4.4. (Two Dimensions 129
Important features of the phase portrait
(1) Location of the xed points.
(2) Regions where :i: and 3} have a constant sign.
(3) Behavior near the xed point. (Emphasis of the next section.) This
includes whether the xed point is stable or unstable, and how tra-
jectories may pass near the xed point and then leave.
(4) Location of periodic orbits. This is much harder and is the focus of
Chapter 6.
Phase portraits using a computer program
There are general purpose programs such as Maple, Mathematica, or Matlab,
as well as programs more speci cally written to draw phase portraits. These
programs can usually draw the vector eld at a grid of points in a selected region.
These are also called direction elds. Also, by picking good representative initial
conditions, a fairly complete phase portrait can be drawn. The phase portraits in
this book are drawn using Maple and the Runge-Kutta numerical method. For
example, see Figures 12, 15, and 16.
A trajectory is a curve in the phase plane that is tangent to the direction eld
at each point. It is often possible to nd the basic properties of the trajectories by
using these direction elds.
(1) If the xed points are known, then pick a window in which to draw the phase
portrait, which includes some or all the xed points. (If there is a nite number
of xed points, then include all the xed points. If there are in nitely many
xed points as is the case for the system zt = y and g = — sin(:c), then include
a representative sample of the xed points.)
(2) Experiment with initial conditions to determine the behavior of trajectories
near the xed points.
(3) In Section 4.5, we discuss the stable and unstable manifolds for saddle xed
points. When these are present, try to include initial conditions near the
unstable manifold and follow these trajectories to nd out where they go.
Also, include initial conditions near the stable manifold very close to the xed
point and follow them backward, or nd initial conditions away from the xed
point which closely follow the stable manifold near the xed point. See Section
4.5.
(4) Try to show the behavior of representative trajectories in all parts of the
window in the phase plane.
Example 4.19. Consider the system of differential equations
i=x—y
3] = 1 — 2 3/2.
The sets where zt = 0 and 37 = 0 are given by y = :1: and :1: = % respectively. See
Figure 11. The xed points are the intersections of these two curves. Algebraically,
130 4. Phase Portraits
the xed points are found by solving the pair of equations
0=x—%
0=1—zf.
The only solution of this system of equations is (:r,y) = (1,1), and so this is the
only xed point.
\ / ..R2 ylR1
\
\ .7____'.rc
/ \-
R3 as
Figure 11. Nullclines for Example -1.19
3 ll I
-\ | 2'8
-2 -1 0
I-
12
Figure 12. Computer plot of solution for Example 4.19
4.4. Two Dimensions 131
These nullclines divide themselves into ve curves a.nd divide the plane into
ve regions:
(N1) y = a: with :1: >1.
(N2) y = :2 with a: <1.
(N3) z =1/g2 with y >1.
(N4) 1 =1/yz with 0 < y <1.
(N5) I =1/ya with y < 0.
(R1) 0<y'2<x<y.
(R2) :0 < y and :2 < y'2 (including the part of a: < y where :1: 5 0).
(R3) y < :1: < y_2.
(R4) 0<y<:|:andy“2<a:.
(R5) y<0andy_2<a:.
To determine the direction arrows, rst note that they are vertical on the nullcline
:5: = O (y = :i:), and horizontal on the nullcline 7,7 = 0 ( :1: = 1/y2 ). Moving up off
y = 2:, causes :i: = a:—y to become negative. Therefore, i: is negative in regions (R1)
and (R2) and on curve (N3). Similarly, :i: = :0 —y becomes positive as :1: is increased
(or as y decreased) off the line y = a: and the point enters regions (R3), (R4), and
(R5), and on curves (N4) and (N5). Turning to the other component, 11] = 1 — zyz
becomes negative as y is increased off (N3) and (N4). Thus, g is negative in regions
(R1) and (R4) and on the curve (N1). Similarly, it is negative in region (R5). On
the other hand, as |y| is decreased off any part of the nullcline made up of (N3),
(N4), and (N5), 3] becomes positive. Thus, 3] is positive in regions (R2) and (R3)
and on (N1). Note that, in the argument we are using, the sign of 3} can change
only on the nullcline. This information is summarized in the following list.
(N1) y = :1: with a: > 1: vertical and 3} < 0.
(N2) y = a: with 2: < 1: vertical and y > 0.
(N3) a: = 1/yz with y > 1: horizontal and :i: < 0.
(N4) as = 1/y2 with 0 < y <1: horizontal and rt > 0.
(N5) :c = 1/yz with y < 0: horizontal and :i: > 0.
(R1) 0<y'2<a:<y: :i:<Oandg}<0.
(R2) :1: < y and a: < y'2 (including :c 3 0): :i: < 0 and y > 0.
(R3) y<a:<y"2: :b>Oand;i]>0.
(R4) 0<y<:randy'2<a:: wherei'>0a.nd3]<0.
(R5) y<Oandy‘2<:r: :i:>0and7)<0.
Solutions are leaving region (R1) on its boundary. However, any solution that
stays in the region for all time has x(t) and y(t) both decreasing. By arguments like
those used for one-dimensional equations, they must approach points where :i: = 0
and 3) = 0 (i.e., the trajectory must approach a xed point, which must be (1,1)).
Because some solutions leave on the left side and some leave on the right side, there
must be at least one such trajectory that is asymptotic to the xed point.
132 4. Phase Portraits
Region (R3) is similar to region (R1) except that :i: > 0 and y > 0, so there
must again be at least one trajectory that approaches the xed point.
Regions (R2), (R4), and (R5) are positively invariant because points on the
boundary either enter the region or remain on the lJOU.l1C13.l'y. Therefore, trajec-
tories starting in region (R2) stay in region (R2), the :1: variable is monotonically
decreasing, and the y variable is monotonically decreasing. Since there are no xed
points in these regions, I|(a:(t),y(t))|| goes to oo as t goes to in nity. Similarly,
trajectories starting in region (R4) and (R5) have |l(a:(t),y(t))|| going to oo as t
goes to oo.
Because trajectories that leave regions (R1) and (R3) must enter regions (R2),
(R4), or (R5), this describes the behavior of all the trajectories as t goes to oo.
Considering t going to —oo, trajectories in regions (R2) and (R4) either leave
or go to the xed point (1, 1) as t goes to —oo, ‘Ii-ajectories starting in region (R5)
must leave and enter region (R3) as t decreases. On the other hand, trajectories in
regions (R1) and (R3) must have ]|(:c(t),y(t)l| going to oo as t goes to —oo.
Using linearization at the xed point, which is discussed in the next section,
we see that the xed point is a saddle point. There is only one orbit in region (R1)
and one orbit in region (R3) going to (1,1) as t goes to oo. Similarly, there is only
one orbit in region (R2) and one orbit in region (R4) that goes to (1,1) as t goes
to —oo. See Figure 11 for the sketch of the nullclines.
Figure 12 shows a computer plot of some trajectories. To draw the unstable
manifolds of the xed point (1, 1), initial conditions where taken slightly off the xed
point in directions of expansion, (1.01, 1.003) and (0.99, 0.997). To draw the stable
manifold of the xed point (1, 1), initial conditions where taken slightly off the xed
point in directions "of contraction, (1.01, 1.033) and (0.99, 0.967), and trajectories
were followed backward in time. Other representative initial conditions were taken
to display the trajectories moving past the xed point.
Example 4.20. Consider the example
i = -2 + y,
1] = my — 1.
The nullclines are
{i=°}={y===} and
{ii = 0} = {:11 = 1/1}-
Solving the nonlinear equations shows that the xed points are (1,1) and (-1, -1).
Al'01111d the xed point at (-1, -1), trajectories move in a clockwise fashion, en-
tering and leaving each of the four regions nearby de ned by the nullclines. See
Figure 13. Thus, by itself, the method of nullclines and directions of ow does not
determine whether the xed point is stable or unstable. In the next section, we will
use the method of linearization at the xed point to show that it is asymptotically
stable.
Exercises 4.4 133
1/
23
Figure 13. Nullclines for Example 4.20
Exercises 4.4
1. Consider the system of differential equations
:i:=—2a:+y—a:3,
1] = —y + $2.
a. Determine the xed points. '
b. Determine the nullclines and the signs of rt a11d 1} in the various regions of
the plane.
c. Using the information from parts (a) and (b), sketch by hand a rough
phase portrait. Explain and justify your sketch.
2. Consider the system of differential equations
:i;=y—a:2,
g=@-9
a. Determine the xed points.
b. Determine the nullclines and the signs of :i: a.nd 11] in the various regions of
the plane.
c. Using the information from parts (a) and (b), sketch by hand a rough
phase portrait. Explain and justify your sketch.
3. Consider the system of differential equations
:i: = y — 1:3,
3) = —y + $2.
a. Determine the xed points.
b. Determine the nullclines and the signs of 2': and -g in the various regions of
the plane.
/
134 4. Phase Portraits
c. Using the information from parts (a) and (b), sketch by hand a rough
phase portrait. Explain and justify your sketch.
d. Using the computer, sketch the direction eld and trajectories for enough
initial conditions to show the important features of the phase portrait.
4. Consider a competitive market of two commodities whose prices are pl and
P2. Let E¢(p1,p2) be the excess demand (demand minus supply) of the i‘l‘
commodity. For 'i = 1, 2, assume that the price adjustment is given by pi’ =
k;E; (111,112) for constants k, > 0. Based on economic assumptions, Shone on
pages 353-357 of [Sho02] gives the following information about the nullclines
and vector eld in the (p1,p;)-space: Both nullclines E.,- = 0 have positive slope
.. in the (phpg)-space and cross at one point p‘ = (p{,p§) with both p: > 0. The
nullcline E1 = 0 crosses from below .E2 = 0 to above at p‘ as pl increases. The
excess demand E1 is: (i). positive to the left of E1 = 0 and (ii) negative to the
right of E1 = 0. The excess demand E2 is: (i) positive below E2 = 0 and (ii)
negative above E2 = O. '
a. Draw a qualitative picture of the nullclines and give the signs of p1’ and
pg in the different regions of the rst quadrant of the (pl , pg)-space.
b. Using the type of argument given for two competitive populations, argue
that for any initial conditions (p1,°,p;,°) for prices with pm, , pm > 0 have
solutions that converge to p‘.
4.5. Linearized Stability of Fixed Points
Often, the knowledge,of the locations of the nullclines and of the signs of it and 3)
in different regions- is not enough to completely understand the properties of the
trajectories. This is especially true when solutions cycle tluough a sequence of these
regions as illustrated in Example 4.20. Another important technique is to linearize
the vector eld near the xed points. The linear terms often dominate near the xed
point, and in that case, they determine the behavior of the nonlinear equations near
the xed point; in particular, the linearized system usually determines the stability
of the fixed point. .
In Example 4.18, the linearized system at the origin is obtained by dropping
the 22:3 term from the 3} equation,
:i: = -2:
-y = -—2y.
The origin is a stable node for this linear system. For an attracting xed point, the
linear terms dominate the higher order terms near the xed point, so the solutions
of the system of nonlinear equations behave similarly to those of the linear system.
In particular, the nonlinear system of di ’erential equations is asymptotically stable.
See Theorem 4.6.
Example 4.21. Consider the system of di erential equations
i=4%
1,7 = :0 + 4 a:3.
4.5. Linearized Stability 135
The linearized system at the origin is obtained by dropping the 42:3 term:
i=4u
1] = 2:.
The origin is a saddle with eigenvalues :l:2. The eigenvector for —2 is (31), and
the eigenvector for 2 is
The linear system has two trajectories that tend toward the origin from opposite
sides along straight lines:
ie'2‘ .
Theorem 4.7 shows that the nonlinear equations have two nonlinear trajectories
that tend to the xed point from opposite sides and are tangent to the vectors:
. (3,).
The points of these two trajectories ma.ke up what is called the stable manifold of
the xed point W"(0) (i.e., the set of all points pg such that the solution ¢(t; pg)
tends to the xed point as t goes to in nity and w(p0) = See Figure 14.
WWW
W%®
Figure 14. Stable and Unstable Manifolds
In the same way, the linear equation has two trajectories that tend to the xed
point as t goes to —oo,
:l:e2‘ .
Again, the nonlinear equations have two nonlinear trajectories that tend to the
xed point from opposite sides as t goes to —oo a.nd are tangent at the origin to
the vectors
i (12) .
The points of these two trajectories make up what is called the unstable manifold of
the xed point W“(0) (i.e., the set of all points pg such that the solution ¢>(t;p0)
tends to the xed point as t goes to —oo and a(p0) =
136 4. Phase Portraits
Example 4.22. As another example, consider the system of differential equations
it = —:|: + y,
y = 2 — 2:123/2.
