Chap-04 B.V.

Ramana August 30, 2006 10:13

Chapter 4
Calculus of Variations

4.1 INTRODUCTION function of the independent variable y(x), which is

a function. Thus
Calculus of variations deals with certain kinds of ! x2 "
“external problems” in which expressions involving L{y(x)} = 1 + y ′2 dx
integrals are optimized (maximized or minimized). x1
Euler and Lagrange in the 18th century laid the foun- Y
dations, with the classical problems of determining cn
B
a closed curve in the plane enclosing maximum area c2
subject to fixed length and the brachistochrone prob-
lem of determining the path between two points in c1
A
minimum time. The present day problems include
the maximization of the entropy integral in third law X
O
of thermodynamics, minimization of potential and Fig. 4.1
kinetic energies integral in Hamilton’s principle in defines a functional which associates a real number
mechanics, the minimization of energy integral in L uniquely to each y(x) (the independent variable).
the problems in elastic behaviour of beams, plates Further suppose we wish to determine the curve hav-
and shells. Thus calculus of variations deals with the ing shortest (least) distance between the two given
study of extrema of “functionals”. points A and B, i.e., curve with minimum length L.
Functional: A real valued function f whose This is a classical example of a variational problem
domain is the set of real functions {y(x)} is known in which we wish to determine, the particular curve
as a functional (or functional of a single independent y = y(x) which minimizes the functional L{y(x)}
variable). Thus the domain of definition of a func- given by (1). Here the two conditions y(x1 ) = y1 and
tional is a set of admissible functions. In ordinary y(x2 ) = y2 , which are imposed on the curve y(x) are
functions the values of the independent variables are known as end conditions of the problem. Thus varia-
numbers. Whereas with functionals, the values of the tional problems involves determination of maximum
independent variables are functions. or minimum or stationary values of a functional. The
Example: The length L of a curve, c whose equa- term extremum is used to include maximum or min-
tion is y = f (x), passing through two given points imum or stationary values.
A(x1 , y1 ) and B(x2 , y2 ) is given by
! x2 " 4.2 VARIATIONAL PROBLEM
L= 1 + y ′2 dx Consider the general integral (a functional)
x1
′
! x2
where y denotes derivative of y w.r.t. x. I {y(x)} = f (x, y, y ′ )dx (1)
Now the length L of the curve passing through x1
A and B depends on y(x) (the curve). Than L is a
4.1
Chap-04 B.V.Ramana August 30, 2006 10:13

4.2 MATHEMATICAL METHODS

Extremal: A function y = y(x) which extremizes

' (
f1 f2 δf1 −f1 δf2
(d) δ f2
= f22
(1) and satisfies the end conditions y(x1 ) = y1 and
dη
y(x2 ) = y2 is known as an extremal or extremizing (e) d
dx
(δy) = dx (ϵη) = ϵ dx
d
= ϵη′ =
function of the functional I (given by (1)). A
' (
dy
δy ′ = δ dx .
variational problem is to find such an extremal
Thus taking the variation of a functional and differ-
function y(x).
entiating w.r.t. the independent variable x are com-
mutative operations.
Variation of a Function and a Functional
When the independent variable x changes to x + !x Result: The necessary condition for the functional
then the dependent variable y of the function y = I to attain an extremum is that its variation vanish
f (x) changes to y + !y. Thus !y is the change of i.e., δI = 0.
the function, the differential dy provides the varia-
tion in y. Consider a function f (x, y, y ′ ) which for 4.3 EULER’S EQUATION
a fixed x, becomes a functional defined on a set of
functions {y(x)}. A necessary condition for the integral
For a fixed value of x, if y(x) is changed to y(x) + x2
!
I= f (x, y, y ′ )dx (1)
ϵη(x), where ϵ is independent of x, then ϵη(x) is x1
known as the variation of y and is denotd by δy.
Similarly, variation of y ′ is ϵη′ (x) and is denoted by to attain an extreme value is that the extremizing
δy ′ . Now the change in f is given by function y(x) should satisfy
# $
!f = f (x, y + ϵη, y ′ + ϵη′ ) − f (x, y, y ′ ) ∂f d ∂f
− =0 (2)
∂y dx ∂y ′
Expanding the first term on R.H.S. by Maclaurins
series in powers of ϵ, we get for x1 ≤ x ≤ x2 .
Note 1: The second order differential equation (2)
# $
∂F ∂F
!f = f (x, y, y ′ ) + η + ′ η′ ϵ +
∂y ∂y is known as Euler-Lagrange or simply Euler’s equa-
tion for the integral (1).
∂ 2 F 2 2∂ 2 F ′ ∂ 2 F ′2 ϵ 2
% &
+ η + ηη + ′2 η +
∂y 2 ∂yy ′ ∂y 2! Note 2: The solutions (integral curves) of Euler’s
equation are known as extremals (or stationary func-
+ · · · − F (x, y, y ′ )
tions) of the functional. Extremum for a functional
or approximately, neglecting higher powers of ϵ. can occur only on extremals.
∂f ∂f ∂f ∂f Proof: Assume that the function y = y(x), is
!f = ηϵ + ′ η′ ϵϵ = δy + ′ δy ′
∂y ∂y ∂y ∂y twice-differentiable on [x1 , x2 ], satisfies the end
(boundary) conditions y(x1 ) = y1 and y(x2 ) = y2
Thus the variation of a functional f is denoted by δf
and is given by and extremizes (maximizes or minimizes) the inte-
gral I given by (1). To determine such a function
∂f ∂f y(x), construct the class of comparison functions
δf = δy + ′ δy ′
∂y ∂y Y (x) defined by
which is analogous to the differential of a function. Y (x) = y(x) + ϵη(x) (2)
on the interval [x1 , x2 ]. For any function η(x), y(x) is
Result: (a) δ(f1 ± f2 ) = δf1 ± δf2
a member of this class of functions {Y (x)} for ϵ = 0.
(b) δ(f1 f2 ) = f1 δf2 + f2 δf1 Assume that
(c) δ(f η ) = ηf η−1 δf
η(x1 ) = η(x2 ) = 0 (3)
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.3

