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1. Some mathematical tools

1. Differential equations

Definition.

A differential equation is one that involves an unknown function and its

derivatives. Intuitively is an equation that relates the rate of change of some process tothe process itself and other processes that are changing over time. For instance

( ) ( )3 4 ' 0y t y t t+ + = (1)

y(t) is the dependent variable and tthe independent variable

The solution to (1) is not a point but rather a function ( )y t . In general this type of

equations, in which a function is the solution, are referred as functional equations1.

Typology.

- Ordinary differential equation, ODE. There is only one independent variable.(1) is an ODE where t is the only independent variable.

- Order, the highest derivative. (1) is of order 1.- Degree, the highest power attained by any derivative in (1). (1) is of the degree 1.- If a differential equation is of degree 1, ( )y t and its derivatives enter linearly and

there are no products between ( )y t and its derivatives then we have a linear

differential equation.- A differential equation is said to be autonomous, if the independent variable only

enters the equation through the dependent variable and not by itself. (1) is not an

autonomous differential equation.

Genesis of a Differential Equation. Where do they come from?

Widely used in natural sciences and engineering.

1 Notation: Given a function, ( , )y F x t= , I will denote its partial derivative with respect to time as

F x,t( )t

=

y

t= F

tx,t( ) = yt = F x,t( ) = y = F and its partial with respect to any variable x as

( )( )

,

,x x x

F x t yF x t y F

x x

= = = =

. Given this notation, the growth rate of a variable (through time) will be

denoted as F F

F. Finally I will be using * to refer to equilibrium/optimal values.

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Ex. 1 A population of bacteria growing asexually by self-division (binary fission).Suppose that at time t the population contains x individuals and that we observe the

population for a short interval of time, t . Let the instantaneous probability of

reproduction (division) be r . Then we can calculate the increase in population as

x = rxt xt

= rx letting_t0# # # ## dxdt

= rx x t( ) rx t( ) =0

which is a differential equation.

Ex. 2 Many radioactive materials, for instance Carbon-14, disintegrate at a rate

proportional to the amount of it present in other materials. If at time twe have a quantityof Carbon-14, C(t), then its rate of disintegration is given by

( ) ( )C t rC t =

which is another a differential equation.

Carbon is a component of many materials (wood, bone, hair, pollen), carbon is made of

3 isotopes (C12, C13 and C14) which appear in constant proportions in all livingorganisms. C14 atoms are always decaying but as long as the organism is alive they are

always being replaced by new C14 atoms. Once a living organism dies it stops absorbingnew C14 atoms. The radiocarbon dating method measures the C14 concentration of a

sample whose age is not known, and compares it with the concentration of a livingorganism. Then using the decay rate of C14 a date for the death of the sample could be

estimated, since the change in proportion of C14 atoms begins when our undated materialdied.

Ex. 3 A good approximation to evaluate the change in temperature of an object is known

as Newtons law of cooling. This law states that the change of temperature of an objectdepends on the difference between its temperature and the temperature of the surrounding

environment. We can model this as,

T(t) = T t( ) S t( )( ) 0 >

which is another differential equation.

In economics the use of differential equations arises from the explicit modeling ofan additional dimension: time. Most economic processes involve a sequence of actions,think of a firm choosing among different productive processes, where current decisions

affect future choices.

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Graph. A sequential decision process

A firm maximizes profits by hiring labor,*

L , up to the point that its marginal cost

(the wage) and its marginal product coincide, this is a static problem. But if the firm isallowed to modify its productive capacity the problem becomes more complicated

because at any point in time the firm needs to make a decision on how much labor to hire

and how much to invest ( )( )I t in capacity ( )( )c t . This investment will affect the futureproductive capacity of the firm and therefore the decision needs to take into account not

only current but also forecasted future market conditions. The nature of this problem isdynamic, and the evolution of the capacity of the firm could be modeled as

c t( ) = I t( )c t( )

The solution to this differential equation (given the path of investment and the initial

capacity) will define the whole path of capacity, ( )c t .