The nullclines are y = :1; and xyz = 1; there is a unique xed point rs = 1 and y = 1.
Using the signs of st and 3] is not enough to determine the behavior of the solutions
near the xed point. However, using the Taylor expansion about the xed point
for the two differential equations
if = f($1y) = _$+Z/»
-- i = 9(I, 1/) = 2 — 2111’.
letting u = at — 1 and v = y —. 1, we have
ft=Ii=f(1,1)+’tL%(1,1)+U-gy£(1,1)+"'=‘—'l£-l-‘U,
1}=3}=g(1,1)+ugg(l,1)+*ug%(1,1)+---=-2u—411+---.
The coe icient matrix for the linear terms has eigenvalues -2 and -3, with eigen-
vectors and , respectively. The linearized system has a stable node
at the origin. The linear terms dominate near the xed point, so the nonlinear
equations also have an attracting xed point. Also, most solutions approach the
xed point with an asymptote of the line y—1 = —(a: -1) (i.e., with an asymptotic
displacement in the direction of the vector ). See Figure 15. The stable man-
ifold of the xed point (1,1) W*"((1,1)) certainly contains a disk about the xed
point. (The gure indicates that it contains at least all the points with O 3 a: 5 2
and 0 5 y 5 2.) Therefore, W’((1,1)) is called the basin of attraction of (1,1).
i2 y
1
ll ll}
0 12
Figure 15. Nonlinear Sink, Example 4.22
4.5. Linearized Stability 137
For a general equation
Ii: = f(Iv. 1/).
19 = y(I» 11),
the linearized system at a xed point (:z:‘, y‘) is given by
<:>=(2Q_: Q <:>»
8:1: 8y
where all the partial derivatives in the matrix are evaluated at (:z:"‘,y‘). When
comparing the solutions of the linearized system with the nonlinear system, the
coordinates (u,v) for the linearized system must be compared with (:r,y) = (u +
2:‘, v + y‘) for the nonlinear systemu If
Fe1> = (58:51?) »
then, as a shorthand notation, we write
{6<1-,1/-> ;3’<w-.1/-)
'" Que y-) @<w' 11')DF($‘ ') = 1
x y
am I ay 1
for the matriz: of partial derivatives or derivative.
For n variables, if
. F1($1, . . .,.1'Jn)
F(a:1,...,:t,,)= E
Fn(I1, . . . ,$n)
and x‘ = (:c1‘,...,a:;) is a xed point, then we write
DF(=<-) = <%(X'))
J
for the n X n matrix of partial derivatives or derivative. The linearized system is
Ill = .DF(x~)l.l.
De nition 4.23. If x‘ is a xed point of 1': = F(x), then we refer to the eigenvalues
of the matrix of partial derivatives DF(x.) as the eigenvalues of the xed point or
the eigenvalues of x‘.
A xed point x‘ is called hyperbolic provided that the real parts of all the
eigenvalues of the matrix DF(,‘-) are nonzero. A hyperbolic xed point is called a
saddle xed point provided that it has at least one eigenvalue with a positive real
part and at least one other eigenvalue with a negative real part.
If a xed point is hyperbolic, then the stability type of the xed point for the
nonlinear system is the same as that for the linearized system. The following result
states this more precisely.
138 4. Phase Portraits
Theorem 4.6. Consider a differential equation x = F(x) in n variables, with a
. 2.
hyperbolic xed point x‘. Assume that F, -%.(x), and (_ x_ k ) are all con-
tinuous. ‘Then, the stability type of the azed pdint for the nolnlinear system is the
same as that for the linearized system at that xed point. In particular, we have
the following.
a. If the real parts of all the eigenvalues of DF(,,-) are negative, then the
red point is asymptotically stable for the nonlinear equation (i.e., if the origin is
asymptotically stable for the linearized system, then x‘ is asymptotically stable for
the nonlinear equation). In this case, the basin of attraction W”(x‘) contains all
the points within a small distance of the xed point.
b. If at least one eigenvalue of DF(,-) has a positive real part, then the red
point x‘ is unstable for the nonlinear equation. -(The linearized system can be a
saddle, unstable node, unstable focus, etc.)
c. If one of the eigenvalues of DF(,,-) has a zero realpart, then the situation
is more delicate. In particular, for n = 2, if the xed point is an elliptic center
(eigenvalues :l:i,B) or one eigenvalue is 0 of multiplicity one, then the linearized
system does not determine the stability type of the azed point.
Remark 4.24. The solutions of the nonlinear and linear equations are similar
enough that most trajectories in Example 4.22 must approach the asymptotically
stable xed point in a direction that is asymptotic to plus or minus the eigenvector
for the weaker eigenvalue of -2 (i.e., tangent to the line y — 1 = :1: — 1).
We next consider the behavior near a nonlinear saddle xed point.
De nition 4.25. The stable manifold of the azed point W‘(x") is the set of all
the points that tend to the xed point as t goes to plus in nity,
W‘(x") = {pg : ¢(t;pg) tends to x‘ as t —> oo} = {pg :w(pg) = {x‘}
Notice that the orbit ¢(t;pg) converges to the single xed point x‘ as t goes to
in nity if and only if w(pg) = {x'}. .
We also consider all those points in the stable manifold whose whole forward
orbit stays near the xed point. For a small r > 0, the local stable manifold of size
r of the xed point x‘ is the set of points in W"‘(x') whose whole forward orbit
stays within a distance r of x‘,
W,:’(x‘) = {pg G W’(x‘) : ||¢>(t;pg) — x'I| < r for all t Z 0 }.
Example 4.10 shows that there can be points on the stable manifold of a xed
point that start near the xed point, go away from the xed point, and nally return
and converge to the xed point; such points are not on the local stable manifold.
De nition 4.26. The unstable manifold of the xed point W“(x‘) is the set of all
the points that tend to the xed point as t goes to minus in nity,
W“(x‘) = {pg : ¢(t;pg) tends to x‘ as t -—> —oo} = {pg : a(pg) = {x‘} }.
4.5. Linearized Stability 139
For a small r > 0, the local unstable manifold of size r of the red point x‘ is
the set of points in W“(x‘) whose whole backward orbit stays within a distance r
of x‘,
W,I‘(x") = {pg E W‘(x') : ||¢(t;pg) — x’|| < r for all t § O}.
In two dimensions, the stable and unstable manifolds of a saddle xed point
are curves, each of which separates the phase plane into the two sides of the curve.
For this reason, in two dimensions, the stable and unstable manifolds of a saddle
xed point are often called separatrices. For a sink, the stable manifold is a whole
region in the plane and is called the basin of attraction of the xed point. (The set
is open as de ned in Appendix A.2.) The unstable manifold of a sink is only the
xed point itself. For a source, the stable manifold is just the xed point and the
unstable manifold is a whole region in the plane.
The Lorenz system considered in Chapter 7 has a a two-dimensional stable
manifold in IRS. The word manifold is a term in mathematics which includes curves,
smfaces, and higher dimensional objects. The word manifold is used to refer to
W‘ (x") and W“(x') because these sets can have a variety of dimensions. A more
thorough discussion of manifolds is given in Appendix A.2.
The next theorem states that the local stable and unstable manifolds can be
represented as graphs. This graph is expressed in terms of subspaces associated
with the eigenvalues at a. xed point, which we de ne next.
The stable subspace at a xed point is the linear subspace spanned by the set
of all the generalized eigenvectors of the linearized equations at the xed point
associated with eigenvalues having negative rea.l parts,
IE” = spa.n{ v : v is a generalized.eigenvector for DF(,,-)
whose associated eigenvalue has negative real part
In this subspace, we consider all vectors whose length is less than r,
Em-> = {v e 1E'= uvu s T1-
Similarly, the unstable subspace at a xed point is the linear subspace spanned by
the set of all the generalized eigenvectors of the linearized equations at the xed
point associated with eigenvalues having positive real parts,
IE“ = span{ v : v is a generalized eigenvector for DF(,.-)
whose associated eigenvalue has positive real part} and
lE“(r) = {v E 1E" : 5 r}.
If the xed point is hyperbolic, then 1R" = IE’ + 1E“. Thus, we can identify
(y,z) E E‘ >< IE“ with y +z G R".
The next theorem states that the local stable (respectively, unstable manifold)
can be expressed as a graph from 1E’ (r) into lE“(r) (respectively, from lE“(r) into
lE’(r)). In two dimensions, a local stable manifold of a saddle xed point is a curve
segment through the xed point; it can be represented as a graph over an interval
of length 2r in the line generated by the stable eigenvector. In higher dimensions,
it is a graph over a ball in the subspace of all of the contracting directions for the
linearized equations.
140 4. Phase Portraits
Theorem 4.7. Assume that a differential equation x = F(x) has a saddle red
point at x".
a. In two dimensions, the eigenvalues for x‘ are real, and one is positive and
one is negative, /\,, < 0 < /\,,. Let v’ be an eigenvector for A, and v“ be an
eigenvector for ,\,,. Then, for sufficiently small r > 0, the local stable manifold of
size 1* for the xed point x‘, W,i'(x'), is a curve passing through x‘ that is tangent
to v’ at x‘. Similarly, the local unstable manifold of size 1', W“(x'), is a curve
passing through x‘ that is tangent to v" at x‘.
b. In n dimensions, let 1E” and IE“ be the stable and unstable subspaces previ-
ously defined. Then, for su iciently small r > 0, the local stable manifold of size
r"_for the xed point x‘, W,f(x"), is a “surface”, which is a graph of a function cr’
from IE‘ (1') into lE"(r) that is. tangent to'the subspace IE’ at x‘,
W§(=<') = {X' + (y,v’(Y)) =3’ 6 ll-3’(T) }-
Here, the pair (y, o~"(y)) with y E lE’(r) and a‘(y) G lE“(r) is identi ed with
y + v’(y) ‘E R"-
In the same way, for su iciently small 1' > 0, the local unstable manifold of size
1' for the fired point x‘, W,!‘(x‘), is a “surface”, which is a graph of a function
o“ from lE“(r) into lE*"(r) that is tangent to the subspace IE“ "at x‘,
Wj‘(x') = {x' + (a“(z),z) : z E lE“(r)
c. In any dimension, the (global) stable manifold is the set of points whose
forward orbit eventually gets into the local stable manifold of size r,
W‘(x"') = {p :-¢(t;p) 6 W;"(x‘) for some t 2 0} = U ¢(¢; W: (x")).
-" < ego
In the same way, the (global) unstable manifold is the set of pointswhose backward
orbit eventually gets into the local unstable manifold of size r,
W“(x‘) = {p : ¢>(t;p) G W}‘(x') for some t 5 0} = U ¢(t;W}‘(x’)).
CZO
See [Rob99] for a. proof.
The preceding theorem says that stable and unstable manifolds exist, but does
not say how to determine them. In general, there is no good way to get an analytical
formula for these curves. It is possible to represent them as a power series near the
xed point, but this is difficult to calculate and not that useful. To nd a numerical
approximation using a computer, an initial condition is taken a. short distance away
from the xed point in the direction of the eigenvector, X9 = x‘ + av“; then, the
solution {¢(t;xq) : t Z 0} gives an approximation for the unstable manifold. To
nd an approximation for the stable manifold, an initial condition xq = x‘ + cv’
is used, and the solution is calculated for negative time.
The theorem also deals with only the stable and unstable manifolds (curves)
near the xed point. Away from the xed point, the stable manifold cannot cross
itself, but can wind around in the phase space and spiral in toward another xed
point or periodic orbit. It can also connect with the unstable manifold of this or
another fixed point, as occurs for the undamped pendulum discussed in Example
5.3.
4.5. Linearized Stability 141
Example 4.27. Consider the phase portrait for the system given in Example 4.20,
:2: = —.’c + y,
g = my — 1.
First, we nd the xed points: a’: = 0 implies a: = y, and y = 0 implies that
1= my = :22 or zr = :l:1. Thus, there are two xed points: (1,1) and (-1, —1).
The linearization is DF(=.t>=(-y1 1
At (1, 1),
—1 1
DF(l.,1) = <1 n
which has characteristic equation A? — 2 = 0 and eigenvalues /\ = :t\/§, so (1, 1) is
a saddle. The eigenvectors are (1 1:1
At (—1,—l),
. -1 1
which has characteristic equation A2 + 2A + 2 = 0 and eigenvalues A = —1 :|: i, so
(1, -1) is a stable focus.
The nullclines are = 0} = {y = :0} and {y = 0} = {y = 1/2:}. See Figure
13. The orbit coming out of the saddle below goes down to the stable focus and
spirals into it. The saddle has a trajectory from either side going asymptotically
into it. These trajectories extend out to in nity. All initial conditions starting
below the stable manifold of the saddle end up by spiraling into the stable focus.
