Lecture 1 - Georgia State University

MATH 4010/6010: Mathema1cal Biology Instructor : Remus Oşan, Georgia State University Lecture 1 – May 19, 2014 Introduc1on and course overview A ﬁrst example Introduc)on n 
Remus Oşan: [email protected], h8p://www2.gsu.edu/~rosan/math_bio_romania_2014.html n 
Assistant Professor, Dept. Mathema)cs and Sta)s)cs, Neuroscience Ins)tute, Georgia State University n 
Romanian organizer: Assist.Prof.Eng. Ioan Muntean n 
Computer Science, U. Tehnica Cluj n 
[email protected] 2 Introduc)on n 
Remus Oşan: n 
Research: Development of analysis methods for neural data n 
Data-‐driven computa)onal modelling for sensory and memory systems in the brain, neural development and regenera)on 3 Introduc)on n 
Course Hours: n 
10 – 12, 2-‐3 pm Mon/Wed n 
10 – 12, 2-‐4 pm Tue/Th n 
Loca)on n 
Bari)u 26-‐28, sala S4.1 (corp cladire Somes, etaj 4) 4 Textbook •  Edelstein-‐Keshet, Mathema)cal Models in Biology. •  Plan to Cover chapters 1 – 8 5 Evalua)on of course performance •  Final grade will be assigned base on •  Homework: 4 hws, lowest one dropped (~ 2 per week) •  Each one counts as 15 percent of the grade (40 overall) •  Final: last week (June 4, tenta)vely): 35 percent •  Class project: June 4 project presenta)on: 20 percent implement and inves)gate a mathema)cal model OR read and cri)cally evaluate a set of papers 6 Course material •  1 Discrete processes •  Chapter 1: Linear diﬀerence equa)ons •  Nth order diﬀerence equa)ons, systems of diﬀerence equa)ons, real and complex eigenvalues and their qualita)ve meaning. •  Applica)ons: Fibonacci numbers and pa8erns in nature, segmental growth, carbon dioxide and breathing. 7 Finite diﬀerence equa)ons •  Chapters 2,3: Introduc)on to nonlinear diﬀerence equa)ons •  Steady states, stability, bifurca)on values •  Behavior of the logis)c equa)on, chao)c solu)ons •  Stability calcula)ons for second order equa)ons •  Scaling and parameter reduc)on •  Applica)ons: spread of disease, host-‐parasite systems, popula)ons gene)cs 8 II Con1nuous processes and ordinary diﬀeren1al equa1ons (ODEs) •  Chapter 4: Introduc)on to con)nuous models •  Review of linear ODEs, comparison to diﬀerence equa)ons, stability •  Dimensional analysis •  Applica)ons: behavior of chemostat, drugs given by con)nuous infusion, glucose-‐insulin kine)cs, compartmental analysis. 9 II Con1nuous processes and ordinary diﬀeren1al equa1ons (ODEs) •  Chapter 5: Phase-‐plane methods and qualita)ve solu)ons •  Vector ﬁelds, nullclines, heteroclinic and homoclinic orbits, construc)ng global •  Phase planes. Is a whole small course! •  Applica)ons: chemostat, more on control of breathing. 10 II Con1nuous processes and ordinary diﬀeren1al equa1ons (ODEs) •  Chapter 6: Applica)ons to popula)on dynamics •  Predator-‐prey systems •  Popula)ons in compe))on •  Popula)on biology of infec)ous diseases 11 II Con1nuous processes and ordinary diﬀeren1al equa1ons (ODEs) •  Chapter 7: Models of molecular events •  Michaelis-‐Menton kine)cs •  Singular perturba)on ideas and quasi-‐steady-‐
state analysis. •  Coopera)ve reac)ons •  Morphogenesis and a molecular switch •  Ac)vator-‐inhibitor and posi)ve feedback systems 12 II Con1nuous processes and ordinary diﬀeren1al equa1ons (ODEs) •  Chapter 8: Limit cycles, oscilla)on and excitable systems •  Hodgkin-‐Huxley nerve conduc)on equa)ons (deriva)on and some analysis) •  FitzHugh Nagumo simpliﬁca)on •  Hopf bifurca)ons •  Oscilla)ons in chemical systems 13 Addi)onal Textbooks •  A Course in Mathema)cal Biology: Quan)ta)ve Modeling with Mathema)cal and Computa)onal (Monographs on Mathema)cal Modeling and Computa)on) (Paperback) •  Gerda de Vries, Thomas Hillen, Mark Lewis, Johannes Muller, Birgi8 Schonﬁsch 14 Addi)onal Textbooks •  An Introduc)on to Mathema)cal Biology, by Linda Allen 15 Addi)onal Textbooks •  Mathema)cal Founda)ons of Neuroscience (Interdisciplinary Applied Mathema)cs) •  G. Bard Ermentrout, •  David H. Terman 16 Addi)onal Textbooks •  Mathema)cal Biology: I. An Introduc)on (Interdisciplinary Applied Mathema)cs) (Pt. 1) by James D. Murray 17 Course sonware •  Matlab (www.mathworks.com) •  Mathema)ca (www.wolfram.com) •  Not a requirement for the course evalua)ons •  Hw solu)ons make use of these packages •  dﬁeld and pplane (phase ﬁeld visualiza)on) (
h8p://math.rice.edu/~dﬁeld/dfpp.html ) 18 Aims of course •  Learn to formulate mathema)cal descrip)ons for biological systems •  Learn techniques for analyzing such descrip)ons •  Develop ability to be thoughoul about where the mathema)cs can be helpful: what does a model tell us about the system that we didn’t know before. 