Differentiating (2), EQUIVALENT FORMS OF EULER’S

EQUATION:
Y ′ (x) = y ′ (x) + ϵη′ (x) (4) (I) Differentiating f , which is a function of x, y, y ′ ,
w.r.t. x, we get
Replacing y and y ′ in (1) Y and Y ′ from (2) and (4),
we obtain the integral df ∂f ∂f dy ∂f dy ′
= + + ′
dx ∂x ∂y dx ∂y dx
! x2
I (ϵ) = f (x, Y, Y ′ )dx (5) df ∂f ∂f ∂f
x1 = + y′ + y ′′ ′ (8)
dx ∂x ∂y ∂y
which is a function of the parameter ϵ. Thus the Consider
problem of determining y(x) reduces to finding the # $ # $
extremum of I (ϵ) at ϵ = 0 which is obtained by solv- d ∂f d ∂f ∂f
y′ ′ = y′ ′
+ ′ y ′′ (9)
ing I ′ (ϵ = 0) = 0. For this, differentiate (5) w.r.t. ϵ, dx ∂y dx ∂y ∂y
we get Subtracting (9) from (8), we have
∂f ∂Y ′
! x2 # $
dI ∂f ∂Y df d
#
∂f
$
∂f ∂f d
#
∂f
$
= I ′ (ϵ) = + dx − y′ ′ = + y′ − y′
dϵ x1 ∂Y ∂ϵ ∂Y ′ ∂ϵ dx dx ∂y ∂x ∂y dx ∂y ′
! x2 # $
∂f ∂f ′
= η+ η dx Rewriting this
x1 ∂Y ∂Y ′ . / . # $/
d ∂f ∂f ∂f d ∂f
f −y ′ ′ − =y ′ − (10)
putting ϵ = 0, dx ∂y ∂x ∂y dx ∂y ′
x2 Since by Euler’s Equation (2), the R.H.S. of (10) is
! # $
∂f ∂f
I ′ (0) = η + ′ η′ dx (6)
x1 ∂y ∂y zero, we get another form of Euler’s equtaion

because for ϵ = 0, we have from (2) Y = y and Y ′ =

. /
d ∂f ∂f
f − y′ ′ − =0 (11)
y ′ . Integrating the second integral in R.H.S. of (6) by dx ∂y ∂x
parts, we have
∂f ∂f
) $ + (II) Since ∂y ′
′ is also function φ of x, y, y say ∂y ′
=
x2 ∂f **x2
! * ! x2 #
′
′
I (0) =
∂f
η+ η − η
d ∂f
dx φ(x, y, y ). Differentiating w.r.t. x
x1 ∂y ∂y ′ *x1 x1 dx ∂y ′
∂φ dy ′
# $
d ∂f ∂φ ∂φ dy
= + + ′
Since by (3), η(x1 ) = η(x2 ) = 0, the second term dx ∂y ′ ∂x ∂y dx ∂y dx
vanishes and using I ′ (0) = 0, we get
# $ # $ # $
∂ ∂f ′ ∂ ∂f ′′ ∂ ∂f
= +y +y
! x2 , # $- ∂x ∂y ′ ∂y ∂y ′ ∂y ′ ∂y ′
∂f d ∂f
I ′ (0) = − η dx = 0 (7) d
#
∂f
$
∂ 2f ∂ 2f ∂ 2f
x1 ∂y dx ∂y ′ = + y′ + y ′′ ′2 (12)
dx ∂y ′ ∂x∂y ′ ∂y∂y ′ ∂y
Since η(x) is arbitrary, equation (7) holds good only
when the integrand is zero Substituting (12) in the Euler’s equation (2), we have

∂f ∂ 2f ∂ 2f ∂ 2f
− y′ − y ′′ ′2 = 0
# $
∂f d ∂f − (13)
i.e., − =0 (2) ∂y ∂x∂y ′ ∂y∂y ′ ∂y
∂y dx ∂y ′
General case: the necessary condition for the occur-
Note: Equation (2) is not sufficient condition. Solu- rence of extremum of the general integral
tion of (2) may be maximum or minimum or a hori- ! x2
zontal inflexion. Thus y(x) is known as extremizing f (x, y1 , y2 , . . . , yη , y1′ , y2′ , . . . , yη′ )dx
function or extremal and the term extremum includes x1

maximum or minimum or stationary value. involving η functions y1 , y2 , . . . , yη , is given by the

Chap-04 B.V.Ramana August 30, 2006 10:13

4.4 MATHEMATICAL METHODS

set of η Euler’s equations Geodesics: A geodesic on a surface is a curve on

∂f d
#
∂f
$ the surface along which the distance between any
− =0 two points of the surface is a minimum.
∂yi dx ∂yi′
for i = 1, 2, 3, . . . , η.
First integrals of the Euler-Lagrang’s equation: 4.4 STANDARD VARIATIONAL
Degenerate cases: Euler’s equation is readily inte- PROBLEMS
grable in the following cases:
Shortest distance
Case (a): If f is independent of x, then ∂f
∂x
= 0 and
equivalent form of Euler’s Equation (11) reduces to Example 1: Find the shortest smooth plane curve
d
#
∂f
$ joining two distinct points in the plane.
f − y′ ′ = 0
dx ∂y
Integrating, we get the first integral of Euler’s equa-
tion
∂f
f − y′ = constant (14)
∂y ′

