CHAPTER

THE FUNDAMENTAL

PROBLEM OF THECALCULUS OF VARIATIONS

We shall begin the study of the calculus of variationswith the fundamental problem:

Maximize or minimize

V[y] = lTF[f,y( f), yr(f)] dt0

(2.1) subject to

y(0) = A (A given)and y(T) = Z (T,Z

given)

YII

Ip(t) I

,------1-\

II/--- .I'----" .- " --- "">+

T FIGURE 2 .1

In locating an extremum of V[y], one may bethinking of either an absolute (global) extremum or arelative (local) extremum. Since the calcu-lus of variationsis based on classical calculus methods, it can directlydeal only with relative extrema. That is, an extremalyields an extreme value of V only in comparison with theimmediately "neighboring9' y paths.

2.1 THE EULER EQUATION

The basic first-order necessary condition in the calculusof variations is the Euler equation . Although it wasformulated as early as 1744, it remains the mostimportant result in this branch of mathematics. In viewof its impor- tance and the ingenuity of its approach, itis worthwhile to explain its rationale in some detail.

With reference to Fig. 2.1, let the solid pathy*(t) be a known extremal. We seek to find someproperty of the extremal that is absent in the(nonextremal) neighboring paths. Such a propertywould constitute a necessary condition for an extremal.To do this, we need for comparison purposes a family ofneighboring paths which, by specification in (2.11, mustpass through the given endpoints (0, A) and (T,2)A simple way of generating such neighboring paths isby using a perturbing curve, chosen arbitrarily except forthe restrictions that it be smooth and pass through thepoints 0 and T on the horizontal axis in Fig. 2.1, so that

We have chosen for illustration one with relatively smallp values and small slopes throughout. By adding ~p(t)toy*(t), where E is a small number, and by varying themagnitude of E,we can perturb the y*(t) path, displacingit to various neighboring positions, thereby generatingthe desired neighbor- ing paths. The latter paths can bedenoted generally as

(2.3) y(t) =y*(t) + ~p(t) [implying yl(t) =y*'(t)

the pa

This constitutes a defining property of the

extremal. It follows that dV/d E = 0 is a necessarycondition for the extremal.

As written, however, condition (2.4) is notoperational because it involves the use of the arbitraryvariable E as well as the arbitrary perturb-ing functionp(t). What the Euler equation accomplishes is to expressthis necessary condition in a convenient operationalform. To transform (2.4) into an operational form,however, requires a knowledge of how to take thederivatives of a definite integral.

Differentiating a Definite Integral

Consider the definition integral

where F(t, x) is assumed to have a continuousderivative F,(t, x) in the time interval [a, b]. Since anychange in x will affect the value of the F function andhence the definite integral, we may view the integralas a function of x-Z(x). The effect of a change in x onthe integral is given by the derivative formula:

dl(2.6) -dx = p,(t,X)dt[Leibniz 's rule]

In words, to differentiate a definite integral withrespect to a variable x which is neither the variable ofintegration (t) nor a limit of integration (a or b), onecan simply differentiate through the integral sign with

respect to x.The intuition behind Leibniz's rule can be seen

from Fig. 2.212, where the solid curve represents F(t, x),and the dotted curve represents the displaced positionof F(t, x) after the change in x. The vertical distancebetween the two curves (if the- change is infinitesimal)

FIGURE 2.2

derivative Fx(t,x) at each value of t.It follows that theeffect of the change in x on the entire integral, dI/dx,corresponds to the area between the two curves, or,equivalently, the definite integral of Fx(t,x) over theinterval [a,b].This explains the meaning of (2.6)

The value of the definite integral in (2.5) canalso be affected by a change in a limit of integration.

Defining the integral alternatively to be

we have the following pair of derivative formulas:

In words, the derivative of a definite integral with respectof integration b is equal to the

integrand evaluated at t = b; and the derivative withrespect to its lower limit of integration a is the negativeof the integrand evaluated at t = a.

In Fig. 2.2b, an increase in b is reflected in arightward displacement of the right-hand boundary of thearea under the curve. When the displace-ment isinfinitesimal, the effect on the definite integral ismeasured by the value of the F function at theright-hand boundary-F(b, x).This provides the intuitionfor (2.8).For an increase in the lower limit, on the otherhand, the resulting displacement, as illustrated in Fig.2.2c, is a rightward move-ment of the left-hand boundary,which reduces the area under the curve. This is whythere is a negative sign in (2.9)

The preceding derivative formulas can also be usedin combination. If, for instance, the definite integral takesthe form

where x not only enters into the integrand function F,but also affects the upper limit of integration, then we canapply both (2.6)and (2.8), to get the total derivative

The first term on the right, an integral, follows from(2.6);the second term,

dK db(x)representing the chain ------,is based on (2.8)

db(x) dxEXAMPLE 1 The derivative of /;e-$ dt with respect tox is, by Leibniz's rule,

EXAMPLE 2 Similarly

EXAMPLE 3 To differentiate j,2"3t2 dt with respect to xwhich appears in the upper limit of integration, we need

the chain rule as well as (2.8). The result is

For ease of understanding, the Euler equation

will be developed in four steps.Step i Let us first express V in terms of E, andtake its derivative. Substituting (2.3) into the objectivefunctional in (2.1),we have

To obtain the derivative dV/d r,Leibniz's rule tellsus to differentiate through the integral sign:

Breaking the last integral in (2.13) into twoseparate integrals, and setting dV/de = 0, we get amore specific form of the necessary condition for anextremal as follows:

