A Course in Stability and Equilibrium Properties of Differential Equations With Applications

8/22/2019 A Course in Stability and Equilibrium Properties of Differential Equations With Applications

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A Course in Stability and Equilibrium Properties

of Differential Equations with Applications

Final Year Project

January 2011.


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ii

Table of Contents

Abstract.......................................................................................................... iii

Lecture 1: I ntroduction to Dif ferential Equations

What is a differential equation? ........................................................ 1

Lecture 2: Stabil i ty of Linear Systems

What does stability mean?................................................................. 5

Asymptotic Stability.......................................................................... 12

Summary............................................................................................ 19

Lecture 3: Stabil ity of Equil ibri um Solutions

Introduction....................................................................................... 14

Examples............................................................................................ 16

Lecture 4: Appli cation: SIR Model

Application......................................................................................... 23

Basic SIR Model................................................................................ 24

Model with Treatment........................................................................ 27

Model with Quarantine...................................................................... 29

Discussion......................................................................................... 32

Conclusion..................................................................................................... 33

References...................................................................................................... 34

Abstract


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The purpose of this research is to provide explicit, concise learning material,

while including real world example on stability of linear systems and stability of

equilibrium solutions. The project will have the layout of four lectures, each

covering a detailed subgroup of stability. Each lecture will include definitions,

theorems, proofs, examples, and solutions to those examples. Real world

applications will be integrated to highlight its importance. When it is finished, it

should be sufficient to teach the topics I have outlined.

Being a student for the majority of my life, I understand the importance of

lecture notes. Using my knowledge of different styles and techniques of teaching,

I will create quality learning material. Described in a friendly and clear way, the

content will be easy to read and understand. The reader of this material should

understand the information as they go, and the content will be written in an

engaging style. An important aspect of the project is to make it easy for the

reader to understand what is being explained. One method which I personally

find helps understand the notes better is to include examples. Therefore, I will be

giving plenty of examples with solutions so the reader will understand the

material at ease. Throughout the course, real world examples will be provided

when the material can be seen. Knowing why and where the material is useful

informs the students how they could use it in reality, and I feel this is the key to

grabbing attention and creates interest in the topic.


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Lecture 1 - Introduction to Differential Equations

What is a differential equation?

A differential equation states how a rate of change in one variable is related to

other variables.

A differential equation is an equation with some derivatives of an unknown in it.

Differential equations are important in the scientific and technical professions

and they are used to represent rates of change or time-varying phenomena.

For example, they are used to calculate electric currents in electric engineering

and rates of change in chemical reaction in chemistry.

Anotherexample of a differential equation is Newtons Law of Cooling, which

claims that the rate at which the objects temperature decreases is proportional to

the difference between the temperature of the object and the temperature of the

location or surroundings.

Let T(t) denote the temperature of the subject and let Ts denotes the temperature

of the location. , Since (i.e. the surrounding temperature is lower than the objectstemperature) then T decreases.

Solution:

Here is a real life application of this:

Example:

A cup of coffee has temperature and is placed in a room oftemperature. After 5 minutes the coffee drops to.


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How many more minutes until the coffee drops to?Solution: [initial temperature]

[temperature at time t] [ =0 so ] We know [temperature of coffee after 5 minutes is 160] [put the value for c in and we get an equation] [working it out with some algebra]

[value for k]

So temperature at time : [T(t)=130 and put the value for k in] [get log of both sides, removes exp.] [simply solve each value to find t]

Conclusion:

It takes approximately 12 minutes for the coffee to drop in temperature from to .There are two types of differential equations, ordinary differential equations and

partial differential equations. The above example is an ordinary differential

equation (ODE). ODEs depend on one variable, meaning there is only one

independent variable. In the above example, the single independent variable is x.

PDEs have more than one independent variable. PDEs are represented with a rather than a d. An example of a PDE is

. Here we have twoindependent variables, x and y, but only one dependent variable, u.

However, while it is important to know the difference of both types of

differential equations, we will only focus on ODEs in this course.


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Consider the differential equation

Where and is a nonlinear function of .We cannot always solve this equation, but in most applications it is not necessary

to find solutions. We are not usually interested in finding values for orother solutions, rather we are interested in their qualitative properties. Oftennumerical solutions do not exist, but we can observe the behavioural properties.

