SCATTERING AND RADIATION OF SHORT SURFACE WAVES BY PLANE-SIDED OBSTACLES

by

ALI MOHAMMED AYAD

MARCH, 1976

A thesis submitted for the degree of

Doctor of Philosophy of the University

of London.

Department of Mathematics,

Imperial College of Science and Technology,

London S.W.7.

ABSTRACT

In this thesis we employ the method of matched

asymptotic expansions to deal with the linearised two-

dimensional surface wave problem of scattering of a regular

surface wave train by rigid obstacles, with at least one plane

side, partially immersed in an ideal fluid of infinite depth

with the plane side, or in some cases the two plane sides,

vertical. The formulation and ,the method are developed for the

prototype problem of transmission of short surface waves under

a plane barrier immersed to a finite vertical depth a. The

technique is then adopted to deal with scattering by more

general obstacle shapes such as a cylinder of rectangular cross-

section, a wedge and other two-dimensional geometries having at

least one vertical plane side. In particular, it is shown that,

in the short-wave asymptotic limit, the transmission constant

is exponentially small compared to the reflection constant,

i.e. the reflection is nearly total.

In some general cases, the reverse (complementary)

problem is considered and the method is used to verify the

fact that when the body scatterer and the coordinate system

are kept fixed, and the direction of the incident wave train

is reversed, then the transmission constant remains unchanged.

This is done by showing that the leading term in the asymptotic

expansion, for short waves, of the transmission constant

remains the same as in the original problem.

Using the same method, the wave-making (radiation)

problem is treated in a brief manner and the Haskind Relations

for two-dimensional motion are also verified.

4

ACKNOWLEDGEMENTS

It is with the greatest pleasure that I acknowledge

the constant encouragement and help given to me by Dr. F.G.

Leppington during the course of this research. His careful

and critical supervision of the work was of the utmost import-

ance in enabling me to complete it. Special thanks are

expressed to the Government of the Libyan Arab Republic for

their financial assistance, without which I would not have been

able to carry out this piece of research.

I would also like to thank my colleague G. Alker

for the constructive way in which he discussed and criticised

many points in the work. This resulted in a number of improve-

ments and the uncovering and correction of some mistakes.

To the staff of the Mathematics Department Library,

especially Miss J. Pindelska, I offer my sincere thanks for

all the cooperation I received in using the facilities during

my stay in the Department.

Finally, I express my warmest thanks to Mrs. S.

Murdock who typed the manuscript in a most efficient, patient

and also good-humoured way, uncovering many linguistic

inconsistencies in the process.

5

CONTENTS

Page

GENERAL INTRODUCTION

8

1. SCATTERING BY A RIGID PLANE BARRIER IMMERSED TO A FINITE VERTICAL DEPTH 19

1.1 Introduction and formulation 19

1.2 The outer approximation 27

1.3 Formulation and solution of the lower 32 inner problem

1.4 Estimation of the reflection and transmission 39 constants

2. TRANSMISSION OF SHORT SURFACE WAVES UNDER GEOMETRIES WITH A KNIFE EDGE 48

Introduction 48

CASE (i) 49

2.1 Statement of the problem 49

2.2 Solution of the lower inner problem 52

2.3 Estimation of the reflection and transmission constants 57

2.4 The complementary problem of Case (i) 61

CASE (ii) 66

2.5 Introduction and solution 66

2.6 The complementary (reverse) problem of Case (ii) 72

2.7 Solution of an outer problem 77

3. TRANSMISSION OF SHORT SURFACE WAVES UNDER A PARTIALLY IMMERSED CYLINDER OF RECTANGULAR

82

CROSS-SECTION

3.1 Formulation of the problem 82

3.2 .The outer approximation 84

3.3 Solution of the inner problem 86

3.4 Estimation of the transmission constant 93

3.5 Solution of the outer problem (3.2.2) 95

- 6 ,

Lao. 4. SCATTERING BY A CYLINDER WITH A VERTICAL PLANE

SIDE AND A CURVED BOTTOM INTERSECTING AT RIGHT ANGLES 100

4.1 Introduction 100

4.2 The lower-right inner problem 103

4.3 The lower-left inner problem 106

4.4 The reverse problem 110

4.5 The second term in the estimate for R1 115

4.6 The solutions of two outer problems 117

5. TRANSMISSION OF SHORT SURFACE WAVES UNDER A WEDGE 122

5.1 Formulation of the problem 122

5.2 The outer approximation 124

5.3 The lower inner problem 127

5.4 The transmission constant 133

5.5 Solution'of the outer problem 137

6. TRANSMISSION UNDER OBSTACLES WITH A VERTICAL PLANE SIDE AND A CURVED PART INTERSECTING AT AN ARBITRARY ANGLE 142

CASE (i) 142

6.1 Statement of the problem 142

6.2 The outer potential 144

6.3 The lower inner problem 145

6.4 The surface inner region. 147

6.5 The complementary problem 148

CASE (ii) 153

6.6 Introduction and solution 153

7. SOME RADIATION PROBLEMS 159

7.1 Introduction 159

7.2 Horizontal oscillations of a vertical plane 164 barrier

Page

7.3 Swaying motion of a cylinder with a vertical 173 plane side

7.4 On the solution of the outer radiation problem 184

APPENDICES

187

REFERENCES

203

GENERAL INTRODUCTION

The theory of wave motion in liquids with a free

surface and subjected to gravitational forces has interested

a considerable number of mathematicians. The literature conc-

erning surface waves in water is very extensive. In addition

to a host of memoirs and papers in the scientific journals,

there is a number of books which deal with the subject at length.

The subject of surface gravity waves has a great

variety, whether regarded from the point of view of the types

of physical problems which occur, or from the point of view of

the mathematical ideas and methods needed to attack them.

physical problems include those of wave motion over sloping

beaches and past obstacles, motion of ships in a seaway, flood

waves in rivers, tidal waves and oscillations in harbours. The

mathematical tools employed include those of potential theory,

the theory of the linear wave equation, complex variable

methods, the theory of integral transforms, methods employing

a Green's function, integral equations and other techniques

concerned with the problems, both linear and non-linear, of

mathematical physics.

Problems relating to the generation, propagation and

scattering of surface waves on a fluid have received a great

deal of attention. Much of the theory deals with gravity

- 9

waves in perfect fluids, and is appropriate to infinitesimal

wave heights. It is within the framework of the linear theory

(with surface tension effects neglected) that we consider the

two-dimensional problem in which infinitesimal time harmonic

surface waves are incident on plane-sided obstacles partially

immersed (with the plane side, or in some cases two plane sides,

vertical) in an ideal fluid of infinite depth. The incidence

is always considered normal, i.e. the wave crests of the

incident wave train are parallel to the generators of the body

scatterer. Such a wave train will be partially reflected and

partially transmitted, and our objective in the present work

is to evaluate the transmission and reflection "constants"

when the wavelength of the surface waves is small compared to

the dimensions of the obstacle.

A typical scattering problem is one in which the

obstacle S is held fixed with its generators parallel to the

z-axis, where the coordinate system is so chosen that the y-axis ,

is pointing into the fluid and the x and z axes are in the plane

of the undisturbed free surface. A regular surface wave train

of potential Re exp f

[(112F) iWd is incident normally on the

body scatterer $, where E = g/W2, g is the acceleration due

(I) to gravity, Trz is the frequency and t is the time. The induced

velocity potential is Re[tp(x,y;E)exp(-iWt)) (it is assumed that

the induced motion is irrotational), where LP(x;E) is specified

-

by the linearised conditions

2 a2 a . e--7 + --70 q)= 0 in the fluid region, ax 837'

an = 0 on S,

(0.1)

(0.2)

where n is the unit normal from S into the fluid. On the

mean free surface, the potential satisfies the condition

akp + Eay

= 0. (0.3)

Any edge singularities are limited by the condition

r a—kP 0 as r 0, Or

(0.4)

where r is the distance from any point on the body scatterer

S. This means that LP(x,y;E) is finite at all points of S.

A radiation condition is needed to ensure that the

scattered waves travel outwards. Thus

exp(ilf-1) + ) as x -co q)(x,y;E) (0.5)

T exp (212-Si) as x + co

. . where R and T are the reflection and transmission constants

respectively, and are unknowns of the problem. Our aim is to

describe the potential field and, in particular, to estimate

the constants R and T, in the short-wave asymptotic limit

E 0.

An analogous problem in high-frequency diffraction

presents itself in the case of scattering of electromagnetic

waves (or sound waves of short wavelengths) by an obstacle of

given shape, and there we have the well-developed "ray theory"

to determine the scattered field. However, for the scattering

of surface waves no such analogous theory is known, and the

major'aim of this thesis is to develop such a theory for a

particular class of problems.

The method used will be that developed by Leppington

(16,17,18) for the application of matched asymptotic expansions

:Van Dyke (33)] to the radiation and scattering of short

surface waves. In this method, the basic assumption is that

the fluid region can be covered by overlapping domains in

which different asymptotic approximations for the potential hold:

The "outer region" consists of that part of the fluid

that is many (short) wavelengths from the free surface and, in

the case when the surface wave train is incident on a vertical

plane side, from the lowest point of the vertical plane side

under consideration. To obtain a first approximation LP(x;E)--,

P(E) ,0o(x) in this region, where 3(E) is a scale factor to be

found, we formally take E --a 0 in the free surface condition

(0.3); the "outer potential" o (x) then satisfies the homo-

geneous conditions

- 12. -

/ 2

a2

k--7 —7) Yo(x) = 0 in the fluid region, -\ ox ay

= 0 on the mean free surface,

ag) °

on = 0 on S,

where n is as in (0.2) above.

(0.6)

The specifications (0.6) permit no surface waves, of course,

so that the condition at infinity is simply that (Mx,y) should

vanish at great distances from the origin.

It is also important to note that %(x,y) a.- 0 unless

it is singular at some point, or points, of the fluid domain.

Since there are points lying outside the region of validity of

po(x), singularities are quite admissible and are to be smoothed

off by solutions that are valid near these points. It trans-

pires, however, that(x,y) has only one singularity in the

domain, and that this singularity is located at the point of

intersection of S (the one on the illuminated side of S) with

the free surface, except in the case when the waves are

incident on a vertical plane side. In the latter case the sing-

ularity is at the lowest point of the illuminated vertical plane side.

In the "inner regions" :points that are very close

(on some characteristic length scale of the obstacle S) to

the edges where the outer potential %(x,y), of (0.6), is not

- 13

valid the potential will, in the short-wave asymptotic limit

E --4 0, depend primarily on the local shape of the body S

near these edges. The approximations in the vicinities of

these edges involve simplifications in the geometry of the

scatterer S, with solutions that are slowly varying functions

of independent variables scaled with respect to the parameter

E. The formal transformations used are given in the main text.

Since all points at great distances from the edges

lie outside the regions of validity of the "inner approximations",

there is some difficulty in assigning boundary conditions at

infinity for the "inner solutions". A similar difficulty

arises when deciding on the correct edge behaviours for the

outer solution T(x,y), which is not valid near the edges under

consideration. The idea of matched asymptotic expansions

provides the means of completing the specifications for the

inner and outer solutions. The formal machinery for using the

matching principle has been propounded by Van Dyke (33), and a

precise description of it, as used here, is given in Leppington

(16) and Crighton (6). Finally, it is remarked that, although

this matching procedure is not rigorous, it does provide a

systematic method for solving a wide class of problems, to any

order of accuracy in principle, that would be difficult to

deal with by other means.

- 14,-

A plan of this thesis is as follows:

In Chapter 1, the problem of scattering of gravity

surface waves by a vertical plane barrier, immersed to a

finite depth, is considered when the crests are parallel to

the plane of the barrier (i.e. the problem is two-dimensional)

and the fluid is ideal and of infinite depth. An explicit

solution to this problem has already been given by Ursell (28)

and Williams (34). In this thesis no attempt is made to

obtain an exact solution to the problem. Instead, this

particular geometry is used to illustrate the formulation and

the method adopted to deal with more general problems of

scattering of short surface waves by plane-sided obstacles.

Using matched asymptotic expansions, it is shown

that the lowest point of the barrier is the only singularity

of the outer potential (P0(x,y) of (0.6). An inner region is

constructed in the neighbourhood of this point and estimates

for the reflection and transmission constants are obtained

immediately after the inner problem is solved. These estimates

can be reproduced from Ursell's solution by expanding, for

short waves, the functions that occur therein.

Chapter 2 deals with the transmission of short

surface waves under geometries having a vertical plane side

touching a curved part in such a way that a knife edge is

- 15 - ,•

formed at the lowest point of the plane side. The curved

part is so chosen that the uniqueness conditions, as given by

John (15), are fulfilled. Two cases are considered: the

first is when the curved part is locally horizontal at its

intersection point with the free surface; this leads to a

problem involving a semi-infinite dock, and the second case is

when the curved part of the obstacle intersects the free

surface normally giving rise to a potential produced by a plane

vertical wave marker ELeppington (16,17) . The chapter starts

by showing that, when the surface wave train is incident

normally on the plane side, the inner problem for the first

term in the inner expansion, in the proximity of the knife

edge, is exactly the same as that around the tip of the vertical

barrier of Chapter 1. Since the first two terms in the

expansion, for small wavelength, of the reflection constant

depend only on the solution of this inner problem, it is then

concluded that the two-term estimate remains the same as in

the case of scattering by a finite vertical barrier. The

single-term estimate obtained for the transmission constant,

on the other hand, is found to depend on the geometry involved

in each case. The complementary (reverse) problem of each

case is also considered (this is a problem obtained by rever-

sing the direction of the incident wave train in the original

problem keeping the rest of the configuration unchanged) in

16-

which we show that the leading term in the asymptotic expansion,

for small E, of the transmission constant remains as in the

original problem.

Adopting the method and formulation used in the

prototype problem of Chapter 1, the scattering of short gravity

surface waves, on an ideal fluid, by a cylinder of rectangular

cross-section, is tackled in Chapter 3. Again the outer

potential is finite everywhere except at the right-angle corner

formed by the vertical plane side, upon which the surface wave

train is incident normally, and the horizontal bottom of the

cylinder.

An inner problem, whose solution is utilised to find

a two-term asymptotic expansion for the reflection constant

and also to fix the exact behaviour of the outer potential

approaching the corner, is formulated and solved by an

integral-equation method in the manner of Chapter 1. The

leading term in the asymptotic expansion for the transmission

constant is found from the application of Green's theorem to

the total potential and a Green's function due to John (15).

The procedure utilises the finite behaviour of the outer

potential near the right-angle corner formed by the bottom and

and other vertical plane side. The chapter ends with the

solution of the outer problem and the consideration of two

- 17. -

limiting cases: the first is when the cylinder becomes a

vertical barrier of finite length and a thickness (> E), and

the second is when it becomes a finite dock of thickness (>> E).

In Chapter 4 we take up the problem of transmission

of short gravity surface waves under an obstacle having, at

least, one vertical plane side intersecting a curved bottom,

or side, in such a way that a right-angle corner is formed. As

in Chapter 2, it is shown that, when the waves are incident

normally on this vertical plane side and in the limit of short

waves, the inner problem in the neighbourhood of this corner is

exactly the same as in the case of the rectangular cylinder of

Chapter 3. A detailed discussion is given for the case of a

cylinder with two (not necessarily equal) vertical plane sides

and a curved bottom locally horizontal at the lowest point of

one side, and is locally vertical at the lowest point of the

other. The reverse problem is also treated in which we show

that, although the details are different, the estimate

obtained for the transmission constant remains the same.

The two-dimensional problem of scattering by a

wedge, with one side vertical and the other making an angle

art (%2- <10(.<1) with the mean free surface, is considered in

Chapter 5. As in Chapter 3, two limiting cases of this

obstacle (a. 2 and a. ---) 1) are considered at the end

- 18

of the chapter with interesting effects on the leading term

for the transmission constant. The generalisation of the

wedge problem is the subject matter of Chapter 6 where, by

taking appropriate limits, some of the results obtained in

the earlier chapters can be easily recovered.

The final chapter of the thesis deals with the wave-

making problem in which the obstacle is given a forced periodic

motion, of small amplitude and high frequency, about some

fixed position. Utilising the outer solution of the radiation

problem and that of the corresponding scattering one, the

Haskind Relations are verified for the sway and heave cases.

These relations show that the exciting force (the y-component

corresponding to the case of heave and the x-component to that

of sway) is proportional to the amplitude of the radiation

potential at infinity, in the direction from which the waves

are incident in the corresponding scattering problem.

CHAPTER 1 - 19. -

SCATTERING BY A RIGID PLANE BARRIER IMMERSED TO A FINITE VERTICAL DEPTH

1.1 INTRODUCTION AND FORMULATION

The problem of scattering of surface waves by a rigid

plane barrier of infinite length and zero thickness, immersed to

a finite vertical depth, has been solved exactly by Ursell (28).

This is a two-dimensional problem in which the wave crests are

parellel to the barrier. Ursell's formulation hinges on a one-

dimensional integral equation for the normal fluid velocity in

the plane of, and below, the barrier, and by adjoining the

derivative of this equation in the same plane, an explicit

solution is effected. In 1966 Williams (34) presented a new

approach to the same configuration by reducing the boundary

value problem to one of a classical type which is easily soluble

and, in particular, the transmission and reflection coefficients

were found without much detailed analysis. Faulkner (9) treated

the three-dimensional water wave scattering problem in which the

effect of a fixed vertical barrier of finite depth on obliquely

incident surface waves is considered. Using Williams's method,

Faulkner reduced the problem to that of solving Helmholtz's

equation in an infinite region in which there is an infinitely

long strip of finite width. This problem is solved by an

approximate method based on the Wiener-Hopf technique, in order

- 20 -

to derive the reflection and transmission coefficients. Levine

and Rodemich (21) solved the same problem by the integral

equation method in the manner of Ursell. In the same report,

the problem was discussed by employing a mapping technique and then

by an integral equation-transform analysis.

In 1968 Jarvis and Taylor (14) extended Williams's

method to the two-dimensional case where the barrier extends

indefinitely downwards from a finite depth. In the present work

we make no attempt to seek a closed form solution to the problem

considered by Ursell (28), but we use it as a prototype to illus-

trate the method used to describe the potential and derive

estimates for the transmission and reflection "constants" in the

short-wave asymptotic limit. The solutions are then generalised

to deal with scattering by other two-dimensional geometries with

appropriate restrictions. We consider a fluid of semi-infinite

extent, assumed inviscid, incompressible and at rest under gravity,

occupying the region 1(x,y,z):-co<Cx<Cco, y ->. 0, -400<z<001

described in a right-handed Cartesian coordinate system, with the

x and z axes in the plane of the undisturbed surface and with the

y axis vertically downwards into the fluid. Any motion of the

fluid will be assumed irrotational, simple harmonic in time and

independent of z.

The surface disturbances will be taken to be of small

amplitude. Under these conditions it can be shown, Stoker (27),

- 21 -

that a single-valued velocity potential exists, satisfying Laplace'E

equation throughout the fluid domain, and a linearised free

surface condition. A rigid barrier occupying the region [(x,y,z):

x=0, 0=..C: y< a, -0o<z<zaq, is fixed in the fluid. The

prescribed condition on the surface of the barrier is that the

normal derivative of the potential is zero there. This is an

expression of the fact that the normal velocity of the fluid on

the surface of the barrier is zero.

The problem considered here is the scattering by the

barrier of a surface wave incident from x= - 00 at right angles to

the barrier and specified by the potential Re[oPi(x;E)exp(-D-001,

where

P1(x;E) =

and -- 211

is the frequency.

The parameter E (= tt, g is the gravitational acceler-

ation) is a measure of the length of the travelling waves and is

taken to be very small compared to the length a (i.e., a —<=1).

Such a wave will be partially reflected and partially

transmitted by the barrier, and in view of this the total potential

Re iT(x;E)exp(-1.110J induced by the incident wave is specified by

the following linearised conditions:

a2 a2 --2)tp(x;E) = 0 in the fluid, Ox ay

+E—aT = 0 on the mean free surface, ay

= 0 on the barrier. ax

The potentialtP(x;E) is required to satisfy the edge condition

r--ar --4 0 as r —) 0, (1.1.5)

where r is the distance from the edge at B(0,a). This condition

represents the physical assumption that whilst the velocity of

the fluid at such a point may be infinite, there is no source

present.

Except for the incident wave term, Q(x;E) must

represent outgoing waves at infinity. This "radiation" condition

is expressed mathematically by

Ti(x;E) +R exp (-lam ) as x –4 - (.0 y(x;s)._

T (1.1.6)

exp(1F) as x --4 + co

where R and T are the reflection and transmission constants

respectively. The uniqueness of the solution of problem (1.1.2)

- (1.1.6) is ensured by the edge and radiation conditions.

Condition (1.1.6) can be written in the equivalent form

2e.--37 cos (E-() + (11- 1)exp(-1i(- Y) as x --> -

T exp (i--2/) as x -- + co s 1.1.7)

- 23.-

Conditions (1.1.2) - (1.1.4) and the form (1.1.7) of the

radiation condition suggest that we may write

31Ecos(i-c)+T1(x;E) (x<0)

(x; E) T2(2.9 E) (x > 0) (1.1.8)

where the functions )1(x;E) and 42(x;E) satisfy the specifications

(1.1.2) - (1.1.5) in their domains of definition with

qyx;E)—(11- 1)exp(1- ) as x --4 - co

kp 2(s;E)-1" exp (i+X) as x --4 +cc)

Since tp(x;E) and ax

are continuous in the fluid region, it

follows at once that the functions Ti(x;E) and .P2(x;E) satisfy

the continuity conditions

ail k, ax

aq: 2e +q)1 (-0 y-E) = ky+0,y;E)

ax -0 y.E) = (+0,y;E)

Y A= (y> a) -1 (1.1.10)

For the particular geometry of figure (1.1.1) it is easy to show,

using the Green's function of appendix (A.1), as we shall see

later, that

Vx,Y;E) -T1(-x/Y;E) (x>0, y;?- 0) (1.1.11)

and hence the respective use of (1.1.9) and (1.1.10) yields the

following properties:

+ T = 1

(1.1.12)

and

- 24 -

Figure (1.1.1)

-25-

4:12(+0,Y;E) = e YAE

(y > a) (1.1.13)

This symmetry is not used explicitly in the following analysis,

since the method is being developed to deal with more general

geometries when the symmetry is not necessarily present. The

constants R and t are unknowns of the problem, and the objective

of the present work is to describe the potential field and, in

particular, to estimate the constants .11 and T, in the short-wave

asymptotic limit E --4 0. The method used will be that of

matched asymptotic expansions as propounded by Van Dyke (33) and

developed by Leppington (16,17,18) for the application to the

scattering and radiation of short surface waves.

In this method, the basic assumption is that the fluid

region can be divided into overlapping domains in which different

asymptotic approximations for the potential are appropriate:

The main body of the fluid, referred to henceforth as

the "outer" region: this region consists of every point of the

fluid domain whose distances 61 and 62' from the tip of the

barrier and the free surface respectively, are both many (short)

wavelengths. This is to say that oi..>E(j = 1,2). In this outer

region a first approximation T(x;E)--0(0,19o (x) is specified by

setting E = 0 in the free surface condition (1.1.3), where 3(E)

is a scale factor to be determined. It is clear that conditions

(1.1.2) - (1.1.4), with E = 0 in (1.1.3), do not permit any

- 26.-

surface waves and therefore 4;1 o(x) must vanish at large distances

(infinity) from the origin on account of the radiation condition

(1.1.6).

In the regions that are very close (on the length scale

a) to the points A(+0,0) and B(0,a), referrred to hereafter as

the "inner" regions, the approximations involve solutions that

are slowly varying functions of independent variables scaled with

respect to E in the following manner:

x = EX

y = a + EY

Lp(EX, a+EY;E) = .EX,Y;E)

} (1.1.14)

for the lower inner region in the vicinity of B(0,a) Dee

figure (1.1.1)j.

For the two upper (i.e. surface) inner regions, in the

neighbourhoods of A(+0,0), we have the respective transformations:

x = -EX1 y = EY

1 (-EX EY •E) =

1(X

l'Y1;E)

(1.1.15)

and

x = E X2 y = EY

2 (1.1.16)

(P(EX2' EY2' •E) = 2 (X2' Y2' •E)

where (X,Y), (X1'Y1) and (X2'Y2) are the inner coordinates and

-27-

1(X,Y;E), k(X1,Y1;E) and 2(X2,Y2;E) are the inner potentials

in the respective domains.

Since all points lying at great distances from A(+0,0)

and B(o,a) are outside the regions of validity of the inner

approximations, there is some difficulty in assigning boundary

conditions at infinity for the inner solutions. A similar

difficulty arises in deciding on the correct edge conditions for

the outer potential q) (x) which is not valid at B(O,a) and does o —

not hold near the free surface. The' idea of matched asymptotic

expansions provides the means of completing the specifications

for the inner and outer solutions Dee Leppington (16,17,18).

1.2 THE OUTER APPROXIMATION

We have seen that an outer approximation 4%, o(x)

can be obtained by formally putting E = 0 in the free surface

condition (1.1.3), where the scale factor 13(E) is to be deter-

mined by matching considerations and the outer potential 1010(x,y)

is specified by the following homogeneous boundary-value problem:

a2 a2 (7 + --7)To(x) = 0 in the fluid domain, ax ay

ek% ax

= 0 on the barrier,

0

(1.2.1)

= 0 on the mean free surface (y = 0),

, 2 o

---) 0 as k x y

- 28.-

Such a homogeneous problem is known to have the unique solution

o = 0 unless o(x) is singular at some point, or points, within

the fluid region. We therefore seek a solution ),c)(x,y) such that

L (x)--Tcost2(9 -FT-t-2)1 as r -.--4 0, (1.2.2)

r

where (x,y-a) = r(cos9,sin9) Esee figure (1.1.1)] and k is a

constant to be fixed by matching and solving for the inner

problem, as will be seen later.

This behaviour is evidently consistent with the

harmonicity of Lp o(x) and the boundary condition on the barrier.

The square root r-dependence is the least singular form ofo(x)

as r 0, and is the one that appears most likely in the light

of Van Dyke's principle of minimum singularity [Van Dyke (33)];

a more convincing justification for this choice is that it is

found to be the only one that can be matched with an inner

solution. At this stage it is also pointed out that possible

singularities at other points such as A(+0,0)] can be ruled out

on similar grounds - they cannot be matched to any regular inner

solutions. The outer approximation is then given by:

(x;E) 143(6)(F)0(x) as E 0, (1.2.3)

where the outer potential Co(x) is now the solution of the

following boundary-value problem:

- 29 -

a' a'g ,

\ 2+7 okx) = 0 in the fluid, Ox ay

at% = 0 on the barrier,

430 = 0 on the mean free surface,

--) 0 as (x2 + y.

