3307/g307 biomathematicsrmjbale/3307/g307_lectures.pdf · chapter 1. general information and...

University College London

3307/G307 Biomathematics

lecture materials

Alexey Zaikin

2010

Contents

1 General information and reading list. 1

1.1 Topics to be covered in lectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Reading list . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Using scaling arguments 5

2.1 First steps: Building a simple mathematical model . . . . . . . . . . . . . . . . . . . . 5

2.1.1 Example. Falling flea. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Scaling arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2.1 Some basic physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.3 Example: How fast can we walk before breaking into a run? . . . . . . . . . . . 9

2.2.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.5 Models that involve Metabolic Rate . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.6 Class Exercise (10mins): How long can a diving mammal stay under water on

one breath? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2.7 Example: Why do large birds find it harder to fly? . . . . . . . . . . . . . . . . . 12

2.2.8 How to obtain lift law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2.9 Kleibers Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2.10 Example. How does heart-rate scale with mass? . . . . . . . . . . . . . . . . . 15

2.2.11 Example: Thickness of fur . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2.12 Class excercise: How long does it take to starve to death? . . . . . . . . . . . 16

2.2.13 Example: Swimming speed of a filter-feeder . . . . . . . . . . . . . . . . . . . . 16

2.3 Example: Ludwig von Bertalanffys Growth Model (1957) . . . . . . . . . . . . . . . . . 17

2.3.1 Case Study: Incubating Eggs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Oxygen transport and Insect respiration 21

iii

4 Bird flight 49

4.1 Basics of bird flight. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.2 Basics of lift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.3 Energy is required to counter: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.3.1 Parasitic drag Dp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.3.2 Induced drag Di . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.4 Steady level flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.5 Stable gliding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.5.1 Linear stability analysis of a glide . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.5.2 Stable glide speeds and angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.5.3 Some crude estimates for stable glide speeds. . . . . . . . . . . . . . . . . . . 61

4.6 Soaring flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.7 Bounding flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5 Electrophysiology 65

5.1 Electrodiffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

5.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.1.2 Constant field approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.2 Poisson-Nernst-Planck equations (PNP) . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.2.2 Example: Ion flux through long aqueous channel . . . . . . . . . . . . . . . . . 69

5.2.3 Transmembrane currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.2.4 Example: Simple cell model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.3 Excitable cells - basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.4 The Cable Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.4.1 A simple model for action potential propagation . . . . . . . . . . . . . . . . . . 77

iv

Chapter 1

General information and reading list.

1.1 Topics to be covered in lectures

1. Using scaling arguments

2. Oxygen transport and Insect respiration

3. bird flight

4. Simple cell electrophysiology

5. Strength of bones

6. Chemotaxis

7. Brain/Memory

8. Blood flood

1.2 Reading list

Warning. This course is not (much) about learning methods and theorems and applying them tostandard problems. As such, there is no single book that you can read to cover the course. Thereare books that you might find helpful, or enjoy reading to supplement the lectures. A list of books thatgo with the course is the following:

1. Scaling Laws.

1

CHAPTER 1. GENERAL INFORMATION AND READING LIST.

(a) Andrew A Biewener. Animal Locomotion. Oxford Animal Biology Series. CUP, 2003.[Good general reading, but particularly pages 10-14. Chapter 7 has an interesting sectionon jump perfprmance].

(b) Knut Schmidt-Nielsen. Scaling: why is animal size so important? CUP, 1984. [Does notbuild models, but is good background information on scaling in biology].

(c) D’arcy Wentworth Thomson. On growth and form. CUP. First published 1961. [Again, nomodel building, but excellent background and a real classic].

(d) Ludwig von Bertalanffy. General Systems Theory. 1969. george Braziller Inc. New York.[He discusses his growth model in pages 171-184].

(e) http://online.itp.ucsb.edu/online/pattern i03/west/ [For general interest,and also von Bertalanffy’s model].

2. Diffusion/Insect Respiration.

(a) Ove Sten-Knudson. Biological Membranes: Theory of transport, potentials and electricimpulses. Cambridge University Press, 2002.

(b) G.R. Grimmet and D.R. Stirzaker. Probability and random processes. Clarendon Press,Oxford. 1992.

(c) http://www.livescience.com/animals/061011 giant insects.html. [Read-able article on oxygen and insect size limitations].

(d) http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/T/Tracheal Breathing.

html

3. Bone.

(a) I.P. Herman, ”Physics of the human body”, Springer, ISBN-10: 3540296034, (2007).

4. Bird flight.

(a) Rayner.J.M.V. (2001). Mathematical modelling of the avian flight power curve, math. Meth.App.Sci., 24:1485–1514.

(b) Loghthill, M.J. (1974). Aerodynamics aspects of animal flight. Bulletin of the Institute ofmathematics and its applications, 10:369 393.

(c) from 1(a) - see above. Chapter 5. Sections 4.1.4.3 may also be useful background readingon fluids.

5. Electrophysiology.

2

1.2. READING LIST

(a) see 2(a) above.

(b) J. Keener and J. Sneyd. Mathematical Physiology. Interdisciplinary Applied Mathematics8. Springer-Verlag, New York 1998. [Parts of Chapters 2,3 might be usefull].

6. Chemotaxis.

(a) J.D. Murray, Mathematical biology. I. Chapter 11, Springer 2001.

(b) J.D. Murray, Mathematical biology. II. Chapter 5.

7. Brain.

(a) J.D. Murray, Mathematical biology. I. Chapter 7, Springer 2001.

8. Blood.

(a) S.I. Rubinow, Introduction to Mathematical Biology, A Wiley-Interscience publication, NewYork. Chapter 4.

3

CHAPTER 1. GENERAL INFORMATION AND READING LIST.

4

Chapter 2

Using scaling arguments

2.1 First steps: Building a simple mathematical model

Warning: In this part of the course we make very simplistic assumptions about the biology. (Never-theless, our efforts are rewarded with answers that make broad sense.)

2.1.1 Example. Falling flea.

Why would a flea survive a fall from 30-storey building, whereas a human would probably not?

Is it:

1. because the human is much heavier?

2. because the flea has a stronger (exo)skeleton and hence can survive the impact?

3. because the fleas legs can absorb the impact (good shock-absorbers)?

4. some other (sensible) reason?

Galileo (or later Newton) tells us two cannon balls of different sizes reach the ground at the sametime - (by experience) this is not what we expect from fleas and humans, so whats missing?

Answer is friction - the drag on bodies due to air friction acts to decelerate a falling body. Overlong distances, bodies reach terminal velocity, which occurs where the frictional drag force balancesweight. So we need to understand how frictional drag depends on the size and shape of a body.

