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DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS

.

.

.

.

.

.

(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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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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Variational Problems with

Moving Boundaries

1. INTRODUCTION

We know that a functional associates a unique number to each function belonging to a certain class of functions. In the preceding chapter, we studied the methods of finding stationary function of the functional of the form

z

x2

x1

F(x, y(x), y′(x)) dx, where the boundary points (x1, y1)

and (x2, y2) were fixed points. In the present chapter, we shall allow one or both boundary points to move. In this case the class of admissible functions is extended because, in addition to the comparison curves with fixed boundary points, we have to admit functions with variable boundary points.

2. FUNCTIONALS DEPENDENT ON ONE FUNCTION

Theorem. Let J[y(x)] =

z

x2

x1

F(x, y(x), y′(x)) dx be a functional defined on the set of functions

y = y(x) which admits of continuous first order derivative and the end points of which (x1, y1) and (x2, y2) lie on the given curves y = φ(x) and y = ψ(x) respectively so that y1 = φ(x1) and y2 = ψ(x2), where y = φ(x) and y = ψ(x) are functions from the class C1[x1, x2]. Also, F is differentiable three times with respect to all its arguments.

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DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS

∴ The extremals of the functional J[y(x)] can be found by solving Euler’s equation of the transformed functional J1[v(u)]. This is called the principle of invariance of Euler’s equation under coordinates transformations.

In solving practical problems, the change of variables is made directly in the integral representing the functional. We then write and solve the Euler’s equation for the new integral involving new variables. In the extremals so obtained, the new variables are changed to the original variables to get the desired extremals.

Example. Find the extremals of the functional

Sol. Let

J[r(θ)] =

z

z

θ2

θ1

θ2

θ1

r 2 + r ′ 2 dθ , where r = r(θ).

r 2 + r ′ 2 dθ .

Let

x = r cos θ, y = r sin θ.

r=

−1 x 2 + y2 , θ = tan

y x

rx + ry y′ dr dr/ dx

=

= r′ =

= dθ dθ/ dx θ x + θ y y′

x

2

x +y

1 y2

1+ 2 x

2

+

FG − y IJ

H xK

2

y

y′ x + y2

1

1

+ y′

2 x y

1+ 2 x

2

FG IJ

H K

x + yy′ x2 + y2

= xy′ − y x2 + y2

=

Also,

dθ =

xy′ − y dθ dx

. dx = (θx + θy y′)dx = 2 dx x + y2

r2 + r′2 = x2 + y2 +

= (x2 + y2)

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DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS

Remark. It can be proved that a curve is uniquely determined (except for its position in space) if we are given its curvature κ(≠ 0) and torsion τ as continuous functions of arc length s. This result shows the importance of curvature and torsion in the study of differential geometry of space curves.

Example 1. For the helix r = a cos ti + a sin tj + btk, a > 0, b ≠ 0, find the torsion at the point t.

Sol. We have r = a cos ti + a sin tj + btk

r = – a sin ti + a cos tj + bk

a 2 sin 2 t + a 2 cos 2 t + b2 =

|r | =

t=

r

=

|r|

t′ =

dt dt dt dt

=

= ds dt ds dt

2

a + b2

= −

=–

|t′| =

n=

a + b2

1

=

1

2

a +b

a

2

a + b2 a

a 2 + b2

(– a sin ti + a cos tj + bk) ds dt

(– a cos ti – a sin tj) |r|

a

2

a 2 + b2

2

(cos ti + sin tj)

a 2 + b2

(cos ti + sin tj)

(cos2 t + sin2 t)1/2 =

a a 2 + b2

a a 2 + b2 t′

(cos t i

+ sin t j

)

.

=− 2

= – (cos ti + sin tj) a a + b2

|t′| i

b=t×n= −

=–

=–

b′ =

1 a 2 + b2

1

2

a + b2

a a 2 + b2

− cos t

1

2

a + b2

sin t

a

k cos t

a 2 + b2

− sin t

i j k

− a sin t a cos t b cos t sin t 0

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