A Course in Math Bio Sol

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Solutions Manual

to accompany

A Course in Mathematical Biology

Quantitative Modeling with Mathematical and Computational Methods

G. de Vries, T. Hillen, M. Lewis, J. Mulller, and B. Schonfisch,

Society for Industrial and Applied Mathematics, Philadephia, 2006

A. Beltaos G. de Vries T. Hillen

November 20, 2006


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2 Solutions manual for de Vries et al, SIAM 2006


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Contents

1.4 Exercises for Modeling . . . . . . . . . . . . . . . . . . . . . . . 6

2.4 Exercises for Discrete-Time Models . . . . . . . . . . . . . . . . 7

3.9 Exercises for ODEs . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.5 Exercises for PDEs . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.8 Exercises for Stochastic Models . . . . . . . . . . . . . . . . . . 61

6.6 Exercises for Cellular Automata . . . . . . . . . . . . . . . . . . 62

7.7 Exercises for Parameter Estimation . . . . . . . . . . . . . . . . 63

3


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Note to Readers

This document contains solutions to a selection of problems, primarily problemsfrom Chapters 2, 3, and 4.

We welcome submissions to fill in the missing solutions (preferably in latex format)and figures (preferably in postscript or encapsulated postscript format). We will bepleased to credit your authorship in the corresponding solution in updated versionsof this document.

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1.4 Exercises for Modeling

Exercise 1.4.1: Discrete-time versus continuous-time models

(a) ∆t = 10 minutes, and the probability of one cell making one other cell in 10minutes is p = 1, since we are told that each cell effectively makes one copyof itself every 10 minutes.

(b) Letting N(t) be the population at time t, we have:

N(t + ∆t) = N(t) + pN(t) (1.1)

= N(t) + N(t)

= 2N(t)

We can see that the whole population doubles every 10 minutes.

(c) We can define α := p∆t (in this case, α = 1

10min−1), and we can change ourdiscrete model. Dividing (1.1) by ∆t, we get:

N(t + ∆t)

∆t=

N(t)

∆t+

p

∆tN(t) =

N(t)

∆t+ αN(t)

⇒ N(t + ∆t) − N(t)

∆t= αN(t)

If we take the limit as ∆t → 0, we obtain the continuous model:

d

dtN(t) = αN(t)

Exercise 1.4.2: Comparison of discrete and continuous models

Solution not available.

Exercise 1.4.3: Structured populations



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A. Beltaos, G. de Vries, T. Hillen, November 20, 2006 7

2.4 Exercises for Discrete-Time Models

Exercise 2.4.1: German population

Let xn = Germany’s population at the end of year n. The simplest model incorpo-rating the given information is

xn+1 = xn − δxn + µxn

= (1 − δ + µ)xn (2.2)

= rxn,

where r = 1− δ + µ.

At the end of the year 1998, we have x1998 = 82, 037, 000, and

µx1998 = 770, 744, and δx1998 = 846, 330,

which gives the values of µ and δ to be:

µ = 770,74482,037,000 = 0.009395,

δ = 846,33082,037,000 = 0.010316,

(2.3)

and hence r = 1 − 0.010316 + 0.009395 = 0.999079 < 1. That is, the populationwill decay.

To make the model more realistic, we could add in terms for immigration andemigration, and use a logistic equation.

Exercise 2.4.2: Drug prescriptions

(a) The dimensionless parameter k represents the fraction of the drug’s dose whichis used up or broken down by the body. The amount of the drug in a dose isrepresented by b. We know that b ≥ 0 (no negative doses), and k > 0 (the bodymust use up some of the drug). Notice that if b = 0, then an+1 = an−kan, sowe better have k ≤ 1 to avoid negative amounts. Summarizing the conditionson k and b,

k ∈ (0, 1],

andb ≥ 0.

(b) The fixed point (we know there is only one, since it is a linear equation), a, ofthis model can be found if we let f(x) := x − kx + b, and solve for f(a) = a.This gives us a = b/k. Further, f ′(x) = 1 − k. We found in part (a) thatk ∈ (0, 1], so that means that f ′(x) ∈ [0, 1), which means that this fixed point,a = b/k, is stable.


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(c) In every case, we will start with the initial value of 0 (representing no dosegiven yet). Looking at the cobweb diagrams below, we can see that the mapalways settles down to the fixed point. The drug will have a constant concen-tration after sufficient time has passed, unless the dose is changed. The fixedpoint is reached more quickly if k is larger. In the figure below on the left,we have k = 0.1, and we can see many steps leading up to the fixed point.In the figure on the right, we set k = 0.6, and it takes fewer iterations of themap to reach the fixed point. Note that in these figures, we have set b = 1.Also, a larger value of k gives a smaller actual value of the fixed point. Sincethe fixed point is a = b/k, we see that b behaves like a scaling factor, simplychanging the value, but not the quality of the fixed point.

0

2

4

6

8

10

12

y

2 4 6 8 10 12x

0

0.5

1

1.5

2

y

0.5 1 1.5 2x

(d) We assume that there is a minimum amount, E, for which the drug will beeffective, and a maximum amount, T , for which the drug becomes toxic. Tomake sure the drug is effective but not toxic, we need the fixed point to bebetween the values E and T :

E ≤ a < T

Substituting in for a (found in part (b)), and multiplying through by k, weneed the dosage b to be between the values kE and kT :

kE ≤ b < kT.

Exercise 2.4.3: Improving the fit of the logistic model to the data


Exercise 2.4.4: Fluctuations in the population of P. aurelia



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Exercise 2.4.5: Whale population

(a) The given equation,

an+1 = an + k(M − an)(an − m),

describes the population of whales next year in terms of the population thisyear. The last term describes the change in the population, which is dependenton the 3 parameters, k, M , and m, with m < M .

• If 0 < an < m, then k(M−an)(an−m) < 0, so the population will decline(specifically, the population next year is smaller than the population thisyear).

• If m < an < M , then k(M − an)(an − m) > 0, so the population willgrow.

• If an > M , then k(M − an)(an −m) < 0, so the population will decline.

• If an = m or an = M , then k(M−an)(an−m) = 0, so the population willremain the same size (that is, m and M are fixed points of the model).

(b) For fixed points a, we require

a = a + 0.0001(5000− a)(a − 100)⇐⇒ 0.0001(5000− a)(a − 100) = 0⇐⇒ a = M = 5000 or a = m = 100.

To determine their stability, we let

f(x) = x + 0.0001 (5000− x)(x − 100)

= x + 0.0001 (5000x− 500000− x2 + 100x)

= x + 0.0001 (5100x− 500000− x2).

Then

f ′(x) = 1 + 0.0001(5100− 2x).

For the fixed point a = m = 100, we have

f ′(100) = 1 + 0.0001(5100− 200)

= 1.49 > 1,

therefore a = m = 100 is unstable. For the fixed point a = M = 5000 we have

f ′(5000) = 1 + 0.0001(5100− 10000)

= 0.51 < 1 (but greater than −1),

therefore a = M = 5000 is stable.


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(c)

100 a0 5000 a

n

an+1

an+1

= an

As before, we conclude that a = m = 100 is unstable, and a = M = 5000 isstable.

(d)

nm = 100

M = 5000

an

nm = 100

M = 5000

an

nm = 100

M = 5000

an

n

m = 100

an

(e) When a0 < m, the population declines, eventually becoming negative. Whena0 � M (i.e., when a0 is greater than the biggest root of the map), then


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a1 < 0, and the population continues to decline. Both cases are problematic,since it is non-sensical to have a negative number of whales.

(f) If we plot an+1 versus an for the given model, we obtain the following quali-tative graph:

m M an

an+1

stableunstable

an+1

= an

We see that if the whale population ever reaches a value located on the high-lighted (thick) portions of the an-axis, the population will be negative nextyear, since the graph of an+1 versus an lies below the an-axis there.

To fix the problem, we could ensure that the graph of an+1 versus an alwayslies above the an-axis. For example:

m M an

an+1

an+1

= an

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Exercise 2.4.6: Second-iterate map

(a)

f2(x) = r (rx(1 − x)) (1 − (rx(1 − x)))

= r2x(1 − (r + 1)x + 2rx2 − rx3)

(b) We already know the fixed points of f . Since they are fixed points of f , theyalso are fixed points of f2. These two “trivial 2-cycles” are x = 0 and x = r−1

r .Since we know they are fixed points of f 2, i.e., we know they satisfy f 2(x) = x,we know that they are solutions of f 2(x)− x = 0. This means we can rewritef2(x)−x as x(x− r−1

r )q(x), where q(x) is an unknown quadratic polynomial.We then set this new expression equal to what we know for f 2(x) − x:

x(x − r − 1

r)q(x) = r2x(1 − (r + 1)x + 2rx2 − rx3) − x.