Solving Differential Equations

We tend to use calculus applying a sequential method. We identify our problemthen we apply a set of standard rules and lead us to the solution.

Problem Rules Solution

So if you have a system with two equations on two unknowns ( x and y ). You solve the

first equation for x and replace the resulting expression in the second equation. As a

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result you have a single equation on y . You find the value of y that satisfies that

equation and you use that value to recover x .

Most of the time when we work with differential equations we do not have such a

set of rules, in fact some of the most interesting differential equations are still analytically

intractable (and need to be solved numerically or characterized graphically). Withdifferential equations we are going to

Problem Guess Conditions to make the guess valid

In this context practice and good intuition become crucial. In this class we are going to

solve only simple differential equations where there is a standard set of rules that isknown to work.

The general form of a first order linear differential equation is

y t( )+u t( ) y t( ) =w t( ) (2)

we are looking for a functiony(t) that satisfies (2)

CASE I: Constant coefficients ( ) ( ),u t u w t w= =

a) Homogeneous case, ( ) 0w t =

(2) can be written as

y t( )+uy t( ) = 0 y t( )

y t( )= u

int_wrt_time ln y t( )( )+ c1 = ut+ c2

exponentiating ( ) ( ) ( ) ( )exp exp exp uty t ut c ut c Ae= + = =

So ( ) uty t Ae= (2.1) is the general solution, which includes an arbitraryconstant

Notice that (2.1) is a family of exponential curves, therefore we need an additional

condition to choose among those curves. This additional condition usually takes the form

of an initial condition, ( ) 00y y= . In our previous examples initial capacity, initialpopulation of bacteria, initial quantity of C-14, or initial temperature of the body.

Using the initial condition and setting 0t=

We get ( ) 0ut

y t y e

= (2.2) is the definite solution

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b) Non-homogeneous case

(2) becomes y t( )+uy t( ) =w (3)

The solution to this equation will consist of two additive terms; the complementary

solution (c

y ) and the particular solution (p

y ).

The complementary solution is the general solution of the homogeneous version of (3),

and therefore

( ) utcy t Ae

= (3.1)

The particular solution is any solution of the complete equation. Lets try first with the

most simple solution possible, a constant, i.e., ( )y t k=

If ( )y t k=

, then

y t( )= 0

replacing in (3)

( )pw w

uk w k y t u u

= = = (3.2)

Since ( ) ( ) ( ) utc pw

y t y t y t Aeu

= + = + (3.3) (general solution)

Setting 0t= and using the initial condition we have

( )0 0 0so utw w w wy A A y y t y e

u u u u

= + = = +

(3.4) (definite solution)

notice that we need to find first the general solution before pinning down the value of thearbitrary constant.

It was evident that this particular solution would not work if 0u = , in that case the

differential equation (3) will be of the form,

( )

( )

( )

y t w

y t dt wdt

y t wt c

=

=

= +

(general solution)

with the definite solution ( ) 0y t y wt= +

To verify the solution, we need to check that the solution satisfies the differential

equation and that the initial condition is met. In some economic applications instead ofan initial condition we might have a condition at any other point in time, the most

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relevant case is a terminal condition (to be met at the end of the planning problem). Thishas a lot to do with the fact that economic agents react to the future, while natural

systems react to the past. In a sense the position of a planet today depends on where itwas yesterday, but my decision to invest in a stock today is driven by what I expect the

price of that stock tomorrow.

Economic Interpretation of the components of the solution

d

s

Q a bP

Q cP

=

=

in equilibrium *d s

aQ a bP cP Q P

b c= = = =

+

(a, b, c >0)

If the initial price is the equilibrium price, then supply equals demand and themarket clears. But we see all the time that there are certain goods that remain unsold and

others that are not available in the shops and therefore it seems that prices are notadjusting instantaneously. Under this more realistic assumption it will be interesting to

explore the dynamics of this market and its stability, i.e. whether if the initial pricediffers from the (intertemporal) equilibrium one, the market achieves equilibrium.

