# Results for: “Parmananda Gupta”

1 eBook

## 15 |
Parmananda Gupta | Laxmi Publications | |||||

5 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. See All Chapters |
|||||||

## 5 |
Parmananda Gupta | Laxmi Publications | |||||

50 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 See All Chapters |
|||||||

## 11 |
Parmananda Gupta | Laxmi Publications | |||||

4 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)]. See All Chapters |
|||||||

## 7 |
Parmananda Gupta | Laxmi Publications | |||||

73 CURVATURE AND TORSION The centre of the osculating circle at point P is called the centre of curvature of the curve C at the point P. By the definition of contact between curves, the osculating circle to the curve C at point P can be considered as the intersection of a sphere with at least 3-point contact with the curve C at point P and a plane with at least 3-point contact with C at P. If κ ≠ 0 at P, then the osculating plane at P is the unique plane having at least 3-point contact with the curve C at P. In particular, if τ ≠ 0 in addition to κ ≠ 0 at P, then the osculating plane is the unique plane having exactly 3-point contact with C at P. Therefore the osculating circle to a curve at a point always lies on the osculating plane to the curve at that point, provided κ ≠ 0 at the point under consideration. Thus, the osculating circle to a curve at a point can be considered as the intersection of a sphere with at least 3-point contact with the curve at that point and the osculating plane to the curve at the point under consideration, provided κ ≠ 0. See All Chapters |
|||||||

## 14 |
Parmananda Gupta | Laxmi Publications | |||||

160 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) See All Chapters |