T(2.14) iT~yp(t)dt + / Fytpf(t)dt = 0

0

While this form of necessary condition is already freeof the arbitrary variable E ,the arbitrary perturbingcurve p(t) is still present along with its derivative p'(t)To make the necessary condition fully operational, wemust also eliminate p(t) and pl(t)

Step ii To that end, we first integrate the second

integral in (2.14) by parts, by using the formula:

Let v = Fy, and u =- p(t). Then we havedv dFyt

dv=-dt=---dt anddt dt

dudu=-dt=

dt ~'(t)dt

Substitution of these expressions into (2.15)-with a

first term to the right of the first equals sign mustvanish under

on (2.2).Applying (2.16) to (2.14) and combining the twointegrals we obtain another version of the necessarycondition for the

Although pl(t) is no longer present in (2.171, thearbitrary p(t) However, precisely because p(t)enters in an arbitrary way, conclude that condition (2.17can be satisfied only if the bracketed [Fy - dFy,/dt] ismade to vanish for every value of t on theotherwise, the integral may not be equal to zero for someadmissi-rbing curve p(t). Consequently, it is a necessarycondition for an that

dFy --F - 0 for all t E [0,T ] [Euler equation]

dt Y' -

the Euler equation is completely free of arbitraryexpressions, and be applied as soon as one is given adifferentiable F(t,y,y')

Euler equation is sometimes also presented in the form

the result of integrating (2.18) with respect to t.

The nature of the Euler equation (2.18 )can be madeclearer when nd the derivative dFyt/d t into a more explicitform. Because F is a with three arguments (t,y,y'), thepartial derivative F,, should function of the same three

consists of three terms:

arguments. Thetotal derivative dFyl/d

dFy, aFyr aF r dy aF t dy'-+2-+--

dt at a~ dt ayl dt= Fty, + FyyPyf(t)+ Fyry,y"(

Substituting this into (2.18), multiplying through by -1,and rearranging, we arrive at a more explicit version ofthe Euler equation:

(2.19) FY.,,y"(t) + FYyty1(t)+ Ft,. - Fy = 0for all t E [0,T ] [Euler equation]

This expanded version reveals that the Euler

equation is in general a second-order nonlineardifferential equation. Its general solution will thuscontain two arbitrary constants. Since our problem in(2.1) comes with two boundary conditions (one initialand one terminal), we should normally possess

sufficient information to definitize the two arbitraryconstants and obtain the definite solution

EXAMPLE 4 Find the extremal of the functional

with boundary conditions y(0) = 0 and y(2) = 8. SinceF = 12ty + f2, we have the derivatives

Fy = 12t Fyt = 2y' Fytyt= 2 and

Fyy.= F,,* = 0 By (2.191, the Euler equation is2yU(t)- 12t = 0 or y"(t) = 6t

which, upon integration, yields y'(t) = 3t2 + c,, andy*(t) = t3 + clt + c2 [general

solution] To definitize the arbitrary constants c, and c,,we first set t = 0 in the general solution to get y(0) =c,; from the initial condition, it follows that c, = 0. Next,

+setting t = 2 in the general solution, we get y(2) = 8 +2c,; from the terminal condition, it follows that c, = 0.The extremal is thus the cubic function

y* (t) = t3 [definite solution]EXAMPLE 5 Find the extremal of the functional

with boundary conditions y(1) = 3 and y(5) = 7.+Here we have F = 3t + (y')'l2. Thus,

The Euler equation (2.19) now reduces to

The only way to satisfy this equation is to have aconstant y', in order that y" = 0. Thus, we write yl(t) =c,, which integrates to the solution

y* (t) = c,t + c, [general solution]

To definitize the arbitrary constants c, and c,, we first+set t = 1to find y(1) = c, + c, = 3 (by the initial

+condition), and then set t = 5 to find y(5) = 5c, + c2 =7 (by the terminal condition). These two equations give usc, = 1and c, = 2. Therefore, the extremal takes theform of the linear function

y*(t) = t + 2 [definite solution]

EXAMPLE 6 Find the extremal of the functional

with boundary conditions y(O) = 0 and y(5) = 3. Since F= t + y2 + 3y1,we have

Fy = 2y and Fy,= 3

By (2.18), we may write the Euler equation as 2y = 0,with solution

Note, however, that although this solution isconsistent with the initial condition y(0) = 0, it violatesthe terminal condition y(5) = 3. Thus, we mustconclude that there exists no extremal among the set ofcontinuous curves that we consider to be admissible.

This last example is of interest because it servesto illustrate that certain variational problems with givenendpoints may not have a solution. More specifically, itcalls attention to one of two peculiar results that canarise when the integrand function F is linear in y'.One result, as illus-trated in Example 6, is that nosolution exists. The other possibility, shown in Example7, is that the Euler equation is an identity, and sinceit is automatically satisfied, any admissible path isoptimal.

EXAMPLE 7 Find the extremal of the functional

and

It follows that the Euler equation (2.18) is

always satisfied. In this example, it is in fact clear fromstraight integration that

The value of V depends only on the given terminaland initial states, regardless of the path adjoining thetwo given endpoints.

The reason behind these peculiarities is that whenF is linear in y', Fy is a constant, and FyrY r= 0, so thefirst term in the Euler equation (2.19) vanishes. TheEuler equation then loses its status as a second-orderdiffer- ential equation, and will not provide two arbitraryconstants in its general solution to enable us to adaptthe time path to the given boundary condi-tions.Consequently, unless the solution path happens to passthrough the fixed endpoints by coincidence, it cannotqualify as an extremal. The only circumstance underwhich a solution can be guaranteed for such a problem(with F linear in y' and with fixed endpoints) iswhen Fy = 0 as well, which, together with the fact thatFy r= constant (implying dFyr/d t= O), would turn theEuler equation (2.18) into an identity, as in Example 7.

