Panfilov(2001)_Introduction to Differential Equations of One and Two Variables

8/2/2019 Panfilov(2001)_Introduction to Differential Equations of One and Two Variables

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INTRODUCTION TO

DIFFERENTIAL EQUATIONS OF

ONE AND TWO VARIABLES

Alexander Panfilov


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INTRODUCTION TO DIFFERENTIAL EQUATIONS

OF ONE AND TWO VARIABLES

Alexander Panfilov

Theoretical Biology, Utrecht University, Utrecht

c

2001

1 October 2001


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Contents

1 Differential equations of one variable 4

1.1 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Differential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Geometrical image of solution.Phase portrait . . . . . . . . . . . . . . . . . . . . 6

1.4 Equilibria; stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.5 Global and local plan.Basin of attraction . . . . . . . . . . . . . . . . . . . . . . . 9

1.6 Systems with parameters.Bifurcations. . . . . . . . . . . . . . . . . . . . . . . . . 121.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Differential equations of two variables 15

2.1 General ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Vector field of an autonomous system . . . . . . . . . . . . . . . . . . . . . . . . 17

2.4 Null-clines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.5 Equilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Calculus of functions of two variables 24

3.1 Functions of two variables are their graphs . . . . . . . . . . . . . . . . . . . . . . 24

3.2 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Linear approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 Linearization of a system. Jacobian 28

4.1 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.2 Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

5 Linear system. General solution 33

5.1 One-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.2 Two dimensional system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2


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CONTENTS 3

6 Types of equilibria I. Saddle, node 39

6.1 Saddle; , or . . . . . . . . . . . . . . . . . . . . . . 39

6.2 Non-stable node;

. . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3 Stable node;

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

7 Types of equilibria II, Spiral 467.1 Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7.2 General ideas on equilibria with complex eigenvalues . . . . . . . . . . . . . . . . 48

7.3 General solution of a system with complex eigenvalues . . . . . . . . . . . . . . . 49

7.4 Center. " $ & ( . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

7.5 Stable spiral " 0 $ & ( 0

. . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.6 Non-stable spiral " 0 $ & ( 0 . . . . . . . . . . . . . . . . . . . . . . . . 54

7.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8 General properties of equilibria 57

8.1 Stability of equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

8.2 Express method for finding type of equilibrium . . . . . . . . . . . . . . . . . . . 588.3 Drawing phase portrait around equilibra using null-clines . . . . . . . . . . . . . . 61

8.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

9 Plan of qualitative study 64

9.1 Plan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

10 Limit cycle 72

10.1 Stable and non-stable limit cycles . . . . . . . . . . . . . . . . . . . . . . . . . . 72

10.2 Dynamics of a system with a limit cycle. . . . . . . . . . . . . . . . . . . . . . . . 7210.3 How do limit cycles occur? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

10.4 Example of a system with a limit cycle . . . . . . . . . . . . . . . . . . . . . . . . 74

10.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

11 Example of an exam 78

12 Extra exercises 80

13 Dictionary 85


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Chapter 1

Differential equations of one variable

1.1 Derivative

One of the most important notations in mathematics is the derivative of a function.

Definition 1 78 @ B D F " H 7 Q H B " T V X ` b ` D c e

`g

ce

`D g

`` D

Informally 7 q @ B D F shows the rate of change of a function at a point B D .

For example:

1. If x(t) is a distance traveled by a car as the function of time s , then H B Q H s gives the velocity

of the car.

2. If v@ s F

is the size of population as the function of time, then Hv Q H s

gives the rate of growth

of the population.

Geometrically the derivative 7 q @ B D F gives the slope of the tangent line to the graph of the function

at the point BD

.

f(x)

x

Figure 1.1:

4


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6 CHAPTER 1. DIFFERENTIAL EQUATIONS OF ONE VARIABLE

N

t

1 2 3

Figure 1.2:

1.3 Geometrical image of solution.Phase portrait

Let us represent the solution of (1.7) graphically (fig.1.2).

If the initial size of the population was different, for example @ F " or @ F " , etc., we

get other solutions of the equation (1.4) and will get other curves in fig.1.2. Let us analyze them.

An important characteristic of any line is its slope. It turns out that we can easily find the slope for

the solutions of (1.4): " H Q H s "

. We see that the slope depends only on (the size

of population). For example if "

the slope is 4*3=12. If "

, the slope is 8*4=32, etc.

Geometrically this means, that if we fix "

, then the integral curves 1,2,3 will have the same

slope at "

( "

).

We can use this information and present a qualitative picture of solutions of (1.4) using one

axis only, namely . First, let us consider the -axis, and at each point of this axis let us show

the slope of the solution of (1.4). We can do it in such a way: We see that this is not very nice.

N= 4 8 12 16 20

1 2 3 4 5 N

To improve the description let us think what it means, that the slope is 8? This means that thepopulation grows with the rate 8 individuals per day. Here we have the following two aspects:

qualitative, the population grows and quantitative, that the rate of this growth is 8. If we neglect

the quantitative aspect, we can represent the situation in the following way:

N

i.e. we show the growth of the population by an arrow which is directed to the right. Of course,

this is not a complete description of our system, but it gives a good idea about the behavior of

our system. It shows that if we start at some initial value of D , then will grow and the size of

the population will approach infinity. Note, that to obtain this result we have used the direction ofarrows in fig.1.4 only.

Now several definitions. Let us define the class of equations for which we can apply the

developed approach. If we consider a general differential equation `

"7 @ B

sF, then the slope

of the solution curve: "

`

"7 @ B

sF

can depend on B and s , and we cannot apply our

approach in this case, as different curves on a figure similar to fig.1.2 will have different slopes at


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1.4. EQUILIBRIA; STABILITY 7

B" B . But if 7 @ B s F does not depends on s explicitly, we can do the same steps, and get a similar

description.

Definition 2 Equation `

"7 @ B F is called the autonomous differential equation

Example

`

" B s v @ B F s v

`

" & v @ B s F v v s v

`

" } & v @ B F s v

We expect that our method will work for autonomous differential equations.

Definition 3 The problem `

"7 @ B F

B@ F " B D is called the initial value problem; Its solution is

called the orbit or trajectory

Definition 4 The collection of all orbits of a differential equation together with the direction ar-

rows is called the phase portrait

1.4 Equilibria; stability

Let us consider two differential equations

HB Q H s " B (1.9)

and

HB Q H s " B (1.10)

where (1.9) is a generalized equation for population growth (we extend it for negative B ), and(1.10) is the equation for growth of corn in the Netherlands. Let us sketch phase portraits for (1.9)

and (1.10). We can do it without finding a solution. In general, to sketch a phase portrait we need

to draw or arrows on the B -axis. The arrow means growth of B , i.e. positive slope of B @ s F ,

or `

. The arrow means decreasing of B , or `

For equation (1.9) HB Q H s " B

so the

graph of the right hand side of this equation is (fig.1.3) If B then H B Q H s , or we have the

arrow to the right , and if B

then HB Q H s

, or we have the arrow to the left . So, we get

the picture of fig.1.3a.

The interesting point here is B"

. Here HB Q H s "

and we have no velocity, or no direction

for the arrow. However, the behavior of our system at this point is trivial: B @ s F - for all s . This

solution means, that if the initial size of the population was zero, it will be zero forever. Such pointsof a phase portrait are called equilibria. They occur at the points where the rate of change is zero

( `

"

). As we will see later, equilibria are very important for qualitative study of differential

equations.

Definition 5 A point B is called an equilibrium point of H B Q H s " 7 @ B F , if 7 @ B F "


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x

x

x>0x


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1.5. GLOBAL AND LOCAL PLAN.BASIN OF ATTRACTION 9

of B . Therefore the graph of 7 @ B F in this case will be as in fig.1.5. If we sketch a phase portrait of

the equation H B Q H s " 7 @ B F with such 7 @ B F we get the flow as in fig.1.5 and B will be a non-stable

equilibrium.

