Lecture Notes. W. Tang. School of Mathematical & Statistical Sciences Arizona State University

Transcription

1 Applied Math / Dynamical Systems Workshop Lecture Notes W. Tang School of Mathematical & Statistical Sciences Arizona State University May. 23 rd -25 th, 2011 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

2 Foreword: Mathematical tools are often used to obtain qualitative and quantitative information of the evolution (dynamics) of physical systems. These tools are of utmost importance in studies on applied mathematics, science and engineering. In this workshop, we discuss how different mathematical tools, typically studied in several college level math courses, are related and used to understand physical systems. We use simple but illustrative examples, built from physically motivated (word) questions, and introduce necessary concepts that are new but useful in our approach. We hope that this workshop will serve as a proxy for you to understand the various concepts discussed in college level math courses, and help you establish abstract understanding of mathematics on applied problems. This workshop is no substitute for the actual math courses, but is the starting point to understand the connections. Please ask questions if you feel puzzled! Some of the material in this workshop can be found in Nonlinear Dynamics And Chaos: With Applications To Physics, Biology, Chemistry, And Engineering, Steven Strogatz, Westview Press, 2001 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

3 May 23 rd Lecture 1: Review of important concepts Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

4 Continuous functions Take the infinitesimal limit Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

5 Continuous functions Take the infinitesimal limit Slope: rate of change of the function df /dx = f = lim x 0 f / x Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

6 Continuous functions Take the infinitesimal limit Slope: rate of change of the function df /dx = f = lim x 0 f / x If f (t), df /dt = ḟ denotes the instantaneous change of f over time Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

7 Continuous functions Take the infinitesimal limit Slope: rate of change of the function df /dx = f = lim x 0 f / x If f (t), df /dt = ḟ denotes the instantaneous change of f over time Link to real world: in many problems you can only describe ḟ, f, f, f, etc Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

8 Calculus Example Consider the Newton s law of motion Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

9 Calculus Example Consider the Newton s law of motion dx(t)/dt = ẋ = v, dv(t)/dt = v = a, F = ma = mẍ Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

10 Calculus Example Consider the Newton s law of motion dx(t)/dt = ẋ = v, dv(t)/dt = v = a, F = ma = mẍ Finding F? trivial. Given F, find a, then v, then x, can be difficult Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

11 Calculus Example Consider the Newton s law of motion dx(t)/dt = ẋ = v, dv(t)/dt = v = a, F = ma = mẍ Finding F? trivial. Given F, find a, then v, then x, can be difficult Message: We can link quantities of physical interest to mathematical equations Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

12 Calculus Example Consider the Newton s law of motion dx(t)/dt = ẋ = v, dv(t)/dt = v = a, F = ma = mẍ Finding F? trivial. Given F, find a, then v, then x, can be difficult Message: We can link quantities of physical interest to mathematical equations We ll discuss how to solve for a subset of these equations later Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

13 Taylor series expansion Instantaneous change of a function focus on local behavior Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

14 Taylor series expansion Instantaneous change of a function focus on local behavior Taylor series expansion gives local information to certain order Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

15 Taylor series expansion Instantaneous change of a function focus on local behavior Taylor series expansion gives local information to certain order Expanding a function f near a point x 0 or (x 0, y 0 ) f (x) f (x 0 ) + f x=x0 (x x 0 ) f x=x0 (x x 0 ) 2 + f (x, y) f (x 0, y 0 ) + f x x=x0,y=y 0 (x x 0 ) + f y x=x0,y=y 0 (y y 0 ) [f xx x=x0,y=y 0 (x x 0 ) 2 + f yy x=x0,y=y 0 (y y 0 ) 2 + 2f xy x=x0,y=y 0 (x x 0 )(y y 0 )] + Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

16 Taylor series expansion Instantaneous change of a function focus on local behavior Taylor series expansion gives local information to certain order Expanding a function f near a point x 0 or (x 0, y 0 ) f (x) f (x 0 ) + f x=x0 (x x 0 ) f x=x0 (x x 0 ) 2 + f (x, y) f (x 0, y 0 ) + f x x=x0,y=y 0 (x x 0 ) + f y x=x0,y=y 0 (y y 0 ) [f xx x=x0,y=y 0 (x x 0 ) 2 + f yy x=x0,y=y 0 (y y 0 ) 2 + 2f xy x=x0,y=y 0 (x x 0 )(y y 0 )] + Local approximation as plane, quadratic surface, etc. Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

17 Taylor series expansion Instantaneous change of a function focus on local behavior Taylor series expansion gives local information to certain order Expanding a function f near a point x 0 or (x 0, y 0 ) f (x) f (x 0 ) + f x=x0 (x x 0 ) f x=x0 (x x 0 ) 2 + f (x, y) f (x 0, y 0 ) + f x x=x0,y=y 0 (x x 0 ) + f y x=x0,y=y 0 (y y 0 ) [f xx x=x0,y=y 0 (x x 0 ) 2 + f yy x=x0,y=y 0 (y y 0 ) 2 + 2f xy x=x0,y=y 0 (x x 0 )(y y 0 )] + Local approximation as plane, quadratic surface, etc. Are the following functions differentiable? Where do you see trouble? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

18 Euler s formula Euler s formula is given as e ix = cos(x) + i sin(x) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

