Steady Periodic Water Waves Solutions Using Asymptotic Approach

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Matematiska Institutionen

Department of Mathematics

Examensarbete

Steady periodic water waves solutions

using asymptotic approach

Shahid Hasnain Reg Nr: LiTH-MAI-EX--2011/14-SE Linköping 2011 Matematiska institutionen Linköpings universitet 581 83 Linköping

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Steady periodic water waves solutions using

asymptotic approach

Mathematics

Shahid Hasnain

LiTH-MAI-EX--2011/14-SE

Handledare: Vladimir Kozlov

Mai, Linköping universitet

Examinator: Vladimir Kozlov

Mai, Linköping universitet

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Avdelning, Institution

Division, Department

Division of Applied mathematics Department of Mathematics Linköpings universitet SE-581 83 Linköping, Sweden

Datum Date 2011-006-13 Språk Language Svenska/Swedish Engelska/English Rapporttyp Report category Licentiatavhandling Examensarbete C-uppsats D-uppsats Övrig rapport

URL för elektronisk version http://www.mai.liu.se http://www.ep.liu.se ISBN — ISRN LiTH-MAI-EX--2011/14-SE

Serietitel och serienummer

Title of series, numbering

ISSN

—

Titel

Title

Steady periodic water waves solutions using asymptotic approach

Författare

Author

Shahid Hasnain

Sammanfattning

Abstract

The aim of this work is to study the relation between two invariants of water flow in a channel of finite depth. The first invariant is the height of the water wave and the second one is the flow force. We restrict ourselves to water waves of small amplitude. Using asymptotic technique together with the method of separation of variables, we construct all water waves of small amplitude which are parameterized by a small parameter. Then we demonstrate numerically that the flow force depends monotonically on the height.

Nyckelord

Keywords Steady two dimensional water wave, perturbation approach, separation of variable, pressure head and flow force.

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List of Figures

1.1 Periodic stoke’s wave . . . 7 2.1 Overview of R(h) . . . 13 3.1 Overview of Flow Force . . . 34

Symbols

η(x) Free surface. φ Potential function . ψ Stream function . ▽ Del Operator g Gravitational acceleration. Q Flux R _{Pressure head.} S Force flow △ Laplace Operator ǫ A small parameter L Period X _{New variable}

Abbreviations

BVP Boundary value problem ODE Ordinary differential equation

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Acknowledgments

First of all I would like to thank The Almighty Allah who is the great source for knowledge and wisdom, who gave me strength and conviction to under-stand and complete this project.

I would like to show my gratitude to my Professor Vladimir Kozlov, thank you very much for providing me with much needed advice during this project, being patient, answering all the silly questions. I would like to thank Prof.Lars Elden who started my journey of computational mathematics when I came to Sweden. I would like to thank Henrik Branden, to teach me the numerical approach of mathematics. I would like to thank Prof.Iryna Yakymenko who teach me important courses to understand the physics of the problem and also she is my program coordinator, my hearty thanks to her. I would like to thank my program Chairman Sven Stafstrom. I would also like to thank the Swedish educational system, not only for the great ed-ucational experience and learning but also for allowing me to experience the Swedish life style. I would like to thank my parents for unconditional love and support. Without your encouragement this work would not have been possible. Thanks for being there when i needed you. Last but not least, I would like to thank all my friends especially Touqeer Pasha and Syed Asmat Shah for their kind support throughout the project.

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Abstract

The aim of this work is to study the relation between two invariants of water flow in a channel of finite depth. The first invariant is the height of the water wave and the second one is the flow force. We restrict ourselves to water waves of small amplitude. Using asymptotic technique together with the method of separation of variables, we construct all water waves of small amplitude which are parameterized by a small parameter. Then we demonstrate numerically that the flow force depends monotonically on the height.

Keywords:

Steady two dimensional water wave, perturbation approach, separation of variable, pressure head and flow force.

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Chapter 1

Introduction

We consider a two dimensional, steady water waves in the channel of finite depth. We assume that the bottom of the channel is defined by y = 0 and x∈ R. We describe the free surface by y = η(x). Thus, the water domain W is defined by

W = {(x, y) : x ∈ R, 0 < y < η(x)},

where η(x) > 0 is a smooth function. We suppose that the density of the water is constant and equal to one. Since the flow is irrotational, the velocity field (u, v) has a potential φ, i.e.

u= φx, and v= φy

Since the velocity field satisfies ux+ vy = 0

the potential φ is a harmonic function. Let us introduce the conjugate function ψ connected with φ by the Cauchy Riemann equations:

u= −ψy, and v= ψx. (1.1)

Then ψ is also a harmonic function called the stream function. The normal derivative of velocity at the bottom and at the free surface is zero. This implies that at the bottom v = 0. Using (1.1), we obtain that ψx = 0 on

the bottom. On the free surface, we have un1+ vn2 = 0,

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CHAPTER 1. INTRODUCTION

which implies

−ψyn1+ ψxn2= 0,

⇔ ▽ψ.(n2,−n1) = 0,

therefore the tangent derivative of ψ is zero at the free surface ∂ψ

∂τ = 0.

Thus ψ is constant on the bottom and on the free surface. Since ψ is defined up to a constant, we can assume that ψ is zero at the bottom, i.e.

ψ(x, 0) = 0, and

ψ(x, η(x)) = Q,

where the quantity Q is constant, which is called the flux of the channel flow. The positive sign of Q shows that the waves propagate in the positive x− direction. We assume that Q = 1. On the free surface, the velocity and η satisfy the Bernoulli’s equation

1 2 u2+ v2 + gη(x) = R,

where g is the gravitational acceleration and we assume that it is equal to one and R is the Bernoulli constant for the flow. Using the definition of stream function, we arrive at

1

2|▽ψ(x, η(x))|

2_{+η(x) = R.}

Thus the functions ψ and η satisfies the following boundary value problem            △ψ = 0, on W (1.2a) ψ(x, 0) = 0, x ∈ [−∞, ∞] (1.2b) ψ(x, η(x)) = 1, x ∈ R (1.2c) 1 2|▽ψ(x, η(x))| 2_{+η(x) = R, x ∈ R.} _(1.2d) Shahid, 2011 6

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CHAPTER 1. INTRODUCTION

Our aim is to find ψ, η and R in the following form

ψ(x, y, ǫ) = ψ0(y) + ǫψ1(x, y) + ǫ2ψ2(x, y) + O(ǫ3), (1.3)

η(x, ǫ) = η0+ ǫη1(x) + ǫ2η2(x) + O(ǫ3), (1.4)

R_{= R}₀_{+ ǫR}₁_{+ ǫ}2R₂_{+ O(ǫ}3_), _(1.5) where ǫ is a small parameter. In (1.3), we assume that ψ0 depends only on

yand in (1.4) we assume that η0 is constant, denoted by h. In other words,

(1.3-1.5) present the perturbation of a uniform stream. Our goal is to find the small periodic solution asymptotically. In the following figure(1.1), we

Figure 1.1: Periodic stoke’s wave

consider steady periodic Stoke’s wave with fixed period 2L and height h which is drawn in Cartesian coordinates. The dotted line represents the position of the free surface where no wave is present. The point of the wave which is the maximum of the free surface is called the crest and it is located at the origin. That point which is the minimum of the free surface is called the trough.