++++ ++++ x x

x x

Figure 1.5:

The case 78 @C B F

gives the opposite picture, and the equilibriumB

will be stable.

Conclusion 1 The equation HB Q H s " 7 @ B F

has two generic (the most probable) types of equilibria:

the stable equilibrium point and the non-stable equilibrium point. The type of equilibrium can be

found from computation of the derivative of the right hand side of the equation at the equilibrium

point.

Let us apply our theorem for the examples (1.9) and (1.10). In the case of (1.9) the equilibrium

point is B " 7 @ B F " B

. Therefore the derivative of the function at the equilibrium point is

78

@B F " @ B F 8

"

. As this derivative is positive 78

@ F

we have a non-stable equilibrium.

In the case of (1.10) the equilibrium point is B " 7 8

@B F " 7 8

@B F "

,

i.e. we have a stable equilibrium, as we expected.

1.5 Global and local plan.Basin of attraction

Now we can formulate two ways of finding the phase portrait ofH

B Q H s " 7 @ B F

. To obtain fig.1.3we use the following plan.

Global plan.

1. Sketch the graph of 7@ B F

.

2. Draw the phase portrait. For that transform the points where 7 @ B F " to equilibria points,

the regions where 7@ B F

to right headed arrows ( ), and the regions where 7@ B F

to

the left headed arrows ( ).

However, if we know the stability of equilibria we can sketch the local phase portraits close to the

equilibria points without sketching the graph of the function 7@ B F

. We can use it, and formulate

the following local plan.

Local plan.

1. Find equilibria points of H B Q H s " 7 @ B F .

2. Study their stability and draw local phase portraits.


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3. Sketch the global phase portraits.

Note that the local plan is in general more complicated than the global plan. However, for differ-

ential equations with two variables the extension of the global plan is impossible, and we will only

be able to use the local plan.

Let us apply these plans to a logistic equation for population growth

Hv Q H s " v @ v Q F

(1.11)

This equation describes growth of population in a medium with limited resources. We can study

(1.11) for arbitrary values of parameters . However for simplicity let us fix "

and "

.

The equation becomes

Hv Q H s " v @ v Q F " @ Q F v @ v F

(1.12)

Let us find the phase portrait and the dynamics of the solutions of (1.12). First we use the global

plan.

1. The right hand side of our equation is @ Q F v @ v F " v @ Q F v

. The graph of this

function is a parabola, opened below with the roots v" v "

.

n

n

n

t

t

n

ab

c

f(n)

Figure 1.6:

2. The construction of the phase portrait is clear from fig.1.6.

The dynamics of the system for different initial conditions is shown in fig. 1.6b for the initial size

of the population v

, and in fig.1.6c for v

. We see that in the course of time the size

of the population becomesv

"

, i.e. the stable equilibrium pointv

"

is the only attractor ofour system.

Now let us sketch the phase portrait of (1.12) using the local plan.

1. Equilibria. We find them from the equation 7@ v F "

. In our case it yields

v @ v F "

and we have two equilibria v "

and v "


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1.5. GLOBAL AND LOCAL PLAN.BASIN OF ATTRACTION 11

a b

Figure 1.7:

2. Stability is given by the sign of 78 @ v F "

v . At the first equilibrium v "

78 @ F "

, i.e. the equilibrium is non-stable; at the second equilibriumv

"

78

@ F "

,i.e. the equilibrium is stable. The local phase portrait is shown in fig1.7a.

3. Global phase portrait is given in fig1.7b.

Note, that in this example the global plan is easier. However, as we will see later, the local plan

can be easily extended to two-dimensional differential equation.

Later, we will use the following property of a phase portrait which can be easily seen in fig.1.7.

Conclusion 2 The trajectory, which starts close to a non-stable equilibrium will go either to in-

finity or to the stable equilibrium. Also, the trajectory which ends at a stable equilibrium can

originate either at infinity or at a non-stable equilibrium.

Sometimes, differential equations have several stable equilibria (stable attractors). For exam-

ple, the model for the spruce bud-worm population (1.13) has the following phase portrait (fig.1.8).

H Q H s " 7 @ F

(1.13)

f(u)

u

u

u1 u2 u3

A2A1

Figure 1.8:

We see, that there are two attractors: A1 and A2 which correspond to bud-worm populations of

different size. We see that if the initial size of the population is D

, then the population

eventually reaches A1; if D

, then population eventually reaches A2. These intervals

are called basins of attraction.

Definition 6 The basin of attraction of a stable equilibrium point B is the set of values of B such

that, if B is initially somewhere in that set, it will subsequently move to the equilibrium point B .

In the case of fig1.8, the basin of attraction of the equilibrium

is the interval D

; the

basin of attraction of the equilibrium

is the interval D

. It is very important to know

basins of attraction of the system in order to predict its behavior.


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1.6 Systems with parameters.Bifurcations.

One of the aims of modeling in biology is to predict the behavior of a system for different con-

ditions. In that case we obtain differential equations with parameters. An example of such an

equation is the logistic equation for population growth (1.11). This equation depends on two pa-

rameters and , where accounts for the intrinsic relative growth rate and accounts for the

carrying capacity. We can study the dynamics of a population at various growth rates and carryingcapacities by changing the parameters and . Let us consider one example. Consider a slight

modification of equation (1.11) for a population which is subject to harvesting at a constant rate :

Hv Q H s " v @ v Q F

(1.14)

where is the harvesting rate.

Let us fix the parameters "

and "

(the same values as in equation (1.12)), and study

the effect of varying the harvesting on, the dynamics of the population.

Hv Q H s " v @ v Q F (1.15)

When " equation (1.15) coincides with equation (1.12) which was studied in fig.1.6. Now

assume that the harvesting is not zero. Let us plot graphs of v @ v Q F

for " "

" (fig.1.9a). The phase portraits for " " are shown in fig 1.9b. We see that

n

f(n)

a

h=0h=0.8

h=1.6

h=0.8

h=1.6

f(n)

n

b c

Figure 1.9:

at "

the behavior of the system is similar to the behavior of the system without harvesting:

the population eventually approaches the stable non-zero equilibrium. However the final size of

the population in this case is slightly smaller than for the population without harvesting (fig.1.6a)

At " the situation is different. We do not have a stable equilibrium anymore. The flow

is directed to the left and the size of the population decreases. In this simple model this means

the extinction of the population. The important question here is, what is the maximal possibleharvesting rate at which the population still survives. From the previous analysis it is clear that the

critical harvesting is reached when the parabola v @ v Q F

touches the v -axis (fig.1.9c).

To find this critical value we note, that just shifts the parabola v @ v Q F

downwards. For

example to plot the graph v @ v Q F

at "

we just need to shift the solid line in

fig.1.9 down, etc. Therefore, the situation of (fig.1.9c) occurs, when the shift equals the maximum


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1.7. EXERCISES 13

of the parabola v @ v Q F . To find the maximal value we find a point where the derivative

H7 Q H v "

H7 Q H v " Q @ v F @ Q F v " v Q "

v ` " Q 7 @ v ` F " @ Q F @ Q F @ Q F " Q

(1.16)

So the maximal value of @ Q F v @ v F

equals Q

and therefore the maximal harvesting is

" Q

If Q

the population will go extinct. At " Q

we have a qualitative change in the

behavior of our system. Such kinds of changes are called bifurcations. Bifurcations are studied in

the special section of mathematics: theory of dynamical systems.

1.7 Exercises

1. Find the derivative of the following functions. Evaluate its value at the given point.

(a) s s "

(b)

s s "

is a parameter (constant)(c)

" B B

s B "

is a parameter (constant)

(d) " B

`

(e) " `

(f) "`

`

s B "

(g) B

2. Sketch graphs of the following functions:

(a) " B

(b) " B B

(c) B "

(d) " B

B

(e) "

`

(f) "

`

`

, where "

. How the shape of the graph depends on the value of the

parameter ? Qualitatively draw a series of graphs when is increasing; when is

decreasing.