19 Euler s formula Euler s formula is given as e ix = cos(x) + i sin(x) This comes from exponentiating a purely complex number e ix = 1 + ix + (ix)2 + (ix)3 + + (ix)n + 2! 3! n! = 1 x 2 2! + x 4 (x 4! + i x 3 3! + x 5 ) 5! = cos(x) + i sin(x) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

20 Euler s formula Euler s formula is given as e ix = cos(x) + i sin(x) This comes from exponentiating a purely complex number e ix = 1 + ix + (ix)2 + (ix)3 + + (ix)n + 2! 3! n! = 1 x 2 2! + x 4 (x 4! + i x 3 3! + x 5 ) 5! = cos(x) + i sin(x) Hence we can also write cos(x) = eix + e ix, sin(x) = eix e ix 2 2i Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

21 Euler s formula Euler s formula is given as e ix = cos(x) + i sin(x) This comes from exponentiating a purely complex number e ix = 1 + ix + (ix)2 + (ix)3 + + (ix)n + 2! 3! n! = 1 x 2 2! + x 4 (x 4! + i x 3 3! + x 5 ) 5! = cos(x) + i sin(x) Hence we can also write cos(x) = eix + e ix, sin(x) = eix e ix 2 2i Why important? Often analyze oscillatory systems (season, day-night) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

22 Euler s formula Euler s formula is given as e ix = cos(x) + i sin(x) This comes from exponentiating a purely complex number e ix = 1 + ix + (ix)2 + (ix)3 + + (ix)n + 2! 3! n! = 1 x 2 2! + x 4 (x 4! + i x 3 3! + x 5 ) 5! = cos(x) + i sin(x) Hence we can also write cos(x) = eix + e ix, sin(x) = eix e ix 2 2i Why important? Often analyze oscillatory systems (season, day-night) Extension cosh(x) = ex + e x, sinh(x) = ex e x 2 2 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

23 Differential equations We already see one example of differential equation mẍ = F (x, t) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

24 Differential equations We already see one example of differential equation mẍ = F (x, t) For unknown function x, usually in the form F (x, x, x,, t) = 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

25 Differential equations We already see one example of differential equation mẍ = F (x, t) For unknown function x, usually in the form F (x, x, x,, t) = 0 Example x = t, x = x + t, x = x 2 + sin(t), Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

26 Differential equations We already see one example of differential equation mẍ = F (x, t) For unknown function x, usually in the form F (x, x, x,, t) = 0 Example x = t, x = x + t, x = x 2 + sin(t), Can be VERY difficult with nonlinearity and with partial derivatives Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

27 Differential equations We already see one example of differential equation mẍ = F (x, t) For unknown function x, usually in the form Example F (x, x, x,, t) = 0 x = t, x = x + t, x = x 2 + sin(t), Can be VERY difficult with nonlinearity and with partial derivatives We will learn to solve some very simple ones coming from physical problems Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

28 Differential equations We already see one example of differential equation mẍ = F (x, t) For unknown function x, usually in the form Example F (x, x, x,, t) = 0 x = t, x = x + t, x = x 2 + sin(t), Can be VERY difficult with nonlinearity and with partial derivatives We will learn to solve some very simple ones coming from physical problems Focus on what each term means (so you feel how to construct a model and interpret a solution) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

29 Matrix Algebra We can usually write a system of equations into matrix form ( )( ) ( ) ( ) ( )( ) a11 a 12 x b1 ẋ a11 a = or = 12 x + a 21 a 22 y b 2 ẏ a 21 a 22 y ( b1 b 2 ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

30 Matrix Algebra We can usually write a system of equations into matrix form ( )( ) ( ) ( ) ( )( ) a11 a 12 x b1 ẋ a11 a = or = 12 x + a 21 a 22 y b 2 ẏ a 21 a 22 y Example: intersection of lines / 2D vector fields ( b1 b 2 ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

31 Matrix Algebra We can usually write a system of equations into matrix form ( )( ) ( ) ( ) ( )( ) a11 a 12 x b1 ẋ a11 a = or = 12 x + a 21 a 22 y b 2 ẏ a 21 a 22 y Example: intersection of lines / 2D vector fields ( b1 b 2 ) Properties of matrix leads to understandings of the system Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

32 Matrix Algebra We can usually write a system of equations into matrix form ( )( ) ( ) ( ) ( )( ) a11 a 12 x b1 ẋ a11 a = or = 12 x + a 21 a 22 y b 2 ẏ a 21 a 22 y Example: intersection of lines / 2D vector fields ( b1 b 2 ) Properties of matrix leads to understandings of the system We ll learn some important concept in matrix (linear) algebra Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

33 Numerical methods Without knowing analytic solutions, we can solve for differential equations numerically Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

34 Numerical methods Without knowing analytic solutions, we can solve for differential equations numerically Solving for ẋ = x, x(0) = 1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

35 Numerical methods Without knowing analytic solutions, we can solve for differential equations numerically Solving for ẋ = x, x(0) = 1 Idea: Starting from x(0) = 1, estimate the slope ẋ, then advance... Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

