Systems Biology: Mathematics for Biologists. Kirsten ten Tusscher, Theoretical Biology, UU

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1 Systems Biology: Mathematics for Biologists Kirsten ten Tusscher, Theoretical Biology, UU 1

2 Introductie: Wiskunde Maandag 20 febr en maandag 27 febr al 2 keer wiskunde les Onderwerpen nodig voor wiskunde en theoretische biologie deel Werkcolleges vallen onder bonuspunt wiskunde/ theoretische biologie Voor mensen die deze wiskunde best lastig vinden: er bestaan geen domme vragen, we willen je graag helpen! doel is niet om jullie zo moeilijk mogelijke wiskunde te leren doel is laten zien dat wiskunde biologische inzichten oplevert alle antwoorden staan uitgewerkt achterin het dictaat extra materiaal: Voor mensen de deze wiskunde makkelijk vinden: Deze wiskunde is alleen de basis voor het volgende cursus deel. Daarna gaan we dingen doen die ook voor jullie nieuw zullen zijn. 2

3 Introductie: Wiskunde (2) Wat is de rode draad deze 2 dagen? Er lijkt misschien helemaal geen rode draad te zijn. We behandelen een hele serie wiskunde onderwerpen. Maar... Allemaal wiskundige kennis die een bioloog zou moeten hebben. Allemaal wiskunde technieken die we in rest cursus gebruiken. Dus... Doel is wiskunde basis voor rest cursus en studie te leggen. 3

4 Chapter 1 Preliminaries 4

5 Chapter subjects Bioinformatics logarithms summations Chapter 1 Wiskunde: solving equations drawing graphs of functions Chapter 2 Wiskunde: matrices, eigenvectors complex numbers 5

6 Logarithms and Exponentials log a (c) = b is inverse of a b = c log a (c) = b to which power b you need to raise a to get c 2 4 = 16 log 2 (16) = log 2 (2 4 ) = = 1000 log 10 (1000) = log 10 (10 3 ) = 3 e log e (e 4 ) = ln(e 4 ) = 4 Example factor 10 diff.: log 10 (b) = a versus log 10 (b ) = a 10 power 10 diff.: b = 10 a versus b = 10 a 10 = (10 a ) 10 = b 10 6

7 Logarithms and Exponentials (2) Rules for log arithmetic: (if logs have the same base) log(a) + log(b) = log(ab) log(a) log(b) = log(a/b) a log(b) = log(b a ) Rules for exp arithmetic: a b + a c a b+c a b a c = a b+c a b : a c = a b c (a b ) c = a bc Logs and exps cancel each other: log 10 (10 a ) = a e ln(b) = b Example e (a ln(b/c)) Solve with rules for logs or with rules for exps 7

8 Summations A simple summation: = 23 Let us define: n 0 = 5, n 1 = 3, n 2 = 6, n 3 = 9 Now we can write: n 0 + n 1 + n 2 + n 3 = 23 This can be more shortly written as: 3 i=0 n i = 23 Or, if we are being a bit sloppy: n i = 23 Examples 5i=0 i 2 = i=0 i 4 = i=4 i=0 (7 i)i3 =

9 Variables and Parameters Equation: x: unknown variable x = 5: solution of equation 3x = 15 Equation:, with a, b constants x: unknown variable a, b: parameters x = b/a: solution of equation ax = b 9

10 Solving a Single Equation Idea: 1) bring variable to one side 2) factor out the variable 3) check for multiple solutions Examples 3x 9 = 15 (ax b)(x c) = 0 (x + a)(3 3x) = 3 10

11 Solving an Equation with Fractions Rules for working with fractions: a b + c b = a+c a b b+c a b + a a b d c = ac bd ca cb = a b a bc = a c b = ac b but = ac+ab bc Example an 1 + bn cn = 0 ( b 1 + N/h d)n = 0 11

12 Solving a System of Equations Idea: 1) start with solving the easiest equation 2) substitute the solutions in the other equation (in case of multiple solutions do it one by one!) 3) solve the other equation for substituted solution (now you have a solution for both variables) (repeat if first equation had multiple solutions) Example { an an 2 bnp = 0 with a, b, k parameters NP kp = 0 12

13 Derivatives Rules for differentiation: General: f(x) = ax b + c f (x) = abx b 1 Chain: f(x) = g(h(x)) f (x) = g (h(x)) h (x) Product: f(x) = g(x)h(x) f (x) = g (x)h(x) + g(x)h (x) Quotient: f(x) = g(x) h(x) f (x) = g (x)h(x) g(x)h (x) h 2 (x) Special: f(x) = e x f (x) = e x Special: f(x) = cos(x) f (x) = sin(x) Examples Find location and height of maximum of f(x) = 3x 2 12x + 6 Find the derivative of f(x) = cos(3x 2 ) 13

14 Linear approximation f(x) x * x Derivative/Slope of f(x) at point x : f (x ) = df dx = lim f(x) f(x ) x x x x d here means a very small change in lim x x means that x approaches x Linear approximation of f(x) around point x : y = f(x ) + f (x )(x x ) 14

15 Limits and Asymptotes What happens when x approaches a: lim x a f(x) = A Possibilities: trivial case 1: a is a number and A = f(a) trivial case 2: a = ± and A = ± horizontal asymptote: a = ± and A is a number vertical asymptote: a is a number and A = ± Examples lim x ax 4 bx 2 +c dx 4 e lim x 3x 4 5x 6 12x lim x c an 2 b N c 15

16 Drawing of Functions Plan 1) Find intersection points with x-axis: x for which f(x) = 0 2) Find intersection points with y-axis y at x = 0 3) Find horizontal asymptotes: lim x ± f(x) 4) Find vertical asymptotes: only if h(x) = 0 in f(x) = g(x) h(x) 5) Find general shape: fill in x = Examples f(x) = a(x p)(x q)(r x) f(x) = 3 x f(x) = 1 x 2 +a 2 + b ar + b R2 R+c drn = 0, R, N 0 16