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Chapter 2

Small periodic water waves

with fixed period

2.1 Calculation of ψ

0

, η

0

and R

0

2.1.1 Asymptotic relations (1.2a)-(1.2d)

Using the fact that ψ is a harmonic and applying △ to both sides of equation (1.3), we get

△ψ(x, y, ǫ) = △ψ0(y) + ǫ △ ψ1(x, y) + ǫ2△ ψ2(x, y) + O(ǫ3) = 0.

By equating the terms of order ǫ0, ǫ1 and ǫ2, we have

△ψ0(y) = 0, (2.1)

△ψ1(x, y) = 0, (2.2)

△ψ2(x, y) = 0. (2.3)

Putting y = 0 in equation (1.3) and taking into account (1.2b), we obtain ψ(x, 0, ǫ) = ψ0(0) + ǫψ1(x, 0) + ǫ2ψ2(x, 0) + O(ǫ3) = 0, from which ψ0(0) = 0, (2.4) ψ₁(x, 0) = 0, x∈ R, (2.5) ψ2(x, 0) = 0, x∈ R. (2.6) 9

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Calculation of ψ0, η0 and R0

Now (1.3) and (1.2c) give

ψ(x, η(x), ǫ) = ψ0(x, η(x)) + ǫψ1(x, η(x)) + ǫ2ψ2(x, η(x)) + O(ǫ3) = 1.

Replacing η(x) by its expression given in (1.4), we have

ψ0(h+ǫη1+ǫ2η2)+ǫψ1(x, η0+ǫη1+ǫ2η2)+ǫ2ψ2(x, η0+ǫη1+ǫ2η2)+O(ǫ3) = 1.

Applying the Taylor’s expansion, we get

ψ₀(h) + ψ0y(h)(ǫη1+ ǫ2η2) + 1 2ψ0yy(h)(ǫη1+ ǫ 2_η 2)2+ ǫψ1(x, h) + ǫψ1y(x, h)(ǫη1+ ǫ2η2) + ǫ2ψ2(x, h) + O(ǫ3) = 1, from which ψ₀(h) = 1, (2.7) ψ1(x, h) = −ψ0y(h)η1, (2.8) ψ2(x, h) = −ψ0y(h)η2− 1 2ψ0yy(h)η 2 1− ψ1y(x, h)η1. (2.9)

Using (1.2d), together with equations (1.3-1.5), we have 1 2 ψ0x+ ǫψ1x+ ǫ2ψ2x 2 +1 2 ψ0y+ ǫψ1y+ ǫ2ψ2y 2 + h + ǫη1 +ǫ2η1= R0+ ǫR1+ ǫ2R2+ O(ǫ3).

Since we know that ψ0 is a function of y only, its derivative with respect to

x is zero. Therefore 1 2 ǫψ1x(x, η(x)) + ǫ2ψ2x(x, η(x)) 2 +1 2 ψ0y+ ǫψ1y(x, η(x)) + ǫ2ψ2y(x, η(x)) 2 + h + ǫη1+ ǫ2η1 = R0+ ǫR1+ ǫ2R2+ O(ǫ3),

expanding the squares, we get 1 2ǫ 2_ψ2 1x(x, η(x)) + 1 2 (ψ0y)2+ ǫ2ψ21y(x, η(x)) + 2ǫψ0yψ1y(x, η(x)) + 2ψ0yǫ2ψ2y(x, η(x)) + h + ǫη1+ ǫ2η1= R0+ ǫR1+ ǫ2R2+ O(ǫ3). Shahid, 2011 10

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Using (1.4) in the above expression, we have 1 2ǫ 2_ψ2 1x(x, h + ǫη1+ ǫ2η2) +1 2 (ψ0y)2+ ǫ2ψ1y2 (x, h + ǫη1+ ǫ2η2) + 2ǫψ0yψ1y(x, h + ǫη1+ ǫ2η2) + 2ǫ2ψ0yψ2y(x, h + ǫη1+ ǫ2η2) + h + ǫη₁(x) + ǫ2η₂(x) = R0+ ǫR1+ ǫ2R2+ O(ǫ3).

Applying Taylor’s expansion, we get 1 2ǫ 2 ψ1x(x, h) + ψ1xy(x, h)(ǫη1+ ǫ2η2) 2 +1 2 (ψ0y)2+ ǫ2 ψ1y(x, h) + ψ1yy(x, h)(ǫη1+ ǫ2η2) 2 + 2ǫψ0y ψ1y(x, h) + ψ1yy(x, h)(ǫη1+ ǫ2η2) + 2ǫ2ψ0y ψ2y(x, h) + ψ2yy(x, h)(ǫη1+ ǫ2η2) + h + ǫη1(x) + ǫ2η2(x) = R₀_{+ ǫR}₁_{+ ǫ}2R₂_{+ O(ǫ}3_), from which R₀₌ 1 2ψ 2 0y(h) + h, (2.10) R₁_{= ψ}_0y_ψ_1y_{(x, h) + η}₁_(x), _(2.11) R₂ ₌ 1 2ψ 2 1x(x, h)+ 1 2ψ 2 1y(x, h)+ψ0yψ1yy(x, h)η1+ψoyψ2y(x, h)+η2. (2.12)

2.1.2 Boundary value problem for ψ0, η0 and R0

Now combining equations (2.1), (2.4), (2.7) and (2.10), we formulate the following boundary value problem

             △ψ0(y) = 0, (2.13a) ψ₀(0) = 0, y ∈ [0, h] (2.13b) ψ0(h) = 1, y ∈ [0, h] (2.13c) 1 2 ψ0y(h) 2 + h = R0. (2.13d) Shahid, 2011 11

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2.1.3 Solution to (2.13)

Solving (2.13) for ψ0 with first two boundary conditions, we obtain

ψ₀(y) = y

h. (2.14)

Now substituting the obtained ψ0 in (2.13d), we get

1

2h2 + h = R0. (2.15)

Let us consider a function R(h) defined by 1

2h2 + h = R(h),

see in graph at figure (2.1). We observe that R(h) has its minimum at h = 1. Therefore equation (2.15) has solutions only when R0 ≥ 3₂. We assume that

R₀ _> 3

2, under this condition, the equation (2.15) has two solutions. If we

choose one of them, then the function ψ0 given by (2.14), η0 = h and R0

solve problem (2.13a-2.13d).