EXERCISE 2.1

1 In discussing the differentiation of definite integrals, nomention was made

of the derivative with respect to the variable t.Is that a justifiable

omission?

Find the derivatives of the following definite integrals withrespect to x:

Find the extremals, if any, of the following fundionals:6 V[yl = \$ty + 2yt2)dt,with y(0) = 1and y(1) = 27 V[yl = \,'tyyf dt, with y(0) = 0 and y(1) = 18 V[yl = /;(2yet + y2+ yr2)dt, with y(0) = 2 and y(2) = 2e2 -t e-' 9 V[yl= j:(y2 +t2yr)dt, with y(O) = 0 and y(2) = 2

se.

AMPLE 1 Find the extremal of the functional

boundary conditions y(0) = y(1) = 1. Since

F = tyr + Y'2 and Fy, = t + 2y'20) gives us t + 2y1(t) = constant, or

direct integration, we obtain

y* (t)= - it2 + clt + c2 [generalsolution] the help of the boundary conditions y(0) =y(1) = 1, it is easily verified c1= f and c, = 1. Theextremal is therefore the quadratic path

y* (t)= - :t2 + f t + 1 [definite solution]pecial case 11: F = F(y,y') Since F is free of t inthis case, we have = 0, so the Euler equation (2.19)simplifies to

solution to this equation is by no means obvious, but it

turns out that ifmultiply through by y', the left-hand-side expression

new equation will be exactly thederivative d(y1FY,- F)/dt, for

d d d-(y1FYt-F) = -(y1FYt)--F(y,y')dt dt dt

Consequently, the Euler equation can be written asd(yfFYt- F)/dt = 0, with the solution y1FYt-F =constant, or, what amounts to be the same thing,

(2.21) F -y'Fy, = constant

This result-the simplified Euler equation alreadyintegrated once-is a first-order differential equation,which may under some circumstances be easier tohandle than the original Euler equation (2.19).Moreover, in analytical (as against computational)applications, (2.21) may yield results that would not bediscernible from (2.19), as will be illustrated in Sec. 2.4.

EXAMPLE 2 Find the extremal of the functional

with boundary conditions y(0) = 1and y(i-r/2) = 0. Since

F = ~ ~ - ~ ' ~and F,,=-2y'

direct substitution in (2.21) gives us

Y'2 + y2 = constant [ = a2, say]This last constant is nonnegative because theleft-hand-side terms are all squares; thus we canjustifiably denote it by a2,where a is a nonnegative realnumber.

The reader will recognize that the equationy'2 + y2 = a2 can be plotted as a circle, as in Fig. 2.3,with radius a and with center at the point of origin. Sincey' is plotted against y in the diagram, this circleconstitutes the phase line for the differential equation.Moreover, the circular loop shape of this phase linesuggests that it gives rise to a cyclical time path,'

'phase diagrams are explained in Alpha C. Chiang, FundamentalMethods of Mathematical Economics, 3d ed., McGraw-Hill, New York,1984, Sec. 14.6. The phase line here is similar to phase line C in Fig.14.3 in that section; the time path it implies is shown in Fig. 14.4CHAPTER 2: THE FUNDAMENTAL PROBLEM OF CALCULUS OF VARIATIONS

1 FIGURE 2.3

with the y values bounded in the closed interval[-a,a], as in a cos function with amplitude a andperiod 2i-r .Such a cosine function can represented ingeneral by the equation

where, aside from the amplitude parameter a, wealso have two ot parameters b and c which relate tothe period and the phase of function, respectively. Inour case, the period should be 2i-r; but since cosinefunction shows a period of 2~/ b(obtained by setting thebt term 2~1,we infer that b = 1. But the values of a andc are still unknown, they must be determined from thegiven boundary conditions

At t = 0, we havey(0) = a cos c = 1 [by the initial

condition]

When t = ~/2,we havey (%) = a cos(i+ c) = 0

curve AZ.

Since this equation has complex characteristic roots r,, r,'= + i,the general solution takes the form of '

y*(t) = a cos t + p sin tThe boundary conditions fur the arbitrary constants a

and 63 at the values of 1and 0, respectively. Thus, we endup with the same definite solution:

y*(t) = cos t

For this example, the original Euler equation (2.18)or (2.19) turns out to be easier to use than the special onein (2.21). We illustrate the latter, not only to present itas an alternative, but also to illustrate some othertechniques (such as the circular phase diagram in Fig.2.3). The reader will have to choose the appropriateversion of the Euler equation to apply to any particularproblem.

EXAMPLE 3 Among the curves that pass through twofixed points A and 2, which curve will generate thesmallest surface of revolution when rotated about thehorizontal axis? This is a problem in physics, but itmay be of interest because it is one of the earliestproblems in the development of the calculus of variations

To construct the objective functional, it may behelpful to consider the curve AZ in Fig. 2.4 as acandidate for the desired extremal. When rotated aboutthe t axis in the stipulated manner, each point on curveAZ traces out a circle parallel to the xy plane, with itscenter on the t axis, and with radius R equal to the valueof y at that point. Since the circumference of such a circle

is 2aR (in our present case, 2ay),all we have to do tocalculate the surface of revolution is to sum (integrate)the circumference over the entire length of the curve AZ.