For this purpose the first thing we need to do is to postulate a process for theadjustment of prices. It seems plausible to assume that the price responds to imbalancesin supply/demand according to a constant factor 0> ,

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )d sP t Q t Q t a bP t cP t a b c P t = = = +

we have a ODE, that we can rewrite as

P t( )+ b+ c( )u

P t( ) = w

a

applying our solution method we reach, given that ( ) 00P P= ,

P t( ) = P0 a

b+ c

w/u

!

"

###

$

%

&&&

e

b+c( )u

t

+

a

b+ c

w/u

= P0 P

*!"

#$e

t

complementaty

+ P*

particular

u b+ c( ) < 0

It turns out that the two components of the solution have a nice economicinterpretation. The particular solution,

py , is the equilibrium price, i.e. the solution to the

static problem, and in this dynamic context we refer to it as the intertemporal

equilibrium level also known as steady state, and the complementary solution,c

y , is a

measure of the deviation from equilibrium.Now we turn to the issue ofstability or whether the system from any initial price

level will converge to the intertemporal equilibrium. Three cases to study:

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* *

0 0,P P P P< = and

*

0P P> . Given 0< the deviation from equilibrium vanishes as

t , and therefore the equilibrium is stable in the dynamic sense. So (known as the

eigenvalue or characteristic root) is the crucial parameter for the stability of the system.

has another interesting interpretation, if we take the time derivative of our solution

P t( ) = P0 P*

e

t= P t( ) P*

P t( )P t( ) P*

=

so is the ratio between the change in prices in a given period and the distance between

the current price and the equilibrium price, in a sense is a measure of the speedconvergence of the system.

What is the meaning of 0.05= ? Each period 5% of the gap ( )( )*P t P is closed.That implies that the half-life of the deviation is

P t( ) P*( ) = P0 P*( )et =1

2P

0 P

*( ) ! ! t= ln2 t.69

.05 14periods.

Using our solution we can draw the time path for the price as follows

Qualitative solution: The phase diagram

In many circumstances analytical solutions are not available or difficult to obtain

An alternative way to infer some properties about the solution of a differential equation is

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to draw a graph. Using our previous example, the dynamics of the price level are givenby,

P t( ) = Qd t( ) Qs t( )( ) = abP t( ) cP t( )( ) = a b+ c( ) P t( )where b+ c( )< 0 .

Lets plot the change in prices, P t( ) , against the level of prices, ( )P t .

The equilibrium condition is satisfied when P= 0 , at that point the price level is

at its intertemporal equilibrium, *P . Since the price level decreases when*( )P t P> and

increases when *( )P t P< , the system is stable, as we already know from the analytical

solution. This graphical approach, known as phase diagram, will be the one that we willbe using with most of our models2.

Most of the time we are going to be evaluating our differential equationsgraphically, since analytical solutions are more difficult to obtain, and therefore we

should look at other examples.

Given y t( ) = ay t( ) b ; , 0a b > . The first step is to find the steady state of theprocess. Since by definition a steady state is a situation in which the process, ( )y t , is not

changing, we can find the steady state by imposing the condition y t( ) = 0

y t( ) = 0 ay t( )b= 0 y* =b

a

2 Notice that time is absent from the graphical analysis, since our differential equation is autonomous.

7

u) r\a

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In this case the steady state is unstable, if the process begins away fromb

a, it will

never go to that point.

y t( )

=

y t( )

y t( )( )

2

Imposing the steady state condition y t( ) = 0 y t( ) y t( )( )2

= 0 y1

*= 0, y

2

*=1 , in

this case we have two steady states, the process has two equilibriums.

The equilibrium located at the origin is unstable, the other one is stable.

Example: 495 as a stable steady state (fix point). Take any 3 digit number where thedigits are not the same, using those digits compose the largest (L) and smallest number

(S), subtract them, L-S, iterate.For instance, take 128, then L=821, S=128, L-S=693.Again L=963, S=369, L-S=594. Again L=954, S=459, L-S=495. Again L-S=495.

