Homework 2 5115: Dynamical systems in biological engineering (Fall 2022)
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Homework 2
5115: Dynamical systems in biological engineering (Fall 2022)
Sep 20, 2022 (due by the end of Sep 27)
Problem 2.1 (Analytical) Considering the logistic equation = f(N) = rN(1 − ).
(a) using the formula N(t) = for the solution of the logistic equation, show that N → B as
t → ∞ .
For parts (b) - (d), you should not use the above formula for the solution. Instead, use the following formula derived from the chain rule:
= = = f ′ (N)f(N)
(Here,f ′ (N) means the derivative off(N) with respect to N .)
(b) The graph of N(t) is concave up at times t where N(t) < .
(c) The graph of N(t) is concave down at times t where < N(t) < B .
(d) The graph of N(t) is concave up at times t where N(t) > B .
Problem 2.2 (Numerical) We will work on Problem 2.1 again, but this time do so computationally. (a) Simulate the logistic equation with an ODE solver. Here, we set r = 0.1,B = 100, and N(t = 0) = N0 = 1.
(b) With the time trajectory of N(t), you can now compute numerically using the following formula (think why; you don’t need to answer this)
d2N N(t + Δt) + N(t − Δt) − 2N(t)
Use the above formula to compute function of t. |
d2N dt2 |
as the function of t using the time trajectory from (a). Plot |
as the |
Problem 2.3 (Analytical) Consider a growth model governed by the diferential equation:
= rN + qN2
Show that there are no solutions defined for all t > 0 , if q > 0.
Hint: check the solution of the logistic equation.
Problem 2.4 (Numerical) Consider two co-existing bacterial species X and Y . We can generalize the logistic equation to model both type of bacteria by considering the cost of nutrient during cell growth. (Check slide #4 from Lecture 0920)
{ Y(Y)
Now, simulate the above ODEs and plot time trajectories of X(t) and Y(t) . We set B = 100. But you will need to try a few combinations of rX , rY , X(t = 0) = X0 and Y(t = 0) = Y0 for (a - c).
(a) Find a situation, i.e., a set of (rX , rY , X0 and Y0 ), where X(t) equals to Y(t) for large t; (b) Find a situation where X(t) is much larger than Y(t) for large t;
(c) Find a situation where X(t) is much smaller than Y(t) for large t;
(d) For each case above, compute X(t) + Y(t) for large t (you can take the values from the last time step)
2022-09-27