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Calculation of ψ1, η1 and R1 0.5 1 1.5 2 2.5 3 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 Pressure Head h R(h) Critical Value Figure 2.1: Overview of R(h)

2.2 Calculation of ψ

1

, η

1

and R

1

Using (2.14) in the equation (2.8), we get

ψ₁(x, h) = −η1(x) h . (2.16) Using (2.14) in (2.11), we have R₁₌ 1 hψ1y(x, h) + η1(x). (2.17) Shahid, 2011 13

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             △ψ1(x, y) = 0, (2.18a) ψ1(x, 0) = 0, x∈ [−L, L] (2.18b) ψ1(x, h) = − η1 h, x∈ [−L, L] (2.18c) 1 hψ1y(x, h) + η1= R1, x∈ [−L, L]. (2.18d) 2.2.2 Solution to (2.18)

We seek a solution to (2.18) in the following form

ψ1(x, y) = γ(y) cos(

xπ

L ), (2.19)

η₁(x) = cos(xπ

L ), (2.20)

where γ(y) is unknown function which we need to find and L is also unknown with half period. Applying △ to both sides of (2.19) and using the fact that ψ1 is harmonic, we have

△ψ1(x, y) = γ′′(y) −

π2

L2γ(y) = 0. (2.21)

Making use of (2.19) in (2.18b), we have

ψ1(x, 0) = γ(0) cos(

xπ L ) = 0, from which

γ(0) = 0. (2.22)

Making use of (2.19) in (2.18c), we have

ψ1(x, h) = γ(h) cos(

xπ L ) = −

η1

h.

Replacing η1 by its expression given in (2.20), we have

γ(h) = −1

h. (2.23)

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Now combining equations (2.21)-(2.23), we have the following ordinary dif-ferential equation                  γ′′_{(y) −} π 2 L2γ(y) = 0, γ(0) = 0, y = 0 γ(h) = −_h1, y= h. (2.24)

Solving the above system of equations, we get

γ(y) = −1 h

sinh(yπ_L) sinh(hπ_L).

Substituting γ(y) in (2.19), we obtain

ψ1(x, y) = − 1 h sinh(πy_L) sinh(πh_L)cos( xπ L ), (2.25) therefore from (2.18d) R₁₌ 1 h −π Lhcos( xπ L) 1 tanh(hπ L) + cos(xπ L ), or R₁₌ −π h2_L 1 tanh(hπ_L) + 1 cos(xπ L ). (2.26)

This equation can be satisfied only when −π

h2_L

1

tanh(hπ_L) + 1 = 0, (2.27) and hence R1 = 0. Relation (2.27) is called the dispersion relation and the

half period L satisfies it. Putting τ = hπ

L , in equation (2.27), we have tanh τ τ = 1 h3. (2.28) Since tanh τ τ <1, Shahid, 2011 15

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then 1 h3 <1,

which implies that h should be greater than 1. Thus, in order to satisfy (2.27) we must choose the root of equation (2.15) which is greater than one. Therefore h > 1 and the equation (2.28) has a solution τ and then L is given by

L= hπ

τ . (2.29)

Thus we have found η1, ψ1, R1 and L which are given by equations (2.20),

(2.25), (2.26) and (2.29) respectively.

2.3 Calculation of ψ

2

, η

2

and R

2

Using (2.14) and (2.25) in equation (2.9), we get

ψ2(x, h) = − η₂ h + h 2 + h 2cos( 2πx L ). (2.30)

Using (2.14) and (2.25) in equation (2.12), we obtain

π2 2h2_L2sin 2₍πx L ) + h2 2 cos 2₍πx L ) − π2 h2_L2 cos 2₍πx L ) + 1 hψ2y(x, h) + η2 = R2. (2.31) Since sin2(πx L ) = 1 2 1 − cos(2πx L ) ,

using above relation in equation (2.31), we have

π2 4h2_L2 1−cos(2πx L ) + h 2 4 − π2 2h2_L2 1+cos(2πx L ) +1 hψ2y(x, h)+η2 = R2. (2.32) Shahid, 2011 16

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                                           △ψ2(x, y) = 0, ψ₂(x, 0) = 0, x ∈ [−L, L] ψ₂(x, h) = −η2 h + h 2 + h 2cos( 2πx L ), x ∈ [−L, L] π2 4h2_L2 1 − cos(2πx_L ) + h 2 4 − π2 2h2_L2 1 + cos(2πx_L ) +1 hψ2y(x, h) + η2 = R2. (2.33) 2.3.2 Solution to (2.33)

We seek a solution ψ2(x, y) in the following form

ψ2(x, y) = ψ(1)(y) + ψ(2)cos(2πx

L ), (2.34)

η₂(x) = η(1)+ η(2)cos(2πx

L ). (2.35)

Applying △ to both sides of equation (2.34) and using the fact that ψ2 is

harmonic, we have △ψ2(x, h) = ψyy(1)+ ψ_yy(2)− ψ(2)(4π 2 L2 ) cos(2πx L ) = 0, from which ψ_yy(2)− (4π 2 L2 )ψ (2) _{= 0,} _(2.36) and ψ_yy(1)= 0. (2.37)

Making use of equation (2.34) in 2nd equation of (2.33), we have

ψ2(x, 0) = ψ(1)(0) + ψ(2)cos(

2πx L ) = 0, Shahid, 2011 17

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from which

ψ(1)(0) = 0, (2.38)

and

ψ(2)(0) = 0. (2.39)

Making use of equation (2.34) in 3rd equation of (2.33), we have ψ2(x, h) = ψ(1)(h) + ψ(2)cos( 2πx L ) = − η₂ h + h 2 + h 2cos( 2πx L ). Replacing η2 by its expression given in (2.35), we have