2 --)00,

o— r 2cos[ 2(9 +1.1)3 as r 0. 2

(r,19) are as defined in (1.2.2).

ox (1.2.4)

Using a conformal transformation, problem (1.2.4) is

found to have the solution

1)0(x,y)=(a).2.1te -iz , 2 va + z2

with z = x + iy and -72-;?. 31I

(1.2.5)

In order to remedy the defect due to the absence of

waves in the potential o(x), the radiation condition (1.1.6)

shows that the addition to the outer approximation of surface

wave trains whose amplitudes are to be determined is an essential

requirement to cover the region close to the free surface.

In the earlier work, Leppington (16,17,18), matching

of the outer field has been done with the wave-free part of the

inner solution. Additionally, the wave terms that are launched

from within the inner regions were matched in a trivial way by

simply adding surface wave terms of the form

- 30-

q>s(x;E) = A+ e xp , (1.2.6)

in the outer region. Although these extra terms in the outer

region adversely affected the boundary condition on the scattering

surface S, their effect was nevertheless negligible in the outer

region (many wavelengths from the free surface). This was

because the surface wave terms s(x;E) of (1.2.6) were clearly —

og such that their normal derivatives -77 were exponentially small

for all points of S in the outer region (y.-- E).

The analysis in our present problem is crucially

different. For it transpires here that the first order outer

approximation is itself exponentially small; we shall find in

fact that the scaling factor p(E) of formula (1.2.3) is equal to

tEexp(--M •

Obviously the potential of (1.2.5) satisfies the

690 boundary condition ox = 0 on the barrier.

If we wish to add free surface waves to 4)(x;E) in

(1.2.3) we must clearly ensure that they do not destroy this

boundary condition on the barrier. Thus any additional wave

terms must also have zero normal derivative on the barrier, or

we must at least ensure that they, and their derivatives, must

vanish to the same order of magnitude as our outer approximation

(1.2.3).

- 31 -

Now the form of the solution suggested by (1.1.8) does

have wave terms with this property, and this suggests that we

write

A: - i__—Zc 2e y cos(0+(R-)exp( -2 6 )+9(EY,a0 (x<O)

T exp(12 )+kf3(E) 0 (x>0)

as E 0. (1.2.7)

But T and 1), as we shall see later, are exponentially small

compared to 3(E). Thus, to our first order approximation, we

have

2e 3r cos ( ) + kP(E) 1130(x) (x<O)

(x;E) 10(E)(60(x) (x>0)

as E --4 0. (1.2.8)

Comparing expressions (1.1.8) and (1.2.8) we see that the

approximations

Lp.(x;E)-0(E)%(14) as E --4 0 (j=1,2) (1.2.9) j —

hold in the outer region with '130(x,y) given by (1.2.5).

According to (1.1.10) and (1.2.9) there is apparently a discon-

tinuity in 430(x,y) across the plane (x = 0, y>a); it transpires, a

however, that P(E) = E1/2exp(7), and the discontinuity as given

by that condition is exponentially negligible in the outer

region (y - a)> E. Thus this discontinuity would appear in a

- 32-

higher order outer approximation but does not appear in our

outer potential ;ID; thus o

(x) is continuous across (x=0, y>a).

1.3 FORMULATION AND SOLUTION OF THE LOWER INNER PROBLEM

Before we proceed to formulate and solve the inner

problem in the neighbourhood of B(O,a) we must point out that the

consideration of either inner problem, on either side of A(0,0),

does not give rise to an outgoing wave train. This is because

the resulting inner problem is, in either case, a vertical beach

one with no incoming waves because of the formulation (1.1.0]

whose solution satisfies the homogeneous boundary conditions,

is large at infinity and hence is wave-free [Alker (2)].

It is clear, therefore, that in order to realise our

objective of estimating the reflection and transmission constants,

in the short-wave asymptotic limit E --4 0, we must focus our

attention on the lower inner region in the vicinity of B(O,a).

Our starting point, to this end, is the outer approximation

(1.2.9).

Because of the behaviour of the outer potential 5 (x), o —

of (1.2.5), near B(0,a) we arrive at the following asymptotic

expression:

LP.(x;E)--kilal(E)r 2cos{-1-(9+TI)1 as E 0, 2 as r 0 (j= 1,2)

(1.3.1)

- 33 -

with (x,y-a) = r(cose,sine) and the limits are taken such that

--4 0. Defining the lower inner coordinates (X,Y), as in

(1.1.14) with q)(EX,a+EY;E) = tjj(X,Y;E) (j = 1,2), rescaling

(1.3.1) and using the matching principle Evan Dyke (33),

Leppington (16,17,10] we obtain:

, ki3(E)

t[5.(x,Y;E)-- 1cosp2-(9 -Fri)i as E 0, as R ----> co 3 R2 (1.3.2)

where (j =1,2), (X,Y)=R(cose,sine) and the limits are taken

such that (ER) 0.

Formula (1.3.2) suggests the inner approximations:

LI1j(X,Y: )—E13,(E) tlijo(X,Y)asE—) 0 ( j =1,2),(1.3.3)

where the harmonic functions

aq110 8X

(X,Y) satisfy the conditions

= 0 on (X=--0, Y < 0)

a 20 0 on (X= +0, Y< 0) (1.3.4)

ax

and at large R they have the behaviour:

€1. jo 2 (X,Y)—kR -cos {.1/2(9 +7-)1 as R —4co

(1.3.5)

with j = 1,2.

The continuity conditions (1.1.10), when rescaled in

the same manner as above, furnish the relations:

2exp(--2--Y)+11.11(-0,Y;E)=02(+0,Y;E)

6112 8X(- YE = ax

(Y>0) (Y>0) (1.3.6)

- 34 -

On substituting from (1.3.3) into (1.3.6) we get:

0 (E) = E%xp(4) (1.3.7)

and

- 2e Y +v210(-0,Y) = 4120(+0,Y)

410 a46 (-o,Y)=(+o,y) = v(Y). ax ax

(Y>o) (1.3.8)

V(Y) is the (unknown) normal velocity distribution on the Y-axis.

Now, on physical grounds, V(Y) does not vanish at any point on

that axis except possibly at infinity.

Differentiating (1.3.5), we obtain a formula for the

behaviour of V(Y) at large distances. Thus

V(Y) kY-3/2 as Y --00, (1.3.9)

Now, we assume that the nature of the singularity in the velocity

field is such that

V(Y)--d XY-G0 as Y 0

(1.3.10)

where Xo is a constant and -1-2 < o <1 in accordance with the

edge condition (1.1.5).

This means that the velocity is infinite at the tip of

the barrier, but there is no source present there.

We shall see later that the assumption (So <:%2. leads

to a behaviour, at infinity, of V(Y) different from (1.3.9) and

hence contradicting our earlier choice for the behaviour of the

outer potential near the tip of the barrier.

- 35 -

Integral equation for the velocity distribution V(Y):

The application of Green's formula to the Green's

function

G(X;X )=— 41Tclog{[(X-Xo) 2+( y.yo ) [(X+Xo)2+(11...Y0)21}

(1.3.11)

and each of the harmonic functions MY) (j = 1,2), and the Jo

usage of (1.3.4) and the second of conditions (1.3.8) yield cc)

10(-0Yo) V(Y)loglY-YoldY (1.3.12)

and td

41200-02Y0) f V(Y)loglY-Yol dY (1.3.13)

Subtracting (1.3.12) from (1.3.13) and using the first of

conditions (1.3.8) we arrive at:

-Y o 1 TC e = – "V(Y)loglY-Y IdY 0

(1.3.14)

The integral equation (1.3.14) is a singular one since, as was

pointed out earlier, V(Yo) / 0 (0 Yo< co).

The formal differentiation of both sides of (1.3.14)

with respect to Yo leads to the new integral equation: 05

e –

-Yo = 1 V(Y)dY 7 0 Y-Yo

where the symbol "I" means the Cauchy Principal Value of the

integral.

(1.3.15)

- 36 -

Solution of the integral equation:

Equation (1.3.15) for V(Y) can be solved by taking the

Mellin transform of both sides EErdelyi (7), Sneddon (26)".]. Thus

V(s) = r(s)tanlTs for clo< Res = G < 1 . (1.3.16) roo,

Note that i7(s) = V(Y)Ys-idY exists for Go< Res =c5< -3- because 0

2 pots-1 of (1.3.9) and (1.3.10), the integral

0 „ ..7-77dt exists for

L

0 <-11es = 0<-1 and r(s) exists for Res =0 > 0.

Writing c/(s) as the product of two functions 1(s) and

g(s) where

1(s) = s2r(s)

g(s) = s-2tanTts < Go < Res =0 <1)

(1.3.17)

we immediately see that

f(u) = u(u-l)e-u (1.3.18)

Using the inversion formula for the Mellin transform [Sneddon

(26)] we obtain: 0+ipol

u-s

g(u) =7 f ---2-tanlIs ds; 2<Go<CS<1 (1.3.19) U-ico s

Using the residue theorem we arrive at the result:

g(u)=

(1.3.20)

(0 < u 1)

V(Y) = f(u)g(u)du U U 0

(1.3.21)

- 37 -

The continuity of g(u) at u= 1 is ensured by the fact that

1 TT2

2 +1)2 8 • n=0

Applying the convolution theorem for the Mellin trans-

form ESneddon (26):1 we arrive at the following solution: co

where f(u) and g(u) are given by (1.3.18) and (1.3.20) respect-

ively. Substituting from (1.3.18) into (1.3.21) and putting

1 u = _T we get:

0

For large values of Y, most of the contribution to V(Y) comes

from the neighbourhood of r= 0.

Using the first of the series (1.3.20), with u = 1 f,

in (1.3.22) and applying Watson's Lemwa ECopson (5)] we get: CO

vV(Y)^- 1

P(n+1/2) as Y % Yn+ 2

(1.3.23)

n=

In particular

V(Y)' — 1 Y-3/2 as Y co (1.3.24)

which is the same as the predicted form (1.3.9) with

k =

(1.3.25) vt

co

V(Y) = ors (Y2T - Y)e-YT g(i)dt ' (1.3.22)

At this point we stop to remind ourselves that if Go, of (1.3.10),

-38 -

is strictly less than .1-2. then the contribution from the pole at

s = 2 to g(u), of (1.3.19), for u .,-;?-1 must be taken into account

and hence will affect the behaviour of V(Y) at large Y. In fact

if ao< z then V(Y) is of the order of Y 2 as Y a) contradicting

the fact that the outer potential is singular at the tip of the

barrier. Therefore our earlier assumption about Go being larger

than or equal to 2 was, in fact, the correct one to make.

It can also be shown see appendix (A.2)] that the

general form of the velocity distribution is given by co

i V(Y) =..r1t ir ii2.2770.dt (1.3.26)

It is clear that 00 00

- Y 2 1 -YT 1 Zee d

V(Y) +— = T e dT +— TL 1 T __t= _1 e

YT dI

ift 0 T1 o T2 % (1--c)

(1.3.27) co

Since , dx _ 0 (this can be seen by making the substitution 0 x'(1-x)

x = 1/I) then (1.3.27) gives:

Hence

[V(Y) + Y = 0. Y=0

(1.3.28)

V(Y) - --Y 2 as Y 0. (1.3.29) vt

Comparing (1.3.10) with (1.3.29) we see that:

Go =

X 0 =

1 (1.3.30)

- 39 -

The fact that (Yo = z can be confirmed by the harmonicity and the

boundary conditions on the barrier.

1.4 ESTIMATION OF THE REFLECTION AND TRANSMISSION CONSTANTS

To find the asymptotic values of the constants R and T

as E --4 0 we need the use of a Green's function due to John (15),

which we may call "fundamental" for two-dimensional wave motion

in water of infinite depth. This Green's function G(x; ;E) is

t-01 the potential at (x,y) co<< due to two oscillatory line

sources of unit strength situated at the points = ,11) and

,11) co < < co ,11> 0 beneath a free surface. The

solution for G(x; 0) is well known see appendix (A.1)11. On

applying Green's theorem to G(x; ;E) and each of the functions

1(x;E) and 4) (x;E) Dee figure (1.1.1)=1 we obtain:

;E) =-

00 ZikP 1 f Go,y;,-71;E)--„,dy,

a co z4)2

(1.4.1)

where

k-P 2( ;E) = a

> 0, 11> 0.

G(0,y; ,T1 ;E).--6-7dY (1.4.2)

Using (1.1.10), (1.4.1) and (1.4.2), formula (1.1.11)

follows at once.

The property (1.1.11) also shows that the scattered

field

(x;E) (x,y,E) - exp(b—p) (x >0)

(x,y;E) + (x< 0)

is odd in x. It can also be shown from (1.4.3) and the first of

equations (1.1.10) that(1)(-0,y;E) =(P(+0,y;E) for y > a, and

therefore is continuous across (x = 0, y> a) proving that

(1)(o,y; E) = 0 (y > a).

Putting the values of the constant k and the scale

factor 3(E), as given by (1.3.25) and (1.3.7) respectively, in

(1.2.9) we get:

E%xp(i) (x" (x,y) as E —40 (x > 0) (1.4 2

.4)

where (00(x,y) is given by (1.2.5).

From (1.1.9) it is seen thatT2(x;E) is an outgoing

wave at infinity, and hence it is appropriate to take the wave-

part of G(0,y; 111 ;E) see appendix (A.1)] in (1.4.2) to obtain

rc°6 2 LP2w( ,11 ;E) = -2itje Y 771c Y ) > 0), (1.4.5)

where q)2w ,T;E) is the transmitted-wave part of the potential

q31 2( ,-11 ;E) (> 0).

Asymptotic evaluation of the integral in (1.4.5):

Choosing a number 0. such that E<=5:(ct<<" a and writing

(1.4.5) in the form:

a

LP2w(. ; E) = -2i(I1 + I2)exp(1411-),

— 41 — (1.4.6)

where

and

a+a

I1 = e 37/ 2 —dv

a

(1.4.7)

sc").-yAE 25, 12 = aa

bx`AY' (1.4.8)

-l-

then, since E <<a << a, the range of integration in 1. is within

the inner region, and that in 12 is within the outer one.

Now we proceed to show that, in the limit E 0, 12

is exponentially small compared to

Using (1.1.14) with LP2(EX, a+EY;E) = 412(X,Y;E), Il

of (1.4.7) becomes:

a/E = j e

-Y aX2dY

Now, the use of formulae (1.3.3) and (1.3.7) and the exploit-

ation of the fact that, by definition, E. co as E —I, 0, yield:

co

- e-Y

ax

0 , 2a, dYlexpk- 77) as E --> 0. (1.4.10)

0

Using the second of equations (1.3.8) and (1.3.26) we arrive at:

I e 2a

2 xP(-1. ) as E 0. (1.4.11)

Formulae (1.2.5), (1.4.4), (1.4.8) and the use of Watson's

Letiuna ECopson (5)] give

E 3/2 (a)

-2a-a, I2 ti exp(----E--) as E 0 (since CC<<a).(1.4.12)

21/FT

(1.4.9)

- 42 -

It is easy to see now that 12 is exponentially small compared

to I1

because of the presence of the factor (--.E )3/2

exp(-2) with

_-4o0 as E --4 0, and therefore (1.4.6) now reads:

4)214( ;0, exp(..2.e+ ) as E 0, (1.4.13)

giving the asymptotic estimate

4) as E 0

for the transmission constant.

Using (1.1.12) we obtain

R exp(_2e) as E ---> 0

as an asymptotic value for the reflection constant.

(1.4.14)

(1.4.15)

Higher order terms:

To improve the above estimates for T and R, we note

that the contribution of I2' as given in (1.4.8), to '.E) 2w --

of (1.4.6) is always asymptotically negligible compared to that

of I1,

as given by (1.4.7), no matter how many terms of the outer

expansion are included in the integral (1.4.8). It is, therefore,

imperative that we concentrate our attention on the integral

as given in (1.4.9). From the behaviour of the outer potential

p(), of (1.2.5), near the tip of the barrier [i.e., for I z-ial<<lj

it is easy to see, after rescaling, that the inner field in the

neighbourhood of B(C,a) has the following asymptotic development:

- 43 -

U2 " (X Y•E),--e-a E j 2 j(x'Y) as E --410, (1.4.16) j=0

where the harmonic functionsc32j satisfy the condition

at/ OX (+c,Y) = 0 (Y <0),

for j = 0,1,2,...

(1.4.17)

Rescaling (1.1.13), substituting from (1.4.16) and

using the fact that 020(+0,Y) = e-Y 0) we get:

T2i(+0,Y) = 0 (Y >0, j 1) (1.4.18)

Formulae (1.4.9) and (1.4.16) give

00 co v

ax a°9

I Ej e-- dY} exp(

--2-i2) as E 0 (1.4.19)

j=0 °

It is clear from (1.4.19) that in order to determine the jth

term in the asymptotic behaviour of I1 we must find the term

2j(X,Y) in the inner expansion (1.4.16).

However, in our present work we outline the method by

(x Y) and noting that more terms can be found by

solving for the outer potentials terms in the outer expansion

as given by (1.4.23) below and using matching techniques.

It is shown in appendix (A.3) that, at large R, the

first term in the inner expansion has the behaviour:

co

20

( -1 )1i n+ P(n+1/2) cost (n+1-) (9 +T-t- 2)1 as R co / % ) n=0

R

where (X,Y) = R(cosA,sine). (1.4.20)

The inner approximation, then, has the behaviour:

e-a/E (-1)11 r(n+k-) () cos t (9 + .11)1

TL n=0

2(X;E)-- % 2 J n+ 2

as E --> 0, as R —> c° and (E R) ----> 0 (1.4.21)

Resealing (1.4.21), bearing in mind thatLP2(EX, a+EY;E) =f12(X;E)

and using the matching principle we arrive at the expansion:

--- -a 2e q31 (x;E) 2 — TL

(-1)1/En Pg n(n+i) cos f(n+) (0 +TI)

r 2

n=

as E 0, as r 0 ana r (1.4.22)

where (x,y-a) = r(cos@, sine).

This suggests the following outer expansion:

2 -e a/E n T2(x; E)— c.

E(x,y) as E--4 0, (1.4.23) vri n=0,

where the harmonic function Lip (x) satisfies the specifications:

ax on the barrier,

LI)n by n-1 = 0 on the mean free surface,

(1311 0 as (x2 + y2)1/2 co 1.4.24)

(-1)n(2n):.. cos [(n+-1/2)(9 +i) as r 0 4n(ni.)rni

with L given by (1.2.5), n >1 and (r,9) as in (1.4.22).

-45-

Now we proceed to find the behaviour, at large R, of

the second term U21(X,Y) in the inner expansion (1.4.16) and

prove that (1.4.24) determines the harmonic function ./13 n(x)

completely:

To this end, we assume that near the tip of the barrier

the harmonic function(x,y) has the most general behaviour I.

consistent with the harmonicity and the boundary conditions. Thus

Lpn(x,y) Y(ea)ri 1/2cost(j4)(91)1 as r --4 0,

j=-n (1.4.25)

where211)(a) is a function of the geometry only with Y(n)(a) = , j -n

(-1)"(2n)! from (1.4.24). 41/(n!)

Substituting from (1.4.25) into (1.4.23) we obtain:

ait

LPi''. vqt y (11 )(a)rH c osH)(9 +TI-2 )il J

E 2e TEn

n=0 j=-n

as E --4 0, as r -- 0 and ,. --4 co. (1.4.26)

Rescaling (1.4.26) by putting r =ER and using the matching

principle we obtain

2 —-)

(X;E)--- e a Y(nn(a)Em R cos (m-n-;-)((g+-N

n=0 m=0

as E --> 0, as R co with (E R) <<1. (1.4.27)

Using (1.4.16) and (1.4.27) it is found that

at121 -3Y 1 ax (+0,Y)

2 (Y > 0).

8a /TT (1.4.33)

-46 -

4121(X'Y (n) R-1/4-1/2 'y (a) cos [(n-2)(9 +7) as R —> co

(1.4.28)

With the help of (1.4.17), (1.4.18) and (1.4.28) we obtain the

following eigensolution:

(n) R-n+ Y(a) cos' t(n-.1-)(0 +II• )) (1.4.29) 2 -n+1 rrc 21(x,Y) =

Because all the inner potentials are finite at the tip of the

barrier, we then conclude that:

y (n+1n) (a) = 0 (n > 1) (1.4.30)

and (0)(a)

.021

.1_

- (X) 1 R2 COS{1/2(0 +.1)1. - — (1.4.31)

Putting n = 0 in (1.4.25) and using (1.2.5) it is found that

and hence

(o),a, _ T 1 ) - 4a (1.4.32)

Formulae (1.4.6), (1.4.19) and (1.4.33) give the improved

estimate for the transmission constant as:

+ 41)exp(-26 a) as E 0.

Using (1.1.12) we get

- i(1+E)exp(-4) as E 0. (1.4.35)

In general

(n) . 'y (a) R3-11-1 cos [(j-n-2)(0+T-t-) (1.4.36) 2 n+j

n=

(1.4.34)

have the exact formulae:

K () 1 E

K (2)— ircI1 E (2)

ilTIi (2) 1 E

11I (2) K (2) lE lE

T - (1.4.37)

(1.4.38)

(n) where j _> 1 and 'Y (a) = 0 (n -->- j),

-n+j

from which we conclude that the harmonic outer potential q3n(x)

is completely determined by (1.4.24). Note that the first term

in the asymptotic estimate for T does not depend on the solutions

of the outer problems and can be obtained hmuediately after V(Y)

of (1.3.8) is found from the solution of the first inner problem

as in §1.3. In fact, the first and second terms in the

asymptotic expansion for R are also independent of the solutions

of the outer problems even in situations when the relation

R + T = 1 does not hold, as we shall see in the next chapter.

Comparison with Ursell's exact solution:

It has already been stated in the introduction that

an exact solution has been obtained by Ursell (28) for this

prototype geometry. In particular, it was found that T and R

On using well-known asymptotic formulae for the modified Bessel

a a a E functions Il(T) and K1(7), for large , the approximate formulae

(1.4.34) and (1.4.35) are reproduced.

- 48 - CHAPTER 2

TRANSMISSION OF SHORT SURFACE WAVES UNDER GEOMETRIES WITH A KNIFE EDGE

INTRODUCTION

The method developed in the prototype problem of

Chapter 1 can now be extended to obtain corresponding results for

a more general class of obstacle shapes, in which the geometry is

plane vertical at one or other end. In the present chapter we

discuss two particular cases in which we utilise the solution

of the inner problem of §1.3 to find a one-term asymptotic

estimate for the transmission constant in terms of a "geometric"

constant depending only on the solution of the outer prdblem for

the particular case under consideration. The complementary

(reverse) problem of each case is also considered briefly, in

which we show that this estimate for the transmission constant

Y remains unchanged when, keeping the rest of the configuration

fixed, the direction of the incident surface wave train is

reversed see appendix (A.4)..].

We shall also show that, in each case, the first two

terms in the asymptotic expansion for the reflection constant

R are the same as those obtained in the prototype problem of

Chapter 1 for the case of scattering of the same surface wave

train by a vertical plane barrier.

- 49 - Case (i)

2.1 STATEMENT OF THE PROBLEM

First we consider the problem of scattering of short

surface waves by a cylinder with a vertical plane side AB,

A(0,0) is on the mean free surface and B(0,a) is vertically below

it [see figure (2.1.1), and an arc BC, C(b,0) is on the mean

free surface, having an equation f(x,y) = 0 and satisfying the

conditions:

(i) The arc is continuous, piecewise smooth and is such

that y i8 a single-valued, finite and non-negative function of x.

(ii) The arc touches the side AB at B(0,a) in such a manner

that a knife edge is created there see figure (2.1.1)] and

and = 0 at (0,a) for n = 1,2, . . . , (M-1) with M.;>2. ay"

(iii)t1-111

0 at (b,0) for n= 1,2, . ., (N-1) with N;?-2. axn

Tho integers M and N give a measure of the steepness and flatness

of the arc at B(0,a) and C(b,0) respectively and they both can

be infinite.

In the case when M =co, condition (i) above applies

to the part DC of the arc, where D is a point on AB lying

between A and B.

As in Chapter 1, we take a fluid of infinite extent

- 50 -

assumed to be incompressible, inviscid and at rest under gravity.

Any motion of the fluid will be considered irrotational and

having simple harmonic time dependence.

The cylinder is held fixed in the manner shown in

figure (2.1.1) with its generators parallel to the z-axis. A

regular train of surface waves of potential ReM(x;E)exp(-iWt)1

is incident from x=- co normally on the vertical side AB, where

-Pi(x;E) is given by (1.1.1) and Wand E are as defined in §1.1

with 1 --1.--C1 and1:3<.<1.

The velocity potential induced by the incident waves

is RqP(x;IE)exp(-iWt)1 where, suppressing the time factor, L1P(x;E)

is given by the linearised conditions that follow:

a2 Z•2 (-7- + = 0 in the fluid, ox ay,

ak-p + E— = 0 on the mean free surface

a = 0 on the cylinder, On

11 --- an denotes differentiation in the direction of the

outward normal to the body scatterer S.

The potential p(x;E) is required to be finite at all

points of S. Thus

lim aq3, r-90 Cray) = 0, (2.1.4)

Yl

tp+ELpy =0

exp EE =0 akP

ax

- 51 -

E cpy =

Texp(b_SLY)

C(ID,O)

aq) =0 e - B(O,a)

X

4'11

R exp (-P17111 E

Figure (211)

- 52 -

where r is the distance from any point on S.

This means that the singularity in the velocity field

is not logarithmic (i.e., source) or worse.

The scattered field is required to obey the radiation

condition at large distances from the body scatterer S. Thus

exp ) as x co

LP(x;E), (2.1.5)

+ 11' exp(1125-) as x - co.

II' and T are the reflection and transmission constants respectively

whose estimation, as E 0, is our main aim.

Condition (i) given at the beginning of this section

is needed to ensure that the solution to the boundary-value

problem (2.1.1) - (2.1.5) is indeed unique John (15a Although

there does not appear to be any a priori reason why uniqueness

should not hold for profiles which do not satisfy this condition,

we shall, nevertheless, assume that the profile is one of this

restricted class.

2.2 SOLUTION OF THE LOWER INNER PROBLEM

Now we proceed to show that the single-term inner

expansion in the neighbourhood of B(O,a) is exactly the same as

- 53 -

that in the vicinity of the tip of the barrier of Chapter 1. To

this end we start by adopting the same formulation as that used

in Chapter 1 by writing the total potential as:

f 2eY/E cos(i) + q)1(2s;E) (x<o)

k-P2(x;E) (x>0)

4)(x;E) =

(2.2.1)

where the harmonic functions q). and L2 satisfy the conditions:

43'1 +Wly = 0 on y = 0, x< 0,

<a,

) as x co -

(2.2.2)

(2.2.3)

= 0 on x = 0, 0 < y ox

(x - 1"

T2 +E-P2y = 0 on y = 0, OLP2

x > b,

BC, = 0 on the arc on

2 (x;E), -texp(1-21) as x +co

In addition the potentials 101 and L 2 satisfy the edge conditions

(2.1.4) and hence are finite on S.

