TEL AVIV UNIVERSITY DISCONTINUITIES AND

TEL AVIV UNIVERSITY

THE IBY AND ALADAR FLEISHMAN FACULTY OF ENGINEERING Department of Solid Mechanics, Materials and Structures

Subject

DISCONTINUITIES AND SELF-SIMILAR WAVES IN THE PLAIN MOTION OF AN INEXTENSIBLE STRING Thesis submitted toward the degree of Master of Science in Mechanical Engineering in Tel-Aviv University by

Michael Slavutin May 1997

TEL AVIV UNIVERSITY

THE IBY AND ALADAR FLEISHMAN FACULTY OF ENGINEERING Department of Solid Mechanics, Materials and Structures

Subject

DISCONTINUITIES AND SELF-SIMILAR WAVES IN THE PLAIN MOTION OF AN INEXTENSIBLE STRING Thesis submitted toward the degree of Master of Science in Mechanical Engineering in Tel-Aviv University by

Michael Slavutin This research work was carried out at Tel-Aviv University in the Department of Solid Mechanics, Materials and Structures Faculty of Engineering under supervision of Prof. Leonid I. Slepyan May 1997

Acknowledgments Grateful acknowledgment to Professor Leonid I. Slepyan for his guidance, great patience and support in preparing this thesis. This research was supported by grants No. 9673-1-96 from the Ministry of Science, Israel, and No. 94-00349 from the United States | Israel Binational Science Foundation (BSF), Jerusalem, Israel.

Abstract Flexible elements are widely used in the engineering structures. In particular, they are seen to be very perspective in space development. The exible structures can be considered as strings in engineering analysis, i. e. onedimensional continuum that can support only tension. The problem of the motion of a string is considered for more than 250 years but the phenomena of the movement of the string have not been completely understood yet. In this thesis the dierent formulations of the string motion are developed, an analysis of propagation of discontinuities and especially the propagation of bends, namely, the discontinuities in tangent vector is presented and nally a new family of special solutions that are self-similar waves is presented and these solutions are connected with the bends. The plain motion of an inextensible string is considered. As far as author knows the formulations of the equations of the motion of the string in terms of integral variables and natural parameters of the string motion, namely, tangent and normal velocities, angular velocity, curvature and tension force, that are developed in this thesis could not be found in literature. An approximation for magnetic induction from the cross-section of the string that shows a good compatibility with the exact expression is proposed. The connection between the motion of the string and the propagation of the discontinuities in it is shown using an analysis of characteristic lines. Necessary conditions of the propagation of the discontinuities are developed. The motion of the bends is investigated using the laws of conservation of mass and momentum. A problem of collision of

two bends is studied. A apping eect is shown and the limits of the assumption of the inextensibility are established. A problem for the self-similar waves is formulated and a system of the ordinary dierential equations that describes this kind of motion is derived and investigated using standard mathematical techniques of investigation of critical points. It is shown that a constant force solution is a special case of such a wave. A few special cases that have a simple analytical solutions are described. Conditions for developing of such a motion in the bends are derived and model problems of the self-similar waves that appear in the semi-in nite initially straight string and in the semi-in nite initially bended string are solved numerically.

Contents 1 Introduction

1

2 Derivation of equations of motion

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1.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Types of mathematical formulations . . . . . . . . . . . . . . . . . . . . . 1.3 Innovation and contribution of this thesis . . . . . . . . . . . . . . . . . .

2.1 General case in three dimensions . . . . . . . . . . . . . . . . . . . . . . 2.2 Relations for inextensible string . . . . . . . . . . . . . . . . . . . . . . . 2.3 Formulation of problem in derivatives . . . . . . . . . . . . . . . . . . . . 2.3.1 Formulation in derivatives for plane motion. Characteristics . . . 2.3.2 Deriving equation for tangent angle . . . . . . . . . . . . . . . . . 2.4 Formulation in integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Formulation for plane motion of inextensible string in natural parameters 2.6 Formulation for disturbance from initial motion . . . . . . . . . . . . . .

1 4 7

8 19 21 25 28 29 30 32

3 Investigation of discontinuities

34

4 Investigation of bends

44

3.1 Case of constant tension (D'Alembert solution) . . . . . . . . . . . . . . 34 3.2 Necessary conditions on discontinuities . . . . . . . . . . . . . . . . . . . 37 3.3 Bend with non-constant force . . . . . . . . . . . . . . . . . . . . . . . . 40 4.1 General laws of mechanics for bends . . . . . . . . . . . . . . . . . . . . . 44 I

CONTENTS

II

4.2 Collision of bends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.3 Investigation of collision . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 Formulation of problem for self-similar waves

5.1 Self-similar rotating wave . . . . . . . . . . . . . 5.2 Non-rotating self-similar waves . . . . . . . . . . 5.3 Analysis of equation for radius . . . . . . . . . . 5.3.1 Special solutions . . . . . . . . . . . . . 5.3.2 Cases that can be solved analytically . . 5.3.3 Critical points of an equation for radius . 5.4 Building of solution with Taylor expansion . . .

6 Self-similar solutions in bends

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6.1 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Energy relations for self-similar wave . . . . . . . . . . . . . . . 6.3 Examples of problems that lead to self-similar solutions . . . . . 6.3.1 Semi-in nite initially straight string drawn from free end 6.3.2 Semi-in nite initially bended string drawn from free end

Summary

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61 61 66 68 69 76 79 85

87

87 95 98 98 101

106

Nomenclature A Auxiliary matrix. A Scaling coecient of a self-similar wave. A0 Area of a cross-section of a string.

B Magnetic induction. B Scaling coecient of a self-similar wave.

B1 External magnetic induction. C Auxiliary matrix. Ci Arbitrary vector constant. Ci; Cij Arbitrary scalar constant. CD Drag coecient. D Length of an interval between two bends. E Modulus of elasticity of material of a string. E (m) Complete elliptic integral of the second kind. F (s; t) Tension force. G Universal gravity constant. III

NOMENCLATURE Gi (t) Arbitrary function of time. H (x) Heaviside step function.

I Identity matrix. I Electric current. K Arbitrary constant. K (m) Complete elliptical integral of the rst kind. L Length of a string. L0 Scaling coecient. L1; L2 Constant and linear scaling coecients. M Mass of the massive point. Na; Nb Auxiliary relations involving ends of a self-similar wave.

P External concentrated force. P (z) Auxiliary function.

Q(s; t) Distributed load. Q0 Acceleration of a mass moving along a string. Qt; Qn; Qb Tangent, normal and binormal projections of distributed load.

R(s; t) Position vector of a point of a string. R0 Reference point on the string. R0 Scaling length parameter.

R0(s; t) Initial motion of string.

IV

NOMENCLATURE

V

R1(s; t) Additional motion of string. Ra; Rb Auxiliary relations involving ends and end radii of a self-similar wave. S Coordinate of a fold. S0 A ow section of a exible pipe. T Scaling time parameter. T (t) Speci ed tension force at an endpoint of a string.

U(t) Speci ed motion of an endpoint of a string. U x-coordinate of a center of gravity of a string in rotated coordinate system.

V Vector for calculating the parameters of a collision of bends. V y-coordinate of a center of gravity of a string in rotated coordinate system. V1; V2; V Particle velocities. W (t) Phase shift of a self-similar wave. W0 Phase shift coecient of a self-similar wave. X; Y; Z Integrals of components of position vector with respect to Lagrangian coordinate. Xi (; z) Functions for determination of a type of critical points of an ordinary dierential equation. a Endpoint of a self-similar wave. a0 Quadratic coecient for determination of a new tangent velocity as a result of collision of bends. a1 Linear coecient for determination of a new tangent velocity as a result of collision of bends.

NOMENCLATURE

VI

a2 Constant coecient for determination of a new tangent velocity as a result of collision of bends. ai() Auxiliary function.

b Binormal unit vector. b Endpoint of a self-similar wave. b0; b1 Quadratic and constant coecients of a self-similar solution of a type of square root of a quadratic polynomial. c Velocity of propagation of discontinuities (\wave velocity"). d Tangent velocity of a bend. d0 Diameter of a cross-section of a string.

e; e1; e2 Unit vectors. f Tension force divided by mass per unit length. f0(t) Arbitrary time-dependent function | constant of integration of tension with respect to Lagrangian coordinate. g Gravity acceleration.

i Unit vector of a x-coordinate axis. j Unit vector of a y-coordinate axis. k Unit vector of a z-coordinate axis. l(t) Scale function of a self-similar wave. m Power of an external force that generates a self-similar wave.

NOMENCLATURE

VII

mi Coecients for determination of a type of critical points of an ordinary dierential equation.

n Normal unit vector. p Pressure (in sense of force per unit length). p0 Linear coecient of pressure.

q Distributed load divided by mass per unit length. q(P ) Auxiliary function. q Derivative of tension with respect to Lagrangian coordinate. q~ Derivative of tension with respect to time. r() Scaled radius.

r() Scaled position vector. r0 Radius of a cross-section of a string. s Lagrangian coordinate along a string. s0 Starting point of a self-similar wave. s0(t) Lagrangian coordinate of a mass or a force moving along a string. sE Euler coordinate along a string.

t Tangent unit vector. t Time. t0 Initial time. u Tangent velocity.

NOMENCLATURE

VIII

v Normal velocity.

v1 Fluid velocity. v? Projection of velocity of a string onto normal plane. v0(t) Euler velocity of a mass or uid along a string. vx0 ; vy0 Velocity components of an origin of a self-similar wave. w(r) Auxiliary equation. x; y; z Coordinates. x1; y1 Rotated coordinates. z() Squared self-similar radius. Quadratic coecient of a self-similar wave. 0 Direction of a force in an endpoint. 2r0 Angle from a reference cross-section on a curved element of a string where a distance between the projection of the center of the cross-section onto the reference crosssection and the center of the reference cross-section is twice a radius of the crosssection of the string. Linear coecient of a self-similar wave.

Angle between a particle velocity and a tangent vector. (x) Dirac delta-function. Complex variable. Self-similar non-dimensional variable. cr Critical point of a self-similar wave.

NOMENCLATURE Pitching angle. Curvature. Stretch of the string. 0 Initial stretch. 1 Additional stretch.

Derivative of a stretch with respect to a Lagrangian coordinate. Vector of space rotation.

Magnetic permeability of medium. 0 Force per unit length. Derivative of a stretch with respect to time. Auxiliary variable. Mass per unit length of string. 0 Density of medium. %k Radius of curvature. Lagrangian coordinate in a coordinate system moving with a discontinuity. Angle of a new bend as a result of a collision. 1(x); 2(x) Arbitrary functions of one variable.

Constant coecient of a self-similar wave. ' Azimuth angle, tangent angle in a plane motion. () Polar angle.

IX

NOMENCLATURE

(t) Rotation of a self-similar wave.

! Angular velocity. ! Vector of angular velocity.

`(t) Lagrangian coordinate of a discontinuity.

Q Integral of distributed load with respect to Lagrangian coordinate. R~ Position vector of a mass or a force moving along a string.

R Integral of position vector with respect to Lagrangian coordinate. T Kinetic energy. W Work of external forces. ( )0 Derivative with respect to Lagrangian coordinate s. (_) Derivative with respect to time. ( )x Derivative with respect to variable x.

r Gradient operator. ( ) Condition right before a discontinuity. ( )+ Condition right after a discontinuity.

a b Scalar product of vectors. a b Cross product of vectors. ab Direct product of vectors.

X

List of Figures 2.1 Element ds of the string. . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The projection of the point onto current cross-section lying outside. The whole cross-section produces magnetic induction. . . . . . . . . . . . . . 2.3 The projection of the point onto current cross-section lying inside. The shaded area does not produce induction. . . . . . . . . . . . . . . . . . . 2.4 Comparison between the exact and the approximated induction at the bend with radius of curvature %k . . . . . . . . . . . . . . . . . . . . . . . 2.5 Comparison between the exact and the approximated induction at the bend %k = 100r0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 De nition of azimuth and pitching angles. . . . . . . . . . . . . . . . . . 2.7 Transformation to the normal and tangent velocities. . . . . . . . . . . .

9 14 14 16 17 22 30

3.1 Supposed solution for semi-in nite initially straight string drawn with constant force. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.2 Supposed solution for nite initially straight string drawn with constant force. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.3 The motion of the string in the time interval dt. . . . . . . . . . . . . . 43 4.1 4.2 4.3 4.4

Motion of the bend. . . . . . . . . . . . . . . . . . . . . Two new bends from the bend impact. . . . . . . . . . . The tangent velocity of the new interval. . . . . . . . . The tangent velocity of the new interval | contour lines. XI

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46 48 54 54

LIST OF FIGURES 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12

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56 56 57 57 58 58 60 60

5.1 Self-similar solution of constant radius. . . . . . . . . . . . . . . . . . . 5.2 Self-similar solution of square root type. . . . . . . . . . . . . . . . . . . 5.3 Regions of solutions of type of square root of a quadratic polynomial at various b0 and b1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Dierent solutions of type of square root of a quadratic polynomial. . . 5.5 Dierent solutions of type = = 0 . . . . . . . . . . . . . . . . . . . .

70 71

6.1 6.2 6.3 6.4

The normal velocity of the new interval. . . . . . . . . . . . . The normal velocity of the new interval | contour lines. . . . The velocity of propagation of the new bend. . . . . . . . . . The velocity of propagation of the new bend | contour lines. The angle of the new bend. . . . . . . . . . . . . . . . . . . . The angle of the new bend | contour lines. . . . . . . . . . . The apping in the normal velocity. . . . . . . . . . . . . . . The apping in the bend propagation velocity. . . . . . . . .

XII . . . . . . . .

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A problem that leads to a self-similar wave. . . . . . . . . . . . . . . . . Problem of drawing of semi-in nite straight string. . . . . . . . . . . . . Solutions for a semi-in nite initially straight string. . . . . . . . . . . . . Non-dimensional length of a self-similar wave for the semi-in nite initially straight string. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Solutions for the semi-in nite initially bended string. . . . . . . . . . . . 6.6 Non-dimensional length of a self-similar wave for the semi-in nite initially bended string. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 The second coecient of an end force in the non-dimensional form for the semi-in nite initially bended string. . . . . . . . . . . . . . . . . . . . .

74 75 78 93 99 100 101 102 104 104

Chapter 1 Introduction 1.1 Applications In this section the term \ exible structures" is used for engineering and natural systems behavior of which can be described by equations of motion of exible string. A string for us is an one-dimensional continuum that can bear only tension forces. Any load that is not the tension is an external load applied on the string. There is a great variety of such structures. We will describe here the basic classes and state on the mathematical treatment required for each one. The engineering structures where cables are the basic elements are suspended and cable stayed bridges, antennas, oating sea platforms and so on. The use of cables instead of rods or beams as supporting elements allows to reduce the total mass of the structure. Here the treatment is static and linearized dynamic since the motion of the cables are small transverse perturbations and the inner tension is large enough in comparison with the other loads. When the external load may be compared with the tension but the motions are still small, the formulation of Kirchho (Kirchho, 1883) and Carrier (Carrier, 1945) that is the formulation for relatively small motions of the string with weak tension force can be used. Electric power lines and funicular railways subjected to gravity load and could be 1

CHAPTER 1. INTRODUCTION

2

forced to make relatively large transverse motions due to the external loads. These could be wind that produces the air drag on power line or the concentrated mass or force on the funicular railway. Still if the transverse motion can be assumed relatively small and the tension is large one can integrate linearized equations as it was performed by Smith in (Smith, 1964). In other cases the numerical modal analysis for large de ections performed by Takahashi and Konishi in (Takahashi and Konishi, 1987a) and (Takahashi and Konishi, 1987b) is necessary. Flexible pipelines are used widely, for example, in refueling of planes during the ight. Here we have to consider the inertia forces produced by the moving liquid, gravity, air drag and variations in tension due to variation in the distance between the aircrafts. This is an example of system where the stable stationary movement is required. It could be noticed that the ordinary \rigid" pipelines might be considered as exible strings for very long waves, since the bending stiness is insigni cant for them in comparison with the wavelength. In the recent time much attention is paid on tethered satellites. These are the systems of satellites connected by a long exible ber (tether). This is a very promising eld of development of space structures. Several experiments were already conducted by NASA. Using of tethered systems allows to perform a complicated orbital manoeuvres without spending fuel, the conducting tether in the magnetic eld of the planet can be used to produce electric power, rotating tethers can be used to establish the economically ecient inter-planet communication: (Mackenzie, 1996) gives, for example, the economically ecient and feasible project of Earth-Moon communication by using a series of rotating tethers that will \catch" the payload, accelerate it and throw like a sling, and so on. The ideas of using the tethers in space are very old and could be found in the early works of Tsiolkovsky in (Tsiolkovsky, 1959). The most perspective tendencies in this eld were described in (Bekey, 1983). Here are the few possible applications (the whole spectrum of ideas and plans of using tethers in space technology could be found in (Penzo and Ammann, 1996)):


3

1. Tethered recovery system | a small payload is deployed from the space station into the upper atmosphere were it would be released into a reentry trajectory. 2. Electrodynamics power generation | a long tether in the Earth magnetic eld acts as a power generator. There is also a possibility of the reverse use | an electric current from the satellite interacts with the Earth magnetic eld to produce a propulsive force on the spacecraft/tether system. 3. Satellite boost from orbiter | a satellite is deployed along a tether \upward" (away from the Earth) from the Shuttle Orbiter. The operation requires less fuel for both satellite and the Orbiter than in the existing procedure. All the application above are scheduled to the near-term (i. e. 5 years or less). From the dynamical point of view these systems could be subjected to all kind of motions (large perturbations) while the tension force could vary a lot even along the length of the string, thus the coupled nonlinear equations of motion are to be considered. The use of tethered systems is, of course, not limited for space systems. A plenty of automatic devices are controlled using cables on which the electric power necessary for the operation is translated. These are tethered robots like the one described in (Wettergreen et al., 1993), underwater vehicles | the example can be found in (Grosenbaugh et al., 1993), kites | (Balsey et al., 1994) describes such a kite used for atmosphere pro ling, and many others. There are many systems that their behavior could be described as the plain motion of string (i. e. it does not depends on the third coordinate). In analysis of such systems (for example conveyor belts) mostly the steady motions were studied. The most interesting natural systems, that are strings as a matter of fact, are molecular chains: polymers, proteins or DNA. The properties of these materials in macro level depend on the dynamical behavior of the molecular chains in micro level. These molecules are described generally as the chains of masses connected by elastic rods with or without bending stiness (as it was described by Harnau, Winkler and Reineker in


4

(Harnau et al., 1995)) | just like the string in part of numerical methods. In such a way viscoelastic and other mechanical properties of plastics and other polymer materials could be predicted: as for example in (Bueche, 1954).