17 Chapter 2 Vectors, Matrices and Complex numbers 17

18 Scalars, vectors and matrices Scalar: single number: λ = 5 e.g expression level of a gene Vector: column or row of n numbers: V = e.g expression levels of two genes ( 2 1 ) or V = ( 2 1 ) Matrix: ( block of n rows and m columns of numbers: A = e.g expression levels of two genes in 3 experiments ) 18

19 Scalar arithmetic This all of you of course know: Addition a + b = b + a = c e.g = 9 and = 9 Multiplication a b = b a = d e.g. 5 4 = 20 and 4 5 = 20 To illustrate differences with vector and matrix arithmetic 19

20 Vector arithmetic Length v = a 2 + b n 2 e.g. v = ( 2 1 ) so v = = 5 Scaling a v = u, ( with ) u i ( = a ) v i 2 6 e.g. 3 =

21 Vector arithmetic (2) Addition v + w ( = w ) + v ( = u, ) with( u i = ) v i + w( i e.g. + = and ) + ( 3 7 ) = ( 5 12 ) Multiplication v w = w v = a, with a = v i w i e.g. ( 3 7 ) ( ) 2 = = 41, ( 2 5 ) ( ) = = 41 21

22 Matrix arithmetic Determinant ( ) a b A = so det(a) = det c d e.g. det a b c d = = 2 = ad cb (only need to know det for 2 2 matrices in this course) Scaling c A = B, with B i,j = c A ( ) ( i,j e.g. 4 = ) 22

23 Matrix arithmetic (2) Addition A + B ( = B + A = ) C, ( with c i,j = ) a i,j + ( b i,j, only if ) A en B same sizes e.g. + = Multiplication A B = C, with c ij = p ( ) ( ) k=1 a ( ikb kj, if #columns ) A=# rows B e.g. = B A = D C, with d ij = p ( ) ( ) ( k=1 b ika kj,if ) #columnsb=#rows A e.g. =

24 Matrices, vectors and systems of linear equations The system: { ax + by = p cx + dy = q can be written as: A X = V with A = ( a b c d ) ; X = ( x y ) ; V = ( p q ) 24

25 Vector transformations Multiplying a vector with a matrix transforms the vector: A v = w usually the new vector has a new direction, length or both Rotation ( matrix ) ( ) ( ) cosφ sinφ vx cosφvx sinφv A v = = y, sinφ cosφ v y sinφv x + cosφv y vector has same length different direction. 25

26 Vector transformations (2) Scaling and shearing matrices new vectors have different lengths and directions. 26

27 D Arcy Thompson Modeling growth, development, evolution: Apply series of scaling, shearing and rotation transforms. Note that A B B A so order of transforms matters. 27

28 Eigenvalues and Eigenvectors Usually Av = w Matrix A transforms vector v into different vector w For special v (not zero) Av = λv Matrix A only rescales v into λv We call v an eigenvector, and the factor λ by which it is rescaled its corresponding eigenvalue. 28

29 Non-uniqueness of eigenvectors Eigenvectors can be scaled by an arbitrary constant k: kav = kλv or A(kv) = λ(kv) kv is also an eigenvector corresponding to eigenvalue λ 29

30 Eigenvectors as dominant directions y v2 A= ( ) eigenvalues: 3, -1 eigenvectors ( 1 1 ) ( 1-1 ) v0= ( 0 v1= 1 ) ( 2 v2= 1 ) ( 4 v3= 5 ) 13) (12 eigenvector 1 v1 v0 eigenvector 2 x The major effect of a transformation matrix is determined by the eigenvector with the largest eigenvalue. 30

31 Finding eigenvalues Can be written as: ( ) ( ) a b vx = λ c d Or: v y ( vx v y Av = λv ) ( a λ b ; c d λ ) ( vx { { avx + bv y = λv x (a λ)vx + bv ; y = 0 cv x + dv y = λv y cv x + (d λ)v y = 0 Gives the so-called characteristic equation: Which can be written as: det λ 2 (a + d)λ + (ad cb) = 0 a λ c v y ) = ( 0 0 b d λ = (a λ)(d λ) bc = 0 31 )

32 Finding eigenvalues (2) Example: Find eigenvalues of: ( ) 32

33 Finding eigenvectors Substitute eigenvalues (λ 1 = 3, λ2 = 1) in equations: For λ 1 = 3: { { (1 3)vx + 2v y = 0 2vx = 2v or y 2v x + (1 3)v y = 0 2v x = 2v y ( ) 1 so v x = v y, so corresponding eigenvector is k 1 Note that the first and second equation give the same info. Example: Do the same for λ2 = 1 33

34 Express method for finding eigenvectors First and second equation give the same info. Using only the first equation we would get: ( v1x ) ( b ) ( v2x ) ( b ) v 1 = v 1y = a λ 1 v 2 = v 2y = a λ 2 Using only the second equation we would get: v 1 = ( v1x v 1y ) = ( d λ1 c ) v 2 = ( v2x v 2y ) = ( d λ2 c ) If one of two methods gives a zero vector use the other one! 34

35 Why complex numbers? Aλ 2 + Bλ + C = 0 From abc formula: where D = B 2 4AC λ 12 = B ± B 2 4AC 2A = B ± D 2A What if D < 0? Use i (i 2 = 1): λ 12 = B ± i D 2A 35

36 The basic complex number i is basic complex number just as 1 is basic real number it is defined as: i 2 = 1 which means the same as: i = 1 36

37 General complex numbers z = α ± iβ Rez = alpha real part Imz = iβ imaginary part Can be displayed on the complex plane: The complex plane imaginary part complex number z real part Looks similar to 2D vector! 37

38 Complex conjugate z 2 = a ib is complex conjugate of z 1 = a + ib: z 1 Complex solutions for eigenvalues are complex conjugates: λ 1 = B + i D 2A λ 2 = B i D 2A Example: x 2 + x + 2 = 0 38