19 Short overview •  Success stories in mathema)cal biology •  S)ll very much an evolving ﬁeld •  Mathema)cal methods used in the life science models •  Keep it rela)vely simple 20 Short overview •  One fundamental property of the biological systems is their high level of complexity, with interac)ons spanning diﬀerent spa)al and temporal )mescales. •  In order to conduct meaningful modeling, the system almost always need to be simpliﬁed, otherwise the mathema)cs become too complex 21 Short overview •  Renewed interest in mathema)cal models in life sciences •  More experimental data: genomics, neural recordings, large-‐scale medical/social studies •  Limita)ons of things one can do in an experimental seqng: mathema)cal models (more powerful computa)onal technologies) 22 Short overview •  Techniques used in this class –  linear diﬀerence equa)ons, –  non-‐linear diﬀerence equa)ons, –  single and systems of ordinary diﬀeren)al equa)ons –  phase plane analysis •  Capture the qualita)ve behavior exhibited by the biological systems 23 Short overview •  The modeling process –  Posing a ques)on –  Determine the important components that aﬀect the behavior of a system. •  A model is then constructed •  Interac)ons between all model components 24 Short overview •  Why bother doing mathema)cal modeling? Framework for understanding the data Account for missing data Conﬁrm/reject hypothesis Make predic)on for untested condi)ons and diﬀerent parameters •  Formulate new hypotheses • 
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25 Short overview •  Models can be simple in nature •  Ignore small details, focus on the large picture •  One cannot have a perfect model (think that it results from ﬁqng data) •  Models can be made increasingly complex at the expense of large-‐scale simula)ons 26 Some simple examples •  Simplest equa)on •  xn+1= xn + 1 – reduces to coun)ng •  Earning a salary •  xn+1= xn + 10k –more elaborate coun)ng •  Investment banking: •  xn+1= xn + 0.05 * xn 27 Some simple examples •  Xn+1 = x0*1.05n+1 •  applies to banking, Ponzi schemes, vampires (see Transylvania) and zombies •  Assumes unlimited growth (right) •  Xn+1 = x0*2n+1 – the chess legend •  h8p://en.wikipedia.org/wiki/Wheat_and_chessboard_problem 28 Success stories in math bio •  Popula)on ecology •  Cycles in the predator-‐prey popula)ons •  Write a simple model for each popula)on 29 Predator-‐prey systems •  Rabbits mul)ply (depends the overall popula)on) •  Rabbits die •  Rabbits are hunted by lynxes •  Lynxes mul)ply (depend on the popula)on size and the available food – rabbits) •  Lynxes die too 30 Neural dynamics •  Hodgkin and Huxley (1950) •  First model that captured the dynamics of cell membrane ac)on poten)als •  Nobel prize, created a new ﬁeld of mathema)cs 31 Development of neural maps •  Sur Lab (MIT) – 2000 •  Rewiring Cortex: Func)onal Plas)city of the Auditory Cortex •  Experiment Model 32 Reac)on-‐diﬀusion pa8erns •  Tiger, Leopard, Giraﬀe, Zebra •  Spa)ally distributed models that reproduce the pa8erns seen in nature (Alan Turing, 1952, morphogenesis) •  Is this the real mechanism? •  Useful to have models that can be tuned to obtain all the observed pa8erns. 33 Epidemiology models • 
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The SIR model, S (for suscep)ble), I (for infec)ous) R (for recovered) 34 Introduc)on •  Let us introduce some simple example as well •  Start at the basics and aim to cover the more complicated examples… •  1.1 BIOLOGICAL MODELS USING DIFFERENCE EQUATIONS -‐ Cell Division 35 1.1 Biological models using diﬀerence equa)ons •  Cell Division – for example bacteria •  Do all cells divide at the same )me? •  Probably not •  To keep things simple we will assume that all cells divide synchronously producing daughter cells 36 1.1 Diﬀerence equa)ons: cell division •  Ques%ons: how many cells exist at a certain moment in )me? •  Answer: Observa)onal study, just do the experiment •  Q: What if the experiment is expensive? •  A: Building a model for guidance is advised. 37 1.1 Diﬀerence equa)ons: cell division •  Some deﬁni%ons: •  Deﬁne the number of cells in each genera)on with a subscript •  M0 ini)al popula)on •  M1 , M2 , … , Mn are respec)vely the number of cells in the ﬁrst, second, ... , nth genera)ons •  How fast is the popula)on increasing? 38 1.1 Diﬀerence equa)ons: cell division •  Mathema)cal model: •  Mn+ 1 = a Mn •  Q: Why do we choose this model? •  A: Simple •  A: Agrees with our understanding of the data 39 1.1 Diﬀerence equa)ons: cell division •  Mathema)cal model: Mn+ 1 = a Mn •  Q: What are the implica)on of the model? •  A: The number of cells in the new genera)ons is completely determined. We can ﬁnd out the general formula by looking at the equa)on in more detail. •  M1 = a M0 (we know M0, so now we know M1) 40 1.1 Diﬀerence equa)ons: cell division •  M1 = a M0; M2 = a M1= a (a M0) = a2 M0 •  By now the procedure is clear •  M3 = a M2= a (a2 M0) = a3 M0 •  In general: Mn = an M0 •  Proof by induc)on 41 1.1 Diﬀerence equa)ons: cell division •  Ini)al popula)on size: M0 •  Parameter of the models: a – number of daughters per cell •  We can compute Mn: the number of cells at genera)on n •  Much faster than doing the experiment 42 1.1 Diﬀerence equa)ons: cell division •  Example: M0 = 100, a = 2 •  M10 = a10 100 = 1024 ·∙ 100 = 102,400 •  What are the long-‐term prospects for this popula)on ? –  Decline –  Explode –  Stay the same 43 1.1 Diﬀerence equa)ons: cell division •  |a|>1 – Explodes (we ignored limi)ng factors) •  Mn increases over successive genera)ons •  |a|<1 (Ex)nct: an → 0 as n→∞) •  Mn decreases over successive genera)ons, •  a = 1 (an = 1, equilibrium, in balance) •  Mn is constant. 