5

CHAPTER 2. USING SCALING ARGUMENTS

Figure 2.1:

What is Drag?- It is a force due to friction (air/water/soil)We know:Force = change momentum / time,so we can find the drag force from

1. the bodys area in contact with the air

2. how much air is moved from standstill to (terminal)speed v in a given time t (which gives the momentumchange in time t)

force × time = changemomentum (of still air)force × t = air mass × speed= (density) × volume × v= ρ × (S × h) × v= ρ ×(S × v × t) × v= ρ S v2 thence drag force ∝ S v2

6

2.2. SCALING ARGUMENTS

Assuming terminal velocity is reached by the flea and human:

weight = drag force = Mg ∝ Sv2 ⇒ v ∝

√MS

(2.1)

So how doesmass/area = M/S

differ for the flea and human?

Approximately:

Flea = 3mm long

Human = 2000mm long

Make the simplest assumption that there is a linear scale L such that

M ∝ L3, S ∝ L2 (2.2)

Then for each bodyM/S ∝ L (2.3)

Thus the terminal velocity varies with the bodys linear scale L as

v ∝√

L (2.4)

We say that the velocity scales as the square root of the body linear scale

For a flea and human we have (very approximately!)

Lflea/Lhuman ≈ 3mm/2000mm = 0.0015 (2.5)

and the terminal velocity of a human is approx 100mph, so

vflea = vhuman×√

Lflea/Lhuman ≈ 4mph (2.6)

This, combined with the strong exoskeleton of the flea, gives it a much better chance of survival!

2.2 Scaling arguments

The previous example is an example of a scaling argument -by making very simple assumptions wewere able to model how terminal velocity scales with linear scale L.

7


The scaling argument summarizes:

weight = Mg ∝ L3 (2.7)

drag ∝ projected area × v2∝ L2v2 (2.8)

weight = drag ⇒ L3∝ L2v2 ⇒ v ∝

√L (2.9)

Notice that we dropped all boring constants to reach the essential point: v scales as square rootof L.

Inherent in our assumptions were that for the linear scale L that distinguishes bodies in the simi-larity class:

• length ∝ L1

• area ∝ L2 (so projected area ∝ L2)

• volume ∝ L3 (so mass ∝ L3)

In applying our model, we were also assuming that the model is being applied to two bodies ofthe same shape (not exactly the case for the flea and human, but this is a first approximation model!).

In the following models, we will assume that we are comparing between families of animals ofsimilar shape, i.e. are isometric, parameterised by linear scale L.

We will be interested in how the size of the animal effects its functions. If L is a length scale,then area scales as L2 and volume scales as L3. Since many of life’s processes depend on trans-port of substances across a surface area (e.g. lung surface), and that transport supplies a volume(e.g. blood), it is intriguing to ask how the fact that volume increases faster than area effects (limits)function.

2.2.1 Some basic physics

(M=mass, L=length, T=time)

1. force = mass × accn : force = MLT−2

2. work done = force × distance = ML2T−2, kinetic energy = M(LT−1)2

3. power = work done/time = ML2T−3

4. flux = amount/area/timee.g. mass flux = mass/area/time = ML−2T−1

heat flux = heat energy/area/time = MT−3

8


5. heat is energy transferred down a temperature gradient

2.2.2 Example 2

How high can an animal jump? Or more precisely: How does the height that an animal from the samesimilarity class vary with linear scale L?

Assumptions:

1. work done by leg muscles = gain in potential energy

2. muscle force ∝ cross-sectional area of muscle ∝ L2

(this is not obvious, but there are models that justifythis experimentally demonstrated fact).

3. height jumped = height gained h by centre of mass(good approx)

Potential energy gained = Mgh ∝ L3×h

Work done by muscles =

muscle force × vertical distance C.O.M. displaced

∝ (L2)× (L) = L3.

Hence, equating PE gained to Wk. Done by muscle

hL3∝ L3

Thus h ∝ L0. That is, the simple model predicts that, for animals in the same isometric class, theheight they can jump is independent of their size.

Is this a good model? How high can a flea jump?

2.2.3 Example: How fast can we walk before breaking into a run?

Consider this (very) simplified picture of the human gait (figure 2.2). The human walks with straightlegs, so the the COM moves in a series of circular arcs. The front foot leaves the ground if the

9


component of weight is not strong enough to provide the centripetal acceleration which increases asv2. This gives us a limit on the walking speed.

10/7/10 17

€

a =v 2

Rcentripetalacceleration

€

v

€

L

€

mg

friction for grip

€

To keep foot on groundduring arc :

mg cosθ > m × v 2

R, so vmax = gR

€

θ

Figure 2.2:

10


2.2.4 Example

Minimum nerve speed required to make it possible for a animal to balance (e.g. flamingo )

toppling of animal scales like free fall :

∝ t =

√2hg

∝√

L

nerve speed required = distance from brain to muscle / timetaken

∝ L/t

Thusnerve speed scales as L√

L=√

L

2.2.5 Models that involve Metabolic Rate

Metabolism is the set of chemical reactions that occur in living organisms in order to maintain life.Animals use ATP to fuel their metabolic demands, e.g. in growth, locomotion, maintenance, immuno-logical defence, etc. The cells power plants” are organelles called mitochondria which generate mostof the cell’s supply of ATP.

food + oxygen ⇒ ATP⇒ ADP + Phosphate + ENERGY ⇒ muscle force.

The metabolic rate (rate of energy metabolism) of an organism (using aerobic respiration) can beassumed to be equated with the rate of oxygen consumption.

A simple (isometric) scaling argument for variation of metabolic rate (assuming a resting state andafter a period of fasting) with size is as follows:

Metabolic rate B = rate of oxygen consumption ∝ area of lungs supplying oxygen to mitochondria∝ L2

Body Mass M ∝ L3

Thus B ∝ L2 = (L3)2/3 ∝ M2/3 (Rubners Law).

Another argument, at least for warm-blooded animals, put forward by Rubner, is that a warm-blooded animal maintains a constant body temperature, and so their metabolism runs at a rate suchthat this temperature is maintained. Since the body loses heat energy at a rate proportional to theirbody surface area, which scales as L2, the metabolic rate ought to scale as L2.

11


Whatever the argument we (might) accept for the L2 law, we take the law as fact for now. (However,we will see later that it can be improved upon with the experimentally determined B ∝ M3/4.)Thus until stated otherwise, we assume that

B ∝ L2∝ M2/3.

2.2.6 Class Exercise (10mins): How long can a diving mammal stay under water onone breath?

A diving mammal (e.g. whale) stores oxygen in it blood before diving. When that oxygen is exhaustedit must surface for more air.

amount of stored oxygen ∝ lung volume ∝ blood volume ∝ L3

Metabolic Rate - rate at which mammal uses stored oxygen is ∝ L2

Thus duration of a dive scales as L3/L2 = L.

Thus the larger you are, the longer you can dive.

NB: We have ignored any specialisation that makes it more efficient for the animal to dive. Thusfor example, whales slow their heart beat and blood flow to their muscles is reduced; these factorsenable whales to dive for longer. When we build our simple models, we keep them simple by ignoringsuch specialisations. We are interested in broad statements about how things typically vary withscale.

