The Partial Diﬀerential Equations of Biology: …keener/classes/math5120/course_notes.pdfThe Partial Diﬀerential Equations of Biology: Signals and Patterns or Biology in Time andSpace

The Partial Differential Equations of Biology: Signals and Patterns

or

Biology in Time and Space

James P. Keener

University of Utah

14 April, 2015

2

Contents

1 Introduction 7

2 Background Material 9

2.1 Review of multivariable calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.1.1 Different Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 A Brief Overview of Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 First Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.2 Systems of first order equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Conservation 17

3.1 The Conservation Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2 Examples of Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 The Diffusion Equation 21

4.1 Discrete Boxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.2 A Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 Realizations of a diffusion process 27

5.1 Following a single particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3

4 CONTENTS

5.2 Other features of Brownian particle motion - Escape times and splitting probabilities . . . . . 28

5.3 Following several particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.3.1 Particle Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

6 Solutions of the Diffusion Equation 33

6.1 On the Infinite Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.2 On the semi-infinite line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

6.3 With Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.4 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

6.5 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.5.1 Method of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.5.2 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.5.3 Crank-Nickolson Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

7 Transport of Oxygen 39

7.1 Transport across Membranes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.2 Delivery of oxygen to tissue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.3 Oxygen exchange between capillaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7.4 Facilitated Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

7.5 Facilitated Diffusion in Muscle Respiration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

8 Diffusion and Reaction 49

8.1 Birth-Death with Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

8.2 Growth with a Carrying capacity - Fisher’s Equation . . . . . . . . . . . . . . . . . . . . . . . 50

8.2.1 On a Bounded Domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

8.3 Resource Consumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

CONTENTS 5

8.4 Spread of Rabies - SIR with Diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

8.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

9 The Bistable Equation 61

9.1 The Spruce Budworm Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

9.2 The Cable Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

9.3 Traveling Waves for the Bistable Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.3.1 Propagation Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

9.4 The Fire-Diffuse-Fire Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

10 Advection 81

10.1 Simple Advection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

10.2 Structured Populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

10.3 Red Blood Cells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

10.4 Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

10.4.1 Delay Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

10.4.2 The Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

10.4.3 Method of Lines; Upwinding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

10.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

11 Advection with diffusion 91

11.1 Axonal transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

11.2 Transport with Switching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

11.3 Ornstein-Uhlenbeck Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

12 Chemotaxis 97

6 CONTENTS

13 Pattern Formation - The Turing Mechanism 101

13.0.1 The Turing Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

13.1 Cell Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

13.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

14 Agent-Based Modeling 105

15 Appendices 107

15.1 Matlab Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

15.1.1 A Matlab Primer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

15.1.2 A.1: Method of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

15.1.3 A.2: Discrete Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

15.1.4 A.3: Motion of a Brownian Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

15.1.5 A.4: Gillespie algorithm for Particle Decay . . . . . . . . . . . . . . . . . . . . . . . . 110

15.1.6 A.5: Diffusion Equation solution via Crank Nicolson . . . . . . . . . . . . . . . . . . . 111

15.1.7 A.6: Diffusion Equation with growth/decay solution via Crank Nicolson . . . . . . . . 112

15.1.8 A.7: Solution of Delay differential equation using the Method of Lines (MOL) . . . . . 114

15.1.9 A.8: Solution of the Ornstein-Uhlenbeck equation . . . . . . . . . . . . . . . . . . . . 114

15.2 Constants and Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

15.2.1 Diffusion Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

15.2.2 Physical constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Chapter 1

Introduction

Mathematical biology is, broadly speaking, about how biological objects move and interact. The interactionpart of biology is often studied using ordinary differential equations. Indeed, there is a lot of insight thatcan be gained from the study of ordinary differential equation descriptions of biological processes. However,this is quite limited when it comes to understanding real biological situations, because biological objects areessentially never homogeneously distributed in space, or well-mixed, even in a test tube or on a Petrie dish.In fact, it is a primary feature of biological objects that there are spatial differences and correspondinglymovement of objects from one place to another. So, whether one is studying the spread of an infectiousdisease or an invasive species, or the movement of an action potential along a nerve axon, it is cruciallyimportant to include the effects of spatial differences.

This book is about the dynamics of biological objects in time and space. Consequently, it is about partialdifferential equations. It is intended for an undergraduate audience, and no previous background in partialdifferential equations is required. In the next chapters you will find very cursory summaries of what isneeded from multivariable calculus, and from ordinary differential equations, because these things are useda lot. Also, this material is presented from a heavily computational perspective, using Matlab to computesolutions. Consequently, one needs to know, or learn Matlab, for this material to be best assimilated.

7

8 CHAPTER 1. INTRODUCTION

Chapter 2

Background Material

2.1 Review of multivariable calculus

For this endeavor, we will be dealing with functions of space and time. The region of space of interest couldbe the inside of a cell, the inside of blood vessels in a human body or a lake in the mountains. Eitherway, we usually describe position in this space by the Cartesian coordinates x, y, and z, and time by thevariable t. The quantity of interest may be the concentration of calcium ions in a cell or the concentrationof microorganisms in the lake, but it is typically denoted by some function u = u(x, y, z, t).

Several quantities of interest for this function include its the rate of change at some particular point, denotedby the partial derivative ∂u

∂t, and its gradient, a vector valued function, ∇u = (∂u

∂x, ∂u∂y, ∂u∂z

).

One important use of the gradient is the object

n =∇u|∇u| , (2.1)

which, if |∇u| 6= 0, is a unit vector pointing in the direction of the greatest increase of the function u. Theimportance of this to a skier or snowboarder is obvious, pointing in the direction of the ”fall-line”. It is alsonoteworthy that n is perpendicular (orthogonal) to level surfaces of the function u.

It could also be that the quantity of interest is a vector valued function, for example, the velocity of the waterin the lake or the velocity of the blood in an artery, given by v = (v1, v2, v3) where each of the componentsof the vector v is a function of x, y, z, and t. Of course, this vector valued function could be the gradientof some other function, u. Either way, one important quantity for a vector valued function is its divergence,denoted

∇ · v =∂v1∂x

+∂v2∂y

+∂v3∂z

. (2.2)

The most important theorem regarding the divergence operator also gives an understanding to its physicalmeaning, called the divergence theorem,

∫

Ω

∇ · vdV =

∫

∂Ω

v · ndS, (2.3)

where Ω is a region of interest, ∂Ω is its boundary, and n is the unit outward normal to the boundary.Consequently,

∫

Ω∇ · vdV represents the net flux of some material with velocity v out of the region Ω.

9

10 CHAPTER 2. BACKGROUND MATERIAL

In one-dimension, the divergence theorem is the same as the fundamental theorem of calculus

v(b)− v(a) =

∫ b

a

dv

dxdx. (2.4)

2.1.1 Different Coordinate Systems

There are some other facts from vector calculus needed for this book, namely what gradient, divergence andLaplacian operators look like in different coordinate systems. The two most important coordinate systemshere are polar and spherical coordinates.

Polar Coordinates

The relationship between polar and Cartesian coordinates is given by

x = r cos θ, y = r sin θ. (2.5)

In polar coordinates, the Laplacian operator is

∇2u =1

r

∂

∂r

(

r∂u

∂r

)

+1

r2∂2u

∂θ2. (2.6)

Spherical Coordinates

The relationship between spherical and Cartesian coordinates is given by

x = r cos θ sinφ, y = r sin theta sinφ, z = r cosφ. (2.7)

In spherical coordinates, the Laplacian operator is

∇2u =1

r2∂

∂r

(

r2∂u

∂r

)

+1

r2 sin2 φ

∂2u

∂θ2+

1

r2 sinφ

∂

∂θ

(

sinφ∂u

∂φ

)

. (2.8)

2.2 A Brief Overview of Ordinary Differential Equations

2.2.1 First Order Equations

An ordinary differential equation specifies a relationship between the (time) derivative of some quantity uand its values through, say,

du

dt= f(u, t). (2.9)

This equation is autonomous if f is independent of t, so that

du

dt= f(u). (2.10)

Many of the problems discussed in this book are autonomous in time.

2.2. A BRIEF OVERVIEW OF ORDINARY DIFFERENTIAL EQUATIONS 11

u-0.2 0 0.2 0.4 0.6 0.8 1 1.2

du/d

t

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Figure 2.1: Plot of dudt

vs. u for the bistable function f(u) = au(1− u)(u− α) with α = 0.25, a = 10.

If u is a scalar quantity, the solution of equation (2.10) can be readily understood using graphical means,i.e., by simply plotting du

dtvs. u. An example is shown in Fig. 2.1.

The first thing to notice are the zeros of f(u), i.e., the equilibria. For the example f(u) = au(u− 1)(α− u),shown in Fig. 2.1, these are at u0 = 0, u0 = α, and u0 = 1. Next, one can determine the direction ofmovement if u is not at an equilibrium. These are shown with arrows in Fig. 2.1. For example, if 0 < u < α,dudt< 0 indicating that u is decreasing there, while if α < u < 1, du

dt> 0 so that u is increasing there. This

is our first indication that u0 = 0 and u0 = 1 are stable equilibria, while u0 = α is unstable.

The next thing to do is to linearize the equations about the equilibria. Linearization is a very importantprocedure; it is a good idea to understand it thoroughly.

The linearization of any (differentiable) object G(u) about u0 is defined as

limǫ→0

∂

∂ǫG(u0 + ǫU), (2.11)

so the linearization of the differential equation (2.10) about any of its equilibria is

limǫ→0

∂

∂ǫ

(

d

dt(u0 + ǫU)− f(u0 + ǫU)

)

, (2.12)

which reduces todU

dt= f ′(u0)U. (2.13)

The solution of the linearized problem is the exponential function

U(t) = U0 exp(f′(u0)t), (2.14)

and it is now obvious that U(t) grows if f ′(u0) > 0 and decays if f ′(u0) < 0. Hence, for our example, theequilibria u0 = 0 and u0 = 1 are linearly stable while the equilibrium u0 = α is unstable. This agrees withour graphical stability analysis.

Finally, it is noteworthy that the equation (2.10) is separable and can be rewritten as

du

f(u)= dt, (2.15)

which, after integrating both sides of the equation, enables us to write

F (u)− F (u(0)) = t, (2.16)


where F (u) =∫ u du

f(u) and u = u(0) at t = 0. In most situations, this is not a particularly useful represen-

tation of the solution, since analytical inversion of the function F (u) to find u(t) explicitly is almost alwaysimpossible. However, through the miracles of Matlab, it is easy to graph this solution. That is, plot t as afunction of u and then reverse the axes.

As an example, for the function f(u) = au(1− u)(u− α),

F (u) =1

aα(α− 1)(α ln(1 − u)− ln(|u− α|) + (1− α) ln(u)) , (2.17)

a plot of which is seen in Fig. 2.2a, and then, reversing the axes gives the plot of u(t) as a function of t,shown in Fig. 2.2b.

u-0.2 0 0.2 0.4 0.6 0.8 1 1.2

F(u

)

-4

-3

-2

-1

0

1

2

3

t-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2

u(t)

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Figure 2.2: Left: Plot of F (u) vs. u, and Right: plot of u(t) as a function of t, for the function (2.17).

This plot illustrates the fact that the solution has a different outcome as t → ∞ depending on the initialcondition. Clearly, (as we already knew), if 0 < u(0) < α, u(t) → 0 as t → ∞, whereas, if α < u(0) < 1,then u(t) → 1 as t→ ∞.

2.2.2 Systems of first order equations

We now turn our attention to systems of first order equations, which can still be written in the form of (2.9)provided we recognize that u is a vector, rather than a scalar, quantity. The most important example forthis text is when there are two unknown scalar functions u(t) and v(t) and the equations describing theirevolution is in the form

du

dt= f(u, v), (2.18)

dv

dt= g(u, v). (2.19)

As with first order equations, a useful way to proceed is with a graphical, or phase plane, analysis. The firststep of this analysis is to plot the nullclines, the curves in the u − v plane along which either u or v do notchange, i.e., du

dt= 0 or dv

dt= 0.

There are many examples of this procedure in this book, however, for purposes of illustration, let’s look atsolutions of the second order differential equation

d2u

dt2+ f(u) = 0, (2.20)

where f(u) = au(1 − u)(u − α), the same function as used above. To write this equation as a first ordersystem, we set v = du

dt, and then the equations are

du

dt= v, (2.21)

2.2. A BRIEF OVERVIEW OF ORDINARY DIFFERENTIAL EQUATIONS 13

dv

dt= −f(u). (2.22)

The nullclines for this system are easily determined, being the line v = 0 for the u nullcline, and f(u) = 0for the v nullclines, i.e., the lines u = 0, u = α, and u = 1. These are shown plotted in Fig. 2.3 as dashedlines.

u-0.2 0 0.2 0.4 0.6 0.8 1 1.2

v

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

u-0.2 0 0.2 0.4 0.6 0.8 1 1.2

v

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Figure 2.3: Phase portrait for the equation (2.21-2.22).

The next step is to identify the critical points, i.e., the points at which dudt

and dvdt

are both zero, hence,points of equilibrium. These are, of course, all the intersections of the u and v nullclines. For this example,they are the points with v = 0 and u = 0, α and 1.

Next, we determine the direction of the flow in regions bounded by the nullclines. For this example, uincreases if v > 0 and decreases if v < 0, while v increases if 0 < u < α and decreases if α < u < 1. It isalso possible at this point to sketch a few typical trajectories by following the vector of flow directions. Itquickly becomes apparent that the equilibria at u = 0 and u = 1 are saddle points, but the nature of thecritical point at u = α is not decided by graphical means alone.

To classify the critical points completely it is necessary to do a linear stability analysis. For the generalsystem (2.18-2.19), the linearized system is

d

dt

(

UV

)

=

(

∂f0∂u

∂f0∂v

∂g0∂u

∂g0∂v

)(

UV

)

= A

(

UV

)

, (2.23)

where the matrix A is the Jacobian for this system, and f0, g0 denote evaluation at the equilibria u0, andv0.

The classification of an equilibrium is made based on the eigenvalues of the Jacobian matrix A, i.e., rootsof the polynomial λ2 + Tr(A)λ + det(A) = 0, with Tr and det representing the trace and determinant of Arespectively. If det(A) < 0, then there are two real roots, of opposite sign; the equilibrium is a saddle point.If det(A) > 0, there are four possible outcomes, depending on the sign of the discriminant, disc=Tr - 4 det,and the sign of Tr. If disc> 0, the two roots are real both with the same sign as Tr, and if disc¡0, the tworoots are complex conjugate pair with the sign of the real part the same as the sign of Tr. If the roots arereal, the equilibrium is called a node, and if the roots are complex, it is called a spiral point. If the realparts are positive, the equilibrium is unstable, while if they are negative, the equilibrium is unstable. Theintermediate case with Tr= 0 has neutral stability and is called a center. Thus, the four cases with det> 0are stable node, stable spiral, unstable node, unstable spiral. These four are summarized in the Table 2.1.

Typical phase portraits for a saddle point, stable spiral and stable node are shown in Fig. 2.4


Table 2.1: Summary of stability criteria.> 0 < 0

> 0 stable spiral stable node< 0 unstable spiral unstable node

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Figure 2.4: Typical phase portraits for (left to right) a saddle point, stable spiral, and stable node.

For the example problem (2.21-2.22), the Jacobian matrix is

A =

(

0 1−f ′(u0) 0

)

, (2.24)

and since Tr = 0, the determining feature is the sign of f ′(u0). In particular, the critical points at u0 = 0and u0 = 1 are both saddle points, while the critical point at u0 = α is a neutral center.

With this information it is now usually possible to get an understanding of the behavior of the solutions.However, there are situations, where this information is not sufficient to tell the whole story, and one suchcase is when there are isolated, closed orbits, i.e., limit cycles. We do not discuss limit cycles any furtherhere.

Another important feature of these equations is that sometimes, but not often, it is possible to find expressionsfor some of the solution curves. For the current example problem, the slope of trajectories in the phase planeis given by

dv

du= −f(u)

v, (2.25)

which is separable, yielding−f(u)du = vdv. (2.26)

Integrating both sides of this equation, we find that

F (u) +1

2v2 = F (u0), (2.27)

where F (u) =∫ u

f(u)du, and u0, v0 = 0 is a point on the trajectory. These trajectories can easily be plottedin the u− v phase plane; Some examples are shown in Fig. 2.3.

2.3 Exercises

1. For the functions

(a) f = x2 + y2

2.3. EXERCISES 15

(b) f = xy

(a) Find ∂f∂x

, ∂2f∂x∂y

, ∇f ;(b) Determine if there are critical points. Which, if any are local maxima?

(c) Visualize the surface and the level curves for the function z = f(x, y) using Matlab. Exploredifferent Matlab functions for plotting: mesh, mesh, mesh, surface, contourplot and see whatthey do. Type

help mesh

or similar to see the available options.

2. Determine which, if any, of the following vector fields are gradient fields. If it is a gradient field, findφ such that F = ∇φ.

(a) F = (x+ y, x− y),

(b) F = (x2y, xy2).

3. Sketch phase portraits for the system

d

dt

(

uv

)

= A

(

uv

)

, (2.28)

with

(a)

A =

(

0 11 0

)

, (2.29)

(b)

A =

(

0.1 1−1 0.1

)

, (2.30)

(c)

A =

(

−0.1 1−1 −0.1

)

, (2.31)

(d)

A =

(

−1 0.20.3 −0.3

)

. (2.32)


Chapter 3

Conservation

3.1 The Conservation Law

Suppose we have some quantity u which is changing in time, and we wish to track how it changes. It mustbe that

d

dt

∫

Ω

udV = −∫

∂Ω

J · ndS +

∫

Ω

fdV, (3.1)

where Ω is any closed region in space, J is the (vector-valued, positive if outward) flux of the quantity u, andf is the rate of production (or destruction, when f is negative) of u. One can put words to this equation bysaying that the way in which the amount of stuff inside Ω changes is determined by the net rate of materialflux across the boundary and the net rate of production of material in the domain.

Now we do two things: First, we apply the divergence theorem and then we pass the time derivative throughthe integral on the left side to get

0 =

∫

Ω

(∂u

∂t+∇ · J − f

)

dV. (3.2)

There are some technical issues associated with these steps that we do not describe here, but the upshot ofthis is that since the region Ω is arbitrary, the integrand is zero, or

∂u

∂t= −∇ · J + f. (3.3)

This equation is called a conservation equation, and it is inviolable.

Where modeling comes into play, and where much of the fun of this book is to be found, is in determinationof the flux J and the production term f .

JFlux

production

f(u)

Ω

17

18 CHAPTER 3. CONSERVATION

3.2 Examples of Flux

There are three examples of flux that are important in biology, these are advection, diffusion and taxis.

Advection: Suppose particles with concentration u are dissolved in water and the water is moving withvelocity v and that the particles are also moving with the same velocity. The flux of concentration at anypoint is the velocity times the concentration

J = uv. (3.4)

Fick’s law: If individual particles have a velocity that is different than that of the water in which they aredissolved, for example, a random motion, then we might reasonably expect that they would tend to move, onaverage down their concentration gradient. This is certainly what happens in our ordinary experience. Forexample, if you put a drop of ink into water, it will very quickly disperse, or diffuse, away, and eventually theink will be uniformly distributed throughout the water, with no regions with higher of lower concentration.In math, this is

J = −D∇u, (3.5)

and is called Fick’s law, and D is called the diffusion coefficient.

Fick’s law is not actually a law, but a model, hence appropriate only in certain contexts. A rough derivationof Fick’s law can be given using ideas from statistical mechanics. For a dilute chemical species the Gibb’sfree energy is approximately G = kbTu lnu and the chemical potential is µ = ∂G

∂u= kBT (lnu+ 1) where kB

is Boltzmann’s constant and T is temperature in degrees Kelvin. (In the chemical literature, the chemicalpotential is taken to be µ = kBT lnu, since additive constants are unimportant for potentials.) Now, forceis the negative gradient of potential, F = −∇µ = −kBT ∇u

uand in a viscous environment, velocity is

proportional to force, ξv = F and flux is velocity times concentration, so that

J = −kBTξ

∇u, (3.6)

which is Fick’s law, and the diffusion coefficient is D = kBTξ

.

In his theory of Brownian motion, Einstein gave a quantitative understanding of diffusion by showing thatif a spherical solute molecule is large compared to the solvent molecule, then the diffusion coefficient is

D =kBT

6πµa, (3.7)

where µ is the coefficient of viscosity for the solute, and a is the radius of the solute molecule.

Fick’s law shows up in other contexts, and is known as Newton’s law of cooling if u is heat, while it is knownas Ohm’s law if u is voltage and J is ionic current.

Taxis: It is often the case that objects respond to the gradient of some other quantity, say ψ, so that

J = ξu∇ψ. (3.8)

When ψ is a chemical gradient and u is cellular concentration, this is called chemotaxis, while when ψ isthe voltage potential and u is the concentration of (charged) ions, this describes the response of ions to anelectric field.

Since ions both diffuse and move along voltage potential gradients, their movement of an ion is described bya combination of terms,

J = −D(∇u+zF

RT∇ψ), (3.9)

3.2. EXAMPLES OF FLUX 19

and is called the Poisson-Nernst law. Here, z is the (integer) charge of the ion, F is Faraday’s constant,R = kBT is the universal gas constant, and NA is Avogadro’s number.

20 CHAPTER 3. CONSERVATION

Chapter 4

The Diffusion Equation

The diffusion equation results from the conservation equation when we assume that Fick’s law holds, andthere is no production or destruction, so that

∂u

∂t= ∇(D∇u), (4.1)

or in one spatial dimension∂u

∂t= D

∂2u

∂x2. (4.2)

In this section we give two different derivations, and corresponding interpretations, of this equation.

4.1 Discrete Boxes

Suppose there are a number of boxes connected side-by-side along a one-dimensional line, with concentrationof some chemical species uj in box j, −∞ < j <∞. Now suppose that the chemical leaves box j at rate 2α,so that the concentration in box j is governed by

dujdt

= −2αuj, (4.3)

provided there is no inflow. However, we also assume that the flow out of box j is evenly split to go into theneighboring boxes j − 1 and j + 1. Consequently,

dujdt

= αuj−1 − 2αuj + αuj+1, (4.4)

since half of the flow out of cells j − 1 and j + 1 is into cell j.

A slightly different way to write this is as

dujdt

= −α(uj − uj−1) + α(uj+1 − uj), (4.5)

and we notice that this represents the discrete Fick’s law, since the terms α(uj−1 − uj) and α(uj+1 − uj)represent the net flux into box j from box j − 1 and j + 1, respectively.

21

22 CHAPTER 4. THE DIFFUSION EQUATION

It is a straightforward matter to simulate this system of ordinary differential equations. The Matlab codeto do so is given in the Appendix A.1.

Now suppose that uj is a sample of a smooth function u(x, t) at points x = j∆x, i.e., uj = u(j∆x, t). UsingTaylor’s theorem,

uj±1 = u(xj ±∆x, t) = u(xj , t)±∆x∂

∂xu(xj , t) +

1

2∆x2

∂2

∂x2u(xj , t)±

1

6∆x3

∂3

∂x3u(xj , t) +O(∆x4). (4.6)

It follows that∂u

∂t= α∆x2

∂2u

∂x2+O(∆x4), (4.7)

which to leading order in ∆x is the diffusion equation with diffusion constant D = α∆x2.

