In applications, we are usually interested in stable equilibrium solutions that are not destroyed by small perturbations.
To determine whether an equilibrium is stable, we need to know the behaviour of a solution x(t), y(t) that is initially very close to the equilibrium. Therefore, just as for first-order and second-first-order differential equations, we can use a linearised stability analysis to investigate whether an equilibrium solution is asymptotically stable. We define a time-dependent deviation from the equilibrium
u(t) = x(t) − x0, ˙u = ˙x, v(t) = y(t) − y0, ˙v = ˙y,
and we expand f (x, y) and g(x, y) in Taylor series as functions of two variables:
f (x, y) = f (x0, y0) + ∂f we approximate the original nonlinear system by a linear system with constant coefficients:
˙u = au + bv,
We have already derived the asymptotic stability criterion for equilibrium solutions of linear systems:
a + d < 0, ad − bc > 0.
Now we can determine whether the solution of a nonlinear system is asymptotically stable by applying this criterion to a linearised system.
We can only use this approach to investigate asymptotic stability. As the following example shows, the linearised system cannot be used to establish whether an equilibrium solution is neutrally stable.
Example. The solution x0 = 0, y0 = 0 is the only equilibrium of the system
˙x = −y + x(x2+ y2),
˙y = x + y(x2+ y2).
The linearised system
˙u = −v, ˙v = u is equivalent to
¨
u + u = 0, v + v = 0.¨
Because its solutions are oscillatory and hence remain small for all t, we might draw the wrong conclusion that x0 = 0, y0 = 0 is a neutrally stable equilibrium of the nonlinear system as well. However, if we consider the function s(t) = x2+ y2, we see that
1 2
ds
dt = x ˙x + y ˙y = x[−y + x(x2+ y2)] + y[x + y(x2+ y2)] = (x2+ y2)(x2+ y2) or
˙s = 2s2.
On integrating this separable first-order equation, we find s(t) = s(0)
1 − 2s(0)t.
It is clear that s = x2 + y2 → ∞ as t → [2s(0)]−1, which can only happen if either x or y or both become infinite as t → [2s(0)]−1. Therefore, x0 = 0, y0 = 0 is an unstable equilibrium.
Example. Previously we used the logistic equation ˙N = a(1 − N/K)N, where a and K are positive constants, to describe the limiting effect of finite resources on the size of a population. Now consider two species that are competing for the same limited food supply. Introducing a parameter α > 0 that measures the degree of competition between the two species for the resources, we can write for the two populations N1 and N2
N˙1 = a1
1 − N1
K1 − αN2
N1,
N˙2 = a2
1 − N2
K2 − αN1
N2.
For simplicity, we assume K1 = K2 = ∞. The resulting nonlinear system N˙1 = f (N1, N2) = a1(1 − αN2) N1,
N˙2 = g(N1, N2) = a2(1 − αN1) N2
has two equilibrium solutions: N10= N20= 0 and N10= N20= α−1.
To investigate the stability of the equilibria, we introduce the perturbations u(t) = N1(t) − N10, v(t) = N2(t) − N20 and linearise the system about an equilibrium. For the solution N10 = N20= 0, the linearised system is
˙u = a1u,
˙v = a2v, and we conclude that the equilibrium is unstable.
For the nonzero equilibrium solution N10 = N20 = α−1, we calculate the partial derivatives
where the subscript ‘eq’ indicates equilibrium values. The linearised system is
˙u = 0u − a1v,
˙v = −a2u + 0v.
To apply our stability criterion, we calculate a + d = 0 + 0 = 0 and ad − bc = 0 × 0 − (−a1) × (−a2) = −a1a2. We see that neither a + d < 0 nor ad − bc > 0 is satisfied.
Hence the model predicts that the equilibrium is unstable: two similar species in the same habitat cannot coexist. The same result is obtained for finite K1 and K2. Therefore we should expect that the struggle for existence between two similar species nearly always leads to the complete extinction of one of them. This result is known as the principle of competitive exclusion in population biology.