For example, take and to denote two populations of competingspecies, at time t. The rates at which and grow are dependent on thedifferential equation

.

We want to find:

(a) If there exists values and at which both species coexist in a steadystate, where and are called equilibrium points of .

(b)If both species coexist in equilibrium, what impact would a suddenincrease in the population of one species have on the population of the

other? Perhaps both will remain close to the equilibrium values for any

time t or would a larger

cause a decrease in the size of

, or vice

versa?

(c)Suppose and have arbitrary values at time t=0, what patterns willemerge as t approached infinity? Will one species prevail over another or

will the species cancel one another out and end in a tie?

Throughout the next four lectures we will be studying and answering these three

questions.


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Example (1.1):

Find all equilibrium solutions of the following systems of differential equations.

Solutions (1.1):

For , is an equilibrium value if and only if c is a solution to .

Using the zero product rule, or, Multiplying both sides by 3, Computing the discriminant now,

Since is less than zero, there are no real solutions.Thus, is the only equilibrium solution.


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Lecture 2 Stability of Linear Systems

What does stability mean?

Stability is hugely important in all physical applications. A stable equilibrium

represents a behaviour which usually cannot be changed. Since we cannot

measure initial conditions, knowing if the system is stable can tell us what we

want to know. Numerical solutions are not always necessary. However, stable

equilibrium can help us predict a solution, since lots of solutions eventually tend

to a certain stable equilibrium point as the time goes to infinity.

Part (b), above, asks if both populations will remain close to their equilibrium

values for all future times. We want to know if each population will diverge

towards, converge away from or remain close to equilibrium as t approaches

infinity.

Let

be a solution of the differential equation

. We are

interested in learning whether is stable or unstable.If every initial state close to equilibrium leads to states consistently close to

equilibrium, then the equilibrium is stable.

Figure 2.1: Stable


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Similarly, if every initial state close to equilibrium does not lead to a state close

to equilibrium, then the equilibrium is unstable.

Figure 2.2: Unstable

Here is an example most people can relate to:

Consider a pendulum, like in an old pendulum clock, which is naturally pointing

downwards. If the pendulum was moved in any direction and it will swing back

and forth until it eventually returns to the starting point again. The pendulum is

an example of a stable equilibrium.

Figure 2.3: Pendulum with stable equilibrium

In these lectures, any mention of stability is a reference to the stability of the

system of differential equations. The system of equations is a model of the

physical behaviour of the objects of the simulation.
http://images-mediawiki-sites.thefullwiki.org/11/1/0/1/33244352464086306.png


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Definition:

Let be the state at time , and be the equilibrium at time . If(t) at (i.e. the initial state) starts close to and remains close to for all

, then the equilibrium is stable.

The solution is unstable if there exists an initial solution which startsnearat but does not remain close to for all future times.The solution is stable if for every there exists such that| | if | | , For every solution

of

.

Or in other words, for every there exists a such that whenever . If at least one solution does not remainclose, then is unstable.The value of depends on the value of, and it tells us how close to wemust start in order to remain within the error.

Before proceeding further we will look at the concept of length of a vector.

Let be a vector with n components, which can be real or complex. Thelength of x can be denoted as , where .

For example, if

,

then || . And if ,

then || . Why? Since , , and so wechoose the largest length, which is .

It is obvious that the length of a vector is based on the length of a number.

|| holds for any vector x and || only if .


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Next, observe that|| Also,

|| This highlights the meaning of length of a vector.

Theorem 2.1:

(a) Every solution of is stable if all eigenvalues of A havenegative real parts.(b) Every solution of is unstable if at least one eigenvalue

of A has a positive real part.

(c) Suppose all the eigenvalues of A have real parts and have no real parts. Let have multiplicity . Thismeans that the characteristic polynomial of A can be factored into the

form where all of the roots of have negative real parts. Then, everysolution of is stable if A has linearly independenteigenvectors for each eigenvalue . Otherwise, every solution

is unstable.

Proof 2.1:

Want will show that every solution is stable if the equilibrium solution is stable and that every solution is unstable if is unstable.Let be any solution of . Note that is a solutionof

. Therefore, if the equilibrium solution is stable, then

will be small if

is sufficiently small.

Consequently, every solution of is stable.