4.2 A Random Walk

Consider the problem where we take a number of random steps, and for each step we make a decision totake a step of length m = -1, 0, or 1, with probability α, 1 − 2α and α, respectively. Let xn be the sum ofall steps, xn =

∑nj=1mj , that is, the total distance traveled after n steps.

The first thing to do here is to simulate this process. It is an easy thing to do, and the Matlab code is foundin the Appendix A.2.

Let’s now calculate the probability that xn has the value k, denoted P (xn = k),

P (xn = k) = αP (xn−1 = k − 1) + (1− 2α)P (xn−1 = k) + αP (xn−1 = k + 1). (4.8)

In words, the probability that xn is k is the sum of three terms, α times the probability that xn−1 is k− 1, αtimes the probability that xn−1 is k + 1, and 1− 2α times the probability that xn−1 is k. Now, we supposethat P (xn = k) is a sampling of a smooth function p(x, t), P (xn = k) = p(k∆x, n∆t). Again, using Taylorseries, it follows that, to leading order in ∆t and ∆x,

∂p

∂t= α

∆x2

∆t

∂2p

∂x2, (4.9)

which is, once again, the diffusion equation, with diffusion coefficient D = α∆x2

∆t.

The variables m and xn are said to be random variables. For any random variable y which can take on kdifferent values, say yj , j = 1, · · · , k, we define the expected value of y to be

E(y) =

k∑

j=1

yjpj , (4.10)

where pj is the probability of the event yj . Some obvious identities are that E(ay) = aE(y) for any scalara, and E(E(y)) = E(y), since E(y) is not random and so occurs with probability one.

Clearly, the expected value of m is

E(m) = −1 · α+ 0 · (1 − 2α) + 1 · α = 0. (4.11)

Furthermore, the expected value of xn is

E(xn) = E(

n∑

i=1

mi) =

n∑

i=1

E(mi) = 0. (4.12)

4.2. A RANDOM WALK 23

A second important statistic for any random variable y is its variance, defined as

var(y) = E((y − E(y))2) = E(y2 − 2yE(y) + E(y)2) = E(y2)− (E(y))2. (4.13)

For example, the variance of m is

var(m) = E(m2) = (−1)2α+ (1)2α = 2α. (4.14)

The variance of xn is a little more complicated to calculate. It is

var(xn) = E(x2n) = E((

n∑

i=1

mi)2) =

n∑

i=1

n∑

j=1

E(mimj). (4.15)

Now, we need to calculate E(mimj) for i 6= j. Since there are three values that mi and mj can each takeon, there are nine ways that mimj can be arranged, but that ends up with three different possible values,namely 1, 0, and -1. It is 1 if mi and mj are the same and nonzero, with probability 2α2, and similarly itis -1 if mi and mj are different and nonzero, also with probability 2α2. It is zero if either of mi or mj arezero. Consequently, E(mimj) = 0. Now, it follows directly that

var(xn) =

n∑

i=1

E(m2i ) = 2αn. (4.16)

It is a general fact that if y and z are independent random variables, meaning that P (yz) = P (y)P (z), then

var(y + z) = var(y) + var(z). (4.17)

(This is a statement that you, the reader, should verify.) This explains why E(mimj) = 0 with i 6= j, sincemi and mj are independent (the outcome of the ith trial is independent of the outcome of the jth trial).

It is interesting to compare the statistics of the random variable xn with those of the probability densityfunction described by the diffusion equation (4.9). Suppose x is a random variable whose probability densityis given by p(x, t) satisfying the diffusion equation (4.2). That p(x, t) is a probability density means thatat time t, the probability of finding the random variable in the interval between x and x + dx is p(x, t)dx.Consequently, since the random variable is always somewhere,

∫ ∞

−∞p(x, t)dx = 1 (4.18)

for all time. This is actually a consequence of the diffusion equation, since if we integrate both sides of theequation with respect to x, we find

d

dt

∫ ∞

−∞p(x, t)dx = D

∫ ∞

−∞

∂2p

∂x2dx = D

∂p

∂x

∣

∣

∣

∞

−∞= 0, (4.19)

so that∫∞−∞ p(x, t)dx is constant in time.

Now let’s determine E(x) by multiplying both sides of the diffusion equation (4.2) by x and integrating, tofind

d

dtE(x) =

d

dt

∫ ∞

−∞xp(x, t)dx = D

∫ ∞

−∞x∂2p

∂x2dx (4.20)

= Dx∂p

∂x

∣

∣

∣

∞

−∞−D

∫ ∞

−∞

∂p

∂xdx (4.21)

= Dp∣

∣

∣

∞

−∞= 0, (4.22)


assuming p(x, t) and its spatial derivatives decay at ±∞ sufficiently fast (which they do). In other words,E(x) is a constant independent of time.

The variance can be calculated in similar fashion

d

dtE(x2) =

d

dt

∫ ∞

−∞x2p(x, t)dx = D

∫ ∞

−∞x2∂2p

∂x2dx (4.23)

= Dx2∂p

∂x

∣

∣

∣

∞

−∞− 2D

∫ ∞

−∞x∂p

∂xdx (4.24)

= −2Dxp∣

∣

∣

∞

−∞+ 2D

∫ ∞

−∞p(x, t)dx = 2D, (4.25)

so that E(x2) grows linearly in time at rate 2D. If we start this process with a particle located exactlyat the origin x = 0, then E(x) = 0 for all time and var(x) = 2Dt. E(x2) is also called the mean-squared

displacement for this process.

What might this distribution be?, you might ask. We get a clue from a fundamental theorem from probabilitytheory, called the

Central Limit Theorem: Suppose m1,m2, · · · , are independent, identically distributed, random variables,with mean µ = E(mi) and variance σ2 = var(mi). Then, the random variable

xn =

n∑

j=1

mj (4.26)

is approximately normally distributed with mean µn = nµ and variance σ2n = nσ2. In other words, the

distribution for xn is well approximated by

fn(x) =1√2πσn

exp(

− (x− µn)2

2σ2n

)

. (4.27)

This most famous of all probability distributions is called the normal distribution or Gaussian distribution

and is often denoted as N(µn, σ2n), meaning a normal distribution with mean µn and variance σ2

n.

For the random walk here, this means that the probability P (xn = k) is approximately (substitute µn = 0and σn = 2αn)

P (xn = k) =1√

4πnαexp(− k2

4αn). (4.28)

Now if we set n = t∆t

and k = x∆x

, this becomes

P (x = k∆x) =1

√

4π t∆tαexp(− x2∆t

4αt∆x2), (4.29)

=∆x√4πDt

exp(− x2

4Dt), (4.30)

where D = α∆x2

∆t. This suggests, but does not prove, that the function

p(x, t) =1√4πDt

exp(− x2

4Dt), (4.31)

is a solution of the diffusion equation.

4.3. EXERCISES 25

To see if this is correct, we try a solution of the diffusion equation of the form

p(x, t) = a(t) exp(b(t)x2), (4.32)

substituting this into the diffusion equation. After factoring out exp(bx2), one is left with the quadraticequation

da

dt− 2Dab+ (a

db

dt− 4Dab2)x2 = 0, (4.33)

which is satisfied for all x if and only if

da

dt= 2Dab, a

db

dt= 4Dab2. (4.34)

Since this is nontrivial only if a 6= 0, we find

b(t) = − 1

4Dt, (4.35)

(taking 1b(0) = 0), and then we can solve da

dt= 2Dab for a to find

da

a= −1

2

dt

t, (4.36)

or

a(t) =a(0)√t. (4.37)

a(0) should be chosen so that∫∞−∞ p(x, t)dx = 1, in which case this is exactly the same as (4.31).

4.3 Exercises

1. Suppose a particle moves to the right with probability α and to the left with probability β and stayput with probability 1 − α − β. Following the above arguments, formulate this as a discrete randomwalk process and determine

(a) The limiting partial differential equation,

(b) The solution of the limiting partial differential equation.


Chapter 5

Realizations of a diffusion process

What we have seen is that the diffusion equation can be viewed as describing the movement of a largenumber of molecules, or the probability of finding a single particle at a particular position and time. But, ofcourse, these are the same. Because, if we follow a large number of molecules once, or follow a single particlemany times (many repeated experiments), the answer will be the same, as long as the many particles arenon-interacting, i.e., independent. This is a reasonable assumption if the particles are dilute, but if they arenot dilute, then there are likely to be interactions so that the diffusion equation is no longer valid. But thisis the topic for another time and place.

The question to be addressed here is how to follow (i.e., simulate) a diffusion process. The answer to thisquestion depends on the context.

5.1 Following a single particle

It may be that one is interested in following only a single particle, or cell. To do this, we make use of a factabout Brownian motion (which is the name for this process). Suppose one were able to precisely follow aparticle and collect large amounts of data on its change of position, denoted dx, in a fixed time incrementdt. Recall from above that the solution of the diffusion equation has the feature that the expected value ofposition of a particle is unchanging in time while the variance of position grows linearly in time with rate2D. What this means for a fixed (small) time increment dt is that dx is random but distributed accordingto

dx =√2DdtN(0, 1). (5.1)

In other words, dx is a continuous random variable that is normally distributed with mean zero and variance2Ddt.

This, then, gives a formula for how to simulate a diffusion process. Specifically, let the position of the particleafter n time steps be denoted by xn. Then, xn is updated by the formula

xn+1 = xn + dxn, (5.2)

where dxn is a random number chosen according to (5.1). The matlab code that carries this out is inAppendix A.3.

Interesting Question: How fast does a molecule of oxygen move, on average?

27

28 CHAPTER 5. REALIZATIONS OF A DIFFUSION PROCESS

The answer is found by understanding the Boltzman equation

1

2mv2 =

1

2kBT, (5.3)

which, in words, says that the kinetic energy of a particle is kBT . Now, you can look up in any book onchemical physics (or Wikipedia) that oxygen (O2) has a molecular weight of 32 g/mole or about 5×10−23

g/molecule and diffusion coefficient (in water) of 2×10−5cm2/s. It follows that, at room temperature (about300 degrees Kelvin), the velocity is

v =

√

kBT

m≈ 3× 104cm/s. (5.4)

Now since v = dxdt

and D = dx2

2dt ,

dx =2D

v= 1.4× 10−9cm, dt =

2D

v2= 2.5× 10−14s. (5.5)

In other words, on average, an oxygen molecule has a velocity of 30 m/s, but runs into something else andswitches direction every 2× 10−14 seconds, and between collisions moves about 10−9cm.

5.2 Other features of Brownian particle motion - Escape timesand splitting probabilities

Now that we know a little bit about how a diffusing particle moves we can ask several other questions. Thefirst is to determine escape times. The question is as follows: How long, on average does it take for a diffusingparticle to escape from some region? In biological terms, how long does it take, on average, for a moleculethat is made in the nucleus of a cell to diffuse to the boundary of the cell? A second question is, if there aretwo different places that a particle can escape from a region, what are the probabilities of escape from eachparticular region, called the splitting probability.

We start with the splitting probability problem. Suppose that a particle is diffusing on a one dimensionalline of length L. We let πL(x) denote the probability that the particle starting at point x will hit point x = Lbefore it hits the point x = 0, and vice versa, let π0(x) denote the probability that the particle starting atpoint x will hit point x = 0 before it hits the point x = L. Obviously, π0(x) + πL(x) = 1 for all x, andπL(0) = 0 and πL(L) = 1.

To derive the relevant equation, let’s start with a line that is discretized into boxes of size ∆x. For unbiaseddiffusion, the probability of moving left or right is equal and so

πL(x) =1

2πL(x +∆x) +

1

2πL(x−∆x). (5.6)

In words, the probability of escaping at x = L from x is half the probability of escaping from x + ∆x plushalf the probability of escaping from x −∆x. Now, we make our usual assumption that πL(x) is a smoothfunction with a Taylor series expansion, and find that (5.6) approaches

d2πLdx2

= 0, (5.7)

in the limit of small ∆x.

The solution of this problem is easy to find and not surprising, being

πL(x) =x

L. (5.8)

5.3. FOLLOWING SEVERAL PARTICLES 29

Now, let T (x) denote the expected time to hit either of the boundaries starting from position x. ClearlyT (0) = T (L) = 0. Also, since steps to the left or right are with equal probability,

T (x) =1

2T (x+∆x) +

1

2T (x−∆x) + ∆t, (5.9)

where ∆t is the time, on average, it takes to move distance ∆x. Again, we use Taylor series for T (x) andfind

Dd2T

dx2= −1, (5.10)

where D = ∆x2

2∆tis the diffusion coefficient.

We do not prove the statement here, but it can be shown that for any region Ω with boundary ∂Ω, theexpected escape time satisfies the boundary value problem

D∇2T = −1, x ∈ Ω, T∣

∣

∣

∂Ω= 0. (5.11)

The solution of this problem is interestingly different in different spatial dimensions. For example, in onedimension, suppose the boundary at x = 0 is reflecting (particles bounce off and cannot escape). Then,

(without proof) T ′(0) = 0, and we must solve D d2Tdx2 = −1, subject to boundary conditions T ′(0) = 0,

T (L) = 0. The solution is

T (x) =L2 − x2

2D. (5.12)

Compare this to the solution for a circle of radius R, which must satisfy

D

r

d

dr(rdT

dr) = −1, (5.13)

with T (R) = 0. The solution of this problem is

T (r) =R2 − r2

4D. (5.14)

For a sphere or radius R, T must satisfy

D

r2d

dr(r2

dT

dr) = −1, (5.15)

with T (R) = 0. The solution is

T (r) =R2 − r2

6D. (5.16)

The obvious conclusion of this is that escape from the interior of a circle or from the interior of a sphere istwo times, or three times, respectively, slower than along a line.

5.3 Following several particles

It is possible, but not practical, to follow a small number of particles using the above method. It is not,however, practical if the particle numbers grow. To follow the diffusion of a medium number (whatever thatmeans) of particles, we adopt the model (4.4) and do a stochastic simulation of it.


5.3.1 Particle Decay

The direct stochastic simulation of (4.4) can be done using the Gillespie algorithm. To describe this algorithm,we start with the simple example of exponential decay, modeled by the equation

du

dt= −αu. (5.17)

The Gillespie simulation for this equation is straightforward. This equation is a continuous representationof a discrete process in which molecules are lost one at a time through the reaction (a Poisson process)

Snnα−→ Sn−1, (5.18)

where Sn is the state in which there are n molecules. What this means is the following: Suppose that attime zero there is one particle. If the particle has a decay rate α, the probability that the particle has notdecayed by time t is p1(t) and it satisfies the differential equation

dp1dt

= −αp1, (5.19)

and consequently,p1(t) = exp(−αt). (5.20)

Similarly, if there are initially n particles, the probability that none of them has decayed by time t is denotedpn(t) and satisfies the differential equation

dpndt

= −nαpn, (5.21)

and has solutionpn(t) = exp(−nαt). (5.22)

This is simply another way of saying that the decay time is exponentially distributed.

Now, we want to pick the next reaction time increment so that it is exponentially distributed and the easiestway to do this is to take the time increment to the next reaction to be

δtn =−1

αnlnR, (5.23)

where R is a uniformly distributed random number 0 < R < 1.

Matlab code that carries this out is in Appendix A.4, and a figure showing a simulation starting with 50initial particles is shown in Fig. 5.1.

Before we go on to simulate the full diffusion process, there is an interesting and important observation tomake about the simple decay process we just simulated, and that is that the process always terminates infinite time, although the exponential curve that approximates it does not. It is interesting to further analyzethe discrete decay process (5.18). To do so we write down the equation governing pk(t), the probability ofhaving k particles at time t. This is

dpkdt

= α(k + 1)pk+1 − αkpk, (5.24)

for k = 0, 1, · · ·N , where pk(0) = 0 for k 6= N and pN(0) = 1. This equation follows since there are twoevents that can cause pk to change, one if one of k + 1 particles decays, leaving k particles (and increasingpk), or if one of k particles decays, leaving k − 1 particles (and decreasing pk).

5.3. FOLLOWING SEVERAL PARTICLES 31

α t0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

Par

ticle

Num

ber

0

5

10

15

20

25

30

35

40

45

50

Figure 5.1: Result of Gillespie simulation of decay (solid curve) compared to the function N exp(−αt) (dashedcurve), plotted as a function of αt, for N = 50 initial particles.

It is useful to write this system of equations in matrix form as

dp0dt

= αp1,dP

dt= AP, (5.25)

where P is the vector P = (p1, p2, · · · , pN)Tand where the matrix A is

A = α

−1 2 0 · · · 00 −2 3 0 · · ·

...0 · · · 0 0 −N

. (5.26)

Now, let’s find the expected extinction time, or the expected time to have zero particles, specified by

E(t) =

∫ ∞

0

tdp0dtdt. (5.27)

Here is the relevant calculation:

E(t) =

∫ ∞

0

tdp0dtdt = α

∫ ∞

0

tp1dt = αeT1

∫ ∞

0

tPdt (5.28)

= αeT1A−1

∫ ∞

0

tdP

dtdt = −αeT

1A−1

∫ ∞

0

Pdt (5.29)

= −αeT1A−2

∫ ∞

0

dP

dt= αeT

1A−2P (0) = αeT

1A−2eN . (5.30)

Here, ek is the column vector whose only nonzero entry is the kth component, which is one. A useful identityis that

1TA = −αeT1, (5.31)

where 1 is a column vector with all the entries one, so that

−αeT1A−1 = 1T , (5.32)

and the above reduces to

E(t) = −1TA−1eN = −1Tx ≡ −N∑

j=1

xj , (5.33)

whereAx = eN . (5.34)


α t0 2 4 6 8 10 12 14

0

100

200

300

400

500

600

700

800

900

Figure 5.2: Histogram of extinction times for a simulation of 10,000 trials for decay of N = 50 initial particles.

It is a straightforward calculation to find the components of x in a sequential fashion starting with xN = − 1αN

,then finding xN−1, then xN−2, and so on. The result of this is

xj =j + 1

jxj+1, (5.35)

leading to

xj = − 1

αj, (5.36)

and

E(t) =1

α

N∑

j=1

1

j. (5.37)

Unfortunately, there is no simple closed form formula for this expression, but it does demonstrate a techniquethat is quite important.

The validity of this formula can be verified by numerical simulation using the code in Appendix A.4. Anexample of this verification is shown in Fig. 5.2 where a histogram of extinction times is shown following10,000 trials starting with 50 particles. For this plot, the mean extinction time was computed to be 4.4939and the theoretical value from (5.37) is αE(t) = 4.4992.

Now that we know how to track a single compartment, we can think about multiple compartments. Supposethere are a total of N particles that are distributed among K boxes, arranged in a row. Let uj , j = 1, · · · ,Krepresent the integer number of particles in box j. We assume that each particle can leave its box and moveto one of its nearest neighbors by a Poisson process with rate 2α if it is an interior box, and rate α if it is aboundary box. Consequently, the ”rate of reaction”, where by reaction we mean leaving its box, is rj = 2αujfor j = 2, · · · , N − 1, and rj = αuj for j = 1, N . Now pick three uniformly distributed random numbersbetween zero and one; the first, R1, we use to determine when the next reaction occurs, and the second two,R2 and R3, we use to determine which of the possible reactions it is. As above, the time increment to thenth reaction, δtn, is taken to be

δtn =−1

RΣlnR1, (5.38)

where RΣ = α∑K

j=1 rj . Then, take j to be the smallest integer for which R2 < ρj = 1RΣ

∑ji=1 ri, and if

2 ≤ j ≤ N − 1, take the particle in the jth box to move to the right if R3 >12 and to the left if R3 ≤ 1

2 . Ifj = 1, the particle moves to the right, and if j = N it moves to the left.

Matlab code to simulate this process is in Appendix A.5. One thing worth noting is that the processbecomes less and less noisy as more particles are included in the system, suggesting that for a sufficientlylarge number of particles we should not be using a Gillespie algorithm, but rather direct simulation of thediffusion equation.

Chapter 6

Solutions of the Diffusion Equation

Now we attack the question of how to solve the diffusion equation on a one dimensional line. We start withanalytical solutions because they give interesting and important insights as to what is happening. However,analytical solutions are too limited in their usefulness for most modern, biological applications, so we alsodescribe numerical solutions.

6.1 On the Infinite Line

When the domain is the infinite line, a solution is the normal distribution found above and given by (4.31).

6.2 On the semi-infinite line

Suppose that a long capillary, open at one end, with uniform cross-sectional area A and filled with water,is inserted into a solution of known chemical concentration u0, and the chemical species is free to diffuseinto the capillary through the open end. Since the concentration of the chemical species depends only onthe distance along the tube and time, it is governed by the diffusion equation (4.2) and for conveniencewe assume that the capillary is infinitely long, so that 0 < x < ∞ . Because the solute bath in whichthe capillary sits is large, it is reasonable to assume that the chemical concentration at the tip is fixed atu(0, t) = u0, and since the tube is initially filled with pure water, u(x, 0) = 0 for all x, 0 < x <∞.

There are (at least) two ways to find the solution of this problem. One is to use the Fourier-Sine transform,a technique which is beyond the scope of this text (but you can learn about it in [?]). The second is tomake a lucky (or semi-informed) guess. Here, we make the guess that the solution should be of the formu(x, t) = f(ξ), where ξ = x√

2Dt. Substitute this guess into the diffusion equation and find

f ′ξ + f ′′ = 0. (6.1)

This is a separable equation for f ′, and can be written as

df ′

f ′= −ξdξ, (6.2)

33

34 CHAPTER 6. SOLUTIONS OF THE DIFFUSION EQUATION

so thatdf

dξ= A exp(−ξ

2

2). (6.3)

From this we determine that a solution of the diffusion equation is given by

u(x, t) = 2u0

(

1− 1√2π

∫ z

−∞exp

(

−s2

2

)

ds

)

, z =x√2Dt

, (6.4)

and the various constants were chosen so that u(0, t) = u0, and u(x, 0) = 0.

From this solution, one can readily calculate that the total number of molecules that enter the capillary ina fixed time T is

N = A

∫ ∞

0

u(x, T )dx = 2u0A

√

TD

π. (6.5)

From this equation, it is also possible to determine the diffusion coefficient by solving (6.5) for D, yielding

D =πN2

4u20A2T. (6.6)

A second useful piece of information is found from (6.4) by observing that u(x, t)/u0 is constant on any curvefor which z is constant. Thus, the curve t = x2/D is a level curve for the concentration, and gives a measureof how fast the substance is moving into the capillary. The time t = x2/D is called the diffusion time forthe process. To give some idea of the effectiveness of diffusion in various cellular contexts, in Table 6.1 isshown typical diffusion times for a variety of cellular structures. Clearly, diffusion is effective for transportwhen distances are short, but totally inadequate for longer distances, such as along a nerve axon. Obviously,biological systems must employ other transport mechanisms in these situations in order to survive.