Example. Suppose x(t) is the number of males infected with gonorrhoea, y(t) is the number of infected females, a1and a2are the rates with which the infectives are cured, c1 = const is the number of promiscuous males, and c2 = const is the number of promiscuous females. We assume that new male infectives are added at a rate b1(c1 − x)y that is proportional to the number of the male susceptibles c1 − x and to the number of the female infectives y. Similarly, we assume that new female infectives are added at a rate b2(c2 − y)x that is proportional to the number of the female susceptibles c2 − y and to
the number of the male infectives x. Then a model for the spread of gonorrhoea due to heterosexual contacts is
˙x = −a1x + b1(c1− x)y,
˙y = −a2y + b2(c2− y)x.
For practical reasons, we are interested in the equilibrium solution x0 = 0, y0 = 0. Is it stable? The linearised system is
˙x = −a1x + b1c1y,
˙y = b2c2x − a2y.
We observe that a + d = −(a1+ a2) < 0, and thus the first part of our stability criterion is always satisfied. Now, ad − bc = (−a1)(−a2) − (b1c1)(b2c2), and the equilibrium is stable if ad − bc > 0 or
b1b2c1c2 < a1a2. This is when the disease disappears from the population.
Otherwise, if b1b2c1c2 > a1a2, it can be checked by a tedious calculation that x0 = b1b2c1c2− a1a2
a1b2+ b1b2c2
, y0 = b1b2c1c2− a1a2 a2b1+ b1b2c1
is a stable equilibrium. In this case the total number of infective males and females ultimately levels off. It should not surprise us that public health data indicate that b1b2c1c2 > a1a2.
Example. Consider now two species, one of which feeds on the other. Suppose x(t) is the prey population, say small fish, and y(t) is the predator population, say sharks. If the species did not interact, we could write ˙x = rx, where r > 0 implies abundant food supply for the prey population, say algae for the small fish. By contrast, the predators would die off without food, and so ˙y = −ky. In reality the two species do interact—the small fish are eaten by the sharks—and the number of contacts between them is proportional to both x and y. Hence we arrive at the following system describing predator–prey interactions:
˙x = f (x, y) = rx − αxy,
˙y = g(x, y) = −ky + βxy.
These equations are called the Lotka–Volterra equations (A. J. Lotka, 1925; V. Volterra, 1926).
We find equilibrium solutions by solving
f (x0, y0) = x0(r − αy0) = 0, g(x0, y0) = (βx0− k)y0 = 0.
The system has two equilibrium solution. The solution x0 = 0, y0 = 0 is unstable since the corresponding linearised system is
˙x = rx, ˙y = −ky,
and so a small nonzero x will deviate strongly from the equilibrium. For the nonzero equilibrium solution x0 = k/β, y0 = r/α, we calculate the partial derivatives
a = ∂f and we use them to obtain the linearised system
˙u = 0u + (−αk/β)v,
˙v = (βr/α)u + 0v.
To apply our stability criterion, we calculate a + d = 0 + 0 = 0 and ad − bc = 0 × 0 − (−αk/β)(βr/α) = rk > 0. We see that a+d < 0 is not satisfied, and thus the equilibrium is not asymptotically stable. More detailed analysis, however, shows that the equilibrium is neutrally stable. On differentiating the linearised equations, we obtain
¨
u + kru = 0, v + krv = 0.¨
Thus the key prediction of the Lotka–Volterra model is that the predator and prey popu-lations should oscillate with period T = 2π/√
kr. In the real world, of course, interaction between species are a lot more complicated than in this one-predator-one-prey model.
Example. The Italian biologist Umberto D’Ancona noticed that the total catch in the Mediterranean had a larger fraction of sharks during World War I when the level of fishing was greatly reduced. To explain this phenomenon, the mathematician Vito Volterra incorporated the effect of fishing into the Lotka–Volterra model. If h is the fraction of sharks and small fish being caught, we can write
˙x = rx − αxy − hx,
˙y = −ky + βxy − hy.
It follows that the equilibrium solution is x0 = k + h
β , y0 = r − h α .
