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
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
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.
Contents
1 Introduction 5
2 Small periodic water waves with fixed period 9
2.1 Calculation of ψ0, η0 and R0 . . . 9
2.1.1 Asymptotic relations (1.2a)-(1.2d) . . . 9
2.1.2 Boundary value problem for ψ0, η0 and R0 . . . 11
2.1.3 Solution to (2.13) . . . 12
2.2 Calculation of ψ1, η1 and R1 . . . 13
2.2.1 Boundary value problem for ψ1, η1 and R1 . . . 13
2.2.2 Solution to (2.18) . . . 14
2.3 Calculation of ψ2, η2 and R2 . . . 16
2.3.1 Boundary value problem for ψ2, η2 and R2 . . . 16
2.3.2 Solution to (2.33) . . . 17
3 Small periodic water waves with fixed Bernoulli constant 21 3.1 Calculation of ψ0, η0 and R . . . 22
3.1.1 Asymptotic relations (3.4a)-(3.4d) . . . 22
3.1.2 Boundary value problem for ψ0, η0 and R . . . 24
3.1.3 Solution to (3.17) . . . 24
3.2 Calculation of ψ1, η1 and L0 . . . 25
3.2.1 Boundary value problem for ψ1, η1 and L0. . . 25
3.2.2 Solution to (3.21) . . . 25
3.3 Calculation of ψ2, η2 and L1 . . . 27
3.3.1 Boundary value problem for ψ2, η2 and L1. . . 27
3.3.2 Solution to (3.37) . . . 28
3.4 Flow Force . . . 33
4 Results 37
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 variableAbbreviations
BVP Boundary value problem ODE Ordinary differential equation
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.
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.
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,
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
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= R0+ ǫR1+ ǫ2R2+ O(ǫ3), (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.
Chapter 2
Small periodic water waves
with fixed period
2.1
Calculation of ψ
0, η
0and R
02.1.1 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
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 2η 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 2ψ2 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
Calculation of ψ0, η0 and R0
Using (1.4) in the above expression, we have 1 2ǫ 2ψ2 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+ ǫR1+ ǫ2R2+ O(ǫ3), from which R0= 1 2ψ 2 0y(h) + h, (2.10) R1= ψ0yψ1y(x, h) + η1(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
Calculation of ψ0, η0 and R0
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).
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
1and R
12.2.1 Boundary value problem for ψ1, η1 and R1
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
Calculation of ψ1, η1 and R1
Now combining equations (2.2), (2.5), (2.16) and (2.17), we formulate the following boundary value problem
△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)
Calculation of ψ1, η1 and R1
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(yπL) sinh(hπL).
Substituting γ(y) in (2.19), we obtain
ψ1(x, y) = − 1 h sinh(πyL) sinh(πhL)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 h2L 1 tanh(hπL) + 1 cos(xπ L ). (2.26)
This equation can be satisfied only when −π
h2L
1
tanh(hπL) + 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
Calculation of ψ2, η2 and R2
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
2and R
22.3.1 Boundary value problem for ψ2, η2 and R2
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 2h2L2sin 2(πx L ) + h2 2 cos 2(πx L ) − π2 h2L2 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 4h2L2 1−cos(2πx L ) + h 2 4 − π2 2h2L2 1+cos(2πx L ) +1 hψ2y(x, h)+η2 = R2. (2.32) Shahid, 2011 16
Calculation of ψ2, η2 and R2
Now combining equations (2.3), (2.6), (2.30) and (2.32), we formulate the following boundary value problem
△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 4h2L2 1 − cos(2πxL ) + h 2 4 − π2 2h2L2 1 + cos(2πxL ) +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)+ ψyy(2)− ψ(2)(4π 2 L2 ) cos(2πx L ) = 0, from which ψyy(2)− (4π 2 L2 )ψ (2) = 0, (2.36) and ψyy(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