44 1.1 Diﬀerence equa)ons: Rabbits!!! •  Nothing reproduces faster than rabbits (this is an exaggera)on) •  Assume one wants to determine the total number of unchecked rabbits raisingrabbits.org 45 1.1 Diﬀerence equa)ons: Rabbits!!! •  The reproduc)on of rabbits captured the Fibonacci’s imagina)on in the year 1202 •  Presumably he could not inves)gate the natural system in detail •  He created a set of rules describing the ideal condi)ons for rabbits breeding •  Then Fibonacci determined the expected number of rabbits as a func)on of )me 46 1.1 Diﬀerence equa)ons: Rabbits!!! •  How many pairs of rabbits can be produced from one pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes produc)ve? •  Rabbits do not start producing un)l they are 2 months old, and then keep on producing forever (or at least beyond the year.) 47 1.1 Diﬀerence equa)ons: Rabbits!!! •  1. You begin with one male rabbit and one female rabbit. These rabbits have just been born. •  2. A rabbit will reach sexual maturity aner one month. •  3. The gesta)on period of a rabbit is one month. •  4. Once it has reached sexual maturity, a female rabbit will give birth every month. •  5. A female rabbit will always give birth to one male rabbit and one female rabbit. •  6. Rabbits never die. 48 1.1 Diﬀerence equa)ons: Rabbits!!! •  Coun)ng the rabbits •  1 1 2 3 5 8 49 1.1 Diﬀerence equa)ons: Rabbits!!! • 
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Deﬁne Rn0 as # of newborn pairs in month n. Rn1 = # 1-‐month old in month n etc. Superscript = age of the rabbits in months Subscript = month in ques)on. •  Note that Rn1 = Rn-‐10. •  number of rabbit pairs of age 1 month at some )me is the number just born the month before. 50 1.1 Diﬀerence equa)ons: Rabbits!!! •  Similarly Rn2 = Rn-‐20 •  the number of rabbit pairs of age 2 month at some )me is the number just born two months before. •  Assump)on: Rn0 = Rn2 + Rn3 + Rn4 + …… un)l the numbers on the RHS are zero. 51 1.1 Diﬀerence equa)ons: Rabbits!!! •  Rn0 = Rn2 + Rn3 + Rn4 + …… •  We seek something that looks like Fibonacci series (since he got famous with this equa)on in the mean)me) •  Rn0 = Rn-‐10 + Rn-‐20 •  The ﬁrst equa)on is very diﬀerent from the second! •  Where does the Fibonacci equa)on come from? 52 1.1 Diﬀerence equa)ons: Rabbits!!! •  Try rewri)ng the equa)on above just with changes indices, using the fact that the above is correct for any n: •  Rn0 = Rn2 + Rn3 + Rn4 + … •  Rn-‐10 = Rn-‐12+Rn-‐13+Rn-‐14 + ... = Rn3 + Rn4 + Rn5 + ... •  Subtract the two equa)ons: Rn0 – Rn-‐10 = Rn2 53 1.1 Diﬀerence equa)ons: Rabbits!!! •  We obtained: Rn0 – Rn-‐10 = Rn2 •  Now use the fact that Rn2 = Rn-‐11 = Rn-‐20 •  Voila: Rn0 = Rn-‐10 + Rn-‐20 •  The Fibonacci series 54 1.1 Diﬀerence equa)ons: Rabbits!!! •  This problem has an alterna)ve formula)on: Problem 14 from pages 32-‐33 • 
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Rabbits reproduce only twice At ages 1 mo and 2 months Produce only one pair of rabbits. None of them die. In genera)on 1, there is one pair. 55 1.1 Diﬀerence equa)ons: Rabbits!!! •  At ﬁrst glance there are fewer rabbits •  Since the rabbits do not reproduce forever •  However these rabbits have a head start •  They start reproducing at an earlier age, namely at month 1 •  Who has the ‘advantage’? 56 1.1 Diﬀerence equa)ons: Rabbits!!! •  Use the same deﬁni)ons •  Deﬁne Rn0 as # of newborn pairs in month n. •  Rn1 = # 1-‐month old in month n etc. •  Superscript stands for the age of the rabbits in months •  Subscript for the month in ques)on. 57 1.1 Diﬀerence equa)ons: Rabbits!!! •  It is easier to write the solu)on for this problem •  The number of newborn pairs equals the number of 1-‐month old and 2 month olds. •  Rn0 = Rn1 + Rn2 = Rn-‐10 + Rn-‐20. •  One adds up the previous numbers of newborns from the past 2 months to get the number of newborns for the current month. •  Done. Fibonacci series 58 1.1 Diﬀerence equa)ons: Rabbits!!! •  We can s)ll count them in a formal way: •  Month 1: Star)ng with one pair of rabbits •  R00 = 1. •  Month 2: The rabbits have mated, but there are no newborns yet: •  R11 = 1, R10 = 0 59 1.1 Diﬀerence equa)ons: Rabbits!!! •  Month 3: A new pair of rabbits is born; the old pair is 2 months old: •  R22 = 1, •  R21 = 0, •  R20 = 1 60 1.1 Diﬀerence equa)ons: Rabbits!!! •  Month 4: A new pair of rabbits is born; the oldest pair is 3 months old, second genera)on is ma)ng: •  R33 = 1, R32 = 0, R31 = 1, R30 = 1 61 1.1 Diﬀerence equa)ons: Rabbits!!! •  Month 5: Two new pairs of rabbits are born •  R44 = 1, •  R43 = 0, •  R42 = 1, •  R41 = 1, •  R40 = 2 62 1.1 Diﬀerence equa)ons: Rabbits!!! •  Month 3: A new pair of rabbits is born; the old pair is 2 months old: •  R22 = 1, R21 = 0, R20 = 1 •  Month 4: A new pair of rabbits is born; the oldest pair is 3 months old, second genera)on is ma)ng: •  R33 = 1, R32 = 0, R31 = 1, R30 = 1 •  Month 5: Two new pairs of rabbits are born •  R44 = 1, R43 = 0, R42 = 1, R41 = 1, R40 = 2 63 1.1 Diﬀerence equa)ons: Rabbits!!! •  Summing up the total number of rabbits gives us the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, … •  Can we solve and obtain an analy)cal formula for this series? •  Formally we have this equa)on: xn+2 = xn+1 + xn •  Where xn = Rn0 64 1.1 Linear diﬀerence equa)ons •  Actually let’s take this one step further and consider the more general equa)on: •  xn+2 + a xn+1 + b xn = 0 •  Solu)on for this equa)on depends on the eigenvalues, which are the roots of the characteris)c equa)on: •  x2 + a x + b = 0 65 ,
