The Calculus of Variations

Jim Emery

Edited: 12/2/2013

Contents

1 About the History of the Calculus of Variations. 2

2 The Simplest Problem 3

3 A Necessary Condition for an Extremum, Eulers’s Differen-

tial Equation 8

4 The Multiple Variable Case 9

5 The Catenary 10

6 Example: A Catenary Solution for A Minimum Surface Prob-

lem. 11

7 Optics, Path of Minimum Time 14

8 Dispersion 14

9 Mechanics, The Principle of Least Action and Lagrange’s

Equations 14

10 Geodesics 14

11 Example: The Brachystochrone, the Curve of Minimum De-

scent Time 15

12 The Finite Element Method 15

1

13 Plateau’s Problem, Minimal Surfaces 15

14 Emmy Noether’s Theorem. 16

15 Problems 16

15.1 Coupled Oscillators Solved Using the Lagrangian . . . . . . . 16

16 Dirichlet’s Principle 17

17 Bibliography 17

18 Appendix A, Uniform Continuity, Compactness, and Differ-

entiating Under the Integral Sign 19

1 About the History of the Calculus of Vari-

ations.

The calculus of variations has a very long history stretching back to Fermat(1601 or 1607 to 1665) and Newton (1642 to 1727).

See the book by Goldstine, A History of the Calculus of Variations:

from the 17th through the 19th Century, which is listed in the bibli-ography and is in Linda Hall Library.

Goldstine takes as the beginning of the Calculus of Variations Fermat’sPrinciple of least time in optics. So according to Fermat a light ray takesa path through a medium where the velocity of light varies, in such a way,that the time of travel of the ray is minimized. Fermat deduced what is nowcalled Snell’s law of refraction for the bending or refraction of light, as a raypasses through an interface between two media. Fermat’s Principle can bededuced mathematically from Huygens’ Principle of secondary waves.

According to Goldstine Newton has a certain variational minimizationproblem that occurs in his principia, although the solution of this statedproblem was given without any derivation (I should look this up). In thehistory of mathematics, often problems are posed and then sometimes solved,but often not quite correctly. The solutions are improved over time as newconcepts and methods are introduced into mathematics. This makes suchhistory important and illuminating, showing how important mathematicalideas were created.

2

2 The Simplest Problem

The most common problem of the Calculus of Variations is the one dimen-sional problem of finding the extreme value of a functional consisting of theintegral of an expression involving a variable t, a function of f(t), and thederivative of the function f ′(t). The variation in the title is the variation ofthe function over the interval of integration. This has some similarity to theconcept of the differential of a dependent variable considered as a function ofsome independent variable about some fixed value of the independent vari-able. So the minimum or maximum value of such a single function occurs atthe point where the differential or variation vanishes. This is the same pointwhere the derivative of the function vanishes. In the calculus of variationsit is a ”function” acting as the independent variable, rather than a ”point”as independent variable in the case of elementary calculus. Such problemsoccur in determining shortest path or geodesic in geometry, or least time inoptics, or the path of motion in mechanics.

Let us define the problem. Consider a class of functions defined on theclosed interval [t1, t2] of the real line. These functions are to be in the classC2[t1, t2], meaning that the second derivatives are continuous. Let F bea function of three variables so that its partial derivatives of order 2 arecontinuous. We write DiF for the first partial derivative of F with respectto the ith coordinate. That is for example if i = 2, then

D2F (x1, x2, x3) =∂F (x1, x2, x3)

∂x2

.

We are given a functional G(f) defined by the integral

G(f) =∫ t2

t1F (t, f(t), f ′(t))dt

This functional maps a function f to a real number, which is the value ofthe integral.

The problem is to find a function f in the class C2[t1, t2] that producesan extreme value of the functional, either a minimum or a maximum.

We shall show below that a necessary condition for an extremum is thatthe function f satisfy Euler’s differential equation

D2F (t, f(t), f ′(t)) − d

dt[D3F (t, f(t), f ′(t))] = 0.