36 Numerical methods Without knowing analytic solutions, we can solve for differential equations numerically Solving for ẋ = x, x(0) = 1 Idea: Starting from x(0) = 1, estimate the slope ẋ, then advance... Pseudo code: (specify dx, say, 0.1) x(0) = 1, ẋ(0) = x(0) = 1, x(dx) = x(0) + dx ẋ(0) = 1.1 x(dx) = 1.1, ẋ(dx) = 1.1, x(2dx) = x(dx) + dx ẋ(dx) = 1.21 x[(n 1)dx] = x n 1, ẋ(n 1) = x n 1, x(ndx) = x n 1 + dx x n 1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

37 Numerical methods Without knowing analytic solutions, we can solve for differential equations numerically Solving for ẋ = x, x(0) = 1 Idea: Starting from x(0) = 1, estimate the slope ẋ, then advance... Pseudo code: (specify dx, say, 0.1) x(0) = 1, ẋ(0) = x(0) = 1, x(dx) = x(0) + dx ẋ(0) = 1.1 x(dx) = 1.1, ẋ(dx) = 1.1, x(2dx) = x(dx) + dx ẋ(dx) = 1.21 x[(n 1)dx] = x n 1, ẋ(n 1) = x n 1, x(ndx) = x n 1 + dx x n 1 Some caveat: accuracy, stability Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

38 May 23 rd Lecture 2: Examples in 1-D Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

39 Fishery in Tempe Town Lake Consider fish population in TTL. Suppose there s no fishing, food resource is abundant, no lag in maturation, what can we write down as an equation to describe the change in fish population? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

40 Fishery in Tempe Town Lake Consider fish population in TTL. Suppose there s no fishing, food resource is abundant, no lag in maturation, what can we write down as an equation to describe the change in fish population? Now consider there is limitation of food resource, so growth is actually slowing down due to increasing population, what equation can we write down? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

41 Fishery in Tempe Town Lake Consider fish population in TTL. Suppose there s no fishing, food resource is abundant, no lag in maturation, what can we write down as an equation to describe the change in fish population? Now consider there is limitation of food resource, so growth is actually slowing down due to increasing population, what equation can we write down? The logistic equation ḟ = kf (1 f /C), where k is the growth rate at no competition, C the carrying capacity indicating availability of resources Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

42 Fishery in Tempe Town Lake Consider fish population in TTL. Suppose there s no fishing, food resource is abundant, no lag in maturation, what can we write down as an equation to describe the change in fish population? Now consider there is limitation of food resource, so growth is actually slowing down due to increasing population, what equation can we write down? The logistic equation ḟ = kf (1 f /C), where k is the growth rate at no competition, C the carrying capacity indicating availability of resources Qualitative understanding: Is there anything you can say about the population without solving for the differential equations? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

43 Quantitative analysis (ODE) Let s consider the simpler part df /dt = kf. Physical meaning? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

44 Quantitative analysis (ODE) Let s consider the simpler part df /dt = kf. Physical meaning? Separation of variables df /f = kdt df /f = kdt + c 0 ln f = kt + c 0 f = e c 0 e kt Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

45 Quantitative analysis (ODE) Let s consider the simpler part df /dt = kf. Physical meaning? Separation of variables df /f = kdt df /f = kdt + c 0 The actual logistic equation? ln f = kt + c 0 f = e c 0 e kt Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

46 Quantitative analysis (ODE) Let s consider the simpler part df /dt = kf. Physical meaning? Separation of variables df /f = kdt df /f = kdt + c 0 ln f = kt + c 0 f = e c 0 e kt The actual logistic equation? Separation of variable, partial fraction, integrate [ 1 f + 1 C f ]df = kt + c 0, f = Cc 0e kt 1 + c 0 e kt Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

47 Solution behavior Solution of the logistic equation subject to different initial conditions: Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

48 Solution behavior Solution of the logistic equation subject to different initial conditions: Qualitative understanding: Is there anything you can say about the population without solving for the differential equations? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

49 Qualitative analysis (Dynamical System) How to describe behavior? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

50 Qualitative analysis (Dynamical System) How to describe behavior? Phase portrait Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

51 Qualitative analysis (Dynamical System) How to describe behavior? Phase portrait Long term behavior? Fixed points? Stability? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

52 Consider fishery How shall we control fishery? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

53 Consider fishery How shall we control fishery? Concept: maintain stable population Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

54 Consider fishery How shall we control fishery? Concept: maintain stable population Interpret each curve Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

55 Numerics Let s try solve the example equations with numerical methods Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

56 Synchronization Firefly sync: dθ/dt = Ω + K 1 sin(θ Θ) dθ/dt = ω + K 2 sin(θ θ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

57 Synchronization Firefly sync: dθ/dt = Ω + K 1 sin(θ Θ) dθ/dt = ω + K 2 sin(θ θ) Subtract the first equation from the second by writing φ = Θ θ: dφ dt = Ω ω (K 1 + K 2 ) sin φ Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

58 Synchronization Firefly sync: dθ/dt = Ω + K 1 sin(θ Θ) dθ/dt = ω + K 2 sin(θ θ) Subtract the first equation from the second by writing φ = Θ θ: dφ dt = Ω ω (K 1 + K 2 ) sin φ If (Ω ω)/(k 1 + K 2 ) 1, φ = arcsin( Ω ω K 1 +K 2 ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

