A rational Strip Theory of ship motions. Part II

ARCHIEF

nische Hogeschool, DOCUMENTATI D Al U M: 1811

1edin1sche Hogesço13

DOIItember

1971

THE UNIVERSITY OF MICHIGAN

COLLEGE OF ENGINEERING

Odd Faltinsen

This research was carried out in part under theNaval Ship SystemsCommand General Hydromechanics Research Program Subprolect SR 009 01 01, administered by the Naval Ship Research and Development Center. Contract No. N00O1467A01810O16

Reproduction in whole or in part is permitted for any purpose of the United States Government.

This document has been approved far public release and sale; its distribution is unlimited.

StTVO,

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A RATIONAL STRIP THEORY

OF SHIP MOTIONS: PART II

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=

TIlE

DEP4RTMEN

0

_{MARINE ENGINEERING}

Onderafdelin

sbouwkwiO

liotheek van d

December 1971

A RATIONAL STRIP THEORY OF SHIP MOTIONS: PART II

Odd Faltinsen

This research was carried out under the Naval Ship Systems Command

General Hydrome chani cs Research Program,

Subproject SR 009 01 01, administered by the

Naval Ship Research and Development Center. Contract No. N00014-67-A-018l-0016

Reproduction in whole or in part is permitted

for any purpose of the United States Government.

This document has been approved for public release

and sale; its distribution is unlimited.

Department of Naval Architecture

and Marine Engineering

College of Engineering The University of Michigan Ann Arbor, Michigan 48104

The exact ideal-fluid boundary-value problem is formulated for the

diffraction of head-sea regular waves by a restrained ship. The problem is then simplified by applying four restrictions: 1) the body must be slender;

2) the wave amplitude is small; 3) the wave length of the incoming waves is of the order of magnitude of the transverse dimensions of the ship; 4) the

forward speed is zero or it is O(Ch/2a) , 0 < a 1/2 , where E is

the slenderness parameter.

The problem is solved by using matched asymptotic expansions. The re-suit shows that the wave is attenuated as it propagates along the ship. The result is not expected to be valid near the bow or stern of the ship.

The pressure distribution and force distribution along a ship model

with circular cross-sections have been calculated. The total force on the ship has been compared with the value predicted by the Khaskind relation The agreement is good.

The experimental and theoretical pressure distribution a1oig a pro-late spheroid have beén compared. The predicted attenuatiOn of the peak pressure is very well confirmed by the experiments. In addition, theory and experiment agree that the peak pressure near the ship enerally leads the Froude-Kriloff pressure peak by 45°.

I thank the members of my doctoral conmittee for their supervision

of this thesis. My foremost acknowledgement goes to the chairman of the

committee, Professor T. Francis Ogilvie.

My thanks go to Professor R. Timman, Technical University of Deift.

The discussions with him were a break-through in my thesis work.

I thank Mr. Arthur N. Reed, University of Michigan, for giving

valuable suggestions in the computational part of my thesis, and Dr. Choung Nook Lee, Naval Ship Research and Development Center, for providing me with

experimental data.

I thank Dr. Nils Salvesen, Naval Ship Research and Development Center, for all his encouragement and help. Further, I am grateful to.my employer, Det norske Verjtas,for making my stay in the U.S.A. possible, and finally,

my thanks go to The Royal Norwegian Council for Scientific and Industrial

Research (NTNF) for supporting my stay.

iv

NOTATION

I INTRODUCTION

II GENERAL FORMULATION 2

III THE ZERO-SPEED PROBLEM 5

Far-field source solution 6

Inner expansion of far-field source solution 14

Comparisonwith another method 22

The near-field problem and the matching 24

IV THE FORWARD-SPEED PROBLEM 33

Far-field source solution 35

Inner expansion of far-field source solution 47

Comparison with another approach 58

The near-field problem and the matching 60

V NUMERICAL CALCULATIONS 67

Theoretical background 67

Numerical example 78

Comparison with experiments 81

a used in the description of the order of magnitude of tue

1/2-a

velocity U; U = O( ), 0 < a 1/2

B beam at midships when used in the figures

c=

U)0

Fn =

U/VFroude number

g = acceleration of gravity G(kY,kZ;k,kT1) See (88)

wave axt1itude of the incoming wave.

function defining the wetted surface of the ship

See (19) for zero-speed problem. See (112) for forward-speed problem.

equal to \)e in the sections about the near-field problem and the matching. Otherwise integration variable in.the Fourier transform.

0

- _{(2w U/g+1)}

0

length of ship

coordinate-axis in the direction of the outward normal on the wetted surface of the ship.

See (66)

coordinate-axis normal to and out of a cylinder with the same cross-section as the ship at a given section.

i = 1,2,3 : the x-,y-,z- component of the unit normal vector to the wetted surface of the ship.

r radial coordinate used in the chapter: "Numerical Calculations" (see Fig. 8). V k k2 = L n N n. 1

U forward speed of the ship

x,y,z Cartesian coordinates (see Fig. 1).( The ship moves in the direc-t-ion of the negative x-axis , z is measured upwards, y to

starboard.).

x,Y,z =

sl

Stretched coordinates (see (66) or (169)). V-k

where k is an integration variable very small positive number

very small positive number

very small positive number

(x,y,z,t) velocity potential in forward-speed problem and inzero-speed problem *

1(x,y,z,t) velocity potential of the incoming wave D(x,Y,z,t) velocity potential of the diffracted wave

*

Note, however, that it means the time dependent part of the velocity

poten-tial in the Chapter "Numerical calculations".

vi

C slenderness parameter

(x,y,t) free-surface displacement

0 angular coordinate used in the chapter: "Numerical Calculatiàns" (see Fig. 8). 0 = 0 is a point on the centerplane.

x _{wave length of the incoming wave}

11 fictitious (Rayleigh) viscosity. (Note that 1(arg) has another

mean-ing.)

0 2W

p density of water (mass per unit volume)

x)e source density per unit length in line distribution of sources.

=

I decy(x)

-wU

T=

(x,y,z) see (16) and (17) for zero-speed probleixt. See (109) and (110) for forward-speed problem.

ip(x,y,z) see (62) in zero-speed problem. See (164) in forward-speed problem.

N

i=l,2,...N

,

w wave frequency of the incoming wave

() +vU 0

The intention in this thesis is to derive a method to find the pressure distribution along a ship due to head-sea regular waves. The ship is res-trained from oscillating,, and the ship meets, the waves with a high frequency.

The wave length of the incoming waves is assumed to be of the order of Inagni-tude of the transverse dimensions of the ship. For the zero-speed case, the frequency of the.wave is the same order of magnitude as the frequency of

os-ciliation used in Ogilvie & Tuck (1969). Ogilvie & Tuck considered the forced

heave and pitch oscillation of a ship when there were no incoming waves. Due

to linearity, the forces obtained,in Ogilvie & Tuck and the forces obtained in this thesis can be superimposed to give the hydrodynamical forces on a

'ship which oscillates in, a. steady-state condition in. regular head-sea waves. In the forward-speed case, the assumptions in Ogilvie & Tuck (1969) and in this thesis are different.

Ogilvie & Tuck got a strip theory result and it is well-known that strip theory gives good results for a wider range of wave lengths than Ogilvie &

Tuck restricted themselves to (see Saivesen, Tuck, & Faltinsen (1970)). So

it is the hope that the theory presented in this thesis also will cover a

wider range of wave lengths. But it is only our experience that 'is going to

tell us for how large wave lengths our theory is capable of predicting the

pressure distribution along the ,ship. The theory predicts that head-sea

waves of small wave length are deformed as they propagate along the ship.

Abels (1959) observed this fact for a wave length which was half of the

length 'of the ship, but he did not observe, it for a wave length which was

three-fourths of the length of the ship.

