if 2 < 1-.
Therefore, if ro is an initial condition with 0 < ro < 2, then r(t;ro) converges to 1
as t goes to in nity. If 2 < ro, then r(t;r0) goes to in nity as t goes to in nity, and
converges to 2 as t goes to minus in nity. Considering the differential equations in
Cartesian coordinates, there is an attracting limit cycle at r = 1 and a. repelling
limit cycle at r = 2. See Figure 1
216 6. Periodic Orbits
ll
2
{D
-2 2
-2
1
Figure 1. Phase portrait for Example 6.2
Example 6.3 (Van der Pol). A more complicated example is the Van der Pol
equation given by 55 — p(1 — :z2)a': + zv = 0, which con'esponds to the system of
di erential equations
:i: = 1),
1'1 = —a: + ,u(1— a:2)v.
For $2 < 1, [1(1 — a:2)n> 0 and the effect is a force in the direction of the velocity,
or an “antifriction"__. Using polar coordinates,
'r1'-=:z:b+m)=a:v—v:c+p.(1-—:c2)v2 or
Therefore, +=v2(—1 — 1:i2.)
T
1">O for —1<:r:<1,
1'~<0 for |:r|>1.
This shows that no periodic orbit can be contained entirely in the strip —1 < a: < 1.
We show in Section 6.3, that there is a periodic orbit that passes through the region
|a:| > 1 as well as the strip |a:| 5 1. See Figure 7.
_Several times in this chapter and later in the book, we use the differential
equation to de ne a function which takes a point on a line in the phase plane
and assigns to it the point on the line obtained by following the trajectory until
it returns again to the same line. This function is called the rst retum map, or
Poincaré map. We discuss this construction more completely in Section 6.7, but
give an introduction to the idea at this time. For the Van der Pol equation, we take
an initial condition on the positive y-axis, {(0, y) : y > 0 }, and follow the trajectory
¢(t; (0, y)) until it returns again to the positive y-axis at a later time T(y). The point
of this rst return can be de ned as the Poincaré map, ¢(T(y), (0,y)) = (0, P(y))-
Although this construction is discussed more completely in Section 6.3, the following
example illustrates this analysis for the simple equations given in Example 4.2.
6.1. Introduction to Periodic Orbits 217
Example 6.4 (Poincaré map for Example 4.2). The di erential equations are
or in polar coordinates :i:=y+a:(1—:z:2 —y2),
9 = —=1=+:1/(1 —r’ -2/2).
1'- = 'I‘ (1 - T2),
é= -1.
We discuss the rst return of trajectories from the half-line {(:c, 0) : :1: > 0} to itself.
In polar coordinates, this amounts to following solutions from 0 = 0 to 0 = —21r.
After separation of variables, a functional relation between 1' and t is obtained
by integrals:
t r(1) 1 r(t) r(t) 1 1'(£)
/O2dt=2/,0 —-i12‘)dr=2/,0 l1‘d'r+/,0 i1d-1‘r—/To L1+d1‘ r,
2‘-'=l!'l *3 /-\_; —l.I1 *1 .
QM
e“;/K Ne‘s-'\/ no
Solving for 'r(t), we have
t2=_r°2e2i!,._
To 1—'r§+r§e2‘
1
_ 1 + @—2=(1-5’ -1)’
rm = 1~@h~%+ 6-“<1 - Tar‘/2
= [1 + e'%‘(r52 —1)]'1/2.
The solution for 0 is
0(t) = 00 — t.
Thus, it takes a length of time of 21r to go once around the origin from 00 = 0 to
0(t) = —21r. So, evaluating 'r(t) at 21r gives the radius after one revolution in terms
of the original radius as
1'1 = 1-(21r) = {1+ e'4”(r§2 - 1)]‘1/2.
We can think of this as taking a point 1'0 on the half-line {6 = 0, 1'0 > 0} and
following its trajectory until it retm'ns to this same half-line at the point 1', =
P(1-0) = r(21r). This rst return map to this half-line is called the Poincaré map.