1.2 Types of mathematical formulations The motion of a exible string has been considered since the works of Jas. Bernoulli and Euler in seventeenth and eighteenth centuries. The static and linearized dynamic problems were studied by the classics. There are ve basic static problems that were formulated and studied since then. Antman in (Antman, 1995) gives a very detailed review with an accent on strings with arbitrary constitutive relations. We will remind the problem of the string suspended at both ends and subjected to the gravity load distributed uniformly along the length of the string (the catenary problem | from the Latin catena, meaning chain ). The equation of the equilibrium state of an inextensible uniform catenary was found by Leibniz and Joh. Bernoulli and integrated by Leibniz in terms of hyperbolic cosine. The suspension bridge problem describes the equilibrium of the string suspended at both ends and subjected to the distributed load that is uniform along the horizontal coordinate. The two problems that are related to the equilibrium of sails and cylindrical membranes describe the equilibrium of the string subjected to the normal loads. These are the so-called velaria problem that determines the equilibrium state of the string under constant pressure (from the Latin velarium | sail ) and the lintearia problem | describing the equilibrium state of a string loaded with pressure that varies linearly with depth (horizontal cylindrical surface holding water). These problems were formulated by Jas. Bernoulli and were re-studied recently in connection with the equilibrium of wires in magnetic eld since the load on the conductor is normal. The fth basic static problem is the equilibrium of the string attracted to a xed point. This problem is connected with the very long strings in the eld of gravity of the planets or other space bodies.


5

The equations of motion of an extensible string were formulated by Euler in 1744. These equations are probably the simplest nonlinear equations of motion since they are dealing with the one-dimensional continuum. Yet we do not have a close-form solution in general case even in a planar motion of an inextensible string. The D'Alembert solution describes only the motion of the string subjected to the constant tension. The solution of the problem of small oscillations of the string pendulum that was found by Jas. Bernoulli deals with the linearized problem where division of variables can be applied. Since the end of nineteenth century we observe attempts to include the nonlinear eects in the description. In 1876, Kirchho introduced in (Kirchho, 1883) the equation that describes the motion of the string clamped at both ends and subjected to a weak inner tension. This equation was also developed later in a more rigorous manner by Carrier in (Carrier, 1945) and (Carrier, 1949). Narasimha in (Narasimha, 1968) obtained the same equation using another approach. These equations were solved by Oplinger in the case when one end is clamped and the other subjected to the unspeci ed periodical load in (Oplinger, 1960). Still this formulation is valid only for the small perturbations of the string and does not include the coupling between the longitudinal and transverse motions. Leissa and Saad in (Leissa and Saad, 1994) derived the similar equations and solved them numerically by use of modi ed Galerkin method. Another way of the research in this eld is studying of the stationary motions: (Antman and Reeken, 1987), for example, considered the problem of whirling and drawing of string and (O'Reilly, 1996) studies the problem of drawing of string in gravity eld, and solitary waves: Slepyan, Krylov and Parnes in (Slepyan et al., 1995b; Krylov et al., 1998) and (Slepyan et al., 1995a) presented the exact solution for solitary waves in the inextensible helicoidal rotating string; Slepyan, Krylov and Rosenau in (Slepyan et al., 1998) | for the arbitrary elastic helicoidal rotating string. There is a lot of practical problems that could be described by the stationary motions of the string. The examples of such problems are the movement of conveyor belts, chain-saws, the various motions in the manufacturing processes (textile manufacturing, ber manufacturing and so on). The


6

study of solitons is interesting since the exible string is probably the simplest waveguide and the motion of the string is deeply connected with the wave propagation. Various special solutions can also be found by assuming some additional conditions | (Rosenau and Rubin, 1986) have pointed, for example, on some possible three-dimensional special motions. A numerical investigation of the string dynamics is modal as, for example, in (Takahashi and Konishi, 1987a) or by the direct numerical integration as in (Perlov and Alesov, 1996). The wave propagation in string is studied generally by means of nite dierences or nite element methods: the example of numerical investigation of such a problem can be found in (Rakowski, 1992). Kuhn et al. compared various methods with the experimental results on the problem of string pendulum in (Kuhn et al., 1995). The problem of the dynamics of string during the translation of mass or force along it arises in investigation of such a systems as funicular railway (Esudri, 1995) or train pantograph (Manabe, 1994). It could be noticed here that many recent authors are formulating the equations of motion in terms of displacements from the initial position of the string. These equations are, generally, much more complicated than the original set. Such a formulation seems to be suitable only in consideration of the small perturbations of the string near the initial curve since the highly nonlinear terms are negligible then. It can be concluded that although the dynamics of the various string-like structures have been studied for more than two hundred years, the motion of the string yet has not been understood completely. The propagation of the waves in string is studied as a special case of the motion of the string in spite of the fact that the motion of the string is a wave motion. The study of propagation of discontinuities in string that leads to the wave propagation still has not been conducted.


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1.3 Innovation and contribution of this thesis In this thesis we develop dierent formulations of the string motion, present the analysis of the propagation of discontinuities and especially the propagation of the bends, namely, the discontinuities in the tangent vector and nally present a new family of special solutions that are self-similar waves and connect these solutions with the bends. The plain motion of an inextensible string is considered. As far as we know the equations of the motion of the string in terms of integral variables and natural parameters of the motion of the string, namely, tangent and normal velocities, angular velocity, curvature and tension force could not be found in literature. The approximation for the magnetic induction from a cross-section of the string that shows a good compatibility with an exact expression is developed. The connection between the motion of the string and the propagation of the discontinuities in it is shown using the analysis of the characteristic lines. The necessary conditions of the propagation of the discontinuities are developed. The motion of the bends is investigated using the laws of conservation of mass and momentum. The problem of collision of two bends is studied and the limits of the assumption of the inextensibility are shown. The problem for the self-similar waves is formulated and the system of the ordinary dierential equations that describes this kind of motion is derived and investigated. It is shown that a constant force solution is a special case of such a wave. The conditions for the developing of such a motion in the bends are derived and the model problems of the self-similar waves that appear in the semi-in nite initially straight string and in the semi-in nite initially bended string are solved numerically.

Chapter 2 Derivation of equations of motion 2.1 General case in three dimensions Here the general equations of motion for string with general elastic constitutive law will be derived. Several cases of distributed external loads and boundary conditions will be considered. Some of the following equations can be found in literature, others are derived in this thesis for the rst time. We de ne string as one-dimensional continuum that can support only tension force, but no compression or a bending moment. To describe the string we will use the Lagrangian coordinate s along the string. We consider the string that is subjected to distributed external load Q(s). Let us choose an element of length ds. The position of the point s is R1, and of the point s + ds | R2. The element is subjected to the forces Qds, F1t1 and F2t2 where t is the tangent unit vector directed along s. The mass of the element is ds where is mass per unit length and thus the second law of Newton can be written in the form: 2 (2.1) ds @ R2 = F2t2 F1t1 + Qds @t and after the division by ds and taking ds ! 0 we receive:

R = (F t)0 + Q 8

(2.2)

CHAPTER 2. DERIVATION OF EQUATIONS OF MOTION

9

F2 t 2 t2

Q

R2

-F 1 t 1

ds

-t 1 R1

Figure 2.1: Element ds of the string. Here and further a dot denotes the partial derivative with respect to time and prime | with respect to the Lagrangian coordinate s. The position vector R is continuous in every point since the string is continuous. The tangent unit vector t and its derivatives can be discontinuous in s. In such a point the derivatives are understood as the directional derivatives (right or left from the point of the discontinuity) and the connection between the two sides of the discontinuity follows from the laws of conservation of mass and momentum as it will be discussed in chapter 4. Note, that the initial length of the element was ds and the current length is ds = jR0j ds. Thus we obtain that R0 = t since the derivative of the position vector R with respect to the Lagrangian coordinate s coincides with the tangent vector. It is assumed that only isolated singular points can exist in which the derivatives in (2.2) does not exist in classical sense, but exist as generalized derivatives (as derivatives of a discontinuity).

CHAPTER 2. DERIVATION OF EQUATIONS OF MOTION Finally we obtain the well-known equation of motion of the string: 0 !0 R R = F + Q

10 (2.3)

This equation is valid for any constitutive law of the material. In the case of an elastic material, that is considered below, F = F (). Strain of the string in point s is (s) 1. Hence in the case of the model of an inextensible string, where the elongation is zero, we obtain: R = (F R0)0 + Q and jR0j = 1 (2.4) Here are some important classes of the external loads: 1. String's own weight that could be treated as the constant gravity acceleration g = const in the direction e = const.

Q = ge Note, that here we can present a new variable 2 t R1 = R g 2 e and thus obtain an original equation without any external loads.

R 1 = (F R01)0

(2.5) (2.6) (2.7)

This substitution is, in fact, a moving to an accelerating coordinate system. 2. The load is distributed uniformly along x-axis with unit vector e1 = const and acts in the direction e2 = const (the suspension bridge problem, where the force is the weight of the bridge in y-direction which is normal to x, is a special case of this kind of loading). Of course, in the general case e1 and e2 are not obliged to be orthogonal. Let 0 be the load per unit length along x-coordinate. The force acting on element ds thus is dP = e20dx. But dx = tds e1 where t is again the tangent unit vector to the string at point s. Finally after substitution of t and we obtain: Q = e2 0 (R0 e1) (2.8)


11

3. The gravity due to the presence of a massive point M is an example of the force directed to a central point. It is natural to choose the origin of the coordinate system in this point. The load is directed towards this point, and if we denote the universal gravity constant as G we obtain

Q = G jM Rj3 R

(2.9)

4. The load is convenient to be described in the natural triedr projections: the normal projection Qn, the tangent projection Qt and the binormal projection Qb. The particular case is the distributed load that is normal to the string (Qt = 0, for example lintearia and velaria problems, electro-magnetic force, air drag: in the rst two cases we have Qt = 0 and Qb = 0). In those cases, one has to consider the natural triedr ft; n; bg. Note, that the Frenet formulae are written for the actual position of the curve, i. e. using Euler coordinate sE : 8 dt > > > dsE = n > > < dn (2.10) > dsE = t + b > > > > : db = n dsE where is a curvature and is a torsion. E The Euler and Lagrange coordinates are connected by the equality: @s @s = , and dt = dt = n (since the derivative of the tangent vector with hence t0 ds dsE respect to the Euler coordinate dsdt = n | the rst equation of (2.10)). In this E way the Frenet formulae in terms of the Lagrange coordinate s are obtained: 8 dt > > > ds = n > < dn (2.11) > ds = t + b > > > : db = n ds


12

Further R00 = (R0)0 = t0 + t0. Finally knowing that R0 = t, R00 = n2 + t0 and b = t n, we obtain the parameters of the natural triedr: 8 R0 = R0 > > t = > j R0 j > q 2 > q 00 j jR0j0 2 > j R 1 2 > 2 00 0 = 2 jR j = > > jR0j2 0 0 > > > 00 R0 jR j ! R > 0 0 < n = 1 2 R00 R0 = q 2 jR j02 (2.12) > 00j jR0j j R > > > R0 (R00 R000) = R0 (R00 R000) > > = > 2 jR00j2 02 jR0j2 jR00j2 jR0j02 > > > > R0 R00 1 R0 R00 = > q b = > : 3 jR0j jR00j2 jR0j02

where is de ned as nonnegative and in the points where = 0 we determine n from the considerations of the continuity if it is possible. Here are some examples:

Constant pressure p (in the meaning of force per unit length, not per unit area and given in Èuler' sense, i. e. constant in space) (the velaria problem): 8 > > > < Qt = 0 (2.13) > Qn = p > : Qb = 0

Pressure that varies linearly along x-coordinate (again, the Èuler' coordi-

nate) p = p0x (the lintearia problem). We will put the origin at the zero of the x-axis: 8 > > > < Qt = 0 (2.14) Qn = p0 (R e) > > > : Qb = 0 where e is the unit vector of the x-axis.


13

The electro-magnetic force. Suppose that the conductive string is found in magnetic eld characterized by magnetic induction B1. The electric current I in the string (positive, if its direction coincides with t) gives us vector di = Ids t. An element R of the string itself thus produces in a reference point R0 additional magnetic induction ) ds (2.15) dB = 4I t (R0 R 3 jR0 Rj where is magnetic permeability of the medium (by the Law of Biot and Savart ). The use of formula (2.15) meets, however, a problem: in the regular case of a nite curvature there is a singularity of the type 1=s (s ! 0) when R ! R0. In fact, we have to look at the cross-section and integrate the magnetic induction, produced by distributed currents on it. The author realizes that this fact has been studied and known, but he did not found the sucient coverage of it in a sources known to him, such as in (Tamm, 1989) or in (Joos, 1934), so he had to derive the equations given below by himself. We have two cases, as it is shown in Figs. 2.2 and 2.3. In Fig. 2.2 the projection of the point R0 where we want to calculate the induction onto the plane of the current cross-section lies outside the cross-section and all the crosssection in uences on the point R0. In the case of Fig. 2.3, the projection is inside the cross-section, and the shaded area does not produce induction since every point in it has the symmetrical point that produces the same induction directed against the rst one, and these two points cancel the in uence of each other. The result of integration in both cases is: m Ids t ( R R ) 1 0 p 1 2 K (m) E (m) (2.16) dB = 2r pr 0 0 jt (R0 R)j3=2 m where 4r0 jt (R0 R)j m= ; jR0 Rj2 + 2r0 jt (R0 R)j + r02 and K (m) is the complete elliptical integral of the rst kind, E (m) is the complete elliptical integral of the second kind and r0 is the radius of the cross-


14

r0

r0

| t x (R 0- R )|

Figure 2.2: The projection of the point onto current cross-section lying outside. The whole cross-section produces magnetic induction.

| t x (R 0 - R )|

Figure 2.3: The projection of the point onto current cross-section lying inside. The shaded area does not produce induction.

section. When the distance jR0 Rj is large enough in comparison with the radius r this formula is reduced to the previous one (2.15). In the case of Fig. 2.3 one can verify that ddsB ! 0 when R ! R0 using the exact formula (2.16). A simpli cation of the formula (2.16) can be derived based on the formula (2.15) but with the current I de ned by the above considerations re ected in the Fig. 2.3. Namely, in the formula (2.15) we take into account only the \white" part of current on the Fig. 2.3 and place it in the center of gravity of the \white" part of the cross-section. The expression obtained is: t (R0 R) dB = I !3=2 (2.17) 4 m (2 m ) 1 1 jR0 Rj2 + jt (R0 R)j2 (1 m )2 H (1 ) 1 where q 2 2 m1 = arccos 1


15

is the ratio of the \white" area to the whole cross-section,

= jt (Rr0 R)j 0

and H (x) is Heaviside step function. In contrast to (2.15), this expression is regular at any point. The accuracy of this approximation can be seen in Figs. 2.4 and 2.5 where the results obtained from (2.16) and (2.17) are compared at the curved part of the string with the radius of curvature %k expressed in terms of the radius of the cross-section of the string r0, and they are shown as function of the angle measured in the curvature circle from R0 to R on the range 0 < < 2r0 where 2r0 is the angle on which jt (R0 R)j = 2r0 . The Figure 2.4 shows that the approximation is as better as the bend is less sharp. Note that at very sharp bends (%k < 10r0) the assumption of the perfect exibility of the string cannot be valid any more. From the Fig. 2.5 where the exact and the approximated results are compared at %k = 100r0 one concludes that the main deviation of our approximation is concentrated in a rather small area near the point R0. The whole string produces, of course: ZL (2.18) B = 0 dB where dB can be taken from (2.16) or (2.17) due to the desired accuracy. We are interested in the force that acts on the string, so, for us R0 is the appropriate point on it. The total induction is B0 = B + B1, the force that acts on element ds of the string is Qds = Idst B0 (by the Ampere's theorem) and the distributed electro-magnetic load on the conducting string in magnetic eld is: Q = I t (B1 + B) (2.19) Note, that since t t = 0, in the case of the electro-magnetic forces, we have Qt = 0 and the load lies in the normal plane.


16

1 0.9

|d B|exact |d B|approx.

0.8 ρk = 200r0 ρk = 100r0 ρk = 20r0 ρk = 10r0

0.7 0.6 0.5 0.2

0.4

α/α 2r

0.6

0.8

1

0

Figure 2.4: Comparison between the exact and the approximated induction at the bend with radius of curvature %k .

The aero- or hydrodynamic drag. Here Qn and Qb are the drag components

that are assumed to be dependent on the projection of the velocity onto normal plane: v? = n n R_ v1 + b b R_ v1 where v1 is the velocity of the uid and are calculated as: 8 > < Qn = 0d0CD jv?j R_ v1 n (2.20) > : Qb = 0d0CD jv?j R_ v1 b where 0 is density of the medium, d0 is the diameter of the cross-section of the string and CD is the drag coecient.

5. Concentrated force P. The representation of a concentrated force in terms of a distributed load is based on Dirac delta-function:

Q = P (s s0 (t))

(2.21)


17

18

2 4 πr 10 µ I0 |ddsB|

16

exact

14

approximation

12 10 8 6 4 2 0

0.05

0.1 α

0.15

0.2

Figure 2.5: Comparison between the exact and the approximated induction at the bend %k = 100r0 . where s0 (t) is the point on the string in which the force is applied. 6. Concentrated mass that can move along the string according to a law s = s0 (t). If we work in Lagrangian description, then the position of the mass is R~ = R (s0(t); t). The dierentiating with respect to time gives us the velocity of the mass R~_ = R_ + R0s_0 and the second derivative which is the acceleration, is Q0 = R~ = R + R0s0 + R00s_02 +2R_ 0s_0 (the acceleration of the point of the string plus the relative acceleration of the mass along the string plus the centrifugal and Coriolis accelerations). The load will result in the concentrated inertia force (the D'Alambert principle) P = M R~ and thus we obtain:

Q = M (s s0 (t)) Q0 = = M (s s0 (t)) R + R0s0 + R00s_02 + 2R_ 0s_0

(2.22)


18

This relation seems relatively simple but the movement of the mass is speci ed usually by velocity v0(t) along the deformed string (for example in the movement of the wagon of the funicular railway). The `Lagrangian' velocity will be then R s_0 = v0= and the position of the mass is R~ = R 0t v0= dt; t . The àbsolute' 0 velocity will be R~_ = R_ + R v0 = R_ + v0t. To obtain acceleration we have to dierentiate this relation with respect to time again and that gives Q0 = R~ = R + R_ 0v0= + v_ 0t + v0t_ + t0v02=. Substituting t and gives, nally: Z t v0 ! Z t v0 ! dt Q = M s dt (2.23) Q = M s 0 jR0j 0 j R0 j ! 0 0 0 00 0R 0 !! _ _ R R R R R 2 0 R + v_0 jR0j + v0 0 2 + v0 2 jR0j R 0 3 jR j jR0j3 jR j 0

This exact but complex expression can be simpli ed: extension of the string causes motion R_ 0 which is negligible in comparison with the total motion of the particle. 7. A owing liquid inside the string (a exible pipe). Let us take the liquid that ows with constant velocity v0 (the same as above it is the velocity in Euler sense). The element of liquid 0S0ds (0 is the density of the liquid and S0 is the ow section) will give us the same force as the concentrated mass, so the load is calculated as in the equation (2.23): Q = 0S0 jR0j Q0 (2.24) where the acceleration Q0 is de ned from the equation (2.23). Initial conditions for the equation (2.3) are the initial con guration of the string R (s; 0) and the initial velocity R_ (s; 0). Boundary conditions could be of the following three types: 1. Speci ed motion at an endpoint. Here R (0; t) = U (t). The particular case is the xed end R (0; t) = 0. 2. Speci ed tension force at an endpoint F (0; t) = T (t). The particular case is the free end F (0; t) = 0. For the elastic string this is the condition on , since f = f () is given.


