Phase Plane for Linear Control System and Design

7/25/2019 Phase Plane for Linear Control System and Design

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Ch 9.1: The Phase Plane: Linear Systems

There are many differential equations, especially nonlinear

ones, that are not susceptible to analytical solution in anyreasonably convenient manner.

Numerical methods provide one means of dealing with theseequations.

Another approach, presented in this chapter, is geometrical incharacter and leads to a qualitative understanding of thesolutions rather than to detailed quantitative information.


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Solutions of Second Order Linear Systems

Consider a second order linear homogeneous system with

constant coefficients of the form x'= Ax, where A is a 2 x 2constant matrix and x is a 2 x 1 vector.

Recall from Chapter 7 that if we assume x = ert, then

Therefore x = ertis a solution of x' = Axprovided that ris aneigenvalue and is an eigenvector of the coefficient matrix A.

The eigenvalues are the roots of the polynomial equationdet(A-rI) = 0, and the eigenvectors are determined up to anarbitrary constant from the equation (A-rI) = 0.

( ) 0IAAA === rreer rtrt


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Equilibrium Solution, Phase Portrait

Solutions x for which Ax = 0 correspond to equilibrium

solutions, and are called critical points.We assume A is nonsingular, or detA 0, and hence x = 0 isthe only critical point for the system x'= Ax.

A solution of x'= Ax is a vector function x = (t) that satisfiesthe differential equation, and can be viewed as a parametricrepresentation for a curve in thex1x2-plane.

This curve can be regarded as a trajectory traversed by a

moving particle whose velocity dx/dtis specified by thedifferential equation.

Thex1x2-plane is called the phase plane, and a representative

set of trajectories is a phase portrait.


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Characterizing Equation by Trajectory Pattern

In analyzing the system x'= Ax, we must consider several

cases, depending on the nature of the eigenvalues of A.These cases also occurred in Sections 7.5 7.8, where wewere primarily interested in finding a convenient formula forthe general solution.

Now our main goal is to characterize the differential equationaccording to the geometric pattern formed by its trajectories.

In each case we discuss the behavior of the trajectories in

general an illustrate it with an example.

It is important to become familiar with the types of behaviorthat the trajectories have for each case, as they are the basic

ingredients of the qualitative theory of differential equations.


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Case 1: Real Unequal Eigenvaluesof the Same Sign (1 of 3)

When the eigenvalues r1 and r2 are both positive or both

negative, the general solution forx'= Ax is

Suppose first that r1 < r2 < 0, and that the eigenvectors(1)

and(2)

are as shown below.It follows that x 0 as t for all solutions x, regardlessof the values of c1 and c2.

trtrecec 21

)2(2

)1(1 x +=


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Case 1: Nodal Sink (2 of 3)

If the solution starts at an initial point on the line through(1)

, then c2 = 0 and the solution remains on this line for all t.Similarly if the initial point is on the line through (2).

The solution can be rewritten as

Since r1 -r2 < 0, for c2 0 the term c1(1)e(r1 -r2)t is negligible

compared to c2(2), for large t.

Thus all solutions are tangent to (2) at the

critical point x = 0 except for solutions

that start exactly on the line through (1).

This type of critical point is called a node

or nodal sink.

( ) )2(

2

)1(

1

)2(

2

)1(

1

21221

x ceceecec trrtrtrtr +=+=


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Case 1: Nodal Source (3 of 3)

The phase portrait along with several graphs ofx1 versus t

are given below. The behavior ofx2 versus tis similar.If 0 < r2 < r1, then the trajectories will have the same patternas in figure (a) below, but the direction will be away fromthe critical point at the origin. In this case the critical pointis again called a node or a nodal source.


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Case 2: Real Eigenvaluesof Opposite Sign (1 of 3)

Suppose now that r1 > 0 and r2 < 0, with general solution

and corresponding eigenvectors (1) and (2) as shown below.

If the solution starts at an initial point on the line through (1),

then c2 = 0 and the solution remains on this line for all t.Also, since r1 > 0, it follows that ||x|| as t .

Similarly if the initial point is on the line through (2), then||x|| 0 as t since r

2

< 0.

Solutions starting at other initial points

have trajectories as shown.

,21 )2(2)1(1 trtr ecec x +=


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Case 2: Saddle Point (2 of 3)

For our general solution

the positive exponential term is dominant for large t, so allsolutions approach infinity asymptotic to the line determined

by the eigenvector(1)

corresponding to r1 > 0.The only solutions that approach the critical point at the originare those that start on the line determined by (2).

This type of critical point is called a saddle point.

,0,0, 21)2(2)1(1 21 += rrecec trtr x


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Case 2: Graphs ofx1 versus t (3 of 3)

The phase portrait along with several graphs ofx1 versus t

are given below.For certain initial conditions, the positive exponential term isabsent from the solution, so x1 0 as t .

For all other initial conditions the positive exponential termeventually dominates and causesx1 to become unbounded.

The behavior ofx2 versus tis similar.


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Case 3: Equal Eigenvalues (1 of 5)

Suppose now that r1 = r2 = r. We consider the case in which

the repeated eigenvalue ris negative. If r is positive, then thetrajectories are similar but direction of motion is reversed.

There are two subcases, depending on whether rhas twolinearly independent eigenvectors or only one.

If the two eigenvectors (1) and (2) are linearly independent,then the general solution is

The ratiox2/x1 is independent of t, but depends on thecomponents of (1) and (2) and onc1 and c2.

A phase portrait is given on the next slide.

rtrtecec

)2(2

)1(1 x +=


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Case 3: Star Node (2 of 5)

The general solution is

Thus every trajectory lies on a line through the origin, as seenin the phase portrait below. Several graphs ofx1 versus tare

given below as well, with the case ofx2 versus tsimilar.The critical point at the origin is called a proper node, or astar point.

0,)2(2)1(1


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Case 3: Equal Eigenvalues (3 of 5)

If the repeated eigenvalue rhas only one linearly independent

eigenvector , then from Section 7.8 the general solution is

For large t, the dominant term is c2tert. Thus every trajectory

approaches origin tangent to line through the eigenvector .Similarly, for large negative tthe dominant term is againc2te

rt, and hence every trajectory is asymptotic to a lineparallel to the eigenvector .

The orientation of the trajectories

depends on the relative positions

of and , as we will see.

rtrtrt etecec x ++= 21


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Case 3: Improper Node (4 of 5)

We can rewrite the general solution as

Note that y determines the direction of x, whereas the scalarquantity ert affects only the magnitude of x.

For fixed values of c1

and c2, the expression for y is a vector

equation of line through the point c1 + c2 and parallel to .

Using this fact, solution trajectories can be sketched for givencoefficients c1 and c2. See phase portrait below.

When a double eigenvalue has onlyone linearly independent eigenvalue,

the critical point is called an improper

or degenerate node.

( ) tccceetccc rtrt yyx 221221 , ++==++=


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Case 3: Phase Portraits (5 of 5)

The phase portrait is given in figure (a) along with several

graphs ofx1 versus tare given below in figure (b).When the relative orientation of and are reversed, the

phase portrait given in figure (c) is obtained.


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Case 4: Complex Eigenvalues (1 of 5)

Suppose the eigenvalues are i, where and are real,

with 0 and > 0.It is possible to write down the general solution in terms ofeigenvalues and eigenvectors, as shown in Section 7.6.

However, we proceed in a different way here.Systems having eigenvalues iare typified by

We introduce the polar coordinates r, given by

212

211

xxx

xxx

+=

+=

= xx

1222

21

2 tan, xxxxr =+=


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Case 4: Polar Equations (2 of 5)

Differentiating the polar equations

with respect tot, we have

or

Substituting

into these derivative equations, we obtain

212211 , xxxxxx +=+=

1222

21

2

tan, xxxxr =+=

( ) ( )2

1

1221222

11

//sec,222

x

dtdxxdtdxx

dt

d

dt

dxx

dt

dxx

dt

drr

=

+

=

) ( ) 21122122211 sec, xxxxxxxxxrr =+=

== ,rr


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Case 4: Spiral Point (3 of 5)

Solving the differential equations

we have

These equations are parametric equations in polar coordinatesof the solution trajectories to our system x'= Ax.