3

Let us put this in a somewhat more intuitive form. So suppose we usethe variables x, y, y′ in the function F . So that we have F (x, y, y′), where xis our independent variable, y is a function of x and y′ is the derivative of y,also a function of x. Then Euler’s equation for an extremum of the problemabove is

∂F

∂y− d

dx

[

∂F

∂y′

]

= 0.

In a certain sense this is more intuitive, but also is potentially mathemat-ically confusing, because it contains a derivative with respect to a variabley′, but which appears to be a derivative of y, and so not an independentvariable. But the first Euler’s equation clarifies the meaning.

Let us pause here to solve maybe the most simple problem of this typeusing Euler’s equation.

Problem . Find the curve that minimizes the distance between two pointsin the plane.

We know of course, that the solution to this problem is the straight linejoining the two points. However, we shall find the solution as a solution toEuler’s equation in order to illustrate the method.

So suppose the curve is to pass through the points (x1, y1) and (x2, y2).Suppose x1 is not equal to x2. We calculate the distance between the pointswith the integral

∫ x2

x1

√

1 + y′2dx.

Note that the arclength increment we are using is

ds =√

dx2 + dy2 =√

1 + (dy/dx)2dx.

Our function F (x, y, y′) =√

1 + y′2.So the Euler differential equation for this problem is

∂F

∂y− d

dx

[

∂F

∂y′

]

= 0,

which reduces to

d

dx

[

∂F

∂y′

]

= 0,

4

because here F is not a function of y.So

∂F

∂y′= c

for some arbitrary constant c.We have

∂F

∂y′=

y′

√1 + y′2

Soy′

√1 + y′2

= c.

Squaring and simplifying we have

y′ =c√

1 − c2.

Then integrating we have

y =c√

1 − c2x + b,

ory = mx + b,

which is the equation of a straight line, where constants c and b are cho-sen to make the line pass through the given points (x1, y1) and (x2, y2).

Notice that m = c/√

1 − c2 takes on all real values for −1 < c < 1.Problem . Find the equation of motion for a particle under a central forcefield, using the Lagrangian.

First let us write down the differential equation for this problem usingNewton’s second law. We assume that there is a central force directed to theorigin of the coordinate system. Consider the motion of a particle of massm located at

r = xi + yj + zk.

The norm or length of this vector is

‖r‖ =√

x2 + y2 + z2

5

We also write this as r. A unit vector directed toward the particle is

ur =r

r

We have for our differential equation of motion

d2r

dt2= − ur

x2 + y2 + z2,

which also can be written as

d2r

dt2= − r

(x2 + y2 + z2)3/2,

where we have assumed the constant multiplying the force on the left is 1 andthat the particle has mass 1. We do this in order to avoid carrying constantsalong in all of the equations.

The component equations are

d2x

dt2= − x

(x2 + y2 + z2)3/2,

d2y

dt2= − y

(x2 + y2 + z2)3/2,

andd2z

dt2= − z

(x2 + y2 + z2)3/2.

The Lagrangian in mechanics is

L = T − V

where T is the Kinetic Energy and V is the potential energy. The curve ofmotion is given as the curve that minimizes the integral of time of L. Thesolution of this variational problem is a solution of the Euler equations, whichare called Lagrange’s equations in mechanics.

Again let a particle have mass 1. Let the position vector of the particlebe

r = xi + yj + zk

6

Suppose there is a central force inversely proportional to the distance to theorigin r = ‖r‖ Then the potential energy is

V = −1

r= − 1√

x2 + y2 + z2.

The difference in potential energy between two locations is computed bycomputing the work done in moving a particle between the locations. Thekinetic energy is

T =1

2(x′2 + y′2 + z′2).

So

L =1

2(x′2 + y′2 + z′2) +

1√x2 + y2 + z2

Lagrange’s equation for coordinate x is

∂L

∂x− d

dt

[

∂L

∂x′

]

= 0.