Initial conditions above the stable manifold have trajectories that eventually go off
to in nity.
Much of this phase portrait can be determined by analyzing the stability of
the xed points and nding the nullclines. However, computers are very helpful in
imderstanding the behavior. In drawing a phase portrait using a computer program,
the rst question is to determine the window in which to display the phase portrait.
Certainly, it should include the xed points and should be big enough to show how
solutions approach the xed points. Often, it is best not to make the window too
big so that the i.mportant features are apparent.
Once the window is chosen, various initial conditions must be picked to display
important behavior of the phase portrait. This takes some experimentation, espe-
cially if not much is known about the equations. By varying the initial conditions,
it is possible to nd solutions that go near the stable manifold of a saddle and then
progress onward to another part of the phase portrait. See Figure 16 for the choices
made for the present example.
Example 4.28 (Linear Center). Consider the system of nonlinear differential
equations
2': = —y + aa:(a:2 + (92),
g = 2: + ay(a:2 + yz).
142 4. Phase Portraits
.y
NU
-1
_ 0i 2m
-1
-'2 -1 '
-2
Figure 16. Phase plane for Example 4.27
The eigenvalues of the matrix of partial derivatives at the origin are ii, no matter
what the value is of the parameter a. Therefore, the origin is a center for the
linearized differential equations. We show that by varying the parameter a, the xed
point of the nonlinear system of differential equations can be attracting, repelling,
or L-stable. ‘
As we have rotation, it is convenient to change to polar coordinates. We have
1'2 = 2:2 + g2, so
1"!" = :|::i: + yy = :|:(—y + a:|:r2) + y(:z + ayrz) = a1'2(:c2 + 1/2) = arr‘ or
1" = 0.1-3. -
Similarly, tan(0) = V/Z, and
sec2(0) 0- = —my —Tvi
__ 2:2 + aaayrz $+2 1/2 — aaiyrz
1.2 or
= F,
6= 1.
If a < 0, then 1‘ < 0 for 1' > O, and the origin is attracting. Thus, the origjn
is not linearly attracting, but it attracts due to the cubic terms in the differential
equations. Similarly, if a > 0, then 1" > 0 for r > 0, and the origin is repelling due
to the cubic terms. Finally, the equations with a = 0 give a linear center. This
shows that, when the linear terms give a center, they are not enough to determine
the stability type of the xed point.
Exercises 4.5 143
Exercises 4.5
1 Consider the system of differential equations
:i:=—a:—y+4,
y=3—a:y.
Find the xed points.
Determine the type of the linearized system at each xed point (saddle,
stable focus, etc.).
Determine the nu.l.lclines and the signs of :i: and 1) on the nullclines and in
the various regions determined by them.
Draw the phase portrait by hand for the system using the information
from parts (a) through (c). Explain and justify your sketch.
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.
2 Consider the system of differential equations
:i:=y—:z3,
y=a:y—1.
Find the (two) xed points.
Determine the type of the linearized system at each xed point (saddle,
stable focus, etc.). .
Determine the nullclines and the signs of :i: and 3} on the nullclines and in
the various regions determined by them.
Draw the phase portraits for the system using the information from parts
(a) through (c). Explain and justify your sketch. _
Using the computer, trajectories for enough initial conditions to show the
important features of the phase portrait.
3 Consider the damped pendulum equation §+ cl? + sin(0) = 0, where c > 0.
Find and classify the xed points of the related system of rst-order equations.
4 Consider the system of di erential equations
:i:=—2:r+y,
§l]=4—2:1:y.
Find the xed points.
Determine the type of the linearized system at each xed point (saddle,
stable focus, etc.).
Determine the nullclines and the signs of :i: and 3'1 on the nullclines and in
the various regions determined by them.
Draw the phase portrait for the system using the information from parts
(a) through (c). Explain and justify your sketch.
144 4. Phase Portraits
5. Consider the system of differential equations
¢—'_ -52 + ya
t=1—¢-
a. Determine the xed points and linear type of each xed point (saddle,
stable focus, etc.).
b. Determine the nullclines.
c. Draw the phase portrait for the system using the information from parts
(a) and (b). Explain and justify your sketch.
6. Consider the system of differential equations
':i:=a:y, » .
3}=1—y—(1+y):r.
a. Determine the xed points and linear type of each xed point (saddle,
stable focus, etc.).
b. Determine the nullclines.
c. Draw the phase portrait for the system using the information from parts
(a) and (b). Explain a.nd justify your sketch.
7. Consider the system of differential equations
i=($_1)(y_]-)1
'!7=3-$21/»
which has xed points at (1,3) and (3,1).
a. Determine the type of the linearized equations at each xed point (saddle,
stable node, etc.).
b. Determine the nullclines and the signs of :i: and 1) in the regions determined
by the nullclines.
c. Draw the phase portrait for the system using the information from parts
(a) and (b). Explain your sketch of the phase portrait.
8. Consider the system of differential equations
:1: = $2 — y — 1,
@=v—Mw—%-
a. Show that there are three xed points and classify them. eld does
b. Show that there are no closed orbits (look at what the vector
along the three lines which go through a pair of xed points).
c. Sketch the phase portrait of the system.
9. Consider the system of di erential equations
a':=a:—y+a:2—a:y,
1] = —y + 1:2.
a. Show that there are three xed points and classify them.
b. Sketch the phase portrait of the system using the nullclines and the infor-
mation of part (a).
4.6. Competitive Populations 145
10. Find the xed points and classify them for the system of equations
,/ :-c=v,
1) = —a: + w 1:3,
Ll.) = —w.
11. Find the xed points and classify them for the Lorenz system of equations
:i: = —10a: + 10y,
1] = 28:1: — y — 1:2,
2—i —3§z+a':y.
12. Prove that a xed point source of a system of differential equations in the plane
is unstable. Hint: Compare with Theorem 4.12 in Section 4.8.
4.6. Competitive Populations
In this section, we discuss two populations which interact and have either positive
or negative effects on the growth of the other population. Just as in the case of one
population, i/¢ and 17/y are the rates of growth per unit population. A general set
of equations would have these rates of growth to be functions of both populations,
=‘f($’y)’ i: mf(Iay)1
E-H!-I =g($,y), O!‘ -
U
y=yg(xIy)'
One classical model takes these dependencies to be linear functions of :1: and y,
yielding the following equations which are called the Lotka— Volterra equations:
ii = $(B1 +l41,1 11? + 141,211)»
17 = y (B2 + 142.1 1? + 142,2 11)-
We consider two cases. In the rst case, each population has a negative e 'ect
on the growth rate of the other population, and A1_2,A2_1 < 0. This case is called
competitive species. (We often also assume that the population has a negative effect
on its own growth rate, so AM, A2,; < 0.)
In the second case, one of the populations has a positive effect on the growth
rate of the other (it is the prey), and the second population has a negative effect on
the growth rate of the rst (it is the predator): A“ > 0 and Am < 0. This case
is called a. predator—prey system, and is discussed in the next chapter, because the
solution method is more similar to those discussed in the next chapter than those
of the present chapter. See Section 5.1.
Competitive Populations
We rst consider competitive systems in which each population has a negative
e 'ect on the growth rate of the other and the dependency is linear:
E = K — :2 — ay,
zr
—ti = L — bar — y,
ll
146 4. Phase Portraits
or
:i:=:r:(K——a:—ay),
zJ=y(L—bI—y)-
We consider only cases in which all the parameters are positive in the form given
here. The constants K and L measure the growth rates of :1: and y, respectively,
when the populations are small; we assume that these growth rates are positive.
Each population has a negative e ect on its own growth rate and the growth rate
of the other population. We have assumed (or scaled the variables so) that the
coe icient of :1: in the rst equation and of y in the second equation is -1.
'" The nullclines are 2: = 0 or K — :0 — ay = 0 for the 2': = 0 equation, and y = 0
or L—ba:—y=0fory=0. Fora xed point, ifa:=0, theneithery=0ory=L.
There is also a xed point with y = 0 and :0 = K .' Finally, there is the xed point
(:v',y"), which is the solution of the two equations
0 = K — a:' — ay‘,
0 = L — bat‘ — y‘.
We are interested in nonnegative populations, so this preceding xed point is of
interest only if both :|:"' > 0 and y‘ > 0. This occurs if the lines cross in the rst
quadrant. Since the 2: and y intercepts of the lines are K , K/0, L/4,, and L, there is
a xed point inside the rst quadrant if either
%<K and L>—, or
—>K and L<-—PNPN.
Thus, there are four xed points, (0, 0), (K, 0), (0, L), and (1:', y‘), that satisfy the
two preceding equations.
The matrix of partial derivatives is
((K — :1: — ay) —- 2: —a.:z:
-by (L—b=v—y)—y '
I. The rst case with an interior xed point has
L<bK and aL>K.
Then, the nullclines cross in the rst quadrant, so rs‘ > 0 and y‘ > 0. The matrix
of partial derivatives at the xed point (K, 0) is
—K —aK
0 L ~ bK '
Since L—bK < 0, (K, 0) is a stable node. Similarly, the matrix of partial derivatives
at (0, L) is
K — q.L 0
—b L —L ’
and (0, L) is a stable node. Finally, at (:1:",y‘),
< -2‘ -0.1‘)
__byI _yI
4.6. Competitive Populations 147
has determinant :c"y"(1 — ab). Since L < bK < b(aL), 1 — ab < 0 and the xed
— “point is a saddle.
The nullclines create four regions in the rst quadrant. See Figure 17. The
bottom region abutting on (K,0), to the right of 1) = 0 and below :i: = 0 has
solutions moving down and to the right. Since they cannot leave the region, they
must go to the attracting xed point at (K, 0).
ll
L
K\
a
\ (E
LK
"5
Figure 17. Competitive Exclusion
Similarly, the points in the region abutting on (0, L) above 5: = 0 and below
3) = 0 are moving up and to the left and they converge to the xed point at (0, L).
For any points in the region abutting on (0,0) below :i: = 0 and 3'1 = 0, both
2': and 3'1 are positive. The solutions eitlier leave and enter one of the two regions
just analyzed, or they must stay in the region forever and tend toward the xed
point (a:',y‘) (i.e., they must be on the stable manifold of (a:‘,y‘)). Similarly,
points above and to the right of both :i: = 0 and 3} = 0 must either be on the stable
manifold of (z*, y‘), or enter the region where trajectories go to (K, 0) or the region
where they go to (0, L). Thus, the stable manifold W‘(a:',y') separates the rst
quadrant into the points that are in the basin of (K, 0) and those in the basin of
(0, L). The points that are on W‘(:1:', y"‘) tend to the saddle point (a:',y"). Thus,
we have competitive exclusion, with most initial conditions leading to one of the
populations dying out as t goes to in nity.
II. The second case with an interior xed point has
L>bK and aL<K.
Again, the nullclines cross in the rst quadrant, so 1:" > 0 and y‘ > 0. See
Figure 18. Since L > bK > baL, 1 — ab > 0, the determinant of the matrix of
partial derivatives at (a:',y“) is positive and the trace is negative. Thus, (:c',y') is
asymptotically stable. In fact, both eigenvalues are real. Each of (K, 0) and (0, L) is
8- saddle, which attracts orbits along the respective axis and repels into the interior
of the rst quadrant. The two regions between the two lines :i: = 0 and 1] = 0 are
invariant, with trajectories tending monotonically toward (a:‘, y‘). Solutions in the
Other two regions either tend directly toward (:c", y’) or enter one of the other two
148 4. Phase Portraits
regions. Therefore, eventually, all trajectories tend toward (:z:‘,y") (i.e., the basin
of attraction of (a:',y') is the entire open rst quadrant). The stable manifold of
(K, 0) is just the positive 1:-axis. Similarly, W'°(0, L) is the positive y-axis. Thus,
we have competitive coexistence of all initial conditions which contain both species.
3/
ii
a.
L\
\ :0
Kt
Figure 18. Competitive Coexistence
4.6.1. Three Competitive Populations. In this section, we do not consider the
general case of three competitive populations. Instead, we consider one example
of the Lotka—Volterra_,system for which the w-limit set of most points is not a
single xed point but a qycle of xed points. Based on what can happen with two
populations, a conjecture might be that the w-limit set of any trajectory is a single
xed point. However, the next example from the book [Hof88] by Hofbauer and
Sigmund shows that this conjecture is false.