Since we are interested in nonzero solutions, we can drop the common x factoron each side. We don’t want the (r − 1)/r root either, so we can drop that,by diving both sides by x − r−1

r . Hence,

q(x) =r2(1 − (r + 1)x + 2rx2 − rx3) − 1

x − r−1r

.

Carrying out the long division, we end up with

q(x) = −r3x2 + (r3 + r2)x − r2 − r.

The roots of this polynomial are the last two roots of f 2. Now we can findthem by simply invoking the quadratic formula:

x =−(r3 + r2) ±

√

(r3 + r2)2 − 4r3(r2 − r)

−2r3.

This can be simplified to

x =r + 1 ±

√

(r − 3)(r + 1)

2r.

It is these two roots we are interested in, because the first two are trivial2-cycles. Recall the restriction of r ∈ [0, 4] which was discussed in section(later), so the term (r + 1) in the radical poses no problem for us. It is the(r − 3) term which tells us when this 2-cycle exists. Obviously, this 2-cycledoes not exist if r < 3 (imaginary roots). If r = 3, this ceases to be a 2-cycle,because the roots which make up the 2-cycle become the same (the radicalevaluates to zero). Hence, a nontrivial 2-cycle exists only for r > 3.


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(c) After a bit of simplification, we find that

d

dxf2(x) = −4r3x3 + 6r3x2 − 2(r2 + r3)x + r2.

(d) This can be done with a few MAPLE commands. One is complicated. Whatwe want to do is turn the derivative, with the fixed points substituted inalready, into a function of r:

> d:=q-> subs( r=q, abs( subs( x=sol[3], diff(f2(x), x)

) ) ):

The explanation of this command is as follows (from the inside out):

We substitute the 3rd solution that MAPLE gave for the fixed points of f 2(x)into the derivative of f2(x), and then take the absolute value. We make it afunction by replacing the r’s that appear with the dummy variable, q, whichwe are using to define the function in terms of. Now if we call d(r0), it willgive us the absolute value of the derivative of f 2 at the fixed point with r = r0.

Next, we want to plot d(r), between 2.8 and 3.6 (to include the points 3 and1 +

√6 ≈ 3.449). A horizontal line has been added to see where the graph

crosses d(r) = 1.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

2.9 3 3.1 3.2 3.3 3.4 3.5 3.6r

Looking at this, we can see that d is above 1 for r < 3 and below 1 for r > r∗,where r∗ ≈ 3.5. This means that the 2-cycle is unstable and stable in thoseregions, respectively. To find the exact values at which the stability changes,we use the solve command:

> solve(d(r)=1, r);

1 +√

6, 1−√

6, 3,−1

Hence, the 2-cycle is stable for 3 < r < 1 +√

6 and unstable for r > 1 +√

6.Note that 1 +

√6 ≈ 3.449, confirming our observation from the graph above.


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Exercise 2.4.7: Fourth-iterate map


Exercise 2.4.8: Exact solution for the Beverton-Holt model

We havexn+1 =

r

1 + r−1K xn

xn

Let un = 1xn

. Then

1

un+1=

r

1 + r−1K

1un

1

un

un+1 =1

r

(

1 +r − 1

K

1

un

)

un

=1

run +

r − 1

rK

This equation is linear, of the form

un+1 = Aun + B,

where A = 1r and B = r−1

rK .

We can find a solution to this equation as follows:

un+1 = Aun + B

= A[Aun−1 + B] + B

= A2un−1 + B(A + 1)

= A2[Aun−2 + B] + B(A + 1)

= A3un−2 + B(A2 + A + 1)

...

= An+1u0 + B(An + An−1 + · · · + A + 1)

= An+1u0 + BAn+1 − 1

A − 1

Returning to the original variables, we have

1

xn+1=

1

rn+1

1

x0+

r − 1

rK

1rn+1 − 1

1r − 1

=1

x0rn+1+

rn+1 − 1

rn+1K

=K + x0(r

n+1 − 1)

x0Krn+1

xn+1 =rn+1x0

1 + x0

K (rn+1 − 1


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Exercise 2.4.9: Fitting the Beverton-Holt model to Gause’s data


Exercise 2.4.10: The tent map

(a) We distinguish three cases: Case 1: µ < 1, Case 2: µ = 1, Case 3: µ > 1.Graphs of f are:

11/20 x

µ/21/2

1/2 10 x

µ/2 = 1/2

11/20 x

µ/2

1/2

Case 1: µ < 1 Case 2: µ = 1 Case 3: µ > 1

(b) In case 1, there is one fixed point, x = 0. In case 2, there are infinitely manyfixed points 0 ≤ x ≤ 1

2 . In case 3, there are two fixed points, x1 = 0 andx2 = µ

1+µ .

To determine their stability, we look at the derivative of f(x) at each of thefixed points:

f ′(x) =

{µ for 0 ≤ x < 0.5

−µ for 0.5 < x ≤ 1

In case 1, f ′(0) = µ < 1, therefore x = 0 is stable. In case 2, f ′(x) = µ = 1,and the stability of these fixed points is undetermined. In case 3, f ′(0) =

µ > 1, therefore x1 = 0 is unstable. Similarly, f ′(

µ1+µ

)

= −µ < −1, and so

x2 = µ1+µ is unstable as well.

(c) We’ll define two functions, f1(x) = µx, and f2(x) = µ(1−x), and loosely definethe terms “the first region” and “the second region” to mean {x : 0 ≤ x < 0.5},and {x : 0.5 < x ≤ 1}, respectively. There are three possibilities for any 2-cycle in this situation, namely that both iterates lie in the first region, theyboth lie in the second region, or one lies in the first region and the second inthe second region (there’s really one more with the opposite condition of thelast one, but this one can be found by simply applying f to the one foundin the last case, and those are the two iterates of the same 2-cycle). To find2-cycles, we set x = f(f(x)).


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• Case (i): Both lie in the first region. Then f(f(x)) = f 21 (x) = µ(µx) =

µ2x, and the fixed point x must satisfy

x = µ2x,

which has only the trivial solution, x = 0, which we already know to be1-cycle, so we are not interested in this case.

• Case (ii): Both lie in the second region. Then f(f(x)) = f 22 (x) = µ(1 −

[µ(1 − x)]) = µ − µ2 + µ2x. The fixed point must satisfy

x = µ − µ2 + µ2x,

which has solution

x =µ(1 − µ)

(1 − µ)(1 + µ)=

µ

1 + µ.

Since x must lie in the second region, we require µ > 1. This is the samesolution as for the 1-cycle, so again, we are not interested in this case.

• Case (iii): The first lies in the first region, and the second lies in thesecond region. Here, applying f twice will give f(f(x)) = f2(f1(x)) =µ(1 − µx) = µ − µ2x. The fixed point must satisfy

x = µ − µ2x,

which has solution x = µ/(1 + µ2). Note that this point lies in the firstregion no matter what µ is. The proof lies in the fact that for any µ ∈ R,we have (µ − 1)2 ≥ 0. With a little rearranging, the inequality turnsinto µ/(µ2 + 1) ≤ 1/2, for all µ ∈ R. As described above, we can findthe other orbit of this 2-cycle by applying f to x. Renaming x to p,the 2-cycle will be p and q, where q = f(p) = µ2/(1 + µ2). Since wefound this 2-cycle by assuming that the first iterate is in the first region(which it is) and the second is in the second region, we must restrictµ such that q is in the second region. That is, we need to make surethat 1/2 < µ2/(µ2 + 1) ≤ 1. The right-hand side of the inequality istrivial, but the left-hand side forces us to put µ > 1 (keeping in mindthat µ > 0). Therefore, a nontrivial 2-cycle exists only for µ > 1, and itis:

{

p =µ

µ2 + 1, q =

µ2

µ2 + 1

}

.


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(d)

0 1/2 1 x

1y = f(x) y = x

As in (c), we think of the different cases of where the iterates of a 3-cycle couldreside. There seem to be eight possibilities. We denote each possibility witha 3-tuple, where 1 means “the iterate lies in the first region”, and 2 means“the iterate lies in the second region”. The eight possibilities thus are

(1, 1, 1),(1, 1, 2),(1, 2, 1),(1, 2, 2),(2, 1, 1),(2, 1, 2),(2, 2, 1),(2, 2, 2).

Since a 3-cycle is literally a cycle, any cyclic permutation of the above 3-tupleswill be the same 3-cycle (it doesn’t matter which is the “first” iterate of thethree, the important thing is the order of the three). This means that we havesome repetition in the above list, for example (2, 1, 1) is the same as (1, 1, 2),which is the same as (1, 2, 1). Eliminating the repeated ones, we obtain thefollowing four essential cases:

Case (i) : (1, 1, 1),Case (ii) : (1, 1, 2),Case (iii) : (1, 2, 2),Case (iv) : (2, 2, 2).

To find a first point of a 3-cycle, we need x = f(f(f(x))).