59 Synchronization Firefly sync: dθ/dt = Ω + K 1 sin(θ Θ) dθ/dt = ω + K 2 sin(θ θ) Subtract the first equation from the second by writing φ = Θ θ: dφ dt = Ω ω (K 1 + K 2 ) sin φ If (Ω ω)/(k 1 + K 2 ) 1, φ = arcsin( Ω ω K 1 +K 2 ) Two equilibrium points, one stable, corresponding to phase lock Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

60 Exercise for the day Suppose the population equation for fish in TTL is ḟ = f 2 f 4, how should we regulate fishery? Hint: look for maximum of ḟ = f 2 f 4 p, when the function reaches max at ḟ = 0, stable structure with max profit Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

61 Exercise for the day Suppose the population equation for fish in TTL is ḟ = f 2 f 4, how should we regulate fishery? Hint: look for maximum of ḟ = f 2 f 4 p, when the function reaches max at ḟ = 0, stable structure with max profit Use numerical methods to solve for the above equation, with several choices of fishery control p Hint: you should see that at maximum fishing rate solution converge from above fixed point and diverge from below Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

62 Exercise for the day Suppose the population equation for fish in TTL is ḟ = f 2 f 4, how should we regulate fishery? Hint: look for maximum of ḟ = f 2 f 4 p, when the function reaches max at ḟ = 0, stable structure with max profit Use numerical methods to solve for the above equation, with several choices of fishery control p Hint: you should see that at maximum fishing rate solution converge from above fixed point and diverge from below The moth population dynamics in a forest obeys the logistic equation. With the introduction of one species of wood pecker, the moth decreases by the predation function p(f ) = Af /(B + f ). Discuss the dynamics. Hint: look for fixed points and discuss stability based on the touching of the curves Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

63 May 24 th Lecture 3: 2D example: Romeo - Juliet Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

64 Statement of the problem A surreal love affair with a twisted story Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

65 Statement of the problem A surreal love affair with a twisted story Verbal statement: Romeo loves Juliet when she loves him, he backs off when she doesn t love him; Juliet is protective when Romeo loves her, but will approach him when he walks away. Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

66 Statement of the problem A surreal love affair with a twisted story Verbal statement: Romeo loves Juliet when she loves him, he backs off when she doesn t love him; Juliet is protective when Romeo loves her, but will approach him when he walks away. Mathematical model: dr dt = aj, dj dt = br, a, b > 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

67 Statement of the problem A surreal love affair with a twisted story Verbal statement: Romeo loves Juliet when she loves him, he backs off when she doesn t love him; Juliet is protective when Romeo loves her, but will approach him when he walks away. Mathematical model: dr dt = aj, dj dt = br, a, b > 0 Differentiate left eq. w.r.t. t and use the right eq. we get d 2 R dt 2 + abr = 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

68 Statement of the problem A surreal love affair with a twisted story Verbal statement: Romeo loves Juliet when she loves him, he backs off when she doesn t love him; Juliet is protective when Romeo loves her, but will approach him when he walks away. Mathematical model: dr dt = aj, dj dt = br, a, b > 0 Differentiate left eq. w.r.t. t and use the right eq. we get d 2 R dt 2 + abr = 0 Close analogy with the spring-mass problem Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

69 Statement of the problem A surreal love affair with a twisted story Verbal statement: Romeo loves Juliet when she loves him, he backs off when she doesn t love him; Juliet is protective when Romeo loves her, but will approach him when he walks away. Mathematical model: dr dt = aj, dj dt = br, a, b > 0 Differentiate left eq. w.r.t. t and use the right eq. we get d 2 R dt 2 + abr = 0 Close analogy with the spring-mass problem Indeed, many physically motivated problems have analogy, so we study a class of problems Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

70 Analytic solutions From physics, solutions are sine and cosine functions Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

71 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

72 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

73 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other The problem has two degrees of freedom Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

74 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other The problem has two degrees of freedom Physical interpretation: Romeo and Juliet will NEVER be together, unless initially together Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

75 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other The problem has two degrees of freedom Physical interpretation: Romeo and Juliet will NEVER be together, unless initially together What s worse: unstable or, marginally stable Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

76 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other The problem has two degrees of freedom Physical interpretation: Romeo and Juliet will NEVER be together, unless initially together What s worse: unstable or, marginally stable Let s handle this with complex exponentials Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

77 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other The problem has two degrees of freedom Physical interpretation: Romeo and Juliet will NEVER be together, unless initially together What s worse: unstable or, marginally stable Let s handle this with complex exponentials Let R = ce kt k 2 + ab = 0 k = ±i ab R = c 1 e i abt + c 2 e i abt Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

78 Analytic solutions From physics, solutions are sine and cosine functions Let R = c 1 sin(kt) + c 2 cos(kt), plugging into the differential equation, k = ab c 1, c 2 are the initial conditions that prescribes the individual love with each other The problem has two degrees of freedom Physical interpretation: Romeo and Juliet will NEVER be together, unless initially together What s worse: unstable or, marginally stable Let s handle this with complex exponentials Let R = ce kt k 2 + ab = 0 k = ±i ab R = c 1 e i abt + c 2 e i abt Same solution, but advantageous (in a moment we see this) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

79 Let s try to make them happily ever after What indicates that R and J are TOGETHER? (R = J = 0) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