39 Calculus with complex numbers Very similar to working with vectors: Addition of two complex numbers: z 1 = i, z 2 = 5 + 4i z 1 +z 2 = (3+10i)+( 5+4i) = 3+10i+ 5+4i = 2+14i Multiplication by real number: z 1 = i 10z 1 = 10(3 + 10i) = i Dissimilar to working with vectors: (uitvermenigvuldingen): Multiplication of two complex numbers: z 1 = i, z 2 = 5 + 4i z 1 z 2 = (3+10i)( 5+4i) = 3( 5)+3 4i+10i( 5)+10i4i = i 50i + 40i 2 = 55 38i 39

40 Modulus z = a 2 + b 2 is absolute value / modulus of z = a + ib (Just like length of vector) Note that z 2 = z z. Turns out usefull if dividing two complex numbers: z 1 z 2 = z 1 z 2 z 2 z 2 Multiply numerator and denominator by complex conjugate to denominator, produces a non-complex denominator and new complex numerator. Example: 1+3i 1 4i 40

41 Chapter 3 Differential equations of one variable 41

42 Differential equations and their solutions Differential equation: dx dt =... describes change of variable x over time Solution: x(t) =... describes the amount of variable x as a function of time 42

43 Why are differential equations used Much easier to write down equations for the change of a variable over time as a function of processes causing changes: dn/dt = bn dn = kn then to write down equations for its amount as a function of time in terms of these same processes: N(t) = Ae (b d)t obtained by solving differential equation, we can not directly write it! 43

44 Why are differential equations used 2 Once we have a differential equation we can use it to: Find out about the long term behaviour of the variable: increase to? decrease to 0? steady state? oscillations? and how this depends on the initial values of variables, and on the particular conditions i.e. the parameter settings of the system. In a forest with rabbits and foxes, do they coexist, do the foxes die out and the rabbit population blows up, do both populations die out? How much fish can we catch before the fish die out? 44

45 Simple differential equations and their solution Simplest possible equation: dx/dt = a eg position change of a car travelling at constant speed a. General solution: (from graph or simple integration) x(t) = at + b, with x(0) = b Solution of initial value problem: given x(0) = 10 solution is x(t) = at

46 Simple differential equations and their solution (2) A bit less simple equation: dx/dt = at eg position change of a car travelling at constant accelleration a. General solution: (from graph or simple integration) x(t) = 1 2 at2 + b, with x(0) = b Solution of initial value problem: given x(0) = 30 solution is x(t) = 1 2 at

47 Simple differential equations and their solution (3) A simple biological equation: dn/dt = bn dn = (b d)n = kn population changes size due to birth and death processes General solution: (less easy to find, only need to proof) N(t) = Ae kt, with N(0) = A Solution of initial value problem: given N(0) = 30 solution is N(t) = 30e kt Characteristic time: time in which e-fold change of variable occurs: τ = 1 k Doubling time: time in which 2-fold change of variable occurs: τ = ln(2) k 47

48 Simple differential equations and their solution (4) Another biological equation: dn/dt = k dn population changes size due to immigration and death processes General solution: (less easy to find, only need to proof) N(t) = k d (1 e dt ) + N(0)e dt Equilibrium: for t we get e dt 0 so N(t) k d indeed dn/dt = k dn = 0 gives N = k d Characteristic time: time in which distance to eq. k d 1 e (N(0) k d ) = N(t) k d = k d (1 e dt ) + N(0)e dt ) k d 1 e (N(0) k d ) = (N(0) k d )e dt τ = 1 d changes e-fold: 48

49 Qualitative analysis Up till now understand the dynamics of a differential equation by solving it. However, a lot of differential equations cannot be (easily) solved. Therefore, qualitative analysis to understand dynamics without solving equation. 49

50 Phase portrait For dn dt = kn with k = 4 and different N(0) we can draw: N a b c t Observations: change of N: slope of N(t): derivative N (t) derivative N (t) is given by dn dt! autonomous equation: dn dt only depends on N 50

51 Phase portrait For qualitative overview of dynamics we only need N-axis: indicate the size of increase, decrease or steady state: N = N or just increase ( ), decrease ( ) or steady state ( ): N we call this a phase portrait 51

52 Phase portrait How to draw a phase portrait of dx dt = f(x): draw if dx dt > 0: f(x) above x-axis draw if dx dt < 0: f(x) below x-axis draw if dx dt = 0: f(x) crosses x-axis So if you can draw f(x) you can find the phase portrait, we do not need solution F (x(t)) of dx dt = f(x)! 52

53 Equilibria What happens if dx dt = f(x) = 0 for x = x : at this point the graph of f(x) crosses the x-axis dx dt = 0 means that x does not change over time so if x = x x remains (unless perturbed) at x we call x = x an equilibrium point of dx dt = f(x) 53

54 Stability of Equilibria Consider dx/dt = 4x (a) and dx/dt = x (b) x <0 x f(x) x x >0 x f(x) x a b both systems have an equilibrium point. 54

55 Stability of Equilibria For dx/dt = f(x) = 4x: in the equilibrium point f (x) > 0 arrows point away from equilibrium perturbation causes divergence from eq. An equilibrium x is unstable if f (x ) > 0 55

56 Stability of Equilibria For dx/dt = f(x) = x: in the equilibrium point f (x) < 0 arrows point towards the equilibrium after perturbation convergence to eq. An equilibrium x is stable if f (x ) < 0 Stable equilibria are called attractors of the system 56

57 Basins of Attraction A differential equation can have multiple equilibria: f(u) u1 u2 u3 A1 A2 u u The basin of attraction of an attractor is the range of x-values for which convergence to that equilibrium occurs. Boundaries are often formed by unstable equilibria. 57

58 Global plan General plan of qualitative analysis: 1. Sketch the graph of f(x). 2. Determine where f(x) = 0 and draw equilibria points. 3. Determine where f(x) > 0 and draw there. 4. Determine where f(x) < 0 and draw there. 5. Together 2, 3 and 4 produce the phase portrait. 6. Determine attractors and their basin of attraction. Now we can predict the systems long term behaviour if we know the initial conditions. 58