(g) B

"

(h)

B

"

3. Draw phase portraits of the following differential equation, using the local plan. Check the

result using the global plan.

(a) HB Q H s " B


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(b) H Q H s "

V

4. A model for the fast inward current in an excitable cell can be written as the following differ-

ential equation: HB Q H s " B @ B F @ B F

. Here B denotes normalized transmembrane

potential of the excitable cell.

Draw the phase portraits of this differential equation, using the local plan. Check the result

using the global plan. Show attractors and their basins of attraction. What is the value

of threshold of excitation of this system (minimal initial value of transmembrane potential

which causes activation of the system).

5. Consider a model population with logistic growth which is subject to harvesting at a constant

rate

Hv Q H s " v @ v Q F (1.17)

Find the maximal yield.

6. Consider a model where the harvesting @ v F is proportional to the size of the population:

Hv Q H s " v @ v Q F v

(1.18)

Find the maximal yield.

7. Compare the harvesting strategies (1.17) and (1.18). Which strategy is better. Why?

8. (Adler 1996) Suppose some species of organism cannot bread successfully when numbers

are too small or too large. Graph a possible function describing the rate of change of this

population as a function of population size, and draw the corresponding phase portrait.

9. Show that the model of population size:

HB Q H s "

B

B

B

satisfies the hypotheses in previous exercise.

Answer the following questions:

(a) Draw the phase portrait.

(b) How many equilibria do we have here? At which B ?

(c) For each equilibrium tell is it stable or non-stable?

(d) Describe the possible types of the asymptotic dynamics of the solution of this system

at s

. (e.g. B converges to ***, or B goes to infinity, etc.)

(e) List attractor/attractors and determine their basin/basins of attraction.


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Chapter 2

Differential equations of two variables

2.1 General ideas

Many biological systems are described by several differential equations. One of the most famous

examples of such models is the Lotka-Volterra system. It describes the dynamics of the predator(P) and prey ( ) populations in the following way:

H Q H s "

H Q H s " } H

(2.1)

where } H are the parameters.

Here the prey growth is given by

. The effect of the predator on the prey population is given

by

, i.e. the predator reduces the growth of prey. The effect of this reduction is proportional

to the prey and predator populations. The death rate of the predator is H

. The growth rate of

the predator is proportional to the available prey and to the size of the predator population.

We will also consider a modification of the Lotka-Volterra model, which includes competition

in the prey population (

).

H Q H s "

H Q H s " } H

(2.2)

Systems (2.1) and (2.2) have many parameters, which account for the specific properties of the

populations. Let us study the system (2.2) at some fixed parameter values ( "

"

"

}"

H" ). Let us also denote as B and as

HB Q H s " B B

B

H Q H s " B

(2.3)

System (2.3) has two unknown functions of time: the size of the prey population B@ s F

and the

size of the predator population @ s F

. What behavior can we expect here? In the case of an one

dimensional differential equation (1.4) we found that there are two types of solution: general, i.e.

for arbitrary initial size of population (which depends on one arbitrary constant) and the unique

solution of the initial value problem.

15


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16 CHAPTER 2. DIFFERENTIAL EQUATIONS OF TWO VARIABLES

In the two dimensional case (2.3) we will also have a general solution and the unique (par-

ticular) solution of the initial value problem. Because in the two dimensional case we have two

populations B@ s F

@ s F

, to find a unique solution we need to know the initial size of each of these

populations, i.e. B@ F " B D

, and @ F " D

. It is also reasonable to assume that the general

solution of (2.3) will depend of two arbitrary constants.

Next, how can we represent the solutions graphically? First, as in one dimension we can solve

an initial value problem (e.g. for B @ F " @ F " ) and plot the size of the prey and thepredator populations as functions of time (fig.2.1).

x

t

y

t

1 1

a b

Figure 2.1:

We see, that in the course of time, B and approach the stationary values B" "

In one dimension we presented the dynamics in terms of a phase portrait using only one axis.

In two dimensions we have two variables, therefore we need to use two axes to represent the

dynamics. To achieve this, let us consider a two dimensional coordinate system B

, and let us

use the B -axis for the variable B and the -axis for the variable y. Such a coordinate system is called

a phase space. Let us represent the trajectory from fig.2.1 on the B -plane. The initial size of

the populations were B @ F " @ F " . Let us put this point ( ) on the B -plane. At

the next moment of time we get other values for B and and we also put them on the B -plane,

etc. Finally, we get the following line (fig.2.2a) To show the dynamics of our system, let us draw

arrows which show the direction of the change of variables in the course of time. As the system

approaches the point (

), the arrows will be as in fig.2.2a. This trajectory is the first elementof the phase portrait. If we start many trajectories from different initial conditions we will get the

complete phase portrait of (2.3) (fig.2.2b). Note, that each trajectory represents a certain type of

0 1 2 x

0

1

2

y

0 1 2 x

0

1

2

y

Figure 2.2:

dynamics of B@ s F

@ s F

, which can be easily shown on time plots similar to fig.2.1.


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2.2. DEFINITIONS 17

The main aim of our course is to develop the procedure of drawing a phase portrait of a gen-

eral system of two differential equations. We will derive a plan similar to the local plan for one

dimensional equation, i.e. find equilibria, find the type of equilibria, draw the local phase portraits

and extend it to the global phase portrait. However, to achieve it we need to consider several im-

portant questions on the theory of two dimensional differential equations and elements of calculus

of functions of two variables.

2.2 Definitions

Consider a general system of two differential equations:

HB Q H s " 7 @ B

sF

H Q H s " @ B

sF

(2.4)

If the right hand side of (2.4) does not depend on time explicitly, the system is called autonomous:

HB Q H s " 7 @ B

F

H Q H s " @ B

F (2.5)

Initial conditions specify the values of B and at some moment of time (usually at s D " ):

B@ F " B D @ F " D

(2.6)

The system (2.5) with initial conditions (2.6) is called the initial value problem. It usually has a

unique solution.

The general solution of (2.5) depends on two arbitrary constants.

2.3 Vector field of an autonomous system

The two variables B in (2.5) can account for many different things such as sizes of population,

concentrations of chemical species, transmembrane potential, etc. Let us consider one important

physical interpretation of these equations. Let B and be the B coordinates of moving body. In

this case the phase space @B

F

is the same as physical space and the trajectory from fig.2.2 is just

the trajectory of motion of the body. What is the physical meaning of HB Q H s

and H Q H s

in (2.5)?

The time derivative gives the velocity, hence HB Q H s

is the rate of change in the B -coordinate of the

body, or the velocity of motion in the B -direction. Similarly, H Q H s

gives the velocity of motion in

the -direction. In other words we describe the total motion of a body as projections onto the B and

the -axes. You have had several examples of such motion in physics. For example the motion of a

body in the gravity field (ballistics)(fig.2.3a). In fig.2.3a the motion is considered as the motion inthe B -direction with the velocity

`and in the -direction with the velocity

. The total velocity

is the sum of these two vectors. Note, that the vector

shows the direction of the tangent line to

the trajectory of the body.

Now let us consider a similar interpretation of the solution of the general system (2.5) (fig.2.3b).

The solution of (2.5),(2.6) is a trajectory, similar to the trajectory of a cannon ball from fig.2.3a.


19/87


y

x x

y

Vx

Vy V

y

x

V

A

a b

Figure 2.3:

The velocity of motion at each point A will also have two components ` " H B Q H s and " H Q H s .

If we add these two vectors we will get the total velocity, or a vector tangent to the trajectory.

An important fact here is that the system (2.5) is autonomous, hence the velocities ` "

`

"

7@ B

F and "

" @ B

F are determined by the values of B only and do not depend on time.

