Autonomous equationsAn autonomous first-order ordinary differential equation is a first-order ordinary differential equation in which the value of x'(t) depends only on x(t), not independently on the value of t.
A phase diagram indicates the sign of x'(t) for a representative collection of values of x. To construct such a diagram, plot the function F, which gives the value of x'. For values of x at which the graph of F is above the x-axis we have x'(t) > 0, so that x is increasing; for values of x at which the graph is below the x-axis we have x'(t) < 0, so that x is decreasing. A value of x for which F(x) = 0 is an equilibrium state.
If x* is an equilibrium and F'(x*) < 0, then if x is slightly less than x* it increases, whereas if x is slightly greater than x* it decreases. If, on the other hand, F'(x*) > 0, then if x is slightly less than x* it decreases, moving further from the equilibrium, and if x is slightly greater than x* it increases, again moving away from the equilibrium. That is, we have the following result.
- For an autonomous first-order ordinary differential equation x'(t) = F(x(t)), x* is an equilibrium if and only if F(x*) = 0. An equilibrium x* is locally stable if F'(x*) < 0 and unstable if F'(x*) > 0.
We see that the equilibrium b is locally stable, whereas the equilibria a and c are unstable.
- Example (Solow's model of economic growth)
The following model generalizes the one in an earlier example. Suppose that the production function is a strictly increasing and strictly concave function F that is homogeneous of degree 1 (i.e. has “constant returns to scale”), rather than taking the specific form assumed in the earlier example.
We now haveK'(t) = sF(K(t), L(t)).As before, the labor force grows at the constant rate λ, so thatL'(t)/L(t) = λ.We may study the behavior of the capital-labor ratio K(t)/L(t) as follows. Let k = K/L and define the function f of a single variable byf(k) = F(k, 1) for all k.Since F is increasing and concave we have f' > 0 and f" < 0, and since it is homogeneous of degree 1 we haveF(K, L) = LF(K/L, 1) = Lf(k).ThusK'(t) = sL(t)f(k(t)).Now,
k'(t) = [K'(t)L(t) − K(t)L'(t)]/(L(t))2 = [K'(t) − k(t)L'(t)]/L(t)k'(t) = sf(k(t)) − λk(t).The phase diagram of this equation is shown in the following figure.
We see that the equation has two equilibria, k* = 0, which is unstable, and the value of k* for which sf(k*) = λk*, which is locally stable.
The stable equilibrium value of k depends on s: the equationsf(k*) = λk*implicitly defines k* as a function of s. Which value of s maximizes the equilibrium value of per capita consumption? Per capita consumption is equal to (1 − s)F(K, L)/L, or, given our previous calculations, (1 − s)f(k). Thus in an equilibrium per capita consumption is equal to (1 − s)f(k*(s)). A stationary point of this function satisfies(1 − s)f'(k*(s))k*'(s) = f(k*(s)).Now, differentiating the equation defining k* and rearranging the terms we obtaink*'(s) = f(k*(s))/[λ − sf'(k*(s))].Combining the last two equations we deduce that if s maximizes per capita consumption thenf'(k*(s)) = λ.That is, the marginal product of capital is equal to the rate of population growth.