The half-line used to construct the Poincaré map is called the transversal or cross
section to the ow. In the current example,
P(TQ) = T(27l')
= wt + 6-“<1 - Ta]-1"
= roll — (1 — e-")<1 — 1%)!“/’
= 11+ e"‘*"<1~a’ - 1>1-*/*-
In Section 6.7, we present a complete discussion of the properties of the graph
of P. However, it should not be too dif cult to see that the graph of P given in
218 6. Periodic Orbits
Figure 2 is basically the correct shape. The points 1' = O and 1' = l return to
themselves, P(O) = 0 and P(1) = 1, as can be seen from the rst equation for
P(-ro). For other initial values, P(r0) 75 1'0. We show in Section 6.7, using the
Poincaré map, that the periodic orbit 1' = 1 is orbitally asymptotically stable and
attracts all orbits other than the xed point at the orig-in.
P(1‘o)
0 8-47’ _-¢_-__-
To
1
Figure 2. Poincaré map for Example 6.4
Limit Cycles for Polynomial Systems
D. Hilbert conjectured i.n 1900 that there was a bound on the number of limit
cycles in terms of the degree of a system of polynomial differential equations in the
plane. This conjecture has not been proved or disproved yet. Yu. Ilyahenko has
proved that any system of polynomial differential equations in the plane can have
only a nite number of limit cycles. Song Ling Shi in 1980 gave an example of a
quadratic system with four limit cycles. The following example of Chin Chu is a
simpler example with two limit cycles.
Example 6.5. Consider the system of differential equations
:i:=—y— no ..
y=51l—l '——y+rvy—;cza /’-
U\-"C
The xed points are (0,0) and (-2, -1). Figure 3 shows an attracting limit cycle
around (-2, -1) and a repelling limit cycle around (0,0). Chin Chu proved that
these limit cycles exist.
Exercises 6.1
1. Show that the system
i = 1/.
z2= —rv+z/(1—r=2—z/2).
has a periodic orbit. Hint: Find the equation for 1" and then nd a value of r
for which 1" = 0.
6.2. Poinca.ré~Bendixson Theorem 219
l as
.i- . .-.
-4 -2 0 2
Figure 3. Two limit cycles for Example 6.5.
6.2. Poincaré—Bendixson Theorem
In Section 6.1, we considered some simple equations for which the limit cycles
could be found by using polar coordinates. In this section, we will give a slightly
more complicated example for which the limit cycle is not a circle and it cannot
be determ.ined as easily: There is a.n annular region for which the trajectories
are entering into the region on both boundaries, and a lim.it cycle is forced to
exist somewhere inside. Indeed, a general theorem, called the Poincaré—Bendixson
theorem, implies the existence of a periodic orbit in this and similar examples.
Example 6.6. Consider the equations
It = y. ,
ti = -11 +1/(4 — =2 — 492),
These equations contain the terms y(4 —- 1:2 411,/2) in the Q equation, which has
the same sign as y for small ||(a:,y)|| (“antifriction”) and which has the opposite
sign as y for large ||(a:,'y)|| (“friction”); thus, this factor y(4 — 1:2 — 41,12) acts as a
generalized nonlinear “friction” in the system. We could use the polar radius to
measure the behavior of the system, but instead, we use the corresponding test or
bounding function L(:c,y) = %(:z:2 + ya). The time derivative is
L1 :y2(4_$2_4y2)= Z0 ‘ 2L(z,y) 1 a2:2+y225l,
$0 if 2L(:c,y)=a: +y 24.
These inequalities imply that the annulus
= {(a:,y) : g 5 L(:|:,y) 3 2 }
is positively invariant. The only xed point is at the origin, which is not in M.
The next theorem then implies that there is a periodic orbit in 4%. The idea is that
this system has a term that is damping on the the outer boundary L‘1(2), and an
antidamping term on the inner boundary L'1(1/2), so that the orbits are trapped
in the annulus where there are no xed points. See Figure 4.
