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Lecture Notes on Differential Equations
Peter Thompson Carnegie Mellon University This version: January 2003.
A differential equation is an equation of the form
x t ( )^ dx t ( ) f x y t ( , , ) = (^) dt =
usually with an associated boundary condition, such as
x (0)= x 0.
The solution to the differential equation,
x t ( ) = g y t x ( , , 0 ),
contains no differential in x. The techniques for solving such equations can a fill a year's course. In this part of the course, we study some basic types, with special emphasis on
- economic interpretation
- types of equations common in the study of economic growth
- characterization of arbitrary equations
- stability of systems
E XAMPLE 0.1 ( Capital accumulation by a firm ). The capital stock of a firm evolves ac- cording to the linear equation of motion
k t ( )^ = i t ( ) − δk t ( ),
gross investment
net depreciation investment
where i ( t ) is the rate of investment at time t and δ is the instantaneous rate of deprecia-
tion. To find the capital stock at any time t given an initial stock k (0)= k 0 requires that we solve the differential equation. •
E XAMPLE 0.2 ( Capital accumulation by a country ). Let GDP per capita be given by the intensive-form production function
y t ( ) = f k t ( ( )),
and let investment satisfy the national income identity for a closed economy,
i t ( ) = sf k t ( ( )).
The equation for motion of capital is then
k t ( )^ = sf k t ( ( )) − δk t ( ),
a nonlinear differential equation. •
E XAMPLE 0.3 ( Labor market matching ). Let L denote the number of workers in the labor force, u ( t ) the unemployment rate, and v ( t ) the vacancy rate (expressed as a fraction of L ). Workers and vacancies are assumed to find each other by a random matching process whereby the total number of matches made in an interval of time ∆ t is given by the matching function
M ∆ t = M uL vL ( , ) ∆ t.
The matching function is increasing and concave in each of its arguments, and homoge-
neous of degree one. Let θ= v / u. The rate at which vacancies are filled, expressed as a
fraction of the unemployed is M (^) t M uL vL ( , ) t uL^ ∆^ =^ uL ∆ vL M ( ,1) t uL = ⋅^ θ ∆
= θM (^) ( θ −^1 ,1)∆ t = θm ( ) θ ∆ t ,
If we know a past value of x ( t ), say x 0 , and the past values of a , a ( s ), s ∈ [0, ] t , we can use them to find the current value of x ( t ):
( ) 0 bt^ t ( ) b t (^^ s ) o
x t = x e + (^) ∫ a s e − ds.
We can verify this is the solution by differentiating with respect to t :
0 (^ ) 0
bt t b t s x t = bx e + a t + (^) ∫ a s be − ds
( ) 0
bt t b t s b x (^) ot e a s e − ds a t
∫ = bx t ( ) + a t ( ).
There is an intuitive interpretation to the backward solution. The current value of x can be decomposed into the sum of the contribution of the initial value, which is x 0 com- pounded at the rate b , and all the individual increments a ( s ), each of which is also com- pounded at the rate b for the interval t − s.
E XERCISE 1.1 Verify that (1.4) is a solution to (1.1).
E XAMPLE 1.1 ( Capital accumulation by the firm ). Recall from Example 0.1 the equation of motion for the capital stock of a firm:
k t ( )^ = i t ( ) − δk t ( ).
The backward solution is
( ) 0
t t t s k t = k e − δ^^ + (^) ∫ i s e − δ^ − ds ,
sum of components of x ( t ), all adjusted for exponential growth
Current value
These two terms come from applying Leibnitz' Rule of differ- entiation.
which states that the current capital stock equals the initial capital stock plus the entire time path of investment, both adjusted for depreciation. The initial capital stock has de-
preciated at the rate δ for the interval of time t , while each addition to the capital stock
from investment at time s has depreciated for the amount of time t − s. •
E XAMPLE 1.2 ( Labor market matching ). From Example 0.3, we have
u t ( ) = λ (1 − u t ( )) − θm ( ) ( ) θu t = λ − ( λ + θm ( ) θ ) u t ( ), u (0)= u 0.
If m ( θ) were a constant, this would be a straightforward linear differential equation. But
θ= v ( t )/ u ( t ). However, when we study matching models later in the course, we will find
that θ is solved by a firm optimization problem, and the solution is a constant that does
not depend on u ( t )! Hence, anticipating this result, let θ m ( θ) be treated as a constant
here. Then, the backward solution is
0 (^ ( ))^ (^ ( ) ()^ ) 0
m t t m t s
u t = u e^ −^^ λ^ + θ^^ θ^ + ∫ λe −^ λ^ + θ^^ θ − ds
0 (^ ( ))^ (^ ( )) ( ) 0
m t^ t u e m^ t^ e m
λ θ θ λ^ λ^ θ^ θ λ θ θ
− +^ −^ + = + (^) +
0 (^ ( ))^ (^1 (^ ( )))
u e m^ t^ e m^ t m
λ θ θ λ λ θ θ λ θ θ
= −^ +^ + −−^ +
which is a weighted average of the initial unemployment rate and the steady-state unem- ployment rate derived in Example 0.3. Note that u t ( ) → λ / (^) ( λ + θm ( ) θ )= u as t → ∞.