An expression for the length of the curve AZ can beobtained with the help of Fig. 2.5. Let us imagine that Mand N represent two points that are located very close toeach other on curve AZ. Because of the extremeproximity of M and N, arc MN can be approximatedby a straight line, with its length equal to the differentialds. In order to express ds in terms of the variables y and

+t,we resort to Pythagoras' theorem to write (ds)' =(dy)' +(dt)'. It follows that (d~)~/(d t)'= (dy)'/(dt)' + 1.Taking the

oror an explanation of complex roots, see Alpha C. Chiang,Fundamental Methods of Mathe-matzcal Econom~cs, 3d ed., McGraw-HillNew York, 1984, See. 15.3. Here we have h = 0, and v = 1; thus, eh' = 1and vt = t.

\x FIGURE 2.4

1D(Yt; 811 both sides, we get

dt

yields the desired expression for ds in terms of y and tas follows:

ds = (1+ yt2)1/2 dt [arc length]

entire length of curve AZ must then be the integral of+ds, namely + y'2)1/2 dt. TO sum the circumference 2~rover the length of curve ill therefore result in thefunctional

The reader should note that, while it isc,, at a nylegitimate to "factor out" the "2~ (constant) part of theexpression ~TR ,the "y" (variable) part must remainwithin the integral sign.

For purposes of minimization, we may actually+disregard the constant 2ar and take y(1 + y'2)1/2 to.be the

expression for the F function, with

Since F is free of t, we may apply (2.21) to get

This equation can be simplified by taking the followingthrough by (1+ y'2)1/2, (2) cancel out yyt2 and -yy'2 on(3) square both sides and solve for yt2 in terms of y andsteps: (1) multiplythe left-hand side,c, and (4) take the

square root. The result is

In this last equation, we note that the variables yand t have been separated, so that each side can beintegrated by itself. The right-hand side poses no problem,

+the integral of dt being in the simple form t + k, where kis an arbitrary constant. But the left-hand side is morecomplicated. In fact, to attempt to integrate it "in theraw" would take too much effort; hence, one shouldconsult prepared tables of integration formulas tofind the re~ul

Equating this result with t + k (and subsuming theconstant c, under the

3~e e,for example, CRC Standard Mathematical Tables ,28th ed., ed. byWilliam H. Beyer, CRC Press, Boca Raton, FL, 1987, Formula 157. Thec in the denominator of the log expression on the right may be omittedwithout affecting the validity of the formula, because its presence or

absence merely makes a difference amounting to the constant -c In c,which can be absorbed into the arbitrary constant of integration c,, atany rate. However, by inserting the c in the denominator, our resultwill come out in a more symmetrical form

CHAPTER 2 THE FUNDAMENTAL PROBLEM OF CALCULUS OF VARIATIONS

43

constant k),we have

e("k)/c =y + ,ly2- c2

orC

[by definition of natural log]

Multiplying both sides by c, subtracting y from bothsides, squaring both sides and canceling out the y2term, and solving for y in terms of t, we finally find thedesired extremal to be

Cy* (t) = -[e(t+h)/c+ e-"+k'/c]

2

[general solution]

where the two constants c and k are to be definitized byusing the boundary conditions.

This extremal is a modified form of the so-called

catenary curve, (2.23) y=i(et+e-')[catenary]

whose distinguishing characteristic is that it involvesthe average of two exponential terms, in which theexponent of one term is the exact negative of theexponent of the other. Since the positive-exponent termgives rise to a curve that increases at an increasing rate,whereas the negative-exponent term produces anever-decreasing curve, the average of the two has ageneral shape that portrays the way a flexible rope willhang on two fixed pegs. (In fact, the name catenarycomes from the Latin word catena, meaning "chain.")This general shape is illustrated by curve AZ in Fig. 2.4.

Even though we have found our extremal in acurve of the catenary family, it is not certain whetherthe resulting surface of revolution (known as a catenoid)has been maximized or minimized. However, geometricaland intuitive considerations should make it clear thatthe surface is indeed minimized. With reference to Fig.2.4, if we replace the AZ curve already drawn by, say, anew AZ curve with the opposite curvature, then alarger surface of revolution can be generated. Hence, thecatenoid cannot possibly be a maximum.

Special case 111: F = F(yf) When the F functiondepends on y' alone, many of the derivatives in (2.19) will

3''.disappear, including Fyy,,F,,,,and 3''. In fact, only thefirst term remains, so the Euler equation becomes (2.24)FypPyw(t)= 0

To satisfy this equation, we must either have y"(t)

= 0 or F,.,, = 0. If y"(t) = 0, then obviously y'(t) = c,and y(t) = c,t f c2, indicating that the

EXAMPLE 4 Find the extremal of the functional

with boundary conditions y(0) = A and y(T) = 2. Theastute reader will note that this functional has beenencountered in a different guise in Example 3.Recalling the discussion of arc length leading to (2.22), weknow that (2.25) measures the total length of a curvepassing through two given points. The problem of findingthe extremal of this functional is therefore that offinding the curve with the shortest distance between

those twopoints.