59 Parameters and Bifurcations Consider a population with logistic growth subject to harvesting: r, k and h are all parameters dn/dt = rn(1 n/k) h Different system behaviour for different parameter values? Focus on h, how much can we harvest without extinction? 59

60 Parameters and Bifurcations Increasing h shifts down graph equilibria converge then disappear: 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 A bifurcation is a qualitative change in system behaviour due to a small change in parameter value 60

61 Chapter 4 Introduction to 2D systems 61

62 Systems of two differential equations { dx dt = f(x, y) = g(x, y) dy dt Note that both functions f(x, y) and g(x, y) depend on x and y Thus rate of change of x and y depends on themselves and each other 62

63 Two example systems Linear system: decaying and converting chemicals: { dx dt = ax + by dy dt = cx + dy,with a = 2, b = 1, c = 1, d = 2 Non-linear system: predator-prey system: { dx dt = rx(1 x/k) bxy dy dt = cxy dy,with r = 3, K = 1, b = 1.5, c = 0.5, d =

64 Solution of linear systems For 1D linear systems the solution is given by with x(0) = A Similarly, for 2D linear systems { dx the solution is given by x(t) = y(t) = dx dt = kx x(t) = Ae kt dt dy dt = ax + by = cx + dy C 1 v 1x e λ1t + C 2 v 2x e λ 2t C 1 v 1y e λ1t + C 2 v 2y e λ 2t with x(0) = C 1 v 1x + C 2 v 2x and y(0) = C 1 v 1y + C 2 v 2y and λ 1, λ 2 eigenvalues and v 1, v 2 eigenvectors of A = ( a b c d ) 64

65 Finding a solution Example find the solution of the following system: { dx dt = 2x + y dy dt = x 2y Note that, just as for 1D systems, is not possible to find an analytical solution for non-linear systems. 65

66 Trajectories Given the general solution and initial values x(0) = 1 and y(0) = 2 We find the particular solution x(t) = C 1 e 1t + C 2 e 3t (1) y(t) = C 1 e 1t + C 2 e 3t (2) x(t) = 1 2 e 1t e 3t (3) y(t) = 1 2 e 1t e 3t (4) What dynamics do we expect x and y to follow? 2 2 x y t t Both eigenvalues are negative decay to 0 66

67 Phase portrait We need 2D phase portrait to depict x and y dynamics together: y y x x Not so easy to draw without a computing device But general behaviour, convergence to (0, 0) follows from λ 1 < 0, λ 2 < 0 Implies that (0, 0) is stable equilibrium 67

68 Equilibria For a 2D system variables x and y are co-dependent { dx dt = f(x, y) dy dt = g(x, y) Therefore (x, y ) is an equilibrium if both dx dt = f(x, y ) = 0 and dy dt = g(x, y ) = 0 If only dx dt = f(x, y ) = 0 or dy dt = g(x, y ) = 0 one variable changes, also causing the other variable to change. 68

69 Equilibria Example Find equilibria of the following two systems: { dx dt = 2x + y dy dt = x + 2y { dx dt = 3x(1 x) 1.5xy = 0.5xy 0.25y dy dt 69

70 Vector field Qualitative phase portrait: only look at derivatives, depict with arrows If dx dt If dx dt If dy dt If dy dt = f(x, y) > 0 x increases: = f(x, y) < 0 x decreases: = g(x, y) > 0 y increases: = g(x, y) < 0 y increases: y x Fill in x,y values to get v = (f(x, y), g(x, y)) Hard to see what long term dynamics will be! 70

71 Null clines Vectorfield: lot of work, and still unclear Can we do this in a smarter, more insightfull way? Only 4 qualitatively different vectors possible: I dy/dt>0 dx/dt>0 II dy/dt>0 dx/dt>0 III dy/dt<0 dx/dt<0 dx/dt<0 IV dy/dt<0 Smarter approach: 1) Divide vectorfield in different regions by finding boundaries. 2) Assign all vectorfield regions to one of these 4 categories. 71

72 Null clines I dy/dt>0 dx/dt>0 II dy/dt>0 dx/dt>0 III dy/dt<0 dx/dt<0 dx/dt<0 IV dy/dt<0 From I to II / III to IV direction dx dt changes. The x null clines are the boundary lines at which dx dt = 0 and the horizontal part of the vector field changes direction. At x null cline vector field has only vertical component. 72

73 Null clines I dy/dt>0 dx/dt>0 II dy/dt>0 dx/dt>0 III dy/dt<0 dx/dt<0 dx/dt<0 IV dy/dt<0 From I to III / II to IV direction dy dt changes. The y null clines are the boundary lines at which dy dt = 0 and the vertical part of the vector field changes direction. At y null cline vector field has only horizontal component 73

74 Combining nullclines and vectorfield Let us draw phase portrait using nullclines and vectorfield for: { dx dt = 2x + y dy dt = x + 2y 2 2 y y x x Together, nullclines and vectorfield give more information! 74

75 Null clines and equilibria At x null cline dx dt At y null cline dy dt = f(x, y) = 0 = g(x, y) = 0 If x and y nullcline intersect f(x, y) = 0 and g(x, y) = 0: equilibrium! Note that intersections of two nullclines of the same type are no equilibria. 75

76 Global plan General plan of qualitative analysis: 1. Draw dx dy dt = f(x, y) = 0 and dt = g(x, y) = 0 null-clines 2. Choose a region of the x, y plane and find v = (f(x, y), g(x, y)) 3. Find the vector field in the adjacent regions using: a. Flip the horizontal component when crossing x null-cline b. Flip the vertical component when cossing y null-cline 4. Show the direction of the vector field on the null-clines. 5. Try Determine the dynamics around an equilibrium. 6. Try to derive the global dynamics of the system. 76

77 Example Example Find vectorfield of following system { dx dt = 3x(1 x) 1.5xy dy dt = 0.5xy 0.25y 77

78 Parameters and bifurcations Consider the following system dx dt = ax + by dy dt = c x 2 x 2 +d 2 ey a, b, c, d and e are all parameters. Let us fix b, c, d and e and study influence of a Note that x and y both cause their own decrease furthermore they both cause each others increase 78