• Case (i): All of the iterates of our 3-cycle lie in the first region. If this isthe case, then the equation we get for the fixed point x of the third-iteratefunction is (remember that µ = 2 in this problem)

x = f3(x) = f31 (x) = µ3x = 8x.

As in part (c), we have the 1-cycle x = 0 again, so this is a trivial 3-cycle.


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• Case (ii): The first two iterates are in the first region and the third is inthe second region. Now we would have for x:

x = f3(x) = f2(f1(f1(x))) = f2(f1(2x)) = f2(4x) = 2(1 − 4x) = 2− 8x.

Hence the fixed point of the third-iterate function is x = 2/9. We canfind the rest of this 3-cycle by applying f to this point, and applyingf again. Let p = 2/9, and then q and r will be the other two iteratesof the 3-cycle. q = f(p) = f(2/9) = f1(2/9) = 2 · 2/9 = 4/9, andthen r = f(q) = f(4/9) = f1(4/9) = 2 · 4/9 = 8/9. We can check thatp = f(r): f(r) = f(8/9) = f2(8/9) = 2(1− 8/9) = 2/9 = p, as expected.Hence, our 3-cycle is

{

p =2

9, q =

4

9, r =

8

9

}

.

We could stop here, since the question asks to find an orbit of period 3 (i.e.,a 3-cycle), but we’ll continue to check the cases, in case we can find anotherone.

• Case (iii): The first iterate lies in the first region, and the other two arein the second region. Our equation for the fixed point of f 3(x) is then

x = f3(x) = f2(f2(f1(x))) = f2(f2(2x)) = f2(2(1 − 2x))

= f2(2 − 4x) = 2(1 − (2 − 4x)) = 2 − 4 + 8x

= −2 + 8x.

This time we get x = 2/7. Doing the same thing as for case (ii), we putp = 2/7, and then q = f(p) = f(2/7) = f1(2/7) = 2 · 2/7 = 4/7, andr = f(q) = f(4/7) = f2(4/7) = 2(1− 4/7) = 2 · 3/7 = 6/7. We check thesolution by finding f(r) = f(6/7) = f2(6/7) = 2(1−6/7) = 2 ·1/7 = 2/7,and we see that this is correct. Hence, another 3-cycle is

{

p =2

7, q =

4

7, r =

6

7

}

.

• Case (iv): All iterates of the 3-cycle are in the second region. Then theequation for a fixed point of f 3(x) is

x = f3(x) = f32 (x) = f2(f2(2(1 − x))) = f2(f2(2 − 2x)) = f2(2(1 − (2 − 2x)))

= f2(2 − 4 + 4x) = f2(−2 + 4x) = 2(1− (−2 + 4x)) = 2(3− 4x)

= 6 − 8x.

This gives us the solution x = 2/3. Again, we set p = 2/3, and letq = f(2/3) = f2(2/3) = 2(1 − 2/3) = 2/3. We need not go any further,because we can see that this is a 1-cycle, a trivial 3-cycle. This is the1-cycle we found in part (b), with µ = 2 (we can verify: µ/(1 + µ) =2/(1 + 2) = 2/3).


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Exercise 2.4.11: Blood cell population

(a) First we graph the function on Maple for the three cases of m: m < 1, m = 1,and m > 1, for a fixed θ:

0

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

32

34

20 40 60 80 100x

0

2

4

6

8

20 40 60 80 100x

0

1

2

3

4

5

20 40 60 80 100x

For m < 1, p2(x) → ∞ as x → ∞. For m = 1, p2(x) approaches bθ forx → ∞. If m > 1, then p2(x) will go to zero as x → ∞. It is actually thiscase where p2(x) takes the shape of p1(x). The graph of p1(x) is shown below,with a = 1 and b = 1.

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.22

0.24

0.26

0.28

0.3

0.32

0.34

0.36

2 4 6 8 10x

If we look at the case m � 1, we see that the peak gets sharper. In the limitof m → ∞ the graph looks linear with a sudden drop to zero, as shown below.


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0

2

4

6

8

20 40 60 80 100x

Changing the value of θ causes the horizontal scale to change, but does notaffect the shape of the curve. Large θ gives larger p2(x) at larger x. Finally,changing the value of b simply changes the vertical scale.

(b) Assume it is q days that red blood cell production is delayed. Then insteadof the original model,

xn+1 = xn − d(xn) + p(xn),

we take the number of cells gained q days earlier, so the model becomes

xn+1 = xn − d(xn) + p(xn−q).

Exercise 2.4.12: Population genetics


Exercise 2.4.13: Competition


Exercise 2.4.14: Spread of infectious disease

(a) For fixed points I , we require

I = I + kI(N − I)⇐⇒ kI(N − I) = 0⇐⇒ I = 0 or I = N.

To determine the stability of the fixed points, we let

f(I) = (1 + kN)I − kI2.


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Differentiating givesf ′(I) = 1 + kN − 2kI.

Sincef ′(0) = 1 + kN > 1,

the trivial fixed point I = 0 is unstable. The stability of the nontrivial fixedpoint I = N is given by

f ′(N) = 1 + kN − 2kN = 1 − kN.

Since kN < 2, we have −1 < 1 − kN < 1, and therefore I = N is stable.

Cobwebbing gives the following picture:

0 N In

In+1

Note that iterates may

In that case, we may

it implies that everyone

In+1

= In

be greater than N.

stop the iteration, since

has caught the disease.

The model predicts that as time progresses, everyone catches the disease.

(b) We introduce a new class of individuals, namely those who have been ill, butnow have recovered with immunity. Let the number of such individuals onday n be Rn. Then the number of newly infected individuals on day n + 1 is

k

people who are sick︷︸︸︷

In (N − In − Rn)︸︷︷︸

people available to catch disease

.

The number of people who move from the infected to the recovered class onday n are those that became newly infected d days ago. That is,

kIn−d (N − In−d − Rn−d).

Putting things together, we obtain{

In+1 = In + kIn(N − In − Rn) − kIn−d (N − In−d − Rn−d)Rn+1 = Rn + kIn−d (N − In−d − Rn−d).


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withI−d = · · · = I−2 = I−1 = 0

I0 = 1

R−d = · · · = R−1 = R0 = R1 = · · · = Rd−1 = 0

Rd = 1

The additional assumption we made is that people remain sick for precisely ddays. That is, all people who become ill at the same time recover at the sametime.

Exercise 2.4.15: Jury conditions

We have

J =

[aR pR

pJ aJ

]

.

(a) The characteristic polynomial of J is

det

[aR − λ pR

pJ aJ − λ

]

= 0

⇐⇒ (λ − aR)(λ − aJ ) − pRpJ = 0

⇐⇒ λ2 − (aR + aJ)λ + aRaJ − pRpJ = 0

⇐⇒ λ2 − tr Jλ + det J = 0,

as required.

(b) The eigenvalues of J are

λ =β ±

√

β2 − 4 γ

2,

where β := tr J and γ := det J .

Thus, we need to prove

|β| < 1 + γ < 2 ⇐⇒ |λ| < 1. (2.4)

We will show both directions of the implication separately.

“⇒”: Suppose the first part of (2.4) is satisfied. We then need to show that botheigenvalues will have magnitude less than 1. The form of the eigenvaluestells us that they could be complex with nonzero imaginary part. If thisis the case, we know that the eigenvalues can be rewritten in the form

λ =β ± i

√

4γ − β2

2, (2.5)


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with 4γ > β2.

Thus, for complex eigenvalues,

|λ|2 =β2

4+

4γ − β2

4= γ < 1,

since 1 + γ < 2 by assumption. Of course, once we have that |λ|2 < 1,we get automatically that |λ| < 1.

Now that we have the complex case covered, we can assume β2 ≥ 4γ, sothat the eigenvalues are real, and proceed the following way:

|λ| =

∣∣∣∣∣

β ±√

β2 − 4γ

2

∣∣∣∣∣

≤ |β|2

+|√

β2 − 4γ|2

(triangle inequality, regardless of the sign of the radical)

<1 + γ +

∣∣∣

√

(1 + γ)2 − 4γ∣∣∣

2(left-hand inequality relating β and γ)

=1 + γ +

∣∣∣

√

(1 − γ)2∣∣∣

2

=1 + γ + |1− γ|

2

=1 + γ + 1 − γ

2(since γ < 1, |1− γ| = 1 − γ)

= 1.

Therefore, |λ| < 1 for real or complex eigenvalues, and we have shownthis direction of the implication.

“⇐”: Suppose now the right hand side of the implication in (2.4) is satisfied(that is, |λ| < 1 for both eigenvalues λ). We then need to show that|β| < 1 + γ < 2.

Again we will break this up into the cases of complex with nonzero imag-inary part, and purely real eigenvalues.