The continuity ofq)(x;E) and tf. across (x=0, y>a)

yields:

/E -y 2e +LP1 (-0 y.c) = 2 "

(+0 y.E)

1 ap2

(y>a) (2.2.4)

bx(-0,Y;E) = 77(+0,Y;E)

- 54 -

As in Chapter 1, the outer approximations are given by:

yx;E) k(3(E) kk)(x) as E --4 0, (2.2.5)

where the constant k and the scale factor 0(E) are to be deter-

mined and the function kis)o (x) is given by the following problem:

v o = 0 in the fluid domain,

k130 = 0 on (y = 0, x<0 and x>b)

(2.2.6) = 0 on S (the body scatterer),

0(x,y) --4 0 as (x2 + y2)1 --•-) 03

O r 2cose as r 0,

ado on

where "02" is the Laplacian operator and (x,y-a) = r(sine,-cose).

Because of condition (ii) of §2.1, we see that the singular

behaviour of Lr)o (x) near B(0,a) is consistent with the harmonicity

and the condition of zero normal velocity on the scatterer.

The outer potential CP0(x,y) is known in principle: in

particular, near C(b,0)

Li/6 o *1 r 2sin1-81 as r1 --) 0, 2 (2.2.7)

where k* is a constant dependent on the geometry of S and

(x-b,y) = r1(cose1 , sine

1 ).

Note that we have located the singularity of the

potential 4-O at the point B(0,a)+ because any singular

- 55 -

behaviour of this function at A(-0,0) cannot be matched to any

inner solution there. Moreover, if (ps (x) is singular at C(b,0) o —

then this implies the existence of standing waves in the region

x > 0. This is impossible since the incident wave train is

coming from x = -00.

Defining a pair (X,Y) of inner coordinates according

to the formulae: )( = EX, y = a + EY with q)j .(EX, a +EY;E) =

4/3. ,(K l y;0 (j = 1,2) and using the same argument as in §1.3 we

arrive at:

P(E) = E%xp(--2-), (2.2.8)

111i (X,Y;E )— e-a/E

jo(X,Y) as E 0 (j= 1,2),(2.2.9)

o (X,Y) k R 2cos1/20 as R co ( j = 1,2), (2.2.10)

J

r = ER.

The continuity conditions (2.2.4) furnish the relations:

-Y 2e +10(-0,Y) = 120

(+0,Y) - (Y> 0) (2.2.11)

-20 410(-0,Y) = —a(+o,y) V(Y)

In the inner region around B(0,a), our intention is to transfer

the boundary condition on the arc to a new boundary condition

satisfied by the inner potential T20 on (X = +0, Y <0). As we

are confined to points in the immediate neighbourhood of B(0,a)

(distances <<a), the deviation of the arc from (X = +0, Y4( 0) is

small.

-56-

By expanding f(x,y) as a Taylor series about the

point B(0,a) we can see that, in the vicinity of B(0,a), the

equation of the arc can be written as:

co Yn(y-a)n

x = (N‹ c° )2 n: n=M

(2.2.12)

M 1 a f

m (M< co ). (2.2.13)

x ay B(o , a) where f

This means that the boundary condition of zero normal velocity

on the arc becomes:

OT2 n(Y-a)11-1

- Ox ( (n-1): n=

= 0 on x =

co Yn(y-a)11

n.1

n=M (2.2.14)

where Vx,y;E) is given by (2.2.3).

Rescaling as above we arrive at the condition:

n-I n d

2 ,aL12 Y-nE Y yil 0: y)n-1

- - 0 on X- - 1 (M <a)) OX OY (n -1): n.

(2.2.15)

Note that in the case M = 00, the boundary condition is simply OU 2

Assuming that 1J2(X,Y;E) (when analytically continued

across the arc) is analytic on the line (X = +0, Y< 0), so that

U2X and 2Y

can be expanded as Taylor series with respect to X

we then obtain:

n=M

ax (+0 'YE) = 0 (Y‹ 0).

- 57 -

y 04-1

(+0, Y; M fliM(l (+0 Y; E) MYM-1U (+0,Y;0.1 -2X L 2XX ' 2Y

+ terms of order E = 0 (M<oo , Y <0). (2.2.16)

Using (2.2.9) and remembering that MI->2, formula (2.2.16) yields:

84520

Since (-0,Y) = 0 (Y<O), formula (2.2.17) completes the

specifications of our inner problem and shows that it is exactly

the same as that in §1.3 with k given by (1.3.25). Using (2.2.5),

(2.2.8) and (1.3.25) we arrive at the following single-term outer

expansion:

q31 E2e

y• 2(xE) 0 (x,y) as E --4 0, Iffc ,

where the potential q50(x) is specified by (2.2.6).

(2.2.18)

2.3 ESTIMATION OF THE REFLECTION AND TRANSMISSION CONSTANTS

From the preceding section we see that the velocity

distribution V(Y) of (2.2.11) is given by the Cauchy Principal

Value integral (1.3.26) Dee appendix (A.2)]. The application

of Green's theorem to the harmonic function(x'y;E), of

(2.2.2), and G(x; ;E), of appendix (A.1), in the space (x< 0,

y2> 0) leads to an estimate for R exactly the same as that given

by (1.4.15).

This shows that as long as the arc f(x,y) = 0 is

locally vertical at B( 0,a) and makes a knife edge there, this

ax (+0,Y) = 0 (Y `C 0). (2.2.17)

- 58 -

two-term asymptotic expansion for R remains the same no matter

what geometry is present to the left of the plane side AB Dee

figure (2.1.1)]. As we shall see, the transmission constant, on

the other hand, depends on the solution of the outer problem

(2.2.6) and hence on the shape of the arc f(x,y) = 0.

The surface inner region:

Because of the general nature of the geometry involved,

it seems reasonable to start looking for an asymptotic estimate

for V by constructing an inner problem in the neighbourhood of

C(b,0) with the help of the formulae x = b + EXl, y = EY1 and

-2(b +EX CY -E) = H(X1,Y1;E). Using (2.2.7) and (2.2.18) we

arrive at:

* -VE L.P2(x,Y;E)'' k E r ;i112-9 as E --4 0, 2 r

as r1 --4 0 and E/r (2.3.1)

where (x-b,y) = ri(cosel,sinel) and the constant k* is defined

implicitly by (2.2.7).

Rescaling and using the matching principle it is found

that

k*C e 1

H(X1,111;C)-- I R1sink-A

1 + possibly outgoing

IFE waves

as E 0, as R1 ---)00and (ER1) 0, (2.3.2)

where RI2 = X12 + Y1

2.

- 59 -

This suggests that the inner potential H has the one-

term asymptotic development:

ke7,e-a/8 H(Xi,Y1;0— Ho(X1,Y1) asE-4 0, (2.3.3)

ITT

where the harmonic function H o (X Y1 ) satisfies the conditions:

H + — = 0 on (Y1 = 0, X1 > 0) Ho 6Y1 (2.3.4)

1 Ho R12 s in-2-91 + possibly outgoing waves as R1 --400.

(2.3.5)

The boundary condition on the arc:

To complete the specifications for Ho(X1,Y1)-we

observe that, in the neighbourhood of C(b,0), the equation of the

arc can be put in the form:

R(x-b)n (N<00), (2.3.6)

where

1 Nf P N ff y bxN C(b,0)

(N <°°) (2.3.7)

Using (2.3.6) and rescaling as above, the boundary condition oLP 2 = 0 on the arc becomes:

CO \/;0 pnEn-l . 13n (E x1)1/-1 q

0 on Y1

n=N

OH( 8H \

6Y1 1

n=

(N<oo) (2.3.8)

Note that in the case N =co, the boundary condition is simply

811 6 ' (X +0•E) = 0 (X1< 0). Y1

3r =

an

-60-

Assuming that H(X1,Y1;E) when analytically continued

across the arc 3 is analytic on the line (Xl < 0, Y1 = +0) so

that Hx

and Hyl

can be expanded as Taylor series with respect 1

to Yl' and substituting from (2.3.3) we obtain

OH 0(x

+0) = 0 ex <o) a 1, 1 Y1 (2.3.9)

This condition completes the specifications required to deter-

mine a unique solution Ho(X1,Y1).

Solution of the surface inner problem:

The problem for Ho(X1,Y1) is, in fact, the same one

met by Leppington Esee Leppington (16); formulae (2.19), (2.20),

(2.27) and (2.28).] when treating the problem of radiation of

short surface waves by a finite dock in a fluid of infinite

depth, in which he utilised the solution, due to Holford (11),

of a related problem.

The estimate for the transmission constant is realised

by extending the outer approximation (2.2.18) up to the free

surface, adding the regular wave train Yexp(1F) Dee (2.2.3)J

and matching with (2.3.3). Thus

T ik E e

irC a ib) as E 0. ' (2.3.10)

fi ------xp(- 8 - C E

The constant k is as in (2.2.7) and is fixed as soon as the

arc f(x,y) = 0 is chosen.

- 61 -

2.4 THE COMPLEMENTARY PROBLEM OF CASE (i)

In this section we show that the single-term estimate

(2.3.10) remains the same when, in figure (2.1.1), the direction

of the incident wave train is reversed keeping the rest of the

configuration unaltered [see appendix (A.03. To this end, we

start by specifying an outer potential q)0, valid many wavelengths

from the free surface, by setting (pc) = 0 on the mean free surface.

Evidently (Po is an eigensolution of the problem and has overall

scale constants that are as yet unknown. Thus we write

q) ()s; -, ki a (E) q:10(x,y) as E ---4 0, (2.4.1)

where (1)(x;E) is the total potential in the complementary problem,

4)0(x) is the solution of the outer problem

02o = 0 in the fluid,

= 0 on the mean free surface, 9

aq) ° = 0 on S, (2.4.2) an

o --4 0 at large distances from S,

()c'")r

1 sin 1 n-p9

1 as r

1 0,

and the constant k1 and the scale factor X(E) are to be found

by matching.

Because the waves are incident from x = +co, the

singularity of 4)0(x,y) is at C(b,0) and its existence elsewhere

cannot be matched to any inner solutions.

- 62-

The inner problem:

Defining a pair of inner coordinates (X 1 ,Y1 ) according

to the formulae: x = b E Xl, y = EY1 with (1) (b+Exi , E Yi; E) =

(10(1,Y1;EX] and using the matching principle we obtain:

NX1,Yi;E)--1 k1E-AX(Ego(Xi,Y1) as E --BO, (2.4.3)

where the harmonic functiono satisfies the conditions:

°I)o + - 0 on Yl = 0, X1 >0, 1

o R12sin1/291 + standing waves as R

1 ,

(2.4.4) and, as in §2.3; formula (2.3.9), the condition

oY°(X+0) = 0 (X1<= 0).

1

The problem for k has a non-trivial solution only if k has the

same order of magnitude as the incident wave, in order that it

may possess both incoming and outgoing waves at infinity.

The solution to the problem was obtained by Holford

(11), and was used by Leppington (16) when considering the

scattering of short surface waves by a finite dock Dee

Leppington (16); formulae (4.15) and (4.16)3.

In particular, the solution for large R1 is seen to be

% 1 o

R12sin%e + (27)2e

-Y Icos(X

1 8 -II) as R1

-400 (2.4.5) 2

7x- y Since the incident wave is given by LLi= exp(

i E ), then

(2.4.3) and (2.4.5) yield

(/(E)

kl =

% -ib = E2exp(-7)

2 1/2 (T) expl-TV

-63 -

(2.4.6)

Formulae (2.4.3) and (2.4.5) also enable us to find an estimate

for the reflection constant -R'1 of the reverse (complementary)

problem. Thus

R1 e xP' 4 - 2i1r)

as E 0. (2.4.7)

The transmission constant T :

Using (2.4.6), formula (2.4.1) becomes:

4)(25;E)--/ 1)0 (2s) as E 0 (2.4.8)

where CT)o is given by (2.4.2).

Since the potential $0(x,y) is finite at B(0,a) see

figure (2.1.1)], then in order to satisfy the boundary

conditions and the harmonicity, it must have the behaviour:

(I)o(x,y)-, p1 + pr2cos228 as r -3 0, ( 2.4.9)

where (x,y-a) = r(sin9,-cos9) and pl. and 11 are constants

dependent on the geometry only.

Differentiating (2.4.9) we get:

a(T) (-e-)x=0 7(y-a) 2

as (y-a) +0.

(2.4.10)

Combining (2.4.8) and (2.4.10) we obtain:

(ax)x,0^-(2Ti) (y-a) 2exp(-44) as E-40, (2.4.11)

as (y -a) --4 +0.

- 64.-

Applying Green's theorem to the total potential (1)(x;E) and

G(x; ;E), of appendix (A.1), in the space (x<0, 0<y<°0), we

arrive at: CO

( ;E) = - J. G(01M;TI;E)t-,-cdy (2.4.12) a

where <0,11 >0.

Taking the wave part of G(0,y;, q ;E), the transmitted

wave ikl,(11;E) has the form: co

g.)14(,"11;E) = 2i {f eY/E ox y.1J exp( E ). (2.4.13)

In (2.4.13) we make the substitution y = a +Et to get:

a

Co r

cr5exp( -11 E ) (2.4.14) QV]; = 2iE

0 bx y=a+ET

(2.4.11) and (2.4.14) give the following asymptotic value for

the transmission constant of the reverse (complementary) problem

of case (i):

iEp. if ex '-a-ib (2.4.15) 1 Pk E 8 ) as E 0.

Comparing (2.4.15) with (2.3.10), we immediately see that the

estimates are the same provided that

= (2.4.16)

To find the relation between the "geometric" constants:

To establish the validity of (2.4.16), we apply Green's

formula to the outer potentials '1)0 and %, of (2.2.6) and (2.4.2)

- 65 -

respectively, in the fluid region except for a small circle of

radius r (Cs---1) centred on B(0,a) and a small semi-circle of

radius r1 (<<l) centred on C(b,0) and then letting r and r1

both

tend to zero remembering that: near B(0,a) the potentials are

such that

3- 0

2cos49

(1)0 + µr 2 cos1e as r 0 (2.4.17)

and in the vicinity of C(b,0) they have the behaviours:

4)0 k r21sin%9 2

1

o r1 2sin

as r1 -9 0.

(2.4.18)

aq) infinity, and since the normal derivatives --2 and --2 on the an an

cylinder both vanish, then

ZM a-N13' rc aCP ji - lif)c) --e)rd9+ ( Jr dr T 0 ar°'r'dg" = 0 0 0 1 1 1

(2.4.19)

Substituting from (2.4.17) and (2.4.18), integrating and then

letting r and r1 tend to zero, we get (2.4.16). This completes

the verification that when the body scatterer and the coordinate

system of figure (2.1.1) are kept fixed and the direction of the

incident wave is reversed, then the leading term in the asymptotic

expansion for the transmission constant remains unchanged :see

appendix (A.4)].

Since qCI and 4)0 both vanish on the free surface and also at

- 66 - Case (ii)

2.5 INTRODUCTION AND SOLUTION

The solution of the inner problem of §1.3 can also be

used in the problem of scattering of short surface waves by a

cylinder with one vertical plane side AB and an arc f(x,y) = 0,

with B(O,a) and C(b,0) as end points, satisfying conditions (i)

and (ii) of §2.1. In place of the horizontal tangency at C(b,0)

it has the property that:

yanf = 0 for n = 1,2,...(N-1), Layn x=b y=0

where N (;=2) is a finite integer giving a measure of the steep-

ness of the arc at C(b,0) on the free surface. When the regular

wave train (1.1.1) is incident normally on AB, the formulation

follows exactly the same lines as those in §2.1 and the

beginning of §2.2. The lower inner problem around B(O,a) is

exactly the same as in §1.3 and §2.2 since the arc is locally

vertical there see figure (2.5.1) and condition (ii) of §2.1].

This means that the reflection constant in this case is still

given by (1.4.15) no matter what shape the arc BC takes. The

geometry of the arc is, of course, expected to make its presence

felt if higher order terms, in the asymptotic expansion for ii,

are to be found.

The outer potential:

Since the problem for the first term in the lower inner

Xi C(b,0)

(P2 4- E (P2 y= 0 a ...1 . 4-1

vi

- 67 -

4,_-tp, x .0

44------- 1.t.

Texp (i ::)/1 aP2 , 0 .... (P-1)exp (-b___

% E i an r e e")

X -s--- B(O,a)

+y

Figure (2.51)

A x.4.

Y (P1 + 6 cpiy=0

- 68 -

expansion around B(0,a) is common to figures (1.1.1), (2.1.1)

and (2.5.1), we then conclude that, in our present case, the

outer approximation is also given by:

-a/E

2 (x1E)-/` e o(x) as E -.".'") 0, '

where q)0(x) is given by the conditions:

o = 0 in the fluid domain,

on = 0 on S,

= 0 on the mean free surface,

o vanishes at large distances from S,

0^1r 2 cos 1e as r --4 0,

where (x,y-a) = r(sinQ, -cos9).

(2.5.1)

(2.5.2)

Again the singularity of L0 (x) is located at B(0,a), since the

existence of such a singularity at A(-0,0) or C(b 0) leads to

an inner problem with a singular solution.

The upper left surface inner region and the transmission constant:

Since the potential (k)(x,y) of (2.5.2) is finite at

C(b,0), its behaviour near that point must be of the form:

Cp‘0,Xr1sine1 as r1 ---"4 0, (2.5.3)

where X is a constant known in principle and (x-b,y) = ri(cos91,

sine). Defining a pair (X1,Y1) of inner coordinates according

- 69

to the formulae: x=b+EXi, y=EY1 with 4)2(b+EXi, EY1;E) =

2(X1,Y1;E) as the upper left surface inner potential, combining

and rescaling (2.5.1) and (2.5.3) and using the matching

principle we, with the help of the free surface condition (2.1.2),

obtain:

, 3/2 - A E e a/E

(X Y -E)— (Rein91 - 1) as E 0, 2 l'

where R12 = X1

2 + Y.

Because the arc is locally vertical at C(b,0), its

equation in the neighbourhood of that point can be put in the 00 n

nY form: P x-b = - nts, (2.5.5)

11=

where

with ON 0.

aNf

f x dyN x=b' (2.5.6)

Rescaling, the boundary condition on the arc becomes:

ao n-1 n-1 a 2 °2 PnE Y1 ax1

+ (ay1 (n - 1); - 0 =iv n-1n

on X1 = - n 1 n. 1

n=

Following Leppington (17) we expand $ 2)( y. and 12 as Taylor 3. 1

series about X1 = 0. Thus

■.g

as R1 -->c0 and (ER1 ) --4 0,

(2.5.4)

(2.5.7)

- 70 -

N-142 N

a22 a°2 13 1‘1 EN1

ax1(°'Yi;E) + NI. filY1 .572 (0'Yl'E)-111 --2°'Yl;

1 dX1

+ terms of order EN = 0 (Y

1 >0) (2.5.8)

Referring to (2.5.4), we assume an expansion of the form:

3/2 -ani:c

°2(X1'Y .E)-\ji‘E 0

20 +6(Egl

21+EN-4

22-

2.5.9) as E --4 0.

The gauge function 0(E) lies asymptotically between unity and

EN-1 and is to be found. There could be several such terms,

and G(E421

is taken to represent a typical term.

Substituting from (2.5.9) into (2.5.8) we get:

as

ax1 x1=o =0

(j = °' 1) Y 1 >0

2 a A r422?

= X1

X1=0 . 1

ax2

20 N-1 ac)

(2.5.11) 1 OY

1 =0 ei

- Y1>0 1 YO

The harmonic functions!2j (j = 0, 1, 2) also satisfy the free

surface condition

0 j + a0, 7T71 = 0 (y = 0, 0) (2.5.12) 2

1

1

With the incident wave train coming from x = -co, any wave-like

part of the inner potential §20

must be outgoing and therefore,

referring to (2.5.4) and (2.5.9), we pose the behaviour

'20 r-sj R1sin91 - 1 + outgoing waves as R1 4°0(2.5.13)

Because of the homogeneous boundary conditions satisfied by 20'

(2.5.10)

and

- 71.-

it is impossible for this potential to contain any outgoing

waves with no incoming waves, Alker (2) proved that the eigen-

solutions of a vertical beach problem are wave-free. Thus

020 = Y1 1. (2.5.14)

The specifications for 21 and 0

22 are completed in principle

by matching with the outer expansion to ascertain their behaviours

at large R1. This will not be carried out here since the outer

solution is not known and our main objective is the wave train

associated with the inner potential 02 of (2.5.9); it is noted

here that the eigenfunction 021 is wave-free Leppington (17),

Alker (2)] and is large for large R1. As for the potential

it is found that:

{°22 = -13N v N-1 (Y > 0) (2.5.15)

oX1

X1=0 (N-1): 3-1 1

and the surface wave train 022w is given by 00

2113N N-1 - *

f Y1 _ ) °22w(xl'Yl) (N-1)! {0 Y1 e dYilexp( Xi Yl

= 20N exp(iXi Yi) (2.5.16)

Formulae (2.5.9), (2.5.16) and matching yield the estimate:

21.X13 N E e

xp(-aib

) as E 0 (2.5.17)

for the transmission constant, where X is determined by (2.5.3).

In addition to the outgoing wave train, 022 contains wave-free

terms which are large at infinity. A detailed treatment of the

function 022 is given in Leppington (17).

- 72' -

2.6 THE COMPLEMENTARY (REVERSE) PROBLEM OF CASE (ii)

If the direction of the incident surface wave train

in figure (2.5.1) is now reversed, with the rest of the config-

uration kept fixed, then the single-term asymptotic estimate for

the new transmission constant will be the same as (2.5.17)

consistent with the fact that the transmission constant remains

unchanged see appendix (A.4)].

The surface inner region:

In the inner region neighbouring C(b,0) we replace the

boundary condition on the scatterer by the one on (X1=0, Y1).0)

as in §2.5: thus we have

N-1 I N-1 0 NE N a2stl N: INY1 -j17(°'Yl'E)-1117(°'Yl;°1

g(0,Y1;E) + 1 1 ax1

+ terms of order EN = 0 (Y1 > 0) , (2.6.1)

where x = b+EX1, y = EY1, q)(b+EXi, EY1;E) = 1(X1,111;E) and

13'(x;E) is the total potential in the reverse problem.

Our starting point is the observation that the incident

wave will be almost totally reflected as E --4 0, since the

obstacle appears almost like a vertical wall,in this limit. Thus

we immediately pose the first inner approximation:

()(1'Y1;E)"^Y(E)60(X1,Y1) as E --4 0, (2.6.2)

where the scale factorY(E) is to be determined.

- 73.-

Since the harmonic function satisfies the free

surface condition o ay1 + = 0 on (Y1 = 0, X1> 0), and the a o

condition 7c.

= 0 on (X1 = 0, Y

1> 0), then we conclude that:

NXI.,Yi;E) 2Y(E)e-Y1cosX1 as 6 0 . (2.6.3)

Formula (2.6.3) and the order of the incident wave furnish the

following value for the scale factor N(E):

'Y (E) = exp( - lb , )E • (2.6.4)

The reflection constant R1 has the asymptotic value:

R1 exp( - 2ib) as E ----> 0. (2.6.5)

Leppington (17) has shown that the asymptotic development for

§(X1,Y1;E) is as follows:

e

[60 +EN"161 + as E --4 0. (2:6.6)

Formulae (2.6.1), (2.6.3) and (2.6.6) give the boundary condition:

a

1 X1

2PN d (y Ne-Y1) (y>0) (2.6.7) ' dYi 1 i

The behaviour of1(Xl'Y1)

at infinity:

Since the harmonic function 1 satisfies the condition

+ = 0 on (Y1 = 0, X1> 0), then the application of Green'saYl 1

theorem to 1 and the Green's function of appendix (A.1) with

E = 1, yields:

213, § =

0

rt2 Re

( > o ;r1 > 0)

CO

-2iexp(i-Ti-Y1)+Vo

expL-t( + iY1 )3dt t +

(2.6.8)

and co 413N e)cpE-t( + j

-TtN: Re (i + t)

4P t expE-t( + rcN Re 0 (1 + t2)(1 + it)N

(yNe -Y1 dY 1 -i-)dY1 dt 0 1

Now we write

if3N 01= -2N -1 exp(g + I1 + I2 (2.6.9)

where co R N I1. TEN dY (YN d -Y1 / : 0 1 _ log )

+ (Y1 -TDI

+ (Y1 +19) 2 1 dY (2.6.10)

0( 1 ) as I + ill I --3c0 • (2.6.11) 1 +11112

To find the behaviour of II as I + iT11---)00 we consider the

integral co I 11.)e-I

Jn = I +G di ' 0 where n is a positive integer and I argal<TE.

(2.6.12)

It is clear that

Jn = n-1 +

(2.6.13)

Therefore

- 75. -

(-1)kok(n-1-k)! (2.6.14)

c° as 101 . (2.6.15)

1 151 (2.6.16) as --*co

Jn = (-1)noPJo +

k=0 But

(-1)j.". a1+j

J 0

Hence

(-1)k(n+k):

al+k J n

k=0

Integrating (2.6.10) by parts and applying (2.6.16) we find that

1 I N as 1+ gl (2.6.17) 1 TL ili2

Referring to (2.6.9) we have

4PN sinel k(X1,Y1;E)", -Ti7rexp(iX1-Y1) + TL

2 R1

as R, co

(2.6.18)

where (X1,Y1) = Ri(cosevsin91).

Formulae (2.6.18) and (2.6.6) lead to the improved

estimate for the reflection constant

if3N N-1 2ib ;_ r\j (1- 1771E

)exp(-) as E ;---) 0, (2.6.19) 2

where the first term is given by (2.6.5).

The transmission constant:

The wave-free term of (2.6.18) together with (2.6.6)

and the matching principle give the following expression for the

outer approximation:

-76-

10(x;EN e 40N,N-ibit o(x,y) as E 0, Tc

where the outer potential satisfies the problem

2^ V o

= 0 in the fluid domain,

(I) = 0 on the mean free surface,

64O = 0 on S of figure (2.5.1),

0 --40 at infinity,

0 r 1 —sine1 as r1 --4 0, 1

where (x-b,y) = r1(cos91,sine1).

on

(2.6.20)

(2.6.21)

In view of the finiteness of 4 (x,y) at B(0,a) Dee fig. (2.5.1)]

and the local property of the arc there, it is readily seen that:

(1) o 2 p1 pr cos-i9 as r 0,

T (2.6.22)

where p1 and p are "geometric" constants known in principle and

.(x,y-a) = r(sin9,-cos9).

Proceeding as at the end of §2.4, we find that:

y 4ippN EN-F1/2e_ ,-a-ib, 1

xpk ) as E--) 0, (2.6.23) Vqi

which is the same estimate as (2.5.17) provided that

X = 2p. (2.6.24)

Relation (2.6.24) is proved by applying Green's theorem to the

outer potentials C00(x,y) of (2.5.2) and q)c, of (2.6.21) in the

fluid region excluding a circle of radius r (<<l) centred on

-77.-

B(O,a) and a quadrant of a circle of radius r1 (<< 1) centred on

C(b,0) and then letting r and r1 tend to zero as in the end of

§2.4.

