HW1_Solutions.pdf

Problem 4.1. Further Exercise 1 Exercise 4.1.FE 1 Is a saddle point an attractor? Justify your answer. For something to be an attractor, it has to attract other trajectories in space. However, for a saddle point, it generally does not. The trajectories can approach the point from an angle, and then almost always run away in the long run. Thus, a saddle point is NOT an attractor. Problem 4.1. Further Exercise 2 Exercise 4.1.FE 2 Does a trajectory that approaches a limit cycle attractor ever reach the attractor? Explain. So let’s say that the trajectory will actually reach the attractor. Let’s think about this... 1. If the trajectory reaches the attractor, that means that the attractor is actually a part of the trajectory. 2. However, we know that all trajectories will approach the attractor. So if #1 is true, it means that the attractor will belong to ALL trajectories. 3. If #2 is true, then the trajectories will cross each other at least at the region of the limit cycle attractor. 4. However, we know that trajectories can NEVER cross. As a result, the limit cycle attractor can never belong to any trajectory. In other words, A trajectory that approaches a limit cycle attractor will NEVER reach the attractor. Problem 4.1. Further Exercise 4 Exercise 4.1.FE 4 Describe jet lag and recovery from it in dynamical terms. The body controls sleep cycle via the circadian rhythm, which oscillates throughout the day. After a long flight, the circadian rhythm witnesses a drastic switch, where the levels of biomarkers do not math with the (expected) normal levels at certain hours. However, the body is not able to adapt immediately since it takes time to control levels of biomarkers in the entire body. Rather, the system slowly adjusts and approaches the stable limit cycle where the circadian rhythm is normally at. 1

Problem 4.1. Further Exercise 6 Exercise 4.1.FE 6 Suppose a 2D system has a stable equilibrium point that is located somewhere outside a limit cycle. Can a trajectory starting inside the limit cycle reach this point? Justify your answer. (Hint: It may help to draw the situation.) As we can see, if a trajectory starts inside the limit cycle, it will try to approach the limit cycle but can never reach it (see Further Exercise 2 ). However, if it can never reach the cycle, it can never cross it to reach the stable equilibrium point on the other side of the cycle. Thus, A trajectory starting inside the limit cycle CANNOT reach this point. Problem 4.2.3 Exercise 4.2.3 Verify this assertion. We are looking to verify the assertion: “This negative feedback model will not oscillate, no matter how steep the feedback.” . To do this, we will need to look at the trajectories and/or time series for the system when it has: (1) a low value of n (not-steep feedback), and (2) a high value of n (steep feedback). There are a few different ways you can accomplish this: animation, interactive, assign-directly, etc. We will show the codes for the most simple method, which is to assign a definite value to n . Recommended Initial Condition: H = 0 . 25 and G = 2 2

1 var( ‘H, G’ ) # define H and G as symbolic CoCalc variables 2 n = 5 # assign a value to n 3 time list = srange( 0 , 100 , 0.1 ) # define the list of time points to obtain solutions 4 5 Hprime(H, G) = 1 / ( 1 + Gˆn ) - 0.2 * H # define H’ 6 Gprime(H, G) = H - 0.2 * G # define G’ 7 8 sol = desolve odeint( [ Hprime(H, G), Gprime(H, G)], dvars = [H, G], ics = [ 0.25 , 2 ], times = time list) # simulate the model 9 10 list plot( sol ) # plot the trajectory Small n Value (e.g. n = 5 ) Large n Value (e.g. n = 50 ) As observed, regardless of the steepness of the feedback (i.e. how large the value of n is), the system always has one stable equilibrium point, instead of a stable oscillation/limit cycle attractor. Problem 4.2.4 Exercise 4.2.4 In the delay differential equation Y ′ ( t ) = 16 Y ( t − 2) + 8 Y ( t ), what does Y ( t − 2) refer to? What does Y ( t ) refer to? We have: • Y ( t ) ≡ the value of Y at time t (i.e. right now) • Y ( t − 2) ≡ the value of Y at time t − 2 (i.e. two time units prior ) Problem 4.2.5 3