Calculation of ψ2, η2 and R2
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 4h2L2 1 − cos(2πx L ) + h 2 4 − π2 2h2L2 1 + cos(2πx L ) + 1 h ψy(1)+ ψ(2)y cos(2πx L ) + η(1)+ η(2)cos(2πx L ) = R2, from which π2 4h2L2 + ( h2 4 − π2 2h2L2) + 1 hψ (1) y + η(1) = R2, (2.42) and − π 2 4h2L2 + ( h2 4 − π2 2h2L2) + 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 4h2L2 + (h 2 4 − π2 2h2L2) + 1 hψ (1) y (x, h) + η(1)= R 2. (2.44) Shahid, 2011 18
Calculation of ψ2, η2 and R2
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 4h2L2 + 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 4h2L2 + ( h2 4 − π2 2h2L2) + 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(2πhL ) , after some simplification,
ψ(2)= (−η (2) h + h 2) sinh(2πyL ) sinh(2πhL ). (2.47) Using (2.47) in forth equation of (2.46), we have
− π 2 4h2L2 + ( h2 4 − π2 2h2L2) + 1 h(− η(2) h + h 2) 2π L cos h(2πhL ) sinh(2πhL ) + η (2)= 0, Shahid, 2011 19
Calculation of ψ2, η2 and R2 or − π 2 4h2L2+ ( h2 4 − π2 2h2L2) + 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 4L2h2 + h2 4 + −η (2) h + h 2 h(1 + π 2 L2h4) + η (2) = 0, or −3π2 4L2h2 + h2 4 + η (2) 1 − 1 − π 2 L2h4 + h 2 2 1 + π 2 L2h4 = 0, therefore η(2) = h 2 4 3L2h4 π2 − 1 . (2.49)
Substituting η(1) and η(2) in (2.35), we get
η2(x) = h2 4 3L2h4 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) sinh2πyL sinh(2πhL ) cos( 2πx L ), or ψ2(x, y) = 3h 4 1 −h 4L2 π2 sinh(2πy L ) sinh(2πhL )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.
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= L0+ ǫL1+ ǫ2L2+ O(ǫ3), (3.3) where X = xL0
L . One can check that if x∈ [−L(ǫ), L(ǫ)]
then
X∈ [−L0, L0].
The boundary value problem (1.2) can be written in new variable as (L0 L) 2ψ 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
Calculation of ψ0, η0 and R
Our aim is to find ψ, η and R asymptotically.
3.1
Calculation of ψ
0, η
0and R
3.1.1 Asymptotic relations (3.4a)-(3.4d)
Using (3.1) in (3.4a), we get
ψ0yy(y) + ǫ (L0 L) 2ψ 1XX(X, y) + ψ1yy + ǫ2 (L0 L) 2ψ 2XX(X, y) + ψ2yy + O(ǫ3) = 0. Using (3.3), we obtain ψ0yy + ǫ (1 + ǫL1 L0 + ǫ 2L2 L0) −2ψ 1XX+ ψ1yy + ǫ2 (1 + ǫL1 L0 + ǫ 2L2 L0) −2ψ 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, ǫ1 and ǫ2, we have
ψ0yy = 0, (3.5)
ψ1XX+ 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,
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 2η 2)2+ ǫ ψ1(X, h) +ψ1y(X, h)(ǫη1+ ǫ2η2) + ǫ2ψ2(X, h) + O(ǫ3) = 1, from which ψ0(h) = 1, (3.11) ψ0y(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) ǫ2ψ1X2 +1 2 ψ0y+ ǫψ1y+ ǫ2ψ2y 2 + h + ǫη1+ ǫ2η1+ O(ǫ3) = R. Shahid, 2011 23
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 ψ0y2 2 + h = R, (3.14) 1 hψ1y(X, h) + η1 = 0, (3.15) 1 2ψ 2 1X(X, h) + 1 2 ψ1y2 (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
Now combining equations (3.5), (3.8), (3.11) and (3.14), we formulate the following boundary value problem
ψ0yy(y) = 0, (3.17a)
ψ0(0) = 0, X ∈ [−L0, L0] (3.17b)
ψ0(h) = 1, (3.17c)
ψ0y2
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
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
1and L
03.2.1 Boundary value problem for ψ1, η1 and L0
Using (3.18) in (3.12), we get
ψ1(X, h) = −η1(X)
h . (3.20)
Now combining equations (3.6), (3.9), (3.15) and (3.20), we formulate the following boundary value problem
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(
Xπ
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( Xπ L0) = 0, (3.25) Shahid, 2011 25
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( Xπ 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(πyL 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( Xπ L0) = 0, or − π h2L0 cosh(πhL0) sinh(πhL0) + 1 = 0, Shahid, 2011 26
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
2and L
13.3.1 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
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(yπL 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
Calculation of ψ2, η2 and L1 or A= 2L1 hπ2L0sinh(πh L0) .
