QUALITATIVE ANALYSIS OF DIFFERENTIAL EQUATIONS. Alexander Panfilov

Transcription

1 QUALITATIVE ANALYSIS OF DIFFERENTIAL EQUATIONS Aleander Panfilov

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3 QUALITATIVE ANALYSIS OF DIFFERENTIAL EQUATIONS Aleander Panfilov Theoretical Biolog, Utrecht Universit, Utrecht c 010 Februar 3, 010

4 Tentative plan for tutorials for 7 da course Da 1 Chapter 1: 1ab,ab,3acd,4c,5abcd Optional: Chapter 1: 3b,4a Da Chapter 1: 6ab,7abc,8abcdef Chapter : 1ab Chapter 1: 3e,4de Da 3 Chapter : ab,3abc,4abcdefgh,8ab,6ab Chapter 1: 8g Optional: Chapter 1: 15f Da 4 Chapter 3: 3abcde,4 Chapter 1: 8h Optional: Chapter 3: 3f,5 Da 5 Chapter 5: 5,6,7,8 Optional: Chapter 5: Da 6 Chapter 6: 1abcd,,3,4 Da 7 Chapter 7: 1,ab,3,4 Chapter 8: 1,

5 Contents 1 Preliminaries Basic algebra Algebraic epressions Limits Equations Sstems of equations Functions of one variable Graphs of functions of one variable Implicit function graphs Eercises Selected topics of calculus 3.1 Comple numbers Matrices Eigenvalues and eigenvectors Functions of two variables Eercises Differential equations of one variable Differential equations of one variable and their solutions Definitions Solution of a differential equation Qualitative methods of analsis of differential equations of one variable Phase portrait Equilibria, stabilit, global plan Sstems with parameters. Bifurcations Eercises Sstem of two linear differential equations Phase portraits and equilibria General solution of linear sstem Real eigen values. Saddle, node Saddle; λ 1 < 0;λ > 0, or λ 1 > 0;λ < Non-stable node; λ 1 > 0;λ > Stable node; λ 1 < 0;λ < Phase portraits for comple eigen values: spiral, center General ideas on equilibria with comple eigenvalues Center, spiral

6 4 CONTENTS 4.5 Stabilit of equilibrium Eercises Additional concepts (appendi General solution for comple eigen values Sstem of two non-linear differential equations Introduction and first definitions Phase portrait Equilibria Vector field Linearization of a sstem: Jacobian Determinant-trace method for finding the tpe of equilibrium Eercises Graphical methods to stud sstems of differential equations Null-clines Graphical Jacobian Eercises Plan of qualitative analsis and eamples Plan Eamples Eercises Limit ccle Stable and non-stable limit ccles Dnamics of a sstem with a limit ccle How do limit ccles occur? Eample of a sstem with a limit ccle Eercises Dictionar Hints: Solution of the initial value problem for a linear sstem Equilibria/derivatives Answers for selected eercises and Formulas lists 107

7 Chapter 1 Preliminaries 1.1 Basic algebra Algebraic epressions Algebraic epressions are formed from numbers, letters and arithmetic operations. The letters ma represent unknown variables, which should be found from solutions of equations, or parameters (unknown numbers on which the solutions depend. Below, we review eamples several basic operations which help us to work with algebraic epress of ions. One of the most basic algebraic operations is opening of parentheses, or simplification of epressions. For that we use the following rule: (a + b (c + d = ac + ad + bc + bd note, that here ac means a c, etc., as in algebra the multiplication is often omitted. Eample (open parenthesis: (4 + a( 3 = a 6a Sometimes we use parentheses to factor epressions: Eample (factor epression: a 4 = ( a 4 = 3 (3 + 1 a 4 In man cases we also need to work with fractions: Eample (the same denominator: a ca 3 = a 3 ca 3 Eample (different denominators: a b + a b = a b a a + b a b b = a a ab + b b ab = a3 +b ab Eample (fractions simplifications: a ca 3a = a c 3 To divide a fraction a b b another fraction c d we just need to multipl it b its inverse d c : a b : c d = a b d c = ad bc, Eample (division: 4 7 : a 5 = a = 0 14a = 10 7a Also note that: + z+d z + d 5

8 6 CHAPTER 1. PRELIMINARIES 1.1. Limits We call A a limit of the function f ( when approaches a, if the value of f ( get closer and closer to A when takes values closer and closer to a. We write it formall as: lim f ( = A (1.1 a In man cases, finding the limit is trivial: we just need to substitute the value of = a into our function: Eample: lim 3 = 3 = 8 lim f ( = f (a (1. a Functions which have such propert are called continuous and most of the functions used in biolog are continuous. However, there are several important eception. The first case, which will be the most important for us, is finding of limit of the function when. Finding such limits is important as it gives an asmptotic behaviour of our sstem when the size of a population becomes ver large. Unfortunatel, there is no such number which we can substitute into our function to find a limit using formula (1.. For functions without parameters, we can guess the limit b substituting large values to (1., e.g. = 10000, 0000, etc, but what to do for functions with parameters? Let us discuss this problem for a special class of functions, which are the most relevant to our course, the so called rational functions f ( = p( g(, where p( and g( are polnomials. In that case we can alwas find the limit using the following propert of the power function: where C is an arbitrar constant and α > 0. C lim α = 0 (1.3 To prove it note that if approaches (becomes larger and larger, the power function α with α > 0 also becomes larger and larger and therefore C α will be closer and closer to zero, thus in accordance with the definition lim C α = 0 To find the limit using this rule we need to do the following: (1 find the highest power of in our epression p( g(, ( divide each term in our function b in that power, and (3 find the limit of each term using propert (1.3. Let us consider three tpical eamples: an Eample (find the limit: lim 3N. N 3 N The highest power is N, division gives: a 0 0 = a Eample (find the limit: lim P ap 3bP 3 cp dp, a,b,c,d 0. an N 3 N N = a 3 N 3 N N 3. N N The limits of the individual terms are:

9 1.1. BASIC ALGEBRA 7 Similar steps give us: ap 3bP3 cp dp = ap P 3 3bP3 P 3 = cp P 3 dp P 3 a P 3b c = 0 3b P P d 0 0 = 3b 0 as we cannot divide b zero and we do not have a finite limit for this function. Eample (find the limit: lim a 3 b +c a 4 b = a 3 4 b 4 + c 4 = a 4 4 b 4 a 3 b +c a 4 b, a,b,c 0. a b + c 4 a b 4 = a 0 = 0 a = 0.. This epression does not have sense Another non-trivial situation occurs when the denominator of our function f ( = p( g( becomes zero for some value of, for eample f ( = 3 for = 3. In this case the formula (1. for limit cannot be used and other more careful analsis is necessar. If using of calculator we substitute some numbers into our function around point 3 we will find the following: if becomes closer and closer to 3 from the left, e.g. = 3.1;3.05;3.01,3.005;etc the function value becomes larger and larger, while if becomes closer and closer to 3 from the right, e.g. =.9;.95;.99,.995;etc the function value is negative and its absolute value also becomes larger and larger. We can formall write it as lim = +, while lim =. However, in a strict sense, as there is no real number for which f ( approaches for 3 3 close to 3 (from either side thus the limit here does not eist. We will use limits for drawing our functions and will see that limits at infinit give us horizontal asmptotes of our graphs, while blow up of functions for some (as in the last eample give us vertical asmptotes Equations An equation is a mathematical relationship involving unknown variables. These unknowns are usuall epressed b letters,, however in biolog we use man other letters (e.g. N, P. T, V, etc., which mabe somewhat related to the name of the species the describe. Solving equations means finding unknown(s such that after substitution in the equation the left and right hand sides will be equal to each other. For eample: equation 16 = 10 has a solution = 3, as 3 16 = 6 16 = 10. The usual wa to solve equations which have unknown variables in the first power onl (linear equations, is to isolate the unknowns: = [known numbers] We can achieve that b using the following rules of equation algebra: (1 we can multipl, or divide both sides of the equation b the same number, and ( we can move numbers/epressions from one to the other side of the equation, b changing their sign. The proof of these rules is trivial. Indeed if two epressions are the same X = Y, then if we multipl (or divide both of them b the same number a, the still will be the same ax = ay. Similarl if X = Y + a, we can add a to the both sides of the equation, which will not change the equalit, but we get: X a = Y + a a, or X a = Y. Thus we see, that we were able to move a from the right hand side to the left hand side of our equation, but it changed its sign as a result. Eample(solve equation: 4 = 4. Solution: + 4 = 4 = = 1

10 8 CHAPTER 1. PRELIMINARIES For a quadratic equation a + b + c = 0 we use the abc formula, which gives us the solutions as 1, = b± b 4ac a. Eample (solve equation: + 1 = Solution: (1+( +1 = = 4; = 0; + 3 = 0 from the abc formula 1, = ± ( 3 = ± 16 = ±4 1 = 1; = 3. Equation ma also contain parameters. A parameter is an unknown number (constant that ma have an value. It is different from the unknown variable, as the parameter is just a constant on which our solution depends. Eample (solve equation: a k P dp = 0, where P in unknown variable and a,k,d > 0 are parameters. Solution: B multipling both sides b P we get ak dp P = 0, or ak = dp P 3, or P 3 = ak d, thus P = 3 ak d. We see that the solution depends on 3 parameters and if someone provides us with their values we will be able to find the solution b substituting the parameter values into the final formula Sstems of equations To solve a sstem of two linear equations we epress one variable via the other and substitute it into the other equation. { + = 5 Eample (solve the sstem of equations: + = 3 Solution: From the second equation we find = 3, so we substitute into the first equation: (3 + = 5; 6 + = 5; = 1; = 1, now substitute this value to = 3 and find = 3 1, thus the solution is =, = 1. Unfortunatel, there are no general rules to solve a sstem of nonlinear equations. The usual practical wa is to start with a more simple equation, tr to obtain from it as much information as possible and then substitute it to the other equation. It is also ver helpful to factor epressions in order to simplif them. { n n Eample (solve the sstem of equations: np = 0 np p = 0 Solution: From the second equation b factoring we find np p = p(n = 0. The product is zero onl if one of the multipliers is zero, thus we have two possibilities p = 0 or n =. If we substitute p = 0 into the first equation we find n n 0 = 0, or n(1 n = 0, thus for p = 0 we have two solutions n = 0, or n = 1; now substitute n = into the first equation: 4 p = 0, 4 8 4p = 0, 4p = 4, thus for n = we found p = 1. Overall, we found the following three solutions of the given sstem (n = 0, p = 0,(n = 1, p = 0,(n =, p = 1. Sstems ma also contain parameters. Eample (solve the sstem: parameters. { an an bnp = 0 np kp = 0, where n, p are variables and a,b,k > 0, are the

11 1.. FUNCTIONS OF ONE VARIABLE 9 Solution: We proceed similarl as in the previous case. From the second equation: np kp = p(n k = 0, thus we have two cases p = 0 or n = k. After substituting p = 0 into the first equation we get: an an 0 = 0, an(1 n 0, thus n = 0, or n = 1; after substituting n = k into the first equation we get: ak ak bkp = 0, ak(1 k = bkp, a(1 k = bp, thus p = a(1 k b. Therefore, we found three solutions: (n = 0, p = 0,(n = 1, p = 0,(n = k, p = a(1 k b. It is eas to see that if we substitute the parameter values a =, b =, k = to these formulas we obtain the solution of the previous problem. Note also, that for sstems with parameters we need to be careful as not all operations are allowed for arbitrar parameter values. In our eample in order to obtain the solution we had to make several divisions b parameters a,b, and k. However we can alwas do that as the parameters are positive numbers (a,b,k > 0 and thus the cannot be equal to zero. Finall note, that we can solve sstems of three and more equations similarl, b subsequent substitutions from one equation to another, etc.. 1. Functions of one variable In science the relationships between quantities are normall epressed using functions. The simplest tpe of functions are functions of one variable. The function of one variable f is a rule that allows us to find the value of a variable (number f from a single variable (number. We denote it as f (. Below are eamples of the most important functions: power functions a, for eample f ( = 1 = ; f ( = = 1. (1.4 polnomials, a c + d, for eample: f ( = (1.5 rational functions f ( = p( g( : trigonometric functions sin, cos, tan: f ( = + 1 (1.6 f ( = sin(; f ( = cos( 1; f ( = tan( (1.7 eponential a and logarithmic function a log( f ( = e +1 ; f ( = 10 log(; f ( = (1.8 etc.

12 10 CHAPTER 1. PRELIMINARIES f( * Figure 1.1: The derivative of a function f ( at point is given b the following limit: f ( = d f d = lim f ( f ( (1.9 The derivative f ( shows the rate of change of a function f ( at a point and has man important applications: If (t is the distance traveled b a car as a function of time t, then d/ gives the velocit of the car. If n(t is the size of a population as the function of time, then dn/ gives the rate of growth of the population. Geometricall, the derivative f ( gives the slope of the tangent line to the graph of the function at the point (fig.1.1. A graph of a line tangent to the function f ( at point (fig.1.1 is given b the following equation: = f ( + f ( ( (1.10 Equation (1.10 is also known as a linear approimation of function f ( at point : Let us check formula (1.10 b approimating the function = + 1 at = 1. We find: f ( = = 3, f = 4, f (1 = 4, hence f ( ( 1. At = 1.1 this approimate formula gives f ( (1.1 1 = 3.4. The eact value is f (1.1 = = 3.4. So the error is just 0.6%. However, if = 0, f ( (0 1 = 1 while the eact value is f (0 = 1. So we see, that the approimate formula works good if is close to onl. Functions with parameters. Functions ma depend not onl on variable(s but also on parameters. We have alread seen the following eample of the function f ( that depends on three parameters a,b,c: f ( = a + b + c (1.11 Equation (1.11 describes a general quadratic polnomial. If we choose, for eample a = 3,b =,c = 1 we will get the function given b equation (1.5. Studing functions with parameters allows us to obtain results for whole classes of functions. We will frequentl use functions with parameters in our course.

13 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 11 This is because biological models usuall depend on man (up to hundreds parameters and in man situations the eact values of these parameters are unknown. One of practical difficulties in working with parameters is that use of calculators is ver limited, because calculators cannot do calculations with unknown quantities. The most valuable methods to stud functions with parameters are direct algebraic computations and analsis of the obtained formulas. In this course we will widel use the graphical methods of representation of function. Let us start with review of the basic function graphs. 1.3 Graphs of functions of one variable Eample of graphs. We usuall represent functions using graphs. To do that we plot the value of the variable along the -ais and the value of the function f ( along the -ais. Let us start first b listing tpical graph shapes that are important in this course. = =p p = 1 a b c Figure 1.: Equation = p produces a horizontal line at the level p (fig.1.. = = 1 =p p a b c Figure 1.3: Equation = p produces a vertical line shifted b p from the -ais (fig.1.3

14 1 CHAPTER 1. PRELIMINARIES = =a (a>0 =3 =0.5 a a =a (a<0 = 1.1 = 0. a b c Figure 1.4: Equation = a + p (linear function produces a straight line with the slope defined b the parameter a: the larger the absolute value of a, the steeper is the slope (fig.1.4. =+ = 1 p =a+p 1 a b c Figure 1.5: The parameter p in = a + p accounts for the vertical shift of the graph fig.1.4. a =a (a>0 = =1. =0.7 = 0.5 = 1.5 a b a c =a (a<0 Figure 1.6: Equation = a produces a parabola, if a > 0 the parabola is opened upward (fig.a,b, and if a < 0 the parabola is opened downward (fig.c. The larger the absolute value of a is, the steeper is the parabola.

15 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 13 =a +c =a +b+c c = b a a b c p =a( p( q q = v b a Figure 1.7: Equation = a + b + c also produces a parabola. Parameter c (fig.1.7a accounts for the vertical shift of the graph. Parameter b accounts for a horizontal shift of the parabola. It is possible to show that the horizontal shift of the parabola is given b a b (fig.1.7b. We can calculate this shift b determining the location of the verte of the parabola which is a point of etremum (maimum or minimum of the function. At this point the derivative of the function to zero (a + b + c = a + b = 0, Thus the coordinate of the verte is given b v = a b, or in other words the (verte of parabola is shifted b v = a b from its central location in (fig.1.7a. Note also, that a parabola ma have up to two points of intersection of the graph with the -ais (zeros of the function. The can be found from the abc formula for roots of the equation a + b + c = 0, and if these roots (p,q are known, the graph can easil be depicted using them (fig.1.7c. Note, that in this case the verte of the parabola is alwas located at the middle between these two roots. = 3 = 3 =a 3 +b +c +d =a( p( q( r p q r a b c Figure 1.8: For a general cubic function = a 3 + b + c + d we have much more possibilities and we will not discuss all of them here. The two basic forms are given b the functions = 3 and = 3 depicted in Fig.1.8a. Important here is the asmptotic behavior of the function at ±. For = 3 we see that goes to + when increases and to when decreases; for = 3 we have the opposite situation. A general graph of = a 3 + b + c + d ma have up to three zeros that can be found from the solution of the equation a 3 + b + c + d = 0, and up to two etrema (fig.1.8b. The etrema are points where the derivative of the function is zero, which in this case results in the following quadratic equation: (a 3 +b +c +d = 3a +b +c = 0. If the zeros of the function (p,q,r are known, the graph can easil be drawn as shown in Fig.1.8c.

16 14 CHAPTER 1. PRELIMINARIES = = 3 = a b c Figure 1.9: Three eamples of graphs of the power function a involving fractional powers are shown in Fig.1.9. If 0 < a < 1 than the graph growth is slower than the function = and is concave downward (in the first quadrant. To draw graph = let us use the graph of parabola = discussed in Fig.g1d5a. If in function = we switch the and we will get =, which is equivalent to = ±. The graph = can be found b switching the and the -ais for the graph of the parabola = in Fig.1.6a and we get a curve depicted in fig.1.9a in which the upper branch corresponds to = (fig.1.9b and the lower branch corresponds to =. Similarl, the graph of the function = 3 (Fig.1.9c can be found b a 90 o rotation of the graph of the function = 3 from Fig.1.8a. = 1 = 1 + b +a a 1 slope a 1 = +a b a b c Figure 1.10: Rational functions p( q( are ver important in theoretical biolog. The graph of the function = 1 (Fig.1.10a has the vertical asmptote ( = 0 and the horizontal asmptote ( = 0. The graph of function = +a 1 + b can be obtained b a shift of the graph = 1 b b units in the (vertical direction and b a units in the (horizontal direction. In this case the vertical asmptote ( at which function goes to infinit is = a, as at this point the denominator in +a 1 equals zero. The horizontal asmptote of this graph is = b, given b lim + b = b. Another rational function = occurs in the classical 1 +a Michaelis-Menten kinetics. Fig.1.10c shows the graph of this function. Because for biological applications and a are alwas considered non-negative ( 0,a > 0, we show the graph in the first quadrant onl. We see that independent of the value of the parameter a the horizontal asmptote is alwas located at = 1, as lim +a +a = 1. The slope of this function at = 0 is given b the function derivative f ( = ( +a = a at = 0, which gives a slope of f (0 = 1 (+a a.

17 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 15 1 λ e t a λ<0 t 1 e λt λ>0 b t π 1 1 c sin( Figure 1.1: Finall in fig.1.1 we show graphs of two other functions that are important in this course e λt and sin(. Note that if t grows the function e λt approaches zero if λ < 0 and diverges to infinit if λ > 0. The function sin( oscillates with a period of π between 1 and +1. π 1 = = 1 + b +a 1 = + a b slope=0 a b c Figure 1.11: Graphs of similar functions involving a second power: = 1 and = 1, are shown in Fig.1.11a,b. +a We see that function = 1 has a graph similar to that of = 1 but located in the first and second quadrants, rather than first and third. One more difference is that function = 1 does not have +a a vertical asmptote, and alwas reaches a maimum at = 0. Function = famous for its ecological applications Hill function = n n +a n +a is an eample of with n =. Its graph (Fig.1.11c has a horizontal asmptote at = 1 (similar to = +a, however, the rate of growth of = for small +a is slower than for = +a : the slope of the tangent line at = 0 here is 0, which can be found from the derivative of this function. Tips on graphs Let us list important rules that ma help to plot graphs of function = f ( with parameters. The graph of the function = f ( + p can be obtained b a vertical shift b p units of the graph of = f (. Eample: In function N+b N N + c, parameter c just shifts the graph of N+b b c units above. Important points of the graph are points at which the graph crosses the -ais (-intercept, given b = f (0, and points where the graph crosses the -ais (zeros of the function, given b f ( = 0. Note, that some graphs do not cross the or the ais and thus do not have -intercepts or zeros. For eample graph of function f ( = 1 (Fig.1.10a does not have finite zeros or -intercepts.

18 16 CHAPTER 1. PRELIMINARIES Eample: For function f (N = N+b N +c, b,c > 0, the -intercept is 0 0+b +c = c. Zeros can be found from N N+b + c = 0, which gives N + c(n + b = 0, or N + cn + cb = 0, or N(1 + c = cb, thus zero is given b the formula N = 1+c cb, which is alwas valid as c > 0. Another important graph feature are asmptotes. To find a horizontal asmptote we need to compute the lim f (. For functions without parameters, ou can tr to compute this limit using calculator b filling in a large numbers 10000, 0000, etc and looking if the function approaches some constant value. For functions with parameter, ou can tr to fill in some reasonable parameter value and tr to find similarl if the asmptote eists, however the best wa here is to find the limit using our plan from section A vertical asmptote is usuall a point where a denominator of a fraction is zero. Not all graphs have asmptotes, for eample graph of function f ( = does not have an vertical or horizontal asmptotes. However, even if the asmptotes are absent it is still useful to understand behavior of the functions at large and show it in the graph. N N N N + N b Eample: For function N+b N + c we can find lim as N N N+b + c = + c = 1 + c = 1 1+ N b c = 1 + c, thus this graph has a horizontal asmptote = 1 + c. The vertical asmptote here is at point where N + b = 0, or line N = b. Several features of the graph can be found from the derivative of the function: a function grows if its derivative f ( > 0, decreases if f ( < 0 and has a local etremum (maimum or minimum if f ( = 0. We do not necessaril need to compute these feature for each graph, but it ma be helpful for some functions. In man applications we will be interested in points of intersection of graphs of two functions f ( and g(. Because at the intersection point functions are equal to each other, such points can be found from the equation f ( = g(. The above mentioned tips are represented graphicall in fig vertical asmptote q(*=0 intercept =f(0 zeros f(*=0 =f(= p( q( horizontal asmptote =lim f( > 8 Figure 1.13: relative etrema df/d=0 Finall let us formulate the main rules for graphing functions with parameters.