ψ(1)(h) = −η (1) h + h 2, (2.40) and ψ(2)(h) = −η (2) h + h 2. (2.41)

Making use of equations (2.34) and (2.35) in 4th equation of (2.33), we have π2 4h2_L2 1 − cos(2πx L ) + h 2 4 − π2 2h2_L2 1 + cos(2πx L ) + 1 h ψ_y(1)+ ψ(2)_y cos(2πx L ) + η(1)+ η(2)cos(2πx L ) = R2, from which π2 4h2_L2 + ( h2 4 − π2 2h2_L2) + 1 hψ (1) y + η(1) = R2, (2.42) and − π 2 4h2_L2 + ( h2 4 − π2 2h2_L2) + 1 hψ (2) y + η(2) = 0. (2.43)

Now combining equations (2.37), (2.38), (2.40) and (2.42), we formulate the following ordinary differential equation

                               ψ(1)_yy = 0, ψ(1)(0) = 0, ψ(1)(h) = −η (1) h + h 2, π2 4h2_L2 + (h 2 4 − π2 2h2_L2) + 1 hψ (1) y (x, h) + η(1)_{= R} 2. (2.44) Shahid, 2011 18

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We assume that η(1) is zero. Solving (2.44) for ψ(1) with first two boundary conditions, we obtain

ψ(1)(y) = −y 2.

Substituting above value in forth equation of (2.44), we get

− π 2 4h2_L2 + h2 4 + 1 h 1 2 = R2, or R₂₌ h 2 4 1 − tanh2(πh L ) + 1 2h. (2.45)

Now combining equations (2.36), (2.39), (2.41) and (2.43), we formulate the following ordinary differential equation

                                 ψ(2)_yy −4π 2 L2 ψ (2) _{= 0,} ψ(2)(0) = 0, ψ(2)(h) = −η (2) h + h 2, − π 2 4h2_L2 + ( h2 4 − π2 2h2_L2) + 1 hψ (2) y (x, h) + η(2)= 0. (2.46)

Solving (2.46) for ψ(2) with first two boundary conditions, we obtain

ψ(2)= (−η (2) h + h 2) 1 2 (e2πyL − e −2πy L ) sinh(2πh_L ) , after some simplification,

ψ(2)= (−η (2) h + h 2) sinh(2πy_L ) sinh(2πh_L ). (2.47) Using (2.47) in forth equation of (2.46), we have

− π 2 4h2_L2 + ( h2 4 − π2 2h2_L2) + 1 h(− η(2) h + h 2) 2π L cos h(2πh_L ) sinh(2πh_L ) + η (2)_{= 0,} Shahid, 2011 19

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Calculation of ψ2, η2 and R2 or − π 2 4h2_L2+ ( h2 4 − π2 2h2_L2) + 2π hL −η (2) h + h 2 coth(2πh L ) + η (2) _{= 0. (2.48)} Since tanh(2πh L ) = 2tanh(πh L) 1 + tanh2₍πh L) ,

and also from (2.27), we have equation (2.48) in the following form −3π2 4L2_h2 + h2 4 + −η (2) h + h 2 h(1 + π 2 L2_h4) + η (2) _{= 0,} or −3π2 4L2_h2 + h2 4 + η (2) 1 − 1 − π 2 L2h4 + h 2 2 1 + π 2 L2h4 = 0, therefore η(2) = h 2 4 3L2_h4 π2 − 1 . (2.49)

Substituting η(1) and η(2) in (2.35), we get

η2(x) = h2 4 3L2_h4 π2 − 1 cos(2πx L ). (2.50)

Now substituting ψ(1) and ψ(2) in equation (2.34), we get

ψ2(x, y) = y 2 + h 2 − h 4( 3h4L2 π2 − 1) sinh2πy_L sinh(2πh_L ) cos( 2πx L ), or ψ2(x, y) = 3h 4 1 −h 4_L2 π2 _sinh(2πy L ) sinh(2πh_L )cos( 2πx L ) + y 2. (2.51) Thus we have found R2, η2(x) and ψ2(x, y) which are given by equations

(2.45), (2.50) and (2.51) respectively. We compare our results with [2] which are mentioned in equations (2.45), (2.50) and (2.51) and we get the required results.

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Chapter 3

Small periodic water waves

with fixed Bernoulli constant

Our goal is to find the small amplitude solution with fixed Bernoulli constant R_{, but the free surface, stream function and the period are unknown. In this} case it is more convenient to look at the solution to (1.2) in the following form

ψ(X, y, ǫ) = ψ0(y) + ǫψ1(X, y) + ǫ2ψ2(X, y) + O(ǫ3), (3.1)

η(X, ǫ) = η0+ ǫη1(X) + ǫ2η2(X) + O(ǫ3), (3.2)

L_{= L}₀_{+ ǫL}₁_{+ ǫ}2L₂_{+ O(ǫ}3_), _(3.3) where X = xL0

L . One can check that if x∈ [−L(ǫ), L(ǫ)]

then

X_{∈ [−L}₀, L₀].

The boundary value problem (1.2) can be written in new variable as                        (L0 L) 2_ψ XX+ ψyy = 0, on W (3.4a) ψ(XL L₀,0) = 0, X ∈ [−L0, L0] (3.4b) ψ(XL L₀, η( XL L₀)) = 1, (3.4c) 1 2|( L₀ L )ψX+ ψy| 2_{+η(X) = R X ∈ [−L} 0, L0]. (3.4d) 21

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Calculation of ψ0, η0 and R

Our aim is to find ψ, η and R asymptotically.

3.1 Calculation of ψ

0

, η

0

and R

3.1.1 Asymptotic relations (3.4a)-(3.4d)

Using (3.1) in (3.4a), we get

ψ0yy(y) + ǫ (L0 L) 2_ψ 1XX(X, y) + ψ1yy + ǫ2 (L0 L) 2_ψ 2XX(X, y) + ψ2yy + O(ǫ3) = 0. Using (3.3), we obtain ψ0yy + ǫ (1 + ǫL1 L₀ + ǫ 2L2 L₀) −2_ψ 1XX+ ψ1yy + ǫ2 (1 + ǫL1 L₀ + ǫ 2L2 L₀) −2_ψ 2XX +ψ2yy + O(ǫ3) = 0, or ψ0yy + ǫ (1 − 2(ǫL1 L₀ + ǫ 2L2 L₀))ψ1XX+ ψ1yy + ǫ2 (1 − 2(ǫL1 L₀ + ǫ 2L2 L₀))ψ2XX +ψ2yy + O(ǫ3) = 0, therefore ψ0yy + ǫ ψ1XX− 2(ǫ L₁ L₀ + ǫ 2L2 L₀)ψ1XX+ ψ1yy + ǫ2 ψ2XX− 2(ǫ L₁ L₀ + ǫ 2L2 L₀)ψ2XX +ψ2yy + O(ǫ3) = 0.