Exercise 4.2.5 What aspect of the function does V max control? Let’s graph V max · X n 1+ X n for a few different values of V max . We also pick: n = 5. X Ventilation Rate V max = 1 V max = 2 V max = 3 As we observed, the V max controls the maximum value for the Ventilation Rate, and graphically, it controls the height of the function. Problem 4.2.6 Exercise 4.2.6 Let X ′ = 6 − 16 · X ( t − 0 . 2) 5 1 + X ( t − 0 . 5) 5 · X Assume that for all t ≤ 0, X ( t ) = 0 . 5. Use Euler’s method with a step size of 0.1 to approximate X (0 . 3). A Note on Our Notation: “ X ( t = 3) ” vs “ X 3 ” In this case, the first notation means the value of X at t = 3. Meanwhile, the second notation means the value of X on its third iteration . This may or may not mean at t = 3. If the step size is ∆ t = 1, then by the third step, you will reach t = 3, hence X ( t = 3) = X 3 for ∆ t = 1. However, let’s say that ∆ t = 0 . 01, then by the third step, you only reach t = 0 . 03, and clearly X 3 = X ( t = 0 . 03) ̸ = X ( t = 3) for ∆ t ̸ = 1. To reach t = 0 . 3 in step size of ∆ t = 0 . 1, we need to take 3 steps . We also have the following equations: X ′ ( t ) = 6 − 16 · X ( t − 0 . 2) 5 1 + X ( t − 0 . 5) 5 · X ( t ) (1) X t new = X t old + X ′ ( t old ) · ∆ t (2) 4

Step #0: Find X at t = 0 As t = 0 ≤ 0: X (0) = 0 . 5 Step #1: Find X at t = 0 . 1 We have t new = 0 . 1 and t old = 0. It is important to remember that we are evaluating X ′ at t old (as seen in Equation (2)). X ( t − 0 . 2) X ( t − 0 . 5) X ≡ X old X (0 − 0 . 2) = X ( − 0 . 2) = 0 . 5 X (0 − 0 . 5) = X ( − 0 . 5) = 0 . 5 X old = 0 . 5 Inputting to Equations (1) and (2), we have: X (0 . 1) = X (0) + X ′ (0) · 0 . 1 = 0 . 5 + " 6 − 16 · (0 . 5) 5 1 + (0 . 5) 5 · (0 . 5) # · (0 . 1) = 1 . 07576 5

Step #2: Find X at t = 0 . 2 We have t new = 0 . 2 and t old = 0 . 1. It is important to remember that we are evaluating X ′ at t old (as seen in Equation (2)). X ( t − 0 . 2) X ( t − 0 . 5) X ≡ X old X (0 . 1 − 0 . 2) = X ( − 0 . 1) = 0 . 5 X (0 . 1 − 0 . 5) = X ( − 0 . 4) = 0 . 5 X old = 1 . 07576 Inputting to Equations (1) and (2), we have: X (0 . 2) = X (0 . 1) + X ′ (0 . 1) · 0 . 1 = 1 . 07576 + " 6 − 16 · (0 . 5) 5 1 + (0 . 5) 5 · (1 . 07576) # · (0 . 1) = 1 . 62360 Step #3: Find X at t = 0 . 3 We have t new = 0 . 3 and t old = 0 . 2. It is important to remember that we are evaluating X ′ at t old (as seen in Equation (2)). X ( t − 0 . 2) X ( t − 0 . 5) X ≡ X old X (0 . 2 − 0 . 2) = X (0) = 0 . 5 X (0 . 2 − 0 . 5) = X ( − 0 . 3) = 0 . 5 X old = 1 . 62360 Inputting to Equations (1) and (2), we have: X (0 . 3) = X (0 . 2) + X ′ (0 . 2) · 0 . 1 = 1 . 62360 + " 6 − 16 · (0 . 5) 5 1 + (0 . 5) 5 · (1 . 62360) # · (0 . 1) = 2 . 14488 In Summary, we have: Step Starting X n X ( t − 0 . 2) X ( t − 0 . 5) Compute X ′ Final X n +1 0 - - - - X (0) = 0 . 5 1 X 0 = 0 . 5 0.5 0.5 X ′ = 5 . 75758 X 1 = 1 . 07576 2 X 1 = 1 . 07576 0.5 0.5 X ′ = 5 . 47842 X 2 = 1 . 62360 3 X 2 = 1 . 62360 0.5 0.5 X ′ = 5 . 21280 X 3 = 2 . 14488 Problem 4.2. Further Exercise 1 Exercise 4.2.FE 1 Some people have difficulty maintaining a stable weight. Instead, they gain a lot of weight, go on a diet, lose the weight, but then gain it back. This pattern is sometimes referred to as yo-yo dieting. 6

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