19 1.3. GRAPHS OF FUNCTIONS OF ONE VARIABLE 17 Plan for graphing functions with parameters 1 Tr to simplif the function and determine if it belongs to a known class of functions with graphs from (Fig Computer trail: (a Put parameters to reasonable values and plot the graph using a calculator. (b Collect qualitative information such as : number of zeros, eistence of vertical and horizontal asmptotes. (c Var parameter values to see how this changes the shape of the graph. 3 Algebraic approach (note, not all steps ma be possible: (a Find -intercept ( f (0, and zeros of the function ( f ( = 0. (b Find horizontal asmptote from the limit = lim f ( and vertical asmptote(s (for rational function p( q( the are at the points where the denominator becomes zero (q( = 0 ( fig (c Find other special points (e.g. maimum, minimum, etc, if the are important determinants of the graph shape. (d Draw the graph and indicate how the graph shape changes for different parameter values. Eample Plot the graph of the function f ( = a 0 a > 0 c > 0. Find how the graph depends +c on the parameters a and c Solution. 1 We do not need to simplif the function. The function equation has some similarities with graph classes listed above, but does not coincide eactl with an of them. a Let us put a = c = 1 and plot the graph using calculator (fig.1.14a. a The graph (fig.1.14a has the following characteristic features: the -intercept here is = 0, we see one zero of the function = 0. If we fill in large values of, we find that f (1000 = , f (5000 = and f (10000 = , thus we epect to have a horizontal asmptote = 0, we do not see an vertical asmptotes and function has an etremum point (maimum. However, will these features persist for other parameter values? In order to answer that let us perform an algebraic stud. 3a -intercept is f (0 = a 0 0 +c = 0. Zeros of the function are given b a +c = 0, which has onl one solution = 0, as + c 0 for all. 3b The function does not have vertical asmptotes as the denominator cannot be zero ( + c > 0 for all and c > 0. To find the horizontal asmptote let us compute = lim = 0, thus the horizontal asmptote is the -ais. a +c = lim a + c =

20 18 CHAPTER 1. PRELIMINARIES 3c Important point here is the location of the maimum of the function. Let us find it. For that let us find the points where the derivative of the function is zero. The derivative of the function is f ( = a ( +c a = ac a = 0, thus the epression is zero if ac a = 0, or = ±c. For ( +c ( +c 0,c > 0 we have just one solution = c. The value of the function at this etreme point is f (c = ac = a c +c c, thus the maimum is at (c, c a. 3d Let us draw the graph now. Because f (0 = 0 the graph alwas goes through the origin. Then the graph will reach the maimum at (c, c a (Fig.1.14a, smbol 1 and then approaches the ais. If we put all this information together we obtain a qualitative graph shown in fig.1.14b. We see that it qualitativel coincides with the calculator sketch. Now let us find how the graph shape depends on the parameter values. We see that the graph has a bell-shape, with a single maimum at (c, c a. The location of this maimum depends on the parameter c onl, but the maimal value of the function increases if a increases. The solid and the dashed line in Fig.1.14c illustrate how the graph shape changes if a increases while we keep c constant. Alternativel, if we keep the a value constant but increase the value of the parameter c (the solid and the dot-dashed line in fig.1.14c, the location of the maimum shifts to the right, and the maimal value of the function a c decreases. Figure 1.14: 1.4 Implicit function graphs As we know the relation between two variables and can be epressed eplicitl in terms of a function = f ( that gives us the value of if we know the value of. It is also possible that the relation between and is given implicitl as an equation. Such relations are called implicit functions, and their graphs are implicit function graphs. One of the most effective methods to plot such graphs is to tr to solve that implicit equation and rewrite it as one or several eplicit functions. In some cases the relations between and can be plotted directl. Let us consider two eamples: Eample: Draw a graph of the function(s given b equation: + = C Solution: We can either rewrite it as two eplicit functions = ± C and draw the two graphs given b this equation. Alternativel, we can note that + gives a square of the distance from the point (0,0 to the point (,, thus equation + = C gives the points located at a distance C from the origin. That is a circle with a radius C with the center at (0,0 (Fig.1.15a. We will use this graph later in our course in chapter 4 to plot fig.4.8a. Eample: Draw a graph of the function(s given b equation: ar + b R are the variables and a,b,c,d 0 are the parameters. R+c drn = 0, where R,N 0

21 1.5. EXERCISES 19 (, N a d a+b d R Figure 1.15: Solution: Let us factor the equation: ar + b R R + c drn = R(a + b R R + c dn = 0 The product of two numbers is zero if one of these numbers is zero, therefore this equation is equivalent to: R = 0 or, a + b R R + c dn = 0 Graphing of R = 0 is trivial. In order to graph a + b R R+c dn = 0 let us rewrite it as dn = a + b R R+c, or N = d a + d b R a R+c. The horizontal asmptote of this graph is: N = lim R d + d b R R+c = d a vertical asmptote occurs if the denominator of the fraction is zero, i.e. at R = c. However, because a+b + b/d = d. The R 0 and c > 0 this asmptote will be outside the range of our function. Additionall note that this function is similar to the graph of Fig.1.10c, but it is shifted upward b d a. Thus the function graph here contains two branches R = 0 and N = d a + b/dr R+c that are plotted in Fig.1.15b 1.5 Eercises Eercises for section Perform the indicated operations: (a (a b (3 4b + b (a + 3 8b (b 6 r. Find limits: 5r 30r+5 (a lim a+q c + (b lim N an +q b N +c +dn, d 0 3. Solve the equation for the specified variable:

22 0 CHAPTER 1. PRELIMINARIES (a find r in: 3r + 5(r + 1 = 6r + 4 (b find in: + 4 = 4 (c find N in: (b N k N = 0 (d find N in: (b d(1 + N k N = 0, d 0; k 0 (e find N in: ( b 1+N/h dn = 0, b 0; 4. Solve the sstem of equations for the specified variables: { = 5 (a find, in: + = 10 { a + b = 0 (b find, in: c + d = b { (1 + = 0 (c find, in: 4 = 0 { 4 (d find, in: = = 0 { b(1 R (e find R,N in: k d anr = 0 (R δn = 0, a,b,d,k,δ 0; Eercises for section Find the derivative of f (: (a f ( = 1 3 (b f ( = e 5 ; (c = (4 ( + 3 (d = a Eercises for section Without plotting the function find the following information about their graph: find the -intercept and zeros; find horizontal or/and vertical asmptotes (if the eist. (Proof of non-eistence of asmptotes is not required. (a function (n given b = an b n +c, a,b,c > 0 (b function N(R given b N = a r (h + R(1 K R, r,a,h,k 0 7. Sketch graphs of the following functions: (a = 3 6 (b = 3 (c = +a Find how the shape of the graph for, 0 depends on the value of the parameter a > 0.

23 1.5. EXERCISES 1 Eercises for section (a Sketch qualitative graphs of the following implicit functions. (b Find how special points of these graphs (intercepts, zeros, asmptotes depend on parameters. (c If graph contains several lines find their intersection points. Note, all parameters represent positive numbers. (a + 3 = 0 on the plane (b + = 9 on the plane (c = 0 on the plane (d dn(a P = 0 on the NP plane (e dn + N+a NP = 0 on the NP plane (f dr(b R = R+a crn on the RN plane (g br(1 R k dr anr = 0 on the RN plane (h an + P(1 ep + bp = 0 on the NP plane Additional eercises 9. Perform the indicated operations: (a (( ( + 1 (b a b p : 4b a p 10. Solve the sstem of equations for the specified variables: ra(1 K A AB = 0 (a find A,B,C in: AB db BC = 0 r,k,d, f > 0 BC fc = Find the derivative of f (: (a f ( = (b f ( = 1 3 (c f (g = cos( ; (d f = (cos( ; (e = a e b a,b > 0 (f = 5 3 (g = b c a, a,b,c > 0 (h =, d > 0 (i = 1+ d n n +a n, a > 0 1. Solutions of differential equations

24 CHAPTER 1. PRELIMINARIES (a Show that function N(t = A e 3t, where A is an arbitrar constant, is a solution of the differential equation: dn(t = 3N. For that, compute derivative of this function and substitute this derivative and the function itself to the equation and show that the left hand side of the equation equals to the right hand side. (b Show, using the same steps that the function N(t = s(1 e t + Ae t, where a is an arbitrar constant and s is a parameters, is a solution of the differential equation dn(t = s N 13. Assume that (t is an unknown function of t. For f ( listed below find the following derivatives: d f d and d f. (a f ( = 3 (b f ( = e a (c Find the epression for d f for an arbitrar f ((t. 14. Without plotting the function find the following information about their graph: find the -intercept and zeros; find horizontal or/and vertical asmptotes (if the eist. (a function ( given b = (b function ( given b = a : Sketch graphs of the following functions: (a = 4e 3t (b = + 3 (c = +3 b 3 c a,b,c > 0 (d = b + 4. Find how the shape of the graph depends on the value of the parameters a > 0 +a and b > 0. (e = b, Find how the shape of the graph depends on the value of the parameter c > 0 and 3 +c 3 b > 0. (f f (n = rn (1 n k h, k 0 find for which values of the parameter h > 0 the graph touches the n-ais. Tip: draw graph for h = 0 and think about how h affects this graph. 16. Sketch qualitative graphs of the following pairs of implicit functions on the same graph. Find all intersection points. (a an P(1 + ep bp = 0 and bp cn = 0 parameters a,b,c,e > 0 (b rr(1 R/K h+r NR = 0 and NR h+r dn = 0 parameters r,k,h,d > 0

25 Chapter Selected topics of calculus In this chapter we introduce several new notions on calculus and algebra which are important for our course..1 Comple numbers Comple numbers were introduced for the solution of algebraic equations. It turns out that in man cases we can not find the solution of even ver simple quadratic equations. Consider the general quadratic equation: λ + Bλ +C = 0 (.1 The roots of (.1 are given b the well known abc formula: where λ 1 = B ± B 4C = B ± D (. D = B 4C (.3 What happens with this equation if D < 0? Does the equation have roots in this case? Comple numbers help to solve such kind of problems. The first step is to consider the equation Let us claim that (.4 has a solution and denote it in the following wa: where λ = 1 (.4 λ 1 = ±i (.5 i = 1 (.6 Here i is the basic comple number which is similar to 1 for real numbers. Using it we can denote solutions of other similar equations. For eample if λ = 4, λ = 1 4 = 1 4 = i (± = ±i. 3

26 4 CHAPTER. SELECTED TOPICS OF CALCULUS Similarl the equation λ = a, has solutions λ = ±ai. Although we call ai a comple number, it is quite different from usual real numbers. Using comple numbers ai we cannot count how man books we have in the librar, for eample. The onl meaning of i is that i = 1, and ai is just a designation of a root of the equation λ = a. Now we can solve equation (.1 for the case D < 0. If D < 0, then D = i D and λ 1 = B ± i D (.7 Eample. Solve the equation λ + λ + 10 = 0 Solution. λ 1 = ± or λ 1 = 1 + 3i, λ = 1 3i. = ± 36 = ± 6i, (.8 We see, that solution of this equation λ 1, has two parts, one part is just a real number -1, which is the same for λ 1 and λ and the other part, is i times another real number 3 which has opposite signs for λ 1 and λ. This is a general form of representation of comple number. An comple number can be represented in the form: z = α + iβ (.9 where α is called the real part of the comple number z, and β is called the imaginar part of z. The notation for the real part is Rez and for the imaginar part is Imz. In our eample Reλ 1 = 1;Imλ 1 = 3. and Reλ = 1;Imλ = 3. We can work with comple numbers in the same wa as with usual real numbers and epressions. The onl thing which we need to remember, is that i = 1. To add two comple numbers we need to add their real and imaginar parts. For eample z 1 = i,z = 5 + 4i, z 1 + z = (3 + 10i + ( 5 + 4i = i i = + 14i. Similarl, multiplication b a real number results in multiplication of the real and imaginar part b this number z 1 = i; 10z 1 = 10 (3 + 10i = i. Multiplication of two comple numbers is just an eercise in multiplication of two epressions z 1 = i,z = 5 + 4i; z 1 z = (3 + 10i ( 5 + 4i = 3 ( i + 10i ( i 4i = i 50i + 40i = 15 38i 40 (as i = 1 = 55 38i. Similarl (z 1 = (3 + 10i = i + (10i = i + 100i = i 100 = i. Now we can check that λ 1 = 1+3i is a solution of the equation in eample (.8. In fact: λ +λ+10 = ( 1+3i + ( 1+3i+10 = ( 1 + ( 1 3i+(3i +6i+10 = 1 6i 9 +6i+10 = ( i+6i = 0 0i = 0, i.e. left hand side of this equation after substitution of λ 1 = 1+3i equals zero and thus λ 1 = 1 + 3i is the root of this equation.

27 .. MATRICES 5 One more definition. The number z = a ib is called the comple conjugate to the number z 1 = a+ib and is denoted as z 1 = z = a ib. Comple conjugate numbers have the same real parts, but their imaginar parts have opposite signs. Roots of a quadratic equation with negative discriminant D < 0 are comple conjugate to each other. It follows from the formula (.7 λ 1 = B + i D λ = B i D (.10 hence: Reλ 1 = Reλ = B D D ; Imλ 1 = ; Imλ =. (.11 Finall consider two more basic operations. If z = a + ib, then, z = a + b is called the absolute value, or modulus of z. Note, that z = z z, as (a + ib (a ib = a (ib = a + b. We use this trick to introduce division of two comple numbers z z z 1 = z 1 z z So, to divide two comple numbers we multipl the numerator and the denominator b a number which is the comple conjugate to the denominator, and we get the answer in the usual form. Eample 1 + 3i 1 4i = 1 + 3i 1 4i 1 + 4i (1 + 3i(1 + 4i 1 + 3i + 4i + 1i i = 1 + 4i = = = i. Matrices From a ver general point of view a matri is a representation of data in the form of a rectangular table. An eample of a matri composed of numbers is given below: ( A = ( This matri A has two rows and three columns. We will call this a matri of the size 3. In general matri size is defined as number o f rows number o f columns. Even if ou did not have matri algebra in school, ou probabl know at least one matri object, that is a vector. Indeed, a vector is an object which is characterized b its components: two numbers in two dimensions or three numbers in three dimensions. In matri algebra vectors can be represented in two forms: as a column vector, i.e. as n 1 matri (preferred representation, or as a row vector, i.e. as 1 n matri. For eample a vector V with the -component V = and the -component V = 1 can be represented a column or as a row vector as: ( V = or V = ( 1 1 Using matrices we can perform the same operations on large blocks of data simultaneousl. For eample, if we need to multipl all 6 numbers of matri A in (.1 b 4, we can write it as 4A which will mean: ( ( ( A = 4 = = (

28 6 CHAPTER. SELECTED TOPICS OF CALCULUS This operation is called multiplication of a matri b a number. For a general matri this can be written as: ( ( a b λa λb λ = c d λc λd Similarl, addition of matrices is adding the numbers that have the same location. This operation is defined onl for two matrices of the same size: ( ( ( ( A + B = + = = ( For general matrices it can be written as: ( ( a b + c d z w ( a + b + = c + z d + w (.15 Multiplication of matrices is not so trivial. In general matri multiplication is defined as the products of the rows of the first matri with the columns of the second matri. Thus, to fund the element in row i and column j of the resulting matri we need to multipl the ith row of the first matri b the jth column of the second matri. Thus we can multipl two matrices A B onl if the number of columns in matri A equals the number of rows in matri B. For a product of two matrices this gives: ( ( a b c d z w ( a + bz a + bw = c + dz c + dw (.16 From this it follows that multiplication of a matri b a column vector is given b: ( ( ( a b v av + bv = c d v cv + dv (.17 The last equation is useful for representation of linear sstems as can be seen from the following eample. Assume we have a sstem of linear equations: { = 5 (.18 + = 10 we can write the coefficients at and in the left hand side as a square matri: ( 1 A =. 1 We also have two numbers in the right hand side which we can write as a column vector: ( 5 V =. 10 Now if we write and as a column vector: X = (.

29 .3. EIGENVALUES AND EIGENVECTORS 7 we can represent sstem (.18 using matri multiplication (.17 as: ( ( ( 1 5 A X = V ; or = 1 10 ( ( ( 1 1 Indeed, from (.17 we get =, that proves this result (.19 Another ( important matri operation is the determinant of a square matri, which for the matri a b A = is defined as: c d det a b c d = ad cb (.0 The determinant of a matri has man important applications in algebra. For eample using determinants it is possible to find solution of sstem of linear equations (e.g. sstem (.18 in the form of so-called Cramer s rule, which was published b Gabriel Cramer as earl as in mid-18th centur. Cramer s rule is briefl formulated in eercise 7 at the end of this chapter. Now let us consider one of the most important problems in matri algebra: the eigen value problem..3 Eigenvalues and eigenvectors Let us start with a definition: Definition 1 A nonzero vector v and number λ are called an eigen vector and an eigen value of a square matri A if the satisf equation: Av = λv (.1 Eigen vectors are not unique, and it is eas to see that if we multipl it b an arbitrar constant k we get another eigen vector corresponding to the same eigen value. Indeed b multipling (.1 b k we get: kav = kλv or A(kv = λ(kv (. therefore, we can sa that kv is also an eigen vector of (.1 corresponding to eigen value λ. ( ( 1 1 For eample, for matri A =, number λ = 3 and vector v = are an eigen value and 1 1 eigen vector as: ( ( ( ( ( Av = = = = 3 = 3v ( ( ( 1 If we multipl v = b an number, e.g., 5, or etc., we will get new eigen vectors v =, ( 1 5 v = of this matri for λ = 3. You can check it in the same wa as we did in (.3 for a vector ( 5 1 v =. 1

30 8 CHAPTER. SELECTED TOPICS OF CALCULUS Finding eigen values and eigen vectors is one of the most important problems in applied mathematics. It arises in man biological applications, such as population dnamics, biostatistics, bioinformatics, image processing and man others. In our course we will appl it for the solution of sstems of differential equations, which we will consider in chapter 4. ( a b Let us consider how to solve the eigen value problem for a matri A =. For that we need ( c d v to find λ and satisfing: v ( a b Av = c d ( v v ( v = λ v. (.4 We can rewrite it as a sstem of two equations with three unknowns λ,v,v : { a v + b v = λv c v + d v = λv (.5 If we collect all unknowns at the left hand side we will get the following sstem: { ( ( (a λ v + b v = 0 a λ b v or in matri f orm = c v + (d λ v = 0 c d λ v ( 0 0. (.6 This sstem alwas has a solution v = v = 0, however it is not an eigen vector, as in accordance with the definition the eigen vector should be nonzero. In order to find non-zero solutions let us multipl the first equation b d λ, the second equation b b and subtract them. Multiplication gives: Subtraction of the equations results in: { (d λ [(a λ v + b v ] = 0 b [c v + (d λ v ] = 0 (.7 (d λ (a λ v + (d λ b v = 0 b c v + b (d λ v = 0 gives (d λ (a λ v b c v + (d λ b v b (d λ v = 0 or as v 0 we get: [(d λ (a λ b c] v = 0 (.8 (d λ (a λ b c = λ (a + dλ + (ad cb = 0 (.9 This is a quadratic equation with unknown λ and for each particular coefficients a,b,c,d we can find two solutions: λ 1 and λ using the abc formula. Thus we found that the eigenvalue problem for a matri (.4 has solutions for the eigen values λ. In general, for a nn matri that the eigen value problem has n solutions for λ. Equation (.9 is ver important in our course and it has a special name: characteristic equation. In most of the courses on mathematics this equation, however, is written in a slightl different matri form.

31 .3. EIGENVALUES AND EIGENVECTORS 9 To derive it let us recall the definition of the determinant of a matri given in section.: det a b ( a λ b c d = ad cb. Similarl the determinant of matri is: c d λ det a λ c b d λ = (a λ(d λ bc (.30 which coincides with the left hand side of characteristic equation (.9 and thus the characteristic equation can be rewritten as: det a λ b c d λ = 0 (.31 Let us use this approach to find the eigen values of matri A from eample (.3. We get the following characteristic equation: Det 1 λ 1 λ = (1 λ(1 λ = 1 λ λ + λ 4 = λ λ 3 = 0 From the abc formula: λ 1, = ± therefore we found two eigen values λ 1 = 3 and λ = 1. = ± 16 ; λ 1 = 3 λ = 1 Now, let us find eigen vectors. For that let us substitute the found eigen values to the original equation (.6 and solve it for v and v. Let us do it first for a particular eample (.3 for which we have found eigen values λ 1 = 3 and λ = 1. For eigen vector corresponding to eigen value λ 1 = 3 we obtain: { (1 3v + v = 0 v + (1 3v = 0 or { v + v = 0 v v = 0 or { v = v v = v (.3 ( 1 Both equations give the same solution v = v. This means that if v = 1, then v = 1 and a pair 1 satisfies the sstem and thus gives an eigen vector of problem (.3. We can also use an( other value for v. For eample, if we use v = then v will be v = and we get another eigen vector, etc. In general an v = k, and v = k give an eigen vector. We can epress it b the following formula: ( v v ( 1 = k 1 (.33 where k is an arbitrar number. Formula (.33 gives all possible solutions of eq.(.3. It also illustrates a general propert of eigen vectors which we have proven in (., that if we multipl an eigen vector b an arbitrar number k will get also an eigen vector of our matri. Using this propert we can formulate an eas wa to write a formula for all eigen vectors. For that we take an found eigen vector and multipl ( it b an arbitrar number k. ( Note, that( if for problem (.3 we use another found eigen vector v, we can write an answer as = k. At the first glance this formula is different from v (.33. However, it is eas to see that both formulas give the same result: this is because k in (.33

32 30 CHAPTER. SELECTED TOPICS OF CALCULUS ( 1 is an arbitrar constant and an vector given b the formula (.33 with can be obtained using ( 1 the formula k for another value of k. Thus the answer to our problem: to find eigen vectors of ( ( ( ( v 1 v matri (.3 for eigen value λ 1 = 3, can be written as =, or =, or etc. v 1 v These ( vectors give ( particular ( solutions ( of this problem. We can also write a formula for all solutions v 1 v as = k, or = k, or etc. As we discussed above all these answers will be v 1 v correct and equivalent. Similarl we find the eigen vector corresponding to the other eigen value λ = 1: 1. Substitution: { (1 ( 1v + v = 0 v + (1 ( 1v = 0 or { v + v = 0 v + v = 0 or { v = v v = v (.34. Relation between v and v : v = v 3. Eigen vector: use e.g. v = 1, thus v = 1 ( 1 The general form is v = k 1 ( 1 v = 1, where k is an arbitrar number. Note, that in both cases in order to find eigen vectors we could use the first equation onl (see equations (.3 and (.34, and the second equation in both cases did not provide us an new information. It is not a coincidence, and this propert is the basis for the following epress method for finding eigen vectors: Epress method for finding eigen vectors Let us derive a formula for finding the eigen vectors of a general sstem (.5. We assume that we have found eigen values λ 1 and λ from the characteristic equation (.31. To find the corresponding eigen vectors we need to substitute the found eigen values into the matri and solve the following sstem of linear equations (.6: { (a λ1 v + bv = 0 (.35 cv + (d λ 1 v = 0 It is eas to check that if we use for v and v the values v = b and v = a λ 1 it gives the solution of the first equation: (a λ 1 v + bv = (a λ 1 ( b + b(a λ 1 = b(a λ 1 + b(a λ 1 = 0

33 .3. EIGENVALUES AND EIGENVECTORS 31 If we substitute these epressions into the second equation we get: cv + (d λ 1 v = cb + (d λ 1 (a λ 1 = 0 To prove that this epression is also zero, note that (d λ 1 (a λ 1 cb is zero in accordance with the characteristic equation (.9. Therefore v = b and v = a λ 1 give a solution of (.35 which is an eigen vector corresponding to the eigen value λ 1. Similarl we find the eigen vector corresponding to the the eigen value λ. However, this approach does not work if in (.6 both b = 0 and a λ = 0. In this case we can use the second equation cv + (d λ 1 v = 0 and find an eigen vector as v = d λ 1 and v = c. Indeed: cv + (d λ 1 v = c(d λ 1 + (d λ 1 ( c = 0 As in the previous case it is eas to show that this vector satisfies the other (first equation as: (a λ 1 v + bv = (a λ 1 (d λ 1 + b( c = 0 due to (.35. The final formulas are: v 1 = ( v1 v 1 ( b = a λ 1 v = ( v v ( b = a λ (.36 or ( ( v1 d λ1 v 1 = = v v 1 c = ( a b where a,b are the elements of the matri A =. c d ( v v = ( d λ c (.37 Either (.36 or (.37 can be used to find eigen vectors. (Both answers will be valid. If, however, one of the formulas gives a zero eigen vector, we should use the other one to obtain a non-zero vector. ( 1 Let us appl these formulas for the sstem (.3 with matri A = and eigen values λ 1 1 = 3;λ = 1. The eigen vectors can be found from (.36 as: ( v1 λ 1 = 3; v 1 ( = 1 (3 ( = ( v λ = 1; v ( = 1 ( 1 ( = (.38 and from (.37 as: ( v1 λ 1 = 3; = v 1 ( 1 3 = ( ( v λ = 1; v = ( 1 ( 1 ( = (.39 We see that the vectors differ from the vectors found earlier, but it is( eas to find( that the are equivalent. For eample, if we multipl the first vector b 1 we find 1 1 =, thus the same 1 vector which we found earlier in (.33. We also see that formulas (.38 and (.39 give equivalent result. ( Indeed, first ( vectors obtained form (.38 and (.39 are the same. For second vectors note that: 1 =.