By equating the terms of order ǫ0_{, ǫ}1 _{and ǫ}2_{, we have}

ψ0yy = 0, (3.5)

ψ_1XX+ ψ1yy = 0, (3.6)

−2L1

L₀ ψ1XX+ ψ2XX+ ψ2yy = 0. (3.7) Putting y = 0 in equation (3.1) and taking into account (3.4b), we obtain

ψ(X, 0, ǫ) = ψ0(0) + ǫψ1(X, 0) + ǫ2ψ2(X, 0) + O(ǫ3) = 0,

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Calculation of ψ0, η0 and R from which ψ0(0) = 0, (3.8) ψ₁(X, 0) = 0, (3.9) ψ2(X, 0) = 0. (3.10) Using(3.1-3.3) in (3.4c), we have ψ0(h + ǫη1+ ǫ2η2) + ǫψ1(X, h + ǫη1+ ǫ2η2) + ǫ2ψ2(X, h + ǫη1+ ǫ2η2) + O(ǫ3) = 1.

Applying Taylor’s expansion, we get

ψ0(h) + ψ0y(ǫη1+ ǫ2η2) + 1 2ψ0yy(ǫη1+ ǫ 2_η 2)2+ ǫ ψ1(X, h) +ψ1y(X, h)(ǫη1+ ǫ2η2) + ǫ2ψ2(X, h) + O(ǫ3) = 1, from which ψ₀(h) = 1, (3.11) ψ_0y(h)η1+ ψ1(X, h) = 0, (3.12) ψ0yη2+ 1 2ψ0yyη 2 1 + ψ1y(X, h)η2+ ψ2(X, h) = 0. (3.13)

Using (3.4d), together with (3.1-3.3), we have 1 2 L₀ L 2 ψ0X+ ǫψ1X+ ǫ2ψ2X 2 +1 2 ψ0y+ ǫψ1y+ ǫ2ψ2y 2 + h+ ǫη1+ ǫ2η1+ O(ǫ3) = R.

Since we know that ψ0 is a function of y only, its derivative with respect to

x is zero, 1 2 1 − 2(ǫL1 L₀ + ǫ 2L2 L₀) ǫ2ψ_1X2 +1 2 ψ0y+ ǫψ1y+ ǫ2ψ2y 2 + h + ǫη1+ ǫ2η1+ O(ǫ3) = R. Shahid, 2011 23

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Calculation of ψ0, η0 and R

Applying Taylor’s expansion, we have ǫ2 2ψ 2 1X− ǫ2ψ21X(ǫ L₁ L₀ + ǫ 2L2 L₀) + ψ 2 0y 2 + ǫ2 2 ψ1y(X, h) + ψ1yy(X, h)(ǫη1+ ǫ2η2) 2 +ǫ h ψ1y(X, h) + ψ1yy(X, h)(ǫη1+ ǫ2η2) +ǫ 2 h ψ2y(X, h) + ψ2yy(X, h)(ǫη1+ ǫ2η2) +h + ǫη1+ ǫ2η2+ O(ǫ3) = R, from which ψ_0y2 2 + h = R, (3.14) 1 hψ1y(X, h) + η1 = 0, (3.15) 1 2ψ 2 1X(X, h) + 1 2 ψ_1y2 (X, h) +2 hψ1yy(X, h) + 2 hψ2y(X, h) + η2 = 0. (3.16)

3.1.2 Boundary value problem for ψ0, η0 and R

            

ψ0yy(y) = 0, (3.17a)

ψ0(0) = 0, X ∈ [−L0, L0] (3.17b)

ψ0(h) = 1, (3.17c)

ψ_0y2

2 + h = R, X ∈ [−L0, L0]. (3.17d)

3.1.3 Solution to (3.17)

Solving (3.17a) for ψ0 with first two boundary conditions, we obtain

ψ0(y) =

y

h. (3.18)

Now substituting the obtained ψ0 in (3.17d), we get

1

2h2 + h = R, (3.19)

see in graph at figure (2.1). R has minimum at h = 1, therefore (3.19) has solutions only when R ≥ 3

2. We assume that R > 3

2, under this condition, Shahid, 2011 24

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Calculation of ψ1, η1 and L0

the equation (3.19) has two solutions. If we choose one of them, then the function ψ0 given by (3.18), η0= h and R solve the problem (3.17a)-(3.17d).

Thus we have found ψ0and R which are given by equations (3.18) and (3.19)

respectively.

3.2 Calculation of ψ

1

, η

1

and L

0

3.2.1 Boundary value problem for ψ1, η1 and L0

Using (3.18) in (3.12), we get

ψ₁(X, h) = −η1(X)

h . (3.20)

             ψ1XX(X, y) + ψ1yy(X, y) = 0, (3.21a) ψ1(X, 0) = 0, X∈ [L0, L0] (3.21b) ψ₁(X, h) = −η1 h, (3.21c) 1 hψ1y(X, h) + η1 = 0, X∈ [L0, L0]. (3.21d) 3.2.2 Solution to (3.21)

We seek a solution to (3.21) in the following form

ψ₁(X, y) = γ(y) cos(Xπ

L₀), (3.22)

η1(X) = cos(

X_π

L₀), (3.23)

where γ(y) and L0 are unknown functions which we need to find. Using

(3.22) in (3.21a), we have ψ₁(X, y) = γ′′_{(y) −}π 2 L2 0 γ(y) = 0. (3.24) Using (3.22) in (3.21b), we get ψ1(X, 0) = γ(0) cos( X_π L₀) = 0, (3.25) Shahid, 2011 25

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Calculation of ψ1, η1 and L0 from which γ(0) = 0. (3.26) Using (3.22) in (3.21c), we have ψ1(X, h) = γ(h) cos( X_π L₀) = − η₁(X) h , (3.27)

Replacing η1 by its expression given in (3.23), we have

γ(h) = −1

h (3.28)

Now combining equations (3.24), (3.26) and (3.28), we have the following ordinary differential equation