Of course, the solution of (2.5) depends on time, but the velocity does not. This means, that if the

system arrives at some point A, it knows the velocity of its motion just from the coordinates of

this point. It is not necessary to know any previous dynamics of the system. For many biologicalsystems it looks quite reasonable. For example, for the system (2.3) this means that the rate of

growth of population depends only on its current size.

Conclusion 3 At any point @ B F of a phase space for an autonomous system (2.5), we can define

the vector

with the components @7 @ B

F

@ B

F F

. Such vectors will be tangent to the trajectories

of our system. We can find this vector field without a solution of our system, just from the right

hand sides of our system.

Let us draw the vector field for the system (2.3). The procedure is simple. 7 @ B F " B

B

B @ B

F " B Hence at B " " 7 @ B F " @ B F " ,

hence we get a vector with components (-1.5,0.25). If we draw it on a graph, we get the followingvector (fig.2.4a). However we do not need such a long vector. The most important thing for us

y

x

y

a bx

Figure 2.4:

is its direction. So, we just need to make the vector shorter, keeping the direction the same (i.e.

keeping the ratio ` Q

). We will get fig.2.4b. In fig 2.4b we have two representations of a vector:

(1) as an arrow (2) as two intervals in B and -directions. They are absolutely equivalent.

If we apply the same procedure in many points, we will get the following vector field of a

system (2.3) (fig.2.5)


20/87

2.4. NULL-CLINES 19

0 1 2 x

0

1

2

y

0 1 2 x

0

1

2

y

Figure 2.5:

2.4 Null-clines

To draw the detailed vector field of a system we usually need a computer. However, in many cases

we can sketch a qualitative picture using the simple method of null-clines.

Each vector of the vector field has two components

"@ `

F " @

`

F . Qualitatively

these components `

can be positive, or negative. So, we can have the following four cases

(fig2.6) Here we denote the horizontal component as a dashed line and the vertical component as

a solid line.

I IIIII IV

x0 x>0 y


21/87


and the direction of the vector field at this line is horizontal.

The boundary between cases I and III is given by equation (2.8) and between cases II and IV by

equation (2.8). In general equations (2.7),(2.8) give two (or more) lines on the B

-plane, which

separate the plane into several regions with different directions of vectors (cases I-IV).

These lines are called null-clines.

Definition 7 TheB

-null-cline (or

`

"

null-cline) is the set of points satisfying the condition7

@ B

F " . The -null-cline (or

" null-cline) is the set of points satisfying the condition

@ B

F "

.

To use the method of null-clines it is useful to note that at the B -null-cline the B -component of

the vector changes its sign and at the -null-cline the -component of the vector changes its sign.

Let us use these ideas and formulate the plan for finding the vector field using null-clines.

Plan of null-cline analysis for system (2.5)

We assume that on the B

-plane the B -axis is the horizontal axis and the -axis is the vertical

axis.

1. Draw the `

"

null-cline from the equation 7@ B

F "

using a dashed line and the

"

null-cline from the equation @ B

F "

using a solid line.

2. Choose a point in one of the regions in the B plane and find the B and the -components

of the vector field. Use the dashed line for the B component and the solid line for the

component. For finding the directions use the following rule: if 7 @ B F the B component

is directed as , if 7 @ B F it is directed as ; if @ B F the -component is directed

as , @ B F it is directed as .

3. Find the vector field in the adjacent regions using the following rule:

(a) change the direction of the dashed component of the vector field if to get to the adjacent

region you cross the dashed null-cline(b) change the direction of the solid component of the vector field if to get to the adjacent

region you cross the solid null-cline

(c) show the direction of the vector field on the null-clines.

Note, that instead of dashed and solid lines you can use lines of different colors. Then the last

step of this plan would be: change the direction of the component of the same color as the color of

the null-cline which we cross to get to the adjacent region.

Note, that although this plan works in most cases, there are cases when components of the

vector field do not change their sign at the corresponding null-cline. These are special so-called

degenerate cases (exceptions). We will not consider them in our course.

Example Find the vector field of the following system using null-clines:

HB Q H s " B

H Q H s " B

(2.9)

Solution. We follow our plan as follows


22/87

2.5. EQUILIBRIA 21

1. The `

" null-cline is given by the equation 7 @ B F " , i.e. B

" , or this is

equation for the parabola on the B -plane " B

. The graph is shown using the dashed

line in fig.2.7 The

"

null-cline is given by the equation @ B

F "

, i.e. B "

, or

" B

. The graph is shown by the solid line in fig.2.7

2. The null-clines separate the plane into 5 regions. Let us find the directions in region 2.

Let us choose a point from this region, for example exampleB

"

"

. At this point7

@ B

F " B

" " , hence the direction of the dashed arrow is to the

left . At the same point @ B F " B " , hence the direction of the solid

arrow is downwards .

3. Now we complete the picture. For example region 1 is separated from region 2 by the dashed

null-cline. Therefore we change the direction of the dashed component here. Similarly for

the other regions. We get the picture as in fig.2.7. Finally we show the vector field on the

null-clines.

x

y

2

3

4

51

Figure 2.7:

2.5 Equilibria

In the one dimensional case we first found the equilibria as stationary points of the vector field.

They were very important for our analysis, as they determine the phase portrait of our system.

In two dimensions we also have such important points. The main property of equilibria in one

dimension is they have zero velocity. In two dimensions the velocity has two components `

,

hence, it is zero if both components are zero, i.e. ` "

`

"

and "

"

. This property is

used as the definition of equilibria in two dimensions.

Definition 8 A point @ B F is called an equilibrium point of a system (2.5) if

7@ B

F"

@ B

F" (2.10)

Equilibria in two dimensions are also stationary points, i.e. the trajectory which starts at the equi-

librium will be there forever. In other words this trajectory contains just one point.


23/87


Example. Find the equilibria of the system (2.9):

HB Q H s " B

H Q H s " B

Solution To find the equilibria we need to solve a system of algebraic equations (2.10). By

equating the right hand sides of (2.9) to zero we get: B

" B "

. From the second

equation we see, that " B . We substitute it into the first equation and get B

B "

B@ B F " . The last equation has two solutions B " B " , therefore, as " B we get

two equilibria of (2.9): B " " and B " " .

Let us find a geometrical representation of the solution of the system of algebraic equations

(2.10). The geometrical image of the first equation is the line 7 @ B F " , which, as we know, is

the B -null-cline. The geometrical image of the second equation @ B F " , is the -null-cline.

The solution of system (2.10) gives the points satisfying both equations, i.e. it gives the points

which belongs to both null-clines, i.e. the points of intersection of the null-clines. So we found the

important fact.

Conclusion 4 Equilibria, are the points of intersection of the B and -null-clines.

Note, that this definition implies points of intersection ofdifferent null-clines. However, some-

times one null-cline can consist of several lines which intersect each other. To distinguish thepoints of intersection of different null-clines from the points of self-intersection of one null-cline

we introduced a special step in our plan of null-cline analysis for the system (2.5). In p.5(a) drew

the B -null-cline as the dashed line and the -null-cline as the solid line. In our approach the inter-

sections of the solid line with the dashed line give the equilibria of our system.

In one dimension our next step was to find local phase portraits close to the equilibria. We

will also do this in the two dimensional case. However, to achieve it we need to know more

about functions of two variables and their derivatives. Our next chapter deals with elements of

multi-variable calculus.

2.6 Exercises

1. Find equilibria of the system:

(a) `

"

" B B

(b) `

" B

"B

(c) `

" B B

"

B

(d) `

"@ B F B

"@ B F

2. Find the vector fields of the system using null-clines:

(a) `

"B

"B

(b)

`

"

" B

(c) `

" B

"B


24/87

2.6. EXERCISES 23

(d) `

"

B

"

3. Study the Lotka Volterra model:

H Q H s "

H Q H s " } H

(a) Find the vector field for " " } " H " using null-clines.