220 6. Periodic Orbits
’__.. ____
1a \\
I \
1 \
/ \
\ /
\ /
\\ 1
\ gr
\“~_ --"T
u
|
Figure 4. Positively invariant region for Example 6.6
Theorem 6.1 (Poincaré—Bendixson Theorem). Consider o di erential equation
x = F(x) on IR2.
a. Assume that F is de ned on all of IR2 and that a forward orbit {¢(t; q) : t Z
0} is bounded. Then, w(q) either: (i) contains a. xed point or (ii) is a periodic
orbit.
b. Assume that M is a closed (includes its boundary) and bounded subset of
R2 that is positively invariant for the di eiential equation. We assume that F(x)
is de ned at all points of M and has no xed point in M. Then, given any xo in
M, the orbit d>(t;x0) is either: (i) periodic or (ii) tends toward a periodic orbit as
t goes to oo, and w(xQ) equals this periodic orbit.
Remark 6.7. In order for a connected region M in the plane to be positively
invariant, but not contain any xed points, it must be “annular” with just one hole.
Thus, it has two boundaries, which are each closed curves (but are not necessarily
circles).
Remark 6.8. ‘To get the region M to be positively invariant, it is enough for the
vector eld of the system to be coming into the region on both boundaries.
Remark 6.9. With appropriate modi cations of Theorem 6.1, the region M could
just as easily be negatively invariant, with the trajectories going out on both bound-
aries.
Idea of the Proof. We indicate the key ideas of the proof here. In Section 6.9,
we include a complete proof. The proof de nitely uses the continuity of the flow
with respect to initial conditions.
The trajectory ¢(t; X0) stays inside M, so the trajectory needs to repeatedly
come near some point z, and there is a sequence of times t,, which go to in nity such
that ¢(t,,;x0) converges to z. This is the idea called compactness in mathematics.
Just as an increasing sequence of points that are bounded in the line must converge,
so a boimded sequence in R2 needs to repeatedly come near some point. The point
z is not xed, so nearby trajectories all go in roughly the same direction. Let S be
a short line segment through z such that the other trajectories are crossing in the
6.2. Poincaré~Bendixson Theorem 221
same direction. Then, for sufficiently large n, we can adjust the times tn so that
each ¢(t,,; xo) is in S. Taking the piece of the trajectory
{¢(t§x0) 3 tn S t S tn+1l’»
together with the part S’ of S between qb(t,,;x0) and ¢(t,,.,.1;x@), we get a. closed
curve which separates the plane into two pieces. See Figure 5. The trajectory
going out from ¢(t,,.,.1;x@) enters either the region inside or the outside of 1". All
the trajectories starting in S’ enter the same region as that starting at ¢(t,,.H; xo).
Therefore, ¢(t;x0) can never reenter the other region for t > t,,+1. This means that
the intersections of the trajectory with S occur in a monotone manner. They must
converge to z from one side. A further argument shows that, if z were not periodic,
then these nearby trajectories could not return. Therefore, z must be periodic. El
X0 xn
gout xn+1 ‘
IF
Fig-ure 5. Transversal at z
The proof actually shows more about the convergence to a periodic orbit. If
one orbit accumulates on a limit cycle from one side, then the Poincaré map is
monotone and attracting on that side. It follows that all initial conditions which
are nearby the limit cycle on that side also have to have the limit cycle as their
w-limit, and the limit cycle is orbitally asymptotically stable from that one side.
Thus, if there are points on both sides of a periodic orbit which have the periodic
orbit as their w-limit set, then the periodic orbit is orbitally asymptotically stable
(from both sides). We summarize this reasoning in the next corollary.
Corollary 6.2. Consider a di erential equation 2': = F(x) on 1R2 with an isolated
periodic orbit 'y.
a. Assume that a point p not on 'y has w(p,d>) = 'y. Then, all points q near
enough to '7 on the same side of '1 as p also have w(q,¢) = *1. In particular, 7 is
orbitally asymptotically stable from that one side.
b. Assume that w(p1,¢) = 7 = w(p2,d>) for points pl and P2 that are on
di erent sides of '1. Then, '1 is orbitally asymptotically stable (from both sides).
222 6. Periodic Orbits
Example 6.10. As a somewhat more complicated example, consider
i = 2/,
1'1 = —:r:3 + y(4 — 2:2 — 41,/2).