79 Parameters and bifurcations Depending on value of a one or three equilibria: y b a y c e c 2e X d So a determines if there is a non-trivial stable equilibrium. 79

80 Chapter 5 Equilibrium types in 2D systems 80

81 Equilibria types 1D systems: two regions, left and right from equilibrium arrows can point away or toward equilibrium so 2 equilibria types, stable and unstable 2D systems: total of four regions around equilibrium point arrows point away or toward equilibrium or both in total 6 different equilibrium types, 2 of which are stable Equilibrium types can be found from: eigenvalues (only works for linear systems) ( x(t) y(t) ) = C 1 ( v1x v 1y ) e λ 1t + C 2 ( v2x v 2y ) e λ 2t vectorfield (works for most but not all situations) 81

82 Equilibria types: stable node { dx dt = 2x y = x 2y dy dt 2 2 y y y 1 y 1 0 x x 0 x x Eigenvalues: λ 1 = 1, λ 2 = 3 stable node Vectorfield: all arrows point to equilibrium stable node 82

83 Equilibria types: unstable node { dx dt = 2x + y dy dt = x + 2y 2 2 y y y 1 y 1 0 x x 0 x x Eigenvalues: λ 1 = 1, λ 2 = 3 unstable node Vectorfield: all arrows point away from eq. unstable node 83

84 Equilibria types: saddle point { dx dt = x y dy dt = 2x y 1 1 y y y 0.5 y x x 0 x x Eigenvalues: λ 1 = 1 + 2, λ 2 = 1 2 saddle Vectorfield: one direction to, one away from eq. saddle 84

85 Complex eigenvalues with eigenvalues { dx dt dy dt = ax + by = cx + dy λ 1,2 = (a + d) ± (a + d) 2 4(ad bc) 2 if D = (a + d) 2 4(ad bc) we can write λ 1,2 = (a + d) ± D 2 assuming D < 0 we can write if (a + d)/2 = α and D/2 = β λ 1,2 = (a + d) ± i D 2 λ 1,2 = α ± iβ 85

86 Complex eigenvectors Because eigenvalues are complex so are the eigenvectors: ( ) ( ) ( ) ( b b b 0 v 1 = = = + i a λ 1 a (α + iβ) a α β where v 1r = ( b a is the real part of the eigenvector and ( 0 i v 1i = i β is the imaginary part of the eigenvector ) ) ) 86

87 General solution? General solution follows from eigenvalues and vectors: ( x(t) y(t) ) = C 1 (v 1r + iv 1i )e (α+iβ)t + C 2 (v 2r + iv 2i )e (α iβ)t But how can we tell the behavior of the system from this? Rewrite the general solution using Euler s formula: From this we find e ix = cosx + isinx e (α±iβ)t = e αt±iβt = e αt e ±iβt = e αt (cos(βt) ± isin(βt)) So we can rewrite the general solution as: ( ) x(t) = C y(t) 1 (v 1r +iv 1i )e αt (cosβt+isinβt)+c 2 (v 2r +iv 2i )e αt (cosβt isinβt) or even: ( ) x(t) = e y(t) αt {C 1 (v 1r +iv 1i )(cosβt+isinβt)+c 2 (v 2r +iv 2i )(cosβt isinβt)} 87

88 General solution What does this general solution tell us? ( ) x(t) = e y(t) αt {C 1 v 1 (cosβt + isinβt) + C 2 v 2 (cosβt isinβt)} Equations for x(t), y(t) consist of: sum of sine and cosine functions: x and y values show complex oscillatory dynamics over time multiplied with e αt : amplitude of oscillations in- (α > 0) or decreases (α < 0) over time So, in x, y phase plane: x and y show outward (α > 0) or inward (α < 0) spiraling trajectories 88

89 Equilibrium types: stable spiral { dx dt = x + 2y dy dt = 2x y y y y 0 x, y 0 y xy 0-2 x x -3 t t -2 x x Eigenvalues: λ 1 = 1 + i2, λ 2 = 1 i2 stable spiral Vectorfield: arrows only suggest rotation 89

90 Equilibrium types: unstable spiral { dx dt = x + 2y dy dt = 2x + y y y x y 0 x, y 0 y 0 y -2 x x -500 t t -2 x x Eigenvalues: λ 1 = 1 + i2, λ 2 = 1 i2 unstable spiral Vectorfield: arrows only suggest rotation 90

91 Equilibrium types: center point { dx dt = x + 2y dy dt = 2x y y y y 0 x, y 0 y y x 0-2 x x -2 t t -2 x x Eigenvalues: λ 1 = i 3, λ 2 = i 3 unstable spiral Vectorfield: arrows only suggest rotation 91

92 Vector field insufficient Sometimes vector field does not give enough information: vector field only suggests rotation, but may not even be spiral! If all feedback is negative equilibrium is stable, but type unknown! If one of feedbacks is positive equilibrium is unstable spiral! 92

93 Stable equilibria An equilibrium is only stable if from all directions trajectories converge on it. One or more diverging directions means that the equilibrium is unstable. Stable equilibria: stable node: real eigenvalues λ 1 < 0, λ 2 < 0 stable spiral: complex eigenvalules α < 0 Unstable equilibria: unstable node: real eigenvalues λ 1 > 0, λ 2 > 0 saddle node: real eigenvalues λ 1 < 0, λ 2 > 0 or vice versa unstable spiral: complex eigenvalules α > 0 A center point is a neutrally stable equilibrium: divergence! no convergence or So an equilibrium is stable if the real part of both eigenvalues is < 0. 93

94 Partial derivatives General function f(x, y) of two variables, example: The function depends on both x and y It may depend in different ways on x and y Not a single derivative, but two partial derivatives 94

95 Partial derivatives Geometrical representation f/ x slope of f in x direction f/ y slope of f in y direction 95

96 Partial derivatives f/ x: 1) Treat y as a constant (derivative of constant is zero) 2) Now compute f/ x as you would df/dx Example f(x, y) = 3x 3x 2 1.5xy y = a is constant f(x) = 3x 3x 2 1.5ax f/ x = 3 6x 1.5a or f/ x = 3 6x 1.5y 96

