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DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS

F(u) = u3 + lu2 + mu + n

Let

(F(u)) = 3u2 + 2lu + m

∂u

The equation of the envelope of the given family is obtained by eliminating u from the equations

(F(u)) = 0.

F(u) = 0 and

∂u

∴ u3 + lu2 + mu + n = 0

...(2)

2

3u + 2lu + m = 0

...(3)

Multiplying (2) by 3 and (3) by u, we get

3u3 + 3lu2 + 3mu + 3n = 0

...(4)

3u3 + 2lu2 + mu = 0

...(5)

2

(4) – (5) ⇒ lu + 2mu + 3n = 0

...(6)

Solving (3) and (6), we get

u2

6ln − 2m

2

=

u2 =

Eliminating u, we get

F ml − 9n I

GH 6m − 2l JK

2

2

=

u

1

= ml − 9n

6m − 2l 2

6ln − 2m 2

6m − 2l

2

=

3ln − m 2

3m − l

2

and

u=

ml − 9n

6m − 2l 2

3ln − m 2

3m − l 2

(ml – 9n)2 = 4(3ln – m2)(3m – l2).

This is the equation of the required equation of the envelope, where l, m and n are as given above.

6. IMPORTANT RESULTS

I. The envelope of the two-parameter family of surfaces F(x, y, z, a, b) = 0, where a, b are parameters is found by eliminating a and b from the equations:

∂f

∂f

= 0,

= 0.

∂a

∂b

The envelope of the two-parameter family of surfaces F(x, y, z, a, b) = 0, where a, b are parameters connected by the equation φ(a, b) = 0 is found by eliminating a and b from the equations:

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y = c1 cosh

x + c2 c1

minimises the functional i.e., the area of the surface of

revolutions by the curve between the given points.

Example 14. Show that the geodesics* on a sphere of radius a are its great circles.

Sol. Let A and B be any two points on the given sphere with origin at the centre of the sphere. Let (a, θ1, φ1) and (a, θ2, φ2) be their spherical coordinates** respectively. Let φ(θ) be a function whose curve passes through A and B and lying itself on the surface of the given sphere.

∴ The length of the arc between A and B is given by

z

θ2

θ1

ds .

Since, we are dealing with spherical coordinates, we have

ds =

(dr) 2 + (r dθ) 2 + (r sin θ dφ) 2

ds =

0 + a 2 ( dθ) 2 + a 2 sin 2 θ . φ′ 2 ( dθ) 2

(Note this step)

(∵ r = a ⇒ dr = 0)

= a 1 + φ ′ 2 sin 2 θ dθ

∴ Length of arc

=

z

θ2

θ1

a 1 + φ ′ 2 sin 2 θ dθ

Let

F(θ, φ, φ′) = a 1 + φ ′ 2 sin 2 θ

Let the length of the curve between A and B has a minimum value for this curve. In other words, let φ(θ) be a geodesic between A and B.

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Variational Problems with

Fixed Boundaries

1. INTRODUCTION

By a function we mean a correspondence between the elements of sets A and B such that to each element of A there correspond exactly one element of the set B. We have also studied the methods of finding maximum and minimum values of real functions of one or more variables. In calculus of variations, we shall deal with functionals and their maximum and minimum values. A functional associates a unique real number to each function belonging to a certain class of functions.

2. FUNCTIONAL

Let A be a class of functions. A correspondence between the functions of class A and the set of real numbers such that to each function belonging to A, there correspond exactly one real number is called a functional. In other words, a functional is a correspondence which assigns a unique real number to each function belonging to some class. Thus a functional is a function, where the independent variable takes functions as values.

A functional is denoted by capital letter I (or J). If y(x) represents a function in the class of functions of a functional J, then we write J = J[y(x)].

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.

.

.

.

.

.

(Fx) x + (Fy) y + (Fz) z = 0

(Gx) x + (Gy) y + (Gz) z = 0

and

.

.

.

Solving these equations for x, y and z , we get

.

.

.

x y z

=

=

Fy G z − Fz G y Fz G x − Fx G z Fx G y − Fy G x

FyGz – FzGy , FzGx – FxGz , FxGy – FyGx are also d.r.’s of the tangent to the curve given by the equations F(x, y, z) = 0, G(x, y, z) = 0 at the point t.

Example 7. Show that the equation of the tangent to the curve of intersection of the ellipsoid

x2 a2

+

y2

+

b2

z2 c2

x( X − x )

2

x2

= 1 and the confocal

2

2

a (b − c )(a − λ )

y2 b2 − λ

y(Y − y)

=

2

a2 − λ

+

2

2

2

2

b (c − a ) (b − λ )

+

=

z2 c2 − λ

= 1 is

z(Z − z)

2

2

c (a − b 2 ) (c 2 − λ )

,

where (x, y, z) is an arbitrary point on the curve.

Sol. The given surfaces are x2 a2

x2

and

+

y2

y2

z2

+

b2

–1=0

c2

...(1)

z2

–1=0

...(2) a 2 − λ b2 − λ c 2 − λ

Let the equation of the curve of intersection of given surfaces be r = r(t), where t is an arbitrary parameter.

Differentiating the equations (1) and (2) w.r.t. t, we get

+

2x

2

+

.

x+

2y

2

.

y+

2z

2

.

z=0

a b c

2x .

2y .

2z . y+ 2 z =0 x+ 2 a2 − λ b −λ c −λ

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Example 4. Find the extremal of the functional

z

a

−a

(λy +

1

2

µy″2) dx which satisfy the

boundary conditions y(– a) = 0, y′(– a) = 0, y(a) = 0, y′(a) = 0.

z FGH

IJ

K

1

µy″ 2 dx .

−a

2

1

2

Let

F(x, y, y′, y″) = λy + µy″

2 d d2

Fy ′ +

F =0

The Euler-Poisson equation is Fy − dx dx 2 y ″

Here Fy = λ, Fy′ = 0, Fy″ = µy″

Sol. Given functional is

∴ (1)

λ−

a

λy +

d d2

(0) +

(µy″ ) = 0 ⇒ dx dx 2

µ

d4 y dx 4

+λ =0 ⇒

...(1)

d4 y

λ

=–

4

µ dx

d3 y

λ

= − x + c1

3

µ dx

d2y

dx 2

=−

λ x2

+ c1x + c2

µ 2

dy

λ x3 x2

=−

+ c1

+ c2x + c3

2 dx

µ 6

λ x4 x3 x2

+ c3x + c4

+ c1

+ c2

6

2

µ 24

This is the equation of the extremals. The boundary conditions are y(– a) = 0, y′(– a) = 0, y(a) = 0, y′(a) = 0

y= −

y(– a) = 0

λa4 c1a 3 c2 a2

– c3a + c4 = 0

+

24µ

6

2

...(4)

λa4 c1 a3 c2 a2

+

+

+ c3a + c4 = 0

24µ

6

2

...(5)

λa3 c1a 2

+

+ c2a + c3 = 0

2

...(6)

y′(a) = 0 ⇒

c1 a3

– 2c3a = 0 ⇒ c1a2 + 6c3 = 0

3

(4) + (6) ⇒ c1a2 + 2c3 = 0

Solving (7) and (8), we get c1 = 0, c3 = 0

(3) – (5) ⇒

(4)

(3) ⇒

λa 3

+ 0 – c2a + 0 = 0

λa 4 a2

−0+

24µ

2

...(3)

λa 3 c1 a2

– c2a + c3 = 0

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