Thus the extremizing function y is obtained as the

solution of a first-order differential equation (14)
involving y and y ′ only.
∂f
Case (b): If f is independent of y, then ∂y
= 0,
and the Euler’s Equation (2) reduces to
d
#
∂f
$ Fig. 4.2
=0
dx ∂y ′
Solution: Assume that the two distinct points be
Integrating, we get the first integral of the Euler’s
P1 (x1 , y1 ) and P2 (x2 , y2 ) lie in the XY -Plane. If y =
equation as,
f (x) is the equation of any plane curve c in XY -
∂f
= constant (15) Plane and passing through the points P1 and P2 , then
∂y ′ the length L of curve c is given by
which is a first order differential equation involving x2
! "
L[y(x)] = 1 + (y ′ )2 dx (1)
y ′ and x only. x1
Case (c): If f is independent of x and y then the The variational problem is to find the plane curve
∂f
partial derivative ∂y ′ is independent of x and y and is whose length is shortest i.e., to determine the func-
therefore function of y ′ alone. Now (15) of case (b) tion y(x) which minimizes the functional (1). The
∂f
= constant has the solution. condition for extrema is the Euler’s equation
∂y ′
y ′ = constant = c1
# $
∂f d ∂f
− =0
Integrating, the extremizing function is a linear func- ∂y dx ∂y ′
tion of x given by 0 ∂f ∂f 1 2y ′
Here f = 1 + y ′2 so ∂y
= 0, ∂y ′ = 2 √
y = c1 x + c 2 1+y ′2

∂f
Then
Case (d): If f is independent of y ′ , then ∂y ′
=0 d
%
y′
&
and the Euler’s Equation (2) reduces to 0− 0 =0
dx 1 + y ′2
∂f
=0 "
∂y or y ′ = k 1 + y ′2 where k = constant
Integrating, we get f = f (x) , i.e., function of x
Squaring y ′2 = k 2 (1 + y ′2 )
alone.
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.5

subject to the boundary conditions y(0) = 0 and

1
k2
i.e., y′ = = m = constant. y(x2 ) = y2 . Integral (1) is convergent although it is
1 − k2
improper. Here
Integrating, y = mx + c, where c is the constant of 0
integration. Thus the straight line joining the two 1 + y ′2
f = √
points P1 and P2 is the curve with shortest length y
(distance). which is independent of x. Now

Brachistochrone (shortest time) problem ∂f 1 1 1

= √ 0 · · 2y ′
∂y ′ y 1 + y ′2 2
Example 2: Determine the plane curve down
which a particle will slide without friction from the The Euler’s equation
point A(x1 , y1 ) to B(x2 , y2 ) in the shortest time. d
,
∂f
-
f − y′ ′ = 0
dx ∂y

reduces to
)0 +
d 1 + y ′2 y ′2
√ −√ 0 =0
dx y y 1 + y ′2

Integrating
1 + y ′2 1 + y ′2 − y ′2
0 0
√ 0 = k1 = constant
y 1 + y ′2
or y(1 + y ′2 ) = k2 (1)
' (2
Fig. 4.3 where k2 = k1 , put y ′ = cotθ where θ is a
1
parameter. Then from (1)
Solution: Assume the positive direction of the y-
axis is vertically downward and let x1 < x2 . Let y=
k2
=
k2 k2
= k2 sin2 θ = (1 − cos 2θ )
P (x, y) be the position of the particle at any time t, 1 + y ′2 1 + cot2 θ 2
on the curve c. Since energy is conserved, the speed (2)
v of the particle sliding along any curve is given by Now
k2
2 (+2 · sin 2θ )dθ
0
v = 2g(y − y ∗ ) dx =
dy
=
# $ y′ cotθ
∗ v12
where y = y1 − 2g
. Here g is acceleration due k2 2 · sin θ · cos θ dθ
= = 2k2 sin2 θ dθ
to gravity, v1 is the initial speed. Choose the origin cotθ
at A so that x1 = 0, y1 = 0 and assume that v1 = 0. dx = k2 · (1 − cos 2θ )dθ.
Then
Integrating, x = k2 θ − sin22θ + k3 , where k3 is
2 3
ds 0
= v = 2gy constant of integration. So
dt
Integrating this, we get the time taken by the particle k2
x − k3 = (2θ − sin 2θ ) (2)
moving under gravity (and neglecting friction along 2
the curve and neglecting resistance of the medium) Since y = 0 at x = 0, we have k3 = 0. Put 2θ = φ
from A(0, 0) to B(x2 , y2 ) is in (1) and (2), then
0
1 x=x2 1 + y ′2 k2 k2
! !
ds x= (φ − sin φ), y = (1 − cos 2φ) (3)
t[y(x)] = √ = √ √ dx (1)
2gy 2g x=0 y 2 2
Chap-04 B.V.Ramana August 30, 2006 10:13

4.6 MATHEMATICAL METHODS

Equation (3) represents a one parameter family of Substituting f and ∂f

, we have
∂y ′
cycloids with k22 as the radius of the rolling circle.
Using the condition that the curve (cycloid) passes y 1
"
y 1 + y ′2 − y ′ 0 · 2y ′ = c1
through B(x2 , y2 ), the value of the constant k2 can 2 1 + y ′2
be determined. y{(1 + y ′2 ) − y ′2 } y
0 = 0 = c1 (2)
1+y ′2 1 + y ′2
Note: A curve having this property of shortest time
is known as “brachistochrone” with Greek words Put y ′ = sinh t, then from (2)
‘brachistos’ meaning shortest and ‘chronos’ meaning y y
time. In 1696 John Bernoulli advanced this ‘brachis- = = c1 or y = c1 cosh t (3)
1 + sin2 ht cosh t
0
tochrone’ problem, although it was also studied by
Leibnitz, Newton and L’Hospital. dy c1 sinh t dt
So dx = ′ = = c1 dt
y sinh t
Minimal surface area Integrating x = c1 t + c2 (4)

where c2 is the constant of integration. Eliminating

Example 3: Find the curve c passing through two
‘t’ between (3) and (4)
given points A(x1 , y1 ) and B(x2 , y2 ) such that the
rotation of the curve c about x-axis generates a sur- x − c2
t=
face of revolution having minimum surface area. c1
x − c2
# $
therefore y = c1 cosh t = c1 cosh (5)
c1
Equation (5) represents a two parameter family of
catenaries. The two constants C1 and C2 are deter-
mined using the end (boundary) conditions y(x1 ) =
y1 and y(x2 ) = y2 .