97 Partial derivatives f/ y: 1) Treat x as a constant (derivative of constant is zero) 2) Now compute f/ y as you would df/dy Example f(x, y) = 3x 3x 2 1.5xy x = a constant f(y) = 3a 3a 2 1.5ay f/ y = 1.5a or f/ y = 1.5x 97

98 Linear approximation In 1D we can linearly approximate f(x) using f (x): f(x) f(x ) + f (x )(x x ) 98

99 Linear approximation In 2D we can approximate f(x, y) using f/ x, f/ y: Move in two steps from (x, y ) to (x, y) f 1 = f(x, y ) f(x, y ) = ( f/ x)(x x ) f 2 = f(x, y) f(x, y ) = ( f/ y)(y y ) Together this gives: f(x, y) f(x, y ) + ( f/ x)(x x ) + ( f/ y)(y y ) 99

100 Linear approximation Example: Find a linear approximation of e x+2y at x = 0, y = 0 100

101 Jacobian Consider { a general 2D system: dx dt = f(x, y) dy dt = g(x, y) Assume that (x, y ) is an equilibrium point Approximate the system close to the equilibrium: f(x, y) f(x, y ) + ( f/ x)(x x ) + ( f/ y)(y y ) g(x, y) g(x, y ) + ( g/ x)(x x ) + ( g/ y)(y y ) 101

102 Jacobian So: { dx dt = ( f/ x)(x x ) + ( f/ y)(y y ) dy dt = ( g/ x)(x x ) + ( g/ y)(y y ) Or: { dx dt = a(x x ) + b(y y ) dy dt = c(x x ) + d(y y ) where a = f/ x; b = f/ y; c = g/ x, d = g/ y 102

103 Jacobian Change { of variables u = x x and v = y y : du dt = au + bv dv dt = cu + dv. Or ( du dt dv dt ) ( a b = c d ( a b With J = c d original system ) ( u ) ) v ( f/ x f/ y = g/ x g/ yd ) the Jacobian of the 103

104 General solution In 1D systems we learned that linear equations can be easily solved: dx dt = ax has as solution x(t) = x(0)eat For non-linear equations we solve the linear approximation instead: dx dt = f(x), f(x ) = 0, dx dt f (x )(x x ) has as a solution for x close to x x(t) = x(0)e f (x )t We can do something similar for 2D systems. 104

105 General solution Solve dx dt = ax using method of substitution: 1) We know that solutions are of form Ce λt 2) We substitute x = Ce λt in the equation dx dt = ax dce λt = dt ace λt λce λt = ace λt λ = a so x(t) = Ce at, similar to before 105

106 General solution Now ( ) we want ( to) use ( the ) same method to solve dx dt a b x dy = c d y dt 1) Again solutions are of form Ce λt 2) We substitute x = C x e λt and y = C y e λt Note that C x C y { λcx e λt = ac x e λt + bc y e λt λc y e λt = cc x e λt + dc y e λt 106

107 General solution { λcx = ac x + bc y ( Cx ) ( a b ) ( Cx ) or λc y = cc x + dc y or λ C y = c d C y ( Cx ) ( a b ) So λ is eigenvalue and we need to solve Det C y a λ c is eigenvector of b d λ = 0 c d and 107

108 General Solution So ( we ) get two particular ( ) solutions: x = v y 1 e λ 1 t x = v y 2 e λ 2 t If v 1 is an eigenvector, so is C 1 v 1 and if v 2 is an eigenvector, so is C 2 v 2 Thus the following are a whole series of solutions: ( x y ) = C 1 v 1 e λ 1 t ( x y ) = C 2 v 2 e λ 2 t But we want a single, general solution! 108

109 General solution If x 1, y 1 and x 2, y 2 are two solutions, then x 1 + x 2, y 1 + y 2 is also a solution: dx 1 dt = ax 1 + by 1 dy 1 dt = cx 1 + dy 1 If we add them up we get: dx 2 dt = ax 2 + by 2 dy 2 dt = cx 2 + dy 2 (5) d(x 1 +x 2 ) dt = a(x 1 + x 2 ) + b(y 1 + y 2 ) d(y 1 +y 2 ) dt = c(x 1 + x 2 ) + d(y 1 + y 2 ) rewriting shows that x 1 + x 2, y 1 + y 2 is a solution (6) We can similarly proof that Ax 1 + Bx 2, Ay 1 + By 2 (with A and B having a specific value) is also a solution. 109

110 General solution General solution should contains all possible solutions: all combinations of the two particular solutions we first found: ( x ) = C y 1 v 1 e λ1 t + C 2 v 2 e λ 2 t for all possible values of C 1 and C 2 110

111 General solution Example Find general solution of: ( dx dt dy dt ) = ( ) ( x y ) 111

112 Eigenvalues and equilibria: saddle λ 1 and λ 2 are real numbers λ 1 and λ 2 have opposite signs: one < 0 one > 0 Example: consider system with following general solution: ( ) ( ) ( ) x 4 = C y 1 e 1t 4 + C 2 2 e 3t. 2 C 1 and C 2 arbitrary constants depending on initial conditions What type of phase plane dynamics does this describe? 112

113 Eigenvalues and equilibria: saddle 1) If C 1 = 0 and C 2 = 0 x and y remain zero: equilibrium ( ) ( ) x 2 2) If C 1 = 0 and C 2 = A: = A e y 1 3t e 3t grows over time A determines the sign/direction v 2 determines ratio x,y ( ) x 3) If C 1 = B and C 2 = 0: = B y e t decreases over time B determines the sign/direction v 1 determines ratio x,y ( ) x 4) If C 1 = B and C 2 = A: = B y B part decreases, A part increases ( 4 2 ( 4 2 so over time A part determines dynamics ) ) e t e t + A ( 2 1 ) e 3t 113

114 Manifolds Manifolds: direction given by eigenvector Not the same as null clines, but may coincide! Unstable manifold: λ > 0: diverging dynamics Stable manifold: λ < 0: converging dynamics 114