We use the corresponding “energy” function
L(w,y) = iv’ + 21"
as a bounding function. The time derivative is
L = y3)+ a:3:i: = —y:r3 +y2(4 — 1:2 -41/2) +a:3y
_ =1/’(4—=v’ -424’)-
It is zero on the ellipse 1'2 + 43/2 = 4, In/'2 0 inside the ellipse, and L $ 0 outside
the ellipse. Letting W(z,y) = 2:2 + 4;/2 be the function determined by L, we can
nd the maximum and minimum of L on W(a:,y) = 4 using Lagrange multipliers.
Solving
$3 = A2:c,
1/=»\8y.
4=:t’+4y2,
we get the critical points (:t2, 0), (0, :h1), and (:t%, $143); the values of L at these
points are 4, 1/2, and 31/64, respectively. Thus, the largest invariant annulus given
by level curves of L is
‘ i_=1= {(1.y) : 2 5 L(r.y) s4}.
The trajectories are coming into the annulus on the outer and the inner boundaries,
so .121 is positively invariant and contains a periodic orbit.
What makes this example work is the fact that the equations have a term
y(4 — :22 — yz), which has a damping e ect on a large level curve of L like L“ (4),
and has an antidamping effect on a small level curve like L’1(3l/64). The region
between is trapped and contains no xed points, so it must contain a periodic orbit.
We cannot completely analyze the phase portrait just using the Poincaré-
Bendixson theorem. If, however, we assume that there is a unique periodic orbit
for this system (which is, in fact, true), then we can state more about the 0z- and
w-limit sets of points. Let p be an initial condition in the region bounded by the
periodic orbit 1". Since the xed point at the origin is repelling, w(p) cannot con-
tain only xed points. The orbit is bounded because it is inside I‘. Therefore, the
only possibility is that w(p) = I‘. On the other hand, if p is outside I‘, the orbit
cannot escape to in nity because L is decreasing for values greater than 4. The
orbit cannot approach 0, so the only possibility is that its w-limit set is a periodic
orbit. If there is only one periodic orbit, it must be 1". In any case, any initial
condition other than the origin has an w-limit set equal to a periodic orbit.
'Ilurning to a-limit sets, notice that there are trajectories on both the inside
and outside that approach F; so, no trajectory can be moving away from 1" as time
increases or can be approaching it as time goes to minus in nity. Therefore, no
point can have I‘ as its Or-limit set. Therefore, if p is inside 1", its cu-limit set must
6.2. P0incaré—Bendixson Theorem 223
equal On the other hand, if p is outside I‘, there is no other xed point or
periodic orbit, so the backward trajectory must leave the annulus and enter the
region where L(x) > 4. Once in this region, L < 0, so the value of L goes to
in nity as t goes to minus in nity, and a(p) = (ll.
Example 5.11 in the section on limit sets has an w-limit set equal to a xed
point, together with two orbits that are on both the stable and unstable manifold
of the xed point. This example illustrates a more general form of the Poincaré-
Bendixson theorem given in the following result.
Theorem 6.3. Consider a di erential equation x = F(x) "on R2. Assume that
.41 is a closed (includes its boundary) and bounded subset 0_f1R2, which is positively
invariant for the di erential equation. We assume that F(x) is de ned at all points
of .41 and has a nite number of red points in M. Then, given any xo in .21, w(xo)
is of one of the following three types:
i. The w-limit set w(x0) is a periodic orbit,
ii. The w-limit set w(x0) is a single xed point.
iii. The w-limit set w(x0) is a. nite number of xed points q1,. . . ,qm, together
with a ( nite or denumerable) collection of orbits {'y,- }, such that for each
orbit 1,-, oz(7_.;) is a single red point q,-(_,-_,,) and w('y,-) is a single xed point
‘Bo.-»)-'
w(Xo)={q1.---,qm}UU'r,-. with
1'
<1('n) =<1¢o".<:> and WW1) =<1¢o".~)-
Moreover, given any two fired points q,-, 94 q,-2, there must be a directed
choice of connecting orbits 'y,-,,....'yj,, with a('y,-,) = q,-,, w('y_,-P) = a('y,-,,+,)
for 1 3 k— 1, and w('y,-,,) = q,-,.