(b) Find equilibria of the model for arbitrary non-negative } H .

(c) Does the vector field qualitatively change if we change the parameters } H .

4. Find equilibria of the following Lotka Volterra model with competition in the prey popula-

tion. Determine for which parameter values both equilibria are non-negative.

H Q H s "

H Q H s " } H

}

H

5. Find equilibria of the following model of the protein-mRNA interaction.

H Q H s "

H Q H s " } H

}

H

6. Find equilibria of the following model of mathematical epidemiology. Determine for which

parameter values both equilibria are non-negative.

H Q H s " (

H Q H s " ( 0

0

(


25/87

Chapter 3

Calculus of functions of two variables

Let us study functions of two variables by comparison to functions of one variable.

3.1 Functions of two variables are their graphs

A function of two variables 7@ B

F

gives the rule of finding the value of function 7 , if we know the

values of the variables B and . For example, the the area of a right-angled triangle with the sides

B , and is given by the following function of two variables: 7@ B

F " B Q

. The other example

is the rate of growth of prey population in system (2.3): 7@ B

F " B B

B

. The graph

of the function of one variable " 7 @ B F

is a line on the B

-plane. To sketch the graph of the

function of two variables 7@ B

F, we must use a three dimensional space @

B

F : the B

-plane

for the values of independent variables B , and the third axis for the function value "7 @ B

F.

In such a representation the graph will be a surface in a three dimensional space. Fig.3.1a shows a

graph of the function 7@ B

F " B B

B

plotted by a computer.

Figure 3.1:

24


26/87

3.2. PARTIAL DERIVATIVES 25

3.2 Partial derivatives

The next step is the derivative of 7 @ B F . The main idea of finding the derivative of 7 @ B F is to

fix one variable, say B " B D . After that we will get a function of one variable only ( 7 @ B D F ).

Now, we can find the derivative of 7 @ B D F , as usual derivative of a function of one variable. For

example, 7 @ B F " B B

B . Let us fix B " B D " . We get the following function of

one variable: 7 @ F "

" . We can easily find the derivative now:

H7 @

F Q H " H @ F Q H " .

This type of derivative is called the partial derivative of 7 @ B F with respect to at B " .

We denote it as

7Q

` "

We can easily find such a derivative at B " , or at any other value of B . So, let us do it for an

arbitrary B" B D

. At B" B D

, 7@ B D

F " B D B D

B D

, and

7Q

` ` D "

@ B D B D

B D F Q

" B D

Here

@ B D F Q

" as we replaced B by a constant B D and the derivative of a constant is zero.

Similarly,

@ B D

FQ

"

, and

@ B D F Q

" B D

, as B D

is a constant

and the derivative of @ F 8 "

. Finally, let us make all these differentiations without replacing B

by BD

. We just keep in mind, that for our differentiation we treat B as a constant. So, we can write

the following

7Q

"

@ B B

B F Q

" B

This expression gives a partial derivative at any point B . It is denoted just as

7Q

.

Similarly, we can introduce a partial derivative of 7 with respect to B :

7Q

B . To compute

it, we fix (treat as a constant) and make the usual differentiations with respect to B . In our

example it gives:

7Q

B"

@ B B

B F Q

B" B

Here

@ B F Q

"

,

@ B

FQ

" B

, and

@ B F Q

"

as we fix .

Example. Find

Q

B and

Q

for "

V B

Solution

Q

B"

B

, as for

Q

B we fix , and

@ V B F Q

B" B

. Similarly,

Q

"

@

V B F Q

"

V B

, as B and hence V B

is treated as a constant.

The geometrical representation of a partial derivative is clear from fig 3.2. To compute

7Q

B we fix , i.e. we choose a curve on a surface representing the graph of our function. This

line will be in the direction of the B -axis (see fig.3.1b).

7Q

B gives a tangent line to this curve, or

it gives a slope of tangent line in the direction of theB

-axis. Note, that in general at each point ofa surface we can draw a tangent line in any direction

Other explanation. Let us imagine a surface 7@ B

F

as a mountain. Assume that we are at some

point on this mountain and we study its steepness. It is obvious that the slope will depend on the

direction in which we are moving. The

7Q

B gives the slope of the mountain in the B -direction

and

7Q

gives a slope for the mountain in the -direction.


27/87

26 CHAPTER 3. CALCULUS OF FUNCTIONS OF TWO VARIABLES

3.3 Linear approximation

The next question is the linear approximation of a function of two variables. For a function of one

variable linear approximation is based on the formula (1.1): 7 8 @ B D F " T V X ` b ` D c e`

g

ce

` g

``

. If we

write this formula without the limit, we get an approximate equality, instead of the exact equality:

78 @ B D F c e

`g

ce

`

g

``

. This formula will be more accurate if B will be closer to B D . Further we find:

7@ B F 7 @ B D F 7 8 @ B D F @ B B D F or

7@ B F 7 @ B D F 7

8

@B D F @ B B D F

(3.1)

Let us check this formula by approximating the function " B

at B D " . 7 @ B D F "

"

78 " B

78 @ F "

, hence 7@ B F @ B F

. At B"

this approximate formula gives

7@ F @ F "

. The exact value is 7@ F "

"

. So the error is

just 0.6%. However, if B"

7@ F @ F "

while the exact value is 7@ F "

. So

we see, that the approximate formula works good if B is close to BD

only.

Let us derive a similar formula for a function of two variables 7@ B

F. Let us assume, that

we know 7@ B

F

and its partial derivatives at some point BD

D

and we want to find the value of

a function at the close point B (fig.3.2). Let us move to the point B in two steps. Let us first

(xo,yo)(x,yo)

(x,y)y

x

Figure 3.2:

move from the point BD

D

to the point B D

, i.e. in the B -direction, and then from B D

to B , i.e.

in the -direction. We see, that on the first part of our motion is fixed at " D

, so the function is

7@ B

D F

and it depends just on one variable B . Let us denote @ B F " 7 @ B

D F

and use the formula

(3.1) to find the approximate value of the function @ B F

:

@ B F " 7 @ B

D F @ B D F

8

@B D F @ B B D F " 7 @ B D

D F @

7Q

BF @ B B D F

(3.2)

as

@ B D F " 7 @ B D

D F

and

8@

B D F

equals the partial derivative of7

with respect toB

,

7Q

B

, at thepoint B D D . So, we found an approximation for our function of two variables for the first part of

our motion fig.3.3. Now let us travel from B D

to B . Here B is fixed and changes from D

to

. We can easily make the same steps, and find the following formula similar to (3.2):

7@ B

F 7 @ B

D F @

7Q

F @ D F

(3.3)


28/87

3.4. EXERCISES 27

where

7Q

is the partial derivative of 7 with respect to at the point B D . Let us assume that

7Q

at ( B D ) is approximately equal to

7Q

at the initial point ( B D D ). (The validity of such

an approximation can be confirmed by a detailed analysis). The last step is to replace 7@ B

D F

in

(3.3) by its approximation (3.2). After that, we get the following approximation for the function of

two variables:

7@ B

F 7 @ B D

D F @

7Q

BF @ B B D F @

7Q

F @ D F (3.4)

This approximation is called linear, as the independent variables B are in the first power only

and we do not have the terms like B

, or B

.

Example Find the linear approximation for the function `

at the point BD "

D "

Solution. We use the formula (3.4) with 7 @ B F " `

and BD "

D "

.

7@ B D

D F "

e

D

Dg

"

7Q

B"

`

; at B"

"

7Q

B"

D

D

"

7Q

"

@

`

FQ

"

`

@B F Q

"

`

, at B"

"

7Q

"

D

D

"

Finally, `

B

.

