PABNA

UNIVERSITY OF
SCIENCE &
TECHNOLOGY Department of Mathematics
Faculty of Science

5th year 1st semester Final Examination 2023

Assignment on – Water Wave Mechanics

Course Tittle: Water Wave Mechanics

Course Code: Math-5110

Submitted By: Submitted To:

Roll:
S. M. Rayhanul Islam
190301,190314,190317,190320,190325,190
329,190330,190333,190335,190336,190337 Assistant Professor
,180305.
Department of Mathematics
Session: 2022-23
Pabna University of Science
Year: 5th Semester: 1st
and Technology, Pabna.
Department of Mathematics
h
Pabna University of Science and
Technology, Pabna.
 WATER WAVE MECHANICS

Question: Define Wave, Hydrostatic Pressure.

Solution:
Definition of Wave
In wave mechanics, a wave is a disturbance or oscillation that travels through space & matter,
transferring energy from one point to another without the net movement of particles of the
medium.
Definition of Hydrostatic Pressure
At any given point, the pressure generated due to the force of gravity is called hydrostatic
pressure.

Question: Define Types of water wave with examples.

Solution:
i) Linear Wave/small amplitude wave
A linear wave is a type of wave where the wave amplitude is small compared to the wavelength
and water depth, allowing the wave motion to be described by linear equations.
A
≪ 1 →Linear Wave
h
A
And ≪1
L
A = Amplitude of the wave
L = wavelength, h = water depth
Examples: sound waves in air, light waves, and water waves.
ii) Long wave
In water wave mechanics, a long wave (also called a solo wave) is a wave whose wavelength is
much greater than the water depth.
A wave is considered long if: λ ≫ h
λ = wavelength
h = water depth
Examples: Tsunamis, Tidal waves, Storm Surges.

Question: Derive Governing equation for water waves.

Solution:
Derivation of Governing equation for water waves:

The motion is defined with respect to two axis x & z in Cartesian coordinate. Here the water
depth z= -h assumed to the constant & viscosity forces are neglected.
Continuity Equation:
For an irrational & incompressible fluid, the continuity equation can be written as,
∇ .u=0 ………… (1)

There exists the velocity potential function Φ . Such that u=−∇ ϕ

Equation (1) becomes
∇ . ( ∇ ϕ )=0
2
∇ ϕ=0
Since the velocity along to the axis x & z, we have
2 2
∂ φ ∂ φ
2
+ 2 =0 ……………. (2)
∂x ∂z
Which is known as Laplace equation.
For an non- divergent & irrotational flow
∂φ ∂φ
w= ∧u= …………….. (3)
∂x ∂z
w, u velocity component w.r.to x & z
Then in 2D irrotational condition is
∂w ∂u
− =0
∂x ∂z
[ φ=Stream function ¿
The velocity must be Zero
By using equation (3), we get

∂ ∂φ
( ¿−
∂x ∂x
∂ −∂ φ
∂z ∂z
=0 ( )
2 2
∂ φ ∂ φ
2
+ 2 =0 ; Which is also known as Laplace equation in term of stream function.
∂x ∂z
Momentum equation:
Momentum equation for irrotational flow is given by the Bernaulli equation
2
∂φ v p
- + + + gz=0 ……………… (4)
∂t 2 p

[( ) ( ) ]
2 2 2
v 1 ∂φ ∂φ
Equation (4) is non-linear and = +
2 2 ∂x ∂z

In the case very slow motion, the convective term neglected, then
−∂ φ p
+ + gz=0 …………….. (5)
∂t p
Finally, it can be say that the equation (2) & (5) are the governing equation of surface wave.

Question: Discuss several types of Boundary conditions.

Solution: several types of Boundary conditions are,
1. Bottom boundary condition (BBC) – at the bottom of the sea.
2. Kinematic surface boundary condition(KSBC)
3. Dynamic surface boundary condition(DSBC)
4. Lateral boundary condition (LBC)
1) Bottom boundary condition:
At the bottom of the Sea, the number of component of the water particle velocity must vanish.
That is no flow normal to the bottom.
−∂ φ
The condition is z=0= …………………….. (1)
∂z
At z=−h where φ is velocity potential
Kinematic surface boundary condition:
Assume that the free surface is z=η (x , t), then the variation of z w.r.to t
∂ z ∂η ∂η ∂x
= + . ……………… (2)
∂t ∂t ∂x ∂t
We know,
∂ x −∂ ϕ
Introducing u = =
∂t ∂x
∂ z −∂ ϕ
w= =
∂t ∂z
At z=ϕ
−∂ ϕ ∂η ∂η ∂ ϕ
= − . ………………. (3)
∂z ∂t ∂ x ∂ x
Which is non-linear equation also known as a Kinematic surface boundary condition

Dynamic surface boundary condition:

The dynamic free surface boundary condition is a requirement that the pressure on the free
surface be uniform along the waveform, the Bernoulli equation with Pn= Constant is applied on
the force surface, z=ƞ (x , t)

( )
2
−∂ φ 1 ∂φ P
+ ¿+ ¿ + η + gz = F (t)
∂t 2 ∂z ρ

Where Pƞis a constant & usually taken gage pressure, Pƞ= 0

( )
2
−∂ φ 1 ∂φ P
So, + ¿+ ¿ + η + gz = f (t) at z=η
∂t 2 ∂z ρ
Lateral Boundary Condition:
At the stage, boundary conditions have been discussed for the bottom & upper surfaces.
For waves that are periodic in space & time, the boundary condition is expressed as a
periodicity condition,
Φ (x , t)=Φ(x + L , t)
Φ ( x , t)=Φ(x , t+T ); where L is the wavelength & T is the wave period.