∂L

∂x= −x(x2 + y2 + z2)−3/2

∂L

∂x′= x′

d

dt

[

∂L

∂x′

]

= x′′

So our differential equation is

x′′ = − x

(x2 + y2 + z2)3/2.

This can also be written as

x′′ = −cos(θx)

r2,

where cos(θx) is the direction cosine of r with respect to the x-axis.

7

That is the acceleration in the x direction is the component of the centralforce in the direction of the x axis, which is just Newton’s law.

The equations for y and z are similar. So we have just demonstrated herethat Lagrange’s equations are the Euler-Lagrange equations for the varia-tional problem of minimizing the integral of L, and that these equations areequivalent to Newton’s second law fob this particular problem. In this specialcase the differential equation is of course already known. In general the vari-ational method can result in huge simplification of the mechanics problemand often works so that forces of constraint need not be computed.

3 A Necessary Condition for an Extremum,

Eulers’s Differential Equation

So consider our problem of finding the extremum (minimum or maximum)of the functional

G(f) =∫ t2

t1F (t, f(t), f ′(t))dt.

Suppose that f does give the extremum. Let η(t) be an arbitrary fixedC2 function that vanishes at t1 and at t2. Consider the function

H(ε) = G(f + εη) =∫ t2

t1F (t, f(t) + εη(t), f ′(t) + εη′(t))dt.

The derivative of H isdH(ε)

dε=

∫ t2

t1[D2F (t, f(t)+εη(t), f ′(t)+εη′(t))η(t)+D3F (t, f(t)+εη(t), f ′(t)+εη′(t))η′(t)]dt.

When ε is zero we get

dH(0)

dε=

∫ t2

t1[D2F (t, f(t), f ′(t))η(t) + D3F (t, f(t), f ′(t))η′(t)]dt.

But at ε = 0 we have the original functional G(f) at its extremal value,because we have assumed the extremal value occurs for the function f . Sothe derivative of H is zero at 0. Thus

∫ t2

t1[D2F (t, f(t), f ′(t))η(t) + D3F (t, f(t), f ′(t))η′(t)]dt = 0.

8

We apply integration by parts to the second part of the integral

∫ t2

t1[D3F (t, f(t), f ′(t))η′(t)]dt

= D3F (t, f(t), f ′(t))η(t2) − D3F (t, f(t), f ′(t))η(t1)

−∫ t2

t1[d

dtD3F (t, f(t), f ′(t))η(t)]dt

= −∫ t2

t1[d

dtD3F (t, f(t), f ′(t))η(t)]dt,

because η vanishes at t1 and t2.So we have

∫ t2

t1[D2F (t, f(t), f ′(t)) − d

dtD3F (t, f(t), f ′(t))]η(t)dt = 0.

But this forces

D2F (t, f(t), f ′(t)) − d

dtD3F (t, f(t), f ′(t)) = 0,

because η(t) is an arbitrary function which could be selected to make theintegral nonzero if

D2F (t, f(t), f ′(t)) − d

dtD3F (t, f(t), f ′(t))

were not identically zero. This is Euler’s equation.

4 The Multiple Variable Case

The problem of finding the extremum of

G(u1, u2, ..., un) =∫ t2

t1F (t, u1(t), u

′

1(t), ..., un(t), u′

n(t))dt.

is similar to the one dimensional problem given above.We look at the variation

Ui(t) = ui(t) + εηi(t), i = 1, ...n,

9

where each ηi is an arbitrary C2 function that vanishes at t1 and t2. Thenwe calculate the derivative

H(ε) = G(U1, .., Un) =∫ t2

t1F (t, U1(t), U

′

1, ..., Un(t), U ′

n)dt,

in a manner similar to the one dimensional case above. We arrive at n Eulerequations

∂F

∂ui− d

dt

[

∂F

∂u′

i

]

= 0, i = 1, ..., n.

by setting H(0) = 0 as above in the one dimensional case.To isolate the ith Euler Differential Equation we are free to choose each

function ηj to be identically zero when j is not equal to i.