The integrand function, F = (1+ y'2)1/2

is dependent on y' alone. We can therefore concludeimmediately that the extremal is a straight line. But if itis desired to examine this particular example explicitlyby the Euler equation, we can use (2.18).with F , = 0,the Euler equation is simply dF,,/dt = 0, and its

+solution is Fy r= constant. In view of the fact thatF,, = yl/(l + y'2)1/2, we can write (after squaring)

Yt2

Multiplying botb sides byrn yf in hrms of

(1+ yt2), rearranging, and factoring out y', we

c as follows: yt2 = c2/(1 - c2). Equivalently,

,,y1= ,/, = constant

(1- c2)

Inasmuch as yl(t) ,the slope of y(t) ,is a constant ,thedesired extremal y*(t) must be a straight line

Strictly speaking ,we have only found an "extremal"which may either maximize or minimize the givenfunctional .However ,it is intuitively obvious that thedistance between the two given points is indeedminimized rather than maximized by the straight-lineextremal ,because there is no such thing as "the longesdistance" between two given points

XERCISE 2.2Find the extremals of the following functionals1 V[yl = /i(t2 + y'2)dt, with y(0) = 0 and y(1) = 22 V[y] = /,27y'3 dt, with y(0) = 9 and y(2) = 113

V[yl = /,'(y + yy' + y' + iYr2)dt, with y(0) = 2 and y(1) = 54 V[yl = /,bt-3y'2 dt (Find the general solution only.)5 V[yl = /,'(y2 + 4yy' + 4y'2) dt, with y(O) = 2e1/2 and

y(1) = 1+ e6 V[yl = \,"'2(y2 -Yr2)dt, with y(O) = 0 and y(~r/2)= 1

TWO GENERALIZATIONS OF THEEULER EQUATION

previous discussion of the Euler equation is based

on an integral ctional with the integrand F(t, y, y').Simple generalizations can be however, to the caseof several state variables and to the case where her-orderderivatives appear as arguments in the F function.

Case of Several State Variablesn > 1state variables in a given problem, the

functional becomes

46

PART 2. THE CALCULUS OF VARIATIONS

and there will be a pair of initial condition and terminalcondition for each of the n state variables.

Any extremal y,*(t), (4 = 1,...,n), must bydefinition yield the ex-treme path value relative to all theneighboring paths. One type of neighbor-ing path arises from varying only one of the functions yJ(t)at a time, say,y,(t), while all the other yJ(t) functions are kept fixed. Then the functionaldepends only on the variation in y,(t) as if we are dealing with the case of asingle state variable. Consequently, the Euler equation(2.18) must still hold as a necessary condition, providedwe change the y symbol to y, to reflect the new problem.

Moreover, this procedure can be similarly used to vary theother yj functions, one at a time, to generate otherEuler equations. Thus, for the case of n state variables,the single Euler equation (2.18) should be replaced by aset of n simultaneous Euler equations:

[Euler equations]

These n equations will, along with the boundaryconditions, enable us to determine the solutions y:(t), ..,y,*(t).

Although (2.27 )is a straightforward generalizationof the Euler equa-tion (2.18)-replacing the symbol y byyj-the same procedure cannot be used to generalize(2.19).To see why not, assume for simplicity that thereare only two state variables, y and z, in our newproblem. The F function will then be a function with fivearguments, F(t,y, z,y', z'), and the partial derivatives Fyand F,,, will be, too. Therefore, the total derivativeof Fyr( t,y,z, y', 2') with respect to t will include fiveterms:

d-Fyv(t,y, Z,y', z') = Fty z+ Fyy'yf( t)+ FZypzt (t)+dt

FyFy'yn(t)+ Fiy,zn(t

with a similar five-term expression for dF,,/dt. The

expanded version of simultaneous Euler equationscorresponding to (2.19) thus looks much morecomplicated than (2.19) itself:

Fytypy"(t)+ F,#ytz"(t) + FyyFy1(t) + F,.z'(t) + Fty- Fy = 0

(2.28)Fytz,yn(

+ FztzI~"(t)+ FyZryl(t)+ FZz,zr(t)+ FtZt-F, = 0forall t E [O,T

EXAMPLE 1 Find the extremal of

kom the integrand, we find thatCHAPTER 2: THE FUNDAMENTAL PROBLEM OF CALCULUS OF VARIATIONS

47

Thus, by (2.27), we have two simultaneous Eulerequations

The same result can also be obtained from (2.28)In this particular case, it happens that each of

the Euler equations contains one variable exclusively.Upon integration, the first yields y' =it + c,, and hence,

Analogously, the extremal of z is

The four arbitrary constants (c,,...,c4) can bedefinitized with the help of four boundary conditionsrelating to y(O), z(0), y(T ), and z(T ).

The Case of Higher-Order Derivatives

As another generalization, consider a functional thatcontains high-order derivatives of y(t). Generally, thiscan be written as

Since many derivatives are present, the boundaryconditions should in this case prescribe the initial and

.terminal values not only of y, but also of the derivativesy', y", ,. ., up to and including the derivative y(*-'),making a total of 2n boundary conditions.

To tackle this case, we first note that an F functionwith a single state variable y and derivatives of y up tothe nth order can be transformed into an equivalent formcontaining n state variables and their first-order deriva-tives only. In other words, the functional in (2.29 )can betransformed into the form in (2.26) .Consequently, theEuler equation (2.27) or (2.28) can again be applied.Moreover, the said transformation can automaticallytake care of the boundary conditions as well.

EXAMPLE 2 Transform the functional

V[y]= lT(ty2 + yy' + yr'2) dt0

with boundary conditions y(0) = A, y(T) = 2, ~'(0)=a, and yr(T) = P, into the format of (2.26) .To achievethis task, we only have to introduce a new variable

which now contains two state variables y and z, with noderivatives higher than the first order. Substitwing thenew F into the functional will turn the latter into theformat of (2.26)

What about the boundary conditions? For theoriginal state variable y, the conditions y(0) = A andy(T) = Z can be kept intact. The other two conditionsfor y' can be directly rewritten as the conditions forthe new state variable z: z(0) = a and z(T) = /3. Thiscomplete the transformation.