Question: Solve the 2D Laplace equation using boundary condition

Or,
For a blow motion φ can be determined from∇ 2 φ=0 , −h ≤ z ≤ η;−∞ ≤ x ≤ ∞ With the
2
∂φ ∂ φ ∂φ
boundary condition =0 where z=-h and 2 + g =0 at z=0.
∂z ∂z ∂z

Solution:
Let us consider the solution of φ (x , z t ) as the form

φ ( x , z , t )=f ( x , z ) cos ⁡(σt + ∈)………………..(1)

2π
Where σ = and T is the time period.
T

Another form,

i ( σt +e )
φ ( x , z , t )=ℜf ( x , z ) e

Let φ ( x , z , t )=u ( x ) p ( z ) f (t ) …………………(2)

Putting (2) in Laplace equation ∇ 2 φ=0

∴ ∇ {u ( x ) φ ( z ) f ( t ) } =0
2

2 2
∂ ( ∂
2
upf )+ 2 ( upf )=0
∂x ∂z

2 2
∂u ∂ p
pf 2
+uf 2
=0
∂x ∂z
2 2
1∂ u 1∂ p
+ =0 …………………(3)
u ∂ x2 p ∂ z 2

2 2
−1 ∂ u 1 ∂ p 2
Assume that = =k … … … … ….. ( 4 )
u ∂ x2 p ∂ z2

Where k is a constant

The equation (4) reduces to

2
d u 2
2
+ k u=0 … … … … … … … … ( 5 )
dx
2
d p 2
And 2
−k p=0 ……………………… (6)
dz

Solving (6) we get,

−kz
p= A e +B e
kz
; where A and B are arbitrary constant

Form (2) , we get

φ=( A e kz + B e−kz ) uf ……………………… (7)

∂φ
For boundary condition =0 at z=−h .
δz

k¿

This equation is true for any x and t,

Then A=B e 2 kh where k ≠ 0

Putting this value of A in equation (7) ,we get

φ=( B e2 kh e kz + B e−kz ) uf

kh
¿ 2 B e cosh k ( h+ z ) uf

∴ φ=D cosh k ( h+ k ) uf ,

Put D=2 B e kh is a constant term .

According to another boundary condition,

2
∂ φ ∂φ
2
+g =0 at z=0 ………………… (9)
∂t ∂z

Differentiating (8) w. r. to z, we get

∂φ
=Dk sinh k ( h+ z ) uf
∂z

∂φ
=Dk sin ( kh ) uf at z=0
∂z

Again diff (8) w.r.to t, we get ,

∂φ df
=D cosh k ( h+ z ) u
∂t dt

2 2
∂φ d f
∴ 2
=D cosh kh u 2 at z=0
∂t dt

Putting these values in equation (9) , we get , D cosh ( kh ) uf’’ + g Dk sin h ( kh ) uf =0

ksin ( kh )
f '' + g f =0
cosh ( kh )

∴ f +σ f =0 … ………………….(10)
'' 2

Where σ 2=g k tanh(kh) (Dispersion relation)

Solution of the equation (10) we get,

f = A1 cos σt + B1 sinσt

Or A1 e iσt + B1 e−iσt

Where A1∧B1 are arbitrary constant

When B1=0 then A1 include in D

iσt
φ=D cos k ( h+ z ) e u ………………..(11)

H
For monochromatic wave, D can be expressed as a function of wave height, a=
2

For the free surface dynamic equation for slow motion.

1 ∂φ
η= z=0
g ∂z

1 iσt
¿ iσD cosh k ( h+ z ) e u at z=0
g

σD
¿− cos h ( kh ) . sinσt .u
g

The expression for φ∧η becomes more convenient

If we get au sinσt for η

−σD
au sin σt = cos h ( kh ) sinσt u
g

−ag 1
∴ D=
σ cosh ( hk )

Equation (11) becomes,

k ( h+ z )
−ag iσt
φ= cosh cosh ( kh ) u e … … … … … ( 12 ) ¿
σ
¿

Again from equation (5) ,we get ,

 2
d u 2
2
+ k u=0
dx

Solving this,

u=∝cosh ⁡(kx)+ B sin ⁡(kx)

ikx −ikx
¿ ∝ e +B e

In particular if u=e−ikx . when ∝=0 , β=1.