           γ′′_{(y) −} π2 L2 0 γ(y) = 0 (3.29a) γ(0) = 0, y= 0 (3.29b) γ(h) = −1 h, y = h. (3.29c)

Solving above system of equations, we get

γ(y) = −1 h sinh(yπL0) sinh(hπ L0) . (3.30)

Substituting γ(y) in (3.22), we have

ψ1(X, y) = − 1 h sinh(πy_L 0) sinh(πh L0) cos(Xπ L₀). (3.31) Using (3.31) in (3.21d), we have − π h2L₀ cosh(πh L0) sinh(πhL0) cos(Xπ L₀) + cos( X_π L₀) = 0, or − π h2L₀ cosh(πhL0) sinh(πhL0) + 1 = 0, Shahid, 2011 26

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Calculation of ψ2, η2 and L1 therefore tanhhπ L₀ = π h2L₀. (3.32)

Relation (3.32) is called the dispersion relation and it is the same that we got in the previous chapter at (2.27). Putting τ = hπ

L0 in above relation, we have tanh τ τ = 1 h3, (3.33) since tanh τ τ <1, then 1 h3 <1,

which implies that h should be greater than 1. Thus in order to satisfy (3.32), we must choose the root of equation (3.19), which is greater than one. Therefore at h > 1, equation (3.33) has a solution τ and then L0 is

given by

L0=

hπ

τ . (3.34)

Thus we have found η1, ψ1 and L0 which are given by equations (3.23),

(3.31) and (3.34) respectively.

3.3 Calculation of ψ

2

, η

2

and L

1

3.3.1 Boundary value problem for ψ2, η2 and L1

Using (3.18) and (3.31) in equation (3.13), we get

ψ2(X, h) = − η2 h + h 2 + h 2 cos( 2πX L₀ ). (3.35)

Using (3.18) and (3.31) in equation (3.16), we obtain

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Calculation of ψ2, η2 and L1 π2 4h2L2 0 (1−cos(2πX L₀ ))+( h2 4 − π2 2h2L2 0 )(1+cos(2πX L₀ ))+ 1 hψ2y(X, h)+η2 = 0. (3.36) Now combining equations (3.7), (3.10), (3.35) and (3.36), we formulate the following boundary value problem

                                               ψ2XX+ ψ2yy = 2L1 hL3₀ sinh(yπ_L 0) sinh(hπL0) , ψ2(X, 0) = 0, X∈ [L0, L0] ψ2(X, h) = − η2 h + h 2 + h 2cos( 2πX L0 ), X ∈ [L0, L0] π2 4h2L2 0 (1 − cos(2πXL0 )) + ( h2 4 − π2 2h2L2 0 )(1 + cos(2πXL0 )) +1 hψ2y(X, h) + η2= 0. (3.37) 3.3.2 Solution to (3.37)

We seek a solution to the above boundary value problem in the following form

ψ2(X, y) = ψ2(P )(X, y) + ψ (H)

2 (X, y), (3.38)

where P and H are particular and homogeneous part of the solution respec-tively[3]. We are looking for the solution ψ(P )₂ in the following form

ψ₂(P )= A sinh(πy

L₀), (3.39)

where A is an arbitrary constant. Substituting ψ(P )₂ in first equation of (3.37), we have Aπ 2 L2 0 sinh(yπ L₀) = 0 + 2 L₁ hL3₀ sinh(yπL0) sinh(hπL0) , Shahid, 2011 28

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Calculation of ψ2, η2 and L1 or A= 2L1 hπ2L₀_sinh(πh L0) .

Substituting the value of A in equation (3.39), we have

ψ₂(P )(X, y) = 2L1sinh( πy L0) hπ2L₀_sinh(πh_L 0) . (3.40)

Function ψ₂(H) satisfies the first equation in (3.37) with zero right hand side ψ_2XX(H) + ψ_2yy(H)= 0. (3.41) Using (3.38) and (3.40) in the 2nd equation of (3.37), we have

ψ₂(H)(X, 0) = 0. (3.42) Using (3.38) and (3.40) in the 3rd equation of (3.37), we have

ψ₂(H)(X, h) = −η2 h + h 2 + h 2cos( 2πX L₀ ) − 2L1 hπ2L₀. (3.43) Using (3.38) and (3.40) in the 4th equation of (3.37), we have

π2 4h2L2 0 (1−cos(2πX L₀ ))+( h2 4 − π2 2h2L2 0 )(1+cos(2πX L₀ ))+ 1 hψ (H) 2y (X, h)+ 2L1 π2L₀+η2 = 0. (3.44) Now combining equations (3.41-3.44), we formulate the following homoge-neous boundary value problem

                                           ψ(H)_2XX+ ψ(H)_2yy = 0, ψ(H)₂ (X, 0) = 0, X_{∈ [−L}₀_{, L}₀_], ψ(H)₂ (X, h) = −η2 h + h 2 + h 2 cos( 2πX L0 ) − 2L1 hπ2L₀, π2 4h2L2 0 (1 − cos(2πX_L 0 )) + ( h2 4 − π2 2h2L2 0 )(1 + cos(2πX_L 0 )) +1 hψ (H) 2y (X, h) + 2L1 π2L₀ + η2 = 0. (3.45) Shahid, 2011 29

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We are looking for a solution ψ₂(H) in the following form

ψ₂(H)(X, y) = ψ(1)(y) + ψ(2)cos(2πX

L₀ ), (3.46)

η2= η(1)+ η(2)cos(

2πX

L₀ ). (3.47)

Using (3.46) in the first equation of (3.45), we have

ψ_yy(1)+ (ψ(2)_yy)2− ψ(2)(4π 2 L2 0 ) cos(2πX L₀ ) = 0, from which (ψ(2)_yy)2− ψ(2)(4π 2 L2 0 ) = 0, (3.48) and ψ_yy(1)= 0. (3.49)

Using (3.46) in 2nd equation of (3.45), we have

ψ₂(H)(X, 0) = ψ(1)(0) + ψ(2)cos(2πX L₀ ) = 0, from which ψ(2)(0) = 0, (3.50) and ψ(1)(0) = 0. (3.51)

Using (3.46) in 3rd equation of (3.45), we have

ψ(1)(h)+ψ(2)cos(2πX L₀ ) = − 1 h η(1)+η(2)cos(2πX L₀ ) +h 2+ h 2 cos( 2πX L₀ )− 2L1 hπ2L₀, from which ψ(2)(h) = h 2 − η(2) h , (3.52) and ψ(1)(h) = −η (1) h + h 2 − 2L1 hπ2L₀. (3.53) Shahid, 2011 30

