Example 1. Applications of eigenvalues/eigenvectors Predator-Prey models. Foxes and Rabbits share a large forest, with the foxes eating rabbits and rabbits eating the abundant vegetation. The sizes of both populations evolve over time according to the following rules: F k+1 = 0.5F k + 0.3R k R k+1 = pf k + 1.R k where F j = the size of the fox population in time period j and R j = the size of the rabbit population in time period j. The parameter p is assumed to be positive and is called the predation parameter. Natural questions to ask are: Will the fox and rabbit populations survive? How does the survival of these populations depend, if at all, on the parameter p and on the initial population sizes, F 0 and R 0? If they do survive, how big do they become over time? 1
To make the study of these questions easier, we express the evolution of the system using matrix multiplication: F k+1 0.5 0.3 = F k. R k+1 p 1. R k The matrix on the right is called the transition matrix, and writing x k = F k 0.5 0.3 and T p =, R k p 1. It follows that x 1 = T p x 0, x = T p x 1 = T p x 0, x 3 = T p x = T 3 p x 0, and so forth. Our questions are ultimately concerned with how depends on p and x 0. x k = T k p x 0
Case study 1. Suppose that p = 0., F 0 = 10 and R 0 = 40, so that x 0 = 10 0.5 0.3 and T p =. 40 0. 1. To get a sense of what s happening to the populations over time, we can do some quick calculations, e.g., 0.5 0.3 x 1 = T p x 0 = 10 = 17, 0. 1. 40 46 and x = T p x 0 = x 3 = T 3 p x 0.3 51.8 6.69 57.7 5 7 58,. 3
It appears that both populations are growing in the short term, but this is no guarantee of future growth. It also appears that the foxes are catching up with the rabbits. I.e., that the ratio F k /R k is getting bigger (from 1/4 to more than 1/3). But to see what will happen in the long run, we need to have a way to estimate x k accurately when k is large. This can be done by considering the eigenvalues of T p their associated eigenvalues. and Eigenvalues: The characteristic equation for T p is λ 0.5 0.3 det(λi T p ) = 0 0. λ 1. = 0 λ 1.7λ + 0.66 = 0 so the eigenvalues of T p are λ 1 = 1.7 +.89.64 = 1.1 and λ = 1.7.89.64 = 0.6. 4
Eigenvectors: To find eigenvectors associated with λ 1 = 1.1 we solve the system (T p (1.1)I) x = 0: 0.6 0.3 0. 0.1 x = 0 y 0 = 0.6x + 0.3y = 0 = y = x, in other words, the eigenvectors associated with λ 1 = 1.1 are all scalar multiples of v 1 = 1. Observe that T k p v 1 = (1.1) k v 1. 5
Likewise, to find eigenvectors associated with λ = 0.6 we solve the system (T p (0.6)I) x = 0: 0.1 0.3 x = 0 0. 0.6 y 0 = 0.1x + 0.3y = 0 = x = 3y. This means that the eigenvectors associated with λ = 0.6 are all scalar multiples of v = 3 1. Observe that T k p v = (0.6) k v. 6
Note that v 1 and v form a basis of R, so we can express the initial state vector v 0 (uniquely) as a linear combination of v 1 and v : x 0 = c 1 v 1 + c v = 1 3 c 1 = 10 1 c 40 = c 1 c = 1 3 1 1 10 40 = 4 In other words x 0 = v 1 4 v 10 40 = 1 4 3 1 7
We are now in position to get very useful (and accurate) estimates for x k (when k is large). Specifically... x k = T k p x 0 = T k p( v 1 4 v ) = (T k p v 1) 4(T k p v ) = (1.1) k v 1 4(0.6) k v = (1.1) k (0.6) k 1 44 4 (1.1) k 44, since (0.6) k 0 as k grows large. In fact, this estimate is already quite accurate for small k, e.g., when k = 5: 0.5 0.3 0. 1. 5 44 = 34.4981 70.5514 and (1.1) 5 44 = 35.431 70.864. 8
