Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 106 ppt

Residing in a large susceptible population is a small group of people with a fatal infectious disease... We make the assumption that the population remains fixed during the time interval

Wheredxdt = 0 but dydt = 0, only y is changing with t so the trajectory’s tangent line is vertical The sign of dydt indicates how y changes with t; we display this information

by orienting the vertical tangents up or down

Wheredydt = 0 butdxdt = 0, only x is changing with t so the trajectory’s tangent line is horizontal The sign ofdxdt indicates how x changes with t; we display this information

by orienting the horizontal tangents right or left

Where dx

dt = 0 and dydt = 0 simultaneously, the system is at equilibrium Equilibrium points are the points of intersection of nullclines for whichdxdt = 0 and those for which dy

dt = 0

The nullclines partition the phase plane into regions in which neitherdxdt nordydt changes sign In each region we’ll have one of the following cases:

dx

dt >0, dy

dt >0

as t increases, both x and y increase

dx

dt >0,dy

dt <0

as t increases,

x increases and y decreases dx

dt <0, dy

dt >0

as t increases,

x decreases and y increases

dx

dt <0,dy

dt <0

as t increases, both x and y decrease Sketch possible trajectories using the information gathered

FactThe limit point of a trajectory must be an equilibrium point By this we mean the following Suppose limt →∞x(t )and limt →∞y(t )both exist and are finite Denote the limits by A and B, respectively Then (A, B) must be an equilibrium point (Verify that this is indeed the case in Examples 31.24 and 31.25.)

Note the analogy to the case of solutions to autonomous differential equations If

y1(t )is a solution to an autonomous equation and limt →∞y1(t ) = L, where L is finite, then y(t) = L is an equilibrium solution

The slope of a trajectory at any point is dydx =dy/dtdx/dt =f (x,y)g(x,y) evaluated at that point (provided f (x, y) = 0) Sometimes explicitly solving for a relationship between x and

yis enlightening

EXAMPLE 31.26 Analyze the following system of differential equations, sketching solution trajectories in

the xy-plane

dx

dt = y dy

dt = −x

The trajectories are horizontal wheredydt = −x = 0, i.e., along the y-axis

The trajectories are vertical wheredxdt = y = 0, i.e., along the x-axis

We can figure out the direction in which the trajectories are traveled along the nullclines by looking back at the original equations For example, dxdt = y, so on the part of the y-axis with y > 0 we know dx

dt >0; the trajectories are traveled from left to right, x increasing with t On the section of the y-axis with y < 0,dx<0 so the trajectories are traveled from

right to left, x decreasing with t Vertical and horizontal tangents are oriented as shown in Figure 31.23

The only equilibrium point is the origin

y

x

Figure 31.23

The nullclines partition the plane into the four regions labeled In each region we look

at the signs ofdydt anddxdt to determine the basic direction of the trajectory

Region I: dx

dt = y > 0

x increases

,dy

dt = −x < 0

y decreases

Region III: dx

dt = y < 0

x decreases

,dy

dt = −x > 0

y increases

Region II: dx

dt = y > 0

x increases

,dy

dt = −x > 0

y increases

Region IV: dx

dt = y < 0

x decreases

,dy

dt = −x < 0

y decreases This information is collected in Figure 31.24

y

x

Figure 31.24

At this point we have some idea of what the trajectories look like However, we are not sure, for instance, if they form closed circles, or ellipses, or if they spiral in or spiral out

We can look atdxdyand try to solve the resulting differential equation to uncover the shapes

of the trajectories

dy

dx =

dy dt dx dt

=−xy , y = 0 so

dx =−xy Separate variables and solve

y dy = −x dx



y dy =



−x dx

y2

2 = −x

2

2 + C1⇒ x2+ y2= C Therefore, the trajectories in the xy-plane are circles centered at the origin; the picture in the phase plane is given in Figure 31.25

y

x

Figure 31.25

Through every point in the xy-plane there is one trajectory passing through that point Each trajectory is a circle, except for the origin, which can be thought of as a degenerate circle with radius 0 As t increases, the point (x(t), y(t)) traces out a circle in the clockwise direction