By integrating the pressure in an appropriate way over the submerged part of the ship, we are able topredictthe exciting force and moment on the

ship. For the zero-speed case 'there is another way to obtain the exciting force and moment on a ship, namely to use the Khaskind relation (see Newman

(1962)). The disadvantage of the Khaskind relation is that it cannot predict

-1-the pressure distribution along -1-the. ship. _{Further it is a formula derived}

on the basis of a general mathematical relationship,and so it does _{not give}

us much insight into the physical problem.

The method of matched asymptotic expansions has been used _{in solving}

our problem, and an important part of our solution: in the near-field _problem

isUrse].l's solution (1968 a) of a closely relatediroblem: _{He obtained a} general expression describing wave motions which can exist in the presence

of an infinitelylong horizontal cylinder, the wave motion being periodic

along the cylinder. Our solution for the total potential in the near-field can be written as Ursell's solution multiplied with a function of _{x ( x is}

the longitudinal coordinate. See Fig. 1). _{The function of x contains the}

factor (x + L/2)1"2, where -L/2 is the x-coordinate of the forward

perpendi-cular. _{Our solution is not assumed tO be valid near the bow or stern. Ursell}

was somewhat discouraged with his solution because it became unbounded laterally

at infinity (when y + ± . See Fig. 1). But in our case this does not matter

because Ursell's solution is only a part of our near-field solution, _{which is}

not assumed to be valid at infinity. The only important thing is that _our near-field solution should match with the far-field solution, which it _does.

II. GENERAL FORMULATION

The coordinate system which is going to be used is shown in Figure 1.

FIGURE 1

The coordinate system is fixed to the sp The plane .z = 0 represents the undisturbed free surface. The z-axis is positive upwards.and the

posi-tive y-direction is in the starboard direction. It is assumed that the ship

moves with a constant velocity U in the direction of the negative x-axis. Since we will refer everything to the coordinate system in the ship it will

look as if 'there is an incident, undisturbed flow with velocity U in the

direction of the psitive x-axis.

It is assumed that the fluid is incompressible and the flow irrotational, so.that there exists a velocity potential which satisfies the three-dimen-sional Laplace equation,

a2 +

ax2

ay2 az2

in the fluid domain. The ship is restrained from performing any, oscillatory rnotions, and so the boundary condition on the wetted surface of the ship will be

- _{0 on z = h(x,y)}

. .' (2)

Here z = h(x,y) is the mathematical description of the wetted surface of

the ship. a/an denotes the derivative in the direction of the outwards nor-mal on the surface of the ship.

The conditions on the free surface, z = _{(x,y,t), are,neglecting surface} tension,

the dynamic free surface condition

g +

'1/2

[2

+

= 1/2 r.i2

on z = (x,y,t) (3)

the kinematic free surface condition

xCx

₊

_yy

- + 0 on z = C(x,y,t). (4)

g is the acceleration of gravity.

0

We must also satisfy a radiation condition. _{We will be more specific} about that later.

It is assumed that the fluid has infinite _{depth, the free surface has}

infi-nite extent, and there are no bodies other than the ship.

We will assume that there are incoming, regular gravity waves propagating

along the positive x-axis. _{The wave amplitude is assumed to be}

small so that

the classical linear free-surface theory is _{applicable. We will later} linear-ize the problem with respect to the wave amplitude. _{The potential of the}

incoming waves will be given by

=

{

e

elt - Vx)

] (5)

Here Re means the real part. _{As is usual we are going to drop}

the notation

Re.

We will write the potential in complex _{form, and it should be understood}

that we should take. the real part-. _{This is only a matter of}

convenience.

h is the wave amplitude,

is the wave frequency, V is _{the wave number,} w the frequency of encounter and t is the time _{variable. The wave number '.}

can be written as

2

A _g (6)

where A is the wave length. _{The relation between}

_V,U,w,

and U for head-sea

waves is

w = w + VU

0

We will assume that the ship is slender, _{and we will introduce the} slenderness parameter . It is a measure of the transverse dimensions

of the ship compared with the length of the ship. _{So C is a small quantity. If we} denote the x,y,z-coinponents of the normal _{n on the wetted surface of the ship}

by n1, n2, n3 respectively, then we _{can set}

2

-w +

g

=

x2

n1 =

0(c),

"2 =

0(1),

n3 = 0(1) (8) We will assume that the frequency of the wave has the following

asyinp-totic behavior

U)0 =

o(c1"2)

(9)

Using (6) this means that

= 0(c) (10)

III. THE ZERO-SPEED PROBLEM

The frequency of encounter, w, is the same as the frequency of the waves, for the zero-speed case (see (7)). The time dependence of the incident

wave is given by

et

(see (5)). It is expected that the time dependence for

the total potential is also given by e1Wt. This means that

/t is

equiva-lent to multiplying by iw. We will use this fact and now write down the equa-tions to determine the velocity potential. FrOm (1), (2), (3), (4) and the

assumption about linearity it follows that satisfies

- 0 in the fluid domain,

0 on z = 0 outside the body.

(1)

In addition, the diffraction part of the potential must satisfy a radiation

condition. We will write as + D = (12) -0 on z =

h(x,y),

Here

D denotes the.dif fraction potential. To find _D we are going to

use the method of matched asymptotic expansions (Van Dyke (1964), Ogilvie

(1970)). As is usual, we introduce a far-field description and a near-field

descr-iption. _{The far-field description is expected to be valid at distances} which are 0(1) and larger from the ship. The near-field description is valid

near the ship at distances which are 0(E).

There are four parts in this chapter: (1) derivation of the far-field

source solution due to a line of pulsating sources located on the x-axis

between -L/2 and L/2 (see Fig. 1); (2) derivation of a two-term inner expansion of the far-field source solution; (3) comparison of the expression found in part (2) with the result obtained by another method; (4) formulation

of the near-field problem, and the matching of a two-term near-field solution

with the far-field solution.

1. Far-field source solution

In the far-field description, we expect to have waves. In order to have

waves, we must satisfy te condition (11). This means that the two terms in (11) must be of the same order of.magnitude in the far-field, and so

=

0(E1).

The existence of a surface wave implies that a/az and,say, are the same order of magnitude, where s _{is measured normal to wave}

fronts. 1n the far-field, we cannot in general say that the normal to the wave fronts should be neither along the x-a.xis nor along the. y-axis. _This

im-plies that

/x and

/y must also be of order

E 1 in the far-field.

From a far-field point of view, one cannot see the shape of the hull.

As C -'- 0 the disturbance from the hull to the far-field seems to emanate from a line of singularities located on the x-axis between -L/2 and L/2. The

dominant far-away effect is expected to appear to be due to a line of sources.

i(wt-Vx)

Since the incoming waves vary as e , it is expected that the line of

sources has a source density of the form i (wt-vx)

Here 00

r

iiy + .lim

de

!ook+

£2 1 --iVx

Fi(x)e

D + + - a(x)e

Here _{is the Dirac-delta function, and initially we take z0< 0. When the} solution of is found, z will be set equal to zero. If we set

z0

= 0 first, we would be in difficulties solving the problem.

We cannot expect that the far-field solution will satisfy the boundary con-dition

on

the hull given by (2), but it must satisfy a proper radiation condition.

We mustbe sure that the diffraction potential _{does not contain an incoming}

wave. This is most easily taken care of by introducing the artificial _Rayleigh viscosity i (see Ogilvie & Tuck (1969)). _{The free-surface condition (11) will}

then be modified to

(iw+M) + g - 0

on z=0

₍₁₄₎

At an appropriate later point, we will let 1. go to zero.

The solution to (13) and (14) with

z0

= 0 can be found in Ogilvie & Tuck (1969) and is

DYz1t) =

1 iwt

_fdk

F {a(x)e"}

-00 i (wt-vx) -ikx

-iUx

dxe cl(x)e = (13) (15)

Expression (15) can be rewritten in a form which is more convenient

to handle for our purposes.. I will follow a procedure described by Ogilvie (1969), and what I am going to do is based on his work.