34 3 CHAPTER. SELECTED TOPICS OF CALCULUS.4 Functions of two variables A function of two variables f (, describes the rule of finding the value of function f, if we know the values of the variables and. For eample, the area of a right-angled triangle with the sides, and is given b the following function of two variables: f (, = /. Another eample is the rate of growth of a pre population in a tpical ecological predator-pre model: f (, = , where is the pre population and is the predator population. The graph of the function of one variable = f ( is a line on the O-plane. To sketch the graph of the function of two variables f (,, we must use a three dimensional space (,,z: the O-plane for the values of the independent input variables,, and the third ais z for the function output value z = f (,. In such a representation the graph will be a surface in a three dimensional space. Fig..1 shows a graph of the function f (, = plotted b a computer. Figure.1: Derivatives. The net step is the definition of the derivative of f (,. The main idea of finding the derivative of f (, is to fi one variable at a constant value, sa =. After that we will get a function of one variable onl ( f (,. Now, we can find the derivative of f (,, as the usual derivative of a function of one variable. For eample, f (, = Let us fi = =. We get the following function of one variable: f (, = = 6 3. We can easil find the derivative now: d f (,/d = d( 6 3/d = 3. This tpe of derivative is called the partial derivative of f (, with respect to at =. We denote it as f / = = 3 We can find such a derivative at = 3, or at an other value of. In fact for an arbitrar =, f (, = , and f / = = ( / = Here (3 / = 0 as we replaced b a constant and the derivative of a constant is zero. Similarl, ( 3 / = 0, and ( 1.5 / = 1.5, as 1.5 is a constant and the derivative of (k = k. It is generall accepted to make all these differentiations without eplicitl replacing b. We just should keep in mind, that for such a differentiation we treat as a constant. Thus, to find the derivative of f with respect to we just write: f / = ( / = 1.5.

35 .4. FUNCTIONS OF TWO VARIABLES 33 keeping in mind that is considered as a constant and not a variable during this differentiation. This epression is called the partial derivative of f (, with respect to and is denoted as f /. Similarl, we can introduce a partial derivative of f with respect to : f /. To compute it, we fi (treat as a constant and make the usual differentiations with respect to. In our eample it gives: f / = ( / = Here (3/ = 3, ( 3 / = 3, and ( 1.5/ = 1.5 as is fied. Eample. Find z/ and z/ for z = 3 sin Solution z/ = 3 cos, as for / we fi, and (sin/ = cos. Similarl, z/ = ( 3 sin/ = 3 sin, as and hence sin is treated as a constant. * =* =* * Figure.: The geometrical representation of a partial derivative is clear from fig... To compute f / we fi, i.e. assume that has some value =. The condition = geometricall gives a horizontal line on the O plane fig..a, or a line parallel to the -ais. In 3D this line gives a curve on the 3D surface in graph fig..b, which is a 1D function. The partial derivative with respect to for this particular will give us the slope of the tangent line to this 1D function. Thus (see fig.. f / gives the slope of the tangent line in the direction of the -ais or the rate of change of f (, in the direction. Similarl, computing f / we fi, i.e. assume that has some value =. It gives us a vertical line on the O plane fig..a, or a line parallel to the -ais. Thus f / gives the slope of the tangent line in the direction of the -ais, or the rate of change of f (, in the direction. If we consider f (, as a mountain f / gives the slope of the mountain if we climb in the -direction and f / gives the slope of the mountain if we climb in the -direction. Note, that in general at each point on a surface we can draw a tangent line in an direction, and partial derivatives f / and f / give the slopes of two of these possible tangent lines. Note, that the slope of a tangent line an direction can be obtained as a combination of these two slopes. Linear approimation Let us derive a formula for approimating a functions of two variables f (,. Let us assume that we know f (, and its partial derivatives at some point, and that we want to find the value of a function at the close point, (fig..3. Let us move to the point, in two steps. Let us first move from the point, to the point,, i.e. in the -direction, and then from, to

36 34 CHAPTER. SELECTED TOPICS OF CALCULUS (*,* f 1 Figure.3: f (, (,*,, i.e. in the -direction. Let us appl the formula for approimation of a function of one variable in formulation (1.10 at each part of this motion. Because on the first part we move along the direction the change of the function ( f 1 will be given as the product of the rate of change of the function in the direction ( f / times the distance between the points ( : f 1 = f (, f (, = ( f / ( (.40 Similarl, on the second part of our motion, we move along the -ais, and the change of the function here ( f will be given as the product of the rate of change of the function in the direction ( f / times the distance between the points ( : f = f (, f (, = ( f / ( (.41 If we add equations (.40 and (.41 and solve it for f (, we find the following formula which gives the approimation for a function of two variables: f (, f (, + ( f / ( + ( f / ( (.4 This epression is called a linear approimation, as the independent variables, are in the first power onl, we do not have terms,, or, or etc. Eample Find the linear approimation for the function e + at the point = 0, = 0 Solution. We use the formula (.4 with f (, = e + and = 0, = 0. f (, = e (0+0 = 1; f / = e + ; at = 0, = 0, f / = e 0+ 0 = 1 f / = (e + / = e + ( + / = e +, at = 0, = 0, f / = e 0+ 0 = Finall, e At = 0.1, = 0.1 the approimate formula gives: e = 1.3. The eact value of e + = e 0.3 =

37 .5. EXERCISES 35.5 Eercises Eercises for section.1 1. Find all roots of the given equations (a = 0 (b = 0 Eercises for section.. Write the following linear sstems in a matri form A X = V. Find the determinant of matri A. (a (b { 4 = 3 + = 1 { a + b = 0 c + d = b Eercises for section.3 3. Find eigen values and eigen vectors of the following matrices: ( 1 (a 1 ( 1 4 (b 1 1 ( 1 5 (c 1 3 Eercises for section.4 4. Find partial derivatives of these functions. After fining derivatives evaluate their value at the given point (if asked. (a z z and for z(, = + 4; at = 1; = (b z for z(, = (5 ; at = 3; = 4 (c z N z z (d P and M z (e (f (g N z M (h z N and z R for z(n,r = N(bR d at R = 0,N = 0; and at R = d b,n = 1 : for z(p,m = a 1+P bm. and z P for z(n,p = an en bnp z vma and A for z(m,a = ML δa h+a z P 1 and z P for z(p 1,P = ap h+p 1 +P and z T for z(n,t = b N T 1+cN+bT N

38 36 CHAPTER. SELECTED TOPICS OF CALCULUS Additional Eercises 5. Perform the indicated operations: (a 3 90 (b ( 1 + i + (4 + 7i (c (4 + 5i (7 + i (d 1 i 6. Find all roots of the given equations (a + 11 = 0 (b = 0 7. Cramer s rule on an eample. Cramer s rule for sstem of two linear equations: { a + b = E c + d = F allows us to find solutions from determinants of matrices. First we need to find a determinant of the main matri A deta = det a b c d. Then we need to find the determinant of a ( matri formed b replacing the -column values of the E matri A with the answer-column values as detd F = det E b F d and similarl for the -column: detd = det a E c F. The solutions of the sstem will be given b the ratios of these determinants as: = detd deta ; = detd deta. Find solutions of the following sstem using the Cramer s rule: { + = 5 + = 4 (.43 Find also solution of (.43 b usual method. Show that Cramer s rule gives a correct result. 8. Find eigen values and eigen vectors of the following matrices: ( 1 6 (a ( 1 (b Find a linear approimation for the function at the given point. (a f (, = + ; at = 1, = 1

39 Chapter 3 Differential equations of one variable Differential equations are equations that contain a derivative of an unknown function. As we know derivative gives a velocit of a process and differential equations occur when we describe various processes via their velocities. Differential equations are widel used for modeling in a variet of disciplines: in mathematics, phsics, chemistr, economics, engineering, medicine, life sciences, etc. Development of methods of stud of differential equations is the main subject of this course. In this chapter we will introduce differential equations, give first definitions, show how to solve simple differential equations analticall and consider a few eamples. Then we will develop qualitative methods for the analsis of differential equations of one variable and will appl them for biological models. 3.1 Differential equations of one variable and their solutions Definitions Let us construct a first differential equation. Consider a motion of a car with a contact velocit, for eample v = 10m/sec. If we denote the distance traveled b the car at time t as l(t we can write velocit = v = dl, as the velocit is the derivative of the distance with respect to time. Thus we can write the following differential equations for this process: dl/ = 10 (3.1 If the car travels with an acceleration, then the velocit will linearl increase with time. If we assume that the acceleration is a = 1.m/sec, then the velocit in the course of time will be given b v = 1.t and we get a differential equation as: dl/ = 1.t (3. Let us consider a biological eample. If N(t is the population size of a species at time t, then the rate of change of the population size is: dn/ = births deaths (3.3 37

40 38 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE Let us assume that the birth and the death terms are proportional to N. This assumption is quite reasonable, as it means, that if, for eample, we know the growth rate of a population of some insects on one tree, then the total growth rate of the whole population of insects in the forest will be proportional to the total number of trees. Thus each term in equation (3.3 will be proportional to N and we get the following famous Malthus equation for population dnamics: dn/ = (α βn = kn (3.4 where α and β are the rate constants for the birth and death processes, and we see that k > 0 if α > β, and k < 0, if α < β. Another model assumes that there is a maimum population size K (called the carring capacit and that the the rate of growth of population depends on how close the population is to this maimum size. This ields the following differential equations: dn/ = k(k N (3.5 Now note, that in mathematical sense equations of the tpe (3. can be written as d = f (t (3.6 as the variable t with respect to which we differentiate function is also present in the right hand side of our equation. Alternativel the equations of the tpe (3.4 and (3.5 can be written as: d = f ( (3.7 as here the variable t is not present at the right hand side and we have onl the unknown function (for eqns(3.4,(3.5 N there. The latter equations are the most important for us in this course and the have a special name an autonomous differential equations : Definition Equation is called an autonomous differential equation d = f ( (3.8 Eample d = tan( d = 3sin( t d = arcsin( autonomous non autonomous autonomous Before we find how to solve differential equations, let us discuss from a ver general point of view what kind of solutions can we epect here. If we assume that a differential equation describes how the size of

41 3.1. DIFFERENTIAL EQUATIONS OF ONE VARIABLE AND THEIR SOLUTIONS 39 a population will change in time, then we ma think about two tpes of problems. The first one it to find this size for a particular population at each time moment. For that we obviousl would need to know the initial size of a population. We can also ask a more general question: to find the population size for an arbitrar initial size. This solution will be called the general solution of a differential equation. Because the general solution contains information on solutions for arbitrar initial conditions, it normall depends on an arbitrar constant. The differential equation with given initial condition is called an initial value problem: Definition 3 The problem d = f (,(0 = is called the initial value problem; Its solution is called the orbit or trajector. The initial value problem for most f ( has a unique solution. Now let us consider how we can solve differential equation Solution of a differential equation e can easil solve an equations of tpe (3.6 using the method of separation of variables. The main idea of this method is to think about the derivative of an unknown function in d = f (t as a fraction d over. If we multipl both sides of this equation b we will get: If we integrate both sides of this equation and get: d = f (t R d = R f (t or = F(t +C (3.9 where F(t is an anti-derivative of f and C an arbitrar constant. This is the general solution of differential equation (3.6. We see that this solution contains an arbitrar constant, as epected from the discussion in the previous section. Let us appl this method to a few eamples. For the simplest differential equation d = 0 we get: d = 0 or d = 0 or R d = R 0 (3.10 or = C Therefore, the solution of this equation is equals an constant C.

42 40 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE For the equation of motion of a car with a velocit 10m/sec (3.1 we get: d = 10 or R R d = 10 or = 10t +C (3.11 Thus this solution shows us the position of a car as a function of time t, and the arbitrar constant C here accounts for the initial position of the car. Finall, for the motion from the rest with an acceleration a = 1.m/sec (3. we get: d = 1.t or R R d = 1.t or = 1. t +C (3.1 We obtained a formula which is well known to ou from our school phsics and C here also accounts for the initial position of a car. It turns out that we can appl the method of separation of variables also for an autonomous equation (3.8 ( d = f (. However, here we will need to separate variables as d f ( = and as a result we will not usuall get the eplicit formula for (t. However, in man cases we can do it after some transformations. Let us solve equation for population dnamics (3.4. dn = kn or R dnn = R k or ln(n = kt +C (3.13 To find N note that equation ln( = a has a solution = e a, thus we find N = e kt+c = e C e kt and if we denote e C = A we will get: N(t = Ae kt, (3.14 where A is an arbitrar constant. Let us appl it for the following initial value problem with k = 4 and the initial population size of 10: dn/ = 4N N(0 = 10 (3.15 The general solution here is given b N = A e 4t. To find the solution of the initial value problem we note that at t = 0 the population size was N(0 = 10, i.e. we can write: N(0 = Ae 4 0 = 10, or we find that A = 10. Hence, we have the following solution of the initial value problem (3.15: N = 10e 4t. Similarl, we can solve the general initial value problem (3.4 for an arbitrar initial size of N(0 and find: N = N(0e kt (3.16

43 3.1. DIFFERENTIAL EQUATIONS OF ONE VARIABLE AND THEIR SOLUTIONS 41 thus A here gives the initial size of the population. Finall let us solve equation (3.5, for specific parameter values K = 0,k = 4: dn = 4(0 N = 4(N 0 or R dn N 0 = R 4 or ln(n 0 = 4t +C or N 0 = Ae 4t or N = 0 + Ae 4t (3.17 This is a general solution. Let us find a particular solution, corresponding to the initial size of the population N(0 = 10, for eample. For that we find: N(0 = 0 + Ae 4 0 = 10, or A = 10 0 = 10, thus the population dnamics in the course of time will be given: N = 0 10e 4t. The solutions of equation (3.4 and equation (3.5 are shown in fig.3.1. N N a t b t Figure 3.1: a- Size of population obtained form the solution of the initial values problem (3.4 with k = 4, N(0 = 10; b-the same for equation (3.5 with K = 0, k = 4, N(0 = 10; We see that the size of the population for eq.(3.4 goes to infinit (fig.3.1a. Quantitativel the size of population increases in e.73 times each 1 4 = 0.5 seconds. Indeed, from the particular solution N = 10e 4t we find N(0 = 10; N(0.5 = 10 e; N(0.5 = 10 e ; etc. In general, for arbitrar k in (3.4 this characteristic time of change is given b τ = 1 k, which follows from equation (3.16. For equation (3.5, the population approaches its carring capacit value of K = 0 (see fig.3.1b and the characteristic time is also determined b the value of k in the following sense: the difference between the current population size and its stationar value decreases in e times over the time period τ = 1 4. This follows from solution (3.17, which gives for this difference N 0 = Ae 4t. Similar epression for an arbitrar value of k is given b N K = Ae kt, which gives for the characteristic time τ = 1 k. This concludes our analtical stud of differential equations. In the net chapter we will formulate another method of analsis of differential equations that does not require direct integration of these equations.

44 4 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE 3. Qualitative methods of analsis of differential equations of one variable In this section we will consider a general non-linear differential equation d = f ( and develop an effective method for qualitative analsis of this equation without finding solutions analticall. In the net chapters this method will be etended to the sstems of two differential equations Phase portrait Let us start with equation (3.18 which we considered in section 3.1. dn/ = kn (3.18 We found that N = 10e 4t is a solution for this equation for k = 4 and the initial population size of N(0 = 10 (see (3.15. This solution was represented graphicall in fig.3.1. N a b c t Figure 3.: Three solutions of equation (3.18 together with tangent lines for these solutions at N = N. If the initial size of the population was different, for eample N(0 = 5, or N(0 = 3, N(0 = 0.1 etc., we get other solutions of equation (3.18 and if we represent these solutions graphicall we will obtain the following curves (a,b,c shown in fig3.. Let us analze them. An important characteristic of an line is its slope. It turns out that we can easil find the slope for the solutions of (3.18: slope = dn/ = 4N. We see that the slope depends onl on N (the size of the population and does not depend on other factors, for eample on the initial conditions. For eample, if N = 3 the slope of the line representing solution at point N = 3 is 4*3=1 for an initial condition. Geometricall this means if we graph several solutions (as in fig3. and determine slopes of these functions for given N (at points of intersection of the dotted line N = N with the solution curves we will find that all slopes are the same (slope = 4 N. We can use this information and represent a qualitative picture of solutions of (3.18 using onl one N-ais. For that let us denote the slope of the solution on the N-ais for each value of N (see numbers 4,8,1,16, etc.. We see, however, that this is not ver helpful for representation of the solution of our equation. To improve it let us think about biological interpretation of the slope of the curve in fig3.. N = N The slope of the curve gives us the rate of change of the function and because fig3. shows how the size of population depends on time, the slope values (represented above the N-ais show the growth rate

45 3.. QUALITATIVE METHODS OF ANALYSIS OF DIFFERENTIAL EQUATIONS OF ONE VARIABLE 43 of population at given N. The most important qualitative aspect of the dnamics here is the gowth of population. We can represent it in the following qualitative wa: N i.e. we show the growth of the population b an arrow which is directed to the right. Of course, this is not a complete description of our sstem, but it gives a good idea about the behavior of our sstem. It shows that if we start at some initial value of N, then N will grow and the size of the population will be continuousl increasing. Note that to obtain this result we have onl used the direction of the arrows in the figure. As we will see in the net section, such representation can be easil obtained for an autonomous differential equation ( d = f ( from the graph of the function f ( at the right hand size. Such representation is called the phase portrait: Definition 4 The collection of all possible orbits of a differential equation together with the direction arrows is called the phase portrait. 3.. Equilibria, stabilit, global plan Let us consider two differential equations that arise in population ecolog: d/ = 4 (3.19 and d/ = (3.0 Let us sketch phase portraits for (3.19 and (3.0. We can do it without finding a solution. In general, <0 f( >0 f( a b Figure 3.3: to sketch a phase portrait of an equation ( d = f ( we need to draw or arrows on the -ais. The arrow means growth of, i.e. d > 0. The arrow means decreasing of, or d < 0. Because d = f ( we will graph function f ( and then assign to that regions where the graph is above the

46 44 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE -ais and to that regions where the graph is below the -ais. For equation (3.19 d/ = 4, the graph of f ( = 4 is shown at the top panel of fig.3.3a and we draw the right arrow for > 0, and the left arrow for < 0 (fig.3.3a bottom. The interesting point here is = 0. Here d/ = 4 = 0, thus the rate of change of here is zero and we cannot assign an direction for the arrow at this point. However, the dnamics of our sstem here is trivial: do not change in the course of time, or (t = 0 for all t. This means, that if the initial size of the population was zero, it will be zero forever. Such points of a phase portrait are called equilibria. The occur at points where the rate of change is zero ( d d = 0 For equation ( = f ( equilibria occur if f ( = 0, which is also used as a definition of an equilibrium. Definition 5 A point is called an equilibrium point of d/ = f (, if f ( = 0 Finall the phase portrait of eq.(3.19 in fig.3.3a gives the following dnamics of : if the initial value of is to the right or to the left left of the equilibrium point = 0, it will go to plus or minus infinit respectivel. Let us stud equation (3.0. Again, our plan is: f ( graph phase portrait (fig.3.3b. Here we have an equilibrium point = 4000 which is the root of the function , and the arrows (flow for this case are shown in fig.3.3b. So, the dnamics of solutions of our equation will be as in the following figure: equilibrium start a start b start b equilibrium start a t Figure 3.4: i.e. for an initial condition, will eventuall approach the equilibrium point. If we compare the equilibria of equations (3.19 and (3.0, we can see that the are different. The variable diverges from the equilibrium point of equation (3.19. Such equilibrium points are called non-stable equilibria. On the contrar, the variable converges to the equilibrium point of equation (3.0. Such equilibria points of are called stable equilibria or attractors. Now we can formulate a general plan for finding the phase portrait of d/ = f (. Global plan. 1. Sketch the graph of f (.. Draw the phase portrait. For that transform the points where f ( = 0 to equilibria points, the regions where f ( > 0 to right headed arrows (, and the regions where f ( < 0 to left headed arrows (. This gives the overall phase portrait. Let us appl it to the logistic equation for population growth dn/ = rn(1 n/k n 0 (3.1

47 3.. QUALITATIVE METHODS OF ANALYSIS OF DIFFERENTIAL EQUATIONS OF ONE VARIABLE 45 This equation describes growth of a population in a medium with limited resources. We can stud (3.1 for arbitrar values of parameters r,k. However for simplicit let us fi r = and k = 3. The equation becomes dn/ = n(1 n/3 = (/3 n (3 n (3. Let us find the phase portrait and the dnamics of the solutions of (3.. First we use the global plan. 1. The right hand side of our equation is (/3 n (3 n = n (/3n. The graph of this function is a parabola, opened below with the roots n = 0;n = 3. f(n a n b t k n n c A B n t Figure 3.5:. The construction of the phase portrait is clear from fig.3.5a. The dnamics of the sstem for different initial conditions is shown in fig. 3.5b, for an initial size of the population 0 < n < 3, and in fig.3.5c for n > 3. We see that in the course of time the size of the population becomes n = 3, i.e. the stable equilibrium point n = 3 is the onl attractor of our sstem. Sometimes, differential equations have several stable equilibria (attractors. For eample, the model for the spruce bud-worm population (3.3 has the following phase portrait (fig.3.6. du/ = f (u (3.3 f(u u1 u u3 A1 Figure 3.6: A u u We see that there are two attractors: A1 and A which correspond to bud-worm populations of different size. We see that if the initial size of the population is 0 < u 0 < u, then the population eventuall

48 46 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE reaches A1; if u < u 0 <, then population eventuall reaches A. These intervals are called basins of attraction. Definition 6 The basin of attraction of a stable equilibrium point is the set of values of such that, if is initiall somewhere in that set, it will subsequentl move to the equilibrium point. In the case of fig3.6, the basin of attraction of the equilibrium u 1 (A1 is the interval 0 < u 0 < u ; the basin of attraction of the equilibrium u 3 (A is the interval u < u 0 <. It is ver important to know basins of attraction of a sstem in order to predict its behavior. 3.3 Sstems with parameters. Bifurcations. One of the aims of modeling in biolog is to predict the behavior of a sstem for different conditions. In that case we differential equations will contain parameters. Let us consider two eamples. The first is a general linear equation with one parameter k dn/ = k n (3.4 If we draw the graph of the right hand side function = kn for different values of the parameter k we find that we can have two possibilities depending on the sign of the parameter k (fig.3.7: a non-stable equilibrium point at n = 0 for k > 0 (fig.3.7a and a stable equilibrium point at n = 0 for k < 0 (fig.3.7b. =kn (k>0 n =kn (k<0 n a b Figure 3.7: Another eample of an equation with parameters is the logistic equation for population growth (3.1. This equation depends on two parameters r and k, where r accounts for the growth rate and k accounts for the carring capacit. Let us consider a slight modification of equation (3.1 for a population which is subject to harvesting at a constant rate h: dn/ = r n (1 n/k h (3.5

49 3.3. SYSTEMS WITH PARAMETERS. BIFURCATIONS. 47 where h is the harvesting rate (an etra parameter. Let us fi the parameters r = and k = 3 (the same values as in equation (3., and stud onl the effect of varing the harvesting parameter h on, the dnamics of the population. dn/ = n (1 n/3 h (3.6 When h = 0 equation (3.6 coincides with equation (3. which was studied in fig.3.5. Now assume that the harvesting h is not zero. Let us plot graphs of n (1 n/3 h for h = 0;h = 0.8;h = 1.6 (fig.3.8a. The phase portraits for h = 0.0;h = 0.8;h = 1.6 are shown in fig3.8b. We see that at h = 0.8 the f(n h=0.0 f(n h=0.8 n n a h=0 h=0.8 h=1.6 h=1.6 b c Figure 3.8: behavior of the sstem is qualitativel similar to the behavior of the sstem without harvesting (h = 0.0: the population eventuall approaches the stable non-zero equilibrium. However the final size of the population in this case is slightl smaller than for the population without harvesting (fig.3.5a At h = 1.6 the situation is different. We do not have a stable and non-stable equilibrium anmore. The flow is alwas directed to the left and the size of the population decreases. In this simple model this means the etinction of the population. The important question here is, what is the maimal possible harvesting rate at which the population still survives. From the previous analsis it is clear that the critical harvesting is reached when the parabola n (1 n/3 h touches the n-ais (fig.3.8c. To find this critical value we note, that h just shifts the parabola n(1 n/3 downward. Therefore, the situation of (fig.3.8c occurs, when the shift equals the maimum of the parabola n (1 n/3. To find the maimal value we find a point where the derivative d f /dn = 0 d f /dn = /3(3 n (/3 n 3 = 4n/3 = 0 n ma = 3/; f (n ma = (/3 (3/ (3 3/ = 3/ (3.7 So the maimal value of (/3 n (3 n equals 3/ and therefore the maimal harvesting is h = 3/ If h > 3/ there are no equilibria and the population will go etinct. If h < 3/ there is a stable and nonstable equilibrium and the population will go to the stable equilibrium. At h = 3/ we are at a boundar between these two qualitativel different cases. Such a qualitative change in sstem behavior is called a bifurcation. Bifurcations are studied in a special section of mathematics: theor of dnamical sstems.