Table 6.1: Estimates of diffusion times for cellular structures of typical dimensions, computed from therelation t = x2/D using D = 10−5cm2/s (typical for molecules the size of oxygen or carbon dioxide).

x t Example10 nm 100 ns Thickness of cell membrane1 µm 1 ms size of mitochondrion10 µm 100 ms Radius of small mammalian cell100 µm 10s Diameter of a large muscle fiber250 µm 60 s Radius of squid giant axon1 mm 16.7 min Half-thickness of frog sartorius muscle2 mm 1.1 h Half-thickness of lens in the eye5 mm 6.9 h Radius of mature ovarian follicle2 cm 2.6 d Thickness of ventricular myocardium1 m 31.7 yrs Length of a (long, e.g. sciatic) nerve axon

6.3 With Boundary Conditions

Up to this point we have not discussed much about boundary conditions, but these can be avoided no longer.Any physical domain is finite in size and so we must describe what is happening at the domain boundaries.There are three possibilities:

6.4. SEPARATION OF VARIABLES 35

• Dirichlet condition is when the value of the unknown u is specified at the boundary. If u is a probability,the condition u = 0 is said to be an absorbing boundary condition.

• Neumann condition is when the flux of the unknown u is specified. In a biological context, the fluxacross a boundary is zero if the boundary is impermeable to the particles, and is often called a no-flux

condition. If u is a probability, ∇u · n = 0 is called a reflecting boundary condition.

• Robin condition is a mixture between Dirichlet and Neumann, typically of the form au+bu′ = c (in onespatial dimension) and is often appropriate when the diffusing species can undergo a chemical reactionat the boundary.

In biology applications, the most common boundary condition is the no-flux (homogeneous Neumann) con-dition, and for the remainder of this section, this is the condition we apply.

An important feature of the no-flux boundary condition is that the total amount of the quantity u isconserved; this follows immediately from the conservation law as stated in (3.1).

6.4 Separation of Variables

When solving any differential equation with constant coefficients, it is reasonable to try an exponentialsolution. For the diffusion equation, we try a solution of the form

u(x, t) = U(x) exp(λt), (6.7)

and upon substituting into the diffusion equation (4.2), we find

Dd2U

dx2− λU = 0. (6.8)

This equation must be solved subject to the no-flux boundary condition U ′(0) = U ′(L) = 0.

There are an infinite number of possible solutions, but they are all of the same form, namely

Un(x) = an cos(nπx

L), (6.9)

with the important restriction that

λ = λn ≡ −n2π2D

L2, (6.10)

with n = 0, 1, 2, · · ·.

Since there are an infinite number of possible solutions, and the diffusion equation is linear, the full solutionis a linear combination of the possible solutions, namely

u(x, t) =

∞∑

n=0

an exp(−Dn2π2t

L2) cos(

nπx

L). (6.11)

At time t = 0, u(x, 0) is specified to be some function U0(x), so for consistency it must be that

U0(x) =

∞∑

n=0

an cos(nπx

L). (6.12)


Now we use the fact that∫ L

0

cos(nπx

L) cos(

mπx

L)dx = δmn

L

2, (6.13)

provided n and m are not both zero. We multiply the equation (6.12) by cos(mπxL

) and integrate from zeroto L, to determine that

a0 =1

L

∫ L

0

U0(x)dx, an =2

L

∫ L

0

U0(x) cos(nπx

L)dx, n 6= 0. (6.14)

One of the important consequences of expressing the solution in this form is that it shows how different modes

behave. Clearly the average value of u, expressed as a0, does not change, since λ0 = 0. Other componentsof the solution decay, and their rate of decay is proportional to the square of the mode number, n. In otherwords, the solution smooths out its ripples very rapidly while variations that are more gradual (i.e., smallern) smooth out less rapidly.

6.5 Numerical Methods

6.5.1 Method of Lines

The first method to numerically simulate the diffusion equation is actually one that we have already done,namely the Method of Lines. With this method, we discretize the spatial region into a grid with points atxj = j∆x, j = 0, 1, · · · , N , and then write the diffusion equation approximately as

dujdt

=D

∆x2(uj+1 − 2uj + uj−1). (6.15)

At the endpoints we take the equations to be

du0dt

=2D

∆x2(u1 − u0),

duNdt

=2D

∆x2(uj−1 − uN ). (6.16)

This system of equation is then simulated using a numerical ordinary differential equation solver, for examplein Matlab. The code for this is in Appendix A.1.

It is convenient for future discussions to represent u(j∆x, t) as a vector u(t) = (uj), and then to rewrite(6.15) using vector/matrix notation as

du

dt=

D

∆x2Au, (6.17)

where the matrix A has diagonal elements -2, and first upper and lower off-diagonal elements 1, except thefirst element of the upper diagonal and last element of the lower diagonal are both 2.

6.5.2 Euler’s Method

The Euler Method constitutes solving the equation (6.15) by doing a forward time step discretization. If thetime step is ∆t, and we set un to be the vector u(n∆t) we have

1

∆t(un+1 − un

j ) =D

∆x2Aun. (6.18)

6.5. NUMERICAL METHODS 37

We can rewrite this to show its iterative nature, as

un+1 = (I +D∆t

∆x2A)un, (6.19)

and it is easy to write code that carries this out. However, one drawback of this method quickly becomesapparent.

Notice that since A is diagonally semi-dominant, its eigenvalues are all non-positive. It is relatively easy toverify (use the Gersch-Gorin Theorem) that the eigenvalues of A all lie between -4 and 0. In fact, using thetrigonometric identity,

cos(x + y)− 2 cos(x) + cos(x− y) = 2(cos(y)− 1) cos(x), (6.20)

one can verify that the vector with components uj = cos( (j−1)kπN

), j = 1, 2, · · · , N + 1, is an eigenvector of

the matrix A corresponding to eigenvalue λk = 2(cos(kπN)− 1), k = 0, 1, 2, · · · , N .

With this information in hand, it is apparent that if D∆t∆x2 is bigger than 1

2 , the matrix I+ D∆t∆x2 A has eigenvalues

which are negative with magnitude larger than one, so that the iterations of (6.19) oscillate with growingamplitude, hence they are unstable.

What this means from a practical point of view is that to have a stable simulation using the Euler method,it is necessary to keep D∆t

∆x2 <12 , and if one wants a high spatial resolution solution (∆x small), this places

a significant restriction on the time step that can be used for the simulation.

6.5.3 Crank-Nickolson Method

A way to overcome this restriction is with a scheme called the Crank-Nickolson algorithm. The idea of thisis to split the spatial difference between time steps, as

un+1 − un

∆t=

D

∆x2Aun+1 + un

2, (6.21)

which leads to the iteration

(I − D∆t

2∆x2A)un+1 = (I +

D∆t

2∆x2A)un. (6.22)

The stability of this iteration is determined by the eigenvalues of the matrix

B = (I − D∆t

2∆x2A)−1(I +

D∆t

2∆x2A). (6.23)

Clearly, if λ is an eigenvalue of A, then the corresponding eigenvalue of B is

1 + D∆t2∆x2λ

1− D∆t2∆x2λ

, (6.24)

but since −4 ≤ λ ≤ 0, this quantity is always between 0 and one. Hence, this iteration is stable for allchoices of ∆x and ∆t, a significant improvement over the Euler time step. Of course, this says nothing atall about the accuracy of the simulation, but at least stability is assured.

Matlab code to solve the diffusion equation using the Crank-Nickolson method is in Appendix A.5.


6.6 Exercises

1. Many species of ant use pheromones as a danger signal. Consider an experiment in which ants arereleased into a long tube and one ant is stimulated until it releases a quantity of pheromone. Use theone-dimensional diffusion equation as a model for the spread of the pheromones in the tube. Assumethat at time t = 0 a bolus of pheromone with total amount α is released, and that other ants react tothe stimulus if the level of pheromone reaches 0.1α. For this exercise assume that D = 1.

(a) Plot the time course of the pheromone level at several different values of position x > 0.

(b) Find the region in the tube 0 ≤ x ≤ X(t) where the other ants react to the stimulus (i.e., theregion of influence).

(c) Sketch the time evolution of the boundary X(t);

(d) Find the region of space that is outside the domain of influence for all time.

Chapter 7

Transport of Oxygen

With this background on the diffusion equation, we can begin to “spice it up” to study problems of physio-logical interest. We begin with some problems related to the delivery of oxygen by the blood to tissues.

7.1 Transport across Membranes

Suppose there is a fixed amount of oxygen on either side of a membrane of width L. What is the flux ofoxygen across the membrane?

We answer this question by solving the (one-dimensional) diffusion equation in steady state, with Dirichletboundary conditions. For this section we denote the concentration of oxygen by c(x, t), so require c(0, t) = c0and c(L, t) = cL. Steady state means that the concentration c(x, t) is not changing in time so that ∂c

∂t= 0

(but this does not mean that there is no movement or flux of oxygen). This does mean that D d2cdx2 = 0 (note

that we can use ordinary rather than partial derivatives here since c is a function of only one variable x.)and consequently

Ddc

dx= −J, (7.1)

where J is the yet to be determined, constant, flux. Integrate this equation again and apply the boundarycondition at x = 0 to find

c(x) = c0 −J

Dx, (7.2)

and then apply the boundary condition at x = L to find

J =D

L(c0 − cL). (7.3)

This is, as expected, Fick’s law. The constant DL

is the membrane permeability.

7.2 Delivery of oxygen to tissue

Now suppose that oxygen is transported by the blood in a capillary, it diffuses across a membrane intotissue where it is consumed by metabolic processes in the tissues. For this problem, we take a simple

39

40 CHAPTER 7. TRANSPORT OF OXYGEN

model of both the tissue and the capillary, assuming that both are one-dimensional with variations onlyin the direction along the capillary, with homogeneous oxygen concentration in the orthogonal direction.Denote the concentration of oxygen in the capillary and the tissue by c and ct, respectively, and write theconservation equations for both as

∂c

∂t= − ∂

∂x(vc) + d(ct − c), (7.4)

∂ct∂t

= −m+ d(c− ct), (7.5)

where v is the velocity of blood in the capillary, hence the rate of transport, m is the rate of metabolism,and d is the membrane permeability. For this problem, we assume that diffusion across the membrane is theonly diffusion that matters.

In steady state, it must be that d(c− ct) = m so that − ∂∂x

(vc) = m, in which case

c = c0 −m

vx (7.6)

where c0 is the inlet oxygen concentration. Consequently, all tissue with x > c0vm

is in oxygen debt. This, ofcourse, illustrates a weakness of the model, namely that oxygen can be consumed even when there is none.

7.3 Oxygen exchange between capillaries

Solutes are exchanged between liquids by diffusion across their separating membranes. Since the rate ofexchange is affected by the concentration difference across the membrane, the exchange rate is increased iflarge concentration differences can be maintained. One important way that large concentration differencescan be maintained is by the countercurrent mechanism. The countercurrent mechanism is important forrenal function, the exchange of oxygen from water to blood through fish gills and the exchange of oxygen inthe placenta between mother and fetus.

Suppose that two gases or liquids containing a solute flow along parallel tubes of length L, separated by apermeable membrane. We model this in the simplest possible way as a one-dimensional problem, and weassume that solute transport is a linear function of the concentration difference. Then the concentrations inthe two one-dimensional tubes are given by

∂c1∂t

+ v1∂c1∂x

= d(c2 − c1), (7.7)

∂c2∂t

+ v2∂c2∂x

= d(c1 − c2). (7.8)

The mathematical problem is to find the outflow concentrations, given that the inflow concentrations, thelength of the exchange chamber, and the flow velocities are known.

We assume that the flows are in steady state and that the input concentrations are c01 and c02. Then, if weadd the two governing equations and integrate, we find that

v1c1 + v2c2 = k (a constant). (7.9)

Pretending that k is known, we eliminate c2 from (7.8) and find the differential equation for c1,

dc1dx

=d

v1v2(k − (v1 + v2)c1) , (7.10)

7.4. FACILITATED DIFFUSION 41

from which we can solve to learn that

c1(x) = κ+ (c1(0)− κ)e−λx, (7.11)

where κ = kv1+v2

and λ = d(

v1+v2v1v2

)

.

There are two cases to consider, namely when v1 and v2 are of the same sign and when they have differentsigns. If they have the same signs, say positive, then the input is at x = 0, and it must be that c1(0) =c01, c2(0) = c02, from which, using (7.9), it follows that

c1(L)

c01=

1 + γρ

1 + ρ+ ρ

1− γ

1 + ρe−λL, (7.12)

where γ = c02/c01, ρ = v2/v1, λ = d

v1(1 + 1

ρ).

Suppose that the goal is to transfer material from vessel 1 to vessel 2, so that γ < 1. We learn from (7.12)that the output concentration from vessel 1 is an exponentially decreasing function of the residence lengthdL/v1. Furthermore, the best that can be done (i.e., as dL/v1 → ∞) is 1+γρ

1+ρ.

In the case that v1 and v2 are of opposite sign, say v1 > 0, v2 < 0, the inflow for vessel 1 is at x = 0, but theinflow for vessel 2 is at x = L. In this case we calculate that

c1(L)

c01=

−γρ+ (1− ρ+ γρ)e−λL

e−λL − ρ, (7.13)

where γ = c2(L)/c01 = c02/c

01, ρ = −v2/v1 > 0, λ = d

v1(1 − 1

ρ), provided that ρ 6= 1. In the special case ρ = 1,

we havec1(L)

c01=v1 + γdL

v1 + dL. (7.14)

Now we can see the substantial difference between a cocurrent (v1 and v2 of the same sign) and a counter-current (v1 and v2 with the opposite sign). At fixed parameter values, if γ < 1, the expression for c1(L)/c

01 in

(7.12) is always larger than that in (7.13), implying that the total transfer of solute is always more efficientwith a countercurrent than with a cocurrent.

In Fig. 7.1 is shown a comparison between a countercurrent and a cocurrent. The dashed curves show thetransfer fraction c1(L)/c

01 for a cocurrent, plotted as a function of the residence length dL/v1, with input

in tube 2, c02 = c2(0) = 0. The solid curves show the same quantity for a countercurrent, with inputconcentration c02 = c2(L) = 0.

In the limit of a long residence time (large dL/v1), the transfer fraction becomes 1 − ρ+ γρ if ρ < 1, and γif ρ > 1. Indeed, this is always smaller than the result for a cocurrent, 1+γρ

1+ρ.

7.4 Facilitated Diffusion

A second important example in which both diffusion and reaction play a role is known as facilitated diffusion.Facilitated diffusion occurs when the flux of a chemical is amplified by a reaction that takes place in thediffusing medium. An example of facilitated diffusion occurs with the flux of oxygen in muscle fibers. Inmuscle fibers, oxygen is bound to myoglobin and is transported as oxymyoglobin, and this transport is greatlyenhanced above the flow of oxygen in the absence of myoglobin (Wyman, 1966; Murray, 1971; Murray andWyman, 1971; Rubinow and Dembo, 1977).


1.0

0.8

0.6

0.4

0.2

0.0

Tra

nsfe

r fr

action

543210

Residence length

ρ = 0.5

ρ = 2.0

Figure 7.1: Transfer fraction for a cocurrent (dashed) and a countercurrent (solid) when γ = 0 plotted as afunction of residence length.

This well-documented observation needs further explanation, because at first glance it seems counterintuitive.Myoglobin molecules are much larger (molecular weight M=16,890) than oxygen molecules (molecular weightM=32) and therefore have a much smaller diffusion coefficient (D = 4.4×10−7 and D = 1.2×10−5cm2/s formyoglobin and oxygen, respectively). The diffusion of oxymyoglobin would therefore seem to be much slowerthan the diffusion of free oxygen. Further, the diffusion of free oxygen is much slower when it is buffered bymyoglobin since the effective diffusion coefficient of oxygen is lowered substantially by buffering.

To anticipate slightly, the answer is that, at steady state, the total transport of oxygen is the sum of thefree oxygen transport and additional oxygen that is transported by the diffusing buffer. If there is a lot ofbuffer, with a lot of oxygen bound, this additional transport due to the buffer can be substantial.

A simple model of this phenomenon is as follows. Suppose we have a slab reactor containing diffusingmyoglobin. On the left (at x = 0) the oxygen concentration is held fixed at s0, and on the right (at x = L)it is held fixed at sL, which is assumed to be less than s0.

If f is the rate of uptake of oxygen into oxymyoglobin, then equations governing the concentrations ofs = [O2], e = [Mb], c = [MbO2] are

∂s

∂t= Ds

∂2s

∂x2− f, (7.15)

∂e

∂t= De

∂2e

∂x2− f, (7.16)

∂c

∂t= Dc

∂2c

∂x2+ f. (7.17)

It is reasonable to take De = Dc, since myoglobin and oxymyoglobin are nearly identical in molecular weightand structure. Since myoglobin and oxymyoglobin remain inside the slab, it is also reasonable to specify theboundary conditions ∂e/∂x = ∂c/∂x = 0 at x = 0 and x = L. Because it reproduces the oxygen saturationcurve (discussed in Chapter ??), we assume that the reaction of oxygen with myoglobin is governed by theelementary reaction

O2 +Mbk+−→←−k−

MbO2,

so that (from the law of mass action) f = −k−c+ k+se. The total amount of myoglobin is conserved by thereaction, so that at steady state e+ c = e0 and (7.16) is superfluous.

7.4. FACILITATED DIFFUSION 43

At steady state,0 = st + ct = Dssxx +Dccxx, (7.18)

and thus there is a second conserved quantity, namely

Ds

ds

dx+Dc

dc

dx= −J, (7.19)

which follows by integrating (7.18) once with respect to x. The constant J (which is yet to be determined)is the sum of the flux of free oxygen and the flux of oxygen in the complex oxymyoglobin, and thereforerepresents the total flux of oxygen. Integrating (7.19) with respect to x between x = 0 and x = L, we canexpress the total flux J in terms of boundary values of the two concentrations as

J =Ds

L(s0 − sL) +

Dc

L(c0 − cL), (7.20)

although the values c0 and cL are as yet unknown.

To further understand this system of equations, we introduce dimensionless variables, σ = k+

k−

s, u = c/e0,

and x = Ly, in terms of which (7.15) and (7.17) become

ǫ1σyy = σ(1 − u)− u = −ǫ2uyy, (7.21)

where ǫ1 = Ds

e0k+L2 , ǫ2 = Dc

k−L2 .

Reasonable numbers for the uptake of oxygen by myoglobin (Wittenberg, 1966) are k+ = 1.4×1010cm3 M−1s−1,

k− = 11 s−1, and L = 0.022 cm in a solution with e0 = 1.2 × 10−5 M/cm3. (These numbers are for an

experimental setup in which the concentration of myoglobin was substantially higher than what naturallyoccurs in living tissue.) With these numbers we estimate that ǫ1 = 1.5× 10−7, and ǫ2 = 8.2× 10−5. Clearly,both of these numbers are small, suggesting that oxygen and myoglobin are at quasi-steady state throughoutthe medium, with

c = e0s

K + s, (7.22)

where K = k−/k+. Now we substitute (7.22) into (7.20) to find the flux

J =Ds

L(s0 − sL) +

Dc

Le0

(

s0K + s0

− sLK + sL

)

=Ds

L(s0 − sL)

(

1 +Dc

Ds

e0K

(s0 +K)(sL +K)

)

=Ds

L(1 + µρ)(s0 − sL), (7.23)

where ρ = Dc

Ds

e0K, µ = K2

(s0+K)(sL+K) .

In terms of dimensionless variables the full solution is given by

σ(y) + ρu(y) = y[σ(1) + ρu(1)] + (1− y)[σ(0) + ρu(0)], (7.24)

u(y) =σ(y)

1 + σ(y). (7.25)

Now we see how diffusion can be facilitated by an enzymatic reaction. In the absence of a diffusing carrier,ρ = 0 and the flux is purely Fickian. However, in the presence of carrier, diffusion is enhanced by the factorµρ. The maximum enhancement possible is at zero concentration, when µ = 1. With the above numbersfor myoglobin, this maximum enhancement is substantial, being ρ = 560. If the oxygen supply is sufficiently


7

6

5

4

3

2

1

0

Oxyg

en

flu

x

1.00.80.60.40.20.0

y

Free oxygen flux

Bound oxygen flux

3.0

2.5

2.0

1.5

1.0

0.5

0.0

Oxygen

concentr

atio

n

1.00.80.60.40.20.0

y

Free

Bound

Total

A B

Figure 7.2: A: Free oxygen content σ(y) and bound oxygen content u(y) as functions of y. B: Free oxygenflux −σ′(y) and bound oxygen flux −ρu′(y) plotted as functions of y.

high on the left side (near x = 0), then oxygen is stored as oxymyoglobin. Moving to the right, as the totaloxygen content drops, oxygen is released by the myoglobin. Thus, even though the bound oxygen diffusesslowly compared to free oxygen, the quantity of bound oxygen is high (provided that e0 is large comparedto the half saturation level K), so that lots of oxygen is transported. We can also understand that to takeadvantage of the myoglobin-bound oxygen, the concentration of oxygen must drop to sufficiently low levelsso that myoglobin releases its stored oxygen.

To explain it another way, note from (7.23) that J is the sum of two terms, the usual ohmic flux term and anadditional term that depends on the diffusion coefficient of MbO2. The total oxygen flux is the sum of theflux of free oxygen and the flux of oxygen bound to myoglobin. Clearly, if oxymyoglobin is free to diffuse, thetotal flux is thereby increased. But since oxymyoglobin can only diffuse down its gradient, the concentrationof oxymyoglobin must be higher on one side than the other.

In Fig. 7.2A are shown the dimensionless free oxygen concentration σ and the dimensionless bound oxygenconcentration u plotted as functions of position. Notice that the free oxygen content falls at first, indicatinghigher free oxygen flux, and the bound oxygen decreases more rapidly at larger y. Perhaps easier to interpretis Fig. 7.2B, where the dimensionless flux of free oxygen and the dimensionless flux of bound oxygen areshown as functions of position. Here we can see that as the free oxygen concentration drops, the flux of freeoxygen also drops, but the flux of bound oxygen increases. For large y, most of the flux is due to the boundoxygen. For these figures, ρ = 10, σ(0) = 2.0, σ(1) = 0.1.

One mathematical detail that was ignored in this discussion is the validity of the quasi-steady-state solution(7.22) as an approximation of (7.21). Usually, when one makes an approximation to boundary value problemsin which the order of the system is reduced (as here where the order is four, and drops by two when ǫ1 andǫ2 are ignored), there are difficulties with the solution at the boundary, because the boundary conditionscannot, in general, be met. Such problems are called singular perturbation problems, because the behaviorof the solutions as functions of the small parameters is not regular, but singular (certain derivatives becomeinfinitely large as the parameters approach zero). In this problem, however, there are no boundary layers,and the quasi-steady-state solution is a uniformly valid approximation to the solution. This occurs becausethe boundary conditions on c are of no-flux (Neumann) type, rather than of fixed (Dirichlet) type. That is,since the value of c is not specified by the boundary conditions, c is readily adjusted so that there are noboundary layers. Only a slight correction to the quasi-steady-state solution is needed to meet the no-fluxboundary conditions, but this correction affects only the derivative, not the value, of c in a small region nearthe boundaries.

7.5. FACILITATED DIFFUSION IN MUSCLE RESPIRATION 45

7.5 Facilitated Diffusion in Muscle Respiration

Even at rest, muscle fibers consume oxygen. This is because ATP is constantly consumed to maintain anonzero membrane potential across a muscle cell wall, and this consumption of energy requires constantmetabolizing of sugar, which consumes oxygen. Although sugar can be metabolized anaerobically, the wasteproduct of this reaction is lactic acid, which is toxic to the cell. In humans, the oxygen consumption of livemuscle tissue at rest is about 5 × 10−8 mol/cm3s, and the concentration of myoglobin is about 2.8 × 10−7

mol/cm3. Thus, when myoglobin is fully saturated, it contains only about a 5 s supply of oxygen. Further,the oxygen at the exterior of the muscle cell must penetrate to the center of the cell to prevent the oxygenconcentration at the center falling to zero, a condition called oxygen debt.