Since > 0, it follows that decreases as tincreases, so thedirection of motion on a trajectory is clockwise.

If < 0, then r 0 as t, while r if > 0.Thus the trajectories are spirals, which approach or recedefrom the origin depending on the sign of , and the critical

point is called a spiral point in this case.

== ,rr

)0(,, 00 =+== tcer t


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The phase portrait along with several graphs ofx1 versus tare

given below.Frequently the terms spiral sinkand spiral source are used torefer to spiral points whose trajectories approach, or departfrom, the critical point.

0

1222

21

2

2

1

2

1

,

tan,

+==

=+=

=

=

tcer

xxxxrx

x

x

x

t

Axx


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Case 4: General System (5 of 5)

It can be shown that for any system with complex eigenvalues

i, where 0, the trajectories are always spirals.They are directed inward or outward, respectively, dependingon whether is negative or positive.

The spirals may be elongated and skewed with respect to thecoordinate axes, and the direction may be either clockwise orcounterclockwise. See text for more details.


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Case 5: Pure Imaginary Eigenvalues (1 of 2)

Suppose the eigenvalues are i, where = 0 and real.

Systems having eigenvalues iare typified by

As in Case 4, using polar coordinates r, leads to

The trajectories are circles with center at the origin, which are

traversed clockwise if > 0 and counterclockwise if < 0.A complete circuit about the origin occurs in a time interval oflength 2/, so all solutions are periodic with period 2/.

The critical point is called a center.

12

21

0

0

xx

xx

=

=

= xx

0, +== tcr


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In general, when the eigenvalues are pure imaginary, it is

possible to show that the trajectories are ellipses centered atthe origin.

The phase portrait along with several graphs ofx1 versus taregiven below.


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Behavior of Individual Trajectories

As t, each trajectory does one of the following:

approaches infinity;approaches the critical point x = 0;

repeatedly traverses a closed curve, corresponding to a periodicsolution, that surrounds the critical point.

The trajectories never intersect, and exactly one trajectorypasses through each point (x0,y0) in the phase plane.

The only solution passing through the origin is x = 0. Other

solutions may approach (0, 0), but never reach it.


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Behavior of Trajectory Sets

As t, one of the following cases holds:

All trajectories approach the critical point x = 0. This is the case whenthe eigenvalues are real and negative or complex with negative real

part. The origin is either a nodal or spiral sink.

All trajectories remain bounded but do not approach the origin, and

occurs when eigenvalues are pure imaginary. The origin is a center.Some trajectories, and possibly all trajectories except x = 0, tend toinfinity. This occurs when at least one of the eigenvalues is positive orhas a positive real part. The origin is a nodal source, a spiral source, ora saddle point.


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Summary Table

The following table summarizes the information we have

derived about our 2 x 2 system x'= Ax, as well as the stabilityof the equilibrium solution x = 0.

Eigenvalues Type of Critical Point Stability

021 >>rr Node Unstable

021


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Ch 9.2: Autonomous Systems and Stability

In this section we draw together and expand on geometrical

ideas introduced in Section 2.5 for certain first order equationsand Section 9.1 for second order linear homogeneous systemswith constant coefficients.

These ideas concern the qualitative study of differential

equations and the concept of stability.


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Initial Value Problem

We are concerned with systems of two simultaneous

differential equations of the form

We assume that the functions Fand G are continuous and havecontinuous partial derivatives in some domain D of xy-plane.

If (x0, y0) is a point in D, then by Theorem 7.1.1 there exists aunique solution x = (t), y = (t), defined in some interval Icontaining t0, satisfying the initial conditions

),(/),,(/ yxGdtdyyxFdtdx ==

0000 )(,)( ytyxtx ==


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Vector Form

We can write the initial value problem

in vector form:

or

where x = xi + yj, f(x)=F(x,y)i + G(x,y)j, x0 = x0i + y0j, and

In vector form, the solution x = (t) = (t)i + (t)j is a curve

traced out by a point in the xy-plane (phase plane).

0000 )(,)(),,(/),,(/ ytyxtxyxGdtdyyxFdtdx ====

00 )(),(/ xxxfx == tdtd

=

=

=

0

00,

),(

),(

y

x

yxG

yxF

y

xxx

=

=

1

0,

0

1ji


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Autonomous Systems

For our initial value problem

note that the functions Fand G depend on x and y, but not t.

Such a system is said to be autonomous.

The system x'= Ax, where A is a constant matrix, is anexample of an autonomous system. However, if one or moreof the elements of the coefficient matrix A is a function of t,then the system is nonautonomous.

The geometrical qualitative analysis of Section 9.1 can beextended to two-dimensional autonomous systems in general,

but is not as useful for nonautonomous systems.



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Phase Portraits for Autonomous Systems

Our autonomous system

has a direction field that is independent of time.

It follows that only one trajectory passes through each point

(x0, y0) in the phase plane.Thus all solutions to an initial value problem of the form

lie on the same trajectory, regardless of the time t0 at whichthey pass through (x0, y0).

Hence a single phase portrait displays important qualitativeinformation about all solutions of the system.




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Stability and Instability

For the following definitions, we consider the autonomous

system x'= f(x) and denote the magnitude of xby ||x||.The points, if any, where f(x) = 0 are called critical points. Atthese points x'= 0 also, and hence critical points correspond toconstant, or equilibrium, solutions of the system of equations.

A critical point x0 is said to be stable if, for all > 0 there is a> 0 such that every solution x = (t) satisfying ||(0) - x0|| < exists for all positive tand satisfies ||(t) - x0|| < for all t 0.

Otherwise, x0 is unstable.


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Asymptotic Stability

A critical point x0 is said to be asymptotically stable if it is

stable and if there exists a 0 > 0 such that if a solution x = (t)satisfies ||(0) - x0|| < 0, then (t) x0 as t.

Thus trajectories that start sufficiently close to x0 not only

stay close to x0but must eventually approach x0 as t.

This is the case for the trajectory in figure (a) below but not forthe one in figure (b) below.

Thus asymptotic stability is a stronger property than stability.

However, note that

does not imply stability.

0)(lim x=

tt


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The Oscillating Pendulum

The concepts of asymptotic stability, stability, and instability

can be easily visualized in terms of an oscillating pendulum.Suppose a mass m is attached to one end of a weightless rigidrod of length L, with the other end of the rod supported at theorigin O. See figure below.

The position of the pendulum is described by the angle , withthe counterclockwise direction taken to be positive.

The gravitational force mg acts downward,

with damping force c|d/dt|, c > 0, alwaysopposite to the direction of motion.


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Pendulum Equation

The principle of angular momentum states that the time rate of

change of angular momentum about any point is equal to themoment of the resultant force about that point.

The angular momentum about the origin is mL2(d/dt), andhence the governing equation is

Here, L and Lsin are the moment arms

of the resistive and gravitational forces.

This equation is valid for all four sign

possibilities of and d/dt.

sin2

22

mgLdt

dcL

dt

dmL =


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Autonomous Pendulum System

Rewriting our equation in standard form, we obtain

To convert this equation into a system of two first orderequations, we let x = and y = d/dt. Then

To find the critical points of this autonomous system, solve

These points correspond to two physical equilibrium positions,one with the mass directly below point of support (= 0), andthe other with the mass directly above point of support (= ).

Intuitively, the first is stable and the second is unstable.

LgmLc ===++

22

,,0sin

yxyyx

==sin,

2

0,0sin,0 2 ==== ynxyxy


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Stability of Critical Points: Damped Case

If mass is slightly displaced from lower equilibrium position, it

will oscillate with decreasing amplitude, and slowly approachequilibrium position as damping force dissipates initial energy.This type of motion illustrates asymptotic stability.

If mass is slightly displaced from upper equilibrium position, it

will rapidly fall, and then approach lower equilibrium position.This type of motion illustrates instability

See figures (a) and (b) below.