AtB

"

" the approximate formula gives:

`

" . The

exact value of `

"

D

"

3.4 Exercises

1. Find partial derivatives of the functions. Evaluate the value of derivative at the given point.

(a)

`

for @B

F " B

at B" "

(b)

`

for @B

F " @ B

F

at B" "

(c) c

`

and c

for 7@ B

F " B

at B" "

(d) c

`

and c

for T @ B

F

2. Find a linear approximation for the function at the given point.

(a) 7 @ B F " B

at B " "

(b) 7@ B

F "

`

at B"

"

(c) 7@ B

F " B @ B F

at B"

"


29/87

Chapter 4

Linearization of a system. Jacobian

4.1 Linearization

Consider a general system of two differential equations:

HB Q H s " 7 @ B

F

H Q H s " @ B

F

(4.1)

We want to study it close to its equilibria points. For that let us approximate the functions of two

variables 7 @ B F and @ B F , using the linear approximation (3.4) and later solve the approximated

system and find the phase portrait close to equilibrium.

Assume that the system (4.1) has an equilibrium point at ( B ). This means (see (2.10)) that:

7@ B

F "

@ B

F "

(4.2)

Let us approximate 7@ B

F

close to the equilibrium using the formula (3.4):

7@ B

F 7 @ B

F @

7Q

BF @ B B

F @

7Q

F @

F

As we assumed that ( B ) is an equilibrium, 7@ B

F"

and we get

7@ B

F @

7Q

BF @ B B

F @

7Q

F @

F (4.3)

A similar approach for @ B

F

yields:

@ B

F @

Q

BF @ B B

F @

Q

F @

F

(4.4)

If we replace the right hand sides of (4.1) by their approximations (4.3), (4.4), we get the following

system:

`

"@

7Q

BF @ B B F @

7Q

F @ F

"@

Q

BF @ B B F @

Q

F @ F

(4.5)

28


30/87

4.1. LINEARIZATION 29

The system (4.5) is more simple than the original system (4.1), as the partial derivatives in (4.5)

are constants (numbers). So we can rewrite (4.5) as :

`

" @ B B F @ F

"} @ B B F H @ F

(4.6)

where

"

7Q

B "

7Q

} "

Q

B

H"

Q

. The partial derivatives here are constants,not functions, as they are evaluated at the equilibrium point B . We can simplify (4.6) even

more. For that let us introduce new variables:

" B B

"

(4.7)

Here are new unknown functions of s . Because they are connected to the old functions of time

B

in a very simple way, we can find equations for using the equation for B . For that, let

us first substitute (4.7) into the right hand side of (4.6)

`

"

"} H

(4.8)

Now let us find an expression for `

and

in terms of . As @ s F " B @ s F B

, then

H Q H s " H B Q H s

(here HB

Q

H s " because B is a constant). Similarly, H

Q H s " H Q H s. Finally,

we replace `

by

and

by

and get

"

"} H

(4.9)

System (4.9) is a linearization of the system (4.1). It is a system oflineardifferential equations.

The next question is, how is the system (4.9) connected to the original system (4.1)? To derive

(4.9) we made two steps: 1). We used the approximate formula for linearization (3.4). 2). Wechanged variables B to . Let us analyze each of these two steps.

As we discussed in the previous section, formulae (4.3),(4.4) work well if B are close to

B

. So we expect that the solution of (4.1) and the phase portrait of (4.1) will be similar to the

phase portrait of (4.6) close to the equilibrium point @B

F

. So, we expect that in the shaded area

in fig4.1a the system (4.1) will be equivalent to a system (4.6).

Next, let us compare systems (4.6) and (4.9). They are basically the same. We obtained (4.9) by

substituting (4.7) into (4.6). That is for any solution @ s F

@ s F

satisfying (4.9), there is the solution

B@ s F " @ s F B

@ s F " @ s F satisfying (4.6). Therefore, if we know the phase portrait of

(4.9) we can draw the phase portrait of (4.6) without any computation by just using simple shifts

in theB

and

-directions. Assume, for example, that we found a trajectory on the phase portrait ofthe system (4.9) which starts from the point

"

"

(fig.4.2a). Then we will have similar

line on a phase portrait of (4.6). As B@ s F " @ s F B

, the B -coordinate of this line will have the

same time dynamics as @ s F

, but the value of this B -coordinate will be larger (by the value B ).

Hence, the initial point of the trajectory will be B" B

, i.e. it is shifted by B units to the

right.


31/87

30 CHAPTER 4. LINEARIZATION OF A SYSTEM. JACOBIAN

x

y

x*

y* y*

x*

y

x

Figure 4.1:

y

xu

v

0.1

0.1

0.1

0.1

x*

y*

Figure 4.2:

Similarly, @ s F " @ s F

gives the shift of the trajectory by

units upwards. So geometri-

cally, if we know the trajectory @ s F

@ s F

of (4.9) we can draw a similar trajectory of (4.6) simply

by shifting @ s F @ s F by B units to the right and by units upwards (fig.4.2b). Finally, if it works

for one trajectory, it will work for all trajectories, so the phase portraits of (4.9) and (4.6) will be

the same in the shaded areas shown in fig.4.3.

Conclusion 5 System (4.9) close to the origin ( " " ) has a phase portrait similar to the

phase portrait of system (4.1) close the to equilibrium point @ B F . To find the phase portrait

of (4.1) close to equilibrium, we can first find a phase portrait of (4.9), and then shift it to the

neighborhood of the equilibrium @ B F .

4.2 Jacobian

To find the linearized system (4.9) we need to find the equilibrium point ( B ) and compute the

following derivatives of right hand sides of our system at this equilibrium:

"

7Q

B "

7Q

} "

Q

BH "

Q


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4.2. JACOBIAN 31

y*

x*

y

xu

v

Figure 4.3:

So, the system (4.9) can be written in a matrix form in the following way

"

(4.10)

where J is the matrix called the Jacobian

"

c

`

c

`

(4.11)

Example Find the linearization of the following system at the nontrivial equilibrium point (i.e.

at the equilibrium where B

"

"

).

HB Q H s " B

H Q H s " B

Solution In section 2.5 we found that this system has two equilibria @ F and @ F . The non-

trivial equilibrium is @

F. To find the Jacobian of our system we compute the partial derivatives

(4.11) and evaluate them at the equilibrium point. In our case 7@ B

F " B

@ B

F "

B .

7Q

B" B at the point @ F this derivative equals

7Q

B" " . Similarly:

7Q

"

Q

B"

Q

" , hence the linearization of our system at the point @ F

is

"

"

or in matrix form:

"

and the Jacobian is:

"


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32 CHAPTER 4. LINEARIZATION OF A SYSTEM. JACOBIAN

Our next step is to find the phase portrait of a linear system (4.9). After that we will be able

to draw a phase portrait of a system (4.1) close to the equilibria and then to extend it to the whole

space.

4.3 Exercises

1. Consider the system: `

"B

`

"

`

.

(a) Check that ( ) is an equilibrium point of the system

(b) Find the general expression for the Jacobian of this system

(c) Find the Jacobian at the point ( )

(d) Write the linearization of the system close to the equilibrium ( )

2. Find the equilibria of the following systems. Compute the Jacobian at the equilibria points.

(a)

`

"

B

"B

(b) `

"

" B B

3. Consider a modification of the Lotka-Volterra model, which includes competition in the prey

population (

).

H Q H s "

H Q H s " } H

where the parameters } H

and the variables

,

.

(a) Find all equilibria of this system.

(b) At which parameter values do we have a non-trivial equilibrium (i.e. an equilibrium atwhich both and are positive.

(c) Compute the Jacobian at each equilibrium point.


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35/87

34 CHAPTER 5. LINEAR SYSTEM. GENERAL SOLUTION

for a solution in some known class of functions which is known in advance from some preliminary

analysis. It is known, that such a class of functions for linear equations are exponential functions.