Equation (12) becomes,

ag
− cosh k ( h+ z )
σ
φ= e−ikx . eiσt
cosh ( kh )

ag
cosh k ( h+ z )
σ ……………… (13)
¿− e i (−kx +σt )
cosh ( kh )

Since the dispersion relation,

2
σ =gk tanh( kh)

Equation (13) becomes,

φ=−a ¿

−aσ k ( h+ z ) i (−kx+σt )
= cosh e
k cosh ( kh )

Taking real part we get,

−aσ coshk ( h+ z )
φ= . cos ( σt −kx ) ………….. (14)
k sinh ( kh )

Which is velocity potential function of progressing wave travelling in the ox- direction

1
Again in particular if ∝=β= , then
2
 1
u= ¿
2

Then (12) ⇒

−ag k ( h+k )
φ= cosh cos ( kx ) e¿ σ
σ cosh ( kh )

Taking real part,

−aσ k ( h+k )
φ= cosh cos ( kx ) cos ⁡( σt )…………… (14)
k cosh ( kh )

When is the velocity potential function for standing wave travelling in the ox directio n.

Question: Derive the free surface boundary condition in the case of very slow motion.
Or,

Derive the Cauchy-Poission condition at the free surface.

Solution: We know that for small amplitude wave at the free surface p=0
Then in the case of slow motion Bernoulli’s equation becomes

∂φ p
+ + gz=f ( t )
∂t ρ

These, p=0

∂φ
+g η=0 at z=η ………….(1)
∂t

∂φ
Provided that the function f(t) and any additive constant can be included in the value of .
∂t

Since the motion is infinity small, η may be written as

1 ∂φ
η= at z =0 …………(2)
g ∂t

Which is the linearized DSBC.

Now for the free surface z= η(x , t)

dz ∂η ∂ η dx
= + ……….. (3)
dt ∂ t ∂ x dt

Since,

dx −∂ φ
=u=
dt ∂x

dz −∂ φ
And =w at z=η= at z=η
dt ∂z

For (3) we get,

∂φ −∂ η ∂ η ∂ φ
at z=η = + …..…………. (4)
∂z ∂t ∂ x ∂ x

∂η ∂φ
The non -linear term form (4) can be neglected due to the solve motion.
∂x ∂x

This gives with sufficient approximation

∂ η −∂ φ
= at z=0………… (5)
∂t ∂z

This is known as the linearized KSBC .

Diff (2) w.r.to t, we get

2
∂η 1 ∂ φ
= at z =0 ………… (6)
∂ t g ∂ z2

Equation (5) and (6), we get,

2
∂φ 1 ∂ φ
+ =0 at z=0
∂ z g ∂ t2

Which is known as Cauchy – Poission slow motion.

Question: Derive the dispersion relation and also find the wave length, wave velocity from
it and also discuss it for both deep and shallow water.
Solution: If we combine dynamic free surface boundary condition and kinematic free surface
boundary condition, then we get
2
∂∅ ∂∅
2
+g =0 at z = 0 ………….(1)
∂t ∂z

For progressive wave, we know

−Hg coshk ( h+ z )
∅= sin ⁡(kx −σt ) ………… (2)
2 σ cosh ( kh )

Differentiating with respect to t, we get

∂ ∅ Hg coshk ( h+ z )
= σ cos(kx−σt ) .
∂t 2σ cosh ( kh )

Again, Differentiating with respect to t, we get

∂ ∅ Hg 2 coshk ( h+ z )
2
= σ sin ( kx −σt )
∂t
2
2σ cosh ( kh )

Hg cosh ( kh )
¿ σ sin ( kx−σt ) at z =0
2 cosh ( kh )
Hg
¿ σ sin ( kx−σt ) .
2
Differentiating with respect to z, we get

∂ ∅ −Hg sinhk ( h+ z )
= k sin ⁡(kx−σt )
∂z 2σ cosh ( kh )

Hgk sinh ( kh )
¿− sin ( kx−σt ) at z=0
2 σ cosh ( kh )
Hgk
¿− tanh(kh)sin ( kx−σt ) .
2σ
Equation (1) becomes
Hg Hgk
σ sin ( kx−σt ) −g tan h ( kh ) sin ( kx−σt )=0
2 2σ

⇒
Hg
2 {
gk
sin ( kx−σt ) σ− tanh ( kh ) =0
σ }
2
⇒ σ −gk tanh( kh)=0
2
∴ σ =gk tanh ( kh ) . ……….(3)
Which is the required equation of progressive wave.
For standing wave, we know

Hg coshk ( h+ z )
∅= coskx . sin σt . ………(4)
2 σ cosh ( kh )

Differentiating with respect to t, we get

∂ ∅ Hg coshk ( h+ z )
= σ cos ( kx ) . cosσt
∂t 2σ cosh ( kh )

Hg coshk ( h+ z )
¿ cos ( kx ) . cosσt .
2 cosh ( kh )

Again, Differentiating with respect to t, we get

∂ ∅ −Hg 2 coshk ( h+ z )
2
= σ coskx . sinσt
∂t
2
2σ cosh ( kh )
Hg
¿− σ coskx . sinσt at z=0 .
2
Differentiating with respect to z, we get