To explain how myoglobin aids in providing oxygen to a muscle cell and helps to prevent oxygen debt, weexamine a model of oxygen consumption that includes the effects of diffusion of oxygen and myoglobin. Wesuppose that a muscle fiber is a long circular cylinder (radius a = 2.5 × 10−3 cm) and that diffusion takesplace only in the radial direction. We suppose that the oxygen concentration at the boundary of the fiber isa fixed constant and that the distribution of chemical species is radially symmetric. With these assumptions,the steady-state equations governing the diffusion of oxygen and oxymyoglobin are

Ds

1

r

d

dr

(

rds

dr

)

− f − g = 0, (7.26)

Dc

1

r

d

dr

(

rdc

dr

)

+ f = 0, (7.27)

where, as before, s = [O2], c = [MbO2], and f = −k−c + k+se. The coordinate r is in the radial direction.The new term in these equations is the constant g, corresponding to the constant consumption of oxygen.The boundary conditions are s = sa, dc/dr = 0 at r = a, and ds/dr = dc/dr = 0 at r = 0. For muscle,

sa is typically 3.5× 10−8 mol/cm3(corresponding to the partial pressure 20 mm Hg). Numerical values for

the parameters in this model are difficult to obtain, but reasonable numbers are Ds = 10−5 cm2/s, Dc =5× 10−7 cm2/s, k+ = 2.4× 1010 cm3/mol · s, and k− = 65/s (Wyman, 1966).

Introducing nondimensional variables σ = k+

k−

s, u = c/e0, and r = ay, we obtain the differential equations

ǫ11

y

d

dy

(

ydσ

dy

)

− γ = σ(1 − u)− u = −ǫ21

y

d

dy

(

ydu

dy

)

, (7.28)

where ǫ1 = Ds

e0k+a2 , ǫ2 = Dc

k−a2 , γ = g/k−. Using the parameters appropriate for muscle, we estimate that ǫ1 =

2.3×10−4, ǫ2 = 1.2×10−3, γ = 3.3×10−3. While these numbers are not as small as for the experimental slabdescribed earlier, they are small enough to warrant the assumption that the quasi-steady state approximation(7.22) holds in the interior of the muscle fiber.

It also follows from (7.28) that

ǫ11

y

d

dy

(

ydσ

dy

)

+ ǫ21

y

d

dy

(

ydu

dy

)

= γ. (7.29)

We integrate (7.29) twice with respect to y to find

ǫ1σ + ǫ2u = A ln y +B +γ

4y2. (7.30)

The constants A and B are determined by boundary conditions. Since we want the solution to be boundedat the origin, A = 0, and B is related to the concentration at the origin.

Now suppose that there is just enough oxygen at the boundary to prevent oxygen debt. In this model,oxygen debt occurs if σ falls to zero. Marginal oxygen debt occurs if σ = u = 0 at y = 0. For this boundary


10

8

6

4

2

0Exte

rna

l o

xyg

en

co

nce

ntr

atio

n

1086420

Oxygen consumption

ρ = 0

ρ = 5

Figure 7.3: Critical concentration σ0 plotted as a function of oxygen consumption γ4ǫ1

. The dashed curve isthe critical concentration with no facilitated diffusion.

condition, we take A = B = 0. Then the concentration at the boundary must be at least as large as σ0,where, using the quasi-steady state σ(1− u) = u,

σ0 + ρσ0

σ0 + 1=

γ

4ǫ1, (7.31)

and where ρ = ǫ2/ǫ1. Otherwise, the center of the muscle is in oxygen debt. Note also that σ0 is a decreasingfunction of ρ, indicating a reduced need for external oxygen because of facilitated diffusion.

A plot of this critical concentration σ0 as a function of the scaled consumption γ4ǫ1

is shown in Fig. 7.3. Forthis plot ρ = 5, which is a reasonable estimate for muscle. The dashed curve is the critical concentrationwhen there is no facilitated diffusion (ρ = 0). The easy lesson from this plot is that facilitated diffusiondecreases the likelihood of oxygen debt, since the external oxygen concentration necessary to prevent oxygendebt is smaller in the presence of myoglobin than without.

A similar lesson comes from Fig. 7.4, where the internal free oxygen content σ is shown, plotted as a functionof radius y. The solid curves show the internal free oxygen with facilitated diffusion, and the dashed curve iswithout. The smaller of the two solid curves and the dashed curve have exactly the critical external oxygenconcentration, showing clearly that in the presence of myoglobin, oxygen debt is less likely at a given externaloxygen concentration. The larger of the two solid curves has the same external oxygen concentration as thedashed curve, showing again the contribution of facilitation toward preventing oxygen debt. For this figure,ρ = 5, γ/ǫ1 = 14.

7.5. FACILITATED DIFFUSION IN MUSCLE RESPIRATION 47

14

12

10

8

6

4

2

0

Fre

e o

xygen

1.00.80.60.40.20.0

Radius

Figure 7.4: Free oxygen σ as a function of radius y. Solid curves show oxygen concentration in the presenceof myoglobin (ρ = 5), the lower of the two having the critical external oxygen concentration. The dashedcurve shows the oxygen concentration without facilitation at the critical external concentration level.


Chapter 8

Diffusion and Reaction

8.1 Birth-Death with Diffusion

Suppose that there is some population (or chemical species) U that diffuses and experiences either decay, asin

Uk−→ φ, (8.1)

or birth via (asexual) duplication as in

Uk−→ 2U, (8.2)

Setting u = [U ], the equation describing the evolution of this population is

∂u

∂t= D

∂2u

∂x2+ ku, (8.3)

where k is negative for decay and positive for growth.

The first thing to do with this equation is to simulate it numerically. A reasonable numerical scheme is touse the Crank-Nicholson method for the diffusion part and the Euler method (explicit forward time step)for the reaction term. Symbolically, this is

un+1 − un

∆t=

D

2∆x2A(un+1 +Aun) + kun, (8.4)

which leads to the time step algorithm

(I − D∆t

2∆x2A)un+1 = (I +

D∆t

2∆x2A+ k∆t)un. (8.5)

Fig. 8.1 shows snapshots of the solution for several times for the two cases with k < 0 and k > 0.

An interesting feature of these solutions is that their ratios are functions of time, independent of space. InFig. 8.2, where the natural logarithm of the ratios of the solution to the solution of the diffusion equation(i.e., set k = 0) are shown.

This observation gives a clue as to how to solve the equation (8.3) analytically. That is, we try a solution ofthe form u(x, t) = f(t)v(x, t) where v(x, t) satisfies the diffusion equation. Obviously,

∂u

∂t=df

dtv + f

∂v

∂t,

∂2u

∂x2=∂2v

∂x2, (8.6)

49

50 CHAPTER 8. DIFFUSION AND REACTION

t0 2 4 6 8 10 12 14 16 18 20

U(x

,t)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t0 2 4 6 8 10 12 14 16 18 20

U(x

,t)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 8.1: Solution of the equation (8.3) with k = −0.05 (left) and k = 0.05 (right), and D = 0.01. Initialdata profile is shown as a dashed curve.

t0 10 20 30 40 50 60 70 80 90 100

ln(U

(x,t)

/u0(x

,t)

100

k=0.05

k=-0.05

Figure 8.2: Natural logarithm of the ratio uk(x,t)u0(x,t)

plottted as a function of time, with k = −0.05 and k = 0.05

and with u0(x, t) the solution of the (8.3) with k = 0.

and the equation (8.3) is satisfied provideddf

dt= kf, (8.7)

and the solution is

f(t) = exp(kt), (8.8)

which is not surprising, given the plots in Fig. 8.2.

8.2 Growth with a Carrying capacity - Fisher’s Equation

Now suppose the population is reproducing but there is population size dependent death. We can representthis by the two chemical reactions

Ukb−→2U, U + U

kd−→U. (8.9)

For these reactions, the conservation equation is

∂u

∂t= D

∂2u

∂x2+ kbu− kdu

2, (8.10)

8.2. GROWTH WITH A CARRYING CAPACITY - FISHER’S EQUATION 51

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

Figure 8.3: Solution of the Fisher equation plotted as a function of position for equal time increments.

Rescaling the variables by setting u = kb

kdv, t = τ

kb, and x =

√

Dkby, the equation simplifies to

∂v

∂τ=∂2v

∂y2+ v − v2, (8.11)

This equation is quite well known, and often referred to as Fisher’s equation (named after the geneticist R.A. Fisher, 1890-1962.)

As usual, the first step to getting an understanding of the behavior of solutions of this equation is bynumerical simulation. For this the Crank Nicholson scheme for diffusion and explicit Euler step for thereaction are the preferred (mostly for reasons of simplicity) methods. The code in Appendix A.6 is set upto integrate this with appropriate choice of parameters.

Solutions of this equations are shown in Fig. 8.3. There we see that following an initial transient, the solutionappears to have a fixed shape moving with constant speed to the left and right.

Following this observation, we try to find solutions of this equation of the form v(y, τ) = F (y− cτ) with thefeature that F (−∞) = 1, F (∞) = 0. To see if this is possible, we substitute this into the governing equation(8.11 ) and find the ordinary differential equation for F (ξ),

d2F

dξ+ c

dF

dξ+ F (1− F ) = 0. (8.12)

The challenge is to see if this differential equation has trajectories that connect F = 1 with F = 0, keepingF positive, and the most natural way to approach this problem is using a phase portrait analysis. To do so,we write the equation (8.12) as the first order system

dF

dξ=G

c,

dG

dξ= −G− F (1 − F ). (8.13)

The phase portrait in Fig. 8.4, shows the nullcline dGdξ

= 0 as a dashed curve.

It is clear that there are exactly two critical points, namely, G = 0 and F = 0 or F = 1. The local linearizedsystem, linearized around F0 = 0 or 1 is

df

dτ=g

c,

dg

dτ= −g − (1− 2F0)f. (8.14)


F-0.2 0 0.2 0.4 0.6 0.8 1

G

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

c = 2.5

c = 0.5

Figure 8.4: Trajectories in the phase plane connecting the saddle point at F = 1 with the critical point atF = 0 with c = 0.5 and c = 2.5;

ξ-25 -20 -15 -10 -5 0 5 10 15 20 25

F(ξ

)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 8.5: Traveling wave solution profile F (ξ) plotted as a function of ξ, for c = 2.5..

The nature of each critical point is determined by the eigenvalues of the Jacobian matrix

J =

(

0 1c

−1 + 2F0 −1

)

, (8.15)

and these are the roots of the quadratic polynomial

λ2 + λ+1

c(1− 2F0) = 0, (8.16)

which are

λ =1

2

(

− 1±√

1− 41− 2F0

c

)

. (8.17)

Clearly, with F0 = 1, there are two real eigenvalues of opposite sign, so that the critical point is a saddlepoint. The critical point F0 = 0 is a node if c > 2, whereas it is a spiral point if c < 2. We conclude thatthere is a trajectory connecting the saddle point at F0 = 1 with the critical point at F +0 = 1 for all valuesof c > 0. However, this trajectory has F > 0 along the entire trajectory only if c ≥ 2.

The conclusion of this analysis is that Fisher’s equation has traveling wave solutions for all values of c ≥ 2.It is also known (using analysis that is beyond the scope of these notes) that if one starts with initial datav(y, 0) for which limy=−∞ v(y, 0) = 1 and for which v(y, 0) = 0 for all large y, then the solution v(y, τ)approaches the traveling wave solution F (y − 2τ) for large τ . In terms of original variables, the solution

u(x, t) approaches kd

kdF (√

k1

D(x − 2

√Dk1t for large t, a traveling wave solution with speed 2

√Dk1t.

8.2. GROWTH WITH A CARRYING CAPACITY - FISHER’S EQUATION 53

8.2.1 On a Bounded Domain

Now we consider a slightly different problem, namely Fisher’s equation on the bounded one-dimensionaldomain 0 < x < L subject to (Dirichlet) boundary conditions u(0, t) = u(L, t) = 0. A biological situation forwhich this may be a somewhat reasonable assumption is a game preserve with no constraints (i.e., fences)at the boundary and where animal occupants are killed whenever they leave the preserve.

As usual, a preliminary analysis of this problem can be gained by numerical simulation. What is shown inFig. 8.6 are the solutions for two different domains, one with L = 2 and the other for L = 5, starting withinitial data v = 1. What is seen is that there are two different outcomes depending on the size of the domain.These are eventual extinction (v → 0) if L is small, and long-term survival if L is large enough. To make

y0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

v(y,

t)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

v(y,

t)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 8.6: Time dependent solutions of Fisher’s equation with Dirichlet boundary data on a domain withL = 2 (left) and L = 5 (right). Initial data profile shown as a dashed curve.

the observation more precise, we look for steady solutions of the Fisher’s equation, satisfying

d2v

dy2+ v(1 − v) = 0, v(0) = v(L) = 0. (8.18)

The analysis of this problem is done by examining the equation in its phase portrait (shown in Fig.8.7),

dv

dy= w,

dw

dy= −v(1− v). (8.19)

Clearly, there are two critical points, with w = 0 and v = v0 where v0 = 0 or v0 = 1. The linearized systemis

dV

dy=W,

dW

dy= (2v0 − 1)V, (8.20)

and the eigenvalues of the Jacobian areλ = ±

√2v0 − 1. (8.21)

Consequently, the critical point at v0 = 0 is a center and the critical point at v0 = 1 is a saddle point. Thismeans that there is a family trajectories that encircle the origin, each having two intersections with v = 0.This family of trajectories is enclosed by a trajectory that intersects the saddle point exactly. These are thestable and unstable manifolds of the saddle point.

Close to the origin, these nearly circular trajectories are approximately described by

v(y) = sin(y), (8.22)


v0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

w

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

Figure 8.7: .

so the distance in y required to traverse between zeros is y = π. This suggests that at L = π there is abifurcation of a nontrivial solution of the boundary value problem from the trivial (i.e., identically zero)solution of the boundary value problem.

The phase portrait trajectories can be found explicitly. Multiply the equation (8.18) by dvdx

and integratewith respect to y once to find

1

2w2 + F (v) = F (v), (8.23)

where F (v) = v2

2 − v3

3 , and v is the intercept of the trajectory on the v axis. The w intercept on the w-axisis

w =√

2F (v). (8.24)

We solve (8.23) for w and find

dv

dx=√

2F (v)− 2F (v), (8.25)

so that

L = 2

∫ v

0

dv√

2F (v)− 2F (v). (8.26)

The curve L = L(v) is depicted in Fig. 8.8.

There is a practical matter associated with calculating this curve numerically that needs to be mentioned.As can be seen, the integral (8.26) is singular and therefore difficult to compute numerically. A much easiernumerical algorithm is to compute phase plane trajectories by solving (8.20) as an initial value problem,starting at v = 0, w = w and terminating the integration when w hits zero, then recording the value of vand y at this v-axis intersection. This is how the curve in Fig. 8.8 was calculated.

The conclusion of this analysis is interesting. It tells us that the organisms cannot survive on a preserve for

which L <√

Dkbπ. Said another way, organisms for which D is large or kb (their reproductive rate) is small

require proportionately more terrain in order to survive.

8.3. RESOURCE CONSUMPTION 55

L0 1 2 3 4 5 6 7 8 9 10

Vm

ax

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 8.8: Plot of the maximum value of the steady solution of Fisher’s equation as a function of domainsize L.

8.3 Resource Consumption

Now consider the situation in which organisms, say bacteria, consume a resource, say glucose, of which thereis a finite supply. For example, suppose bacteria are grown on an agar gel on a Petrie dish. The chemicalreaction describing this is

U +Gk−→ 2U. (8.27)

The units on U and G are such that one unit of G converts into one unit of U . We assume that both theglucose and the bacteria move by diffusion. Consequently the differential equations describing this evolutionare (in one spatial dimension)

∂u

∂t= Du

∂2u

∂x2+ kug, (8.28)

∂g

∂t= Dg

∂2g

∂x2− kug. (8.29)

Numerical simulation of this system of equations is shown in Fig. 8.9.

This simulation again suggests that there should be a traveling wave solution. The first step of the analysis

is to simplify the equations by introducing scaled variables t = τk, x =

√

Du

ky, in terms of which the equation

become

∂u

∂τ=

∂2u

∂y2+ ug, (8.30)

∂g

∂τ= D

∂2g

∂y2− ug. (8.31)

where D =Dg

Du.

Now, to examine the possibility of traveling wave solutions, we look for a solution of the form u(y, τ) =U(y − cτ), g(y, τ) = G(y − cτ), and find the system of ordinary differential equations

0 =d2U

dξ2+ c

dU

dξ+ UG, (8.32)

0 = Dd2G

dξ2+ c

dG

dξ− kUG. (8.33)


x0 20 40 60 80 100 120 140 160 180 200

v(x,

t)

0

0.1

0.2

0.3

0.4

0.5

x0 20 40 60 80 100 120 140 160 180 200

g(x,

t)

0

0.1

0.2

0.3

0.4

0.5

Figure 8.9: Solution of the bacteria (top) -glucose (bottom) system, with initial profile for each shown dashed.For this simulation Du = Dg = 0.01, k = 0.01.

Before we dive into a full-fledged analysis, let’s examine a special case, D = 1. Then, adding the twoequations together yields

d2

dξ2(U +G) + c

d

dξ(U +G) = 0 (8.34)

This can be integrated once to getd

dξ(U +G) + c(U +G) = c (8.35)

where, without loss of generality, we assume that limξ→±∞(U +G) = 1. In fact, the only bounded solutionof (8.35) is U +G = 1. Substituting this into the equation for U , we find

d2U

dξ2+ c

dU

dξ+ U(1− U) = 0, (8.36)

which is Fisher’s equation in traveling wave form. It follows that traveling wave solutions exist for c ≥ 2.

Now let’s try the general case. As before, add the two equations together and integrate once to get

d

dξ(U +DG) + c(U +G) = c (8.37)

We cannot solve this equation as before, so we write the two governing equations as a third order system byintroducing the variable W

dU

dξ= W, (8.38)

DdG

dξ= c−W − c(U +G), (8.39)

dW

dξ= −cW − UG, (8.40)

The two critical points are at W = 0 with U = 0, G = 1, or U = 1, G = 0. The stability of these two criticalpoints is found by examining the eigenvalues of the Jacobian matrix,

J =

0 0 1− c

D− c

D− 1

D

−G0 −U0 −c

. (8.41)

8.4. SPREAD OF RABIES - SIR WITH DIFFUSION 57

The eigenvalues of J are the roots of the cubic polynomial

p0(λ) = −(λ2 +c

Dλ− 1

D)(λ + c), (8.42)

when G = 0, U = 1, and

p1(λ) = −(λ2 + cλ+ 1)(λ+c

D), (8.43)

when G = 1, U = 0. The polynomial p0 has two negative roots (assuming c > 0) and one positive root,so that the critical point at U = 1, G = 0 is a saddle point. On the other hand, the polynomial p1 hasthree roots with negative real part, one of which is λ = − c

D. The other two roots are complex if c < 2 and

real if c ≥ 2. It takes a little more work to show that there is a trajectory connecting U = 1, G = 0 toU = 0, G = 1 for which U and G are always positive and W is always negative. However, such is the case ifc ≥ 2. Consequently, we conclude that there is a traveling wave solution for c ≥ 2, as with Fisher’s equation.Therefore, as with Fisher’s equation, the wave speed is

√2Duk in original dimensional variables.

8.4 Spread of Rabies - SIR with Diffusion

It has been observed in England that rabid foxes tend to travel across much larger distances than rabies freeanimals. This observation has led to consideration of the spread of an infectious disease where the infectedanimals diffuse, but susceptible animal do not. For this we consider the standard SIR disease dynamics

S + Ik−→ 2I, I

µ−→ R. (8.44)

and the corresponding differential equations

∂s

∂t= −ksi, (8.45)

∂i

∂t= ksi− µi+D

∂2i

∂x2. (8.46)

x0 5 10 15 20 25 30

I(x,

t)

0

0.05

0.1

0.15

0.2

0.25

x0 5 10 15 20 25 30

S(x

,t)

0

0.2

0.4

0.6

0.8

1

Figure 8.10: Solution of the SIR model system, with initial profile for each shown dashed. For this simulationD = 0.01, k = 1, µ = 0.05.

A simulation is shown in Fig. 8.11.

As before, we look for traveling wave solutions of the form

s = S(x− ct), i = I(x− ct), (8.47)


and find the system of equations

dS

dξ=

k

cSI, (8.48)

dI

dξ= U, (8.49)

dU

dξ= − c

DU +

µ

DI − k

DSI. (8.50)

The trick to solving this system of equations is to find a conserved quantity, i.e., a first integral of the motion.To do so we set

H = S + a lnS + bI + dU, (8.51)

and try to pick constants a, b, and d so that dHdξ

= 0. Since

dH

dξ=k

cSI + a

k

cI + bU + d(− c

DU +

µ

DI − k

DSI), (8.52)

we pick

d =D

c, a = −µ

k, b = 1, (8.53)

so that

H = S − µ

klnS + I +

D

cU, (8.54)

is a constant of the motion. Suppose that the initial state has S = S0, I = 0, then (8.54) implies that

dS

dξ=

k

cSI, (8.55)

D

c

dI

dξ= S0 − S +

µ

kln

S

S0− I. (8.56)

Now, scale the problem, setting σ = SS0, I = ηS0 so that

dσ

dξ=

kS0

cση, (8.57)

D

c

dη

dξ= 1− σ +

µ

kS0lnσ − η, (8.58)

a system whose phase portrait can be readily studied. Once again we determine the behavior of this systemby examining the phase portrait. First, the critical points have η = 0 and are roots of the equation1− σ + µ

kS0lnσ = 0. One of these roots is σ = 1, and there is a second root, 0 < σ∗ < 1 provided µ

kS0< 1.

At this root, µkS0σ∗

> 1.

For a linear stability analysis, we examine the linearized system

dσ

dξ=

kS0

cσ0η, (8.59)

D

c

dη

dξ= −η + (

µ

kS0− 1)

σ

σ0, (8.60)

where σ = σ∗ or 1. The characteristic polynomial is

p(λ) = λ2 +c

Dλ+

1

D(kS0σ0 − µ). (8.61)

8.5. EXERCISES 59

σ0 0.2 0.4 0.6 0.8 1 1.2 1.4

η

-0.05

0

0.05

0.1

0.15

0.2

Figure 8.11: Phase portrait for the system with µcS0

= 0.5, kS0

c= 1, the dη

dξ= 0 null cline shown as a dashed

curve, with two trajectories leaving the saddle point at σ∗ and approaching σ = 1, for parameter Dc= 0.8

and 1.8.

so that the eigenvalues are

λ = − c

2D± 1

2D

√

c2 − 4D(kS0σ0 − µ). (8.62)

Clearly, since µ > kS0σ∗, the critical point at σ0 = σ∗ is a saddle point, since the eigenvalues are of opposite

sign. On the other hand, the critical point at σ0 = 1 is either a stable node or a stable spiral, a stable nodeif c2 ≥ 4D(kS0 − µ). It follows that there is a trajectory connecting σ = σ∗ with σ = 1 for which η ≥ 0 andσ∗ ≤ σ ≤ 1, i.e., a traveling wave solution.