For a proof, see [Hal69].
Another consequence of the Poincaré—Bendixson theorem is that any periodic
orbit of a di erential equation in the plane has to surround a xed point.
Theorem 6.4. Consider a di erential equation >2 = F(x) on 1R2. Let 7 be a
periodic orbit which encloses the open set U. Then, F(x) has a xed point in U
or there are points in U at which F(x) is unde ned.
Proof. We argue by contradiction. Assume that '7 is a periodic orbit which encloses
the open set U, U contains no xed point, and F(x) is de ned at all points of U.
We can take such a '7 of minimal area, so there can be no other periodic orbits
inside '7. Take an initial condition xo in U. We argue that the limit sets w(x0)
and a(xo) carmot be both equal to '7 as follows. Assume w(x0) = '7. By the proof
of the Poincaré—Bendixson theorem, the Poincaré map is monotone on the inside
of 'y and all nearby points yq inside 'y have w(y0) = 'y and a(y0) cannot equal 7.
Then, a(x0) is nonempty and must be either a periodic orbit or a xed point in
U: Either conclusion contradicts the assumptions, since U contains no xed points
and the minimality of the areas implies there can be no periodic orbit inside U.
This contradiction proves the theorem. El
224 6. Periodic Orbits
6.2.1. Chemical Reaction Model. The Brusselator, which is a system of equa-
tions that model a hypothetical chemical reaction, has the equations
:i:=a——b:r+a:2y—:z:,
y=ba:-a:2y.
We show, using the Poincaré—Bendixson theorem, that if b > 1+a2 and a. > 0, then
the system has a periodic orbit. We include this example in the main sections of the
chapters rather in the applications, because the method of nding the positively
invariant region is an interesting one that is di erent from the other examples
considered so far.
.. For another system of differential equations that model chemical reactions,
which have periodic limiting behavior, see the oregonator system of di erential
equations in Section 6.8.1. '
To nd the xed points, adding the 3'1 = 0 equation to the :i: = 0 equation,
we get 0 = a — a:, or :1: = a. Substituting a: = a. into the 11) = 0 equation yields
0 = ba — a2y, or y = 5/a. Thus, the only xed point is (:c‘,y‘) = (a,"/,1).
The matrix of partial derivatives at the xed point is
(Zzy-b—1 1:2) _(b—1 (12)
b — 2a:y -12 :=a‘y=b/Q —b —a2 '
This matrix has its determinant equal to a2, which is positive, and has trace
b - 1 — a2, which we assumed is positive. Therefore, the xed point is either an
unstable node or an unstable focus. If we remove a small elliptical disk D of the
correct shape about the xed point, then the trajectories will leave this elliptical
disk and enter the region outside. Thus, the boundary of D is the inside boundary
of the annular region .41 we are seeking to nd.
To get the outside boundary of the region 42¢’, we take a quadrilateral with sides
(1) :|: = 0, (2) y = 0, (3) y = A —- zr, for a value of A to be determined, and (4)
y = B + 0:, for a value of B to be determined. Let the region .2! be between the
boundary of D _a.nd the boundary formed by these four sides. See Figure 6
Along side (1), = a. > 0, so trajectories are entering into the region M.
Along side (2), 3) = ba: > 0 (except at :1: = 0), so trajectories are entering 12/
along this boundary.
Side (3) is in the zero level set of the function L3(:|:, y) = y + a: — A, with .2! on
the side where L3 is negative. The time derivative is
L3 = e + 1;
=a—;
which is negative if :1: > a. Therefore, we need to take A and B so that the top
vertex of .2!/, where the sides (3) and (4) come together, has :1: 2 a. In fact, in the
following, we take t-hem so that :1: = a at this top vertex.
Side (4) is in the zero level set of the function L.;(a:,y) = y — zr — B, with .13’
on the side where L4 is negative. The derivative is
L4=y—:i:=2ba:—2:i:2y-a+:c.