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2. Systems of differential equations

The systems of differential equations are generated through the interaction of

two or more processes, each of them changing through time.

Ex. 1 Imagine a psychologist trying to model a love-hate relationship. Janes love forGeorge increases with Georges love (and decreases with his disdain), the opposite is true

for George. Defining Janes love for George as J t( ) , and Georges love for Jane as G t( ) ,a way to model the process could be as follows

J t( ) =G t( )G t( ) = J t( )

, 0 >

Ex. 2. A Malthusian model of population growth. In Ex. 1 of the previous section, we

introduced a model of growth for a population of bacteria ( )( )x t which was exponential

x t( ) rx t( ) = 0 x t( ) = Aert = x0e

rt( ) . Malthus suggested that sooner or later, ashortage of resources must bring this increase to a halt. The fact that resources are limited

to a level ( )K leads to the logistic equation,

x t( ) = rx t( ) 1x t( )

K

(4)

So, we can use our graphical analysis to see what happens with the population of bacteria

under this new process. We can determine the steady state as,

x t( ) = 0 rx t( ) 1x t( )

K

= 0 x1

*= 0,x

2

*= K

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Imagine that we have two populations that behave according to some version of

(4) and that share the same environment (and therefore resources). Think of rabbits

( )( )x t and mice

( )( )y t living in the same area and (partially) competing for the same

resources. This implies that the rate of growth of one of the populations is inhibited by itsown level and by the level of the other population.

x t( ) = r1x t( ) 1 ay t( ) x t( )( )y t( ) = r2 y t( ) 1 y t( ) bx t( )( )0 < a,b

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Following with our second example, we define an equilibrium or steady-state

(or a fix-point) for (5) as a combination of values of ( )x t and ( )y t , such that thepopulation of x and y are stationary (i.e. do not change). Since in steady state the two

populations have to be constant, this implies that their rates of change have to be equal to

0, sox =

y = 0 . Now we can solve both equations simultaneously

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( )

*

*

10 1 1 1

1

10 1 1

1

aay t x t x t ay t x t a abx t x

ab

by t bx t y t bx t y

ab

= = = + =

= = =

Besides there are other 3 equilibriums ((0,0),(1,0),(0,1)), one with neither rabbitsnor mice, one with only rabbits, and a third with only mice (we abstract from these threetrivial equilibriums where there is no interaction between species).

b) Qualitative solutions and stability.

As we already know the phase diagram is a graphical representation of thedynamic behavior of a differential equation, but we can also use it to analyze systems of

two autonomous (do not depend on time explicitly) differential equations. Its constructioninvolves 2 steps

1. Construction ofequilibrium loci.2. Determination of the arrows of motion and streamlines.

We define the equilibrium locus for ( )x t , as the set of pairs ( ) ( )( ),x t y t such thatx t( ) = 0

is satisfied. To construct the equilibrium locus for ( )x t

Equilibrium/ steadystateforx t( )x t( )=0

1 ay t( ) x t( )( ) = 0 (5.1)

Since we want to graph it in an x y space, it is useful to rewrite (5.1) as3

( ) ( )( ) ( )( )1

1 0x t

ay t x t y t a

= =

3 When we cannot solve explicitly for y in terms of x , we can still apply the implicit function theorem to

sign the slope. Totally differentiating (5.1) we reach

ady dx = 0 dy

dx

x=0

= 1

a< 0

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So the line we have just plotted is the set of combinations of ( )x t and ( )y t such

that x t( ) = 0 , and we call it the x = 0 line . Notice that it divides the first quadrant

(positive values) in two areas. The next step is to analyze the behavior of ( )x t in thoseareas, i.e. to draw the arrows of motion (drawn in the direction of x ). There are several

(equivalent) ways to determine this behavior.