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Stability of Critical Point: Undamped Case

Now consider the ideal situation in which the damping

coefficient c (or ) is zero.In this case, if the mass is displaced slightly from the lowerequilibrium position, then it will oscillate indefinitely withconstant amplitude about the equilibrium position.

Since there is no dissipation in the system, the mass willremain near the equilibrium position but will not approach itasymptotically. This motion is stable but not asymptotically

stable. See figure (c) below.


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Determination of Trajectories

Consider the autonomous system

It follows that

which is a first order equation in the variables x and y.If we can solve this equation using methods from Chapter 2,then the implicit expression for the solution, H(x,y) = c, givesan equation for the trajectories of

Thus the trajectories lie on the level curves of H(x,y).

Note this approach is applicable only in special cases.


),,(/),(/ yxFyxGdxdy =

),(/),(/ yxFyxGdxdy =


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Example 1

Consider the system

It follows that

The solution of this separable equation is

Thus the trajectories are hyperbolas, as shown below.

The direction of motion can by inferredfrom the signs of dx/dtand dy/dtin the

four quadrants.

xdtdyydtdx == /,/

dxxydyyxdxdy == //

cxyyxH == 22),(


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Example 2: Separable Equation (1 of 2)

Consider the system

It follows that

The solution of this separable equation is

Note that the equilibrium points are found by solving

and hence (-2, 2) and (2, 2) are the equilibrium points.

2

312/,24/ xdtdyydtdx ==

( ) ( )dxxdyyy

x

dx

dy 22

31224

24

312=

=

cxxyyyxH =+= 32 124),(

2,20312,024 2 ==== yxxy


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Example 2: Phase Portrait (2 of 2)

We have

A graph of some level curves of Hare given below.

Note that (-2, 2) is a center and (2, 2) is a saddle point.

Also, one trajectory leaves the saddle point (at t= -), loopsaround the center, and returns to the saddle point (at t= ).

cxxyyyxH =+=

32

124),(


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Ch 9.3: Almost Linear Systems

In Section 9.1 we gave an informal description of the stability

properties of the equilibrium solution x = 0 of the 2 x 2 systemx'= Ax. The results are summarized in Table 9.1.1.

We required detA 0, and hence x = 0 is the only critical pointof the system x'= Ax.

Now that we have precisely defined the concepts of asymptoticstability, stability, and instability, we can restate these resultsin the Theorem 9.3.1, on the next slide.


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Theorem 9.3.1

The critical point x = 0 of the 2 x 2 linear system x'= Ax is

(1) asymptotically stable if the eigenvalues r1 and r2 are realand negative or have negative real part;

(2) stable, but not asymptotically stable, if r1 and r2 are pureimaginary;

(3) unstable if r1 and r2 are real and either is positive, or if theyhave positive real part.


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Perturbations

Thus by Theorem 9.3.1, or Table 9.1.1, the eigenvalues r1, r2

of A determine the type of critical point at x = 0 and itsstability characteristics.

Now r1, r2 depend on the coefficients in the system x'= Ax,which in turn may depend on physical measurements.

Since these measurements are typically subject to smalluncertainties, it is of interest to investigate whether the smallchanges (perturbations) in the coefficients can affect the

stability or instability of a critical point and/or significantlyalter the pattern of the trajectories.


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Perturbations in Pure Imaginary Eigenvalues

Recall that the eigenvalues r1 and r2 of A are the roots of the

polynomial equation det(A

-rI

) = 0.It can be shown that small perturbations in some or all of thecoefficients are reflected in small changes in the eigenvalues.

The most sensitive situation occurs whenr1 = iand r2 = -i,that is, when the critical point is a center.

Small changes to the coefficients results in r1, r2 taking on newvalues: r1 = '+ i' and r2 = '-i', where ' 0 and ' .


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Perturbations and Centers

Thus r1, r2become r1 = '+ i' and r2 = '-i'.

If ' 0, which almost always happens, then trajectories ofperturbed system are spirals rather than closed curves.

The system is asymptotically stable if '< 0; unstable if '> 0.

Thus small perturbations in coefficients may change a stablesystem into an unstable one, and in any case may be expectedto alter radically the trajectories in the phase plane.


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Perturbations in Equal Eigenvalues

Another slightly less sensitive case occurs whenr1 = r2, that is,

when the critical point is a node.Small perturbations in the coefficients will normally cause thetwo equal roots to separate (bifurcate).

If the separated roots are real, then the critical point remains a

node, but if the separated roots are complex conjugates, thenthe critical point becomes a spiral point.

Here, the stability or instability of the system is not affected by

small changes in the coefficients, but the trajectories may bealtered considerably.


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Perturbations in Other Cases

In all other cases the stability or instability of the system is notchanged, nor is the type of critical point altered, by sufficientlysmall perturbations in the coefficients of the system.

For example, if r1 and r2 are real, negative, and unequal, then asmall change in the coefficients will neither change the sign of

r1 and r2 nor allow them to coalesce. Thus the critical pointremains an asymptotically stable node.


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Nonlinear Systems

Consider a nonlinear two-dimensional autonomous system

x'= f(x)Our main object is to investigate the behavior of trajectories ofthis system near a critical point x0.

We do this by approximating the nonlinear system near acritical point x0by an appropriate linear system, whosetrajectories are easy to describe.

The crucial question is whether the trajectories of the linear

system are good approximations to those of nonlinear system.For convenience, assume critical point is at the origin, x0 = 0.This involves no loss of generality, since in general the

substitution u = xx0

shifts the critical point to the origin.


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Nonlinear and Nearby Linear Systems

First, we consider what it means for a nonlinear system to beclose to a linear system.

Accordingly, suppose that x'= f(x) = Ax + g(x).

Assume that x = 0 is an isolated critical point of this system.This means that there is some circle about the origin within

which there are no other critical points.

In addition, assume that detA 0, and hence x = 0 is also anisolated critical point of the system x'= Ax.

For the nonlinear system x'= f(x) to be close to the linearsystem x'= Ax, we must assume that g(x) is small.


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Almost Linear Systems: Vector Form

Recall: x'= f(x) = Ax + g(x), where g(x) is small.

More precisely, assume the components of g have continuousfirst partial derivatives and satisfy the limit condition

Thus ||g(x)|| is small compared to ||x|| near the origin.

Such a system is called an almost linear system in the

neighborhood of the critical point x = 0.

0)(

lim =

x

xg

0x


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Almost Linear Systems: Scalar Form

Recall: x'= f(x) = Ax + g(x), where

In scalar form, if we let

then

and the corresponding limit condition is

0)(lim = xxg

0x

=

=

),(),()(,

2

1

yxgyxg

yx xgx

,),(),()(, 222122 yxgyxgryx +==+= xgx

0),(

limand0),(

lim 2

0

1

0

== r

yxg

r

yxg

rr


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Example 1: Nonlinear System (1 of 2)

Consider the system

Note that (0, 0) is a critical point and detA 0.

It can be shown that the other critical points are (0, 2), (1, 0)and (0.5, 0.5). Thus (0, 0) is an isolated critical point.

The limit condition

is more readily verified if we use polar coordinatesx = rcos

andy = rsin.

+

=

2

2

25.075.05.0001

yxyxyx

yx

yx

0)(lim = xxg

0x


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Example 1: Almost Linear System (2 of 2)

Using polar coordinatesx = rcosandy = rsin, we have

Thus

and hence our system is almost linear near the origin.

( )

( )

2

2222

2

2

22221

sin25.0sincos75.0

sin25.0sincos75.025.075.0),(

,sincoscos

sincoscos),(

+=

==

+=

==

r

rrr

ryxy

ryxg

r

r

rr

r

xyx

r

yxg

0),(

limand0),(

lim 20

1

0==

r

yxg

r

yxg

rr


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Example 2: Pendulum System

The motion of a pendulum system is given by

or

Note detA 0, and critical points are (n, 0) for n = 1, 2,.Thus (0, 0) is an isolated critical point, with g1(x,y) = 0 and

where, from the Taylor series representation of sinx,

Thus this system is almost linear.

yxyyx ==

sin,

2

=

xxy

x

y

x

sin

010 22

,0cos

lim)(sin

lim),(

lim332

0

2

0

2

0

=

=

= r

r

r

xx

r

yxg

rrr

L++= 523

1sin xaxaxx


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General Nonlinear System (1 of 3)

Consider the general nonlinear system x'= f(x), or

We will show that this system is almost linear near a criticalpoint (x0,y0) whenever the functions Fand G have continuouspartial derivatives up to order 2.