∂ ∅ Hg sinhk ( h+ z )
= k coskx . sin ⁡(σt )
∂ z 2σ cosh ( kh )

Hg sinh ( kh )
¿ k coskx . sin ( σt ) at z=0
2 σ cosh ( kh )

Hg
¿ k tanh (kh) coskx .sin ( σt ) .
2σ
Equation (1) becomes,
−Hg Hg
σ coskx . sinσt + g k tan h ( kh ) coskx . sin ( σt ) =0
2 2σ

⇒−
Hg
2 { gk
coskx . sinσt σ− tanh ⁡(kh) =0
σ }
2
⇒ σ −gk tanh ⁡(kh)=0
2
⇒ σ =gk tanh ( kh ) . ………(5)

Which is the required equation of standing wave.

Wave length and wave velocity:
2π 2π
We know that σ = ∧K= .
T L
Where, σ is the frequency, T is the wave period, K is the wave number, and L is the wave length.
Putting the values of σ and K in (3) , we have
2
2π 2π 2 πh
( ) =g . . tanh ⁡( )
T L L
2
4 π 2 gπ 2 πh
⇒ 2
= . tanh
T L L

2 2 2 πh
⇒ 4 π L=2 πg T tanh
L
2
gT 2 πh
⇒ L= tanh . ……… (6)
2π L
Again, we know that
L
C=
T
2
1 gT 2 πh
¿ tanh
T 2π L
2 πh
¿ ¿ tanh ….….. (7)
2π L
This is also a relation for wave length, wave speed and water depth.
L σ
Again, = . Then
T K
σ 1
⇒ C= = √ gk tanh kh
K K

¿
√ g
K
tanh kh

¿
√ gL
2π
tanh
2 πh
L
h
For shallow water, ≪1
L
 2 πh 2 πh
Then tanh ≈
L L
2
g T 2 πh
Wave length, L=
2π L
2 2
⇒ L =gh T
∴ L=T √ gh
L
& wave velocity, C=
T
T √ gh
¿ = √ gh
T
For shallow water wave, wave length depends on water depth and wave period. And Wave
velocity depends on water depth.

h 2 πh
For deep water, ≫ 1, then tanh ≈1
L L
2 2
gT gT
Wave length, L= .1=
2π 2π
L
And wave velocity, C=¿
T
2
gT
¿ = ¿
2 πT 2 π
For deep water, wave length and wave velocity depend on wave period.

Question: Find the particle velocity components and acceleration components for
progressive wave.

Solution:
We know, the velocity potential function for progressive wave,

−Hg coshk ( h+ z )
∅= sin ⁡(kx −σt ) ……….. (1)
2 σ cosh ( kh )

Now, the horizontal velocity under the wave is given by

 −∂∅
u=
∂x
Hgk coshk ( h+ z )
¿ cos (kx−σt )
2 σ cosh ( kh )

Hσ coshk ( h+ z ) σ
2
gk
¿ cos ( kx−σt ) [∵ = ]
2 sinh ( kh ) sinh ( kh ) cosh ( kh )
And the horizontal acceleration
∂u
a x=
∂t

H σ coshk ( h+ z )
2
¿− sin ( kx−σt )
2 sinh ( kh )
The vertical velocity
−∂ ∅
w=
∂z
Hgk sinhk ( h+ z )
¿ sin ( kx−σt )
2 σ cosh ( kh )

Hσ sinhk ( h+ z )
¿ sin ( kx−σt )
2 sinh ( kh )

And the vertical acceleration is

∂w
az=
∂t

H σ sinhk ( h+ z )
2
¿ cos ( kx −σt ) .
2 sinh ( kh )

Question: Find the pressure field under progressive wave & discuss the case for deep water
and shallow water.
Solution:
The pressure field associated with a progressive is determined from the unsteady Bernoulli equation

p 1 2 2 ∂φ
+ gz + ( u +v ) − =C ( t ) … … … … …(1)
ρ 2 ∂t
p ∂φ 1 2 2
+ gz− =C ( t )− ( u + v )
ρ ∂t 2
All the free surface z=η , the pressure is taken zero, then the Bernaulli equation becomes

1 2 2 ∂φ
C ( t )− ( u + v )=gη−
2 ∂t
p ∂φ ∂φ
Then, + gz− =gη− … … … … .(2)
ρ ∂t ∂t
We know that from linearized Dynamic free force boundary condition,

1 ∂φ
η= . At z=0
g ∂t
Then equation (2) becomes

p ∂φ
+ gz− =0
ρ ∂t
p ∂φ
Now, = −gz
ρ ∂t
∂φ
p=− ρgz+ ρ
∂t
hg csch k ( h+ z )
p=− ρgz+ ρ . .σ . . cos ( kx −σt )
2σ csch ( kh )

hg csch k ( h+ z )
p=− ρgz+ ρ . . .cos ( kx−σt )
2 csch ( kh )
Which is required pressure field.

There 1st term, −ρgz is the hydrostatic pressure & 2nd term is the dynamic pressure.