8.5 Exercises

1. A particle diffuses in a region −L < x < L starting from x = 0. The particle can degrade withdegradation rate γ. The goal of this exercise is to determine the probability that the particle reachesone of the boundaries before it decays.

(a) The answer should be a function of what single non-dimensional parameter?

(b) Numerically simulate this problem to get a numerical estimate of the answer for several parametervalues.

(c) The probability that the particle decays before reaching the boundary is π(0) where π(x) satisfiesthe differential equation

Dπxx − γπ = −γ, π(−L) = 0, π(L) = 0 (8.63)

Calculate the probability that it reaches the boundary before decaying and compare with whatyou found from your numerical simulation.

2. A dingo population which lives in the eastern parts of Australia is prevented from invasion to the westby a fence that runs north-south. Imagine that the fence breaks somewhere at time t = 0. A farmis located on the west side of the fence, exactly 100 miles west of the hole in the fence. The farmerswould like to know how long it will take the dingoes to reach their farm. Model the spread of dingoesas:

ut = Duxx + ku(1− u

K) (8.64)

with k = 1 (1/month), and K = 1 (in units of u).


(a) The region between the fence and farm is flat and the diffusion constantD1 = 100 (miles2/month).When does the dingo population reach the farm? (consider a traveling wave and calculate thewave speed)

(b) How will the time change if the diffusion constant is D2 = 50 miles2/month due to some rock andslope in the region?

Chapter 9

The Bistable Equation

9.1 The Spruce Budworm Problem

The spruce budworm is an insect that attacks spruce and fir trees in the forests of eastern North America.Most years its population level is relatively low, but in some years it exhibits outbreaks in which the popu-lation may increase by a factor of 1000. During an outbreak, budworms may eat enough new needles in anevergreen forest to kill 80% of the trees in the forest and effectively destroy the forest. The destruction ofthe forest also eliminates the budworms’ food supply, causing a collapse of the budworm population, afterwhich the forests can begin a slow recovery. A full model of this behavior would include both the sprucebudworm and the forest, and they operate on quite different time scales.

Let u denote the budworm population size. The equation governing u, proposed by Ludwig (1987), is

∂u

∂t= D

∂2u

∂x2+ ru(1− u

A)− βu2

K2 + u2. (9.1)

where r is the intrinsic growth rate, A is the carrying capacity. Further, we assume that the spruce budwormis subject to predation by birds and parasites, predation which is modeled by a Hill function of order 2(without much mechanistic motivation).

First, we rescale variables setting u = Av, x =√

Dry, τ = rt in terms of which we find

∂v

∂τ=∂2v

∂y2+ v(1 − v)−R

v2

κ2 + v2. (9.2)

where R = βrA

, κ = KA.

Nontrivial steady solutions of this equation satisfy

R =(1 − v)(κ2 + v2)

v, (9.3)

a plot of which is shown in Fig. 9.4. Here we see bistable behavior has a function of the parameter R.

Let’s explore the question of when there are steady solutions with Dirichlet boundary conditions u(0) =u(L) = 0.

61

62 CHAPTER 9. THE BISTABLE EQUATION

v0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

f(v)

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

Figure 9.1: f(u) for (9.2) for κ = 0.05, r = 0.11.

9.2 The Cable Equation

The next equation we consider is the bistable equation

∂u

∂t= D

∂u

∂x2+ f(u) (9.4)

where f(u) is a function with three zeros, i.e., without loss of generality, f(0) = f(α) = f(1) = 0, 0 < α < 1and f(u), 0 for u > 1. It is called bistable because the two equilibria U0 = 0 and u0 = 1 are stable equilibriaof the differential equation du

dt= f(u).

There are several biological situations in which this equation is relevant. The first is in population dynamicswhere the logistic growth function u(1 − u) is multiplied by the factor (u − α) due to something called theAllee effect, yielding f(u) = u(1−u)(u−α). The second occurs in population genetics when there are allelesfor which the heterozygote has a fitness disadvantage to either of the two homozygous traits. The third,which we have already seen, is the spuce budworm problem. The fourth, and the one we describe in moredetail, is the propagation of action potentials in nerve axons.

The membrane of a cell is a barrier to the movement of ions between the intracellular and extracellularspaces. Indeed, the movement of ions across a membrane is carefully regulated and is through a variety ofion channels. Consequently, the electrical activity of a cell can be described by a capacitor (the membrane)and resistors (the ion channels) in parallel, as shown in Fig. 9.5, and modeled by the equation

Cm

dV

dt+ Iion = Iin, (9.5)

where Cm is the membrane capacitance, and V = Vi −Ve is the transmembrane voltage potential, Vi and Veare the intracellular and extracellular voltage potentials, respectively.

Nerve cells typically have three types of ion channels, namely sodium, potassium, and chloride ion channels.The current-voltage relationships for all of these are of the form

Ij = gj(V − Vj), (9.6)

9.2. THE CABLE EQUATION 63

R0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

v

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 9.2: Steady state solution for (9.2) for κ = 0.05.

with j =Na, K, or Cl. The quantity Vj is called the Nernst potential, and gj is the conductance. Onefascinating aspect of many ions channels is that gj is not constant but varies in a time dependent fashion inresponse to the transmembrane potential V .

Now to make life a bit easier, we scale V so that VK = 0. Then, simple models for the sodium and potassiumcurrents are

INa = gnaV2(V − Vna), IK = gkV. (9.7)

A word of warning is important here. These models for sodium and potassium currents are reasonable onlyfor the initial phase of the action potential because they both ignore the fact that these ion channels becomeinactive at high potentials.

This model applies only for a small homogeneous patch of membrane. To incorporate the effects of anelongated membrane, as an axon, consider the diagram in Fig. 9.6 showing two patched of membrane oflength dx.

We view the cell as a long cylindrical piece of membrane surrounding an interior of cytoplasm (called acable). We suppose that everywhere along its length, the potential depends only on the length variable andnot on radial or angular variables, so that the cable can be viewed as one-dimensional. This assumption iscalled the core conductor assumption (Rall, 1977). We now divide the cable into a number of short piecesof isopotential membrane each of length dx. In any cable section, all currents must balance, and there areonly two types of current, namely, transmembrane current and axial current (Fig. ??). The axial currenthas intracellular and extracellular components, both of which ared assume to be ohmic, i.e., linear functionsof the voltage. Hence,

Vi(x+ dx) − Vi(x) = −Ii(x)ridx, (9.8)

Ve(x+ dx) − Ve(x) = −Ie(x)redx, (9.9)

where Ii and Ie are the intracellular and extracellular axial currents respectively. The minus sign on theright-hand side appears because of the convention that positive current is a flow of positive charges fromleft to right (i.e., in the direction of increasing x). If Vi(x + dx) > Vi(x), then positive charges flow in thedirection of decreasing x, giving a negative current. In the limit dx→ 0,

Ii = − 1

ri

∂Vi∂x

, (9.10)


v0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

w

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Figure 9.3: Phase portrait for (9.2) for κ = 0.05, r = 0.11.

Ie = − 1

re

∂Ve∂x

. (9.11)

The numbers ri and re are the resistances per unit length of the intracellular and extracellular media,respectively. In general,

ri =Rc

Ai

, (9.12)

where Rc is the cytoplasmic resistivity, measured in units of Ohms-length, and Ai is the cross-sectional areaof the cylindrical cable. A similar expression holds for the extracellular space, so if the cable is in a bathwith large (effectively infinite) cross-sectional area, the extracellular resistance re is nearly zero.

Next, from Kirchhoff’s laws, any change in extracellular or intracellular axial current must be due to atransmembrane current, and thus

Ii(x)− Ii(x + dx) = Itdx = Ie(x + dx)− Ie(x), (9.13)

where It is the total transmembrane current (positive outward) per unit length of membrane. In the limitas dx→ 0, this becomes

It = −∂Ii∂x

=∂Ie∂x

. (9.14)

In a cable with no additional current sources, the total axial current is IT = Ii+Ie, so using that V = Vi−Ve,we find

−IT =ri + rerire

∂Vi∂x

− 1

re

∂V

∂x, (9.15)

from which it follows that1

ri

∂Vi∂x

=1

ri + re

∂V

∂x− reri + re

IT . (9.16)

On substituting (9.16) into (9.14), we obtain

It =∂

∂x

(

1

ri + re

∂V

∂x

)

, (9.17)

where we have used (9.10) and the fact that IT is constant. Finally, recall that the transmembrane currentIt is a sum of the capacitive and ionic currents, and thus

It = p

(

Cm

∂V

∂t+ Iion

)

=∂

∂x

(

1

ri + re

∂V

∂x

)

, (9.18)

9.2. THE CABLE EQUATION 65

L0 1 2 3 4 5 6 7 8 9 10

Vm

ax

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 9.4: Bifurcation diagram for (9.2) for κ = 0.05, r = 0.11.

C

Outside

Inside

Iion C dV/dtmm

Figure 9.5: Electrical circuit diagram for a cell.

where p is the perimeter of the axon. Equation (9.18) is usually referred to as the cable equation. Notethat Cm has units of capacitance per unit area of membrane, and Iion has units of current per unit areaof membrane. If a current Iapplied, with units of current per unit area, is applied across the membrane (asbefore, taken positive in the outward direction), then the cable equation becomes

It = p

(

Cm

∂V

∂t+ Iion + Iapplied

)

=∂

∂x

(

1

ri + re

∂V

∂x

)

. (9.19)

It is useful to nondimensionalize the cable equation. To do so we define the membrane resistivity Rm as theresistance of a unit square area of membrane, having units of ΩC2

m. For any fixed V0, Rm is determined bymeasuring the change in membrane current when V is perturbed slightly from V0. In mathematical terms,

1

Rm

=dIiondV

∣

∣

∣

∣

V=V0

. (9.20)

Although the value of Rm depends on the chosen value of V0, it is typical to take V0 to be the restingmembrane potential to define Rm. Note that if the membrane is an ohmic resistor, then Iion = V/Rm, inwhich case Rm is independent of V0.


Ve(x)

It dx

Cm dxIion dx

Vi (x)

Ie(x)

Ii(x)

It dx

Cm dxIion dx

Ve(x+dx)

Vi (x+dx)

Extracellularspace

Intracellularspace

Cellmembrane

re dx

ri dx

Figure 9.6: Electrical circuit diagram for several cell membrane patches.

Assuming that ri and re are constant, the cable equation (9.18) can now be written in the form

τm∂V

∂t+ RmIion = λ2m

∂2V

∂x2, (9.21)

where

λm =

√

Rm

p(ri + re)(9.22)

has units of distance and is called the cable space constant, and where

τm = RmCm (9.23)

has units of time and is called the membrane time constant. If we ignore the extracellular resistance, then

λm =

√

Rmd

4Rc

, (9.24)

where d is the diameter of the axon (assuming circular cross-section). Finally, we rescale the ionic currentby defining Iion = −f(V, t)/Rm for some f , which, in general, is a function of both voltage and time and hasunits of voltage, and nondimensionalize space and time by defining new variables X = x/λm and T = t/τm.In the new variables the cable equation is

∂V

∂T=∂2V

∂X2+ f(V, T ). (9.25)

Although f is written as a function of voltage and time, in many of the simpler versions of the cable equation,f is a function of V only (for example, (??) below).

9.3 Traveling Waves for the Bistable Equation

The bistable equation is a specific version of the cable equation (9.25), namely

∂V

∂t=∂2V

∂x2+ f(V ), (9.26)

9.3. TRAVELING WAVES FOR THE BISTABLE EQUATION 67

where f(V ) has three zeros at 0, α, and 1, where 0 < α < 1. The values V = 0 and V = 1 are stable steadysolutions of the ordinary differential equation dV/dt = f(V ). Notice that the variable V has been scaledso that 0 and 1 are zeros of f(V ). In the standard nondimensional form, f ′(0) = −1. (Recall from (9.20)that the passive cable resistance was defined so that the ionic current has slope 1 at rest.) However, thisrestriction on f(V ) is often ignored.

An example of such a function can be found in the Hodgkin–Huxley fast–slow phase plane. When therecovery variable n is held fixed at its steady state, the Hodgkin–Huxley fast–slow model is bistable. Twoother examples of functions that are often used in this context are the cubic polynomial

f(V ) = aV (V − 1)(α− V ), 0 < α < 1, (9.27)

and the piecewise-linear function

f(V ) = −V +H(V − α), 0 < α < 1. (9.28)

where H(V ) is the Heaviside function (Mckean, 1970). This piecewise-linear function is not continuous, nordoes it have three zeros, yet it is useful in the study of traveling wave solutions of the bistable equationbecause it is an analytically tractable model that retains many important qualitative features.

By a traveling wave solution, we mean a translation-invariant solution of (9.26) that provides a transitionbetween the two stable rest states (zeros of the nonlinear function f(V )) and travels with constant speed.That is, we seek a solution of (9.26) of the form

V (x, t) = U(x+ ct) = U(ξ) (9.29)

for some (yet to be determined) value of c. The new variable ξ, called the traveling wave coordinate, hasthe property that fixed values move in space–time with fixed speed c. When written as a function of ξ, thewave appears stationary. Note that, because we use ξ = x+ ct as the traveling wave coordinate, a solutionwith c positive corresponds to a wave moving from right to left. We could equally well have used x − ct asthe traveling wave coordinate, to obtain waves moving from left to right (for positive c).

By substituting (9.29) into (9.26) it can be seen that any traveling wave solution must satisfy

Uξξ − cUξ + f(U) = 0, (9.30)

and this, being an ordinary differential equation, should be easier to analyze than the original partial dif-ferential equation. For U(ξ) to provide a transition between rest points, it must be that f(U(ξ)) → 0 asξ → ±∞.

To study (9.30) it is convenient to write it as two first order equations,

Uξ = W, (9.31)

Wξ = cW − f(U). (9.32)

To find traveling front solutions for the bistable equation, we look for a solution of (9.31) and (9.32) thatconnects the rest points (U,W ) = (0, 0) and (U,W ) = (1, 0) in the (U,W ) phase plane. Such a trajectory,connecting two different steady states, is called a heteroclinic trajectory, and in this case is parameterizedby ξ; the trajectory approaches (0, 0) as ξ → −∞ and approaches (1, 0) as ξ → +∞ (see the dashed line inFig. 9.7A). The steady states at U = 0 and U = 1 are both saddle points, while for the steady state U = α,the real part of both eigenvalues have the same sign, negative if c is positive and positive if c is negative,so that this is a node or a spiral point. Since the points at U = 0 and U = 1 are saddle points, the goal isto determine whether the parameter c can be chosen so that the trajectory that leaves U = 0 at ξ = −∞connects with the saddle point U = 1 at ξ = +∞. This mathematical procedure is called shooting, and somesample trajectories are shown in Fig. 9.7A.


First, we can determine the sign of c. Supposing a monotone increasing (Uξ > 0) connecting trajectoryexists, we multiply (9.30) by Uξ and integrate from ξ = −∞ to ξ = ∞ with the result that

c

∫ ∞

−∞W 2dξ =

∫ 1

0

f(u)du. (9.33)

In other words, if a traveling wave solution exists, then the sign of c is the same as the sign of the area underthe curve f(u) between u = 0 and u = 1. If this area is positive, then the traveling solutions move the statevariable U from U = 0 to U = 1, and the state at U = 1 is said to be dominant. In both of the special cases(9.27) and (9.28), the state U = 1 is dominant if α < 1/2.

Suppose∫ 1

0f(u)du > 0. We want to determine what happens to the unstable trajectory that leaves the

saddle point U = 0, Uξ = 0 for different values of c. With c = 0, an explicit expression for this trajectory isfound by multiplying (9.30) by Uξ and integrating to get

W 2

2+

∫ U

0

f(u)du = 0. (9.34)

If this trajectory were to reach U = 1 for some value of W , then

W 2

2+

∫ 1

0

f(u)du = 0, (9.35)

which can occur only if∫ 1

0f(u)du ≤ 0. Since this contradicts our assumption that

∫ 1

0f(u)du ≥ 0, we

conclude that this trajectory cannot reach U = 1. Neither can this trajectory remain in the first quadrant,as W > 0 implies that U is increasing. Thus, this trajectory must intersect the W = 0 axis at some value ofU < 1 (Fig. 9.7A). It cannot be the connecting trajectory.

Next, suppose c is large. In the (U,W ) phase plane, the slope of the unstable trajectory leaving the restpoint at U = 0 is the positive root of λ2 − cλ+ f ′(0) = 0, which is always larger than c (Exercise ??). Let Kbe the smallest positive number for which f(u)/u ≤ K for all u on the interval 0 < u ≤ 1 (Exercise: How dowe know K exists?), and let σ be any fixed positive number. On the line W = σU the slope of trajectoriessatisfies

dW

dU= c− f(U)

W= c− f(U)

σU≥ c− K

σ. (9.36)

By picking c large enough, we are assured that c − K/σ > σ, so that once trajectories are above the lineW = σU , they stay above it. We know that for large enough c, the trajectory leaving the saddle point U = 0starts out above this curve. Thus, this trajectory always stays above the line W = σU , and therefore passesabove the rest point at (U,W ) = (1, 0).

Now we have two trajectories, one with c = 0, which misses the rest point at U = 1 by crossing the W = 0axis at some point U < 1, and one with c large, which misses this rest point by staying above it at U = 1.Since trajectories depend continuously on the parameters of the problem, there is a continuous family oftrajectories depending on the parameter c between these two special trajectories, and therefore there is atleast one trajectory that hits the point U = 1,W = 0 exactly.

The value of c for which this heteroclinic connection occurs is unique. To verify this, notice from (9.36) thatthe slope dW/dU of trajectories in the (U,W ) plane is a monotone increasing function of the parameter c.Suppose at some value of c = c0 there is known to be a connecting trajectory. For any value of c that islarger than c0, the trajectory leaving the saddle point at U = 0 must lie above the connecting curve for c0.For the same reason, with c > c0, the trajectory approaching the saddle point at U = 1 as ξ → ∞ must liebelow the connecting curve with c = c0. A single curve cannot simultaneously lie above and below anothercurve, so there cannot be a connecting trajectory for c > c0. By a similar argument, there cannot be aconnecting trajectory for a smaller value of c, so the value c0, and hence the connecting trajectory, is unique.


0.20

0.15

0.10

0.05

0.00

W

1.00.80.60.40.20.0U

c = 1

c = 0.57

c = 0.56c = 0

1.0

0.8

0.6

0.4

0.2

0.0

-10 -5 0 5 10ξ

U(ξ)

U'(ξ)

A B

Figure 9.7: A: Trajectories in the (U,W ) phase plane leaving the rest point U = 0, W = 0 for the equationUξξ − cUξ + U(U − 0.1)(1 − U) = 0, with c = 0.0, 0.56, 0.57, and 1.0. Dashed curve shows the connectingheteroclinic trajectory. B: Profile of the traveling wave solution of panel A. In the original coordinates, thisfront moves to the left with speed c.

For most functions f(V ), it is necessary to calculate the speed of propagation of the traveling front solutionnumerically. However, in the two special cases (9.27) and (9.28) the speed of propagation can be calculatedexplicitly. In the piecewise linear case (9.28) one calculates directly that

c =1− 2α√α− α2

(9.37)

(see Exercise ??).

Suppose f(u) is the cubic polynomial

f(u) = −a2(u− u0)(u − u1)(u − u2), (9.38)

where the zeros of the cubic are ordered u0 < u1 < u2. We want to find a heteroclinic connection betweenthe smallest zero u0, and the largest zero u2, so we guess that

W = −b(U − u0)(U − u2). (9.39)

We substitute this guess into the governing equation (9.30), and find that we must have

b2(2U − u0 − u2)− cb− a2(U − u1) = 0. (9.40)

This is a linear function of U that can be made identically zero only if we choose b = a/√2 and

c =a√2(u2 − 2u1 + u0). (9.41)

It follows from (9.39) that

U(ξ) =u0 + u2

2+u2 − u0

2tanh

(

a√2

u2 − u02

ξ

)

, (9.42)

which is independent of u1. In the case that u0 = 0, u1 = α, and u2 = 1, the speed reduces to

c =a√2(1− 2α), (9.43)


showing that the speed is a decreasing function of α and the direction of propagation changes at α = 1/2.The profile of the traveling wave in this case is

U(ξ) =1

2

[

1 + tanh

(

a

2√2ξ

)]

. (9.44)

A plot of this traveling wave profile is shown in Fig. 9.7b.

Once the solution of the nondimensional cable equation (9.26) is known, it is a simple matter to express thesolution in terms of physical parameters as

V (x, t) = U

(

x

λm+ c

t

τm

)

, (9.45)

where λm and τm are, respectively, the space and time constants of the cable, as described in Chapter ??.The speed of the traveling wave is

s =cλmτm

=c

2Cm

√

d

RmRc

, (9.46)

which shows how the wave speed depends on capacitance, membrane resistance, cytoplasmic resistance, andaxonal diameter. The dependence of the speed on ionic channel conductances is contained (but hidden) inc. According to empirical measurements, a good estimate of the speed for an axon is given by

s =

√

d

10−6mm/sec. (9.47)

Using d = 500mm for squid axon, this estimate gives s = 22.4 mm/ms, which compares favorably to themeasured value of s = 21.2 mm/ms.

Scaling arguments can also be used to find the dependence of speed on certain other parameters. Suppose,for example, that a drug is applied to the membrane that blocks a percentage of all ion channels, irrespectiveof type. If ρ is the fraction of remaining operational channels, then the speed of propagation is reduced bythe factor

√ρ. This follows directly by noting that the bistable equation with reduced ion channels

V ′′ − sV ′ + ρf(V ) = 0 (9.48)

can be related to the original bistable equation (9.30) by taking V (ξ) = U(√ρξ), s = c

√ρ.

A second example of the nonlinear bistable function is the piecewise linear function

f(u) = k(H(u− α)− u). (9.49)

This function is nice because there are many explicit calculations that can be done with it. For example, tofind traveling wave solutions, we solve

DU ′′ + cU ′ − kU = 0, ξ > 0 (9.50)

andDU ′′ + cU ′ − k(U − 1) = 0 ξ < 0 (9.51)

with boundary conditions U(−∞) = 1, U(∞) = 0 and matching conditions u(0) = α and u′(0) continuous.It follows that

U(ξ) =

α exp(λ−ξ ξ > 01 + (α− 1) exp(λ+ξ ξ < 0

(9.52)

subject to the condition αλ− = (α− 1)λ+, where λ± are roots of the quadratic equation Dλ2 + cλ− k = 0,that is,

λ± =−c±

√c2 + 4Dk

2D(9.53)

Consequently,

c =1− 2α√α− α2

√Dk. (9.54)


9.3.1 Propagation Failure

There are several important biological situations where the homogeneous cable equation fails to be theapronriate model for signal propagation. These include cardiac cells, myelinated axons, and calcium releasein cardiac cells.