First, if β2 < 4γ, the eigenvalues will be complex and we’ll get:

|λ| < 1 ⇒ |λ|2 < 1

⇒ β2

4+

4γ − β2

4< 1

⇒ γ < 1

⇒ 1 + γ < 2,

which implies one of the Jury conditions. To show the other condition,


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we start with

γ2 − 2γ + 1 > 0

⇒ γ2 + 2γ + 1 > 4γ

⇒ (γ + 1)2 > 4γ

⇒ (γ + 1)2 > β2 (4γ > β2, by assumption)

⇒ γ + 1 > |β|.

Note that the last step is allowed because 4γ > β2 ≥ 0 ⇒ γ > −1 (ifγ ≤ −1, we would need to put |γ + 1| > |β|). We have now shown thedesired result in the case of complex eigenvalues when 4γ > β2.

In the real case, we substitute in for both of the eigenvalues in the in-equality |λ| < 1:

−1 <β +

√

β2 − 4γ

2< 1,

−1 <β −

√

β2 − 4γ

2< 1.

We can then isolate β/2:

−1 −√

β2 − 4γ

2<

β

2< 1 −

√

β2 − 4γ

2,

−1 +

√

β2 − 4γ

2<

β

2< 1 +

√

β2 − 4γ

2.

Now we can mix these together,

−(

1 +

√

β2 − 4γ

2

)

<β

2< 1 +

√

β2 − 4γ

2,

−(

1 −√

β2 − 4γ

2

)

<β

2< 1 −

√

β2 − 4γ

2.

so as to put it into the following more compact form:

|β|2

< 1 −√

β2 − 4γ

2.

It follows that

1 − |β|2

>

√

β2 − 4γ

2, (2.6)

and since√

β2 − 4γ ≥ 0, we get

|β|2

< 1. (2.7)


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In addition, we can square both sides of (2.6), since they are both posi-tive, and obtain

1 +β2

4− |β| >

β2

4− γ

⇒ 1 − |β| > −γ

⇒ |β| < 1 + γ,

(2.8)

which is one of the Jury conditions. To obtain the other Jury condition,note that since β2 ≥ 4γ, we can write

( |β|2

)2

≥ γ. (2.9)

Combining (2.7) and (2.9) gives γ ≤(

|β|2

)2

< 12 = 1, and hence γ +1 <

2, which is the second Jury condition. The two Jury conditions togethergive

|β| < 1 + γ < 2,

which is what we were trying to prove.

Now putting together the results from “⇒” and “⇐”, we have shown (2.4):

|β| < 1 + γ < 2 ⇐⇒ |λ| < 1.

Exercise 2.4.16: Romeo and Juliet in love/hate-preserving mode

(a)

det A = det

[aR − 1 pR

pJ aJ − 1

]

= (aR − 1)(aJ − 1) − pRpJ

= (−pJ)(−pR) − pRpJ

= 0, as required

(b)

J =

[aR pR

pJ aJ

]

The eigenvalues are given by

det

[aR − λ pR

pJ aJ − λ

]

= (λ − aR)(λ − aJ) − pRpJ

= λ2 − (aR + aJ)λ + aRaJ − pRpJ

= λ2 − (aR + aJ)λ + aR + aJ − 1 (since det A = (aR − 1)(aJ − 1) − pRpJ = 0)

= [λ − 1][λ − (aR + aJ − 1)] = 0,


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That is, λ = 1 or λ = aR + aJ − 1, as required.

Exercise 2.4.17: Host-parasitoid systems: the Poisson distribution

∞∑

i=0

P (i) =∞∑

i=0

νie−ν

i!

= e−ν∞∑

i=0

νi

i!

= e−νeν (Taylor expansion)

= 1

Exercise 2.4.18: Host-parasitoid systems: the Nicholson-Bailey model

(a) Fixed points (H, P ) must satisfy

H = kHe−aP

P = cH [1 − e−aP ]

The first equationgives H = 0 or P = ln ka .

Substituting H = 0 into the second equation gives P = 0, thus yielding thetrivial fixed point,

(H∗1 , P ∗

1 ) = (0, 0).

Substituting P = ln ka into the second equation gives H = k ln k

ac(k−1) , yielding

the nontrivial fixed point,

(H∗2 , P ∗

2 ) =

(k ln k

ac(k − 1),ln k

a

)

.

To ensure P ∗2 > 0, we require k > 1.

(b)

J(H, P ) =

[ke−aP −akHe−aP

c[1 − e−aP ] acHe−aP

]

,

so

J(H∗1 , P ∗

1 ) = J(0, 0) =

[k 00 0

]

.

The eigenvalues are λ1 = k and λ2 = 0. Thus, the trivial fixed point is stablewhen 0 < k < 1, and unstable when k > 1.


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(c)

ln k +ln k

k − 1> 1

⇐⇒ ln k

(k

k − 1

)

> 1

⇐⇒ k ln k > k − 1

⇐⇒ k ln k − k + 1 > 0.

Let f(k) = k ln k − k + 1. Since f ′(k) = 1 + ln k − 1 = ln k > 0, we see thatf(k) is a monotonically increasing function for k > 1. Further, f(1) = 0.

thusk ln k − k + 1 > 0,

or

ln k +ln k

k − 1> 1,

as required.

Exercise 2.4.19: Host-parasitoid systems: the Beddington model

Note that the problem as stated in the text is very challenging. As such, this solution

is not complete, but reflects approximately what we would expect a student to be able

to do.

(a) We can find two trivial fixed points by inspection, namely

(H1, P1) = (0, 0)

and(H2, P2) = (K, 0).

The first of these fixed points represents the situation that both hosts and par-asitoids are extinct. The second represents the situation that the parasitoidsare extinct, and the host population is at its carrying capacity.

There is a third fixed point, representing co-existence. This fixed point cannotbe solved for explicitly (our apologies for the misleading wording in the prob-lem statement). Please consult the paper by Beddington, Free and Lawtonfor details on how they handled this case (see text for the complete referenceto the paper).

(b) We can determine the stability of the two trivial fixed points found above.

The Jacobian matrix of the Beddington model is

J(H, P ) =

exp[

r(

1 − HK

)

− aP] (

1 − HrK

)

−aH exp[

r(

1 − HK

)

− aP]

c(1 − e−aP ) acHe−aP

.


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The stability of (H1, P1) = (0, 0) is determined by

J(0, 0) =

(er 00 0

)

.

Here, tr J = er and det J = 0. Since r > 0 (why?), tr J > 1, and hence thefixed point (H1, P1) = (0, 0) is unstable.

The stability of (H2, P2) = (K, 0) is determined by

J(K, 0) =

(1 − r −aK

0 acK

)

.

Here, tr J = 1− r + acK and det J = acK(1− r). Depending on the value ofthe model parameters, this fixed point may be stable or unstable. Stability isguaranteed by the Jury conditions, which are

|1 − r + acK| < 1 + acK(1 − r) < 2.

Determining the conditions that guarantee stability of the steady state rep-resenting co-existence is beyond the scope of this book and solution manual.The reader is encouraged to consult the paper by Beddington, Free and Law-ton for details. It turns out that the co-existence fixed point is stable in alarge region of parameter space, especially for small values of r. For largevalues of r, the fixed point usually is unstable (depending on the values of theother model parameters).

(c) The numerical experiment is left to the reader. Lots of interesting patternsare possible depending on the choice of parameters. It is especially revealingto plot the orbits in the (H, P ) phase plane. We refer the reader to the bookMathematical Models in Biology by Leah Edelstein-Keshet for a nice discussionof a variety of numerical solutions.


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3.9 Exercises for ODEs

Exercise 3.9.1: The C14-method

Let C(t) be the concentration of carbon at any time t. We use the equation describedin the chapter on ODEs,

C ′(t) = −k C(t),

where k > 0 is a constant. This equation has the solution

C(t) = Ae−kt,

where A is a constant of integration. From the initial condition, C(0) = c0, we findthat A = c0, so the equation we get for C is

C(t) = c0e−kt.

We know by definition that C(T1/2) = 12c0 = c0e

−kT1/2 . Solving for k, we find

k =log(2)

T1/2.

The wood now has 75% of its original concentration. Call the time T3/4. Currently,

we have C(T3/4) = 34c0. We set this value for the function and solve for T3/4:

3

4c0 = c0e

−kT3/4

⇒ log(4/3) = kT3/4

⇒ T3/4 = log(4/3) · 1

k

= T1/2 ·log(4/3)

log(2)

= 2390 a

The model tells us that this piece of wood is about 2390 years old. This is about1000 years after Tutankhamen’s time (about 3300 years ago), so Tutankhamen couldnot have sat in a boat made of wood from the same tree that this piece came from.

Exercise 3.9.2: Learning curves

(a) dP/dt represents the rate of change of the performance over time, or how fastsomeone picks up a skill.

