Chapter 1: First Order Diﬀerential Equations 12 Long Term ... · Chapter 1: First Order Diﬀerential Equations T = .001. You may want to program the RK2(3) method into a calculator

Chapter 1: First Order Differential Equations

T = .001. You may want to program the RK2(3) method into a calculator or computer.Unfortunately, unlike the first extra credit assignment, this method isn’t well suited to aspreadsheet because you keep having to test the step-size and possibly change it (perhapsmore than once) at each step, though some students have used a spreadsheet successfullyanyway.

§12 Long Term Behavior of Solutions

Discussion: Now we will examine the qualitative behavior of solutions to first order initialvalue problems of the form

dP

dt= f(P ), P (0) = P0

with particular emphasis on their long term behavior. A first order equation of the aboveform, where the right hand side depends only on the dependent variable, P , and not onthe independent variable, t, is called autonomous. Our goal is to be able to draw thesolution curves just from looking at the function f(P ) in the equation

dP

dt= f(P ).

Then in the next couple of sections we will develop and analyze models of populationgrowth using the techniques we develop here.

We start by looking for equilibrium points for the differential equation. An equilibrium

point for the differential equation dP/dt = f(P ) is a value P0 which satisfies f(P0) = 0.

Suppose P0 is an equilibrium point for our differential equation. Then P = P0 is a solutionto the differential equation. We quickly check that dP/dt = 0 = f(P ) since P0 is a constantand f(P0) = 0. So if our initial value is P (0) = P0, our solution is just a straight line(and P is unchanging, hence in equilibrium). Now since the solution curves don’t crosseach other (as long as f(P ) is continuous), we know that if P (0) < P0, then P (t) < P0

for all t. Similarly if P (0) > P0, then P (t) > P0 for all t. Next by examining the sign ofdP/dt at P (0) we can determine whether P (t) is increasing or decreasing. For autonomousequations P (t) will continue to increase or decrease until it encounters an equilibrium point,which will be a barrier since P (t) can’t cross an equilibrium point. If no equilibrium pointis encountered, then P (t) will increase or decrease without bound. This information isenough to let us determine the long term behavior of P (t).

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§12 Long Term Behavior of Solutions

Paradigm: What is the long term behavior ofdP

dt= 1

4 (P 2 + P − 2) ?

STEP 1: Find the equilibrium points.

We solve 14 (P 2 + P − 2) = 0 and find the equilibrium points are P0 = −2 and P0 = 1.

STEP 2: Find the sign of dP/dt in the regions between equilibrium points and classify P

as increasing or decreasing in these regions.

If P < −2 then 14 (P 2+P −2) > 0 so dP/dt > 0 and so P is increasing. If −2 < P < 1 then

14 (P 2 + P − 2) < 0 so dP/dt < 0 and so P is decreasing. If P > 1 then 1

4 (P 2 + P − 2) > 0so dP/dt > 0 and P is increasing.

STEP 3: The limiting value is calculated by using the rule that P will increase or decreasein a given region until it encounters a “barrier” formed by an equilibrium point. If no suchbarrier is encountered, P increases or decreases without bound (to ±∞).

If P < −2 then P will increase until it hits the barrier formed by the solution curvecorresponding to the equilibrium point at P = −2, so in the limit as t → ∞, P → −2. If−2 < P < 1 then P is decreasing, but P can only decrease until it hits the barrier formedby the solution curve corresponding to the equilibrium point at P = −2, so in the limit ast → ∞, P → −2. If P > 1 then P is increasing and there is no barrier so P → ∞. Thegraph of the vector field and solution curves for this problem is in figure 3.

More Discussion: Note that I didn’t say that for P > 1, P → ∞ as t → ∞. Thereason I didn’t say this is that it isn’t true. It doesn’t take infinitely long for P to getto ∞, it explodes to ∞ in finite time. When you have a differential equation of the formdP/dt = f(P ) where f(P ) is a polynomial, and you have a solution to the differentialequation which tends to ±∞, then the solution explodes to ±∞ in finite time if thepolynomial f(P ) is of degree 2 or greater. If f(P ) is a first degree polynomial then thesolution politely tends to ±∞ in infinite time. This is one reason for doing a geometricrather than an analytic analysis of the long term behavior of solutions. It is clear fromthe geometric analysis what is happening. But if you solve the equation analytically andthen take the limit as t → ∞, you will get the wrong answer because you will cross thesingularity at the explosion. There are examples like this in the exercises.