Deep water Kh >>1 i.e. Kh > ∞

csch k ( h+ z ) e k (h +z )+ e−k (h+ z )

= kh −kh
csch ( kh ) e +e
kz
¿e
hg kz
p=− ρgz+ ρ . e . cos ( kx−σt ) ; Term for deep water
2
Shallow water, Kh <<1, Kh → 0

csch k ( h+ z ) e k (h +z )+ e−k (h+ z )

= kh −kh
csch ( kh ) e +e
kz −kz
e +e
¿
2
 ¿ cosh ( kz )
hg
p=− ρgz+ ρ . cosh ( kz ) . cos ( kx−σt ) ; Term for shallow water.
2

Question: Show that the particle in water wave travels in a closed path. Also discuss this
path in shallow and deep water.
Or,
Particles are elliptical for deep and shallow water.

Solution: Let δ x and δ z be the displacement of the particle in the horizontal and vertical
direction respectively. We know for the progressive wave,

−Hg cosh k ( h+ z )
ϕ= sin ( kx −σt ) … … … … … …(1)
2σ cosh kh
2
σ
And σ 2=gk tanh kh ⇒ g= … … … … … … …( 2)
k tanh kh
Putting these values of g in equation (1)

−Hσ cosh k ( h+ z )
ϕ= sin ( kx−σt )
2 k tanh kh cosh kh
Hσ cosh k ( h+ z )
¿− sin ( kx −σt )
2 k sinh kh
d −∂ ϕ cosh k ( h+ z )
Now, (δ ¿¿ x)=u= =aσ cos ( kx−σt ) ¿
dt ∂x sinh kh
d −∂ ϕ sinh k ( h+ z )
(δ ¿¿ z)=w= =aσ sin ( kx−σt ) ¿
dt ∂z sinh kh
Integrating,

cosh k ( h+ z ) sin ( kx−σt ) δx

δ x =aσ ⇒ =−sin ( kx−σt ) … … … … … … (3)
sinh kh −σ cosh k ( h+ z )
a
sinh kh
δz
=co s ( kx−σt ) … … … … … … … …(4)
Similarly, sinh k ( h+ z )
a
sinh kh
 2 2
δx δz
+ =1
Now, (3)2 +( 4)2 ⇒ cosh k ( h+ z )
2
sinh k ( h+ z )
2
…………………… (5)
(a ) (a )
sinh kh sinh kh
Equation (5) shows that , Particle travels in closed path .This path are elliptical with major axis
horizontal. Since, cosh k ( h+ z ) >sinh k ( h+ z ) .
cosh k ( h+ z ) sinh k ( h+ z )
Let, A=a ∧B=a
sinh kh sinh kh
2 2
δx δz
(5)⇒ 2
+ 2
=1
( A ) (B)

For shallow water, when z=−h ,

cosh k ( h−h ) a sinh k ( h−h )
Let, A=a = =constant and B=a =0
sinh kh sinh kh sinh kh
⇒ δ x =constant ∧δ z=0

This means that parallel to the horizontal show that the same of such ellipse the minor axes
vanishes at the bottom so degenerate ellipse into a straight line.
h
For deep water, when ≫ 1i . e . kh →0
L
cosh k ( h+ z¿ )
A=a
sinh kh
k (h+ z) −k (h+z )
H e +e
¿ kh −kh
2 e +e
k ( h+ z) −2 k ( h+ z)
H e (1+ e )
¿ kh −2 kh
2 e (1−e )
k ( h+ z) −2 kh −2 kz
H e (1+ e e )
¿
e (1−e−2 kh )
kh
2

H kz
¿ e at kh →0=D(say)
2
Again,
sinh k ( h+ z )
B=a
sinh kh
k (h+ z) −k(h+ z)
H e −e
¿ kh −kh
2 e −e
 k ( h+ z) −2 k ( h+ z )
H e (1−e )
¿ kh −2 kh
2 e (1−e )
k ( h+ z) −2 kh −2 kz
H e (1−e e )
¿ kh −2kh
2 e (1−e )
H kz
¿ e =D(say )
2
2 2
δx δz 2 2 2
( 5) ⇒ 2
+ 2
=1⇒ δ x + δ z =D
( D) ( D)
Which is the equation of circle .Thus the particle path are circle with exponentially decreasing
radius as the path increasing.

Question: What do you mean by velocity potential? Classify the ocean wave according to
their water depth.

Solution: The velocity potential ϕ (x , y , z ,t ) is a scalar function such that the fluid velocity
∂ϕ ∂ϕ ∂ϕ
V =⃗
V is the gradient ofϕ : ⃗
vector ⃗ ∇ ϕ=( , , ).
∂x ∂y ∂z
This means that the fluid particles move in the direction of increasing ϕ and the velocity at any
point is given by the spatial rate of change of ϕ .
Classification of the ocean wave according to their water depth:
d
Ocean waves are classified according to their relative water depth using the ratio , where
L
d=water depth and L=wavelength of the wave .