50 48 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE 3.4 Eercises Eercises for section Assume that a population grows in accordance with the following equation: dn = 1.5n If the initial size of the population was n(0 = 30, find what will be size at time t = 4. Find at what time the population will double its initial size.. A bacterial population doubles its size each 0 minutes. The growth of this population N satisfies the differential equation dn = kn. Find the value of k in sec 1. Eercises for section Stud the listed differential equations b answering the following questions: Draw the phase portrait. How man equilibria do we have here? At which? For each equilibrium tell whether is it stable or non-stable What will be the final value of if t. (e.g. converges to equilibrium, or goes to infinit, etc. List attractor/attractors and determine their basin/basins of attraction. (a d = (b d = (c d = ( + 6 (d d = 8 3 (e d = f ( with the following graphs of f (: f( f( (f d = 3 (this equation (Adler 1996 describes the dnamics of population of species + which cannot bread successfull when numbers are too small or too large 4. The following equation describes the production of a gene product with concentration : d/ = ( 0.( 1 + s Here s is the parameter accounting for a chemical which produces the gene product. The initial state of the sstem was: = 0 and the value of s = 0. At some moment of time the value of s was slowl increased from s = 0 to s = s ma and then slowl decreased back to the value s = 0.

51 3.4. EXERCISES 49 (a The value of function ( 0.( 1 at its local minimum is (Optional Show this from function derivative, or using calculator. (b What will be the value of the concentration of the gene product at the end of the described process for s ma = 0.005? (c The same for s ma = 0.0? (d Show that there is a critical value of s ma that separates different outcomes of this process. Find this critical value of s ma. 5. Consider a model population with logistic growth which is subject to harvesting at a constant rate h dn/ = r n (1 n/k h (3.8 Find the maimal ield h. Additional Eercises 6. Assume that the growth of a mass of an animal can given b the following differential equation: dw = W where W is the weight in grams and t time in weeks. (a Find the solution for the initial W(0 = 10. (b At what time the mass will reach half of the saturated value. (c If we assume that the linear size of the animal is proportional to the cubic root of the mass, find at what time the object will reach half of its saturated linear size. 7. The dnamics of the ionic channels in the famous Hodgkin-Hule model for a nerve cell is described b the following tpe of equations: dm = α(1 m βm where m is a gating variable and α,β are the parameters. Find the stead state values of the gating variable m and the characteristic time of approaching this stead state. 8. Consider a model where the harvesting (hn is proportional to the size of the population: Find the maimal ield. dn/ = r n (1 n/k hn ( Compare the harvesting strategies (3.8 and (3.9. Which strateg is better. Wh?

52 50 CHAPTER 3. DIFFERENTIAL EQUATIONS OF ONE VARIABLE

53 Chapter 4 Sstem of two linear differential equations Man biological sstems are described b several differential equations. One of the most simple tpes of such sstems is a sstem of two linear differential equations that on a general form can be written as: { d = a + b d = c + d here (t and (t are unknown functions of time t, and a,b,c,d are constants (parameters. (4.1 Sstem (4.1 b itself has man practicall important applications, for eample compartmental models in biolog and pharmacolog, electrical circuits in phsics, models in economics, etc. Sstem (4.1 will also be ver important for stud so-called nonlinear sstem of differential equations which is widel used in theoretical biolog and will be considered in chapter 5. This chapter we will introduce main definitions for linear sstems (phase portrait and equilibria points we will derive a formula for general solution of this sstem and classif possible solutions of this sstem and their phase portraits. These results will be used later to stud models of biological processes. 4.1 Phase portraits and equilibria Let us consider an eample of sstem (4.1 with particular values for the parameters a,b,c,d: { d = + d = Let us first solve this sstem on a computer. For that we need to choose initial values for and and let the computer find their dnamics in the course of time. Solutions for (0 = 1,(0 = are shown in fig.4.1. We see, that in the course of time, and approach the stationar values = 0., = 0. Let us represent this solution graphicall. For a differential equation with one variable ( d = f ( we presented the solutions in terms of a one-dimensional phase portrait using the -ais. For sstem with two variables, we need to use two aes to represent the dnamics. Let us consider a two dimensional coordinate sstem O with the -ais for the variable and the -ais for the variable. Such a coordinate sstem is called a phase space. Let us represent the trajector from fig.4.1 on the O-plane. The initial sizes 51 (4.

54 5 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS t t Figure 4.1: of the populations were (0 = 1,(0 =, thus we put this point (1, on the O-plane. At the net moment of time we get other values for and and we also put them on the O-plane and the and the coordinates of the net point, etc. Finall, we will get the line shown in fig.4.a that starts at (,1 and ends at (0,0. To show how and change in the course of time we draw an arrow as in fig.4.a. This trajector is the first element of the phase portrait. If we start trajectories from man different initial conditions we will get the complete phase portrait of sstem (4. (fig.4.b. Each trajector represents a certain tpe of dnamics of (t,(t, which can be easil shown on time plots similar to fig.4.1. The phase portrait in fig.4.b give us the overall qualitative dnamics of our sstem: the variables and approach 0,0 from all possible initial conditions Figure 4.: Phase portrait of sstem (4.: ( d = + ; d = found numericall. The main aim of our course is to develop a procedure of drawing a phase portrait of a general sstem of two differential equations without using a computer which will allow us to stud models of biological processes. In the 1D case the phase portrait consisted of two main elements: equilibria points and flows (trajectories between them. Similar elements also compose the phase portrait of a sstem of two differential equations. Let us start with the first main element and define equilibria of the sstem. In the 1D case equilibria were points where our sstem is stationar: placed at an equilibrium point the sstem will sta there forever. Mathematicall equilibria for eq. d = f ( (3.8 were determined as

55 4.. GENERAL SOLUTION OF LINEAR SYSTEM 53 the points where d = 0, i.e. where f ( = 0. In the D case it is required that at the equilibrium point both variables and do not change their values, i.e. both d = 0 and d = 0. For sstem (4. these conditions give the following the sstem of two algebraic equations for finding equilibria: t { d = 0 = + d = 0 = { = = 0 { = = 0 { = 3 = 0 that have the onl solution = 0, = 0. Therefore sstem (4. has an equilibrium point (0,0. As we see in fig.4.b this equilibrium is an attractor for all trajectories. (4.3 For a general linear sstem (4.1 the equilibria will be given b: { a + b = 0 c + d = 0 (4.4 This sstem alwas has a solution = 0, = 0 and thus the general linear sstem (4.1 alwas has an equilibrium at the point = 0, = 0. In the net sections we will find out how to sketch a phase portrait of (4.1 around this equilibrium. Our plan will be the following. We will first find the general analtical solution of this sstem and then will use it to draw the phase portraits. 4. General solution of linear sstem Consider a general sstem of two differential equations with constant coefficients: { d = a + b d = c + d The general solution of (4.5 can be written in the following form ( ( ( v1 = C 1 e λ1t v +C v 1 v ( a b where λ 1,λ are eigen values of the matri A = c d eigen vectors. ( v1, and v 1 e λ t ( v, v (4.5 (4.6 are the corresponding We will not derive the formula (4.6, we will show the main ideas behind the real derivation. For that let us consider first the one dimensional case, and then etend it to a two dimensional sstem. The easiest wa to find solutions of sstem (4.5 is b the method of substitution. Let us illustrate this method on eample of 1D analog of sstem (4.5 which is 1D linear differential equation: d = a (4.7 We can easil solve (4.7 using the direct method of separation of variables and subsequent integration. However, let us find the solution using another method, the method of substitution. The main idea of this

56 54 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS method is to look for a solution in some known class of functions which should be chosen in advance from some preliminar analsis. It was found that for linear sstems this class is the class of eponential functions Ce λt. Important questions such as: how was this class found and is this class unique etc, will not be discussed. Our aim here will be an illustration of the main components of the solution rather than comprehensive analsis of linear sstems, which is a large special section of mathematics. Once the class of functions is given (in our case the class of eponential functions, we need to check under which circumstances it will satisf the equations we are solving. Thus we will look for a solution of (4.7 of the form = Ce λt, where C and λ are unknown coefficients. The main idea of the method of substitution is to find these unknown coefficients for a particular sstem. Let us substitute = Ce λt into (4.7. We find: d = (Ce λt = λce λt, or: We can cancel e λt and C, and we get: λce λt = ace λt Hence we found the following the solutions of (4.7: where C is an arbitrar constant. λ = a (4.8 = Ce at (4.9 Now, let us use the same approach for the two dimensional sstem (4.5. It turns out that the class of functions in two dimensions will be the same as in one dimension, but because we have two variables, we need to introduce different constants for and, so our substitution will be = C e λt ; = C e λt or ( = ( C C e λt (4.10 where C,C,λ are unknown coefficients. Let us make this substitution for a particular eample: ( d d = ( ( (4.11 Substitution for d = λc e λt, d = λc e λt, = C e λt, = C e λt gives: we can cancel e λt, (but not C,C as in one dimensional case, and get: { λc e λt = C e λt + 4C e λt λc e λt = C e λt +C e λt (4.1 or in the matri form: ( C λ C { λc = C + 4C λc = C +C (4.13 = ( ( C C (4.14 We see that to find the solution of (4.11 we need to solve the problem (4.14, which is the eigen value problem considered in section.3. To solve it we find eigen values from the characteristic equation

57 4.. GENERAL SOLUTION OF LINEAR SYSTEM 55 (.31: Det 1 λ λ = (1 λ (1 λ 4 = λ λ 3 = 0 λ 1 = ± +3 4 = 1 ± 1 16 = 1 ±. Hence λ 1 = 1,λ = 3 We use the epress formula (.36 for eigen vectors and get: ( v1 λ 1 = 1; v 1 ( = 4 1 ( 1 = ( 4 ( v λ = 3; v = ( 4. (4.15 Formula (4.15 gives just one eigen vector for each eigen value. We also know (see formula (.33 in section.3 that all eigen vectors can be found b multiplication of this eigen vector b an arbitrar constant. If we denote b C 1 the constant for the first eigen vector and b C the constant for the second eigen vector we will get the following solution of the eigen value problem (4.14: ( v1 λ 1 = 1; v 1 = C 1 ( 4 ( v λ = 3; v = C ( 4. (4.16 If we substitute these eigen vectors into the formula (4.10 we find the following solutions of (4.11 ( = C 1 ( 4 e 1 t ( = C ( 4 e 3 t (4.17 Now let us prove the following propert of sstem (4.5: If 1, 1 and, are two solutions of (4.5, then 1 +, 1 + is also a solution of (4.5. Proof: As 1, 1 and, are the solution this means that the satisf (4.5, i.e. { d1 = a 1 + b 1 d 1 = c 1 + d 1 { d = a + b d = c + d (4.18 If we add equations for d 1 and d we get: d 1 + d = a 1 + b 1 + a + b, which can be re-written as: d( 1+ = a( 1 + +b( 1 +. If we do the same for equations for d 1 and d we will finall get: { d(1 + = a( b( 1 + d( 1 + = c( d( 1 + (4.19 which eplicitl shows that 1 +, 1 + is a solution of (4.5. If we appl this result for two found solutions (4.17 of (4.11 we can conclude that the sum of these two solutions is also a solution of (4.11: ( = C 1 ( 4 e 1 t +C ( 4 e 3 t (4.0 We proved formula (4.6 for a particular sstem. If we appl the same steps for a general sstem (4.5 we will get the general solution in the form (4.6. So, we solved sstem (4.5. In the net sections we will find out how to draw its phase portraits.

58 56 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS 4.3 Real eigen values. Saddle, node. As we know, the general solution of (4.1 is given b (4.6. We can use this formula to sketch a phase portrait of this sstem. It turns out, that we can have several different tpes of equilibria depending on the eigen values λ 1,λ. As we know λ 1,λ are the roots of the characteristic equation (.31, which is a general quadratic equation. Therefore, the roots can be real or comple numbers. In this section we consider the case of real roots. So, assume that the eigen values λ 1 and λ are real. This ields the following three cases: 1. Eigen values have different signs (λ 1 < 0;λ > 0, or λ 1 > 0;λ < 0.. Both eigen values are positive (λ 1 > 0;λ > 0 3. Both eigen values are negative (λ 1 < 0;λ < 0 Note, that we do not consider the case when λ = 0. This situation is quite rare and is not considered in this course Saddle; λ 1 < 0;λ > 0, or λ 1 > 0;λ < 0 Sstem (4.11, which we solved in section 4., had eigen values λ 1 = 1,λ = 3. Let us draw its phase portrait. The general solution of this sstem is given b (4.0 ( ( ( 4 = C 1 e 1 t 4 +C e 3 t. (4.1 Because C 1,C are arbitrar constants let us consider three simple cases, in which one of these constants, or both of them are zero. 1 If C 1 = 0,C = 0, then = 0, = 0 and do not depend on time. The trajector is just one point (0,0, which is the equilibrium point of the sstem (4.11. ( ( If C 1 = 0,C = an number, then = C e 1 3 t. Because e 3 t can change from 1 (at t = 0 to an infinitel large number and C is an arbitrar positive or negative number, this epression can be rewritten as: ( ( ( 4 = C e 3 t 4 = K = V K (4. ( 4 where K is an arbitrar number from < K < and V is a vector. Thus epression (4. means multipling of the eigen vector V b an arbitrar number K. In general, if we multipl a vector b a positive number K we get a vector with the same direction but the length will be increased K times. If we multipl the vector b a negative number, the direction of the vector will be changed to the opposite and the length will be changed b a factor K. Because ( in (4. K assumes all values from < K <, this will give a straight line along this vector (fig.4.3a. To complete drawing the trajector we 1 need to show arrows indicating the motion of a point along the trajector in the course of time. Because

59 4.3. REAL EIGEN VALUES. SADDLE, NODE. 57 time dnamics is given b e 3t, K in (4. will grow in the course of time, i.e. it will become either more positive, or more negative depending on its initial sign. Geometricall this means that a point will move apart from the origin of the O coordinate sstem and we will get a picture as in fig.4.3b. 3 The 4 a Figure 4.3: b third case is C 1 = an number, C = 0. The solution in this case is ( ( 4 = C 1 e t. (4.3 As in previous case ( we conclude, that all trajectories in this case will be located on a straight line 4 along the vector and we just need to show the direction of flow along this line. In this case, time dnamics is given b the function e t, which approaches zero when t goes to infinit. Therefore, independent of initial conditions (independent of the value of C 1 we will approach the point = 0, = 0, and the arrows will have the following direction (fig.4.4. Finall let us draw the phase portrait for 4 a a b Figure 4.4: arbitrar C 1 and C (fig.4.4b. Let us consider one trajector for which C 1 0;C 0, for eample: C 1 = 0.1, C = 0.1. The solution in this case is given b: ( ( ( 4 = 0.1 e t e 3t (4.4 This trajector starts as point = = 0.; = = 0.1. In the course ( of time e t will become smaller and smaller, while e 3t 4 will grow. So, the first term in ( e t will

60 58 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS ( 4 be small compared to the second term 0.1 e 3t, and the dnamics at large t will be described ( ( 4 b an approimate formula : 0.1 e 3t, thus the trajector will approach the line (4. presented in fig.4.3b. There will be similar behavior for an other trajectories: independentl on staring points the will approach line of fig.4.3b from various directions. The qualitative picture will be as in fig.4.4b. Such a phase portrait is called a saddle point. It has the following important features: (1 There is an equilibrium point at = 0, = 0. ( There are two lines associated with eigen vectors of our sstem (fig.4.3b,fig.4.4a. These lines are called manifolds. The manifolds in fig.4.3b and fig.4.4a are different. If we follow the trajector along the manifold in fig.4.3b the distance to the equilibrium increases (see fig.4.4. On the contrar, if we follow the trajector along the manifold in fig.4.4a the distance to the equilibrium decreases. The manifold from fig.4.3b is called a non-stable manifold. The manifold from fig.4.4a is called a stable manifold. Conclusions of this stud can be easil generalized. If we consider an epression: ( ( v = C e λt, (4.5 it will obviousl determine a manifold (straight line along the vector v ( v v and the stabilit of this manifold will be determined b e λt. There are two main tpes of behavior of the function e λt (fig.4.5. If λ < 0,e λt approaches zero, when t increases. If λ > 0,e λt grows to infinit with increasing t. Hence, if λ < 0, eq.(4.5 will determine a stable manifold:, will approach 0 in the course of time (as in fig.4.4a. If λ > 0, then, will diverge to infinit and we will get a non-stable manifold. 1 λ<0 λ>0 a t 1 b t Figure 4.5: Conclusion 1 The equation (4.5 on a phase portrait gives a manifold ( in the form of a straight line. v This line goes through the origin and is directed along the vector. This manifold is stable if λ < 0 and non-stable if λ > 0. Finall the formal definition of a saddle point: v Conclusion Fig.4.4 shows the phase portrait of a saddle point. It occurs close to equilibrium, at which eigen values of the sstem are real and have different signs, i.e. λ 1 < 0;λ > 0, or λ 1 > 0;λ < 0.

61 4.3. REAL EIGEN VALUES. SADDLE, NODE. 59 The phase portrait of a saddle point has two manifolds directed along the eigen vectors. One manifold is stable (corresponding to the negative eigen value of the sstem. The other manifold is non-stable (corresponding to the positive eigen value of the sstem Non-stable node; λ 1 > 0;λ > 0 Let us draw the phase portrait for the case when eigen values are real and are both positive. The general solution of the sstem is given b (4.6: ( ( ( v1 = C 1 e λ1t v +C v e λ t (4.6 1 v From the previous analsis we immediatel conclude, that the phase ( portrait in this ( case has the equilibrium point at (0,0 and two unstable manifolds along the vectors and (fig.4.6a. v1 v Let us put that on the graph (fig.4.6b. v 1 v V V1 a b Figure 4.6: To complete the picture we need to add several trajectories which start between the manifolds. For such trajectories C 1 0 C 0 and both terms in (4.6 will diverge to plus or minus infinit. So, we get trajectories as in fig.4.6b. Such an equilibrium is called a non-stable node. Conclusion 3 If the eigen values of sstem (4.1 are real and both positive (λ 1 > 0,λ > 0 we have an equilibrium point called a non-stable node. To draw a phase portrait at this equilibrium we need to show two non-stable manifolds along the eigen vectors of sstem (4.1 and add several diverging trajectories between the manifolds Stable node; λ 1 < 0;λ < 0 The general solution in this case has the same form (4.6. The phase portrait will be similar to fig.4.6, but because λ 1 < 0;λ < 0 both manifolds will be stable. So we get a picture fig.4.7a If the trajector starts between the manifolds (C 1 0 C 0 it will also approach equilibrium as both terms in (4.6 will converge to 0, because λ 1 < 0;λ < 0 (fig.4.7b. Such an equilibrium is called a stable node.