Note that if P = −2 but then gets “bumped” a little bit away, P will return to -2. Wesay P = −2 is a stable equilibrium point. On the other hand, if P = 1 gets “bumped” a

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Figure 3

little bit, P will tend to move away from 1, toward either ∞ or −2. We say P = 1 is anunstable equilibrium point. The rule is that in the limit as t →∞, P will tend to a stableequilibrium point, or ±∞. This rule is only true for autonomous equations; the situationis much more complicated for general equations where the right hand side depends on t aswell as P . But this rule will suffice for our purposes.

Exercises:

(1) SolvedP

dt= −P 2. Explicitly compute the limit of P (t) as t →∞. Check that the rules

given in this section work. How can you identify the explosion from the analytic solution?

(2) SolvedP

dt= −P 2. Explicitly compute the limit of P (t) as t → −∞. You can generalize

the rules given in this section to the case of t → −∞. Write out a paragraph explaininghow to compute limt→−∞ P (t) using geometric reasoning without solving the equation.Do you run into problems with explosions here?

(3) SolvedP

dt= 5P − P 2. Explicitly compute the limit of P (t) as t → ∞. Check that

the rules given in this section work. How can you identify the explosion from the analytic

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§13 Population Models

solution?

(4) SolvedP

dt= 5P − P 2. Explicitly compute the limit of P (t) as t → −∞. You can

generalize the rules given in this section to the case of t → −∞. Write out a paragraphexplaining how to compute limt→−∞ P (t) using geometric reasoning without solving theequation. Do you run into problems with explosions here?

(5) SolvedP

dt= P 2 + 4P + 3. Explicitly compute the limit of P (t) as t →∞. Check that

the rules given in this section work. How can you identify the explosion from the analyticsolution?

(6) SolvedP

dt= P 2 + 4P + 3. Explicitly compute the limit of P (t) as t → −∞. You can

generalize the rules given in this section to the case of t → −∞. Write out a paragraphexplaining how to compute limt→−∞ P (t) using geometric reasoning without solving theequation. Do you run into problems with explosions here?

Graph several solution curves for the following equations using the techniques of thissection. Find the equilibrium points and classify them as stable or unstable. You don’thave to actually solve the equations.

(7)dP

dt= (P − 2)(P + 3)

(9)dy

dx= (y − 1)(y − 2)(y − 3)

(11)dP

dt= P 3 − P

(13)dP

dt= P 3 − 105P 2 + 500P

(15)dP

dt= −P (P − 5)(P − 50)

(17)dP

dt= −P 3 + 4P 2

(19)dy

dx= y3 + 2y2 + 4y

(8)dP

dt= (P + 2)(P + 1)

(10)dP

dt= (3P − 1)(P + 4)(P + 2)(P − 5)

(12)ds

dr= s2 + 2s + 1

(14)dP

dt= −P 4 + 4P 3 − P 2

(16)dP

dt= −(P − 1)(P − 2)(P − 3)

(18)dy

dx= y3 + 2y2 + 4y

(20)dP

dt= P 4 − 10P 3 + 35P 2 − 50P + 24


Discussion: My favorite application of first order differential equations is population

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dynamics. Several different models of population growth can be easily derived from whatwe know so far.

EXPONENTIAL GROWTH

Consider a bacterium. It reproduces by fission, that is every so often it splits into twonew bacteria. The two progeny then split into four bacteria sometime later. This processcontinues indefinitely. We represent this situation by saying that if there are p bacterianow, we will add about rp new bacteria every hour where r is a constant of proportionalitycalled the growth rate. Or in the standard calculus phrasing, the rate of change of p isabout rp per hour, and if we start with p0 bacteria we get the initial value problem

dp

dt= rp, p(0) = p0

where p is the population of bacteria and t is time in hours with the present time beingset to t = 0. This is a separable first order differential equation and we can solve it to find

p = p0ert.

Hence the name exponential growth. The graph of population vs. time for this model isgiven in Figure 4.

Figure 4

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LOGISTIC GROWTH

You may have noticed that in the preceeding model, the population explodes to ∞. Givena single bacterium, you should grow a colony the size of the earth in a matter of days. Ofcourse this doesn’t happen. The previous model provides a model of the “birth” of newindividuals but makes no allowance for the death of individuals. Bacteria do not usuallydie of old age, but they do die from competition for resources. We add this to the modelby saying the rate of deaths in the population is proportional to the likelihood of contactsbetween two individuals (which is usually assumed to be proportional to p2). This leadsto the model

dp

dt= rp− kp2 = rp(1− (k/r)p), p(0) = p0

where k is a constant of proportionality. This is also a separable equation (as all theequations in this section will be) but it is easiest to analyze it using the ideas of theprevious section rather than trying to solve the equation directly. The equilibrium pointsof the equation are at p = 0, and p = r/k. The equilibrium point at p = 0 is unstable whilethe equilibrium point at p = r/k is stable. r/k is called the carrying capacity of themodel. So the population should increase to a level p = r/k and then stay there. This ismuch more in line with what we expect to happen to a population of bacteria. The graphof population vs. time for this model is given in Figure 5.