5 The Catenary

The catenary is the curve that gives the shape of a hanging cable. Thecatenary satisfies the differential equation

d2y

dx2=

w

T0

ds

dx=

1

a

√

1 + (dy/dx)2,

where the origin is at the lowest point of the catenary, the horizontal tensionthere is T0, w is the weight per unit length of the cable, and where thecatenary constant is

a =T0

w,

One finds that the solution of this equation is

y(x) = a cosh(x/a).

See Estimating the Parameters of a Catenary, by James Emery,catenary.tex, catenary.pdf, 2003.

http://www.stem2.org/je/catenary.pdf

10

6 Example: A Catenary Solution for A Min-

imum Surface Problem.

Let a curve be rotated about the x axis generating a surface. With a fixedstarting point and a fixed ending point for the curve, what is the surface ofminimum area?

This is a problem that is given as an example in many books. The solutionis a catenary passing through the two fixed points.

The area of a surface of revolution about the x axis is

S = 2π∫ x2

x1

yds = 2π∫ x2

x1

y√

1 + y′2dx.

Ignoring the 2π we have

S =∫ x2

x1

y√

1 + y′2dx,

whereF (x, y, y′) = y

√

1 + y′2

∂F

∂y=

√

1 + y′2

∂F

∂y′=

yy′

√1 + y′2

.

Euler’s equation is

√

1 + y′2 − d

dx

[

yy′

√1 + y′2

]

= 0.

The expanded second term is

− y′2

√1 + y′2

− yy′′

√1 + y′2

+yy′2y′′

(1 + y′2)3/2.

So Euler’s equation becomes

√

1 + y′2 − y′2

√1 + y′2

− yy′′

√1 + y′2

+yy′2y′′

(1 + y′2)3/2= 0.

11

Multiplying by√

1 + y′2, we have

1 + y′2 − y′2 − yy′′ +yy′2y′′

1 + y′2= 0,

so

1 − yy′′ +yy′2y′′

1 + y′2= 0.

Multiplying by 1 + y′2 we have

1 + y′2 − yy′′(1 + y′2) + yy′2y′′ = 0.

Then1 + y′2 − yy′′ = 0.

Let

y′ =dy

dx

y′′ =dy′

dx=

dy′

dy

dy

dx=

dy′

dyy′

y is a function of x, and y′ is a function of x, so assuming that y has aninverse, x is a function of y. Then y′ is a function of y, so that

dy′

dy

makes sense.Substituting in the simplified Euler equation we found above we get

yy′dy′

dy− y′2 − 1 = 0

we have

yy′dy′

dy= y′2 + 1

y′dy′

1 + y′2=

dy

y.

12

Integrating both sides we have

1

2ln(1 + y′2) = ln(y) + c,

for some constant c. Then

ln(1 + y′2) = 2 ln(y) + 2c.

So defining a new constant c1 by c = − ln(c1), we get

ln(1 + y′2) = 2 ln(y) − 2 ln(c1) = ln

[

y2

c21

]

.

Then

1 + y′2 =y2

c21

.

y′2 =y2 − c2

1

c21

so

y′ =

√

y2 − c21

c1

ordy

√

y2 − c21

=dx

c1

.

Thencosh−1(y/c1) =

x

c1

+ c2

and so

y = c1 cosh(x/c1 + c2).

Now constants c1 and c2 need to be determined numerically to make thecurve pass through the given points (x1, y1) and (x2, y2), if possible.

13

7 Optics, Path of Minimum Time

Fermat’s principle states that the path of a ray of light is the path thatminimizes the travel time. Fermat applied this idea to prove Snell’s law ofrefraction, which does not of course require the Calculus of variations becausethere is just a noncontinuous change in velocity at the interface between twooptical media. Huygens principle, which allows the construction of a newwave front by assuming that each point of the previous front is a source ofspherical waves, implies Fermat’s principle.