We first introduce

k' k + -V

in (15). We can then write

= -urn 31' +0 1 i(wt-vx) 4rr2

I

iiy

+

zv'(k'-)2

+ £2 :00 1/(kt_V)2 + £2 _-

(-if

Here 3i' = We drop the primes and write

i (wt-vx) = (x,y,z) e where 00 (x,y,z) =

42

_L

dk 00 (17) r iiy + z/2+ (v-k)2

lim

die

-

v+(v-k)

- (v-i31)

We will let y = 0(1) and we are going to assume that y > 0. The derivation for y < 0 will be quite similar, and we are not going to _go

through that. Since we are operating with sources, we can later _{use the}

fact that

dk' elk'X G*(k')

(16)

We now define

1(k) = lixn

iO

- (ViJJ)

The poles of the integrand are important in the evaluation of I (k). They are given in the limit i -'O by:

£2 _{= (2\.-k)k.}

t.

QVIc-_k

Let us first study the case in which these. singularities _{are imaginary, which} means k < 0 or k > 2v. Then we study case II, in which 0 < k < 2v.

Case I: k < 0 or k > 2\.). We define

£ = iv'kk-2V)

0

We introduce a closed curve ABCDEA in the complex _{i-plane, as shown in}

Figure 2.

d2,

i9y +

zv'2'l-(v-k)2

B FIGURE 2

COMPLEX INTEGRATION PATH

The integrand in (19) has a branch point at ijv-kI , and we choose the branch

cut along the positive imaginary axis. _{By using the residue theorem, letting}

C+-Fi, E-'+i, A+oo, wewill get

1(k) 2Tti Res

r

izv'2,2-(v-k).2 +

I

idLe

y1e

V-kI

L j/L2-(v-k)2-v

a

f

dL ± i (v-k)2

-u/i?.2- (v-k)

2-v

dLe

This can be written as

rdL

_-

_[

'

e/'m22

=1

UdK

(22)

J

/p,,22

v'm+a

₀

a

See Abramowitz and Stegun (1964). K0 _{is a modified Bessel function of the} second kind. Since V =

0(C1),

_y

=

o(1)

and

a > 1, the argument of Kis

large. By using the asymptotic expansion of K0 for large _{arguments, we can} show that the integral term in (20) is

bounded by a quantity which is

O(e1).

So we can

write

- yvk(k-2v)

1(k)

2TrVe

+ 0(e1)

&(k-2v)

for k < 0

or

k

> 2v, and

y=0(l)

(20)

(21)

(23)

27tVe

- yVk(k-2V)

reivzi.2c2

+

e/'L2

V'k(k-2)

j

_{/L2a2 -}

i

a

where a _{a will be greater than 1, since we are considering the case}

k < 0 or k > 2v.

We want to show that the integral in (20) is exponentially small _as

case.II: 0 < k < 2V. The poles of the integrand of (19) are now real.

The Rayleigh viscosity will help us to determine how to indent the _integration

path of 1(k) around the poles. If we define

the poles will be at ± _when i = 0. _{The integration path will be as shown} in Figure 3.

-0

FIGURE 3

COMPLEX INTEGRATION PATH

In the same way as we did for Case I, we introduce a closed _{curve ABCDEA in} the complex £-plane and use the residue theorem to evaluate 1(k). The closed curve is shown in Figure 4.

FIGURE 4

COMPLEX INTEGRATION PATH

Noting that

ve - iyv'k (2v-k)

Res(-L)

-0

-/k(2v-k)

we get by use of the residue theorem in the same way as for Case I that

a v-k

Here a=

-.

<1

1(k) = iyvk(2v-k) 2rtive V'k (2v-k) ivzv'2-a2

-ivzVQ,2-a2

d2, e' e + i V -. (2v-k)

+ ... for 0<k<2v

(25) (26)

As long as a is not of order C or less, we can use the same argument as for Case I to show that the integral in (26) is exponentially small with

respect to C as C + 0*. But when

a

6(c)

we see from (22) that the

argu-ment of the bounding function K0 is of order 1. We therefore have to use another procedure. We can write 1

I'

d2,

eV7

e±ia

dJ e'

e7C

= 0(c) J _V'5t2-a2

±i

a a

When a = 0 it can be shown that the integral term is of 0(c). Since k is

0(u) = 0() when

a = 0(c) or less, it is easily seen that the first term

in (26) is 0(1). So the integral will be of higher order than the first term in (26) when

a

= 0(e), as well as when

a.

0(1).

We now have the re-suits

(0<y

0(1)). 211\)e - yv(k-2V) k< 0

+...

for lk(k-2v)

k>2v

I(k)= (27) 27riVe - iyv'k(2V-k)

* _{It will be evident later why k is related to}

c,

_{and therefore why}

a

By using equation (17) and (19), we can write

f

(

ic

eX -

ylk(k-2V)

G*k)

'(x,y,z)

_-

t i 27r

_V'k(k-2v)

e1 -

iyi/k(2v-k) a*(k)

We rewrite

(28)

as (x,y,z) =

-3-I.

2\

ikx dk e ,'k(2v-k) -

yv'k(k-2\)

/k(k-2v)

+ higher order terms

Ve _r

_i

_dk

- yifk (k-2v)

_{a* (k)}

-

27r

/k(k-2v)

0

I-..

+ -

2v)

cy*(k)

r

dk

e1

---(1--)

Vc(k-2v)

I

dk a

ikx - iy/k(2V-k)

/k(2v-k)

2'J

_ik

- iyVk(2v-k)

dke

-2..J

_6(1(51)

_Vk(2v-k)

+

f

dk -

yv'k(k-2v)

00 _/1.. Il..

-.

2v

VA.

Here i is some very small positive nuiriber. _{It will be evident in the next}

-(l-)

section why we have introduced C

It can easily be shown that the first integral in

(29) is

_{exponentially}

small. So we drop that term.

a* (k)

* (k)

+

£

y* (k)

]

+..

:i

+

I.)

This is the final form of the far-field source solution. It is valid for y = 0(1).

2. Inner expansion of far-field source solution

We are now going to find a two-term inner expansion of the far-field

source solution. The result is given by equation (54).

We now let y be of order . , and we reorder the terms in (30).

In the first integrand we want an expansion of

e1k (k-2v)

(31) i/k (k-2V)

It will now be evident why we introduced c in the previous section. Since k canriot.be greater than yV'k(k-2\)) will be o(1). We expand (31) and keep two terms

We prefer to write the last two terms in another way by introducing the new variable u = k-2v. After introducing the new variable in the last

integral, we drop the contribution from integrating from to

the argument is. the same as that used above in dropping the first integral in (29). Equation (29) can now be written as

ve

_f

ãj - y/k(k-2v) 27T j

J_(l_s1)

V(k-2v)

-(l-cS) öj - iy/k(2\)-k) -i (2v-k) . - (30) du e1 + i2Vx - iYV'(U+2V*U),*(+2) i/(u+2V)(-u)

r

_due

iux + ivx - y/(u+2v)u

_*(u+2)

1

= y +... (32)

/k(k-2v) v"k.(k-2v)

The omitted, terms are higher order. Further, we expand

1 0

-f

ikx (k)1 1 2ir

dke

-(l-i)

L/2V1k1 -c

In a similar way we will find that the second integral in (30) can be

written as

/k(k-2v)

We can then write (31) as follows

1

k

=

Il+-- f...

-y+...