Solid of revolution with least resistance

Example 4: Determine the shape of solid of revo-

lution moving in a flow of gas with least resistance.

Fig. 4.4

Solution: The surface area S generated by revolv-

ing the curve c defined by y(x) about x-axis is
! B ! x2 "
S[y(x)] = 2πy ds = 2πy 1 + y ′2 dx (1)
A x=x1
Fig. 4.5
To find the0extremal y(x) which minimizes (1).
Here f = y 1 + y ′2 which is independent of x. The Solution: The total resistance experienced by the
Euler’s equation is body is
d
#
∂f
$
∂f
! L
f − y′ ′ = 0 or f − y′ = constant = c1 F [y(x)] = 4πρv 2 yy ′3 dx
dx ∂y ∂y ′ 0
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.7

with boundary conditions y(0) = 0, y(L) = R. Here

4 ' (2
ρ is the density, v is the velocity of gas relative to Here f = a 1 + sin2 θ · dφ
dθ
is independent of φ,
solid. Here f = yy ′3 is independent of x. The Euler’s but is a function of θ and dφ
. Denoting dφ = φ ′ , the
dθ dθ
equation is Euler’s equation reduces to
# $
∂f d ∂f d
= y ′3 − (3yy ′2 ) = 0
# $
− (1) d ∂f ∂f
∂y dx ∂y ′ dx =0 or = constant.
dθ ∂φ ′ ∂φ ′
Multiplying (1) by y ′ , we get
1 1
d i.e., a · " · 2 · sin2 θ · φ ′ = k = constant
(yy ′3 ) = 0 2 2
1 + sin θ φ ′2
dx
Integrating Squaring a 2 sin4 θ · φ ′2 = k 2 (1 + sin2 θ · φ ′2 )
c1
yy ′3 = c13 or y′ = 1 dφ k kcosec2 θ
y3 or = φ′ = = 0
sin θ · sin2 θ − k 2
0
dθ 1 − c2 cosec2 θ
1
Integrating y 3 dy = c1 dx yields
Integrating, we get
4
y3
kcosec2 θ dθ
!
= c1 x + c2
4 φ(θ ) = + c2
3 (1 − k 2 ) − (kcotθ )2
0
3
or y(x) = (c3 x + c4 ) 4 (2) 5 6
−1 kcotθ
Using boundary conditions φ(θ ) = cos 0 + c2
1 − k2
0 = y(0) = 0 + c4 ... c4 = 0
where c2 is constant of integration. Rewriting
4
3 R3
R = y(L) = (c3 L) 4 ... c3 = kcotθ
L 0 = cos(φ − c2 ) = cos φ · cos c2 + sin φ · sin c2
1 − k2
The the required function y(x) is given by
' x (3 or cotθ = A cos φ + B sin φ
4
y(x) = R . 0
L (cos c2 )( 1 − k 2 )
where A = ,
k
Geodesics
0
( 1 − k2 )
Example 5: Find the geodesics on a sphere of B = (sin c2 )
k
radius ‘a’.
Multiplying by a sin θ , we have
Solution: In spherical coordinates r, θ, φ, the dif-
ferential of arc length on a sphere is given by a cos θ = A · a · sin θ · cos φ + B · a · sin θ · sin φ

(ds)2 = (dr)2 + (rdθ )2 + (r sin θ dφ)2 Since r = a, the spherical coordinates are x =
a sin θ cos φ, y = a sin θ sin φ, z = a cos θ , so
Since r = a = constant, dr = 0. So
# $2 # $2 z = Ax + By
ds 2 2 2 dφ
= a + a sin θ
dθ dθ which is the equation of plane, passing through ori-
Integrating w.r.t. θ between θ1 and θ2 , gin (0, 0, 0) (since no constant term) the centre of
1 sphere. This plane cuts the sphere along a great cir-
$2
θ2 cle. Hence the great circle is the geodesic on the
! #
dφ
s= a 1 + sin2 θ dθ
θ1 dθ sphere.
Chap-04 B.V.Ramana August 30, 2006 10:13

4.8 MATHEMATICAL METHODS

2
Solution: Here f = 1+y y ′2
which is independent of
WORKED OUT EXAMPLES x. So the Euler’s equation becomes
# $
d ∂f
f − y′ ′ = 0 (1)
Variational problems. dx ∂y
f is dependent on x, y, y ′ 2 2(1 + y 2 )
% &
∂f ∂ 1+y
Here = = − (2)
Example 1: Find a complete solution of the Euler- ∂y ′ ∂y ′ y ′2 y ′3
Lagrange equation for Substituting (2) in (1), we have
! x2 , -
2 ′ 2
1 + y2 2 1 + y2
% & % &
y − (y ) − 2y cosh x dx (1) d ′ (−2)(1 + y ) d
x1 −y =3 =0
dx y ′2 y ′3 dx y ′2
Solution: Here f (x, y, y ′ ) = y 2 − (y ′ )2 − y ′2 (2yy ′ ) − (1 + y 2 )2y ′ y ′′
2y cosh x, which is a function of x, y, y ′ . The =0
y ′4
Euler-Lagrange equation is
# $ or (1 + y 2 )y ′′ − yy ′2 = 0 (3)
∂f d ∂f
− =0 (2) ′
Put y = p, then y = ′′ d ′
y = d
p = dp dy
=
∂y dx ∂y ′ dx dx dy dx
y ′ dp
dy
= p dp
dy
. Putting these values in (3),
Differentiating (1) partially w.r.t. y and y ′ , we get
dp dp py
∂f (1 + y 2 )p − yp 2 = 0 or =
= 2y − 2 cosh x (3) dy dy 1 + y2
∂y
dp y dy 1 d(1 + y 2 )
∂f Integrating = =
= −2y ′ (4) p 1 + y2 2 (1 + y 2 )
∂y ′
p 2 = c12 (1 + y 2 ).
Substituting (3) and (4) in (2), we have
dy
" "
d so p = c1 (1 + y 2 ) or = c 1 + y2
2y − 2 cosh x − (−2y ′ ) = 0 dx
dx dy
Integrating = c1 dx we get
y ′′ + y = cosh x (5) 1 + y2
0