8 Dispersion

In the optics case in a variable dispersive medium, the velocity of light mayvary continuously and so the calculus of variations is required to computeray paths

9 Mechanics, The Principle of Least Action

and Lagrange’s Equations

See James Emery Mechanics, mechanics.pdf, mechanics.tex.

http://www.stem2.org/je/mechanics.pdf

10 Geodesics

A geodesic is a curve of minimum length connecting two points. Geodesicson a sphere are great circles. Geodesics are determined using variational ar-guments. In general relativity, geometry is bent by mass-energy. Light takesthe path that is a geodesic in this bent Riemannian space-time geometry. Sofor example one of the early verifications of general relativity was to observethe bent path of light as it passed near the gravitational field of the sun.This was done famously during a solar eclipse.

14

11 Example: The Brachystochrone, the Curve

of Minimum Descent Time

See James Emery, Brachystochrone, bra.pdf, bra.tex.

http://www.stem2.org/je/bra.pdf

12 The Finite Element Method

The Finite Element Method is a technique for solving certain partial differ-ential equations numerically. It consists in breaking up the solution domaininto small elements which have node points that define an interpolation func-tion inside the element. In a simple example the elements might be triangleswith the vertices as nodes. An interpolation function can be defined in theelement using values at nodes. An approximation to the solution of the par-tial differential equation is computed inside the element using a variationaltechnique. Then the solutions in the elements are glued together giving asolution to the entire boundary value problem.

See James Emery A Finite Element Current Flow Program, fec.tex,fec.pdf

http://www.stem2.org/je/fec.pdf

(This document was written in 1980 originally, and has only partially beenconverted to tex. The finite element program itself ran on a CDC computerwith limited memory, using an old version of a Fortran compiler that did notsupport dynamic memory allocation. If I can find the time I will convert itto a modern program. )

13 Plateau’s Problem, Minimal Surfaces

Plateau’s Problem is to find the surface of minimal area that has a givenclosed curve as a boundary. Such surfaces are the shape of soap films obtainedby dipping a wire loop into a soap solution. See the entertaining lectures byBoys given to children in the late nineteenth century Soap Bubbles and

the Forces that Mold Them available at the Project Gutenberg site.

15

14 Emmy Noether’s Theorem.

(Emmy Noether 1882-1935, German pronunciation: em ee no ter, englishno ther, or, no eh ther) According to Noether’s Theorem, symmetries ofthe Lagrangian in physics imply certain conservation laws. This is widelyapplied in particle physics. Noether’s Theorem shows how symmetries ofthe Lagrangian can be used to construct constants of the motion from theLagrangian. See chapter 5 of Analytical Mechanics by Hand and Finch.This chapter is titled Noether’s Theorem and Hamiltonian Dynamics. Themathematics used is very nonrigorus in this treatment, which might makea mathematician quite uncomfortable. For a more advanced treatment seeChapter 4, of Mariano Giaquinta, Stefan Hildebrandt, Calculus of Varia-

tions I: The Lagrangian Formalism Volume I. The arguments used inNoether’s theorem are perhaps the source of the entry of Lie Groups intophysics. Emmy Noether is one of the great mathematicians of the 20thcentury and pioneered much of modern Commutative Algebra. Noetheriangroups are named after her. Most mathematicians know her for the algebraicwork, and do not know her big influence on physics.

For example using this theorem applied to Quantum Mechanics, one canestablish the conservation of charge in particle physics.

15 Problems

15.1 Coupled Oscillators Solved Using the Lagrangian

See the section on coupled oscillators in Vibrations by James Emery (vi-bra.pdf, vibra.tex).

http://www.stem2.org/je/vibra.pdf

Coupled oscillators are physically demonstrated in the Wilberforce Pen-dulum, and in a pair of lightly coupled ordinary pendulums. The Wilberforcependulum consists of a weight suspended from a spring. The weight has twomodes of oscillation, an up and down motion, and a rotary motion. If thespring and the mass are chosen carefully so that the moment of inertia ofthe weight and its mass give nearly equal periods of vibration in the upand down, and in the rotary motion, the pendulum will repeatedly switchoscillation modes and energy, between translational and rotory motion.