1. Looking at the graph, fix a ( )x t and move above/below the x = 0 line .2. Finding x t( )

y t( )= ar

1x t( ) < 0

3. x t( )x t( )

= r1 r

1ay t( )+ 2x t( )( ) . Since we cannot sign it, this approach is not useful

in this case.

We repeat the same analysis for y t( ) = 01 y t( )bx t( ) = 0 y t( ) =1bx t( ) , plotthe y = 0 lineand the arrows of motion.

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Combining both figures the point of intersection is our intertemporalequilibrium and the arrows of motion we have just drawn determine the evolution of the

system through time. In this system, like in most of our applications, chronological timedoes not matter, again as a consequence of the system being autonomous. Think about the

problems that might arise if chronological time matters.

Now we can turn to evaluate its stability. Given any initial value of the two

populations (except 0 for any or both of them) all the arrows of motion point towardssteady state and therefore the system eventually converges to the equilibrium level. Our

equilibrium with the two species co-existing is stable. The advantage of the graphical

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analysis is that the system does not need to be linear, and we still can infer someinteresting qualitative information about the variables of interest.

c) Analytical solution; linear approximation and stability

Consider a system of non-linear differential equations such as

x = f x, y( )y = g x, y( )

(6) with an equilibrium at*

*

x x

y y

=

=

In general, we dont know how to solve systems of non-linear differential equations andtherefore we will resort to approximation (linear) techniques, since the solutions to linear

systems of differential equations are well understood The solution to the linearcounterpart of (6) will provide a good approximation to the dynamics of (6), at least in

the neighborhood of the approximation point.

Linear approximation

We are going to find a linear approximation to (6) in the neighborhood of itssteady-state. Using a first order Taylor series expansion we obtain the following linear

approximation to the original system around its steady-state,

x f x*, y*( )+ fx x*, y*( ) x x*( )+ fy x*, y*( ) y y*( )y g x

*, y

*( )+ gx x*, y*( ) x x*( )+ gy x*, y*( ) y y*( )(7)

since by definition, in steady-state

x =y = 0 , then (7) simplifies to

x fx

x*, y*( ) x x*( )+ fy x*, y*( ) y y*( )y g

xx

*, y

*( ) x x*( )+ gy x*, y*( ) y y*( )

since all the starred terms are just numbers, we can rearrange and denoting

( )* * *,x xf x y f we have

x

fx*

x+

fy*

y

fx*

x

*

fy*

y

*=

fx*

x+

fy*

y

kx

y gx

*x+ gy

* y gx

*x* gy

*y* = gx

*x+ gy

* y ky

(8)

which is a linear system.Like with an ODE the solution will be made of the sum of twoterms, the complementary solution (deviation from equilibrium) and the particular

solution (intertemporal equilibrium level). But in this case we will have two particular

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solutions and two complementary ones. Besides with two dependent variables we will

have two time paths, i.e. one for ( )x t and one for ( )y t .Complementary solution

The technique applied to (linear) systems of differential equations parallels the

one used to solve ordinary differential equations. First we solve the homogeneouscounterpart of our linear system (8)

x fx

*x fy

*y x a11x a

12y = 0

y gx

*x gy

*y y a21x a

22y = 0

(8a)

Our solution method will propose a solution, we replace it in (8a) and find the conditionsunder which (8a) is satisfied. Given our experience with differential equations an

educated guess would be,

x= Aet then x= Aet

y = Betthen

y = Bet

we are looking for values of , ,A B that would make our proposed solution a solution

for (8a), therefore replacing our proposed solutions in (8a)

11 12

21 22

0a a A

a a B

=

(8b)

If we set 0A B= = , the system would be satisfied but we want to use those constants tosatisfy the initial conditions. Therefore we are looking for the values of which satisfy

(8b)