(b) When M ≥ P , dP/dt ≥ 0, so P (t) is increasing or staying constant in time.If M < P , then dP/dt < 0, which means that P (t) is decreasing in time.We expect that with more and more training, a person will never have a


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decrease in performance. Notice that if a we start with P below M , P cannever get larger than M . If P = M , P will remain constant. This model isreasonable. We interpret M as the level when someone has mastered the skill(M for master). A reasonable initial condition could be P (0) = 0; no previousknowledge.

0

0

graph of dP/dt versus P

P

dP/d

t

M

kM

(c)

00

graph of P versus t

t

P(t

)

P(0) = 0P(0) > M

M

Exercise 3.9.3: Harvesting

The differential equation we have is

u′(t) = g(u),


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where g(u) = au(1 − u/K)− cu. Rewriting, we get

g(u) = u(

(a − c) − a

Ku)

.

We can see that the graph of g(u) vs u is a parabola. Since a, K > 0, the parabolaopens downwards. It has intersections with the u axis at u = 0 and at u = K

a (a−c).

We refer to u = 0 as the trivial steady state, and u = Ka (a − c) as the nontrivial

steady state.

There are 3 different scenarios which will give three different qualitative sketches.

• Case (i): a < c. This means that the nontrivial steady state is negative. Thereis only one biologically realistic steady state, namely the trivial steady stateu = 0.

0

0

case(i): a < c

u

du/d

t

The phase portrait shows that the trivial steady state u = 0 is stable.

• Case (ii): a = c. Now there is just one intersection, at the origin. Hence thereis only one steady state, namely the trivial steady state u = 0.


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0

0

case(ii): a = c

u

du/d

t

The phase portrait shows that u = 0 is stable (for u > 0).

• Case (iii): a > c. Both steady states are biologically realistic.

0

0

case(iii): a > c

u

du/d

t

The phase portrait shows that the nontrivial steady state at u = Ka (a − c) is

stable, and the trivial one at u = 0 is unstable.

In summary, the nontrivial steady state exists only for a > c, and it is stable. Ifa ≤ c, there is only the trivial steady state, and it is stable.

Biologically, this means that when the population is not reproducing faster than itis being harvested, it will die out. If it is reproducing fast enough, it settles down


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to a steady state, determined by how much faster it is reproducing than it is beingharvested (the larger the difference, the larger the population at the steady state).

Exercise 3.9.4: Fishing

(a) 1: The fishing term says that there is a constant number of fish per unittime being removed due to fishing. This number is H1, in the units offish per unit time.

2: Here the fishing term says that the number of fish which are fished outper unit time is proportional to the current population. H2, which is thefraction of fish caught per unit time, must be between 0 and 1, since youcannot catch more than all of the fish, or fewer than none of them. Thismodel reflects the fact that fish must be found to be caught. A certainfraction of them can be found.

3: In the third model, the limits of small and large values for N reduce tomodels 2 and 1, respecively. As N → 0, the fishing term, H3N/(A + N),goes to H3N/A. This is the same situation as model 2, but with H2

“=” H3/A. That is, for small amounts of fish, the number of fish caughtper unit time it proportional to the population. More fish implies morefishing, and fewer fish implies less fishing. As N → ∞, the fishing termwill go to H3, a constant, which works the same way as H1 in model 1.That is, when there are more fish than can be caught, or more fish thanare wanted, there is just a constant amount of fish being caught per unittime. A determines how fast this limit of “too many” fish is effectivelyreached. When A is higher, model 3 stays at model 2 for a longer time.When A is lower, model 3 turns into model 1 sooner. We can see thatA is actually in the units of fish. When N = A, we see that the fishingterm becomes H3/2, giving an idea of how fast we are approaching thelimit of H3, in terms of N .

(b) Model 1 is not biologically realistic because a constant number of fish caughtper unit time regardless of how many fish there are makes no sense. For exam-ple, suppose there are 10 fish in the pond, and H1 = 3 fish/hour. Neglectingthe first term in the model equation, there is no way that this model can workafter 3 hours, because there will be only 1 fish left. Model 1 would work onlyif there was an abundance of fish. When the abundance becomes less, and thefish population becomes closer to the value H1, the model breaks down.

(c) Model 3 is a better model, because, as explained in part (a), it takes thebest parts of both models. We see that model 3 is precisely the modificationto model 1 desired in part (b). The problem with model 2 is that there isa limited number of fish which can actually be caught. Model 3 says whathappens when this limit is reached. For example, say you have a boat whichcan catch 50 fish per hour. We use model 2. Say that H2 = 0.5, and thatinitially, N = 100. After an hour, you will have caught 50 fish. After 2 hours,


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you will have caught 25 more fish, etc. If, on the other hand, you start fishingat N = 1000, then model 2 says you will have caought 500 fish after an hour,even though your boat can catch only 50 fish per hour. Model 3 correctsthis and says that there is a constant number of fish that will be caught,when the population is high. To a close approximation, H3 represents thecombined effects of people who fish with limited boat capacities, or limiteddesired quantities of fish, and puts it into a single term; one big “boat” whichcan catch at the most ||H3|| fish per hour, where ||H3|| means the magnitudeof H3, without the units.

Exercise 3.9.5: A metapopulation model

Thanks to Pandora Lam, University of Alberta, for providing this solution.

(a) We first rewrite P ′:

P ′ = cP (h − P ) − µP

= chP − cP 2 − µP

= (ch − µ)P − cP 2

= P ((ch − µ) − cP )

and then set P ′ = 0 to find the steady states, namely P = 0 and P = ch−µc =

h − µc .

Note that the steady state P = 0 always exists. The steady state P = h − µc

exists only if h − µc > 0 or h > µ

c .

(b) Let f(P ) = cP (h − P ) − µP . Then f ′(P ) = ch − µ − 2cP .

The stability of the steady state P = 0 is determined by f ′(0) = ch − µ.

If h > µc , f ′(0) > 0, and P = 0 is a stable steady state.

If h < µc , f ′(0) < 0, and P = 0 is an unstable steady state.

The stability of the steady state P = h − µc is determined by

f ′(h − µ

c) = ch − µ − 2c(h − µ

c) = −(ch − µ).

Recall that this steady state exists only if h > µc , so that f ′(h− µ

c ) < 0, whichimplies that the steady state is stable.

The bifurcation diagram follows:


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00

bifurcation diagram

h

P

P = 0

P = h − µ/c

µ/c

In conclusion, the population dies out as soon as the number of habitablepatches falls below µ

c .

Note that the horizontal line in the bifurcation diagram should be dotted for

h > µ/c.

Exercise 3.9.6: Gene activation

(a) The first term is a constant growth term. The growth of g depends linearlyon the concentration of S, with a growth rate k1. In this case, it is a constantgrowth, because s0 is constant.

The second term is the gene’s natural decay term. With nothing else, thegene would decay exponentially at a rate k2.

The third term is a self-production term, with a limited rate of reproductionof k3. When g gets large, this term becomes approximately constant, keeping

a the k3 factor (limg→∞g2

k24+g2 = 1). The parameter k4 determines how large

g has to be before this term starts behaving as in the large g limit.

(b) We start with the first equation,

dg

dt= k1s0 − k2g +

k3g2

k24 + g2

.

In order to get the last term to look like x2/(1+x2), we must divide everything

by k3, and then cleverly mulitply the last term by 1, as in1/k2

4

1/k24

= 1. We then

get:1

k3

dg

dt=

k1s0

k3− k2

k3g +

(g/k4)2

1 + (g/k4)2


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It seems logical to define x := g/k4, so that the last term is correct. We nowmust think about how we can define τ in order to make the dx/dτ = 1

k3dg/dt.

To get just a single 1k3

as the difference, we’ll need τ = k3

k4t, to cancel out the

k4’s and to bring in the 1k3

. Since x and τ are just scalar mulitples of g andt, respectively, we can pull the scalars out of the differential operator, so that

dx

dτ=

d(g/k4)

d(k3t/k4)=

1/k4

k3/k4

dg

dt=

1

k3

dg

dt.

Hence,

dx

dτ=

1

k3

dg

dt

=k1s0

k3− k2

k3g +

(g/k4)2

1 + (g/k4)2

=k1s0

k3− k2k4

k3x +

x2

1 + x2

= s − rx +x2

1 + x2,

with s := k1s0/k3, and r := k2k4/k3.

(c) We really need to graph only two more cases; when s is such that there aretwo intersections of the horizontal axis, and when s is high enough that thereare no intersections of the horizontal axis.