Figure 5

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EXPONENTIAL DECAY

Large creatures (rabbits, whales, us, etc.) do not reproduce by fission. If you take a rabbit,it doesn’t split into two as time passes. It sits there quietly for a few years and then itdies. Mathematically, you could write that some proportion of rabbits dies every year sothe negative rate of change of the population of rabbits is proportional to the population.Assuming an initial population p0 this gives rise to the model

dp

dt= −cp, p(0) = p0

where p is the population of rabbits, t is time (in years) and c is a constant of proportion-ality. We can solve this initial value problem to find the population is

p(t) = p0e−ct

This is clearly not a reasonable model for rabbits. But it is a reasonable model for thepopulation of Carbon-14 molecules in fossil materials. One of the nice features of popula-tion dynamics is that it can be applied to many different sorts of populations, from rabbitsto U-235 nuclei. The graph of population vs. time for this model is given in Figure 6.

Figure 6

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EXPONENTIAL GROWTH WITH THRESHOLD

Returning to rabbits, the population of rabbits doesn’t just fade away to nothing in reallife. Our problem earlier with bacteria was failing to include death in our model. Ourcurrent problem is failing to include birth. We may assume that a certain percentage ofinteractions between two rabbits will lead to the birth of several new baby rabbits. Therevised model is then

dp

dt= −cp + rp2 = −cp(1− (r/c)p), p(0) = p0

The equilibrium points for this equation are p = 0 and p = c/r. c/r is called the threshold.p = 0 is stable while p = c/r is unstable. So if the population falls below the threshold,c/r, needed to sustain itself, it will die off. If the population starts above the threshold, itwill explode to ∞. The graph of population vs. time for this model is given in Figure 7.

Figure 7

LOGISTIC GROWTH WITH THRESHOLD

Of course the population of rabbits doesn’t tend to ∞ (though you couldn’t prove it bymy garden). Rabbits also experience competition for scarce resources. Since interactions

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of 2 rabbits are presumed to have positive effects, we subtract a term proportional to theinteractions of 3 rabbits (p3) to model competitive pressures. This gives the model

dp

dt= −cp + rp2 − kp3, p(0) = p0.

Now we have three equilibrium points (assuming r2 > 4kc), 0, pu = (r −√

r2 − 4kc)/2k

and ps = (r+√

r2 − 4kc)/2k. 0 is a stable equilibrium point. pu is an unstable equilibriumpoint and is called the threshold as in the previous example. ps is a stable equilibriumpoint, called the carrying capacity as in logistic growth. So if the populations fallsbelow the threshold pu it dies out but otherwise it tends to the level ps. Looking at thegraphs of the population vs. time, you should note that the behavior above the thresholdclosely resembles the behavior of the logistic growth model. For established populations,the logistic growth model we established for bacteria is often used because it is simplerand has the right general properties (and population dynamics is by no means as accuratea subject as physics). Logistic growth with threshold is used in studying populations indanger of extinction, such as blue whales. The graph of population vs. time for this modelis given in Figure 8.

Figure 8

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§14 Harvesting Models

§14 Harvesting Models

Discussion: In this lab you will analyze two harvesting models. In both cases, assume youhave a population, traditionally fish, which you wish to harvest. You must decide how manyfish to harvest to obtain the maximum yield. Start by assuming the population undergoeslogistic growth (see section 13) with growth rate .07 (7%), the carrying capacity of theenvironment is 10,000 and that the population is currently 8,000. The initial population isusually below the carrying capacity in harvesting models because people usually start toharvest species before they work out mathematical models to predict the best harvestingstrategy. There are two strategies that can be easily modeled. In the first, proportionalharvesting, a fixed proportion of the fish are harvested every year. This is the model

dp

dt= .07p(1− p/10000)− Ep

where E is the proportion of fish harvested, called the harvesting constant. In the secondmodel, constant harvesting, a constant number of fish are harvested every year. This isthe model

dp

dt= .07p(1− p/10000)− h

where h is the number of fish harvested every year. In this lab you will experiment withdifferent levels of harvesting and determine the yields. You will then explain your resultsin terms of the mathematics involved.