80 Let s try to make them happily ever after What indicates that R and J are TOGETHER? (R = J = 0) What is needed from the exponential functions? (negative real part) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

81 Let s try to make them happily ever after What indicates that R and J are TOGETHER? (R = J = 0) What is needed from the exponential functions? (negative real part) Let s still work on the linear model but introduce dr dt = cr + aj, dj dt = br + dj, a, b > 0, c, d arbitrary Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

82 Let s try to make them happily ever after What indicates that R and J are TOGETHER? (R = J = 0) What is needed from the exponential functions? (negative real part) Let s still work on the linear model but introduce dr dt = cr + aj, dj dt The second order equation: = br + dj, R (c + d)ṙ + (ab + cd)r = 0 a, b > 0, c, d arbitrary Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

83 Let s try to make them happily ever after What indicates that R and J are TOGETHER? (R = J = 0) What is needed from the exponential functions? (negative real part) Let s still work on the linear model but introduce dr dt = cr + aj, dj dt The second order equation: = br + dj, R (c + d)ṙ + (ab + cd)r = 0 Let R = ce kt k 2 (c + d)k + (ab + cd) = 0 k = [(c + d) ± (c + d) 2 4(ab + cd)]/2 a, b > 0, c, d arbitrary Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

84 Let s try to make them happily ever after What indicates that R and J are TOGETHER? (R = J = 0) What is needed from the exponential functions? (negative real part) Let s still work on the linear model but introduce dr dt = cr + aj, dj dt The second order equation: = br + dj, R (c + d)ṙ + (ab + cd)r = 0 Let R = ce kt k 2 (c + d)k + (ab + cd) = 0 k = [(c + d) ± (c + d) 2 4(ab + cd)]/2 At least need c + d < 0 to damp. Physical meaning? a, b > 0, c, d arbitrary Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

85 Perspective from linear algebra Let s forget about the love story at the moment but think about population dynamics (T ) ( )( ) n T n P n+1 = P n with T 0 = 50000, P 0 = Use the computer to check the long term behavior of T n+1, P n+1. Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

86 Perspective from linear algebra Let s forget about the love story at the moment but think about population dynamics (T ) ( )( ) n T n P n+1 = P n with T 0 = 50000, P 0 = Use the computer to check the long term behavior of T n+1, P n+1. There is a convergence to some T, P. Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

87 Perspective from linear algebra Let s forget about the love story at the moment but think about population dynamics (T ) ( )( ) n T n P n+1 = P n with T 0 = 50000, P 0 = Use the computer to check the long term behavior of T n+1, P n+1. There is a convergence to some T, P. The above equation then becomes ( ) ( )( ) T T P = P But how to find them? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

88 Eigenvalues and eigenvectors Let v n = [T n, P n ] T, v n+1 = k[t n, P n ] T (because of alignment) ( ) ( )( ) T n T n k P n = P n Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

89 Eigenvalues and eigenvectors Let v n = [T n, P n ] T, v n+1 = k[t n, P n ] T (because of alignment) ( ) ( )( ) T n T n k P n = P n So (A ki )v n = [0, 0] T, or ( 0.3 k k But we need [T n, P n ] to be non-trivial )( ) T n = P n ( ) 0 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

90 Eigenvalues and eigenvectors Let v n = [T n, P n ] T, v n+1 = k[t n, P n ] T (because of alignment) ( ) ( )( ) T n T n k P n = P n So (A ki )v n = [0, 0] T, or ( 0.3 k k But we need [T n, P n ] to be non-trivial )( ) T n = P n Hence (A ki ) = 0 (explain determinant) ( ) 0 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

91 Eigenvalues and eigenvectors Let v n = [T n, P n ] T, v n+1 = k[t n, P n ] T (because of alignment) ( ) ( )( ) T n T n k P n = P n So (A ki )v n = [0, 0] T, or ( 0.3 k k But we need [T n, P n ] to be non-trivial )( ) T n = P n Hence (A ki ) = 0 (explain determinant) ( ) 0 0 For this case k 1 = 1, v 1 = [8, 7] T, k 2 = 0.5, v 1 = [ 1, 1] T Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

92 Same idea on differential equations For the love circle, (Ṙ ) = J ( c a b d )( ) R J Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

93 Same idea on differential equations For the love circle, (Ṙ ) = J ( c a b d )( ) R J We characterize rates and directions with eigenvalues/vectors Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

94 Same idea on differential equations For the love circle, (Ṙ ) = J ( c a b d )( ) R J We characterize rates and directions with eigenvalues/vectors Special directions are just those where velocity vector has the same direction as the position vector. (eg: ẋ = x, ẏ = y; ẋ = y, ẏ = x) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

95 Same idea on differential equations For the love circle, (Ṙ ) = J ( c a b d )( ) R J We characterize rates and directions with eigenvalues/vectors Special directions are just those where velocity vector has the same direction as the position vector. (eg: ẋ = x, ẏ = y; ẋ = y, ẏ = x) Find the eigenvalues and eigenvectors for the love circle problem Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

96 May 24 th Lecture 4: Nonlinear 2D problems Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

97 Rabbit Vs. Sheep Rabbit and sheep both eat grass in a closed lawn, so they form a competition Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