This completes the verification that when, in figure

(2.5.1), the direction of the incident surface wave train is

reversed keeping the rest of the configuration unchanged, the

leading term in the asymptotic expansion for fremains unchanged

Dee appendix (A.4)J.

2.7 SOLUTION OF AN OUTER PROBLEM

Referring to case (ii) of §2.5 and figure (2.5.1) we

now solve the outer problem (2.5.2) in the special case when the

arc f(x,y) = 0 is as shown in figure (2.7.1). The transformation

= log(ib z) iz - b (2.7.1)

with a cut in the z-plane from z=ib to z=ia, and the Schwarz-

Cristoffel transformation

= k

(w-e)dw (2.7.2) o uNw-1)2(w-d)

with a cut in the -plane along lImc = 11/2, -log(n)1

transform the fluid region onto the upper half of the w-plane

with e, d and k as constants to be determined. From the nature

of the Schwarz-Cristoffel transformation it is found that

e > d >1.

0

C(b,0)

Cp.°

8 (P° =0 ax

A x

x +y2 b2

D (0,b)

- 78 -

B (0,a)

Figure (2.7.1)

k(e - d) '

(d2 - d) 2 e a+b (e u)du

1 (u-d)(u2-u)V

- 79.-

Note that in (2.7.2) the path of Integration lies in

Imw > 0 and does not pass through the pole of the integrand at

w = d.

When z = 0, it is found that = 3111 and w = 1 and 2

hence

_ 31T k ir (e-u)du

7 0 (d-u)(u-u2)2'

which shows that k< 0.

(2.7.3)

When z = is we have C= - log(a + b) and w = e and 2 a- b

hence

0 w1/2(w-0(w-d)'

where the path of integration in (2.7.4) is the real line except

for an arc about w=d in the upper half of the w-plane.

Integrating along the above described path we obtain:

(2.7.5)

(2.7.6)

where the integral now is a Cauchy Principal Value. From

(2.7.3) and (2.7.5) we get:

K = -1/2, (2.7.7)

e - d = 2(d2 - d)2. (2.7.8)

e DT , l - og1/4a

-1-b\) b k ir (w-e)dw (2.7.4) 2

- 80.-

The constants e and d are given by equations (2.7.6) and

(2.7.8).

The solution to the outer problem (2.5.2) in the

special case of figure (2.7.1) is then given by:

LP(x,y) =.YR4w k(w.-1)2 , (2.7.9) o w

where w(z), z = x+iy, is given by (2.7.1) and (2.7.2) and Y is

a real constant to be found. Now

(15o (e2 - e)2Re(W- e) as lw - el '-'••••4 0 .

(2.7.10)

But

8b(e -d)(e2 - a2-b2

w- e

- is

as lz - ial 0. (2.7.11)

The behaviour of k./. o near z = is combined with (2.7.10) and

(2.7.11) yields:

8b(e-d)

(a2 b2 )(e2 0 =2- •

To find the behaviour of the potential near the point z =

we proceed as follows. The transformation C(w) and C= C(z)

give:

2.(iwk) as I WI 0, (2.7.13)

and

•-•-1-13--(z-b) as z - 131 0. (2.7.14)

Near w = 0 the potential has the behaviour:

(2.7.12)

-81 -

"11 f31:::• , - —Re(iw2) as I w I ---4 0, e

and therefore

(2.7.15)

)( yd qi0 d 2Reli(z-b)1 = --7risinel as r1 --4-0, (2.7.16) be be

where z-b = r1ei91 .

Comparing (2.5.3) and (2.7.16) we see that:

)( 1 -1-

d d 8(e - d) 2

A = 2 2 1 2 2 2 • . be e b(a - b )(e - e) 2

(2.7.17)

Estimate (2.5.17) is now completely determined with N = 2 and

1 P2 = it.

CHAPTER 3 - 82 -

TRANSMISSION OF SHORT SURFACE WAVES UNDER A PARTIALLY IMMERSED CYLINDER OF RECTANGULAR

CROSS-SECTION

3.1 FORMULATION OF THE PROBLEM

An infinitely long rectangular cylinder, of breadth 2b,

is immersed to a vertical deptha in a fluid of infinite extent. A

regular wave train of potential Rel(yx;E)exp(-2.00/, )1.(x;E) is

given by (1.1.1) with a and b both large compared to the wave-

length 2TCE and CO is the angular frequency, is incident from

x = -co with the wave crests parallel to the generators of the

cylinder and the problem is a two-dimensional one. Our

objective is to estimate the reflection and transmission

constants, in the short-wave asymptotic limit E --4 0, by

employing the method developed in Chapter 1 to deal with the

problem of scattering of the same wave train by a finite vertical

plane barrier.

A Cartesian coordinate system is chosen as shown in

figure (3.1.1) with the generators of the cylinder parallel to

the z-direction and y is pointing into the fluid.

The formulation of the problem follows the same lines

as in Chapter 1. Thus

q)(x;E) = f- 2e-Y/Ecos(x/E) + qyx,y;E) (x<O)

Li) 2 (x>0) (x y.E)

(3.1.1)

A x-ae D (2b,0)

(A +E tply =0 (1)2 +6 (p2y =o Y

..0.4----- 'I exp(1):-Y) E

`p2.=0,,,. (F1-1)exp(-__Ix-y)

i C(2b,a) X

- 83 -

el 3. 1 -- i ,n f

2y=0 ...- r 1 sr ,, I r 1

I tY

Figure (all)

- 84. -

In addition to the free surface condition T. +E-1, = 0 and the ay

condition of zero normal velocity on the cylinder .-.R;1 = 0 on S,

the harmonic function T.J

(x;E) satisfies a condition limiting the

edge singularities. Thus

r --1 0 as r --4 0, ar

(3.1.2)

where r is the distance from any sharp edge of S. The radiation

condition ensures that the potential !pj is purely an outgoing

wave at large distances from S. Thus

1 -- ("I - 1)exp( 1-1.1) as x --4 -co

q)2 T exp(41;I) as x --4 +co 3.1.3)

The edge and radiation conditions ensure that the potential

T.

J

(x;E) is bounded everywhere in its domain of definition and is

unique.

As in Chapter 1, the continuity of the total potential

T(x;E) and 22 across (x=0, y2> a) gives: ax

2e YA: + TI(-0,y; E) = 4)2(+0,y;E)

aql aLp2 ax (-0,y;E) = ox (+0,y;E)

3.2 THE OUTER APPROXIMATION

Reasoning in exactly the same way as in Chapter 1, we

see that the outer estimates are given by:

Ti(x;E) k0(.(E)450(x) as E--4 0 (j = 1,2) (3.2.1)

(y>a)(3.1.4)

- 85. -

where k is a constant independent of E, c(E) is a scale factor,

and they are to be found from the consideration of the inner

problem in the vicinity of B(0,a).

The outer potential %(x,y) is the solution of the

following problem:

00 = 0 in the fluid domain

'160 = 0 on the mean free surface, bq

= 0 on the cylinder S, an ,

O -- 0 as kx

2 + y )

0 r :3coA9 as r -"*-4 0,

(3.2.2)

where (x,y-a) = r(cos9,sin9) and — an indicates differentiation

in the direction of the outward normal to S.

Formula (3.2.1) and the first of conditions (3.1.4)

show that the outer potential L?)0(x,y) is apparently discontinuous

across the plane (x=0, y>a). Because it transpires that

a(E) = E 3exp(- i!), this apparent discontinuity is exponentially

small in the outer region (y-a)»E and kk)(x,y) is indeed

continuous everywhere.

There is found to be no singularity in Lk.(x,y) at

points other than B(0,a); for any other singularity would lead

to a singular inner problem whose solution does not satisfy the

edge condition at a point of the inner domain under consider-

-3- ation. Again it is found that the r 3 behaviour indicated in

- 86- -

(3.2.2) is the only one which matches with the inner solution

in the vicinity of B(0,a). This is consistent with the

"principle of minimum singularity" of Van Dyke (33).

3.3 SOLUTION OF THE INNER PROBLEM

Formula (3.2.1) and the behaviour of x,y) near B(0,a)

lead to the expression:

LP.(x;E)--/ ka.(E)r%os%-9 as E 0, as r --4 0, (3.3.1) J —

where (x,y-a) = r(cosel,sine) and the limits are taken such that

r/ E --) co.

Defining inner coordinates (X,Y) according to the

equations

= EX

y = a +EY (3.3.2)

with ip.(EX,a+EY;E)=0.(X,Y;E) and using the matching principle

[Van Dyke (33), Leppington (16,17,18),we are led to the

following form for the inner approximation near B(0,a):

J(X,Y; E)-- E- CO:2)4J. (X,Y) as E --a0, (3.3.3)

where the harmonic function U. 0 (1.7=1,2) has the behaviour 3

jS-cosg3-9 as R co jo

where (X,Y) = R(cosQ,sinQ).

(3.3.4)

The boundary condition on the cylinder S gives:

420 ay (x,+0) = 0 (x >0), }

ato( 0 Y) = 0 (3.3.5)

(y < 0), ax '

Rescaling the continuity conditions (3.1.4) and substituting

from (3.3.3), we obtain:

0.(E) = E 3exp( (3.3.6)

EvidentlyT.Jo satisfy the boundary conditions:

2e-11 +II,10(-0,y) . TI20(+0,Y) 4 (Y> 0

410( o,Y) = 20(+0,Y) = V(Y) ) ax ax

(3.3.7)

The unknown velocity distribution V(Y) has the property that it

does not vanish except possibly at infinity. Formula (3.3.4)

and the definition of V(Y) give:

k -5/3 V(Y) — Y as Y co .

Near Y = 0, V(Y) has the behaviour

V(Y)-- X0Y- as Y --4 0,

where Xo is a constant.

(3.3.8)

(3.3.9)

The behaviour (3.3.9) can either be obtained from the

behaviour. --constant + (21.o/5)R3cosgse as R --4 0, or by an Jo

argument similar to that in §1.3.

- 88.-

The integral equation for V(Y):

Because of the conditions satisfied by the harmonic

function U20, the application of Green's formula to it and the

Green's function

G2 " (X Y.X Y1 ) = \ 2}

41- log f(X-Xii2,

1 (X-X1)2+

(Y+Y1 )2/ [(X+X1 )2 +(Y-Y1 )2

) f(x+x 1)2÷

(3.3.10)

yields

20(Xl'Y 20 (0 Y-X Y )4 ----dY 2 " 1 ax (3.3.11)

In particular: co

(+0 y = — v(Y) {log(Y+Y1) + log 1 Y-Y11 dY(3.3.12) 20 1 TC 0

In a similar manner, the application of Green's formula to the

Green's function

r Gi(X,Y; Xj.,Y1) = ziT . log (X-Xi.)2 +(Y-Yi)2 lt(X+Xi)

2 +(Y-Yi)2d

(3.3.13)

and the harmonic function '00 gives: ives. -1

15 (-0,1) =

V Y log I Y-Y dY. 10 I0

1 (3.3.14)

Formulae (3.3.12), (3.3.14) and the continuity conditions (3.3.7)

lead to cc

2e-Y1 =TT j

- V(Y)tlog (Y+Y1 g ) + 210 1Y- Y11/ dY, (3.3.15)

0

- 89. -

as the singular integral equation to determine the unknown

function V(Y). We are especially interested in the determination

of the constant k specified by formula (3.3.8).

Solution of the integral equation:

The formal differentiation of both sides of (3.3.15)

yields the new integral equation co

-2rce-Y1 = V(Y)dY + 2 fv(Y)dy 0 Y1+Y

0 1 Y Y (3.3.16)

where the last integral is a Cauchy Principal Value one. To

solve the integral equation (3. 3 . 16) , we take the Mellin transform

of both sides [Sneddon (26), Erdelyi (7)] and obtain:

f7O oo s-1

-27r(s)=V(s)f t dt - 2 r---dtl, (3.3.17) i+t 01 - - t

dt

.11(s) = fV(Y)Ys-idY for 3 <Res < 5/3. (3.3.18)

0

The restriction 3 <Res < 5/3 is made because of the behaviour

of V(Y) for large and small Y [cf. formulae (3.3.8 and (3.3.9)].

Because the integrals in (3.3.17) are defined for 0<Res<1 and

r(s) exists for Res >0, then

F(s)sinTts V(s) a 2-• 1- <Res < 1. (coslis - ) (3.3.19)

Writing V(s) as the product of two functions f(s) and "g(s) where

f(s) = s2 r(s)

sinTCs

<Res < 1), (3.3.20) g(s)

s2(cosits -

- 90 -

we immediately see that

f(u) = (u2 u) e

-u (3.3.21)

and by using the inversion formula for the Mellin transform we

obtain:

- s 1 I u sinus g(u) = ds,

CI 2i1T 2

(cosTts - 1/2) - ico

s

where .1/2 <Res =Ci < 1.

(3.3.22)

Integrating (3.3.22) with the help of the residue

theorem we arrive at:

co 1 1 g(u) =

TL .. (2n+1/2)2u2n+-15 (u._>-.1) 5)2u2n-1/2 +

n=1 (3.3.23)

2n- u2n+1/2 g(u) = 2TC-.-

f 1/2 2+ (0<u<1)(3.3.24)

2 11=0 (2n-1/2) (2n+1/2)2

Since 1 the continuity of g(u) at u = 1 is (2n+1)2 8 n=

ensured.

From the knowledge of f(u) and g(u) and the convolution

theorem for the Mellin transform Dneddon (26)] we arrive at the

formal solution: co r

v(Y) = Jf(I)g(u)du. 0 u u

Substituting from (3.3.21) and making the change of variable

u = T1, we obtain:

co

V(Y) = Y J (YT - 1)e-Ilig(1)dT. (3.3.26)

(3.3.25)

0

- 91. -

By Watson's Lemma ECopson (5)], the leading contribution to

V(Y) for large Y comes from the neighbourhood of T = 0. Using

(3.3.23) with u = it it is found that:

R5/3) V(Y)--, as Y --4 , EY5/3

which is of the predicted form (3.3.8), and the constant k of

that formula has the value:

k = FX5/3). (3.3.28)

The general form of the velocity distribution V(Y):

By differentiating formulae (3.3.23) and (3.3.24) we

get:

and

1/u %.4 4/31/u

ug'(u) (t--I- t 2)dt = -1 1

(1 J (1)(t)dt T, 0 _ t - 0

for u >1, (3.3.29)

u 4/3 3 % 1 Jr (t A- t )dt ug e (u) +Tt(u - u 3) = - 7. ;

16 0 1 - t2

1 =— 1 log(1-u) +--ft fq)(t)dt for 0<u<1,(3.3.30) 0 where

+ tl + t — 3- (1 + t)(1 + t 3 + t3)

' (3.3.31)

Using (3.3.21), (3.3.25) and following the same steps as those

outlined in appendix (A.2), then utilising (3.3.29)-(3.3.31), we

obtain the solution:

(3.3.27)

3- 4/3 -VT V(Y) = 1 (t 3 + T )e

160 1 - T (3.3.32) 2 dl,

- 92- -

co

where we have utilised the fact that

- fug' (0} u ÷u 35 du - 7.11(u2 - 1)

Now consider the quantity:

(u 0,1). (3.3.33)

v(Y) Ea/ - Y 3 5 3 jr-3- . e-- d +1 dt (t3+t4/3 )e-t

rc iTy 3 0 0 1 - t2 co

= 1 (ji + 0 1 - t2

Putting Y = 0 we obtain:

V(Y) La- = At (t3±e 31,dt = 0.

3- -3-

II Y-6 Y=0 1- t2

(3.3.34)

(3.3.35)

1 This can be seen by making the substitution t = 1- in the

integral of (3.3.35) and hence

-FO)Y-3 TL as Y 0,

which is of the assumed form (3.3.9) with the constant

x _ -F(3) O IC

(3.3.36)

The reflection constant:

Referring to figure (3.1.1), we apply Green's formula

to the Green's function G(x;;E) of appendix (A.1) and the

harmonic function k-P1 (x;E) of (3.1.1) to get:

aq; 1 (;r1;E) = - jr c(o,y;',11;E)- o=x"dy ( <0). (3.3.37)

a

CO

- 93.-

Taking the wave part of G(0,y;,11 ;E) leads to:

cc

X;E)= 2i f eY/E y_lexpCiE 11 ) (< 0) a (3.3.38)

where Lplw

is the wave part of P1.

Using (3.1.3) and remembering that, in the limit E --4 0,

most of the contribution to the integral in (3.3.38) comes from

the inner region in the neighbourhood of B(O,a), we arrive at: co

2a 1+12i i'V(Y)e-YdY/ exp(-- 2Ea.). 1 .. 4i exp( ) 0 3/5

as E --4 0, (3.3.39)

where V(Y) is given by (3.3.32) and the integrL is evaluated

in appendix (A.6).

Again, note that the above estimate for R is

independent of the solution of the outer problem and hence does

not depend on the other dimension of the cylinder. The length

scale b would, of course, be expected to enter at higher order

approximations.

3.4 ESTIMATION OF THE TRANSMISSION CONSTANT

Formulae (3.2.1), (3.3.6) and (3.3.28) give:

j3". 5 -aA^ (x;E)-- F(-5)E 33- e = Lpo(x,y) as E 0, (3.4.1)

where (00 is given by (3.2.2).

-94-

Since q (x,y) is finite at C(2b,a) its behaviour near o

that point must be of the form:

C IP0-,,p1-2.ricosC3-(914)} as r1 --4 0, (3.4.2)

where (x-2b, y-a) = ri(cosel,sin91) and p and g are constants

determined from the solution of the outer problem (3.2.2), and

can be considered as known in principle.

From (3.4.1) and (3.4.2) it is found that:

P2 ar-1(5/3)E g- 3e-a/E

( ox }x=21;j 3- as E --4 0, ft (y-,-a)3 as (y-a) --,4 +O. (3.4.3)

As before, referring to figure (3.1.1) and applying Green's

formula to the function P2(x;E) of (3.1.1) and the Green's

function of appendix (A.1), we get the following expression for

,, aq)., Lp2wQ,-(1;E) = -2i{ fe-YR-(-L') dv}exiia -

Dx x=2b ' --x- 2b) -ell

a E

for > 2b. (3.4.4)

Using (3.4.3) and (3.1.3), formula (3.4.4) gives

- 16 iTIR E4/3expe-2a-2i b) asE-4 0. 9tF(

2 E (3.4.5

It will become clear in the next chapter that if the geometry of

figure (3.1.1) is replaced by the more general configuration of

two vertical plane sides AB and DC having lengths al and a2

respectively, and an arc BC, satisfying condition (i) of §2.1,

the transmitted wave LP • 2w. co

- 95 -

which is locally horizontal at both ends B(0,11) and C(b,a2),

then (3.3.39) will remain valid with a replaced by al, and the

estimate (3.4.5) will read

- 16 6 xpk 4/3e , -a -a -ib) as E 0, (3.4.6) 9 / n(31)12

E

where a is the "geometric" constant found from the solution of

the new outer problem.

3.5 SOLUTION OF THE OUTER PROBLEM (3.2.2)

Using the method of conformal transformations, problem

(3.2.2) is found to have the solution: 2 2 32-

k Y qi0(x,y) =1(R4 1 + (3.5.1)

where)/ and k2 (>1) are real constants and is a function of

z = x+iy given by the implicit relation:

2 OQ z=b+ia+k1 Q _ k2

22cI)1/2

; arg TE),(3.5.2) d ((Q

where k1 is a constant.

From the nature of the transformation the constants k1

and k2 are given by:

1

b = k 1 2 2 % 0 (k2 - )2

/2 62 a = k1j 2 2 %

1 (k2 )2

from which we see that k1 2> O.

(3.5.3)

8k1

The point z = 2b + ia corresponds to the point = 1 in the

+ f9(k22 -

2 (z- ia)31 as z ""'") ia. (3.5.4)

-96-

Using the fact that z = ia corresponds to = -1

(3.5.2) gives:

-plane and hence:

19(k22-1T r

1 , 2 <i(z - 2b- ia) as z 2b+ia.

Formulae (3.5.1), (3.5.4) and the behaviour of kPo near B(0,a)

furnish the value:

9 3

(3.5.6)

•

Similarly (3.5.1), (3.5.5) and (3.5.6) give the following

behaviour near z = 2b + ia:

(1 + k22)

% ,., 1 3/77A 3 4/3 ,-71.(7-1(2 -1) , , 2 1 16(k2 - 1)3

as z 2b + ia

i(z - 2b -

(3.5.7)

Comparing (3.5.7) and (3.4.2) we immediately see that:

3 4/3 (k2

2 4. 1) - (---) (3.5.8)

2 2 1 16(k -1)3

Making the substitutions = sin Land = k2[1- (1 ---17)sin2tr

k2 in the first and second of equations (3.5.3) respectively, and

eliminating k1 we arrive at:

a {EGO - 2K(I -1-)i =jE(fa') - µ2K(., )}, (3.5.9)

8k1 (3.5.5)

- 97 -

where 10.1) and 01) are the elliptic integrals of the first 1 and second kinds respectively, p.= IT <1 is the modulus and 2 2 % = (1 - p. ) 2 is the complementary modulus [Erdelyi (8)].

The general solution to equation (3.5.9) is difficult,

if not impossible, to find because of the transcendental nature

of the equation. However, in the case a= b, it possesses the

solution p.= p! 1 1 = = In this case if 2

ki = a -1(.3i-)-12, (3.5.10) VOT

and from (3.5.8) we have

2 = - 4/3

37 1/7 ? f 3

16

a n()4)] 2 •

With this value of a, (3.4.5) reads:

4 ilT3 (3E)4/3 e - 2a(1+ •[ ti 3 (

. 2 1 -V3/3 2a xp

as E --4 0 (,<.=<].). (3.5.12)

The examination of equation (3.5.9) reveals that two limiting

cases can be considered:

(1)

The case b/a<1:-

Inspecting the equation we immediately see that the

smaller the ratio (b/a), the smaller the modulus . Hence

(3.5.11)

—) 0 as a— . (3.5.13)

- 98 -

From the power series expansions for E(11) and K(p) we obtain

1 bb P- = 2( ---)2 as --4 0, k2 II a a (3.5.14)

since µ2K(µ') --) 0 as p, 0 and E(µ' ) 1 as P- 0.

The constant k1 of (3.5.3) has the asymptotic value:

kl 2010 ) as la --4 0.

In this case (3.5.8) yields:

9. 16(41)4/3 a154/3 as 1.2 --4 0. (3.5.16)

The asymptotic expression (3.4.5) now reads:

iTI2 ATJJ,E.4/3 ,-2a- k 9 ) kb) expk 2ib, ) as E 0, 4(r(1)1

as a — 0, such that .- <<1. (3.5.17)

ii) The case a/b<-:‹1:

In a similar manner to case (0 above, we see that

equation (3.5.9) and the power series expansions for E(W) and

K(1.1!) give

4a a ,2 = 1 - 2 Tlb as b— -- 0,

since p! 0 as 0.

(3.5.18)

In this case the asymptotic values of k1 and k2 are

(3.5.15)

given by:

-99 -

4a Ttb

a as 1-; 0. (3.5.19)

The constant a of (3.5.8) is found to have the asymptotic value:

-a 3TEN4/3 a as 0. N2 TT b(4a ) (3.5.20)

Using (3.4.5) we obtain the asymptotic value:

jail (270(EN4/3exp(-2172i b) as E 0, btr(1)}2‘ 9 / \al E

as a — --4 0. (3.5.21)

The limits are taken such that.<(1.

In the following chapter we generalise the results

obtained for the rectangular cylinder by considering the problem

of scattering of the same regular wave train by shapes having, at

least, one vertical plane side, of length al, intersecting a

curved bottom at right angles. First, we show that, when the

wave train is incident normally on this plane side, the problem

for the first term in the inner expansion in the proximity of

the right-angle corner is the same as that considered in §3.3,

and hence the estimate (3.3.39) for the reflection constant

holds in the general case with a replaced by al. The transmission

constant, on the other hand, will be estimated in terms of a

"geometric" constant depending on the solution of the outer

problem.

- 100 - CHAPTER 4

SCATTERING BY A CYLINDER WITH A VERTICAL PLANE SIDE ANDA CURVED BOTTOM INTERSECTING

AT RIGHT ANGLES

4.1

INTRODUCTION

To generalise the results obtained in Chapter 3, we

consider the particular problem of scattering of a regular

surface wave train, of the type (1.1.1), by a cylinder with two

vertical plane sides AB and DC, having lengths al and 22

respectively, and an arc BC having an equation f(x,y) =0 and

satisfying condition (i) of §2.1 together with:

(i) (___) a f ax n B(O,a1)

n = 0 for n = 1, 2, ..., M-1 co).

This expresses the property that the arc is locally horizontal

at B(O,a1).

(ii) The arc is locally vertical at C(b,a2), in the manner

shown in Y2 figure (4.1.1), and (1)C(b,a)= 0 for n = 1, 2, ..., a n

37. N-1 N<co).

The cylinder is held fixed in the fluid and is

irradiated by the wave train which is incident from x = - co

normally on AB, i.e. with the wave crests parallel to the

generators of the cylinder. The surface wave train is partially

reflected and partially transmitted and our objective, as in the

Lp +cipy=o

exp ( xe-Y) yr

oe. x =o

B(0,al) exp E I

- 101 -

D(b,O)

X1 <- — --1C(ba 81/4P = 0 .1,1

4,

Figure (4.1.1)

tp+E (Ty

• -44----11 i-exp fIX-y )

- 102 -

preceding chapters, is to find the values of the reflection and

transmission constants in the short-wave asymptotic limit E --4 0.

As before, the total potential is R4Y(x,y; E)exp(- iWt)1 ,

with kP(x;E) satisfying the, by now, familiar mixed boundary-

value problem, and W is the angular frequency of the travelling

waves.

Proceeding as before, we suppress the time factor

and write:

x y ; E

x 2e-y/E cosT + yx,y;E) (x< 0),

(4.1.1)

2(x r "E) (x > 0)

with the harmonic functions LG1 and!2 satisfying the following

radiation conditions:

Th -

LPI(x,y;E)—(R-1)exp( i x-y

) as x -co 7 (4.1.2)

LP2(x)37;E)-,Texp(LiEY) as x --4 +co 7

ti

where R and T are the reflection and transmission constants

respectively.

The continuities of the potentialY(x,y;E) and its

ati) derivative TT, across the plane (x=0, y> al) provide the

conditions:

-F-P (-0 37-E) = (+0 37-E) 1 2 " aq) (y >al) (4.1.3) 1

02

-103 -

The outer potential:

An argument similar to that in Chapter 1 can be used

to justify that the outer approximations are given by:

Lpj(x; - ki3(E)430(x) as E --> 0 (j= 1,2) (4.1.4)

with the constant k and the scale factor p(E) to be determined.

The outer potential 0o (x)is the solution of the problem given

by the following specifications:

V245 = 0 in the domain occupied by the fluid,

LPo = 0 0

an = 0

%(x0r)

q30(x9Y)

n the mean free surface y = 0,

on the cylinder,

--4 0 at large distances from the cylinder,

--r 3 cos 39 as r -9 0,

(4.1.5)

where ( x, y-a1) = r( cose , sine ) and — an denotes differentiation

along the outward normal to the cylinder.