Given the functional (2.291,it is also possible todeal with it directly instead of transforming it into theformat of (2.26) .By a procedure similar to that used inderiving the Euler equation, a necessary condition foran extremal can be found for (2.29) .The condition,known as the Euler-Poisson equation, is

for all t E [0,T ] [Euler-Poissonequation]

This equation is in general a differential equation oforder 2n. Thus its solution will involve 2n arbitrary

constants, which can be definitized with the help of the2n boundary conditions.

EXAMPLE 3 Find an extremal of the functional inExample 2. Since we have

Fy = 2ty + y' Fyr= y Ff = 2y"the Euler-Poisson equation is

which is a fourth-order differential equation.

EXERCISE 2.3

1Find the extremal of V[y] = /,'(I + yr2) dt, with y(0) = 0and yf(0) =

y(1) = ~'(1)= 12 Find the extremal of

only).

the first quadrant, giving us an upward-sl oping

.4 DYNAMIC OPTIMIZATIONOF A MONOPOLIST

et us turn now to economic applications of the Euler

equation. As the first xample, we shall discuss the classicEvans model of a monopolistic firm, ne of the earliestapplications of variational calculus to economics.*

Dynamic Profit Functiononsider a monopolistic firm that produces a single

commodity with a uadratic total cost function5

ince no inventory is considered, the output Q is alwaysset equal to the uantity demanded. Hence, we shall use asingle symbol Q(t)to denote both uantities. The quantitydemanded is assumed to depend not only on price (t),butalso on the rate of change of price P1(t)2.32) Q = a - bP(t) + hP1(t(a,b > 0; h =+ 0)he firm's profit is thus

?T=PQ-C= P(a - bP + hP') -a(a - bP + hP')' - /3(a -

bP + bP)- ywhich is a function of P and P'.Multiplying out the above expression and ollecting terms,we have the dynamic profit function

2.33) r(P,P') = -b(l + ab)P2 + (a + 2aab + pb)P- ah2P'2 - h(2aa + @)PI + h(l + 2ab)PP' - (aa2 + pa

50in the general format of

PART 2: THE CALCULUS OF VARIATIONS

The ProblemThe objective of the firm is to find an optimal path

of price P that maximizes the total profit rI over afinite time period [0,TI. This period is assumed to besufficiently short to justify the assumption of fixeddemand and cost functions, as well as,the omission of adiscount factor. Besides, as the first approach to theproblem, both the initial price Po and the terminal priceP, are assumed to be given

The objective of the monopolist is therefore to

Maximize II[P] = LTT(p, P') dt

(2.34) subject to P(0) = Po (Pgiven)

and P(T)=P, (T,PTgivenThe Optimal Price PathAlthough (2.34) pertains to Special Case 11, it turns outthat for computa tion with specific functions it is simplerto use the original Euler equation (2.19), where weshould obviously substitute .rr for F, and P for y. Fromthe profit function (2.339, it is easily found that

and n-,,=-2ah2 rrP,=h(l+2ab) a,,=OThese expression turn (2.19)-after normalizing-intothe specific form

b(l + ab)p= - a + 2aab + pb(2.35) P " -ah2 2ah2

[Euler equation]This is a second-order linear differential equation

with constant coef- ficients and a constant term, in thegeneral format of

Its general solution is known to be6

his type of differential equation is discussed in Alpha C. Chiang,Fundamental Methods ofMathematical Economics, 3d ed., McGraw-Hill, New York, 1984, Sec.

where the characteristic roots r, and r, take the

values

r,, r2 = +(-a, * J-)and the particular integral 3 is

Thus, for the Euler equation of the present model

(where a, = 0), we have

b(l + ab)where rl,r2 = 5~

[characteristic roots] anda + 2aab + fib

F= [particular integral]2b(l + ab) Note that

the two characteristic roots are real and distinct under

our sign specifications. Moreover, they are the exactnegatives of each other. We may therefore let r denotethe common absolute value of the two roots, andrewrite the solution as

(2.36' ) P*(t) = Aler' + A2e-rt + F [generalsolution]

The two arbitrary constants A, and A, in (2.36'can be definitized via the boundary conditions P(0 )= Poand P(T )= P,. When we set t = 0 and t = Tsuccessively ,in (2.36'1 ,we get two simultaneousequations in the two variables A, and A2

P,=A,+A,+PP, = AlerT + A2eWrT+ F

with solution values

(2.37P, - H - (P, - P)erT

A, = 1- e2rT

P, - H - (P, - F)e-rTA, = 1- e-2rT

The determination of these two constants completes

the solution of the problem , for the entire price pathP*(t ) is now specified in terms of the knownparameters T ,Po,PT ,a,P,y ,a,b,and h.Of theseparameters, al have specified signs excep t h .Butinasmuch as the parameter h enters into the solutionpath (2.36' )only through r,and only in the form of asquared term h2,we can see that its algebraic signwill not affect our results, although its numerical valuewill

A More General View of the Problem

Evans' formulation specifies a quadratic cost functionand a linear demand function. In a more general studyof the dynamic-monopolist problem by Tintner, thesefunctions are left ~nspecified .~Thus, the profit functionis simply written as dP ,P'). In such a generalformulation, it turns out that formula (2.21) [for SpecialCase I11 can be used to good advantage. It directly yields asimple necessary condition

that can be given a clear-cut economic interpretation.To see this, first look into the economic meaning of

the constant c. If profit rr does not depend on the rate of

change of price P'-that is, if we are dealing with thestatic monopoly problem as a special case of thedynamic model-then arr/aP' = 0, and (2.38) reduces torr = c. So the constant c represents the static monopolyprofit. Let us therefore denote it by rr, (subscript s forstatic). Next, we note that if (2.38) is divided through byrr, the second term on the left-hand side will involve

which represents the partial elasticity of rr with respectto P'. Indeed, after performing the indicated division, theequation can be rearranged into the

7~erhardTintner, "Monopoly over Time," Econornetnca, 1937, pp.160-170. Tintner also tried to generalize the Evans model to the casewhere T depends on higher-order derivatives of P, but the economicmeaning of the result is difficult to interpret.