6.2. Poincaré—Bendixson Theorem 225
,U
8 _.
\
/I \\
1' \
6
4 ' \\\\\\
\\
\\
2 -‘ \
\\\
\\ I
\\\
I I I ‘F I
0 2 4 6 s 10
Figure 6. Positively invariant region and trajectories for Bruelator
This is negative if
2b 1 -
Let h(:c) equal the right-hand side of the preceding inequality. Then, (i) h(a:)
goes to minus in nity as a: goes to zero, h(:z:) goes to zero as :2: goes to plus
in nity, (iii) h(:c) equals zero at a/(2b +'1), and (iv) h(a:) has a unique maximum
at rc = 2a/(2b + 1). The maximum of h.(z:) is as indicated because
"'<“> = i2 a.=— (2 b»+ 1i) a's
where h’(a:) is zero at rs = 20./(2b + 1), h'(a:) is positive for a: < 2a/(2b + 1), and
h'(:1:) is negative for a: > 20,/(2b + 1). The value at the maximum is
h 2.1 =(2b+l)2
2b+1 80, '
If we take B = (2 b + 1)2/(8 a), then certainly y = B + :1: is greater than h(:c) along
side (4), since y Z B = max{h(a:) : a: > O}. Having chosen B, set A = B + 2a, so
the two edges intersect when :r = a. and y = B + a = A — a..
With these choices for the boundary of .91, all the points on the boundary have
trajectories that are entering the region .21/. Since the only xed point is not in the
region .21, there must be a periodic orbit in the region .21 by the Poincaré—Bendixson
theorem.
226 6. Periodic Orbits
Exercises 6.2
1. Consider the system of di erential equations
i=3/0
y=-4a:+y(l —a:2—y2).
Show that the system has a periodic orbit. Hint: Use a bounding mction.
2. Consider the system of equations
1i: = a: — 2y — :r(:r2 + 3y2),
19-= 2w + 1/ —' 2/(I2 +311’)-
a. Classify the xed point at the origin.
b. Rewrite the system of differential equations in polar coordinates.
c. Find the maximum radius 1'1 for the circle on which all the solutions are
crossing outward across it.
d. Find the minimum radius 1‘; for the circle on which all the solutions are
crossing inward across it.
e. Prove that there is a periodic orbit somewhere in the armulus 1'1 3 r 3 T2.
f. Use a computer program to plot the periodic solution.
3. Consider the system of equations
:i: =3:z:+2y —:i:(a:2 +3/2),
zi= —-1=+y—z/(¢’+y’)-
8 Classify the xed point at the origin.
b. Show that (0, 0) is the only xed point.
c Calculate 1'1‘ in terms of :|: and y.
d Show that 1" is positive for small 1' and negative for la.rge 'r. Hint: To show
the quadratic terms are positive de nite (positive for all (a:, y) 76 (0,0)),
either'(i) complete the square or (ii) use the test for a minimum of a
function.
B Prove that the system has a periodic orbit.
4. Assume that the system of di erential equations in the plane x = F(x) has an
annulus
{X113 S llxll 5"2l=
where the trajectories are coming in on the boundary and there are no fixed
points in the annulus.
3. Assume that there is exactly one periodic orbit in the annulus. Prove that
it is orbitally asymptotically stable. Hint: Use Corollary 6.2, as well as
the Poincaré—Bendixson theorem.
b Assume that there are exactly two periodic orbits in the annulus. Prove
that one of the periodic orbits is orbitally asymptotically stable.
c Prove that there is an orbitally asymptotically stable periodic orbit in the
annulus if there are exactly n periodic orbits, where n = 3 or 4. (The
result is true for any positive integer n.)
Exercises 6.2 227
5. Consider the system of equations
:i: = 3:: — 2y — :r(a:2 + 4y2),
ii = 41 -1/—:1/(12 +411’)-
a. Show that the linearization of this system at the origin is an 11I1SI38.l)l6
spiral.
b. Show that (0,0) is the only xed point. Hint: One way to do this is to
consider the equations which say that (:c.,y.) is a xed point. Express
them as a matrix equation. This matrix equation says something that
contradicts what you discovered in (a).