4.2 THE LOWER-RIGHT INNER PROBLEM

Defining inner coordinates (X,Y) according to the

formulae

x = EX (4.2.1)

y = al +EY

withy.3(EX,a1-1EY;E)=C).

.3(X,Y;E), then using the same argument

as in §3.3, it is found that the potentials

that:

i(X;E) are such J

fy bxM B(0,a1). where

n. 4.2.8)

-104-

45.(X;E)— 13(V (X) as E o (j= 1,2) (4.2.2) 3 E 3 30 —

with

and jo

?5 kR -cos?je as R ∎4 °O

io ax (-000 = 0 (Y< o).

As in §3.3, the rescaling of (4.1.3) and the use of (4.2.2)

furnish the following value for the scale factor 13(E) of (4.1.4):

0(6) = E%xP( (4.2.5)

The harmonic fanctdcms30(x,y) = 1,2) then satisfy the

continuity conditions:

11510(-°Mai 4120(+°'Y) L10 20 (Y > 0) (4.2.6)

ax (-°,Y) = ax (4)'Y) v(Y)

Condition (i) of §4.1 enables us to write the equation of the

arc in the neighbourhood of B(0,a1) as:

(4.2.7) an

Xn (M < co) , n. y - a1

n=M

This means that the boundary condition of zero normal velocity

on the arc can be expressed locally as:

n-1 a x

(n-1): 0 on y-a1 =

n=M n=

Rescaling as above we arrive at:

an(EX)11-1 (n 1): = 0 on Y-

(4.2.9)

dir5 42

dY ax n=M

-105-

Note that in the case M =co, the boundary condition is simply

ay 2(X,+0;E) = 0 (x> 0).

Assuming that Cf2(X,Y;E) when the arc touches the

X-axis from above, T2 analytically continued across the arc] is

analytic on the line (Y = 7-0, X> 0), so that 2X and 17.5

2Y can be

expanded as Taylor series with respect to Y and substituting

from (4.2.2) we obtain:

1120 = 0 (X > 0). (4.2.10) aY

This completes the specifications for the lower right inner

problem and shows that it is exactly the same as that considered

in §3.3, and hence the value of the constant k of (4.1.4) and

(4.2.3) is given by (3.3.28). The velocity distribution V(Y)

of (4.2.6) is, therefore, given by (3.3.32).

Since (3.3.39) was obtained by utilising the inner

solution, then we conclude that the same estimate holds true,

with a = a1, for our present problem no matter what geometry exists

to the left of the plane side AB as long as the arc is locally

horizontal at B(0,a1).

as r1 -4 0 and the transmission constant, in this case, has the

asymptotic value:

-isp/7 7/6 -al-a2-ib T F(5/3)E exp( E` ) as E --4 0. ITE (4.3.3)

- 106 -

4.3 THE LOWER-LEFT INNER PROBLEM

Putting the values of k and 13(E) in the expression

(4.1.4), for the outer approximation, we obtain:

kp2(?c;E) ,_, r(5,3)E3-3-e-aiALp0(s) as E --4 0,(4.3.1)

where the potential %(x,y) is given by (4.1.5).

Because of condition (ii) of §4.1 and the finiteness

of LUx,y) at C(b,a2) we have

coo x xlr1sinel as r1 0, (4.3.2)

where (x-b, y-a2) = ri(cosel,singi) and the constants Xand X1

are known in principle :see (4.1.5)].

At this stage we remark that if the arc BC is locally

.e; horizontal at C(b,a2), then p + s ri3cos3(01+2) as r1 --4 0

and the estimate (3.4.6) predicted at the end of §3.4 can easily

be obtained. Moreover, if the configuration is such that a

knife edge is created at C(b,a2), then +.air(2cos2(e1 +7)

To arrive at an estimate for the transmission constant in our

present problem we start by defining inner coordinates (X1,Y1)

as follows:

- 107 -

x = b + EX

1 y = a2 + EY1, (4.3.4)

with kyb+EX1, a2+EY1;E) = k(X1,Y1;E) as the lower-left inner

potential.

Formulae (4.3.1) and (4.3.2) furnish the following

expression for the behaviour of k.P2(x;E) for small E and ri:

a1 k.1/5 -a1/E X/7 5 3- kP2(x,y; g)—-TI-R7)E3exp(---E-)+----R)E e (rising].)

as E 0, as r1 --4 0 such that --4 0. (4.3.5) ri

Resealing by using (4.3.4) and using the matching principle

ELeppington (16,17,18)] we arrive at: 5

XI5 5 % al X1/5 5 5 -a /E k (X/ ,Yi; E)— xp(- -T)+ - -R-5)E e (Risinel)

as E --40, as R1 --4°0 such that (ER1) -4 0,(4.3.6)

where (X1,Y1) = Ri(cosAi, sinQ1).

Because of condition (ii) of §4.1, the equation of

the arc, in the neighbourhood of C(b,a2) can be expressed as:

x-b = - -Tn (37 - a2)n

. (4.3.7)

where n=N

f1 aNf RN fx ayNLtb,a

1 2

(4.3.8)

As before, the boundary condition on the arc is then locally

given by:

-108 -

aq) aq)2 131-1(Y-a2)n-1 co ,

Pn 2+ ( =0 on x - b=-> (y-a )1/. ax ay (n - 1): n—T 2

n=N n=N (4.3.9)

Rescaling as above and expanding ky and as Taylor series about 1

X1 = 0 ELeppington (17)] we obtain the new boundary condition:

Now, formula (4.3.6) suggests that we pose the asymptotic

development:

5 3' al qi(24; E) X15 r1(7)E 3exP( E )

i, 1 \-5/- e Y10+a(E)H(xy)

+ EN-1 11+ ***-

as E --4 0, (4.3.11)

where 010 --/ R1sinO1 as R1 ---4°O and 5(E) is a typical coeff-

icient (there could be more than one) lying asymptotically

between unity and EN-1.

Substitution from (4.3.11) in (4.3.10) yields:

a 10(o,v ) = 0 (y >0), aX1

-1 1 (4.3.12)

aX 1 N. 1 1 (0,Y ) = Y RNI N

ax2 ' 1 (0 Y ) -NY 1 W(0'111) N-1 lo

(4.3.14)

Because ako -57(0,Y1) = 0 (Y1 <( 0) then, from the above, we see

1 that:

N-1- E N

-INY1--111Y (°I Yl ; " k X1(°9 Yl ; E) N

1 Y51X X(°'111;E)1 1 1 + terms of order E N = 0 (Y1 > 0). (4.3.10)

ax (0,Y1) = 0 (Y1 > 0) , (4.3.13) 1

and

- 109 -

=

(4.3.15)

Formulae (4.3.14) and (4.3.15) give:

411(o ) N Y1N-1 (Y

1> 0). (4.3.16)

ax

1 (N-1):

We also have:

all)11(°`/ ) = ax '1

and

ax (0'1/1)

To complete the specifications for H(X1,Y1) and k1(X1,Y1), a

knowledge of the outer solutions (terms in the outer expansion)

is essential to fix the behaviours of these functions at large

R1. Since our main aim is to find an estimate for the trans-

mission constant, and since all the terms in (4.3.11) are wave-

free on account of the boundary conditions, then an alternative

to solving the inner problems must be sought. This is provided

by the application of Green's formula to the Green's function

G(x;:E) of appendix (A.1) and the function kyx;E) of (4.1.1).

The transmission constant:

In order to find an asymptotic value for the transmission

constant, we apply Green's formula to the two functions mentioned

above to obtain: co

2w b J --r (E ; = -2if e-316(a) df eyn ax x= E a2

(4.3.19)

0 (Y1 <0),

0 (yi < o).

(4.3.17)

(4.3.18)

- 110 -

where)2w

is the transmitted-wave part of (f2.

Using (4.1.2) and (4.3.19) we obtain: co oT

2 = -2i e y (-7c-)x.bdy exp ( - iEE), a

(4.3.20)

2 Rescaling as in (4.3.4) and remembering that most of the

contribution to T comes from the inner region around C(b,a2),

then using (4.3.11)-(4.3.13) and (4.3.16) we obtain:

y -a -a

2 -ib

__Zi TC `: R ./5 FY

3/AEN-1-%exp( 1 ) as E __4 O.

(4.3.21)

4.4 THE REVERSE PROBLEM

If, in figure (4.1.1) the configuration is kept fixed

except that the direction of the incident wave train is now

reversed, then we predict that the leading term in the asymptotic

expansion, as E --4 0, of the transmission constant, in this

new scattering problem, will be the same as (4.3.21) Dee

appendix (A.4)]. This we now proceed to verify.

The lower-left inner problem:

The lower-left inner region is defined as in (4.3.4)

with LP(b+EXi, a2+EY1;E)= §(X/,Yi;E), where LP(x;E) is now the

total potential.

As in §4.3, formula (4.3.10), we have:

a a (0,Y •E)+PnNE

N-1 a N ,-, .1 + terms

N: ay1 1 al/. Xi 1

of order EN = 0 (Y1 > 0).

Next, we pose an expansion of the form:

N-1A Nxi;E) "-NJ Y(E){ +E Ti+•••1 as E 0,(4.4.2)

where Y(E) is a scale factor to be found.

Note that the exclusion of terms with coefficients

lying asymptotically between unity and EN-1, in (4.4.2), is now

necessary to avoid the location of the singularity of the outer

potential at a point other than C(b,a2) which would, otherwise,

lead to singular inner problems.

Substitution from (4.4.2) in (4.4.1) yields:

(o Y1 ) = 0 (Y

1 0), (4.4.3)

X 'i

al(O,Y1) =-.°4 erlY1N ;:2(0,Y1)} 1 1 1

(Y1 > 0).(4.4.4)

The harmonic functions 00 and 01 also satisfy:

--1(o,y ) = 0 (Y < 0, j = 0,1). ax 1 (4.4.5)

Since the wave train is incident from x = +co, then the existence

of standing waves in (x>b, y;?0) is admissible and

-Y1 o = 2e cosX1

. (4.4.6)

Since the incoming wave is of order unity, then (4.4.6) together

with the first term in (4.4.2) yield:

(4.4.1)

and

Y(E) = exp( -a2 - ib

E )

- 112 -

(4.4.7)

and the reflection constant l'

then, has the single-term

estimate:

---'exp(- 2r) as E --4 0. (4.4.8)

Formulae (4.4.6) and (4.4.4) give:

43N a N -Y1 ax (°'Y l) V-(Y1 e ) (Y1 >0). 1 1

Considering the problem for 01(X1'Y1)' we now proceed to find

the behaviour of this function at large R1 = (X12 +Y12 ), where (X1,Y1) are given by (4.3.4). Writing the resulting behaviour

in outer variables (x,y) and using the matching principle will

enable us to find the behaviour of the outer solution when

approaching the point C(b,a2). In fact, as we shall see, this

point turns out to be the only singularity of the outer solution.

The outer approximation:

111(X1'Y1) of (4.4.2) and the Green's function

Applying Green's formula to the harmonic function

G(i)=4-:-Ttlog{(X3.-X1)+(Yi-Yi) H(XileXi + * 2 * 2 * 2

Lyic Y1-4)2i]

(4.4.10)

and using (4.4.5) and (4.4.9) we arrive at the solution:

(X") (YNe 1 roN:

l)llogrX*2+ - N -Y

(4> 0). (4.4.11)

(4.4.9)

ay 0 1

2 (Y -Y") 1 1 1 1 )

Integration by parts yields:

— 113 — co -Y

s;N

i (X )= -- 21; (Y1-Yi)Yie

1 -I TUN! dY 2+(Y1- c51 1. 0 f(ii

Utilising (2.6.12) and (2.6.16) we arrive at the algebraic

behaviour:

(4.4.12)

) 2ON sine].

1X1 --JTc R1

* * * * where (X

1 ,Y1 ) = R1(cosel'sine1).

(4.4.13) as R1

Substitution in (4.4.2) and the use of (4.4.6) and

(4.4.7) give:

-Y 1 2pN N-1 sine1 -a2 NXI.,Yi;E)- {2e cosXi+ E

as E --4 0, as R1 --4 c° such that (ER1) --4 0, (4.4.14)

where (X1,Y1) = R1(cosei,sin91).

Writing (4.4.14) in outer variables and using the

matching principle we obtain the following outer approximation:

2PN N -a2-ib up(x;E) 1-E fexp( ) (1)0(x,y) as E —40, (4.4.15)

where the outer (wave-free) potential 10(x) is given by the

conditions:

2^ V = 0 in the fluid region,

(4.4.16)

(I)9 = 0 on the mean free surface,

of an = 0 on the body scatterer,

p , ( c)

k 0 as x2 +y2 ;i ) —4 CO

,fr11 sine1 as r1 -- 0,

where (x-b, y-a2) = ri(cose1,sin01).

114 -

Note that the singular behaviour of 4)0 near C(b,a2)

was obtained purely from matching considerations, confirming,

again, that the least singular form is the only one which matches

with the inner solution in the proximity of C(b,a2).

The transmission constant:

Since the potential (Po of (4.4.16) is finite at B(0,a1)

[See figure (4.1.1), then, to be consistent with the boundary

conditions with condition. (i) of §4.1 taken into account: and

the harmonicity, its behaviour near there must be of the form:

(Po' P, + 1-t1r.cos?-3-9 as r 0, (4.4.17)

where (x,y-a1 ) = r(cosA,sin9) and !land 111 are constants

obtained from the solution of (4.4.16). Formulae (4.4.15) and

(4.4.17) furnish:

11: (y - a1)1. (4.4.18)

The application of Green's theorem to the total potentialT(x;E)

and the Green's function G(x; ;E) of appendix (A.1) in the

space (x <0, y> 0), leads to the result:

cc)

k.P w( ;E) = 2i e-Y/E(4 ( ), ax x=0d y exp a1

where Tu, is the transmitted wave and < 0. (4.4.19)

Formulae (4.4.18) and (4.4.19) give the following estimate for

the transmission constant:

–a2-ib aLP E 2p,PNENexp( )

(8x)x=0 as E --4 0, as y-a1 —4 +0.

- 115 -

-a1-a2 -ib

11 rsj

TC ,

v vTn

N r (5/3)EN-05exp( ) as E 0. J

(4.4.20)

This estimate is the same as (4.3.21) provided that Xi of (4.3.2)

is the same as pl of (4.4.17).

To prove that 'Ai = X1, we apply Green's formula to

the potentials, of (4.1.5), and qi, of (4.4.16), in the fluid

3TC domain excluding a sector 0 — of a circle of radius r 2

(<.<1) centred on B(0,a1), and a semi-circle of radius ri (<<l)

centred on C(b,a2). In the limit (r, r1) 0 we obtain the

desired result.

As in Chapter 2, this verifies that when a certain

configuration is kept fixed and the direction of the incident

wave train is reversed, then the leading term in the estimate

for the transmission constant remains the same see appendix

(A.4)]. As we have seen, and as we shall see, the same thing

cannot be said about the reflection constant.

4.5 THE SECOND TERM IN THE ESTIMATE FOR i 1

In this section we outline the steps which lead to the

improvement of the estimate (4.4.8) for the reflection constant

R1

of the reverse problem.

Writing P(x,y;E) = 2{exp( -1. )1cos(4)) +(j)(x,y; E)

rescaling as in (4.3.4) and using (4.4.2), (4.4.6) and

- 116 -

(4.4.7), we arrive at:

-a2 A

1

-ib t1)(X •E)—EN-lt E exp( )} (Xl' Y1 as E 0,

(4.5.1)

where qi(b+EX1,a2 +EY1 ;E) = 41(X Y .E).

Using (4.4.9) we get: -a2-ib -Y1

( '

y .E 2(3N EN-Itexp( 2E Y,N e ) OX1 N. 1

as E --4 0 (Y1> 0). (4.5.2)

It is clear that the harmonic function (1)(x,y;E) defined above

satisfies:

(1) +Ell) = 0 (y = 0, x > b),

ax = 0 (x = b, 0<y<a2),

q)N (Rl- e-2ibiE i2saY - co )exp( E ) as x -4 +

The application of Green's formula to the function (1)(x;E) and

the Green's function of appendix (A.1) yields:

;E) = - 2i e-YA=0) dyl xpfi( - b) - y. ax x=b - )e la2

b), (4.5.4)

where q)w is the wave-part of the potential 4)(x;E).

Rescaling as above and using (4.5.2) and (4.5.3) we

obtain: -2a2 1 -2ib E N-1 - 43N(7) exp( as E-- 0.E

(4.5.5)

- 117 -

This shows that the second term is exponentially small compared

to the first, confirming that the surface wave train is nearly

totally reflected in this short-wave limit.

4.6 THE SOLUTIONS OF TWO OUTER PROBLEMS

We conclude this chapter by giving the solutions to

two outer problems. In each case, we quote the estimate

obtained for the transmission constant associated with the

corresponding problem of scattering of a regular wave train, of

the type (1.1.1), incident from x = - 00 , bearing in mind that

if the direction of the incident wave train is reversed, then

the estimate for the transmission constant remains unchanged

['see appendix (A.4)].

First problem:

First we seek a solution to the problem:

0(1. = 0 in the region>0 exterior to the

act quadrant ABC Lysee figure (4.6.1)],

- 0 on the side AB and on the circular an

arc BC,

(I)0 = 0 on the mean free surface (EA + CD), (4.6.1)

- o —40 as (x2 + y2 )2 —4 oo,

r1

s ing1 as r

1 -4 0,

„ a „ where -- On

indicates differentiation along the outward normal

Figure (4'61)

E(too)

00 .o D(-co)

O. . o

- 118 -

- 119 -

and (x+a, y) = r1(-cos91, sinel).

The method of conformal transformations shows that

problem (4.6.1) has the solution:

) (i) 0 = - 3a + 1 (0 <argc < TL),

(c - i)2-

where

(4.6.2)

= cosh(3 Q)(ReQ> 0, 0 < ImQ 3TE 2 )' (4.6.3)

Q = log(2aein w)(- -rc<argw (4.6.4)

1 a - iz w 2a(z - is) (z = x+ iy). (4.6.5)

The behaviour of (1Jlo near B(0,a) is obtained by expanding (4.6.2)

for large Icl.

The transmission constant of the corresponding

scattering problem has the estimate

-i(2)13/3 Tti E 9 n (,3) (a)8/3 exp -E(1+ i)}. as E ---> 0.(4.6.6)

Second problem:

If, in figure (4.6.1), the circular arc x2+y2 = a2

x2 2 is replaced by the elliptic one -7 + x7 = 1 (b >a), then

b a problem (4.6.1) is found to have the solution:

(1)o - 4-C Re ) (0 < arg TO, (4.6.7) 1

where CC, p and K1 (r3› X > 1) are constants and is a function

of z = x+ iy given by:

- 120 -

z = (b2 - a2)coshw , (4.6.8)

where

with

(S2 — 1) 2ch 1 , (4.6.9) w = DT+ 2-log(112)j.7- aa") +1(1. 0 Q1/2(Q_ a) (Q- 13)

0 <: Imw.( for 0 < Rew < ;log(iin) and 0 < Imw

for Rew _> 21og(n).

The constants a, p and K1 are given by the equations

1 TC (1 - u) 'du

K 2 1 9

0 u2(0C- u) 2([3 -

) L1 u1/20, u)2-

(u-i) 2du log( bb+...aa —

_ iss

TC (u-1)2du = u2 (u - (X) % 2(p - U) 2%.

From the above we see that K1> 0 and is a dimensionless number.

Equations (4.6.10)-(4.6.12) are transcendental, since

the integrals can be expressed in terms of the hypergeometric

function, and, therefore, are difficult, if not impossible, to

solve for a, 1 and K1.

In terms of the constants 01,p and K1, the trans-

mission constant of the corresponding scattering problem has

the asymptotic value:

(4.6.10)

(4.6.11)

(4.6.12)

T -, 6ib g' 68/3 exp(-a-ib) as E --4 0, (4.6.13) 3a M)

-121-

where the "geometric" constant s. has the value:

p z i3/(cf,.-1)(B-1) (4.6.14) = 4aK1

`a.- 11 2bK1

If the elliptic arc is such that b <a, then the transformation

z = (a2 -b

2 )2sinhw followed by a Schwarz-Christoffel transform-

ation leads to the solution of the problem.

CHAPTER 5 - 122 -

TRANSMISSION OF SHORT SURFACE WAVES UNDER A WEDGE

5.1 FORMULATION OF THE PROBLEM

The scattering of time-periodic surface waves, by a

partially immersed wedge, is investigated in the short-wave

asymptotic limit. The wedge CAB is held fixed with C(0,0) and

B(b,0) on the horizontal mean free surface. The point A(O,a) is

vertically below, and at a distance a from, C(0,0) and the side

BA makes an angle aT1 (1/2<a<l) with the mean free surface.

A regular wave train of potential Repi(x;E)exp(-iWt)},

1P(x;E) is given by (1.1.1) and Wand E are as defined in §1.1,

is incident from x = -co normally on the vertical side CA Dee

figure (5.1.1) . We shall assume that both a and b (=- a cotaTt)

are large compared to the wavelength 27E of the travelling waves.

As in the previous chapters, the existence of a velocity potential

follows from the assumption that any induced motion of the fluid

is irrotational in nature. The total velocity potential is of

the form RetT(x;E)exp(-iWt)1, where the function T(x;E) satisfies

the following linearised conditions:

+ tp =0 in the domain occupied by the fluid, (5.1.1) xx yy

Ly = 0 on the mean free surface,

dq) = 0 on the wedge, On

- A(O,a)

- 123 -

XI B(:1,0) X •• i- -

(p +apy=o el 1 no-co (p+E (py =

.., 1 ./ ri , ■-• V

Yi

to. Texp (P

e4) () § exp (--ii-i) e =0

an

C

(ID, =0 -.0-4-------

exp ril LY } ` 1

Figure (5.1.1)

kl)(x;E)

- 124 -

where n is the unit vector in the direction of the outward

akf) normal to the wedge and 171 = n .740.

The edge singularities are limited by the condition

r -4 0 as r --4 0, or

where r is the distance from any point on the wedge.

(5.1.4)

The scattered waves are required to travel outwards

at large distances from the wedge. Thus

R exp ( 1 ) as x —4 - 00

( 5 . 1 . 5 ) T—T exp ( i)—f-X) as x + co

11 and Y are constants whose asymptotic evaluation for small E is

our main objective.

Conditions (5.1.4) and (5.1.5) ensure that P(x;E) is

bounded everywhere and is unique.

5.2 THE OUTER APPROXIMATION

Using the formulation adopted in Chapter 1, we write

the total potential LP(x,y;E) as follows:

+ LPI(x,y;E) (x <0)

Vx,Y;E) (x > 0),

where, from the continuity conditions on the plane (x=0, y>a),

(5.2.1)

we have:

- 125 -

2e Y/E -FY1( -0,y;E) = (P2(+01y;E)

64)1f n '4)21 n (y>a) (5.2.2)

ax \-vy3r;Ei = 7;70v237;Ei

Formulae (5.1.5) and (5.2.1) give:

4)1(2E; E) — CA - 1 )exp( - i---=L-i ) as x --4 - CO ,

Y2(X;E) -- I' exp(-1- iX) as x --4 + co . 5.2.3)

Using the same argument as that used in Chapter 1, we arrive at

the outer approximations

q).(X;E)-,k13(E)(RD(X) aSE-4 0 ( j =1-y 2)y (5.2.4)

where, as before, the constant k and the scale factor P(E) are

to be found by matching.

Formula (5.2.4) and the first of conditions (5.2.2)

indicate the existence of an apparent discontinuity in the

potential q30(x,y) on (x = 0, y>a). This apparent discontinuity

is exponentially small in the outer region (y-a)>> E since it

a transpires that p(E) (E)1/(5/2)-a exp(-0. The potential

(x,y) is, therefore, continuous everywhere and is given by

the conditions:

V2 = 0 in the fluid domain,

aq)0 = 0 on the wedge, an

k-f3 = 0 on the mean free surface,

0 as (x2 +y2)-1/2 oo , 5

r -1/2 cos `` 4- (x ) as r — 0. 0

) 1I(1 — 5

(5.2.5)

- 126 -

where (x,y- a) = r(cosQ,sinQ).

Conditions (5.2.5) specify %(x) completely and the

function can therefore be regarded as known in principle. In

particular its behaviour near B(b,0) is of the form:

kr)o(x,y), p r Ilsinp.91 as r1 —* 0, (5.2.6)

• where r(cose,sine 1 i l l) = (x—b,y), 1-1, = TRand p is a "geometric"

constant known in principle.

Note that we have located the singularity of L.P10(x,y)

at the point A(0,a) because its location at C(-0,0) leads to an

inner problem whose solution is singular. Moreover, the

existence at B(b,0) of a singularity of gives rise to a

sloping beach problem, in the inner region around B(b,0), whose

solution is bounded everywhere including infinity and hence has

a standing-wave part [Peters (23), Isaacson (13) and Holford

(12)]. This is impossible in our present problem since the

incident wave train is coming from x = - CO

Note also that we have taken the least singular form

consistent with the harmonicity and the boundary conditions,

because it is the only one which matches with the solution of

the lower inner problem in the neighbourhood of A(0,a).

- 127 -

5.3 THE LOWER INNER PROBLEM

The single-term outer expansion (5.2.4) and the

behaviour of qO(x,y) near A(O,a) give: _ 1 5 ,

4) i (x;E)—kp(E)r T"'L cos

e +5 O_ -a.) as E --4 0,

2 'A- as r --4 0,(5.3.1)

where j = 1,2 and the limits are taken such that -C;,. --4 0.

Now, we define a pair of inner coordinates (X,Y)

according to the equations

x =EX 5.3.2)

y= a + EY

withq).(EX,a+EY:0=45.(X,Y;E), rescaling (5.3.1) and using

matching we obtain the inner approximations

-1 5

.(X;E)--(E)2 (3(E) 1. (X) as E --4 0 0=1,2) (5.3.3) o —

where the harmonic functions qjo (j=1,2) have the behaviour:

-1 5

U. o (X,Y)---k{R1 2 cos 5 - a) as R j

where k has to be determined and R(cosG,sinG) = (X,Y).

Rescaling conditions (5.2.2) and substituting from

(5.3.3) we arrive at:

= 0 on 9 = -TE(1 -a)

0 on 9 = -- ae 2

a 20 ae

atli10 (5.3.7)

2e-11 + U10(-0

ax

- 128 -

,y)a7 T200.0,y) > 0), (5.3.5) 20 TR-(+0,Y)

and 1 _ ct

R (E) = (E)2 exp( (5.3.6)

Results (1.3.7) and (3.3.6) can, now, be recovered by, respect-

ively, letting a -' k and O. 1 in (5.3.6). The boundary

condition on the wedge yields:

Solution of the inner problem:

Applying the Mellin transform to the equations

at3 2LT. R--a j(R ) + = 0 (j = 1,2), OR OR

o

ae2o

we obtain:

s2 d2 — Tio(s,9) +

U. o(s,9) = 0 (O<Res<5 2-75c) (5.3.9) d92 j

T since R U.