19

3. The connection between force and motion at an end is speci ed. The particular case is some concentrated mass M attached at the endpoint. 0 F R (2.25) M R (0; t) = jR0j 0;t

2.2 Relations for inextensible string Here some of the relations of the previous section are written in the case of inextensible string, that is in the case when extensibility can be neglected. In the case of the inextensible string there is no elongation, = jR0j = 1, and therefore many of the above equations are simpli ed a lot. First, the equation of motion will be written in form (2.4). Secondly, the geometrical relations (2.12) are written as: 8 > t = R0 > > > > = jR00j > > > 00 > < n = R00 jR j (2.26) > 0 00 000 > R ( R R ) > > = > 00j2 j R > > 0 00 > > : b = RjR00R j For acceleration of the point mass M moving on the string we obtain Q0 = R~ = R + s0t + s_20n + 2_s0 t_ and the equation (2.23) can be re-written as: (2.27) Q = M (s s0 (t)) R + s0R0 + s_20R00 + 2_s0R_ 0 The equation (2.24) for the liquid ow of uniform velocity v0 for the inextensible string takes the form: (2.28) Q = 0S0 R + v02R00 + 2v0R_ 0 where 0 is density of the liquid and S0 is the ow section of the pipe.


20

If we take the equation of inextensibility R0 R0 = 1 and dierentiate it with respect to time and with respect to the Lagrange coordinate s we obtain the following relations: 8 > < R_ 0 R0 = 0 (2.29) > R00 R0 = 0 : The meaning of the second equation is obvious | the normal n and the tangent vector t are perpendicular to each other. This property allows us to express the tension F from (2.4). After opening the parentheses we obtain R = F 0R0 + F R00 (we suppose that there is no external loads Q = 0). By scalar multiplication of this equation on R0 (projection on tangent) we obtain: F 0 = R0 R (2.30) and by scalar multiplication on R00 (projection on normal): 00 F = RR00 RR00

(2.31)

Let us perform some further analysis of (2.29). The rst equation of (2.29) means that the vector of the derivative of velocity is normal to the tangent vector. From this relation we can make some conclusions about the point on the string where velocity is maximal (or minimal). Really, in the coordinate notation we can write x0x_ 0 + y0y_ 0 + z0z_ 0 = 0. Now suppose that two of the velocity projections have reached their extremal value (without loss of generality we consider x_ 0 = 0 and y_ 0 = 0). From the rst equation of (2.29) it yields z0z_ 0 = 0. There are two possibilities: z_ 0 = 0 or z0 = 0. The latter means that the tangent t lies in the x{y plane. But the rst means that if the tangent to the string is not lying in the x{y plane then all the velocity projections and thus the whole velocity vector reach their extremum at the same point. In the case of plane motion the conclusion is even stronger. Really in this case z 0 and thus the system (2.30) can be written in the form of system of linear equations: 2 38 9 8 9 > > > > 0 0 64 x_ y_ 75 < x0 = = < 0 = (2.32) ; > :0> ; : y0 > x00 y00 >


21

that always has nontrivial solution since x02 + y02 = 1. The conclusion is that the determinant of the system is zero, or x_ 0y00 = y_ 0x00. Suppose that the string in the point s is not straight, i.e. x002 + y002 6= 0. From the determinant yields that if x_ 0 = 0 then x00 = 0 or y_ 0 = 0. If x00 = 0 then we have y0y00 = 0 and y0 = 0 since we have supposed that the string is not straight (x00 = 0 and x002 + y002 6= 0 mean y00 6= 0). But that means that the tangent coincides with x and the conclusion is that the tangent velocity reached its extremum. Otherwise we have to conclude that in the plain motion of the string all the velocity projections reach their extremum in the same point if in this point the coordinate axes are not coincide with the tangent and normal axes.

2.3 Formulation of problem in derivatives Now we return to the extensible string. Beginning from this section we consider the string of uniform mass per unit length = const. Let us divide our equations by and move to the relative force f = F= and to the relative distributed load q = Q=. Note, that the dimension of f is velocity squared and the dimension of q is acceleration. It will be shown that f= is, in fact, the velocity of propagation of discontinuities squared. In this section we derive the equations for the derivatives and discuss the characteristics of the equations in the case of the plane motion. Consider the equation (2.3) after dividing by : 0 !0 R R = f + q (2.33) Derive it with respect to s and substitute R0 = t where t is the tangent vector. We receive: (t)= (f t)00 + q0 (2.34) when f = f () is speci ed. After opening the parentheses we obtain the following vector equation when we are


22

working with the homogeneous case q = 0: t + 2_ t_ + t = f 00t + 2f 0t0 + f t00

(2.35)

We present below two dierent formulations based on various representations of the unit tangent vector and its derivatives. The rst one describes the tangent vector in the xed (non-rotating) frame fi; j; kg by means of two angles: the pitching angle that is the deviation of the tangent vector from the \vertical" axis z and the azimuth angle ' that is the rotation of the t{k plane from the i{k plane (Note, that ' is not de ned uniquely in the case when = 0). The two angles are presented at Fig. 2.6. k θ

t

j

i

ϕ

Figure 2.6: De nition of azimuth and pitching angles. In these parameters we can write t = i sin cos ' + j sin sin ' + k cos . Now let us derive the latter expression and substitute to the equation (2.35). After dividing of the vector equation into three scalar components and combining them in the appropriate way we obtain the following system: 8 _2 2 2 00 02 02 2 > > + '_ sin = f f + ' sin > < (2.36) 2_ _ + = 2f 0 0 + f00 > > > : 2_ '_ + ' + 2'_ _ cot = 2f 0 '0 + f'00 + 2f'00 cot


23

and since f = f () is speci ed we can substitute also: 8 df > > < f 0 = d 0 (2.37) 2 > > : f 00 = d f2 02 + df 00 d d The above representation presents, however, a problem that the coordinates present in the equations directly (the pitching angle is written explicitly in the equations) while the original formulation (2.33) is written in derivatives only (the position vector R is not presented explicitly). To avoid this, another formulation based on the natural triedr ft; n; bg can be derived. Let us present two vectors to describe the rotation of the natural triedr: the vector of angular velocity ! such, that t_ = ! t, n_ = ! n and b_ = ! b and the vector of space rotation such, that t0 = t, n0 = n and b0 = b. Now we can write ! = !t t + !n n + !b b and in the similar way = t t + n n + b b. The Frenet formulae in the Lagrangian coordinates (2.11) imply t = , n = 0 and b = and it can be easily veri ed by direct dierentiation that !_ = !_ t t + !_ n n + !_ b b and = t 0t + b 0b. We have to establish the connection between the angular velocity and the space rotation. Recall, that the mixed second derivatives are independent of the order of 0 dierentiation, or (t0)_ = t_ and the same can be written for n and b. Now, that means, that ( t)_= (! t)0, or: 0

_ t + (! t) = ! 0 t + ! ( t)

and after some simpli cation we obtain: _ !0 + ! t = 0 and since similar expressions has to be valid for normal unit vector n and binormal unit vector b we can write nally: _ ! 0 = ! (2.38)

CHAPTER 2. DERIVATION OF EQUATIONS OF MOTION In these terms formula (2.35) becomes: t + 2_ ! t + !_ t + (! (! t) !2t) = = f 00t + 2f 0 t + f 0 t + f ( ( t) 2t)

24 (2.39)

Now let us write the equations (2.39) and (2.38) in projections on tangent, normal and binormal axis: 8 2 2 2 00 > > > (!n + !b ) = f fb > 2_ !b + (!_ b + !t!n ) = 2f 0b + fb 0 > > > < 2_ !n + (!_n !t!b) = ftb (2.40) > 0t _ t = !n b > ! > > > !n0 = t!b !t b > > > : _ ! 0 = ! b

b

t n

and the above system is complemented by the representation of tension in terms of stretch (2.37). For the model of the inextensible string we have no elongation 1, so equations (2.36) take the form: 8 > f 00 = f 02 + '02 sin2 _2 + '_ 2 sin2 > < (2.41) = 2f 00 + f00 > > > : ' + 2'_ _ cot = 2f 0 '0 + f'00 + 2f'00 cot and the equations (2.40) become: 8 > f 00 = fb 2 (!n 2 + !b2) > > > > 2f 0b + fb 0 = !_ b + !t !n > > > < !_ n !t!b = ftb (2.42) > 0 _ t = !n b > ! t > > > !n0 = t!b !t b > > : _ b !b 0 = t!n Initial conditions for this problem are the initial curvature, torsion and angular velocity. Boundary conditions de ned in section 2.1 for the regular formulation in coordinates can be rewritten partly in terms of derivatives.


25

1. When the motion at the endpoint is speci ed R (0; t) = U(t) we have to use the equation (2.33) at this endpoint:

U = f 0t + f t0

(2.43)

If U(t) = 0 then we have f 0(0; t) = 0, '0(0; t) = 0 and 0(0; t) = 0 or b = 0. If U(t) 6= 0 we cannot write the boundary conditions for the derivatives and to enforce those conditions we have to integrate our equations completely. 2. The boundary conditions on force enforced in this formulation in the same way as in the original formulation since the tension force presents in this system. 3. The boundary condition of mass at the endpoint (2.25) cannot be rewritten in terms of derivatives and demands the complete integration.

2.3.1 Formulation in derivatives for plane motion. Characteristics The plane motion is a special case of the system (2.36) when the pitching is zero ( = =2). It is also the special case of (2.40) when t = !t = !n = 0 and b = '0 and !b = '_ . The angle ' in the case of the plane motion is tangent angle of the string so that x0 = cos ' and y0 = sin '. The equations that will be obtained in this section could be found yet in (Routh, 1955) but here we present, probably, the most elegant way of the derivation of the equations. Consider our vector written in complex notation R = x + iy. Then 0 = ei' and equation (2.34) could be re-written: i' i'00 (2.44) e = fe where we work again with the case q0 = 0. Now after performing dierentiation, dividing both parts by ei', splitting the equation into real and imaginary parts and substituting


26

f = f () we obtain:

8 2 > < f'02 '_ 2 = df 00 + d f2 02 d d > df 0 0 00 : 2 ' + f' = 2_ '_ + ' d and for the inextensible string: 8 > < f'02 '_ 2 = f 00 > : 2f 0'0 + f'00 = '

(2.45)

(2.46)

Now we derive equations of characteristics. Note that the rst equation in system (2.45) (and in (2.46) as well) is of the second order in (or f ) but of the rst order in ' while the second equation is of the second order in ' but of rst order in (f ). We denote '0 = since this is the curvature (although curvature was de ned non-negative in the previous sections, it is more convenient here to de ne it to be positive or negative as well, so in this section we rede ne it), '_ = ! since this is the angular velocity at point s, 0 = , _ = , f 0 = q and f_ = q~. In system (2.45) the second equation gives us the system: 9 9 8 2 38 > > > > 00 d > > > ' 66 ds dt 0 77 > > > > > = 66 0 ds dt 77 < '_ 0 = = < (2.47) d! 64 > 75 > > > > > > > ; : 2! 2 df > : ' > ; > f 0 > d On the characteristic lines the determinant of this system has to be zero, and since the system still has the solution on this lines the rank of the expanded matrix (the left-hand side matrix and the right-hand side vector) has to be equal 2. These conditions give us two equations of characteristic curves and equations for the variables on this curves: 8 s s ! > ds f f df d! 2 d > = d ! > < dt = and dt dt s s (2.48) ! > f f ds d df d! 2 > > : dt = and dt + dt = d ! These equations de ne hyperbolic equation which is the equation of propagation of transverse discontinuities (in '). So it can be seen here that the tension-to-stretch ratio f= is


27

the velocity of the propagation of the transverse discontinuities squared. The rst equation of (2.45) gives:

9 9 8 2 38 > > > > 00 d ds dt 0 > > > 66 77 > > > > > = < = < 66 0 ds dt 77 _ 0 = d (2.49) > 64 df > > 75 > 2 > > > > > ; : f2 !2 d f2 2 > ; > : > d 0 1 d which gives the characteristics: s s 8 > ds df df d = d2f 2 + !2 f2 d > = and < dt d dt s d dt d2 s (2.50) > 2f df df d d d ds > 2 2 2 : = dt d and dt + d dt = d2 + ! f These two equations de ne hyperbolic equation which describes propagation of longitudinal discontinuities (in ). It can be seen that the derivative ddf is the velocity of the propagation of the longitudinal discontinuities squared. In the case of linear elasticity f = EA0( 1) and ddf = EA0 / that is the sound velocity squared. For the inextensible string we have from the second equation of (2.46): 9 38 9 8 2 > > > > 00 > > > ' d 66 ds dt 0 77 > > > > > < < = 66 0 ds dt 77 '_ 0 = d! = (2.51) 75 > > > 64 > : ' > ; : 2q > ; > f 0 1 > which gives us: 8 ds q d! qf d = 2q > > f and = < dt dt dt (2.52) q q > > : ds = f and d! + f d = 2q dt dt dt and the rst equation of (2.46) gives: 9 8 9 2 38 > > > > 00 > > > > f d q 66 ds dt 0 77 > > > > < = < = 66 0 ds dt 77 f_0 = (2.53) d q ~ 64 75 > > > > > > > > : f > ; > : f2 !2 > ; 0 0 1 > and that gives only one characteristic:

dq = f2 !2 dt = 0 and ds

(2.54)


28

It can be seen that in the case of inextensible string the equation for tension is of parabolic kind. In, fact this single characteristic yields from the two characteristic lines of the elastic string when ddf ! 1, namely, the case when the modulus of elasticity tends to in nity.

2.3.2 Deriving equation for tangent angle Here we derive an equation for the tangent angle ' for the plane motion of the inextensible string by eliminating f from the system (2.46). To eliminate f we derive the second equation in (2.46) with respect to s: 2f 00'0 + 3f 0'00 + f'000 = '0

(2.55)

now we substitute f 00 from the rst equation of (2.46), f 0 from the second equation of (2.46) and obtain the tension f explicitly: 0 '0 3'' 00 + 4'_ 2 '0 2 2 ' f = 2'000'0 3'002 + 4'04

(2.56)

To obtain the equation for ' one has now to substitute this expression for f into the second equation of (2.46) and after performing simpli cations we receive: (2.57) ('00 + 4'0'_ '_ 0) 2'0'000 3'002 + 4'04 IV ' + 3'02'00 2'0'0 3'00' + 4'_ 2'02 + +2'0'00 2'000 5'03 + 3'_ 2'00 6'0'000 5'002 + +2' 9'02'002 2'0002 5'03'000 2'06 = 0 It can be seen that this equation is homogeneous, i. e. each term has total of 6 derivatives with respect to s and 2 derivatives with respect to t. In the next sections we will see that the solutions of the kind F (st 31 ) are very important and wide class.


29

2.4 Formulation in integrals In the previous section we obtained equations in derivatives by deriving the original equation with respect to s. Now let us take the original equation (2.33): 0 !0 R R = f + q R R R and integrate it with respect to s. We denote R = R ds, Q = q ds, X = x ds, Y = R y ds and Z = R z ds and obtain:

R = f R00 + Q

where = jR00j

(2.58)

It is clearly seen that the tension-to-stretch ratio f= is the velocity of the propagation of the transverse discontinuities, as it was stated in section 2.3.1. Again to complete the formulation, f = f () has to be given. We work with the case Q = 0. Since p = R00 R00 we perform the scalar multiplication of (2.58) by R00 and obtain the relation for f : R00 f = 1 R (2.59)

Now let us move to the plain motion of inextensible string. In this case 1 and we obtain: 8 > 00 > > < X = fX (2.60) Y = fY 00 > > > X 002 + Y 002 = 1 : 00 = Y =Y 00 while from the third equation of the above system and from here f = X=X ' = arccos X 00 = arcsin Y 00. The system (2.60) can be transformed into an equation of the fourth order for a single function X or Y by substitution of the relations from (2.60) into the rst equation of (2.46): 8 !00 X X 0002 _ 002 X X > > < X 00 = X 00 1 X 002 1 X 002 (2.61) > !00 Y Y 0002 _ 002 Y Y > > : Y 00 = Y 00 1 Y 002 1 Y 002


30

or after performing the dierentiation: 00 2 X 002 X_ 002 IV = X 0002 3X 0022 X00 + 2X 000X 0 X 00X 00 XX (2.62) 1 X X 1 X 002 and similar equation for Y . These are the simplest equations for single variable that can be obtained from the original system. In particular, they seem to be much simpler than the equation (2.57) for the tangent angle.

2.5 Formulation for plane motion of inextensible string in natural parameters The \natural parameters" are those that are independent on the coordinate system. For the planar motion of inextensible string these are the normal component of velocity v, the tangent velocity u, the tension f , the curvature and the angular velocity. The latter two can be expressed in terms of the tangent angle ': the curvature = '0 and the angular velocity ! = '_ . Here we de ne the tangent velocity to be positive if it coincides with the tangent unit vector t = R0 and the normal velocity to be positive if it is directed to the left from the tangent unit vector. y . y

u v ϕ . x

x

Figure 2.7: Transformation to the normal and tangent velocities.

CHAPTER 2. DERIVATION OF EQUATIONS OF MOTION From the gure 2.7 one can conclude that: 8 > < x_ = u cos ' v sin ' > : y_ = u sin ' + v cos '

31

(2.63)

After substituting of (2.63) into the rst equation of (2.29) and simplifying we obtain u0 v'0 = 0. In section 4.1 we will show that this equation expresses the conservation of mass on the string. It is rather obvious that (x_ )0 = (x_0) and (y_ )0 = (y_0). After substituting of (2.63) into the left-hand side of these equations and of the de nitions x0 = cos ' and y0 = sin ' into the right-hand side and performing the simpli cations we obtain that '_ = v0 + u'0. Next we substitute the relations (2.63) into the equations of motion (2.4) (we remind that = const and we divide the original equations by it to obtain the relative tension f ) in homogeneous case q = 0 and after projecting on the tangent and normal, obtain the equation of momentum on the tangent f 0 = u_ v'_ and the equation of momentum on the normal f'0 = v_ + u'_ . Summarizing the equations, we obtain the following system of 4 equations of the rst order: 8 > u0 v'0 = 0 > > > < v0 + u'0 = '_ (2.64) > 0 > u _ v ' _ = f > > : v_ + u'_ = f'0 To obtain the equations of characteristic lines for this formulation we have to write

CHAPTER 2. DERIVATION OF EQUATIONS OF MOTION our system (2.64) in the following matrix form: 2 66 1 0 0 0 v 0 0 66 0 0 1 0 u 1 0 66 66 0 1 0 0 0 v 1 66 66 0 0 0 1 f u 0 66 66 ds dt 0 0 0 0 0 66 66 0 0 ds dt 0 0 0 66 0 0 0 0 ds dt 0 64 0 0 0 0 0 0 ds

38 > u0 0 7> > 7> 0 777 > > u_ 77 > > v0 0 77 > > 7> < v_ 0 77 > 77 > 0 ' 0 77 > > 77 > '_ 0 77 > > > 7 f0 0 77 > > > 5: _ f dt >

9 8 9 > > > > > 0 > > > > > > > > > > 0 > > > > > > > > > > > > 0 > > > > > > > > = = < 0 > => > > > du > > > > > > > > > > > dv > > > > > > > > > > d' > > > > > > > > ; ; : df >

32

(2.65)

and analyze it by the method described in section 2.3.1. We obtain the following relations that are the analogs of the equations (2.52) and (2.54) in the variables de ned in this section: 8 > > > < ds = cdt and dv + (u c)d' = 0 (2.66) ds = cdt and dv + (u + c)d' = 0 > > > : dt = 0 and du vd' = 0 p Where c = f is the wave velocity. The rst two equations of the above system are the equations of propagation of the transverse discontinuities written in the terms of natural parameters and the third equation is the parabolic \double-characteristic".