The complimentary function of (5) is sinh−1 y = c1 x + c2

yc = c1 cos x + c2 sin x Thus the required extremal function is
and particular integral of (5) is y(x) = sinh(c1 x + c2 )
1 where c1 and c2 are two arbitrary constant.
y= cosh x.
2
Thus the complete solution Euler-Lagrange Equation f is independent of y
(5) is
Example 3: If the rate of motion v = ds dt
is equal to
1 x then the time t spent on translation along the curve
y(x) = c1 cos x + c2 sin x + cosh x.
2 y = y(x) from one point P1 (x1 , y1 ) to another point
P2 (x2 , y2 ) is a functional. Find the extremal of this
f is independent of x functional, when P (1, 0) and P2 (2, 1).
Example 1: Find the extremals of the functional ds ds
Solution: Given = x or = dt.
! x2 (1 + y 2 ) dt x
I [y(x)] = dx 0 0
x1 y ′2 But ds = 1 + y ′2 dx so 1 + y ′2 dx
x
= dt.
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.9

Integrating from P1 to P2 Squaring (1) and (3) and adding

x 2 + (y − c3 )2 = (c2 sin v)2 + (−c2 cos v)2
0
x2 x2 1 + y ′2
! !
dt = dx. The functional is
x1 x1 x = c22 = c4 (4)
0
x2 1 + y ′2 Equation (4) represents a two-parameter family of
!
t[y(x)] = dx
x1 x circles. If (4) passes through P1 (1, 0) Then y(0) = 1.
√ Then (4) becomes
1+y ′2
Here f = which is independent of y. Euler’s
x7 8 1 + (0 − c3 )2 = c4 or 1 + c32 = c4
∂f
equation is d
dx ∂y ′
=0
If (4) passes through P2 (2, 1) then y(2) = 1.
So from (4),
5 6
d 1 1 1 ′
· ·0 · 2y = 0
dx x 2 1 + y ′2 4 + (1 − c3 )2 = c4 = 1 + c32 ... c3 = −2
. /
x (1 + y ′2 )y ′′ − y ′ (1 + y ′2 ) + xy ′ y ′′ and c4 = 5. Thus the required extremal satisfying the
0
end points P1 and P2 is
3
x 2 (1 + y ′2 ) 2
x 2 + (y + 2)2 = 5.
xy ′′ − y ′ (1 + y ′2 )
= 3
=0
x 2 (1 + y ′2 ) 2 Invalid variational problem
′′ ′ ′2
or xy − y (1 + y ) = 0. Example 4: Test for an extremum of the functional
′
Put y = u, then x dx
du
− u(1 + u2 ) = 0 ! 1
I [y(x)]= (xy + y 2 − 2y 2 y ′ )dx, with y(0)=1, y(1)=2.
du du udu dx 0
2
= − 2
=
u(1 + u ) u 1+u x
2 u 32 Solution: Here f = xy + y 2 − 2y 2 y ′ . Differenti-
Integrating x
= c12 (1 + u2 ) ating partially w.r.t. y and y ′ , we have
y = c12 x 2 (1 + y ′2 )
′2 ∂f
= x + 2y − 4yy ′ and
∂f
= −2y 2 .
" ∂y ∂y ′
or y ′ = c1 x (1 + y ′2 ).
Substituting these in the Euler’s equation
√
1 + y ′2 = 1 + tan2 v =
0
′
Put
√ y = tan v, then ∂f d
#
∂f
$
d
− = (x + 2y − 4yy ′ ) − (−2y 2 ) = 0
sec2 v ∂y dx ∂y ′ dx
y′ 1 tan v 1
so x= = = sin v (1) = x + 2y − 4yy ′ + 4yy ′ = 0
c1 (1 + y ′2 ) c1 secv c1
x
1 or x + 2y = 0 i.e., y = − .
and dx = cos v dv 2
c1
However, this function y = f (x) does not satisfy the
dy
Now = y ′ = tan v given boundary conditions y(0) = 1 and y(1) = 2
dx
i.e., 1 = y(0) ̸= 0 and 2 = y(1) ̸== − 21 . Thus an
1
dy = tan v dx = tan v · · cos v dv = extremum can not be achieved on the class of con-
c1
tinuous functions.
1
= sin v dv
c1 Geodesics
Integrating y = −c2 cos v + c3 (2)
1 Example 5: Determine the equation of the
where c2 = or y − c3 = −c2 cos v (3) geodesics on a right circular cylinder of radius ‘a’.
c1
Chap-04 B.V.Ramana August 30, 2006 10:13