1.1 Linear diﬀerence equa)ons •  Compute the roots of the characteris)c equa)on x2 + a x + b = 0 (
µ = (− a +
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− 4b )/ 2
φ = − a − a 2 − 4b / 2
a2
•  Turns out that the complete solu)on is: •  xn = c1 ϕn + c2 µn 66 ,
1.1 Linear diﬀerence equa)ons •  For the rabbit problem a = 1 and b = -‐1, therefore ( )
µ = (1 − 5 )/ 2 ≈ -0.6180
φ = 1 + 5 / 2 ≈ 1.6180
•  Hence the number of rabbits in month n is: •  xn = c1 φn + c2 μn = c1 1.618n + c2 (-‐0.618)n 67 1.1 Linear diﬀerence equa)ons ,
•  The number of rabbits in month n is: •  xn = c1 φn + c2 μn = c1 1.618n + c2 (-‐0.618)n •  Are we done here? •  No, we s)ll need to determine the coeﬃcients c1 and c2 •  Need to use what we know about the ini)al condi)ons: x0 = 1, x1 = 1, x2 = 2 68 ,
1.1 Linear diﬀerence equa)ons •  How to ﬁnd speciﬁc solu)ons, given ini)al condi)ons. •  Get two equa)ons for unknowns c1 and c2 by plugging in for x0 with n = 0 and x1 for n = 1. •  Note that need exactly two terms in the series to get all the terms aner it. 69 ,
1.1 Linear diﬀerence equa)ons •  Write down the system of equa)ons and solve it. ⎧x 0 = c1 µ 0 + c 2 φ 0 ⎧1 = c1 µ 0 + c 2 φ 0
⎨
⎨
1
1
1
1
x
=
c
µ
+
c
φ
1
=
c
µ
+
c
φ
⎩ 1 1
⎩
2
1
2
•  We obtain: ⎧ c1 = (φ − 1) /(φ − µ ) ≈ 0.2764
⎨
⎩c2 = (1 − µ ) /(φ − µ ) ≈ 0.7236
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1.1 Linear diﬀerence equa)ons •  We ﬁnally obtain the following formula: xn = 0.7236 ⋅ 1.618 n + 0.2736 ⋅ ( −0.618)n
1− µ n φ −1 n
xn =
⋅φ +
µ
φ−µ
φ−µ
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1.1 Linear diﬀerence equa)ons •  Can we make addi)onal use of this formula? xn = 0.7236 ⋅ 1.618 n + 0.2736 ⋅ ( −0.618)n
1− µ n φ −1 n
xn =
⋅φ +
µ
φ−µ
φ−µ
•  Yes, we can use it for a quick es)mate for the TOTAL NUMBER of rabbits when n (the number of months) is large 72 ,
1.1 Linear diﬀerence equa)ons •  Quick es)mate at large numbers xn = 0.7236 ⋅ 1.618 n + 0.2736 ⋅ ( −0.618)n
•  Idea: solu)on is c1 ϕn + c2 µn, where µ is nega)ve and < 1 in absolute value. •  Hence, for n large, the second term hardly ma8ers. 73 ,
1.1 Linear diﬀerence equa)ons •  So we have to add up the number of newborns from month 0 thru month N •  Approximately c1(1 + ϕ + ϕ 2+….. ϕ n) = c1 S. •  Geometrical series = (ϕn – 1)/(ϕ -‐ 1) •  Total number of rabbits = c1(ϕn – 1)/(ϕ -‐ 1) 74 ,
1.1 Linear diﬀerence equa)ons •  Also, the exact formula is the sum of two geometrical series. •  The problem is that the second term oscillates between posi)ve and nega)ve values, but decays to zero •  Hence is dominated by the ﬁrst term which grows exponen)ally. 75 ,
1.1 Linear diﬀerence equa)ons •  Other problems with the formula)ons (e. g. future direc)ons) •  Where does the ﬁrst pair come? Wrong ques)on for this class, we just assume it. •  There are gene)cs concerns if we start with one pair for breeding. Let’s assume a larger ini)al popula)on, and not immediate family breeding and this can be addressed. 76 ,
1.1 Linear diﬀerence equa)ons •  Other problems with the formula)ons (e. g. future direc)ons) •  How about the ra)o of male/females in each genera)on being equal to 1? •  This is probably not right, and also the number of oﬀsprings is generally greater than two and varies. •  Puqng randomness in the formulas is a big problem for obtaining analy)cal results. 77 ,
1.1 Linear diﬀerence equa)ons •  According to wiki, •  Humans have a Fisherian sex ra)o. •  In humans the secondary sex ra)o is commonly assumed to be 105 boys to 100 girls 78 ,
1.1 Linear diﬀerence equa)ons •  Tradi)onally, farmers have discovered that the most economically eﬃcient community of animals will have a large number of females and a very small number of males. •  A herd of cows and a few prize bulls or a ﬂock of chickens and one rooster are the most economical sex ra)os for domes)cated livestock. •  h8p://en.allexperts.com/e/s/se/sex_ra)o.htm 79 ,
1.1 Linear diﬀerence equa)ons •  Elephants reproduce well and in approximately even numbers in the wild •  Skewed birth sex ra)o in cap)vity (0.71) •  Skewed birth sex ra)o and premature mortality in elephants, Animal Reproduc;on Science, Volume 115, Issue 1, Pages 247-‐254, J. Saragusty, R. Hermes, F. Göritz, D. Schmi8, T. Hildebrandt 80 ,
1.1 Linear diﬀerence equa)ons •  The Fibonacci Spiral is a geometric spiral whose growth is regulated by the Fibonacci Series. •  Its sudden, almost exponen)al growth parallels the rapid growth of the series itself. •  The spiral itself is a series of connected quarter-‐
circles drawn inside an array of squares with Fibonacci numbers for dimensions. 81 ,