Forward Solutions
If we know a future value of x ( t ), say x ( T ), for some T > t , and the future values of a , a ( s ), s ∈ [ , t T ], we can use them to find the current value x ( t ):
( ) ( ) b T (^^ t^ )^^ T ( ) b s (^^ t ) t
x t = x T e −^ −^ − ∫ a s e −^ − ds.
By differentiating, you can verify that this is a solution.
As these assets could be used to increase consumption, lim (^) t →∞ a t ( ) > 0 cannot be opti- mal. Thus, an economic solution, as opposed to simply a mathematical solution to the budget constraint problem, allows us to impose a priori the condition lim (^) t →∞ a t ( ) =0. But this condition implies
lim ( ) r T (^^ t ) 0 T a T e
− − →∞^ =^.^ (1.2) Substituting (1.1) into (1.2), we have
( ) r s (^^ t^ )^ ( ) ( ) r s (^^ t ) t t
c s e ds a t y s e ds
∞ (^) − − ∞ − − ∫ =^ +∫ ,
which has the nice interpretation that the discounted present value of the family's life- time consumption is equal to the sum of its initial wealth and the discounted present value of its lifetime labor income. This is the solution to the family's budget constraint when it is behaving optimally ( lim (^) t →∞ a t ( ) ≤ 0 ) and feasibly ( lim (^) t →∞ a t ( ) ≥ 0 ). •
E XERCISE 1.2. Solve the following differential equations (a) x t ( ) = e − t −2 ( ) x t , x (0) =3 / 4. (b) x t ( ) = te −^2 t −2 ( ) x t , x (1) = 0. (c) x t e ( )^ t = t + (1 − x t ( )) et , x ^ (0) = 0.
If x ( t ) is a continuous function of time (i.e. it does not have any jumps), the back- ward and forward solutions are merely alternative ways of representing the same solution (although one may allow us to define constraints more readily, as in Example 1.2). The choice between them depends only on the information you have – the future or past val- ues of a ( s ) and a future or initial value for x. However, if x ( t ) can make a discrete jump at any given point in time, these expressions will not be equal, and we must use the logic of the model to which the equations apply to decide between the forward and backward solutions. In economic and financial problems, the variable of interest frequently is able to jump at a point in time, and so the distinction between forward and backward solu-
tions is often an important matter. We will illustrate this idea with a classic model about hyperinflation.
E XAMPLE 1.4 ( Hyperinflation ). Cagan (1956) developed a model of hyperinflation de- scribed by the following equations:
m t ( ) − p t ( ) = − απ ( ) t , (1.3)
π ( ) t = γ (^) ( p t ( ) − π ( ) t ). (1.4) Equation (1.3) states the demand for real money balances depends negatively on the
expected inflation rate; α is the semi-elasticity of real money demand with respect to in-
flation expectations. Equation (1.4) was the first statement of the theory of adaptive ex- pectations. The change in expected inflation is proportional to the current mistake made
in expectations. The parameter γ is the speed of adjustment of expectations.
Assume m ( t ) is constant except possibly for a one-time jump. Then, differentiating (1.3) gives p t ^ ( ) = απ ^ ( ) t. Using this in (1.4) and substituting into (1.3), we get
p t^ ^ ( ) = (^1) −^ γαγ ( m t ( ) − p t ( )). (1.5)
Suppose now there is a once and for all rise in m , beginning from a position where m ( t )= p ( t ) and p t ^ ( ) = 0. The quantity theory of money leads one to expect that a posi-
tive jump in m would induce p t ( )^ > 0. But this will only be true in (1.5) if αγ<1. This is
Cagan’s famous condition for monetary stability. The stability condition states that a sensitive money demand is consistent with monetary stability only if expectations adapt sufficiently slowly. Adaptive expectations are backward looking, so one way to think about p ( t ) is that it can be obtained from the backward solution to the differential equation.
p t^ ( ) = − (^1) −^ γ αγ^ p t ( ) + 1 − γαγm
ln(price level)
expected inflation rate
m held constant
ln(money demand)
and if m t ( ) = m for all t , we have p t ( ) = m. Rational expectations requires that the price today jump to accommodate whatever people think the path of m will be. Such jumps are not possible in a backward solution, because the backward solution assumes that, once p (0) is given, its subsequent path is always differentiable (and you can't be differentiable if you’re not continuous!). The lesson here is that different economic as- sumptions (i.e. adaptive versus rational expectations) impose very different boundary conditions on the differential equation, and this can have important consequences for the way a model behaves. •
E XERCISE 1.3 (An asset market model). Let p ( t ) denote the price of equity, let d ( t ) denote the dividend paid at time t, and let r denote the yield on a risk free bond. (a) What equation of motion yields the following forward solution for the price of equity : ( ) lim ( ) r T (^^ t^ )^ ( ) r s (^^ t ) p t (^) T p T e (^) td s e ds − − ∞ − − = (^) →∞ + (^) ∫?