To see this, we use Taylor expansions about the point (x0,y0)to write F(x,y) and G(x,y) in the form

where

),(),,( yxGyyxFx =

=

),())(,())(,(),(),(

),,())(,())(,(),(),(

200000000

100000000

yxyyyxGxxyxGyxGyxG

yxyyyxFxxyxFyxFyxF

yx

yx

+++=

+++=

0

)()(

),(lim

)()(

),(lim

20

20

2

),(),(20

20

1

),(),(0000

=

+

=

+

yyxx

yx

yyxx

yx

yxyxyxyx


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Rewriting General Nonlinear System (2 of 3)

We have F(x,y) and G(x,y) in the form

Since (x0,y0) is a critical point, F(x0,y0) = G(x0,y0) = 0. Also,

note that dx /dt = d(x -x0)/dt and dy /dt = d(y -y0)/dt.Thus the original system of equations

reduces to

),())(,())(,(),(),(),,())(,())(,(),(),(

200000000

100000000

yxyyyxGxxyxGyxGyxGyxyyyxFxxyxFyxFyxF

yx

yx

+++=

+++=

),(),,( yxGyyxFx ==

+

=

),(

),(

),(),(

),(),(

2

1

0

0

0000

0000

0

0

yx

yx

yy

xx

yxGyxG

yxFyxF

yy

xx

dt

d

yx

yx


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General Nonlinear System: Almost Linear(3 of 3)

Thus our system of equations can be written as

In vector notation,

Thus if Fand G are twice differentiable, then the nonlinear

system of equations x'= f(x) is almost linear, and the linearsystem that approximates the nonlinear system is given by

+

=

),(),(

),(),(),(),(

2

1

0

0

0000

0000

0

0

yxyx

yyxx

yxGyxGyxFyxF

yyxx

dtd

yx

yx

( ) ( ) ( ) ( )

=

=+=

),(),(,where,

2

1

0

00

yx

yx

yy

xx

d

d

dt

d

xxuxux

x

fu

=

=

0

0

2

1

2

1

0000

0000

2

1 where,

),(),(

),(),(

yy

xx

u

u

u

u

yxGyxG

yxFyxF

u

u

dt

d

yx

yx


59/206

Example 3: Pendulum System (1 of 2)

The motion of a pendulum system is given by

Thus

The critical points are (n, 0) for n = 0, 1, 2,.Since Fand G are twice differentiable, the system of equationsis almost linear near each critical point. We have

To find the approximating linear system at (x0,y0), we use

yxyyx

==sin,

2

yxyxGyyxF == sin),(,),( 2

==== yxyx GxGFF ,cos,1,02

=

=

0

0

2

1

2

1

0000

0000

2

1 where,),(),(

),(),(

yy

xx

u

u

u

u

yxGyxG

yxFyxF

u

u

dt

d

yx

yx

Example 3:


60/206

Example 3:

Approximating Linear Systems (2 of 2)

To find the approximating linear system at (x0,y0), we use

with

At the origin, the approximating linear system is

At (, 0), the approximating linear system is

=

y

x

y

x

dt

d

210

=

=

0

0

2

1

2

1

0000

0000

2

1 where,),(),(),(),(

yyxx

uu

uu

yxGyxGyxFyxF

uu

dtd

yx

yx

=

y

x

y

x

dt

d

2

10

==== yxyx GxGFF ,cos,1,0

2


61/206

Theorem 9.3.2

Consider the almost linear system x'= Ax + g(x). Let r1 and r2be the eigenvalues of A. Then the type and stability of thecritical point (0,0) of the linear system x'= Ax and the almostlinear system x'= Ax + g(x) are as given in the table below.

Linear System Almost Linear System

Eigenvalues Type Stability Type Stability

021 >>rr Node Unstable Node Unstable

021


62/206

Theorem 9.3.2 Discussion (1 of 2)

Since nonlinear term g(x) is small compared to the linear termAx when x is small, we hope that the trajectories of the linearsystem x'= Ax are good approximations to those of nonlinearsystem, at least near the origin.

Theorem 9.3.2 states that this is true in many but not all cases.

For small x, the nonlinear terms are small and do not affect thestability and type of critical point as determined by linear term,except in two sensitive cases: r1 and r2pure imaginary, and r1and r2 real and equal.

As we have seen earlier, small changes in eigenvalues can alterthe type and stability of the critical point for a linear system,

but only in these two cases. Thus a small nonlinear term might

have a similar effect for these two cases as well.


63/206

Theorem 9.3.2 Discussion (2 of 2)

Even if the critical point is of the same type as that of thelinear system, the trajectories of the almost linear system may

be much different in appearance than for the linear system,except very near the critical point.

However, it can be shown that the slopes at which the

trajectories enter or leave the critical point are givencorrectly by the linear equation.


64/206

Damped Pendulum System (1 of 2)

Recall that the motion of a pendulum system is given by

Near the origin these nonlinear equations are approximated by

whose eigenvalues are

The nature of the solutions to the linear and nonlinear systemsdepends on the sign of 2 42, as examined on the next slide.

yxyyx == sin, 2

=

y

x

y

x

210

2

4,

22

21

=rr

2

4,

22

21

=rr


65/206

Damped Pendulum System (2 of 2)

Thus we have the following cases:If 2 42 > 0, then the eigenvalues are real, unequal, and negative.The critical point (0,0) is an asymptotically stable node of the linearsystem, and of the almost linear system.

If 2 42 = 0, then the eigenvalues are real, equal, and negative. Thecritical point (0,0) is an asymptotically stable (proper or improper) node

of the linear system. It may be either an asymptotically stable node or aspiral point of the almost linear system.

If 2 42 < 0, then the eigenvalues are complex with negative realpart. The critical point (0,0) is an asymptotically stable spiral point of

the linear system, and of the almost linear system.

Hence the origin is a spiral point of the nonlinear system if thedamping is small, and a node if damping is large enough. Ineither case, the origin is asymptotically stable.

2


66/206

Pendulum System: Small Damping (1 of 5)

Consider the small damping case, 2 42 < 0.

The direction of motion of the spirals near (0,0) can beobtained directly from the equations

For the point at which a spiral intersects the positivey-axis, atx = 0 andy > 0, it follows that dx/dt> 0. Thus the point (x,y)on the trajectory is moving to the right, so the direction ofmotion spirals clockwise.

yxdtdyydtdx == sin/,/ 2


67/206


The direction of motion of the spirals near (2n, 0) is thesame as near the origin.

As before, this can be obtained directly from the equations

We can expect this on physical grounds, since all these criticalpoints correspond to the downward equilibrium position of thependulum.

yxdtdyydtdx == sin/,/ 2


68/206


Next, consider the critical point (, 0). Here, the nonlinearequations are approximated by the linear system

whose eigenvalues are

One eigenvalue (r1) is positive and the other (r

2) is negative.

Therefore, regardless of the amount of damping, the criticalpoint (, 0) is an unstable saddle point both of the linearsystem and of the almost linear system.

,10

2

=

y

x

y

x

dt

d

2

4,

22

21

+=rr


69/206


To examine the trajectories near the saddle point (, 0) in moredetail, consider the general solution of the linear system:

Sincer1 > 0 and r2 < 0, it follows that the solution that

approaches zero as t corresponds to C1 = 0.

For this solution v/u = r2, thus slope of entering trajectories isnegative; one lies in second quadrant (C2 < 0) and the other

lies in the fourth quadrant (C2 > 0).For C2 = 0, we obtain a pair of trajectories exiting from thesaddle point. They have slope r1 > 0; one lies in first quadrant(C

1> 0) and the other lies in the third quadrant (C

1< 0).