Consider the competitive system of three populations given by
ii = $1(1— 101- £1912 -13%),
i2 = $2(1— $1 - $2 - 01113),
$3 = I3 (I. -— (ICE-1 — 1l!2 — 233),
with 0 < £3 < 1 < a and 0: + )6 > 2. We consider only points with nonnegative
coordinates,
{($1,1I2,$a) 11111 Z 0, $2 Z 0, $3 2 0 }-
The xed points are u1 = (1,0,0), 112 = (0,1,0), u;-, = (0,0,1), and x‘, where
0:; = as; = 1:5 = 1 / (1 + 0' + B). The matrix of partial derivatives at x‘ is
_ -0, _
DF(x~> = - - -I-5
.—ow=»- — — »-so
Because of the symmetry in the problem, it is possible to show that the eigenvalues
are -1 and
/\ 2 = _ = 1 (_ 1 _ a e1. 211'/3 __ ,6 e2. 4-Ir/3) ,
A3 -—+ a +i1B
4.6. Competitive Populations 149
dthe real parts of /\2 and /\3 are
__1__ (-1, %) = ail > 0
1+a+6 1+a+ '
The eigenvector for -1 is (1, 1, 1). Thus, x" has a one-dimensional stable manifold
and a two-dimensional unstable manifold. The diagonal A are the points x for
which $1 = :2 = 1:3. Since :i:; = 5:2 = :i:3 at points of A, it is invariant and the
stable manifold of x‘ consists of the points on the diagonal with positive entries,
W8(X) = {(£L‘-1,132,583) ZI1 =32 =$3 >
The matrix of partial derivatives at u; is
-1 -0: —,B
DF(,,,) = 0 1 — 0.
1—a
0O
The the eigenvalues are —1 < 0, 1-0: < 0, and 1 — > 0. Thus, this xed point has
a two-dimensional stable manifold and a one-dimensional unstable manifold. The
other two xed points 112 and u3 have similar eigenvalues and the same dimensional
stable and unstable manifolds.
We start by showing that most orbits tend to the union of the coordinate planes,
{$1 = 0}, {$2 = 0}, and {$3 = 0}. Let
P = $1 022 $3 and
S = :21 + :02 + :03.
Then ,
S = 2:; + 2:2 + :23 — [zvf + $3 + :1:§~+(a + )(a:1:|:2 + 1:22;; + $311)]
5 2:1 + 2:2 + :03 — [sf + + :n§ + 2 ($112 + a:2a:3 +:r3:n1)]
= s - $2
= S (1 — S).
Therefore, any orbit must be asymptotic to the set for which S = 1 (unless S = O)
and has to enter and remain in the set where S $ 2. This shows that all orbits are
bounded. Turning to P,
P = 531 $21123 -l-IE1i)2£B3 +181 1:2i3
=P[3-(1+a+ )S].
To see that solutions go to the set for which P(x) = 0, a computation shows that
Ed (PS_3) = PS_4 (1 — 1+;) [(IB1—$2)2 + (:22 - a:3)2 + (:03 —:c1)2]
3 0.
The preceding derivative is strictly negative if P(x) > 0 and x is not on the diagonal
A. Since the orbits are bounded, we show in Section 5.3 that, in such a situation,
PS4 is a strict Lyapunov function and the trajectories rnust go to the minimum
of PS'3, which has P(x) S’3(x) = 0 (i.e., to the coordinate planes where at least
one coordinate variable is zero).
150 4. Phase Portraits
On the coordinate plane {$3 = 0}, the nullcline $2 = 0 is strictly above the
nullcline $1 = 0, and all orbits having initial conditions with $1 > 0 and $2 > 0
tend to the xed point U2. Thus,
W‘(ug) D {($1,$2,$3):$3 =0, $12 0, $2 > 0} C {$3 =0}.
We saw from the eigenvalues that the xed point u; has a one-dimensional unstable
manifold, and it has a one-dimensional unstable manifold in the coordinate plane
{$3 = 0}. Therefore, W“(u1) is contained in this plane and
W“(u1) C W’(ug) C {$3 = O}.
.. The coordinate planes {$1 = 0} and {$2 = 0} are similar, with
W“(ug)"C W"’(u3) C {$1 ='0} and
W"(u3) C W’(u1) C {$2 = 0}.
Thus, the three xed points ul, 112, and 113 form a cycle of xed points connected
by stable and unstable manifolds.
Take any initial condition ($9, $2, $3) with all three coordinates positive. Then,
its w-limit set w($§’,$g,:1:g) must be in the set
P‘1(0)|'lS_1(1) ={(11=1»I2,=Fa) =P($1,1=2,$a)= 0 and -5'(fB1.$2,$s) = 1}-
Since the stable manifolds of the three xed points do not enter the positive octant
where all coordinates are positive, w($‘{, rug, mg) cannot be just a single xed point.
It must be positively invariant. Indeed, the orbit cycles near each xed point in
succession and then repeats the process. Therefore, the w-limit set is the whole
triangular set ’
~(w‘l.:3.¢S> = P"(0> n S*‘<1)-
Thus, the w-limit set is not a single xed point but a cycle of unstable manifolds
running from one xed point to the next.
,.
Exercises 4.6
1. Consider the system of differential equations for a competitive pair of popula-
tions
:i:=$(K—$-ay),
1J=u(L—b¢v—u)-
Assume that K > L/b and K/a. > L.
a. Find the xed points for :1: Z 0 and y Z 0, and determine the type of the
linearized equations at each xed point (e.g., stable node, stable focus,
saddle, unstable node, etc.).
b. Sketch the phase portrait for $ Z 0 and y 2 0, including the nullclines.
c. Do most solutions with $0 > 0 and yo > 0 tend to a xed point? Which
xed point?
Exercises 4.6 151
2 Consider the system of di erential equations for a competitive pair of popula-
tions
1'1 = =v(2 — 1 — 1/).
ii = 1/(3 — 11 — 1/)-
a. Find the xed points for this system a.nd classify them.
b. Use the nullclines and the stability type of the xed points to sketch the
phase portrait. Indicate the basin of attraction of any attracting xed
points.
Consider the system of differential equations for a competitive pair of popula-
tions
$ = $ (K — ay),
1) =' y (L — b$).
Note that there is no limit to the growth of each population separately.
a. Find the xed points for $ Z 0 and y 2 0, and determine the type of the
linearized equations at each xed point (e.g., stable node, stable focus,
saddle, unstable node, etc.).
b. Sketch the phase portrait for $ Z 0 and y Z 0, including the nullclines.
Consider a competitive market of two commodities whose prices (against a nu-
meraire) are p1 and P3. Let E,-(_p1,p2) be the excess demand (demand minus
supply) of the i"" commodity. For i = 1,2, assume that the price adjustment
is given by p,- = k,E1(p1,pg) for constants k,- > 0. Based on economic as-
sumptions, {Sho02] (pages 353-357) gives the following information about the
nullclines and vector eld in the (p1,p2)-space: Both nullclines E, = 0 have
positive slope in the (p1,p2)-space ahd cross at one point p‘ = (p§,p§) with
both pf > 0. The nullcline E1 = 0 crosses from below E2 = 0 to above at p"
as p1 increases. The partial derivatives satisfy -68E-ii < 0 and 56E2 < 0.
12
a. Draw a qualitative picture of the nullclines and give the signs of p1 and [lg
in the different regions of the rst quadrant of the (p1, P2)-SPBCB.
b. Argue that for any initial conditions (p1,o, p2_11) for prices with p1_11, , pm >
0 have solutions that converge to p‘.
Consider the situation in which both species have a positive e ect on the growth
of the other, given by the system of differential equations
:i:=$(K+a.y—c$),
1):?/(L-l_ba:_ey)a
where all the parameters are positive. K, a, c, L, b,e > 0. Also assume that
c/a > b/e.
a. Show that there is a xed point ($‘,y") with $‘ > 0 and y‘ > O.
b. Show that the xed point ($',y') from part (a) is asymptotically stable.
Hint: For the matrix of partial derivatives, show that the trace 'r < 0 and
the determinant A = (ce — ab) $'y" > 0.
c. Draw the phase portrait for $ 2 0 and y 2 0, including the nullclines.
Discuss the basin of attraction of ($',y") a.nd the stable and unstable
manifolds of other xed points.
152 4. Phase Portraits
4.7. Applications
4.7.1. Chemostats.
A simple chemostat
A chemostat is a laboratory apparatus used to grow micro-organisms under
controlled experimental conditions. In this apparatus, a constant supply of nutri-
ents is pumped into a vessel containing the micro-organism and the same amount
of uid containing the cells, nutrients, and byproducts ows out. Because the ow
of uid into and out of the vessel is the same, the volume inside the chemostat
remains constant.
'- In the rst model of a simple chemostat, the concentration of the micro-
organism in the effluent tends to a constant value (an equilibrium), as would be
expected in the chemical reaction. This source of an effluent with a constant con-
centration of the micro-organism could be used in a further experiment, or the
equilibrium concentration could be used to study the rate of cell growth or the rate
of metabolic product formation.
Sections 6.2.1 (Brusselator) and 6.8.1 (Field—Noyes model for the Belousov—
Zhabotinsky chemical reaction) discuss models of chemical processes that tend to
periodic levels of concentrations in the e fluent. These examples need at least three
chemicals, even though the example in Section 6.2 contains only two variables. In
fact, people have found situations in which the concentrations are chaotic.
Let S be the concentration of the nutrient and a: the concentration of the micro-
organism. The concentration of the nutrient pumped in is C, if the nutrient were
not consumed by the micro-organism, the rate of change of S would be proportional
to C — S, S = (C — S)D. -:The nutrient is consumed by the micro-organism at a rate
that is a function of the concentration of the nutrient 'r(S) times the concentration
of the micro-organism $. A model commonly used is
“”"zmISs
with m > 0 and a > 0, which was introduced by Monod in 1950: it has the property
that 1'(0) = 0 and 1'(S) approaches a limit value of m as S approaches in nity. Using
this model, if C is the concentration of the nutrient pumped into the vessel, the
differential equations become
S-_-(C S)D am+.SS'$,
”_ ‘EmIS§$_D”
where D is a constant determined by the rate of ow of uid in and out of the
vessel and so can be adjusted. We assume that all of the parameters are positive
and m > D. This last condition implies that the rate 1-(S) for large S is greater than
the amount removed by the ow of uid out of the vessel. The initial conditions
4.7. Applications 153
satisfy 12(0) > 0 and S(0) Z 0. The nullclines are
{:i=0}={S=—m—i&D } and
-_ _ __nw-sns+@
The second nullcli.ne has :2 = 0 when S = C. Also, the slope of the nullcline is
dz dS( BD 5+ D(C' — a.) + D a C )
dS d m Bm mS
_ D DaC
_ _mi " imiS
<0,
and a: goes to oo as S goes to 0. Therefore, there is a xed point in the interior of
the rst quadrant if and only if
* - ;;Ta_D-5 < C-
Theorem 4.8. For the preceding system of di 'e'rential equations, assume that m >
D and /\ < C. Then, the interior xed point (S',a:') is given by
S"=/\ and a:"=i l,
and it is asymptotically stable. The only other xed point is given by S = C and
:0 = 0, and it is a saddle point. In fact, all initial conditions with :00 > 0 and S0 > 0
tend to the interior rced point S‘ and 1;‘.
Proof. A little algebra shows that the unique xed point in the interior has the
form given in the theorem. The matrix of partial derivatives at this xed point is
_D_Q@:£§£ _@2$
a+$‘ a+S‘_
(m — D) :12’ 0
o. + S‘
The trace is negative and the determinant is positive. By Theorem 4.3, this interior
xed point is asymptotically stable.
The matrix of partial derivatives at the second xed point (C, 0) is
mC'
D a+C
0 —m—C——D
Since A < C, the eigenvalues are a+C
-0<0 wd 151-D>0, U
a+C
$0 this xed point is an unstable saddle point.
154 4. Phase Portraits
The next result gives the results when there are no interior xed points. We
leave the details to the reader.
Theorem 4.9. For the preceding system of dijferential equations, assume that m >
D and X > C. Then, there is no interior xed point in the rst quadrant and the
red point (C,0) is asymptotically stable.
For references, see the books on population biology [Bra01] and [VVa.l83].
Two micro-organisms competing for one nutrient
We next consider two micro-organisms in a vessel, with concentrations :0 and y,
that are competing for the same nutrient supplied as the simple chemostat, but are
not directly a ecting each other. Using ,D for the rate of ow through the vessel,
the system of equations becomes -
S: (C_S) D_ »@1m131l? _ l52’m2-5'3/,
G1 + S G2 + S
it = ml S :1: _ Dir,
G1 + S
. = —mg—S—y » D .
y G2 + S y ‘
We are interested only in values with S 2 0, as 2 0, and y Z 0.
Rather than considering these equations with all three variables, we use an
argument to reduce the problem to two variables. We de ne the new variable
i z=C—$— w—&a
whose derivative is ’
. =—C—S.D B-1 m-i 5-I -B-2 "1~2 5'11
2( ) + (Z1 + S + (12 + S
m1Sa: mgSy
l31a1+S+51D$-52 a2+s+32Dy
" =—Dz.