16

16 Dirichlet’s Principle

Roughly speaking A solution φ of Laplace’s equation in a bounded region Cis a minimum of the Dirichlet Integral. For example, in two dimensions asolution φ is the function that minimizes the Dirichlet integral

∫

C

[

∂φ

∂x

]

2

+

[

∂φ

∂y

]

2

dxdy.

This is a two dimensional variational problem.

17 Bibliography

[1] Weinstock Robert, Calculus of Variations, Dover reprint, 1974.

[2] Wylie C. Ray, Advanced Engineering Mathematics, 4th Edition,1975, McGraw-Hill.

[3] Sagan Hans, Introduction to the Calculus of Variations, McGraw-Hill, 1969, Dover reprint, 1992.

[4] Hildebrand F. B. , Methods of Applied Mathematics, Prentice-Hall,1952, 9th printing 1963.

[5] Young L.C. , Lectures on the Calculus of Variations and Optimal

Control Theory, 1969, W. B. Saunders Company.

[6] Caratheodory C., Variationsrechnung Teubner, Berliin and Leipzig,1935. (English Translation, R. B. Dean and J. J. Brandstatter: Calculus

of Variations and Partial Differential Equations of the First Order,Holden-Day, 1965.)

[7] Mariano Giaquinta, Stefan Hildebrandt, Calculus of Variations I:

The Lagrangian Formalism, Springer, 1996, 2nd printing 2004, Chap-ter 4, Emmy Noether’s Theorem. (QA315 .G467 Linda Hall). (Volume 310,Grundlehren Der Mathematischen Wissenschaften.)

17

[8] Goldstine Herman H, A History of the Calculus of Variations: from

the 17th through the 19th Century, Springer-Verlag, 1980, (QA315 .G58Linda Hall).

[9] Almgren F J, Plateau’s Problem, W A Benjamin 1966.

[10] Osserman Robert, A Survey of Minimal Surfaces, Van NostrandReinhold, 1969.

[11] Courant Richard and Hilbert David, Methods of Mathematical Physics,Volumes I and II, 1937 Springer and 1953 Interscience, Chapt 4, Vol. I, Cal-culus of Variations.

[12] Boys C V, Soap Bubbles and the Forces Which Mold Them,Doubleday Anchor 1959, Originally three Lectures delivered at the LondonInstitution in December 1889 and January 1890. Available from ProjectGutenberg. The HTML file has very nice illustrations.

[13] Hand Louis N and Finch Janet D, Analytical Mechanics, CambridgeUniversity Press, 1998.

[14] Emery James, Optimization, (optimization.pdf, optimization.tex,) at

http://stem2.org/je/optimization.pdf

[15] Kellogg Oliver Dimon, Foundations of Potential Theory, Dover 1953.

18

[16] Oden J T, Reddy J N, Variational Methods in Theoretical Me-

chanics, Springer-Verlag, 1970.

[17] Lanczos Cornelius, The Variational Principles of Mechanics, FourthEdition, Toronto University Press, 1970, Dover reprint 1986.

[18] Goldstein Herbert, Classical Mechanics, Addison-Wesley, 1950, 7thprinting, 1965, (there is a second and third edition).

[19] Bradbury T C, Theoretical Mechanics, John Wiley, 1968.

[20] Abraham Ralph and Marsden Jerrold E, Foundations of Mechanics,Benjamin/Cummings 1978.

[21] Apostol Tom M, Mathematical Analysis, Second Edition, 1975, Addison-Wesley.

[22] Widder David V, Advanced Calculus, 2nd Edition, 1961, Pentice-Hall.

[23] Flanders, Harley (JuneJuly 1973). Differentiation under the inte-

gral sign, American Mathematical Monthly 80 (6): 615627.