2.2.7 Example: Why do large birds find it harder to fly?

Facts/assumptions:

1. Drag ∝ L2v2 (see flea/human model)

2. (Not obvious!) Maximum lift during gliding and wingflapping ∝ Awv2 where Aw = wing area ∝ L2

3. Metabolic rate ∝ L2 = rate at which energy is available.

12


2.2.8 How to obtain lift law

Bernoulli’s Theorem (Sketch)

For steady flow of inviscid incompressible fluid

ρv2

2+ p+ρgz = const

along streamline. Here p = pressure, z = fluid depth, v = fluid speed, ρ = density.Proof.

F = MA :

ρdvdt

=−ρgk−∇p

(Here d/dt means rate of change following a fluid particle, the so-called material derivative.)

ρ

2d |v|2

dt=−ρgk · v−∇p · v =− d

dt(ρgz+ p)⇒ integrate

(use :ddt

p(x(t)) = ∇p · ddt

x(t) = ∇p · v etc)

Figure 2.3:

Back to wing lift now. Air particles moving around the wing profile start and end at same time, sotop particle must move faster. By Bernoulli, this generates a lift ∝ v2, where v is the wind speed, thatis the speed of the air relative to the wing.

Max lift must just overcome gravity, so minimum flying speed v is given by

Awv2∝ Mg ∝ L3,

13


so that since Aw ∝ L2, v ∝ L1/2.

The required power (for flapping wings to get lift) is Power (work done /time = force x distance/time)

Flying power = drag× v ∝ L2v3∝ L2+3/2 = L7/2

Metabolic power ∝ L2, so that required power exceeds supplied power for larger L (L7/2 > L2 for Llarge enough), and hence there is an upper limit on bird size.

2.2.9 Kleibers Law

But experimentally B ∝ M2/3 is not observed!

Figure 2.4:

Instead one finds experimentally (by measuring oxygen consumption of animals in a resting state

14


and after they have fasted for sufficient period) that

metabolic rate B ∝ m3/4.

Figure 5.2 illustrates the remarkable range of scales over which the 3/4 law holds.Aside: An new argument for 3/4 power law was recently published by West (1997) under the

following assumptions:

• mammalian energy distribution networks (circulatory system, lungs) are fractal-like in structure;

• systems have evolved to maximise their metabolic capacity by maintaining networks that occupya fixed percentage of the volume of the body.

So from now on we will acknowledge experimental data and assume the allometric scaling (asopposed to isometric scaling) law for metabolic rate with mass:

B = B0m3/4. (Klieber’s Law).

2.2.10 Example. How does heart-rate scale with mass?

Assume that the heart beats fast enough to supply enough oxygen for the organism’s metabolism.Facts:

1. Metabolic rate ∝ m3/4.

2. Blood volume ∝ L3 ∝ m.

The rate of oxygen is transport around body is ∝ r×L3 ∝ r×m where r = heart rate (assume pumpvolume ∝ body volume). Thus r×m ∝ m3/4 giving r ∝ m−1/4. Smaller bodies have faster heart rates:

e.g. masked shrew (0.003kg) has r = 600, whereas elephant (4000kg) has r = 30. A human(80kg) has r = 80.

2.2.11 Example: Thickness of fur

Consider a class of similar animals in a cold environment. How does their fur thickness scale withmass?

Recall: Heat is energy transferred down a temperature gradient ∆T/∆x. heat flux = heat en-ergy/area/time = k(∆T/∆x), k = thermal conductivity of material (independent of scale).

To maintain body temperature (in surrounding of constant temperature, so that ∆T is constant) wethus need metabolic rate ∝ heat flux ∝ surface area × (temp difference / fur thickness h)

m3/4∝ L2/h ∝ m2/3/h,

15


so that fur thickness h ∝ m2/3−3/4 = m−1/12. Hence larger animals tend to have thinner fur.

2.2.12 Class excercise: How long does it take to starve to death?

Using Kliebers law, power ∝ m3/4

And energy reserves ∝ mass m, energy used up to starvation ∝ power × time to starve = m3/4× t.Thus t ∝ m×m−3/4 = m1/4.

2.2.13 Example: Swimming speed of a filter-feeder

rate of gain of stored energy ∝ food energy input rate - metabolism of stores

food input rate = F0U ,metabolism = basal rate + power to overcome (speed-dependent) drag= P0 + drag ×U

= P0 +(P1U2)×U = P0 +P1U3.

Thus need to look at the functionG(U) = F0U−P0−P1U3.G′(U) = F0−3P1U2 = 0 where U =

√F0/3P1 .

Not viable if Gmax < 0, i.e. (after some algebra) P0 > F0

√F03P1

(1− 1

3P1

).

Figure 2.5:

16

2.3. EXAMPLE: LUDWIG VON BERTALANFFYS GROWTH MODEL (1957)

2.3 Example: Ludwig von Bertalanffys Growth Model (1957)

von Bertalanfy was one of the founders of ”General Systems Theory” (http://en.wikipedia.org/wiki/Ludwig von Bertalanffy) Here is a (very) simple model he developed to study growthof an organism.

He assumed that all an organism’s available energy is channeled into :

1. Growth of the organism - building new cells, all taking the same energy to generate

2. Maintenance of the existing cells - keeping existing cells alive by supplying resources and re-moving waste products.

We have:

incoming power (metabolic rate [B]) = number of cells in body Nc(t)× metabolic rate of one cell [= Bc]+ energy required to create new cell [= Ec]× rate of increase in number of cells Nc(t)

⇒ B = NcBc︸︷︷︸maintenance

+EcdNc

dt︸︷︷︸growth

Now:body mass m = Ncmc, where mc =mass of 1 cell (assumed identical for all cells). Take B = B0m2/3

(isometric scaling, i.e. ∝ L2 [see exercise sheet 1 for the m3/4 case]). Thus

B0m2/3 =Bcmmc

+Ec

mc

dmdt

rearrange ⇒ dmdt

= αm2/3−βm,

where α = mcB0/Ec, β = Bc/Ec.Solve:

dmdt

= αm2/3−βm, m(0) = m0 (small, since organism starts small!).

Write asdmdt

= m2/3(α−βm1/3)

and substitute u = m1/3. Then 3du/dt = m−2/3dm/dt which gives

dudt

=13(α−βu), u(0) = m1/3

0 .

This has general solution u(t) = α

β+Aexp−βt/3. To find A, use initial data:

m1/30 =

α

β+A.

17


Hence we obtain:u(t) =

α

β(1− exp−βt/3)+m1/3

0 exp−βt/3,

and finally in terms of m:

m(t) =(

α

β(1− exp−βt/3)+m1/3

0 exp−βt/3)3

.

(Sketch similar in Q2, sheet 1).

2.3.1 Case Study: Incubating Eggs

• An egg is a self-contained unit. It has all the nutrients it needs for the embryo to develop -except for oxygen.

• But must also be rid of waste products, such as carbon dioxide and water. The shell is mainlycalcium carbonate with pores that allow influx and outflux of nutrients and waste products(gases and water).

• The shell must be strong enough to withstand roosting, but weak enough to allow chick to breakout when hatching.

Consider a spherical Egg!