4.10 MATHEMATICAL METHODS

Solution: In cylindrical coordinates (r, θ, z), the Then

differential of arc ds on a cylinder is given by # $2 # $2
ds dr
2 2 2 2 = + r 2 sin2 α
(ds) = (dr) + (rdθ ) + (dz) dφ dφ
Since radius r = a = constant, dr = 0. Then Integrating w.r.t., φ
1
# $2 # $2 # $2 φ2
1
$2
ds dz ds dz
! #
2 dr
=a + or = a2 + s= + r 2 sin2 α · dφ
dθ dθ dθ dθ φ1 dφ
Integrating The arc
4'length
(2
s of the curve is to be minimized. Here
1
+ r 2 sin2 α is independent of φ. Then
$2 dr
θ2 f =
! #
dz
s= a2 + dθ. dφ
θ1 dθ the integral of Euler’s equation is
Since geodesic is curve with minimum" length, we ∂f
2 dz 32 f − r ′ ′ = constant = k
2
have to find minimum of s. Here f = a + dθ ∂r
1 2r ′
"
which is independent of the variable z. Now the or r ′2 + r 2 sin2 α − r ′ · 0 =k
Euler’s equation is 2 r ′2 + r 2 sin2 α
# $ "
d ∂f ∂f
= 0 or
= constant = k r ′2 + r 2 sin2 α − r ′2 = k r ′2 + r 2 sin2 α
dθ ∂z′ ∂z′
5 # $2 6 squaring, r 4 sin4 α = k 2 (r ′2 + r 2 sin2 α)
∂f 0
2
dz 1 2 · z′
so = a + = 0 =k
∂z ′ dθ 2 a 2 + z′2 r 2 sin2 α(r 2 sin2 α − k 2 )
r ′2 =
k2
or z′2 = k 2 (a 2 + z′2 )
r sin α
"
dr
k2 a2 or = · r 2 sin2 α − k 2
z′2 = dφ k
1 − k2 kdr
dz ka i.e., = sin α · dφ.
r r 2 sin2 α − k 2
0
i.e., z′ = = 0
dθ 1 − k2 !
dr
Integrating k · = sin α · φ + c1
Integrating z(θ ) = √kaθ + c1 . Thus the equation of r r 2 sin2 α − k 2
0
1−k 2
the geodesics which is a circular helix is where c1 is the constant of integration. Introducing
∗
z = k θ + c1 and r=a
r = 1t , dr = − t12 dt, t = 1r , the L.H.S. integral trans-
forms to
ka
where k∗ = 0 . !
1
#
dt
$ !
dt
1 − k2 k· ·t " · − 2 = −k 0
sin2 α t 2
sin α − k 2 t 2
t2
− k2
Example 6: Find the geodesics on a right circular # $
cone of semivertical angle α. = cos−1
kt
.
sin α
Solution: In spherical coordinates (r, θ, φ) the dif- #
kt
$
ferential of an arc ds on a right circular cone is given Then cos−1 = φ sin α + c1
sin α
by
kt
(ds)2 = (dr)2 + (rdθ )2 + (r sin α dφ)2 . = cos(φ sin α + c1 )
sin α
With vertex of the cone at the origin and z-axis as k
Thus = cos(φ sin α + c1 )
the axis of the cone, the polar equation of cone is r sin α
θ = α = constant so dθ = 0. and θ = α are the equations of the geodesics.
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.11

Hint: EE: y ′′ + y = 2 cos x, y = c1 cos x +

EXERCISE c2 sin x + x sin x, c1 = 0, c2 = arbitrary
Ans. y = (C + x) sin x.
Variational problems
1. Test for extremum of the functional 4.5 ISOPERIMETRIC PROBLEMS
π
In calculus, in problems of extrema with constraints
! 'π (
2
I [y(x)] = (y ′2 − y 2 )dx, y(0) = 0, y = 1.
0 2 it is required to find the maximum or minimum of a
Hint: Euler’s Equation (EE): y ′′ + y = 0, y = function of several variably g(x1 , x2 , . . . , xη ) where
c1 cos x + c2 sin x using B.C, c1 = 0, c2 = 1 the variables x1 , x2 , . . . , xη are connected by some
given relation or condition known as a constraint.
Ans. y = sin x The variational problems considered so far find
Find the extremal of the following functionals the extremum of a functional in which the argument
9x
2. x12 (y 2 + y ′2 − 2y sin x)dx functions could be chosen arbitrarily except for pos-
sible end (boundary) conditions. However, the class
Hint: EE: 2y − 2 sin x − 2y ′′ = 0 of variational problems with subsidiary conditions
Ans. y = c1 ex + c2 e−x + sin2 x or constraints imposed on the argument functions,
91 apart from the end conditions, are branded as isoperi-
3. 0 (y ′12 + 12xy)dx, y(0) = 0, y(1) = 1. metric problems. In the original isoperimetric (“iso”
Hint: EE: y ′′ = 6x, y = x 3 + c1 x + C2 , C = for same, “perimetric” for perimeter) problem it is
0, c2 = 0 required to find a closed curve of given length which
Ans. y = x 3 enclose maximum area. It is known even in ancient
9π Greece that the solution to this problem is circle. This
4. 02 (y ′2 − y 2 + 2xy)dy, y(0) = 0, y π2 = 0
2 3
is an example of the extrema of integrals under con-
Hint: EE: y ′′ + y = x, y = c1 cos x + straint consists of maximumizing the area subject to
c2 sin x + x the constraint (condition) that the length of the curve
is fixed.
Ans. y = x − π2 sin x The simplest isoperimetric problem consists
9x
5. x12 (y 2 + 2xyy ′ )dx, y(x1 ) = y1 , y(x2 ) = y2 of finding a function f (x) which extremizes the
functional
Hint: EE: 2y + 2xy ′ − 2(xy ′ + y) = 0 i.e., ! x2
0=0 I [y(x)] = f (x, y, y ′ )dx (1)
x1
Ans. Invalid problem
92 3 subject to the constraint (condition) that the second
6. 1 yx′2 dx, y(1) = 1, y(2) = 4 integral
! x2
Ans. y = x 2 J [y(x)] = g(x, y, y ′ )dx (2)
9 3 ′2 x1
7. 2 yx 3 dx, y(2) = 1, y(3) = 16
assumes a given prescribed value and satisfying the
y ′′
Hint: EE: y′
= x3 , y ′ = cx 3 , y = c1 x 4 + c2 prescribed end conditions y(x1 ) = y1 and y(x2 ) =
3 4 y2 . To solve this problem, use the method of
Ans. y = 13 x − 35
9 x1 2
13 Lagrange’s multipliers and form a new function
8. x0 (y + y ′2 + 2yex ) dx
H (x, y, y ′ ) = f (x, y, y ′ ) + λg(x, y, y ′ ) (3)
−x 1
Ans. y = Ae + Be + x
2
xex where λ is an arbitrary constant known as the
9π
9. 0 (4y cos x − y 2 + y )dx, y(0) = 0, y(π ) =
′2
Lagrange multiplier. Now the problem is to find the
0 extremal of the new functional,
Chap-04 B.V.Ramana August 30, 2006 10:13