So we will look for a solution of (5.4) of the form B"

, where

and are unknown

coefficients. The main idea of the method of substitution is to find these unknown coefficients for

a particular system. Let us substitute B"

into (5.4). We find: `

"@

F

8 "

, or:

"

"

We can cancel

and

, and we get:

" "

(5.5)

So, we conclude, that if "

(or in the general case "

), then the left hand side of (5.4) equals

the right hand side, or that B"

with "

, (in general "

) is the solution of (5.4). Hence

the solutions of (5.4) are:

B"

B"

(5.6)

where

is an arbitrary constant. So, we got the same solution as in chapter 1, but using another

method.

5.2 Two dimensional system

Now, let us use the same approach for the two dimensional system (5.1). It turns out that the class

of functions in two dimensions will be the same as in one dimension, but because we have two

variables, we need to introduce different constants for B and , so our substitution will be

B"

`

"

B

"

`

(5.7)

where

`

are unknown coefficients. Let us make this substitution for a particular example:

`

"

B

(5.8)

Substitution for `

"

`

"

B"

`

"

gives:

`

"

`

"

`

(5.9)

we can cancel

, (but not

`

as in one dimensional case), and get:

`"

`

"

`

(5.10)

or in the matrix form:

`

"

`

(5.11)


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5.2. TWO DIMENSIONAL SYSTEM 35

We see that to find the solution of (5.8) we need to solve the problem (5.11). But (5.11) is the well

known problem of finding eigen values and eigen vectors of a matrix. You know how to solve it:

(a)We need to find eigen values from the characteristic equation:

s @ F "

s

} H

"

(5.12)

In our case it gives:

s

"@ F @ F "

" "

" $

" $ . Hence " "

(b)We need to find eigen vectors of . For that we need to solve the following system of

equations:

} H

`

"

(5.13)

where are the eigen values of the matrix . Let us substitute the first eigen value "

:

@ F

@ F

`

"

or

` "

` "

Then both equations give the same solution

` "

. To find an eigen vector we just put

to some number. For example if " , then ` " and we get the eigen vector

.

We can also use any other value for

. For example "

yields the eigen vector

.

In general, if

is an eigen vector, then

(where

is an arbitrary constant) is also an eigen

vector for the same eigenvalue . So, the general expression for the eigen vector of

" is:

`

"

(5.14)

Note, that if we use the other found eigen vector

in (5.14) we get the same result. This

is because

in (5.14) is an arbitrary constant and any vector given by the formula (5.14) with

can be obtained using the formula with the other eigen vector

`

"

at

some other value of

.


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Similarly we find the eigen vector corresponding to the other eigen value " :

`

"

or

` "

` "

` "

`

"

If we substitute the eigen vectors into the initial formula (5.7) we find the following solutions

of (5.8)

B

"

B

"

(5.15)

It turns out that the general solution of (5.8) is just the sum of these two solutions:

B

"

(5.16)

If we apply the same steps for a general system (5.1) we will get the solution in the form (5.2).

So, we were able to solve system (5.1). In the next two sections we will analyze solutions and

draw their phase portraits. However let us first consider a simplification of the procedure of finding

eigen vectors.

Express method for finding eigen vectorsLet us derive a formula for finding the eigen vectors of a system (5.12). As we know, to find the

eigen vectors we first need to solve equation (5.12) to find the eigen values. Then substitute the

found eigen values into the matrix and solve the system of linear equations. For the first eigen

vector this system is the following:

} H

`

"

or

@ F ` "

} ` @ H F "

(5.17)

It is easy to see that ` "

and "

gives the solution of the first equation:

@ F ` " @ F @ F @ F "


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5.3. EXERCISES 37

If we substitute these expressions into the second equation we get:

} @ H F @ F "

To prove that this expression is also zero, note that @H F @ F }

is just a determinant

s

} H

which is zero in accordance with the equation (5.12). Therefore ` "

and " give a solution of (5.17) and therefore an eigen vector corresponding to the eigen

value . Similarly we find the eigen vector corresponding to the the eigen value . The final

formulae are:

`

"

`

"

(5.18)

where are the elements of the matrix A (see equation(5.1)). Let us apply these formulae for

the system (5.8). It gives:

"

`

"

@ F

"

"

`

"

(5.19)

We see, that the first vector is the same as obtained earlier in (5.14). The second vector differs

from the vector

found earlier , but it is easy to see that if we multiply

by

we get

the vector

`

"

. So, in accordance with our previous discussion these vectors are

the same.

Note the formulae (5.18) do not work if both "

and "

. In this case we need to use

the standard method for finding the eigen vectors.

5.3 Exercises

1. Find the general solution of the following systems of ordinary differential equations:

(a)

`

"

B

(b)

`

"

B

2. Find the solution for the following initial value problem:

`

"

B

B@ F

@ F

"


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3. Two different concentrations of a solution are separated by a membrane through which the

solute can diffuse. The rate at which the solute diffuses is proportional to the difference in

concentrations between two solutions. The differential equations governing the process are:

H

Q H s "

@

F

H

Q H s "

@

F

where

and

are the two concentrations,

and

are the volumes of the respective

compartments, and is a constant of proportionality. If

" & s ,

" & s , and

" & s Q & v and if initially

" Q & s and

" , find

and

as

functions of time.


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Chapter 6

Types of equilibria I. Saddle, node

Consider a general system of two differential equations with constant coefficients:

`

" B

"} B H

(6.1)

As we know, the general solution of (6.1) is given by (5.2). We can use it and sketch a phase

portrait of this system. It turns out, that we can have several different (generic) types of equilibria

depending on the eigen values . As we know are the roots of the characteristic equation

(5.12), which is a general quadratic equation. Therefore, the roots can be real or complex numbers.

In this chapter we consider the case of real roots.

So, assume that the eigen values and are real. This yields the following three cases:

1. Eigen values have different signs ( , or ).

2. Both eigen values are positive ( )

3. Both eigen values are negative (

)

Note, that we do not consider the case when "

. This situation is quite rare and is not interesting

for us in this course.

6.1 Saddle;

, or

The system (5.8), which we solved in sec.5.2, had eigen values "

"

. Let us draw its

phase portrait. The general solution of this system is given by (5.16)

B

"

(6.2)

Because

are arbitrary constants, let us consider three simple cases, when one of these con-

stants or both of them are zero.

39


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40 CHAPTER 6. TYPES OF EQUILIBRIA I. SADDLE, NODE

1) If

"

" , then B " " and does not depend on time. So the trajectory is just

one point (0,0), which is the equilibrium point of the system (5.8).

2) If

"

"

any number, then B"

"

. This expression will give us a

line on the B

-plane. To see it, let us fix

"

. Then B"

"

. This is a trajectory

which starts at s"

at the point B"

D

"

"

. In the course of time B"

and "

will become more and more positive. We can easily find a shape of a trajectory for any

. On this

trajectory B "

"

, so we can exclude

from these expressions, and express

through B . To do it, we find from the first expression that

" B Q

. Then we substitute it to

the second expression and get: " B Q

. That is the trajectory for all

will be on a straight line

" @ Q F B

. Let us draw the trajectory for

"

. We know that it starts at the point @

F. In the

course of time B become more and more positive, but we also know that the trajectory always

stays on the straight line " B

. So we get the trajectory as in fig.6.1

y y y

x x

a b c

x 22

1 1

Figure 6.1:

Let us represent trajectories for other values of

. If

"

, our trajectory will start at

B" " "

. However, the dynamics will be the same: it will stay on the same line

" @ Q F B

and will move to the positiveB

. At

"

, or

"

we will get similartrajectories, but they will start closer and closer to the equilibrium point (0,0). So, we represent

them all as a left branch of a straight line " @ Q F B with the direction arrows to the positive

B

. Next, if

is negative, e.g.

"

we have the trajectory B"

"

. We see

that both B and decrease in the course of time. However, as we know from previous analysis

this trajectory will also stay on the line " @ Q F B

. So to draw it we need to draw arrows on

the line " @ Q F B pointing to the direction of negative B . Similar consideration for small

negative

" , or

" shows that all these trajectories can be represented as the

right branch of the straight line " @ Q F B

with arrows directed to the right. So, finally the case

"

" any numbergives the following part of a phase portrait (fig.6.1b).