Figure 2.6:

18

2.3. EXAMPLE: LUDWIG VON BERTALANFFYS GROWTH MODEL (1957)

Questions:

1. How does the egg incubation time scale with the mass of the egg?

2. How does the shell thickness vary with egg size?

Assumptions/Notes

1. Loss of water is the limiting effect - must not be too rapid else the embryo dehydrates. Oxy-gen and CO2 diffuse across embryo and egg shell faster than water (so can be consideredinstantaneous on the time scale of water movement).

2. Rate of water production ∝ metabolic rate

3. Would expect shell thickness to increase with shell size, since shell has to contain and protectyolk.

4. Total water loss ∝ size of egg

5. Water is lost via pores length = shell thickness d and with total area Apores over the shell.

Total water lost in Tinc = daily water loss ×Tinc

Tinc ∝Megg

daily water loss

and daily water loss ∝ metabolic rate ∝ M3/4egg . Hence

Tinc ∝Megg

M3/4egg

= M1/4egg ,

which agrees quite well with the experimentally observed M0.217.But how might the shell thickness d change with egg size?

water flux ∝ pore area × concentration gradientpore length

(see later lectures on diffusion).

∝density of pores in shell ×area

shell thickness×∆C

Since the area of the shell is proportional to R2, and ∆C is constant,

⇒M3/4egg ∝

density pores×R2

d∝

density pores×M2/3egg

d

19


density poresd

∝ M3/4−2/3egg = M1/12

egg .

Hence if pore density is a constant, independent of egg size,

d ∝ M−1/12egg

which would mean that eggs get thinner with increasing egg size!So the density of pores must be size dependent. In fact, experimentally it is observed pore density

∝ M4/3 and the shell thickness scales as d ∝ M1/2egg .

PS: It is also experimentally established that for birds Megg ∝ M3/4.

20

Chapter 3

Oxygen transport and Insect respiration

21

CHAPTER 3. OXYGEN TRANSPORT AND INSECT RESPIRATION

22


24


26


28


30


32


34


38


40


42


44


46


48

Chapter 4

Bird flight

4.1 Basics of bird flight.

Figure 4.1:

In planes: wings produce lift. Propeller or jet engine produces thrust. In birds: wings must produceboth thrust and lift.

Birds have evolved so that they

• have light weight skeleton - have smaller porous bones, hollow bones with strengthening struts,skull ≈ 1% body weight

• efficient respiratory system to provide for high metabolic rate required for flight. Anterior airsacs lungs, posterior air sacs.

• eat beries and other high energy foods

49

CHAPTER 4. BIRD FLIGHT

4.2 Basics of lift

The lift force is always perpendicular to the forward motion and the drag is along the line of forwardmotion.

Figure 4.2: Vu > Vl ⇒ pressure difference by Bernoulli ⇒ lift force.

Figure 4.3: Angle of attack.

Why evolve an airfoil shape? Why not just use the angle of attack for lift?Asymmetry of wing

1. produces lift at zero angle of attack

2. produces more lift than any symmetric wing at any angle of attack

3. produces less (pressure) drag.

50

4.2. BASICS OF LIFT

Stalling: large angle of attack leads to large drag.

Figure 4.4: Turbulent flow ⇒ less lift, more drag.

The Kutta-Joukowski theorem is a fundamental theorem of aerodynamics. It states that the lift perunit of the wing length is a product of density of the fluid, velocity at some distance, and the circulationaround the wing.

Figure 4.5: Calculation of the velocity circulation.

L =−ρu×Γ (4.1)

where Γ is a velocity circulation, i.e.Γ =

Iγ

V cosθdθ

51


Figure 4.6: Actual flow around wing is the mathematical sum of flow yielding no lift and a flow of purecirculation

As the wing starts from rest in stationary air, a vortex with circulation −Γ appears behind the wingand is matched by a circulation Γ around the wing itself (conservation of vorticity). The circulationcomes off the tips of the wings and leaves as vortices that trail behind the wings. The circulationaround the wings, in the form a bound vortex, gives rise to lift given by equation (4.1). However, theenergy used to create the trailing vortices manifests as “induced” drag on the wing. If there is no liftthen there is no induced drag.

4.3 Energy is required to counter:

• weight

• parasitic drag = frictional drag (body surface drag) + pressure drag (low pressure ”suck” behindwing)

• induced drag ∝ L2, where L2 =lift generated. This is from the energy in the trailing vortices ordisturbances in a large region of air in the wake of the wing

52

4.3. ENERGY IS REQUIRED TO COUNTER:

Figure 4.7: Vortex wake leads to a trailling vortex behind the wing. The vortices dissipate eventually,but may persist for some time.

4.3.1 Parasitic drag Dp.

We assume that the pressure drag is small in comparison to the frictional drag . Dp = rate of transferof momentum from the surface of the bird to the thin mass of air (boundary layer) which is draggedforward with speed u.

Figure 4.8: Boundary layer. Zero velocity on surface ⇒ air speed in a small distance.

frictional drag force =ddt

(momentum of air displaced by wing) ∝ u2

Define dimensionless drag coefficient

Cd =D

12 ρSu2

,

where u = air speed of wing.

Dp =12

CdρSu2,

where ρ is air density and S wing area.

53


4.3.2 Induced drag Di

Induced drag = rate of transfer of momentum to the trailing vortices

= Kinetic Energy per unit length in the trailing vortex system.

mass M in vortex region×air speed UT

=MU2

TU=

MU2

d

Also we have

Momentum transferred per unit length to vortices = Lift force L× (1/U)

HenceLU

=momentum of vorticeslength of vortex system

mass per unit length of vortex system ∝ ρ× wing area ∝ ρb2,

where b is wing semi-span. So the induced drag

Di = K. E. per unit length

=1/2(momentum)2/masslength d of vortex system

= 1/2(momentum/length d)2/mass/length d

∝ (L/u)2/(ρ×b2) =L2

ρb2u2 .

Define

Di =KL2

12 ρb2u2

,

where K is some factor for wing shape, etc. Hence total drag

D(u) =12

CdρSu2 +KL2

12 ρb2u2

4.4 Steady level flight

We have L = Mg

drag =KL2

12 ρb2u2

+12

ρSu2Cd (4.2)

54

4.5. STABLE GLIDING

Figure 4.9:

In level flight L = Mg so in equation (4.2)

D =KM2g2

12 ρb2u2

+12

ρSu2Cd

is the drag under these conditions. The bird must produce a thrust T = D to counteract the drag.The rate at which the bird does work to counter the drag, the power=P = thrust × velocity = force×distance

time , i.e.

Power = P = U×D(U) =KM2g2

12 ρb2u

+12

ρSCdu3.

4.5 Stable gliding

This is where the bird glides in a downward straight line without flapping its wings (at least betweenwing movements that allow it to change direction). Birds with suitable wing characteristics, e.g.buzzards, albatrosses, can travel large distances using this mode of flying. Here the power to counterthe drag and provide lift comes from the gravitational pull on the bird, and we are interested in a glidedownwards at a constant angle θ to the horizontal.