Substituting the value of A in equation (3.39), we have
ψ2(P )(X, y) = 2L1sinh( πy L0) hπ2L0sinh(πhL 0) . (3.40)
Function ψ2(H) satisfies the first equation in (3.37) with zero right hand side ψ2XX(H) + ψ2yy(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∈ [−L0, L0], ψ(H)2 (X, h) = −η2 h + h 2 + h 2 cos( 2πX L0 ) − 2L1 hπ2L0, π2 4h2L2 0 (1 − cos(2πXL 0 )) + ( h2 4 − π2 2h2L2 0 )(1 + cos(2πXL 0 )) +1 hψ (H) 2y (X, h) + 2L1 π2L0 + η2 = 0. (3.45) Shahid, 2011 29
Calculation of ψ2, η2 and L1
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
ψyy(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 ψyy(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
Calculation of ψ2, η2 and L1
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
Calculation of ψ2, η2 and L1
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
h2π2L0 + y
2. (3.59)
Using (3.59) in forth equation of (3.58), we have
L1= π 2L 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 h2π2L0 + y 2 + − η (2) h + h 2 2π L0 sinh(2πyL0 ) sinh(2πhL 0 ) cos(2πX L0 ). (3.62) Using (3.40) and (3.62) in equation (3.38), we have
ψ2(X, y) = 2L1 hπ2L 0sinh(πhL0) sinh(πy L0) − 2L1y h2π2L0 + 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.
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) 2ψ2 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 2ψ 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 L0h2 π L0 + ǫ2 1 4( 3L2 0h4 π2 − 1) + π L0hsinh(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 L0hsinh(πh L0) ( 2L1 π2L0− 1 h)+ π2 2L20hsinh2(πh L0) + (−η (2) h + h 2) 2π L0 1 sinh(2πhL 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
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.
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).
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 Rare 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(πyL) sinh(πhL)cos( xπ L ) + ǫ2 3h 4 (1 − h4L2 π2 ) sinh(2πyL ) sinh(2πhL )cos( 2πx L ) + y 2 + O(ǫ3), (3.70) η(x, ǫ) = h + ǫcos(πx L ) + ǫ 2 h2 4 ( 3L2h4 π2 − 1) cos( 2πx L ) + O(ǫ3), (3.71) and R= R0+ ǫ2 h 2 4 (1 − tanh 2(πh L )) + 1 2h + O(ǫ3), (3.72) where R0= h + 1 2h2 (3.73)
Rhas 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
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) hπ2L0sinh(πh L0) − 2L1y h2π2L0 + y 2 + −η (2) h + h 2 sinh(2πy L0 ) sinh(2πhL 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= L0+ 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
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 L0hsinh(πh L0) ( 2L1 π2L 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
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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[11] Lokenath Debnath., Nonlinear Water Waves ISBN: 0122084373. Ch.3
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2011, Shahid-Hasnain