=

+

=

y

x

v

ue

rCe

rC

v

utrtr

where,

1121

2

2

1

1


70/206


The analysis at (, 0) can be repeated to show that the criticalpoints (n, 0), n odd, are all saddle points oriented in the sameway as the one at (, 0).

These critical points all correspond to the upward equilibriumposition of the pendulum, so we expect them to be unstable.

Diagrams showing the trajectories in the neighborhood of twosaddles points are given below.


71/206

Example 4: Pendulum System (1 of 6)

The motion of a certain pendulum system is given by

wherex = andy = d/dt.

Note that 2 = 9, and thus the damping coefficient, = 1/5, isrelatively small. It follows that 2 42 < 0 here.

The phase portrait for this system is given below.

5/sin9, yxyyx ==


72/206

Example 4: Critical Points (2 of 6)

The critical points are (n, 0) for n = 0, 1, 2,.

Even values of n, including (0,0), correspond to the downwardposition of the pendulum, while odd values correspond to theupward position.

Near each of the asymptotically stable critical points, the

trajectories are clockwise spirals that represent a decayingoscillation about the equilibrium solution.


73/206

Example 4: Whirling (3 of 6)

The wavy horizontal portions of the trajectories for largervalues of |y| represent whirling motions of the pendulum.

A whirling motion cannot continue indefinitely, sinceeventually the angular velocityy is so much reduced bydamping that the pendulum can no longer go over the top, and

instead begins to oscillate about its downward position.


74/206

Example 4: Separatrix (4 of 6)

The trajectories that enter the saddle points separate the phaseplane into regions. Such a trajectory is called a separatrix.

Each region contains exactly one of the spiral points.

The initial conditions onx = andy = d/dtdetermine theposition of an initial point (x, y) in the phase plane.

The subsequent motion of thependulum is represented by thetrajectory passing through theinitial point as it spirals towardthe asymptotically stablecritical point in that region.


75/206

Example 4: Basins of Attraction (5 of 6)

The set of all initial points from which the trajectoriesapproach an asymptotically stable critical point is called thebasin of attraction, or the region of asymptotic stability, forthat critical point.

Each asymptotically stable critical point has its own basin of

attraction, which is bounded by the separatrices through theneighboring unstable saddle points.

The basin of attraction is shownin blue on the graph.


76/206

Example 4: Asymptotic Stability (6 of 6)

An important difference between nonlinear autonomoussystems and the linear systems x'= Ax discussed in Section

9.1 is illustrated by the pendulum equations.

Recall that x'= Ax has only the single critical point x = 0 ifdetA 0. Thus if the origin is asymptotically stable, then not

only do the trajectories that start close to the origin approachthe origin, but every trajectory approaches the origin. In thiscase the critical point x = 0 is globally asymptotically stable.

This property is not true in general for nonlinear systems, andthus it is important to determine, or estimate, the basins ofattraction for each asymptotically stable critical point.


77/206

Ch 9.4: Competing Species

In this section we explore the application of phase plane

analysis to some problems in population dynamics.

These problems involve two interacting populations and are

extensions of those discussed in Section 2.5, which dealt with

a single population.

Although the relationships discussed here are overly simplecompared to the complex relationships in nature, it is still

possible to acquire some insight into ecological principles

from a study of these model problems.


78/206

Logistic Equations

Suppose that in some closed environment there are two similar

species competing for a limited food supply.

For example, two species of fish in a pond that do not prey on

each other but do compete for the available food.

Let x and ybe the populations of the two species at time t.

As in Section 2.5, assume that the population of each species,in the absence of the other, is modeled by the logistic equation.

Thus

where 1 and 2 are the growth rates of the two populations,

and 1/1 and 2 /2 are their saturation levels.

),(/),(/ 2211 yydtdyxxdtdx ==


79/206

Competing Species Equations

However, when both species are present, each will impinge on

the available food supply for the other. In effect, they reduce

each others growth rates and saturation populations.

The simplest expression for reducing growth rate of species x

due to the presence of species y is to replace the growth factor

1 - 1xby 1 - 1x - 1y, where 1 is a measure of the degreeto which species y interferes with species x.

Similarly, we model the reduced growth rate of species y due

to presence of species xby replacing the growth factor 2 - 2y

by 2 - 2y - 2x.

Thus we have the system of equations

)(/),(/ 222111 xyydtdyyxxdtdx ==


80/206

Example 1: Population Equations (1 of 8)

Consider the system of equations

To find the critical points, we solve

obtaining (0,0), (0,0.75), (1,0), and (0.5,0.5). These critical

points correspond to equilibrium solutions.

The first three points involve the extinction of one or both

species. Only the fourth critical point corresponds to the long

term survival of both species.Other solutions are represented as trajectories in the xy-plane

that describe the evolution of the populations over time.

)5.075.0(/),1(/ xyydtdyyxxdtdx ==

,0)5.075.0(,0)1( == xyyyxx


81/206

Example 1: Direction Field (2 of 8)

A direction field for our system of equations is given below.

Only the first quadrant is depicted, as this corresponds to

positive population size.

The heavy dots in the figure correspond to the critical points.

Based on the direction field, (0.5,0.5) attracts other solutions

and therefore appears to be asymptotically stable.

The other three critical points

appear to be unstable.

To confirm these observations,we can examine the linear

approximations to each point.


82/206

Example 1: Linearization (3 of 8)

Our system of equations,

is almost linear, sinceFand G are twice differentiable.

To obtain the linear system near a critical point (x0, y0), we use

the results of Section 9.3, given below.

Thus

=

=

0

0

0000

0000where,

),(),(

),(),(

yy

xx

v

u

v

u

yxGyxG

yxFyxF

v

u

dt

d

yx

yx


=

v

u

xyy

xyx

v

u

dt

d

000

000

5.0275.05.0

21


83/206

Example 1: Critical Point at (0,0) (4 of 8)

For the critical point (0,0), the approximating linear system is

The eigenvalues and eigenvectors are

and hence the general solution for this linear system is

Thus the origin is an unstable node of both the linear andnonlinear systems. The trajectories near origin are all tangentto y-axis, except for one trajectory that lies along the x-axis.

=

y

x

y

x

dt

d

75.00

01

==

== 1

0

,75.0;0

1

,1

)2(

2

)1(

1 rr

ttecec

y

x75.0

21

1

0

0

1

+

=


84/206





Thus (1,0) is an unstable saddle point of both the linear andnonlinear systems. One pair of trajectories approach (1,0)along the x-axis, while all other trajectories depart from (1,0).

=

v

u

v

u

dt

d

25.00

11

==

==

5

4

,25.0;0

1

,1

)2(

2

)1(

1 rr

ttecec

v

u25.0

21

5

4

0

1

+

=


85/206

Example 1: Critical Point at (0,0.75) (6 of 8)

For the critical point (0,0.75), the linear system is



Thus (0,0.75) is an unstable saddle point of both the linear andnonlinear systems. One pair of trajectories approach (0,0.75)along y-axis, while all other trajectories depart from (0,0.75).

=

v

u

v

u

dt

d

75.0375.0

025.0

==

==

1

0

,75.0;3

8

,25.0

)2(

2

)1(

1 rr

ttecec

v

u75.0

2

25.0

1

1

0

3

8

+

=


86/206

E l 1 Ph P i (8 f 8)


87/206


A phase portrait is given below, along with the direction field,

for our nonlinear system

Note that the quadratic terms in these equations are all

negative, and hence x'< 0 and y'< 0 for large x and y.

Thus the trajectories are directed inward towards (0.5,0.5).


E l 2 P l i E i (1 f 9)


88/206


Consider the system of equations

The critical points are (0,0), (1,0), (0,2), and (0.5,0.5). These

critical points correspond to equilibrium solutions.

The first three points involve the extinction of one or both

species. Only the fourth critical point corresponds to the longterm survival of both species.

Other solutions are represented as trajectories in the xy-plane

that describe the evolution of the populations over time.