The Matter of the Terminal Price

The foregoing discussion is based on the assumption thatthe terminal price P(T) is given. In reality, however, thefirm is likely to have discretionary control over P(T) eventhough the terminal time T has been preset. If so, it willface the variable-terminal-point situation depicted inFig. 1.5a, where the boundary condition P(T) = P,must be replaced by an appropriate transversalitycondition. We shall develop such transversality

conditions in the next chapter.

EXERCISE 2.4

1 If the monopolistic firm in the Evans model faces a staticlinear demand

(h = O) ,what price will the firm charge for profitmaximization? Call that

price P,, and checkthat it has the correct algebraic signThen compare the

values of P, and P, and give the particular integral in(2.36) an economic

interpretation2 Verify that A, and A, should indeed have the values shownin (2.37)3 Show that the extremal P*(t )will not involve a reversal inthe direction of

price movement in the time interval (0,T 1unless there is a value 0< to < T

such that A,erLo = A,eCr'0 [i.e. ,satisfying the conditionthat the first two

terms on the right-hand side of (2.36') are equal at t = to]4 If the coefficients A, and A, in (2.36') are both positivethe P*(t) curve

will take the shape of a catenary .Compare the location ofthe lowest point

on the price curve in the following three cases :(a)A, > A,,(b)A, = A,, and (c)A, < A,. Which of these cases can possiblyinvolve a price reversal in the interval [O,T] ?What type ofprice movement characterizes the remaining cases?5 Find the Euler equation for the Evans problem by usingformula (2.18)6 If a discount rate p > D is used to convert the profit dPP') at any point of time to its present value, then theintegrand in (2.34 )will take the general form F = dPP')e-pf(a) In that case, is formula (2.21) still applicable?(b)Apply formula (2.19 )to this F expression to derive thenew Euler

equation(el Apply formula (2.18 )to derive the Euler equation ,andexpress the

result as a rule regarding the rate of growth of q,

2.5 TRADING OFF INFLATION ANDUNEMPLOYMENT

The economic maladies of both inflation andunemployment inflict social losses. When a Phillipstradeoff exists between the two, what would be the bestcombination of inflation and unemployment over time?The answer to this question may be sought through thecalculus of variations. In this section we present asimple formulation of such a problem adapted from apaper by Taylor.' In this formulation, the unemploymentvariable as such is not included; instead, it is proxied by

(Yf - Y)-the shortfall of current national income Yfrom its full-employment level Yf.

The Socid Loss Function

Let the economic ideal consist of the income level Y fcoupled with the inflation rate 0. Any deviation, positiveor negative, of actual income Y from Y f is consideredundesirable, and so is any deviation of actual rate ofinflation p from zero. Then we can write the social lossfunction as follows:

Because the deviation expressions are squared, positiveand negative devia-tions are counted the same way (cf.Exercise 1.3, Prob. 2). However, Y deviations and pdeviations do enter the loss function with differentweights, in the ratio of 1to a, reflecting the differentdegrees of distaste for the two types of deviations.

"~eanTaylor, "Stopping Inflation in the Dornbusch Model: OptimalMonetary Policiea with Alternate Price-Adjustment Equations," JournalofMacroeconomics, Spring 1989, pp. 199-216. In our adapted version, we

CHAPTER 2 THE

FUNDAMENTAL PROBLEM OF CALCULUS OF VARIATIONS

55

The expectations-augmented Phillips tradeoffbetween (Yf - Y) and p

is captured in the equation

p = -p(Yf - Y) + a (p > 0)

where n, unlike its usage in the preceding section, nowmeans the expected tion. The formation ofinflation expectations is assumed to be

dPi(=z)=j(p-n)

(O

When substituted into (2.401, (2.42) yields

n'

p=_ +nJ

And, by plugging (2.42) and (2.43) into (2.39), we are ableto express the loss function entirely in terms of n and nl

2 2

(2.44) A(P, n') = ($) + a(; + P)[loss function]

The Problem

The problem for the government is then to find anoptimal path of n that minimizes the total social lossover the time interval [O, TI.The initial (current) valueof P is given at no,and the terminal value of n,apolicy target ,is assumed to be set at 0.To recognize theimportance of the present over the future,all social lossesare discounted to their present values via a positivediscount rate p

Minimiz

A[rr] = kT~(rr,lip)e-@ d

(2.45) subject to - rr(0) = rr( rr, > 0 given

and a(T) = 0 (T given

The Solution PathOn the basis of the loss function (2.44), the integrandfunction A(P, .rrr)e-fi yields the first derivatives

with madderivatives

Fvri = 2(%) e-pt

2aF,, = -e-~t

J

and

1+ ap2J

Consequently, formula (2.19) gives us (aftersimplification) the specific

necessary condition(2.46) rr" - prr' - fir = 0

[Euler equation], ap2j(p +j),where R =1+ apZ

Since this differential equation is homogeneous, its

particular integralis zero, and its general solution is simply its

complementary function:

(2.47) rr*(t) = Aler'' + A2er2' [generalsolution]

where r,, r, = f (p k dm.The characteristic roots are real and distinct.