The biology:

Cardiac Cells

gap junctions

Figure 9.8: Gap junctional coupling

Figure 9.9: Schematic diagram of the myelin sheath. (Guyton and Hall, 1996, Fig. 5-16, p. 69.)

dundt

= d(un−1 − 2un + un+1) + f(un). (9.55)

Numerically simulate this equation to find speed as a function of d and observe that for small enough D,propagation fails. Here we attempt to decide if this is correct, and if so, when. To do so, we look forstationary (i.e., not moving) profiles. This means solving the difference equation

d(un−1 − 2un + un+1) + f(un) = 0 (9.56)

subject to the conditions, limn→−∞ un = 1, limn→∞ un = 0. This is not easy to do for general f(u), but itis possible for the piecewise linear function (9.49). In particular, we take

un =

1−Aλ−n n < 0Bλn n ≥ 0

(9.57)

where 0 < λ < 1 is a root of the polynomial λ2 − (2 + kd)λ + 1 = 0. Assuming B < α and 1 − Aλ > α, we

have two consistency conditions to satisfy, namely

n = 0 : 1−Aλ− 2B +Bλ− k

dB = 0, (9.58)

n = −1 : 1−Aλ2 − 2(1−Aλ) +B +k

dAλ = 0, (9.59)


α0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

d/k

0

1

2

3

4

5

6

7

8

9

10

Figure 9.10: Plot of equation (9.63).

0.8

0.6

0.4

0.2

c*/c

e

43210β

traveling waves

standing waves

Figure 9.11: Plot of ace

(lower curve) and Cce

− bceµf (upper curve) plotted as functions of β =

√

kL2

Dfor

Ak= 5.

which we solve for A and B to find

1− λ = B(λ+1

λ), (9.60)

1−A−B = 0. (9.61)

or

B =λ− λ2

λ2 + 1, A = 1−B, (9.62)

and then, the condition B < α leads tod

k≤ α(1− α)

(2α− 1)2. (9.63)

A plot of this function is shown in Fig. 9.10.

A second example of the effects of discreteness is that of calcium release. Most models of calcium releasefrom the ER assume that release is homogeneous in space. In fact, this is not the case. In Xenopus oocytes,for example, IPR are arranged in clusters with a density of about 1 per 30 mm2, with each cluster containingabout 25 IPR. Furthermore, the propagation of Ca2+ waves is saltatory, with release “jumping” from onecluster of release sites to another.

To explore the properties of such saltatory waves, and to see the effect of this discrete structure, we assumethat calcium is released from discrete release sites but is removed continuously. We consider the equation


Cardiac Cell

Calcium release sites

Figure 9.12: Calcium Release through CICR Receptors

for calcium concentration∂c

∂t= Dc

∂2c

∂x2− ksc+ L

∑

n

δ(x− nL)f(c). (9.64)

The function f is your favorite description of calcium release, and L is the spatial separation between releasesites.

Two examples of the release function f(c) that are useful are

f(c) = Ac2(ce − c), (9.65)

andf(c) = AH(c− c∗)(ce − c), (9.66)

where H is the usual Heaviside function. Both of these release functions have calcium dependent release,representing CICR.

The corresponding continuous space model can be found by homogenization (See Exercise ??). It is not

difficult to show (and it makes intuitive sense) that in the limit that L2kDc

≪ 1, the spatially inhomogeneousproblem

∂c

∂t= Dc

∂2c

∂x2− ksc+ g(x)f(c), (9.67)

where g(x) is periodic with period L, can be replaced by its average

∂c

∂t= Dc

∂2c

∂x2− ksc+Gf(c), (9.68)

where G = 1L

∫ L

0g(x)dx.

Since the effective release function F (c) = Gf(c)− ksc is of bistable type, with three zeros 0 < c1 < c2, onewould expect there to be traveling waves of calcium release, provided the medium is sufficiently excitable,i.e., provided

∫ c2

0

F (c)dc > 0. (9.69)

However, this condition fails to be correct if L2ks

Dc6= 0.

Waves fail to propagate if there is a standing wave solution. Standing waves are stationary solutions of(9.64), i.e., solutions of

0 = Dc

∂2c

∂x2− ksc+ L

∑

n

δ(x − nL)f(c). (9.70)

On the intervals nL < x < (n+ 1)L, this becomes

0 = Dc

∂2c

∂x2− ksc. (9.71)


We find jump conditions at x = nL by (formally) integrating from nL− to nL+ to obtain

Dccx|nL+

nL− + Lf(cn) = 0, (9.72)

where cn = c(nL).

Now we solve (9.71) to obtain

c(x) = (cn+1 − cn coshβ)sinh

(

βL(x− nL)

)

sinhβ+ cn cosh

(

β

L(x− nL)

)

, (9.73)

for nL < x < (n+ 1)L, where cn = c(nL), β2 = ksL2

Dc, so that

cx(nL+) = (cn+1 − cn coshβ)

β

L sinhβ. (9.74)

Similarly,

cx(nL−) = −(cn−1 − cn coshβ)

β

L sinhβ. (9.75)

It follows that (9.72) is the difference equation

k

β sinhβ(cn+1 − 2cncoshβ + cn−1) + f(cn) = 0, (9.76)

which is a difference equation for cn.

Finding solutions of nonlinear difference equations is non-trivial, in general. However, if f(c) is piecewiselinear, as it is with release function (9.66), analytical solutions can be found. We seek solutions of (9.76) ofthe form

cn =

aµ−n0 n ≤ 0C − bµn

f n > 0, (9.77)

where the steady solution C satisfies

k

β sinhβ(C − 2Ccoshβ + C) +A(ce − C) = 0, (9.78)

so that

C = ceAf

2coshβ − 2 +Af

, Af = Aβ sinhβ

ks. (9.79)

The numbers µ0 and µf are both less than one and satisfy the quadratic equations

µj − 2λj +1

µj

= 0, j = 0, f (9.80)

where

λ0 = coshβ, λf = coshβ +Af

2, (9.81)

so that

µj = λj −√

λ2j − 1. (9.82)

It is easy to determine that µ0 = exp(−β); the expression for µf is more complicated.

We determine the scalars a and b by examining the difference equation for n = 0

k

β sinhβ(c1 − 2c0 coshβ + c−1) + f(c0) = 0, (9.83)

9.4. THE FIRE-DIFFUSE-FIRE MODEL 75

and for n = 1ks

β sinhβ(c2 − 2c1 coshβ + c0) + f(c1) = 0. (9.84)

Since c0 < c∗ and c1 > c∗ by assumption, f(c0) = 0 and f(c1) = A(ce − C + bµf ). After some algebraicmanipulation we find that

a = C

(

µf − 1

µf − 1µ0

)

, b = C

(

1− 1µ0

µf − 1µ0

)

. (9.85)

Now, the condition for the existence of these standing waves is that

a ≤ c∗, C − bµf ≥ c∗, (9.86)

so a plot of a and C − bµf is revealing. In Fig. 9.11 are the two curves ace

(lower curve) and Cce

− bceµf

(upper curve) plotted as functions of β =√

ksL2

Dcfor A

k= 5. The interpretation of this plot is that for a fixed

value of β, if c∗

celies in the region below the curve a

ce, then there is a traveling wave, whereas, if c∗

ceand β

lies between the two curves, there is a standing wave, which precludes the possibility of propagation. Thus,in general, the larger β the more excitable the release sites must be ( c

∗

cemust be smaller), for propagation

to occur. Said another way, discrete release from clumped receptors makes propagation less likely than ifthe same amount of calcium were released in a spatially continuous and homogeneous fashion. Notice alsothat for fixed c∗

ce, increasing β sufficiently always leads to propagation failure. Increasing β corresponds to

increasing the distance between release sites L or the rate of uptake ks, or decreasing the calcium diffusioncoefficient Dc.

9.4 The Fire-Diffuse-Fire Model

It is difficult to determine when propagation failure occurs and it is even more difficult to find the speed ofpropagation in a discrete release model. One approach, which uses techniques that are beyond the scopeof this text, was used in Keener (2000b). Another approach is to use a different model. One model thathas attracted a lot of attention is the fire-diffuse-fire model (Pearson and Ponce-Dawson, 1998, Keizer et al.(1998), Dawson et al., 1999; Coombes, 2001; Coombes and Bressloff, 2003; Coombes et al., 2004).

In this model, once [Ca++] reaches a threshold value, c∗, at a release site, that site fires, instantaneouslyreleasing a fixed amount, σ, of Ca2+. Thus, a Ca2+ wave is propagated by the sequential firing of releasesites, each responding to the Ca2+ diffusing from neighboring release sites. Hence the name fire-diffuse-fire.

We assume that Ca2+ obeys the reaction diffusion equation

∂c

∂t= Dc

∂2c

∂x2+ σ

∑

n

δ(x− nL)δ(t− tn), (9.87)

where, as before, L is the spacing between release sites. Although this equation looks linear, appearances aredeceptive. Here, tn is the time at which c first reaches the threshold value c∗ at the nth release site. Whenthis happens, the nth release site releases the amount σ. Thus, tn depends in a complicated way on c.

The Ca2+ profile resulting from the firing of a single site, site i say, is

ci(x, t) = σH(t− ti)

√

4πDc(t− ti)exp

(

− (x − iL)2

4Dc(t− ti)

)

, (9.88)

where H is the Heaviside function. This is the fundamental solution of the diffusion equation with a deltafunction input at x = i, t = ti, and can be found in any standard book on analytical solutions to partial


50

40

30

20

10

0

η

0.80.60.40.20.0c*L/σ

1.0

0.8

0.6

0.4

0.2

0.0

g(η)

403020100η

A B

Figure 9.13: A: Plot of g(η), where g(η) is defined by (9.92). B: Plot of the delay η using (9.92).

differential equations (see, for example, Keener, 1998, or Kevorkian, 2000). If we superimpose the solutionsfrom each site, we get

c(x, t) =∑

i

ci(x, t) = σ∑

i

H(t− ti)√

4πDc(t− ti)exp

(

− (x− iL)2

4Dc(t− ti)

)

. (9.89)

Notice that because of the instantaneous release, c(x, t) is not a continuous function of time at any releasesite.

Now, suppose that sites i = N,N − 1, · · · have fired at known times tN > tN−1 > · · ·. The next firing timetN+1 is determined by when c at xN+1 first reaches the threshold, c∗, that is,

c((N + 1)L, t−N+1) = c∗,∂

∂tc((N + 1)L, t−N+1) > 0. (9.90)

Thus, tN+1 must satisfy

c∗ = σ∑

i≤N

1√

4πDc(tN+1 − ti)exp

(

−L2(N + 1− i)2

4Dc(tN+1 − ti)

)

. (9.91)

A steadily propagating wave corresponds to having ti − ti−1 = constant = τ for all i, i.e., each site fires afixed time after its leftward neighbor fired. Note that the resulting wave does not propagate with a constantprofile, but has a well-defined wave speed L/τ . If such a τ exists, then tN+1 − ti = τ(N + 1− i) and τ is asolution of the equation

c∗L

σ=

∞∑

n=1

1√4πnη

exp

(

− n

4η

)

≡ g(η), (9.92)

where η = DcτL2 is the dimensionless delay.

To find η we need to invert this equation. A plot of g(η) is shown in Fig. 9.13. It can be shown that0 ≤ g(η) ≤ 1 and that g is monotonic with g → 0 as η → 0 and g → 1 as η → ∞. It follows that a solutionto (9.92) exists only if c∗L

σ< 1. Thus, when the intercluster distance or the threshold is too large, or the

amount of release is too small, there is no propagation. However, when c∗Lσ

< 1 a unique solution of (9.92)is guaranteed, and thus there is a propagating wave.

It is an easy matter to plot the delay as a function of c∗Lσ

by appropriately reversing the axes of Fig. 9.13A.

Thus, in Fig. 9.13B is plotted the dimensionless delay η as a function of c∗Lσ

. It is equally easy to plot the

dimensionless velocity 1ηas a function of c∗L

σ(not shown). The result is that the velocity is infinitely large

as c∗Lσ

→ 0 and is zero for c∗Lσ

≥ 1.

9.4. THE FIRE-DIFFUSE-FIRE MODEL 77

If a solution of (9.92) exists, then the speed of the wave is proportional to Dc. This is because the velocityis L

τ= D

ηL. This result is disconcerting for two reasons. First, the speed of propagation of waves in spatially

homogeneous reaction diffusion systems usually scales with the square root of D. It follows from simplescaling arguments that since the units of the diffusion coefficient, Dc, are (length)2 per time, the distancevariable can be scaled by

√Dck, where k is a typical time constant, to remove all distance units, so that the

wave speed scales with√Dck. Second, this disagrees with the result of the previous section showing that if

Dc is sufficiently small, there is propagation failure. For the fire-diffuse-fire model, propagation success orfailure is independent of Dc.

This mismatch is explained by the fact that the fire-diffuse-fire model has no typical time constants, and isnot translationally invariant. Furthermore, since the fire-diffuse-fire model allows only for release of calcium,but no uptake, the calcium transient is unrealistically monotone increasing.

This deficiency of the fire-diffuse-fire model is easily remedied by adding a linear uptake term that is homo-geneous in space (Coombes, 2001), so that the model becomes

∂c

∂t= Dc

∂2c

∂x2− ksc+ σ

∑

n

δ(x− nL)δ(t− tn), (9.93)

The analysis of this modified model is almost identical to the case with ks = 0. The fundamental solution ismodified slightly by ks, with the Ca2+ profile resulting from the firing at site i given by

ci(x, t) = σH(t− ti)

√

4πDc(t− ti)exp

(

− (x− iL)2

4Dc(t− ti)− ks(t− ti)

)

. (9.94)

Following the previous arguments, we learn that a propagating solution exists if there is a solution of theequation

c∗L

σ=

∞∑

n=1

1√4πnη

exp

(

− n

4η− β2n

)

≡ gβ(η), (9.95)

where η = DcτL2 is the dimensionless delay, and β2 = ksL

2

Dcas in the previous section.

Estimates of the size of β2 can vary substantially. For example, using ks = 143/s (= 1/α2 from Table ??),L2 = 30µm2 and Dc = 20µm2/s, we find β2 = 172, whereas with ks = 3.78/s (= k4 from the Friel modelTable ??), L2 = 4µm2 (appropriate for calcium release in cardiac cells) and Dc = 25µm2/s, we find β2 = 0.6.Regardless, the effect of β is significant. Plots of gβ(η) are shown for several values of β in Fig. 9.14. Inparticular, if β 6= 0, the function gβ(η) is not monotone increasing, but has a maximal value, say gmax(β),

which is a decreasing function of β. Furthermore, gβ(η) → 0 as η → 0 and as η → ∞. If c∗Lσ

> gmax(β),

then no solution of (9.95) exists; there is propagation failure. On the other hand, if c∗Lσ

< gmax(β), thenthere are two solutions of (9.95), although, the physically meaningful solution is the smallest of the two,corresponding to the first time that c(x, t) reaches c∗.

Denote the maximal value of gβ(η) by gmax(β). For larger values of β, the curve gmax(β) agrees very wellwith the exponential function exp(−β). This provides us with an approximate criterion for propagation

failure, namely, if c∗Lσ

> exp(−β), propagation fails. Recall that β =√

ksL2

Dcto see the

√Dc dependence in

this criterion.

By reversing the axes on Fig. 9.14, one obtains a plot of the dimensionless delay as a function of c∗Lσ

. Noticethe significant qualitative difference when β 6= 0 compared to β = 0. With β 6= 0, propagation ceases at afinite, not infinite delay. This implies that propagation fails at a positive, not zero, velocity.


0.8

0.6

0.4

0.2

0.0

g β(η

)

403020100η

β=0.1

β=0.5

β=0.01

Figure 9.14: Plots of gβ(η) for three different values of β. Note that gβ is a decreasing function of β for allpositive values of η.

9.5 Exercises

1. The purpose of this exercise is to demonstrate that the bistable equation

∂u

∂t=∂2u

∂x2+ f(u), (9.96)

with f(u) = u(u − α)(1 − u) and −∞ < x < ∞ exhibits threshold behavior. For this exercise takeα = 0.1.

(a) Find a nontrivial steady solution of this problem, i.e., a solution of ∂2u∂x2 + f(u) = 0 for which

limx→±∞ u(x) = 0. Plot it both in the phase plane and as a function of x.

(b) Solve (i.e., simulate numerically) the bistable equation starting from initial data which are scalarmultiples of the steady solution found in the previous part. Take the scalar multiple to be slightlylarger than one and slightly smaller than one. What is the eventual fate of the different solutions?

2. Morris and Lecar (1981) proposed the following two-variable model of membrane potential for a bar-nacle muscle fiber:

Cm

∂V

∂t+ Iion(V,W ) = Iapp +D

∂2V

∂x2, (9.97)

∂W

∂t= φΛ(V )[W∞(V )−W ], (9.98)

where V = membrane potential, W = fraction of open K+ channels, Cm = membrane capacitance,Iapp = externally applied current, φ = maximum rate for closing K+ channels, and

Iion(V,W ) = gCaM∞(V )(V − V 0Ca) + gKW (V − V 0

K) + gL(V − V 0L ), (9.99)

M∞(V ) =1

2

(

1 + tanh

(

V − V1V2

))

, (9.100)

W∞(V ) =1

2

(

1 + tanh

(

V − V3V4

))

, (9.101)

Λ(V ) = cosh

(

V − V32V4

)

. (9.102)

Typical rate constants in these equations are shown in Table 5.1.

9.5. EXERCISES 79

Cm = 20 µF/cm2

Iapp = 0.06 mA/cm2

gCa = 4.4 mS/cm2

gK = 8 mS/cm2

gL = 2 mS/cm2

φ = 0.04 (ms)−1

V1 = −1.2 mV V2 = 18 mV

V3 = 2 V4 = 30 mV

V 0Ca = 120 mV V 0

K = −84 mV

VL = −60 mV

Table 9.1: Typical parameter values for the Morris–Lecar model.

(a) Make a phase portrait for the Morris–Lecar differential equations (with D = 0). Plot the nullclinesand show some typical trajectories, demonstrating that the model is excitable.

(b) Find travelling wave solutions of the Morris-Lecar equations numerically. Plot these solutions inthe V −W phase plane. Estimate the wave speed as a function of D. Verify (numerically) thatthe speed scales like

√D.


Chapter 10

Advection

Now we begin to study problems in which transport (or advection) play an important role.

10.1 Simple Advection

The first example we study is the simplest possible case, in which the material u is being carried at constantvelocity v, so that the flux is J = vu, and the conservation law reads

∂u

∂t= − ∂

∂x(vu). (10.1)

It is reasonable to believe that the quantity u is simply translated at velocity v and so it is a reasonableguess that u(x, t) = f(x− vt). In fact, substituting this form into the equation (10.1) verifies this.

10.2 Structured Populations

The idea pursued here is that when following the dynamics of some population, it may be important to keeptrack of some additional feature of the population, such as age. Such considerations give rise to so-calledstructured population models. To begin with a simple example, suppose u(a, t) is a population density, sothat u(a, t)da represents the number of individuals in the population with ages between a and a+ da. Howcan u change?

We derive a conservation law for u by noting that

∂

∂t

∫ b

a

u(x, t)dx = u(a, t)− u(b, t)−∫ b

a

µ(x)u(x, t)dx, (10.2)

= −∫ b

a

∂u

∂ada−

∫ b

a

µ(x)u(x, t)dx, (10.3)

since the rate of aging is the same as the rate of passage of time, where µ(a) is the death rate, possibly a

81

82 CHAPTER 10. ADVECTION

function of age a. Removing the integrals yields

∂u

∂t= −∂u

∂a− µ(a)u. (10.4)

In addition, we must specify the birthrate, that is, the rate of production of individuals of age a = 0. Wespecify,

u(0, t) =

∫ ∞

0

β(a)u(a, t)da. (10.5)

Here β(a) is the age-dependent birthrate, in units of number of births per individual.

The solution of this problem employs an important technique called the method of characteristics. The ideais to look for solutions of the equation along curves, called characteristic curves, of the form a = X(t) Alongsuch a curve,

du

dλ=∂u

∂a

dX

dt+∂u

∂t, (10.6)

and if we pick dXdt

= 1, then

du

dt= −µu. (10.7)

In other words, by this ”trick”, we have reduced the solution of the partial differential equation to thesolution of the two ordinary differential equations

dX

dt= 1,

du

dt= −µ(X)u. (10.8)

This is not difficult to solve. Setting a = X = t− t0, we find

u(a, t) = U0(t− a) exp

(

−∫ a

0

µ(x)dx

)

, (10.9)

where u(0, t) = U0(t). Now, we can close this equation by using (10.5) to find

U0(t) =

∫ ∞

0

β(x)u(x, t)dx =

∫ ∞

0

β(x)S(x)U0(t− x)dx, (10.10)

where

S(a) = exp(

−∫ a

0

µ(x)dx)

(10.11)

is the survival probability to age a. This is an integral equation for the birthrate U0 as a function of time.To get an idea of its solution we try an exponential solution, U0(t) = exp(λt) and find upon substitutinginto (10.10),

1 =

∫ ∞

0

β(x)S(x) exp(−λx)dx, (10.12)

which is a nonlinear equation for λ. Clearly, the function F (λ) is a monotone decreasing function of λ withF (0) =

∫∞0β(x)S(x)dx. Therefore, if F (0) > 1 the solution has λ > 0, whereas, if F (0) < 1 the solution

has λ < 0. In other words, the population grows exponentially if F (0) > 1 whereas it decays exponentiallyif F (0) < 1.

This all makes sense when one realizes that the quantity F (0) represents the fitness of the population, i.e.,the true, age structure dependent, reproductive rate.

10.3. RED BLOOD CELLS 83

10.3 Red Blood Cells

Growth and differentiation of blood cells are controlled by an intricate array of soluble factors, or cytokines,that include hematopoietic growth and differentiation factors. However, most cytokines are known to havemultiple effects and interactions, and there is no clear distinction between those that control growth andthose that control differentiation. Formation of cytokines is itself controlled by factors outside the bonemarrow such as, in the case of red blood cells, low oxygen concentration for an extended period of time.

The feedback system that controls erythrocyte production is relatively well understood. The principal factorstimulating red blood cell production is the hormone erythropoietin. About 90% of the erythropoietin issecreted by renal tubular epithelial cells when blood is unable to deliver sufficient oxygen. The remainder isproduced by other tissues (mostly the liver). When both kidneys are removed or destroyed by renal disease,the person invariably becomes anemic because of insufficient production of erythropoietin.

The role of erythropoietin in bone marrow is twofold. First, it stimulates the production of pre-erythrocytes,called proerythroblasts, and it also controls the speed at which the developing cells pass through the differentstages. Normal production of red blood cells from stem cells takes 5–7 days, with no appearance of new cellsbefore 5 days, even at high levels of erythropoietin. At high erythropoietin levels the rate of red blood cellproduction (number per unit time) can be as much as ten times normal, even though the maturation rateof an individual red blood cell varies much less.

Thus, in response to a drop of oxygen pressure in the tissues, an increased production of erythropoietincauses an increase in the rate of production of red blood cells, thus tending to restore oxygen levels. Thecontrol mechanisms that operate when the red blood cell count is too high (a condition called polycythemiaor erythrocytosis) are less clear. Details of the regulatory system governing red blood cells can be found inWilliams (1990), while an excellent review of much of the material discussed in this section can be found inHaurie et al. (1998).

