1. The growth of a cancerous tumor can be modeled by the Gompertz Law: dn. = an ln ( )

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1 1. The growth of a cancerous tumor can be modeled by the Gompertz Law: ( ) dn b = an ln, (1) dt N where N measures the size of the tumor. (a) Interpret the parameters a and b (both non-negative) biologically. Potential Answer: The model parameters a and b are analogues to those found in the logistic model for population growth. a is a tumor growth rate, while b is a carrying capacity coming from external factors such as limited space and nutrients. A growth rate or a carrying capacity of 0 produces a model with uninteresting (no) dynamics, so we will assume both are positive. (b) Define a new variable x = ln(n). Derive an equation for the evolution of x. Answer: Taking the derivative with respect to time on both sides of x = ln(n) and applying the chain rule gives dx dt = 1 dn N dt. (2) Substituting in dn dt = an ln ( ) b N and using N = e x (again from x = ln(n)) gives us a differential equation for x(t): dx dt = a(ln(b) x), (3) where we have used ln ( b N ) = ln(b) ln(n). (c) Solve your equation from part (b). Your solution should be the variable x(t) as a function of time. Answer: Rewriting equation (3) as dx dt + ax = a ln(b), (4) we can see that this is a linear differential equation with a homogeneous part x h (t) satisfying dx h dt + ax h = 0 (5) and a particular part x p (t) satisfying dx p dt + ax p = a ln(b). (6) Substituting x h (t) = x 0 e λt, we find that λ = a. Since the equation for x p (t) has a constant inhomogeneous part a ln(b), we will try a constant test solution x p (t) = C. Substituting this into (6) yields x p (t) = ln(b). Therefore our solution x(t) is x(t) = x h (t) + x p (t) = x 0 e at + ln(b). (7) (d) Use your answer to part (c) to solve for the tumor size N(t) as a function of time. What happens to the tumor? Answer: We need to undo the change of variables x = ln(n). Since the inverse transformation is e x = N, we have N(t) = e x 0e at +ln(b) 1 = be x 0e at.

2 Using that the intial value x 0 and the initial tumor size N 0 at time t = 0 satisfy x 0 + ln(b) = ln(n 0 ), we arrive at N(t) = be (ln(n 0) ln(b))e at = be ln(n 0/b)e at = b (e ) ln(n e 0/b) at = b ( ) e at N0 b, (8) which is as far as the equation simplifies. Note that N(0) = N 0. If N 0 = 0, then N(t) = 0 for all time; however, if N 0 > 0, then in the limit that t, the exponent e at goes to 0, so we have that N(t) b, analogous to the population in a logistic model growing or shrinking to a carrying capacity. 2. Suppose the population size of some species follows the model (a) Find the Equilibrium populations. Answer: At equilibrium we have the rate of change dn dt equilibrium population is or The three solutions are N = 0, N = 1, and N = 2. dn dt = 3N 2 N. (9) 2 + N 2 = 0, so the equation for the N = 3N N 2, (10) N(N 2 3N + 2) = 0. (11) (b) Draw the phase-line diagram. Answer: Using MATLAB to plot 3N 2 2+N 2 and N, we get the following figure that matches the result above. Figure 1: Plots of both terms in equation (9) up to N = 3. Equilibria (crossing points) are marked with a green star. 2

3 Looking at the difference between these two terms, it is clear graphically that the difference (the rate of change of N) is positive on (, 0) (1, 2) and negative on (0, 1) (2, ). We can then create a stability or phase line diagram (shown below). dn dt N (c) Which equilibria are stable? Answer: Based on the diagram above, N = 0 and N = 2 are stable equilibria, while N = 1 is unstable, as N(t) moves toward one of the other equilibria even if it starts close to N = 1. (d) Interprest these results in biological terms. Why might this a population behave as it does for small values? Potential Answer: One of the (qualitative) purposes of this model is to account for realistic population dynamics at low population levels. If a biological population pass below a critical threshold (N = 1 in this model), the population will most likely die off, possibly from random events that hit a small population harder, or more likely from a loss of genetic viability or communally driven self-sustainability. This is something often not captured in more basic population models, for example, the logistic equation N (t) = r(1 N/K)N, which predicts that even if there is a biologically unrealistic population (i.e. one individual or even a fraction of an individual), as long as the population is not 0, it can grow to its carrying capacity K. Of course, the model parameters in this more complex model are unrealistic, and are chosen to make the model solvable largely by hand. See the next problem for a case where parameters come into play more generally. 3. Consider the following model for the population of a single species (denoted u): du (1 dt = r u ) u u2 k 1 + u 2. (12) We may interpret γ(u) = r ( 1 u k ) as the (population dependent) growth rate. It is worth noting that this comes from a population model for N(t) that has a similar term VmaxN 2 N 2 1/2 +N 2. The current model is derived by rescaling time and the population so that the parameters V max and N 1/2 are both 1. (a) Use phase-line analysis to determine the behavior of this population for various values of r and k. 3