115 Eigenvalues and equilibria: unstable node λ 1 and λ 2 are real numbers λ 1 and λ 2 both > 0 Two unstable manifolds: unstable node 115

116 Eigenvalues and equilibria: stable node λ 1 and λ 2 are real numbers λ 1 and λ 2 both < 0 Two stable manifolds: stable node 116

117 Complex eigenvalues The Jacobian matrix may also have complex eigenvalues: J = ( a b c d ) det(j) = λ 2 + (a + d)λ + (ad bc) = 0 λ 1,2 = B± D 2 with B = a + d, C = ad bc and D = B 2 4C if D < 0: λ 1,2 = B±i D 2 117

118 Real part and convergence and divergence Real part determines convergence or divergence As complex eigenvalues are conjugates only two possibilities: α < 0: both real parts < 0: convergence from 2 directions α > 0: both real parts > 0: divergence from 2 directions (Proof of this is outside the scope of this course) 118

119 Imaginary part and rotation Imaginary part adds rotating dynamics to this How and why??? We will demonstrate this with an example: ) ( ) ( ) 0 2 x = or 2 0 y ( dx dt dy dt with λ 1,2 = ±2i: no real, only imaginary part { dx dt = 2y dy dt = 2x Should produce only rotation, no convergence or divergence 119

120 Imaginary part and rotation Apply the following trick: multiply dx dt = 2y with x multiply dy dt = 2x with y add up the two new equations This produces: x dx dt + ydy = 2xy 2xy = 0 dt one equation describing dynamics of x and y 120

121 Imaginary part and rotation Now substitute x dx dt = 1 2 dx2 dt and y dy dt = 1 dy 2 2 dt This produces: d(x 2 + y 2 ) dt = 0 If we integrate this over time we get: x 2 + y 2 = C with C an arbitrary integration constant 121

122 Imaginary part and rotation Over time x and y vary, but as x 2 + y 2 = C they remain at constant distance from (0, 0). This distance depends on initial conditions. Direction of rotation can be found from vector field. 122

123 Eigenvalues and equilibria: center λ 1,2 = ±iβ (α = 0): center The imaginary part produces rotation. No real part to produce convergence or divergence Initial condition determines distance from equilibrium 123

124 Eigenvalues and equilibria: stable spiral λ 1,2 = α ± iβ and α < 0: stable spiral The imaginary part produces rotation (as for center). The negative real part causes convergence (as for stable node) Together a dynamics spiralling in towards equilibrium results. 124

125 Eigenvalues and equilibria: unstable spiral λ 1,2 = α ± iβ and α > 0: unstable spiral The imaginary part produces rotation (as for center). The positive real part causes divergence (as for unst. node) Together a dynamics spiralling out from equilibrium results. 125

126 Eigenvalues and stability of equilibria An equilibrium is stable only if all trajectories starting in the neighborhood of the equilibrium converge to the equilibrium. If there are one or more diverging trajectories the equilibrium point is unstable. 126

127 Eigenvalues and stability of equilibria Stable: stable node: λ 1 < 0 and λ 2 < 0 stable spiral: λ 1,2 = α ± iβ and α < 0 Unstable: unstable node: λ 1 > 0 and λ 2 > 0 unstable spiral: λ 1,2 = α ± iβ and α > 0 saddle: λ 1 > 0; λ 2 < 0 or λ 1 < 0; λ 2 > 0 Inconclusive / Neutral: center: λ 1,2 = ±iβ (α = 0) 127

128 Eigenvalues and stability of equilibria Summary: If all eigenvalues have negative real parts the equilibrium is stable. In contrast to the null cline approach, this approach always works to determine the equilibrium type and stability. 128

129 Chapter 6 Efficient analysis of systems of two differential equations 129

130 Summary of previous chapter General { 2D system: dx dt = f(x, y) dy dt = g(x, y) equilibrium at (x, y ) Linearize ) ( to: a b = c d ( dx dt dy dt ) ( x y ) Determine eigenvalues and eigenvectors: a λ b Det c d λ = 0, Gives equilibrium type and stability 130

131 Trace and Determinant Consider ( ) a matrix A: a b c d tra = a + d is called the trace of a matrix deta = ad bc is the determinant of a matrix 131

132 Rewriting the characteristic equation With the determinant and trace Det a λ c b d λ = λ2 (a + d)λ + (ad bc) = 0 can be rewritten into λ 2 traλ + deta = 0 132

133 Rewriting the eigenvalues So also λ 1,2 = (a+d)± (a+d) 2 4(ad bc) 2 can be rewritten as λ 1,2 = tra± D 2 with D = tra 2 4detA 133

134 Some other relations Furthermore λ 1 + λ 2 = tra λ 1 λ 2 = deta 134

135 Determinant Trace Method Find equilibrium type without finding eigenvalues: the determinant trace method 1) deta < 0 gives D = tra 2 4detA > 0 so real roots as λ 1 λ 2 = deta and deta < 0 so saddle 2) deta > 0 and D > 0 so real roots λ 1 λ 2 = deta and deta > 0 so stable or unstable node assume tra > 0, as λ 1 + λ 2 = tra unstable node 3) deta > 0, D > 0 and tra < 0 stable node 4) deta > 0 and D < 0 so complex roots tra is real part eigenvalues assume tra > 0 unstable spiral 5) deta > 0, D < 0 and tra < 0 stable spiral 6) deta > 0, D < 0 and tra = 0 center point 135

136 Determinant Trace Method Graphical representation: 136

137 Determinant Trace Method Example 1 ( dx dt dy dt ) = ( ) ( x y ) Example 2 ( dx dt dy dt ) = ( ) ( x y ) 137

138 Graphical Jacobian Consider again: sufficient to know sign of tra and deta to find whether equilibrium is saddle, stable spiral/node or unstable spiral/node 138

139 Graphical Jacobian Use vector field to determine sign of elements Jacobian: f x f y J = g g x y f(x +h,y ) f(x,y +h) J h h g(x +h,y ) f(x,y +h) ( h h / x J +h / y +h / x +h / y +h ) 139