D =12

ρu2SCd +KL2

12 ρu2b2

Write asD = αu2 +β/u2

where α = 12 ρSCd and β = 2KL2

ρb2 . To find Dmin, differentiate w.r.t. u and set the gradient of D to zero:

2αu− 2β

u3 = 0 ⇒ u4 = β/α, umd = (β/α)1/4

55


Figure 4.10: Forces by gliding. D as a function of u.

Dmin = α

√β

α+β

√α

β= 2√

αβ

So needmgsinθ≥ 2

√αβ

But at critical angle θmin,

12

mgsinθmin =(

1/2ρu2SCdKL2

1/2ρu2b2

)1/2

=(

SCdKL2

b2

)1/2

12

mgsinθmin = (SCdK)1/2 Lb

=(SCdK)1/2

bmgcosθmin

tanθmin =2b(SCdK)1/2

θmin = tan−1(

2b(SCdK)1/2

)θmin is the minimum glide angle possible.

What is the air speed umd (md=minimum drag) at this minimum angle?

56

4.5. STABLE GLIDING

umd =(

β

α

)1/4

=

(KL2

12 ρb2

/12

ρSCD f

)1/4

=(

4KL2

ρ2b2SCD f

)1/4

=(

4Km2g2 cos2 θmin

ρ2b2SCD f

)1/4

Figure 4.11: But this is not a stable glide.

Figure 4.12:

For steeper angles, θ > θmin, a stable glide is possible. mgsinθ = D(u) has more than one root u∗

(see figure 4.10) but only one of them is stable, namely u > umd . To see why consider how the birdresponds to a small perturbation in its speed when at the higher speed uhigh where D f dominates. Ifthe speed slightly increases, the drag increases and reduces the speed. If the speed decreases, thenthe drag is less than the weight component mgsinθ and so there is an acceleration and the speedincreases back to uhigh. On the other hand, if the speed is ulow where Di dominates, then a slightincrease in speed will decrease the drag, and lead to an acceleration, increased speed and so on, sothat the speed would build up to uhigh where it is stable. If at ulow the speed slightly decreased then itwould continue to decrease until the lift becomes too small and the bird would stall.

To confirm these ideas we may do the linear stability analysis:

57


Figure 4.13:

4.5.1 Linear stability analysis of a glide

Consider a small perturbation ξ away from the glide speed u∗: u = u∗+ξ.

mu = mgsinθ−D(u∗+ξ)

m(ξ) = mgsinθ− [D(u∗)+D′(u∗)ξ+ ...] = [(mgsinθ−D(u∗)) = 0(glide)]−D′(u∗)ξ+ ...

mξ =−D′(u∗)ξ, to first order in ξ.

Henceξ =−D′(u∗)

mξ

giving

ξ(t) = ξ(0)exp(−D′(u∗)

mt

)for growth of the perturbation.

Hence the glide is stable only if D′(u∗) > 0, i.e. the bird needs the larger velocity uhigh of the twofor a stable glide at a glide angle θ.

Thus for a stable glide with air speed u, we need D′(u) > 0, which is equivalent to Dp = D f > Di,i.e. at speed u,

u2SCD f >4KL2

ρ2u2b2

b2 >4KL2

ρ2SCD f u4

b >

(KSCD f

)1/2 L12 ρu2S

58

4.5. STABLE GLIDING

Hence at lower speeds, need b larger. That is the bird spreads its wings.

Figure 4.14:

See fig. 4.15

Figure 4.15:

L = mgcosθ, D = mgsinθ. Need D f > Di so see fig 4.15 (middle and right).

4.5.2 Stable glide speeds and angles

For a stable glide, we have the force balance

mgsinθ =KL2

12 ρb2u2

+12

SCdρu2

Hence

mgsinθ =2Km2g2 cos2 θ

ρb2u2 +12

SCdρu2

59


Figure 4.16:

D(u) =α

u2 +βu2

α

β=

4Km2g2 cos2 θ

ρ2b2SCd

We solve for u:α

u2 +βu2 = γ(= mgsinθ)

βu4− γu2 +α = 0

u2 =γ

2β± 1

2β(γ2−4αβ)1/2

For stability we take the larger root:

u2 =mgsinθ

2 12 ρSCd

+mgsinθ

ρSCd

(1−4cot2 θ

KSCd

b2

)1/2

u2 =mgsinθ

ρSCd

1+(

1−4cot2 θKSCd

b2

)1/2

Notice that since θ > θmin,tanθ > tanθmin =

2b

√KSCd

the speed u is real.

60

4.6. SOARING FLIGHT

4.5.3 Some crude estimates for stable glide speeds.

As we have seen, for stable glide we need D f > Di. This gives D = D f + Di < 2D f ⇒ D f > 12 D and

D = D f +Di ≥ D f > 12 D. Since D = mgsinθ, we then have mgsinθ≥ D f > 1

2 mgsinθ which gives

mgsinθ≥ 12

ρu2SC f >12

mgsinθ

√2mgsinθ

ρSC f≥ u >

√mgsinθ

ρSC f(4.3)

which is a crude estimate for the stable glide speed at a glide angle θ > θmin.

4.6 Soaring flight

This is where the bird glides, but takes advantage of upward air currents to maintain or gain theirheight, such as from the windward side of cliffs, or from a thermal). All that is required is that verticalrate of descent is less than the upward component Uair of the rising air:

Uair sinφ≥U sinθ. (4.4)

For a seagull to remain stationary with respect to the cliff, then, it needs to adjust its stable glidespeed U and angle θ (see figure 4.17) to satisfy

U sinθ = Uair sinφ

U cosθ = Uair cosφ.

Hence the seagull chooses θ = φ and U = Uair if it is remain stationary. That is, the seagull alignsitself head on into the wind and adjusts its wings so that it remains stationary. The balance betweenthe air speed U , angle θ satisfies equation (4.5), provided that u exceeds the stalling speed of theseagull and the full-span (maximised b) minimum drag speed.

Some birds, such as vultures and buzzards, use rising air currents known as thermals for lift.They soar in circles using the energy in the thermal to provide lift. If the upward air speed is v (soφ = π/2 here) then they can fly in a circle at the same altitude and at speed u when v = usinθ. Fromequation (4.5) this gives that u satisfies√

2mg(v/u)ρSC f

≥ u >

√mg(v/u)

ρSC f(4.5)

61


Figure 4.17: Soaring flight. The bird takes advantage of rising air currents to remain aloft for longperiods of time.

so that (mgvρSCd

) 13

< u≤(

2mgvρSCd

) 13

. (4.6)

4.7 Bounding flight

This is the mode of flying used by many small birds that do not have the wing area or span for glid-ing. The bounding flight consists of periodic change between essentially parabolic projectile motion,where the wings are folded, and flapping in order to regain height. Here we are concerned with theefficiency of flapping fight over pure glides.