1.1 Linear diﬀerence equa)ons •  The Fibonacci Spiral 82 ,
1.1 Linear diﬀerence equa)ons •  The Fibonacci number: golden ra)o 83 ,
1.1 Linear diﬀerence equa)ons •  Another notable example is human body. •  In human body, the ra)o of the length of forearm to the length of the hand is equal to 1.618, that is, Golden Ra)o. 84 ,
1.1 Linear diﬀerence equa)ons •  Another well-‐known examples on human body are: –  The ra)o between the length and width of face –  Ra)o of the distance between the lips and where the eyebrows meet to the length of nose –  Ra)o of the length of mouth to the width of nose 85 ,
1.1 Linear diﬀerence equa)ons •  Another well-‐known examples on human body are: –  Ra)o of the distance between the shoulder line and the top of the head to the head length –  Ra)o of the distance between the navel and knee to the distance between the knee and the end of the foot –  Ra)o of the distance between the ﬁnger )p and the elbow to the distance between the wrist and the elbow 86 1.1 Linear diﬀerence equa)ons ,
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Fibonacci numbers or pa8erns are found in: Sea Shells Petals on Flowers Sunﬂower seed Heads Pine Cones, Palms Pineapple and other Bromeliads Plant Growth or leaf/petal arrangements in 90% of plants. 87 ,
1.1 Linear diﬀerence equa)ons •  The same happens in many seed and ﬂower heads in nature. •  The reason seems to be that this arrangement forms an op)mal packing of the seeds so that, no ma8er how large the seed head, they are uniformly packed at any stage, all the seeds being the same size, no crowding in the centre and not too sparse at the edges. •  hZp://milan.milanovic.org/math/english/ﬁbon/index.html 88 ,
1.1 Linear diﬀerence equa)ons •  h8p://bri8on.disted.camosun.bc.ca/ﬁbslide/jbﬁbslide.htm 89 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  Linear diﬀerence equa)ons are characterized by the following proper)es •  1. An mth-‐order equa)on: a0 xn + a1 xn-‐1 +… + am xn-‐m = bm •  2. The order m of the equa)on is the number of previous genera)ons that directly inﬂuence the value of x in a given genera)on. 90 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  3. When a0 , a1 , …, am = constant, bm = 0 •  constant-‐coeﬃcient homogeneous linear diﬀerence equa;on •  Solu)ons are of linear combina)ons of: •  xn = C λn 91 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  4. We can solve for λ using the following equa)on: •  a0 λm + a1 λm-‐1 + … + am = 0 •  5. First-‐order equa)on has one solu)on •  Second-‐order equa)on has two solu)ons. •  In general, an mth-‐order equa)on has m basic solu)ons. 92 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  6. The general solu)on is a linear superposi)on of the m basic solu)ons of the equa)on •  7. For real values of λ the qualita)ve behavior of a basic solu)on depends on whether λ falls into one of four possible ranges: (provided all values of λ are dis)nct). •  λ ≥ 1, λ ≤ -‐1, 0 < λ < 1, -‐1 < λ < O. 93 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  (a) For λ > 1, λn grows as n increases; •  thus xn = C λn grows without bound. 94 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  (b) For 0 < λ < 1, λn decreases to zero with increasing n; thus xn decreases to zero 95 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  (c) For -‐1 < λ < 0, λn oscillates between posi)ve and nega)ve values while declining in magnitude to zero. 96 ,
1.6 QUALITATIVE BEHAVIOR OF SOLUTIONS TO LINEAR DIFFERENCE EQUATIONS •  (d) For λ < -‐1, λn oscillates as in (c) but with increasing magnitude. 97 ,
1.1 Linear diﬀerence – segmental growth •  Problem 1: Growth of Segmental Organisms •  Page 26 from Edelstein-‐Keshet book (also Problem 15/page 33) •  Hypothe)cal situa)on arises in organisms such as certain ﬁlamentous algae and fungi that propagate by addi)on of segments. •  The rates of growth and branching may be complicated func)ons of densi)es, nutrient availability, and internal reserves. 98 ,
1.1 Linear diﬀerence – segmental growth •  Simpliﬁed version of this phenomenon for illustra)on purposes •  Prove versa)lity of diﬀerence equa)on models. •  A segmental organism grows by adding new segments at intervals of 24 hours in several possible ways 99 ,
1.1 Linear diﬀerence – segmental growth •  1. A terminal segment can produce a single daughter with frequency p, thereby elonga)ng its branch. •  2. A terminal segment can produce a pair of daughters (dichotomous branching) with frequency q. •  3. A next-‐to-‐terminal segment can produce a single daughter (lateral branching with frequency r. 100 ,
1.1 Linear diﬀerence – segmental growth •  The ques)on to be addressed is how the numbers of segments change as this organism grows. •  In approaching the problem, it is best to deﬁne the variables depic)ng the number of segments of each type (terminal and next-‐to-‐terminal), to make several assump)ons and to account for each variable in a separate equa)on. 101 ,
1.1 Linear diﬀerence – segmental growth •  Let us use the following nota)ons •  an = number of terminal segments, •  bn = number of next-‐to-‐terminal segments, •  Sn = total number of segments. 102 ,