62 60 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS a b Figure 4.7: Conclusion 4 If the eigen values of sstem (4.1 are real and both negative (λ 1 < 0,λ < 0 we have an equilibrium point called stable node. To draw a phase portrait at this equilibrium we need to show two stable manifolds along the eigen vectors of sstem (4.1 and add several trajectories converging to the equilibrium (0, Phase portraits for comple eigen values: spiral, center In the previous section we have studied the case when the roots of the characteristic equation (.31 are real, and found three possible tpes of phase portrait (equilibria: saddle, stable node and unstable node. Here we will stud the case when the roots of the characteristic equation (.31 are comple General ideas on equilibria with comple eigenvalues From section 4. we know that in order to find the tpe of phase portrait of linear sstem we need to solve the characteristic equation (.31 which is a general quadratic equation, and because the parameters a, b, c, d of the linear sstem are arbitrar the coefficients of the characteristic equation are also arbitrar. Therefore, it ma happen that the discriminant of this quadratic equation will be negative and we will have comple eigen values. From equation (.7 we know that these eigen values will be given b λ 1 = B + i D λ = B i D (4.7 or if we denote the real part of these comple numbers as : α = B and the imaginar part as β = D we can rewrite (4.7 as λ 1, = α ± iβ (4.8 Which tpe of dnamics do we epect here. If we drop for a while the imaginar part iβ in (4.8, we get that λ 1 = λ = α, i.e. both eigen values will be the same and hence the will have the same sign. Which tpe of equilibria do we have for two real eigen values which have the same sign? We can have

63 4.4. PHASE PORTRAITS FOR COMPLEX EIGEN VALUES: SPIRAL, CENTER 61 either a non-stable node if λ 1 > 0,λ > 0, or a stable node, if λ 1 < 0,λ < 0 (see section 4.. In the case of a non-stable node all the trajectories diverge from the equilibrium (fig.4.6b, while in the case of a stable node all the trajectories converge to it (fig.4.7b. It turns out that we will get similar behavior for comple eigen values: if Reλ 1, > 0 we will get divergence of the trajectories, if Reλ 1, < 0 we will get convergence of the trajectories to the equilibrium. However, the picture will be slightl different from fig.4.6b and fig.4.7b, as the imaginar part of the eigen values Imλ 1, will add rotation to the trajectories. The complete derivation of the formula for the general real solution of sstem (4.1 with comple eigen values is given in the appendi at the end of this chapter for etra reading. Here we will just demonstrate that Imλ 1, adds rotation Center, spiral Let illustrate that the imaginar part of the comple eigen value of the characteristic equation results in a rotation of the trajectories. For that let us consider a sstem: ( d ( ( { 0 d d = or = 0 d (4.9 = In this case the eigen values are given b: Det λ λ = λ + 4 = 0 λ 1, = ±i. Thus we have purel imaginar eigen values. It turns out that we can draw a phase portrait of this sstem using the following trick. Let us multipl the first equation in (4.9 b, the second equation b and let us add them together. We get Now note that d = + d = gives d + d = = 0 d = 1 d (just check this b appling the chain rule for d, and similarl hence eq.(4.30 can be rewritten as: d = 1 d ( d + 1 d = 0 d( + = 0 We know that if the derivative of the function f is zero ( d f = f ( = 0 then the function f is a constant, thus the above equation implies: + = Const (4.31

64 6 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS Epression (4.31 gives a so-called first integral of our sstem: combination of variables which are preserved in time. It is not equivalent to the solution of our sstem, but using it we can draw the phase portrait of sstem (4.9. Because Const in (4.31 is an arbitrar positive number, let us denote it as Const = A, where A is just another arbitrar constant. Thus we will get the equation + = A, which represents a graph of a circle with radius A and with the center at the origin (fig.4.8a A t Figure 4.8: To find the dnamics on this circle let us find d at a point = A, = 0 (the bold point in fig.4.8a. From eq.(4.9 d = = A < 0, i.e. the coordinate decreases at this point in the course of time. Thus we conclude that motions of trajector along the circle will be in the clockwise direction. If we draw a series of such circles for different A we will get the phase portrait as shown in fig.4.8b. The dnamics of the and variables can be found b considering motion along trajectories of the sstem: motion here is a rotation of a point along the circle. During this rotation the variable changes periodicall (between the values +A and A, and thus we will see periodic oscillations. The same is valid for the variable. An eample of such dnamics is represented in fig.4.8c. The equilibrium point (0,0 in such sstem is called a center. The phase portrait in Fig.4.8 is a set of circles, which is a consequence of the smmetr of sstem (4.9. In a general case, if λ 1, = ±iβ, we will also get a center point, similar to that in fig.4.8, but instead of circles we can get a series of embedded ellipses. The dnamics of the variables will alwas be oscillations. Conclusion 5 If the eigen values of sstem (4.1 are λ 1, = ±iβ, we have an equilibrium point called a center. The dnamics of variables, are oscillations. The phase portrait is a set of embedded ellipses. A computer generated phase portrait of the sstem d = ; d = + is shown in fig.4.9a. The eigen values in this case are λ 1, = ±i0.1. The time-plot for the and variables is shown in fig.4.10(left. Conclusion 6 The imaginar part of the eigen values results in the rotation of trajectories on a phase portrait. Now let us consider the net two cases: λ 1, = α ± iβ. As we discussed in section 4.3, Reλ 1, determines convergence or divergence of the trajectories to the equilibrium. In the case λ 1, = α ± iβ;α < 0 we epect that the real part of the eigenvalue will give

65 4.4. PHASE PORTRAITS FOR COMPLEX EIGEN VALUES: SPIRAL, CENTER 63 a behavior similar to the case of both negative real eigenvalues, hence we epect the convergence of trajectories to the equilibrium, as for a stable node. In addition to this, as we saw in the previous section, the imaginar part of the eigen values causes the rotation of the trajector. If we add these two processes together we will get convergence to the equilibrium with rotation, hence trajectories will have the form of spirals. A computer generated phase portrait for this case is shown in fig.4.9b. The sstem is d = ; d = The eigen values in this case are λ 1, = 0.15 ± i0.13. The time-plot for the and variables is shown in fig.4.10(middle. The dnamics of the sstem are oscillations with graduall decreasing amplitude. Conclusion 7 If the eigen values of sstem (4.1 are λ 1, = α ± iβ;α < 0, we have an equilibrium point called a stable spiral, fig.4.9b. The last case occurs if λ 1, = α ± iβ;α > 0. This case is similar to the previous one. The onl difference is that because Reλ 1, = α > 0, the real part gives motion equivalent to a non-stable node, or divergence of trajectories from the equilibrium. So, together with rotation from the imaginar part of λ we get the following phase portrait (fig.4.9c. This phase portrait is generated b computer for the sstem d = ; d = The eigen values in this case are λ 1, = 0.1 ± i0.89. The time-plot for the and variables is shown in fig.4.10(right. The dnamics of the sstem are oscillations with graduall increasing amplitude Figure 4.9: Figure 4.10: The dnamics of variables and for the phase portraits shown in fig.4.9. The left picture corresponds to the equilibrium point center, the middle picture corresponds to the equilibrium point stable spiral and the right picture corresponds to the equilibrium point non-stable spiral. Conclusion 8 If the eigen values of sstem (4.1 are λ 1, = α ± iβ;α > 0, we have an equilibrium point called a non-stable spiral, fig.4.9c.

66 64 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS We have found all possible tpes of equilibria which can occur in D sstems: saddle, non-stable node, stable node, center, non-stable spiral and stable spiral. The net question which we will discuss here is the stabilit of these equilibria. 4.5 Stabilit of equilibrium We will call an equilibrium point stable, if there is a neighborhood of this equilibrium, such that all trajectories which start in this neighborhood will converge to the equilibrium (Fig.4.11a. We will call the equilibrium point non-stable, if there is at least one diverging trajector from the close neighborhood of this equilibrium (fig.4.11b. a b Figure 4.11: If we analze the stabilit of the 6 tpes of equilibria studied in the previous section we find the following: Stable equilibria 1. Stable node λ 1 < 0;λ < 0 real. Stable spiral λ 1, = α ± iβ;α < 0 Non-stable equilibria 1. Non-stable node λ 1 > 0;λ > 0 real. Non-stable spiral λ 1, = α ± iβ;α > 0 3. Saddle point λ 1 < 0;λ > 0; or λ 1 > 0;λ < 0 real In case of spirals and nodes the stabilit and non-stabilit is obvious. In case of a saddle point we have a converging trajector, however the eistence of the diverging trajectories (fig.4.4b implies that this equilibrium point is non-stable. The last case, λ 1, = ±iβ (center point, is non conclusive. Trajectories do not converge and do not diverge from the equilibrium. Usuall this case is treated as neutrall stable. All these cases can be formulated in the following theorem.

67 4.6. EXERCISES 65 Theorem 1 If all eigenvalues of the linear sstem (4.1 have negative real parts, then the equilibrium point = 0, = 0 is stable. It is eas to see, that this theorem includes all listed stable equilibria. It obviousl works for a stable spiral, but it also works for a stable node, because an real number can be considered as a comple number with imaginar part equal to zero. For eample: 3 can be represented as z = 3 = 3 + i0, and Rez = 3;Imz = Eercises Eercises for section Find the general solution of the following sstems of ordinar differential equations: ( d ( ( (a 1 d = 1 ( ( 3 1 (b = 4 ( d d. Find the solution for the following initial value problem: ( d d = ( ( (Hint: See an eample of solution in section ( (0 (0 = ( Two different concentrations of a solution are separated b a membrane through which the solute can diffuse. The rate at which the solute diffuses is proportional to the difference in concentrations between two solutions. The differential equations governing the process are: { dc1 / = k V 1 (C 1 C dc / = k V (C 1 C where C 1 and C are the two concentrations, V 1 and V are the volumes of the respective compartments, and k is a constant of proportionalit. If V 1 = 0liters, V = 5liters, and k = 0. liters/min and if initiall C 1 = 3 moles/liter and C = 0, find C 1 and C as functions of time. Eercises for sections 4.3 and Find eigen values and eigen vectors (for real eigen values onl of the following sstems. Determine equilibrium tpe and sketch phase portraits. For real eigen values show non-stable, stable manifolds and several trajectories between the manifolds. (a { d = + 4 d = + 3

68 66 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS (b (c (d (e (f (g { d = 5 d = 3 + { d = 3 5 d = { d = + d = { d = d = { d = d = + { d = d = Stud the following linear sstem with a parameter a: { d = a d = 3 Find the tpes of equilibrium which are possible for different values of < a <. Give the parameter region for each equilibrium and draw qualitative phase portraits. For which parameter values is the equilibrium stable? 6. For which values of parameters a and b does the following sstem has periodic oscillations (i.e. a center equilibrium point: { d = a + d = (a 3 b 7. Compartmental models pla an important role in different parts of population biolog, pharmacolog and biochemistr. The describe the interaction between several processes, which ma be interactions of populations, chemical reactions, etc. A two compartment model is schematicall shown in fig.4.1. It represents two interacting species and. The concentration of the species can be changed either due to a transition to species with the rate given b a, or can die with the rate c. Similar transitions eist also for the species. The rates of these processes are specified in the figure. a b c e Figure 4.1: (a Derive a sstem of two differential equations for species and. (b If a = 0.5,b =,c = 4.5,e = 3 find equilibrium tpe. Is it stable? (c Determine the stabilit of the equilibrium in a general case when a > 0, b > 0, c > 0, e > 0

69 4.7. ADDITIONAL CONCEPTS (APPENDIX Additional concepts (appendi General solution for comple eigen values Notes about general solution of a sstem with comple eigenvalues It turns out that if λ 1, = α ± iβ formula (4.6 is still valid, so the solution can be represented in the form: ( = C 1 ( v1 v 1 e λ 1t +C ( v v e λ t (4.3 but as λ 1 = α + iβ;λ = α iβ we get ( = C 1 ( v1 v 1 e (α+iβt +C ( v v e (α iβt (4.33 This epression gives the correct solutions ( of (4.5. ( However, the form of the solution is not good. First, because v1 v λ 1, are comple, the eigenvectors and will also be comple. Net, we should consider C 1,C v 1 v as general comple constants. Therefore (4.33 is a quite complicated epression which gives and as comple valued functions of time t. Mathematicall it is correct and if we substitute (4.33 into the original equation (4.5 we get an equalit. However, we need to draw a phase portrait of (4.5 on the O-plane, i.e. we onl need the real solutions of our sstem. Such real solutions are present in the general epression (4.33, i.e. the are a subset of all the possible solutions. However, etracting them from (4.33 is not a simple task. We will highlight the main idea behind this derivation below. To derive the general epressions we will need the following Euler formula. Euler formula The Euler formula gives a representation of e iβt in terms of trigonometric functions. It is quite unepected: or in another representation: e iβt = cosβt + isinβt (4.34 e iφ = cosφ + isinφ (4.35 When ou see this formula for the first time it looks quite craz. We know that sinφ and cosφ come from the simple geometr of triangles, i = 1 and e is a special eponential function. Wh are these functions connected together in such a simple wa (4.35? To prove this formula, one should use Talor series. However, here I will present another simpler derivation of (4.35 on the basis of differential equations. At the beginning of this chapter, in order to find a solution of (4.5, we first considered a one dimensional differential equation d = a, and we found its solution Ce at. Consider the following initial value problem for this equation: d = a (0 = 1 (4.36 This initial value problem has the unique solution (t = e at. So we sa, that the solution of (4.36 is e at. But we can also sa it vice versa: we can define the eponential function e at as the function which satisfies the initial value problem (4.36. For eample, if we give to a person just this equation and a computer, he will be able to

70 68 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS solve it and to draw the graph of e at, even without knowledge about eponential functions. The advantage of such a definition is that it can be easil etended to comple numbers. So, let us define e it as a function which satisfies the initial value problem (4.36 with a = i d = i (0 = 1 (4.37 In other words: e it must be the function (t, such that (0 = 1, and the derivative of this function d (t must be equal to this function times i, i.e. d (t = i(t. Let us find an epression which satisfies these conditions. It turns out that it will be (t = cost + isint. Let us check it. The first condition is satisfied: To check the second condition we write: if we replace 1 b i we get: (0 = cos0 + isin0 = 1 + 0i = 1, d (t = (cost + isint = cos t + isin t = sint + icost d (t = sint + icost = i sint + icost = i(cost + isint = i(t, i.e. the second condition is also satisfied. So the function (t = cost + isint gives the solution of (4.37, hence it is the same as e it or e it = cost + isint and we get the Euler formula (4.35. The formula (4.34 is just the formula (4.35 in which instead of φ the letters βt are used. To find e (α+iβt we write: e (α+iβt = e αt e iβt = e αt (cosβt + isinβt (4.38 General solution Now let us find a solution of a sstem with imaginar eigenvalues. As we know the general solution is given b the formula (4.33 and because of the Euler formula we can rewrite it in the following wa: ( = C 1 ( v1 v 1 where C 1,C are arbitrar comple constants and e αt (cosβt + isinβt +C ( v v ( v1 v 1 and ( v v e αt (cosβt isinβt, (4.39 are comple eigen vectors. Now we should find a real part of this complicated epression and get a general real solution of our sstem in this case. A general solution of a sstem of two differential equations should depend on two arbitrar constants as the initial value of each of the variables can be arbitrar. We will use this fact to find the general solution. Our idea is instead of etracting all real solutions from (4.39 we will find just two real solutions. B multipling them b two arbitrar constants we will get a general solution. We will be able to find these two solutions from the first term of (4.39: ( v1 Y 1 = e αt (cosβt + isinβt = v 1 e αt (cosβt + isinβt (4.40 v 1

71 4.7. ADDITIONAL CONCEPTS (APPENDIX 69 Let us etract real and an imaginar parts of this term. If we use the formula (.36 for the eigen value λ = α + iβ we find the eigen vector v 1 : It has the real and imaginar parts: ( ( ( ( ( v1 b b b 0 v 1 = = = = + i = v v 1 a λ 1 a α iβ a α β r + iv i (4.41 The vector v 1 has the real part v r and imaginar part v i. So, the term (4.40 can be written as: If we denote: the term (4.40 can be rewritten as Y 1 = (v r + iv i e αt (cosβt + isinβt = e αt (v r cosβt + iv r sinβt + iv i cosβt v i sinβt = e αt (v r cosβt v i sinβt + ie αt (v r sinβt + v i cosβt 1 = e αt (v r cosβt v i sinβt = e αt (v r sinβt + v i cosβt Y 1 = 1 + i. (4.4 Let us prove that both 1 and are the real solutions of (4.5. For that we will use the fact that (4.39 gives a solution of (4.5 and hence Y 1 is a comple solution of (4.5 as it is a part of (4.39. Sstem (4.5 in a matri form can be written as: As Y 1 is a solution, it satisfies (4.43: dx = AX (4.43 Equating the real and imaginar parts ields dy 1 = AY 1 d 1 + id = A( 1 + i d 1 + id = A 1 + ia d 1 = A 1 d = A Hence 1 and are real solutions of (4.5. Finall, because 1 and are real solutions of (4.5 the general solution is given b the formula: ( = C 1 1 +C (4.44 where C 1 and C are arbitrar constants and 1 and are given b (4.4. Eample Find the general solution of the following sstem. ( ( 0 = 0 ( d d v 1 or { d = d = (4.45 Solution. The eigen values are given b: Det λ λ = λ +4 = 0, λ 1, = ± 4 = ±i. So the eigen vector v 1 corresponding to the eigen value λ = i is: ( ( ( ( v1 0 v 1 = = = + i = v 0 i 0 r + iv i (4.46

72 70 CHAPTER 4. SYSTEM OF TWO LINEAR DIFFERENTIAL EQUATIONS So, ( 1 = e 0 ( 0 ( = e 0 ( 0 ( 0 cost sint + ( 0 ( cost sint = sint cost = ( sint cost Therefore the solution from the formulas (4.44 is: ( ( cost = C 1 sint +C ( sint cost (4.47 Or if we denote C 1 = A and C = B we get: { = Acost + Bsint = Asint + Bcost Let us check that (4.48 does gives a solution of (4.45. Substitution of (4.48 into equation (4.45 ields: { (Acost + Bsint = ( Asint + Bcost ( Asint + Bcost or = (Acost + Bsint { Asint + Bcost = ( Asint + Bcost Acost Bsint = (Acost + Bsint (4.48 So, (4.48 is a solution of (4.45. Finall note that formula (4.31 which we got for the same equation (4.45 is of course valid for functions given in (4.48. For that ou need to find + with and given b (4.48. A direct computation will give us + = A + B = Const i.e. the same result as in (4.33. (Note that in order to get the final result ou need to appl several times the well-known formula from trigonometr sin (α + cos (α = 1.

73 Chapter 5 Sstem of two non-linear differential equations 5.1 Introduction and first definitions Phase portrait After analzing linear sstems let us consider a general non-linear sstem which can be written in the following general form: { d = f (, d (5.1 = g(, Man biological sstems are described b such sstems. One of the classical eamples of ecological models (the predator-pre model can be derived as follows. Let us consider the pre population with a logistic growth given b eq.(3.1: d = r(1 /k, which interacts with the predator and let us assume that the effect of the predator on the pre population is given b the term b. Then, if we assume that the growth of the predator population is proportional to the predator pre interaction c and that the death rate of the predator is given b d, we will get the following sstem of differential equations: { d = r(1 /k b d (5. = c d Formall sstem (5. describes the predator-pre interactions with competition in the pre population. It has several parameters, which account for the specific properties of the populations. Let us stud it for r = 3,k = 1,b = 1.5,c = 0.5,d = 0.5: { d = 3(1 1.5 d = If, as in section 4.1, we solve this sstem on a computer, we will get the following phase portrait (fig.5.1b. We see, that in the course of time all trajectories approach the stationar values of = 0.5; = (5.3

74 7 CHAPTER 5. SYSTEM OF TWO NON-LINEAR DIFFERENTIAL EQUATIONS Figure 5.1: A trajector starting at point (0 =,(0 = 1.5 (a and the complete phase portrait of sstem (5.3 generated b a computer The main aim of our course is to develop the procedure of drawing a phase portrait of a general sstem of two non-linear differential equations without using a computer. We epect that as for 1D differential equation (section 3..1 and for D linear sstems (section 4.1 the phase portrait should include two main elements: equilibria points and flows (trajectories between them. Let us define first equilibria points of a general non-linear sstem ( Equilibria In the 1D case and for D linear sstems the equilibria were the points where our sstem is stationar: placed at equilibrium point the sstem will sta there forever. Therefore, for 1D equation d = f ( equilibria were determined as the points where d = 0, i.e. where f ( = 0. For D linear sstem (section 4.1 we required that both variables and are stationar at equilibria points, i.e. both d = 0 and d = 0. For a general non-linear sstem (5.1 these conditions ield the following definition of equilibria: Definition 7 A point (, is called an equilibrium point of a sstem (5.1 if f (, = 0, g(, = 0 (5.4 Equilibria in two dimensions are also stationar points, i.e. if sstem is placed at the equilibrium it will sta there forever. Thus this trajector will contain just one point. Eample. Find the equilibria of the sstem (5.3: Solution To find the equilibria we need to solve a sstem of algebraic equations (5.4 which in our case becomes: { 3(1 1.5 = 0 ( = 0 From the second equation we find ( = 0, which can be either when = 0 or when = 0.5. Substitution of = 0 to the first equation ields 3(1 0 = 0. This equation has two solutions = 0

75 5.1. INTRODUCTION AND FIRST DEFINITIONS 73 and = 1. Substitution of the other case = 0.5 to the first equation gives ( = 0, or = 1. Thus we have found three equilibria points: (0,0,(1,0 and (0.5,1. We see in fig.5.1 that point (0.5,1 is indeed an important attractor of our sstem which determines the final state of the populations. The other two points are not apparent in fig.5.1, however, as we will see later the also account for important changes of trajectories of our sstem. Thus we have defined equilibria for D sstems. Our net step is to understand what is the D analog of flows, which on the 1D phase portrait were represented b the or arrows Vector field In 1D flows, visualizations of the direction of change of the variable were given via the sign of its derivative d d. In D, both variables can change and the rate of their change is given b the derivatives. In 1D we were able to find the direction of flow at an point from the right hand side function of the equation d = f (. Similarl in D we can find d and d at an point (, from the right hand sides of sstem (5.1 ( functions f (, and (g(,. For eample for sstem (5.3 at a point = 1, = 1 we find d = f (, = = = 1.5, and d = g(, = = = 0.5. However, what do these two numbers show? The tell us that if the size of the pre population = 1 and the size of the predator population is = 1, then the pre population decreases with the rate of 1.5 and the predator population grows with the rate of 0.5. On the phase plane, this will result in a shift of a point representing populations from point (1,1 (point A in fig.5.a to some point B which is to the left and upward from point A. Let us make it more quantitative. We know that the rate of change of in our case is 1.5/0.5 times larger than the rate of change of. This determines the direction of shift of point B relative to point A. The easiest wa to represent it is to draw from point (1,1 a horizontal arrow heading to the left with the length of 1.5 and a vertical arrow heading upward with the length of 0.5. The direction of the overall shift will be given b the resultant vector of these two vectors fig.5.b. The resultant vector will give us the direction tangent to the trajector which goes through the given point. We can generalize this result as: and d 1 B A B A 1 a b Figure 5.: Conclusion 9 At an point (, of a phase space for sstem (5.1, we can define the vector v with the components ( f (,,g(,. Such vectors will be tangent to the trajectories of our sstem. We can find this vector field without a solution of our sstem, just from the right hand sides of our sstem. Note, that the length of the vector in fig.5. is not important, as we are interested in the direction, onl. If we appl the same procedure at man points and represent the directions b shorter vectors we

76 74 CHAPTER 5. SYSTEM OF TWO NON-LINEAR DIFFERENTIAL EQUATIONS Figure 5.3: will get the following vector field of the sstem (5.3 (fig.5.3. We see that although the vector field describes qualitativel the direction of trajectories in the phase portrait of our sstem, it does not give us information on convergence/divergence of trajectories, thus we cannot determine the attractors of our sstem, which is crucial for our stud. However, as we will see in chapter 6 we can elaborate the vector field based methods and in man cases will be able to obtain convergence/divergence information using the so-called graphical Jacobian approach. But in order to derive this approach we need to understand how to find attractors of a general non-linear sstem (5.1 using an analtical approach. For that we need to establish the relation of the general non-linear sstem (5.1 and the general linear sstem (4.1 that we studied in chapter Linearization of a sstem: Jacobian Consider a general sstem of two differential equations: { d = f (, d = g(, (5.6 In this section we will show that close to equilibrium point the phase portrait of this general non-linear sstem (5.6 can be found from the solution of the linear sstem (4.1 that we studied in chapter 4. As a consequence we will get si possible tpes of equilibria, that are: saddle, non-stable and stable node, non-stable and stable spiral, and a center. Our main tool here will be a formula (5.7 for the approimation of function of two variables f (, around point (, which we derived in section.4: f (, f (, + ( f / ( + ( f / ( (5.7 where f / and f / are the values of the partial derivatives at the point (,, i.e. the are just numbers. We will appl this formula to approimate the functions f (, and g(, of our sstem (5.6 and later solve the approimated sstem and find the phase portraits close to the equilibrium.