What scope is there for energy saving for birds flying at speeds well in excess of the minimumdrag speed umd? At such speeds the induced drag (∝ 1/U2 = energy cost per unit distance to supportweight with lift) is relatively small, but the parasitic drag from the loss of momentum to the boundarylayer around the outstretched wings and body is large. Some small birds have evolved a mode offlying called “bounding flight” where they reduce the drag cost by fold their wings for part of their flightpattern. Thus bounding flight consists of periods where the bird flaps and glides (so the wings areoutstretched) to gain height alternating with periods when their wings are folded (so no lift) and theirflight path is parabolic (projectile motion). Since on average, for level flight, the lift has to balance theweight, if the wings are to be folded for some of the time, the lift has to be enhanced when the wings

62

4.7. BOUNDING FLIGHT

Figure 4.18: Bounding flight. The bird flies with alternating periods of flapping/gliding (wings out-stretched) and free fall (wings folded).

are outstretched. Thus (as suggested by Mr S. B. Furber [unpublished]) if the wings are outstretchedfor a fraction f of the time, then the enhanced lift must be

L =mgf

.

This increases the induced drag when the wings are outstretched, but they are not outstretched forall of the time. So is there any benefit from bounding flight?

For normal flight ( f = 1) let Di be the induced drag and D f = Db + Dw be the frictional drag splitinto body drag Db (which is always present) and Dw the drag from the outstretched wing. The averagedrag is

D = Db + f

(Dw +

K(mg/ f )2

12 ρb2u2

)= Db + f Dw +

Di

f.

How does this drag vary with f ?dDd f

= Dw−Di

f 2 ,

so that D is extremal at f =√

DiDw

which is a minimum. So bounded flight is less costly when the birdflies at speeds such that Di/Dw < 1. Indeed, at the minimum D = Db + 2

√DwDi and comparing with

the drag for normal flight f = 1 we obtain

Db +Dw +Di− (Db +2√

DwDi) = Dw +Di−2√

DwDi > 0,

63


by the arithmetic-geometric mean inequality.

64

Chapter 5

Electrophysiology

5.1 Electrodiffusion

Electrodiffusion occurs when ions diffusing through a medium are subject to an external force due tothe interaction of the ionic charge with an electric field.

We recall that the diffusive flux J for a substance with concentration C(x, t) in a medium of diffusiveconstant D is given by

J =−D∇C.

The presence of an electric field causes the ions to have an additional migrational flux

Jmig = µqCE,

where q is the charge of the ion, E the electric field strength, and µ the mobility of the ion. Hence thetotal flux is

J =−D∇C +µqCE.

Introducing the electric potential (i.e. the voltage) φ we have E =−∇φ so that

J =−(D∇C +µqC∇φ) .

Finally we use Einstein’s relation D = kBT µ (kB Boltzmann’s constant, T temperature in Kelvin) to give

J =−µ(kBT ∇C +qC∇φ) . (5.1)

65

CHAPTER 5. ELECTROPHYSIOLOGY

Using kbT/qe = RT/F (R= gas constant, qe charge on electron) gives the alternative form:

J =−D(

∇C +FzRT

C∇φ

). (5.2)

where z is valency of the ion (q = qez). Hence, with suitable boundary conditions on a domain Ω wehave

∂C∂t

=−divJ = µ div(kBT ∇C +qC∇φ) .

This equation is highly nonlinear and presents a major challenge! We make two simplifying assump-tions: (i) our domain is an interval of the real line, (ii) we are only interested in the steady stateconcentration (i.e. ∂C

∂t = 0). In this case the steady state condition reads

∂C∂t

=−∂J∂x

= 0,

so that the flux is independent of x.

5.1.1 Example

Consider a membrane with diffusive constant D that separates two phases of the same electrolytewith ions charge q; the inside of the membrane has constant concentration C(0) =Cin, and the outsidehas C(d) = Cout . The voltage is taken to be V at x = 0 and 0 at x = d (see figure 5.1). The phases oneach side of the membrane are so large that movement of ions across the membrane does not altertheir concentrations. We wish to find the steady flux J through the membrane. We have

J =−D(

dCdx

+zFRT

Cdφ

dx

),

is constant (in both x, t). Rewrite this as

− JD

=dCdx

+zFRT

Cdφ

dx.

Using V0 = RT/F and the integrating factor exp(zφ/V0) we obtain

− JD

exp(zφ/V0) =ddx

(C exp(zφ/V0))

66

5.1. ELECTRODIFFUSION

Figure 5.1:

Integrating across the membrane,

−Z d

0

JD

exp(zφ(x)/V0)dx =Z d

0

ddx

(C exp(zφ/V0)) dx

= C(d)exp(zφ(d)/V0)−C(0)exp(zφ(0)/V0)

= Cout −Cin exp(zV/V0).

This gives the membrane flux as

J = D

(Cin exp(zV/V0)−CoutR d

0 exp(zφ(x)/V0)dx

). (5.3)

Notice that there is still the unknown voltage profile φ(x). Since we only have (at present) one equationfor the two unknowns C,φ this is expected. Even so, this formula is useful since it tells us that the fluxvanishes when

V =−RTzF

log(Cin/Cout)

regardless of what the actual voltage profile is. This last voltage is known as the Nernst Equilib-rium potential for that ion. It is the membrane voltage for which, given the inside and outside ionconcentrations Cin,Cout , there is no transmembrane flux.

67


5.1.2 Constant field approximation

In some circumstances, it is appropriate to suppose that the electric field E is a constant, so thatthe voltage is linear through the membrane. The flux through the membrane - equation (5.3) - thensimplifies to

J = D

(Cin exp(zV/V0)−CoutR d

0 exp((zV/V0)1− x/d)dx

)

= DV z(

Cin exp(zV/V0)−Cout

−dV0(1− exp(zV/V0))

)=

DV zdV0

(Cout −Cin exp(zV/V0)

1− exp(zV/V0)

)= Pz

VV0

(Cout −Cin exp(zV/V0)

1− exp(zV/V0)

).

where P = Dd is the permeability of the ion through the membrane.

5.2 Poisson-Nernst-Planck equations (PNP)

The missing information concerns how the electric potential φ depends on the local density of chargefrom the ions. This is given by Poisson’s equation:

∇2φ =− ρ

ε0K, (5.4)

where ε0 = 8.85× 10−12 farad per metre, and K is the relative dielectric constant. ρ is the chargedensity given by

ρ = F ∑k

(z+kC+k−|z−k|C−k) . (5.5)

Here F is Faraday’s constant (96485 Coulombs per mole), z+k is the valency of the kth cation ofconcentration C+k and similarly for the anions −k.

Thus in one dimension, we have

d2φ

dx2 =− Fε0K ∑

k(z+kC+k−|z−k|C−k) .

The Poisson-Nernst-Planck equations consist of the flux equations (5.2), one for each ion, plusPoisson’s equation (5.4) for the potential.

68

5.2. POISSON-NERNST-PLANCK EQUATIONS (PNP)

5.2.1 Example

The salt HCL dissociates into H+ cations (actually protons) and Cl− anions. Writing ψ = φ/V0 (V0 istypically around 50-100 millivolts) we have

d2ψ

dx2 = − Fε0K

(CH −CCl)

= −1.37×1014 (CH −CCl) .