Therefore, we find that the line on fig.6.1b is a graphical representation of the expression

B

"

(6.3)

What is direction of this line? In fig.6.1 we found it by excluding

from the expression (6.3).

However, it is easy to see, that the direction of this line is the same as the direction of the eigen


42/87

6.1. SADDLE;

, OR

41

vector

(fig.6.1c). (Formally it follows, from the fact, that the line given by (6.3) goes

through the origin, at

"

, and through the point @

F, at

" s "

). So we conclude,

that the graphical representation of (6.3) is a straight line which goes through the origin, and has a

direction determined by the vector

.

3)The third case is

" any number,

" . The solution in this case is

B

"

(6.4)

As in previous case we conclude, that all trajectories in this case will be located on a straight line

along the vector

and we just need to show the direction of flow along this line. In this

case, time dynamics is given by the function , which approaches zero, when s goes to infinity.

Therefore, independent of initial conditions (value of

) we will approach the point B"

"

,

hence the arrows will have the following direction (Fig.6.2).

x

y

a

2

4

y

x

a bFigure 6.2:

(here shows the head of the vector

which was used to draw the line).

Finally let us draw the phase portrait for arbitrary

and

. First, we put the lines from

fig.6.1 and fig.6.2a to the fig.6.2b. Next, let us consider one trajectory for which

"

"

,

for example:

"

"

. The solution in this case is given by:

B

"

(6.5)

The trajectory starts as the point B" " " "

(i.e. at (-0.2,0.1)).

In the course of time will become smaller and smaller, while

will grow. So the first term

will be small compared to the second term

, and the dynamics at


43/87


large s will be described by an approximate formula :

B

, so it will be close

to the flow (6.3) presented in fig.6.1b. The trajectory will be as shown in fig.6.2b. There will be

similar behavior for any other trajectories. The qualitative picture will be as in fig.6.2b.

Such a phase portrait is called a saddle point. It has the following important features: (1).There

is an equilibrium point at B " " . (2)There are two lines associated with eigen vectors of

our system (fig.6.1b,6.2a).These lines are called manifolds. The word manifold is widely usedin abstract mathematics, but in our case it just means the line. The manifolds in fig.6.1b and

fig.6.2a are different. If we follow the trajectory along the manifold in fig.6.1b the distance to the

equilibrium increases (see fig.6.1). On the contrary, if we follow the trajectory along the manifold

in fig.6.2a the distance to the equilibrium decreases. The manifold from fig.6.1b is called a non-

stable manifold. The manifold from fig.6.2a is called a stable manifold.

There is a very simple method to find stability of the manifold. The general equation for a

manifold is

B

"

`

(6.6)

where

`

is the eigen vector, which determines the direction of the manifold and is the

eigen value, which determines the flow on this manifold. There are two main types of behavior of

the function

fig.6.3. If

approaches zero, when s increases. If

grows to

infinity with increasing s .

a bt t

0

Figure 6.3:

The solution (6.6) in coordinate form is: B "

`

and "

. Hence, if , both

B

in the course of time approach , or the trajectory approaches an equilibrium point. In this

case we will have the case of fig.6.2a, or a stable manifold. If , then B grow to infinity and

we will have a non-stable manifold.

Conclusion 6 The equation (6.6) on a phase portrait gives a manifold in the form of a straight

line. This line goes through the origin and is directed along the vector

`

. This manifold is

stable if

and non-stable if


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6.2. NON-STABLE NODE;

43

Conclusion 7 Fig.6.2 shows the phase portrait of a saddle point. Such types of phase portrait

occur close to equilibrium, at which eigen values of the system are real and have different signs,

i.e.

, or

. The phase portrait of a saddle point has two manifolds

directed along the eigen vectors of the linearized system. One manifold is stable ( corresponding

to the negative eigen value of the system). The other manifold is non-stable (corresponding to the

positive eigen value of the system).

6.2 Non-stable node;

Let us draw the phase portrait for the case when eigen values are real and are both positive. The

general solution of the system is given by (5.2):

B

"

`

`

(6.7)

From the previous analysis we immediately conclude, that the phase portrait in this case hasthe equilibrium point at @ F and two unstable manifolds along the vectors

`

and

`

(fig.6.4a). Let us put that on the graph (fig.6.4b).

x

y y

x

a b

V1V2

Figure 6.4:

To complete the picture we need to add several trajectories which start between the manifolds.

For such trajectories

"

"

and both terms in (6.7) will diverge to plus or minus infinity.

So, we get trajectories as in fig.6.4b. Such an equilibrium is called a non-stable node.

Conclusion 8 If the eigen values of system (6.1) are real and both positive we

have an equilibrium point called a non-stable node. To draw a phase portrait at this equilibrium

we need to show two non-stable manifolds along the eigen vectors of system (6.1) and add several

diverging trajectories between the manifolds.


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6.3 Stable node;

The general solution in this case has the same form (6.8). The phase portrait will be similar to

fig.6.4, but because both manifolds will be stable. So we get a picture fig.6.5a

x

y y

x

a b

Figure 6.5:

If the trajectory starts between the manifolds @

"

" F

it will also approach equilibrium

as both terms in (6.7) will converge to 0, because

(fig.6.5b). Such an equilibrium

is called a stable node.

Conclusion 9 If the eigen values of system (6.1) are real and both negative we

have an equilibrium point calledstable node. To draw a phase portrait at this equilibrium we need

to show two stable manifolds along the eigen vectors of system (6.1) and add several trajectories

converging to the equilibrium (0,0).

6.4 Exercises

IFind general solution of the system. Sketch phase portrait. Show non-stable, stable manifolds and

several trajectories between the manifolds.

1.

`

"

B

2.

`

"

B

3.

`

"

B

4.

`

"

B


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6.4. EXERCISES 45

II. The following model describes the dynamics of HIV virus (

) and of the infected cells ( )

under the treatment by the drug lamivudine (Perelson s 1996):

H

QH s "

H Q H s "

(6.8)

here

are positive constants (parameters).

1. Find the general solution of (6.8)

2. Which type of equilibrium do we have here?

3. Put "

"

" and draw the phase portrait of (6.8)

4. On this phase portrait qualitatively show a trajectory starting from the point

"

"

.

Draw a qualitative graph of

@s F

for this trajectory using the

s axes and also using the

v @

F

s axes.

5. *Find the particular solution of (6.8) corresponding the following initial conditions:

@ F "

D

@ F " @

Q F

D .

6. *Using this particular solution propose a method for finding the parameters from the

experimental recordings of

@s F .


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Chapter 7

Types of equilibria II, Spiral

We start with a short review on complex numbers.

7.1 Complex numbers

Complex numbers were introduced for a solution of algebraic equations. It turns out that in many

cases we can not find the solution of even very simple quadratic equations. Consider the general

quadratic equation:

"

(7.1)

The roots of (7.1) are given by the well known formula:

"

$

"

$

(7.2)

where

"

(7.3)What happens with this equation if

? Does the equation have roots in this case? If (7.1) is a

characteristic equation of a system of differential equations (5.1), does the system has solutions if

?

Complex numbers help to solve such kind of problems. The first step is to consider the equation

"

(7.4)

Let us claim that (7.4) has the solution and denote it in the following way:

" $ & (7.5)

where&

"

(7.6)

Here & is the basic complex number which is similar to 8 8 for real numbers. Now we can solve

other similar equations. For example if

"

"

"

" & @ $ F " $ &

46


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7.1. COMPLEX NUMBERS 47

Similarly the equation

"

, has solutions " $ & . Note, that although we call & a number,

or a complex number, it is quite diffe

Panfilov(2001)_Introduction to Differential Equations of One and Two Variables

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