77 5.. LINEARIZATION OF A SYSTEM: JACOBIAN 75 Let us start the derivation. Assume that sstem (5.6 has an equilibrium point at (,. This means (see (5.4 that: { f (, = 0 g(, (5.8 = 0 Let us approimate f (, close to the equilibrium using the formula (5.7: f (, f (, + ( f / ( + ( f / ( As we assumed (, is an equilibrium, i.e. f (, = 0 and we get A similar approach for g(, ields: f (, ( f / ( + ( f / ( (5.9 g(, ( g/ ( + ( g/ ( (5.10 If we replace the right hand sides of (5.6 b their approimations (5.9, (5.10, we get the following sstem: { d = ( f / ( + ( f / ( d = ( g/ ( + ( g/ ( (5.11 The sstem (5.11 is simpler than the original sstem (5.6, as the partial derivatives in (5.11 are constants (numbers, as the are evaluated at the equilibrium point,. So we can rewrite (5.11 as : { d = a( + b( d = c( + d( (5.1 where a = f / ;b = f / ;c = g/,d = g/. We can simplif (5.1 even more. For that let us introduce new variables: u = v = (5.13 were u,v are new unknown functions of t. If we substitute them into the right hand side of (5.1,we get: { d = au + bv d = cu + dv (5.14 In order to substitute u and v into the left hand side of (5.14, note that u(t = (t, i.e. du = d 0 (here d = 0 because is a constant. Similarl, dv = d d. After replacing b du and d b dv in (5.14 we get: { du = au + bv dv (5.15 = cu + dv. Sstem (5.15 coincides with a general linear sstem (4.1 that we studied in chapter 4 and for which we found si possible tpes of solutions resulting in si possible phase portraits: saddle, non-stable and stable node, non-stable and stable spiral, and a center. However, how can we use the results of stud of sstem (5.15 for the stud of the original sstem (5.6? In order to derive (5.15 we made two steps: (1 we used formula (5.7 for function approimation; ( we changed variables, to u,v. Let us analze each of these two steps. (1 As we discussed earlier, formula (5.7 gives a good approimation of the function f (, onl if, is close to the point of approimation,. So we can use the linear sstem (5.15 for approimating the solutions of the non-linear sstem (5.6 onl close to the equilibrium point

78 76 CHAPTER 5. SYSTEM OF TWO NON-LINEAR DIFFERENTIAL EQUATIONS (,. ( about change of variables. Equation (5.13 gives the variables u,v via the variables,. However, we can also solve the equations and find how, will be epressed in terms of u,v: = u + = v + (5.16 Using this epression we can draw the trajectories for our original variables (t, (t if we know the trajectories u(t,v(t of the linearized sstem (5.15. Indeed, as the coordinate of the trajector equals u plus number, and the coordinate equals v plus number, the onl thing what we need to do is just to draw the trajector u(t,v(t and shift its -coordinate b and the -coordinate b units. As we know in a general linear sstem (5.15 the phase portrait is centered around a point (0,0, therefore all what we would need to do in order to draw the phase portrait of the non-linear sstem (5.6 close to its equilibrium (, is just to shift the phase portrait of the linear sstem (5.15 to a location of equilibrium in non-linear sstem (5.6. If, for eample, the linearized sstem (5.15 will have a stable node tpe phase portrait, then a non-linear sstem (5.6 will have the same stable node point but around an equilibrium (, as shown in fig.5.4. v * u * Figure 5.4: Conclusion 10 The phase portrait of liner sstem (5.15 close to the origin (u = 0, v = 0 is similar to the phase portrait of non-linear sstem (5.6 close the to equilibrium point (,. To draw the phase portrait of non-linear sstem (5.6 close to equilibrium, we need to shift a phase portrait of linear sstem (5.15 from the origin the equilibrium point (,. To find the linearized sstem (5.15 we need to find the equilibrium point (, and compute the following four numbers: the values of derivatives of right hand sides of our sstem at this equilibrium: So, sstem (5.15 can be written as: a = f / b = f / c = g/ d = g/ { du = f u + f v dv = g u + g v. (5.17 From coefficients of this sstem we can construct a matri J that is called the Jacobian J = ( f g f g (5.18

79 5.3. DETERMINANT-TRACE METHOD FOR FINDING THE TYPE OF EQUILIBRIUM 77 Eample Find qualitative phase portrait of sstem (5.3 close to a nontrivial equilibrium point (i.e. at the equilibrium where 0, 0. { d = 3(1 1.5 d = Solution In section 5.1. we found that this sstem has three equilibria (0,0, (1,0, and (0.5,1. The nontrivial equilibrium is (0.5, 1. To find the Jacobian of our sstem we compute the partial derivatives (5.18 and evaluate them at the equilibrium point. In our case f (, = 3(1 1.5; g(, = f / = at point (0.5,1 this derivative equals f / = = 1.5. Similarl: f / = 0.75; g/ = 0.5; g/ = 0, hence the linearization of our sstem at point (0.5, 1 is { du = 1.5u 0.75v dv = 0.5u or the Jacobian is: J = ( In order to find a phase portrait of this linear sstem we need to find eigen values of the Jacobian matri from the following characteristic equation (.31: Det 1.5 λ λ = ( 1.5 λ( λ Using abc formula we find that:. = λ + 1.5λ = 0 λ 1 = 1.5 ± = 1.5 ± 0.87 (5.19 or, λ 1 = 1.36 and λ 1 = Because both eigen values are real and negative we will have a stable node and a phase portrait qualitativel similar to that in fig.5.4b. 5.3 Determinant-trace method for finding the tpe of equilibrium In this section we derive a simple method for finding signs of eigen values and the tpe of equilibrium of the linear sstem (4.1. The results will also be applicable for a general non-linear sstem (5.6, because, as discussed in the previous section, the equilibrium tpe of a non-linear sstem can be found from its linearization. The eigen values of sstem (4.1 are the roots of the characteristic equation (.31: Det a λ b c d λ = 0, (5.0 or:

80 78 CHAPTER 5. SYSTEM OF TWO NON-LINEAR DIFFERENTIAL EQUATIONS Det a λ b c d λ = (a λ(d λ cb = λ λ(a + d + ad cb = 0 Let us epress the last equation in a slightl different form using the following definition: ( a b Definition 8 The trace of the matri c d is tra = a + d. The determinant of the matri A is deta = ad cb So we can rewrite the characteristic equation using the definition of the trace and the determinant as follows: λ traλ + deta = 0 (5.1 We see that although the original sstem depends on four parameters a, b, c, d the characteristic equation depends onl on two parameters tra and deta, thus if we know the determinant and the trace of our sstem we can find the eigen values and the tpe of the equilibrium of the sstem. Indeed, from (5.1 we can easil find that the eigen values are: λ 1, = tra ± D where D = (tra 4detA (5. Roots of the equation (5. are as the roots of an quadratic equation, connected in the following wa to the coefficients of the equation: λ 1 + λ = tra (5.3 λ 1 λ = deta (5.4 To prove the properties (5.3 and (5.4, just note that if λ 1 and λ are the roots of a quadratic (5.1, then it can be written as: λ traλ + deta = (λ λ 1 ((λ λ. The direct computation ields: λ traλ + deta = (λ λ 1 ((λ λ or λ traλ + deta = λ λ 1 λ λ λ + λ 1 λ or λ traλ + deta = λ (λ 1 + λ λ + λ 1 λ If we now compare the left and the right hand sides of the last equation we will get both properties (5.3 and (5.4. Let us start classification. 1. If deta < 0, then D = (tra 4detA > 0, so we have real roots. From (5.4 we conclude that their product is negative, i.e. roots have different signs, i.e. λ 1 < 0,λ > 0, or λ 1 > 0,λ < 0 and we have a saddle point.. If deta > 0, then D = (tra 4detA can be negative as well as positive. This means the roots can be real, or comple. Let us consider the case of real roots first, i.e. D = (tra 4detA 0 (5.5

81 5.3. DETERMINANT-TRACE METHOD FOR FINDING THE TYPE OF EQUILIBRIUM 79 If (5.5 holds, the roots are real. Net, let us use the propert λ 1 λ = deta. In the case of deta > 0, the product of the roots is positive, i.e. the roots have the same sign. The can be both positive, or both negative. The sign of the roots can be found from the trace of the matri (λ 1 + λ = tra. When tra > 0, then λ 1 > 0, and λ > 0 and we have a non-stable node. When tra < 0, λ 1 < 0 and λ < 0 and we have a stable node. Let us formulate it as a separate case: 3. If deta > 0, D > 0 and tra < 0 the equilibrium is a stable node. Let us put this information into a graph (fig.5.5. On this graph let us use tra as the -ais and deta as the -ais. Case 1 of a saddle point then corresponds to the lower half plane (region 1. The line (5.5, which separates real and comple roots, is the parabola given b deta = (tra /4, or = /4. Real roots are below this line. Therefore, in region, where tra > 0, we have case of a non-stable node. In region 3 tra < 0 and we have case 3 of a stable node. 4. If deta > 0, and D < 0, we have comple roots. In accordance with (5.3 and (4.7 the are: λ 1, = tra D ± i Hence, Reλ 1, = tra/. From this we immediatel see that if tra > 0, we have a non-stable spiral (region If deta > 0, D < 0 and tra < 0, we have a stable spiral (region The last case of a center point appears when Reλ 1, = tra/ = 0, or when tra = 0 and deta > 0. In our graph is it the upper part of the deta ais (region 6 det A stable spiral non stable spiral D=0 stable node center non stable node saddle 1 tr A Let us appl this method for several eamples: Figure 5.5: Eample. Find the equilibrium tpe for the following sstems: ( d ( ( ( a 1 d ( ( d = ; b d = ; 1 ( d ( ( ( 3 d ( ( 1 1 c = ; d = ; 1 1 d d

82 80 CHAPTER 5. SYSTEM OF TWO NON-LINEAR DIFFERENTIAL EQUATIONS Solution Our plan is to find deta,tra for the corresponding matrices and make a conclusion. ( 1 a The matri is A = ; tra = = 5; deta = =. Therefore we have case 1, 3 4 hence the sstem has a saddle point. ( 4 1 b The corresponding matri is A = ; tra = 4 + = 6; deta = = 7. As deta > 0 1 we need to check that the roots are real. We have D = (tra 4detA = = 36 8 > 0, hence the roots are real and we have case, hence the sstem has a non-stable node. c tra = +1 = 1; deta = +6 = 4; D = (tra 4detA = = 15 < 0, we have comple roots and we are in region 5 and have a stable spiral d tra = 1 1 = 0; deta = 1 + = 1, so we have case 6 and we have a center point, or oscillation in our sstem. 5.4 Eercises Eercises sec (A Find equilibria of the following sstems { d = f (, d = g(, (B Find the following partial derivatives at each equilibrium point ( f { d { (a = 4 d d = (b = 9 + d = { d { d (c = d = + (d = (1 3 d = (1 (Hint: See an eample of solution in section (see definition in section 5.1.., f, g, g.. Find equilibria of the following Lotka Volterra model with competition in the pre population. Determine for which parameter values all equilibria are non-negative. { dn/ = an en bnp dp/ = cnp dp a,b,c,d,e > 0 3. Protein snthesis depends on DNA transcription (a making mrna molecules (M and translation (c of mrna into proteins (P. Some proteins inhibit the transcription of their own mrna ( 1+P 1. mrna and proteins are degraded at rates b and d, respectivel. This process gives the following of two differential equations. Find equilibria of this model. { dm/ = a 1+P bm P,M 0 dp/ = cm dp a,b,c,d > 0 4. Mathematical epidemiolog also makes use of simple ODE models. One of these models describes the number of susceptible individuals (S and infected individuals (I. Individuals are born at rate (B, and die at rate (µ. Susceptible individuals can become infected when the come into contact

83 5.4. EXERCISES 81 with infected individuals ( βsi. Once infected, an individual has a certain death rate (α; this ma be different from the death rate{ of non-infected individuals. This process can therefore be ds/ = B βsi µs modeled b the following equations: di/ = βsi αi B,α,β,µ > 0 Find equilibria of this model. Determine for which parameter values all equilibria are nonnegative. Eercises sec. 5. and Find { the tpe of equilibria using the det-tr method. Determine the stabilit of the equilibrium. d { (a = 3 + d { d = (b = + d { d = 10 (c = + d d = 5 (d = + 10 d = Consider the following model for the algae population: (a Find equilibria { d (b Find the general epression for the Jacobian of this sstem (c Determine tpe of each equilibrium using det-tr method = (1 0; d = 0. (d Draw qualitative local phase portraits around each equilibrium point 7. Consider the following biological model: { de/ = b e e 3 g dg/ = e g b 0 (5.6 For which values of parameter b the sstem has onl one equilibrium? Determine stabilit and tpe of this equilibrium in found parameter range (in which sstem (5.6 has onl one equilibrium. { dn 8. Stud the Lotka-Volterra model for a predator-pre sstem: = an bnp dp = cnp dp here N denotes the pre population, P denotes the predator population and a > 0,b > 0,c > 0,d > 0 are parameters (a Find the nontrivial equilibrium of the sstem (i.e. an equilibrium where N 0,P 0. (b Find the linearization of the sstem at this point (i.e. the Jacobian matri (c Determine the tpe of the equilibrium (d Sketch the phase portrait around this equilibrium. Which kind of dnamics do we epect here? Additional eercises 9. Consider the sstem: d = e 1; d = e (a Check that (0,0 is an equilibrium point of the sstem

84 8 CHAPTER 5. SYSTEM OF TWO NON-LINEAR DIFFERENTIAL EQUATIONS (b Find the general epression for the Jacobian of this sstem (c Find the Jacobian at the point (0,0 (d Write the linearization of the sstem close to the equilibrium (0,0 10. Find the equilibria of the following sstems. Compute the Jacobian at the equilibria points. Determine tpe of each equilibrium using det-tr method. Draw qualitative local phase portraits around each equilibrium point. (a { d = 3 + d = (b { d = d = Revisit solutions of problems 1. Use found equilibria and partial derivatives at these equilibria points to determine tpe of each equilibrium using det-tr method. Draw qualitative local phase portraits around each equilibrium point. 1. Consider a modification { of the Lotka-Volterra model, which includes competition in the pre population ( en : dn = an en bnp dp, where the parameters a,b,c,d,e > 0 and the variables = cnp dp N 0, P 0. (a Find all equilibria of this sstem. (b Compute the Jacobian at each equilibrium point. (c At which parameter values do we have a non-trivial equilibrium (i.e. an equilibrium at which both N and P are positive. Find stabilit of this equilibrium. 13. Consider the following model for cardiac tissue: { de/ = e(e a(e 1 g 0 < a < 1 dg/ = εe ε > 0 (5.7 Here the variable e accounts for the transmembrane potential, the variable g accounts for the refractor period and a,ε are the parameters. The shape of the action potential in cardiac tissue is an important characteristic of mocardium. If the recover of the transmembrane potential shows oscillation as in fig.5.6b. it can cause dangerous cardiac arrhthmias. From a mathematical point of view the oscillations in fig5.6b occur when sstem (5.7 has an equilibrium point which is a stable spiral. Monotonous recover occurs when this equilibrium is a stable node. a b Figure 5.6: The monotonous (a and oscillator recover (b in an ecitable medium Determine for which parameter values we will have situation fig.5.6a and for which parameter values we will have situation fig.5.6b.

85 Chapter 6 Graphical methods to stud sstems of differential equations In the first section of this chapter we will start from the vector field of a general non-linear sstem introduced in section 5.1.3, and find how we can approimate the vector field b the so-called null-cline method, without using a computer. Then, in section 6., we will show that the null-clines can be used not onl for vector field approimation but also for determining the tpe of an equilibrium point without eplicit computation of the Jacobian of the sstem. 6.1 Null-clines We introduced the vector field in section 5.1.3, where we showed that for a general D sstem { d = f (, d = g(, the direction of the trajectories on the phase portrait will be along the vector v with the components ( f (,,g(,. Thus, to draw the vector field of a particular sstem we need to evaluate the values of the functions ( f (,,g(, in man points which usuall requires using a computer. In this section we will develop a so-called method of null-clines, which will allow us to sketch a qualitative picture of the vector field analticall. The main idea here is similar to what we did in the 1D case, where we have represented the derivative d b arrows of two tpes: for d > 0 and for d < 0. In D the vector field has two components V = (v,v = ( d d. Each of these components can be positive, or negative. Therefore, we can have the following four cases shown fig.6.1. (Note, that we use different line tpes there that will be important in the future., d, d (6.1 I d/>0 d/>0 d/>0 d/>0 II III d/<0 d/<0 Figure 6.1: d/<0 IV d/<0 83

86 84CHAPTER 6. GRAPHICAL METHODS TO STUDY SYSTEMS OF DIFFERENTIAL EQUATIONS The method of null-clines represents the vector field using these four cases. The main idea behind this method is the following. If we compare cases I and II we see that the differ b the sign of the d derivative: for case I d > 0 and for case II d < 0. Therefore these cases are separated b the boundar where d = 0 (fig.6.a. We know that for sstem (6.1 d = f (,, therefore the boundar between cases I and II is given b the condition f (, = 0 (6. I I >0 >0 >0 >0 III >0 <0 II =f(,=0 =g(,=0 <0 >0 a Figure 6.: b Geometricall equation (6. gives a graph of one or more lines in the O-plane (see section 1.4. Note that at this line the horizontal component of the vector field is zero, therefore the vector is vertical. Similarl, the transition from case I to case III occurs when d = 0 (fig.6.b. As we know for the sstem (6.1 d = g(,, hence the separation line in this case is given b the equation: and the direction of the vector field at this line is horizontal. g(, = 0 (6.3 Equation (6.3 will give us not onl the boundar between cases I and III but also a boundar between cases II and IV, as the transition between cases II and IV also occurs when d = 0. In general equations (6.,(6.3 give two (or more lines on the O-plane, which separate the plane into several regions with different directions of vectors (cases I-IV. These lines are called null-clines. Definition 9 The -null-cline (or d = 0 null-cline is the set of points satisfing the condition f (, = 0. The -null-cline (or d = 0 null-cline is the set of points satisfing the condition g(, = 0. To use the method of null-clines it is useful to note that at the -null-cline the -component of the vector changes its sign, and at the -null-cline the -component of the vector changes its sign. To use this rule effectivel we will alwas denote the vector field components using lines of different tpes: the horizontal component as a dashed line and the vertical component as a solid line. Let us use these ideas and formulate a plan for finding the vector field using null-clines.

87 6.1. NULL-CLINES 85 Plan of null-cline analsis for sstem (6.1 We assume that on the O-plane the -ais is the horizontal ais and the -ais is the vertical ais. 1. Draw d = 0 null-clines from the equation f (, = 0 using a dashed line and d = 0 null-clines from the equation g(, = 0 using a solid line.. Choose a point in one of the regions in the, plane and find the and the -components of the vector field. Use the dashed line for the component and the solid line for the component. For finding the directions use the following rule: if f (, > 0 the component is directed as, if f (, < 0 it is directed as ; if g(, > 0 the -component is directed as, g(, < 0 it is directed as. 3. Find the vector field in the adjacent regions using the following rule: (a change the direction of the dashed component of the vector field if in order to get to the adjacent region ou cross the dashed null-cline (b change the direction of the solid component of the vector field if in order to get to the adjacent region ou cross the solid null-cline (c show the direction of the vector field on the null-clines. Note, that instead of dashed and solid lines ou can use lines of different colors. Then the last step of this plan would be: change the direction of the component of the same color as the color of the null-cline which we cross to get to the adjacent region. Note, that although this plan works good in most of cases, there are some situations when components of the vector field do not change their sign at the corresponding null-cline. These are special so-called degenerate cases (eceptions. We will not consider them in our course. Eample Find the vector field of the following sstem using null-clines. { d = 3(1 1.5 d = (6.4 Solution. We follow our plan as follows 1. The d = 0 null-clines are given b the equation f (, = 0, i.e. 3(1 1.5 = ( = 0. This equation has two solutions: = 0 (the vertical line which coincides with the - ais and = which is a straight line with the negative slope which goes through the point (,0. The graphs are shown using dashed lines in fig.6.3. The d = 0 null-clines are given b the equation g(, = 0, i.e = ( = 0, which also has two solutions: = 0 (horizontal line which coincides with the -ais and = 0.5 (vertical line through the point = 0.5. Graphs are shown b solid lines in fig We find that at point (, d d = 3 (1 1.5 = 1 < 0 and = = 1 > 0, hence the direction of the dashed arrow is to the left and of the solid arrow is upward.

88 86CHAPTER 6. GRAPHICAL METHODS TO STUDY SYSTEMS OF DIFFERENTIAL EQUATIONS Figure 6.3: 3. Now we complete the picture. For eample, to get into the region to the left from point (, we cross the solid line, thus we change the direction of the solid component here. Similarl for the other regions. We get the picture as in fig.6.3. Finall we show the vector field on the null-clines. We see that the vector field in fig.6.3 is a good approimation of the flow in fig.5.3. We also see that attractor (0.5,1 is a special point in fig.6.3: the point of intersection of the and null-clines. This is not a coincidence. If we compare the conditions for finding the equilibrium point (5.4 and equations for null clines (6., (6.3, we see that the first equation for finding equilibria f (, = 0 is also the equation for null-cline and the second equation for finding the equilibrium point g(, = 0, is the equation for the -null-cline. Thus, the solution of sstem (5.4, which gives the points satisfing both equations, gives the points which belong to both null-clines, i.e. the points of intersection of the null-clines. So we found that: Conclusion 11 Equilibria are the points of intersection of the and -null-clines. Note that this definition applies points of intersection of different null-clines onl. For eample intersection of null clines of the same tpe in fig.6.3 at points (0.5,0 and (0, do not give equilibria of sstem (5.3, while intersection of the different null-clines at points (0,0,(1,0, (0.5,1 give the equilibria of this sstem. Our net step will be to stud how can we appl the null-clines for finding the tpes of equilibrium. 6. Graphical Jacobian The method which we present here allows, in a number of cases, to find the tpe of an equilibrium from the null-clines of the sstem, i.e., even without computation of partial derivatives of the Jacobian in the equilibrium. The main idea behind this method can be seen from the scheme in Fig.6.4. Let us consider an equilibrium point (, and two close points: one located to the right, with the coordinates ( + h,, the other upward with the coordinates (, + h. Because we assume that (, is an

89 6.. GRAPHICAL JACOBIAN 87 equilibrium point, f (, = and g(, = 0 (see (5.4. We can approimate the partial derivative f at (, using a formula similar to (1.9 as : f f ( +h, f (, h, but because f (, = 0, we get f f ( +h, h. If we appl the same approach for all derivatives constituting the Jacobian at the equilibrium point (, we get: J = ( f f ( +h, h g g( +h, h f f (, +h h g g(, +h h This approimation will be better if points ( +h, and (, +h are closer to the equilibrium point, i.e., if h is small. It turns out that in man cases the eact values of the derivative will not be important for us and we will be able to find the equilibrium tpe from just the sign of the components of the Jacobian. From (6.5 it is clear that the sign of the Jacobian components is the same as the sign of the functions at the appropriate points, e.g. the sign of f is the same as the sign of f ( + h,, etc. Let us now recall, that the sign of the functions f (,,g(, is represented on the vector field of our sstem. In fact, means that d d > 0 and occurs at the points where f (, > 0, the means that > 0 and occurs at the points where g(, > 0 (see fig.6., etc. Thus from the vector field we can easil determine the sign of the components of the Jacobian matri. For eample, the negative direction of the -component in fig.6.4a will mean that f ( + h, < 0 and hence f < 0, the positive direction of the -component in fig.6.4a will mean that g( + h, > 0 and thus g > 0. Similarl, in Fig.6.4b the vector field at point ( + h, is vertical, thus g > 0 and f = 0. (6.5 d/=g(*,*+h (*,* a g(*+h,* >0 f(*+h,* <0 (*+h,* (*,* f(*+h,* =0 (*+h,* b d/ (*,*+h d/ d/=f(*+h,* (*,* (*+h,* Figure 6.4: We will formulate this result as the following conclusion: Conclusion 1 The sign of the and vector field components to the right from the equilibrium point give the sign of f g and components of the Jacobian. The sign of the and vector field components upward from the equilibrium point give the sign of f g and components of the Jacobian matri J = ( f g f g at this equilibrium point (fig.6.4c. Three notes on the application of this rule: The testing points must be eactl horizontal ( + h, and eactl vertical (, + h with respect to the equilibrium point.