Thus we have

εd2ψ

dx2 =−σ,

where ε 1 and σ = CH −CCl. Look for solutions ψ = ψ0 + εψ1 + ε2ψ2 + · · · and similarly for σ. Thenwe have

ε(ψ′′0 + εψ′′1 + ε

2ψ′′2 + · · ·) =−σ0 + εσ1 + ε

2σ2 + · · ·

Reading off powers of ε, assuming that there are no large gradients of the electric field, we obtain,for the first two terms,

0 = σ0

ψ′′0 = −σ1

The first equation expresses that the HCL solution is approximately (to first order in ε) electricallyneutral: CH ≈ CCl. (NB: This electroneutrality may break down close to the boundaries of phaseswhere large gradients in the electric field can be observed).

5.2.2 Example: Ion flux through long aqueous channel

(Based on Sneyed and Keener, Mathematical Physiology.) The cell membrane separates two phaseswhere there are monovalent ionic species 1,2, say species 1 is a cation, species 2 an anion. At eachend of the channel, in the two phases the concentrations of the two ions are equal (electroneutrality):

c1(0) = c2(0) = ci, c1(L) = c2(L) = ce,

and as is convention we takeφ(0) = V,φ(L) = 0,

so that the membrane potential is V = φ(0)−φ(L).

69


Figure 5.2:

In the channel, the Poisson-Nernst-Planck (PNP) equations read

εd2φ

dx2 = −(c1− c2)

J1 = −D1

(dc1

dx+

FRT

c1dφ

dx

)J2 = −D2

(dc2

dx− F

RTc2

dφ

dx

)where ε = ε0K

F 1.

It is hard to solve this system without some form of approximation. Here we will assume that thechannel is long: L 1. To see what this implies for the solution, we rescale the equations:

x =xL

, c1 =c1

C, c2 =

c2

C, ψ =

Fφ

RT, C = ci + ce.

70

5.2. POISSON-NERNST-PLANCK EQUATIONS (PNP)

We obtain:

d2ψ

dx2 = − L2CRT ε

(c1− c2)

− J1LD1C

=dc1

dx+ c1

dψ

dx

− J2LD2C

=dc2

dx− c2

dψ

dx

with boundary conditions

c1(0) = c2(0) = ci/C

c1(1) = c2(1) = ce/C

ψ(0) = V,ψ(1) = 0

Now we set η = RT ε/L2C, Jk =−JkL/CDk (k = 1,2) and drop the tildes to obtain

ηd2ψ

dx2 = −(c1− c2)

−J1 =dc1

dx+ c1

dψ

dx

−J2 =dc2

dx− c2

dψ

dx

Now we make the assumption that the channel is long, so that η 1. We end up with potentiallya singular perturbation, but first let’s try the first order approximation c1(x) = c2(x) = c(x) (say). Thissatisfies the boundary conditions, so to first order we have that the concentrations are equal in thechannel. Thus to first order in η we have

−J1− J2 = 2dcdx

.

Hence c(x) is linear in the channel and so, since we know the end concentrations,

c(x) =ci

C+(ce− ci)

xC

.

To find the potential we use the flux equations again:

2cdψ

dx= J2− J1,

so thatdψ

dx=

C(J2− J1)ci +(ce− ci)x

,

71


which may be integrated to yield

ψ(x) =C(J2− J1)

ce− cilog(ci +(ce− ci)x)+B,

where B is the constant determined from the boundary condition ψ(1) = 0:

B =−C(J2− J1)ce− ci

logce.

Hence we obtainψ(x) =

C(J2− J1)ce− ci

log(

ci

ce+(

1− ci

ce

)x)

.

Now ψ(0) = V , so that we obtain

ψ(x) =V

log(ci/ce)log(

ci

ce+(

1− ci

ce

)x)

.

We may also calculate the flux, say, of ion 1 through the channel. This is

J1 =−dc1

dx− c1

dψ

dx

But, setting V1 =− log(ci/ce) we have

dψ

dx=−V

V1

ce− ci

ci +(ce− ci)x.

HenceJ1 =−ce− ci

C+

ci

C+(

ce− ci

C

)x

VV1

ce− ci

ci +(ce− ci)x

so that the flux of ion 1 isJ1 =

(ce− ci)(V −V1)CV1

.

All these results can now be written back in the original unscaled coordinates.

5.2.3 Transmembrane currents

The current I carried by an ion of valency z is given by I = zFJ where J is the flux of the ion and F isFaraday’s constant.

72

5.3. EXCITABLE CELLS - BASICS

5.2.4 Example: Simple cell model

In the cell the main ionic species are Na+, Cl− and K+. The fluxes of these ions across a membrane(assuming the constant field approximation) read

JK = PKVV0

(Kout −Kin exp(V/V0)

1− exp(V/V0)

)JNa = PNa

VV0

(Naout −Nain exp(V/V0)

1− exp(V/V0)

)JCl = −PCl

VV0

(Clout −Clin exp(−V/V0)

1− exp(−V/V0)

)The corresponding currents are

IK = FJK , INa = FJNa, ICl =−FJCl.

The total current across the membrane is thus I = IK + INa + ICl which comes out to be

I =VV0

(PKKout +PNaNaout +PClClin− [PKKin +PNaNain +PClClout ]exp(V/V0)

1− exp(V/V0)

). (5.6)

Hence in this scenario, there is net zero current flow across the membrane when the membranepotential is

V =RTF

log(

PKKout +PNaNaout +PClClinPKKin +PNaNain +PClClout

). (5.7)

This last equation is known as the Goldman-Hodgkin-Katz voltage equation. If the membrane isimpermeable to sodium and chloride, the equation is known as the Nernst equation.

5.3 Excitable cells - basics

Many cells, such as for example, nerve axons, heart pacemaker cells, muscle cells are capable ofgenerating action potentials - sudden depolarizations of the membrane (from a negative potential ofaround -70 millivolts to a positive potential of around +55 millivolts). Surprisingly such depolarizationsare brought about by the transfer of relatively small numbers of ions. The potential changes occurdue to changes in the permeability of the cell membrane to the ions K+ and Na+.

Let us suppose that (in the resting state) the membrane in the simple cell model above is per-meable to potassium ions only (the membrane in this state is relatively impermeable, but the smallpermeability is due to potassium channels). For a typical cell, there are strong chemical gradientsof Na and K across the cell membrane. Extracellular K (20mmol/litre) is much lower than intracellu-

73


lar K (400 mmol/litre). (We will ignore the role of chloride ions in what follows.) On the other handextracellular sodium (440mmol/litre) is much higher than intracellular sodium (50mmol/litre).

Then the Nernst equilibrium potential for potassium EK is given by setting PNa = PCl = 0 in equation(5.7). We obtain, for a typical cell, RT

F = 58 and

EK = 58log(20/400)≈−75mV.

SimiilarlyENa = 58log(440/50)≈ 55mV.