(b) Explain the economic intuition behind this equation of motion. (c) What as- sumption about the forward solution implies that the price of equity is equal to the present value of current and future dividends? (d) What are the economic justifications for this assumption?
General Linear Differential Equations
So far we have discussed differential equations of the form
x t ^ ( ) = a t ( ) + bx t ( ).
We would also like to be able to solve more complicated equations of the form
x t ^ ( ) + p t x t ( ) ( ) = g t ( ).
The following exercise will lead you to discover the technique for solving this variable coefficient problem.
E XERCISE 1.4 (Solving general linear equations). Consider equations of the form , x t ( )^ + p t x t ( ) ( ) = g t ( ). (a) If g ( t ) is identically zero, show that the solution is
0
( ) exp ( )
t x t A p s ds
∫^ , where A is a constant determined by boundary conditions. (b) If g ( t ) is not iden- tically zero, assume a solution of the form
0
( ) ( ) exp ( )
t x t A t p s ds
∫^ , where A is now a function of t. Show that A ( t ) must satisfy the condition
0
( ) ( ) exp ( )
t A t g t p s ds
∫
(c) Find A ( t ) from this expression and then use your answer to write an expres- sion for x ( t ). (d) Using the formulae just obtained, solve x t ( )^ x t^ ( ) 3 + (^) t =.
2. Nonlinear Differential Equations
A single linear differential equation can always be solved (we may not be able to evaluate the integral, but then the solution stated with the integral expression is the solu- tion). However, most economic problems are nonlinear:
x t ^ ( ) = a x t ( , ) − f x t ( , ). Special types of nonlinear equations have known explicit solutions. However, most economic problems that exploit these special cases tend to be quite contrived. For this reason, and also to save time, we will look at only a couple of examples, a little later in this section.
Note also that any initial value of k other than k* is always associated with a movement toward k*. That is, the steady state is also globally stable. •
E XAMPLE 2.2 ( Multiple equilibria in the Solow model ).
For most applications, of course, a purely graphical analysis is insufficient. We can usu- ally obtain a useful analytical approximation to the differential equations by linearizing around the steady-state value. From a first-order Taylor expansion , we have:
k t^ ( ) ≈ sf k ( *^ ) − δk *^ + sf^ /^ ( k *^ ) − δ ( k t ( ) − k *)
= sf^ /^ ( k *^ ) − δ ( k t ( ) − k *)
Now let sf /^ ( k )− δ = b. Then, the linear approximation to the differential equation is k t ( )^ = − bk + bk t ( ). (2.2) This is a constant coefficient linear differential equation, which we can solve in our sleep. The obvious question, however, is how useful this equation will be. That depends upon the researcher’s purpose. The nonlinear equation (2.1) and its linear approximation (2,2) give very similar answers about how k behaves for values of k close to _k_ , but this similarity weakens the fur- ther we move away from _k._ Figure 2.2 illustrates by plotting k t ( )^ as a function of k ( t ) for the nonlinear model and its linear approximation.
F IGURE 2.2. Exact and linear approxima- tions to k t ( )^.
=0 in the steady state.
Implies k is increasing when k ( t )< k* and A negative constant: the slope of the decreasing when^ k ( t )> k* curve in Figure 1, sf /( k* ), must be less than the slope of the straight line, δ, at the steady state.
(^0) k * k
k^ Nonlinear equation (negative slopecomes from f // (^) <0. The exact shape depends on f///^ about which no assumptions were made).