REMARK The system of equations in Example 31.26 can arise when modeling the vibrations

of a frictionless spring Consider a spring and attached block positioned as shown in Figure 31.26 Let x give the position of the block (of negligible weight) attached to the end of the spring Suppose we stretch the spring by pulling the block from its original position (x = 0) out to x0and release the block

x0

x = 0

Figure 31.26

The block will oscillate back and forth about x = 0 and, in the absence of friction, will

repeat its motion ad infinitum We expect the graph of x versus t to look like Figure 31.27

on the following page

x0 x

t

Figure 31.27

Let y = the velocity of the block Then y =dxdt; velocity is the derivative of position with respect to time By Newton’s second law we know that force = (mass) · (acceleration)

We can denote the mass of the block by m and its acceleration byd2x

dt 2 ordydt Hooke’s law, an experimentally derived law, tells us that the force exerted by the spring is proportional to the displacement from its equilibrium length Putting the two laws together gives − ∝ x = mdydt

or dydt = −kx for some k > 0 Acceleration, dydt, is proportional to x but opposite in sign The proportionality constant is determined by the mass of the block and the nature of the spring (Think through some special cases to assure yourself that the equationdydt = −kx makes sense in terms of the spring.) The frictionless block and spring system can therefore

be modeled by the system of differential equations

dx

dt = y dy

dt = −kx

For convenience let’s assume that k = 1 Then the equations describing the motion become

dx

dt = −x

Look back at the phase-plane diagram (Figure 31.25) to see how the motion of the spring is reflected in this picture If we pull the block out a distance x0and simply let it go, giving it no initial velocity, we should look at the point (x0, 0) in the phase plane Follow the trajectory around in the direction of the arrow and note how the position of the block (the x-coordinate) and the velocity of the block (the y-coordinate) change with time

 x(0) = x0

x(0) = 0 (a) Sketch a possible graph of x versus t for the spring example, considering the force of friction and assuming the block vibrates back and forth several times

(b) Now, letting y =dxdt = velocity, sketch the trajectory in the xy-plane (In the next section

we will deal analytically with the friction issue.)

The answer to Exercise 31.8 appears at the end of the section.

Returning to Epidemic Model B: A Contagious, Fatal Disease

EXAMPLE 31.27 Consider the epidemic model that was used to motivate this discussion Residing in a large

susceptible population is a small group of people with a fatal infectious disease An infected

individual can immediately infect others, and any person not infected is susceptible We make the assumption that the population remains fixed during the time interval in question except for deaths due to the disease I(t) = the number of infected people at time t, and S(t ) = the number of susceptible people at time t

dS

dt = −rSI , where r > 0

As the susceptible class loses members, the infected class gains members but it also loses people due to death from the disease

dI

dt = rSI − kI , where r, k > 0 Analyze the system of equations

dS

dt = −rSI dI

dt = I(rS − k) r, k > 0, sketching solution trajectories in the SI -plane

the vertical axis Then the slopes of the trajectories are given bydSdI =dS/dtdI/dt

The trajectories are horizontal wheredIdt = I(rS − k) = 0, that is, for I = 0 or S =kr The trajectories are vertical wheredSdt = −rSI = 0, that is, for S = 0 or I = 0 The system is at equilibrium if and only ifdIdt = 0 anddSdt = 0 simultaneously S cannot simultaneously be 0 and kr, so the equilibria for the system are at I = 0 Every point

on the S-axis is an equilibrium point

The nullclines are S = 0, I = 0, and S =k

r These partition the first quadrant into two regions, I and II, as indicated in Figure 31.28

I

S

k r

Figure 31.28

sign ofdS

dt: − −

We know that the solution curves go in the direction of the arrows drawn in Figure 31.28 Further information about the shape of the trajectories in the SI -plane can be obtained by looking at dIdS

dI

dS =dI/dt dS/dt =I(rS − k)

rS − k

k

rS

The slope is a function of S; the trajectories are vertical translates (As an exercise, solve for I in terms of S.) The second derivative of I with respect to S is given by d2I

dS 2 = −rSk2, where r and k are positive constants d2I

dS 2 <0, so the trajectories are concave down See Figure 31.29

(S0, I0)

I

S k

r

Figure 31.29

REMARKS Observe that if S0<kr, then I immediately decreases toward zero; the disease leaves the population without becoming an epidemic On the other hand, if S0>kr, then the disease will spread and the number of infected people will increase until S drops tokr Only when S falls below the threshold value ofkr does I begin to decrease The disease does not die out completely for lack of a susceptible population but rather for lack of infected people Generally, some individuals will remain who have not caught the disease

The same system of differential equations that were set up in the previous example can also be used to model an epidemic such as measles or chicken pox, where instead of the disease being fatal, sickness confers immunity upon those who recover Once an individual

is infected he cannot become reinfected, so after leaving the “infected” class he does not rejoin the “susceptible” class but belongs to a “recovered” class

EXAMPLE 31.28 In an isolated community of 800 susceptible children, one child is diagnosed with chicken

pox Suppose the spread of the disease can be modeled by the system of differential equations

dS

dt = −0.001SI dI

dt = 0.001SI − 0.3I (a) What is the maximum number of children sick at any one moment?