( )11 12 2 11 22 11 22 21 1221 22

0 0a a

a a a a a a

a a

= + + =

Charact. equation

We will find two roots1

and2

known as characteristic roots or eigenvalues, which

are going to be crucial for the dynamic stability of our system and the speed at which it

converges. Furthermore, the eigenvalues satisfy the following relations

1 2 11 22 21 12* a a a a = = determinant of the jacobian

and

1 2 11 22a a + = + = trace of the jacobian

since we have two eigenvalues, the complementary solution takes the following form

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1 2

1 2

1 2

1 2

t t

t t

x A e A e

y B e B e

= +

= +

Remember that we are solving a system of two differential equations on two unknowns,

( )x t

and ( )y t , as a result we will have only two initial conditions ( )0x

and ( )0

y . Butnow it seems that we have four indeterminate constants,

1 2 1 2, , ,A A B B and only two initial

conditions to pin them down, but it turns out, that given the properties of the

eigenvalues the constants are not independent. Using (8b) for the first eigenvalue,1

,

111 1 12

21 22 1 1

0Aa a

a a B

=

we can find that 1 11 211 1 1 1 1

12 1 22

a aB A A A

a a

= = =

Then combining (8b) with2

, we find a similar relation for the second eigenvalue,

2 11 21

2 2 2 2 2

12 2 22

a aB A A A

a a

= = =

And therefore two initial conditions are just enough to pin down the four constants.

Particular and general solution solution

Using our knowledge from solving differential equations, we know that the particular

solution is the intertemporal equilibrium level, and therefore we have the followinggeneral solution, which is just the sum of the complementary solution and the particular

solution.

1 2

1 2

*

1 2

*

1 1 2 2

t t

t t

x x A e A e

y y A e A e

= + +

= + +

(9 general solution)

with the constants being determined by initial/terminal conditions ( ) ( )0 00 , 0x x y y= = .

Stability

As before the complementary solutions represent deviations from equilibrium,

i.e the 'A s and 'B s are these deviations, and therefore the sign of the eigenvalues will be

crucial to determine the dynamic stability of the system. If the eigenvalue is positive any

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deviation from equilibrium will grow as time goes by, therefore positive eigenvalues areassociated with unstable behavior, the opposite being true for negative eigenvalues.

Assuming the eigenvalues are distinct (and real), we have 3 possibilities

a)

Both eigenvalues are negative, stable equilibrium. (tr0)b) Both eigenvalues are positive, unstable equilibrium. (tr>0, det>0)c) One is positive and the other negative, saddle point behavior. (tr>=< 0, det

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Ex. 3 (from the previous section). A household can either consume or save andaccumulate wealth, according to the following set of linear (or linearized) equations,

c= g r A( )( );gA < 0

A=

f c,A( ); fA>

0, fc , that captures the fact that present is

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preferred to future) of an instantaneous objective function with the general form

( ) ( )( ),V s t c t , i.e. she aims to

max

c

J = ( ) ( )( )0

,t

V s t c t e dt

(1)

s.t. s t( ) = f s t( ),c t( )( )s 0( ) = s0

(2)

and possibly some non-negativity constraints

where we have one state variable, ( )s t , and one control variable, ( )c t (both could bevectors). The solution to this problem amounts to find the time-path for the control,

( )optc t , such that the value ofJ is maximized and the constraint is satisfied.

One approach will be to propose a sequence ,

find the associated values of the state variable , replace them in

V(.) and calculate the value ofJassociated with this proposal, repeat this process untilyou checked all (the infinite) possible paths and choose the one with highest value

An alternative way to solve (1) s.t. (2) implies the application of what is known asthe maximum principle. To solve a problem using the maximum principle we define the

following function, the Hamiltonian (it could be defined in alternative ways, you adjustthe conditions to the way you define, same results)

H s t( ),c t( ), t( )( ) =V s t( ),c t( )( )et+ t( )et f s t( ),c t( )( ) s t( )"#$% (3)

where ( )t is called the co-state variable and measures the sensitivity of the objectivefunction (1) to changes in the constraint