-4 0 as R 0, a (OM is finite and aR -jo -jo co

IJj (s'9) = r.(13-._(R,9)Rs-ldR (5.3.10) 0 -3°

Solving (5.3.9) and using the transforms of (5.3.5) and (5.3.7)

we arrive at:

(5.3.8)

:rii ts,(3\ _ 2r(s)sin(Us) cos 19 +11(1 - 00 is T20‘ I sikts(4 -a)}

(O<Res< 2 ). 5-2Ck (5.3.11)

- 129 -

The differentiation of both sides of (5.3.11) with respect to

9 yields:

420(s,9).m{611120(R,e) -2s I-1(s )sin us sin s +nu co} -69 sin{TCs - a)1

(0 < Res < 5 -2

20(.) , (5.3.12)

where MR20(R,9)3 =

Putting

f(s ) = -2s3 r(s),

and gs,e) sin(Tts) sinf9 +TE(1 -a)ls

s2sintits(2 - CL) 2 where 0 < Res < 5 then -

22CC'

f(u) = -2(u3 -3u2 + u)e-u,

(5.3.13)

(5.3.14)

(5.3.15)

and by the inversion formula for the Mellin transform we have a-H.«)

u-ss in 0-1 s) sinfe +1T( g(u, A) - f ct- sds Q-ico s 2s in-{TC s( 2 - a)1

o<C3= Res < 2 5 - 2a.

Using the residue theorem, formula (5.3.16) yields:

(5.3.16)

2n TC [A +TC(1 - 0.)} (-1)n+lsin(7.. 72 sin 2n 5_ 2a 2n

n 2(u)5-20( g(u, 9) —(5 2c) 2T-C,

11=

(u > 1) , (5.3.17)

4T2o(st9).

and

42(31 8 ti

de 1T(5-2a) ( n=1

as R co.

In particular:

- 130 -

g(u,0) = 2[9 +TE(1 Qt)1 5-2a

2n

2 (u) sin(5.2a)sin2n -1- 12-a° (u 1) (5-200 5-2CX, 21.1 TI

9

n=1 5.3.18)

The continuity of g(u,9) at u= 1 is ensured by the identity:

2Th 9 +TC(1 - (X) _ ( -1)n+1 2nTC . 2n 9 +1-1(1 - cx) n2 sin(5-2a)sin 5 - 2a

(5 - 2002 n= (5.3.19)

remembering that -no. in the definition of T20(R,9).

420 a9 (R,9) whose Mellin transform is given by (5.3.12). Referring

to formulae (5.3.12)-(5.3.14) and using the convolution theorem

for the Mellin transform [Sneddon (26)] we get:

61)20 -co

du co

69 = f()g(u,9)-- = f(RT)g01T, • (5.3.20) 0 u u 0

Using formula (5.3.15) we see that the behaviour of the

integral in (5.3.20), for large R, is dominated by the contrib-

ution from the neighbourhood of ti = 0.

Substitution from (5.3.17) and the application of

Watson's Lemma ECopson (5)] leads to the behaviour:

Now we proceed to find the behaviour, at large R, of

nFl 2n 2nTi. te-g1(1-0) TT 57 1 (-1) (Osin(- 2a)sin2n

2n

R5-2a

(5.3.21)

- 131 -

2O -8 2 . 211 . {9 +11(1 - ct)

1-"(5.--7E)sin(sin2 5_ 2ct 'NJ

59 2 TL(5-2C()

2

R5 -2a

as R --4 co (5.3.22)

Putting j = 2 in (5.3.4) and differentiating with respect to 9

we obtain the behaviour (5.3.22) with

4 2 k = 211

I"( ----)sin( 11( 5- 2 co 5-2a 5 2CC (5.3.23)

Letting OL--4 1/2 in (5.3.23) leads to (1.3.25) obtained previously

for the vertical barrier. When OL--4 1, the constant (3.3.28),

obtained for the right-angle corner, is immediately recovered.

The reflection constant:

The application of Green's formula to the function

q(x,y;E)' of (5.2.1), and the Green's function G(x..E) of

appendix (A.1), yields co

6'1 (1--)1(,11;E) = - J- G(0,y;;11;E)-6-7-dc y (< 0). (5.3.24) a

As before, the (outgoing) wave part of the potential LP'1 is

realised by taking the wave-part of G(0,y;,0 ;E). Thus co tp

q):1_14 ( ;1 ; E ) = 2i e-Y1E a y exp( rl ) C < 0).

a

Using (5.2.3) we arrive at: co atP

R - 1 = 2i 5 e YAE --idy. (5.3.26) ox a

Because most of the contribution to the integral comes from the

inner region, then using (5.3.3), (5.3.5) and (5.3.6) we get:

(5.3.25)

6U -Yrd3 I = jre ----dY = 20

0 -Y 20

OX o Y ae

Letting co

TEdY. =-2. (5.3.28)

(5.3.31)

- 132 -

cc° al'? 2a ii — 1 + 2i i e-Y --1(2dY.lexp( ---E-) as E --) 0. (5.3.27) 1 0 ax

Now we proceed to evaluate the integral in (5.3.27) in order to

determine the estimate for R completely.

Evaluation of the integral:

Using the convolution theorem for the Mellin transform we get: co m

Y 1 = ... f e ,- f f(y.)g(u,i)duldy,

(5.3.29) 0 Y ,0

where f(u) and g(u) are given by (5.3.15) and (5.4.16) respectively.

Using formula (5.3.15) we obtain: co

I = 2 f(u 1) TC g(u,-)du.

0 (1 + u)3 2

Substituting from (5.3.16) and interchanging the order of integ-

(5.3.30)

ration we arrive at:

1 (--141)° sin(PCs) sinlCs 3 (g

ill 0-Ipo s S inEsq- 2

I =

0< 0 = Res <- 5-2°C

Since 0 <Res < 1 <1, then the u-integration yields:

r sir) Tcs(73 -a] 2 I = ds (0 < Res = Ci< 5-2a -) . (5.3.32)

0-loo 2 ••• a]

Now consider the integral:

-. 133 -

11-k° J = I cosXs ds, i 0-1)0 s sinEs(2 - Cc2

where 0 < CI < 2 — and 0 <X < 7(-5 - a). 5-2a 2

(5.3.33)

(-I) n+1 27■n J = 2 n cos(5727.) = 2 logt2cos(5=2-6-c$ (5.3.34)

n=1

Now, (5.3.32) and (5.3.33) give:

I 4 x=/ 3 _a) - 2 / 5

-2 bx ru(_2

77-275cotk 0!' (5.3.35)

This completes the estimate (5.3.27). Thus

(5420) cot(5i4exp(4) as E --4 0. (5.3.36)

Letting E we recover the estimate (1.4.15) obtained in the

case of scattering by a vertical barrier. The asymptotic

estimate (3.3.39) of chapter 3 can be obtained by letting CC

in (5.3.36).

5.4 THE TRANSMISSION CONSTANT

Using formulae (5.2.4), (5.3.6) and (5.3.23) we arrive

at the one-term outer asymptotic expansion:

2 e-a/'

LP2(x;E)- ‘4.( -/ q)(x,y) as E 0, (5.4.1) 3-21a (5-2c) F(----) 0

where the potential %(x) is specified by (5.2.5). Formulae

(5.2.6) and (5.4.1) yield:

By the residue theorem we have: co

- 134 - 2

5-2a. -ah: x;E)_,4pfEl e r sinP.A as E--4 0, 5-2(a)

as r --4 0 such that --- -- 0. 1 rl

The surface inner region:

23-20. 1 1 (5-20)

(5.4.2)

An inner region in the neighbourhood of B(b,0), see

figure (5.1.1), is obtained by rescaling the outer variables

(x,y) with respect to the parameter E in the following manner:

x = b +EXl' (5.4.3) y = EY1,

with ki)2(b+EX1, EY1;E) = 0(X ,Yi;E)•

Rescaling (5.4.2) and using the matching principle

[Leppington (16,17,18) and Van Dyke (33)] it is found that

2

4p(E}11+5-2ae-a/E ti (5 - 2a)1A3-2a) o(X Y 1) asE--4

0, 5- 20.

(5.4.4)

where the harmonic inner potential 00 satisfies the boundary

conditions:

t,iRisinP91 + possibly outgoing waves as R1 (5.4.5)

where R (cosOl'sine1) = (X1'Y1).

4_ o + „u = o (y1 = ol xi >o),

ull

= o (el = arc 691

- 135-

Writingo = H + R1 sinp0l'

we see that the harmonic function

H(X1,Y1) is given by:

ae = o (9 =COT>:), a9

5H H + = (Y1 = 0, X1>0), .L1

H(X1,Y1)-,possibly outgoing waves as RI

Now we tentatively pose the behaviour

H(X1,Y1)-,Tlexp(iX1-Y1) as IX1+iYil , (5.4.9)

where T1 is a constant to be found.

To determine the constant T1 we utilise the solution

ED(X1,Y ) of the sloping beach

`2 2

-=-7 + = o in bx OY 1 1

C), +

ei1 - 0 on Yl = .1

problem:

R11->-0 0<-19 <CITC

1

0, X1> 0,

I

as. = 0 on =CM> 2-' 1 be1 e(xi ,y,) is bounded everywhere including

infinity,

e(o,o) = 1. J

(5.4.10)

Referring to Peters (23), Isaacson (13) and Holford (12),

problem (5.4.10) has a unique solution with the following

-Y1 e

sin1101 C)(X1 , Y1) s Xl - 2T( 1 - R

1 r(1 _ II)

as R1

co , (5.4.11)

behaviour at large R1:

1 where 0 Q1 an , <11 = -2-cz <1 and (Xi,Y1)=Ri(cosei,singi).

rco

J ti-r-1(t)dt 2 211' 0 l+t (5.4.17)

- 136 -

We now apply Green's theorem to the harmonic function

H(X1,Y1), of (5.4.8), and e(X1,Y1), of (5.4.10), throughout

the sector 0 < 91<:COT, R1 => 0 to give

T1 =3/2Iat)exp - 4(1 - Id)} , (5.4.12)

where co co roo p.-1 1 'Oa) = f (a , 0 )2. dg= –7—; eate 0)daldt

0 R1-11)0 0 (5.4.13

Holford (12) shows that:

e-'at0(g.,0)da = tf(t

1 + t)2' 0

where the function f(t) is given by:

(Co f(t) = exp[k J ,t+ 1 -

log(1 - . 2 F21-1

Formulae (5.4.13) and (5.4.14) yield:

(5.4.14)

(5.4.15)

I(P) — 1 dt. (5.4.16) PJ 0 1+t2

Holford (12) also gives the value of the integral in (5.4.16) as:

Formulae (5.4.12), (5.4.16) and (5.4.17) give:

0 irc T1 = ro1 exp 7(1 - P.)1 .

The inner potential ks(X1,Y1) then has the behaviour:

2 - /ko- Risinpf9i + iTc

4(1

as R1 = I X1+iY1j co .

Using (5.4.4) we arrive at:

(5.4.18)

(5.4.19)

- 137- 2

5-20(.-aiE 4p (E1 R sin1191 (5-2a)11:7,E

1 )

iTE 4(1-1.1)]-} as E --"-4 0,

as R1 --400 such that (ER1

) --4 0. (5.4.20)

Extending the outer approximation (5.4.1) up to the free surface

by adding the second of the regular wave trains (5.2.3) and using

matching, we arrive at the asymptotic estimate: 5+20C

4iaplE)20,(5-20) [ a . - 0 (1 .1)1 exp - E-cia tan in 4 - 20. T--

(2a) 2(5 - - 0V )1-Y3 2‘5 - 20V

as E 0, (5.4.21)

where 1/2<:(1<1 and p is defined implicitly by (5.2.6).

In order to determine the estimate (5.4.21) completely,

it is necessary to solve the outer problem (5.2.5) to find the

value of the constant p. This we now proceed to do:

5.5 SOLUTION OF THE OUTER PROBLEM

Using the method of conformal transformations, problem

(5.2.5) is found to have the solution:

LPID(X'371

2

R (1); (5.5.1)

(5-2aN5-2a -‘ 28 1 - 1+2a g- 1

(d-1)2(5-2a)

where

- 138 - 3 2-C1

z = x+iy = A J (9-1) dQ (0-c_arg (5.5.2) 0 0(c2- d)1-u-

The constants A and d (31) are given by the equations: 1 _ a

a =461 q -E) 2 0 - E)1-a

d 3-a (2<la<a). (5.5.3)

, a (g - 1) cK

sinarc

7

g 2(d.. )i-cc

Investigating the behaviour of >0(x,y) near = d E = d

corresponds to z=-acotaTO we arrive at the following value for

p, of (5.2.6), in terms of A and d: 1+2U. 1

2 5 (d) ita(coax(7-ct)s-za

P 5+20. 3(5 + 2a)• (5.5.4) (6)200-200 (d_1)400-200

It is easy to see that A has the dimensions of length and it

occurs in the expression for p in such a way as to make (5.4.21)

dimensionless. This should be the case since the incident wave

train has the dimensionless expression (1:1.1).

Equations (5.5.3) can be written in terms of the

hypergeometric function. Thus

a B(;-, 5 -00 1 _ F(1 -CI, ;i; 3 -a•' d-) A a (01-

a (d-1)32 B(a , 5 7 - a)Fa-,Gt; 5 7; l -1 7.) AsinCtTI - ve- J

(5.5.5)

Eliminating A we arrive at the transcendental equation for

d (>1):

- 139 - 3

a' 1 4(1 -c)(2 -a)(d-1)2 5 •) - F(--2-,a;-11 - d

3(d) a'- 2

(2<a‹.1). (5.5.6)

Because of the transcendental nature of equation (5.5.6), its

general solution is difficult, if not impossible, to find. It is,

however, easy to see that, when a= 2, d = 2 giving the mapping

when the obstacle is a vertical barrier.

Now we proceed to consider two limiting cases of

equation (5.5.6).

Case (1): -

First we consider the case a 1/2 such that HcotaTC1

>> 1. In this case

d 2

,a

as a (5.5.7)

Formulae (5.5.4) and (5.4.21) give: 3

i( 2a-1 ) ( 2 TL)(Ea) 2exp t- E+.4cotaTI -141--C(1 - Ac)i

as E 0, as a k such that (4cotaTT)>> 1.

(5.5.8)

Result (5.5.8) shows that the leading term in the asymptotic

expansion for T vanishes when a = 2. This is consistent with the

fact that the transmitted wave cannot be obtained by considering

the inner region in the vicinity of the right-angle corner where

- 140 -

a vertical plane barrier meets the free surface [Alker (2)].

Case (ii):-

In this case we consider what happens when the para-

meter CC tends to unity (i.e. the limit when the angle BAC

approaches a right angle) Dee figure (5.1.1)]. It is obvious

that when a tends to unity the value of the number d becomes indefinitely large. When expressed mathematically, this last

statement reads:

d co as a —4 1. (5.5.9)

From equation (5.5.6) we then obtain:.

1-- 4(1 - 3 C)d 5

' F(1/2,1.'2-1) as a —4 1.

Hence

(5.5.10)

1 d 2(1 - as CC —* 1. (5.5.11)

Either of formulae (5.5.5) then yields:

A, 2a as a-4 1. (5.5.12)

Again (5.5.4) and (5.4.21) give:

TC2 3,,2, E ,T 1 a is i(1 - a) ( 2 ) exp - + T(1- 2(x} 3 a iTC 1

as E --4 0, as a --4 1, such that E >2.> 1. (5.5.13)

This shows that the leading term in the asymptotic expansion

for T vanishes when a = 1. This is consistent with the non-

7

- 141 -

existence of transmitted waves when the wedge of figure (5.1.1)

is replaced by the rigid obstacle consisting of the two planes

Ex' 0, 0<ya]andEy=a, 0 <x<cop.

- 142 - CHAPTER 6

TRANSMISSION UNDER OBSTACLES WITH A VERTICAL PLANE SIDE AND A CURVED PART INTERSECTING AT

AN ARBITRARY ANGLE

In this chapter a brief description is given of two

cases in which the regular wave train (1.1.1) is scattered by

shapes leading to the same lower inner problem as that consid-

ered in §5.3.

Case (i)

6.1 STATEMENT OF THE PROBLEM

We start by generalising the results obtained in

Chapter 5 to scattering by an obstacle with a vertical plane side

CA, C(0,0) is on the mean free surface and A(O,a) is vertically

below it, and an arc AB, B(b,0) is on the mean free surface,

having an equation f(x,y) = 0 and satisfying condition (i) of

§2.1. In addition the arc AB has the properties:

(i) The tangent at A(O,a) is such that:

t(cosorC ax a —6 + sin57 -8y)11f(x,Y)/A(0,a) 0 for n = 1,2,..., N-1.

(ii) The arc AB intersects the mean free surface in such

a way that:

t(cosa . a ,n , ,

a11—+sinaTC—) “x,y) ax ay 1B(b,0)-

- 0 for n = 1,2,..., M-1

where -4 < 5 < 1/2 , 1/2<a < 1, M _>, 2 and N > 2.

- 143 -

Figure (6.1.1)

-144 -

Both integers N and M can be infinite.

In the case when N is infinite and b = -k, condition

(i) of §2.1 applies to the part DB of the arc where D is a point

on CA lying between C(0,0) and A(0,a) Dee figure (6.1.1)].

Note that when the arc touches the line y-a= xtanoTt

from below, then -2<5< 1/2. Similarly, if the arc touches the

line y = (x-b)tanajt from below, then 2<C(<1.

The wave train is incident from x = - co normally on

the plane side CA. As before, the problem is two-dimensional

and the formulation follows exactly the same steps as those

outlined in Chapter 5.

6.2 THE OUTER POTENTIAL

As in the previous chapters, we write the total

potential QP(x,y; E) as:

q)(x,E)- Vx7Y;E) (x >0)

2e-Y/Ecos(f) + 1 (x y•E) (x <0)i (6.2.1)

In the outer region [distances many (small) wavelengths from

the free surface and from the point A(0,a)] L.(x;E) = 1,2)

is approximated by:

qyx;E)--'kf3(E)4)(x) (j = 1,2) as E 0, (6.2.2)

- 145 -

where the constant k and the scale factor R(E) are to be deter-

mined from the solution of the lower inner problem around A(0,a)

as in Chapter 5, and the potential -Pc)(x,y) is the solution of

the problem:

= 0 on the scatterer, an

q3 --4 0 as (x2 + y2 )1/2 co ,

430 —j r2/(3-2 o) 9 -67 cos2(3 25) as r 0, j

where (x,y-a) = r(cose,sine) and, in our case, -4 sz: b < 2. The

potential o(x) is known in principle. In particular near B(b,0)

it has the behaviour:

LiVx,y) --fprill sinj-l.91 as r1 —* 0, (6.2.4)

where (x-b,y) = ri(cosAl,sinel), 11= 2a-- iL (=.;<:cx.<1 in our case)

and p is the constant of geometry.

The behaviours of k-P(x) near A(0,a) and B(b,0), as o —

given above, are consistent with the boundary conditions and

the harmonicity of the function.

6.3 THE LOWER INNER PROBLEM

Defining the lower inner coordinates as in (5.3.2),

with Lp.(EX, a-FEY;E) .(X,Y;E) = 1,2), and assuming that

, V 2T0 = 0 in the fluid domain,

4O = 0 on the mean free surface, 4i°

1

( 6 . 2 . 3 )

- 146 -

the harmonic function 1112(X,Y;E) [when continued analytically

across the arc] is analytic on the line Y = XtanoTC, then the

boundary condition of zero normal velocity on the arc can be

applied on this line when the problem for the first term 4120(X'Y)

in the asymptotic expansion of VX,Y;E) is considered Dee

Chapters 2 and 4].

Working through the solution as in §5.3 we find that

the constant k and the scale factor P(E) of (6.2.2) are given by

(5.3.23) and (5.3.6) respectively with a replaced by (1+5). Thus

4

(6.3.1) -

(3 - 25) n(13 _ 2625)'

2

R(E) = E3-25 expc

where —1/2 <1/2-.

The reflection constant:

(6.3.2)

The first two terms in the asymptotic expansion for

the reflection constant R are obtained in exactly the same way

as in Chapter 5. Therefore putting a= 1 +5 in (5.3.36) we

arrive at:

2a —1 (3 4i_ 26)[cot(3.---.7-6)exp( -7) as E --4 0, (6.3.3) where < <1/2.

- 147 -

6.4 THE SURFACE INNER REGION

Formulae (6.2.2), (6.2.4), (6.3.1) and (6.3.2) yield:

2

4PE3-25 e-a'E k

2 — (x; E)- 1-25 r1 sinlIA1 as E 0 (3-2or(5:75)

as r1 --4 0 and the limits are taken such that

co

(6.4.1)

Defining a pair (X1,Y1) of surface inner coordinates as in

(5.4.3), with q)2(b+EX1, EYi;E) = (X1,1/1;E), then (6.4.1), when

rescaled, together with matching yields:

2

4M1 3-215e

-a/E

1-25 §0(X1,Y1) as E --4 0, (31-26) (776)

(6.4.2)

where

%(X1,Y1)--R1 sintlei + possibly outgoing waves

as , R1 -•-...4 CO (6.4.3)

where (X1:Y.1) = R1(cos91,sine1).

Analytically continuing the potential §(X1,Y1;E)

across the arc and assuming that it is analytic on the line

Y1 = X1tanaT1, we replace the boundary condition of zero normal

-L speed on the arc by the condition -

a5-6jo = 0 on the line Al =CUT>

T17,

1 and the problem for the potential 00(X1,Y1) is then the same

one as that treated in §5.4.

- 148 -

The transmission constant:

The use of formulae (5.4.19) and (6.4.2) gives:

, 2 3-25e-a/t

()( Y •E)— 1-1- ") p)r., \r,11 25 1R1 sinIA 91 (3-4-vii .3 _ 25/

ri(4 exP [iX1-111- 4

(1 -141

as E --4 0, as R1 co such that (ER1) O. (6.4.4)

Extending the outer approximation (6.2.2) [With j = 2] up to

the free surface by addilig the outgoing wave train Texp(41),

writing (6.4.4) in outer variables and using matching we arrive

at the asymptotic estimate: 1 c a ( .?

r , exp 2a+ 32

5 2

4i1TPtE) - ib E 4 -

1 )

ill v 201

where -2 < 6 < 2 and 11 < a < 1.

It is noted that the estimate (2.3.10) can be recovered

by putting S = -%-2. and a, =1.

6.5 THE COMPLEMENTARY PROBLEM

If in figure (6.1.1) we reverse the direction of the

incident wave train, keeping the rest of the configuration

unchanged, then the resulting leading term in the estimate for

the transmission constant will be the same as that given by

(6.4.5) above in consistency with the result of appendix (A.4).

as E —4 0, 1 (2a) 2(3 - 25)F(1 - -)F( ) 2a 3 - 26 (6.4.5)

- 149 -

To verify this we start by considering the following

approximation in the outer region (many short wavelengths from

the free surface):

(x;E)-J\0110EY$0(x,y) asE--4 0, (6.5.1)

where (1)(x;E) is the total potential in this case, the constant

V and the scale factorY(E) are to be found and the function

(x,y) is given by the specifications:

V2(1. = 0 in the fluid region, 0

(Po = 0 on the mean free surface,

Oq = 0 on the scattering body,

(po 0 as (x2 + y) - 4 C°

---Lsinide as r --4-0, o r 1

an (6.5.2)

1 where, as before, (x-b,y) = ri(cosel,sin91) and p, 20C

Formula (6.5.1) and the behaviour of (1)0(x,y) near B(b,0) yield:

tp(x;E)---/ "(E P-)sinP-A as E --- 0 , as r1 --4 0, (6.5.3) r1

where the limits are taken in such a way that -- --4 0. 1

Using (5.4.3), with(1)(b+EXi, EY1;E) = ()(1,Y1;E) we

rescale (6.5.3) and use the matching principle Van Dyke (33)

and Leppington (16,17,18)] to get:

■().C1,Y*1;E) (E (X1,Y1) as E -s 0, (6.5.4)

where

R 11. V

sinp.91 + standing waves as R

1 --co , L(xl ,Y1)--

1 (6.5.5)

- 150-

where, as before, (X1,Y1) = Ri(cosel,sin91).

The harmonic function §o(Xl'Y1) is therefore bounded

everywhere (note that k satisfies the edge condition) and

satisfies the free surface condition:

(Yi °' 0). (6.5.6)

Assuming that 1()C1,Y1;E) (after analytic continuation across

the arc) is analytic on the line Yi = Xitanall, then the

boundary condition on the arc yields:

69 = 0 on el =OLTI>7. (6.5.7) 1

It is clear from (5.4.10), (5.4.11), (6.5.5) and the conditions

satisfied by lk)(X1,Y1) that:

00(x1,Y1) =VP(1-11) ED(X1,Y1). (6.5.8)

Using formulae (5.4.11) and (6.5.4) we obtain:

00(1/Y1; E) V r(1-1-1.)\(iCA ;irill:191_111) ;:--;.cos - zn(1- I-)11

as E 0, as Ri co (6.5.9)

where the limits are taken such that (ER1) --4 0.

Since the incoming wave is of order unity, we then

conclude that:

- 151 -

R12 v- R exp 1- 4111(1 P)1

Y ( E) = E - lEb).

Formula (6.5.9) also provides the estimate

2ib it exp {- -E ------2-(1-1-) as E --4 0,

for the reflection constant in the reverse problem.

(6.5.10)

(6.5.11)

Using (6.5.1) and (6.5.10) we arrive at the approximation

i h fx.c..\ 2 p. 2E [ ib in, ‘-' I 11(1-p) exP L- TO- - t-L) ($0(x,Y) as E -- 0

(6.5.12)

for the outer potential.

The finiteness of 10(x,y),

2A-P' ib (I)(x;E)—F(1 - P.) [exP E

of (6.5.2), at A(0,a) yields:

2

IT--1(1-1,q[gi+ar3-25cos 2 4 (3-2b

as E —4 0 as r --4 o such that

(6.5.13)

The "geometric" constants .91 and a are found from the solution

of the outer problem (6.5.2) and (x,y-a) = r(cose,sin9).

The transmission constant:

Formula (6.5.13), when differentiated, gives:

4g µ2Eµ [exp {- E- 1--z7(1 - sin(A)

2

(3- 2 )1-- ci p) - 1 3-25

as E --+ 0 as (y-a) --4 +0 (y-a› E).

'6x x=0-'-'

(6.5.14)

- 152 -

The application of Green's theorem to the total potential qi(x; E)

and the Green's function G(x; ;E), of appendix (A.1), in the

space (x <0, y > 0) yields:

qiw( 1 11 ; E) = 2i/ E e-Y/E( Ox x=0

(6.5.15)

where <0 and (w( ;E) is the transmitted wave. Formulae

(6.5.14) and (6.5.15) and the use of Watson's Lemma [Copson (5)]

give the estimate:

1 2 2CC+ 3-26 I a ib 1 ;1

ti 8iTES.IE3 exP t- E 7(

1 - Tdi

as 0 1 (20) 2(3 - 215) I' - — 2a.n3 - 25) (6.5.16)

for the transmission constant of the complementary problem.