Moreover, inasmuch as thesquare root has a numerical value greater than p, we

know that

(2.48) r, > 0 and r,

To definitize the arbitrary constants A, and A,, wesuccessively set= 0 and t = T in (2.47), and use theboundary conditions to get the pair of lations

A, +Az = rr,

AlerlT + A2erzT = 0olved simultaneously, these relations give us

-rroerzT rroerlT.49) A, =

e IT - er~T A2 = erlT -er

verified Prom the derivative

rr*'(t) = r,A,erlt + r2A2erz' < 0Having found rr*(t) and rr*'(t), we can also derive

some conclusions regarding p and (Yf- Y).For these,(2.43) and (2.42), respectively.

we can simply use the relations in

EXERCISE 2.51 Verify the result in (2.46) by using Euler equation

(2.18).2 Let the objective functional in problem (2

.45) be changed toj:ih(~, rr)e-pt dt.(a) Do you think the solution of the problem will be

different?

(b) Can you think of any advantage in including acoefficient in the

integrand?S Let the terminal condition in problem (2.45) be

changed to T(T)= rT0 < TT < TO(a9 What would be the values of A, and A,?(b) Can you ambiguously evaluate the signs of A,

and A,?

COVERCONTENTS PREFACEPART1 INTRODUCTIONCHAPTER 1 THE NATURE OF DYNAMIC OPTIMIZATION 1.1 SALIENT FEATURES OF DYNAMIC OPTIMIZATION PROBLEMS 1.2 VARIABLE ENDPOINTS AND TRANSVERSALITY CONDITIONS 1.3 THE OBJECTIVE FUNCTIONAL1.4 ALTERNATIVE APPROACHES TO DYNAMIC OPTIMIZATION

PART2 THE CALCULUS OF VARIATIONS CHAPTER 2 THE FUNDAMENTAL PROBLEM OF THE CALCULUS OF VARIATIONS 2.1 THE EULER EQUATION 2.2 SOME SPECIAL CASES 2.3 TWO GENERALIZATIONS OF THE EULER EQUATION 2.4 DYNAMIC OPTIMIZATION OF A MONOPOLIST 2.5 TRADING OFF INFLATION AND UNEMPLOYMENT

CHAPTER 3 TRANSVERSALITY CONDITIONS FOR VARIABLE-ENDPOINT PROBLEMS 3.1 THE GENERAL TRANSVERSALITY CONDITION 3.2 SPECIALIZED TRANSVERSALITY CONDITIONS 3.3 THREE GENERALIZATIONS 3.4 THE OPTIMAL ADJUSTMENT OF LABOR DEMAND

CHAPTER 4 SECOND-ORDER CONDITIONS 4.1 SECOND-ORDER CONDITIONS 4.2 THE CONCAVITY/CONVEXITY SUFFICIENT CONDITION 4.3 THE LEGENDRE NECESSARY CONDITION 4.4 FIRST AND SECOND VARIATIONS

CHAPTER 5 INFINITE PLANNING HORIZON 5.1 METHODOLOGICAL ISSUES OF INFINITE HORIZON 5.2 THE OPTIMAL INVESTMENT PATH OF A FIRM 5.3 THE OPTIMAL SOCIAL SAVING BEHAVIOR 5.4 PHASE-DIAGRAM ANALYSIS 5.5 THE CONCAVITY/CONVEXITY SUFFICIENT CONDITION AGAIN

CHAPTER 6 CONSTRAINED PROBLEMS 6.1 FOUR BASIC TYPES OF CONSTRAINTS 6.2 SOME ECONOMIC APPLICATIONS REFORMULATED 6.3 THE ECONOMICS OF EXHAUSTIBLE RESOURCES

PART3 OPTIMAL CONTROL THEORY CHAPTER 7 OPTIMAL CONTROL:THE MAXIMUM PRINCIPLE 7.1 THE SIMPLEST PROBLEM OF OPTIMAL CONTROL 7.2 THE MAXIMUM PRINCIPLE 7.3 THE RATIONALE OF THE MAXIMUM PRINCIPLE 7.4 ALTERNATrVE TERMINAL CONDITIONS 7.5 THE CALCULUS OF VARIATIONS AND OPTIMAL CONTROL THEORY COMPARED 7.6 THE POLITICAL BUSINESS CYCLE 7.7 ENERGY USE AND ENVIRONMENTAL QUALITY

CHAPTER 8 MORE ON OPTIMAL CONTROL 8.1 AN ECONOMIC INTERPRETATION OF THE MAXIMUM PRINCIPLE 8.2 THE CURRENT-VALUE HAMILTONIAN 8.3 SUFFICIENT CONDITIONS 8.4 PROBLEMS WITH SEVERAL STATE AND CONTROL VARIABLES 8.5 ANTIPOLLUTION POLICY

CHAPTER 9 INFINITE-HORIZON PROBLEMS 9.1 TRANSVERSALITY CONDITIONS 9.2 SOME COUNTEREXAMPLES REEXAMINED 9.3 THE NEOCLASSICAL THEORY OF OPTIMAL GROWTH 9.4 EXOGENOUS AND ENDOGENOUS TECHNOLOGICAL PROGRESS

CHAPTER 10 OPTIMAL CONTROL WITH CONSTRAINTS 10.1 CONSTRATNTS INVOLVING CONTROL VARLABLES 10.2 THE DYNAMICS OF A REVENUE-MAXIMIZING FIRM 10.3 STATE-SPACE CONSTRAINTS 10.4 ECONOMIC EXAMPLES OF STATE-SPACE CONSTRAINTS 10.5 LIMITATIONS OF DYNAMIC OPTIMIZATION

ANSWERS TO SELECTED EXERCISE PROBLEMS INDEX