4. It is usually interpreted as the shadow price of

4 The best way to understand the economic interpretation of is to look at a simple example of static

constrained optimization. Imagine that you can allocate a given amount of land ( )T between two uses,grow bananas or apples, which have the following given prices,

bP

and

aP

and that are produced according

to the following technologies ( )b bF T and ( )a aF T . Assuming you want to maximize revenue, your

objective could be stated as ( ) ( ) ( )max , max s.t.b a b b b a a a b aR T T P F T P F T T T T= + + = . Under standardconditions of concavity of the production functions the FOC are necessary and sufficient for a maximum:

( ) ( )* * ' * * ' * *; ;a b b b b a a aT T T P F T P F T + = = = . We can already see that the last two conditions look like theusual conditions for profit maximization with playing the role of the price of the input. Now imagine that

they give us an small amount of extra land dT , how should we allocate it between bananas and apples?

Since we are using a small increase in land the increase in revenues from the extra land will be* ' ' sin the FOC * * *u g

b b b a a a b adR P F dT P F dT dR dT dT = + = + and since it is optimal to use all the additional

land, thenb a

dT dT dT = + and* *

dR dT = so * *dR dT = . The multiplier is the total derivative of the

objective function with respect to the constrained resource, in this case land. It measures how revenue

changes when we change land (assuming we are allocating it optimally). Intuitively one extra unit of land

( ) ( ) ( ) ( ){ }0 , 1 , 2 ,...c c c c ( ) ( ) ( ){ }1 , 2 ,...s s s

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the constrained resource ( )s t , i.e. the amount the agent is be willing to pay for anadditional unit of the constrained resource.

The First Order Conditions for this problem are

( )( )

( ) ( )( ) ( ) ( ) ( )( )0 , , 0c cH t

V c t s t t f c t s t c t

= + =

(4.1)

H t( )s t( )

=

t

H t( )s t( )

"

#

$$

%

&

''V

sc t( ),s t( )( )+ t( ) fs c t( ),s t( )( ) = t( )+ t( ) (4.2)

H t( ) t( )

= 0 s t( ) = f s t( ),c t( )( ) (4.3) = (2)

In addition, we also require the following transversality condition:

( ) ( ) 0tt

Lim t s t e

= (4.4)

So we have can solve (4.1), (4.2) and (4.3) to reach a system of differential equations of

the following form,

c t( ) = g c t( ),s t( )( )s t

( )=

f c t( ),s t

( )( )

we use the fact that (4.1) yields ( ) ( ) ( )( ),t c t s t =

that we can evaluate it using a phase diagram.

For an economic interpretation of these conditions it is helpful to go back to our

fishery example, assume that we are in charge of exploiting the fishery forever, where we

denote the stock of fish at any point in time as . We want to find the optimal path of

captures, i.e. the amount of captures at each point in time, , that maximizes the

present value of intertemporal profits, discounted at the market interest rate, we need tosolve something like

maxPc

= c t( )( )ert0

dt where c > 0;cc < 0 (1)

will generate * units of revenue and therefore * is the value we attach to an extra unit of land or its

shadow price.

( )f t

( )c t

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s.t.

f t( ) = gf t( ) c t( )f 0( ) = f0

(2)

where g is the constant rate of reproduction and is a concave function representing

instantaneous profits, its concavity reflects that the price of fish (the cost of fishing) is

decreasing (increasing) in the amount captured5

.

Setting the Hamiltonian,

H= c t( )( )ert+ t( ) ert gf t( ) c t( ) f t( )"# $%

we get the following optimality conditions,

0c

H

c

= =

(4.1)

Hs

=t

H s

g = + r g+

= r (4.2)

H

= 0 f = fg c (4.3) = (2)

In addition, we also require the following transversality condition:

0rt

t

Lim fe

= (4.4)

(4.1) equates at the margin the contribution of a fish in its two alternative uses, capture it

and increase current profits, leave in the lake (shadow-price of the constrained resource).This is a static condition.