Putting a= 1 and 5 = in (6.5.16) we immediately recover the

estimate (2.4.15).

Comparing (6.4.5) and (6.5.16) we see that the two

estimates are the same provided that p = 2a. To show this, we

apply Green's theorem to the potential L0(x,y) of (6.2.3) and the

function (1-(x,y) of (6.5.2) in the fluid region excluding the

3T1 sector — 67 of a circle of radius r ( << 1) centred on 2

A(0,a), and the sector an 91 > 0 of a circle of radius (<< 1)

centred on B(b,0). In the limit (r,ri) ---) 0 we get the desired

result.

From the above it is seen that when, in figure (6.1.1),

the coordinate system and the obstacle are kept fixed and the

- 153 -

direction of the incident wave train is reversed, then the

leading term in the asymptotic expansion of the transmission

constant remains the same. In fact, every term in the asymptotic

expansion of the transmission constant should remain the same

because the latter is the same in both problems [see appendix

(A.4): .

Case (ii)

6.6 INTRODUCTION AND SOLUTION

In what remains of this chapter we discuss the_ problem

of scattering of a regular wave train, of the type (1.1.1), by

an obstacle having two vertical plane sides CA and BD and an arc

AD. A coordinate system is so chosen that C(0,0) and B(b,0) are

on the mean free surface, and the points A(O,a1) and D(b,a2) are

vertically below C(0,0) and B(b,0) respectively. The arc AD has

an equation f(x,y) = 0 and satisfies condition (i) of §2.1

together with the property that the tangents at A(O,a1) and

D(b,a2) are such that:

(cosoT12-+sinOTE ay )re

'f(x,y)y)1A(0,ai)0 for n=1,2,..,N-1 a

and

[(sin CC1-437 - cosaTta)nf(x,y)1,„b \.0 for n=1, 2, . . , M-1 'a2I

where , < 2 [see figure (6.6.1) and N ( > 2) and

M 2) are integers which can be infinite. Note that in the

EI(ID,O) C (0,0)

-4,41------ T exp ( D11)! ) E

- 154 -

7( •All- — Lp.tecpy tp+c tpy =

Figure (6.6.1)

- 155 -

case (N = ap, 6 = and/or (M = op, a= -2), condition (i) of

§2.1 applies to that part of the arc AD with (apart from its two

end ones) points (x,y) having the property that 0 <:x <b.

The wave train is incident from x = -00 normally on

the vertical side CA and our objective is to find asymptotic

estimates for the reflection and transmission constants in the

short-wave limit E---) 0. The formulation and the method are

exactly as in Chapter 5.

The outer approximation:

In the outer region many wavelengths from the mean

free surface and the point A(0,a1)] the approximations

Tj(x;E)-- kP(E)C1:30(x) as E --4 0 (i = 1,2) (6.6.1)

hold, where k and f3(E) are given by (6.3.1) and (6.3.2),

respectively, with a replaced by al, and the potentials p.(x;E)

are defined by the formulation:

q:)(x;E) Yfc cos() + (x.E) (x <0)

E 1 ' (6.6.2) q302(x; E) (x >0),

where -P(x;E) is the total potential.

The function o(x) is the solution of the outer

problem:

- 156 -

..1 V2 46 = 0 in the region of the fluid,

q)0 = 0 on the mean free surface, aLp 0

0 on the body scatterer, an

--+ 0 as (x2 + y2 )2 CO

,, r-2/(3-2.5)cos29 3- 26

511j1 as r --4

where (x,y-a) = r(cos8,sin9).

(6.6.3)

The reflection constant:

Since the solution of the problem for the single-term

inner expansion, in the neighbourhood of A(0,a1), enables us to

find a two-term asymptotic estimate for the reflection constant,

and since this problem in our present case is identical (after

analytically continuing the inner potential across the arc and

assuming that it is analytic on the line Y = XtanEITE) to that

considered in §5.3, we then conclude that the asymptotic value

of the reflection constant 11 in our present problem is given by

(6.3.3) with a replaced by al. Thus

2a1 1 - 3146+ot(3.---7-r 52 )1e)cp( - El) as E 0, (6.6.4)

where < 5 < 2.

The transmission constant:

Using (6.3.1), (6.3.2) with a replaced by al, and

(6.6.1) we obtain the following outer estimate for the potential

2(x;E):

-.157-

2 ..a1 r 13-25 E

tp (x. 0, 4 kE3 e ,,,,, 1 , 2 —' (3-20-1(1-25) 'rokx,Y) as E -4 0,

3-25

6.6.5)

where %(x,y) is given by (6.6.3).

Because Lk)(x,y) is finite at D(b,a2) its behaviour

near there must be of the form:

2 3 °1+1

"Po' P1 p rl -2acos2 1 3 -

where (x-b, y-a2) = ri(cosepsin91).

The constants pl and p are functions of the geometry

only and are known in principle from (6.6.3). Formulae (6.6.5)

and (6,6.6) yield:

dq 2aC e-aI/E

(---) 2 dx x=b 1-25 (3-25)(3-20T'(7.75)[y-aW 3-2a

as E 0 as (y-a2) --4 +0 (y-a2>> E). (6.6.7)

The application of Green's formula to the potential Lyx;E) of

(6.6.2) and the Green's function G(x; ;E) of appendix (A.1) in

the space Dx,y): x > b, y> 0] yields:

q2wa'11 ;E) = -2i

(t > b).

t j e-Yr-(--=) dy exp i( - b) ax x=b a2

(6.6.8)

Formulae (6.6.7) and (6.6.8) and the use of Watson's Lemma

Ecopson (5)] furnish the following asymptotic value for the transmission constant:

as r1 0, (6.6.6)

2

- 158 -

2 2

-16iTIRE?-26+ 3-2C1 exp -1-a2-ibl f„

(3 - 26)(3 - - 25 -•

3- 261 `3- 20Ci

as E 0 , c0 < 21. (6.6.9)

If, in (6.6.9), we put a =5 = 0, then the estimate (3.4.6) will be recovered immediately. The leading term (4.3.3) can be

obtained from (6.6.9) by putting 5 = 0 and a=

on a(P = v(x) (x E S), (7.1.2)

- 159 - CHAPTER 7

SOME RADIATION PROBLEMS

7.1 INTRODUCTION

So far we have been concerned with the two-dimensional

problem of how a regular train of gravity waves, of amplitude and

wavelength very much less than the dimensions of the obstacle, is

scattered by a partially immersed cylindrical obstacle of given

cross-section with its generators parallel to the wave crests:

the so-called transmission (scattering) problem.

The corresponding wave-making (radiation) problem is

one in which the obstacle is given a forced periodic motion of

small amplitude about some fixed position. If the amplitude of

motion of the - obstacle is small enough, then the amplitude of

motion at any point of the fluid is proportional to it and at

large distances from the obstacle the motion on each side of it

is a regular wave train travelling away from the obstacle. It

is to some of these problems that we devote the remainder of

this thesis. A typical radiation problem, in which the normal

velocity on the wave-maker $ has the prescribed value

RetV(x)exp(-iWt)} , is to determine a functionT(x;E) that

satisfies the linearised conditions:

C740 = 0 in the fluid region, (7.1.1)

- 160 -

where n is the unit vector in the direction of the outward MEM

normal from .S into the fluid. On the mean free surface Lx;E)

is required to satisfy:

+ ELP = 0, (7.1.3)

where E= -g-, g is the gravitational acceleration and surface (I)

tension effects are neglected.

Two supplementary conditions are needed; the first

is the radiation condition:

Lp(x;E) -., A + ) as x ± CO , (7.1.4)

where A+ are constants in whose evaluation we shall be particularly

interested, and the second is a condition limiting the edge

singularities. Thus

lim aLP ) = 0 (7.1.5)

r•-•---) kJ or

where r is the distance from any point on S. With S satisfying

certain conditions [John (15)], problem (7.1.1)-(7.1.5) has a

unique solution. The regular wave trains (7.1.4), on the other

hand, are always unique this can be shown as in the end of

appendix (A.5)].

The Haskind Relations:

If (4)(x;E) is the corresponding scattering potential

in the case when the incident wave train is coming from x =

then

021J = 0 in the fluid region,

on . 0 on S,

where n is as in (7.1.2).

1p+E(1) = 0 on the mean free surface, (7.1.8)

ix E -y l'exp( ) as x —4 +co

- (7.1.9)

xp( E ) + R exp ( ) as x --4 -m (1)(x;E)

The behaviour of tkx;E) near any sharp edge is such that

r—ar

—40 as r --4 0, (7.1.10)

where r is the distance from the edge under consideration. Note

that the reflection and transmission constants k and T are

always unique Dee appendix (A.5)J.

Using the contour described in appendix (A.4), applying

Green's theorem to the two potentials 'P(x;E) and (P(x;E) and

taking limits we arrive at the result:

A = 14)(x; E)V(x)ds, (7.1.11) 5

where the integration on S is such that the arc length s is

increasing with x.

This gives the radiation amplitude (constant) at

infinity, in the direction from which the waves are incident

Ex = in this case in the transmission (scattering) problem,

in terms of the scattering potential and the prescribed normal

(7.1.14)

Similarly, if S is swaying [i.e. oscillating horizontally] with

a velocity cosWt, then V(x) = cos p on S and

Fx =fW-3- exp(- iWt)} A_, (7.1.15)

FY = -iW-acf(1)(x;E)sinpds = < w-aexp(-iWt) A...

s /

-162 -

velocity on S. Note that (7.1.11) holds without any restrict-

ions on the profile S.

If the wave-maker S is heaving [i.e. oscillating

vertically] with a velocity cosWt, then V(x) = sinO on S,

where n = (cos3 , sin i3 , 0) is the unit outward normal from S

into the fluid, and

A = f(1)(x/Y;E)sinp ds. (7.1.12)

In the scattering problem, the y-component of the force exerted

on $ (the heave force) is given by:

F = - 1 p sin p ds, (7.1.13)

where p is the pressure and the integration on S is such that

the arc length s is increasing with x on S.

Now, p = at --a a [Cx;E)exp(- = iW-al)(x; E)exp(- iWt), -

where -a is the density. Hence

where Fx is the x-component of the force exerted on S in the

-163-

scattering problem :sway force].

In the case when the wave-maker S is rolling about

the z-axis with an angular velocity Q cosC0t, where Q = (0,0,1),

then

V(x) = a8114:1 = (QAr).n = sing - ycosI3) on S. (7.1.16)

In this case (7.1.11) and (7.1.16) give

=if{y cosp- x sini3}(j)(x;E)ds. (7.1.17) 8

In the scattering problem we have that the roll moment on S in

the z-direction is given by:

M = jr pfycos13 - x sin Plds

(y cos13 x sini3 ) (I) (s;E)dsl exp(- iWt)

qw,a exp(- itot)1A_. (7.1.18)

Similar relations between A+, of (7.1.4), and the exciting

forces and moment can be found when the wave train, in the

transmission problem is incident from x = + co .

Relations (7.1.14), (7.1.15) and (7.1.18) are the

Haskind Relations for two-dimensional motion ['Newman (22)1

In what follows we shall restrict our discussion to

fluid motions induced by the rapid vertical (heave), or

- 164-

horizontal (sway); oscillations executed by some of the rigid

obstacles considered for the scattering (transmission) problems

of the preceding chapters. In each case the vertical, or

horizontal, velocity is taken to be Vcose0t, with V a given

constant. The amplitude of the oscillations is considered to be

very small and the constants A+ of (7.1.4) are estimated in the

high-frequency (i.e. short-wave) limit E O. Since the

problem is linear, we may take V to be equal to unity [as we

have done already in the derivation of the Haskind Relations],

without any loss of generality.

7.2 HORIZONTAL OSCILLATIONS OF A VERTICAL PLANE BARRIER

A plane barrier of zero thickness and infinite length

is immersed to a vertical depth a in an ideal fluid of infinite

extent Dee figure (1.1.1)]. The barrier undergoes horizontal

oscillations, of small amplitude and high frequency, with a

2TC velocity cosWt, where - (<-K1) is the period of the oscillation.

The velocity potential of the induced motion is

RefLP(x;E)exp(- it,001 , where

OLP = 0 in the fluid domain,

ot-P = 1 on the barrier,

ax

+E aq) = 0 on the mean free surface. ay

(7.2.1)

(7.2.2)

(7.2.3)

- 165-

The edge singularities are limited by the condition

r or - -4 0 as r 0, ag)

(7.2.4)

where r is the distance from any point on the immersed part of

the barrier.

At large distance from the barrier, the potential is an

outgoing wave. Thus it is required that

LP(x;E),, A+ exp(-111c11) as x + co • E (7.2.5)

The constants A+ are unknowns of the problem and give a measure

of the energy radiated towards infinity by the motion of the

barrier.

The application of Green's theorem to the function

LP(x;E) of (7.2.1)-(7.2.5) and the Green's function G(x;;E) of

appendix (A.1) readily yields the property:

LP(x,y;E) + q)(-x,y; = 0, (7.2.6)

since aq) is continuous across x = 0, y>a. ox

Formulae (7.2.5) and (7.2.6) give:

A+ + A- = 0. (7.2.7)

The continuity of P(x,y;E) across (x = 0, y>a) and formula

(7.2.6) yield:

)(0,y;E) = 0 (y>a). (7.2.8)

The potential at any point (,11 ) is given by:

aq)0 = 1 on the barrier,

ax

-166 -

a co

q3I q ;E) = G(0,y; ; Ody+(aLP) is7c,x0G,O,y;5 ,71;Ody 0 a

( > 0,11 > 0). (7.2.9)

Formulae (7.2.8) and (7.2.9) lead to an integral equation whose

solution determines the velocity distribution ax 2 on (x=0, y>a)

and hence the potentialLP( ,M;E) at any point (E ). The aim

of the present investigation is not to seek a closed form

solution to problem (7.2.1)-(7.2.5), but is to determine the

asymptotic forms of the constants A+ of (7.2.5) in the limit of

short waves. To this end we proceed as follows:

The outer potential:

A first approximation to the potential, and one that

is presumed to be valid at all points that are many (short)

wavelengths distant from the free surface, is specified by

formally taking the limit E --4 0 in (7.2.3), to get

02q)0 = 0 in the fluid,

(7.2.10)

0 = 0 on the mean free surface.

The specifications (7.2.10) permit no surface waves, of course,

so that the condition at infinity is simply that L0(x,y) should

vanish at great distances from the origin. The outer potential

LiC(x,y) is also required to be finite at all points of the

immersed part of the barrier [i.e. satisfying condition (7.2.4)J,

-167-

otherwise it cannot be matched to any inner solutions.

The problem for t40:(x,y) is a simple one and can be

solved by conformal transformations. Thus

LP(x,y) = -n2Re tiz log( z

if27.)

a+ a + z (7.2.11)

3 where z = x + iy and - 2 arg( z - ia) Ta .

The addition of eigensolutions to (7.2.11) is not

permitted because (since they are singular in the domain) they

cannot be matched to any inner solutions.

With this 4O(x,y), the single-term outer expansion is then given by:

q)(x;E)--, -721 Re iz log( )1 as E —4 0,

a +42 z2 (7.2.12)

where - - TC arg( z - ia) 311. 2 2

When expanded for lz - la' <<1, formula (7. 2.12) gives

27; - r 2 cos?--- (e +I) +rcos9 as E-4 0, 2

as r --4 0, such that —4 co , (7.2.13)

where z - ia = x + i(y - a) = re19.

Differentiation of (7.2.13) with respect to x yields

( B ‘b

Tx)

x=0-' (y-a) 2 as E 0, as (y-a) —) +0,

(7.2.14)

- 168 -

where the limits are taken in such a way that (y-a)»E.

Taking the wave part of G(0,y;71;E) formula (7.2.9)

gives ,c0

kipw( ;E) = —24E(1—e—a-/E)+ e—Y/E0-2)x=0 dylexp(i 11),E a

ax (7.2.15)

where Vg,11;E) is the radiated-wave part of the potential

kp( ,m ;E) in the region > 0,11> 0).

Using (7.2.5), (7.2.14) and Watson's Lemma [Copson (5)J

formula (7.2.15) furnishes the result:

aE 1/2 A -2iiE - (277) exp(-f)) as E --4 0. (7.2.16)

It is noted that the second term in the estimate (7.2.16) is

exponentially small compared to the leading one. In fact the

latter could be obtained by considering the inner region in the

vicinity of the point where the barrier intersects the free

surface.

The surface inner region:

In place of the single-term outer expansion (7.2.12)

we assume the asymptotic development

k-P(x;E).-ik-P0 +Y(E)(1)0 + + ... as E 0, (7.2.17)

where %(x) is given by (7.2.11) and the scale factorY(E) is

such that

and

(7.2.20) ax = 0 on the barrier,

LP1 = boy on the mean free surface,

0 at infinity.

02 1 = 0 in the fluid domain,

aq1

- 169 -

Y(E) --4 0

E Y --4 0 (E) as E -4 0. (7.2.18)

The outer potentials (1) and q°1 are specified as follows:

02q) = 0 in the fluid region, a r

= 0 on the barrier, ax

o = 0 on the mean free surface,

(1)o --4 0 at .infinity,

(7.2.19)

It is clear that the eigensolution %(x,y) of (7.2.19) is singular

in the domain. Next we suppose that its singularity is located

at the point of intersection of the barrier with the free

surface and therefore (because of the boundary conditions) it

must at least have the behaviour

--sin()1 as r1 —4 0, ri

(7.2.21)

where (x,y) = r1(cos9sine1

) and IIo is a constant depending on

the geometry.

/In 2n +1 Note that sin(2n + 1)9 where n is a non-negative

integer, satisfies all the conditions and n = 0 gives the least

singular form.

- 170 -

Appealing to the free surface condition (7.2.3) and

to appendix B, formula (B6), of Crighton (6), we find that

;+ Mlog() + 1 is finite at the origin and hence

2r1 2r1

q)(s; c)— {9icos91+(logrdsing flog(21 sine].

+ .,;'1(o 2 r1 1+log(2 + terms finite at ri = 0

r sineeE - 1

as E --4 0, as r -4 0 such that -- 0. (7.2.22) 1 rl

Defining a pair of inner coordinates (X,Y) by the formulae

x = E X (7.2.23)

y =EY

with LP(EX,EY;E) = §(X,Y;E), resCaling (7.2.22) and using the

matching principle and (7.2.18) we arrive at:

sine1 il(X,Y;E)--11 Y(E) + 2TC (ElogE)(R1sine1 -1) o E R1

+E [

2 2R1 -- a(R1 sine1 -1)log(2a)+7.I Picos91+ (logRdsin91

2, + logy as E --40, as R1 —4 co

such that (ER1) 0

(7.2.24)

where (X,Y) = R1(cos91,sin91).

This suggests that the surface inner potential (X,Y;E)

has the development:

I-1,Y(E) 2 1(X;IE) -6 Ho(X,Y) + ft(ElogE)

as E -4 0,

X,Y) +EyX,Y)

(7.2.25)

- 171 -

with the harmonic functions H o ,

o and 0

1 satisfying the free

surface condition aY

= 0 on (Y=0, X>0) and have the

following behaviours at large R1:

sine1 Ho + outgoing waves as R1 --4°°, R

1 R1sine1 - 1 + possibly outgoing waves as

R1 co,

2R1 (7.2.26) - 1)log(2a)+—fr eicosei

+ (logysinq- + logy +possibly

outgoing waves as R1 —4 op

The boundary condition on the barrier yields

aH (o Y) = 0 ax

ax° ' (o Y) = o

OX (0'Y) = 1

(Y > 0). (7.2.27)

It is clear from the above that the problem for the potential

Ho(X,Y) is a singular one and hence q)(x,y) of (7.2.19) is not o

singular at the point of intersection of the barrier with the

free surface.

A similar argument can be used to show that the eigen-

solution (1)o(x,y) is finite at the tip of the barrier and hence

l!o ta- 0. The problem for 0 (X,Y) is a homogeneous vertical beach -o

one and the eigensolution is given by:

00(X,Y) = Y - 1. (7.2.28)

- 172 -

The potential k:

To find the potential yX,Y) we proceed as follows:

Defining a new potentiallD(X,Y) such that:

4-(R1sing1-l)log(2a) - 4(1+ logy 2R,

,

then the harmonic function tlf(X,Y) is given by the conditions:

' (o Y) = 0 (y>o) ax 11 + tV = 0 oz = o , x> o )

(7.2.30) possibly outgoing waves at infinity,

OP - 14logy is finite at R1 = 0.

It is obvious that

*0(X,Y) = G(X,Y;0,0;1),

where G(x,y;p] ;E) is given in appendix (A.1).

(7.2.31)

Extending the outer approximation up to the free

surface, adding the regular wave train Ai_exp( E y) and matching

with (7.2.25), remembering that ',lb E 0, we arrive at the estimate

A+ti -2iE as E 0, (7.2.32)

which is the leading term in (7.2.16).

The above analysis can be applied to any swaying two-

dimensional obstacle whose normal intersection with the free

(7.2.29)

-173 -

surface is planar for a distance of many (short) wavelengths.

For the case of non-planar normal intersection with the free

surface, the resder is referred to Dilker, G. "Ph.D. Thesis",

Mathematics Department, Imperial College, London (to be

published)J,

7,3 SWAYING MOTION OF A CYLINDER WITH A VERTICAL PLANE SIDE

We start this section by considering briefly the fluid motion

induced by a partially immersed cylinder, of the type shown in

figure (6,1,1), executing horizontal oscillations, of small

Amplitude and high frequency, with a velocity cosWt. The

induced velocity potential Rel(1)(x;E)exp(-:iWt)1 is the solution

to the following linear problem:

= 0 in the fluid region, (7.3.1)

E-- = 0 on the mean free surface, Y

(7.3.2)

where F g is the gravitational acceleration.

P64 on 5, (7.3.3)

where n = (CO5P, SinP, 0) is the unit outward normal from S

into the fluid domain and i. = (1,0,0) is the unit vector in the

direction of x increasing. For uniqueness of solution, (1)(x;E)

is required to satisfy

lim aq) (r) = 0,

r

(7.3.4) --40 ar

-174-

where r is the distance from any sharp edge. Waves are outgoing

at infinity. Thus

(1)(x,y;E) — A+exp(±) as x --4+ co . (7.3.5)

Our objective is to find the leading term in the asymptotic

expansion (for small E) of the constant A_, and verify the

Haskind Relation (7.1.15). The estimation of A+ is as illustrated

for the heaving case at the end of this section.

The outer approximation:

An outer potential, valid throughout the main body of

the fluid (many small wavelengths from the free surface) is

obtained by putting E = 0 in condition (7.3.2). This outer

potential is then specified by

\7= 0 in the fluid region, 2 *0

(7.3.6)

o = 0 on the mean free surface,

a4) cost on S Esee (7.3.3).],

--4 0 as (x2 +y2)

lo is finite on S,

and the outer approximation then reads

q)(x;E)--ilko(x) as E --4 0.

an

(7.3.7)

Near the point A(0,a) Dee figure (6.1.1)] the outer potential

110(x, y) has the behaviour:

- 175 -

gio

2

+ rcos9 as r --4 0,

(7.3.8)

P- + Xr3-25 cos [--)-.5T-C 3-26

where (x,y-a) = r(cos9,sin9), 15 is as defined in §6.1, and p.

and X are constants dependent on the geometry of figure (6.1.1).

The application of Green's theorem to the total

potential 4)(x,y;E), of (7.3.1)-(7.3.5), and the Green'S function

G(x; E ;E), of appendix (A.1), in the space (x <0, y2> 0) yields:

re° C 9 7I;E) = 2i e-Y/E(2i).) d =0 y exp 6 , 7.3.9) i v

0 ax x

where < 0, 11 > 0 and q)w is the radiated wave towards = - 00 .

Referring to (7.3.5), formula (7.3.9) gives

a co Y/E -y/E OLP) A = 2i e dy +21 e (67 _ody.

0 a

Upon using (7.3.7), (7.3.8) and Watson's Lemma ECopson

(7.3.10)

formula (7.3.10) furnishes the asymptotic value

2

2TaJEj 3-2 exp(-)

A'-' 2iE+ 1 -25 as E --4 0. (7.3.11)

\ (3 - 25)L N,3 2S.

Note that the first of the two terms in (7.3.11) could have been

obtained by considering the surface inner region in the proximity

of the point C(-0,0) see figure (6.1.1)] bearing in mind that

near this point the outer potential liJo of (7.3.6) has the

behaviour:

- 176--

2ri (1)0(x,y)^/ - TL

{9icosei + (logri)sin9i +korisin9i

as r1

-4 0, (7.3.12)

where (x,y) = r1(-cos9,sin91) (0 < 91 < 2) and ko is a constant

known in principle. The procedure is as outlined in §7.2.

Next we proceed to find the constant X of (7.3.8) in

terms of the outer potential t.po, as given by (6.2.3), of the

related scattering problem.

To accomplish this we apply Green's formula to the

two potentials (00(x,y) and %(x,y) of (6.2.3) and (7.3.6)

respectively, in the fluid region except for the sector 2-1 .- 9 On of a circle of radius r (<<l) centred on A(0,a). In the

limit r --4 0 we arrive at

B(b,0) 3iT

1: -

C(0,0) cps.cospds = ri2:0 t (

.0 Or - O Or )rd9

(7.3.13)

where 11.. 111 means that the integration is taken over the outer

region and the arc length s of the profile is measured from

C(0,0) to B(b,0) i.e. in the direction of x increasini].

Utilising the behaviours of kpo and (1) near A(0,a) we obtain:

B(b,0)

X = (x,y)cosPds. IL

C(0,0) (7.3.14)

-177 -

The exciting (sway) force:

In the related scattering problem of §6.1 - 06.4 the

incident wave train is coming from x = - co and the x-component

of the force exerted on the obstacle of figure (6.1.1) is given

by: B(b,0) B(b,0) Fx = f (-pn.i)ds = - S p cos 13ds

C(0,0) COM r B(3,0)

= - j_WO. L j. kP(X;E)COSPCIS exp(-iWt), (7.3.15) C(0,0)

where n and i are as defined in (7.3.3), 'P(x;E) is the total a scattering potential, p = ---6aP(x:E)exp(- iWt)1 is the pressure

and is the density.

Using (6.2.1) we obtain: iB(b,0) a

Fx = S .P,(Lc;E)cosPds f ,A(0,a) 0

+(-0 y•E)1dy W exp(-it) 1

(7.3.16)

With the help of (6.2.2), (6.3.1), (6.3.2) and the fact that

most of the contribution to Fx comes from the outer region we

arrive at: 2 4(E)377e-a/E

iCOliexp(- iWt)} 2E (e - 1) + (3-2 ar(2-----25)

-a/E

B(1,?,0) A(0 ) 3-25

body asE O. [

locos (ads A(0,a) C(0,0)

(7.3.17)

-178 -

Since 5 < 1/2 and p = Tt on CA, then using (7.3.14) we obtain 2

2 TIME}3-26exp ( - Fx 2164+ exp(- iwt)

/1 - 26, (3-26)1A777161

as E 0. (7.3.18)

Comparing (7.3.11) with (7.3.18) we iionediately see that the

leading terms in the expansions of A and F- are in consistency

with the Haskind Relation (7.1.15).

