The Growth Model In Continuous Time
Prof. Lutz Hendricks
Econ720
September 27, 2016
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Topics
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We study the standard growth model in continuous time.
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To solve it: Optimal Control
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To characterize it: phase diagrams
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Continuous Time vs. Discrete Time
[Some of you will find the next several slides obvious.]
Continuous time
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So far, time was divided into discrete "periods."
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It is often more convenient to shrink the length of periods to 0.
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Difference equations then become differential equations.
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Continuous time
Example: Law of motion for capital
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Discrete time:
Kt+1 − Kt = It − δ Kt
(1)
Kt+∆t − Kt = [It − δ Kt ] ∆t
(2)
More generally:
Continuous time (∆t → 0):
Kt+∆t − Kt
= K̇t = It − δ Kt
∆t→0
∆t
lim
(3)
Notation: K̇ = dK/dt.
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Growth rates in continuous time
The growth rate of a variable is defined as
g (x) =
ẋ d ln x
=
x
dt
(4)
Growth rate rules (easy to prove):
1. g (xy) = g (x) + g (y).
2. g (x/y) = g (x) − g (y).
3. g (xα ) = αg (x).
4. x (t) = eγt =⇒ g (x) = γ.
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Differential equations
Differential equations
Take a function of time:
x (t) = a + bt
(5)
There is another way of describing this function:
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Take the derivative:
ẋ (t) = dx (t) /dt = b
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Fix x (0) = a.
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The two pieces of information (the derivative and x (0))
completely describe x (t).
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Only one function x (t) satisfies both pieces.
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(6)
But note that infinitely many functions satisfy the derivative!
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Definition: Differential equation
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A differential equation (DE) is a function of the form
ẋ (t) = f (x (t) , t)
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This is actually a "first-order" DE.
Higher order DEs contain higher order derivatives of time.
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E.g.: A second order DE
d2 x (t) /dt2 + dx (t) /dt = a + bt
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(7)
(8)
Together with a boundary condition, the DE can be solved for
x (t).
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Solving DEs
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The bad news: There is no algorithm for solving DEs.
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But one look up solutions in tables.
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It is also easy to verify a solution one may guess.
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Guess + Verify
Consider again
ẋ (t) = b
(9)
x (0) = a
(10)
x (t) = a + bt
(11)
Guess
Verify:
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Take the time derivative and find that it matches ẋ = b.
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Verify that x (0) = a.
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Example: constant growth
ẋ (t) = b x (t)
(12)
x (0) = a
(13)
Guess:
x (t) = a ebt
(14)
ẋ (t) = b a ebt = b x (t)
(15)
Verify: Take the derivative
0
x (0) = a e = a
(16)
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Boundary conditions
Boundary conditions can take many forms:
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Rb
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x (1.7) = 5.
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x (T) − x (T − 2) = 5.
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etc.
a
x (s) ds = 5.
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The Solow Model
The Solow Model - Structure
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Modify the discrete time growth model in two ways:
1. Continuous time.
2. Fixed saving rate.
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This is not an equilibrium model, but can be interpreted as
one.
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Model Elements
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Demographics: households live forever;
Lt = ent
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Preferences:
Z ∞
e−(ρ−n)t u(ct )dt
(17)
(18)
t=0
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Endowments:
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at each moment, the household has 1 unit of work time
at t = 0 he has K0 goods
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Model Elements
Technology:
F (Kt , Lt ) = K̇t + δ Kt + Lt ct
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(19)
F: constant returns to scale
Markets:
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competitive markets for goods (numeraire), labor rental,
capital rental
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Firms
The firm solves a static problem.
The same as in discrete time.
max F (K, L) − wL − qK
(20)
qt = FK
(21)
wt = FL
(22)
FOC
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Firms: Intensive Form
Define kF = K/L and
f kF = F (K, L) /L = F kF , 1
(23)
The first order conditions are then
q = f 0 (kF )
(24)
w = f (kF ) − f 0 (kF )kF
(25)
and
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Households
Budget constraint
K̇t = wt Lt + (qt − δ )Kt − Lt ct
It is convenient to have everything per capita.
Define k = K/L.
Law of motion for k:
k̇/k = K̇/K − n
= w/k + (q − δ ) − c/k
Or
k̇t = wt + (qt − δ − n)kt − ct
(26)
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Constant saving rate
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The modern way: Set up an optimization problem and derive
the saving function.