)75.025.05.0(/),1(/ xyydtdyyxxdtdx ==

E l 2 Di ti Fi ld (2 f 9)


89/206

Example 2: Direction Field (2 of 9)

A direction field for our system of equations is given below,

where the heavy dots correspond to the critical points.

Based on the direction field, (0.5,0.5) appears to be a saddle

point, and hence unstable, while (1,0) and (0,2) appear to be

asymptotically stable.

Thus only one species will eventuallysurvive, and this species is determined

by the initial conditions.

E l 2 C iti l P i t t (0 0) (3 f 9)


90/206





Thus the origin is an unstable node of both the linear andnonlinear systems. All trajectories leave the origin tangent toy-axis, except for one trajectory that lies along the x-axis.

=

y

x

y

x

dt

d

5.00

01

==

==

1

0

,5.0;0

1

,1

)2(

2

)1(

1 rr

ttecec

y

x5.0

21

1

0

0

1

+

=

E l 2 C iti l P i t t (1 0) (4 of 9)


91/206



The general solution for this linear system is

Thus (1,0) is an asymptotically stable node of both the linear

and nonlinear systems.

One pair of trajectories approach (1,0) along the x-axis

All other trajectories approach (1,0) tangent to the line with

slope 3/4, determined by the eigenvector (2).

=

v

u

v

u

dt

d

25.00

11

tt

ececv

u25.0

21 3

4

0

1

+

=

E ample 2: Critical Point at (0 2) (5 of 9)


92/206


For the critical point (0,2), the linear system is

The general solution for this linear system is

Thus (0,2) is an asymptotically stable node of both the linear

and nonlinear systems.

One trajectory approaches (0,2) along the line with slope 3,determined by the eigenvector (1).

All other trajectories approach (0,2) along the y-axis.

=

v

u

v

u

dt

d

5.05.1

01

tt

ececv

u5.0

211

0

3

1

+

=

Example 2: Critical Point at (0 5 0 5) (6 of 9)


93/206

Example 2: Critical Point at (0.5,0.5) (6 of 9)

For the critical point (0.5,0.5), the linear system is



=

v

u

v

u

dt

d

125.0375.0

5.05.0

( )

( )

+=

=

=

+

=

5687.0

1

8/573

1,7844.0

16

575

;3187.1

1

8/573

1

,1594.016

575

)2(

2

)1(

1

r

r

ttecec

v

u7844.0

2

1594.0

15687.0

1

3187.1

1

+

=

Example 2: Critical Point at (0 5 0 5) (7 of 9)


94/206

Example 2: Critical Point at (0.5,0.5) (7 of 9)

Thus for the critical point (0.5,0.5), we have

It follows that (0.5,0.5) is an unstable saddle point of both the

linear and nonlinear systems.

One pair of trajectories approaches (0.5,0.5) as t, whilethe others depart from (0.5,0.5).

The entering trajectories approach (0.5,0.5) tangent to the line

with slope of 0.5687, determined by the eigenvector (2).

tt

ececv

u7844.0

2

1594.0

15687.0

1

3187.1

1

+

=



95/206


A phase portrait is given below, along with the direction field.

Of particular interest is the pair of trajectories that enter the

saddle point. These trajectories form a separatrix that divides

the first quadrant into two basins of attraction.



96/206


Trajectories starting above the separatrix approach the node at

(0,2), while those below approach the node at (1,0).

If initial state lies on separatrix, then the solution will approach

the saddle point, but the slightest perturbation will send the

trajectory to one of the nodes instead.

Thus in practice, one species willsurvive the competition and the

other species will not.

Coexistence Analysis (1 of 7)


97/206

Coexistence Analysis (1 of 7)

Examples 1 and 2 show that in some cases the competition

between two species leads to an equilibrium state of

coexistence, while in other cases the competition results ineventual extinction of one of the species.

We can predict which situation will occur by examining the

governing equations

Note that this system is almost linear, since Fand G are

quadratic polynomials.

There are four cases to be considered, depending on the

relative orientation of the lines


0,0 222111 == xyyx

Coexistence Analysis: Nullclines (2 of 7)


98/206

Coexistence Analysis: Nullclines (2 of 7)

The graphs below show the relative orientation of the lines

The lines are called the x and y nullclines, respectively,

because x'= 0 on the first and y'= 0 on the second.

The heavy dots represent equilibrium solutions.

Thus sustained coexistence is not possible in cases (a) and (b).

We show that sustained coexistence happens only in case (d).

0,0 222111 == xyyx

Coexistence Analysis: Linear System (3 of 7)


99/206

Coexistence Analysis: Linear System (3 of 7)

Let (X,Y) be a critical point, with corresponding linear system

Since (X,Y) is a critical point, we solve

to obtain nonzero values of Xand Y, as given below:

Further, with (X,Y) a critical point, the linear system reduces to

=

v

u

XYY

XYX

v

u

dt

d

2222

1111

2

2

0,0 222111 == XYYX

2121

2112

2121

1221 ,

=

= YX

=

v

u

YY

XX

v

u

dt

d

22

11

Coexistence Analysis: Eigenvalues (4 of 7)


100/206

Coexistence Analysis: Eigenvalues ( )

The eigenvalues of the linear system are

If 1212 < 0, then the eigenvalues are real and of

opposite sign, and hence (X,Y) is an unstable saddle point, and

thus sustained coexistence is not possible.If 1212 > 0, then the eigenvalues are real, negative and

unequal, or complex. It can be shown that the eigenvalues are

not complex, and hence (X,Y) is an asymptotically stable node,and thus sustained coexistence is possible.

In Example 1, 1212 = (1)(1) - (1)(0. 5) = 0.5 > 0.

In Example 2, 1212 = (1)(0.25) - (1)(0.75) = -0.5 < 0.

( )2

)(4)( 21212

2121

2,1

XYYXYX

r

++

=

Coexistence Analysis: Case (c) (5 of 7)


101/206

Coexistence Analysis: Case (c) ( )

In case (c), we have

These inequalities, together with

yield 1212 < 0.

Therefore sustained coexistence is not

possible in case (c).

21121

1

2

2

12212

2

1

1 or,or

>>>>

,0,02121

2112

2121

1221

>

=>

=

YX

Coexistence Analysis: Case (d) (6 of 7)


102/206

Coexistence Analysis: Case (d) ( )

In case (d), we have

These inequalities, together with

yield 1212 > 0.

We can also show that the other critical

points are unstable, and thus the twopopulations approach the equilibrium

state of coexistences in case (d).

2112

1

1

2

2

1221

2

2

1

1

or,or


103/206

Discussion of Coexistence Conditions ( )

For our competitive species equations,

sustained coexistence is possible when 1212 > 0.

The s are a measure of the inhibitory effect that the growth

of each population has on itself.

The s are a measure of the inhibitory effect that the growth

of each population has on the other species.

Thus when 1212 > 0, the competition is weak, and the

species can coexist.When 1212 < 0, the competition is strong, and the

species cannot coexist one must die out.


Ch 9.5: Predator-Prey Systems


104/206

y yIn Section 9.4 we discussed a model of two species that

interact by competing for a common food supply or other

natural resource.In this section we investigate the situation in which one

species (the predator) preys on the other species (the prey),

while the prey lives on some other source of food.For example, foxes and rabbits in a closed forest.

Again we emphasize that a model involving only two species

cannot fully describe the complex relationships among species

that occur in nature.

Nevertheless, the study of simple models is the first step

toward an understanding of more complicated phenomena.

Assumptions


105/206

pLet x and ybe the populations of the prey and predator,

respectively, at time t.

We make the following assumptions:In the absence of the predator, the prey grows at a rate proportional to

the current population; thus dx/dt= ax, a > 0, when y = 0.

In the absence of the prey, the predator dies out at a rate proportional to

the current population; thus dy/dt= -cy, c > 0, when x = 0.

The number of encounters between predator and prey is proportional to

the product of their populations. Each such encounter tends to promote

the growth of the predator and to inhibit the growth of the prey. Thus

the growth rate of the predator is increased by a term of the form xy,while the growth rate of the prey is decreased by a term xy, where

and are positive constants.