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This time, suppose that is unstable. There exists a solution which initially is very small. However, it grows larger and larger as t approaches

infinity. The function

is a solution of

. Also,

is

close to to begin with, but it diverges away from as t increases. As aresult, every solution of is unstable.(a)Every solution of is of the form . Let be the ij element of the matrix , and let be the

components of. Then, the component of is ,Suppose that all the eigenvalues ofA have negative real part. Let be the

largest of the real parts of the eigenvalues ofA. It is a simple matter to show that

for every number, with , there exists a number K such that|| For any a,b and c>b which are all positive numbers,there exists a positive constant K at any time , . is always

positive for

and c a positive number. Therefore,

|| Consequently, ,for some positive constants K and . Now, ||. Hence,|| .Let be given. Choose . Then, ||. Hence,

|| ||

Consequently, the equilibrium solution is stable.(b)Let be an eigenvalue of A with real part and let v be an

eigenvector of A with eigenvalue . Then, is a solution of for any constant c. If is real then v is also real and || .Clearly,

||approaches infinity as t approaches infinity, for any choice of


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, no matter how small. Therefore, is unstable. If iscomplex, then is also complex. In this case

is a complex-valued solution of . Therefore is a real-valued solution of

, for any choice of constant c. Clearly,|| is unbounded as t approaches infinity if c and either or is nonzero.Thus, is unstable.

(c) If A has

linearly independent eigenvectors for each eigenvalue

of multiplicity , then there exists a constant K such that . There, || for every solution of . It nowfollows immediately from the proof of (a) that is stable.On the other hand, if A has fewer than linearly independent eigenvectors witheigenvalue , then has solutions of the form

where . If , then || is unbounded as t approachesinfinity for any choice of . Similarly, both the real and imaginary parts of are unbounded in magnitude for arbitrarily small , if .Therefore, the equilibrium solution is unstable. If all the eigenvalues of A have negative real part, then every solution of approaches zero as t approaches infinity. This follows immediately fromthe estimate which we derived above. Thus, not only isthe equilibrium solution stable, but every solution of approaches it as tapproaches infinity. This type of stability is known as

asymptotic stability (which we will be looking at in the next lecture).


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Example (2.1):

Show that every solution of the differential equation

=

x

is stable.

Solution (2.1):

The characteristic polynomial of the matrix

A = x,

,Thus the eigenvalues of A are , , , .Therefore, by part (c) of Theorem 1.1, every solution of isstable.

Example (2.2):

Show that every solution of the differential equation

xis stable.

Solution (2.2):

The characteristic polynomial of the matrix A is

.So the eigenvalues of A are .Since at least one eigenvalues has positive real part, the solution

of

is unstable.

Note that (A-I) is , .


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Asymptotic Stability

An asymptote of a curve is a line such that the distance between the curve and

line approaches zero as they tend to infinity.

If the line and the curve are extended far enough, it would appear as though they

merge together.

In the below graph on the Cartesian coordinates, the x-axis and y-axis are the two

asymptotes of the function .The chart shows the corresponding values of x and y for different values of each

variable.

As the values of x become smaller and smaller, the values for y become larger

and larger. Alternatively, as the values of x become greater, the values of y

become smaller.

For example, if x is 10, then y is 0.1. If x=100, then y=0.01.

Figure 2.4: Asymptotes

Definition:

A solution of the differential equation is asymptoticallystableif it is stable, and if every solution (t) which starts sufficiently close to

(t) must approach as t tends to infinity. In particular, an equilibriumsolution of is asymptotically stable if every solution of which starts sufficiently close to at time not only

X Y

0.00...001 100...000

0.01 1000.1 10

1 1

10 0.1

100 0.01

100...000 0.00...001


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remains close to for all future times, but ultimately approaches at tapproaches infinity.

The asymptotic stability of any solution of is equivalent tothe asymptotic stability of the equilibrium solution .Example (2.3):

Determine whether each solution of the differential equation (1)

is stable, asymptotically stable, or unstable.

Solution (2.3):

The characteristic polynomial of the matrix is

Solving using the quadratic formula gives eigenvalues and Since they both have negative real part, we conclude that every solution of (1) is

asymptotically stable.