_2 -
c. Let L(z,y) = 5; + and let M(a:,y) = 9; + Show that M > 0
if (a:,y) is close to (0, 0) and L < 0 if (x, y) is sufficiently far from (0,0).
d. Show that there is a trapping region of the form {(:r,y) : 0.2 5 M(_a:, y)
and L(z:,y) 5 A2}, where a is small and A is large.
e. Show that there is a closed orbit in the set de ned in (d).
6. In [Mur89], Murray presents a system of di 'erentia.l equations based on a
trimolecular reaction given by
it = a — u + uz '0,
'0 = b —- uz v,
with a > 0 and b > 0. In this exercise, we consider only the case in which
a = 1/s and b = 1/2. (Notice that the equations are similar to the Brusselator
given in the text.)
a. Find the xed point.
b. Show that the xed point is a source, with the real parts of both eigenvalues
positive.
c. Show that the following region is positively invariant:
§?={(u,v):u21—16, 05115128, u+v5130}.
d. Explain why there is a periodic orbit in the region .92.
7. In [Bra01], Brauer and Castillo—Chavez give the system,
. = "’< 1_s3_o _z-_+U1_0)’
”’
1.’ _r 1’ a: " 51 ) *
as an example of a more general type of predator—prey system.
a. Show that the only xed points are (0, 0), (30.0), and (5, 12.5). Also show
' that (0, 0) and (30, 0) are saddle xed points and that (5, 12.5) is a repelling
xed point (a source).
b. Show that the region
.9Z={(a:,y):05a:, 053/,a:+y550}
is positively invariant. Hint: Show that the line m + y = 50 lies above the
curvewhere:i:+y=0,so:i:+;§1<0on:|:+y=50.
228 6. Periodic Orbits
c. Use Theorem 6.3 to show that, for any point po = (:rq,y0) in .9? with
2:0 > 0, yo > 0, and pg 96 (5,12.5), w(p0) must be a periodic orbit.
Hints: (i) If w(p0) contained either (0,0) or (30,0) then it must contain
both. (ii) Thus, w(p0) would contain an orbit 'y with o:('y) = (30,0) and
w('y) = (0,0), i.e., 'y C W“(30,0) F1 W“(0,0). Since there are no such
orbits, this is impossible and w(p0) must be a periodic orbit.
8. A model by Selkov for the biochemical process of glycolysis is given by
:i:= —:i:+ay+:r2y,
g = b _ ay _ $2 yr
" where a and b are positive constants.
a. Show that the only ‘ xed point is (a:‘,y*'T) = (b, g).
b. Show that the xed point (a:‘,y’) is unstable for a = 0.1 and b = 0.5.
c. To get an invariant region in the next part, we hse one boundary of the
form :1: + y equal to a constant. We want the line to go through the point
(b, I’/G), where b = a;' and 5/a is the y-intercept of 14] = 0. Therefore, we
consider :0 + y = b + I’/Q. We also consider the nullcline zt = 0. Show that
there is a unique point (a:1,y1), such that
b+2a—'_ x+ ya
_m
y_a+ '
Hint: Plot the two curves.
d. Show that the region
.92={(:r,y):051:5z1,05y5bz,a:+y5b+Zb }
is positively invariant, where 2:1 is the value found in the previous part.
e. For a = 0.1, b = 0.5, and (a:0,yo) 94 (a:',y‘) in 5?, show that w((.'c0,y0)) is
a periodic orbit.
9. Consider the system of differential equations
i= = 2/.
g = -¢- 2¢° +y(4 ~ 1:2 - 492).
Show that the system has a periodic orbit. Hint: Find an "energy function”
L(a:,y) by disregarding the terms involving y in the y equation. Find the
maximum and minimum of L on the curve L = 0. Use those values to nd a
region that is positively invariant without any xed point.
10. A special case of the Rowenzweig model for a predator-prey system had the
following system of differential equations
:i: = :i:(6'— :i:) — yzvi,
3] = y (12% — 1).
a. Find all the xed points.
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An Introduction to Dynamical Systems - 2nd Edition
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