2.6 Formulation for disturbance from initial motion Many authors (for example, Takahashi in (Takahashi and Konishi, 1987a) and (Takahashi and Konishi, 1987b) or Narasimha in (Narasimha, 1968)) write the equations of motion in form of displacement from the initial position. Those equations are generally much more complicated than the original set. This approach seems to be very inecient in studying of the large vibrations, although \the computer will eat it all". However, in analyzing of the small disturbances this formulation will be ecient and in this section we will show it.


33

Consider the original set of equations (2.3) and present the solution as sum of two vectors R = R0 + R1. Consider R0 as a known solution that satis es the equations (2.3): F0 0 (2.67) R0 = R00 + Q0 0 where Q0 is the distributed load as function of this known solution R0 and substitute R into the equations (2.3): !0 !0 F () F () 0 0 R0 + R1 = R0 + R1 + Q (R0 + R1) (2.68) q q where = (R00 + R10) (R00 + R10) = 02 + 2R00 R10 + 12. We will proceed with an equation of inextensible string taking F = F0 + F1:

R 0 + R 1 = (F0R00)0 +(F0R10)0 +(F1R00)0 +(F1R10)0 + Q and 2R00 R10 + R10 R10 = 0 (2.69) since R00 R00 = 1. Substituting the equation (2.67) we obtain nally: R 1 = (F0R10)0 + (F1R00)0 + (F1R10)0 + Q Q0 and 2R00 R10 + R10 R10 = 0 (2.70) We now consider a small disturbance from the known motion R0. In this case we can neglect all the quadratic terms of this disturbance and also expand Q in the Taylor series. We obtain:

R 1 = (F0R10)0 + (F1R00)0 + (R0 rQ) R1 and R00 R10 = 0

(2.71)

The above equation is linear in R1, so in this case the process of the solution, generally speaking, will be much simpler. Note also that we can consider that the tangent of the disturbance coincides with normal to the initial solution as yields from the second equation of (2.71). The main eld of application of the equations (2.71) seems to be in the analysis of the stability of motion of the string.

Chapter 3 Investigation of discontinuities In the following chapter we derive necessary conditions that have to be satis ed on discontinuities in the tangent vector (bends) and in curvature. The examples of discontinuities along the lines of constant and non-constant force will be shown.

3.1 Case of constant tension (D'Alembert solution) We consider planar homogeneous motion of inextensible string, i. e. the case of zero external load excluding the endpoints of the string where the tension force is applied and interpreted by us as a boundary condition. Let the internal tension force be constant, i. e. f = const = c2. We rewrite the system (2.46) in section 2.3.1 in derivatives in this case: 8 > < (c'0 + '_ )(c'0 '_ ) = 0 (3.1) > : c2'00 ' = 0 The second equation of this system gives us the well-known D'Alembert solution ' = 1(s + ct) + 2(s ct). But the rst equation gives two equations that only one of them must be zero: c'0 + '_ = 0 or c'0 '_ = 0 (3.2) and that means that ' = 1(s + ct) or ' = 2(s 34

ct). The superposition of such

CHAPTER 3. INVESTIGATION OF DISCONTINUITIES

35

steady-state waves is impossible, and the reason of this phenomenon is the nonlinearity. Let us show on the following example the initiation of a wave of the constant tension that is joined with the rest of the string in the point of discontinuity in tangent vector. y f = c2

l (t) R α0

x

Figure 3.1: Supposed solution for semi-in nite initially straight string drawn with constant force. Consider a straight, semi-in nite, inextensible string that until the moment t = 0 was in rest (R_ (s; 0) = 0) and from this moment is drawn from its free end with constant force f = c2 directed with constant angle 0 6= 0. We seek the solution in the form R = lr (s=l) and also f = f (s=l) where l = l(t) in the interval [0; l] and the initial straight con guration on the rest of the string as shown in Fig. 3.1. The point s = l(t) is a point of discontinuity. The inextensibility equation R0 R0 = 1 gives us r0 r0 = 1 where prime denotes the derivative with respect to the non-dimensional variable = s=l and by _ 2 = l=l _ deriving this relation with respect to we obtain r0 r00 = 0. Note that _ = sl=l and thus R = l (r r0) + l_22r00=l. Substituting this into (2.30) and multiplying by l gives: f 0 = ll (r0 r ) (3.3) This equation can be integrated with respect to . We obtain: f = 21 ll r2 2 + f0(t)

(3.4)


36

The left-hand part of this equation is the function of only while the right-hand part is the sum of products of functions of on functions of t and f0(t) is the constant of integration independent on . Notice that at = 1 we have x = l and y = 0 or, another words, r = 1. Also this is the point where two dierent solutions are joining together and so it has to lie on the characteristic line. The conclusion is that f ( = 1) = l_2 (the equation (2.52) for the characteristics of the inextensible string). Since f 6= f (t), we have l_ = const and f0(t) = f (1) = const. The fact that l_ = const means that l = 0 and f = const = f (0) = c2. Since l(0) = 0 we obtain l(t) = ct. The equation of motion (returning to s, t and R) is R = c2R00 with boundary conditions R0(0; t) = cos 0i + sin 0j and R(ct; t) = cti while the solution has to be R = ctr (s=ct). Substituting the solution into the equation we obtain (s2 c2t2) r00 = 0 that has to be satis ed at each s and t and the conclusion is, obviously, r = C1 + C2 s=ct, or R = C1ct + C2s. Substituting of boundary conditions nally gives us the solution:

R = ict + ( i cos 0 + j sin 0) (s ct)

(3.5)

that is the straight line with the inclination that coincides with the one that is directed along the force on the section s 2 [0; ct) while on the rest of the string s 2 (ct; 1) we still have R = is (the initial straight con guration). Note that we can change the direction of the force at the end at any time and this results in another bend (like the one at s = ct). By changing the direction of the force continuously we can initiate the wave of any desired shape propagating with velocity c along the string. The solution described above is a special case of self-similar waves that will be described in chapter 5. Next we show that from all tensions for which f 00 = 0 the only one that satis es the problem is the constant tension if the solution is not a straight line. Really, if f 00 = 0 then f = G1(t) + sG2(t), where G1(t) and G2(t) are arbitrary functions of time. The equations of motion (2.46) are written then: 8 > < 0 = '02(sG2 + G1) '_ 2 (3.6) > : ' = 2'0G2 + '00(sG2 + G1)


37

We take the rst equation and dierentiate it with respect to t and to s, eliminate '_ 0 and substitute '_ 2 from the rst equation. The result is: _ _ 1 0 s G + G 2 1 1 2 0 00 0 (3.7) 2(sG2 + G1 )' @ 2 ' G2 + 2(sG + G ) '_ + ' (sG2 + G1) 'A = 0 2 1 and now the substitution of the second equation of (3.6) gives: ! _ 2 + G_ 1 s G 2 0 0 ' (sG2 + G1) '_ sG + G 3' G2 = 0 (3.8) 2 1 There are two basic possibilities. The rst is the straight string (no curvature, '0 = 0) and in this case it is possible f = G1 + sG2 . The second case is when the expression in the parentheses vanishes. This one gives after eliminating of '0, combining the coecients at the appropriate powers of s and setting them to zero, the following system: 8 > _2 > > < G2 = 0 (3.9) 2G_ 2 G_ 1 = 9G32 > > > : G_ 21 = 9G22G1 From the rst two equation of the above system we obtain G2 = 0 and from the third G1 = const and that proves our proposition.

3.2 Necessary conditions on discontinuities Let us derive necessary conditions for the discontinuities propagating along the string. Here we mean the discontinuities in the tangent vector or in its derivatives, namely the curvature. Recall again the system of equations (2.46) from section 2.3.1 for the plane motion of the inextensible string in derivatives: 8 > < f 00 = f'02 '_ 2 > : 2f 0'0 + f'00 = ' and suppose that the discontinuity in tangent vector (i. e. in ') or in curvature (i. e. in '0) propagates along the line s = `(t). From the equations of characteristics (2.52) we


38

know that on this line f = `_2 , and that means that the tension force is continuous on the discontinuity of this kind. Now notice that the rst equation of the above system is a non-linear equation of the rst order for '. If we suppose that the tension force f is given, we can derive the equations of the characteristics for this equation: ds = dt = d' = d '0 = d '_ (3.10) _ 02 2f'0 2'_ 2f 00 f 000 f 0'02 f_00 f' and from here: ds = f '0 (3.11) dt '_

It is a well-known fact from the theory of the partial dierential equations that the discontinuities can propagate only along the characteristic lines (it can be found, for example in (Courant and Hilbert, 1962), volume II, (Rozdestvenski and Janenko, 1983); there exists also the speci c theorem proved by Courant and Lax in (Courant and Lax, 1956)). It means that on the discontinuity the characteristic (3.11) must be the same as pf = f'0='_ or, nally: = the characteristic (2.52), or ds dt q (3.12) '0 f + '_ = 0 and the rst equation of (2.46) gives us now: q q f 00 = f'0 '_ f'0 + '_ = 0

(3.13)

Let us move now into the coordinate system connected with the moving discontinuity. We choose new variables: = s `(t) and t, that is the time that describes the events in this moving coordinate system. In this variables '0 = ' and '_ = ' `_ + 't (we denote by subscript the derivative with respect to certain variable) and the equation (3.12) gives us: 't = 0 (3.14) which means that there is no rotation of the string in the moving coordinate system around the point of discontinuity. We rewrite also the equation (2.4) for the motion of the inextensible string: R = (f R0)0


39

in this new variables:

f R + f R = `_2 R `R 2`_ Rt + Rtt

(3.15)

Since R = R0 = t (tangent unit vector) and on the discontinuity f = `_2 and also (as we have just proved) there is no rotation, or Rt = 0, the equation (3.15) transforms on the discontinuity into: f + ` t = Rtt (3.16) Note that in our moving coordinate system the point of discontinuity is R(0; t) and it is the continuous function of t since the motion of the point of the discontinuity itself in space is continuous. This implies that Rt and Rtt are continuous. Considering this we obtain: + + f + ` t = f + ` t (3.17) where the `minus' superscript denotes the parameters right before the point of discontinuity and the `plus' superscript | right after it. Here we distinguish two cases: 1. The discontinuity in tangent vector (bend). In this case t 6= t+ and the equation (3.17) yields: f + = f = ` (3.18) Let us now rewrite the second equation of (2.46) in the moving coordinate system: 2f ' + f' = ' `_2 ' ` 2't `_ + 'tt

(3.19)

Since on a discontinuity f = `_2 and 't = 0 (and thus 'tt = 0) and on the discontinuity of this kind (bend) it is also f = `, the equation (3.19) gives us: _ t `' = 2`'

(3.20)

that is true on both sides of the point of the bend. This equation can be integrated with respect to t and we obtain: q 8 > < ' = C0 l_ q (3.21) > : ' + = C0+ l_


40

where C0+ and C0 are constants of integration. 2. t = t+ | the discontinuity in the curvature. Here the equation (3.17) gives us f + + ` = f + `, or: f = f + (3.22) Let us summarize the results of this section: The necessary conditions on discontinuity in tangent vector or curvature in plane motion of inextensible string, are: 1. No rotation around the point of discontinuity 't = 0. 2. The tension force is continuous and f = `_2 where s = `(t) is the line of discontinuity. 3. The tension gradient f 0 is continuous and on the sharp bend f 0 = `. 4. The second derivative of the tension force is zero f 00 = 0. 5. If the discontinuity is a bend, then the curvature on both sides of the bend is proporq tional to the square root of the velocity of the propagation of the bend '0 = C0 `_.

3.3 Bend with non-constant force In section 3.1 we have presented the discontinuities that propagate with constant velocity. Now we show an example of a solution where the bend propagates along the string with velocity that varies in time. Consider a nite inextensible string of length L drawn from its end with a tension force F directed with an angle 0 starting from the moment t = 0. Let us nd such a tension force that provides the bend, that propagates along the string with non-constant velocity. We suppose that the solution is a straight line the direction of which coincides with the direction of the force before the bend, and the remaining \tail" after the bend that


41

has only tangent velocity. To prove that such a solution is possible we have to show that the normal velocity of the part before the bend is constant along s. y1

y x1

F U V α0 x

Figure 3.2: Supposed solution for nite initially straight string drawn with constant force. Since the solution consists of two straight pieces and we have shown in section 3.1 that the tension force is linear along a straight non-rotating piece and we have also shown in the previous section that the derivative of the tension force is continuous on the bend, the solution for the tension force is:

f = FL (L s)

(3.23)

and if we denote by l the coordinate of the bend that equals to the length of the moving part before the bend, then the tension on it is l_2 = F (L l)=L. We choose the origin of the coordinate system in the middle of the string in its initial position. Now let us rotate the coordinate system by angle 0 so, that the new axis x1 is normal to the direction of the force and the axis y1 is parallel to it (Fig. 3.2 illustrates these designations). We denote by V the y1-coordinate of the center of gravity of the string and by U the x1-coordinate of the bend in this new coordinate system. We can write the equations of motion of the center of gravity nd from them we conclude that the center of gravity does


42

not move in x1-direction since there is no projection of the external force on it and:

F = LV

(3.24)

The equations that determine U and V are the equations of position of the center of gravity in x1: U 1 U U 1 (3.25) U l + 2 U sin = 2 L l sin sin 0 L l sin 0 0 0 and in y1: ! l U tan 0 + 2 V l + 12 U tan 0 V sinU = (3.26) 0 1 U U = 2 L l sin cos 0 + V L l sin 0

0

The equations (3.25) and (3.26) yield: U = 21L (L l)2 sin 0 V = 21L l2

(3.27) (3.28)

and by using (3.28), (3.24) and (3.23) we receive the equation to determine l: (L l)l = l_2 with the initial conditions l(0) = 0 and F (0) = c2. The solution is: q p l = L L2 cLt 2 and the tension required is:

(3.29)

(3.30)

L3 (3.31) p 3=2 L2 cLt 2 Now we have to check that the laws of conservation of mass and momentum are satis ed on the bend by our solution. From the Fig. 3.3 one sees that the part of the _ tan 0)dt while string before the bend receives in the time interval dt the mass (u2 + U= _ sin 0)dt and both mass changes have to be equal to the \tail" looses the mass (u1 + U= F = c2


43

u1 dt

l

. U dt

. l + l dt

α0

u2 dt

Figure 3.3: The motion of the string in the time interval dt.

l_dt. We also have to satisfy the equations of momenta, that are: in the projection onto the direction of the \head" u2l_ u1l_ cos 0 = l_2 l_2 cos 0 and onto the direction of the \tail" u2l_ cos 0 u1l_ = l_2 cos 0 l_2 since the tension force on both sides of the bend is f = l_2. These equations can be easily checked and veri ed. Note also that: (3.32) U_ = p1 c sin 0 = const 2 and that means that all the \head" is moving with the constant normal velocity, hence its straight shape is preserved. It can be checked also that on the bend the condition f 0 = l is satis ed and this is another veri cation of the theorem that we have proved in the previous section. Also, when the string tends to semi-in nite (L ! 1) we can see that the linear term in the denominator of (3.31) vanishes and the expression becomes F = c2 | the solution for semi-in nite string that we have studied in section 3.1. At the moment when the bend reaches the end of the string as we can notice, the L and this leads to in nite force in (3.31). So, the bend that equation (3.30) gives t = p c 2 propagates in the nite straight strings leads to a strong impact (a singularity of type x 3=2 when x ! 0) at the end of the propagation.

Chapter 4 Investigation of bends In the previous chapter we have obtained the necessary conditions that are valid for the discontinuities of general kind. In this section we will investigate the discontinuities in tangent vector, or bends. In particular, we will consider the case of collision of two bends and will show that in some cases it leads to apping, that is the cases when the tension force suddenly grows.

4.1 General laws of mechanics for bends First, let us rewrite the equations of motion formulated in natural parameters (2.64) on the line of discontinuity (of any type) s = `(t) in the variables that have been introduced in section 3.2: 8 > u v' = 0 > > > < v + u' = 't ' `_ (4.1) > _ _ > u u ` v' + v' ` = f t t > > : vt v `_ + u't u' `_ = f' and after performing the substitution f = `_2 on the line of discontinuity and some simpli cation we obtain: 8 > < ut v't = f (4.2) > : vt + u `_ 't = 0 44

CHAPTER 4. INVESTIGATION OF BENDS

45

and if we recall that around the point of discontinuity there is no rotation 't = 0 (it was proven in section 3.2) we obtain: 8 > < ut = f (4.3) > : vt = 0 The main conclusion from the equations (4.3) is, that the normal velocity is constant in time on both sides of the discontinuity. We want to apply the basic laws of mechanics on the bend, namely, the conservation of mass and momentum. First, we apply the law of the conservation of mass. We denote by subscript 1 the parameters of the motion of the string right before the bend and by subscript 2 the parameters of the motion of the string right after the bend. We have proved above that the normal velocities v1 and v2 are constant in time. In the time interval dt the bend moves by v1dt in the direction of the normal vector before the bend and by v2dt in the direction of the normal vector after the bend. The new position of the bend can be calculated from the geometrical considerations as the point of crossing of two straight lines. Note, that we can omit the multiplier dt and work only with the velocities. Let us denote by d1 and d2 the velocities of the bend in tangent directions before and after the bend respectively when the positive d means enlarging of the interval. From the Fig. 4.1 it can be seen that: 8 v1 v2 v1 cos' v2 > > < d1 = tan' sin' = sin' (4.4) > v v cos ' v1 v 1 2 2 > : d2 = tan' sin' = sin' The velocity of loosing of length by the part before the bend is u1 d1 while the velocity of receiving of length by the part after the bend is u2 + d2. These are tangent velocities in the coordinate system that moves with the bend. The law of the conservation of mass implies the equality of the loosen and the received lengths. After some simpli cation we obtain: u1 cos 2' + v1 sin 2' = u2 cos 2' v2 sin 2' (4.5)


d2

46

v2 u2

d1 v1

∆ϕ u1

Figure 4.1: Motion of the bend. or:

u2 u1 = (v1 + v2) tan 2'

(4.6)

The meaning of the equation (4.5) is the equality of the projections of velocities on both sides of the bend on the direction normal to the bisector of the angle of the bend. This is an analog of Snell's law in geometrical optics and ray theory of wave propagation. Really, let us denote by V1 and V2 the total velocities of the parts before and after the bend and by 1 and 2 | the angles between these velocities and tangent directions. Then, from (4.5): V1 cos 1 2' = V2 cos 2 + 2' (4.7)

If we look at the equation (4.6) we can see that the right-hand side of this equation is constant in time (really, we have proved that there is no rotation around the point of the discontinuity and that the normal velocity is constant in time) and, therefore, the left-hand side of this equation have to be also constant in time, or, the dierence of the tangent velocities from the both sides of bend is constant. Note that this is true


47

for the discontinuity of any type. Really, it was already proved in section 3.2 that the tension gradient f 0 is continuous at the discontinuity in tangent angle or curvature and after substitution of this fact into the rst equation of (4.3) we obtain, that (u2 u1)t = 0. Note also, that when ' ! 0 the equation (4.6) takes the form du = v d' and this demonstrates that the rst equation in the system (2.64) expresses the law of the conservation of mass. Next let us examine the equation of momentum. We write it in projections on the tangent direction of the string before the bend and on the tangent direction after the bend. The projection before the bend gives:

f2 cos' f1 = (u1 d1) (u2 cos' v2 sin' u1) = (d1 u1)2 (cos' 1) (4.8) and after the bend:

f2 f1 cos' = (u2 + d2) (u2 u1 cos' v1 sin') = (d2 + u2)2 (1 cos') (4.9) Due to the conservation of mass we obtain:

f2 f1 cos' = f1 f2 cos' or:

(f2 f1)(1 + cos') = 0

(4.10)

and the conclusion is that the force is continuous on the bend as it was demonstrated in section 3.2 except the case of the folded string (' = ) where it can be discontinuous. In fact, the latter case is the case of plastic impact where the energy is not conserved (it was reviewed in (Steiner and Troger, 1995)). In all other cases the equation (4.8) or (4.9) gives: f = c2 (4.11) where c = d1 u1 is the velocity of the propagation of the bend.