98 Rabbit Vs. Sheep Rabbit and sheep both eat grass in a closed lawn, so they form a competition Both species have logistic dynamics in each, but there is also an interaction between the species Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

99 Rabbit Vs. Sheep Rabbit and sheep both eat grass in a closed lawn, so they form a competition Both species have logistic dynamics in each, but there is also an interaction between the species When rabbit and sheep compete, sheep wins very often, but occasionally rabbits can still get something Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

100 Rabbit Vs. Sheep Rabbit and sheep both eat grass in a closed lawn, so they form a competition Both species have logistic dynamics in each, but there is also an interaction between the species When rabbit and sheep compete, sheep wins very often, but occasionally rabbits can still get something A simple model ṙ = r(3 r 2s) F (r, s) ṡ = s(2 r s) G(r, s) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

101 Rabbit Vs. Sheep Rabbit and sheep both eat grass in a closed lawn, so they form a competition Both species have logistic dynamics in each, but there is also an interaction between the species When rabbit and sheep compete, sheep wins very often, but occasionally rabbits can still get something A simple model ṙ = r(3 r 2s) F (r, s) ṡ = s(2 r s) G(r, s) Find all fixed points, analyze the dynamics and interpret the physical meanings Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

102 The Jacobian The Jacobian captures local dynamics by linearization Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

103 The Jacobian The Jacobian captures local dynamics by linearization Concept: on fixed points (r, s ), ṙ r = 0, ṡ s = 0, so locally ξ = r r, η = s s and ṙ = ξ r = ξ, ṡ = η s = η Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

104 The Jacobian The Jacobian captures local dynamics by linearization Concept: on fixed points (r, s ), ṙ r = 0, ṡ s = 0, so locally ξ = r r, η = s s and ṙ = ξ r = ξ, ṡ = η s = η On RHS, linearize F, G around (r, s ) F F (r, s ) + F r (r r ) + F s (s s ) = 0 + F r (r r ) + F s (s s ) G G(r, s ) +G r (r r ) +G s (s s ) = 0 + G r (r r ) + G s (s s ) ( ) ( ) ( ) ξ Fr F = s ξ η η G r G s Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

105 The Jacobian The Jacobian captures local dynamics by linearization Concept: on fixed points (r, s ), ṙ r = 0, ṡ s = 0, so locally ξ = r r, η = s s and ṙ = ξ r = ξ, ṡ = η s = η On RHS, linearize F, G around (r, s ) F F (r, s ) + F r (r r ) + F s (s s ) = 0 + F r (r r ) + F s (s s ) G G(r, s ) +G r (r r ) +G s (s s ) = 0 + G r (r r ) + G s (s s ) ( ) ( ) ( ) ξ Fr F = s ξ η η G r G s Local dynamics governed by the Jacobian matrix ( ) Fr F s G r G s Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

106 Fixed points and linearization From governing equations, r = 0 or r = 3 2s s = 0 or r = 2 s Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

107 Fixed points and linearization From governing equations, r = 0 or r = 3 2s s = 0 or r = 2 s Let x = (r, s), 4 fixed points: (0, 0), (0, 2), (3, 0), (1, 1) where ẋ = 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

108 Fixed points and linearization From governing equations, r = 0 or r = 3 2s s = 0 or r = 2 s Let x = (r, s), 4 fixed points: (0, 0), (0, 2), (3, 0), (1, 1) where ẋ = 0 Local dynamics captured by the Jacobian ( 3 2r 2s 2r s 2 r 2s ) ( ) ξ η Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

109 Fixed points and linearization From governing equations, r = 0 or r = 3 2s s = 0 or r = 2 s Let x = (r, s), 4 fixed points: (0, 0), (0, 2), (3, 0), (1, 1) where ẋ = 0 Local dynamics captured by the Jacobian ( 3 2r 2s 2r s 2 r 2s ) ( ) ξ η Problem becomes eigenvalues/eigenvectors of the Jacobian matrix Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

110 Stability of fixed points At (0, 0), A = [ ], k 1 = 3, v 1 = [ 1 0 ] [ 0, k 2 = 2, v 2 = 1 ] Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

111 Stability of fixed points [ ] [ ] [ ] At (0, 0), A =, k = 3, v 1 =, k 0 2 = 2, v 2 = 1 [ ] [ ] [ ] At (3, 0), A =, k = 3, v 1 =, k 0 2 = 1, v 2 = 1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

112 Stability of fixed points [ ] [ ] [ ] At (0, 0), A =, k = 3, v 1 =, k 0 2 = 2, v 2 = 1 [ ] [ ] [ ] At (3, 0), A =, k = 3, v 1 =, k 0 2 = 1, v 2 = 1 [ ] [ ] [ ] At (0, 2), A =, k = 1, v 1 =, k 2 2 = 2, v 2 = 1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

113 Stability of fixed points [ ] [ ] [ ] At (0, 0), A =, k = 3, v 1 =, k 0 2 = 2, v 2 = 1 [ ] [ ] [ ] At (3, 0), A =, k = 3, v 1 =, k 0 2 = 1, v 2 = 1 [ ] [ ] [ ] At (0, 2), A =, k = 1, v 1 =, k 2 2 = 2, v 2 = 1 [ ] 1 2 At (1, 1), A =, k = [ ] 2 2 1, v 1 =, k 1 2 = 2 1, [ ] 2 v 2 = 1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