Predator-Prey Equations


106/206

y qThus we have the system of equations

The constants a, c,,are all positive, where a, c are the

growth rate of prey and death rate of predator, respectively,

and , are measures of the effect of the interaction betweenthe two species.

The predator-prey equations are known as the Lotka-Volterra

equations. Although they are rather simple equations, they do

characterize a wide class of problems.

Our goal here is to determine the qualitative behavior of the

solutions for arbitrary positive initial values x and y.

( )

( )xcyxycydtdy

yaxxyaxdtdx

+=+=

==

/

,/



107/206

p p qConsider the system of equations

The critical points are (0,0) and (3,2).

Given below is a direction field for this system of equations,

with the critical points indicated as heavy dots.

The trajectories appear to be closed curves surrounding thecritical point (3,2).

0,,25.075.0/,5.0/ >+== yxxyydtdyxyxdtdx



108/206

p ( )For the critical point (0,0), the approximating linear system is



Thus (0,0) is an unstable saddle point of both the linear andnonlinear systems. One trajectory approaches (0,0) along they-axis, while all other trajectories depart from (0,0).

=

y

x

y

x

dt

d

75.00

01

==

==

1

0,75.0;

0

1,1 )2(2

)1(

1 rr

ttecec

y

x75.0

21

1

0

0

1

+

=



109/206



Thus (3,2) is stable center point of the linear system, but is

indeterminate for the nonlinear systems, by Theorem 9.3.2.

To find the trajectories for the linear system, we have

=

=

2

3,

05.0

5.10

y

x

v

u

v

u

v

u

dt

d

==

==

3/

1,

2

3;

3/

1,

2

3 )2(

2

)1(

1 i

ir

i

ir

v

u

v

u

du

dv

dtdu

dtdv

35.1

5.0

/

/===



110/206

Near the critical point (3,2), we thus have

The solution to this separable equation is

Thus the trajectories of the linear system are ellipses centered

at the critical point (3,2), and are elongated horizontally.

Returning to the nonlinear system

we have

vududv 3// =

0,3 22 >=+ kkvu

0,,25.075.0/,5.0/ >+== yxxyydtdyxyxdtdx

xyx

xyy

dx

dy

5.0

25.075.0

+=



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Thus


It can be shown that the graph of this equation, for a fixed c, is

a closed curve about the point (3,2).

Thus the critical point (3,2) is also a center of our nonlinear

system, and hence the predator and prey populations exhibit acyclic variation about the equilibrium solution (3,2).

This behavior is seen in the phase portrait on the next slide.

cxyyx =+ 25.05.0lnln75.0

( )

( ) dx

x

xdy

y

y

yx

xy

dx

dy 25.075.05.01

5.01

25.075.0 +=

+=



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Given below is a phase portrait for our nonlinear system.

For some initial conditions, the trajectories represent small

variations in x and y about (3,2), and are almost elliptical inshape, as the linear analysis suggests.

For other initial points, the oscillations in x and y are more

pronounced, and the shape of the trajectories are significantlydifferent from an ellipse.

Note that the trajectories are

traversed counterclockwise.



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A phase portrait along with population graphs x(t) and y(t), for

a typical set of initial conditions, are given below.

Note from both of these figures that the oscillation of thepredator population lags behind that of the prey.



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Starting with a state in which both populations are relatively

small, the prey first increase because of little predation.

Then the predators, with abundant food, increase in population.

This causes heavy predation, and the prey tend to decrease.

Finally, with a diminished food supply, the predator population

also decreases, and the system returns to original state.

General Predator-Prey Equations (1 of 7)h l f i


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The general system of equations

can be analyzed in the same way as in Example 1.

The critical points are the solutions of the equations

yielding the points (0,0) and (c/,a/).

We next examine the solutions of the corresponding linear

system near each critical point.

( )

( )xcyxycydtdy

yaxxyaxdtdx

+=+=

==

/

,/

( ) ( ) ,0,0 =+= xcyyax

General System: Critical Point at (0,0) (2 of 7)F th iti l i t (0 0) th i ti li t i


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Thus (0,0) is an unstable saddle point of both the linear andnonlinear systems. One trajectory approaches (0,0) along they-axis, while all other trajectories depart from (0,0).

=

y

x

c

a

y

x

dt

d

0

0

==

==

1

0,;

0

1, )2(2

)1(

1 crar

tctaecec

y

x

+

=

1

0

0

121

Critical Point at (c/,a/): Linear System (3 of 7)F th iti l i t ( / / ) th li t i


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For the critical point (c/,a/), the linear system is

The eigenvalues are

Thus (c/,a/) is stable center point of the linear system, but isindeterminate for the nonlinear systems, by Theorem 9.3.2.

To find the trajectories for the linear system, we have

=

=

/

/,

0/

/0

ay

cx

v

u

v

u

a

c

v

u

dt

d

aciracir == 21 ,

vc

ua

vc

ua

du

dv

dtdu

dtdv2

2

)/(

)/(

/

/

===

Critical Point at (c/,a/): Ellipses (4 of 7)N th iti l i t ( / / ) th h


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Near the critical point (c/,a/), we thus have


Thus the trajectories of the linear system are ellipses centeredat the critical point (c/,a/).

Returning to the nonlinear system

we have( )( )yax

xcy

dx

dy

+=

( ) ( ) 022

2

2

=+= dvvcduuavc

ua

du

dv

0,2222 >=+ kkvcua

( ) ( )xcydtdyyaxdtdx +== /,/

Critical Point at (c/,a/):

Nonlinear System (5 of 7)Thus


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Thus


It can be shown that the graph of this equation, for a fixed C, isa closed curve about the point (c/,a/).

Thus the critical point (c/,a/) is also a center of our

nonlinear system, and hence the predator and prey populationsexhibit a cyclic variation about (c/,a/).

Cxxcyya =+ lnln

( )

( ) dx

x

xcdy

y

ya

yax

xcy

dx

dy

+=

+=

Critical Point at (c/,a/):

Small Deviations and Linear System (6 of 7)The cyclic variation of the predator and prey populations can


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The cyclic variation of the predator and prey populations can

be analyzed in more detail when the deviations from the point

(c/,a/) are small and the linear system can be used.The solution of the linear system

can be written in the form

where the constants Kand are determined by the initial

conditions.

=

v

u

a

c

v

u

dt

d

0/

/0

( ) ( )

+=+= tacK

a

catvtacK

ctu sin)(,cos)(

Critical Point at (c/,a/): Small Deviations and

Elliptical Approximation (7 of 7)The equations below are good approximations to the nearly


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The equations below are good approximations to the nearlyelliptical trajectories near (c/,a/).

We can use them to draw several conclusions about the cyclicvariation of the predator and prey on such trajectories:

The predator and prey population sizes vary sinusoidally with period2/(ac). This oscillation period is independent of initial conditions.

The predator and prey populations are out of phase by one-quarter of acycle. The prey leads and the predator lags.

The amplitudes of oscillations are Kc/ for the prey, and (c/a)Ka/for the predator, and hence depend on initial conditions and parameters.

The average predator and prey populations over one complete cycle arec/ and /a, respectively same as equilibrium populations.

( ) ( ) +=+= tacKacatvtacKctu sin)(,cos)(

Modified Predator-Prey EquationsCyclic variations of predator and prey as predicted by our


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Cyclic variations of predator and prey, as predicted by our

equations, have been observed in nature; see text.

One criticism of the Lotka-Volterra equations is that in theabsence of the predator, the prey will grow without bound.

This can be corrected using a logistic model for x when y = 0.

As a result of this modification, the critical point at (c/,a/)moves to (c/,a/- c/ ) and becomes an asymptotically

stable point. It is either a node or a spiral point, depending on

the parameters in the differential equations.

In either case, other trajectories are no longer closed curves but

approach the critical points as t.