(4.2) it is an intertemporal allocation condition equating the return (through time) we

obtain by leaving a fish in the lake, measured as its reproductive power plus the change inits shadow price, and the return to fish it today instead of tomorrow, measured by the

interest rate.

(4.3) it is just a restatement of the constraint.

5Every time I introduce a new variable x , I will denote it as ( )x t if it depends on time, but once I have

introduced it I will drop the ( )t . Nonetheless it is important that you know when some variable is afunction of time.

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(4.4) it is best understood by considering the same maximization problem with a finitehorizon, so imagine that the planning horizon ends at t=T, then (4.4) becomes,

LimtT

fert = T( ) f T( )erT = 0 it implies that at the end of the planning horizon either the amount of fish in the lake,

( )f T , should be zero or the value we place on it, ( )T

, should be zero (from (4.1) thisis equivalent to marginal contribution of a capture to profits of 0) . If there is still fish andit has value, then we would be better off capturing it and increasing profits. (4.4) is the

limit of this condition as Tgoes to infinity.

4. Continuous vs. discrete time

When modeling time there are two potential approaches: time might be treated as

a continuum (continuous time modeling) or the time dimension might be divided intointervals of a constant length (discrete time modeling) and therefore time is only allowed

to take integer values.

The results of both approaches are equivalent and as a matter of pragmatism oneshould use the method that serves ones purposes more satisfactorily. In general

theoretical work is conducted under a continuous-time framework due to its analytical

tractability (notice that since time is only allowed to take integer values the concept ofderivative will no longer be appropriate under discrete time). In the other hand discrete-

time modeling is often used when the researcher plans to contrast the model with

empirical observations since data is always collected in discrete intervals, or for

theoretical work in a stochastic environment.Finally it is worth noticing that the continuous-time model is obtained as the limit

of the underlying discrete-time model, as the unit time interval shrinks to zero.

5. Some Final Remarks.

5.a Working with growth rates

( )X t denotes the growth rate of ( )X t

Z t( ) =dlnZ t( )

dt=

dlnZ t( )dZ t( )

dZ t( )dt

=1

Z t( )Z t( ) =

Z t( )Z t( )

Y t

( ) B t

( )/ A t

( ) Y t

( )= B t

( )

A t

( )X t( ) B t( )* A t( ) X t( ) = B t( )+ A t( )

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Proof:

( )( ) ( ) ( ) ( )

( )

( )

( )( )

( )

( )( )

( )

( )( )

( )

( )

( ) ( )( ) ( )

2

/

then

B t B tA t

B t A t A t B t B t B t A t A tY t A t Y t B t A t

A t A t B t A tA t

= = = =

5.b Graphing exponential growth

Exponential (or geometric growth) occurs when the change in a variable, Z t( ) , is

proportional to the variables current size, Z t( ) . Population and income both followapproximately exponential laws (as a result income per capita also grows exponentially).

Lets take a variable that grows exponentially at a rate g , then after t periods its value

will be Z t( ) =Z0egt . Its graphical representation is difficult to interpret, since its value

grows very fast, a convenient transformation consists in taking logarithms of our variable

of interest, so ln Z t( )( ) = ln Z0( )+gt. As you can see the resulting relation is linear andtherefore much easier to interpret. The following 3 figures present alternativerepresentations of the same data, Dutch per capita GDP from 1820 to 2000.

Dutch per capita GDP

Ln (Dutch per capita GDP)

0

5,000

10,000

15,000

20,000

25,000

1820

1831

1842

1853

1864

1875

1886

1897

1908

1919

1930

1941

1952

1963

1974

1985

1996

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Dutch per capita GDP (logarithmic scale)

6.00

6.50

7.00

7.50

8.00

8.50

9.00

9.50

10.00

10.50

11.00

1820

1831

1842

1853

1864

1875

1886

1897

1908

1919

1930

1941

1952

1963

1974

1985

1996

100

1,000

10,000

100,000

1820

1831

1842

1853

1864

1875

1886

1897

1908

1919

1930

1941

1952

1963

1974

1985

1996