4.12 MATHEMATICAL METHODS

9x
I ∗ [y(x)] = x12 H (x, y, y ′ )dx, subject to no con- where t is the parameter. The area enclosed by curve
straints (except the boundary conditions). Then the C is given by the integral
modified Euler’s equation is given by 1
! t2
. .
I= (xy − x y)dt (2)
2
# $
∂H d ∂H t1
− =0 (4) . .
∂y dx ∂y ′ where x = dx , y = dy . We have x(t1 ) = x(t2 ) = x0
dt dt
The complete solution of the second order Equation and y(t1 ) = y(t2 ) = y0 , since the curve is closed.
(4) contains, in general, two constants of integration Now the total length of the curce C is given by
say c1 , c2 and the unknown Lagrange multiplier λ. ! t2 "
. .
These 3 constants c1 , c2 , λ will be determined using J= x 2 + y 2 dt (3)
t1
the two end conditions y(x1 ) = y1 , y(x2 ) = y2 and 1 . . . .
"
given constraint (2). Form H= (xy − x y) + λ x 2 + y 2 (4)
2
Corollary: Parametric form: To find the Here λ is the unknown Lagrangian multiplier. Prob-
extremal of the functional lem is to find a curve with given perimeter for which
t2 area (2) is maximum. Euler equations are
. .
!
I= f (x, y, x , y , t)dt # $
t1 ∂H d ∂H
− . = 0 (5)
∂x dt ∂x
subject to the constraint # $
∂H d ∂H
! t2
. . and − . =0 (6)
J = g(x, y, x , y , t)dt = constant ∂y dt ∂y
t1 . .
Differentiating H in (4) w.r.t. x, x , y, y and substi-
solve the system of two Euler equations given by tuting them in (5) and (6), we get
.
# $ # $ ⎛ ⎞
∂H d ∂H ∂H d ∂H 1. d ⎝ 1 λx
− . =0 and − . =0 y− − y+" ⎠=0 (7)
∂x dt ∂x ∂y dt ∂y 2 dt 2 . .
x2 + y2
resulting in the solution x = x(t), y = y(t), which
.
⎛ ⎞
is the parametric representation of the required func- 1. d ⎝1 λy
− x− x+" ⎠=0 (8)
tion y = f (x) which is obtained by elimination of t. 2 dt 2 . .
. . x2 + y2
Here x = dxdt
and y = dydt
and
. . . . . . Integrating (7) and (8) w.r.t. ‘t’, we get
H (x, y, x , y , t) = f (x, y, x , y , t) + λ g(x, y, x , y , t)
.
λx
The two arbitrary constants c1 , c2 and λ are deter- y−" = c1 (9)
. .
mined using the end conditions and the constraint. x2 + y2
.
λy
and x+" = c2 (10)
. .
4.6 STANDARD ISOPERIMETRIC x2 + y2
PROBLEMS
where c1 and c2 are arbitrary constants. From (9) and
Circle (10) squaring (y − c1 ) and (x − c2 ) and adding, we
get
Example 1: Isoperimetric problem is to determine ⎛
.
⎞2 ⎛
.
⎞2
−λy λx
a closed curve C of given (fixed) length (perimeter) (x − c2 )2 + (y − c1 )2 = ⎝ " ⎠ +⎝" ⎠
which encloses maximum area. . . . .
x 2 +y 2 x 2 +y 2
. .
Solution: Let the parametric equation of the curve (x 2 + y 2 )
= λ2 . 2 . 2 = λ2
C be (x + y )
x = x(t), y = y(t) (1) i.e., (x − c2 )2 + (y − c1 )2 = λ2
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.13

which is the equation of circle. Thus we have 1 2y ′

"
i.e., (y + λ)( 1 + y ′2 ) − y ′ (y + λ) · 0 = k1
obtained the well-known result that the closed curve 2 1 + y ′2
of given perimeter for which the enclosed area is a . / "
′2 ′2
maximum is a circle. (y + λ) (1 + y ) − y = k1 ( 1 + y ′2 )
"
Catenary or y + λ = k1 1 + y ′2 (4)

Put y ′ = sinh t, where t is a parameter, in (4)

Example 2: Determine the shape an absolutely
flexible, inextensible homogeneous and heavy
"
Then y + λ = k1 1 + sin2 ht = k1 cosh t (5)
rope of given length L suspended at the points A
and B dy k1 sinh t dt
Now dx = = = k1 dt
y′ sinh t
Integrating x = k 1 t + k2 (6)

Eliminating ‘t’ between (5) and (6), we have

x − k2
# $
y + λ = k1 cosh t = k1 cosh (7)
k1
Equation (7) is the desired curve which is a catenary.