Heaving motion of the same cylindrical obstacle:

If the obstacle of figure (6.1.1) is undergoing small

vertical oscillations with a velocity costiit, where W ( the

angular frequency) is large, then, as before, the outer approxim-

ation is given by

LP(x;E)--, Vx) as E 0, (7.3.19)

where LP(x;E) is the total radiation potential and k-P (x) is o -

specified by the linearised conditions:

02LP = 0 in the fluid domain,

LP = 0 on the mean free surface,

oLP

(7.3.20) on = sin p on S, where n = (cos(3 , sin P , 0) is the outward

1 nornal, LPo

--> 0 as (x2 + co ,

q:1o is finite on S.

- 179 -

The requirement that the potential %(x,y) must vanish at

large distances from the heaving obstacle stems from the

radiation condition (7.1.4) and the fact that the outer

potential is wave free. Following Leppington D16),(17),(18)],

the surface wave region is then covered by extending the

potential(x,y) up to the free surface and adding the regular

wave trains that are formed.

Near the point B(b,0) see figure (6.1.1)], the

potential L.c(x,y) has the behaviour:

q)0(x,y)-./krili sinp,91 + risin91 as r1 0, (7.3.21)

1 where 1-1, = — (1 > 1/2) , (x-b, y) = r (cose1 '

sine ) and k is a 2a "geometric" constant determined from the solution of the outer

problem (7.3.20). It can therefore be regarded as known in

principle.

The surface inner region:

Formulae (7.3.19) and (7.3.21) yield

4)(x;E),-,J kri5int,191 + risinei as E 0, asr1 0,

(7.3.22)

where the limits are taken subject to the condition that r

co

Defining a pair (X1,Y1) of inner coordinates according

to the equations

OA = 0 on el = au> 2' 1

10 7.3.25)

- 180-

x = b + EX1

y = EY1

7.3.23)

with 4)(b+EXi, EYi;E) = k(X1,Y1;E), rescaling (7.3.22) and

using the matching principle EVan Dyke (33), Leppington (16,17,18)

and Crighton (6)] we arrive at the inner approximation

k(Xl'Yl; E) ti k kO(Xl'Yl) as E --4 0, (7.3.24)

where [after analytically continuing the potential k(X1,Y1;E)

across the arc] the inner potential is given by the

following sloping beach problem:

a2 a2 2 2110 = 0 in X1+1.Y11=> 0, 0 <91<:OLTI)

ax1 Y1 10

10 aY1 = 0 on Y1 = 0, X1 >0,

Iho is finite at R1 = = 0 f

10 v R1 91 + outgoing waves as R1 --4 1

The function On satisfying conditions (7.3.25) is exactly the

same as the one encountered in §5.4 Esee (5.4.4)-(5.4.9)1

Using (5.4.19) and (7.3.24) we obtain:

§ (X Y k EliRtisint-09 + iT1112 l' 1 1 r(1 -1,),)exP iX-Y 1 1 4

as E.-, 0, as R1 00 such that (CR1 ) 0. (7.3.26)

- 181 -

Extending the outer approximation (7.3.19) up to the free

surface, adding the first of the regular wave trains (7.1.4) and

matching with (7.3.26) we obtain the estimate

ib rm • A ink 1-1 2 E exp{-- — ± r-0.-11) E 4 - as E 0 .

( 7 . 3 . 2 7 )

Now we proceed to find the value of the constant k, of (7.3.21),

in terms of the outer potential ljJo, as given by (6.5.2), of the

related transmission problem considered in @6.5.

Referring to figure (6.1.1), we apply Green's formula

to the functions (1) and q) of (6.5.2) and (7.3.20) respectively,

where the domain of integration is the fluid region excluding

the sector 0 C 9 <an of a circle of radius r1 (<< 1) centred

on B(b,0). In the limit r1 0 we get:

B(b 0) 0,7

4 64)

A(0, )

rosin ds = r1 im 6 J (4O o gio Eto)rldel 1

0 1 1rosin

(7.3.28)

where the arc length s is measured from A(0,a) to B(b,0) [see

figure (6.1.1)].

The behaviours of and (1)o near B(b,0) and formula

(7.3.28) yield the relation:

B(b 0)

k =-- (x,y)sinPds, A(0,a) °

since p. = 1 < 1.

(7.3.29)

- 182 -

The exciting (heave) force:

In the scattering problem considered in §6.5, the

y-component of the force exerted on the obstacle of figure (6.1.1)

is given by:.

B(b,0)

F = S (-pn.j)ds, (7.3.30) A(0,a)

where p.,---aa-4)(2s;E)exp(- iLOt) is the pressure, -a is the density,

kl)(x;E) is the total scattering potential, n = (cos p, sin p , 0)

is the unit outward normal from S into the fluid, J = (0,1,0)

and s is measured from A(0,a) to B(b,0) see figure (6.1.1)].

Substituting for the pressure we obtain:

r B(b,0)

F = J ,(x; Osirads exp( - iW t). (7.3.31) y A(0,a)

Now we proceed to show that most of the contribution to F y

comes from the outer region. We start by writing the integral

(7.3.32)

where t is the total length of the arc AB see figure (6.1.1)]

and 20 is chosen in such a way that E<<( - 2o)<<min(a,b).

In the range (0,2,0) i.e. within the outer region we use the

approximation (6.5.12) to obtain:

of (7.3.31) in the form:

B(b,0) 2

S q)((;E)sin1ds=q1(is;E)sin1ds+SP(x;E)sin1ds, P A(0,a) 0 20

- 183 - ,

2 p.32'El-' r , irt (X2s;E>sinpds—,„ . j To(a)sin exp E 4 0. 0 ikl 1-1-) 0

as E —4 0. (7.3.33)

In the interval (2,o ,2,) i.e. within the inner region around

B(b,O)], we continue the inner potential §(X1,Y1;E)----(1)(b+EX1,EY1;E)

analytically across the arc and, see §6.5, assume that it is

analytic on the line Yi = Xitanan Thus making the substitution

s = g-Er' we get: 0

r2 R(x,y;E)sinPds = (-EcosaTo f §(ncosc(.Ttigincurt, E)dfl

0 (7.3.34)

Using (6.5.4) and (6.5.10) we arrive at: ib co E Jr

jrq)(x,y;E)sinf3ds — (-EcosaTT)e 0o(rara cosTEsinrC)dF 2.o as E —p 0, (7.3.35)

where the sloping beach potential L)( X1,Y1) is dealt with in

§6.5.

Since 11<i, then (7.3.31), (7.3.32), (7.3.33) and

(7.3.35) yield:

2ito,ativj, B(b,0),

sinPds exp[- _ FY (1 - 1-0 A(6,a) °

as E --4 0. (7.3.36)

Using (7.3.29) and comparing with (7.3.27) we see that the

leading terms in the expansions of A+ and F are consistent

- 184-

with the Haskind Relation F = tW,aexp(-iWt)1A+ which is

(7.1.14) with A_ replaced by Al..

7.4 ON THE SOLUTION OF THE OUTER RADIATION PROBLEM

In §7.3 we saw that in order to determine the asymptotic

estimates of the "radiation constants" A+ completely, it is

necessary that integrals like (7.3.14) and (7.3.29) are

evaluated to fix the geometric constants which occur in the

expressions for these estimates. If this proves to be difficult,

an alternative way is to solve the outer radiation problem and

find the constants directly from the behaviours of the outer

solutions near the edges under consideration. We end this

chapter by giving the derivation of a formula which is needed to

transform the given boundary condition on the wave maker S from

one plane to the other when the method of conformal transform-

ation is adopted, and then quote two solutions which were

obtained for a rectangular cylinder in heave and then, for the

same cylinder in sway.

If S is the profile of the wave maker in the z-plane

n is the unit normal from S into the fluid region D and 1 is the

angle made by this normal with the positive real axis, then

- 185 -

% 43 df tclw ) LIE eip an =Re(e = Re c-i-; e dw

dw ei(p -ip* ) '54)O) dz an

(7.4.1)

where Apo = Relf(z)1 , = Re(F(w) , fl-z(w)} = F(w), n is the

unit normal from S* the image of S under the conformal trans-

formation z = z(wfl into the region D" Lthe image of D under the

same transformation] and p is the angle it makes with the

positive real axis in the w-plane.

Examples where the method of conformal transformations

is used and formula (7.4.1) is employed are:

(i) Heaving rectangular cylinder:

6q)0

on Because — = 1 on the side BC Dee figure (3.1.1)]

and is zero everywhere else on the cylinder, then (3.5.2)

followed by the transformation

w = (c-k2)2/(-Fk2)2 (0 < arg w , (7.4.2)

leads to the outer solution

4k, 1/0 2_2)(1-C?v2)] w-iv

(1-Ct) (l+vi2)2 w+iv1 o ' '0 Re f 1 1 - log( . 1)dV e(u v)-

(7.4.3)

2 k2-1 where w = u+iv, 0 =

and k and k2 are given by (3.5.3). k2+1

(ii) Swaying rectangular cylinder:

aLp 0 Because = cos 13 on S in this case, where

- 186 -

aLP n = (cosp, sin 3, 0) is the unit outward normal, then an = 0

aqb on the side BC, see figure (3.1.1), and —67 = 1 everywhere else

on the cylinder. Conformally mapping the region as in (i)

above and making use of formula (7.4.1) we arrive at the

solution:

[(02-v2 )(1-02v2):

(1 + v2)2

w-ivl log(17T7)dvii}

4k1 (u,v) TL(1-02)

R

co

E(.712_02)(v1202..id2 w-ivi

lo(77717—)dv, 1/0 (1 + v

2)2 wriv1

g

1

4k1 Re TE(1-0`)

where w, U, k1 and k2 are as in (i) above.

Note that the field is even in (i) and is odd in (ii)

because if the symmetry about the line x.= b. Similar

solutions can be obtained for the wedge of figure (5.1.1) when

it is either in heave or in sway. The outer solution for the

rolling case can be obtained in a similar way.

- 187 -

APPENDICES

A.1 THE GREEN'S FUNCTION

For two-dimensional motion of a liquid of infinite

depth the Green's function can be defined as a solution

G(x,y; ;372 of the potential equation

a2 a2 1 ( a 2 + ;72 = 5(x- + 5(x - s) in the ax ay

space {(x,y): - oO <x < oND , 0<y <001, (A.1.1)

with the following conditions:

ay cc XG + ay — = 0 for (y=0, - <x < 00 and X>0),

G 1 is finite at x=

1 K*

G x- I is finite at x= ; , 2m – – – –

aG = 0 on (x = 0, 0 <y < co ), ax

G is an outgoing wave at infinity,

((A.1.2)

where = ( 01), x = (x,Y), * = ), 15(x- = 5(x- >5(3r-71),

5(x) is the Dirac 6-function and 11 is chosen such that 0 <1 < co

The solution is given by John (15) in the following form:

G(x;L;k)=.21 log )24-(Y-11)2ffix-i- )21-(Y=1)2..] co

(X+ P)(cosP-k- I )exPE-11(Y+11): ( X ) µ 1-1"

1 (X+P-)(cosidlx+l)expE-P-(Y+V] e-µdl 11( X - IA)

(A.1.3)

- 188 -

where the path of integration, in the 11-plane, is along the

real line except for a small arc round t_1,=X in the lower half

plane.

To simplify the integrals in (A.1.3) we consider

r(X+p.)(coskcpexp[- µ(y+-q)] e-I4 k dg (A.1.4) P- (X - P,)

where ko and the path of integration is as described above.

I + = -2iTC(cosXko)expE-X(y+/1)]+Re(J),

(A.1.5)

where

J } (x+oexppiciko - y-11)] e - P-

µ0l

where t means the Cauchy Principal Value.

Defining s = 11+ iP1, we integrate along a contour

consisting of:

The arc 0 < arg s <3 of a circle of radius R (>> 1)

centred on the origin (s = 0),

(ii) The line (P.= 0, 0 < µl < R),

(iii) The semi-circle 0 < arg(s -X) < II of radius 0 (<<l)

centred on s = X ,

dg, (A.1.6)

(iv) The line (11 = 0, 0 <1.-1, < -0 and X+ Q < µ < R).

- 189 -

In the limit R --4 co , 0 -- 0 we obtain: co OD

I -ot I

t e to -at- -it

HIT J = - exp {X (iko-y-70} 4- 2i j 2-- i—dt + j dt

A- 0 0

where 6= ko + i(y+-1). (A.1.7)

Defining s = t + iZ , we integrate along the contour specified

by:

(1) The line (s = ueiGt ), where (0 <: u < R>>1,

0 > a>- 2 (ii) The line (1: = 0, 0 < t < R),

(iii) The arc 0 > args> a. of a circle of radius R centred

on the origin (s = 0).

Letting R co and using the formula co e-V1

- e x -V2x V2 cc_log(77)foroev.>0), (A.1.8)

0 1 we obtain

co e-St - e-it dt - exp(-uoe a) - exp(-iue)du= log(t). t 0

(A.1.9)

Formulae (A.1.7) and (A.1.9) give CO

edtl Re(J) = 2TC(sinXko )expEX(y+11)}- logl5l+Re{2if 0 X- it

(A.1.10)

Using (A.1.5) and (A.1.10) we arrive at

- 190 -

I = -21 iexp (iko-y-11 - 2 log Eko2 + (y+ )2J

co expf-t(k

+Rer2i cth X- it 0

since El= ko + i(y +11).

(A.1.11)

Utilising (A.1.11) by putting ko = jx-q and I in

turn, (A.1.3) then yields:

G(x;t;E)=-1.{exp(klx-I)+exp(kix+1)}exp( -Yr)

+ Alog 1.--(x_ )2+(y.0

2

i oA)2+(y.1)2

lL2

2iL2 2j(x-D +( Y+7 ) (x+D +CY+10 co

fS EexP( -t I x-U ) + exp(-t Ix+ M] - T-tRe i + Et .

exp E-it(y+.11)]d, where E = --. (A.1.12)

The behaviour at infinity:

Keeping the point ( ,11) fixed, letting the point

(x,y) tend to infinity and using Watson's Lemma we obtain the

behaviour:

G(x,y;,TI; E) -2iexp(L--XxEl 2

rc(x2Ey +y2 )

as lx+iy I —.400. (A.1.13)

This means that G(x,y;,11;E) is purely an outgoing wave at

infinity. It is also seen that

G(x,y;0,0;E) = -2iexp(41)+.2i-fRe(iI), (A.1.14)

- 191 -

where 0= lx1+iy and co

e-Ot ------ 1 - iEt—dt. 0

(A.1.15)

Defining s = t + iZ , we integrate along a contour consisting

of:

(1) The arc 0 > arg s> -arg 0 of a circle of radius R>> 1

centred on s = 0,

(ii) The line (1 = 0, 0 < t < R),

(iii) The line s = u/0 (0 < u< 101R).

In the limit R ---)00 we obtain

I (A.1.16)

since 0 < arg 0 <

Integrating (A.1.16) by parts, we iniwediately see

that (I - . log0) is finite at 0 = 0 and therefore IE

G(x,y;0,0;E) T2c logiCil as 101 0. (A.1.17)

A.2 EVALUATION OF THE VELOCITY DISTRIBUTION V(Y)

We here find the general form of the velocity distrib-

ution V(Y) of §1.3, formula (1.3.8). Using (1.3.21) and (1.3.18)

it is found that:

co

V(Y) = de-Y/u1g(u)du du- 0

(A.2.1)

- 192 -

where g(u) is given by (1.3.20).

Writing (A.2.1) in the form

1-E co --N

{

V(Y) =Ei---K) (ue )g(u)du+ Lc c-rute lim : 1. d Y -Y/u d Y -Y/u)g(u)du? 0 du 1+E.2

(A.2.2)

integrating by parts and utilising the properties that g(u) is

continuous at u = 1 and that g(u) --4 0 as u Dee (1.3.20)]

we obtain:

1-E co S ug, (u\ d ie uNdu c 1

V 0(Y)=. -E1.22-11) ue(u)-c-1--(e u)du + 0 du 1+E2

Iduv u) du

Another integration by parts and the use of the fact that

ug'(u)exp( 0 as u —b 0 and as u --4c° lead to

V(?) = lim (l+E)gi(l+E)exp( -Y )-(1-Ede(1-9)exp(7-Y E _40f 2 2 1.c 1.1 -,

e duEle(u)idu+ e-Y/u TI d P (u) du -Y u d 1+E2

Differentiation of the power series (1.3.20) yields:

du[ig'(u)1 1 3/2 (u 0,1)

TC(u - 1)u

(A.2.4)

(A.2.5)

and hence

- 193 -

V(Y) = limE _40 3

13E1

+ co -Y e-Y/udu 1

I e-Y/udu I E2 log(-- 1 7 1+E

2(u-1)u E ) +- 1 IT 0 (u-1)u

3/2 7 3/2

By choosing El = E2 = Ewe obtain co

V(Y)- -Y/ T 2e

-VT

0 (u-1)u3/2 1-00 (1 - T)cit Tt e udu I

(A.2.6)

(A.2.7)

where the integrals are now Cauchy Principal Value ones and

hence (1.3.26).

A.3 THE ASYMPTOTIC EVALUATION OF THE POTENTIAL U20

To find the asymptotic behaviour of 4520, of §1.3,

formula (1.3.3), at infinity we apply Green's theorem to the

Green's function G(X,Y;Xo,Yo) of (1.3.11) and the harmonic

function 1,j20 as in §1.3. Thus co

(a 0 \ = 1 V(x)loga,2+ (x-R)2

20\ 't'ji 0 (A.3.1)

where X o , Yo and Y are replaced, for convenience, by a, p and x

respectively, and hence the inner coordinates (cc, (3) are such

that a+ if3 = Reie .

Substituting from (A.2.7) for V(x), doing the x- 90

integration once by parts, utilising the fact that 2dT 0 1 - T

:this can easily be shown by putting I = -t-j and then

= 0

interchanging the order of integration we obtain:

- 194-

4120(a" )

where

cc

(x-(3)I(x)dx, [cc2+ (x-09

(A.3.2)

(A.3.3)

co --- e -xi 2 , I(x) = (1-T) ' 0

with I(0) = 0 Dee above] and I(x) --4 0 as x --4 co . It can

easily be shown that

dI I ift +• dx 2.

Now, formula (A.3.2) gives

1

cc

412o ( = + 7-51I(x)dx, 0 x

(A.3.4)

(A.3.5)

where CI = p + is and 5 is its complex conjugate. Next we

consider

J(0) =I(x)dx 0

Using (A.3.4) we obtain

(A.3.6)

JO) = 0

co

dx It(x)dx

(x-ICI) 0 (x (A.3.7)

Integrating the second integral by parts and utilising the

properties of I(x) at x = 0,00 we arrive at:

J(o) = S dx LE/L 0 x--(x-05) 0 (x-05)2.

Repeating the procedure by using (A.3.4) we find that:

(A.3.8)

- 195 -

co co

J(0)--'/TT(-1)nn:[f dx +11 as 101 co • 0 x-2(x-0)fl n=0 (A.3.9)

But co

jr dx B(n-1/2,-1) t , ( l argil <It), 0 x(x+p)n (On1

Dradshteyn (10)]. Hence

(A.3.10)

(-1)nr(n+2) as 101 , (-0)114-1 n=0

(A.3.11)

with a similar result for J(d).

Referring to (A.3.5) and the definition of O, we obtain

the behaviour:

11520(a' P)^'- -1 )nfAn+1/2) cos {(n+) ($9 +a) 2 R 2 n=0

as R = 1 CC + 3131 —) co . (A.3.12)

A.4 THE EFFECT OF THE REVERSAL OF THE INCIDENT WAVE ON THE TRANSMISSION CONSTANT

If S is the submerged part of the obstacle, and

P(x;E) is the scattering potential when the surface wave train

is incident from x = co then

02LP = 0 in the fluid domain,

4)-F E q) = 0 on the mean free surface, aY a4) -- an 0 on S,

(A.4.1)

(A.4.2)

(A.4.3)

- 196 -

where n is the unit outward normal from S into the fluid region.

expfillY\ exp +k(:125.= )as x - OD / E kgx; — (A.4.4)

Texp(ix-y) as x —4 +co

and

aLP uT6 —40 as 6 --4 0,

where 5 is the distance from any point on S.

(A.4.5)

If q)(x;E) is now the corresponding scattering potential

in the case when the direction of the incident surface wave train

is reversed keeping everything else in the configuration

unchanged, then

v2(1). 0 in the fluid region, (A.4.6)

+ E— ay = 0 on the mean free surface

(A.4.7)

on = 0 on S, (A.4.8)

where n is as in (A.4.3).

(x;E)--, ' .{. T

l exp(11. Y) as x —4 - co ()

E

expi:1).1.=Y z . E )+ K e xp (1)=Y- as x 1 E

_.....> + co ?(A.4.9)

and

A 134) --4 0 as 15 --4 0, ab

where b is as in (A.4.5).

(A.4.10)

To prove that Yi = T we start by choosing a control

-197 -

contour C consisting of:

(1)

The two vertical lines Ex=xo, 0 < y < Q land Ex=-xo,

(ii) The profile of 8, the line Ey= 2, -xo x <xo] and

the mean free surface between the points (xo,0) and (-xo,0),

where xo and. are both very large and positive.

With the interior of C as the domain of integration,

the application of Green's theorem to the two potentials

x; E) and (1)(x;E) yields

S aP LP f aLP (p - a q)-) dy + dy 0 ax ax x=xo 0

ax ax x=-xo

+ (T2LA2, dx = 0.

aY aY Y= (A.4.11)

-x

Letting 2 c0 and using the fact that Tandq) are both

exponentially small (and so are their derivatives) at large

depths we obtain: co

S aLP (P (1)7Z-TTdx=-

x )dy = 0.(A.4.12) aq) aq) 0 ax ax x=xo

Since, by choice, xo is very large, then substitution from

(A.4.4) and (A.4.9) into (A.4.12) and integration give the

result:

T1 ti

= T. (A.4.13)

- 198 -

A.5 THE RELATION BETWEEN THE REFLECTION AND TRANSMISSION CONSTANTS

Now we proceed to find an important relation between

the moduli of R and Y of (A.4.4). Using the contour C described in (A.4) and referring to problem (A.4.1)-(A.4.5), we

apply Green's theorem to the vector function ['071, where

(x;E) is the complex conjugate of q)(x;E) to obtain:

2 xo 2 f (ea= ,(41Y + Li (P*- dx - j (P*4 dY 0 o -xo

ay y=.2, 0 ax x=-xo

fl 12 dx = (v q)).(v,)kdxdy, (A.5.1)

where D is the interior of C and F is the part of the free

surface between (xo,0) and (-xo,0).

As in (A.4), letting 2 ---)°0 leads to:

S * .p (''P 67)x=xdY 0

0

co aT

- *) ay + S q2dx ax x=-xo EF

(Vq)).(VLP)*dxdy, (A.5.2)

where D = lim cy and the integration on F is such that x is

increasing.

Since xo is large, then substitution from (A.4.4)

is legitimate and (A.5.2) gives:

-199 -

ie I

exp(21:0) i[-. - 7 T'2 + RI2 - + -2ix

-

+ .E....",,,I2dx = Sf (VP).(774))*dxdy. (A.5.3) F D --- ---

-2ix 1* - +2ix * Since blexp(-7-2) = [-iR exp(-772)1, then by equating the

imaginary part of (A.5.3) to zero we arrive at:

.11 2 + l R ! 2 1. (A.5.4)

This is the equation conserving energy.

Uniqueness of the reflected and transmitted wave trains:

Next we make an attempt to show that R and T of

(A.4.4) are in fact unique no matter what the shape of 8 is.

To this end we suppose that there exist two distinct solutions

1 (x;E) and 4)2(x;E) of problem (A.4.1)-(A.4.5), and let

I-0 = LP1. - LP2, then

V 2r = 0 in the fluid region,

+E —az = 0 on the mean free surface, by

aul= 0 on 8, on

n is as in (A.4.3). (A.5.5)

g u 0 as 5 --4 0,

ti is as in (A.4.5)

T5/ __ (ki-R2)exp(2 S. ) as E x a _.....), co

(T1-"I'2)exp(i)- 1- ) as x + co ,

-200-

whereLandf.are the reflection and transmission constants, 3 3

respectively, associated with the potentials T.(x;E) = 1,2).

In (A.5.1) we putIZ(x;E) in place ofP(x;E) and take

limits as before, to arrive at:

1 ,1 - T2 12 + - k212 = 0,

from which we immediately see that:

(A.5.6)

R1 = 2'

T1 = '12'

(A.5.7)

and hence the wave trains are unique.

A.6 EVALUATION OF THE INTEGRAL IN (3.3.39)

We here use (3.3.32) to evaluate the integral occurring

in the estimate (3.3.39). Thus

I = — (x + x4/

32)e-xY _y dx e dY.

0 0 1 - x

Interchanging the order of integration and making the

substitution x = u3, we arrive at:

3 3 3 I= + -- - 41T, 1 411 2 2T1 3'

,(A.6.1)

(A.6.2)

where

oo 1 u

I1 ----du

0 1 -u3 00

- 201 -

I = 3 ' r 0 1 + u3 2

CO

I3 = f u du. 0 (l+u3)2

(A.6.3)

Making the change of variable u = --c 1 - in I1, it immediately

follows that I1 = 0. To find the values of 12 and I3, we

consider the integral:

J = S (y -13)dY 0 (y3 + X3)

(X > o, p> o). (A.6.4)

Resolving the integrand into partial fractions we get: co co

dy 2 ' 1

J= A ,cf(2(y(22y yX±x)2 ) y+Xjdy + D X 2 3X 0 (y - 2)+ 4

X -f3 where A = +B and D = X

Zk (A.6.5)

It is obvious that the first integral in (A.6.5) is

zero, and therefore

2TI 1 J = 3/-3-(—X - x2)

Putting p = 3 and X = 1 we obtain

1= - 4T1 2 3(5

(A.6.6)

(A.6.7)

Differentiating both sides of (A.6.4) with respect to X and

using (A.6.6)

J 0

Again, with

I

And hence using

I

we obtain: co

(3r - 0)dY 2TE 1 _ ( 22

- 202 -

(A.6.8)

(A.6.9)

(A.6.10)

3 3 2 2 2 Cy +X ) 9X tr5 X

X= 1 = 3, it is found

TT. = _

- ). , A3

that:

3 9V7

(A.6.2) leads to

2 = - . 3V

- 203-

REFERENCES

1. Abramowitz, M. & Stegun, I.A.

"Handbook of Mathematical Functions",

National Bureau of Standards, Washington: 1965.

2. Alker, G.

"On the radiation of short surface waves by a heaving

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