90 88CHAPTER 6. GRAPHICAL METHODS TO STUDY SYSTEMS OF DIFFERENTIAL EQUATIONS Testing points should be as close as possible to the equilibrium and should never cross a null cline when going from the equilibrium. (Putting the testing point to the right from the dashed line in fig.6.4a will be wrong. If a null cline is eactl horizontal or eactl vertical then one of the corresponding derivatives will be zero. ( In fig.6.4b, f g = 0; > 0. =0 =0 =0 =0 =0 =0 a b c Figure 6.5: Eamples. Let us consider several eamples of the application of this graphical approach for null-clines presented in Fig.6.5 and Fig.6.6. On all these figures the testing points are marked b filled circles. ( α β From fig.6.5a we find the following components of the Jacobian: J =, where α,β,γ,δ +γ δ stand for unknown positive numbers. Thus we see that detj = αδ + βγ > 0, and trj = α δ < 0, thus (see fig.5.5 we have a stable equilibrium (stable node or stable spiral. On the basis of these data we cannot sa whether this equilibrium is a node or a spiral as we cannot compute whether the discriminant of the Jacobian is positive or negative, as this depends on the eact values of the partial derivatives. ( α 0 From fig.6.5b we find: J =. Thus detj = αδ > 0, trj = α δ < 0 and D = tr 0 δ 4Det = (α + δ 4αδ = α + αδ + δ 4αδ = α αδ + δ = (α δ > 0, thus we have a stable node. ( +α 0 From fig.6.5c we find: J =. Thus detj = αδ < 0, and we have a saddle. 0 δ From fig.6.6a we find: J = ( +α +β +γ δ Figure 6.6:, hence detj = αδ γβ < 0, and we again have a saddle.

91 6.3. EXERCISES 89 ( +α +β From fig.6.6b we find: J =. detj = αδ + γβ, trj = α δ. Because we do not know the γ δ values of the coefficients α,β,γ,δ, just their signs, we do not know if the detj and trj are positive or negative, thus we cannot determine the equilibrium tpe in this case using the graphical Jacobian. ( +α +β Finall from 6.6c we find: J =. Thus detj = αδ γβ, trj = α + δ > 0. Again we cannot +γ +δ determine the sign of the detj, but positive trj implies that the equilibrium will be unstable. We do not know its tpe: non-stable node, non-stable spiral and a saddle are all possible. We see that the method of graphical Jacobian is a useful tool for finding the equilibrium tpe from the vector field, but sometimes it is not sufficient and we need to know not onl the sign but also the values of the Jacobian coefficients. 6.3 Eercises 1. Sketch the phase portraits of the following sstems using null-clines. Tr to find the tpe of equilibrium using the graphical Jacobian (section 6.. If ou are unable to find the equilibrium tpe using the graphical Jacobian, find it using the method of sec.5.3. Sketch a qualitative phase portrait (without computation of eigen values and eigen vectors. (a (b (c (d { d = 3 + d = { d = + d = { d = d = 3 { d = 4 d = +. The figure below shows the null-clines of three sstems of two non-linear differential equations a b c (a Complete the figures b showing the direction of the vector field in all regions on the plane and on the null-clines. (b Mark the equilibria and find their tpes using the graphical Jacobian (if possible.

92 90CHAPTER 6. GRAPHICAL METHODS TO STUDY SYSTEMS OF DIFFERENTIAL EQUATIONS 3. Algae model using graphical Jacobian (see also problem 6 from section 5.4: { d = (1 0; d = 0. (a Sketch the vector field for the Algae sstem using null-clines. (b Find equilibria. (c Find the tpe of each equilibrium using the graphical Jacobian and draw qualitative local phase portraits. 4. Lotka Volterra model using graphical Jacobian (see also problem 8 from section 5.4: { dn/ = an bnp N 0 a,b,c,d,e > 0 dp/ = cnp dp P 0 (a Sketch the vector field of the sstem using null-clines. (b Find equilibria. (c Find the tpe of each equilibrium using the graphical Jacobian and draw qualitative local phase portraits. (d Does the vector field change if we change the values of parameters a,b,c,d? Additional eercises 5. Find the vector fields of these sstems using null-clines. Find equilibria. Determine the tpe of each equilibrium using the graphical Jacobian and draw qualitative local phase portraits. (a (b (c (d { d = 3 d = + { d = d = { d = 9 + d = { d = d = 1

93 Chapter 7 Plan of qualitative analsis and eamples 7.1 Plan Let us now formulate a plan to qualitativel stud a sstem of two differential equations with two variables and consider several eamples. We stud the sstem: { d = f (, d (7.1 = g(, Our main aim is to plot the phase portrait of this sstem and then predict its dnamics. Based on methods which we have developed we will do it in two steps: I Null-cline analsis II Jacobian analsis Where the Jacobian analsis can be either performed using the determinant-trace method from section 5.3, or using the graphical Jacobian from sec.6.. Note, that the determinant-trace method alwas give us a definitive answer, while the graphical Jacobian method sometimes fails. In more details: Null-cline analsis We assume that on the O-plane the -ais is the horizontal ais and the -ais is the vertical ais. 1. Draw the d = 0 null-clines from the equation f (, = 0 using dashed lines and the d = 0 nullclines from the equation g(, = 0 using solid lines.. Choose a point in one of the regions on the, plane and find the vector field for the -component. Denote the component as a dashed if f (, > 0 and as a dashed if f (, < Find the vector field for the -component at the same point. Denote the component as a solid if g(, > 0 and as a solid if g(, < 0 91

94 9 CHAPTER 7. PLAN OF QUALITATIVE ANALYSIS AND EXAMPLES 4. Find the vector field in the adjacent regions using the following rule: (a change the direction of the dashed component of the vector field if to get to the adjacent region ou cross the dashed null-cline. (b change the direction of the solid component of the vector field if to get to the adjacent region ou cross the solid null-cline. (c show the direction of the vector field on the null-clines close to the equilibrium Jacobian analsis using the determinant-trace method 1. Find equilibria from equations: { f (, = 0 g(, = 0. For each equilibrium (,, find the Jacobian at that equilibrium point ( f / f / J = g/ g/ (, (7. (7.3 Note: Do not forget to substitute =, = into the Jacobian. 3. Determine the tpe of each equilibrium (,. For this compute ( f detj = g f g trj = ( f + g (, (, (7.4 (7.5 To find the tpe of equilibrium use fig.5.5, or the following list: (a If detj < 0; the point is a saddle point (b If detj > 0,trJ > 0, D 0; the point is a non-stable node (c If detj > 0,trJ < 0, D 0; the point is a stable node (d If detj > 0,trJ > 0, D < 0; the point is a non-stable spiral (e If detj > 0,trJ < 0, D < 0; the point is a stable spiral (f If detj > 0,trJ = 0; the point is a center D = (trj 4detJ ( Draw local phase portraits using both knowledge on the equilibrium tpe and the vector fields obtained using null-cline analsis 5. Connect local phase portraits to get the global picture and show attractors and their basins of attraction. Jacobian analsis using the graphical Jacobian

95 7.. EXAMPLES For each equilibrium point (point of intersection of different null-clines choose two points, one of which is located to the right and the other upward from the equilibrium. Find the components of the Jacobian using the following rule: the sign of the component of the vector field to the right of the equilibrium point gives the sign of f right of the equilibrium point gives the sign of g upward of the equilibrium point give the sign of f and the sign of the component of the vector field to the. The sign of the and vector field components g and.. Note that if one of the components is zero, then the corresponding derivative of the Jacobian will be zero. The latter can happen if at a given equilibrium point one or both null-clines are eactl horizontal or eactl vertical. 3. Tr to compute the sign of the determinant and of the trace of the Jacobian and tr to identif the tpe of equilibrium from fig If the tpe of equilibrium cannot be identified, use the analtical determinant-trace method formulated above. 5. Draw local and then global phase portraits of the sstem and make predictions about its dnamics. 7. Eamples Eample 1 Stud sstem (5.3, using null-clines and the determinant-trace method. { d = 3(1 1.5 d = (7.7 Solution We have made studies of different aspects of this model throughout the reader, so let us collect the information. I.Null-cline analsis. The null-clines of this sstem are given in fig.6.3. We repeat it here in fig7.1a: Figure 7.1: II.Jacobian analsis:

96 94 CHAPTER 7. PLAN OF QUALITATIVE ANALYSIS AND EXAMPLES 1. Equilibria. In eample eq.(5.5 we found that this sstem has three equilibria points: (0, 0,(1, 0 and (0.5,1.. Jacobian. We compute( the Jacobian as: f / = ; f / = 1.5; g/ = 0.5; g/ = , thus: J = Let us find equilibria tpes from the Jacobian. ( 3 0 At point (0,0 the Jacobian is: J 1 = point. At the point (1,0 the Jacobian is: J = ( , detj 1 = 3 ( 0.5 < 0, thus this is a saddle, detj = ( < 0, thus this is a saddle point. ( At the point (0.5,1 the Jacobian is: J 3 =, detj = ( ( 0.75 < 0.375, thus we need to find trj 3 = 1.5 and D = ( = 0.75 > 0 thus this is a stable node. 3. Local phase portraits are presented in fig.7.1b. From the null-cline analsis we find the approimate locations of the manifolds of the saddle points and the direction of the trajectories around the stable node. 4. Global picture. There are no general rules to draw the global picture. However in the case of fig.7.1b we can epect that the non-stable manifold of the saddle point at (0,0 will end at the other saddle point (1,0, the non-stable manifold of the saddle point (1,0 as well as most of other trajectories should go to the stable node at (0.5,1 as this is the onl attractor here. Thus we get the phase portrait presented in fig.7.1c. We see that we will have a stable global attractor with the basin of attraction the whole region > 0, > 0 (ecept two aes = 0; = 0. We see that the phase portrait which we got as a result of our stud is qualitativel the same as the phase portrait of this sstem obtained using a computer (see fig.5.1b. Let us tr to appl for this problem the method of the graphical Jacobian Eample Stud the same sstem (7.7, using the graphical Jacobian method. Solution From fig.7.1a we find in the following components of the Jacobian: ( +α 0 1. Point (0,0, J = 0 δ ( α β. Point (1,0, J = 0 +δ ( α β 3. Point (0.5,1, J = +γ 0 equilibrium (stable node or stable spiral.. Thus detj = αδ < 0, thus we have a saddle.. Thus detj = αδ < 0, thus we have a saddle.. Thus detj = γβ > 0, trj = α < 0, thus we have a stable 4. Drawing of the local and global phase portraits is eactl the same as with the previous method

97 7.3. EXERCISES 95 Thus we see that in this case we were able to solve the problem completel using the graphical Jacobian method. The onl difference with the determinant-trace method is that we were not able to sa that the stable equilibrium is a node. However, this has almost no consequences for the phase portrait. 7.3 Eercises 1. Complete the vector field approimations for the null-clines shown in fig7.a,b,c. Mark equilibria and determine their tpe using the graphical Jacobian method. Draw local and then global phase portraits. Tr to describe the dnamics of the sstem, b saing what happens with the variables and in the course of time. a b c Figure 7.:. Draw the phase portrait of the following sstems of differential equations. Eplain their dnamics. (a (b { d = (1 d 0; 0 = (1 { dn = N NP N dp = 3P NP P N 0; P 0 3. In theoretical biolog the following model has been used to stud mrna (M protein (P interaction: { dp/ = bm dp P a,b,d > 0 dm/ = a K b (7.8 d K +P M M P > 0;M > 0 Stud this sstem using the graphical Jacobian approach, i.e. draw null-clines, mark equilibria as points of intersection of the null-clines, determine stabilit of these equilibria and sketch a qualitative phase portrait. 4. The following equations describe the dnamics of predator (P and pre (N populations: here K, b, r are parameters dn/ = rn(1 N K bnp N 0 P 0 dp/ = bnp bp b 0 K 0 r 0 This sstem of differential equations alwas has an equilibrium corresponding to an etinct population of the predator and non-zero population of pre (P = 0;N 0.

98 96 CHAPTER 7. PLAN OF QUALITATIVE ANALYSIS AND EXAMPLES (a Find this (P = 0; N 0 equilibrium. (b Find the Jacobian at this equilibrium point. (c For which values of the parameters the predator population can be driven to etinction (i.e. to that equilibrium. (N.B. Do not use the graphical Jacobian approach for this problem!. Additional eercises 5. Draw the phase portrait of the following sstems of differential equations. Eplain their dnamics. { d (a = d = 0.5 { d (b = + d = + (c (d { d = (5 d 0; 0 = ( 3 { d = d = 4 < < ; < < 6. Lotka Volterra model with competition in the pre population: { dn/ = an en bnp dp/ = cnp dp for a = 3,b = 1.5,c = 0.5,d = 0.5,e = A swimming pool is infested with algae whose population is N(t. The owner attempts to control the infestation with an algicidal chemical, poured into the pool at a constant rate. In the absence of algae, the chemical decas naturall; when algae are present it is metabolized b them and kills them. The equations of the rates of change of N(t and the concentration of the chemical in the pool, C(t, are { dn/ = an bnc dc/ = Q αc βnc where a,b,q,α,β are positive constants. Discuss the meaning of each term in these equations. (a Put a = 1,b = 1,α = 1,β = 1 and show, that the sstem has two non-negative equilibria if Q > 1, find them, draw the phase portrait of this sstem and eplain the dnamics. (b What happens if Q < 1? (c * Find the conditions for control of the infestation for arbitrar positive values of a,b,q,α,β.

99 Chapter 8 Limit ccle 8.1 Stable and non-stable limit ccles In previous chapters we found several possible tpes of equilibria: saddle, node, spiral and center. Some of these equilibria can be attractors which determine the ultimate dnamics of a sstem. However, it turns out that there eists another important attractor for sstems of two differential equations. It is called a limit ccle. The geometrical image of a limit ccle on a phase portrait is a closed curve. The usual phase portrait of a sstem with a limit ccle is shown in fig.8.1a. Here the limit ccle is shown as a bold ellipse. The figure also shows two more trajectories: one inside the limit ccle and one outside the limit ccle. Inside the limit ccle ou can see a spiral. This is not unusual. There is a theorem which states that for sstems of two equations there is alwas at least one equilibrium point inside the limit ccle. In most of the cases such an equilibrium point is a spiral. Outside the limit ccle we can see a trajector which approaches the limit ccle and winds onto it. The limit ccle which is shown in fig.8.1a is called a stable limit ccle. This is because if we start on a trajector close to this limit ccle, this trajector will approach this limit ccle in the course of time. There also eists another tpe of limit ccle called a non-stable limit ccle (fig.8.1b. If we start on a trajector close to a non-stable limit ccle, it will diverge from this limit ccle. In order to distinguish the stable and non-stable limit ccles we will draw the non-stable limit ccle using a dashed line. The main questions regarding the limit ccle are: what will be the dnamics of sstems with limit ccles and how do limit ccles occur in sstems of two differential equations? We will also consider an eample of a biological sstem with a limit ccle. 8. Dnamics of a sstem with a limit ccle. What will be the dnamics of sstems with a limit ccle? In section 4.4. we studied an equilibrium point called a center. The phase portrait of that point (fig.4.8 was a set of embedded ellipses. The corresponding dnamics are oscillations of the and variables. Therefore, the dnamics which corresponds to the trajector which starts on the limit ccle will also be oscillations. Fig.8.a, fig.8.3a shows an eample of dnamics of the variable for the trajector which starts on the limit ccle (at the point C 97

100 98 CHAPTER 8. LIMIT CYCLE B C A C A B a b Figure 8.1: Phase portrait of a sstem of two differential equations with a limit ccle. A C B a time b time c time Figure 8.: Dnamics of a sstem with a stable limit ccle from fig.8.1a. C A B a time b time c time Figure 8.3: Dnamics of a sstem with an non-stable limit ccle from fig.8.1b. in fig.8.1. The dnamics of trajectories originating around the limit ccles will depend on its tpe. If the limit ccle is stable, then a trajector which starts inside the limit ccle will approach it in the course of time and we obtain oscillations with initiall increasing amplitude (fig.8.b. If the trajector starts outside the limit ccle, then we obtain oscillations with an initial decrease of amplitude until the trajector reaches the limit ccle (fig.8.c. For a non-stable limit ccle the dnamics are different. Onl the trajector which starts on a limit ccle will have oscillator dnamics (fig.8.3a. Other trajectories will have different behavior. The trajector which originates inside the limit ccle will approach the stable equilibrium inside the limit ccle and we obtain oscillations with graduall decreasing amplitude (fig.8.3b. The trajector which starts outside the limit ccle blows up to infinit or to another equilibrium (fig.8.3c. The oscillations in fig.8.3a with a non-stable limit ccle are highl improbable in real sstems. This is because for such oscillations the trajector must start eactl on the limit ccle and even small disturbances will switch our sstem either to the behavior of fig.8.3b or fig.8.3c. Therefore, for real sstems the non-stable limit ccle just determines the basin of attraction of the stable equilibrium which is located inside this limit ccle.

101 8.3. HOW DO LIMIT CYCLES OCCUR? How do limit ccles occur? In man cases the limit ccle occurs as a result of the changing of a parameter in a sstem of differential equations. The most usual process of formation of a limit ccle is the following. Assume we have a sstem of two differential equations with a parameter c: { d = f (,,c d = g(,,c (8.1 Assume that at some parameter value c = c 1 sstem (8.1 has a global attractor which is a stable spiral (fig.8.4a. This means that all the trajectories which originate close to or even far from this equilibrium approach it in the course of time. Because our equilibrium point is a stable spiral the Jacobian of the sstem at this point will have two comple eigen values λ 1, = α ± iβ;α < 0. However, because our sstem will now depend on the parameter c the eigen values will also depend on this parameter: λ 1, (c = α(c ± iβ(c; (8. Because we have assumed that at c = c 1 sstem (8.1 has an equilibrium which is a stable spiral, then α(c 1 < 0. When we change the parameter c the value of α(c will change and at some c it can become a positive number α(c > 0. This means that the stable spiral in fig.8.4a will become an unstable spiral. However, as we found in chapter 4 (fig.4.3, the eigenvalues onl give us the dnamics close to the equilibrium point of our sstem. Therefore, it can happen that although close to the equilibrium point our spiral becomes unstable, the global behavior far from the equilibrium remains the same, i.e. far from the equilibrium we still have a converging flow (fig.8.4b. Hence, in our phase portrait we have two tpes of flow: the diverging flow from the unstable spiral around the equilibrium and the converging flow from the peripher. Simple geometrical consideration shows that these two flows must be separated from each other. The line of separation will be the limit ccle in our sstem. Therefore, we will obtain a phase portrait as in fig.8.1a. a b Figure 8.4: Appearance of a limit ccle in a sstem with a parameter. Note, that such a mechanism of limit ccle formation frequentl occurs in sstems of two equations. It is called the Hopf bifurcation. From our analsis it follows that the Hopf bifurcation occurs when α(c changes its sign, i.e. at the parameter value c where: α(c = 0 (8.3

102 100 CHAPTER 8. LIMIT CYCLE 8.4 Eample of a sstem with a limit ccle Let us consider the Holling-Tanner model for predator-pre interactions. { dp/ = rp(1 P K arp d+p dr/ = br(1 R P P > 0;R > 0 (8.4 here P denotes the pre and R the predator population, the term rp(1 P/K describes the growth of arp the pre in absence of predator, d+p accounts for the predator-pre interaction. At a = 1,b = 0.,r = 1.,d = 1.,K = 0.7 sstem (8.4 has one equilibrium point (for P > 0,R > 0 which is a stable spiral. The null-clines, phase portrait and dnamics of this sstem are shown in fig.8.5. r 10 r pr p p Figure 8.5: Dnamics of sstem (8.4 for K = 0.7. a-null-clines; b-an orbit; c-time-plot for the both variables for the orbit from fig.b If we increase the value of the parameter K which accounts for the carring capacit (K = 1.0 the tpe of equilibrium changes and it becomes an unstable spiral. As we discussed in section 8.3 we epect the formation of a stable limit ccle. We can clearl see it in fig.8.6b. The trajector, which starts from the same initial conditions as the trajector from fig.8.5b will now approach some closed curve which is a limit ccle and the dnamics of the sstem will be oscillator (fig.8.6c. r 10 r r p p p Figure 8.6: Dnamics of sstem (8.4 for K = 1.0. a-null-clines; b-an orbit; c-time-plot for both variables for the orbit from fig.b If we start a trajector inside the limit ccle then, as we predicted in fig.8.1a and fig.8.b, the trajector will approach the limit ccle and the dnamics will be oscillation with increasing amplitude fig.8.7a,b. The complete phase portrait of this sstem at K = 1 is shown in fig.8.7c. You will learn more about the biological consequences of this tpe of dnamics in the course Theoretical Biolog b R. de Boer.