Therefore, the solution z(t) decays to zero,
z(t) = 2(0) e_D‘.
Since we are interested only in the long-term behavior in the equations, we can take
z = 0 and use it to solve for S in terrns of :1; and y. Note that, since we consider
only S Z 0, we need to restrict our consideration to
m=+mu=C-S50
On this set with z = O, the di erential equations for a: and y become
¢=x( -D) ’
1+C— 1$-529
. = imn0—im=i-an — D) .
y y(¢12+C'-1311-13231
Since B1 In + z y 3 C, the denominators of the expressions are positive.
4. 7. Applications 155
We want to consider only parameters for which each organism would approach
a nonzero limit concentration if it were present alone. So, based on the simple
chemostat, we assume
ml > D,
mg > D,
,\, = LD < c, and
171,1 — D
A = i(13 D < C.
2 177.2 — D
By renumbering the organisms, we can assume that
)\1 < X2.
We analyze the phase portrait by considering the nullclines for the differential
equations in only the rc and y variables. With a little bit of algebra, we nd that
the two nullclines are the following:
C—/\1= 11-‘+523! and
C-/\2=l31$+l521l/-
These two lines are parallel, and since A, < /\;, the rst line is outside the second.
Therefore, the nullclines and direction arrows are as in Figure 19. The xed points
are (zrI ,0) and (0,y'), where
$*=C_A1 and y:=C—A2
151 Z31 l
By phase plane analysis similar to that in Section 4.6 for competitive systems, any
trajectory ¢(t; (:co,y0)) with 2:0 > 0 a.nd~y0 > 0 tends to the xed point at (:z',0)
as t goes to in nity.
3/
_/
1/ \
7, \ /
2;.
Figure 19. Nullclines for the competitive chemostat
The discussion of this competitive chemostat is based on the treatment in
lW8183]. For further details, see that reference. That book also considers other
related situations such as a. food chain, in which the organism as lives o the rst
nutrient S, and the second organism y lives oif the rst organism a:.
156 4. Phase Portraits
4.7.2. Epidemic Model. We present a model of epidemics, in which S is the
susceptible population, I is the infected population, and R is the recovered popu-
lation. Because of the variables used for different types of populations considered,
the model is called an SIR model of epidemics. The model we present includes
births in the susceptible group at a constant rate /.LK. (If the total population
were a constant K, then this would be the births occu.rring with a rate of ,u.) We
also assume that there is a death rate of -—/.L from each class of population. There
is an infection rate of people in the susceptible group who become infected which
is proportional to the contacts between the two groups BS1. There is a recovery
of '71 from the infected group into the recovered group. Finally, the disease is fa-
ta] to some of the infected group, which results in the removal rate of —aI from
the infected population. Putting this together, the system of differential equations
becomes '.
S=pK— SI—aS
I= SI—1I-pI—aL
R = 7 I — /i R.
Notice that the change in total population is
S+l+R=MK—S—I-m—aL
Thus, without the term —aI, a total population of K would be preserved.
Because the rst two equations do not involve R, we can solve them separately,
and then use the solution to determine the solution for R. Setting 6 = a + '1 + p,
the system of equations becomes
S=pK— SI—pS
i=os1-at
The nullclines for the equations are
I-_g&s _ 25 - _for {S-0} and
S=% orI=0 for{I=0}.
Thus, the xed points a.re
S_-K,I-_0 and S._-5 ,I._-aK6 5#.
The value I‘ <0 when K/6< 1, and I‘ >0 when [iK/6> 1.
To check the stability of the xed points, we consider the matrix of partial
derivatives
(— I—# -55)
,6 I [3 S — 6 '
If BK/6 < 1, then the only xed point with nonnegative values is at S = K
and I = 0. The matrix of partial derivatives is
c *"<>0 ,BK—6'
4. 7. Applications 157
This has eigenvalues —p. and 13K ~ 6, both of which are negative. It is not hard
to see that this xed point attracts all trajectories with S0 Z 0 and I0 Z 0. See
Figure 20. Thus, there is no epidemic, but the disease dies out.
6
I5
m
\ /<-
XX ‘_
I> S
(K30)
Figure 20. Nullclines for the epidemic model with BK/6 < 1
Next, assume that K/6 > 1. The xed point S = K and I = 0 now becomes
a saddle point because K — 6 > 0. The matrix of partial derivatives at (S", I") is
(-5? — u -55‘)
L‘! I" 0 '
This matrix has a trace of —,BI' - p,, which is negative, and has a determinant of
B2I"S" > 0. Thus, this xed point is asymptotically stable, and is either a. stable
focus or a. stable node. By considering the nullclines and. direction arrows, it follows
that any solution for initial conditions with So > 0 and I0 > 0 goes around the
xed point (S", I'). In fact, such a solution tends to the xed point (S“, I‘). Thus,
the epidemic reaches a steady state. See Figure 21
This system of equations is one of the models considered in [B1-a01] (Section
7.4). For more discussion of the interpretation and other models, see this reference.
158 4. Phase Portraits
a
it *3
‘i_
\ /<_\
(5‘»I')
N \
>T'>‘<1<.<>><
Figure 21. Nullclines for the epidemic model with BK/6 > 1
Exercises 4.7
Consider the SIR model for a population with vaccination given by
" ,_9= (1 —P)#K—5-5'I—#-5',
i=as1-11-an
R =p/.zK+'yI—uR.
Here the population R contains: (i) those who have recovered and are immune
and (ii) those who have been vaccinated and are immune. The parameter p,
which satis es 0 < p < 1, is the proportion of those born who are vaccinated
immediately and are transferred to the immune state. Also, /.¢ > 0, > 0,
and 7 > 0. The parameter it is the birth rate and death rate, Li is the rate of
infection, and" 7 is the rate of recovery from infected to recovered.
a. Show that $(.S'+ I + R) = p(K — S — I — R) and so decays to a. total
population S + I + R = K . Also, why can we consider the system of
differential equations for (S, I) rst and then nd R(t)'? The rest of the
problem considers only the system of differential equations for (S, I).
b. Find the two xed points for the system of di erential equations for (S, I),
(§,0) and (S‘, I‘) with I‘ 76 0. For what parameter values is I‘ > 0?
c. Find the matrix of partial derivatives.
d. What is the matrix of partial derivatives at the xed point (5',0)? For
what parameter values is this xed point asymptotically stable?
e. What is the matrix of partial derivatives at the xed point (S‘, I‘)? What
is the trace and determinant? For what parameter values is this xed point
asymptotically stable?
4.8. Theory and Proofs 159
2. For this epidemic model, let a: and y be the healthy and sick populations,
respectively. Assume that the system of differential equations is given by
01' = —k:a:y,
ll = My — ls.
where It and l are both positive.
a. Find and classify all the xed points.
b. Sketch the nullclines and the direction of the vector eld in each region
bounded by the nullclines.
c. Indicate the set of points i.u the rst quadrant at which the epidemic is
getting worse (i.e. where ‘lg/at > 0).
d. Find an expression for 4%’/as by taking the quotient of the equations. Then
separate variables and nd a function of :2 and y that is conserved by the
system. '
e. Sketch the phase portrait.
4.8. Theory and Proofs
Limit sets
Restatement of Theorem 4.1. Assume that ¢(t;x0) is a trajectory. Then, the
following properties of the w-limit set are true:
a. The limit set depends only on the trajectory and not on the particular point,
so w(xg) = w(¢>(t;x0)) for any real time t.
U
b. The w(xo) is invariant: if Z0 E w(xo), then the orbit ¢>(t;z0) is in UJ(XQ) for
all positive and negative t.
c. The w(x0) contains all its limit points (i.e., w(x0) is closed).
d. If yo is a point in w(x0), then w(y0) C w(xo).
In addition, assume that the trajectory ¢(t; xo) stays bounded fort Z 0 (i.e., there
is a constant C > 0 such that ||¢(t;x0)|| 5 C for t Z 0). Then, properties (e) and
(f) are true.
e. The w(x0) is nonempty.
f. The w(xo) is connected; it is not made up of more than one piece.
Similar properties hold for the Oz-limit set if ¢>(t; xo) stays bounded for t 3 0.
Proof. (a) Assume that zo G w(xq), so there exists a sequence of times tn going
to in nity such that ¢(t,,;xo) converges to zo. Then <;b(t,, — t;¢(t;x0)) = d>(t,,;xo)
converges to z, so z is in w(¢(t;x0)), w(x0) C w(<;b(t;xo)). Reversing the roles of xo
and ¢(t; xo) gives the other inclusion, proving (a).
(b) Assume that zo E w(x0), so there eidsts a sequence of times tn going to
in nity such that ¢(t,,; X0) converges to zo. By continuity of the ow, ¢(t+t,,; xo) =
¢(t;¢(t,,;x0)) converges to ¢>(t;z0). This proves ¢(t; zo) is in w(x0).
160 4. Phase Portraits
(c) Assume that zj is a sequence of points in w(xq) converging to a point zoo
in IR". In fact, we can take the sequence so that
“zj - Zwll < ‘/1-
Each of the zj is an w-limit point, so there exist sequences of times t_,-_,, such that
"¢(tj,n§x0) “ zill < 1/11-
(We have taken the times so that the points are within 1/n.) Then
"¢(tn.nix-0) _ zoo" 3 "¢(t1l-,n;x-0) _ Zn" + "Zn _ zoo"
11 2
§ — +*— = ~. .
Tl 77- Tl .
Since 2/1, is arbitrarily small, the points d>(t,,_,,;x0) converge to zw, and zw is in
the w-limit set of xo.
(d) Assu.rne that zo is a point in w(x0). Assume that y is an w-limit point of
zo, so there exist times tn going to in nity such that ¢(t,,, Z0) converges to y. By
(b), the points ¢(t,,;z0) are all contained in w(xo). Because the limit set is closed
by (c) and ¢(t,,; Z0) converges to y, the limit point y must be in w(x0). This shows
that any limit point of Z0 is contained in w(x0), or w(zg) C w(xo).
(e) The sequence of points ¢(n;xq) is bounded, so |d>(n;xo)| 3 C for some
C > 0. It follows from compactness of the set
{xG IR" : |]x|| 3 C}
that there must be ‘a subsequence ¢(nk;xo) which converges to a. point z in lR".
This point z must be an w-limit point of Z9. See a book on real analysis such as
[Lay01], [Lew93], or [Ma.r93] for discussion of compactness and convergence of a
bounded sequence in IR".
(f) Assume that w(x0) is not connected and there are two pieces. Then there
are two disjoint open sets U1 and U; such that w(x0) C U1UU2 and w(x0) U_,- 96 0
for j = 1,2. (The intuition is that U1 and U; are a positive distance apart, but
this is not necessarily the case. We just assume both are open so that a curve going
from one set to the other must pass tin'ough points outside of both sets.) Then, the
trajectory must come close to points in both U1 and U2 for arbitrarily large times,
so there must be times tn going to in nity such that ¢>(t,,; xo) is in neither U1 nor
U2. Since the trajectory is bounded, there must be a subsequence tn, such that
¢(t,,k; xo) converges to a point z that is in neither U1 nor U2. But this contradicts
the fact that w(x0) c U1 u U2. El
Stability of xed points
In this section we use the norm of a matrix introduced in Section 2.5 to prove
that, if all the eigenvalues of A have negative real parts, then the origin is attracting
at an exponential rate determined by the eigenvalues. We start with an example,
which illustrates the fact that the Euclidean norm is not always strictly decreasing.
4.8. Theory and Proofs 161
Example 4.29. Consider the system of linear differential equations with constant
coef cients given by
:b=—:1:—9y,
y=z—y
The eigenvalues are -1 :l: 3i, and the general solution is given by
(x(t)) = e_, ( cos(3t) —3sin(3t) (avg) _
y(t) § sin(3t) cos(3t) yo
The Euclidean norm is not strictly decreasing: its derivative is given by
gs’ + W” = gs” + ya-1/*(2w¢ + no
= (I2 + 1/2)"/’(—=v’ — 1/’ — 812/)-
Along the line y = —a:, the derivative is positive and the norm is increasing, while
along the line y = :1: the derivative is negative and the norm is decreasing. Therefore,
the Euclidean norm does not go to zero monotonically. However, the length of the
solution does go to zero exponentially:
x(t) _, cos(3t) —3sin(3t
(y(t)) 5° (§sin(s¢) cos(3t
33e-
Here, we have used the fact that the length of a matrix times a vector is less than or
equal to the norm of the matrix time the length of the vector. The norm of a matrix
is briefly discussed in Appendix A.3. The exact constant in the preceding inequality
is not very important. The entries of the matrix are periodic, so the norm has a
maximum. In fact, the maximum can be taken to be three. Thus, the Euclidean
norm goes to zero at a rate determined by the real parts of the eigenvalues -1 i3 i,
with a coefficient determined by the eigenvectors.