Based on this ratio, ocean waves are classified into three types:
d 1
1) Deep water wave ( > )
L 2
1 d 1
2) Transitional water wave ( < < )
20 L 2
d 1
3) Shallow water waves ( < )
L 20

Question: Define stream function and find the stream function for small amplitude wave.
Solution: We have known the velocity potential for progressive wave is
−H g cos h k ( h+ z )
∅= sin ( kx−σt )−−−−−− (1 )
2σ cosh ( kh )

The horizontal velocity under this wave is

−∂∅ −δφ
u= and the stream function u=
∂x δz
And the vertical velocity under this wave is
−δ ∅
ω=
δz
And for stream function,
δ∅
ω=
δx
∂∅ δφ
∴ = =+ u−−−−−−−−−−−−−(2)
∂ x δz
And
−δ ∅ δφ
= =w−−−−−−−−−−−−−−(3)
δz δx
Form (1)

∂ ∅ −H gk cos h k ( h+ z )
= cos ( kx−σt )−−−−−(4)
∂x 2σ cosh ( kh )

δ ∅ −H gk sin h k ( h+ z )
And = sin ( kx−σt )−−−(5)
δz 2σ cosh ( kh )

From (2) and (3) we have

δφ −H gk cos h k ( h+ z )
= cos ( kx−σt )
δz 2σ cosh ( kh )

Integrating

−H g sin h k ( h+ z )
φ= cos ( kx −σt )
2σ cosh ( kh )

From (3) and (5) we have

δφ H gk sin h k ( h+ z )
= sin ( kx−σt )
δx 2 σ cosh ( kh )
Integrating,

−H g sin h k ( h+ z )
φ= cos ( kx −σt )
2σ cosh ( kh )

Which is the stream function for progressive wave.

H g cos h k ( h+ z )
For Standing wave, ∅ = cos kx sin σt
2 σ cosh ( kh )

∂ ∅ −H gk cos h k ( h+ z ) δφ
¿> = sin kx sinσt= −−−(6)
∂x 2σ cosh ( kh ) δz

And
δφ H gk sin h k ( h+ z ) δφ
= cos kx sinσt=−¿ −−−−(7)¿
δz 2 σ cosh ( kh ) δx

Now, integrating (6) and (7), we have

−H g sin h k ( h+ z )
φ= sin kx sinσt
2σ cosh ( kh )

Which is the required stream function for standing wave.

Question: Define Potential and kinetic energy. Show that wave energy can be expressed as
a function of wave height.

Solution:

Potential Energy: Potential energy of the wave is defined as the work done to deform a
horizontal free surface into the disturbed state when water is at rest with a uniform free
surface elevation. It can be show readily that the potential energy is a minimum.
The potential energy of a small column of fluid shown in the figure with mess dm relative to the
bottom is, d (PE)=dmg z -------------------------------- (1)
Where, z is the central of gravity of the mess of the column. Can be written as,
h+n
z= −−−−−−−−−−−−−−−−(2)
2
And the differential mass per unit with is,
dm= ρ ( h+n ) dx −−−−−−−−−−−−(3)
The potential overage over one wave height for a progressive wave of wave of height H, then
(PE)T¿ ( PE )wave −(PE)Wlo
Where, wlo means without wave
L L
1 1
PE= ∫ d ( PE )− ∫ d (PE)
L0 L0
L L
1 1
¿ ∫ dmg z − ∫ dmg z
'
L0 L0
L L
1
¿ ∫ ρg ( h 2+ 2hn+ n2 ) dx− 21L ∫ ρgh 2 dx
2L 0 0

L
1
¿ ∫ ρg ( 2 hn+ n2 ) dx
2L 0

H
Inserting surface elevation as a function of x, for n = cos (kx−σt )
2

{ }
L 2
ρg
Then average PE is, PE= ∫ cos ( kx−σt ) + H4 cos 2(kx−σt ) dx
2L 0
L
ρghH
¿
2L 0
∫ ¿¿

[ ] [ ]
L L
ρghH sin ( kx−σt ) ρg H 2 sin2 ( kx −σt )
¿ + X+
2L k 0 16 L 2k 0

¿
2L [
ρghH sin ( kx−σt ) sinσt ρg H 2
k
+
k
+
16 L
L+
] 2K [
sin 2 ( kL−σt ) sin 2 σt
+
2k ]
[ ] [ ]
2
ρghH −sinσt sinσt ρg H sin 2 σt sin 2 σt
¿ + + L− +
2L k k 16 L 2k 2k
 2
ρg H
¿ .L
16 L
1 2
¿ ρg H
16
∴ Potential energy of the waves per unit area depends on the wave height.