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The Solow way: Assume that the saving rate is fixed:
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c = (1 − s) (w + qk)
(27)
k̇ = s(w + qk) − (n + δ )k
(28)
Therefore:
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Market Clearing
Capital rental:
k = kF
(29)
Goods market:
F(Kt , Lt ) = Ct + δ Kt + K̇t
or in per capita terms
k̇ = f (k) − (n + δ )k − c
(30)
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Equilibrium
An equilibrium is a collection of functions (of time)
ct , kt , ktF , wt , qt
that satisfy
1. the firm’s first order conditions (2)
2. the household’s budget constraint and the behavioral equation
k̇ = s(w + qk) − (n + δ )k
3. market clearing (2)
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Law of Motion
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The entire model boils down to to one key equation:
k̇t = sf (kt ) − (n + δ )kt
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(31)
This is simply the household’s behavioral equation after
applying f (k) = w + qk.
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Steady state
Steady
state
The steady
state requires k̇ = 0 or
The steady state requires k̇ = 0 or
(32)
sf (k) = (n + δ ) k
sf (k) = (n + d) k
(34)
With strictly concave f , there is a unique steady state with k > 0.
With strictly concave f , there is a unique steady state with k > 0.
(n + δ) kt
s f(kt)
k0
L. Hendricks ()
k
kSS
Continuous time
August 6, 2009
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Dynamics
With strictly concave f , there is a unique steady state with k >
(n + δ) kt
s f(kt)
k0
k
kSS
L. Hendricks ()
The steady
state is stable.
Continuous time
August 6
Convergence is monotone.
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Adding Technical Change
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The model does not have sustained growth in per capita
income.
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This requires technical change (A grows).
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Assume exogenous growth in A:
A(t) = A(0)eγt
(33)
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Adding Technical Change
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Assume that technical change takes the following form:
Y(t) = F(K(t), A(t)L(t))
(34)
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This type of technical change is called “labor-augmenting” or
“Harrod-neutral.”
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This is the only form of technical change that is consistent
with balanced growth.
Definition
A balanced growth path is a path along which all growth rates are
constant.
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How to analyze a growing model?
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Construct a stationary transformation.
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Divide each variable by its balanced growth factor:
x̃ (t) = x (t) e−gx t
(35)
where gx is the balanced growth rate of x.
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Or take ratios of variables that grow at the same rate.
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The economy in transformed variables (x̃) has a steady state.
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How to find the balanced growth rates?
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For equations that involve sums:
Y (t) = C (t) + I (t) + G (t)
(36)
Constant growth (usually) requires that all summands grow at
the same rate.
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For other equations: Try taking the growth rate of the whole
equation.
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Example:
Y (t) = K (t)α [A (t) L (t)]1−α
(37)
g (Y) = αg (K) + (1 − α) [g (A) + n]
(38)
implies
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Balanced growth path: Solow Model
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Start from
K̇ (t) = sF (K (t) , A (t) L (t)) − δ K (t)
A (t) L (t)
−δ
g (K (t)) = sF 1,
K (t)
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(39)
(40)
Constant growth requires that
k̄ (t) =
K (t)
A (t) L (t)
(41)
be constant over time. Thus, on a balanced growth path:
g (K) = γ + n
(42)
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Balanced growth path
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Production function:
ȳ (t) =
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Y (t)
= F k̄ (t) , 1
A (t) L (t)
(43)
must be constant on a balanced growth path.
Thus: The model has a steady state in k̄,ȳ .
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Law of motion
g k̄
Or
= g (K) − γ − n
A (t) L (t)
= sF 1,
−δ −γ −n
K (t)
= s f k̄ /k̄ − δ − γ − n
˙ = sf (k̄(t)) − (n + δ + γ)k̄(t)
k̄(t)
(44)
Nothing changes, except the constant term in the law of motion.
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Reading
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Acemoglu (2009), ch. 2 covers the Solow model and
stationary transformations of growing economies.
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Barro and Martin (1995), ch. 1
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Romer (2011), ch. 1
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Krusell (2014) ch. 2 discusses some insights that might be
gained from the Solow model (and its limitations).
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References I
Acemoglu, D. (2009): Introduction to modern economic growth,
MIT Press.
Barro, R. and S.-i. Martin (1995): “X., 1995. Economic growth,”
Boston, MA.
Krusell, P. (2014): “Real Macroeconomic Theory,” Unpublished.
Romer, D. (2011): Advanced macroeconomics, McGraw-Hill/Irwin.
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