The next theorem gives the general result.
Theorem 4.10. Let A be a real n xn matrix, and consider the di erential equation
x = Ax. The following conditions are equivalent:
(i) The-re exist constants a > 0 and C‘ Z 1 such that, for an initial condition
xo in IR", the solution satis es
||¢A’Xo|l S Ce-°'||X<>||
for all t 2 0.
(ii) The real parts of all the eigenvalues of A are negative.
Proof. First, we show that (i) implies (ii). Suppose that (ii) is not true and one
of the eigenvalues /\ = a + i has a nonnegative real part a Z 0. (If the eigenvalue
is real, then )6 = 0.) The solution method shows there is a solution of the form
e°“(sin( t)u + cos( t)w). This shows
Since a Z 0, this solution does not go to zero, which contradicts
that if (ii) is false, then (i) is false, or if is true, then (ii) is true.
162 4. Phase Portraits
Next, we show the converse, that (ii) implies Let v1, ..., v" be a basis of
generalized eigenvectors, each of length one (unit vectors). We let A5 = Clj + i j
be the eigenvector associated with vj. Then, the solutions eA’vj involve terms
dependent on t, such as t"e°‘1‘cos(,6,-t) and t"e"-1‘sin( ,-t) for some k less than the
multiplicity of the eigenvalue for vj, so 0 3 k 3 n. Assume Re(/\) < —a < 0 for
all the eigenvalues /\ of A. Let c > 0 be such that R.e()\) < —a — e for all the
eigenvalues /\ of A. By the form of the solutions, there is C’; Z 1 such that
||e**vi|| s <1+»">e“1‘c;||vi||
s (1 +¢")e-~*~<=c;||v1||
3 e_“‘(t"e““)C';-||vj||
5 ¢_°iC1l|vj ll '.
where (t"e‘“) C_;- 3 Cj for all t Z 0. Any initial condition xo can be written as a
combination of the v-7', xo = EL, yjvj, so the solution is
1.
eAtxo -_ E y,-eArv-7'
i=1
and H
Letting C’ = max{C'j},
l|¢”"Xo|| 3 5°‘ Z ll:/J'lC.1' llv’ ll-
J'=1
,n _ n
||e*'x@|| s e-"W Z It/1| ~ uvin s e-"*0 Z lwl
i=1 i=1
n
3 e_“'C"n max{|y,-| :1 3 j 3 n} 3 e"°’C"n 2 |y_,-|2.
.'l=1
Let V be the matrix whose columns are the vectors v-7 and let y be the vector with
entries yj. Then,
xo = J-v-7 = Vy, U
<:M=y= ‘5Q "
TI.
uyu = Z |y.-|= s ||v"‘||-||=<@1|-
J'=1
Combining with the preceding inequality, we have
llemxoll 3 ¢_'“C'" ||V_1|| ' llxoll
= Cequllxollv
where C‘ = C"'n.||V“1|].
4.8. heoryand Proofs 163
One-dimensional differential equations
Restatement of Theorem 4.4. Consider a. di erentiol equation :i: = f (2) on
IR, for which f(a:) has a continuous derivative. Assume that x(t) = d>(t;z0) is the
solution, with initial condition :00. Assume that the maximum interval containing
0 for which it can be de ned is (t',t*').
a. Further assume that the solution ¢(t;a:o) is bounded for 0 3 t < t+ (i.e.,
there is a constant C > 0 such that |¢(t;:r0)| 3 C for 0 3 t < t*). Then, ¢(t;:1:0)
must converge either to a xed point or to a point where f(a:) is undefined as t
converges to t'*.
b. Similarly, if the solution ¢(t;a:Q) is bounded for t‘ < t 3 0, then ¢(t;a:o)
must converge either to a azed point or to a point where f(a:) is unde ned as t
converges to t‘.
c. Assume that f(a:) is de ned for all :1: in IR. (i) If f(z0) > 0, assume that
there is o azed point 2:‘ > 2:0, and in fact, let 2:‘ be the smallest xed point larger
than 0:0. (ii) If f'(a:o) < 0, assume that there is a. azed point as‘ < mo, and in fact,
let 0:‘ be the largest azed point less than $0. Then, t* = oo and ¢(t;a:0) converges
to a:' as t goes to in nity.
Proof. This theorem states that a bounded solution for a differential equation on
the real line must converge to either a xed point or to a point where _f(:|:) is
unde ned as t converges to either t+ or t‘.
We give a proof which treats both parts (a) and (c). The proof of part (b)
is similar. By the uniqueness of solutions, the solution cannot cross points where
f(a:) = 0. Thus, the solution ¢(t;:r0) must stay in a region where f (zv) has one
sign. Assume that f (¢(t;a:Q)) > 0. Tlius, ¢>(t;:cq) is increasing and is bounded
above. By properties of the real numbers, it must converge to a limiting value as t
converges to ti‘, the supremum of times for which the ow is de ned. (The value
ti" = oo if _f(a:) is de ned everywhere on the line.) Call this limiting value 2:00.
Because ¢(t;:c0) converges to zoo and f(d>(t;:z:o)) > 0, f(a:w) Z 0 or is unde ned.
If either f (mm) = 0 or f(:c°,,,) is unde ned, we are done. Otherwise, if f (mm) > 0,
then ¢(t;a;,,<,) goes from 2:- = ¢(—6;a:.,°) < zoo to :c+ = ¢(6;a:°°) > raw as t goes
from -6 to 6. But there is some time t1 such that :r_ < ¢(t1;:cg) < mm. Then,
mm < a:+ = ¢(26;:z:_) < ¢(26+t1;$0), so the solution starting at $0 goes past mm.
This contradicts the fact that f(z.,°) > 0 and we are done. El
A very similar proof shows the next lemma.
Lemma 4.11. Assume that, for allt Z 0, g(t) is de ned, g(t) is bounded, and
g(t) = %g(i) > 0.
a. Then, g(t) approaches o limiting value as t goes to in nity.
b. If g’ (t) is also uniformly continuous, then g’ (t) approaches 0 as t goes to
in nity.
Proof. (a) By the properties of the real numbers, the monotone values, g(t), must
8-pproach a limiting value goo as t goes to in nity.
164 4. Phase Portraits
(b) Since it is increasing, g(t) < goo for all t 3 0. The lim inf,_.oo g’ (t) 2 0
because g’(t) > 0 for all t. If lim sup,_,m g’ (t) = L > 0, then there is a sequence of
times to such that g'(t,,) Z L/2. By uniform continuity, there is a 6 > 0 such that
g’(t) 2 L/4 for tn 3 t 5 to + 6, so g(t,, + 6) Z g(t,,) + 6L/4. For sufficiently large
n, g(t,,) Z goo — 6L/8, so
y(tn + 6) 2 9(t,.) + 6L/4
Z goo — 6L/8+6L/4
= goo + 6 L/8
> goo.
This contradicts the fact that < goo for all t Z 0. El
Restatement of Theorem 4.5. Consider a fixed point 1:‘ for the dijferential
equation :i: = f (:2) on R, where f and f’ are continuous.
a. If f’ (1') < 0, then a:‘ is an attracting xed point. '
b. If f’(a:’) > 0, then :c" is a repelling xed point.
c. If f’ (:i:") = 0, then the derivative does not determine the stability type.
Proof. (a) If f’(:|:") < 0, then there is an interval about :i:‘, (a:' — 6, 1:‘ + 6), such
that f'(a:) < 0 for all :1: in this interval. Then, for any :1: with :2‘ — 6 < 1: < 2:‘ + 6,
f(:i:) = _f(a:') + f’(zo):z: = f’(zo)a: for some zo, between a:' and 2:. Therefore,
f(a:)>0for:v‘-6<:c<a:" and f(:c)>0for:r:"' <:r:<a:'+6.
Now take an initial condition :00 with :1:"‘ — 6 < zro < 2:‘. The solution ¢(t; $0)
is increasing because > 0. By the uniqueness given in Section 3.1, the solution
cannot cross the point Therefore, the solution must stay in the interval (:r‘ —
6, 1:‘), where f’ (:2) > 0. The solution must be increasing and must converge to
:2‘. This shows that 2‘ is attracting from the left. A similar argument shows it is
attracting from the right and so it is an attracting xed point.
(b) The proof for _f'(a:‘) > 0 is similar" to the preceding proof and is left to the
reader.
(c) The examples of Section 4.3 show that, if f’ (:r') =' 0, then the xed point
can be attracting, repelling, or attracting from one side or can have other possibil-
ities. Therefore, the derivative does not determine the stability type. U
Linearized stability of xed points
In this subsection we prove that a xed point whose eigenvalues all have nega-
tive real parts is asymptotically stable.
Restatement of Theorem 4.6a. Let x‘ be a xed point for 2': = F(x), where F
is C2. (The theorem is actually true for F is C‘ but the proof uses C2.) Assume
all the eigenvalues of DF(,o-) have negative real part. Then x‘ is asymptotically
stable.
Proof. We consider a. Taylor expansion of F about x‘. Let A = DF(,,-) and
y = x - x‘, the displacement away from the xed point. Since F(x') = 0,
Y = Ay + s(y).
4.8. Theory and Proofs 165
where ||g(y)|| 3 M||y||2 for ||y|| 3 60. Along a solution y(t), we can think of
G0) = s(Y(¢))
as known. So we get the time-dependent linear equation,
Y = Ay + GU).
which has a solution
E
y(t) = eA‘y0 + / eA(‘_’) G(s) ds.
0
For the matrix em there is a scalar K Z 1 such that ||eA‘|| 3 Ke'°“, so
|l¢‘“Yl| S |l¢A‘l| - llyll S K¢_“‘l|y|l-
If ||y|| is sufficiently small, then M||y]|2 3 m||y||, where m is small. So,
|y<¢>1| s Kc-"*||y.|| + /0E Kc-"<‘-"> Mlly(8)|l’ds
s Ke-"'||y.|| + /t K ~‘=‘°“">m||y<s)|l ds.
o
and multiplying by e°‘, we obtain
e"*r|y<¢>|| s K ||y.|| + /02 m1<e"*||y(@)||d==-
Considering ¢>(t) = e°"||y(t)||, this is like a di erential inequality ¢'(t) 3 mK¢(t),
with ¢(0) = K|lyQ||, so it is reasonable that ¢(t) 3 ¢(0)e’"K‘ = K||y0||e"‘K‘.
Indeed, by Gronwall’s inequality, this is true, so
e"*||y<¢>|| s K||y.||e"“‘ and
||y<t>|| s I<||y.||e<'~“-">*.
If ||yo|| 3 60/ K is sufficiently small and m is sufficiently small so that mK — a < 0,
then
||y<¢>|| s K|ly.||e"""-">' s K 1|:/on s 6..
and ||y(t)|| goes to zero exponentially. Also, it is clearly L-stable, because for any
e > O, if
llyoll S 5 = miI1{¢/K60/K}. El
then the solution stays less than both e and 60 for all t 2 0.
We have said that a saddle point has an unstable manifold; we have not proved
this since it is not easy to do. See [Rob99] for a proof. We can prove that a saddle
point is unstable, which we do in the next theorem.
Theorem 4.12. Assume that x" is a saddle xed point for a system of di erential
equations in the plane. Then, x‘ is unstable.
Remark 4.30. A similar result is true for an unstable xed point. We leave this
result as an exercise. Also, the result is true in any dimension. We restrict to two
variables just to make the notation and details simpler.
166 4. Phase Portraits
Proof. By changing the variable from x to u = x—x", we can assume that the xed
point is at the origin. A saddle xed point in two dimensions has real eigenvalues.
We can also use the coordinates from the eigenvectors to put the equations in the
following form:
:i: = aa: + R(:r,y),
Q = _by + S($1y)I
where a, b > 0. Let m = min{a., b}. We also assume that R(z,y) and S(:r, y) have
zero partial derivatives at (0,0): they contain quadratic terms and higher, so we
can nd constants C > 0 and 6 > 0 such that
" |R(1'. 1/)l + I-5'(1-1. 1/)| 3 C(-‘P2 + 142)
for ||(-"v.1/)|| S 5- "
To show that trajectories move away from the xed point, we use the test
function
L(a:,y) =
If LCM!) 2 0 (1-6-» It/I S IIBI) and ||(I»y)l| S 6, We have
L = 2:5: - yg
= w (M1 + R(I.1/)) - (—by + 5(I»i/))
= aa:2 + by2 +a:R(a:,y) — yS(a:,y)
§m(1=2 +1/2) - l$| lR(==»y)| - I1/I I-5'(M/)|
2 T" (1-'2 + I/2) — III (lR(I,y)l + |$(1‘»1/)1)
zmw+m-m@w+m
= (m - C III) (1-'2 + 112)
z§w+m
>0
for |:|:| 3 m/(ZC). Since this derivative is positive, a trajectory ¢(t; (20, 1/0)) stays
in the region where L(a:,y) Z 0 until either |a:(t)| = m/(2C) or |]¢(t; (:z:0,y0))|] = 6.