The linear approximation has the same slope as the exact function at k* , but may induce large errors a long way from k *
Nonlinear Equations with Exact Solutions
One has to be lucky, but sometimes an economic model gives rise to a nonlinear differen- tial equation that has an exact solution. We provide two examples here. In some special cases, a nonlinear equation can be transformed into a linear differential equation that can always be solved a substitution of variables:
E XAMPLE 2.3 ( A closed-form solution to the Solow Model ).^1 Consider the following ver- sion of the Solow model:
Yt = Kt α ( A Lt t )^1 − α , α ∈(0,1) K ^ t = sYt − δKt , K (^) 0 > 0 , s ∈(0,1) Lt = L e 0 nt , L 0 (^) > 0 At = A e 0 gt , A 0 (^) > 0
where n , g , and δ are non-negative parameters. This model present two difficulties. The
first is that A (^) t and L (^) t are growing over time. Standard practice is to normalize variables by dividing by effective labor, A (^) tL (^) t , and then denoting the normalized variables with lower case letters: yt = Yt / A Lt t and kt = Kt / A Lt t. With this normalization,
y t = kt α , k t^ = skt α − ( n + g + δ ) kt ,
with initial condition k (^) 0 = K (^) 0 / A L 0 0. The second difficulty is that the equation of motion for normalized capital just de- rived is nonlinear. Previously, we solved for the steady state in the general case yt = f k ( (^) t ). But we can do better with the particular functional form yt = kt α. In this case, the equation of motion is known as a Bernoulli equation and can be solved explicitly by a change of variables. Let zt = kt^1 − α. Then,
z (^) t^ = (1 − α ) kt − αk t ,
which can be used to obtain a differential equation in z :
(^1) This section is taken from Jones (2000).
z (^) t^ = (1 − α ) s − λzt ,
where λ = (1 − α )( n + g + δ ). This is a simple linear differential equation, which is readily solved. Moreover, the equation has a nice interpretation. As zt = kt^1 −^ α^ = kt / kt α = kt / yt , the equation represents the evolution of the capital-output ratio. The backward solution is
z (^) t = z e 0^ −^ λ^ t^ + (^) n + sg + δ ( 1 − e − λt ),
yielding
( ) 1 1 t 0 t^^1 t k k e s e n g
α λ λ^ α δ
1 = ^ −^ −^ + − − ^ − (^) + +
and
( ) (1 )/^1 t 0 t^^1 t y y e s e n g
α α α λ λ α δ
= ^ −^ −^ + − − ^ −
Income per effective worker is a weighted average of the initial and steady-state values.
The parameter λ governs the rate at which the economy converges onto its steady state,
and this will be of direct interest in a later application. •
A second class of nonlinear models are separable equations that take the form:
( ) ( ) ( ) x t g t = f x
That is, they can be written in terms of two distinct functions, one involving the endoge- nous variable, the other involving time. These equations are interesting because they can often be solved by writing one side of the equation in terms of x and the other in terms of t. Thus writing the equation in the form
f x ( )^ dxdt = g t ( ),
we can integrate both sides with respect to t
∫ f x^ ( )^^^ dxdt dt^ =∫ g t dt ( ) ,
ing use of this regularity write down a simple model of R&D-driven growth in a firm’s market share, s, that incorporates the following features: (i) market share is bounded between 0 and 1; (ii) if a firm does no R&D it will lose market share due to the R&D efforts of other firms; (iii) there are n firms; (iv) all firms have the same R&D ability. Solve (if possible) and characterize the solution of the model. What is (are) the steady state(s)?
3. Systems of Differential Equations
In many problems relevant to the study of economic dynamics, we will be concerned with solving two or more differential equations simultaneously. Generally, these systems will also be nonlinear, and we will want to characterize their behavior using graphical techniques. However, just as for single nonlinear equations, local approximations around the steady state can be obtained by linearizing the system. It will therefore be easier to understand what is going on if we first consider the behavior of linear systems. We will limit ourselves to systems of two equations, which allow for graphical analy- sis. Systems of three equations can often be reduced to a system of two equations using a technique known as the time-elimination method (Mulligan and Sala-i-Martin [1992]). Systems of more than three equations are, fortunately, almost unheard of in our areas of study, and would usually need to be tackled with numerical methods.
Linear Systems of Two Equations
Consider the following system: x t ^ ( ) = 0.06 ( ) x t − y t ( ) +1.4, (3.1) y t ( )^ = −0.004 ( ) x t +0.04, (3.2)
with boundary conditions x (0)=1 and lim (^) t →∞( e −0.06 tx t ( )) = 0. To analyze this system, we construct a phase diagram , a graphical tool which allows us to visualize the dynamics of the system. The first step in constructing the phase dia- gram is to use the two equations to construct two curves plotting out stationary values
for x ( t ) and y ( t ). These curves plot all possible pairs of x and y such that x ^ = 0 and all possible pairs of x and y such that y ^ = 0. Setting x t ^ ( ) = 0 in (3.1), yields 0 = 0.06 ( ) x t − y t ( ) + 1.4, which gives y t ( ) = 0.06 ( ) x t + 1.4 (3.3)
Setting y t ( )^ = 0 in (3.2) gives x t ( ) = 10. (3.4)
We can then plot these curves:
F IGURE 3.1. Constructing the phase diagram, step 1.
Only one value of x is consistent with an unchanging y.
0^ x
x= 0
Intersection gives the steady state of the system y= 0
10
y
All pairs of { x , y } yielding a constant value of x.