4 Answer: Set k = 30. Plotting the terms γ(u)u and u2 together for 0.05 r 1 1+u 2 using MATLAB gives the following results (note for scale that the blue curve does not change from figure to figure). Initially at r = 0, the growth rate is not enough to overcome the decay rate, and the only equilibrium is u = 0, which is stable. There is actually a transcritical bifurcation at r = 0 as a negative equilibrium passes through u = 0. For small but positive r, there is a small, positive equilibrium that is stable, while the ( zero ) equilibrium is unstable (Figure 2 4). This can be seen analytically since d u 2 du at u = 0 is 0, 1+u 2 while d du γ(u)u at u = 0 is r ( 1 2 ) k > 0. Figure 2: At r = 0.05 the only two equilibrium are u = 0 and a small positive equilibrium (not visible here). For slightly raised values of r (around ), the growth rate become enough to create two additional positive equilibria, both greater than the one already present, in a saddle node bifurcation (Figure 3). The largest positive equilibrium is stable, while the second largest is unstable. The stability for the two previous equilibria remain unchanged. Figure 3: A saddle node bifurcation has occured, resulting in two new positive equilibria 4

5 Figure 4: A zoomed in view at r = 0.39 showing the two smaller positive equilibria after the saddle node bifurcation. The largest equilibrium is around u = 25. A second saddle node bifurcation occurs between the smallest and second smallest positive equilibria, leaving only one positive equilibrium. Both it and the zero equilibrium retain their stability (Figures 5 and 6). Figure 5: A second saddle node bifurcation has occured, leaving only one positive equlibrium. 5

6 Figure 6: The equilibrium near u = k (k is the carrying capacity) is now the only positive equilibrium present. (b) Repeat part (a) in the limit of population independent growth (the limit k ). Answer: In the limit k, the carrying capacity becomes infinite, and we recover a constant growth rate γ(u) = r that no longer depends on the population. Initially at r = 0.05, the growth rate is enough to overcome the decay rate. Three equilibrium are already present. The smaller positive equilibrium is stable and appears after r = 0 through a transcritical bifurcation (just like in part (a)) with the zero equilibrium, which becomes unstable. The larger positive equilibrium also appears for r > 0 at large u values. It is unstable, with the population growing to + for initial populations greater than it, or dropping to the small positive equilibrium if the initial population starts below the larger equilibrium. Figure 7: At r = 0.05 there are two equilibrium, one at u = 0, and one small positive equilibrium. 6

7 As r increases, the two positive equilibrium approach each other (Figure 8). Figure 8: The two positive equilibria steadily approach each other with increasing r. A saddle node bifurcation occurs between the two positive equilibrium, leaving only the zero equilibrium, which is unstable (Figures 9 and 10). The population grows without bound past this critical value of the growth rate (around r = 0.5). Figure 9: A saddle node bifurcation has occured, leaving only the zero equilibrium. 7

8 Figure 10: Unbounded population growth is present. (c) Potential Answer: The results from part (a) and (b) highlight the differences based on the growth term γ(u). In both cases, the saturating self-competition term u2 1+u 2 influences the dynamics more if the growth rate is fairly small, adding in small positive equilibria, but its effects are washed out for large r. In these cases the dynamics qualitatively resemble those of the equation u = γ(u)u. In part (a) this was with γ(u) = r ( 1 u ) k, leading to capped growth based on a carrying capacity k. In part (b), since γ(u) = r, there was unbounded growth due to an infinite carrying capacity. In terms of a population living within a certain environment, if that environment cannot support a large population, there will be a hard cap to the population size. The population will also be more sensitive at low levels, indicated in the model by a much larger level for the unstable smaller positive equilibrium for k = 30 compared to k =. If events occur within the environment to drop the population below this threshold, the population will die off. A finite carrying capacity raises this threshold significantly. 8