Thus, we can nd initial conditions arbitrarily near the xed point which move to
a distance away of at least min{6,m/(2C)}. This proves that the xed point is
unstable. Cl
Conjugacy of phase portraits
To a large extent, the linearized equations at a xed point determine the fea-
tures of the phase portrait near the xed point. In particular, two di erential
equations are conjugate provided there is a change of coordinates which takes the
trajectories of one system into the trajectories of the other, preserving time. Some-
times the change of coordinates preserves the sense of direction along the trajec-
tories, but not the time: Two systems related in this way are called topologically
equivalent. In this section, we make these ideas more precise; in particular, we need
to be more speci c about how many derivatives the change of coordinates has. The
4.8. Theory and Proofs 167
results in this section have proofs beyond the scope of this text. We merely state
the theorems and provide references for the proofs.
De nition 4.31. For an integer 'r Z 1, a system of nonlinear diiferential equations
x = F(x) is said to be C’ provided F is continuous and all partial derivatives up
to order 1' exist and are continuous. If partial derivatives of all orders exist (i.e., F
is C’ for all r), then it is called C°°.
De nition 4.32. Let U and V be two open sets in IR". A homeomorphism h from
U to V is a continuous map onto V, which has a continuous inverse k from V to
U (i.e., k o h is the identity on U and k o h is the identity on V). It follows that
h is one to one (i.e., if h(x1) = h(x-_;), then x1 = x2).
For an integer 1" 2 1, a C’ diffeomorphism h from U to V is a homeomorphism
from U to V such that h and its inverse k each have all partial derivatives up to
order r and they are continuous. It follows that det(Dh(,,)) 76 0 at all points x.
A homeomorphism (respectively, di eomorphism) can be considered a contin-
uous (respectively, differentiable) change of coordinates.
De nition 4.33. Two ows d>(t; -) and 1,/1(t;-) are called topologically conjugate on
open sets U and V provided there is a homeomorphism h from U to V such that
1'1 ° ¢(i; X0) = ‘PU; h(Xo))-
Such a map h is a continuous change of coordinates.
These two ows are called topologically equivalent provided that there is a home-
omorphism h from U to V which takes the trajectories of ¢(l; -) into the trajectories
of 1/1(t; -), preserving the sense of direction of time, but not necwarily the exact
same time. ~
For an integer r Z 1, these ows are called C’ conjugate on open sets U and
V provided there is a C’ diffeomorphism h from U to V such that
1'1 ° ¢(l5;X0) = Wt; h(X0))-
Such a map h is a differentiable change of coordinates.
These two ows are called C’ equivalent on open sets U and V provided that
there is a C’ diffeomorphism h from U to V, which takes the trajectories of ¢(t; -)
into the trajectories of 1/)(t; ~), preserving the sense of direction of time.
As we state shortly, there is a general theorem called the Grobma.n—Hartman
theorem that states that a nonlinear system with a hyperbolic xed point is topo-
logically conjugate to its linearized ow. Unfortunately, this topological conjugacy
does not tell us much about the features of the phase portrait. Any two linear
systems in the same dimension for which the origin is asymptotically stable are
topologically conjugate on all of IR". See [Rob99]. In particular, a linear system
for which the origin is a stable focus is topologically conjugate to one for which the
origin is a stable node. Thus, a continuous change of coordinates does 11ot preserve
the property that trajectories spiral in toward a xed point or do not spiral. A
differentiable change of coordinates does preserve such features. Unfortunately, the
theorems that imply the existence of a di erentiable change of coordinates require
more assumptions.
168 4. Phase Portraits
We now state these results.
Theorem 4.13 (Grobman—Hartman). Let x = F(x) be a C1 nonlinear system
of di erential equations that has a hyperbolic xed point x‘ and ow ¢(t; -). Let
A = DF(,,.) be the matrix of partial derivatives, and 1,/»(t;xo) = eA‘x0 the ow of
the linearized system of equations, with xed point at the origin. Then, there are
open sets U containing x‘ and V containing the origin, such that the ow d>(t; -)
on U is topologically conjugate to the ow itv(t; -) on V.
See [Rob99] for a proof, since the proof of this theorem is more involved than
we want to pursue.
-- To get a differentiable change of coordinates, we need to treat special cases or
add more assumptions. A result with more derivatives was shown by Sternberg,
but requires a “nonresonance” requirement on the eigenvalues.
Theorem 4.14 (Stemberg). Let x = F(x) be a C°° system of di erential equations
with a hyperbolic ared point x‘ and ow ¢(t; -). Let ,\1, . . . , /\,, be the eigenvalues at
the xed point and assume that each /\_,- has algebraic multiplicity one. Moreover,
assume that
ml/\1+m2/\2+"'+mn/\n 94 /\k,
for any k and any nonnegative integers m_,- with 21- m_,- Z 2. (This condition is
called nonresonance of the eigenvalues.) Then, there exist open sets U containing
x‘ and V containing the origin, such that the ow ¢(t; -) on U is C°° conjugate to
the ow for the linearized system on V.
The proof of this» theorem is much harder than the one for the Grobman—
Hartman theorem. *-See [I-Iar82] for a proof. It also requires the nonresonance of
the eigenvalues. The following two theorems give results about a C1 conjugacy
without the nonresonance of the eigenvalues.
Theorem 4.15 (Hartman). Let x = F(x) be a C2 nonlinear system of differential
equations with an asymptotically stable red point x" and flow ¢(t; -). Then, there
exist open sets U containing x" and V containing the origin, such that the ow
¢(t; -) on U is C1 conjugate to the ow for the linearized system on V.
Theorem 4.16 (Belickii). Let :i: = F(x) be a C2 system of di erential equations
with a hyperbolic led point x‘ and ow ¢(t; ~). Assume that the eigenvalues satisfy
a nonresonance assumption that says that /\;, 95 /\¢ + /\_,- for any three eigenvalues.
Then, there ea-ist open sets U containing x‘ and V containing the origin, such that
the ow ¢(t; ~) on U is C1 conjugate to the ow for the linearized system an V.
See [Be172].
i
Chapter 5
Phase Portraits Using Scalar
Functions
This chapter continues our consideration of phase portraits for systems of nonlinear
differential equations. In the previous chapter, the systems considered could be
analyzed using nullclines and the linearization at the xed points. Those considered
in this chapter are analyzed using a real-valued function, an “energy function”. The
rst section treats the predator—prey model for two populations; this continues our
treatment of Lotka—Volterra equations, but for the casein which one population has
a positive effect on the growth of the other. In Section 5.2, we consider nonlinear
oscillators such as the pendulum, for which we can use energy to analyze the phase
plane. When friction is added to these non.linear oscillators, the energy decreases.
The energy in these situations is an example of what is called a Lyapunov function.
The Lyapunov function can be used in certain classes of problems to determine the
set of initial conditions which have trajectories tending to a xed point. Section
5.4 uses a real-valued function (test fimction) to determine the limit sets of some
more complicated examples. These examples are used in the next few chapters to
illustrate concepts which are introduced. Finally, for a system given as the gradient
of a function G, the function G‘ is strictly decreasing along trajectories not at a xed
point, so any trajectory that is bounded tends to a xed point.
5.1. Predator—Prey Systems
In Section 4.6, we introduced the Lotka—Volterra equations, but looked mainly at
the case of two competitive populations. In this section, we consider two popula-
tions for which the rst population has a positive effect on the second population,
but the second population has a negative effect on the rst. This situation is called
a predator—prey system. The simplest system of equations of this type is given by
=i1=1Y(a—lYy),
32:1/(_c+ea:)v
_
169
170 5. Phase Portraits Using Scalar Functions
with all parameters positive, 0., b, c,e > 0. Notice that :2: is the prey population
and y is the predator population. We have not included any limit to growth or
any possible negative effect either population has on its own growth rate. Without
any of the predators around, the prey grows at a linear positive rate of a. As the
predator population increases, the growth rate of :1: decreases linearly in y. Without
any prey, the predator population has a negative growth rate of —c. As the prey
population increases, the growth rate of y increases linearly in :r:.
These equations have xed points at (0,0) and (C/¢,°'/b). Analyzing the signs
of i‘ and 3,‘; in the regions separated by the nullclines leads to the possibility of
cycles, or periodic orbits. The matrix of partial derivatives at (°/e, “/5) is
-- bc
(G _ by _b1 ) ’ r: 0 _.; _
0
ey ea: _ C :=c/e,y=a/b ' Eb
This matrix has characteristic equation /\2 + ac = O, and the eigenvalues ii,/ac are
purely imaginary. Thus, (°/¢, °/5) is an elliptic center.
For this system, there is a way to show that the orbits close when they go once
around the xed point. Notice that
Q : y(—c+ err)
:i: :c(a.—- by) ’
so we can separate variables
, (y ) ( . )-a— — b y = —-—c + e
This equality can be integrated with respect to t to yield
K+a.ln(y)—by=—cln(:r)+e:c or
K = (ea: — c ln(:|:)) + (by— a. ln(y)),
where K is an arbitrary constant. Thus, K is a conserved quantity that can be
considered as_a function of :1: and y on the phase plane. This function is the sum
of a function of :z: and a fimction of y,
f(:r:) = em —- c ln(:r:) and g(y) = by — a ln(y).
To understand the function f, note that f”(a:) = $—c2 > 0.
f'(a:) = e — 5c and
The graph of f is concave up and has a unique minimum at ‘/6. See Figure 1.
Similarly, the graph of g(y) is concave up and has a unique minimum at “/5. When
a trajectory starting on the li.ne y = “/1, goes around the xed point and returns
to this line at time ti, the value of f(:c) must be the same as when it left, so
:c(t1) = 12(0) and y(t1) = y(0) = “/11. Therefore, all these solutions are periodic,
and the xed point is not only a linear center, but is also a nonlinear center that
is surrounded by periodic orbits. '
Since the function f (2) + g(y) is a conserved quantity along trajectories, each
level set of this function corresponds to a trajectory of the system of diiferential
equations. Thus, to analyze the phase portrait more carefully and more completely,
5.1. Predator—Prey Systems 171
f (iv)
0 F F‘ I
2 e
6
Figure'1. Graph of f(:c)
‘Q
0.
B —-— - ,— _ - - - - -- -
1 :1:
D 7' 2
3
Figure 2. Predator—prey system
we need to analyze the level curves in the rst quadrant of the function f(:1:)+g(y).
The minimum value is K0 = f(°/3) + g(°'/5). The level curve f (22) + g(y) = K0 is
just the xed point (C/e, '1/4,). If K < K0, then level curve f (:i:) +g(y) = K is empty.
Finally, we consider K > K0 and let K’ = K — K0 > 0. Therefore, for a given :r,
we need to nd the y that satisfy
sh/) — 9(°/bl = K’ — [f($) — f(°/¢)]-
For :1: with f (as) — f(c/a) > K’ , the right-hand side is negative and there are no
solutions for y. There are two values of av, :21 and $2, for which f(x) — f(°/e) = K’ .
For these :0,-, the right-hand side is zero and the only solution is y = “/5. For
$1 < 1: < a:2, the right-hand side is positive and there are two values of y which
satisfy the equations. One of these values of y will be less than “/1, and the other
will be greater. Therefore, the level curve for K > f(°/e) + g(“/b) is a Closed Curve
that surrounds the xed point. Compare with the trajectories drawn in Figure 2.
5. Phase Portraits Using Scalar Functions
Exercises 5.1
1 Consider the predator—prey system
i=@(1—y).
y=y (—1+:r:§).
a. Find the nullclines and xed point with as‘ > 0 and :0‘ > 0. r.
b. Show that there is a fU.I1C'Cl0I'l L(a:, y) that is constant along trajectories
and has a minimum at (z",y*). Conclude that all trajectories in the first
quadrant other than the xed p_oint are on closed curves that surround the
xed point. ' -
Consider the SIR epidemic model with no births or deaths given by
S = — SI,
I = SI —— '11,
for S Z 0 and I Z 0. (In terms of the equations given in Section 4.7, we are
taking p. = 0 and a = 0, because there are no deaths or fatalities, and ;iK = 0
because there are no births.) We do not include the equation R = 'yI because
it can be solved once the solution I (t) is known.
a. What are the nullclines and xed points’?
b. Show that there is a real-valued function G‘(S, I) that is constant along
trajectories. II-Iint: Consider I/3.
c. Draw the phase portrait for s 3 0 and I Z 0 using the flmction c:. Hint:
Write a level curve of G in the form of I as a function of S.