We model the blood production cycle as follows. We let n(x, t) be the density of blood cells at time t thatare x units old, i.e., that were released into the bloodstream at time t − x. Red blood cells can die at anyage, but for this discussion, we assume that they all die at some age X . Consider the rate of change of thetotal number of cells with age in the interval between x = a and x = b. Since n(b, t) is the rate at whichcells leave the interval [a, b] due to aging, and n(a, t) is the rate at which they enter that interval, the rateof change of total cell number is given by

d

dt

∫ b

a

n(x, t) dx = n(a, t)− n(b, t). (10.13)

Differentiating (10.13) with respect to b, and replacing b by x gives the conservation equation,

∂n

∂t= −∂n

∂x. (10.14)

At any given time the total number of blood cells in circulation is

N(t) =

∫ X

0

n(x, t)dx. (10.15)

Now we suppose that the production of blood cells is controlled by N , and that once a cohort of cells isformed in the bone marrow, it emerges into the bloodstream as mature cells some fixed time d later, about5 days. Here we ignore the fact that at high levels of feedback (for example, at high levels of erythropoietin,which occurs when oxygen levels are low) cells mature more rapidly. Thus,

n(0, t) = F (N(t− d)), (10.16)


1.0

0.8

0.6

0.4

0.2

0.0

Pro

du

ctio

n r

ate

2.01.51.00.50.0

N

β = 0.8

β = 0.5

β = 0.2

Figure 10.1: Plot of left- and right-hand sides of (10.19) for three different values of β = 1X

and for F (N) =1

1+N7 .

where F is some nonlinear production function that is monotone decreasing in its argument. The functionF is related to the rate of secretion of growth inducer (erythropoietin, for example) in response to the bloodcell population size.

The steady-state solution for this model is easy to determine. We set ∂n/∂t = 0 and find that

n(x) =

n(0), x < X,0, x > X,

(10.17)

where n(0) is yet to be determined. Since N0 =∫X

0 n(x) dx is the total number of cells in steady state, itfollows that

N0 = Xn(0). (10.18)

At steady state, F (N0) = n(0), and thus it follows that

F (N0) =N0

X. (10.19)

Since F (N0) is a monotone decreasing function of N0, (10.19) is guaranteed to have a unique solution. Infact, the solution is a monotone decreasing function of the parameter X , indicating that with decreasingdeath age, the cell population drops while the production of cells increases. An illustration of these facts isprovided by the graph in Fig. 10.1, where the two curves F (N) and N

Xare plotted as functions of N . Here

the function F (N) is taken to be F (N) = A1+N7 , as suggested by data from autoimmune-induced hemolytic

anemia in rabbits (Belair et al., 1995).

The next interesting question to ask is whether this steady solution is stable or unstable. It is convenient tointegrate the partial differential equation (10.14) to get an ordinary differential equation. Integrating (10.14)from x = 0 to x = X gives

dN

dt= n(0, t)− n(X, t). (10.20)

Since n(0, t) = F (N(t − d)) and n(X, t) = F (N(t − d −X)), it follows that N(t) is governed by the delaydifferential equation

dN

dt= F (N(t− d)) − F (N(t− d−X)). (10.21)

We now linearize around the steady state, N0, by looking for solutions of the form N(t) = N0 + ǫN1(t).Substituting this form into (10.21), differentiating with respect to ǫ and taking the limit ǫ→ 0 we find

dN1

dt= F ′(N0)

(

N1(t− d))−N1(t− d−X))

, (10.22)

10.3. RED BLOOD CELLS 85

and then we try a solution of the form N1 = exp(λt). It follows that λ must satisfy the characteristicequation

λ+ F ′(N0)e−λ(d+X) − F ′(N0)e

−λd = 0, (10.23)

from which it follows that

F ′(N0) =λeλd

1− e−λX. (10.24)

The roots λ of this equation determine the stability of the linearized solution. If all the roots have negativereal part, then the solution is stable, whereas if there are roots with positive real part, the steady solutionis unstable. Notice that, since F ′(N0) < 0, there are no positive real roots.

There is possibly one negative real root; all other roots are complex. It follows that even if the steady solutionis stable, the return to steady state is oscillatory rather than monotone. Thus, following rapid disruptions ofblood cell population, such as traumatic blood loss or transfusion, or a vacation at a high-altitude ski resort,the blood cell population oscillates about its steady state.

The only possible way to have a root with positive real part is if it is complex. Furthermore, a transitionfrom stable to unstable can occur only if a complex root changes the sign of its real part, leading to a Hopfbifurcation. If a Hopf bifurcation occurs, it does so with λ = iω. We substitute λ = iω into (??) and separatethis into its real and imaginary parts to obtain

F ′(N0)(cos (ωd)− cos(ω(d+X))) = 0, (10.25)

F ′(N0)(sin (ωd)− sin(ω(d+X))) = −ω. (10.26)

There are two ways to solve (10.25). Because cosine is symmetric about any multiple of π, we can takenπ − ωd = nπ + ω(d +X), or ω(2d +X) = 2nπ, for any positive integer n. Because cosine is 2π periodic,we could also take ωX = 2nπ, however, since sine is also 2π periodic, this fails to give a solution of (10.26).With ω(2d+X) = 2nπ, (10.26) becomes

2dF ′(N0) sin(ωd) = −ωd, (10.27)

or

2dF ′(N0) = − 2nπ

2 + Xd

1

sin(

2nπ2+X

d

) . (10.28)

Finally, we use that F (N0) = N0/X to write

N0F′(N0)

F (N0)= −1

2

X

d

2nπ

2 + Xd

1

sin( 2nπ2+X

d

). (10.29)

For each integer n, this equation defines a relationship between N0 and X/d at which there is a change ofstability and thus, a Hopf bifurcation. If we take F to be of the special form

F (x) =A

1 + xp, (10.30)

we find thatN0F

′(N0)

F (N0)= − pNp

0

1 +Np0

, (10.31)

and that

dA =d

XN0(1 +Np

0 ). (10.32)

These two relationships allow us to solve (10.29) for N0 as a function of Xd

and then find dA(= dF (0)) asa function of X/d. Shown in Fig. 10.2 is this curve for n = 1 and p = 7. The case n = 1 is the only curve


5

4

3

2

1

0

d F

(0)

14121086420

X/d

Unstable

Stable

Figure 10.2: Critical stability curve (Hopf bifurcation curve) for cell growth.

of interest, since it is the first instability. That is, the steady state solution is unstable for all the criticalstability curves with n > 1, and so these curves do not lead to physically relevant bifurcations.

The implications of this calculation are interesting. If the nondimensional parameters X/d and dF (0) aresuch that they lie above the curve in Fig. 10.2, then the steady solution is unstable, and a periodic oroscillatory solution is likely (but since we do not know the direction of bifurcation, this is not guaranteed).On the other hand, if these parameters lie below or to the far right of this curve, the steady solution is stable.

From this we learn that there are three mechanisms by which cell production can be destabilized, and theseare by changing the maximal production rate F (0), the expected lifetime X , or the production delay d. IfX/d is sufficiently large (>≈ 14 for these parameter values), the system cannot be destabilized. However,if X/d is small enough, increasing F (0) is destabilizing. Increasing d is also destabilizing. If F (0) and Xare held fixed, then changing d moves y = dF (0) and x = X/d along the hyperbola yx = constant. Thus,decreasing d is stabilizing, as it increases X/d, moving it out of and away from the unstable region.

For normal humans, with d = 5 days and X = 120 days, there is no instability, since X/d = 24. However,any mechanism that substantially shortens X can have a destabilizing effect and can result in oscillatoryproduction of blood cells. Near the bifurcation, the period of oscillation is T = 2π

ω, where ω(2d+X) = 2π,

so that

T = 2d+X. (10.33)

Thus, for example, a disorder that halves the normal lifetime of blood cells to X = 60 days should result inoscillatory blood cell production with a period on the order of 70 days.

10.4 Simulation

It is always nice when problems have solutions that can be found with analytical techniques, but with realapplications this is rarely the case. This raises the question of how to best simulate advection equations.There are two methods that are recommended.

10.4. SIMULATION 87

time0 10 20 30 40 50 60 70 80 90 100

u(t)

0

1

2

3

4

5

6

time0 10 20 30 40 50 60 70 80 90 100

u(t)

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

10.4.1 Delay Differential Equations

Because they are closely related to advection equations, we begin with the example of the delay differentialequation

dN

dt= −βN + F (N(t− d)). (10.34)

This is the equation one finds for a population of cells that has a death rate β, and for which the highestpossible age X → ∞.

The key to simulating this equation is to realize that one must keep track of the function N(t − x) on theinterval 0 < x < d. Obviously, u = N(t− x) satisfies the partial differential equation

∂u

∂t+∂u

∂x= 0. (10.35)

So, we discretize the function N(t − x) for discrete values of x = j∆x, j = 0, 1, · · · , J , setting Nj(t) =N(t− j∆x). If J is such that d = J∆x, then the equation (10.34) can be written as

dN0

dt= −βN0 + F (NJ). (10.36)

and for all other values of j, j = 1, · · · , J ,

dNj

dt=

1

∆x(Nj−1 −Nj), (10.37)

approximately. The system (10.36-10.37) constitutes a Method of lines approximation for the delay differen-tial equation, and can readily be simulated using one’s favorite differential equation solution technique.

In Fig. 10.4.1 are shown the simulation results for the equation (10.34) starting from zero initial population,with β = 1

25 (left), and β = 115 (right). The Matlab code for this simulation is shown in Appendix A.7.

10.4.2 The Method of Characteristics

In the previous sections we used a very important technique to reduce the partial differential equation to asystem of ordinary differential equation, a method called the method of characteristics. In this section wedescribe this method in more detail and show some more example of it usefulness.

The general problem can be stated as follows: we seek a solution of the partial differential equation

∂u

∂t= f(u, x, t)

∂u

∂x+ g(u, x, t). (10.38)


The idea, as stated above, it to look for curves in x− t space along which the equation simplifies. So we lookfor a curve x = X(t), with yet to be determined functions X(t). Of course, along any such curve,

du

dt=∂u

∂tt+

∂u

∂x

dX

dt. (10.39)

Now, notice that if we pickdT

dt= 1,

dX

dt= −f(u, x, t), (10.40)

the partial differential equation reduces to the ordinary differential equation

du

dt= g(u, x, t). (10.41)

Thus, we have reduced the partial differential equation to the two ordinary differential equations

du

dt= g(u, x, t), (10.42)

dx

dt= −f(u, x, t). (10.43)

This formulation lends itself very nicely to numerical simulation, since one can easily solve the system ofdifferential equations numerically for a discretized grid of x values, and be assured that this method isexact (i.e., equations (10.42-10.43) were derived with no approximations). The method has disadvantages,however, that need to be understood. First, unless the characteristic lines are all parallel (i.e., f(u, x, t) isindependent of both u and x), then what might start out as a uniform grid of x-values will compress orexpand to be non-uniform, so that the representation of the function u(x, t) as a function of x at a fixedtime t could become highly non-uniform. What most people do in this situation is ”regrid”, that is, fromtime to time, reinitialize the simulation using a uniform x- discretization, using a smoothed interpolatedrepresentation of the solution.

The second disadvantage to this method is that it cannot be adapted to more difficult equations, such asequations with both advection and diffusion. This is a disadvantage that cannot be overcome, so it encouragesus to look for another method.

10.4.3 Method of Lines; Upwinding

In the section on simulating delay differential equations, it was necessary to simulate the partial differentialequation

∂u

∂t+∂u

∂x= 0, (10.44)

and the method proposed was to discretize u(x, t) setting uj(t) = U(j∆x, t) and then replacing the equation(10.47) with the discretized equation

dujdt

=1

∆x(uj−1 − uj), (10.45)

replacing the derivative ∂u∂x

with its backward finite difference. Notice, that we could just as easily used theforward finite difference to get

dujdt

=1

∆x(uj − uj+1), (10.46)

but we didn’t! Why?

This question is partially answered by a simple observation, and that is that at first glance (10.45) appearsto have exponentially decaying solutions, while (10.46) has exponentially growing solutions. This follows

10.4. SIMULATION 89

simply from the fact that the equation dudt

= −u+ f has exponentially bounded solutions, while dudt

= u+ fexhibits exponentially growing behavior.

A more precise explanation is found by examining the numerical solutions of the advection equation

∂u

∂t+v∂u

∂x= 0, (10.47)

where the velocity v is some constant. The method of lines with backward differences for this gives theequations

dujdt

=v

∆x(uj−1 − uj). (10.48)

If we now choose to simulate this using forward Euler stepping in time, we get

un+1j = unj +

v∆t

∆x(unj−1 − unj ), (10.49)

or in matrix form

un+1 = un +v∆t

∆xAun, (10.50)

where the matrix A has the form

A =

−1 0 0 · · · 01 −1 0 0 · · ·

...0 · · · 0 1 −1

. (10.51)

Because A is a lower diagonal matrix, all of its eigenvalues are -1, so that the eigenvalues of the matrixI + v∆t

∆xA are all 1− v∆t

∆x. Hence, the numerical algorithm is stable only if 0 < v∆t

∆x< 2.

This has important consequences. First, if v < 0 this scheme is always unstable which means that if v < 0one must use forward, rather than backward, differences. This choice is called upwinding referring to thefact that the difference is taken in the upwind direction.

The second consequence is that ∆t must not be too large. This is referred to as a CFL condition (i.e., theCourant-Friedrichs-Levy condition).

The idea of upwinding requires a little extra thought when the velocity is not constant. To be specific,suppose we have the equation, written in conservation form,

∂u

∂t+

∂

∂x

(

v(x)u)

= 0. (10.52)

We define Fj to be the discretized flux at position xj as

Fj = vj− 12

(

(vj− 12> 0)uj−1 + (vj− 1

2< 0)uj

)

, (10.53)

and then the method of lines approximation is

∂uj∂t

= − 1

∆x(Fj+1 − Fj), (10.54)

where vj− 12= v(xj− ∆x

2 ), and (v > 0) represents the logical variable which is one if v > 0 and zero otherwise.

Notice that this discretization correctly does ”upwinding” and is also conserving, since ddt

∑

j uj = 0.


10.5 Exercises

1. The death rate for a population of organisms is a constant µ while the birth-rate is β(a) = B exp(λa),with λ < 0. For what values of parameters µ, B, and λ is the population sustainable?

2. Suppose that the probability that an organism is alive at age a is

p(a) =1

1 + a2. (10.55)

(a) Write an equation relating p(a) to the death rate µ(a);

(b) Find µ(a);

(c) Suppose the birth rate is the constant β. Find the smallest value of β for which the population issustainable;

(d) What is the age distribution for this minimally sustainable population?

3. Consider a population with age and density dependent birth rate

β(a,N) = Ba(1 − N

K), (10.56)

where N is the total population size. Suppose the death rate is constant µ. Find the equilibriumpopulation size.

4. Simulate the differential equation∂u

∂t+

∂

∂x(u2) = 0 (10.57)

on the interval −10 < x < 10, starting from initial data u(x, 0) = exp(−x2), using two numericalmethods:

(a) the method of characterstics;

(b) the method of lines with upwinding.

Describe any differences you find between these two solutions.

Chapter 11

Advection with diffusion

Now let’s consider the case where the particle is both advecting and diffusing. There are two examples weconsider.

11.1 Axonal transport

Consider the situation in which a molecule can attach to or detach from a microtubule. When it is attachedit moves along the microtubule path with a fixed velocity v, and when it is detached it simply diffuses. Thisis the situation for many molecular motors, including kinesis and dynein. The chemical reaction for this isdenoted

Uα

−→←−β

B, (11.1)

where U denotes the unbound state and B denotes the bound state. To study this problem we must keeptrack of the particle’s position in space and its state, bound or unbound. We let pu(x, t) and pb(x, t) be theprobability that the particle is at position x at time t in the unbound or bound state, respectively. Then,

∂pu∂t

= −αpu + βpb +D∂2pu∂x2

, (11.2)

∂pb∂t

= αpu − βpb − v∂pb∂x

. (11.3)

The first equation has terms representing the fact that the unbound particle diffuses and can become boundand the bound particle can become unbound. Similarly, the terms in the second equation reflect the factthat the bound particle can advect with velocity v or become unbound, while the unbound particle maybind.

Now we make an approximation, namely that the binding/unbinding reaction is in equilibrium so that

pu =β

α+ βp, pb =

α

α+ βp. (11.4)

Then, adding the above two equations we find

∂p

∂t= −v α

α+ β

∂p

∂x+D

β

α+ β

∂2p

∂x2. (11.5)

91

92 CHAPTER 11. ADVECTION WITH DIFFUSION

This is an equation, called an advection-diffusion equation, of the form

∂p

∂t= −ve

∂p

∂x+De

∂2p

∂x2. (11.6)

where ve, the effective advection velocity is ve =vαα+β

, and De, the effective diffusion constant, is De =Dβα+β

.

11.2 Transport with Switching

Suppose there is some object that moves in one of two directions along a one-dimensional path (again,one could imagine a molecular motor complex moving along a microtubule), and it randomly switches itsdirection of motion via the process

Lα

−→←−β

R, (11.7)

where L and R refer to the left-moving, and right-moving states, respectively. This is similar to the casedescribe above, except that now the object is moving with a fixed, but different, velocity in each state.

Setting pL and pR to be the probability of being at position x at time t in the left or right-moving state,respectively, we find that

∂pL∂t

= −αpL + βpR + vL∂pL∂x

, (11.8)

∂pR∂t

= αpL − βpR − vR∂pL∂x

. (11.9)

(11.10)

To simplify these equations we could assume, as we did above, that the states are distributed in equilibrium,so that

pL =β

α+ βp, pR =

α

α+ βp. (11.11)

Then, after adding the two equations together we find

∂p

∂t=vLβ − vRα

α+ β

∂p

∂x. (11.12)

This approximation is useful, because it shows that there is an average, or effective, rate of drift, but it failsto include any effects of the randomness of switching.

To include the effects of randomness requires a slightly more sophisticated analysis. To do this we expressthe rates α and β as

α =α

ǫ, β =

β

ǫ, (11.13)

where α+ β = 1, and ǫ is assumed to be a small positive number.

Now, we introduce the exact change of variables(

pLpR

)

=

(

βα

)

p+

(

1−1

)

w, (11.14)

and find the governing equations in these new variables to be

(

βα

)

∂p

∂t+

(

1−1

)

∂w

∂t=

1

ǫ

(

−11

)

w +

(

vLβ∂p∂x

+ vL∂w∂x

−vRα ∂p∂x

+ vR∂w∂x

)

. (11.15)

11.3. ORNSTEIN-UHLENBECK PROCESS 93

Adding the two equations we find

∂p

∂t= (vLβ − vRα)

∂p

∂x+ (vL + vR)

∂w

∂x, (11.16)

which, because it contains terms involving w, is slightly different than (11.12). Now, find an equation for

∂w∂t

by taking the dot product of (11.15) with the vector

(

α

−β

)

, to find

∂w

∂t= −1

ǫw + (vL + vR)βα

∂p

∂x+ (vLα− vRβ)

∂w

∂x. (11.17)

Now, notice that this last equation is of the form

∂w

∂t= −1

ǫw + f(x, t) (11.18)

which, since ǫ is small, implies that w equilibrates rapidly to something of order ǫ in size. In other words, wis approximately

w = ǫ(vL + vR)βα∂p

∂x. (11.19)

Substituting this into (11.21) gives

∂p

∂t= (vLβ − vRα)

∂p

∂x+ ǫ(vL + vR)

2βα∂2p

∂x2, (11.20)

or, expressed in terms of the original turning rates,

∂p

∂t=vLβ − vRα

α+ β

∂p

∂x+ αβ

(vL + vR)2

(α+ β)3∂2p

∂x2, (11.21)

This equation has an interesting interpretation. It tells us that for this process, there is an effective drift

rate ve = vLβ−vRαα+β

, and also an effective diffusion constant De = αβ (vL+vR)2

(α+β)3 . Consequently, in spite of the

fact that the movement of the particles is by deterministic processes, the fact that the switching is by astochastic process leads to diffusion away from the deterministic answer.

There are several ways that we might proceed to find solutions of the advection-diffusion equations. Thefirst is to make a change of variables into a uniformly moving coordinate system, by setting

y = x− vet, s = t. (11.22)

In these new variables, the equation (11.23) is

∂p

∂s= De

∂2p

∂y2, (11.23)

and, of course, this is the diffusion equation. It follows that if U(t, s) is a solution of the diffusion equationequation, then

u(x, t) = U(x− vet, t) (11.24)

is a solution of the advection-diffusion equation (11.23).

11.3 Ornstein-Uhlenbeck Process

Now consider a particle that diffuses, but is also constrained by a restoring force, like a spring, that pulls itback to its zero position. To find the equation of motion for this particle, we first consider the differential


equation for the particle if it were not diffusing. In this case, we have a simple mass-spring system for whichthe governing equation is

mdv

dt+ ξv + kx = 0, (11.25)

where v = dxdt, and x is the particle’s position. This is a statement of Newton’s second law (F = ma), where

ξv is the force due to viscosity. Now we make the approximation that this is a viscosity dominated process(inertia is negligible compared to viscosity) so that

ξv = −kx. (11.26)

Now we can write an equation for the probability that the particle is at position x at time t by includingboth the advection (velocity) term and the diffusive term

∂p

∂t=

∂

∂x

(

kx

ξp

)

+D∂2p

∂x2. (11.27)

The process described here is called an Ornstein-Uhlenbeck process.

There are several things to notice about this process. First, the total integral of p(x, t) is constant (andequal to 1). Second, if there were no diffusion, the position x would satisfy the differential equation

dx

dt= −k

ξx, (11.28)

which has a solution x(t) = x0 exp(−kξt).

This suggests that we should look for a solution of the form

p(x, t) =1

√

b(t)πexp

(

− (x− c(t))2

b(t)

)

. (11.29)

Substituting this into the governing equation, we find that we must require

(b− 2c2)(db

dt− 4D) + 2b2

k

ξ+ 4bc

dc

dt+ 4(bc

k

ξ− b

dc

dt+ c(

db

dt− 4D))x− 2(

db

dt− 4D + 2b

k

ξ)x2 = 0. (11.30)

Since this is a quadratic equation in x, each of the three coefficients of the powers of x must individually bezero, leading to the differential equations

db

dt= 4D − 2

k

ξb, (11.31)

dc

dt= −k

ξc. (11.32)

The solutions of these are also relatively easy to obtain, being

b(t) = 2Dξ

k

(

1− exp(−2k

ξt)

)

, (11.33)

c(t) = x0 exp(−k

ξt). (11.34)

The equation for c(t) is exactly as we expected from the deterministic equation of motion.

The Matlab code for this simulation is shown in Appendix A.8.

11.3. ORNSTEIN-UHLENBECK PROCESS 95

-10 -8 -6 -4 -2 0 2 4 6 8 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 11.1: Simulation of the Ornstein Uhlenbeck equation with parameters k = 0.1, D = 0.1.


Chapter 12

Chemotaxis

Intro - the life cycle of the amoebae Dictyostelium discoideum (dd).

