A 2 B C D2

and when we apply the rule for determinants (i.e. that Eq. (2.41) is equivalent to (A — 2)(D — 2) — BC = 0) we obtain the same equation for 2, Eq. (2.39).

Equation (2.39) is a quadratic equation, so that we know there are two solutions, given by the quadratic formula. We will denote these solutions by 21,2 and they are

where, for convention, we will assume that 1 corresponds to + and 2 to - in the quadratic formula. If we define the discriminant by

D = (A — D)2 + 4BC, then we can write that 21 ,2 = t(A + D)± pD ] /2.

We are now able to classify the steady state (0, 0) of the system given in Eq. (2.37). Before doing that, let's have a brief interlude.

Exercise 2.11 (M)

Show that if l is a solution of Eq. (2.42) and that if we set u = B and v = l — A that Eq. (2.40) is satisfied. Thus, we know how to find the eigenvectors too.

As long as D = 0, which we will assume in this chapter, the exercises up to this point have allowed us to find the general solution of the system given by Eq. (2.37):

Although it is nice to have an explicit form for the solution, what is nicer is that we now know how to classify the steady state.

We begin with the case in which D > 0. Then both of the eigenvalues are real. We conclude that if they are both positive, the origin is an unstable node. Since solutions will grow exponentially, whichever eigenvalue is larger will ultimately dominate the behavior of the solution. If both of the eigenvalues are negative, we conclude that the origin is a stable node. If one of the eigenvalues is positive and the other is negative, we conclude that the origin is a saddle point.

When D < 0, the eigenvalues are complex numbers, so if we set q = vjDf we can rewrite the eigenvalues as l1j2 = [(A + D) ± iq]/2, where i = V—X. Consequently, when we compute solutions given by Eq. (2.43), we will need to consider expressions of the form

From this, we see that if A + D, the real part of the eigenvalues, is negative, then whatever else happens solutions will decline in time. If A + D is positive, they will grow in time. The question then becomes how we interpret the exponential of iqt.

For this interpretation, we need a brief reminder. Recall that the solution of the differential equation d2x/dt2 = —kx involves sines or cosines. (If you do not recall this, confirm that if x = sin^V^t) or x = cos(v/kt) then the differential equation is satisfied.) Since this is a linear equation, the general solution must be of the form cX sm(v/kt)+ c2 cos(vkt), where the are constants. Suppose that we had guessed an exponential solution for this equation, i.e. that x = ceAt. In this case, the second derivative of x(t) is cl2eAt so that we conclude l must satisfy the equation l =— k or that l = iiVk In other words, exponentials involving X lead to oscillations. Our x(t) =cXBe1l( + c2Bel2t y(t) = c1(21 — A)e1l( + c2(l2 — A)el2t

solution is at hand. If D < 0, we now know that the solutions will oscillate. Such a steady state is called a focus or a spiral point. If A + D < 0, the focus is stable and if A + D > 0 the focus is unstable.

But, of course, all this work (and it is hard work) only corresponds to the linear system of equations (2.37) and the equations that we actually encounter in population biology are nonlinear. What do we do about this? That is, in general we will have a pair of differential equations of the form

0 0

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