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Lecture 3Stability of Equilibrium Solutions

Introduction

Previously only the simple differential equation was considered. In thenext lecture, we will look at the equation

where is very small in comparison to x. We assume that

are continuous functions of which vanish for .If then the equilibrium solution of is . Nextquestion is whether this is stable or unstable. This equation can be solvedexplicitly, however if is very small, then is even smaller compared to .Therefore, is seems likely that the stability of the equilibrium solution of

should be determined by the stability of the approximateequation

. This is almost the case as the following theorem will show.Theorem 3.1

Suppose that the vector-valued function iscontinuous function of which vanishes for . Then,

(a)The equilibrium solution of is asymptoticallystable if the equilibrium solution

of the linearized equation

is asymptotically stable. Equivalently, the solution of is asymptotically stable if all the eigenvalues of A havenegative real part.

(b)The equilibrium solution of is unstable if atleast one eigenvalue of A has positive real part.


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Proof 3.1

(a)Any solution x(t) of can be written in the form

(2)

We must show that approaches zero as t goes to infinity. If all thereal parts of the eigenvalues of A are negative, then we can find positiveconstants K and such that|| and || ||.We can find anotherpositive constant such that

if .This follows immediately from our assumption that iscontinuous and vanishes at . Consequently, Equation (2) impliesthat

|| || || || provided || , for .

Then, multiplying both sides of this inequality by gives|| || || . (3)Let || be called represented by , then we have || . (4)

Now differentiate both sides of (4) with respect to t. However, it is not

possible to differentiate both sides explicitly while preserving the sense of

the inequality. We can overcome this by setting .Then || Or


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Multiplying both sides of this inequality by the integrating factor gives

,Or [ ||] .Consequently, [ ||] || ,So that || .Returning to the inequality (4), notice

|| || (5)as long as || , .

Now, if|| , then the inequality (5) ensures that || for all future time t. Consequently, the inequality (5) is true for all if

|| . Finally, we note from (5) that

|| ||and

|| approaches zero as t approaches infinity. Therefore, theequilibrium solution of is asymptotically stable.

From this proof, it is visible that the equilibrium solution of is

asymptotically stable if all the eigenvalues of A have negative real part.

Now, consider some examples of this theorem.

Example (3.1):

Consider the system of differential equations, and determine whether the

equilibrium solution , is stable or unstable.


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Solution (3.1):

Rewrite this system of differential equations in the form where

,

and

.

Assuming that the vector-valued function .is a continuous function of which vanishes for .

Now, find the characteristic polynomial, , of matrix A. Recall from lecture 2,

)Thus the eigenvalues are

and

The function g(x) satisfies the hypotheses of Theorem 3 part (b), since the

eigenvalues of A are 6 and 1. Hence, the equilibrium solution isunstable.

Example (3.2):


equilibrium solution , , is stable or unstable. .


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Solution (3.2):


, and

.The vector-valued function .is a continuous function of which vanishes for .Again, find the characteristic polynomial, , of matrix A.

, (10)Now, factor the quadratic polynomial in order to find the eigenvalues. To find

the eigenvalues, use the factors of the largest number in (10), which is 27, and

calculate if the factor is or is not an eigenvalue.

Taking 1 first;

So 1 is not an eigenvalue, since .

Next take 3:


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So 3 is an eigenvalue. is a factor of (10). So we can rewrite the quadratic equation including . Thus

So the eigenvalues are and The function satisfies the hypotheses of Theorem 3 part (b), since theeigenvalues of A are 3 and -3. Hence, the equilibrium solution isasymptotically unstable.

Example (3.3):


equilibrium solution , , is stable or unstable.

.

Solution (3.3):


, and

.The vector-valued function

.is a continuous function of which vanishes for .First, find the characteristic polynomial, p(), of matrix A.

,

,


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Thus the eigenvalues are , and The function satisfies the hypotheses of Theorem 3 part (a), since theeigenvalues of A are all negative with real part. Hence, the equilibrium solution

is asymptotically stable.Lemma 1:

Let have two continuous partial derivatives with respect to each of thevariables . Then, (11)where

is a continuous function of z which vanishes for

.Proof:

Equation (11) is derived straight from Taylors Theorem which states that each

component of can be written in the form

Where is a continuous function of z which vanishes for . Hence, ,where

(

)

.