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Using (3.46) in 4th equation of (3.45), we obtain

− π 2 4h2L2 0 + h 2 4 − π2 2h2L2 0 +1 hψ (2) y (X, h) + η(2) = 0, (3.54) and π2 4h2L2 0 + (h 2 4 − π2 2h2L2 0 ) + 2L1 π2L₀ + 1 hψ (1) y (X, h) + η(1)= 0. (3.55)

Now combining equations (3.48), (3.50), (3.52) and (3.54), we formulate the following boundary value problem for ψ(2)

                                   ψ(2)yy 2 − ψ(2)₍4π2 L2 0 ) = 0, ψ(2)(0) = 0, ψ(2)(h) = h 2 − η(2) h , − π 2 4h2L2 0 +h 2 4 − π2 2h2L2 0 + 1 hψ (2) y (X, h) + η(2)= 0. (3.56)

Solving (3.56) for ψ(2) with first two boundary conditions, we obtain

ψ(2)= − η (2) h + h 2 _{sin h(}2πy L0 ) sinh(2πhL0 ) . (3.57)

Using (3.57) in forth equation of (3.56), we have −3π2 4L2₀h2 + h2 4 + −η (2) h + h 2 h(1 + π 2 L2 0h4 ) + η(2) = 0, or η(2) = h 2 4 3L2 0h4 π2 − 1 . Shahid, 2011 31

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Now combining equations (3.49), (3.51), (3.53) and (3.55), we formulate the following boundary value problem for ψ(1)

                               ψ(1)_yy = 0, ψ(1)(0) = 0, ψ(1)(h) = −η (1) h + h 2 − 2L1 hπ2L₀, π2 4h2L2 0 + (h 2 4 − π2 2h2L2 0 ) + 2L1 π2L₀ + 1 hψ (1) y (X, h) + η(1) = 0. (3.58)

We assume that η(1) _{is zero. Solving (3.58) for ψ}(1) _{with first two boundary}

conditions, we obtain ψ(1)= − 2L1y

h2_π2L₀ + y

2. (3.59)

Using (3.59) in forth equation of (3.58), we have

L₁₌ π 2_L 0 2(h−3_{− π)} h2 4 + 1 2h− π2 4h2L2 0 . (3.60)

Using η(1) and η(2) in equation (3.47), we have

η₂(X) = h 2 4 3L2 0h4 π2 − 1 cos(2πX L₀ ). (3.61) Using (3.57) and (3.59) in equation (3.46), we have

ψ(H)₂ = − 2L1y h2_π2L₀ + y 2 + − η (2) h + h 2 2π L₀ sinh(2πyL0 ) sinh(2πh_L 0 ) cos(2πX L₀ ). (3.62) Using (3.40) and (3.62) in equation (3.38), we have

ψ₂(X, y) = 2L1 hπ2_L 0sinh(πhL0) sinh(πy L₀) − 2L1y h2_π2L₀ + y 2 + −η (2) h + h 2 _sinh(2πy L0 ) sinh(2πhL0 ) cos(2πX L₀ ). (3.63) Thus we have found L1, η2 and ψ2 which are given by equations (3.60),

(3.61) and (3.63) respectively.

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Flow Force

3.4 Flow Force

Since we assume that the Bernoulli constant is fixed. The flow force for steady waves can be deduced from expression for horizontal momentum flux plus pressure force, see [4]

3 2s= 3 2rη(X, ǫ) − 1 2η(X, ǫ) 2₊ Z η(X,ǫ) 0 1 2ψ 2 y− 1 2( L₀ L) 2_ψ2 X dy, (3.64) where S₌ 3 2s, (3.65) R₌ 3 2r, (3.66)

also see at figure (1.1), we have X = 0. Now using (3.1), (3.65)-(3.66) in (3.64) S_{= Rη(ǫ) −} 1 2η 2_{(ǫ) +}Z η(ǫ) 0 1 2(ψ0y+ ǫψ1y+ ǫ 2_ψ 2y)2 dy+ O(ǫ3),

using (3.2) and doing some simplification, we have

S₌ h2+ 1 2h− 1 2h 2₊ 1 2h + ǫ 1 h2 − π L₀_h2 π L₀ + ǫ2 1 4( 3L2 0h4 π2 − 1) + π L₀_h_sinh(πh L₀) ( 2L1 π2L₀ − 1 h) + π2 2L2 0hsinh2( πh L₀) + (−η (2) h + h 2) 2π L₀ 1 sinh(2πhL0 ) +O(ǫ3), therefore S₌ 1 h+ 1 2h 2 +ǫ2 1 4( 3L2₀h4 π2 −1)+ π L₀_h_sinh(πh L₀) ( 2L1 π2L₀− 1 h)+ π2 2L2₀hsinh2(πh L₀) + (−η (2) h + h 2) 2π L₀ 1 sinh(2πh_L 0 ) + O(ǫ3). (3.67) Using (3.2) at X = 0, we have η(ǫ) = h + ǫ + ǫ2h 2 2 3h4L2 0 π2 − 1 + O(ǫ3), (3.68) Shahid, 2011 33

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Flow Force

The following figure shows the graph of flow force at different value of η which can obtain at different value of ǫ. We have chosen the following values for ǫ, ǫ= (0, ±0.01, ±0.02, ... ± 0.08). (3.69) The following graph shows that the flow force S has minimum at η = 1.5.

1.48 1.5 1.52 1.54 1.56 1.58 1.6 1.62 1.78 1.8 1.82 1.84 1.86 1.88 1.9 Flow force eta S

Figure 3.1: Overview of Flow Force

Using (3.19), we fixed the value of R, which give us h,

R₌ 31 18,

h= 3 2.