Get into Matlab for Windows and give the command xc3. This will get you a window witha plot of the population over time, an editable window where you can change the harvestingfunction, and the final yield, that is the number of fish harvested 200 years later using thatharvesting function. The fish population will reach equilibrium eventually and the numberharvested in the year 200 is usually, though not always, a close approximation to thelimiting value of the yield.

Start by analyzing the proportional harvesting model. Try using the harvesting function.05.*p as your first guess. Then try .1.*p as your second guess. See how this changes thepopulation and the yield. Play around with the harvesting constant until you can answerthe following questions. Note that old population plots are retained for comparison. Youcan always dismiss the window (double-click in the upper left hand corner) and start overwhen things get too messy. Remember to answer all questions in complete sentences.

(1) How does changing the harvesting constant change the population curve?

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(2) How does changing the harvesting constant change the yield?

(3) What value of the harvesting constant gives the greatest final yield?

(4) What value of the harvesting constant results in extinction for the population?

Next consider the constant harvesting model. Try the harvesting function 100. In thiscase the yield will be 100 because you are always harvesting 100 fish. Next try 195. In thiscase the yield is 0, because the population becomes extinct. Play around with differentamounts of constant harvesting until you can answer the following questions. Rememberto answer in complete sentences.

(5) How does changing the amount harvested change the population curve?

(6) What is the maximum possible harvest without extinction? (Note that since thepopulation is only graphed for 200 years, the population may be headed for extinctioneven though it hasn’t reached 0 by the time the graph ends. When you are determiningthe maximum possible harvest you should try to guess whether the population will becomeextinct in the future even if it hasn’t yet. You should be able to judge pretty well fromlooking at the population trends at the end of the graph.)

(7) How does the maximum yield in the constant harvesting model compare to the maxi-mum yield in the proportional harvesting model?

Now compute the theoretical maximum yields as follows.

For the proportional harvesting model, determine the stable fixed point, p1(E), as a func-tion of E. Then the yield as a function of E is y(E) = E × p1(E). Now maximize y(E)to find the theoretic maximum yield. Can you also find the value of E that results inp1(E) = 0 (extinction) theoretically?

For the constant harvesting model, the harvest is h as long as the population doesn’tbecome extinct. So determine the stable fixed point, p2(h), as a function of h. What isthe maximum h with p2(h) > 0?

(8) How do the experimental results compare with the theoretical values?

Most people find their experimental results for optimum harvesting for the proportionalharvesting model are quite accurate but that they are usually a little high on the best

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§15 Theory of First Order Equations

sustainable harvest for the constant harvest model. The trouble is that you have to guesswhether the population curve has leveled off or if it is about to turn down and head toextinction after t = 200. Of course in real life you don’t normally have precise values forthe growth rate or the carrying capacity and you have to guess based on the data whatit will do in the future just given the graph of past behavior. This is one reason whyconservation officials managing wildlife populations generally use a proportional harvestmodel.

§15 Theory of First Order Equations

Discussion So far we have concentrated on finding solutions to first order equations;explicitly when we can and using numerical approximations when we can’t find explicitsolutions. But whenever you can’t find an explicit solution to a mathematical problemand are about to use a numerical approximation instead, you ought to pause and thinkfor a minute. Could it be the reason you can’t find an explicit solution is because no suchsolution exists? How do we know there is any function whose derivative satisfies someequation we have written down? This can be a tricky and dangerous question in somesituations, but fortunately for this class we have the following theorem:

Theorem. Let the functions f and ∂f/∂y be continuous in some rectangle a < x < b and

c < y < d containing the point (p, q). Then in some interval p− h < x < p + h contained

in a < x < b, there is a unique solution y = g(x) of the initial value problem

y′ = f(x, y) y(p) = q

I won’t give a formal proof of this theorem (come see me in my office if you are interested).A heuristic justification is offered by the geometric ideas discussed earlier. Draw thevector field defined by f(x, y). Start at the point (p, q) and follow the arrows to draw anintegral curve. This integral curve must be the graph of the solution to the initial valueproblem. The fact that the solution curves don’t cross means that a solution to the initialvalue problem is unique as claimed above. If there were two solutions to the initial valueproblem then they would give two different solution curves both passing through the point(p, q). The only possible problem is if the arrows change too abruptly or contradict oneanother so we can’t follow them. The continuity conditions in the hypothesis rule out