(b) According to our model, how many susceptible children will avoid getting chicken pox while the epidemic runs its course?

when S =kr, that is, whendSdI = 0 So I is maximum at S = .0010.3 = 300 To find I when

S = 300 we need an explicit relationship between I and S I = 800 − S because some infected children have already recovered

dI

dS =dI/dt dS/dt =0.001SI − 0.3I

0.3 0.001S = −1 +300

S

Separate variables and integrate to obtain I as a function of S.

dI =



−1 +300S

 dS



dI =

 (−1)dS + 300

 dS S

I = −S + 300 ln S + C, S >0, so |S| = S

Use the initial conditions, I = 1 and S = 800 when t = 0, to solve for C

1 = −800 + 300 ln(800) + C, so C = 801 − 300 ln(800) ≈ −1204.38 Then

I(S) ≈ −S + 300 ln S − 1204, and

I (300) ≈ −300 + 300 ln 300 − 1204 ≈ 207

There were at most about 207 children sick at one time This is just over a quarter of the susceptible population

(b) As pointed out in Example 31.27, the epidemic ends not for lack of susceptible children but for lack of infected children Therefore, we set I = 0 and solve for S Because the equation involves both S and ln S we can only approximate the solution Using

a calculator, computer, or numerical methods, we find that S ≈ 70 At the end of the epidemic about 70 children will still be susceptible to chicken pox; 730 children will have caught the disease

In some sense, in the previous two examples it is the ratio kr that governs the course

of the epidemic The likelihood of the disease being passed from one individual to the next is reflected in r For a disease such as chicken pox, communities sometimes attempt

to raise the value of r (to confer immunity before adulthood) whereas for a fatal infectious disease, like AIDs, communities sometimes educate to lower the value of r When dangerous epidemics break out in livestock populations, farmers and ranchers often remove sick animals, effectively raising the value of k, in order to end up with more uninfected animals

Recurrent Epidemics

Many diseases recur in various populations with some regularity For instance, in London

in the early 1900s, measles epidemics recurred approximately every two years In 1929, when the mathematical biologist H E Soper was attempting to model this cyclic measles epidemic, he dropped the assumption that the population remains fixed for the time interval being observed Instead he assumed that the population of susceptibles grows at a constant rate µ and arrived at this system of differential equations:

dS

dt = −rSI + µ dI

dt = rSI − kI for r, k, and µ positive constants

In fact this system of equations does not predict recurrent outbreaks of the epidemic Instead

it predicts that the disease will reach a steady state level since the cycles are heavily damped The problem with this model is the assumption that the susceptible population grows at a

constantrate Try to modify this system of equations If you assume the population grows at

a rate proportional to itself, then the solutions to the equations are cyclic and undamped, as desired Think about the repeating cycle of measles epidemics again after working through Example 31.29

Modeling Population Interactions

The more closely we look at the world the more clearly we see its interconnected nature

We can use the tools we’ve developed in this section to model interactions between distinct populations We can model symbiotic interactions, such as the interaction between sea anemones and clown fish, or competitive interactions, such as the interaction between lion and hyena populations competing for small prey We can model predator-prey interactions, such as that between hyenas and the Thomson’s gazelle, lions and water buffalo, or cats and mice In the examples in this section we’ll simplify our models to focus on just two populations

EXAMPLE 31.29 Consider the predator-prey relationship between hyenas and the Thomson’s gazelle, a small

gazelle native to Africa Let’s make the following simplifying assumptions

i Assume hyenas are the gazelles’ major predator and that in the hyenas’ absence the gazelle population would grow exponentially

ii Assume gazelles are the major food source for hyenas; with no gazelles the hyenas would die off