Vk(k-2V) /2\)JkI L

We want to keep only the two lowest-order terms in the first integral in (30). k in the first integral will be

O(Eahl)

, where ci can be greater than or equal to 5 , depending on k. So the first term in (33) will be

o(clh/'2) _{the second term} o(cl+CL1/2) _{and the third term}

o(E). When

(33) is put into the first integral in (30), the second term in (33) will give the highest-order term. The two lowest-order terms in the first integral in

(30) will be

iT

ikx

dk e *(k)

(33)

We are going to follow the same procedure as above to evaluate the third

and fourth integrals in (30). It will be evident later that the latter will give a contribution which is of the same order of. magnitude as the higher-order terms in (34) and (35). We will therefore keep only the lowest-order

terms in the two last integrals. But we must be careful with the third inte-gral. The lower integration limit is 0(V) = O(c For the integration

variable u of order V we cannot truncate the series expansion of 0

27r

I /2Vk

(35)

and -2V ye _ei2 27r

I

e12 C-u)

i/(u+2V) (-u)

after two terms in the same manner as we did in (32). But let us set the

-(l-6)

lower integration limit in the third integral equal to -E . The

dif-ference between iux _ly du e e /(u+2V) C-u)

*(2)

/(u+2V) C-u) cY(u+2V) ±UX .

r

du e e

J

-(l-)

V(u+2V) (-u) -E

is higher order. As long as we only want the lowest-order term.of (36), we should be safe in changing (36) to (37).

We now do for the third integral as we did for the first integral.

will get 0 .Vz ei2Vx

due

J

iux a*(u+2\)

-(l-5)

/2vuj

And for the fourth integral we will get

iux a*(u+2v) du e

By now putting the expressions (34), (35), (38) and (39) for each of

the four integrals into (30), we can write

(36) (37) (38) 0 --' (39)

and 211 0

I

dk e (x,y,z) = L

2W

I_(l_&i)

ikx a*(k) V2v k

-C

dk e

I

ikx + 2 1

I

0

dk ea*(k)

+ e'

e2

f0 du

eUC

a*(u+2J) 27r v'2\)IuI ye. e'2

+ higher order terms

Note that the first brackets contain the lowest-order terms, the second

brackets the next-lowest-order terms.

Because we want to apply Fourier-transform techniques, we want to set C (1-cSi) equal to °°. _{For the three higher-order terms in (40) we could do}

that; the effect would be only to introduce higher-order, negligible effects. But we must be careful with the lower order terms:

jux du e

dk ei a*(k)

v12\)Ik.I

We will consider especially the integral (42). We can write it as

(40) (41)

/

dic e a*(k) (42) * (u+2V)

I

ve 211 y

r ikx a*(k)

f

ikx I

dke

j

dk e

v'5i

-(l-5)

0

We will try to find, the order of magnitude of the second integral. _{We then}

need to know how a*(k) _{behaves for large k . We will asSume that a(x)} and

a'

(x) are continuous in the interval -L/2 x L/2 [including the end

*

points] (see Fig. 1). Outside -L/2 <x < L/2 , a(x) = 0 . It can then be

shown (see Lighthill (1958)) that. IkLI3O*(k) remains bounded as k + ±

So the second integral in (43) can be bounded by

dk a*(k)

-

---l (1-cSi)5/2

IT

_-(l-1)

k7'12 5V'

C1 _{is a constant determined so that the inequality above is satisfied.} If we now put (43) into (40) and _{use the estimate above of the order of}

mag-nitude of

f

ikx a*(k) dk e

it should be obvious that we can replace

dk a*(k) V2\k

f

e

0

in (40).

In a similar way, we can show that we can replace

(43)

* _{It should be noted specifically that we assume a(±L/2).= 0}

, a'(±L/2) = 0,

and continuity in the neighborhood of x = ±L/2.

I

by

in (40).

We can now write (40) as follows:

0 V \)z

I

r

dk e1 a*(k) (I(x,y,z) =

- - e

I I 2ff

J

/2vJkJ 0

fdke

i/2Vjk 0 1. ikz a*(k) _.L dk e v12VJkJ -y

I

i2Vx -i.e

I

k>0

jux du e V2vfuj +e12'

f

du a*(u+2v) O*(u+2V) ₍₄₄₎ 0 /2Vu

+ higher order terms.

This is now in a form that can be greatly simplified through _{use of the}

pro-perties of Fourier transforms.

We define

k< 0.

F*(k) (45) Co

if

0

F* (k) denotes the Fourier transform of a function F (x). So = dk

e3

i dk 21Ij - 0 So (x,y,z) = -- 1-i 27r

I

-i'rr/4 = k.t/_ii

{l+sgnx}_e

_H(x) 27r 21x1 vrrIxI

where H(x) is Heaviside step function.

We also define 00 ye 2Tr/ Vz i2Vx \)e e 27r/ 0 00 r )UX

G(x)=

__2-_j

due

+ r

due"

211j

By using (45) and (47) we can rewrite (44) as follows:

dk

eC

cy*(k)F*(k) + \)yeyz 0(x) 00 jux

J

du e *(u+2y) G*(u) -00 {cöskx + sinkx} -irr/4 cosux - sinux} e _H(-x) v'ir x (46) (48) (49) 1

u<0

G*(u) =

4T

₍₄₇₎ 1

U>

0

We now apply the convolution theorem to (49) and use (46) and (48). We get -iTr/4 Vz i2Vx

f

-i2V e - ye e dEcY()e x Vz -iir/4

f

d o()

= -Ve e V2iTjx-+ \)y e 0(x) -L/2 L/2

vz i2yx -iir/4 r

da()e12

-ye

e e

In the last expression we have used the definition of the Heaviside. step

function and the fact that the source density 0 _{is zero outside the ship.}

The integral L/2

I

-i2V

dO()e

x

can be further simplified. It is a Fourier integral. The integrand has a

singularity at the lower integration limit, x , and it is zero at the upper integration limit. (See discussion following (43).) _{The theory for finding}

asymptotic expansions of such integrals can be found in Erdélyi (1956). _We find that (51) can be written

L/2

d0()e12

=

i-;;-e2

e'4 0(x)

+ Q(!) (52) x

So, by putting (52) into (50), we get

(x,y,z) = -ye

_{f do()}

el/I4

H(x-) + \)ye 0(x)

-G

D1Y1zt)

et_

e

[e_i1r/4

f

+vIyja(x) -

e'2a

]

3. Comparison with another method

The method applied above by using Fourier transforms is a very good

-method and has been applied to many problems in ship hydrodynamics. (Ogilvie (1973)). Other methods can also be applied to find inner expansions of

far-field expansions, but generally speaking it is very difficult to obtain the inner expansion by other methods than the Fourier transform method as

used above. Sometimes it even seems to be impossible.

Below I will attack the above problem by another method. I cannot

Obtain all three terms in (54), but I will obtain the two terms which remain when y = 0. These two terms will agree with the two relevant terms in (54). This is a good Oheck on our result.

In Ogilvie & Tuck (1969), it is described how a. far-field solution of the velocity potential, c(x,y,z,t) , of a line distribution of sources of

da()

-L/2 v'27rvJx-I

(53)

(54)

(x,y,z) =

en/4

c)

+ vye G(x)

--I/2 V27vIx-J

2

+ higher order terms.

We remember that we have assumed y > 0 . But we can use (18) together

with (16),

= (x,y,z) ei (wt-vx) ₍₁₆₎

so that we can write down a two-term inner expansion of the far-field source

solution

density p(x)e1Wt spread along the line y = z = 0 , -L/2 < x < L/2

in the presence of a free surface, can be written as

iwt

(x,y,z,t) = Re[4(x,y,z)e ] , (55)

where

L/2 00

(x,y,z) = -2 f

_{di() j}

dkke

J0

[kV'(x-)2+ y2)

(56)

(55) is a solution of (1) and (11), and it satisfies a proper radiation condition. The inner integral is a contour integral, indented at the pole

as indicated.

J

is

a Bessel function of the first kind.