Consider the predator—prey system with terms added corresponding to limited
growth,
:i: =:c(a—by—ma:),
1'1 =y(—c+e:|:~'n.y),
where all the constants are positive, a > 0, b > 0, c > 0, e > 0, m > 0, and
n > 0. Also assume that °/m > °/e.
a. Show that there is a xed poi11t (:r:',y') with .'c" > 0 and y‘ > 0.
b. Show that the xed point (a:‘,y") is asymptotically stable.
c. Draw a sketch of the phase portrait for 2: Z 0 and y Z O. (You may use
the fact that there are no periodic orbits, which is true but not trivial to
show. See Example 6.24.)
Consider the system of differential equations given by
i $1/2|
gr] = —4ya:2.
Show that there is a real-valued function G(:c, y) that is constant along trajec-
tories. Hint: Consider F/,3.
5.2. Undamped Forces 173
5.2. Undamped Forces
In this section, we consider a particle of mass m moving with position given by only
one scalar variable, :12. To describe its complete state, we need to also specify its
velocity, zb. We assume the motion is determined by forces F(a:) that depend only on
the position and not on the velocity (i.e., there is no damping or friction). Since the
mass times the acceleration equals the forces acting on the particle, the differential
equation which determines the motion is given by ml} = F (a:). This equation can
be written as a system of differential equations with only rst derivatives by setting
1/ = 531
i = 1/,
1'/ = —m1—F(Iv) .
Such a system is said to have one ‘degree of freedom, even though the system of
equations has the two variables, position and velocity.
We want to solve for a quantity (energy) that is conserved along trajectories by
“integrating” the differential equation. We can multiply both sides of the equation
mzii = F(x) by :i:, obtaining
m:i§:i:= F'(a:):i:,
mitii: - F'(a:)a': = 0,
i lm'L'2—/:F(s)ds =0 or
dt 2 ' Io '
I
1m:i:2—/ F(s)ds=E,
2 am
where E is a. constant along trajectories. The potential energy of the force eld is
de ned to be the integral of the negative of the force eld,
v(¢) = - f:F(s)ds.
Therefore,
E(a:,a.:) = 51 mz.z + V(a:)
is a constant along solutions. The quantity E(a:,:i:) is called the energy or total
energy, and is the sum of the potential energy V(:r) and the kinetic energy %m:i:2.
Because there is a function that is constant along trajectories, these systems are
often called conservative systems. We usually consider E a function of the phase
plane variables :2 and y.
In physics, the potential energy is interpreted as the work (or energy) needed
to move the particle from the position 0:0 to :2. The term %my2 is the kinetic
energy determined by the velocity of the particle. For simplicity, in what follows,
we usually take the mass coefficient to be equal to one, or incorporate it into F.
An equation of the form mi = F(x) is often called an oscillator because,
frequently, the resulting motion is periodic with the particle going back and forth.
See Section 5.6.1 for further discussion of the connection with oscillators.
174 5. Phase Portraits Using Scalar Functions
For a real-valued function L(x), we often use the following notation for the
level set for which L has the value C:
L'1(C)={x:L(x)=C}.
If I is any set of real values (e.g., an interval), then we also denote the set of points
talcing values in I by
L-1(1) = {X = L(x) e I}.
Example 5.1 (Double-Well Potential). The rst example we consider is called
the Duf ng equation, whose potential has two minima. Consider the system
i = 2/.
1) = 3: — :23. I
The xed points are zc = i1, 0 for y = 0. The matrix of partial derivatives is
O1 \
whose evaluation at the fixed points equals
O1 and O 1
DF(g,g) = (1 0) DF(:H‘0) = (_2 0) .
Therefore, the xed point at (0,0) is a saddle and the xed points at (:l:1,0) are
centers.
The potential energy is
_ =1: _ 3 =_a:_2 L4
V(a:)_[0 s+s ds 2+4,
which has a local maximum at as = 0 and rninima at :1: = i1. Since the potential
function has two minima, it is called a double-well potential. The values of the
potential energy at the critical points are V(0) = 0 and V(i1) = '1/4. See the
graph of V in Figure 3. The total energy is
:23 :24 1 2
E($>y)__?+ ?+5y .
We consider values of the energy C2 > V(0) = 0, C1 with V(:t1) = "1/4 <
01 < 0 = v(0), vo = -1/4, and nally v1 = 0 = v(0). See Figure 3.
We take C2 = 2 > 0 so the points come out easily; therefore, we consider the
level set
E"*<2> = {<w.y>=Ea.y> = %2 +V(rv) =2}.
Since P2/2 Z 0, the only allowable as have V(:z) 3 2. Solving for the J: with
V(a:) = 2, we have
24
2=_?_"+ i,
24
O=a:'l -21-’ +8,
0 = (:02 -4)(a:2 +1),
5.2. Undamped Forces 175
2 lV(=v)
0. -1 2
1
_7
64
Figure 3. Double-well potential function V(a:) for Example 5.1 (not drawn to scale)
l\s
\
<\\\\t\\a{\\“-\r\\s\A\l-\\:t\L.l\~\?I\le#-‘-alI.#1I\'l\'I|\:1\:II:\:\I5I:%:I.:;:I,.,,9'I/I¢0I’),If}!///Z///»/7;'/1/l7l \’/
\\§-":-.-.='-'=:-..=.=_=~I~_‘\"is.t=‘¢,17/,}}///
“‘-?‘—"‘-‘:”1'*?""““““~e‘\\.‘—\'¢,’0//
Flgure 4. Level curve E‘ ' (2) on three-dimensional graph of energy function
for Example 5.1
and :1: must be :t2. For |:z:| > 2, V(a:) > 2 and there are no solutions of E(:c,y) =
2. For a: = :l:2, V(:l;2) = 2, and the only allowable value of y must solve 1/2/2 = 2—
V(:|:2) = O, so y must be zero. For —-2 < :1: < 2, V(a:) < 2, so there are two values
of y for which E(:c,y) = 2, one the negative of the other, y = :h[4 — 2 V(:c)]‘/2.
Therefore, the level set E"(2) = {(a:,y) : E(:n,y) = 2} is a closed curve around
the three xed points and it extends from :1: = —2 to :1: = 2. See Figure 4 for a plot
of the level curve on the three-dimensional graph of the energy function E(a:, y).
See Figure 5 for the level curve in the (a:,y)-plane. A similar analysis of E'1(C2)
holds for any C2 > 0 = V(0) with just a change of the endpoints in the 2 values.
Next, for a value of C1 with V(:t1) = "1/4 < C1 < 0 = V(O), we take
C1 = *7/64 = V (;l;1/2) = V <i\/7/2). Again, we need V(:r) 5 "7/64, so
__7 <_$_2 +_$4
s4‘ 2 4’
0Za:4—2:1:2+l.
16
176 5. Phase Portraits Using Scalar Functions
1/
I
-2
Figure 5. Level curve E“l(2) for Exa.mple_5.1
The right-hand side is zero when
2,: 2i\/4-i =4=t,/TsT7
=4i:3r2=7z”dr14
Therefore, the allowable an are ‘*/7/2 5 :1: 5 '1/2 or 1/2 5 a: 5 ‘/7/2. For the range
"\/7/2 5 :1: 5 '1/2, E“ ('7/64) is a closed curve around (-1, 0), and for the range
1/2 5 a: 5 ‘/7/2, E“ (J/sa) is a closed curve around (1, 0). See Figure 6 for the level
curve E" ('7/64) in the (:i:,y)-plane. Similarly, for any '1/4 < C1 < 0, E'1(C1)
is the union of two closed curves, one around (-1, 0) and the other aroiuid (1, 0).
0.4 y :1:
-0.
Figure 6. Level curve E'1 (—%) for Example 5.1
The level set E‘1 (-1/4) is just the two points (—1,0) and (1,0).
Finally, we consider the level E_1(O). The level set has values of :1; with
V(a:) 5 0,
$4 $2
02?‘?
0 2 :04 — 2:22 = a:2(a:2 — 2).
Since 11:2 Z 0, we need 0:2 5 2, or —\/2 5 a: 5 For 2: = O,i\/2', the only
value of y is 0. For —\/2 < an < 0 or 0 < a: < \/2, there are two values of y.
Thus, the level set is a “ gure eight”, which crosses nontangentially at :1: = 0. See
5.2. Undamped Forces 177
Figure 7. For a point pg = (:c0,y0) on E-1(0) with 2:0 95 0, the trajectory ¢(t; pg)
approaches (0,0) as t goes to both :l:0O. Such an orbit, which is asymptotic to
the same xed point as t goes to both :|:oo, is called a homoclinic orbit. Homoclinic
roughly means self-inclined; the orbits are called this because they are inclined to
the same point both forward and backward in time.
9
it
Figure 7. Phase plane for Example 5.1
Thus, the two xed points (:t1,0) are surrounded by periodic orbits. There are
also periodic orbits that surround all three xed points. The two types of periodic
orbits are separated by the level set E',1(0) which is made up of two homoclinic
orbits.
This example indicate the steps needed to analyze a system without damping
forces. See Table 1 on the next page.
Remark 5.2. Computer programs, such as Maple, Mathematica, or Matlab, can
also help with the preceding analysis by using a command to plot the contour curves
of the real-valued function E.
Example 5.3 (Pendulum). Consider the system of equations for a pendulum
i = y,
1) = — sin(:r:).
The potential function is
I
V(a:) =/0‘ sin(s) ds = 1 — cos(a:).
The potential function has a minimum at a: = 0 and multiples of 21r, V(0) =
0 = V(n 21r). It has maxima at :1: = :h‘n' and other odd integer multiples of 1r,
V(1r) = V(1r + n21r) = 2. See Figure 8. A direct check shows that the xed
points (n2'rr,0) are centers, and (1r + n21r,0) are saddles. Therefore, we need to
understand the level sets of the energy for C1 with 0 < C1 < 2, for 2, and for Cg
Wltll 2 < C2.
178 5. Phase Portraits Using Scalar Flinctions
Steps to analyze a system with undamped forces
1. Plot the potential function v.
a. Determine the local maxima and local minima of V.
b. List of all the critical values V0 < V1 < < V1. that are
the values attained by V(a:,-) at critical points :0,-, where
V’ (Tj) = O. (These include the local maximum, local mini-
mum, and possibly in ection points.)
c. Sketch the graph of V.
2. Take values C_,- with C0 < V0, V_,-_1 < Cj < for l 5j 5 lc— 1,
- and C;¢+1 > Vk.
a. For each of these values, determine the allowable range of
as-values, {zc : V(:r:) 5 Cj}, using the graph of V.
b. Using the range of allowable a:-values for C,-, determine the
level set E_1(C,-) in the phase space. Note that if a compo-
nent of {:z: : V(a:) 5 C,-} is a closed interval for one of these
values of the energy which is not a critical value, then the
corresponding component of E'1(C,-) is a periodic orbit.
3. Repeat the last step for the level sets E‘1(VJ-), 1 5j 5 k.
a. If V0 is an absolute minimum, then the level set E'1(Vo) is
just the xed points with this energy.
b. Some of these values correspond to local maxima of V, and
so the level set E‘1(V,-) contains the stable and unstable
manifolds of the saddle xed point.
(i) lf a component of {a: : V(a:) 5 is a closed inter-
val that does not contain any other critical points of
V, then the corresponding component of the energy
level set E'1(V_.;) closes back up on itself, resulting in
homoclinic orbits. This is the same as case 2b.
(ii) If the potential function has the same value at two
local maxima 2:1 and 1:2 of V, V(a:1) = V(:c2) = V,-,
and if < Vj for 1:1 < as < :22, then the energy
level set E‘1(V,-) connects the two saddle xed points
p1 = (:c1,0) and pg = (a:2,0) with one curve with"
positive y and one with negative y. Thus, there are
two orbits each of whose w-limit set is one xed point
and whose a-limit set is the other xed point: such
orbits are called heteroclinic orbits. See Example 5.3
and Figure 9.
Table 1
For a constant 0 < C1 < 2, the level set E_1(C1) in —1r < 2: < 1r is a level curve
surrounding (0,0). This repeats itself in every strip —1r + n21r < a: < 1r + 11.21!-
Thus (0, 0) is surrounded by periodic orbits.
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An Introduction to Dynamical Systems - 2nd Edition
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