Model this interaction with a system of differential equations

Let g(t) = the number of hundreds of gazelles at time t

We can model the interaction by a system of differential equations of the form

dh

dt = −k1h + k2hg dg

dt = k3g − k4hg where k1, k2, k3, and k4are positive constants Just as the rate of transmission of disease

is proportional to interactions between the infected and the susceptible, which is in turn proportional to the product of their numbers, so too is the rate of nourishing/fatal interaction between hyena and gazelle proportional to the product of their population sizes Observe that if h = 0 thendgdt = k3gand if g = 0 thendhdt = −k1h For the sake of concreteness, we’ll work with the values of k1, k2, k3, and k4given below and analyze solutions in the gh-plane

dh

dt = −0.3h + 0.1gh = 0.1h(−3 + g) dg

dt = +0.4g − 0.4gh = 0.4g(1 − h) Nullclines: dhdt = 0 when h = 0 or g = 3 In the gh-plane this is where trajectories have horizontal tangent lines

dg

dt = 0 when g = 0 or h = 1 In the gh-plane this is where trajectories have vertical tangent lines

Equilibrium points: dhdt = 0 at h = 0 or g = 3, so at an equilibrium point either h = 0 or

g = 3

Suppose h = 0 Then in order for dgdt to be zero we must have g = 0 (0, 0) is an equilibrium point

Suppose g = 3 Then in order for dgdt to be zero we must have h = 1 The point (3, 1)

is an equilibrium point

Use the differential equations to orient the vertical and horizontal tangents For instance, when g = 3 we know thatdgdt >0 for 0 < h < 1 anddgdt <0 for h > 1

h

g

1

3

Figure 31.30

The h and g axes are oriented as shown in Figure 31.30 The nullclines divide the relevant first quadrant region into four subregions The direction of trajectories in each region is indicated in Figure 31.30

From Figure 31.30 we see that for g(0) > 0 and h(0) > 0 the trajectories either spiral

in toward (3, 1) or spiral outward, or are closed curves From the slope field it appears that the curves are closed Looking atdhdg =0.1h(−3+0.1g)0.4g(1−h) enables us to distinguish between these options Separating variables, we find that

ln(h) − h = −3 ln(g) + g + C

Suppose we start at the point (5, 1) We can solve for C and show that the trajectory through (5, 1) intersects the line g = 3 exactly twice, once for h > 1 and once for h < 1 Similarly

we can show that it intersects the line h = 1 exactly twice, once for g < 3 and once for

g >3 The trajectory through (5, 1) is a closed curve In fact, all the trajectories in the first quadrant are closed

g h

3 1

Figure 31.31

Our model predicts that the hyena and gazelle populations will oscillate cyclically When there are few gazelle the hyena population decreases due to lack of food The decrease

in the number of hyenas allows the gazelle population to thrive, but as the gazelle population increases the hyenas’ food source is replenished, allowing the hyena population to flourish This flourishing takes its toll on the gazelles, and the cycle repeats

To the extent that a model reflects observed population dynamics, it is a good model Models that don’t reflect observed behavior must be modified There are various ways this predator-prey model can be modified For instance, there may be competition among gazelle for limited grazing land so that in the absence of hyena the gazelle population exhibits logistic growth This can be reflected in the system of differential equations by inserting a

−k5g2term as shown

dh

dt = −k1h + k2hg dg

dt = k3g − k4hg − k5g2, where the −k5g2term (k5>0) reflects competition among gazelles

The predator-prey system of differential equations given in Example 31.29 is sometimes referred to as Volterra’s model after the Italian mathematician Vito Volterra (1860-1940) who was encouraged to analyze the predator-prey relationship between sharks and the fish they prey upon by his son-in-law, the biologist Humberto D’Ancona.10By inserting

a term in each equation to account for fishing, Volterra was able to explain why fishing conducted in the Adriatic Sea was raising the average number of prey and lowering the average number of predators over any cycle Similar analysis has been successfully used

to analyze unexpected results of introducing DDT into a predator-prey system, leaving predator populations lowered and prey populations elevated

Competition between species can be modeled by differential equations of the form

dx

dt = k1x − k2xy dy

dt = k4y − k5xy

or

dx

dt = k1x − k2xy − k3x2 dy

dt = k4y − k5xy − k6y2 where k1, k2, , k6are positive constants The latter set of differential equations takes into account competition between members of the same species in addition to competition between species

Answers to Selected Exercises

Answers to Exercise 31.8

x

x

y

(b) (a)

x0

x0 t

10For more information on a very interesting story see Martin Braun, Differential Equations and their Applications, 2nd ed New York Springer Verlag, 1978 or G F Gause, The Struggle for Existence, New York Hafner, 1964, or Umberto D’Ancona,