114 Stability of fixed points [ ] [ ] [ ] At (0, 0), A =, k = 3, v 1 =, k 0 2 = 2, v 2 = 1 [ ] [ ] [ ] At (3, 0), A =, k = 3, v 1 =, k 0 2 = 1, v 2 = 1 [ ] [ ] [ ] At (0, 2), A =, k = 1, v 1 =, k 2 2 = 2, v 2 = 1 [ ] 1 2 At (1, 1), A =, k = [ ] 2 2 1, v 1 =, k 1 2 = 2 1, [ ] 2 v 2 = 1 Phase portrait Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

115 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

116 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Linearization still shows a marginally stable cycle Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

117 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Linearization still shows a marginally stable cycle Contribution of the cubic terms is going to determine the fate of their love Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

118 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Linearization still shows a marginally stable cycle Contribution of the cubic terms is going to determine the fate of their love R 2 + J 2 polar coordinates! Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

119 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Linearization still shows a marginally stable cycle Contribution of the cubic terms is going to determine the fate of their love R 2 + J 2 polar coordinates! R = r cos θ, J = r sin θ so ṙ = ar 3, θ = 1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

120 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Linearization still shows a marginally stable cycle Contribution of the cubic terms is going to determine the fate of their love R 2 + J 2 polar coordinates! R = r cos θ, J = r sin θ so ṙ = ar 3, θ = 1 We can even make their love circle structurally stable! Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

121 Limit cycles in love affair Let s look at a more complicated love cycle problem Ṙ = J + ar(r 2 + J 2 ) J = R + aj(r 2 + J 2 ) Linearization still shows a marginally stable cycle Contribution of the cubic terms is going to determine the fate of their love R 2 + J 2 polar coordinates! R = r cos θ, J = r sin θ so ṙ = ar 3, θ = 1 We can even make their love circle structurally stable! Try ṙ = ar(1 r 2 ), θ = 1, unstable at r = 0, stable at r = 1. Can you write this under Cartesian coordinate? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

122 Nonlinear pendulum Let s look at the dynamics of a REAL pendulum, not the small angle approximation Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

123 Nonlinear pendulum Let s look at the dynamics of a REAL pendulum, not the small angle approximation Discuss the formulation of problem by analyzing forces Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

124 Nonlinear pendulum Let s look at the dynamics of a REAL pendulum, not the small angle approximation Discuss the formulation of problem by analyzing forces The resulting equation: θ + g/l sin(θ) = 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

125 Nonlinear pendulum Let s look at the dynamics of a REAL pendulum, not the small angle approximation Discuss the formulation of problem by analyzing forces The resulting equation: θ + g/l sin(θ) = 0 Analyze the fixed points (0, 0), (π, 0), their stability (marginally stable, unstable) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

126 Nonlinear pendulum Let s look at the dynamics of a REAL pendulum, not the small angle approximation Discuss the formulation of problem by analyzing forces The resulting equation: θ + g/l sin(θ) = 0 Analyze the fixed points (0, 0), (π, 0), their stability (marginally stable, unstable) Phase portrait, direction of trajectories Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

127 Damped nonlinear pendulum Adding some damping corresponding to friction forces Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

128 Damped nonlinear pendulum Adding some damping corresponding to friction forces The governing equation: θ + γ/m θ + g/l sin(θ) = 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

129 Damped nonlinear pendulum Adding some damping corresponding to friction forces The governing equation: θ + γ/m θ + g/l sin(θ) = 0 Analyze the fixed points by linearization (0, 0), (π, 0), their stability (stable, unstable) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

130 Damped nonlinear pendulum Adding some damping corresponding to friction forces The governing equation: θ + γ/m θ + g/l sin(θ) = 0 Analyze the fixed points by linearization (0, 0), (π, 0), their stability (stable, unstable) Phase portrait Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

131 May 25 th Lecture 5: Sea ice dynamics, a 2D version Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

132 Formulation Assume a simple sea ice recession-progression model based on two parameters, L and C Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

133 Formulation Assume a simple sea ice recession-progression model based on two parameters, L and C The hypothetical cycle: Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

134 Formulation Assume a simple sea ice recession-progression model based on two parameters, L and C The hypothetical cycle: Seems like a simple spring-mass oscillation but on sphere! Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

135 Formulation Assume a simple sea ice recession-progression model based on two parameters, L and C The hypothetical cycle: Seems like a simple spring-mass oscillation but on sphere! Coordinate: Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

136 Formulation Assume North-South symmetry, azimuthal invariance and perfect mixture of carbon Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

137 Formulation Assume North-South symmetry, azimuthal invariance and perfect mixture of carbon Excessive amount of carbon is linearly proportional to the volume of ice melt sea surface area change Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

138 Formulation Assume North-South symmetry, azimuthal invariance and perfect mixture of carbon Excessive amount of carbon is linearly proportional to the volume of ice melt sea surface area change Sea surface area away from equilibrium is linearly proportional to reduction of carbon Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

139 Formulation Assume North-South symmetry, azimuthal invariance and perfect mixture of carbon Excessive amount of carbon is linearly proportional to the volume of ice melt sea surface area change Sea surface area away from equilibrium is linearly proportional to reduction of carbon Assume an equilibrium state where Latitude of ice cap and carbon concentration remain constant Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