Using Theorem 3 and Lemma 1, we can determine whether an equilibrium

solution is stable or unstable using the following method:

1. Let .2. Write in the form Az+g(z) where g(z) is a vector-valued

polynomial in

beginning with terms of order two or more.

3. Compute the eigenvalues of the matrix A. The solution isasymptotically stable if all eigenvalues of A each have negative real part.


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But the solution is unstable if at least one eigenvalues has positive real

part.

Example (3.4):

Find all equilibrium solutions of the system of differential equations (12)And determine (if possible) whether they are stable or unstable.

Solution (3.4):

The equilibrium points of (13) will be determined by two equations

and .The first of these two equations suggests that .Substitute this into the other equations, giving

.Next, find the value for y using this value for x.

.Hence and are the only equilibrium solutions of (13). Set and .


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Then

Also

Rewrite this system in the form

So the matrix Has eigenvalues

and

.

Hence, the equilibrium solution and of (12) is unstable.


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Lecture 4Application: SIR Model

In this lecture previously acquired information will be combined and applied in a

real world example. Differential equations are important in life, as they canpredict patterns of how events unfold.

We are going to study the spreading patterns of the swine influenza. In the winter

of 2009/10, the H1N1 influenza virus was the cause of 98% of all flu cases in

Ireland. This flu spreads from person to person, just like most other seasonal

viruses. It enters the body via the eyes, mouth or nose and is contracted from

being close to, from direct skin to skin contact, from indirect contactsfor

example touching where someone with swine flu previously touched, or from

respiratory droplets belonging to an individual with it, spreading to someone

without it. It can be caught from environmental surfaces which an infected

person may have coughed or sneezed upon.

Swine flu results in death to certain groups of people more so than others.

Babies, the elderly and those already suffering from other illnesses are more

prone to death. If a person has a weak immune system, swine flu can attack the

body more effectively. In health bodies, swine flu can turn to pneumonia

following a few days without treatment. It causes swelling around the lungs,

which reduces the capacity to breathe and possibly ending in death.

In this lecture, some analysis of different variations of SIR models will be

studied and the stability of each will be determined. The Jacobian Matrix will be

used to help decipher if each variation is stable, unstable or asymptotically stable.

Recall, the model is stable if all the real parts of the eigenvalues of the Jacobian

Matrix are less than zero. If one of the eigenvalues real parts is positive, it is an

unstable equilibrium. Also, the equilibrium is asymptotically stable if all the

eigenvalues are negative.


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Basic SIR Model

The population can be grouped into three categories:

S; susceptible: those at risk of catching the virus,

I; infected: those who have contracted the virus, and R; removed: those who have died, both from natural causes and virus

related causes. Also those who have recovered and are now immune.

Firstly, derive the differential equations for this SIR model in a period, where

treatment has yet to be discovered. In this model, assume each infected

individual can recover naturally over time. Since they recovered naturally, their

immune systems can now resist the influenza in the future, providing immunity.

Once a person becomes infected, their next destination is always going to be the

Removed category, from death or immunity.

An average member of the population makes contact, sufficient for contraction,

with others times per unit time, where N is the total number of individuals.The probability that an infected person is in contact, direct or indirect, with a

susceptible person is S/N; thus the number of newly infected individuals is:

The number of susceptible individuals can increase or decrease naturally by birth

(parameter) and death (parameter) rates respectively. However, it can also

decrease by members of the S category becoming infected and moving into the I

category, meaning the effective transmission rate is shown by the contraction

parameter. An ordinary differential equation for this information looks like:

. (13)The change in the number of infected people increases as newly-infected humans

in the S category cross into the I category, (SI). Fatal cases of the virus cause

the I category to decrease in size, (parameter). Infected individuals can also die

of natural causes, (parameter ). Finally, those who have recovered, (parameterr)

over time are also removed from this category. So the ODE is:

. (14)


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Finally, the change in the number of removed people only is always positive. It is

equal to the number of individuals from the S category who have died naturally,

(S), infected people who died from the flu, , and from natural causes, (I),and infected people who have recovered, (rI). Therefore the ODE is:

. (15)

The ordinary differential equation satisfies: Hence

As as long as .Consider the model over a short period of time in a closed population of constant

size. Therefore, let births, immigration and deaths all equal zero, and . This time, .Therefore, the new differential equations are:

(16)

(17) (18)The Jacobian is then:

.From (16), we know that either

(i) or (ii) I


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(i) When , the equilibrium is: Which means everyone alive is infected. Therefore, this equilibrium shows that

in this model, a susceptible-infected coexistence population is impossible.So the Jacobian for this susceptible-free equilibrium is:

.So, Therefore the eigenvalues are:

, and .Note that is always a real number since it represents the probability that asusceptible person becomes infected. It will always lie between 0 and 1.