Therefore, as long as h < r, harvesting increases the equilibrium prey population and decreases the equilibrium predator population. (There is no biologically meaningful equi-librium in the case h > r that corresponds to overfishing.) This is why reduced fishing during the war (smaller h) had increased the shark population (larger y0) and decreased the average fish population (smaller x0). More recently, an unanticipated effect of DDT spraying was to destroy not the harmful insects but rather their natural predators.
Exercises
1. Suppose R(t) measures Romeo’s love for Juliet and J(t) measures Juliet’s love for Romeo. Romeo’s love grows in response to Juliet’s love for him and vice versa. At the same time, both lovers try to rein in their feelings for each other. The resulting system of equations is as follows:
R = −R + J,˙ J = −J + R.˙
Find the general solution of the system. Solve the initial value problem R(0) = 1, J(0) = 0.
2. We wish to model the outbreak of an infectious disease. The number of individuals susceptible to infection at time t is S(t), and the number of infected individuals is I(t). We postulate that the population increases at a constant rate in the absence of an epidemic.
We also postulate that the infection rate is proportional to the product of S and I and that the death rate is proportional to I. Consider the resulting system of coupled differential equations for S(t) and I(t):
S = −rSI + µ,˙ I = rSI − γI.˙
Here r, µ and γ are positive constants. Derive an equilibrium solution of the system, S = S0 and I = I0, and determine its stability.
3. We wish to model the size of an insect population. The population at time t is P (t), and the total food supply is F (t). We postulate that the birth rate is proportional to the product of the population size and the food supply and that the death rate is proportional to the population size. We also postulate that F (t) is described by a linear first-order differential equation in the absence of the insects and that the food consumption rate is proportional to the population size. Consider the resulting system of coupled differential equations for P (t) and F (t) in appropriate units:
P = F P − kP,˙
F = c − F − P.˙
Here k and c are positive constants, and we assume c > k. Derive an equilibrium solution of the system, P = P0 > 0 and F = F0, and determine its stability.
4. We wish to model the size of a fish population in a lake. The fish population at time t is F (t), and the number of fishermen is M(t). We postulate that fish grow logistically in the absence of fishing, and that the presence of fishermen decreases the growth rate by an amount proportional to the product of the fish and fishermen numbers. We also postulate that fishermen are attracted to the lake at a rate proportional to the amount of fish in the lake, and that fishermen are discouraged from the lake at a rate proportional to the number of fishermen already there. Consider the resulting system of coupled differential equations for F (t) and M(t) in appropriate units:
F = aF (1 − F ) − F M,˙ M = bF − M.˙
Here a and b are positive constants. Calculate the equilibrium solutions of the system and determine their stability.
Chapter 5
Difference equations
5.1 Geometric growth
In many applied problems, we are interested in variables that are only known at certain time intervals. For example, economic quantities (income, savings, gross domestic prod-uct, unemployment) can be reported weekly, monthly or annually. Learning results are evaluated by tests and exams. Population sizes of annual plants vary from year to year.
These problems lead to mathematical models that involve difference equations.
The difference equation for geometric growth is xk+1 = axk, a = const,
where xk is some variable at time step k. Given an initial value x0, we can determine x1 = ax0, x2 = ax1 = a2x0, and in general
xk = x0ak.
Example. If a sum of money accumulates at compound interest, S0 is the initial deposit, and Sk is the sum on deposit after k compounding intervals, then Sk+1 = Sk + iSk = (1 + i)Sk, where i is the interest rate per interval. Hence Sk= S0(1 + i)k.
Example. In a model for economic growth, we consider annual national income Yk for year k. The total income is divided into consumer expenditures Ckand private investments Ik,
Yk= Ck+ Ik.
We assume that consumption is proportional to the available income, Ck = mYk, 0 < m < 1,
and also that income growth is proportional to the invested amount, Yk+1− Yk = rIk, r > 0.
We have
Yk+1 = Yk+ rIk
= Yk+ r(Yk− Ck)
= Yk+ r(Yk− mYk)
= (1 + r − rm)Yk or
Yk= Y0(1 + r − rm)k.
The model neglects many important factors, such as unemployment, so its prediction of a robust economic growth is not very realistic.