103 8.5. EXERCISES 101 r r p 5 r p p Figure 8.7: Dnamics of sstem (8.4 for K = 1.0. a-an orbit originating inside the limit ccle; b-timeplot for the both variables for the orbit from fig.a; c-phase portrait of sstem (8.4 Conclusion 13 A limit ccle is a closed trajector on the phase portrait of a sstem of two differential equations. If trajectories around the limit ccle converge onto it, then the limit ccle is called a stable limit ccle. If trajectories around the limit ccle diverge awa from it, then the limit ccle is called a non-stable limit ccle. The dnamics of a sstem with a stable limit ccle is oscillator. The dnamics of a sstem with an non-stable limit ccle is either converging to the equilibrium which is located within the limit ccle or diverging, possibl to infinit. Limit ccles can appear as a result of a Hopf bifurcation, i.e. the process where the real part of the comple eigenvalues change their sign. 8.5 Eercises 1. Assume that a sstem of two differential equations has two equilibria which are a saddle point and a non-stable spiral. The phase portrait of this sstem is partiall shown in fig.8.8a What is missing here? (a Complete the phase portrait of this sstem (b Qualitativel sketch the dnamics of the variable ( dependence of the variable on time for the initial conditions which are shown in fig.8.8 b the point A in fig.8.8a.. Complete the phase portrait of this sstem shown in fig.8.8b,c. Qualitativel sketch the dnamics of the variable ( dependence of the variable on time for the initial conditions which are shown b points A,B,C in fig.8.8b,c. (Note, that in fig.8.8b,c a stable limit ccle is shown b the solid line and a non-stable limit ccle is shown b the dashed line 3. As we discussed in (8.3 the necessar condition for the appearance of a limit ccle via a Hopf bifurcation is that the eigenvalues of the Jacobian matri of a sstem at the equilibrium point are comple and the real part of the eigenvalues are zero α(c = 0. Prove that this condition can be rewritten in the following wa using the det and tr of the Jacobian matri: trj(c = 0; detj(c > 0 (8.5

104 10 CHAPTER 8. LIMIT CYCLE 3 1 a A B C b A C B A c Figure 8.8: 4. One of the classical models for oscillator phenomena in biochemical sstems is a model called the Brusselator. In dimensionless form this model can be written as the following sstem of two differential equations: { d = a (b > 0; > 0 d = b a > 0;b > 0 here and are concentrations of two biochemical species and a and b are parameters. Stud sstem (8.6 for a = 1. (a Find a non-trivial equilibrium. (b Determine the stabilit of this equilibrium as a function of the parameter b. (c Find the value of b when sstem (8.6 undergoes a Hopf bifurcation. (Note: it can be helpful to use equations (8.5. (d For which values of b do ou epect oscillations in sstem (8.6? 5. Stud the following predator-pre model { dp/ = rp(1 P K arp h+p dr/ = carp h+p dr P 0;R 0 (8.7 (a Draw null-clines of this sstem for r = 1,a = 3,h = 0.1,c = 1.,d =.5 and two values of K, K = 0.8 and K = 1.6. (Hint: the maimum of the parabola A( a( b is reached at the middle between its roots (i.e. at = a+b. (b Determine the stabilit of non-trivial equilibrium in both cases using the graphical Jacobian. (c For which value of K do ou epect oscillator behavior? (d Etend our analsis for arbitrar positive values of the parameters (r,a,h,c,d,k > 0 provided ca > d. Find for which values of K the non-trivial equilibrium is stable. When do we epect oscillations? (Note, that critical values of K should be a function of the other parameters of the sstem. (8.6

105 Chapter 9 Dictionar absolute value autonomous sstem attractor basin of attraction bifurcation carring capacit center point comple conjugate number component of the vector determinant derivative differential equation direction field eigen value eigen vector equilibrium general solution harvesting imaginar part of comple number initial value problem Jacobian linear approimation modulus node non-stable manifold absolute waarde autonoom ssteem attractor basin van attractie bifurcatie draagkracht centrumpunt comple toegevoegde component van de vector determinant afgeleide differentiaal vergelijking vectorveld eigenwaarde eigenvector evenwicht algemene oplossing oogsten het imaginaire deel van een comple getal beginwaarde probleem Jacobi-matri lineaire benadering modulus knooppunt instabiele manifold 103

106 104 CHAPTER 9. DICTIONARY null-cline parameter partial derivative particular solution phase space phase portrait real part of comple number saddle spiral stabilit stable manifold sstem of differential equations trajector trace of the matri variable vector isocline parameter partiële afgeleide specifieke oplossing faseruimte faseplaatje het reëele deel van een comple getal zadelpunt spiraalpunt stabiliteit stabiele manifold stelsel differentiaal vergelijkingen trajectorie spoor van de matri variabele vector

107 Chapter 10 Hints: Solution of the initial value problem for a linear sstem Let us illustrate how we can find a solution for the initial value problem, on eample of sstem (4.11. Problem: Find the solution for the following initial value problem: ( d d ( = , ( ( (0 (0 ( = 46 (10.1 Solution: We alread found a general solution of (10.1 as formula (4.0: ( ( ( = C1 4 e 1 t +C 4 e 3 t (10. Using it we can find a particular solution, corresponding to an given initial conditions. We proceed as follows. At time t = 0 (10. gives: ( ( ( = C1 4 e 1 0 +C 4 ( This epression will satisf the initial conditions if: C 4 1 { sstem of equations for unknowns C 1 and C : 4C1 4C = 4 C 1 C = 6 ( ( e 3 0 = C 4 1 +C 4 ( +C 4 =. ( 46. This gives the following We now solve this sstem using the method described in section From the second equation we find: C 1 = 6 + C, or C 1 = 3 +C. After substitution of this epression to the first equation we find: 4(3 +C 4C = 4, i.e. 1 4C 4C = 4, or 8C = = 16 and C = 16/( 8 =. We now find C 1 from our substitution as C 1 = 3 +C = 3 = 1. Thus C 1 = 1,C = give the solution of our sstem satisfing given initial conditions. Let us rewrite it as: ( ( ( ( = 1 4 e 1 t 4 e 3 t = 4e 1 t ( 4e 3 t e 1 t ( e 3 t, thus the particular solution is given b (t = 4e 1 t + 8e 3 t, and (t = e 1 t + 4e 3 t Equilibria/derivatives { d Problem:(A Find equilibria of the following sstems = f (, d = g(, (B Find the following partial derivatives at each equilibrium point ( f (a { d = 4 d = 4 (see definition in section 5.1.., f, g, g.

108 106 CHAPTER 10. HINTS: Solution: (A We rewrite the first equation as 4 = ( = 0. It has solutions = 0 or =. Let us substitute them to the second equation: For = 0, we get 0 4 = 4 = 0, thus = 0. Therefore we found one equilibrium = 0, = 0. Now substitute =. We get 4 = 4 8 = 0, thus =. We found the second equilibrium =, =. (B f (, = 4, thus f = 4 ; f =. g(, = 4, thus g g = ; = 4. At the equilibiria points: Equilibrium = 0, = 0: f = 4 0 = 4; f g g = 0 = 0; = 0 = 0; = 0 4 = 4. Equilibrium =, = : f = 4 = 0; f g g = = 4; = = 4; = 4 = 0.

109 Chapter 11 Answers for selected eercises and Formulas lists 1. (a 3a (b 6 r Eercises Chapter 1 5r 30r+5 = r +36r+6 r(6r+1. (a lim a+q c + = 0 (b lim N an +q b N +c +dn = a d 3. (a 3r + 5(r + 1 = 6r + 4; r = 7 8 (b + 4 = 4; 1, = ± 0 = (c (b N k N = 0; N = 0, or N = bk. (d N = 0, or b d(1 + N k(b d k = 0, thus N = d. (e N = 0, or N = h( b d 1 4. (a From 1st eq. = 5, substitution to nd eq. gives = 4, thus = 3. (b from 1st eq. = b a to nd eq. gives = b a = b ba a da bc = b da bc (c From nd eq. (4 = 0, this = 0 or = 4. Substitution to the 1st eq. gives or = 0 and = 0.5. Now substitute = 4, we get = 7, thus all solutions are given: (0,0,(0.5,0,(4,7. (d From nd eq. (9 3 = 0, thus = 0 or = 9 3. Substitution to the 1st eq. gives: (0,0,(4,0,(0,9,(.5,1.5. (e From nd eq. N = 0, or R = δ. Substitution to the 1st eq. gives: : (0,0,(k(1 d,0,(δ, 1 d δ ak. 5. (a f ( = 1 3 = 3 4 (b f ( = e 5 and f ( = 5e 5 (c ((4 ( + 3 = (d = +a (a b a, horizon- 6. (a -intercept = b, zeros n = ± c tal asmptote: = a a (b -intercept = hr a, zeros R = h and R = K. 7. (a fig.a (b fig.b (c horizontal asmptote = 7, intercept (0 = 4. (fig.c. Dependence on a. If a increases, the horizontal asmptote = 7 will be approached at a slower rate. (fig.c a b c 8. (a (see below,fig.a (b (see below,fig.b a P a 0.33 a b c (c (see above, fig.c (d (see above, fig.d N (e N = 0 and P = d(n + a (see below, fig.a (f R = 0 and d c (b R(R + a = N (see below, fig.b da N a N b b R N c b/a 1 (d+1/k (g R = 0 and b a (1 ( 1 k + dr = N (see above, fig.c. (h N = P a (ep (1 + b (see above, fig.d. 9. (a (( ( + 1 = 5 = 5 (b a b p : 4b a p = 1 4 p R P d N a d (1+b/e P

110 108 CHAPTER 11. ANSWERS FOR SELECTED EXERCISES AND FORMULAS LISTS 10. no answer provided for this eercise 11. (a = e ln(, thus ( = (e ln( = ln((e ln( = ln( 1 (b = 3 = 3 3, thus ( 3 = 3 5 (c f ( = cos( and f ( = sin( (d f ( = cos ( and f ( = sin(cos( (e (a e b = ae b + abe b (f = ( 3 (4 3( 5 = ( 3 ( 3 (g = a(b c ab (b c (h = (1+ d d (1+ d = 1 (1+ d = ab ac (b c (i = nn 1 ( n +a n n n 1 n ( n +a n = nn 1 a n ( n +a n 1. no answer provided for this eercise 13. (a d f d = 3, d f = 3 d (b d f d = ae a, d f = ae a d (c d f = d f d d 14. (a -intercept =,zero = 4,horizontal asmptote = 0, as = lim 3+ = lim = 4 0, vertical asmptote is at 3 + = 0, i.e. = 1, and =. b (b = a : 3 c = a b (3 c, thus -intercept = ac b, one zero: = 3 c 15. (a no answer provided for this eercise (b = + 3 = ( 1(+3, as we have + at the parabola is opened upward. (Graph fig.a (c horizontal asmptote = 0, vertical asmptote = 3, no zeros (fig.b 3 a 1 3 (d (fig.a below. Parameter b shifts the horizontal asmptote. (e no answer provided for this eercise (f h = kr 4. b 4+b a 16. no answer provided for this eercise 1. (a 1, = ± i. Eercises Chapter b (b 1 =, = 3. (. (a 4 ( ( 1 1 = 31, deta = +4 = 6. ( (b a b ( ( c d = 0 b, deta = ad bc. ( ( 3. (a λ 1 = 1, v 1 = k 1 1 ; λ 1 = 3, v = k 1 1, where k is an arbitrar number. ( (b λ 1 = 1, v 1 = k 4 ; λ 1 = 3, v = k where k is an arbitrar number. ( (c λ 1 = 1+i, v 1 = k 5 i k ( 5 + i n ( 4 ; λ 1 = 1 i, v =, where k is an arbitrar number. z z 4. (a = at (1, it is ; = at (1, it is 4; (b z = 5 3, at (3,4 it is -18; (c z N = br d, at 0,0 it is d; at R = d b,n = 1 it is 0; z R = bn, at 0,0 it is 0; at R = d b,n = 1 it is b. z (d P = (e (f (g (h a z, (1+P M = b; z z N = a en bp, P = bn; z M = L νa h+a, z νmh A = δ ; (h+a z P 1 = ap 1P (h+p 1 +P and z P = a(h+p 1 (h+p 1 +P z N = b NT (+cn (1+cN+bT N 5. (a 3 90 = 81 = ±9i z T = b N (1+cN( (1+cN+bT N (b ( 1 + i + (4 + 7i = 3 + 9i (c (4+5i (7+i = 8+8i+35i+10i = i (d 1 i = i i = i 6. (a 1, = ±11i (b 1, = 1 ± i,

111 ( A =, deta = 3; D = ( detd = 3 D = = = 1; = 3 =. ( 5 4 1,, detd = 6, thus B usual method: = 5 thus after subs into nd eq. we get = 4, or = and hence = 5 = 1. ( 8. (a λ 1 =, v 1 = k 6 3 ; λ 1 = 5, v = k where k is an arbitrar number. ( (b λ 1 = 3, v 1 = k 1 1 where k is an arbitrar number. 9. f (, + + ; λ 1 = 5, v = k Eercises Chapter 3 ( 6 4, ( 1 7, 1. at t = 4,n = 30e , The double size at : t = ln( k = ln( sec (a 15+8 = (3 ( 5+ Phase portrait in fig.a. (below, (zeros of parabola 1 = 3 and = 5, attractor = 5, basin > 3. f( (b Phase portrait in fig.a. (below, attractor = 4, basin > 1. (c Phase portrait in fig.c. (below, attractor = 3, basin < 0 and =, basin, > 0. 1 a c (d fig.d. (above, attractor =, basin < 0 and =, basin, > 0. (e fig.a (below attractor = 4 basin 0 < < 10; fig.b (below, attractor = 0, basin < 1 and attractor = 6, basin > 1. a f( 0 1 b 6 d 0 1 (f graph is shown in fig.c (above, attractor = 0, basin < 1 and attractor =, basin > 1. c General consideration. Tpical graphs are shown in the figure. s shifts the graph upward. If the total shift is less then the minimum of the graph, nothing will change. If the shift is more than the minimum (fig.b, the will go to the right equilibrium a For questions (a f min (b final is = 0, for question (c final is = 1. (d s ma = The maimal ield if h equals maimum of r n (1 n/k (see the last section of this chapter, which gives h ma = rk 4 6. (a the general solution W = 400/0.3+Ae 0.3t, W = e 0.3t ; (b t = ln(0.504/0.3 =.9; (c t = ln(0.9/0.3 = 0.4; 7. the stead state value is m = α α+β and the characteristic time τ = 1 α+β. 8. stable equilibrium at n = k(r h r, ield is given b ield = hn = kh(r h r, which has maimal value ield ma = kr the last strateg is better as population is more stable: small decrease in population size is OK for the last strateg, but for the other case small decrease will result in the population etinction. b Eercises chapter 4 1. (a ( ( ( = C1 1 1 e t +C 1 1 e 3 t (b ( ( ( = C1 11 e t +C 1 e 5t (. ( = 3 e 3t { dc1 3. sstem is = 0.01C C dc, the solution: ( ( = 0.04C C ( C1 C = e 0.05t, or C 1 = e 0.05t, and C =.4.4e 0.05t. 4. (a det 1 λ 3 λ 4 = (1 λ(3 λ 8 = λ ( 4λ 5 = 0, λ 1 = 5,v 1 = k 4 4 ; λ = 1,v = ( k 4, thus this is a saddle point (the corresponding phase portrait form the figure (a below.

112 110 CHAPTER 11. ANSWERS FOR SELECTED EXERCISES AND FORMULAS LISTS a (b characteristic ( eq. is λ 6λ + ( 8 = 0, λ 1 =,v 1 = k 13 ; λ = 4,v = k 11, i.e. a non-stable node (phase portrait see fig.b. (c characteristic eq. is λ λ + = 0, λ 1 = 1 + i, λ = 1 i,(no eigen vectors required. A non-stable spiral (qualitative phase portrait see fig.c. b (d characteristic ( eq. is λ + 4λ + 3 = ( 0, λ 1 = 3,v 1 = k 1 1, λ = 1,v1 = k 1 1 i.e. a stable node ( phase portrait see fig.a below. (e characteristic eq. is λ +λ+ = 0, λ 1 = 1+ i, λ = 1 i,(no eigen vectors required. A stable spiral (qualitative phase portrait see fig.b. (f characteristic eq. is λ + 1 = 0, λ 1 = +i, λ = i,(no eigen vectors required. A center (qualitative phase portrait see fig.c. (g characteristic ( eq. is λ 1 ( = 0, λ 1 = +1,v 1 = k 1 3 ; λ = 1,v 1 = k 1 1, i.e. a saddle ( phase portrait see fig.d. c more than 1 1 we will have a stable spiral (stable eq.. Qualitative phase portrait can be plotted as in the previous eercise. 6. In linear sstem oscillation occur if the equilibrium tpe is a center. Characteristic equation is given b λ + (a + bλ + ab + 3 a = 0. The center occurs if the roots comple with a zero real part (b = a. These conditions give a + a 3 < 0, thus oscillations occur, if b = a and 3 < a < (a sstem is { d = (a + c + b d = a (b + e { d (b sstem is = 5 + d = 0.5 5, eigen values λ 1 = 4, λ = 6, stable node, stable equilibrium. (c formula for eigen values in general case are: λ 1, = (a+b+c+e± (a+b+c+e 4 [(a+c(b+e ab]. We see that (a + c(b + e ab = ae + bc + ce > 0, thus epression under is less than (a+b+c+e, thus we can have either stable node or stable spiral, both tpes of equilibria are stable. Eercises chapter 5 1. (a Equilibria: (0,0,(4,0. f, f, g, g are: at (0,0:0, 4,4, 0.5; at (4,0: 0, 4, 4, 0.5. (b Equilibria: (0, 0,( 9, 9. Derivatives are: at (0,0:9,0,1, 1; at ( 9, 9: 9, 18,1, 1. (c Equilibria: (0, 0,(0.5,. Derivatives are: at (0,0:,0,0, 1; at (0.5,: 0, 0.5,4,1., a b c d 5. Characteristic equation is given b det 3 λ 1 a λ = ( λ( 1 λ+3a = λ +3λ++3a = 0, λ 1, = 3± 1 1a. If 1 1a < 0, i.e. a > 1 1 we have comple roots. Because the real part for comple roots is negative we have a stable spiral. If the roots are real (a > 1 1, then λ = 3 1 1a is alwas negative. The first root λ 1 = a will be positive is a > 0, i.e. if the epression under the root we have a value more than 9 ( 9 = 3, this gives us 1 1a > 9, or a < 8 1 = 3. Thus if a < 3, then λ 1 > 0 and we have a saddle point. In the interval 3 < a < 1 1 we have λ 1 < 0, thus both real roots negative, and we have a stable node. Conclusion, if we increase a from, we will first have a saddle point (unstable eq. till a = 3, then the saddle will become stable node (stable eq., until a = 1 1. If a becomes (d Equilibria: (0, 0,(0, 0.5,(1, 0,(0.5, 0.5. Derivatives are: at (0,0:1,0,0,1; at (0,0.5: 0.5,0, 1, 1; at (1,0: 1, 3,0, 1; at (0.5, 0.5: 0.5, 0.75, 0.5, From nd P = 0 or N = d c, which after substitution to 1st equation gives 3 equilibria (0,0, ( a e,0,( d c, a b ed bc. All non-negative if: ac ed 3. From nd M = d c P. Substitution to 1st gives one non-negative equilibrium: P 1 = ac bd and thus M 1 = d c P 1 4. From nd I = 0, or S = α β. Substitution to 1st gives equilibria: (I = 0,S = B µ and (I = B α µ β,s = α β. The first equilibrium is alwas positive, the second one is positive if Bβ > αµ. 5. (a non-stable spiral, non-stable. (b saddle, non-stable. (c saddle, non-stable.

113 111 (d center, neutrall stable. 6. (a equilibria (0,, (1,1 which are stable node and saddle; (d see fig.a below. (d Null-clines are given below in fig.c, the graphical Jacobian gives a non-stable equilibrium (node, spiral. The real Jacobian gives a non-stable spiral. Phase portrait fig.d. a 7. onl one equilibrium (0, 0. At this equilibrium, detj > 0,trJ < 0, thus equilibrium is stable. ( 8. (a ( d c, a dv 0 b, linearization = Jv, where J = (c center point; (d see fig.b above. 9. no answer provided for this eercise 10. no answer provided for this eercise 11. no answer provided for this eercise 1. no answer provided for this eercise b bd ac c b Equilibria of sstem. From nd e = 0, to 1st eq., 0 g = 0, i.e. equilibrium is (0,0. Jacobian at the F equilibrium is: e = ( e3 +(1+ae ae g e = 3e + F (1 + ae a, at (0,0 it is a, g = 1, G ( e = ε G g = 0, thus J = a 1 ε 0 and we get detj = ε > 0,trJ = a,d = a 4ε, thus equilibrium is alwas stable, and it is a node if D > 0, i.e. a > 4ε, and a stable spiral if a < 4ε. a b. For fig.a below we have two equilibria (marked. For the left one the graphical Jacobian gives a stable equilibrium (stable node or stable spiral. The right equi- ; librium gives a saddle. a b c For fig.b we have one equilibrium (marked. The graphical Jacobian gives a stable equilibrium (stable node or stable spiral. For fig.c we have one equilibrium (marked. The graphical Jacobian gives a stable node. 3. (a See fig.a below. (b,c equilibria: (0,,graphical Jacobian gives a stable node; (1, 1: saddle. For phase portrait see solution problem 6 chapter 5. c d Eercises chapter 6 1. (a Null-clines are given below in fig.a, thus the ( graphical Jacobian on basis of black points is: J = α β a b c γ δ. This gives for detj = αδ+βγ, and we do not know 4. (a See fig.b above. (b,c equilibria: (0, 0,graphical what is the sign. Thus graphical Jacobian ( does not Jacobian: saddle; nd equilibrium ( d c, a b, graphical work here. The real Jacobian here is: J = , Jacobian: center. Phase portrait fig.c above. (d no, for positive parameters. that gives detj = < 0, thus we have a saddle point. Phase portrait is in fig.b. 5. (a, (b no answer provided for this eercise a b c d (c vector field ( see fig.a; Equilibria (0, 0,graphical Jacobian: J = α β γ δ, detj = αδ betaγ < 0,saddle; nd ( equilibrium ( 9, 9,graphical Jacobian: J = α β γ δ, detj = αδ+βγ, sign unknown, trj = α δ, equilibrium tpe unknown; phase portrait fig.b. (b Null-clines are given above in fig.c, Graphical Jacobian does not work here. The real Jacobian gives stable spiral. Phase portrait is in fig.d. (c Null-clines are given below in fig.a, the graphical Jacobian gives saddle point. Phase portrait in fig.b. a ( 9, 9 (0,0?? b

114 11 CHAPTER 11. ANSWERS FOR SELECTED EXERCISES AND FORMULAS LISTS (d no answer provided for this eercise M M P r/b P Eercises chapter 7 ab dm 1 a 0 P b P 0. K c N d N 1 1 a d 1 1 b e 1. (a graphical Jacobian: at 1 stable node. At saddle. Null-clines fig.a above, phase portrait fig.d. (b At 1 saddle. At not known, stable node/spiral. At 3 saddle. Null-clines fig.b above, phase portrait fig.e. (c At 1 not known, stable node/spiral. At saddle. At 3 not known, stable node/spiral. Null-clines fig.c above, phase portrait fig.f.. (a Null-clines : = 0, = 1, : = 0, = 1, fig.a below. Equilibria (0,0,(1,0,( 1, 1. Graphical Jacobian: (0,0 unstable node. (1,0: stable node. ( 1, 1 saddle. Phase portrait fig.b. One attractor (1, 0. Basin of attraction shaded c f 3 ( 4. (aequilibrium (K, 0; (b J = r rn K bp bn ( bp bn b r bk 0 bk b ; (c Equilibrium is stable if 0 < K <. 5. (a-(d no answer provided for this eercise 6. Fig c,d above 1. see figure below 3 1 Eercises Chapter 8 A t = P. see figure below 1 B a 1 b 0 c N d B 3 C A 3 1 C A (b Null-clines N: N = 0,P = N, P: P = 0,P = 3 N, fig.c above Equilibria (0, 0,(, 0,(0, 3,(1, 1. Graphical Jacobian: (0,0, unstable node; (,0, stable node; (0, 3, stable node: (1, 1 cannot determine equilibrium tpe using graphical Jacobian. Need real Jacobian, which gives saddle. Phase portrait fig.d. Two attractors (, 0 and (0, 3, basins of attraction shaded C A B B t A C 3. Null-clines P: M = d PP b, M: M = abk d M, fig.a below (K +P One equilibrium. Graphical Jacobian gives stable equilibrium node/spiral. Phase portrait fig.b. One attractor 1. Basin of attraction shaded.

115 113 Main graphs: Real eigen values λ1 λ λ 1>0 λ <0 saddle stable node λ <0 λ <0 non stable node λ >0 λ >0 1 1 Comple eigen values α + i β center stable spiral non stable spiral α=0 α<0 α>0 Equilibrium tpe can be determined form the det-tr of the sstem, as shown in the net figure: det A stable spiral non stable spiral D=0 stable node center non stable node Quadratic equation: Equation Aλ +Bλ+C = 0, has solutions given b the following abc formula: λ 1 = B± B 4AC A saddle 1 tr A Differentiation rules: ( n = n n 1 ; (e = e ; (sin( = cos(; (cos( = sin(; (ln( = 1 ( f g = f g+g f ; ( f g = f g g f g ; f (g( = f (gg. 1D differential equations: Equation dn = kn has the solution: N(t = N 0 e kt ; N 0 is an (arbitrar initial value of N. Characteristic time of change is τ = 1/k. Sstems of linear differential equations: { d For sstem = a+b d = c+d, characteristic equation: det a λ b c d λ = 0, gives eigen values λ 1,, which can be real ( λ 1, λ, or comple (λ 1, = α±iβ. Eigen vectors can be found from sub- ( stitution of( λ 1, λ to: v b = a λ v or ( v The general solution is given b: ( = C 1 ( v1 v 1 v e λ 1t +C ( v v ( d λ = c e λ t Eigen values determine equilibrium tpe as shown in the figure below. For saddle and nodes the manifolds are directed along the eigen vectors. Note, that for the linear sstem deta = ad bc, tra = a+d, D = (tra 4 deta and D < 0 above the parabola on the figure. { d { For general sstem = f(, f(, = 0 d equilibira are = g(, g(, = 0 The -null-cline is given b f(, = 0,the -null-cline is given b g(, = 0. Equilibrium tpe can be found from the Jacobian: J = ( f g evaluated at the equilibrium. The signs of these derivatives can be found using the graphical Jacobian method as shown in the figure below: f g 1

116 114 CHAPTER 11. ANSWERS FOR SELECTED EXERCISES AND FORMULAS LISTS