It is part of the life cycle of dd to respond to starvation by forming aggregates that then form fruiting bodies,which release spores into the wind to colonize new populations. A simple model of this aggregation processis the system of equations

∂a

∂t= − ∂

∂x

(

aχ∂c

∂x− µ

∂a

∂x

)

, (12.1)

∂c

∂t= ρa− kc+D

∂2c

∂x2, (12.2)

where a and c represent the concentrations of amoebae and cAMP, respectively. The interpretation of thismodel is that cAMP is produced by amoebae and naturally degrades, and that amoebae diffuse but alsomove up gradients of cAMP. For this discussion, we assume that the domain has length L and that thereare no-flux boundary conditions for both c and a. Consequently, the total amount of a

∫ L

0

a(x, t)dx = a∗L (12.3)

is a conserved quantity.

First, notice that the steady state solution a∗ and c∗ satisfy fa∗ − kc∗ = 0. Now, we linearize the governingequations about the steady solution, finding

∂δa

∂t= − ∂

∂x

(

a∗χ∂δc

∂x− µ

∂δa

∂x

)

, (12.4)

∂δc

∂t= ρδa− kδc+D

∂2δc

∂x2. (12.5)

Now we look for solutions of the linearized equations of the form

δa = A exp(λt+ iωx), δc = C exp(λt+ iωx), (12.6)

finding the linear system of equations

λ

(

AC

)

=

(

−µω2 a∗χω2

ρ −k − ω2D

)(

AC

)

. (12.7)

97

98 CHAPTER 12. CHEMOTAXIS

x0 1 2 3 4 5 6 7 8 9 10

a

0

5

10

15

Figure 12.1: Numerical solution for a(x, t) for several values of t, with parameter values L = 10, D = 0.25,µ = 0.07, ρ = 0.2, k = 0.3, χ = 0.2.

The characteristic equation is

λ2 − Trλ+ det = 0, (12.8)

where Tr = −k − ω2D − µω2, and det = µω2(ω2D + k) − ω2ρχa∗. Since Tr < 0, the only way that thesolution can be unstable is if det < 0, so neutral stability occurs wherever det = 0. It follows that the steadysolution is unstable if

µω2D < ρχa∗ − µk. (12.9)

Obviously, if ρχa∗ − µk < 0, the homogeneous solution is stable. However, if ρχa∗ − µk > 0, then thehomogeneous solution is unstable if the domain size L satisfies

L2 >π2

ρχa∗ − µk. (12.10)

This follows since, for a domain with no-flux boundary conditions, ωL = π.

There are a few observations to make here. First, if χ < 0, the homogeneous solution is always stable.However, if χ > 0 and if a∗ is large enough, the homogeneous solution is unstable, suggesting that anaggregated pattern forms if the domain is sufficiently large.

Now, we look for steady, inhomogeneous solutions. That is, set time derivatives to zero, to learn that

aχ∂c

∂x− µ

∂a

∂x= 0, (12.11)

so thatχ

µ(c−K) = ln

a

a∗, (12.12)

where the constant K is yet to be determined. Solving for a, we find the single equation for c,

ρa∗ exp(χ

µ(c−K)

)

− kc+D∂2c

∂x2= 0. (12.13)

Now we also require that the homogeneous steady state satisfy ρa∗ − kc∗ = 0. Consequently, K = ρa∗

kand

ρa∗ exp(χ

µ(c− ρa∗

k))

− kc+D∂2c

∂x2= 0. (12.14)

99

Before proceeding any further, it is useful to rescale the problem. We set y = x√

kD

and u = kcρa∗

in terms

of which the governing equation becomes

d2u

dy2+ f(u) = 0, (12.15)

where f(u) = exp(κ(u − 1))− u.

What to do next should be clear. (Answer: phase plane analysis.)

Notice that f(0) > 0, and f ′(0) = κ exp(−κ) − 1 < 0 for all positive κ. Furthermore, f ′(u) > 0 for usufficiently large, and f(1) = 0. Hence, f(u) = 0 has two positive roots. Since f ′(1) = κ − 1, the root atu = 1 is the smaller of the two roots if κ < 1 and is the larger of the two roots if κ > 1.

u0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

du/d

y

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

length l0 5 10 15

um

in,u

max

0

0.5

1

1.5

Figure 12.2: (Left)Phase portrait for solutions of the equation (12.15), and (Right) plot of the maximum and

minimum of the solution plotted as a function of the non-dimensional length l = L√

kD

in the case κ = 2.0.

y0 1 2 3 4 5 6 7 8

u, a

/a*

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Figure 12.3: Typical solution profile for u and aa∗

(dashed).

The phase portrait analysis for this equation is straightforward (and has been done before several times).The larger of the two critical points is a center, and the smaller of the two critical points is a saddle point.Consequently, there is a family of closed orbits surrounding the center, expanding out to the closed saddle–saddle homoclinic trajectory.

100 CHAPTER 12. CHEMOTAXIS

The solutions of interest here are when u = 1 is the larger of the two roots, which is when κ > 1. In thatcase, the family of closed orbits exists and satisfies no-flux boundary conditions provided l > π√

κ−1 . The

phase portrait and bifurcation diagram depicting this result are shown in Fig. 12.2, and an example of asolution profile is shown in Fig. 12.3.

Chapter 13

Pattern Formation - The TuringMechanism

13.0.1 The Turing Instability

There are many chemical and biological systems that might be modeled as reaction-diffusion systems of theform

∂u

∂t= Du∇2u+ f(u, v), (13.1)

∂v

∂t= Dv∇2v + g(u, v), (13.2)

where the reaction kinetics have a stable steady state solution u = u0, v = v0. Here we want to determine ifthis spatially homogeneous stable steady state can be unstable when spatial variations are allowed.

Before we consider this problem, let’s remind ourselves about what happens with only one species. In thiscase we have the diffusion reaction equation

∂u

∂t= Du∇2u+ f(u), (13.3)

where f(u0) = 0. We linearize this equation around u = u0 to find

∂U

∂t= Du∇2U + f ′(u0)U. (13.4)

We know from Section 8.1 that the solution of this problem is given by

U(x, t) = exp(f ′(u0)t)v(x, t), (13.5)

where v(x, t) satisfies the diffusion equation. Obviously, if f ′(u0) < 0, this solution is linearly stable. Inother words, the spatially inhomogeneous solution has the same stability characteristic as the homogeneous(space-independent) solution.

Now, we perform a linear stability analysis for the two-variable system (13.1-13.2). That is, we let u = u0+ǫU ,v = v0 + ǫV , substitute into the governing equations, differentiate with respect to ǫ and then set ǫ = 0. The

101

102 CHAPTER 13. PATTERN FORMATION - THE TURING MECHANISM

consequence of this procedure is the linear system of equations

∂U

∂t= Du∇2U +

∂f0

∂uU +

∂f0

∂vV, (13.6)

∂v

∂t= Dv∇2V +

∂g0

∂uU +

∂g0

∂vV, (13.7)

where 0 refers to evaluation at the steady state solution u = u0, v = v0.

The assumption that this steady state solution is a stable solution of the ordinary differential equation system

implies that Tr ≡ ∂f0

∂u+ ∂g0

∂v< 0 and Det ≡ ∂f0

∂u∂g0

∂v− ∂f0

∂v∂g0

∂u> 0.

The stability is determined by setting(

UV

)

=

(

αβ

)

exp(λt) cos(ωx) (13.8)

and consequently require(

∂f0

∂u− ω2Du

∂f0

∂v∂g0

∂u∂g0

∂v− ω2Dv

)

(

αβ

)

= λ

(

αβ

)

). (13.9)

This implies that λ be a solution of the quadratic equation

λ2 − (Tr − ω2Du − ω2Dv)λ+Det− ω2Du

∂g0

∂v− ω2Dv

∂f0

∂u+ ω4DuDv = 0. (13.10)

Stability is determined by the sign of the real part of the roots of this polynomial, and since

Tr − ω2Du − ω2Dv < 0, (13.11)

the only way that there can be a change of stability is if

Dω ≡ Det− ω2Du

∂g0

∂v− ω2Dv

∂f0

∂u+ ω4DuDv = 0. (13.12)

is au is the state vector and D is a diagonal matrix. If u represents chemical concentrations, then thisequation applies on a bounded domain Ω subject to homogeneous Neumann boundary conditions, (becausethere is no flux across the boundary of the domain Ω).

Suppose there is some non-trivial value u = u0 at which f(u0) = 0. Then there is a steady state solutionthat is uniform in space, and it is of interest to determine the stability of this steady solution. If we linearize(13.1) about the steady state u0, we find the linear system of equations

∂U

∂t= D∇2U +AU, (13.13)

where A is the Jacobian matrix of f(u) at u = u0.

Suppose that the state space is two-dimensional and that the steady state solution is stable in the absenceof diffusion. This means that the trace of A is negative and the determinant of A is positive. An interestingquestion to ask is if it is possible for diffusion to destabilize the steady solution.

To examine this question we take the specific case in which a11 < 0, a22 > 0, a12 > 0 and a21 < 0. In thiscase the variables u are sometimes identified as inhibitor and activator. Notice that u2 activates its owngrowth and the growth of u1 while u1 inhibits the growth of u2.

13.1. CELL POLARITY 103

The linear stability analysis for (13.13) is straightforward. We look for solutions of the form

U(t) = vφneλt (13.14)

where φn is an eigenfunction of the Laplacian on the domain Ω, ∇2φn = −λnφn. Then (13.13) becomes aneigenvalue problem for λ

Anv ≡ (−λnD +A)v = λv. (13.15)

Clearly the matrix An has negative trace. Thus, the sign of the real part of the eigenvalues are determinedby the sign of the determinant of An; if det(An) < 0, one eigenvalue is positive and the solution is unstable,while if det(An) > 0 both eigenvalues are negative and the solution is stable. We calculate

det(An) = (−D1λn + a11)(−D2λn + a22)− a21a12. (13.16)

Several curves on which det(An) = 0 (for several values of n) are depicted in Fig. 13.1, plotted in the D1,D2 parameter plane. Above the nth curve the nth eigenmode is stable while below it the mode is unstable.Notice that any specific mode can be made unstable if the diffusion coefficient D1 is sufficiently large and D2

is sufficiently small. This instability is called a “diffusion-driven instability”, or simply a diffusive instability.It is also called the Turing instability, in recognition of its discovery by A. Turing in 1952 [?].

Figure 13.1: Neutral stability curves for the diffusion reaction system (13.13).

13.1 Cell Polarity

To illustrate a biological situation in which a Turing-like mechanism is involved, we described a simple modelof cell polarization. Cell polarization is the process by which a cell decides where its front should be.

13.2 Exercises

1. Consider the system of equations

∂u

∂t= αf(u, v) +Du∇2u (13.17)

∂v

∂t= αg(u, v) +Dv∇2v (13.18)

with no-flux boundary conditions on the one dimensional interval 0x < π, Du = 1, Dv = 10. Supposethe Jacobian matrix of the linearization of the system about its spatially uniform steady state solutionis given by

J =

(

2 −44 −6

)

. (13.19)

(a) Derive conditions for instability of the nth mode;

(b) For values of α, 0 < α < 10, find all the bifurcation values of α, i.e. the values where one of themodes gains or loses stability.

104 CHAPTER 13. PATTERN FORMATION - THE TURING MECHANISM

Chapter 14

Agent-Based Modeling

105

106 CHAPTER 14. AGENT-BASED MODELING

Chapter 15

Appendices

15.1 Matlab Codes

15.1.1 A Matlab Primer

A primary emphasis of this text is solving equations numerically, and while there are several choices for thelanguage in which to do numerical simulations, the choice for this text is Matlab. Consequently, all thesimulations and plots shown in this text were done using Matlab, and there are a number of Matlab codesthat are included in the following Appendices. The purpose of this section is to give a brief introduction tothose features of Matlab that are used most heavily in this text, with the hope that this will enable you, thereader, to at least understand what these codes do and how they might be modified to do slightly differenttasks.

15.1.2 A.1: Method of Lines

This is matlab code to simulate the system of ode’s (4.4)

% code to simulate diffusion process via Method of lines (MOL)

function Diff_sim

global alpha sc

alpha = 0.1; %The time constant

N = 52; %number of cells

sc = [1;2*ones(N-2,1);1];

u0 = zeros(N,1);

u0(N/2) = 1;

tstep = 1;

t_end = 50;

%specify the output points

107

108 CHAPTER 15. APPENDICES

tspan = [0:tstep:t_end];

[T,S] = ode23(@deRHS,tspan, u0, odeset(’maxstep’,1));

figure(1)

for j = 1:length(T)

plot(S(j,:),’*’)

axis([0 N 0 1])

pause(.1)

end

figure(2)

plot(T,S(:,N/2),T,S(:,fix(N/4)))

xlabel(’time’,’fontsize’,16)

ylabel(’U’,’fontsize’,16)

%the right hand side for ode simulation:

function s_prime=deRHS(t,s)

global alpha sc

Fu = alpha*(-sc.*s+[0;s(1:end-1)]+[s(2:end);0]);

s_prime = Fu;

15.1.3 A.2: Discrete Random Walk

Matlab code to simulate a random walk.

% diffusion on a grid via a random walk

% create a vector of numbers

N = 1000; % number of objects

X = zeros(N,1); % initial position

lambda = 0.1; % Probability of movement

M = 100; % number of time steps

XS = zeros(N,M);

c = [lambda,2*lambda,1];

% Specify the movement vector

xm(1) = 1;

xm(2) = -1;

xm(3) = 0;

for j = 1:M

R = rand(N,1);

for k = 1:N

15.1. MATLAB CODES 109

ndx= min(find(R(k)<=c));

X(k) = X(k) + xm(ndx);

XS(:,j) = X;

end

figure(1)

hist(X,[-20:20])

axis([-25 25 0 300])

pause(0.1)

end

figure(2)

plot(XS’)

15.1.4 A.3: Motion of a Brownian Particle

% This is to illustrate diffusion of particles (i.e., Brownian motion)

clear

N = 1000; %number of particles

K = 100; %number of time steps

y = [-100:100]/5;

D = 1/2;

binc = [-20:20];

ksk = 5;

x = zeros(K,1);

X = x;

dt = .1;

t = 0;

var(1) = 0;

T(1) = 0;

k = 1;

diff = sqrt(dt);

est1 = 0;

est2 = 0;

for j = 2:N

x = x + diff*randn(K,1);

X = [X,x];

var(j) = std(x)^2;

t = t +dt;

est1 = est1 + t^2;

est2 = est2 + t*var(j);

T(j) = t;


if (fix(j/ksk)\frack\xisk==j)

figure(1)

hist(x,binc)

hold on

p = K*exp(-y.^2/(4*D*t))/sqrt(4*pi*D*t);

plot(y,p,’r’,’linewidth’,2)

hold off

axis([-20 20 0 45])

pause(0.01)

end

end

figure(3)

sl = est2/est1;

plot(T,var,T,sl*T,’--’ )

hold off

15.1.5 A.4: Gillespie algorithm for Particle Decay

% exponential decay of particles via Gillespie algorithm

alpha = 1; % rate of decay

N = 100; % initial number of particles

K = 20000 %total number of trials

t = zeros(N+1,K);

nn = reshape([ N:-1:0; N:-1:0],2*(N+1),1);

R = rand(N,K);

for j = 1:N

n = N-j+1;

dt = -log(R(j,:))/(alpha*n);

t(j+1,:) = t(j,:) + dt;

end

s = 1;

tt = reshape([t(:,s)’;t(:,s)’],2*(N+1),1);

figure(1)

plot(tt(2:end),nn(1:end-1),tt,N*exp(-tt),’--’)

%figure(2)

% semilogy(tt,nn)


figure(3)

hist(t(N,:),20)

mean(t(N,:))

E = sum(1./[1:N])

15.1.6 A.5: Diffusion Equation solution via Crank Nicolson

% This file solves the diffusion equation

clear

L = 1000; % length of the domain

d = .01; % diffusion coefficient

N=50; % number of spatial grid points

M = 130;

h = L/N;

dt = h/2;

K = 0.01;

vel = 0.1; % set this to zero to have no transport

vel = 0;

sc = 100;

scal = d*dt/h^2;

X = h*(1:N)’;

V = exp(-(X-h*N/2).^2./(2*h)); % initial gaussian

kstep = 10;

t=0;

% uses Crank Nicolson to solve the diffusion equation

% set up matrix

Atm=(1+scal)*ones(N,1);

Atm(1,1) = 1+scal/2;

Atm(N,1) = 1+scal/2;

Am=diag(Atm)-diag(scal/2*ones(N-1,1),1)-diag(scal/2*ones(N-1,1),-1);

% The algorithm

for n=2:M


for k = 1:kstep

F = vel*(V - [V(1 );V(1:N-1 )]);

V_t = scal/2 * ([V(2:N );V(N )] - 2*V + [V(1 );V(1:N-1 )])-dt*F/h ;

%V_t = scal/2 * ([V(2:N );0] - 2*V + [0;V(1:N-1 )]) ; Dirichlet

%conditions

V = Am\(V+V_t);

t=t + dt

end

figure(1)

plot(X,V )

axis([0 L 0 1])

% axis([0 L -.5 1])

pause(0.05)

end

TL = t;

time=cputime-time % outputs the cpu time taken to solve the equations

time_steps = M;

15.1.7 A.6: Diffusion Equation with growth/decay solution via Crank Nicolson

% This file solves the diffusion equation

clear


d = .01; % diffusion coefficient



h = L/N;

dt = h/2;

K = -1; %pick K <0 for decay, K > 0 for growth

if (K>0) then

kp = K;

km = 0;

else

kp = 0;

km = K;

end

sc = 100;


scal = d*dt/h^2;

X = h*(1:N)’;

V = zeros(N,1);

V(N/2) = 1; %initialize V

kstep = 10;

t=0;


% uses explicit Euler for decay, implicit euler for growth

% set up matrix

Atm=(1+scal)*ones(N,1)-kp;

Atm(1,1) = 1+scal/2-kp;

Atm(N,1) = 1+scal/2-kp;


% The algorithm

for n=2:M

for k = 1:kstep

F = vel*(V - [V(1 );V(1:N-1 )]);

V_t = scal/2 * ([V(2:N );V(N )] - 2*V + [V(1 );V(1:N-1 )])+dt\frack\xim*V ;

%V_t = scal/2 * ([V(2:N );0] - 2*V + [0;V(1:N-1 )]) ; %Dirichlet conditions

V = Am\(V+V_t);

t=t + dt;

end

figure(1)

plot(X,V )

axis([0 L 0 1])

% axis([0 L -.5 1])

pause(0.05)

end

TL = t;


15.1.8 A.7: Solution of Delay differential equation using the Method of Lines(MOL)

% code to solve Mackey-Glass delay differential equation

% using method of lines

function delay_de

global N beta dx

d = 5; % time delay

beta = 1/25; % decay time constant

N = 61;

dx = d/(N-1);

x = zeros(N,1); %initial data

tstep = 0.1;

t_end = 100;

tspan = [0:tstep:t_end];

s = zeros(N,1);

[T,S] = ode23s(@deRHS,tspan,s);

plot(T,S(:,1),’linewidth’,2)

xlabel(’time’,’fontsize’,20)

ylabel(’u(t)’,’fontsize’,20)

function s_prime=deRHS(t,s)

global N beta dx

Fw = (s(1:N-1)-s(2:N))/dx;

Fs = -beta*s(1) + 1/(1+s(N)^7);

s_prime = [Fs; Fw];

15.1.9 A.8: Solution of the Ornstein-Uhlenbeck equation

% This file solves the Fokker-Planck equation for an OU process

% using Crank-Nicholson and upwinding

clear


d = 0.1; % diffusion coefficient

K = 0.1; % restoring force parameter



h = L/((N-1)/2); %grid size


K = 0.1; % restoring force parameter

X = h*((1:N)’-(N+1)/2);% grid points

Xm = (X(1:N-1)+X(2:N))/2; %midpoints

vel = -K*Xm;

vmx = max(abs(vel));

msk = (vel>0);

dt = 0.5*h/vmx; % set time step

scal = d*dt/h^2;

U = zeros(N,1); % specify initial data

U(N-1) = 5; %initial position

kstep = 10;

t=0;


% set up the CN matrix

Atm=(1+scal)*ones(N,1);

Atm(1,1) = 1+scal/2;

Atm(N,1) = 1+scal/2;


% The algorithm

for n=2:M

for k = 1:kstep

F = [0;vel.*(msk.*U(1:N-1)+(1-msk).*U(2:N));0];

Cdiff = (F(2:N+1)-F(1:N))/h;

U_t = scal/2 * ([U(2:N );U(N )] - 2*U + [U(1 );U(1:N-1 )])-dt*Cdiff ;

U = Am\(U+U_t);

t=t + dt

end

figure(1)

plot(X,U,’linewidth’,2 )

axis([-L L 0 1])

% axis([0 L -.5 1])

hold on


Table 15.1: Molecular weight and diffusion coefficients of some biochemical substances in dilute aqueoussolution.Substance Molecular Weight D(cm2/s)hydrogen 1 4.5× 10−5

oxygen 32 2.1× 10−5

carbon dioxide 48 1.92× 10−5

glucose 192 6.60× 10−6

insulin 5,734 2.10× 10−6

Cytochrome c 13,370 1.14× 10−6

Myoglobin 16,900 5.1× 10−7

Serum albumin 66,500 6.03× 10−7

hemoglobin 64,500 6.9× 10−7

Catalase 247,500 4.1× 10−7

Urease 482,700 3.46× 10−7

Fibrinogen 330,000 1.97× 10−7

Myosin 524,800 1.05× 10−7

Tobacco mosaic virus 40,590,000 5.3× 10−8

end

TL = t;

15.2 Constants and Parameters

15.2.1 Diffusion Coefficients

15.2. CONSTANTS AND PARAMETERS 117

15.2.2 Physical constants

Quantity Name Symbol UnitsAmount mole molElectric charge coulomb CMass gram gTemperature kelvin KTime second sLength meter mForce newton N kg ·m · s−2Energy joule J N ·mPressure pascal Pa N ·m−2Capacitance farad F A · s · V−1Resistance ohm Ω V ·A−1Electric current ampere A C · s−1Conductance siemen S A · V−1 = Ω−1

Potential difference volt V N ·m · C−1Concentration Molar M mol · L−1Atomic Mass Dalton D g N−1A

Physical Constant Symbol ValueBoltzmann’s constant k 1.381× 10−23 J ·K−1Planck’s constant h 6.626× 10−34J · sAvogadro’s number NA 6.02257× 1023 mol−1

unit charge q 1.6× 10−19 Cgravitational constant g 9.78049 m/s2

Faraday’s constant F 9.649× 104C ·mol−1

permittivity of free space ǫ0 8.854× 10−12 F/muniversal gas constant R 8.315 J mol−1 ·K−1atmosphere atm 1.01325× 105N ·m−2insulin unit U 1

24000g

Lumen: 1 lm = quantity of light emitted by 160 cm2 surface area of pure platinum at its melting temperature

(1770 C), within a solid angle of 1 steradian.

Liter: 1 L = 10−3m3.

ǫ: dielectric constant for water = 80.4ǫ0

Other Identities1 atm = 760mmHgR = kNA

F = qNA

pH = − log10[H+] with [H+] in moles per liter

273.15 K = 0C (ice point)TKelvin = Tcentigrade − 273.15TFarenheit =

95Tcentigrade + 32


RTF

= 25.8 mV at 27C

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