Note: The three unknowns λ, k1 , k2 will be

determined from the two boundary conditions (curve
passing through A and B) and the constraint (2).
Fig. 4.6

Solution: The rope in equilibrium take a shape such WORKED OUT EXAMPLES
that its centre of gravity occupies the lowest position.
Thus to find minimum of y-coordinate of the centre
Example 1: 9 π Find the extremal of the function
of gravity of the string given by
I [y(x)] = 0 (y ′2 − y 2 )dx with boundary condi-
9 x2 0
′2 tions y(0)
9 π = 0, y(π ) = 1 and subject to the con-
x y 1 + y dx
I [y(x)] = 9 1x 0 (1) straint 0 y dx = 1.
2
x 1 + y ′2 dx
1
Solution: Here f = y ′2 − y 2 and g = y. So choose
subject to the constraint H = f + λg = (y ′2 − y 2 ) + λy where λ is the
! x2 " unknown Lagrange’s multiplier. The Euler’s equa-
J [y(x)] = 1 + y ′2 dx = L = constant (2) tion for H is
x1 # $
∂H d ∂H
Thus to minimize the numerator in R.H.S. of (1) sub- − =0
∂y dx ∂y ′
ject to (2). Form
" " " Using derivatives of H w.r.t. y and y ′ , we get
H = y (1 + y ′2 ) + λ 1 + y ′2 = (y + λ) 1 + y ′2 d
(3) (−2y + λ) − (2y ′ ) = 0
dx
where λ is Lagrangian multiplier. Here H is inde-
pendent of x. So the Euler equation is or y ′′ + y = λ

∂H whose general solution is

H − y′ = constant = k1
∂y ′ y(x) = CF + P I = (c1 cos x + c2 sin x) + (λ) (1)
Chap-04 B.V.Ramana August 30, 2006 10:13

4.14 MATHEMATICAL METHODS

The three unknowns c1 , c2 , λ in (1) will be deter- B.C’s y(1) = 3 and y(4) = 24 (i.e., passing through
mined using the two boundary conditions and the points P1 and P2 ) and the given constraint. From (1)
given constraint. From (1) λ
3 = y(1) = + c1 + c2 (2)
0 = y(0) = c1 + c2 · 0 + λ or c1 + λ = 0 4
Again from (1)
1 = y(π ) = −c1 + c2 · 0 + λ or − c1 + λ = 1
24 = y(4) = 4λ + 4c1 + c2 (3)
Solving λ = 21 , c1 = −λ = − 21
Now from the constraint
Now from the given constraint ! x2 =4
! π y(x)dx = 36
y dx = 1, we have x1 =1
0 ! 4 #λ $
π
or x 2 + c1 x + c2 dx = 36
!
(c1 cos x + c2 sin x + λ)dx = 1 1 4
0 *4
*π λ x3 x2
i.e., + c2 x ** = 36
*
c1 sin x − c2 cos x + λx ** = 1
* · + c1
4 3 2 1
0
(0 + c2 + λπ ) − (0 − c2 + 0) = 1 or 42λ + 60c1 + 24c2 = 288 (4)
π( From (2) & (3):
'
or 2c2 = 1 − π λ = 1 −
2
λ − c2 = 12
Thus the required extremal function y(x) is
and from (3) & (4)
1 1 π 1
# $
y(x) = − cos x + − sin x + . 2λ − c2 = 8
2 2 4 2
Solving λ = −4, c2 = −16, c1 = 20. Thus the spe-
Example 2: Show that the9 extremal of the isoperi- cific parabola satisfying the given B.C.’s (passing
x
metric problem I [y(x)] = x12 y ′2 dx subject to the through P1 and P2 ) is
9 x2
condition J [y(x)] = x1 y dx = constant = k is a 4
y = − x 2 + 20x − 16
parabola. Determine the equation of the parabola 4
passing through the points P1 (1, 3) and P2 (4, 24) and i.e., y = −x 2 + 20x − 16.
k = 36.

Solution: Here f = y ′2 and g = y. So form EXERCISE

′2
H = f + λg = y + λy.

The Euler equation for H is 1. Find the curve of given length L which joins
# $ the points (x1 , 0) and (x2 , 0) and cuts off from
∂H d ∂H the first quadrant the maximum area.
− =0
∂y dx ∂y ′
Ans. (x − c)2 + (y − d)2 = λ2 , c = x1 +x
2
2
,
d (x2 −x1 ) 2 2 2
√
λ− (2y ′ ) = 0 a = 2 ,λ = d + a , d + a 2 2
dx
cot −1 da = L2 .
2 3
λ
or y ′′ − = 0 2. Determine the curve of given length L which
2
Integrating twice, joins the points (−a, b) and (a, b) and gen-
erates the minimum surface area when it is
λ x2 revolved about the x-axis.
y(x) = + c1 x + c2 (1)
2 2
Ans. y = c cosh xc − λ, where c = a' (
, λ=
which is a parabola. Here c1 and c2 are constants of sin h−1 L
2
√
integration. To determine the particular parabola, use c
2
4 + L2 − b
Chap-04 B.V.Ramana August 30, 2006 10:13

CALCULUS OF VARIATIONS 4.15

9π
3. Find ′2
9 π 2the extremal of I = 0 y dx subject to 5. Find the curve of given length L which mini-
0 y dx = 1 and satisfying y(0) = y(π ) = 0
mizes the curved surface area of the solid gen-
Hint: EE: y ′′ − λy = 0 erated by the revolution of the curve about the
" x-axis.
Ans. yη (x) = ± π2 sin ηx, η = 1, 2, 3 . . . Ans. Catenary
4. Show that sphere is the solid of revolution 91
6. Determine y(x) for which 0 (x 2 + y ′2 )dx
which has maximum volume for a given sur- 91 2
is stationary subject to 0 y dx = 2, y(0) =
face area.
0, y(1) = 0.
Hint:√H = πy 2 + λ[(2πy) (1 + y ′2 )], EE:
0
Ans. y = ±2 sin mπ x, where m is an integer.
4λ2 −y 2
y′ = y
, (x − 2λ)2 + y 2 = (2λ)2 ; cir-
cle, solid of revolution sphere.