48

4.2 Collision of bends Suppose that the condition (4.6) is not preserved at the bend starting from the moment t = t0, for example, because of the collision of two bends. In this section we will nd the reaction of the string at the point of impact at the time t0. We suppose that the bend splits into two new bends and seek for a solution that is a straight piece between the two new bends, that the laws described in the previous section are satis ed at each new bend (Fig. 4.2). On the straight piece we use the rst equation of (4.3) while remembering that f = (f2 f1)=D where D is the length of the piece. Substituting the tension force on the bend from (4.11) we obtain:

Du_ = D_ (d2 d1 2u) ∆ϕ

(4.12) u

1

d2 d1

∆ϕ 2

v

u2 v1 ∆ϕ v2

u1

Figure 4.2: Two new bends from the bend impact. The increasing of length D_ = d1 + d2 is constant since d1 and d2 are constant thus D = (d1 + d2)(t t0) (since the initial length of the new piece is zero). The solution of (4.12) is: u = (t C0t )2 + d2 2 d1 0


49

where C0 is a constant of integration, or, since the tangent velocity is bounded at t = t0: u = d2 2 d1 (4.13) Note that since the tangent velocity in the formula (4.13) is constant in time while the velocities u1 and u2 may be non-constant, the solution that we will obtain will be valid only at the moment of the collision. Summarizing the conditions on the point of impact: 8 > u u1 = (v + v1) tan '1 > 2 > < '2 (4.14) u2 u = (v + v2) tan 2 > > > : u = d2 d1 2 The angles '1 and '2 and their functions can be expressed in terms of velocities from the rst two equations of (4.14): 2 uv + uv 1 tan'1 = u u1 2 1 v + v1 1 u u 1 2v +v sin'1 = u u1 2 1 + v + v1 1 u u 2 2 tan'2 = vu+ v2u 2 2 1 v + v2 2 uv2+ vu sin'2 = u 2u 2 1 + v2+ v 2 and using these relations we can also express d1 and d2 in terms of the velocities by substitution of the above formulae into (4.4): 8 2 v 2 (u u1)2 v > 1 > < d1 = 2(u u1) (4.15) 2 v 2 (u2 u)2 > v > 2 : d2 = 2(u2 u)


50

After substitution of (4.15) into the third equation of (4.14) and some transformations, the normal velocity v of the \newborn" piece can be expressed in terms of its tangent velocity u and the velocities of the neighbored parts: 2 2 2 2 u2 u2 ) (u1 u) 2 v2 = (v1 + u1 u ) (u2 u u+) u+ (v22+ (4.16) u 1 2 Further, the angle ' between the two initial parts is known and ' = '1 + '2. We write tangent of sum: tan 2'1 + tan 2'2 ' tan 2 = 1 tan 2'1 tan 2'2

and substitute the tangents of the angles of the rst and the second bends from the rst two equations of (4.14). The substitution results in the following expression: ' v u2 u1 tan 2 (v2 + v1) = (4.17) ' 2 2 = v + u u(u2 + u1) + v1v2 + u1u2 tan 2 v2(u u1) v1(u2 u) or, substituting v2 from (4.16) and simplifying: ' (4.18) v u2 u1 tan 2 (v1 + v2) = = (v1 + v2) (v1(u2 u) + v2(uu1 + uu)) +2u2(u1 + u2)(u1 u)(u2 u) tan 2' 1 2 v2(u u1) v1(u2 u) We square (4.18) and substitute v2 from (4.16) again. That gives an equation of the fourth degree for the tangent velocity u. Two roots of this equation can be guessed: u = u1 and u = u2. The remaining equation of the second order was simpli ed with the program \Mathematica" and has the form: 4u2a0 + 4ua1 + a2 = 0 where a0 = VT IV, a1 = VT AV, a2 = VT CV and: 9 8 ' > = < u2 u1 tan (v1 + v2) > 2 V = > ; : v2 v1 + tan 2' (u1 + u2) >

(4.19)

CHAPTER 4. INVESTIGATION OF BENDS 3 2 I = 64 1 0 75 the 2 2 identity matrix 0 1 2 3 v + v 1 2 0 77 2 A = 664 v1 + v2 5 ( u + u ) 1 2 2 2 3 2 (v v )2 2(u v + u v ) ( u + u ) 2 1 2 1 1 2 7 C = 64 1 2 5 2(u2v1 + u1v2) 4u1u2

51

The equation (4.19) gives two roots:

q a1 a21 a0a2 (4.20) u= 2a0 and we have to choose the \right" one. The condition to be satis ed is that the length of the new interval is increasing, or d1 + d2 0. The substitution of d1 and d2 from (4.15) and simpli cation leads to the condition: v12 + u21 (v22 + u22) 0 (4.21) u1 + u2 2u q which means that if the velocity before the bend (V1 = u21 + v12) was greater than the velocity after the bend, the new tangent velocity u will be less than the mean of the initial tangent velocities u1 and u2, and if the velocity after the bend was greater, the new tangent velocity will be greater than that mean. Let us also determine the velocities of propagation of the new bends, namely, the square root of the new tension force: 8 > < c2 = d2 u = d2 (d2 d1)=2 = (d1 + d2)=2 (4.22) > : c1 = (u + d1) = ((d2 d1)=2 + d1) = (d1 + d2)=2 That means that the result of the collision of the two bends is two new bends \running" toward the both sides from the point of the collision with equal velocities of propagation. By virtue of (4.11) and knowing from section 3.1 that the tension force is linear on a straight piece we obtain that the tension force is constant on the whole interval.


52

To end this section, we have to discuss the case when V1 = V2. In this case after the substitution of (4.15) into the third equation of (4.14) instead of (4.16) we obtain: 2 V1 (u2 + v2) (u1 + u2 2u) = 0 (4.23) The possibility u2 + v2 = V12 and the condition (4.5) implies not only the equality of the values of velocities, but the equality of the vectors of the velocities on both sides. That means that the condition (4.5) is maintained on the initial bend, hence there is no splitting of the bend. Therefore, in this case

u = u1 +2 u2

(4.24)

We substitute the tangent velocity into the rst two equations of (4.14) to express the tangents of the bend angles and substituting them into the tangent of sum we obtain instead of (4.17) the following equation for the normal velocity v: (2v + v1 + v2)2 tan 2' 2(u2 u1)(2v + v1 + v2) tan 2' (v2 v1)2 + (u2 u1)2 = 0 (4.25) with the solution: 0 1 v u B (u2 u1)2 + (v v )2C u2 u1 u u v = 21 B ( v + v ) + (4.26) 1 2 2 1 C @ A ' t ' 2 tan 2 sin 2 or, in terms of the velocity V1 and the angles 1 and 2 that were de ned in the previous section: 0 1 s ' V

+

'

+

v = 1 ' @ sin 1 2 2 sin 2 2 1 + 2 sin 2 2 1 1 cos2 2 cos2 2 2 1 A sin 2 (4.27) The answer on the question which one of the root must be taken comes again from the condition of adding the length into the new interval. In this case instead of (4.21) we obtain: v2 + u2 V12 0 (4.28) u u 2

1

CHAPTER 4. INVESTIGATION OF BENDS or in terms of 1 and 2: 3 s 2 1 ' 2

+

'

+

' 1 2 1 2 sin 2 + 2 4 sin 2 cos 2 1 cos2 2 cos2 2 5 0

53

(4.29)

Since the expression in the square brackets in the condition (4.29) is always positive when the plus sign is chosen and always negative at the minus sign (that can be easily checked), the choice of the sign is dependent on the sign of the rst sine in (4.29). If it is positive, the \plus" root has to be chosen, if negative | the \minus" root.

4.3 Investigation of collision To understand the reaction of the string to the collision of bends we will describe here the model problem. Let us look at the bend that one of its sides was in rest (u1 = 0, v1 = 0). Since we can always choose the coordinate system that moves with one of the sides of the bend, every collision can be treated like this one. We de ne the angle | the angle between the velocity vector and the tangent velocity as it was done in section 4.1. If we also denote by V the velocity after the bend (V 2 = u22 + v22) as in the previous sections, the equation (4.19) takes the form: 4u2 4uV sin 2' sin + 2' V 2 cos2 + 2' = 0 (4.30) The condition (4.21) implies the choice of the root with the plus sign, so: s ' ! ' ' V ' 2 u = 2 sin 2 sin + 2 + 1 cos2 2 sin + 2

(4.31)

The tangent velocity is presented in Fig. 4.3 and its contour lines | in Fig. 4.4 The normal velocity v can be found from (4.18):

V ' 2 cos + 2 s ! ' 2 ' ' ' 2 sin 2 sin + 2 + 1 cos 2 sin + 2

v =

(4.32)


54

1 0.8 0.6 u/V 0.4 0.2 0 0.5 γ/π 1 1.5 2

1.5

0

0.5

1 ∆ϕ/π

Figure 4.3: The tangent velocity of the new interval. 0

0.5

∆ϕ/π 1

1.5

2

1.5

1 γ/π

0.5

0

Figure 4.4: The tangent velocity of the new interval | contour lines.

CHAPTER 4. INVESTIGATION OF BENDS 55 1 0 ' B ' CC sin 2 B B B@ s ' ' sin + 2 CCA ' ' cos cos + 1 cos2 sin2 +

2 2 2 2 | it is presented in Fig. 4.5 and its contour lines | in Fig. 4.6. The velocity of propagation of the bend can be calculated from (4.22): V=2 c=s (4.33) ' ' ' ' 1 cos2 2 sin2 + 2 cos 2 cos + 2 it is presented in Figs. 4.7 and 4.8. One can notice, that in most of the domain the propagation velocity is nearly constant and approximately equals to V=2. The angle of opening of the new bend '1 (in this section we denote it by ) can be calculated from the rst equation of (4.14): (4.34) = 2 arctan uv where u is substituted from (4.31) and v | from (4.32). It is presented in Figs. 4.9 and 4.10. The values of the arctangent are in the interval ( ; ], the \walls" in the Fig. 4.9 as well as the thick lines in the Fig. 4.10 are the lines where the angle reaches . The rst conclusion from the relations (4.31 { 4.34) is that the solution exists for each ' and . Really, it is easy to see, that the discriminant of the equation (4.30), that is the expression under the square root in (4.31 { 4.33) is always positive, hence the existence of the solution is guaranteed. Next, in the interval 2 2 when ' ! 0 the equation (4.32) can be presented as an asymptotic expansion: cos 1 v 4V sin (4.35) 2 ' and the equation (4.33) gives a similar asymptotic expansion: cos 1 c 8V sin (4.36) 2 '2 These singularities present the apping eect, when the tension in the string suddenly becomes unbounded. We visualize it in Figs. 4.11 and 4.12. In these points the tension


3 2 1 v/V 0 -1 -2 -3 0

0.5

γ/π

1

56

0.2 0.4 0.6 0.8 1 ∆ϕ/π 1.2 1.4 1.6 2 1.8

1.5

Figure 4.5: The normal velocity of the new interval. 0.2

0.4

0.6

0.8

∆ϕ/π 1

1.2

1.4

1.6

1.8 2

1.5

1 γ/π

0.5

0

Figure 4.6: The normal velocity of the new interval | contour lines.


57

4 3 c/V 2 1 0.2

0.6

γ/π 1

1.4

1.8

0.6 0.4 0.8 1.2 1 1.4 ∆ϕ/π 1.6

Figure 4.7: The velocity of propagation of the new bend. 0.4

0.6

0.8

∆ϕ/π 1

1.2

1.4

1.6 1.8 1.6 1.4 1.2 1 γ/π 0.8 0.6 0.4 0.2

Figure 4.8: The velocity of propagation of the new bend | contour lines.


58

1 0.5 Φ/π 0 -0.5 -1 0 2

0.5 1.5

∆ϕ/π 1 1.5

1 γ/π

0.5 2 0

Figure 4.9: The angle of the new bend. 0

0.5

∆ϕ/π 1

1.5

2

1.5

1 γ/π

0.5

0

Figure 4.10: The angle of the new bend | contour lines.


59

can easily tear the string, although there were no previous signs of failure. It is clear, that in this case the model of an inextensible string is not valid any more. The tension will not grow unboundedly, but due to the constitutive law of the material of the string.


60

60 50 40 v/V 30 20 10 0 0.04 0.08 ∆ϕ/π 0.12 0.16 0.2

1.6 1.8 1 1.2 1.4 0.8 0.6 0.2 0.4 γ/π

Figure 4.11: The apping in the normal velocity.

800 600 c/V 400 200 0 0.02 0.06 0.1 ∆ϕ/π

0.14 0.18

1.8 1.6 1.4 1.2 1 0.8 γ/π 0.6 0.4 0.2

Figure 4.12: The apping in the bend propagation velocity.

Chapter 5 Formulation of problem for self-similar waves In this chapter we will formulate ordinary dierential equations that describe a special kind of motion of the inextensible string, namely, self-similar waves. The solutions of partial dierential equations in a self-similar form is a widely studied eld in research. The self-similar motion of medium is connected with the dimensional theory described, for example, in (Sedov, 1959) | the monograph that studies the dimensional methods in mechanics. Nevertheless, self-similar waves in the string is a eld that seems not to be touched yet. In this chapter we will formulate the equations of the self-similar waves and in the next chapter we will try to describe the behavior of the string in the bends by means of the self-similar waves.

5.1 Self-similar rotating wave We have seen in the previous sections that the motion of the string is deeply connected with the propagation of waves in it. The regular linear waves are the solutions of form R = R(s ct) where c is a constant wave velocity. This variable presents the shift of the origin with constant velocity. In the previous sections we have studied the behavior of 61

CHAPTER 5. FORMULATION OF PROBLEM FOR SELF-SIMILAR WAVES

62

the discontinuities that are shifted with the non-linear function `(t). Now let us present a solution of a more general kind | we introduce the scale factor l(t), the phase shift W (t) and the rotation shift (t) such that: 8 > < x = lr() cos( () + ) (5.1) > : y = lr() sin( () + ) where = s W is a new non-dimensional variable. In fact, we have introduced here l self-similar non-dimensional functions to describe the position of the point of the string in polar coordinates | the scaled non-dimensional radius r and the polar angle . We can present the derivatives of the variable with respect to the old variables in terms of functions of t and : 8 > 0=1 > > l > > > > < 00 = 0 (5.2) _ > 1 > _ = l l + W_ > > > _ > > : = 1 l + W + 22l l_ + W_ l l Using the relations (5.1) we can rewrite the derivatives of x and y: First derivatives with respect to s: 8 > < x0 = r cos( + ) r sin( + ) > : y0 = r sin( + ) + r cos( + )

(5.3)

here and further the subscript denotes partial (or full) derivative with respect to . Second derivatives with respect to s: 8 i h 1 [2r + r ] > 00 = cos( + ) 1 r r 2 sin( +

) x > < l l i h > > : y00 = sin( + ) 1 r r 2 + cos( + ) 1 [2r + r ] l l

(5.4)


63

First derivatives with respect to time: 8 h i h i > < x_ = cos( + ) l_r r l_ + W_ sin( + ) l _ r r l_ + W_ (5.5) > : y_ = sin( + ) hl_r r l_ + W_ i + cos( + ) hl _ r r l_ + W_ i Second derivatives with respect to time: " 8 2# 2 1 1 > x = cos( + ) lr + r l l_ + W_ r l + W lr _ l l_ + W_ > > > > > 1 l + W + 2l_ _ r 1 l_ + W_ 2 > sin( +

) lr

+ > 2 > l l > > > > 2r l_ + W_ _ 1l l_ + W_ < " 2 # 2 > 1 1 > + y = sin( + ) lr + r l l_ + W_ r l + W lr _ l l_ + W_ > > > > 1 l + W + 2l_ _ r 1 l_ + W_ 2 > > + cos( +

) lr

+ 2 > l l > > > > : 2r l_ + W_ _ 1l l_ + W_ (5.6) The condition of inextensibility takes in the new variables the following form:

r2 + r2 2 = 1

(5.7)

and this allows us to express with r and r . Tangent and normal velocities are expressed with the new variables after substitution of (5.3) and (5.5) into 8 > : v = yx _ 0 xy _ 0 as:

8 > : v = l _ rr l_r2 The particle velocity takes the following form after substitution of the tangent and the normal velocities and using the condition of inextensibility (5.7): 2 V 2 = u2 + v2 = r2 l_2 + l2 _ 2 2r r l_ + r l _ l_ + W_ + l_ + W_ (5.9)


64

To obtain the equation of motion we will express the derivative of tension with respect to s as in (2.30) f 0 = x0x + y0y in the new variables expressing from the condition of inextensibility (5.7): q f 0 = l l _ 2 rr + 2l_ _ + l r 1 r2 l + W (5.10) 00 + y 00y and the tension itself as in (2.31) f = xx00x2 + y002 expressing from the condition of inextensibility (5.7) in the similar manner: _ 2 r2 1 r2 2 (5.11) f = l l l rr + r2 1 + l_ + W_ q q _ _ r2 r 1 r2 _ _ _ _ r 1 r2 l 2l + l rr + r2 1 + 2l l + W rr + r2 1

The nal equation is obtained by dierentiation of the latter equation (5.11) with respect to s and equating it with (5.10): 3 2 2 _ 2 r2 1 r2 r (5.12) l l 4 rr + r2 1 2 5 20 2 q 1 3 q 27 _ _ 6 r r 1 r2 2l + l 4@ rr + r2 1 A + r 1 r 5 + 3 3 2 2 _l2 _ r 1 r2 _l _ _ r 1 r2 + 2 l 4 rr + r2 1 5 + 2 l W 4 rr + r2 1 5 + ! ! _l2 _l _ + 2 +l + 2 W +W =0 l l The equation (5.12) is a sum of products of functions of two independent variables | t and . Let us analyze this relation. If all three functions of time | the scaling l(t), the phase shift W (t) and the rotation (t) are non-zero, there have to exist a system of three ordinary dierential equations to determine these functions. If we take any three dierent values of in the equation (5.12) this will give us a system of three dierential equations for the determination of the three functions of t mentioned above. The condition on this system will be that the rank of the matrix of the system will be three, i. e. that the


65

equations will be independent linearly. Another words, the system of equations for the determination of scaling, phase shift and rotation will have the following form according to (5.12): 9 8 2 > > _ l l

> > > _ _ > > > > 2l + l > > > > > > 2 _ _ = < 2l =l > =0 (5.13) [[A] [C]] > _W_ _ =l > > > 2 l > > > 2 _ > > 2l =l + l > > > > ; : 2l_W=l _ + W >

where A and C are the matrices of size 3 3 with non-zero determinant. Suppose also, that all the functions in the vector of the expressions of t in (5.13) are non-zero. Then we rewrite the rst three expressions in the vector in terms of the last three: 8 8 9 9 > > > > 2 _ _ _ _ > > > > l l

2 l W

=l > > > > < < = = 1 2 _ _ _ (5.14) = A C 2l =l + l > 2l + l > > > > > > > > > _ + W > : 2l_W=l : 2l_2 _ =l > ; ; and rewrite the equation (5.12) using (5.14) in the following matrix form: 8 9 > > _ _ _ > > 2 l W

=l > > = h h 1 i i< 2 fa1() a2() a3() a4() 1g A C [I] > 2l_ =l + l > = 0 (5.15) > > > _ + W > : 2l_W=l ; where I is a 3 3 identity matrix and: 2 2 3 2 2 1 r r r a1() = 4 rr + r2 1 2 5 1 0 2 q q r r 1 r 2 a2() = @ rr + r2 1 A + r 1 r2 3 2 2 r 1 r a3() = 4 rr + r2 1 5 2 2 3 r 1 r a4() = 4 rr + r2 1 5


66

Since the functions in the vector in (5.15) are dependent only on t and are not equal zero identically, the row-matrix product in this equation has to be zero. Really, by taking any xed we present linear combination of functions of t that equal zero while the functions are independent linearly by our presumption. Hence, the coecients of the linear dependency are zero, or: h h i i fa1() a2() a3() a4() 1g A 1 C [I] = 0 (5.16) The problem in the system (5.16) is that this is a system of three ordinary dierential equations for only one function r(). It is clear that this fact limits a lot the class of solutions that can be obtained. Therefore, from the next section we will consider the non-rotating self-similar waves (t) 0.

