Solving the System Back

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(1) Cobb-Douglas Solution

The explicit solution to the fundamental Solowian differential equation, dk/dt = s¦ (k) - nk, is particularly easy if we assume a specific form for the production function. Let us assume that F(K, L) is Cobb-Douglas, so that:

Y = Ka L(1-a )

where a is the elasticity parameter (0 £ a £ 1). In intensive form, the Cobb-Douglas production function can be written as:

y = ka

This permits us to rewrite our fundamental Solowian differential equation (without depreciation) as:

dk/dt = ska - nk

which is a non-linear differential equation (see earlier Figure 3). To resolve this, we can linearize it by defining a new term z = k1-a , which upon differentiation with respect to t yields:

dz/dt = (1-a)(dk/dt)/ka

Now, dividing the original Solowian equation by ka, we obtain:

(dk/dt)/ka = s - nk/ka

so, plugging this into our dz/dt term:

dz/dt = (1-a )[s - nk/ka]

Now, as k/ka = k1-a = z by definition, then this equation can be expressed as:

dz/dt = (1-a)s - (1-a )nz

which is a simple linear first order differential equation in z. As we know from simple mathematics, the solution z(t) is:

z(t) = Ce-(1-a)nt + z*

where z* is the equilibrium value of z and C is a constant, both of which have to be deciphered. The equilibrium z* is defined where dz/dt = 0, i.e.

(1-a)s - (1-a )nz* = 0

so:

z* = s/n.

Thus:

z(t) = Ce-(1-a )nt + s/n

To translate this to obtain the solution k(t), we just re-transform it back, i.e. as z = k1-a , then k = z1/(1-a ). So:

k(t) = {Ce-(1-a )nt + s/n}1/(1-a )

To decipher C, we need to assume some initial value of k, call it k(0) = k0. So, when t = 0:

k(0) = {C + s/n}1/(1-a ) = k0

so, rearranging:

C = k01-a - s/n.

thus, written out in full, the solution to the Solowian differential equation is:

k(t) = {[k01-a - s/n]e-(1-a )nt + s/n}1/(1-a )

Now, as n > 0 and 0 < a < 1 by assumption, then it is evident that this equation is stable, i.e. [k01-a - s/n]e-(1-a)nt ® 0 as as t ® ¥ , so that in the end, k(t) ® k*, where:

k* = (s/n)1/(1-a )

is the steady-state capita-labor ratio.

(2) General Solution

For more general functional forms, the simpler mathematical method of checking the stability of the Solowian differential equation is via Lyapunov's method. For this, let us introduce a Lyapunov function:

V(t) = (k(t) - k*)2

where k* is capital-labor ratio. Notice that if k(t) = k*, when we are in steady-state, then V(t) = 0, but outside of steady-state, k(t) ¹ k*, so V(t) > 0. The system is considered "stable" if dV(t)/dt < 0, i.e. as time progresses, the difference between k(t) and k* is reduced.

Now, let z = k(t) - k*, so that the function becomes V(t) = z2. So differentiating V(t) with respect to time:

dV(t)/dt = 2zｷ(dz/dt)

As dz/dt = dk(t)/dt, then by the fundamental Solow-Swan differential equation, dk/dt = s¦(k) - nk, then:

dV(t)/dt = 2zｷ[s¦ (k(t)) - nk(t)]

or as k(t) = z + k*, then:

dV(t)/dt = 2zｷ[s¦ (z + k*) - n(z +k*)]

Now, the concavity of the production function implies that [¦ (z+k*) - ¦ (k*)]/z £ ¦ ¢ (z), or ¦ (z+k*) £ ¦ (k*) + z¦¢(k*), so:

dV(t)/dt £ 2zｷ[s¦ (k*) + sz¦ ¢ (k*) - n(z +k*)]

Since s¦ (k*) = nk* (by definition of steady-state equilibrium), then this reduces to:

dV(t)/dt £ 2zｷ[sz¦ ¢ (k*) - nz]

Now, as s¦ (k*) = nk*, then s = nk*/¦ (k*), so:

dV(t)/dt £ 2zｷ[znk*¦ ¢ (k*)/¦ (k*) - nz]

factoring out nz/¦ (k*):

dV(t)/dt £ [2nz2/¦ (k*)]ｷ[k*¦ ¢ (k*) - ¦ (k*)]

or factoring out -1:

dV(t)/dt £ -[2nz2/¦ (k*)]ｷ[¦ (k*) - k*¦ ¢ (k*)]

Notice that as we have assumed constant returns to scale, then the term [¦ (k*) - k*¦ ¢ (k*)] is the marginal product of labor, which is positive, while the term [2nz2/¦ (k*)] is unambiguously non-negative, thus:

dV(t)/dt < 0

and thus we know that the steady-state capital-labor ratio k* is globally stable. In other words, beginning from any capital-labor ratio (other than 0), we will converge to the steady-state ratio k*.

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