AMATH 383 Introduction to Continuous Mathematical Modeling Spring, 2020 AMATH 383 Homework Assignment #6 (Due online: 5pm, Thu., May 14) 1. Exercise 2 of Chapter 7 of the textbook. 2. Consider logistic difference equation, which has been widely used as a model for annual population of certain insects: xn+1 = rxn(1− xn) = f(x), 0 ≤ xn ≤ 1. (1) (a) Show that expression f (f(x))− x = 0 can be factorized into rx (rx− r + 1) ( rx2 − (1 + r)x+ 1 + r r ) = 0. (b) Show that x∗1 = 1 + r + √ (1 + r)(r − 3) 2r and x∗2 = 1 + r −√(1 + r)(r − 3) 2r satisfy: f(x∗1) = x ∗ 2 and f(x ∗ 2) = x ∗ 1. They are known as a period-2 cycle solution to Eq. (1). You can find more information on this model in this paper: https://www.nature.com/articles/261459a0 3. Consider a reversible bi-molecular association reaction: A+B k1 GGGBFGG k−1 C, in which k1 is an association rate constant, and k−1 is a dissociation rate constant. (a) Classical chemical kinetics uses ordinary differential equation to describe the con- centrations of A, B, and C at time t, cA(t), cB(t), and cC(t). dcA(t) dt = −kˆ1cAcB + k−1cC , (2a) dcB(t) dt = kˆ1cAcB − k−1cC , (2b) dcC(t) dt = kˆ1cAcB − k−1cC . (2c) Show that d dt ( cA(t) + cC(t) ) = 0 and d dt ( cB(t) + cC(t) ) = 0, Prof. Hong Qian May 9, 2020 AMATH 383 Introduction to Continuous Mathematical Modeling Spring, 2020 and explain the meaning of these two equations, in terms of the molecules and the atoms within. (b) Molecular concentration in chemistry is usually measured in terms of molarity (M): the number of moles per liter of solution, where a mole contains 6.022× 1023 number of molecules. Let us now consider a single biological cell with a tiny volume v, in which the number of A, B, and C are only on the order of tens to hundreds. In this case, the differential equation above is not a realistic model; rather one uses random variable n(t) to count the number of C molecular in the cell at time t. The reaction is stochastic, as shown schematically below: 0 -ff 1 -ff 2 -ff · · ·n-1 -ff n -ff n+1 -ff · · · -ff NA k1NANB k−1 k1(NA − 1)(NB − 1) 2k−1 k1(nA + 1)(nB + 1) nk−1 k1nAnB (n + 1)k−1 k1(nA − 1)(nB − 1) NAk−1 k1(NB −NA + 1) The integers in the circles represent the number of C molecules in the cell. Assuming at the very beinning, nC(0) = 0, and nA(0) = NA, nB(0) = NB. Thus if there are n number of C at time t, the number of A molecule and B molecule are nA = NA− n and nB = NB − n, correspondingly. The rate of forward reaction is k1nAnB = k1(NA − n)(NB − n), and the rate of backward reaction is k−1n. Note that the k1 is diffrent from kˆ1 in the differential equation: k1 = kˆ1v , where v is the volume of the cell: concentration = number v . Write a program to simulate this stochastic chemical kinetics. You should think of k1nAnB = k1(NA − n)(NB − n) as the “birth rate” un, and k−1n as the “death rate” wn. Use k1 = 0.1 and k−1 = 1, and start the reaction with nC(0) = 0, nA(0) = NA = 30, nB(0) = NB = 40, and ∆t = 0.005, simulate this reaction for 1, 000∆t, with the total time of 5 units. Do this 3 times with different random seeds, and plot your results in a graph of nC(t) vs. t. (c) Now running the program in much longer time with 50,000 steps, obtain the his- togram for the steady state probability distribution for the number of C molecules. Com- pare your result from computer simulation, in a graph, with pss(n) = 1 Ξ n∏ j=0 (NA − j + 1)(NB − j + 1)η j = 1 Ξ ( NA!NB!η n n!(NA − n)!(NB − n)! ) , where NA = 30, NB = 40, η = k1k−1 = 0.1, and Ξ is just a constant. To compare, adjust the Ξ to get a rough good fit. (d) According to the differential equations in Eq. 2, the equilibrium of the chemical reaction is when ( nAnB nC ) eq = (NA − n∗)(NB − n∗) n∗ = k−1 k1 . Prof. Hong Qian May 9, 2020 AMATH 383 Introduction to Continuous Mathematical Modeling Spring, 2020 Using k1 = 0.1, k−1 = 1, NA = 30, and NB = 40, solve this quadratic equation for the equilibrium number of C molecule, n∗, and compare it with the histogram you obtained in (c). Prof. Hong Qian May 9, 2020