Kinetic Energy: The kinetic energy is associated with a small parcel of fluid with mass dm is,
2 2
u +w
d (KE )=dm
2
2 2
u +w
¿ ρdxdz .
2
To Find the average KE per unit surface area, d(KE) must be integrated over depth and average
over wave length.
L n 2 2
KE=
1
∫ ∫ u + w ρfx dz −−−−−−−−−−−−(4)
k 0 −h 2

But we know that for the progressive wave is,

−Hg coshk ( h+ Z )
∅= sin(kx−σt )
2σ cosh ( kh )

−δ ∅ Hgk cosh k ( h+ z )
∴ u= = cos(kx−σt )
δx 2σ cosh ( kh )

And

−δ ∅ Hgk sinhk ( h+ z )
w= = sin ( kx−σt )
δz 2 σ cosh ( kh )

Now,
2 2 2
2 2 H g k
u +w = 2 2
¿
4 σ cos h(kh)
2 2 2
H g k
= 2 2
¿
4 σ cos h ( kh )

+ sin2 hk ( h+ z ) .sin 2 (kx−σt )¿ ¿

2 2 2
H g k
= 2 2
¿
4 σ cos h(kh)

Putting this value of u2 +w 2 in equation (4), we get

 L n 2 2 2
ρ H g k
∴ KE= ∫ ∫ ¿¿¿
2 L 0 −h 4 σ 2 cos 2 (kh)
2 2 2 L n
ρH g k
¿ 2 2 ∫∫ ¿ ¿¿
16 σ L Co s h (kh) 0 −h
2 2 2 L n
ρH g k
¿ 2 2 ∫∫ [1+ cosh 2 k ( h+ z )−1+ cosz ( kx−σt ) ] dxdt
16 σ L Co s h (kh) 0 −h

[ ]
L n
ρ H 2 g2 k 2 sin 2 k ( h+ z )
¿ 2 2 ∫
16 σ L Co s h (kh) 0 2k
+ zcos 2(kx −σt ) dx
−h

L
ρ H 2 g2 k 2
¿ ∫¿¿
16 σ 2 L Co s 2 h (kh) 0
2 2 2 L
ρH g k sin 2 k ( h+n )
¿ 2 2 ∫
16 σ L Co s h ( kh ) 0
(
2k
. x +(h+n)hcos 2(kx−σt )¿)dx ¿

[ ]
L
ρ H 2 g2 k 2 sin 2 k ( h+n ) sin 2(kx −σt )
¿ 2 2
+ ( h+ n )
16 σ L Co s h (kh) 2k 2k 0

ρH g k
2
sin 2 k ( h+n )
2 2
sin 2(kL−σt ) sin 2 σt
¿ 2 2
[ + ( h+n ) + ]
16 σ L Co s h (kh) 2k 2k 2k
2 2 2
ρH g k
¿ 2 2
¿¿
16 σ L Co s h ( kh )
2 2 2
ρH g k L
¿ 2 2
. . sinh ⁡(2 kh)
16 σ L Co s h ( kh ) 2 k

ρH g k
2
sin ( 2 kh )
2 2
¿ 2 2
.
16 σ L Co s h ( kh ) 2k

ρH g k
2
2 sinh ( kh ) cosh ⁡(kh)
2 2
¿ 2 2
.
16 σ L Co s h ( kh ) 2k

ρ H g k tanh ( kh )
2 2 2
¿ 2
16 σ k
2 2 2 2
ρH g k σ
¿ 2
.
16 σ gk . k
1 2
¿ ρg H
16
Kinetic energy of the waves also depends on the water height.
So that, the total energy,
E=KE+ PE
1 2 1 2
¿ ρg H + ρg H
16 16
1 2
¿ ρg H
8
Thus, the total energy depends on the wave height.

Question: Show that the energy flux of the waves can be expressed as E f =E C g , where
C g is the group velocity.

Solution: The rate at which the energy is transmitted is called energy flux E f , and for linear
wave theory it is the rate at which work is being done by the fluid on the other side.

Figure: Energy Flux

 η

Mathematically, E f =∫ Pd u dz where Pd is the dynamic pressure. Pd =P+ ρgz .

−η

cosh k ( h+ z)
Pd =ρgη
cosh kh
The average energy flux E f is obtained by averaging over a wave period.
T n
1 cosh k ( h+ z ) gkη cosh k ( h+ z )
Ef= ∫ ∫
T 0 −η
ρgη
cosh kh σ cosh kh
dz dx

T n 2 2
1 ρgσ H 2 cosh k ( h+ z )
¿ ∫∫
2
cos (kx −σt ) dz dt
T 0 −η 4 sinh 2 kh

And we obtain the energy flux,

Ef=
ρg H 2 σ 1
8 k 2
1+
[(
2 kh
sinh 2 kh )]
=Ecn, Where cn is the speed at which the energy is transmitted.

This called group velocity C g , since C g=cn .

( )
Cg 1 2 kh
∴ n= = 1+
C 2 sinh 2 kh

Thus E f =E C g .

Question: Define radiation stress. Shows that for shallow water the RS can be expressed as
3 2
δ xx = ρg H .
16

Solution: A stress is by definition equivalent to a flow of momentum. Thus the radiation stress
may be defined as the excess of flow of momentum due to the pressure of the waves. Thus flux
of momentum is formed by two contributions:
One due to the wave-induced velocities of the particles and another one due to the pressure.
Consider first an undisturbed body of water of uniform depth. The pressure p at any point is
equal to the hydrostatic pressure.
P0=− ρgz … … … … … … … … … … … …(i)

Thus the total flux of horizontal momentum between surface and bottom (z =- h) is
0
I p=∫ P 0 dz … … … … … … … … … … … … .(ii)
−h

Now consider the momentum flux in the pressure of a simple progressive wave motion (Fig-1).
A quite general expression for the instantaneous flux of horizontal momentum across unit area
of a vertical plane in the fluid is P+ ρ u2 .