5.2 Non-rotating self-similar waves We choose the non-rotating wave (t) 0. Thus the equation (5.12) becomes: 2 2 2 3 2 _2 ! _ _ ! 1 r r r l l l 4 5 (5.17) rr + r2 1 2 + 2 l + l + 2 l W + W = 0

and this equation (5.17) can be integrated with respect to : 0 2 2 1 2 1 r r r ll @ rr + r2 1 2 A + l_2 + 12 ll 2 + 2l_W_ + lW + W_ 2 f0 = 0

(5.18)

where f0(t) is a constant of integration. Now we have the sum of the products of the functions of and the functions of t where the three last functions of | 2, and 1 are independent linearly. That allows us to choose three arbitrary numbers , and such that: r2 1 r2 r2 = 2 + + (5.19) rr + r2 1 2

and after substitution of (5.19) into (5.18) we receive the sum of products of functions of t and 2, and 1 respectively that has to be equal zero. Since 1, and 2 are


67

independent linearly, we have to equate to zero their coecients, which gives us three following equations to determine l(t), W (t) and f0(t): 8 > _2 > > < l + ll ( + 1=2) = 0 (5.20) 2l_W_ + lW + ll = 0 > > > ll + W_ 2 f0 = 0 : The system (5.20) has simple analytic solutions for any , and . We distinguish two special cases: = 1=2 and = 3=2.

The case = 12 . The system (5.20) transforms into the following system: 8 > _ > > < l=0 W = 0 > > > : f0 = W_ 2

(5.21)

which leads to the linear D'Alembert solutions: 8 > > > < l = L0 = c onst W = s0 + ct where s0 = c onst and c = c onst > > > : f0 = c2

(5.22)

The case = 23 . The exponential solution:

8 > l = L0et=T > > > < W = s0 L0 et=T + W0e 2t=T 3 > > 2 2! L > 0 > + 9 e2t=T + 43 LT02W0 e : f0 = T

2 t=T + 4 W0 e 4t=T T

(5.23)

where L0, T and W0 are constants of integration. In all other cases the solution for l is as follows: 2+1

l = (L1 + L2t) 2+3

(5.24)


68

where L1 and L2 are constants of integration. The solution for W is obtained after substitution of (5.24) into (5.20) and integration: 2+1 2 1 W = s0 + W0 (L1 + L2t) 2+3 + 2 (L1 + L2t) 2+3 (5.25) where s0 and W0 are constants of integration. From (5.25) we can see that = 0 is a limiting case. In this case the solution for W is logarithmic: 8 p3 > l = L1 + L2t < q (5.26) > : W = s0 + 2 1 ln(L1 + L2t) 1 + W0 3 L1 + L2t 3 The function f0 is obtained after substitution of l and W into the third equation of (5.20). Before we begin the analysis of the equation (5.19) we rewrite some of the expressions from the previous section for the case of the non-rotating wave: The equation for the tension (5.11) transforms after the substitution of (5.19) and the system (5.20) into: + f0 (5.27) f = 12 ll r2 2 lW The equations for the tangent and the normal velocities (5.8) take the form: 8 > : v = l_rq1 r2 And nally the equation for the particle velocity (5.9) becomes: 2 V 2 = r2l_2 2rr l_ l_ + W_ + l_ + W_

5.3 Analysis of equation for radius We have the equation (5.19) for the scaled non-dimensional radius: r2 1 r2 r2 = 2 + + rr + r2 1 2

(5.28)

(5.29)


69

complemented with the equation for the angle that follows from the inextensibility equation (5.7): 1q (5.30) = 1 r2 r

As far as we know, the equation (5.19) has not been investigated by the mathematicians. We could not nd it in the reference books of ordinary dierential equations, such as in (Kamke, 1959) or in (Murphy, 1960). Nevertheless we will obtain here some simple special solutions and the solutions in the less simple cases when the equation still may be solved analytically by methods known from the theory of the ordinary dierential equations.

5.3.1 Special solutions 1. The case = = 0 has a special solution r = c onst. Since we are free to choose an arbitrary constant of scaling, we can choose r = 1. After the substitution into (5.19) that gives = 3=2. Substituting r = 1 into (5.30) gives us the solution for the angle (we choose the plus sign | the choice of minus means counting the angle in a clockwise direction) = 0 + . After the substitution of = = 0 and = 3=2 into (5.20) and solving or using the solutions (5.21 { 5.26) we obtain: 8 q3 > l = R 0 1 + t=T > > q3 < W = s + W R 1 + t=T (5.31) 0 0 0 > 2 > > f0 = 1 R0 3 + W02 (1 + t=T ) 4=3 : 9 T The tangent and normal velocities are obtained after substitution of the solution into (5.28): 8 1 s s0 > > < u= 3 t+T (5.32) > 1 R 0 2 = 3 > : v= 3 T (1 + t=T )


70

The tension force is obtained from (5.27): 2 (1 + t=T )2=3 + (s s0 )2 2 R f= 0 9 (t + T )2

(5.33)

Note, that the constant W0 does not appear in the physical parameters of the motion of the string | the velocities and the tension thus we can set it in an arbitrary convenient manner. We take W0 = 0. The meaning of s0 for this solution is the point where the tangent velocity is zero, as it yields from the rst equation of (5.32). This is the point where the angle maintains constant | (s0; t) = 0. The solution described above arises when the in nite string rolled into a circle of the initial radius R0 starts its motion at time t = 0 with the normal velocity v0. The characteristic time T is connected with the initial normal velocity T = 3Rv0 . 0 Of course, this solution can describe also the behavior of nite piece rolled in an arc of a circle (Fig. 5.1) if the tangent velocities on its ends s1 and s2 are maintained as in the rst equation of (5.32). u

s2

v s0 l

s1

Figure 5.1: Self-similar solution of constant radius.

p

2. The case = 0 has a special solution r = C1 + C2 where C1 and C2 are arbitrary constants. We consider the solution r = p since the constant C1 can be taken


71

arbitrary due to the arbitrarity of the scaling coecient and also C2 can be taken arbitrary since the meaning of this constant is a shift of the `zero' . After substitution of this solution into (5.19) we obtain = 3=2 and = 1=4. Substituting this solution into (5.30) and integrating yields for the angle (here and further we will take the plus sign that means the increasing of angle in counter-clockwise direction): p4 1 p q = 0 (5.34) p + 2 ln 2 + 4 1 = p 2 4r 1 + 2 ln 2r + p4r2 1 = 0 p r2 4r 1 + 2 arccosh(2r) = 0 r

we show this solution on Fig. 5.2. One can notice from (5.34) that this solution implies 1=4 and r 1=2.

0

-2 y/l -4

-6

-8 -2

0

x/l

2

4

6

Figure 5.2: Self-similar solution of square root type. The scaling length l is again of the power 1=3, so the non-dimensional variable


72

is again of the type st 1=3. As we stated in section 2.3.2 the solutions of this kind are a very wide class of solutions. These are the solutions that follow from the homogeneity of the tangent angle. As we can see, these are all the self-similar solutions with = 0. From (5.26) and (5.20): 8 > l = R0 q3 1 + t=T > R0 q q > < 3 W = s0 + 3 1 ln R 1 + t=T R0 3 1 + t=T (5.35) > 1 q R0 2 > 1 4=3 ln2 R0 3 1 + t=T > : f0 = (1 + t=T ) T R1 18 The tangent velocity in this solution is: (5.36) u = 67 RT0 (1 + t=T ) 2=3 31 st + sT0 and the normal velocity: v R0 q u 1 R s s u 0 0 2 = 3 3 t q v = 6 T (1 + t=T ) 13 + 12 ln R 1 + t=T 4 3 (5.37) 1 R0 1 + t=T The tension force in this solution is: R0 2 1 f = 9 T (1 + t=T ) 4=3 2 0 10 q 1 s s 7 R 0 0 3 4 @ A @ q3 4 2 6 ln R 1 + t=T + qs3 s0 1 R0 1 + t=T R0 1 + t=T

(5.38) 13 7 A5 2

The constants of integration R0, T and R1 are determined from the initial conditions. We have to specify the radius R, the tangent velocity u and the normal velocity v at some point at t = 0 at one condition | we can not do it at the point = 1=4. Really, in this point r = 1=2 and thus as it can be seen from (5.19) in this point r = 1 and from the second equation of (5.28) it follows that in this point v = 0, so R1 can not be determined from the conditions that are given in this point. This point is the singular point of the equation (5.19). 3. The case 6= 0 has a special solution of the type of square root of a quadratic polynomial. In this case we cannot set arbitrary the coecient at the rst term


73

of the polynomial as 1 as we did in the previous two cases, since this coecient determines the power of the scaling l and thus it is crucial for the self-similar wave. On the other hand we can set arbitrary the origin of the non-dimensional variable , hence, we can always bring the linear term of the quadratic polynomial to zero. p Another words, we consider the solution of type r = b02 + b1. Substitution of this solution into (5.19) yields: 8 3b > = > > 20 < =0 (5.39) > > b (3 b ) > : = 2(1 b 1)0 0 The rst question that is of interest for us in this solution is the valid combinations of b0 and b1. The rst condition that limit us is that the scaled radius r must (obviously) be real number. The second condition is that the angle is also real and from the equation (5.30) we can see that this condition implies r2 1. Combining the conditions we obtain: 8 > < b02 + b1 0 (5.40) > : b0(1 b0)2 + b1 0 The second condition of (5.40) is obviously stronger than the rst since it can be rewritten in the form b02 + b1 (b0)2 and the right-hand part of this inequality is already greater or equal zero. Solving this inequality we obtain that the valid values of depend on the sign of b0(1 b0) and b1. Using the bounds that are obtained from this inequality, we also can learn about the range of r. Next question that interests us in this kind of solutions is whether the solution is convex or concave towards the origin. To deduce this we have to know when rr 2r2 r2 is greater than zero (the function is concave towards the origin) or less the zero (the function is convex towards the origin) where the subscript denotes the derivative with respect to . In our variables the condition of concaveness results in (after substitution of from (5.30) and simpli cation):

CHAPTER 5. FORMULATION OF PROBLEM FOR SELF-SIMILAR WAVES convex

b1 −

b1 b1 ≤η ≤ b0 (b0 − 1) b0 (b0 − 1)

−

b1 b1 ≤η ≤ b0 (b0 − 1) b0 (b0 − 1)

r ≥ b1 0

b1 ≤ r ≤

b0b1 b0 − 1 b0

1

η≥ prohibited

concave

any η

b0b1 ≤ r ≤ b1 b0 − 1

74

b1 b0 (b0 − 1)

prohibited

b0b1 r≥ b0 − 1

Figure 5.3: Regions of solutions of type of square root of a quadratic polynomial at various b0 and b1.

rr + r2 1 0

(5.41)

and after substitution of the solution into (5.41) we obtain that it is concave if b0 > 1 and convex otherwise. We summarize the regions of existence, concaveness and ranges of the solution on Fig. 5.3. To calculate the angle we will integrate (5.30) from = 0 to the current . The result is: 0 q p 21 s b (1 b ) x 1+x 2 1 0 0 1 b 0 = 0 + b arcsinh x 2 arctan@ b x2 (1 b ) (1 + x2) A (5.42) 0 0 0 s where x = b0(1b b0) . We present dierent solutions in Fig. 5.4. 1

Now we proceed with the other parameters of the solution. The scaling l is: the phase shift W :

1 3b0 l = (L1 + L2t) 3(1 b0)

(5.43)

1+3b0 W = s0 + W0 (L1 + L2t) 3(1 b0)

(5.44)


75

0.6 0.4 0.2 y/l 0 -0.2 -0.4 -0.6 -0.8

-0.2

0

b0 -2 0.5 0.5 2 5 0.2 x/l

b1 0.2 -0.1 0.1 0.2 0.4 0.4 0.6

Figure 5.4: Dierent solutions of type of square root of a quadratic polynomial. The tangent velocity is obtained from the rst equation of (5.28): 2(3b0 1) ! s s 3 b 7 L 0 0 2 u = 3 (3b0 1) L + L t + W0b0 1 b (L1 + L2t) 3(1 b0) (5.45) 1 2 0 and the normal velocity | from the second equation of (5.28) (we write it in terms of for simplicity): 2 q 1 3 b L 0 2 3( b 0 v = 3 1 b (L1 + L2t) 1) b0(1 b0)2 + b1 (5.46) 0 The tension force is obtained from (5.29) (here we use the notation of , l and W ): ! b 1 b 0 1 2 f = ll 2 + b 1 + 2l_W + W_ 2 (5.47) 0 Note again that at the singular points r = 1 and the normal velocity is zero, as it can be seen from (5.28). In these points we cannot de ne the initial conditions, since not all the parameters of the wave can be determined.


76

5.3.2 Cases that can be solved analytically There are four cases when the equation (5.19) can be reduced to the rst-order ordinary dierential equation. In the rst case when the independent variable is not presented explicitly in the equation, it can be integrated in terms of elliptic integrals. In the three other cases it can be reduced to an Abel equation of the second kind. Note that we can de ne a new function z = r2. For this function the equation (5.19) is rewritten as: 2 4z z2 z = 2 + + (5.48) z 2

sometimes it is simpler to work with the equation that is written in the form (5.48).

1. The independent variable is not presented explicitly in the equation, or = = 0. We de ne the derivative z as a new function P (z) where the function z is now an independent variable. The equation (5.48) is rewritten as: 4z P 2 z = 2 PPz 2

(5.49)

where the subscript z denotes the derivative with respect to z. The integration gives: q 4z (z + 2)2 K (5.50) P= z + 2 where K is a constant of integration. Now recalling that P z we can see that an equation of the rst order with separable variables has been obtained. The constant K can be found if the radius r and its derivative r are given as the initial conditions. We can write the integral for : Z r2+4=3 + 23 s d (5.51) = 0 + 0 16 64 2 3 4 3 K 3 27 that is a sum of two elliptic integrals of the rst and second kind in Weierstrass normal form.


77

The angle that is written in (5.30) as a function of can be rewritten as a function of r using the obvious relation r = =r . In our case that allows us to write the explicit integral that determines as a function of r:

p Zr K

dz (5.52) r z 4z (z + 2) 2 K For this equation to be real, the expression under the square root has to be positive. Thus if we allow K < 0, then it must be satis ed 4z (z + 2)2 K < 0. But since z = r2 0, the rst member of the sum is positive or zero, while the second ( K ) is also positive by our presumption thus the whole sum is positive and the expression under the square root is negative. The conclusion is that only K 0 are allowed. In that case we have to satisfy 4z (z + 2)2 K 0. In terms p of r we have the condition 2r (r2 + 2) K 0. Let us look at the function p p w(r) = 2r (r2 + 2) K . w(0) = K 0 and the derivative wr = 6r2 + 4 is always greater than zero at large r, which means that the cubic equation w(r0) = 0 has only one positive root, and w(r) > 0 when r > r0. So, in this case the solution has a lower bound r = r0. The solution is also concave, since rr + r2 1 = 4z z2 4r2 1 r2 (z 2)=2 > 0. Really, z 2 = z + 2 = r2 + 2 . The numerator of the p latter fraction is obviously positive, while the denominator r2 + 2 2K r 0 is also positive, hence, the whole fraction is positive and the solution is concave towards the origin. The only exception is the partial solution r = c onst that we presented in the previous section. In this case r 0 and r r0 where r0 is the positive root of the cubic equation w(r0) = 0. We present two examples of the solutions of this type in Fig. 5.5. = 0+ 2

2 0

2

q

The singular points jrj = 1 are asymptotic for this solution. It can be seen from (5.52) that jrj ! 1 when r ! 1. 2. In the general case 6= 0, 6= 0 and 6= 0 we can always choose the origin of


78

4 K=1 φ=1 2 K=2 φ = 0.1 y/l 0

-2

-4 0.5

1

2

1.5

2.5

3

x/l

Figure 5.5: Dierent solutions of type = = 0 in such a way that the linear term of the quadratic polynomial vanish, or, another words, we can always consider the case = 0: 2 4z z2 z = 2 + (5.53) z 2 The case = 0 is the homogeneous case when the equation can be reduced to a one of the rst order. We perform a change of variables z = 2P () where = ln . That leads to the equation that lacks in term of independent variable: 4P (2P + P )2 P = 2 (5.54) 2P + 3P + P 2 so we can introduce a new function P = q(P ). In terms of this function our original equation can be rewritten as the one of the rst order:

qP q(P + 2) = q2 q(6 + 7P ) + (6P + 4)(1 P )

(5.55)

The equation (5.55) is of a class known as the Abel equations of the second kind.