Similarly, is the probability of an infected person dying from the flu, while r isthe probability of an infected person recovering naturally. Again, these will fall

between 0 and 1. Also, I is the population of infected people, which is a non-

negative whole number. Consequently, those are all real numbers.

Since all the eigenvalues are negative, it follows that the equilibrium is

asymptotically stable. Which means as time goes approaches infinity; infected

people will be the only ones to remain.

(ii). When I=0, we get the equilibrium: In this case, everyone is disease-free and infected people do not exist.

Select N and not S since N is the total population without disease.

So the Jacobian for this equilibrium is:

.and Therefore the eigenvalues are:

and .


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Knowing that , since is those who have caught the disease and is the total people who have died from the virus plus those who have recovered

naturally. Logically, the number of people who die from a disease or recover is

less than the number of people who previously have the disease. It is not possible

to have more people dying and recovering that the number who have it. Also,

from above, it is clear that , and r are real numbers.Since the characteristic equation always has a root with positive real part, it

follows that this infection-free equilibrium is always unstable.

Figure 4.1: Infected people do not exist. The population of the susceptible

community is 500,000 and remains at this since the disease is not present. The

equilibrium is unstable.

Model with Treatment

Suppose now there is a treatment to cure infected bodies.

The treatment would allow infected individuals to return to theirsusceptible state.

Once cured, they are vulnerable; however they are far less likely tocatching the disease again, as the cure does not mean full immunity.

The treatment will ensure the death rate by the virus will lower.


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Again, there will be the same three categories:

S; susceptible: however this category also includes people cured from theinfection by the new treatment,

I; infected, and

R; removed.

As a consequence, new changes to the Susceptible and Infected differential

equations occur.

The new change in the number of susceptible individuals, accounts for the

infected people who have been cured (parameter c). Those individuals return to

the S category following their healing. The ODE is:

. (19)The change in the number of infected allows for the return of infected people to

the S category. So the ODE is: . (20)There are no new changes for the Removed differential equation. Therefore the

ODE remains as:

. (21)This model can be illustrated as:

As in the previous model, if , then .Again, consider our model in the short term, letting and . Therefore,the new differential equations are: (22)

(23)

(24)


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Thus, from (23) it is clear that I=0.

So the Jacobian is then:

.

Then

,so .The root

is positive since

, r and c are subsets of

. There

exists one positive eigenvalue with real part. Therefore the equilibrium is

unstable.

Then

So

Since all the eigenvalues with real parts are negative, it follows that theequilibrium is asymptotically stable.

Model with Quarantine

Next, consider a model where a partial quarantine is introduced. Infected

individuals would be isolated from other humans to contain the spread of the

contagious virus. This is one precaution which would limit the contamination,

and eventually the virus would cease spreading and should result in an endemic.

In this model there would be slight tweaks to the previous model.

The quarantined area would consist of infected peoples own homes, orbedrooms. Only those infected would be allowed to stay in these areas

while they recover.


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People would return to full health following a period in quarantine. Theflu would die off with aid from treatment or medication.

Recovered people who are in quarantine will become removed from thiscycle, and immune to the influenza.

An ordinary differential equation for this information looks like: . (25)Where represents births, represents deaths and is the positive parameter,where susceptibles become infected.

. (26)

This is the change in the infected population. It consists of those infected in the

last time unit (parameter), minus those who have died from natural causes(parameter), minus those who have died from virus related causes (parameter), minus those who have recovered naturally or with help from vaccinations(parameter), and minus those who have gone into quarantine (parameter).

. (27)This is the change in recovered people, and it is found by adding the number ofsusceptible people who have died naturally (), the number of infected peoplewhom have died naturally (), the number of infected people who have died as aresult of becoming infected (), the number of people who have recovered fromthe infection (, and the number of people who have been cured and haveexited quarantine (

), together.

