11.1 Models of Optimal Economic Growth
11.1.4 Solution by the Maximum Principle
H=u(c) +λ[f(k)−c−γk]. (11.12) The adjoint equation is
λ˙ =rλ−∂H
∂k = (r+γ)λ−f(k)λ= (ρ+δ)λ−f(k)λ. (11.13) To obtain the optimal control, we differentiate (11.12) with respect to c, set it to zero, and solve
u(c) =λ. (11.14)
Let c=h(λ) =u−1(λ) denote the solution of (11.14). In Exercise 11.3, you are asked to show that h(λ) < 0. This can be easily shown by
340 11. Applications to Economics inverting the graph of u(c) vs. c.Alternatively you can rewrite (11.14) asu(h(λ)) =λand then take its derivative with respect to λ.
To show that the maximum principle is sufficient for optimality, it is enough to show that the derived Hamiltonian
H0(k, λ) =u(h(λ)) +λ[f(k)−h(λ)−γk] (11.15) is concave in k for any λ satisfying (11.14). The concavity follows im- mediately from the facts thatλis positive from (11.10) and (11.14) and f(k) is concave because of the assumptions on F(K, L).
Equations (11.9), (11.13), and (11.14) now constitute a complete au- tonomous system, since time does not enter explicitly in these equations.
Such systems can be analyzed by the phase diagram method, which is used next.
In Fig.11.1we have drawn a phase diagram for the two equations k˙ = f(k)−h(λ)−γk= 0, (11.16) λ˙ = (r+γ)λ−f(k)λ= 0, (11.17) obtained from (11.9), (11.13), and (11.14). In Exercise11.2you are asked to show that the graphs of ˙k = 0 and ˙λ = 0 are like the dotted curves in Fig.11.1. Given the nature of these graphs, known as isoclines, it is clear that they have a unique point of intersection denoted as (¯k,λ).¯ In
Figure 11.1: Phase diagram for the optimal growth model
11.1. Models of Optimal Economic Growth 341 other words, (¯k,λ) is the unique solution of the equations¯
f(¯k)−h(¯λ)−γ¯k= 0 and(r+γ)−f(¯k) = 0. (11.18) The two isoclines divide the plane into four regions, I, II, III, and IV, as marked in Fig.11.1. To the left of the vertical line ˙λ = 0, we have k <k¯and thereforer+γ < f(k) in view off(k)<0.Thus, ˙λ <0 from (11.13). Therefore, λis decreasing, which is indicated by the downward pointing arrows in Regions I and IV. On the other hand, to the right of the vertical line, in Regions II and III, the arrows are pointed upward because λ is increasing. In Exercise 11.3, you are asked to show that the horizontal arrows, which indicate the direction of change ink,point to the right above the ˙k= 0 isocline, i.e., in Regions I and II, and they point to the left in Regions III and IV which are below the ˙k= 0 isocline.
The point (¯k,λ) represents the optimal long-run stationary equilib-¯ rium. The values of ¯k and ¯λ are obtained in Exercise 11.2. The next important thing is to show that there is a unique path starting from any initial capital stock k0, which satisfies the maximum principle and converges to the steady state (¯k,¯λ).Clearly such a path cannot start in Regions II and IV, because the directions of the arrows in these areas point away from (¯k,¯λ). For k0 < ¯k, the value of λ0 (if any) must be selected so that (k0, λ0) is in Region I. For k0 >k,¯ on the other hand, the point (k0, λ0) must be chosen to be in Region III. We analyze the casek0 <¯konly, and show that there exists a uniqueλ0 associated with the givenk0,and that the optimal path, shown as the solid curve in Re- gion I of Fig.11.1, starts from (k0, λ0) and converges to (¯k,¯λ).It should be obvious that this path also represents the locus of such (k0, λ0) for k0∈[0,¯k].The analysis of the casek0>¯kis left as Exercise 11.4.
In Region I, ˙k(t) >0 and k(t) is an increasing function oft as indi- cated by the horizontal right-directed arrow in Fig.11.1. Therefore, we can replace the independent variable t by k, and then use (11.16) and (11.17) to obtain
λ(k) = dλ dk = dλ
dt 3dk
dt = [f(k)−(r+γ)]λ
h(λ) +γk−f(k). (11.19) Thus, our task of showing that there exists an optimal path starting from any initial k0 < ¯k is equivalent to showing that there exists a solution of the differential equation (11.19) on the interval [0,¯k],beginning with the boundary condition λ(¯k) = ¯λ. For this, we must obtain the value λ(¯k).Since both the numerator and the denominator in (11.19) vanish
342 11. Applications to Economics atk= ¯k,we need to deriveλ(¯k) by a perturbation argument. To do so, we use (11.19) and (11.18) to obtain
λ(k) = [r+γ−f(k)]λ
f(k)−γk−h(λ) = [f(¯k)−f(k)]λ
f(k)−f(¯k)−γk+γ¯k−h(λ) +h(¯λ). We use L’Hˆopital’s rule to take the limit as k→k¯ and obtain
λ(¯k) = −f(¯k)¯λ
f(¯k)−γ−h(¯λ) = −f(¯k)¯λ
f(¯k)−γ−λ(¯k)/u(h(¯λ)), (11.20) or
− (λ(¯k))2
u(h(¯λ))+λ(¯k)[f(¯k)−γ] + ¯λf(¯k) = 0. (11.21) Note that the second equality in (11.20) uses the relation h(¯λ) = 1/u(h(¯λ)) obtained by differentiating u(c) = u(h(λ)) = λ of (11.14) with respect to λatλ= ¯λ.
It is easy to see that (11.21) has one positive solution and one negative solution. We take the negative solution forλ(¯k) because of the following consideration. With the negative solution, we can prove that the differ- ential equation (11.19) has a smooth solution, such that λ(k) <0.For this, let
π(k) =f(k)−kγ−h(λ(k)).
Sincek <k,¯ we haver+γ−f(k)<0.Then from (11.19), sinceλ(¯k)<0, we have λ(¯k−ε) > λ(¯k). Also since ¯λ > 0 and f(¯k) < 0, Eq. (11.20) with λ(¯k) implies
π(¯k) =f(¯k)−γ− λ(¯k) u(h(¯λ)) <0, and thus,
π(¯k−ε) =f(¯k−ε)−γ(¯k−ε)−h(λ(¯k−ε))>0.
Therefore, the derivative at ¯k−εis well defined and λ(¯k−ε)<0. We can proceed as long as
π(k) =f(k)−γ− λ(k)
u(h(λ(k))) <0. (11.22) This implies that f(k)−kγ −h(λ) > 0, and also since r +γ −f(k) remains negative for k <¯k,we have λ(k)<0.