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such nasty behavior. On the other hand, if the continuity conditions fail, then all sortsof things can go wrong. Furthermore, the theorem only guarantees us that we have asolution that exists in some interval around our initial point. We have no guarantee thatthe solution doesn’t stop somewhere. In fact, it is quite possible for a solution to only existfor a finite interval as we saw when we discussed explosions earlier. On the other hand,Euler’s method and the improved Euler’s method will almost always produce results onan infinite interval. It is up to you to watch for signs that the numerical results no longerapproximate the true solution (which may not exist any longer). One easy clue is that thesolution to a differential equation should be a smooth curve. It should not have any spikesin it. Spikes in the curve result from having the slope field changing discontinuously, andthat means we are in the danger zone for existence of solutions. Explosions, unfortunately,don’t usually signal their existence quite as dramatically. To detect explosions you needto recognize an explosion exists in the differential equation and/or use more sophisticatedsoftware that tests for singularities in the result. The messages you got occasionally frommatlab warning that “Singularity likely” were matlab’s warnings that it thinks the solutionhas exploded. Unfortunately, sometimes the numerical method can be fooled into thinkinga solution explodes even when it doesn’t. This is a final reason for understanding thegeometric approach to dealing with autonomous equations.

§16 Review Problems

Discussion: In real life, you are unlikely to be asked if a problem is Bernoulli. In fact youare unlikely to be directly asked to solve any differential equation at all. More commonly,you will have some question that after doing some work can be reduced to solving adifferential equation, without any clue as to what type of equation it is. Given such asituation, I check to see if it is one of the types we have considered in the following order.

Separable

Linear

Bernoulli

Exact

Homogeneous

There is nothing magical about this ordering. I just find it easy to check if an equation isone of the first three types by inspection so I do that first. Equations which are homoge-

60


neous are often solvable using some other method and if they are, it is usually easier touse the other method. So I check if an equation is homogeneous only after I have triedevery other method. This is just the order I use — you are welcome to use a differentorder of techniques to try if you want. Of course, lots of equations can’t be solved usingany of the five methods we have covered. There are a lot of other techniques that aresomewhat less common that we haven’t covered in this class. The standard reference ifyou need to solve an equation that doesn’t fit into any of our paradigms is “Differential-gleichungen; Losungsmethoden und Losungen,” by Erich Kamke (available from ChelseaPublishing Co.). This book contains a vast assortment of different tricks and techniquesfor solving differential equations. Unfortunately it is written in German, but you can’thave everything.

Exercises: Solve the following problems.

(1)dy

dx+ 3y = 2x

(2)dy

dx= x2y

(3)dy

dx=

x2 + x + 2xy

y2 − x2

(4)dy

dx= x− y

(5)dy

dx= xy − y2

(6)dy

dx= xy − 4x + 3y − 12

(7)dy

dx=

2x2y + x3

y3 + 2x3

(8)dy

dx= y − y3

(9) 3x + 4y + (4x− sin(y))dy

dx= 0

(10)dy

dx+ y/x = cos(x)

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(11)dy

dx= xy + 4x, y(0) = 0

(12)dy

dx= y + y2, y(0) = 1

(13)dy

dx= 6

x + y

x− y, y(0) = −1

(14)dy

dx= x2ey, y(−2) = 0

(15)dy

dx=

x + 2y

3x− y, y(1) = −1

(16)dy

dx+ sin(x)y = −2 sin(x), y(0) = 0

(17)dy

dx= y/x + x/y, y(1) = 1

(18) 2x2 + 3xy + ey + ((x + 7)ey +32x2)

dy

dx= 0, y(0) = 0

(19)dy

dx=

3y + x

y + 3x, y(0) = −2

(20)dy

dx− y/x = e2x, y(1) = 2

(21) SupposedP

dt= P (1− P/2)(P/5− 1), P (0) = 3. Compute limt→∞ P (t) and

limt→−∞ P (t).

(22) SupposedP

dt= P (1− P/2)(P/5− 1), P (0) = 8. Compute limt→∞ P (t) and


(23) SupposedP

dt= P (1− P/2)(P/5− 1), P (0) = −2. Compute limt→∞ P (t) and


(24) SupposedP

dt= −25P + 100P 2 − P 3, P (0) = 10. Compute limt→∞ P (t) and


(25) SupposedP



(26) SupposedP



(27) Graph several of the solution curves ofdy

dx= y2 − 6y − 16.

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dx= −y2 + 6y − 16.


dx= y2 + 7y + 14.


dx= −y2 − 7y − 14.

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Documents

Chapter 1: First Order Diﬀerential Equations 12 Long Term ... · Chapter 1: First Order Diﬀerential Equations T = .001. You may want to program the RK2(3) method into a calculator