In our problem we must set

-iV

4Tr1J()

= a() e

, (57)

in accordance with what we did in the previous chapters. The factor 411

is needed: because of the different normalization of source strength here and in Ogilvie & Tuck (1969).

Ogilvie & Tuck (1969) simplified the expression (56) for

_{y =0(1)}

_and

we only state the result here. We get: L/ 2

\)z 111/4

(x,y,z)

e -.

f

d

a()er ei+

z2 -L/2

((x-)2+

2)l/4

We want an inner expansion of (58), which means that y = 0(c). It is

diffi-cult to do anything with (58) when y 0(c) , but, if we simply set y = 0

we can get a special case of the inner expansion of (58). _{By setting y = 0}

we get

_Je"4

[

.Ix

dO()1

L/2 -L/2 _x (58) (59)

By using (52) in the last integral, we can write (59) as -

e" {

X +

e11'4

_0(x) ] (60) -L/2.

By combining this with (55) we get two of the terms in the inner expansion

of the far-field solution:

{_iTr/4

X

d0()

_{e2 o(x)}

]

'L/2

I2wIx-I

We see that (61)agrees with the two terms in (54) that remain when _{y = 0.} Since we set y = 0 , we could not get the y-terin in (54), and, as said above, it is very difficult to get the y-dependent term by this method. I will not

go further with this method, but I think we have a good check on our result.

4. The near-field problem and the matching

We are now going to formulate the near-field prOblem and perform, the matching 'between the near-field and the far-field solutions. A one-term

far-field solution is found to be due to a line of sources with source density i (Wt-Vx)

Gi(x) e

spread along the line

y =z

0, -L/2 x L/2 (see Fig.l), and a

one-term near-field sOlution will be found to be the negative of the incident

wave. The matching between the lowest-order term in the near-field problem and the lowest-order term In the far-field problem gives an integral equation

for 01(x) (see (78)).. 0 is needed in the second order near-field,

solu-tion. The two-term near-field solution is given by (94).

It should be noted that "Near-field"

means

the region near the body,

where the distance from the body is 0(c) . However, we do not expect the

near-field approximations to be valid near the bow and stern.

We will express the potential of the diffracted wave as follows:

i (wt-vx)

= e i(x,y,z)

By putting (62) into the Laplace equation, (1), we get

+ _2iV

-+

--4

= o

y2

z2

x 3x

in the fluid region. _{The free-surface condition, (11), is:}

-- = 0

onz=0

The body boundary condition, (2), together with (12), gives

gh Vz - ivn T _{+ n} --j = (iVn1 - Vn3]

_{- e}

[n

+fl

1

13x

2ay

3 ₀₎ on z = h(x,y).

and n3 have been explained before equation (8). _{A last} con-dition on is that it must match with the far-field solution.

We will assume that ij.i varies very slowly in the x-direction _compared

with the variation of in the transverse plane. We assume that the rate

of change of in the transverse plane is governed by the order of magni-tude of the transverse dimensions of the body and that the _{rate of change}

of ip _{in the x-direction is governed by the order of magnitude}

of the longi-tudinal dimensions of the bOdy.

We therefore stretch the coordinates

y=Y , zZ

_IlDEN

,

x=X

₍₆₆₎

Here _/an2D denotes the directional derivative normal to and out of _a

We assume that, inthe near-field

-= O(ij) , -

O()

, - = 0(ip) ,

-. = 0()

(67)

It should be noted, however, that the rate of change in the x-direction

of the diffraction potential, , as given by (62),is of the same order

of magnitude as the rate of change of in the transverse dimensions.

We will assume an asymptotic expansion of of the form

i(X,Y,Z;c) (68)

where

= o()

as -'- 0 for fixed X,Y,Z. By putting (68) into

(63) we get (69) 2 2 .

v22)

( l4I] + 2 = 2iVE2 l

The free-surface condition, (64), gives

2

'l +

2

onZ=0.

(70)

The body boundary condition, (65), becomes

1 a

...

S a (71) gh Vz e on z = h(x,y).1

+iVn e -Vn

_1w

3 We introduce . =.k =:0(l). (72)

We will set

Since the problem is linear in C , we shall not be bothered with the order

of magnitude of C

The lowest-order equations become:

k2) = 0

(--k)1=0 onZ=0

- Cn3kekZ _{on the body.}

In addition, _{must match with the far-field solution.}

A one-term far-field solution is assumed to be the potential associated with a line distribution of sources of density

i (wt-vx) e

spread along the line y = z = 0 , -L/2 x L/2 . That solution has been

obtained in a previous chapter, and a one-term inner expansion of a one-term

far-field solution can be found from (54)

For any fixed x

_{greater than}

-L/2 , it is. obvious that a one-term inner expansion is

et_

[ c_l/2

eer'4

.rL,2

dc11()

] (77)

and that the second-order term in the inner- expansion is _{of order} l/2 compared with the first-order term. .

i (wt-'x)

(77) should match with a one-term outer expansion of i, e

* J.

as determined from (74), (75) and (76). Ursell (1968a) has given a solu-tion to that problem, but it does not appear to match with (77). However,

if we say that the one-term near-field solution is just the negative Of

the incident wave .(thi. is a special case of Ursell's solution), then (74), (75) and (76) are satisfied, and if we require that

e

e"4

kZ = -Ce

We solve (78) for 1(x) formally by letting it be an equality for

all x . -L/2 . We recognize (78) as Abel's integral equation (see Dettman

1965)), which has the solution

- c

Trk(x+L/2) e C

1/2 _f 2 irr/4

(80)

This solution is singiilar at x = -L/2 , which is a violation of the assump-. tions made earlier. However, this is not a serious difficulty i-f we do not try to use our results very near the bow of, the ship. The near-field expan-sion of which (79) gives the first tert,is. hot uniformly valid near x = -L/2.

In order to examine.the solution precisely in the neighborhood of x = -L/2 ,

we should construct a separate expansion for a region in which x + '12 = O(E'), for some y > 0 . One may xpect then that 1(x) is. not given in that

region by (80); rather, Cx)

will

decrease continuously to zero at x = -L/2,

rX dc1()

J

JIx-I

-L/2

kz = Ce

then we see that a one-term outer expansion of the one-term near-field solution matches with a one-term inner expansion of a one-term far-field solution.

So we have the solution

(X,Y,Z;C) = -C1"2 e

e"4

-L/2

*

Ursell's solution will be needed in the second order tern and will be

dis-cussed then.

(78)

as physical considerations require that it must. Using (80) to express

produces a higher-order, i.e., negligible, error in the velocity po-tential, provided that we restrict our attention to a region in which

= o(x+L/2).

We wish next to find

2 , but first we need to say some more about the far-field.

We expect that a two-term far-field expansion is obtained by a line dis-tribution of sources of density

(x) + 02 (x)) e (wt-Vx) (81)

spread along the line y = z = 0 , -L/2 x L/2 . It is assumed that

02 = o(a)

(82)

A two-term inner expansion of this- two-term far-field expansion can be

ob-tained from (54), and it is

eWt

[

ekZ en'4

do1()

2ir

_l/2ri

e

e1V4

f

da2()

2rr x (83) kZ i -iir/2 01(x) kZ 1 +e kjYIa1(x) -e 2 e j

Let us now look at the second term,

2eWt

, in the near-field. Since = 0 , it follows from (69) that

2 has to satisfy the Helxnholtz.

equation - - -

-2 2

From (70) it follows that satisfies the free-surface condition

In the body boundary condition, (71), we know that

iVn11

cancels

iVn1()e

. Further = 0 ,

and

so the only possibility is

0 on the body

In addition must match with the far-field solution.

Ursefl (1968a) has derived a solution of (84), (85) and (86). It

can

be written as = A2(x) [ e ]41(s;k) [ G(kY, kZ; kg(s), kfl(s)) c(-I-) + G(kY, kZ; kg(s), kr(s)) ] _{ds ]} where

onZQ.

