University of California, Los Angeles LIFESCIENC 30A. LS 30A-1 Homework 9 Solutions. 48 Solutions. Total points: 98

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Homework 9 Solutions Total points: 98 Section 3.2 Further Exercises FE 3.2.5 (4 pts, 2 pts each) FE 3.2.7 (2 pts) It is not possible for a one-dimensional system to have two stable equilibria... without an unstable one between them. This is because in order for an equilibria point to be stable the change equation must go from negative to positive when passing through the equilibrium point. In order for the next point to be stable the change equation would also have to go from positive to negative while passing through the equilibrium point. However, since after the previous equilibria the change equation was negative, it can only become positive again by crossing the X-axis which would indicate an unstable equilibrium point. Simply put, if you had two stable equilibrium points next to each other, the change vectors between them would not know which way to point. FE 3.2.8 (2 pts) 3 Section 3.3 Further Exercises FE 3.3.2 (10 pts, 2 pts each) a) Suppose R0 = 0; this means that J = 0. Now suppose, in addition, that J0 = 0; this means that -R + (0:1)(0) = 0, so that R = 0. Thus, (0; 0) is the only equilibrium point of this model. 4 The equilibrium point at (0; 0) is an unstable spiral. FE 3.3.3 (6pts,2 pts each) a) For values of a less than 0, the equilibrium point at (0,0) is a stable node. Once a becomes equal to or greater than 0, the equilibrium point becomes a saddle point. The saddle point has stability along the J-dimension and instability along the R-dimension. 5 This figure has a negative b parameter and a positive c parameter. If the signs of the b and c were switched, the vector field would rotate in the other direction. The equilibrium point at the origin would be considered a center. This figure shows a hyperbolic-like vector field. The equilibrium point at the center is actually a saddle point. However, instead of being stable along one axis and unstable along the other axis, this vectorfield is stable along the line R = -J and unstable along the R = J. 7 Section 3.4 Exercises 3.4.1 (2 pts) Solving the system of two equations above, we find that the equilibria are (S∗; T∗) = (0; 0) and (S∗; T∗) = (5; 20): 3.4.2 (2 pts) M = 0 and M = -0:5D + 2: 3.4.3 (2 pts) Nullclines are the lines in which one change equation is equal to zero for all values along that line. When they intersect both change equations are equal to 0, which is the condition for an equilibrium point. By definition, they cannot occur anywhere else. 3.4.4 (2 pts) M = 0 and M = -D + 2 3.4.5 (2 pts) (M∗; D∗) = (0; 0) and (M∗; D∗) = (1; 1) 8 3.4.6 (2 pts) 9 3.4.7 (2 pts) The biological significance of the saddle point is that one species will out-compete the other. Depending on where you start in the model, you will end up having only deer or only moose after an extended period of time. 3.4.8 (2 pts) The equilibrium points are (N∗; P ∗) = (0; 0) and (N∗; P ∗) = (5; 20): The point at (0,0) is a saddle point, and the point at (5,20) is a stable spiral. 3.4.9 (2 pts) The M-nullclines exist along the lines M = 0 and M = (rM - kMD)=cM. Section 3.4 Further Exercises FE 3.4.1 (4 pts, 2 pts each) a) There cannot be an equilibria with a nonzero point and a zeropoint in the Lotka-Volterra model because in both of change equations contain a single state multiplied by non zero parameter. In other words, if P = 0 and N 6= 0, then N0 cannot equal zero because it still contains an rN term which cannot be zero. 11 b) There are equilibrium points for the Lotka-Volterra model at (N∗; P ∗) = (0; 0) and (N∗; P ∗) = (δ=ca; r=a): The equilibrium point at (0; 0) is a stable node and the equilibrium point at (1; 9:998) is a stable spiral. FE 3.4.4 (8 pts, 2 pts each) 13 Based on the vector field we can see that the equilibrium point at (0; 0) is unstable and that the equilibrium point at (-4; 4) is a saddle point. FE 3.4.5 (10 pts, 2 pts each) 14 the yellow line flow towards (15; 0), the ones above it towards (0; 12). Section 1.4 Exercises 1.4.21 (2 pts) In the three compartment HIV simulation you can see that there is an initial level of uninfected cells in the body. However, once exposure occurs there is a huge spike in the number of viruses in the body which coincides with a huge spike in the number of infected cells. This leads to an immediate drop off in the number of uninfected cells. The number of infected cells and viruses then drop off shortly after that due to cells dying off. Then as a result of the drop off in viruses and infected cells, the number uninfected cells begin to grow again. After the uninfected cells increase for a time, the virus then spikes again which leads to an increases in the number of infected cells and this is followed by another cell die off. This process continues repeatedly, but with the the spikes becoming smaller and the number of uninfected cells approaches a stable value. 16 1.4.22 (2 pts) This model does not accurately represent the long term development of AIDS as the model reaches nearly it’s equilibrium value as early as several weeks. The model is characterized by a very sharp drop off in the number of uninfected cells and while it still oscillates after that the number of infected cells does not reach even 5% of it’s original value. Section 1.4 Further Exercises FE 1.4.13 (10 pts, 2 pts each) 17 slows the infection of new cells, the healthy T-cell count will stay higher throughout the duration of the simulation. However, the increased activation rate the cell drop off will be similar, but the recovery to normal T-cell levels will occur faster due to wiping out most of the infected (active and latent) cells faster. d) Manipulating α does in fact determine the time of cell recovery. Large values of α lead to quicker recoveries infection rate, infected cells die off before they can infect new ones and with a higher activation rate the latent cells aren’t able to build up and are eliminated quickly. 18 [Show More]

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