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 )

(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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