0 0.5 1 1.5 2 2.5−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25graph of dx/dτ versus x

x

dx/d

τ

s = 0s = 0.02s = 0.1s = 0.04

(d) We note that a bifurcation in the number of steady states occurs at s* ≈ 0.04.For s < s*, there are three steady states; the outer two steady states arestable, and the inner steady state is unstable.For s > s*, there is one stable steady state.The bifurcation diagram follows:


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0 0.05 0.1 0.15 0.2 0.25 0.30

0.5

1

1.5

2

2.5

3qualitative sketch of bifurcation diagram

x

s

X: 0.04183Y: 0.2198

Exercise 3.9.7: Linear systems


(a)

A =

[1 13 −1

]

tr A = a + d = 1 + (−1) = 0

det A = ad − bc = 1 ∗ (−1) − 1 ∗ 3 = −4 < 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A = 0 ± 1

2

√

0 − 4 ∗ (−4) = ±1

2∗ 4 = ±2

Hence, (0, 0) is a saddle point.


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5000

10000

15000

20000

25000

30000

35000

y2

5000 10000 15000 20000 25000 30000 35000

y1

(b)

A =

[2 12 3

]

tr A = 2 + 3 = 5 > 0

det A = ad − bc = 2 ∗ (3) − 1 ∗ 2 = 4 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =5

2± 1

2

√

25 − 4 ∗ (4) =5

2± 3

2= 4, 1

Hence, (0, 0) is an unstable node.

0

1000

2000

3000

4000

5000

y2

500 1000 1500 2000 2500

y1


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(c)

A =

[−1 −2

2 −1

]

tr A = −1 + (−1) = −2 < 0

det A = ad − bc = −1 ∗ (−1) − (−2) ∗ 2 = 5 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =−2

2± 1

2

√

4 − 4 ∗ (5) = −1± 1

2

√−16 = −1 ± 2i

Hence, (0, 0) is a stable spiral.

–0.2

0

0.2

0.4

0.6

0.8

1

1.2

y2

–0.5 0.5 1 1.5 2

y1

(d)

A =

[1 2

−2 1

]

tr A = 1 + 1 = 2 > 0

det A = ad − bc = 1 ∗ 1 − 2 ∗ (−2) = 5 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =2

2± 1

2

√

4− 4 ∗ (5) = 1 ± 1

2

√−16 = 1 ± 2i

Hence, (0, 0) is an unstable spiral.


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–100

–50

0

50

100

150

y2

–300 –250 –200 –150 –100 –50 50

y1

(e)

A =

[0 −22 0

]

tr A = 0 + 0 = 0

det A = ad − bc = 0 ∗ 0 − (−2) ∗ 2 = 4 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A = 0 ± 1

2

√

0 − 4 ∗ (4) = ±1

2

√−16 = ±2i

Hence, (0, 0) is a center.

–2

–1

1

2

y2

–2 –1 1 2

y1


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Exercise 3.9.8: A linear system with complex eigenvalues


We need to show that both x(1)(t) and x(2)(t) satisfy the differential equation

d

dt

[x1

x2

]

=

[α β−β α

] [x1

x2

]

.

If we let[

x1

x2

]

= x(1)(t) =

[eαt cosβt−eαt sinβt

]

,

then

d

dt

[x1

x2

]

=

[αeαt cosβt − βeαt sin βt−αeαt sinβt − βeαt cosβt

]

=

[α(eαt cosβt) + β(−eαt sin βt)−β(eαt cosβt) + α(−eαt sin βt)

]

=

[α β−β α

][eαt cosβt−eαt sin βt

]

=

[α β−β α

][x1

x2

]

,

as required.

Similarly, if we let[

x1

x2

]

= x(2)(t) =

[eαt sin βteαt cosβt

]

,

then

d

dt

[x1

x2

]

=

[αeαt sin βt + βeαt cosβtαeαt cosβt − βeαt sin βt

]

=

[α(eαt sin βt) + β(eαt cosβt)−β(eαt sin βt) + α(eαt cosβt)

]

=

[α β−β α

] [eαt sin βteαt cosβt

]

=

[α β−β α

] [x1

x2

]

,

as required.

We now let x(t) = c1x(1)(t) + c2x

(2)(t), and rewrite x(t) in the required form as


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follows:

x(t) = c1x(1)(t) + c2x

(2)(t)

= c1

[eαt cosβt−eαt sinβt

]

+ c2

[eαt sin βteαt cosβt

]

= eαt

[c1 cosβt + c2 sin βt−c1 sinβt + c2 cosβt

]

.

Introducing a and φ such that c1 = a cos(−φ) and c2 = a sin(−φ), we get

x(t) = aeαt

[cosβt cos(−φ) + sin βt sin(−φ)

−(sin βt cos(−φ) − cosβt sin(−φ))

]

= aeαt

[cos(βt + φ)− sin(βt + φ)

]

.

Note that c1 = a cos(−φ) and c2 = a sin(−φ) imply

c21 + c2

2 = a2 cos2(−φ) + a2 sin2(−φ) = a2,

or

a =√

c21 + c2

2

anda sin(−φ)

a cos(−φ)= tan(−φ) = − tan(φ) =

c2

c1,

orφ = arctan(−c2

c1).

Exercise 3.9.9: The trace-determinant formula

Given a matrix,

A =

(a bc d

)

,

The eigenvalues of A are the λ satisfying |λI −A| = 0, where I is the 2× 2 identitymatrix. Notice that tr(A) = a + d, and det(A) = ad − bc. Hence,

0 = |λI − A|= (λ − a)(λ − d) − bc

= λ2 + λ(−a − d) + ad − bc

= λ2 − tr(A)λ + det(A).

From the quadratic formula, we find

λ1/2 =tr(A) ±

√

(tr(A))2 − 4 det(A)

2.


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Q.E.D.

Exercise 3.9.10: Using the trace-determinant formula


(a)

A =

[1 53 2

]

tr A = a + d = 1 + 2 = 3

det A = ad − bc = 1 ∗ 2 − 5 ∗ 3 = −13 < 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =3

2± 1

2

√

9− 4 ∗ (−13) =3

2± 1

2

√61 ≈ 5.41,−2.41


(b)

A =

[0 −21 −3

]

tr A = a + d = 0 + (−3) = −3 < 0

det A = ad − bc = 0 ∗ (−3) − 1 ∗ (−2) = 2 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =−3

2± 1

2

√

9 − 4 ∗ (2) =−3

2± 1

2= −1,−2

Hence, (0, 0) is a stable node.

(c)

A =

[−2 4−3 4

]

tr A = a + d = −2 + 4 = 2 > 0

det A = ad − bc = −2 ∗ 4 − 4 ∗ (−3) = 4 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =2

2± 1

2

√

4− 4 ∗ (4) = 1 ± 1

2

√−12 = 1 ±

√3i

Hence, (0, 0) is an unstable spiral.

(d)

A =

[2 11 3

]

tr A = a + d = 2 + 3 = 5 > 0

det A = ad − bc = 2 ∗ 3 − 1 ∗ 1 = 5 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A =5

2± 1

2

√

25− 4 ∗ (5) =5

2± 1

2

√5 ≈ 3.62, 1.38

Hence, (0, 0) is an unstable node.


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(e)

A =

[−2 −1

1 2

]

tr A = a + d = −2 + 2 = 0

det A = ad − bc = −2 ∗ 2 − (−1) ∗ 1 = −3 < 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A = 0 ± 1

2

√

0 − 4 ∗ (−3) = ±1

2

√12 = ±

√3


(f)

A =

[−1 −2

2 1

]

tr A = a + d = −1 + 1 = 0

det A = ad − bc = −1 ∗ 1 − (−2) ∗ 2 = 3 > 0

λ1, λ2 =tr A

2± 1

2

√

(tr A)2 − 4 ∗ det A = 0 ± 1

2

√

0 − 4 ∗ (3) = ±1

2

√−12 = ±

√3i

Hence, (0, 0) is a center.

Exercise 3.9.11: Two-population model

Thanks to Pandora Lam, University of Alberta, for providing the outline of this

solution.

The two-population model, (3.8), is

x = αx + βxy,

y = γy + δxy.

There are two steady states, namely P1 = (0, 0) and P2 = (−γδ ,−α

β ).

In the solutions shown below, we determine the stability of any biologically relevantsteady states. Note that P1 always is biologically relevant. However, P2 only isbiologically relevant if α and β as well as γ and δ have opposite signs.

Knowing the stability of the steady states will be helpful in sketching the phaseportraits, not (yet) provided here.

The Jacobian matrix for the system is

J =

[∂f1

∂x∂f1

∂y∂f2

∂x∂f2

∂y

]

=

[α + βy βx

δy γ + δx

]

.


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In general then, the stability of P1 = (0, 0) is determined by

J(0, 0) =

[α 00 γ

]

,

with eigenvalues λ1 = α and λ2 = γ.

Similarly, the stability of P2 = (−γδ , −α

β ) is determined by

J(−γ

δ,−α

β) =

[0 −βγ

δ−αδβ 0

]

,

with tr J = 0 and det J = −αγ.