140 Formulation Assume North-South symmetry, azimuthal invariance and perfect mixture of carbon Excessive amount of carbon is linearly proportional to the volume of ice melt sea surface area change Sea surface area away from equilibrium is linearly proportional to reduction of carbon Assume an equilibrium state where Latitude of ice cap and carbon concentration remain constant The simplest Cartesian coordinate analogy: dl/dt = a(c C 0 ) dc/dt = b(l L 0 ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

141 Formulation Assume North-South symmetry, azimuthal invariance and perfect mixture of carbon Excessive amount of carbon is linearly proportional to the volume of ice melt sea surface area change Sea surface area away from equilibrium is linearly proportional to reduction of carbon Assume an equilibrium state where Latitude of ice cap and carbon concentration remain constant The simplest Cartesian coordinate analogy: dl/dt = a(c C 0 ) dc/dt = b(l L 0 ) Our model is a deviation from this simple formulation Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

142 The Latitude equation Volume change of ice strip: RdL 2πr thickness(100m) = 200πR 2 cos LdL Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

143 The Latitude equation Volume change of ice strip: RdL 2πr thickness(100m) = 200πR 2 cos LdL Change of carbon concentration over time (change of carbon substance in air): a(c C 0 )dt Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

144 The Latitude equation Volume change of ice strip: RdL 2πr thickness(100m) = 200πR 2 cos LdL Change of carbon concentration over time (change of carbon substance in air): a(c C 0 )dt Thus: dl dt = a (C C 0 ) F cos L Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

145 The Carbon equation Sea surface area surplus/deficit: A = L L 2πrRdL = 2πR 2 cos L dl = 2πR 2 (sin L sin L 0 ) L 0 L 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

146 The Carbon equation Sea surface area surplus/deficit: Thus A = L L 0 2πrRdL = L L 0 2πR 2 cos L dl = 2πR 2 (sin L sin L 0 ) dc dt = ba = b (sin L sin L 0 ) G Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

147 Dynamics near fixed point Do we still expect a circle near (L 0, C 0 )? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

148 Dynamics near fixed point Do we still expect a circle near (L 0, C 0 )? Jacobian: A = = [ a (C C 0 )[ sin L/ cos L 2 ] a ] / cos L b cos L 0 L0,C 0 [ 0 a ] / cos L 0 b cos L 0 0 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

149 Dynamics near fixed point Do we still expect a circle near (L 0, C 0 )? Jacobian: A = = [ a (C C 0 )[ sin L/ cos L 2 ] a ] / cos L b cos L 0 L0,C 0 [ 0 a ] / cos L 0 b cos L 0 0 Linearization doesn t tell, need higher order terms maybe we need polar coordinate! Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

150 Dynamics near fixed point Do we still expect a circle near (L 0, C 0 )? Jacobian: A = = [ a (C C 0 )[ sin L/ cos L 2 ] a ] / cos L b cos L 0 L0,C 0 [ 0 a ] / cos L 0 b cos L 0 0 Linearization doesn t tell, need higher order terms maybe we need polar coordinate! Let s try a little in this direction by expanding F, G into higher order terms Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

151 Numerical predictions Use numerical simulations to see what happens, choose L 0 = π/3, C 0 = 1, a = 0.1, b = 0.1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

152 Numerical predictions Use numerical simulations to see what happens, choose L 0 = π/3, C 0 = 1, a = 0.1, b = 0.1 Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

153 Numerical predictions Use numerical simulations to see what happens, choose L 0 = π/3, C 0 = 1, a = 0.1, b = 0.1 Using these parameters to check into our prediction on the dynamics? Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

154 Additional constraints and physical interpretations Physically we don t expect C to drop below 0, L is between equator and the pole Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

155 Additional constraints and physical interpretations Physically we don t expect C to drop below 0, L is between equator and the pole Long term dynamics on phase plane Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

156 Additional constraints and physical interpretations Physically we don t expect C to drop below 0, L is between equator and the pole Long term dynamics on phase plane Physical meaning of the data: ice-free snowball in short time Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

157 May 25 th Lecture 6: Sea-ice dynamics, 3D version Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

158 Formulation The carbon-sea ice interaction is rather indirect Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

159 Formulation The carbon-sea ice interaction is rather indirect Factor in temperature interactions with the rest Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

160 Formulation The carbon-sea ice interaction is rather indirect Factor in temperature interactions with the rest Surface absorption of carbon unchanged dc dt = ba = b (sin L sin L 0 ) Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

161 Formulation The carbon-sea ice interaction is rather indirect Factor in temperature interactions with the rest Surface absorption of carbon unchanged dc dt = ba = b (sin L sin L 0 ) Melting (change in latitude) directly link to temperature dl dt = a (T T 0 ) cos L Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

162 Formulation The carbon-sea ice interaction is rather indirect Factor in temperature interactions with the rest Surface absorption of carbon unchanged dc dt = ba = b (sin L sin L 0 ) Melting (change in latitude) directly link to temperature dl dt = a (T T 0 ) cos L Contributions to temperature change: solar input - reflection of heat from ice/ocean surface Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46

163 The temperature equation Assume constant heat flux straight into earth equator q Tang (ASU) AMDS Workshop Notes May. 23 rd -25 th, / 46