140 Graphical Jacobian Example f x = g x = + f y = g y = 140

141 Graphical Jacobian So J = ( α β +γ δ ) So detj = ( α δ) ( β γ) = αδ + βγ > 0 trj = α δ < 0 So either stable spiral or stable node 141

142 Graphical Jacobian Example 2 f x = + g x = f y = + g y = 142

143 Graphical Jacobian So J = ( +α +β γ δ ) So detj = αδ + γβ trj = α δ So we can not determine whether detj and trj > 0 or < 0 143

144 Summary of Different Methods 144

145 Plan for Analysis Now that we have learned these different methods: when to use which method?: From easy to more difficult! General method for determining equilibrium type: 1) Draw null-clines and vectorfield, find equilibria 2) Look at vectorfield to determine equilibrium type 3) If 2 fails: Find graphical jacobian to determine eq. type (using determinant-trace method) 4) If 3 fails: Find analytical jacobian to determine eq. type (using determinant-trace or eigenvalue method) Finally, connect local phase portrait of equilibria to get global phase portrait and determine basins of attraction. 145

146 Example Let us use this approach for: { dx dt = 3x(1 x) 1.5xy = 0.5xy 0.25y dy dt (We have seen this system in previous chapters) 146

147 Example Null clines and vector field: We see three equilibria: (0, 0), (0.5, 1) and (1, 0) (0, 0): one converging one diverging direction: saddle (1, 0): one converging one diverging direction: saddle (0.5, 1): rotation, equilibrium type unclear 147

148 Example Graphical Jacobian: ( ) +α 0 (0, 0): J =, detj = αδ < 0: saddle 0 δ ( ) α β (1, 0): J =, detj = αδ < 0: saddle 0 +δ ( ) α β (0.5, 1): J = detj = γβ > 0 trj = α < 0: stable +γ 0 spiral or node 148

149 Example Analytical Jacobian: ( ) 3 0 (0, 0): J 1 =, detj = < 0: saddle ( ) (1, 0): J 2 =, detj = < 0: saddle ( ) (0.5, 1): J 3 =, detj = = > 0, trj 3 = 1.5 < 0 and D = ( 1.5) = 0.75 > 0: stable node 149

150 Example Local phase portraits around equilibria and global one: Single large basin of attraction for (0.5, 1) 150

151 A backup plan We know that linearization, Jacobians, etc are hard. No need to panic: nullclines and vectorfield often enough. We can even push the use of vectorfields a bit further: add x: does x grow or decline? growth: unstable, decline: stable add y: does y grow or decline? growth: unstable, decline: stable If 2 stable directions: stable! If 1 or 2 unstable directions: unstable! 151

152 A backup plan Note that approach is close to graphical Jacobian approach! Case 1: add x: decline, add y decline: stable! Case 2: add x: decline, add y increase: unstable! Case 3: add x: decline, add y decline: stable! But...does not work for horizontal/vertical nullclines! 152

153 Summary The discussed mathematics was needed to: understand what is going in systems understand why more simple methods work understand all possible cases But often more simple methods sufficient. 153

154 Chapter 7 Limit cycles 154

155 Parameter change in LV-model Lotka-Volterra predator prey model: F = R h+r dr dt = rr(1 K R ) bnf dn dt = bnf dn with b = 0.5, d = 0.43, h = 0.1, r = 1, variable K R-nullcline: R = 0 and N = r b (1 K R )(h + R) = 2(1 K R )( R) N-nullcline: N = 0 and R = dh b d

156 Parameter change in LV-model K = 2dh b d + h : center: hard to find! 156

157 Parameter change in LV-model So in determinant-trace scheme: K : stable node stable spiral center stable limitcycle 157

158 Dynamics Center point: series of closed loops rotating dynamics in phase plane repeated oscillations of variables which loop depends on initial conditions, and determines amplitude oscillations 158

159 Dynamics Limit cycle: single closed loop so single amplitude of oscillations stability determines if you go and stay there 159

160 Stable limitcycle A stable limitcycle is an attractor consisting of a closed loop of points rather than a single point: 160

161 Stable limitcycle Stable limitcycle: convergence to it: 161

162 Unstable limitcycle An unstable limitcycle is an unstable equilibrium consisting of a closed loop of points that acts as a boundary for the basin of attraction of the attractor it often contains: 162

163 Unstable limitcycle Unstable limitcycle: divergence from it: 163

164 Why and when do limitcycles occur? Assume { general 2D system with parameter c: dx dt = f(x, y, c) dy dt = g(x, y, c) Assume that eigenvalues are complex: λ 1,2 (c) = α(c) ± iβ(c); Assume that as c changes α changes between < 0 to > 0 Locally, close to the equilibrium, we change between a stable and unstable spiral. However, globally nothing changes. 164

165 Why and when do limitcycles occur? Needed to resolve conflict between local and global dynamics: Globally system converges, α changes from < 0 to > 0 Around unstable spiral, stable limitcycle needed. Hopf bifurcation 165

166 Why and when do limitcycles occur? Needed to resolve conflict between local and global dynamics: Globally system diverges, α changes from > 0 to < 0 Around stable equilibrium unstable limitcycle needed Hopf bifurcation 166

167 Why and when do limitcycles occur? 167

168 Example Holling-Tanner model for predator-prey interactions: dp/dt = rp (1 P K ) arp d+p dr/dt = br(1 R P ) P > 0; R > 0 We fix a = 1, b = 0.2, r = 1, d = 1 and vary K 168

169 Example Null clines and dynamics for K = 7: Stable spiral with whole phase plane as basin of attraction 169

170 Example Null clines and dynamics for K = 10: Unstable spiral, stable limit cycle, single basin of attraction 170

171 Example Hopf bifurcation: Intersection of nullclines both times left from top! Close to top: stable spiral, consistent with global dynamics Further from top: unstable spiral, stable limitcycle 171

172 End of Math part This was the end of the Math part! 172