; (85) G (kY, kZ; k, kfl) = K (k V1

(y)

2 (z-)2] + -L

1cosi_±

41J

J

Jcoshii--

-The synbol denotes integration along a contour passing below the double

pole at p = 0 , with a corresponding. meaning for . K0 is a modified

Bessel function of the second kind. In (87)

c(+)

denotes the half of the boundary curve of the submerged cross-section at x for which y > 0 ;

(86)

(87,)

(88)

Y = c(s)

,

z = fl(s)

are the parametric equations of the curve

_c(+).

i(s,k)

_{is determined from satisfaction of the body boundary condition}

_(86).

It should be noted that (84), (85) and (86) will be satisfied for an

arbi-trary

A2 (x)

in (87).

_{The reasons are that 1) none of the equations (84),}

(85),and (86) involve differentiation with respect to

x ,

and

2)

(84),

(85), and (86) are homogeneous.

A2(x)

_{has to be determined by the matching}

procedure.

In order to match, we need

an

outer expansion of (87).

Ursell (1968a) has

done that.

The result is

kZ

- 4k2Tr

jj

kZ

kr(s)

A2 x)

[e

e

J

J(s,k) e

ds]

C(+)

A three-term outer expansion of the two-term near-field solution,

(pl

i (ut-\x)

(90)

can now be written down.

It is

4kTr

e

+A2(x)

ekZ

_{- A2(x)}

2 JYI kZ

J p(s,k) ek5)

ds J

C

C(+)

The last tezm is the lowest-order term and the first term is the

_next

lowest-order term.

(91) should match with (83) and we see that it does if

_{we set}

(x)

A2(x)

=

4kTr f p(s,k)

ds

e

(wt-\x)

_{-C"2 ! ::j-}

ekZ

-iTr/4

l/2f

2

iir/4

1Tk(x+L/2)

e

C 4kTT

f

p(s,k)

ds

f.

-L/2

doj ()

v'jx-j

(89)

(91)

(92)

(a1 is given by (80).), and if also

El/2

e14

£/2

+ = A2(x) + e

=(

11/2 c11(x) 2 1 + 4klr

f

.i(s,k)

ds

1/2 L2

irr/4 lTk(x+L/2) e C

which is a condition to be satisfied by a2 . Equation (93) gives us Abel's

integral equation, and it can be solved in principle. It is to be noted

that the term in the brackets is a function of x , which is determined in practice by numerical computation.

It is the near-field solution that has the primary interest. So let us summarize our result: A two-term near-field solution of the diffraction potential is given by 112 _f 2 irr/4

et

=

et_

_1Ce

rrk(x+W2) e C e_iTt'2 2 (93) 4kTr

f

i(s,k) ds

d+)

(94).

[e -

f

ki(s;k) f G(kY,kZ;k (s) ,kTl(s)) + G(kY,kZ; k Cs) ,k (s))J ds]

C(+)

where k and C are given by (72).and (73), and G is given by (88). The first term in (94) is just the negative of the incident waye and so (94) tells us that the total (incident- plus diffracted-wave). potential near the body

(x+L/2)-1/2 (95)

in the x-direction. But note that L in (94) is also a function of x

and so (95) does not give the total x-dependence. However, i will be

the same for similar cross-sections. So, if the cross-sections are not varying much in the x-direction, the potential will, roughly speaking,

drop off with the factor (x+L/2)1"2 in the lengthwise direction. Note that we have assumed that the wave length is of the order of magnitude of the trans-verse dimensions of the ship.

IV. THE FORWARD-SPEED PROBLEM

We write the total potential as follows:

(x,y,z,t) = Ux + U(x,y,z) + _T(xy,zft)

, (96)

where U is the perturbation velocity potential in the steady motion

problem. satisfies the three-dimensional Laplace equation

+ +

____ -

0 (97)

ax2 az2

in the fluid domain and the body boundary condition

(Ux+U) = 0

on z = h(x,y) (98)

Since satisfies (1) and (2), this implies tat satisfies

a2 a2

T T T

-0

(99)

ax2 ay2 az2

on z h(x,y) (100)

By combininq (3)

and

(4) and

using

the

asurnption about linearity, it can

be shown that

satisfies the free-surface condition

] T + g

on z0

(101)

(see Ogilvie

& Tuck (1969)). Since the time dependence of the incident wave

is given by e1Wt (see (5)), it is expected that the time dependence for the

potential

T

is also given by

e1Wt

. This implies that we can write equation (101) as

+ + g = 0

on z0.

(102)

We will write as

= + = e + (103)

where denotes the, diffraction potential. must satisfy a radiation

condition.

As in the zero-speed problem we are going to use the method of

matched asymptotic expansions to find

We will assume that

U=O(CJ/2a)

_0(al/2

₍₁₀₄₎

In the steady forward-motion problem we know that there is a length scale

in the x-direction which is connected with the wave length 21T132/g . So

(104) implies that this length scale is large compared with the transverse dimensions of the ship, and that it can be o,f the same order of magnitude as the length of the ship. I-n some way, I expect this length scale will enter

our diffraction problem and affect the rate of change of the variables in the

x-direction. But it turns out that it will-not have any influence on the first two approximations of the diffraction potential. The

iiiortant length

a4T

scale in the x-direction will be connected with the wave length of the

incom-ing wave, in the same way as for the zero-speed problem. As we remember

from equation (10), this wave length is assumed to be of order c

If, however, we had assumed that a were zero in (104), we would have been in difficulties finding the second approximation to the diffraction po-tential. The reason must be that there then are two important length scales of order C in the x-direction, one.connected with the wave length of the incoming wave and one connected with the forward speed, and it is difficult

to separate out the effect of one. of the length scales from the other.

Using (7) and (9), we can show that (104) implies that the order of

magnitude of the frequency of encounter, w will be

w = Q(

6l/2a)

, 0 < a l,'2 (105)

We then see that the order of magnitude of

T =

is

g

WU

Q(2a)

, 0 < a < 1/2 (106)

g

It is obvious that T will be larger than 1/4. This. is important,

because the solution will be singular when T = 1/4 (see Ogilvie and Tuck

(1969)).

There are four parts in this chapter; (1) derivation of the far-field

source solution due to a line of pulsating, translating sources located on the x-axis between -L/2 and L/2 (see Fig. 1); (2) a two-term inner expan-sion of the far-field source solution; (3) comparison of the expression found in part (2) with the result obtained by another approach; (4) formulation

of the near-field problem, and the matching of a two-term near-field solution

with the far-field solution.

1. Far-field source solution

In the far-field description we expect to have waves. It is difficult

to say how differentiation changes order of magnitudes in the far-field. So,

aD

ax2 ay2 az2

a 2 D

(iL+U +

have to be sure that we have a systemof equations that describes a wave

motion.

Using arguments similar to those in the section "Far-field source solu-tion" in the chapter on the zero-speed problem, we can find that must

satisfy the Poisson equation,

-. (x) e

(wt-vx)

(y)

(z-Zt)

where z < 0. We write the free-surface condition as follows:

0

= 0

on z=0

where i is the artificial Rayleigh viscosity, which will approach zero at

a proper later point. This equation system does give waves.