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Flow Force

Substituting the value of h in (3.34), we obtain the value for L0 which is

1.398. This L0 solves the dispersion relation at (3.32).

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Results

First part

We have presented the periodic waves solution with the help of perturbation approach. The governing equations were formulated in terms of stream function see [5]. In the first part of the project, we assume that the period of the waves is fixed and stream function ψ, free surface η and pressure head R_{are unknown.}

We are looking for unknown ψ, η and R in the form of (1.3)-(1.5) and asymptotically solving (1.2), we can write the following expressions for

ψ(x, y, ǫ) = y h + ǫ − 1 h sinh(πy_L) sinh(πh_L)cos( xπ L ) + ǫ2 3h 4 (1 − h4L2 π2 ) sinh(2πy_L ) sinh(2πh_L )cos( 2πx L ) + y 2 + O(ǫ3), (3.70) η(x, ǫ) = h + ǫcos(πx L ) + ǫ 2 h2 4 ( 3L2_h4 π2 − 1) cos( 2πx L ) + O(ǫ3), (3.71) and R_{= R}₀_{+ ǫ}2 h 2 4 (1 − tanh 2₍πh L )) + 1 2h + O(ǫ3), (3.72) where R₀_{= h +} 1 2h2 (3.73)

R_{has minimum at h = 1, therefore (3.19) has solutions only when R ≥} 3 2. We assume that R > 3

2, under this condition, the equation (3.19) has two 37

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Flow Force

solutions. which implies that h should be greater than 1. Thus in order to satisfy (3.32), we must choose the root of equation (3.19), which is greater than one. Therefore at h > 1, equation (3.33) has a solution τ and then L0

is given by

L= hπ

τ , (3.74)

therefore L satisfies the dispersion relation (3.32).

Second part

We assume that the pressure head R is fixed and period L(ǫ), free surface η(X, ǫ) and stream function ψ(X, y, ǫ) are unknown. We used the change of variable technique in the 2nd part to find the unknown ψ, η and L.

We are looking for unknown ψ, η and L in the form of (3.1)-(3.3) and asymptotically solving (1.2), we can write the following expressions for

ψ(X, y, ǫ) = y h + ǫ − 1 h sinh(πyL0) sinh(πh L0) cos(Xπ L₀) + ǫ2 _2L 1sinh(πyL0) hπ2L₀_sinh(πh L0) − 2L1y h2_π2L₀ + y 2 + −η (2) h + h 2 _sinh(2πy L0 ) sinh(2πh_L 0 ) cos(2πX L₀ ) + O(ǫ3), (3.75) η(X, ǫ) = h + ǫ cos(πX L₀) + ǫ2 h 2 4 ( 3L2₀h4 π2 − 1) cos( 2πX L₀ ) + O(ǫ3), (3.76) and L_{= L}₀_{+ ǫ} π2L₀ 2(h−3_{− π)}( h2 4 + 1 2h− π2 4h2L2 0 ) + O(ǫ2), (3.77) where L₀₌ hπ τ (3.78)

which implies that h should be greater than 1. Thus in order to satisfy (3.32), we must choose the root of equation (3.19), which is greater than

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Flow Force

one.

Last but not least the expression for flow force

S₌ 1 h+ 1 2h 2 +ǫ2 1 4( 3L2 0h4 π2 −1)+ π L₀_h_sinh(πh L₀) ( 2L1 π2_L 0 −1 h)+ π2 2L2 0hsinh2( πh L₀) + (−η (2) h + h 2) 2π L₀ 1 sinh(2πhL0 ) + O(ǫ3). (3.79) Shahid, 2011 39

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Conclusion

In this thesis, bifurcation approach with a fixed period and Bernoulli con-stant for small periodic water waves in a finite depth channel is analyzed. The governing equations are formulated in term of stream function which satisfy the boundary value problem (1.2) in conjunction with free surface. Our aim to solve (1.2) asymptotically and find the relation for R and S which depend monotonically on height.

This thesis work is divided into two parts. In the first part, we solve the boundary value problem (1.2) asymptotically by fixing the period of the small periodic water waves. We have presented our results at (3.69)-(3.74) and compare with [2]. We observe, these results only exist for the value of h which is greater than one.

In the 2nd part we fixed the Bernoulli constant and obtain the results at (3.75)-(3.78). We found R at (3.19) and observe that it has two roots. We choose one of them which is greater than one, with this condition L0satisfies

the dispersion relation at (3.34). At the end we found the expression for flow force which depend monotonically on height, see at (3.67).

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Bibliography

[1] Vladimir Kozlov, Nikolay Kuznetsov., The Benjamin-Lighthill conjec-ture for near critical values of Bernoulli’s constant. Archive for Rational

Mechanics and Analysis, Volume 197, Issue 2, pp. 433-488.

[2] S. H. Doole and J. Norbury., The bifurcation of steady gravity water waves in (R, S) parameter space. J. Fluid Mech. 1995, Vol. 302,

2877-305.

[3] http://en.wikipedia.org/wiki/Method of undetermined coefficients [4] T. Brooke Benjamin., Verification of the Benjamin-Lighthill conjecture

about steady water waves. J. Fluid Mech. 295(1995), 337-356.

[5] S. H.Doole., The bifurcation of long waves in the parameter space of pressure head and flow force. Q. J. Mech. Appl. Math. Vol. 50, Pt. 1,

1997.

[6] J. J. Stoker., Water Waves (The Mathematical Theory With Applica-tion.) Pure And Applied Mathematics Volume IV Ch. 2, P 19-27. [7] Germund Dahlquist and Ake Bjorck, Numerical Methods. ISBN:

0-486-42807-9 (PBK), Ch.8 P 342-344.

[8] http://en.wikipedia.org/wiki/Hyperbolic function

[9] T. B. Benjamin and M. Lighthill., On Cnoidal waves and Bores. Proc.

Roy. Soc. Lond. A 224(1954), 448-460.

[10] Charles L. Mader., Numerical Modeling of Water Waves. eBook ISBN:

978-0-203-49219-2, Ch.1.

[11] Lokenath Debnath., Nonlinear Water Waves ISBN: 0122084373. Ch.3

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c

2011, Shahid-Hasnain

Steady Periodic Water Waves Solutions Using Asymptotic Approach

Matematiska Institutionen

Department of Mathematics

Examensarbete

Steady periodic water waves solutions

using asymptotic approach

Steady periodic water waves solutions using

asymptotic approach

Contents

List of Figures

Symbols

Abbreviations

Acknowledgments

Abstract

Chapter 1

Introduction

Chapter 2

Small periodic water waves

with fixed period

2.1

Calculation of ψ

, η

and R

2.2

Calculation of ψ

, η

and R

2.3

Calculation of ψ

, η

and R

Chapter 3

Small periodic water waves

with fixed Bernoulli constant

3.1

Calculation of ψ

, η

and R

3.2

Calculation of ψ

, η

and L

3.3

Calculation of ψ

, η

and L

3.4

Flow Force

Results

Conclusion

Bibliography