(a) Case α > 0, β > 0, γ > 0, δ < 0

For P1 = (0, 0):The eigenvalues are λ1,2 > 0, therefore P1 = (0, 0) is an unstable node.

For P2 = (−γδ , −α

β ):Since α and β have the same sign, P2 is not biologically relevant.

INSERT PHASE PORTRAIT HERE

Biological interpretation: We have a predator-prey model . . .

(b) Case α > 0, β > 0, γ < 0, δ < 0

For P1 = (0, 0):The eigenvalues are λ1 = α > 0 and λ2 = γ < 0, therefore P1 = (0, 0) is asaddle point.

For P2 = (−γδ , −α

β ):P2 is not biologically relevant.



(c) Case α < 0, β > 0, γ < 0, δ < 0

For P1 = (0, 0):The eigenvalues are λ1 = α < 0 and λ2 = γ < 0, therefore (0, 0) is a stablenode.

For P2 = (−γδ , −α




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(d) Case α > 0, β > 0, γ > 0, δ > 0

For P1 = (0, 0):The eigenvalues are λ1 = α > 0 and λ2 = γ > 0, therefore (0, 0) is an unstablenode.

For P2 = (−γδ , −α



Biological interpretation: We have a mutualism or symbiosis model . . .

(e) Case α > 0, β > 0, γ < 0, δ > 0

For P1 = (0, 0):The eigenvalues are λ1 = α > 0 and λ2 = γ < 0, therefore (0, 0) is a saddlepoint.

For P2 = (−γδ , −α



Biological interpretation: We have a mutualism or symbiosis model . . .

(f) Case α > 0, β < 0, γ > 0, δ < 0

For P1 = (0, 0):The eigenvalues are λ1 = α > 0 and λ2 = γ > 0, therefore (0, 0) is an unstablenode.

For P2 = (−γδ , −α

β ):P2 IS biologically relevant! Since tr J = 0 and det J = −αγ < 0, P2 =(−γ

δ , −αβ ) is a saddle point.


Biological interpretation: We have a competition model . . .

(g) Case α < 0, β < 0, γ < 0, δ < 0

For P1 = (0, 0):The eigenvalues are λ1 = α < 0 and λ2 = γ < 0, therefore (0, 0) is a stablenode.

For P2 = (−γδ , −α




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Biological interpretation: We have a competition model . . .

Exercise 3.9.12: Predator-prey model


(a) Let x(t) be the prey population, and y(t) be the natural predator population.

Assuming exponential growth for the prey population in the absense of thepredator, and exponential decay for the predator population in the absense ofprey, the 2-species interaction model reads

dx

dt= αx − βxy,

dy

dt= γy + δxy.

(b) Let r1 be the rate that the poison kills the prey population, and r2 be therate that the poison kills the predator population.

The new model then reads

dx

dt= αx − βxy − r1x,

dy

dt= γy + δxy − r2y.

Exercise 3.9.13: Inhibited enzymatic reaction

Let s = [S], e = [E], b1 = [B1], q = [Q], b2 = [B2], and i = [I ].

The first reaction gives the following differential equations:

ds

dt= −k1se + K−1b1,

de

dt= −k1se + K−1b1 + k2b1,

db1

dt= k1se − K−1b1 − k2b1,

dq

dt= k2b1.


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The second gives the following three equations:

db2

dt= k1ei − k−1b2,

de

dt= −k1ei + k−1b2,

di

dt= −k1ei + k−1b2.

Exercise 3.9.14: A feedback mechanism for oscillatory reactions


We are given the following pathway:

Ak1

k−1

Bk2

k−2

Ck3

k−3

A.

Let a = [A], b = [B], and c = [C].

A differential equation model for the above pathway then is

da

dt= k−1b + k3c − k1a − k−3a,

db

dt= k1a + k−2c − k−1b − k2b,

dc

dt= k2b + k−3a − k−2c − k3c.

Exercise 3.9.15: Enzymatic reaction with two intermediate steps


We are given the following reaction:

S + Ek1

k−1

C1

k2

k−2

C2

k3

k−3

E + P.

Let s = [S], e = [E], c1 = [C1], c2 = [C2], and p = [P ].


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A differential equation model for the above reaction then is

ds

dt= k−1c1 − k1se,

de

dt= k−1c1 + k3c2 − k1se − k−3ep,

dc1

dt= k1se + k−2c2 − k−1c1 − k2c1,

dc2

dt= k2c1 + k−3ep − k−2c2 − k3c2,

dp

dt= k3c2 − k−3ep.

Exercise 3.9.16: Self-intoxicating population


We are working with the following system:

n = (α − β − Ky)n,

y = γn − δy.

To avoid having to consider all sorts of special cases in the solution below, we assumeα, β, γ, δ, K > 0 instead of α, β, γ, δ, K ≥ 0.

(a) The term αn represents birth, increasing the population.The term −βn represents natural death, decreasing the population.The term −Kyn represents death due to a toxic environment, decreasing thepopulation.

The term γn represents the production of waste products, proportional to thesize of the population.The term −δy represents natural degradation of the waste products.

(b) We begin with the nullclines.

There are two n-nullclines, given by n = 0, namely the vertical line

n = 0

and the horizontal line

y =α − β

K.

Similarly, there is one y-nullcline, given by y = 0, namely the straight linepassing through the origin (with positive, finite slope γ/δ)

y =γ

δn.


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We now find steady states by looking for all the intersections of an n-nullclinewith a y-nullcline.

The intersection of the first n-nullcline, n = 0, and the y-nullcline is given bythe solution of n = 0 and y = γn/δ, that is, at

P1 := (n, y) = (0, 0).

The intersection of the second n-nullcline, y = (α−β)/K, and the y-nullclineis given by the solution of y = (α − β)/K and y = γn/δ, that is, at

P2 := (n, y) =

(δ

γ

α − β

K,α − β

K

)

.

We will refer to P1 as the trivial steady state and P2 as the nontrivial (co-existence) steady state. Note that P2 is biologically relevant only providedα > β.

We think it doesn’t make sense to sketch a phase portrait here (since there aretoo many cases, and not all information has been determined yet). It shouldcome later, in part (e).

(c) We think it doesn’t make sense to sketch a vector field here (since there aretoo many cases, and not all information has been determined yet). It shouldcome later, in part (e).

(d) The Jacobian matrix of the system is

J(n, y) =

[∂f1

∂n∂f1

∂y∂f2

∂n∂f2

∂y

]

=

[α − β − Ky −Kn

γ −δ

]

The stability of P1 is determined by

J(0, 0) =

[α − β 0

γ −δ

]

.

The eigenvalues of J(0, 0) are λ1 = α − β and λ2 = −δ < 0.

If α < β, then P1 is the only biologically relevant steady state. In this case,λ1 < 0, and P1 is a stable node.

If α > β, then both steady states are relevant. In this case, λ1 > 0, and P1 isa saddle point.

Similarly, the stability of P2 is determined by

J

(δ

γ

α − β

K,α − β

K

)

=

[0 − δ

γ (α − β)

γ −δ

]

.

We have tr J = −δ < 0 and det J = δ(α − β).


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If α < β, then det J < 0, and P2 is a saddle point (but in this case, P2 is notbiologically relevant).

If α > β, then det J > 0, and P2 is either a stable node or a stable spiral.

To summarize what we have so far:

If α < β, then P1 is the only relevant steady state, and it is a stable node.

If α > β, then both P1 and P2 are biologically relevant. In this case, P1 is asaddle point, and P2 is a stable node or a stable spiral.

For P2 to be a stable node, we need (tr J)2−4 detJ > 0, that is δ2−4δ(α−β) >0, or δ > 4(α − β) > 0.

Similarly, for P2 to be a stable spiral, we need (tr J)2 − 4 det J < 0, orδ < 4(α − β).

(e) Here we look at one of the cases determined above, namely when δ < 4(α−β).In this case, P1 is a saddle point, and P2 is a stable spiral.

INSERT VECTOR FIELD AND PHASE PORTRAIT HERE

Interpretation in terms of the biology: Starting from any initial population(other than zero), the population and amount of toxicity eventually reach asteady state. That is, under ideal conditions (no stochasticity), the populationpersists, no matter how much waste it produces. The steady state is reachedin a damped oscillatory fashion. However, depending on the initial conditions,trajectories may pass close to the first n-nullcline, n = 0. When this happens,n is very small. That is, in the presence of stochastic events, the populationcould become extinct.

(f) Solution not available.

Exercise 3.9.17: Fish populations in a pond

(a) Exponential growth:dT

dt= rT T

(b) Growth with competition:

dT

dt= (−mB + rT )T

(c) Exponential growth:dB

dt= rBb

Growth with competition:

dB

dt= (−nT + rB)B


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(d) Solution not available.

(e) We get the system

dT

dt= rT T − mBT,

dB

dt= rBB − nBT.

