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Unformatted text preview: Solutions for the second midterm exam 1. (a) The rate at which people who are not immune get infected is proportional to x (the number of such people) and y (the number of sick people). That is Eq. (1). The number of sick people changes because health people get infected (this is the origin of the term rxy in Eq. (2)) and because sick people recover (this is the origin of the term sy ). People become immune by contracting the disease and recovering, that is Eq. (3). The model is known as the KermackMcKendrick model. (Anderson Gray McKendrick (18761943), an epidemiologist in Schotland, was the first to use mathematics in the study of epidemics. I dont know who Kermack was.) (b) At a fixed point, x = 0 or y = 0 by Eq. (1). But if x = 0 at a fixed point, then y has to be 0 anyway, by Eq. (2). So the fixed points are the point on the xaxis (all of them). (c) The dx / dt = 0 nullcline consists of the x and yaxes. The dy / dt = 0nullcline consists of the xaxis, and the line x = r / s . (d) d dt parenleftBig y + x s r ln x parenrightBig = y + x s r x x = rxy sy rxy s r ( ry ) = . (e) Notice that as long as x > 0 and y > 0, dx / dt < 0. So x is a strictly decreasing function of t , and therefore y , a function of t , can also be viewed as a function of x . Then dy dx = dy / dt dx / dt by the chain rule. So dy dx = rxy sy rxy = s rx 1 and therefore y = s r ln x x + C . This means precisely that y + x s r ln x is a constant. 1 (f) I use Matlab: x =  r x y y = r x y  s y r = 1 s = 1 0.5 1 1.5 2 2.5 3 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x y You can easily see analytically that this is what the functions y = ( s / r ) ln x x + C look like. The maximum is at x = s / r in general. In my plot, s / r = 1. (g) There is an epidemic if x > s / r , that is, if the population of healthy individuals that are not immune is sufficiently large. An epidemic occurs more easily if r is larger (that is, if the disease is more infectious) and if s is smaller (that is, if people take longer to recover)....
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 Spring '09
 Dr.Borgers

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