[24] Goldberg Richard R, Methods of Real Analysis, Blaisdell, 1965.

18 Appendix A, Uniform Continuity, Com-

pactness, and Differentiating Under the

Integral Sign

Definition. A real function defined on set A is uniformly continuous if givenany ε > 0 there exists a δ > 0 independent of any x0 ∈ A and x ∈ A, so thatfor any x ∈ A and x ∈ A, such that

|x − x0| < δ,

then|f(x) − f(x0)| < ε.

19

Example. Consider the function f(x) = −1/x defined on (0, 1]. Notice thatas x approaches 0, the slope of f(x) becomes arbitrarily large. So given anyε > 0, a pair of points x and x0 can be chosen near zero so that for any δ > 0,no matter how small,

|x − x0| < δ,

but|f(x) − f(x0)| > ε.

So this function f(x) is not uniformly continuous on (0, 1], although it iscontinuous everywhere on (0, 1].

However, if f(x) is continuous and A is a compact set then f(x) is uni-formly continuous. We won’t take the time here to introduce the conceptof compactness, except to say that a closed interval on the real line, and aclosed rectangle in the plane are compact sets. The reader is directed to abook on Analysis, such as the book in the bibliography by Goldberg, or to abook on Topology.

Goldberg Richard R, Methods of Real Analysis, Blaisdell, 1965.

In deriving the necessary condition for a minimum above we differentiatedunder the integral sign. We will justify this below. Here are some referencesconcerning differentiating under the integral sign:

Apostol Tom M. , Mathematical Analysis, Second Edition, 1975, Addison-Wesley, section 10.16.

Widder David V, Advanced Calculus, 2nd Edition, 1961, Pentice-Hall,Chapter 10, article 7.2.

Flanders, Harley (June-July 1973). Differentiation under the integral

sign, American Mathematical Monthly 80 (6): 615627.

Note. The section on differentiation under the integral sign in Apostol usesthe Lebesgue Integral.

Theorem. Differentiating Under the Integral Sign. Consider the integral

20

F (x) =∫ b

af(x, t)dt.

If f(x, t) and its derivative

f1(x, t) =∂f

∂x

are continuous on the closed rectangle [a, b] × [c, d], then F is differentiableon the closed interval [a, b]. The derivative is

dF

dx=

∫ b

af1(x, t)dt,

obtained by differentiating under the integral sign.

Proof. The derivative of F is the limit as h → 0 of

F (x + h) − F (x)

h=

∫ b

a

f(x + h, t) − f(x, t)

hdt.

By the mean value theorem the integrand on the right can be written as

f(x + h, t) − f(x, t)

h= f1(x̄, t),

where x̄ is strictly between x and x+h, and depends upon t. x̄ can be writtenas

x̄ = x + αh,

where 0 < α < 1, and again depends on t. So we can write

F (x + h) − F (x)

h−

∫ b

af1(x, t)dt =

∫ b

a[f1(x + αh, t) − f1(x, t)]dt.

Then∣

∣

∣

∣

∣

F (x + h) − F (x)

h−

∫ b

af1(x, t)dt

∣

∣

∣

∣

∣

≤∫ b

a|f1(x + αh, t) − f1(x, t)|dt.

f1 is continuous on the compact rectangle [a, b]× [c, d], so uniformly con-tinuous there.

21

So given ε > 0 there exists a δ > 0 so that if |x1 − x| < δ, then

|f1(x1, t) − f1(x, t)| < ε/(b − a).

In particular, if |h| < δ then α|h| < δ, and so

|f1(x + αh, t) − f1(x, t)| < ε/(b − a).

So if |h| < δ then

∣

∣

∣

∣

∣

F (x + h) − F (x)

h−

∫ b

af1(x, t)dt

∣

∣

∣

∣

∣

≤∫ b

a|f1(x+αh, t)−f1(x, t)|dt <

∫ b

aε/(b−a)dt = ε,

which proves the theorem.

22