CHAPTER 5. FORMULATION OF PROBLEM FOR SELF-SIMILAR WAVES 3. Let us rewrite the equation (5.53) in the form: 2! z 2 2(2 + ) 3z z ( + ) = 4

79 (5.56)

In the case = 3=2 the equation (5.56) is exact, i. e. it is the exact derivative of the equation of the rst order. This equation of the rst order is again the Abel equation of the second kind:

z (z + 32 + 2) = 6z + 23 + 4 + K

(5.57)

where K is a constant of integration. 4. In the case = 1=2 the equation (5.56) can be made exact by multiplying on the integrating factor . The corresponding equation of the rst order is again the Abel equation of the second kind: (5.58) z(z + 3 + 2) = 12 z2 + (32 + 2)z + 21 4 + 22 + K where K is again a constant of integration. Although there are three special cases that can be reduced to the rst-order equation of type of the Abel equation of the second kind, these equations have no analytical solutions known to the author. Thus we have to use more general techniques to investigate the equation for radius and build the solution. First, we will investigate the critical points of the equation (5.53) in its general form ( 6= 0) (5.48).

5.3.3 Critical points of an equation for radius There is a limited class of equations of the second order with xed critical points, i. e. the critical points that their position does not depend on the initial conditions. These are 6 Painleve transcedents and some other classes of equations. To determine if our equation belongs to this type, we have to present it in the form, according to (Murphy, 1960): z = X0 (; z)z2 + X1(; z)z + X2(; z) (5.59)

CHAPTER 5. FORMULATION OF PROBLEM FOR SELF-SIMILAR WAVES where

X0(; z) =

that gives for our equation:

80

4 X

mi i=1 z ai ( )

! 1 4 z z = z + 2(2 + + ) z2 + z + 2(2 + + ) + 2 (5.60) and from (5.60) m1 = 1, m2 = m3 = m4 = 0. That means that our equation is not the one with the xed critical points (from the comparison with the valid mi that the theory gives). Now let us investigate the points = cr where z = 2(2 + + ). In these points 4z z2 = 0 as the equation (5.48) implies. To determine z in this point we rewrite the equation (5.48): 4z z2 z = 2 + z + 2(2 + (5.61) + )

The numerator and the denominator of the fraction in the equation (5.61) are tending to zero when ! cr . To determine z in this point we have to take the limit value: 4z z2 2z (2 z) = 2 + lim (5.62) z = 2 + lim lim ! z + 2(2 + ) ! z + 2( 2 + + ) ! where we have used the L'Hospital rule. We rewrite the last equation in the form: 3z + 2(2 + ) = 0 ( z 2) lim (5.63) ! z + 2(2 + ) If the limit of the numerator is zero, this will mean that our equation has xed critical point since in this case 3z + 2(2cr + ) = 0 and z2 = 4z = 8(cr2 + cr + ) which means that cr depends only on the parameters of the equation and not on the initial conditions, i. e. the critical point is xed. Since this is not true, we have z ! 2 when ! cr . The exception here is the special solution z = C12 + C2 + C3 that we have considered in section 5.3.1 since in this solution 3z + 2(2 + ) 0. To nd the limit value of z3 in this point, where the subscript 3 denotes the third derivative with respect to , we have to dierentiate the original equation: (z 2) z + 2(2 + + ) = 4z z2 (5.64) cr

cr

cr

cr


81

with respect to :

z3 z + 2(2 + + ) = (2 z) (3z + 2(2 + ))

(5.65)

solve it for z3 and take the limit. We have to use the L'Hospital rule again and that gives: z + 2 + = 0 (5.66) 4 z lim 3 ! z + 2(2 + ) cr

which means that z3 ! 0 when ! cr . The dierentiating of the equation (5.65) with respect to gives us: z4 z + 2(2 + + ) + 4z3(z + 2 + ) + (z 2)(3z + 4) = 0 (5.67) and taking the limit with the L'Hospital rule gives the following expression: 5z + 6(2 + ) = 0 z lim 4 ! z + 2(2 + ) cr

(5.68)

Now we suppose that lim! z(k+1) = 0 at every k = 2 : : : n where the subscript (k +1) denotes the (k +1)-th derivative with respect to . Then if we derive the equation (5.64) n times we obtain after some simpli cation: z(n+2) z + 2(2 + + ) + (5.69) + z(n+1) ((n + 2)z + 2n(2 + )) + 1 + zn 2 n(n + 3) + 1 z + 2n(n 1) 6 + nX2 n + 1! z(k+1)z(n k+1) = 0 + k=2 k + 1 cr

and if we solve this recursive relation for z(n+2) and take the limit using L'Hospital rule and our assumption, we obtain: (n + 3)z + 2(n + 1)(2 + ) = 0 z lim ( n +2) ! z + 2(2 + ) cr

(5.70)

from which follows that lim! z(n+2) = 0. Thus we have proved by induction that all the derivatives of z except the rst and the second tend to zero at the critical point of cr


82

the equation. The points that were discussed above are the critical points of the equation since the derivatives above the rst cannot be determined in a unique way in these points and thus there exist non-unique solution of the equation if the initial conditions are given in this point. In any other point of the domain the solution is unique as it follows from the theorem of the existence and uniqueness since the coecients of the equation are rational analytical functions of , z and z. There exist a special solution of the equation (5.64) that starts in the critical point. This is the solution that satis es the equations z = 2 and z2 = 4z simultaneously, namely:

z = ( + C0)2

(5.71)

where C0 is a constant of integration. Hence, the function behaves near the critical point as: z ( + C0)2 + U ()e 1=( ) (5.72) and from here one can conclude that this critical point is a point of the contact of in nite order. Now let us look at all the points where z = 2. From (5.64) it follows that in this point z2 = 4z | this is the critical point that we have seen already or z +2(2 + + ) ! 1 and the dierence 4z z2 cannot be determined. In the latter case the dierential equation has a special solution: cr

z = 2 + C1 + C2

(5.73)

So, we have seen that every point where z = 2 is a critical point. That means that between the critical points can not exist any point with z = 2 which, in virtue of the continuity of all the derivatives of the function in regular points means that between every two neighbourough critical points z > 2 or z < 2 which means that the self-similar wave can be only convex or concave towards the origin between the two neighbourough critical points. Let us take, for example, a concave interval z > 2. Since jr j 1 then 4z z2 and that means that: 4z z 2 z = 2(2 + + ) + z 2 2(2 + + ) (5.74)


83

when the inequality becomes an equality only in the critical points that are the boundaries of the chosen interval. The condition z > 2 implies also that: ZZ (5.75) z = z0 + z1 + z d2 > ZZ 2 d 2 = > z0 + z1 + = z0 + z1 + 2 = ( + C0)2 cr

cr

since the special solution is de ned from the initial conditions z0 and z1 in the point = cr . The conclusion is that for the concave solution the exact solution is always lies above the special solution (from (5.75)) and also above the parabola 2(2 + + ) and all these three functions are equal in the critical point. The analogous result is valid also for the convex solution where the exact solution lies beneath the two parabolae mentioned above. Thus we can see that the parameters that separate between the convex and the concave family give us the equality of two parabolae when each point of the solution is critical, or: 2(2 + + ) = ( + C0)2 (5.76) that gives = 1=2, = C0 and = C02. Now we recall from section 5.2 that = 1=2 is the special case of linear waves when and are not even speci ed. To nish the analysis of the critical points, we rewrite the equations (5.27 { 5.29) for the tension force and the velocities in terms of z = r2 and look on the interaction between these mechanical parameters and the critical points of the equation (5.48): The equation (5.27) takes the following form: f = 21 ll z 2 lW + f0 and its second derivative with respect to s is: f 00 = 21 ll (z 2)

(5.77) (5.78)

We can see that the condition f 00 = 0 is equivalent to z = 2 that is the critical point of the equation (5.48). But as it was shown in section 3.2 one of the necessary


84

conditions on the discontinuity is f 00 = 0. The conclusion is that we can insert the self-similar wave into the moving string between the two critical points of the equation (5.48). The system (5.28) can be rewritten as: 8 > : v = l_q4z z2

(5.79)

We can dierentiate the system (5.79) with respect to s to nd the points where the normal or the tangent velocity is minimal or maximal: 8 _ > 0 = 1 l (z 2) > u > < 2l (5.80) > (z 2) 0 = l_ zq > v > : 4z z2 that means that the extremum of the tangent velocity is reached only at the singular points, that is the ends of interval of the self-similar solution. The normal velocity has its extremum values at the ends of the interval (critical points) and in the point z = 0. Also we can see from (5.79) that at the ends of the interval the normal velocity is zero, since 4z z2 = 0 in the critical points. In virtue of the continuity of the normal velocity we can conclude that in all the interval between the endpoints it has the same sign: if we suppose that the normal velocity changes its sign in the middle of the interval it will mean, that in that point it passes zero value, and that means that it passes the critical point while we have no critical points in the interval interior. The conclusion is that the maximum of the absolute value of the normal velocity is reached in the point z = 0. The squared velocity from (5.29) is rewritten as: 2 V 2 = zl_2 zl_ l_ + W_ + l_ + W_

(5.81)


85

Let us nd also the extremal values of the velocity. We will dierentiate the squared velocity: 8 _ > > < (V 2)0 = l l_ + W_ (2 z) l_ (5.82) > 00 > (V 2) = l l_ (2 z ) z3 l_ + W_ : l2 The rst equation shows that the extremums are in the critical points z = 2 and _ in the point = W_ . From the second equation we can conclude that the critical l points are the saddle points (z = 2 and zk = 0 when k > 2) and the nature of _ the extremum in the point = W_ is dependent on the nature of the solution | l it is a maximum for the concave solutions and a minimum for convex.

5.4 Building of solution with Taylor expansion Since we could not nd the solution of the equation (5.48) by integration we have to use the numerical methods to solve the problem. One of the methods of building of the solution of the equation (5.48) and (5.30) is with the help of the Taylor expansion. The function z and its rst derivative z have to be given as the initial parameters in the point where we wish to build the Taylor expansion | this point, of course, cannot be one of the critical points of the equation (5.48). Then the second derivative in this point z can be found directly from the equation (5.48), the third derivation | from the equation (5.65) that is the derivative of the original equation (5.48), the fourth derivative | from the equation (5.67) that is the second derivative of the equation (5.48), and the next derivatives | from the recursive relation (5.69) until the desired accuracy of the approximation is reached. The Taylor expansion for the angle may be derived in the similar manner from the equation (5.30) by successive dierentiation. Every derivative contains the derivatives of z that are known already. To solve the boundary-value problem we will use the method of initial parameters, i. e. we will assume some initial parameters (that are unknown at the beginning) and


86

by using the approximation with the Taylor expansion we will obtain the values at the boundary. Thus we write the nonlinear equation for the initial parameters. It can be solved by programming or using the existing computer packages.

Chapter 6 Self-similar solutions in bends In the previous chapter we have formulated the problem for the special cases of the motion of the string that are the self-similar waves. Here we will see how these waves may arise from the bends.

6.1 Formulation We seek the solutions of the form: 8 > < x = vx0t + lr cos( ) > : y = vy 0t + lr sin( )

(6.1)

where vx0 and vy 0 are x- and y-projections of the velocity of the origin and l(t), r() and () where = s l W are the parameters of the self-similar non-rotating wave ( 0) as it was de ned in section 5.1. Note that since the motion of the origin is uniform, the second derivatives with respect to time are zero and since it is independent on s the rst and the second derivatives with respect to the Lagrange coordinate also vanish. That means that the formulation of the equation for the self-similar wave will remain the same as in the previous sections. We take the case of the bend between two straight pieces of string. The self-similar wave develops from the zero length (from the point of origin that is the position of the 87

CHAPTER 6. SELF-SIMILAR SOLUTIONS IN BENDS

88

bend at the initial time t = 0). It follows that the linear solution join with the selfsimilar wave. As we recall from the necessary conditions obtained in section 3.2 in this joining points there must be no rotation and the second derivative of the force has to vanish. The last condition immediately gives us that the self-similar solution join with the straight line in its critical point as it was discussed in section 5.3.3 and the rst | that the orientation of the string from both sides of the join remain the same. In the critical points of the self-similar wave the string is pointed towards the origin in the convex solutions and outwards in the concave solutions since in this points r = 1 and = 0. That gives us only two possibilities | the self-similar solution is concave and there is no bend between it and the straight continuation of the string, or it is convex and in the point of the join there is a fold (a bend with the angle of 180 degrees). In the latter case the energy is not conserved and we will not consider it here. Thus we are seeking the self-similar solutions that are concave towards the origin, bounded between two critical points and have the discontinuity in the curvature at the join with the straight pieces of the string. It follows then that the normal and the tangent velocity will be continuous at the join points. The velocities in the self-similar solution of the kind (6.1) are: 8 > < x_ = vx0 + l_r cos + l_ + W_ (r sin r cos ) (6.2) > : y_ = vy 0 + l_r sin l_ + W_ (r sin + r cos ) while the directions of the tangent vector are the same as in the section 5.1: 8 > < x0 = r cos r sin (6.3) > : y0 = r sin + r cos If we calculate the tangent and the normal velocities, we obtain: 8 > : v = vx0( r sin r cos ) + vy 0(r cos r sin ) l_r2 and the squared particle velocity:

V 2 = vx20 + vy 20 + 2l_r(vx0 cos + vy 0 sin ) +

(6.5)

CHAPTER 6. SELF-SIMILAR SOLUTIONS IN BENDS + 2 l_ + W_ vx0(r sin r cos ) vy 0(r cos + r sin ) + 2 + l_2r2 2l_ l_ + W_ rr + l_ + W_

89

To obtain the point of maximum velocity we perform the dierentiation of the last equation and equating the result to zero. We obtain: 20 V = 2 l_ + W_ (1 r2 rr ) (6.6) 0l 1 @ q r 2 (vx0 sin vy 0 cos ) + 1r (vx0 cos + vy 0 sin ) + l_A = 0 r 1 r _ where we can see that here the maximum velocity is obtained also at = W_ as it l was in section 5.3.3. In this point the normal and tangent velocities do not reach their extremum, and that means as we have proved in section 2.2 that both projections of velocity reach their maximum in this point simultaneously. Another words, in the point _ = W_ we have (x_ )0 = 0 and (y_ )0 = 0 or (x_ 0)2 + (y_ 0)2 = 0. Recalling that x0 = cos ' l and y0 = sin ' where ' is the tangent angle, we obtain that in the point of maximum . velocity there is no rotation '_ = 0. Further, in this point _ = l_ + W_ l = 0 or the Lagrangian coordinate s has always the same value while any other is moving on the s-axis. It is convenient to choose this point as the origin of the non-dimensional coordinate = 0. In this case W_ = 0, W = s0 and from (5.20) we have = 0. At the moment of the start t = 0 the length of the interval occupied by the selfsimilar wave was zero since the wave arises from a single point. The ends of the interval s1 = s0 + l1 and s2 = s0 + l2 were thus in the point s0 at t = 0. That means that l(0) = 0 and from the solution for the scaling l(t) (5.24) it follows that: 2+1

l = L0t 2+3

(6.7)

Since l tk where k is a number, there are three possibilities: the rst derivative l_ may be zero, constant or in nity at t = 0. We have to drop the latter case, since from (6.4) it follows immideately that u(t = 0) = 1. That gives us the condition k 1 or: 2 + 1 1 (6.8) 2 + 3


90

The solution of (6.8) is 23 or 0. We recall from section 5.3.3 that the concaveness of the solution demands 12 so nally we obtain the condition 32 (that is the exponential scaling). The ends of the self-similar wave are the critical points of the equation (5.48). In these points jr j = 1 and = 0. At the moment t = 0 there was also l_ = 0. Substituting all these into (6.4) we obtain that in the moment t = 0 there was at the ends of the wave: 8 > u10 = vx0 cos 1 vy 0 sin 1 > > > < v10 = vx0 sin 1 vy 0 cos 1 (6.9) > > cos + v = v u sin 2 y0 x0 20 2 > > : v20 = vx0 sin 2 + vy 0 cos 2 where the subscript 0 means that the initial velocities are under consideration. Since here the discontinuities are the discontinuities in the curvature (' = 0), the normal and the tangent velocities are continuous in these points as it can be seen from the analysis in section 4.1. From the necessary conditions on discontinuities we can also omit the subscripts 0 of the normal velocities since they must maintain their values in the point of the join all the time. Therefore it is logical to nd the origin velocities vx0 and vy 0 in terms of the normal velocities by solving the second and the fourth equations of the system (6.9): 8 1 > = > v x 0 < sin( 1 2) (v1 cos 2 + v2 cos 1) (6.10) > 1 > : vy 0 = sin( 1 2) (v1 sin 2 + v2 sin 1) The dierence of the angles of the ends of the self-similar wave is in fact the dierence of the tangent angles of the straight pieces ' that we have used in section 4.1. The rst and the third equations of (6.9) give us the initial tangent velocities that will produce the self-similar wave: 8 v2 v1 > > u = 1 0 < tan ' sin ' (6.11) > v v 2 1 > : u20 = tan ' + sin '


91

If we compare these conditions on the initial tangent velocities with the tangent velocities of the bend motion d1 and d2 from section 4.1 and calculate the initial tension force with the formulae of that section we see that the condition for the self-similar wave that arises from the bend is zero initial force. The most common case of such a condition is, of course, the movement starting from rest, or zero initial velocities. Now let us look at the motion of the straight piece that is joined with the self-similar solution. We have already noted that the continuity of the tangent vector in the joining point yields the continuity of the normal and the tangent velocities. Let us analyze the conditions on the rst point of join 1 since the conditions on the second are derived in an analogous way. Recall again that in the endpoint r = 1 (minus in the beginning, plus in the end, r 1 = 1), = 0 and = 1 is constant in time. In these conditions the tangent and the normal velocities from (6.4) take the following form at the endpoint 1: 8 > : v = vx0 sin 1 + vy 0 cos 1 The normal velocity is maintained constant as it should be while the tangent velocity is changing in time due to the function l_. We distinguish here three cases: 1. The in nite straight piece. It is clear that in this case the tangent velocity is maintained constant. In the rst formula of (6.12) the only term that is dependent on time is the last term therefore we have to equate it to zero in this case. That means that in such a point r1 = 1. Recalling from section 5.3.3 that in the critical point r2 = 2(2 + + ) and since here = 0 we obtain: 1 2 = 1 + 2 (6.13) 2. The straight piece is between the two neighbouroghed bends. Since the equation (6.12) is valid at the end of the rst self-similar wave and at the beginning of the second, we have to conclude that in this case the scaling parameters l(t) of both


92

waves are the same functions of time with the dierent constants of integration. That means that both waves have the same parameter . Moreover, the projections of the velocities of the origins onto the normal to the joining straight piece have to be equal since these projections equal to the normal velocity of this joining interval. Let us denote by a the end of the rst wave and by b the beginning of the second wave, by Al | the scaling of the rst wave and by Bl | the scaling of the second 2+1 wave (here l = t 2+3 ). From the equality of the tangent velocities we obtain then A(ra a) = B ( rb b). Deriving the tangent velocity with respect to time we obtain at the endpoints: u_ 1;2 = l(r1;2 1;2) (6.14) Now applying the second law of Newton to the straight piece we write: (b a)u_ = b_ 2 a_ 2