1.1 Linear diﬀerence – segmental growth •  Assump%ons: •  All daughters are terminal segments; •  All terminal segments par)cipate in growth (p + q = 1) and thereby become next-‐to-‐terminal segments in a single genera)on; •  All next-‐to-‐terminal segments are thereby displaced and can no longer branch aner each genera)on. 103 ,
1.1 Linear diﬀerence – segmental growth •  Show that equa)ons for an and bn can be combined to give: an+l -‐ (1 + q)an – r an-‐1 = 0, •  Explain the equa)on. •  p = frac)on of end segments that have one bud next genera)on; •  q = frac)on with 2 buds on end segments; p+q = 1. •  r = frac)on of next to end segments that bud 104 ,
1.1 Linear diﬀerence – segmental growth •  The growth equa)ons then becomes: •  an+1 = an p + 2 an q + bn r •  bn+1 = an •  sn+1 = sn + an p + 2 an q + bn r 105 ,
1.1 Linear diﬀerence – segmental growth •  Solving for an yields: an+1 = (p + 2q) an + r an-‐1 •  Using p + q = 1, it follows that the equa)ons for an and bn can be combined to give: •  an+l -‐ (1 + q) an – r an-‐1 = 0 •  Use ini)al condi)ons: a1 = 1, b1 = 0, so a0 = 0. 106 1.1 Linear diﬀerence – segmental growth ,
•  A numerical example for p = ½, q = ½ and r = ½. • 
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 
an 1.0000
1.5000
2.8000
4.9000
8.7000
15.5000
27.5000
49.1000
87.4000
155.6000
bn
Sn 0
1.0000 1.5000 2.8000 4.9000 8.7000 15.5000
27.5000
49.1000
87.4000
1.0000 2.5000 5.3000 10.1000 18.8000 34.3000 61.8000 110.9000 198.2000 353.8000 107 ,
1.1 Linear diﬀerence – segmental growth •  A numerical example for p = ½, q = ½ and r = ½. 108 ,
1.1 Linear diﬀerence – segmental growth •  Example: Simula)ng colonial growth of fungi with the Neighbour-‐Sensing model of hyphal growth, Audrius Meškauskas1, Mark D. Fricker2 and David Moore1, Mycological Research, Volume 108, Issue 11, November 2004, Pages 1241-‐1256 109 Topics covered so far •  Introduc)on in mathema)cal models in life sciences •  Linear diﬀerence equa)on –  Cell division –  Rabbits problem (Fibonacci numbers) –  Segmental growth model •  Matlab overview –  Obtaining and ploqng solu)ons with Matlab 110 1.1 Linear diﬀerence – Red blood cells •  A Schema;c Model of Red Blood Cell Produc;on •  number of red blood cells (RBCs) circula)ng in the blood – approx 2–3 × 1013 according to wiki •  25% of the total human body cell number •  Each circula)on takes about 20 seconds •  Each cell lives approx 100–120 days in the body 111 1.1 Linear diﬀerence – Red blood cells •  A Schema;c Model of Red Blood Cell Produc;on •  number of red blood cells (RBCs) circula)ng in the blood – approx 2–3 × 1013 according to wiki •  25% of the total human body cell number •  Each circula)on takes about 20 seconds •  Each cell lives approx 100–120 days in the body 112 1.1 Linear diﬀerence – Red blood cells •  Re)culoendothelial system (spleen, liver and bone marrow) •  Removes old or defec)ve cells, con)nually replenishing the cell count •  Normally occurs at the same rate of produc)on, balancing the total numbers. h8p://www.nature.com/nature/journal/v420/n6917/ﬁg_tab/nature01321_F1.html 113 1.1 Linear diﬀerence – Red blood cells •  Blood doping – collect blood cells, freeze and reintroduce them into the system, diﬃcult to detect •  Risk of infec)ons, increase viscosity of blood (may lead to heart failure) •  Replaced by gene)c engineering: Athletes may simply inject erythropoe)n (EPO), which causes the body to make the cells •  Iden)cal to natural chemicals made by the body -‐ making sure detec)on diﬃcult or impossible •  h8p://whyﬁles.org/090doping_sport/3.html 114 1.1 Linear diﬀerence – Red blood cells •  In the circulatory system, the red blood cells (RBCs) are constantly being destroyed and replaced. •  Since these cells carry oxygen throughout the body, their number must be maintained at some ﬁxed level. •  Assume that the spleen ﬁlters out and destroys a certain frac)on of the cells daily and that the bone marrow produces a number propor)onal to the number lost on the previous day. 115 1.1 Linear diﬀerence – Red blood cells •  Ques1ons: What is the number of red cells at a certain moment in )me? •  Answer: Build a mathema)cal model •  1) Assume discrete )me •  1 basic unit of )me = 1 day •  Deﬁne variables 116 1.1 Linear diﬀerence – Red blood cells •  2) Variables: •  Rn = # of RBCs in circula)on on day n, •  Mn = # of RBCs produced by marrow on day n •  3) Parameters •  f = frac)on RBCs removed by spleen, •  γ = produc)on constant (# produced per # lost). 117 1.1 Linear diﬀerence – Red blood cells •  4) variables and parameters → equa1ons •  Rn+1 = Rn – f Rn + Mn •  RBC in day n + 1 = RBC in day n – RBC removed by spleen in day n + RBC produced by marrow in day n •  Mn + 1= γ f Rn •  RBC produced by marrow in day n + 1 = frac)on RBCs removed by spleen x marrow RBC in day n 118 1.1 Linear diﬀerence – Red blood cells •  Ader simpliﬁca1ons, we obtain the following system of equa1ons: •  Rn+1 = (1 – f) Rn + Mn •  Mn + 1= γ f Rn •  A coupled (Rn+1 depends on Mn) system of two linear 1st order diﬀerence equa)ons 119 1.1 Linear diﬀerence – Red blood cells •  Q: How do we solve this system of eqs? •  A: One op)on is to use subs)tu)on. •  Rn+1 = (1 – f) Rn + Mn •  Mn + 1= γ f Rn •  Rn+2 = (1 – f) Rn+1 + Mn+1 = (1 – f) Rn+1 + γ f Rn •  Second order linear diﬀerence equa;on 120 1.1 Linear diﬀerence – Red blood cells •  We had a similar example for the segmental growth problem (an+l -‐ (1 + q) an – r an-‐1 = 0) •  How do we solve this type of equa)on: •  xn+l + a xn + b xn-‐1 = 0 •  Guess a solu)on: xn = c λn 121 1.1 Linear diﬀerence – Red blood cells • 