The solution to the equation system with = 0 can be found in

Ogilvie & Tuck (1969). It is

where -iVx F {cy(x) e } = 1 iwt (x,y,z,t) = - e D 47r2

I

I

= ikx -iVx e F{cY(x)e }

/k2+92 -

I

(w+Uk-ii)2 -ikx -iJx e a(x) e

We will rewrite (107) in a way similar to the way we did with (15) in

the zero speed problem. We; introduce:

(107)

We drop the primes and write i

(wt-vx)

DYz1t)

=

(x,y,z)

e where c(x,y,z) =

urn

Ii--0 1 47.2

urn

k' =

(107)

can

then be written:

e 1 i(wt-vx)

f

dk' ik'x = _e

I

I

-iQ.y

+ z(k'-V)

2 +

de

-.o /(k'-v)2+ (w-i-U(k'-v) -

ui)2

dk G*(k) g d2, +

zVV-k)2

+ v"(v-k) 2 (EL) + Uk g 0

r

d

e17

+

zV2,2+ (v-k)

2

1(k)

=

urn

1

ii-o

i/9,+(V-k)2-

-i1j)

2

We will let y = 0(1) > 0 The derivation for y = 0(1) < 0 will be

similar and, instead of going through that, we use the fact that _{is a} source solution, which implies that

DYz1t)

=

D1,z1t)

(111)

We now define (108) (109) (110)

(112)

The poles of the integrand of (112) are- important. They are given in the limit by:

= ± (w0+Uk) - (v-k)2

Vg2

We have to study the sign of the radicand in order to determine the

location of the poles in the complex £.-plane. We study therefore the

equation:

---(u +Uk)

- (v-k)2 =

0

2 0

g

which is the same as

(w

+uk)2

-

(v-k)]

[

1

(w

+Uk)2

+ V-k ] = 0

g 0 g -o

The zeroes of the first factor are:

k=k10

r 2 U .3_I 0

k=k2-L g

+1]

r2WU

_.j_I

0

u21

g j 2 2

But from (106) and the text that followed (106), we know that T = 1/4

This means that the latter pair of zeroes of (114) are imaginary. But k is

a real variable, and so this means that the quantity

-and the zeroes of the second factor are:

r2uu

0

_jI

\J

I 0

-

8---u2I

g g g k

u2

2 I (116)

Jl

4 g

f(k) = (w

+Uk)2 +

\)-k

g 0

can never be zero. And since expression (117) for large k behaves like

1 ₂

(w +Uk) g 0

we see that (117) has to be a positive quantity.

Going back now to expression (113) we can conclude:

,

a) when

k>k1

= 0 or

k<k2

= -

(2uU/g

+ 1)

U2 the integrand of (112) are real;

b) when k2< k < 0 _{the poles of the integrand of (112) are imaginary.}

The integration path in (112) is along the real 9..-axis, and so, when

the poles of the integrand are real, we have to know how to indent the inte-gration path at those poles. The Rayleigh viscosity _{J helps us to do that.}

For

j

0 , the poles of the integrand

are

given by

= _{(+Uk-i.j) - (k-v)2}

=

(k-k1)

(k-k2) fCk)U2/g-4(w0k)3iWg2 + o(p)

where k1, k2, and f(k) are given by (115), (116), and (117), respectively. The poles of the integrand of. (112) are then given by

= _{± /(k-k1)} (k-k2)

f(k)U2/g

[1

2 (w+Uk) _i31 the poles of

gU2 (k-k1)

(k-k2) f(k)

for kk1andkJc

(117) (118)

When k < k is positive.

in Figure 6.

FIGURE 6

COMPLEX INTEGRATION PATH

Let us now study

1(k),

, given by (112), for different ranges of

k

Case

I:

k2

<

k < k1

. This is the case in which the poles are imagi-nary. We define:

For the case k > , the imaginary part of the factor in parenthesis in (118) is negative. This means that the integration path of (112) for

k > k1 will be shown in Figure 5, where

+ I and -j9I are the poles of

the integrand

FIGURE 5

COMPLEX INTEGRATION PATH

= iv'(V-k) 2

-L+/g2

0

We introduce a closed curve ABCDE in the complex 9..-plane in the.same way

as we did for the zero-speed case. See Figure 2. The residue at 2 = is

(Th

1 z 0

k>0

the imaginary part of the term In the parethesis in (118) This means that the integration path of. (112) will be as shown

Res () =

urn (-)

V2+(v-k)2-(w0+Uk)2/g

i2., y \)z(1+Uk/W0)2

\)(1+Uk/W0)2 e 0 e

By using the residue theorem we can write

1(k) = 2i Res( ) +

f

d e 0 1 '2.2-a2 + i I.', a 2. 0 i\)zv19..2-a2 e1 + zv'2,2+

(u-k)

2 -ivzi/P..2-a2 e + 2 i/Z2-cx2 -

i[l+

-J

where a=

> 1 , since k is negative.

The integral terms in (119) can be bounded by

(

d.

e2 ±

ivz/Z2-a2

<

d2, e'

J

)/2,2-a2

±iIl+

J

/9-a

a L w0j a Using 1) equation (22) d - - -J _V22-a2 0 a

2) the fact that

v =

o(1)

, y = 0(1) ,

a >

1 , and 3)the

asympto-tic expansion of K for large arguments, we -see that the integral terms

in (119) are exponentially small with respect to C . So we can write -\)y

[ e

(119)

Case II: 1(k) = Uk'2 i2. y Vz [1 +

-j

1i+

12 e e 2.. 0 > k1 . Let us define for k2 < k < k1 2..

= -/(u

+tJk)4/g2 - (v-k)2 0 0

We have earlier studied how to indent the integration path of the expression

for 1(k) at the poles ±2. . See Fig. 5. In the same way as before, we

introduce a closed curve ABCDEA in the complex 2.-plane We can use Fig. 4,

noting

that - in Fig. 4 is the same as defined above.

+

By the residue theorem

1(k) = 2iri Res (2.) +

[

d.Z e 00

r

iVZ/2.

-vy2. e & L,/2.2...2 + _.j 2_,,,2 + e v12.2c2 -

jfl +

12 I.

WJ.

0 -2

As before,

K(vya)

is a bounding function for the integral terms.

But when k 0(v)

0(c1)

= is of 0(c). This means that the

argument of the

K

function is 0(1), and so we cannot say that the

inte-gral terms are exponentially small when c = 0(c) , and it does not seem

(120)

(121)

We have to evaluate the

Res(2.)

0 residue

VIl+le

at 2.=9 0 r Uk 2

izy

L

wJ

_-0 . ' e It is given by

uk1

2 W0J

probable that there exists any such bounding function when ci. = 0(6) . We

should note that

Uk -a

= 0(6 ) ,

0<a<l/2

w 0

when k =

0(61)

_{(see (104) and (9)). This means that}

r

Uki2

I 1 +

-L

e 0

is exponentially small when k = 0(6_i) _{, which makes the first term in} (121) exponentially small when k = 0(6_i) . So we should be careful to

1(k) =

+ exponentially small terms

for 0 < k 6-(l-a-f-62)

0<a<l/2

is some very small positive number)

For all other values of k in Case II, 1(k) will be exponentially small.

keep the integral term in (121) when k =

e0

0(6_i) {i + We Uk wo J can write 2 27riV [1 +

is some very small positive number) ₍₁₂₂₎

- iz,/2.2-o

I(k)=

I

d9eW 1e

L/Q.2_c2 ₊ UkJ2 _{)'2-ch2 - i [1 +}

+ exponentially small terms

Case III: k < k2 . We define

(Th

9,.. = v"(W

+Uk)k/g2 _{- (V-k)2}

0 0

We have earlier studied how to indent the integration path of 1(k) at the poles ±2.. See Fig. 6. In the same way as before, we introduce

a closed curve ABCDEA in the complex 2.-plane . See Fig. 7.

v-k

FIGURE 7.

COMPLEX INTEGRATION PATH

The residue at £. = Res 2.. 0 (2.) is v + i2. y ]2 e' 0

vz Ii+

Wo 1PI 2

E C

A

B

By means of the residue theorem we get ivzik2-ct2 -vy

_Fe

1(k) = 2lTiRes (9)

+ J

d9, e

[,,i,2_2

+ .