Note that once a person is removed or has recovered from the influenza, they

become immune. Once they contract the virus, their immune system develops

sufficient resistance to avoid future infection of the same microbes.

There is a new ODE for this model; it will be for the quarantine population. It is

calculated simple by subtracting the number of people who exit quarantine (

)

from those who have entered quarantine ().


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.Again, if , then .Consider our model in the short term, letting

and

. Therefore, the

new differential equations are: (28) (29) (30) (31)And again, we see from (28) that either or .The Jacobian is a square 4x4 matric this time:

.When ,

So

.

So the eigenvalues are,

,

,

, and


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Therefore, since all the eigenvalues are negative, the equilibrium of this model is

stable.

Now find the equilibrium status for when

.

So

Therefore, the eigenvalues of this infection-free model is stable since the

eigenvalues are non-positive, that is negative or zero.

Discussion

While explicitly results were calculated regarding the stability of different SIR

models, further inputs are necessary to calculate the effect of such models.

Assigning numerical values to the different parameters would allow the real

solution to be found. Using figures, which could be found of various statistic

websites, would allow the reader to derive their own values for the parameters

and then input them.

However, the motivation behind these lectures is to be capable of deriving the

differential equations and specific models determine their equilibrium stability.

The statements below can be used to decide the stability:

1. Asymptotically stable if all eigenvalues have real part< 0.

2. Stable if all eigenvalues have real part 0.3. Unstable if the real part

> 0 for at least one eigenvalue.


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- 33 -

Conclusion

The lectures provided show how to formulate a system of differential equations

and, using the Jacobian Matrix, find the stability of any given system. They

provide sufficient techniques for working out stability of any sort of system of

ordinary differential equations.

This SIR model on Swine Flu shows the broad flexibility of mathematical

modelling. There exists many variations of the SIR model, such as SIS and SEIS.All these different models can be discussed and written up. As well as variations

in the models, there is an endless list of biological and physical applications

which can also be drafted. It is obvious that stability of equilibrium solutions and

linear systems is highly important and widely used outside of the class room.

The inclusion of real world examples shows the user how the material being

studied is relevant. It is often asked in maths class by sceptical students, why do

we need to know this when we will never use it again? It is obvious that the

answer to this question is that you may need it once again when you enter the

real world of work. An important aspect of these lectures is the relation between

theorems and applications.

Next in the line of lectures, would be to connect this data with a computer class,

developing a MatLab program to solve the different systems of differential

equations. This would allow the user to input value for the parameters and create

graphs to represent the models.


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References

Jordan, D.W., and Smith, P. (1999)Nonlinear Ordinary Differential Equation

An Introduction to Dynamic Systems (Third Edition). Oxford: Oxford University

Press.

Braun, M. (1979)Differential Equations and Their Applications (Second

Edition). New York: Springer-Verlag.

Ellner, S.P., and Guckenheimer, J. (2006)Dynamic Models in Biology. Oxford:

Princeton University Press.

Murray, J.D. (1993)Mathematical Biology (Second, Corrected Edition). Berlin:

Springer.

University of British Columbia Department of Mathematics (2012)Equilibrium

Solutions and Stability [online]

http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/stability.htm

l Accessed 20 October 2011.

Munz, P., Hudea, I., Imad, J., and Smith, R.J. (2009) When Zombies Attack!:

Mathematical Modelling of an Outbreak of Zombie Infection [online].

http://www.mathstat.uottawa.ca/~rsmith/Zombies.pdf Accessed 19 October

2011.

Health Service Executive (2011) Winter Flu Advice [online]. Dublin: Health

Service Executive. http://www.hse.ie/eng/services/swineflu/ Accessed 16

January 2012.

FluCount (2012) Worldwide Statistics of the H1N1 Influenza A Pandemic

[online] http://www.flucount.org/ Accessed 16 January 2012.
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/stability.html%20%20%20Accessed%2020/10/2011http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/stability.html%20%20%20Accessed%2020/10/2011http://www.hse.ie/eng/services/swineflu/http://www.flucount.org/http://www.flucount.org/http://www.hse.ie/eng/services/swineflu/http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/stability.html%20%20%20Accessed%2020/10/2011http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/stability.html%20%20%20Accessed%2020/10/2011

Documents

A Course in Stability and Equilibrium Properties of Differential Equations With Applications