• 
• 
• 
• 
• 
• 
xn+l + a xn + b xn-‐1 = 0 Guess a solu)on: xn = c λn Subs)tute in: c λn+1 + a c λn + b c λn-‐1 = 0 The parameter c cancels out λn+1 + a c λn + b c λn-‐1 = 0 λn-‐1 (λ2 + a λ + b)= 0 The characteris)c equa)on corresponding to the general equa)on is: λ2 + a λ + b= 0 122 1.1 Linear diﬀerence – Red blood cells •  For our problem: •  Rn+2 = (1 – f) Rn+1 + Mn+1 = (1 – f) Rn+1 + γ f Rn •  a = -‐(1 -‐ f), b = -‐ γ f •  We then obtain (1 − f ) ± (1 − f )2 + 4γ f
λ 1,2 =
2
•  The general solu)on is then: Rn = C1 λ1n + C2 λ2n 123 1.1 Linear diﬀerence – Red blood cells •  An analy)cal solu)on! 2
n
2
⎛ (1 − f ) + (1 − f ) + 4γ f ⎞
⎛ (1 − f ) − (1 − f ) + 4γ f ⎞
⎟ + C ⎜
⎟
R = C1 ⎜
2
⎜
⎟
⎜
⎟
2
2
⎝
⎠
⎝
⎠
n
• 
• 
• 
• 
Diﬀerence equa)on is linear C1 λ1n and C2 λ2n are solu)ons Their sum is also a solu)on We can determine C1 and C2 from R0 and R1 124 n
1.1 Linear diﬀerence – Red blood cells •  Let’s determine C1 and C2 from R0 and R1 ⎧R 0 = c1 λ10 + c2 λ02
⎨
1
1
R
=
c
λ
+
c
λ
⎩ 1 1 1 2 2
c2 = R 0 − c1
⎧ R 0 = c1 + c2
⎧
⎨
⎨
⎩R1 = c1 λ1 + c2 λ2 ⎩R1 = c1 λ1 + (R 0 - c1 ) λ2
c2 = c1 - R 0
⎧ c 2 = c1 - R 0
⎧
⎪
⎨
R 1 − R 0λ2
⎨
R
=
c
(
λ
−
λ
)
+
R
λ
2
0 2
⎩ 1 1 1
c1 =
⎪⎩
λ1 − λ2
⎧
R 1 − R 0λ2
⎪⎪c 2 = R 0 - λ − λ
1
2
⎨
⎪ c1 = R 1 − R 0λ2
⎪⎩
λ1 − λ2
125 1.1 Linear diﬀerence – Red blood cells •  Finally the solu)on for C1 and C2 ⎧
- R 1 + R 0λ1
⎪⎪c 2 = λ − λ
1
2
⎨
⎪ c1 = R 1 − R 0λ2
⎪⎩
λ1 − λ2
•  Use numerical example with f = 0.01, γ =1, R0 = 1000, M0 = 10 126 1.1 Linear diﬀerence – Red blood cells •  Numerical example •  f = 0.1, γ =1, R0 = 1000, M0 = 10 •  Run matlab script rbc.m 127 1.1 Linear diﬀerence – Red blood cells •  Let’s look at another way to solve this equa)on •  The new approach is more general and will work for more complicated cases •  We’ll think about the system as two 1st order equa)ons (We don’t combine) •  We’ll use concepts from linear algebra •  Linear algebra review 128 1. 4 Linear algebra review •  Consider the following system of linear algebraic equa)on a11 x +a12 y = 0 ⎫
⎬ (1)
a21 x +a22 y = 0⎭
•  Note we don’t have diﬀerence equa)ons here (that is no n, n+1 indices, etc) •  Rewrite in vector nota)ons 129 1. 4 Linear algebra review •  Two by two matrix mul)plied by 2x1 vector = 2x1 vector ⎛ a11 a11 ⎞⎛ x ⎞ ⎛ 0 ⎞
~
~
~
⎜
⎟
=
⎜
⎟
⎜
⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Re-‐express as: M v = 0
⎝ a21 a22 ⎠⎝ y ⎠ ⎝ 0 ⎠
~ ⎛ a11 a11 ⎞
⎟⎟
M = ⎜⎜
⎝ a21 a22 ⎠
⎛ x ⎞
~
v = ⎜⎜ ⎟⎟
⎝ y ⎠
~ ⎛ 0 ⎞
0 = ⎜⎜ ⎟⎟
⎝ 0 ⎠
130 1. 4 Linear algebra review •  Q: What are the solu)ons? ⎛ 0 ⎞
~
•  A: There is one that is trivial v = ⎜⎜ ⎟⎟
⎝ 0 ⎠
•  Q: Do others exist? •  A: Yes, but the system of equa)ons must be redundant 131 1. 4 Linear algebra review •  a11 x + a12 y = 0
&
a21 x + a22 y = 0 •  y = -‐ a11/a12 x &
y = -‐a21/a22 x •  y = -‐ a11/a12 x = -‐a21/a22 x •  If x ≠ 0, then a11/a12 = -‐a21/a22 •  Or: a11a22 – a12a21 = 0 132 1. 4 Linear algebra review •  This must hold to sa)sfy our equa)ons simultaneously: •  a11a22 – a12a21 = 0 •  Q: Can we reinterpret this result? ⎛ a11 a12 ⎞
~
⎟⎟ = a11a22 − a12a21
•  A: det( M ) = det⎜⎜
⎝ a21 a22 ⎠
133 1. 4 Linear algebra review •  In order to have a NON-‐TRIVIAL solu)on to the system of equa)ons (1), we must have ~
det( M ) = 0
•  Q: Can we apply this to our system of lin eqs? xn +1 = a11 xn + a12 yn
yn +1 = a21 xn + a22 yn
134 1. 4 Linear algebra review •  A: Re-‐write in matrix/vector nota)on ⎛ xn ⎞
~
vn = ⎜⎜ ⎟⎟
⎝ yn ⎠
~~
~
vn +1 = M vn
•  Similar to the one-‐dimensional case let’s guess that n
~ ⎛ A
λ ⎞ , where A and B depend ⎜⎜ n ⎟⎟
v
=
n
B
λ
on the ini)al condi)ons ⎝
⎠
135 1. 4 Linear algebra review •  Now we need to ﬁnd λ •  Subs)tute in: n +1
n
n
⎛ Aλ ⎞ ~ ⎛ Aλ ⎞ ⎛ a11 a12 ⎞⎛ Aλ ⎞
⎜⎜ n +1 ⎟⎟ = M ⎜⎜ n ⎟⎟ = ⎜⎜
⎟⎟⎜⎜ n ⎟⎟
⎝ Bλ ⎠
⎝ Bλ ⎠ ⎝ a21 a22 ⎠⎝ Bλ ⎠
•  A λn+1 = a11 A λn + a12 B λn •  B λn+1 = a21 A λn + a22 B λn 136 1. 4 Linear algebra review •  Aner canceling out λn, we obtain: •  A λ = a11 A + a12 B •  B λ = a21 A + a22 B •  A (a11 – λ) •  A a21 + + B a12 = 0 B (a22 – λ) = 0 137 1. 4 Linear algebra review •  Rewrite & reduce the system linear equa)ons to a system of linear algebraic equa)ons: a12 ⎞⎛ A ⎞
⎛ 0 ⎞ ⎛ a11 − λ
⎟⎟⎜⎜ ⎟⎟
⎜⎜ ⎟⎟ = ⎜⎜
a22 − λ ⎠⎝ B ⎠
⎝ 0 ⎠ ⎝ a21
•  Now we either have the trivial solu)on: a12 ⎞
⎛ a11 − λ
⎛ A ⎞ ⎛ 0 ⎞
⎟⎟ = 0
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ OR det⎜⎜
a21
a22 − λ ⎠
⎝
B
0
⎝ ⎠ ⎝ ⎠
138 1. 4 Linear algebra review •  The zero determinant can be expressed as: •  (a11 -‐ λ)(a22 -‐ λ) – a12 a21 = 0 •  λ2 -‐ (a11 + a22) λ + a11 a22 – a12 a21 = 0 •  λ2 -‐ β λ + γ= 0 β ± β 2 − 4γ