Uk12

woJ

-i\)zVX,-cc

e v'9..2-ct2 -

i Ii +

a12

I

Wj

₀

nentially small terms in 1(k) _{will only give exponentially small terms in}

the expression for , we can write

(x,y,z) -k + (124) dk *(k) Uk 2

i0y vz Ii +

a12

Ii+-1

e e L U)0] L

WJ

z /v

0 As before,

K(vyc)

this case, small. So we 1(k) v-k =

is a bounding functioxi for the

> 1 ,.and so the Integral terms

Uk

Ii +

12 e°' e'

[i + W _J I

wI

L 0J

integral terms. For

are exponentially

can write

2ITiV

=

(123) + exponentially small terms

for k<k2

By using (110), (112), (120), (122), _{(123), and the fact that the}

expo-- (1expo--aexpo--s)

0 + r e (l-a+c52)

+ 1(1-a-)

I

In the first integral,

=

(w01

/g2

-

(v-k)2

.

In the. second and third

integrals, _2.

= iV'(V-k)2 - ((+Uk)

k/g2 . In the fourth and fifth integrals,

=

_{/(u+tJk)'/g2 - (v-k)2}

Further, is some very small number, 0 < a

1/2

, and

This means: We will write

1 =1

+1

o. cxl a2

ikx

k4(cL1) e y*(k) k3 l

Here k3(C&1) < V and

k4(a1) > v

and we choose 1-61

a1

= C

/

d9. e

-vyZ

2

/2a2+i[i +

1

(125)

+

+.J2

_j

is chosen in accordance with

(122),

(126)

where the integration in I is from k =

k3 (a1)

to k

= v

, and the

integration in I is from

k.=

V to

k = k4(cc1)

We introduce

cx-

v-k

as a new integration variable in

(125).

So, for

v-k

v

(127)

1 I

a

42

For I ct2 This means:

jVx

'Je 41r2

I

1-al iVx I ivcLx 2

J

dcxe 0

r

ivz/9.2-cc2

.Ie

-ivcxx

dcx e V cy* (v-cxV) ak(V-I-Vcx) Uw v'9,2-cx2

+i[1 +

(1tcL)]2

v'L2-g d

eV

-iVzV'9..2-cz2 e + ill 1

+ 2-(l+cx)l

2 g J

We note that the second and third integrals in. (124) could be written as one integral. The reason for not doing so will be obvious in the next

sec-tion. The same is true for the fourth and fifth integrals in (124).

2.

Inner expansion of far-field source solution

We are now going to find a two-term inner expansion of the far-field

source solution. The result is given by equation (161).

We now let y be of order , and we reorder the terms in (124).

We will first take a look on the integrand of the fourth integral in (124).

We note that

(128)

r

ivzvt2-cx2

-iVz/2-ct2

.Ie

I Uw UU) +1 + __.2 _(].-cx)1 2

),'22 _i[l

+

_2(l cx)] 2

L g .) g 4 ]L _[i

-:

2 = o(l) (130)

/

d2..

(129)

This .is because 4 is the lowest order term in (130) and that the lowest w

0

Uk

possible order of - i-s

This means we

_can

write

Further

1/2-a

=

o(_1/2

a+-1

₎ =

O(c)

= o(l) (131)

-ivy + -]

-e

/

[1_}2 J[.±]

_k2[+

8

1-J

\J

{i+J

V 0 1 +

O(c1)]

Jk

12w

L + 21JV =

i-ivy4i

+

]_

[i -+ higher Order terms

1

Since o(1) (see (131)) we get by Taylor expansion

1

f

(w2 + 6U +

3u2v2)

k

[2w+

+ higher order terms

2

(132)

Ii

+ I -

Ii

F] ic12

[

vJ

W

0

11 + L

Here .,

> > 0 . Further

21-2UV k

vwO

Thus, the inner expansion of the fourth integral in (124)

_can

be written:

ikx

_r*(1e'

_I

_Ii

_J2

e"

_[l_]2eVz [i+

]2

1

f/2

=

_ole

1

\Jk [2w+2uv]

Ii

+

2.

e =

e'

a +

o(El))

This means that, in the integrand of the inner expansion of the fourth inte-gral in (124), we can write

[i +

!]2

e"

-

[i k]2 e +

evz

[I + Uk14

wJ

0

Uk2

wo_J

VzI1+l2

+

Uk]2

e ₊ wo Vz e 'Jz =

-

lVye

Q(J/2)

l > > 0 i

r

..-

Iwo

d

\J[l+94

[l]2

(l-a-)

ei

2rr (133)

f

dk e1

a* (k)

F 1

iVY]

+

higher order terms

-j

L /2w+2UV k V i)y [i

Let us now study the inner expansion of the third integral in (124).

The derivation is quite similar to that for the fourth integral in (124)

and so we only state the result

0 Fr

k2

r

-vYJI1--I _1.1+

ci e1 a*(k) {i + e % wo Uk 1 z+

1 + -'

\J[l]2

[

woJ (134) dk eikx a*(k) [ 1

-)

I2w+2Uv Iki

We will now study the inner expansion of the fifth integral in (124). We then need an asymptotic formula for C7*(k) as k ' 00 _{We will assume.}

that a(x) and a' (x) are continuous in the interval -L/2 < x < L/2

[including the end points] (see Fig. 1). Outside -L/2

x

L/2 , a(x) 0

It can then be shown (see Lighthill (1958)) that IkLI3a*(k) remains bounded

as

k-'±

The fifth integral in (124) can be bounded by

-(l-a+c52) -(l-a+ó

C C 2

fdkla*(k)I< C

f

-'- (l-a-)

.J (la)

C1 is a constant determined so that the inequality above is satisfied. It is obvious from (133), (134), and (135) that the inner expansion of the

fifth integral in (124) will give a term of higher order of magnitude than the highest-order terms already retained in the inner expansion of the third

and fourth integral in (124). So we can drop the contribution from the fifth integral.in (124).

r

Uk

I 1 +

W

e

vy

+ higher order terms

(135)

Let us now study the inner expansion of in (124). I is given

by (127) as

e

00 _I,.

-C1 \)

2

I

=1

+1

al a2

where I and _Ic2 are given by (128) and (129), respectively. Since the argument of 0* _{is large both in (128) and (129) we can use the} pre-viously stated fact that IkLI3o*(k) remains bounded as kI-00 For the inner integrals in (128) and (129) we can write

Iivzyx,-a -Vy9 e e

f

d

/22

_{±i {i +}

g (1±)] 00 d

eV

e = < ₌ Vy

So the inner expansion of (128) can be bounded by

dc

(\'-cv)

A similar bounding function can be found for the inner expansion of (129). It should be obvious, for the same reasons used to drop the contribution from the inner expansion of the fifth integral in (124), that we can drop

the contribution from the inner expansion of in (124).

Let us now look at the first two integrals in (124). _{We have to be}

careful since the integrands have a pole at

k = k2

where

k = (2w U/g+l)

2

U2

We will study the first integral first. We introduce the new variable

V. = k - k

(127)

(136)

This means that where ik2 x e 2rr dv

eTX

*(v+k)

£

e2X

0 2Tr (1+Uk/w0) = H3 H2 (1-k/v)2 = H3 H12 V H 1 1 \)H32 H2 = 1 - W0H3

We can now write the first integral in (124) as

eH322_H2 +

VzH32H22

We write this integral as a sum of two integrals. The first integral. is from - to

_-(l-a-)

. The integrand is well-behaved, and the

upper limit is a large number. So, since the infinite integral from -to 0 necessarily has to converge, the first integral has to be of higher

-(l-a-)

order than the second integral from

-E

to 0 . It will be

evi-dent later on that the integral from

_-(l-a-)

to 0 will give a term

which is of the same order of magnitude as the highest order terms already retained in the inner expansion of the far-field source solution. So we drop

the integral from - to (l-a-)

We therefore write (141) as

dve1

a*(v+k)

.

ivyH32/H2-Vl-H12/ 2 + \)zH2H (142) (141) H = 1 +

wu

(140)