(6.15)

since the tension forces in the joining points are the squared velocities of the propagation of the discontinuities. Substituting a = Ala + s0a and b = Blb + s0b we obtain: l(bB aA) A(ra a) + + 12 (Bb + Aa) + (6.16) + (s0b s0a) A(ra a) = 0 From the equation (6.16) we can conclude that the self-similar solutions are inapplicable for this case. Really, the rst term of the sum is dependent on time while the second is constant. To satisfy the equality, the rst term has to be zero, but then the second term also has to be equal zero and that is impossible | none of the parentheses in the second term is allowed to vanish. 3. The straight piece is connecting the self-similar wave and the free end where the force is speci ed. Let us look at the valid tension forces. We have to write the second law of Newton similar to (6.15):

au_ = a_ 2 F

(6.17)


93

where a is again the Lagrangian coordinate s of the join and F is the speci ed tension at the free end. Substitution of the appropriate parameters into (6.17) gives: (ala + s0)la( ra a) = a2l_a2 F (6.18) 1 and from the equation (6.18) substituting from (5.20) l_2 = ll + 2 we obtain the valid forces: 1 F = lalaa a 2 ra + las0(ra + a) (6.19) To conclude, the self-similar solutions are describing the motion of the string from the initially piecewise-straight con guration with a single bend with the zero initial force. f1

f2 a v0 y

b

v0

x

Figure 6.1: A problem that leads to a self-similar wave. Suppose that we have such a problem. The forces from the both ends have to be the functions of the same type that is dictated by (6.19). More precisely, we have at the endpoints: 8 > < F1 = C11ll + C12l (6.20) > : F2 = C21ll + C22l


94

where l is one of the valid scaling functions. The equation (6.19) with the conditions at the endpoints give us the following system of 6 equations: 8 2 1 > L0a a 2 ra = C11 > > > > L0s0(ra + a) = C12 > > > < ra2 = 2 (a2 + ) (6.21) > 2 b b 1 + rb = C21 > L 0 2 > > > L0(L s0)(rb b) = C22 > > > : rb2 = 2 (b2 + ) where L is the total length of the string | the Lagrangian coordinate of the endpoint. We have also 6 variables | ra, a, rb , b, and L0. The parameter is speci ed from the appropriate functions presenting in F1 and F2. But although the system (6.21) is a system of 6 equations for 6 variables, not all the 6 can be determined and the coecients Cij are dependent one on the other. Let us show it. The rst two equations of (6.21) present a system of linear equations for a and ra: 8 C 9 8 9 > 2 3 11 > 1 > > > > 2 < = < 64 1 + 2 75 ra = L0a = (6.22) > > > > C : ; 12 > > 1 1 a : sL ; 0 0 and the fourth and the fth equations present the similar system for rb and b : 8 C 9 9 8 > > 21 2 3 1 > > > 2 = < < = L0b > 64 1 + 2 75 rb = (6.23) > > C ; : > > 22 1 1 b > : (L s0)L0 > ; The system (6.22) gives for a: 1 C12 1 C12 C11 2 (6.24) a + 2 = s L L2 or a + 2 s L a + CL112 = 0 0 0 0 0 0 a 0 a and the solution of (6.24) can be written as a = N L0 where Na is independent on L0. For ra we obtain ra = a + sCL12 = RLa where Ra is also independent on L0. In the 0 0

0


95

Rb where N and R are b and r similar way the solution of (6.23) gives b = N b = b b L0 L0 independent on L0. Comparing the third and the last equations of (6.21) we can write = 12 ra2 a2 = 21 rb2 b2 or:

1 R2 + N 2 = 1 R2 + N 2 (6.25) a 2 b b 2 a The last equation does not contain any unknowns | this is the connection between the boundary conditions. On the other hand we can conclude that the scaling constant L0 cannot be determined from (6.21). Thus we cannot specify the boundary-value problem from (6.21), and the last condition that completes all the parameters of the boundaryvalue problem to determine the self-similar wave from the boundary conditions on the string is the angle of the opening of the bend ( 2 1) is given.

6.2 Energy relations for self-similar wave In this section we consider the energy of the self-similar wave. First of all we have to remind the well-known fact that the mechanical energy is conserved during the motion of the inextensible string. Really, the equation (2.4) of the motion of the string after division by the mass per unit length:

R = (f R0)0

(6.26)

may be multiplied with a scalar multiplication by R_ and integrated with respect to s from the rst endpoint to the second which gives us: Z ZL _R R ds = L R_ (f R0)0 ds (6.27) 0

0

The right-hand part can be integrated by parts which gives us the following equation: 1 Z L R_ 2_ds = f R_ R0jL + Z L f R0 R_ 0 ds (6.28) 0 2 0 0 where the last integral is identically zero since its integrand is the derivative of the inextensibility condition R0 R0 = 1. The left-hand side of (6.28) is the derivative of the


96

kinetic energy of the string with respect to time while the remaining rst term in the right-hand side is the derivative with respect to time i of the work applied at the both ends of the string. The last conclusion is not valid in the case when there is a fold in the middle of the interval [0; L] since as we have seen in the section 4.1 the force may be discontinuous on the fold. In this case the equation (6.28) takes the form: 1 Z L R_ 2_ds = f R_ R0jS + f R_ R0jL (6.29) 0 S 2 0 where S is the Lagrangian coordinate s of the fold. First we calculate the work invested into the self-similar wave that arises form the bend. The work on each end is the integral with respect to time of the product of the tension force and the tangent velocity, we can obtain this integral using the equation (6.7) for the scaling factor and thus calculate the total work W : W = L3 2 (2 + 1)2 (r ) 1 + r + (6.30) 2 0 (2 3) (2 + 3)2 2 2 2 2 2 2 3 + 1(r1 + 1) 1 12 r1 t 2+3 + 1 2(2 + 1) 2 + L0 (2 1)(2 + 3) 2 2 2 + r2 (vx0 cos 2 + vy 0 sin 2)+ 2 1 + 1 1 21 r1 (vx0 cos 1 + vy 0 sin 1) t 2+3 + h + L0 2 + 1 (L s0)(r2 2)(vx0 cos 2 + vy 0 sin 2)+ 2 + 3 i 2 + s0(r1 + 1)(vx0 cos 1 + vy 0 sin 1) t 2+3 + 2 h (2 + 1) 2 + s (r + )2 i t 24+3 2 ( L s ) ( r ) + L0 0 2 2 0 1 1 2 (2 + 3)2 The powers of t in (6.30) are the results of the integration so they have to be calculated from the initial zero time to the time t. Since the work must be nite the powers of t in (6.30) have to be positive. The compound of the conditions on the powers in this equation together with the condition of concaveness 12 gives us nally the condition on that already has been obtained in the previous section using another approach: 32 (6.31)


97

which means that the exponential scaling (that is the special case of = 3=2) divides between the waves on the nite intervals that demand the nite work to initialize them and the waves on the in nite string. The kinetic energy of the string can be calculated as the product of the integral of the squared particle velocity along the whole length of the string and the mass per unit length . To apply the law of conservation of energy on the string (6.28) we need to calculate the addition of the kinetic energy T to the string from the moment t = 0. That calculation gives: T = L3 (2 + 1)2 Z 2 r2 d+ (6.32) 0 (2 + 3)2 1 2 3 + 13 23 13 + r11(r1 + 1) r22(r2 2) t 2+3 + Z 2 2 + 1 2 + L0 2 + 3 vx0 2 r cos d + 1(2r1 + 1) cos 1 2(2r2 2) cos 2 + 1 2 1 Z 2 + vy 0 2 r sin d + 1(2r1 + 1) sin 1 2(2r2 2) sin 2 t 2+3 + 1 h 2 + 1 + L0 2 + 3 s0(r1 + 1)(vx0 cos 1 + vy 0 sin 1)+ i 2 + (L s0)(r2 2)(vx0 cos 2 + vy 0 sin 2) t 2+3 + 2 h i 4 + L20 (2 + 1) 2 s0 (r1 + 1)2 + (L s0) (r2 2)2 t 2+3 2 (2 + 3) The last equation have to be equal to (6.30) due to the conservation of the mechanical energy of the string. Comparing the dierent powers of t we can observe that the two last square brackets are equal identically. The equality of the rst pair of square brackets gives the formulae for the integrals on the wave: Z 2 2 d = 2(4 3) 3 3 + 2 1 2 r2 1 r2 r (6.33) 1 3(2 3) 1 2 2 3 2 1 Z 2 (6.34) r cos d = 22 1 (2r2 cos 2 1r1 cos 1) 1 Z 2 r sin d = 22 1 (2r2 sin 2 1r1 sin 1) (6.35) 1

The last two integrals can be also obtained writing the law of momentum in projection on the x- or the y-axis.


98

6.3 Examples of problems that lead to self-similar solutions Let us examine two problems that have the self-similar solutions: the semi-in nite straight string initially in rest drawn from its free end and the semi-in nite initially bended string drawn from its free end.

6.3.1 Semi-in nite initially straight string drawn from free end Consider the semi-in nite string that was initially straight and in rest. From the moment t = 0 it is drawn with the force F = c2tm directed with an angle of 90 degrees to the initial string con guration. Recall that in the point of maximum velocity the Lagrangian coordinate s maintain the same value as well as the non-dimensional coordinate . Here this is obviously the point s = 0 which will be chosen as the origin of . The tangent angle in this point has to coincide with the direction of the force, hence, it has to be 90 degrees. From the relations between the derivatives of the scaled radius r, the polar angle and the tangent angle ' we obtain in the point = 0: 8 > > > < r cos r sin = 1 (6.36) r sin + r cos = 0 > > > r2 + r2 2 = 1 : or r = cos and r = sin in = 0. Also, the tension force in the string must be equal to the external force in = 0 which gives us: 4 c2tm = (2 + 1)2 L20t 2+3 r02 + 2 (6.37) (2 + 3) Since 32 > > 1 the valid forces have m 0. Next, the scaling constant L0 is found from the other parameters of the wave: L0 2 2 1 (2 + 3) (6.38) c = (2 + 1) r02 + 2


99

The other end of the self-similar wave joins the semi-in nite straight piece. We with 1 recall from (6.13) that in this case = 12 + 2 and so we can choose the coordinate of the endpoint arbitrary. We choose it = 1 (Fig. 6.2) which gives us: 1 (6.39) = + 2 F=c 2 t m y/l

η=0 r0

ψ0

η=1 x/l

Figure 6.2: Problem of drawing of semi-in nite straight string. That also means that r(1) = 1 as it follows for the semi-in nite straight piece from section 6.1. The meaning of L0 then is naturally the length of the self-similar wave at the time t = 1. We solve the system of ordinary dierential equations for determination of the radius r and the polar angle by means of the Taylor expansion. The problem is solved by the method of initial parameters when the values of r and at the end of the interval are known (r(1) = 1, (1) = 0) and the initial conditions are the unknowns while the functions z = r2 and are approximated with the Taylor expansion employing 16 terms in z and 10 terms in . That gives the system of two non-linear equations that is solved with the \Maple-V" computer program. The dierent solutions are shown in Fig. 6.3. The solutions are approaching the straight vertical line as ! 1 that is the special solution for the constant force. Really, at ! 1 m = 2 4+ 3 ! 0 that is the constant force and the equation (5.48)


100

transforms then into z = 2 where z = r2. The applying of boundary conditions gives p in that case the solution r = 2 2 + 2 and = arctan(1 ) and the numerical investigation in this section con rms this result. 1 0.8 α F -1.5 exponential y/l 0.4 -1.9 power 5 -2.5 quadratic -3.5 linear 0.2 -5.5 square root −∞ constant 0.6

0 0

0.2

0.4

0.6

0.8

1

x/l

Figure 6.3: Solutions for a semi-in nite initially straight string. The next parameter that interests us in this problem is the length of the self-similar wave L0=c. We determine it from (6.38) using the numerical solution. It is shown in Fig. 6.4 in logarithmic horizontal axis since the range of is very large. As the force tends to constant ( ! 1) the non-dimensional wavelength tends to 1 as it should be. Near the exponential scaling ! 32 the non-dimensional wavelength tends to zero expressing the fact that to initialize the self-similar wave with the large powers of the force one has to invest large work.


101

1

0.8

L 0 /c

0.6

0.4

0.2

0 5 −10

4

−10

3

−10

2

α

−10

1

−10

0

−10

Figure 6.4: Non-dimensional length of a self-similar wave for the semi-in nite initially straight string.

6.3.2 Semi-in nite initially bended string drawn from free end Consider a semi-in nite string with the 90-degree bend at the beginning in the point s = s0 that initially was in rest and from the moment t = 0 is drawn from its free end with the tension force F of the kind that allows the self-similar solution. The force of this kind has only two degrees of freedom that we can choose arbitrary. We write comparing with (6.20) and (6.21): (2 + 3)2 C tm + C t 21 m 1 F = 2(2 (6.40) 1 2 + 1) 8 > 1 C2 = 2(2 + 1) (r + ) > > < s0 L0 (2 + 3)2 a a (6.41) > C1 2(2 + 1) 1 > : L20 = (2 + 3)2 a a 2 ra


102

First of all the equation (6.40) gives us = m2 32 . The other end of the selfsimilar wave is joined with the semi-in nite part and that gives us = 21 and as in the previous section we choose b = rb = 1. The knowledge of the position of at the end plus the condition of the angle of the opening of the bend the critical point a b= 2 is enough to determine the position of the second critical point (a; ra) and knowing the coordinates of the second critical point we can determine from the rst equation of (6.41) the scaling coecient L0. The second equation of (6.29) dictates us then the coecient C1 of the force. Thus the degrees of freedom in this problem are the power of the force m, the length of the bend s0 and one of the coecients of the force C1 or C2. 2

1.5

y/l 1

α −1.56 −1.9 −2.5 −3.5 −5.5 −∞

0.5

0 0 0.2 0.4 0.6 0.8 1 x/l

Figure 6.5: Solutions for the semi-in nite initially bended string. To solve this problem the method of initial parameters was used again while the function was approximated with the Runge-Kutta method of the 5 order. The initial


103

dz (0) was obtained then guess was the initial radius squared z0 = r2(0). The derivative d from the condition z(1) = 1. Next the second critical point was calculated from the condition z(a) = 2 (a2 + ) and at last the condition (a) (1) = 2 was checked. Since the Runge-Kutta method does not allow to calculate a value of function exactly in the critical point, it was approximated with the error of 0:0001 0:005, the larger error at closer to 23 . The solutions at various are shown in Fig. 6.5. As approaches 1 the solution tends to a straight line that can be calculated from the formulae of section 4.2 and 4.3 as it was done in the previous problem | the straight piece that has an angle of 3 with the semi-in nite part. It is described in terms of , z() and () as: 8 > < z = 2 + 1 (6.42) arctan p2 1 > = : 6 2 3 The results for the coecients of the tension force are shown in Fig. 6.6 for C1 and in Fig. 6.7 for C2. Fig. 6.6 presents the graph of the non-dimensional wavelength (the coecient of the scaling function l(t)) analogously to the Fig. 6.4. Although the two graphs look very similar, the numerical comparison shows the dierence up to 10% at close to 32 . Note an interesting paradox that arises in this problem. As it was stated in sections 2.2 and 6.1 the two components of the particle velocity x_ and y_ reach their maximum in the point = 0 (s = s0). That means that the velocity of the string in the vertical direction y_ is maximal in this point for all the string, in particular, it is greater than the vertical velocity of the string in the point s = 0 where the pulling force is applied. We obtain that the point s = s0 (and, in fact, all the points in the interval s 2 [a; s0]) are following the pulling force faster, than the force itself moves. The last question of this section is the duration of the self-similar wave, or in another words, the time when the end of the self-similar wave reaches the end of the string. To answer this we have to solve the equation a = s0 which gives: ! 22+3 +1 s 0 te = L (6.43) a 0


104

1

0.6

L0

q

C1

0.8

0.4

0.2

−10

3

2

−10

α

1

−10

0

−10

Figure 6.6: Non-dimensional length of a self-similar wave for the semi-in nite initially bended string. 4

10

2

C2 /(s0 L0 )

10

0

10

−2

10

−4

10 3 −10

2

−10

1

α

−10

0

−10

Figure 6.7: The second coecient of an end force in the non-dimensional form for the semi-in nite initially bended string.


105

where one has to determine the scaling coecient L0 from the one of the coecients of the force C1 or C2.

Summary In this thesis we developed dierent formulations of the motion of the string, presented the analysis of propagation of discontinuities and especially of propagation of bends, namely, the discontinuities in tangent vector and nally presented new family of special solutions that are self-similar waves and connected these solutions with the bends. Mainly the plain motion of an inextensible string was considered. The formulation of the equations of the motion of the string in terms of integral variables seems to be the simplest possible formulation of the equations of the motion of the string. It expresses in the most clear manner the wave nature of the motion of the string. The future analytical investigation of the dynamics of the string is proposed to be concentrated on this formulation. The formulation of the equations of the motion of the string in terms of natural parameters, namely, tangent and normal velocities, angular velocity, curvature and tension force seems to be convenient for the numerical integration by means of the nite dierences since this formulation presents the equations of the rst order where each variable has a de nite physical meaning. It is proposed to develop a numerical method of the integration of the equations of the motion of the string based on this formulation. Another direction of research based on this formulation is developing of the analogous formulation for three-dimensional motion of the string. In addition it will include torsion as well as the curvature, the binormal projection of the particle velocity and tangent and binormal projections of the angular velocity. The approximation for magnetic induction from a cross-section of the string that 106

SUMMARY

107

shows a good compatibility with the exact expression was developed. The future research in this direction will be concentrated in comparison of results produced using this approximation with exact solutions and evaluating of the quality of the proposed approximation. The connection between the motion of the string and propagation of discontinuities in it was shown using an analysis of characteristic lines. Necessary conditions of the propagation of the discontinuities were developed. It was proved that the string cannot rotate around the point of the discontinuity | the suggestion that follows from the physical fact that the string cannot bear the bending moment. A special solution in which the bend propagates along the string with a non-constant velocity was obtained. In the future research it is proposed to investigate the conditions on the discontinuities in three-dimensional motion. The motion of the bends was investigated using the law of conservation of mass and momentum. The problem of collision of two bends was studied and the limits of the assumption of inextensibility were shown. It was shown how the collision of two bends leads to apping, i. e. sudden increase of the tension force. In the future research the collisions of bends and apping should be investigated in elastic string. Laws of the propagation of the bend present a framework for understanding the motion of the string as wave motion. In the future research the numerical method for the computing of the motion of the string is proposed to be developed using the results of this thesis. The problem for the self-similar waves was formulated and a system of the ordinary dierential equations that describes this kind of motion was derived and investigated. It was shown that a constant force solution is a special case of such a wave. The equation for a shape of the self-similar wave was investigated using the existing analytical techniques. The conditions for developing of such a motion in the bends were derived and the model problems of the self-similar waves that appear in the semi-in nite initially straight string and in the semi-in nite initially bended string were solved numerically.

SUMMARY

108

The self-similar waves seem to be one of the simplest expansions of the D'Alembert solution. They support only a limited number of degrees of freedom in the initial condition although give a rather good understanding of the opening of the bends. In the future it is proposed to enhance this formulation. The nal goal is to solve the problem of the initially bended string that is drawn from its free end with a constant force.

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