•  With solu)ons: λ 1, 2 =
2
139 1. 4 Linear algebra review •  Q: what are the values λ •  A: the eigenvalues of matrix M 2
λ 1, 2 =
β ± β − 4γ
2
~~
~
•  In conclusion, given v n + 1 = M
v n + 1 , there exist n~
~
solu)ons of the form v n = λ
v 0 , where λ is an ~
eigenvalue of matrix M
140 1. 4 Linear algebra review •  Note: each eigenvalue has an associate eigenvector: ~~
~~
~
M v1 = λ1 v1 M v2 = λ2 v~2
⎛ 2 2 ⎞
~
•  Example: M = ⎜⎜
⎟⎟
⎝1 / 2 2 ⎠
•  Compute eigenvalues and eigenvectors 141 1. 4 Linear algebra review ~ ⎛ 2 2 ⎞
⎟⎟
•  Example: M = ⎜⎜
⎝1 / 2 2 ⎠
•  Compute eigenvalues and eigenvectors 2 ⎞
⎛ 2 2 ⎞
⎛ 1 0 ⎞ ⎛ 2 − λ
~
det( M − λI ) = ⎜⎜
⎟⎟ − λ ⎜⎜
⎟⎟ = ⎜⎜
⎟⎟ = 0
⎝1 / 2 2 ⎠
⎝ 0 1 ⎠ ⎝ 1 / 2 2 − λ ⎠
2
2
(2 − λ ) − 1 = λ − 4λ + 3 = 0 ⇒ λ1 = 1, λ2 = 3
142 1. 4 Linear algebra review ~~
•  Q: What are the eigenvectors? M v2 = λ2v~2
•  A: Write the equa)ons and solve for a, b ⎛ 2 2 ⎞⎛ a ⎞ ⎛ a ⎞
⎜⎜
⎟⎟⎜⎜ ⎟⎟ = 3⎜⎜ ⎟⎟
⎝1 / 2 2 ⎠⎝ b ⎠ ⎝ b ⎠
2 a + 2 b = 3 a , ⇒
2 b =
a ⎫ Redundant, just ⎪
1 ⎬ choose a = 2, b = 1 a + 2b = 3b, ⇒ 1 / 2a = b ⎪
2 ⎭
143 1. 4 Linear algebra review ~~
•  Q: What are the eigenvectors? M v1 = λ1v~1
•  A: Write the equa)ons and solve for a, b ⎛ 2 2 ⎞⎛ a ⎞ ⎛ a ⎞
⎜⎜
⎟⎟⎜⎜ ⎟⎟ = 1⎜⎜ ⎟⎟
⎝1 / 2 2 ⎠⎝ b ⎠ ⎝ b ⎠
2 a + 2 b =
a , ⇒
−
2 b = a ⎫⎪ Redundant, just 1 a + 2 b = b , ⇒
1 / 2 a = − b ⎬ choose a = -‐2, b = 1 ⎪⎭
2
144 1. 4 Linear algebra review •  In summary M has the following eigenvectors and eigenvalues: Eigenvalues Eigenvectors 1 λ1 = 1 ⎛ 2 ⎞
⎜⎜ ⎟⎟
⎝ 1 ⎠
2 λ2 = 3 ⎛ − 2 ⎞
⎜⎜ ⎟⎟
⎝ 1 ⎠
145 1. 4 Linear algebra review •  Graphical representa)on •  Any vector in the plane is a linear combina)on of the eigenvectors ⎛ x ⎞
~
x = ⎜⎜ ⎟⎟ = c1v~1 + c2v~2 ; c1 = c2 = 1 / 2
⎝ y ⎠
146 1. 4 Linear algebra review ⎛ 0 ⎞
~
•  E. g. x = ⎜⎜ ⎟⎟ → ⎝ 1 ⎠
⎛ 0 ⎞
⎛ 2 ⎞
⎛ − 2 ⎞
⎜⎜ ⎟⎟ = c1 ⎜⎜ ⎟⎟ + c2 ⎜⎜ ⎟⎟
⎝ 1 ⎠
⎝ 1 ⎠
⎝ 1 ⎠
•  0 = 2c1 – 2c2 => c1 = c2 •  1 = c1 + c2, 1 = 2 c1 => c1 = c2 = ½ ⎛ 2 ⎞
⎛ − 2 ⎞ ⎛ 0 ⎞
•  Check: 1 / 2⎜⎜ ⎟⎟ + 1 / 2⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ 1 ⎠
⎝ 1 ⎠ ⎝ 1 ⎠
147 1. 4 Linear algebra review •  Matlab code: •  [eigenvalues, eigenvectors] = eigs([2 2; 1/2 2]) •  eigenvalues = 0.8944 -‐0.8944 0.4472 0.4472 •  eigenvectors = 3 0 0 1 148 1. 4 Linear algebra review •  Q: Why is this useful? ~~
~
•  A: Remember vn +1 = M vn
•  To solve, one needs to compute the eigenvectors and eigenvalues for matrix M 149 1. 4 Linear algebra review •  Then, given ini)al condi)ons we can write ~
~
~
v0 = C1v− + C2v+
~~
~ ~
~
v1 = M v0 = M (C1v− + C2v~+ ) =
~~
~~
~
~
C Mv + C Mv = C λ v + C λ v
1
−
2
+
1 1 −
2 2 +
150 1. 4 Linear algebra review •  For the second itera)on: ~~
~
~
v2 = M v2 = M (C1λ1v~− + C2λ2v~+ ) =
~~
~~
2~
2~
C λ Mv + C λ Mv = C λ v + C λ v
1 1
−
2 2
+
1 1
−
2 2
+
•  By induc)on: n~
n~
~
vn = C1λ1 v− + C2λ2 v+
151 1. 4 Linear algebra review •  Q: Will the solu)on grow/decay? •  A: Depends on the speciﬁc eigenvalues. •  E. g. 1 < λ1 < λ2. Let n →∞ n~
n~
~
vn = C1λ1 v− + C2λ2 v+ =
n
⎛ λ1 ⎞
n~
n~
~
⎜
⎟
λ2 (C1 ⎜ ⎟ / λ2 v− + C2v+ ) ≈ C2λ2 v+
⎝ λ2 ⎠
n
152 1. 4 Linear algebra review Remember the system of equa)ons: Rn+1 = (1 – f) Rn + Mn Mn + 1= γ f Rn ⎛ 2 ⎞
~
Assume that one of the eigenvectors is v+ = ⎜⎜ ⎟⎟
⎝ 1 ⎠
⎛ Rn ⎞ ~
n~
n ⎛ 2 ⎞
⎜⎜
⎟⎟ = vn ≈ C2λ2 v+ = C2λ2 ⎜⎜ ⎟⎟
⎝ 1 ⎠
⎝ M n ⎠
• 
• 
• 
• 
•  Then Rn/Mn = 2/1 = 1, that is, ﬁxed ra)o 153 1. 4 Linear algebra review •  E. g. 0 < λ1 < λ2< 1. Let n →∞ n~
n~
~
vn = C1λ1 v− + C2λ2 v+ → 0
•  In general if magnitude of both is less than 1, solu)ons decay •  If at least the magnitude of at least one eigenvalues is more than 1 it explodes 154 Red blood cell example revisited •  Rn+1 = (1 – f) Rn + Mn •  Mn + 1= γ f Rn •  Use linear algebra instead of combining the equa)ons ⎛ Rn +1 ⎞ ⎛1 − f
⎜⎜
⎟⎟ = ⎜⎜
⎝ M n +1 ⎠ ⎝ γ f
1 ⎞⎛ Rn ⎞
⎟⎟ =
⎟⎟⎜⎜
0 ⎠⎝ M n ⎠
~⎛ Rn ⎞
⎟⎟
A⎜⎜
⎝ M n ⎠
155 Red blood cell example revisited •  Q: What are the eigenvalues of matrix A? •  A: compute the determinant ⎛1 − f − λ
det ⎜⎜
⎝ γ f
1 ⎞
⎟⎟ = 0
− λ ⎠
•  (1 – f -‐ λ)(-‐λ) – γ f = 0 •  λ2 – λ (f – 1) + γ f= 0 156 Red blood cell example revisited •  Approach 1 and 2 yield iden)cal results 2
1 − f ± (1 − f ) + 4γ f
λ1,2 =
2
•  Can the # RBC be constant in the future? •  Under what condi)ons? 157 Red blood cell example revisited •  The larger eigenvalue is 1 2
1 − f ± (1 − f ) + 4γ f
1=
2
1 + f = (1 − f )2 + 4γ f
1 + 2 f + f 2 = 1 − 2 f + f 2 + 4γ f
•  True if γ = 1 158 Red blood cell example revisited •  Easy to check that: •  λ1 = -‐f, λ2 = 1 •  f is the frac)on of RBC remove (0 < f < 1) •  Remember that solu)on are on the form: n~
n~
~
vn = C1λ1 v− + C2λ2 v+
159 Red blood cell example revisited •  It is easy to see now that: n~
n~
n~
n~
~
vn = C1λ1 v− + C2λ2 v+ = C11 v− + C2 (− f ) v+ ≈ C1v~−
•  The resultant state is now constant, and depends only on the ini)al condi)ons 160 Red blood cell example revisited •  Stable solu)ons requires γ = 1 •  It is true that our body is tuned so that the # RBC produced in the marrow per day is the same lost in the spine (so that γ = 1)? •  Probably not •  What happens in γ < 1, or γ >1 161

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