The steady states are determined by dT/dt = dB/dt = 0. This means thatany steady state (T , B) must satisfy

rT T = mBT ,

rBB = nBT .

Therefore, we get the trivial steady state,

(T , B) = (0, 0),

and the nontrivial steady state,

(T , B) = (rB

n,rT

m).

The jacobian matrix of this system, evaluated at the nontrivial steady state,is

J(rB

n,rT

m

)

=

rT − mB −mT

−nB rB − nT

=

0 −mrB

n

−nrT

m 0

.

Exercise 3.9.18: Exact solution for the logistic equation

(a) We have

N ′ = µN

(

1 − N

K

)

, N(0) = N0.

Solution method 1: We recognize the differential equation as a separableequation, so that we can write

∫ N(t)

N0

dN

N(

1 − NK

) =

∫ t

0

µ dt.

Using partial fractions, we can rewrite the left hand side:

∫ N(t)

N0

[

1

N+

1K

1 − NK

]

dN =

∫ t

0

µ dt.


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We integrate to obtain

[

ln N − ln

(

1 − N

K

)]N(t)

N0

= µt

ln

(

N

1 − NK

)N(t)

N0

= µt

ln

(

N(t)

1 − N(t)K

)

− ln

(

N0

1 − N0

K

)

= µt

ln

(1 − N0

K

)N(t)

(

1 − N(t)K

)

N0

= µt.

Exponentiating both sides and rearranging gives

K − N0

K − N(t)N(t) = N0e

µt

(K − N0)N(t) = N0eµt(K − N(t))

(K − N0 + N0eµt)N(t) = N0Keµt

N(t) =N0Keµt

K − N0 + N0eµt

=eµtN0

1 + N0

K (eµt − 1).

Solution method 2: Let u = 1N . Then N = 1

u and

dN

dt= − 1

u2

du

dt.

Substitution into the logistic equation gives

− 1

u2

du

dt= µ

1

u

(

1 − 1

K

1

u

)

du

dt= µ

(1

K− u

)

.


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We separate variables and integrate, as follows:

∫ u(t)

u0

du1K − u

=

∫ t

0

µ dt

− ln

(1

K− u

)∣∣∣∣

u(t)

u0

= µt

− ln

(1

K− u(t)

)

+ ln

(1

K− u0

)

= µt

ln

( 1K − u0

1K − u(t)

)

= µt

1 − Ku0

1 − Ku(t)= eµt

1 − Ku(t) = (1 − Ku0)e−µt

u(t) =1

K

[1 − (1 − Ku0)e

−µt].

Now we return to original variables, as follows:

1

N(t)=

1

K

[

1 −(

1 − K1

N0

)

e−µt

]

N(t) =K

1 −(

1 − KN0

)

e−µt

=Keµt

eµt − 1 + KN0

=eµtN0

N0

K eµt − N0

K + 1

=eµtN0

1 + N0

K (eµt − 1).

(b) This solution is of the same form as that of the Beverton-Holt model, exceptwe have eµt in place of rn+1.


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4.5 Exercises for PDEs

Exercise 4.5.1: Diffusion through a membrane

ut = Duxx, ut = 0

(a) uxx = 0 ⇒ ux = const = c ⇒ u(x) = cx + d

Boundary conditions:

u(0) = c1 ⇒ d = c1

u(L) = c2 ⇒ cL + c1 = c2 ⇒ c =c2 − c1

L

Solution:

u(x) =c2 − c1

Lx + c1

For c2 > c1:

0 L x

u(x)

outsideinside

u(0) = c1

u(L) = c2

(b) J(x) = −D ∂∂xu(x) = −D c2−c1

L = −DL (c2−c1). The flux is proportional to the

concentration difference. The proportionality factor DL is called permeability.

Exercise 4.5.2: Fundamental solution


Exercise 4.5.3: Signalling in ant populations

ut = Duxx, u(0) = αδ0(x), D = 1 (4.10)

(a) Fundamental solution of {ut = Duxx, u(0) = δ0(x)} is g(x) = 1√2πt

e−x2

4t .

Hence u(x) = αg(x) solves (4.10).


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At x(t): u(x(t)) = 0.1 · α = αg(x)

⇒ g(x) =1

10, e−

x2

4t =

√2πt

10, e

x2

4t =10√2πt

, x2 = 4t ln

(10√2πt

)

⇒ x(t) =

√

4t ln

(10√2πt

)

(b)

0 5 10 15t

0

1

2

3

x(t)

Range of Influence

(c) x(t) defined only for ln(

10√2πt

)

> 0, hence 10√2πt

> 1. So

10 >√

2πt, 100 > 2πt,50

π> t

⇒ t∗ =50

π≈ 15.9

Exercise 4.5.4: Dingos in Australia

(Thanks to Dr. Markus Owen (Nottingham), who used this problem in one of hisMath-bio classes).

ut = Duxx + ku(1 − u), k = 1

(a) D1 = 100, wave speed of a travelling wave,

c∗ = 2√

D1f ′(0), f ′(0) = k = 1

= 2√

D1 = 20

(miles

month

)

distance = 100 miles ⇒ T = 100 milesc∗ = 100

20 = 5 months.


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The decay rate of this wave front is:

λ1 = − c∗

2D1= − 10

100= − 1

10,

The wave looks like e−110 x near farm A.

e (−x/10)

c*

(b) Between A and B: D2 = 50

Decay rate λ1 = − 110 = − c

2D2

⇒ wave speed = c = −λ12D2 =1

10· 2 · 50 = 10

⇒ T2 = 10 months from farm A to B.

c

A B

Exercise 4.5.5: Signal transport in the axon

ut = uxx + u(1 − u)(u − 1

2)

ux(t, 0) = 0, ux(t, l) = 0

0 l

(a) Steady states: ut = 0. Introduce v := ux.

ux = v

vx = −u(1− u)(u − 1

2) = u3 − 3

2u2 +

1

2u

(b) equlibria of (a): v = 0, u = 0, 1, 12 .


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Jacobian:

Df(u, v) =

0 1

3u2 − 3u + 12 0

Df(0, 0) =

0 1

12 0

, tr(Df(0, 0)) = 0, det(Df(0, 0)) < 0 ⇒ saddle

Df(1

2, 0) =

0 1

34 − 3

2 + 12 0

=

0 1

− 14 0

tr(Df(1

2, 0)) = 0, det(Df(

1

2, 0)) > 0 ⇒ center

Df(1, 0) =

0 1

12 0

, tr(Df(0, 0)) = 0, det(Df(0, 0)) < 0 ⇒ saddle

(c) Hamilton function if

d

dxH(u, v) = 0 and ux =

∂H

∂v, vx = −∂H

∂u,

Here H(u, v) = 12 (v)2 − 1

4u4 + 12u3 − 1

4u2.

Let’s check:

∂H

∂v= v = ux X

∂H

∂u= −u3 − 3

2u2 − 1

2u = −vx X

d

dxH(u, v) =

∂H

∂u

du

dx+

∂H

∂v

dv

dx= −vxux + uxvx = 0.

(d)


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0 1/2 1 u

v

(e) Neumann boundary conditions:

v(0) = 0 v(l) = 0

Following candidates in the phase-portrait of (d):

1/2 u

v

I

1/2 u

v

II

1/2 u

v

III

etc.

As functions of x:

x

u(x)

0 lI x

u(x)

0 lII x

u(x)

0 lIII

etc.


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x

u(x)

l0

a = 9

a = 8 + 1

(4.11)

(f) Solution not available.

Exercise 4.5.6: Separation


Exercise 4.5.7: Linear transport


Exercise 4.5.8: Correlated random walk



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5.8 Exercises for Stochastic Models

Exercise 5.8.1: Forest ecosystem succession


Exercise 5.8.2: Princeton forest


Exercise 5.8.3: Mean and variance for a sum of random variables


Exercise 5.8.4: Mean and variance for a negative binomial distribution


Exercise 5.8.5: Random walk derivation of a diffusion-advection equation


Exercise 5.8.6: Spatially varying diffusion model


Exercise 5.8.7: Variance for a branching process


Exercise 5.8.8: The survival of right whales


Exercise 5.8.9: An explicit solution for the pure birth process



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6.6 Exercises for Cellular Automata

Exercise 6.6.1: Wolfram rule 108


Exercise 6.6.2: Monotonous automaton


Exercise 6.6.3: Greenberg-Hastings automata


Exercise 6.6.4: Game of life


Exercise 6.6.5: Boundaries of finite grids


Exercise 6.6.6: Stochastic epidemic automaton



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7.7 Exercises for Parameter Estimation

Exercise 7.7.1: Maximum likelihood estimation for binomial distribution


Exercise 7.7.2: Maximum likelihood estimation for exponential distri-bution


Exercise 7.7.3: Hill-climbing algorithm


A Course in Math Bio Sol

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