The sudden changes in membrane potential are brought about by the changes in membrane perme-ability to sodium and potassium. Prior to a stimulus, when the excitable cell is in its resting state, themembrane potential is at the so-called resting potential, which is about −72mV. Thus the membraneis initially relatively impermeable, but substantially more permeable to potassium than sodium. Whensubject to a sufficiently strong (suprathreshold) stimulus (see figure 5.4), however, a depolarizationcauses voltage-dependent sodium channels in the membrane to open and sodium ions surge in un-der the combined electric and chemical gradients. These sodium channels are sensitive to changesin membrane potential, in that they can open as the cell membrane potential increases. Since theentry of sodium cations into the cell depolarizes the membrane (increases the membrane potential),more sodium channels open. This leads to further depolarization and further opening of sodiumchannels. This positive feedback results in a rapid depolarization of the membrane to a value closeto the sodium equilibrium potential of around 55mV (since now the permeability of the membraneis dominated by sodium). As the peak is approached the sodium channels tend to close under thepositive potential and potassium channels begin to open and potassium ions surge out of the cell andthe membrane repolarizes back to the cell resting potential. After a brief refactory period the cell isready to respond to new stimuli. The spike in the voltage lasts around one millisecond. If the stimulusis not strong enough, the feedback mechanism that leads to rapid opening of the sodium channelsis not triggered and the cell returns to the resting state. Such a stimulus is subthreshold. Thereis a threshold, usually around 15millivolts above resting potential, beyond which the suptrathesholdstimulus triggers an action potential.

Later in the course, the Fitzhugh-Nagumo model for excitable cells will be introduced. Here we areconcerned with the propagation of the action potential from a nerve cell down a long thin nerve axon.The propagation ensues because local action potentials trigger action potentials in neighbouringsections of the axon. We will now develop a simple model for action potential propagation down anaxon modelled as a conducting cylinder covered by a excitable membrane whose voltage-currentcharacteristic Iion(V ) is designed to model the complex membrane conductance changes outlinedabove.

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5.4. THE CABLE EQUATION

Figure 5.3: Membrane voltage changes during generation of an action potential

The above description is a substantial simplification of the whole action potential generation pro-cess, leaving out actions of sodium pumps that pump out the sodium ions that enter the cell ondepolarization, or other ions and ion channels involved. Nevertheless, it suffices for our simple modelof action potential propagation.

5.4 The Cable Equation

We now consider the transmission of an action potential along a nerve axon (see figure 5.5). Thedistance down the centreline of the axon is x, and the currents and voltages are all assumed tobe radially independent. We let Ii(x, t) be the longitudinal current at x, t along the axon. The voltageinside the axon is Vi(x, t) and the extracellular voltage is assumed constant set at zero. The membranepotential is thus Vm(x, t) = Vi(x, t)−Ve(x, t) = Vi(x, t). We let ri denote the resistance of the axon coremeasured in Ohm/cm. Then by Ohm’s law,

Vm(x+∆x, t)−Vm(x, t) =−Ii(x, t)ri∆x,

to first order in ∆x. HenceIi =− 1

ri

∂Vm

∂x. (5.8)

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Figure 5.4: Propagation of an action potential from a nerve cell down a nerve axon.

Now let Im(x, t) be the total membrane current per unit length of the axon. By KCL,

Ii(x, t)− I(x+∆x, t) = Im(x, t)∆x,

(again to first order) so that

Im(x, t) =−∂Ii

∂x. (5.9)

Combining equations (5.8) and (5.9) we obtain

Im =1ri

∂2Vm

∂x2 . (5.10)

Now using KCL again, the total membrane current is the sum of the capacitative current and themembrane ionic current Iion per unit length of the membrane:

Im =(

cm∂Vm

∂t+ Iion

). (5.11)

Here cm is capacitance per unit length of the membrane. Hence putting (5.10) with (5.11) we obtainthe cable equation:

cm∂Vm

∂t=−Iion(Vm)+

1ri

∂2Vm

∂x2 . (5.12)

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Figure 5.5: Small section of nerve axon, thickness ∆x

5.4.1 A simple model for action potential propagation

Let us model the ionic current as

Iion(V ) = gV (V −Vthres)(V −Vp),

where 0 < Vthres < Vp, and g > 0 (see figure 5.6). Then the cable equation reads

cm∂V∂t

=−gV (V −Vthres)(V −Vp)+1ri

∂2V∂x2 .

Now introduce α = VthresVp

, U = VVp

, τ = gVpcmVthres

, so that we obtain

∂u∂τ

= f (u)+D∂2u∂x2 , (5.13)

where f (u) = u(α−u)(u−1) and D = griV 2p . We will now seek travelling front solutions of (5.13). The

travelling front of speed c > 0 is a wave that has constant profile to an observer moving at speed c.Thus we seek a profile U such that u(x,τ) = U(x− cτ). If such a wave exists then it is a solution of(5.13). Now we have, setting ξ = x− cτ,

∂u∂τ

= U ′(ξ)∂ξ

∂τ=−cU ′(ξ),

and∂u∂x

= U ′(ξ)∂ξ

∂x= U ′(ξ),

∂2u∂x2 = U ′′(ξ).

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Figure 5.6: Voltage-current characteristic for membrane in cable equation model. Note how the cur-rent changes correspond to the movement of various ions following membrane permeability changes.

Hence (5.13) transforms into the following second order ordinary differential equation:

DU ′′(ξ)+ cU ′(ξ)+ f (U(ξ)) = 0.

What about boundary conditions? We want our wavefront to propagate from left to right (c > 0)exciting the cells from resting state u = 0(V = 0) to u = 1(V = Vp) as it progresses. Hence for ourprofile U : (−∞,∞)→ (−∞,∞) we would expect u(ξ)→ 1,u′(ξ)→ 0 as ξ→−∞ (see figure 5.8). On theother hand u(ξ)→ 0,u′(ξ)→ 0 as ξ→ ∞. We may represent such a curve in (U,U ′) space (see figure5.7), where the curve is given by the points (U(ξ),U ′(ξ)) as ξ ranges form −∞ to +∞. Let X = U ′.Then we are solving X ′ =− 1

D(cX + f (U)), so that

dXdU

=−cX− f (U)

DX.

We look for a quadratic solution X(U). This must satisfy X(0) = 0 and X(1) = 0, and so X(U) =θU(U−1) for some θ which we must find.

Dθ(2U−1)×θU(U−1) =−cθU(U−1)−U(α−U)(U−1)

Dθ2(2U−1) =−cθ− (α−U)

Equating coefficients leads to

c =

√D2

(1−2α), θ =1√2D

.

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Figure 5.7:

This shows that the profile U satisfies

dUdξ

= θU(U−1).

This last equation may be solved (using partial fractions) to obtain

U(ξ) =e−θξ

K + e−θξ,

where K is an arbitrary constant of integration. Notice that this U satisfies all the boundary conditionsthat we have stated. A sketch of the profile can be seen in figure 5.8.

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Figure 5.8: Profile of travelling wavefront for action potential model (α < 1/2)

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