Conclusions (for Case Study 1): 1. With p = 0. and F 0 = 10 and R 0 = 40, both populations thrive. In fact, in the long run, both populations grow exponentially (by 10% per time period).. In the long run (and even in the relatively short run), the ratio of foxes to rabbits stabilizes at about 1/. 9
Case Study : Staying with p = 0., so that 0.5 0.3 T p =, 0. 1. we ll use the eigenvalues and eigenvectors that we have already found to see the impact of the initial population sizes on the long term survival of the species. Suppose that the initial populations are simply F 0 and R 0, and x 0 = F 0 R 0 As in the first case study, to estimate T k p x 0, we have to express x 0 as a linear combination of v 1 and v :. x 0 = c 1 v 1 + c v. 10
And, as before, we conclude that x k = T k p x 0 = T k p(c 1 v 1 + c v ) = c 1 (1.1) k v 1 + c (0.6) k v (1.1) k (c 1 v 1 ) In terms of the individual populations, we have (1.1) k This means that F k R k c 1 c 1 the two populations survive in the long run if and only if c 1 > 0. To see how c 1 depends on x 0 = F 0 R 0., we can use Cramer s rule... 11
F 0 R 0 = c 1 1 + c 3 1 = 1 3 1 c 1 c = F 0 R 0 = c 1 = F 0 3 R 0 1 1 3 1 = F 0 3R 0 5 = 3R 0 F 0 5. For c 1 to be greater than 0, it must be the case that 3R 0 > F 0. In other words, for the two populations to survive, the initial population of rabbits must be more than a third the size of the initial fox population, which itself must be positive. 1
Case study 3: Now we consider how the survival of the populations depends on the predation parameter p. When we change p, the eigenvalues (and eigenvectors) of T p change. Specifically, if 0.5 0.3 T p = p 1., then the characteristic equation of T p is λ 1.7λ + (0.6 + 0.3p) = 0, with solutions λ 1 (p) = 1.7 + 0.49 1.p and λ 1 (p) = 1.7 0.49 1.p. There are several cases to consider. 13
1. If p > 0.49/1. 0.40833, then λ 1 (p) and λ (p) are complex conjugates. In this case, T p has the effect of rotating vectors in R, and after a finite number of rotations, one of the two populations would cross into negative territory. In practice, this means that the foxes are eating so many rabbits, that the rabbit population dies off quickly, and after that, the fox population dies off.. If p = 0.49/1., then λ 1 (p) = λ (p) = 0.85. In this case, T p does not have two linearly independent eigenvectors, so the analysis we used in the previous case studies does not apply directly. It is possible to show however that both populations die off in this case as well. 3. If 0.49/1. > p > 1/3, then 0 < 0.49 1.p < 0.09, so 0 < 0.49 1.p = α p < 0.3. In this case, 0 < λ (p) = 1.7 α p < 1.7 + α p and once again, both populations die off. = λ 1 (p) < 1, 14
4. If p = 1/3, then 0.49 1.p = 0.09 = 0.3, so 1.7 + 0.3 1.7 0.3 λ 1 (p) = = 1 and λ (p) = The vectors, v 1 = 3 and v = 3 5 = 0.7. are eigenvectors associated with λ 1 and λ, respectively. If x 0 = c 1 = F 0 R 0 F 0 3 R 0 3 3 5, then x 0 = c 1 v 1 + c v, where = 3R 0 F 0 9 and c = 3 F 0 5 R 0 3 3 5 = 5F 0 3R 0 9 15
In this case, x k = T k p x 0 = T k p(c 1 v 1 + c v ) = c 1 v 1 + c (0.7) k v c 1 v 1. It follows that if 3R 0 > F 0 (so that c 1 > 0) both populations will stabilize at the constant levels: F k 3R 0 F 0 3 and R k 15R 0 10F 0 9 If 3R 0 F 0, then both populations die off rapidly. 5. If 0 < p < 1/3, then 0.7 > α p = 0.49 1.p > 0.3 so λ 1 (p) = 1.7 + α p In this case, if > 1 and 0.5 < λ (p) = 1.7 α p R 0 > ( 7 6 5α ) p F 0 3 then both populations survive (and grow exponentially). both populations die off rapidly.. < 0.7. If not, 16