Ch 9.6: Liapunovs Second MethodIn Section 9 3 we showed how the stability of a critical point


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In Section 9.3 we showed how the stability of a critical pointof an almost linear system can usually be determined from a

study of the corresponding linear system.However, no conclusion can be drawn when the critical pointis a center of the corresponding linear system.

Also, for an asymptotically stable critical point, we may wantto investigate the basin of attraction, for which the localizedalmost linear theory provides no information.

In this section we discuss Liapunovs second method, or direct

method, in which no knowledge of the solution is required.Instead, conclusions about the stability of a critical point areobtained by constructing a suitable auxiliary function.

Physical PrinciplesLiapunovs second method is a generalization of two physical


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Liapunov s second method is a generalization of two physicalprinciples for conservative systems.

The first principle is that a rest position is stable if thepotential energy is a local minimum, otherwise it is unstable.

The second principle states that the total energy is a constant

during any motion.To illustrate these concepts, we again consider the undamped

pendulum, which is a conservative system.

Undamped Pendulum Equation (1 of 5)The governing equation for the undamped pendulum is


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The governing equation for the undamped pendulum is

To convert this equation into a system of two first orderequations, we letx = andy = d/dt, obtaining

The potential energy Uis the work done in

lifting pendulum above its lowest position:

0sin2

2

=+

L

g

dt

d

xLg

dtdyy

dtdx sin, ==

)cos1(),( xmgLyxU =

Undamped Pendulum System:

Potential Energy (2 of 5)The critical points of our system


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p y

arex= n,y = 0, for n = 0, 1, 2,.

Physically, we expect the points (2n, 0) to be stable, sincethe pendulum bob is vertical with the weight down, and the

points ((2n+1), 0) to be unstable, since the pendulum bobis vertical with the weight up.

Comparing this with the potential energy U,

we see that Uis a minimum equal to zero at (2n, 0) and Uis a maximum equal to 2mgL at ((2n+1), 0).

x

L

g

dt

dyy

dt

dxsin, ==

),cos1(),( xmgLyxU =


Total Energy (3 of 5)The total energy Vis the sum of potential and kinetic energy:


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gy p gy

On a solution trajectoryx = (t),y = (t), Vis a function of t.The derivative of V((t), (t)) with respect to tis called therate of change of Vfollowing the trajectory.

Forx = (t),y = (t), and using the chain rule, we obtain

Sincex andy satisfy the differential equations

it follows that dV(, )/dt= 0, and hence Vis constant.

22)2/1()cos1(),( ymLxmgLyxV +=

( )dt

dyymL

dt

dxxmgL

dt

dV

dt

dV

dt

dVyx

2sin),(),(,

+=+=

,/sin/,/ Lxgdtdyydtdx ==


Small Energy Trajectories (4 of 5)Observe that we computed the rate of change dV(, )/dtof


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p g (, )the total energy Vwithout solving the system of equations.

It is this fact that enables us to use Liapunovs second methodfor systems whose solution we do not know.

Note that V= 0 at the stable critical points (2n, 0), where

we recall

If the initial state (x1,y1) of the pendulum is sufficiently near astable critical point, then the energy V(x1,y1) is small, and thecorresponding trajectory will stay close to the critical point.

It can be shown that if V(x1,y1) is sufficiently small, then thetrajectory is closed and contains the critical point.

22)2/1()cos1(),( ymLxmgLyxV +=


Small Energy Elliptical Trajectories (5 of 5)Suppose (x1,y1) is near (0,0), and that V(x1,y1) is very small.


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The energy equation of the corresponding trajectory is

From the Taylor series expansion of cosx aboutx = 0, we have

Thus the equation of the trajectory is approximately

This is an ellipse enclosing the origin. The smaller V(x1,y1) is,

the smaller the axes of the ellipse are.Physically, this trajectory corresponds to a periodic solution,whose motion is a small oscillation about equilibrium point.

22

11 )2/1()cos1(),( ymLxmgLyxV +=

2/)!4/!2/1(1cos1 242 xxxx = K

1/),(2/),(2 211

2

11

2

=+mLyxV

y

mgLyxV

x


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Damped Pendulum System:

Nonincreasing Total Energy (2 of 2)Thus dV/dt= -cLy2 0, and hence the energy is nonincreasing


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along any trajectory, and except for the liney = 0, the motion

is such that the energy decreases.Therefore each trajectory must approach a point of minimumenergy, or a stable equilibrium point.

IfdV

/dt

< 0 instead ofdV

/dt

0, we can expect this to hold forall trajectories that start sufficiently close to the origin.

General Autonomous System:

Total EnergyTo pursue these ideas further, consider the autonomous system


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and suppose (0,0) is an asymptotically stable critical point.Then there exists a domainD containing (0,0) such that every

trajectory that starts inD must approach (0,0) as t .

Suppose there is an energy function Vsuch that V(x,y) 0for (x,y) inD with V = 0 only at (0,0).

Since each trajectory inD approaches (0,0) as t , then

following any particular trajectory, Vapproaches 0 as t

.The result we want is the converse: If Vdecreases to zero onevery trajectory as t , then the trajectories approach (0,0)

as t , and hence (0,0) is asymptotically stable.

),,(/),,(/ yxGdtdyyxFdtdx ==

Definitions: DefinitenessLet Vbe defined on a domainD containing the origin. Then


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we make the following definitions.

Vis positive definite onD if V(0,0) = 0 and V(x,y) > 0 for allother points (x,y) inD.

Vis negative definite onD if V(0,0) = 0 and V(x,y) < 0 for allother points (x,y) inD.

Vis positive semi-definite onD if V(0,0) = 0 and V(x,y) 0for all other points (x,y) inD.

Vis negative semi-definite onD if V(0,0) = 0 and V(x,y) 0

for all other points (x,y) inD.

Example 1Consider the function

( )


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Then Vis positive definite on the domain

since V(0,0) = 0 and V(x,y) > 0 for all other points (x,y) inD.

( )22sin),( yxyxV +=

2/:),( 22


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Then Vis only positive semi-definite on the domain

since V(x,y) = 0 on the liney = -x.

( )2),( yxyxV +=

2/:),( 22


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Theorem 9.6.1Suppose that the origin is an isolated critical point of the

t t


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autonomous system

If there is a function Vthat is continuous and has continuousfirst partial derivatives, is positive definite, and for which

is negative definite on a domainD in thexy-plane containing(0,0), then the origin is an asymptotically stable critical point.

If negative semidefinite, then (0,0) is a stable critical point.

See the text for an outline of the proof for this theorem.

),,(/),,(/ yxGdtdyyxFdtdx ==

),(),(),(),( yxGyxVyxFyxVV yx +=&

V&

Theorem 9.6.2Suppose that the origin is an isolated critical point of theautonomous system


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autonomous system

Let Vbe a function that is continuous and has continuous firstpartial derivatives.

Suppose V(0,0) = 0 and that in every neighborhood of (0,0)there is at least one point for which Vis positive (negative).

If there is a domainD containing the origin such that

is positive definite (negative definite) onD, then the origin isan unstable critical point.

See the text for an outline of the proof for this theorem.


),(),(),(),( yxGyxVyxFyxVV yx +=&

Liapunov FunctionThe function V in Theorems 9.6.1 and 9.6.2 is called aLiapunov function


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Liapunov function.

The difficulty in using these theorems is that they tell usnothing about how to construct a Liapunov function, assumingthat one exists.

In the case where the autonomous system represents a physicalproblem, it is natural to consider first the actual total energy ofthe system as a possible Liapunov function.

However, Theorems 9.6.1 and 9.6.2 are applicable in cases

where the concept of physical energy is not pertinent.In these cases, a trial-and-error approach may be necessary.

Example 3: Undamped Pendulum (1 of 3)For the undamped pendulum system

( )sin/// uLgdtdvvdtdu


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use Theorem 9.6.1 show that (0,0) is a stable critical point, anduse Theorem 9.6.2 to show (, 0) is an unstable critical point.

Let Vbe the total energy function

and let

Thus Vis positive definite onD, since V> 0 onD, except atthe origi

Phase Plane for Linear Control System and Design

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