Thus the total flux of momentum across a plane, x=constant , we have integrate between
z=−h to z=η is,
n
I m=∫ ( P+ ρ u ) dz
2

−h

The components of the radiation stress in the direction of wave propagation is defined as the
time averaged total momentum flux due to the presence of waves, minus the mean flux in the
absence of waves:
n 0
i .e S xx =∫ ( P+ ρ u ) dz−∫ P0 dz … … … … … … .(iv)
2

−h −h

Where the subscript xx denotes the x directed momentum flux across a plane define by
x=conestrant ,
Now,
 n n 0
S xx =∫ ρu dz +∫ Pdz−∫ P0 dz
2

−h −h −h

0 n 0 n 0
¿ ∫ ρu dz +∫ ρ u dz+ ∫ Pdz+∫ Pdz−∫ P0 dz
2 2

−h 0 −h 0 −h

0 n 0 n

¿ ∫ ρu dz +∫ ρ u dz+ ∫ (P−P¿¿ 0)dz +∫ Pdz ¿

2 2

−h 0 −h 0

1 2 3 4
¿ S xx + S xx + S xx +S xx … … … … … … … … … … …(5)

Consider the contribution S2xx . Since it contributes only third order term of H.

So neglecting the higher order term of H.

0 0
S xx =∫ ρu dz=∫ ρ u dz
1 2 2

−h −h

0 0
S =∫ (P−P 0)dz=∫ (−ρ ω)dz
3
xx
−h −h

We know that mean flux of vertical momentum across any horizontal plane, which is P+ ρ ω 2 dz
must be just sufficient to support the weight of the water about it. Since the mean water level
is z=0 have then,
2
P+ ρ ω =−ρgz=P0
2
⇒ P−P0=−ρ ω
0
∴ S =∫ −ρ ω dz
3 2
xx
−h

For S4xx we can write the pressure term in the range of 0to η as

P= ρg(η−z )
0 0 η
∴ S xx =∫ ρ u dz +∫ − ρω dz+∫ ρg(η−z )dz
2 2

−h −h 0

0 η
¿ ∫ ρ ( u −ω ) dz+∫ ρg(η−z )dz
2 2

−h 0

0 η
¿ ∫ ρ ( u −ω ) dz+∫ ρg(η−z )dz ……………………….(vi)
2 2

−h 0

Now we have
0

∫ ρ ( u2−ω2 ) dz
−h

0 T
ρ
¿ ∫ ∫ ( u −ω ) dz dt
2 2

−h T 0

0 T
ρ
¿ ∫∫ {a 2 σ 2 ¿ ¿ ¿
T −h 0
2 0 T
a σρ 1
¿ 2 ∫ ∫ {cosh2 k ( h+ z)cos 2 (kx−σt ¿)−sinh 2 k ( h+ z ) sin2 (kx−σt ) }dtdz ¿
sinh kh T −h 0
M
¿ (t −t )
T 1 2
Where,
0 T
t 2=∫ ∫ cosh k ( h+ z)cos (kx−σt ¿)dtdz ¿
2 2

−h 0

0 T
t 2=∫ ∫ sinh k ( h+ z ) sin (kx−σt ) dtdz
2 2

−h 0

2
a σρ
And M = 2
sinh kh
Now,
0 T
t 1=∫ cosh k (h+ z)dz ∫ cos ( kx−σt ¿) dt ¿
2 2

−h 0

0 T
1
¿ ∫ cosh k (h+ z)dz ∫ 2cos (kx−σt ¿)dt ¿
2 2
2 −h 0

0 T
1
¿ ∫ cosh 2 k (h+ z)dz ∫ {1+cos (2 kx−2 σt ¿)}dt ¿
2 −h 0

0
1
¿ ∫ cosh k (h+ z)dz ¿ ¿ ¿
2
2 −h
0
1
¿ ∫ cosh k (h+ z)dz ¿ ¿
2
2 −h
0 0
1 sin ⁡(2 kx−4 π) sin ⁡(2 kx ) 2π T
¿ ∫ cosh k (h+ z)dz [T −¿ ]¿ ¿ ∫ cosh k (h+ z )dz
2 2
+ ][σ=
2 −h 2σ 2σ T 2 −h
 0
T
¿ ∫ {1+cosh 2 k (h+ z)}dz
4 −h
0
T sin ( 2 kh+2 kz )
¿ [ z+ ]
4 2k −h

T sin 2kh
¿ [0+ + h−0]
4 2k
T 1
= [ sin 2 kh+h]
4 2k
Similarly,
T
t 2= ¿
4
0 M
∴ ∫ ρ ( u −ω ) dz = T ( t 2−t 1 )
2 2

−h
.
M T
¿ { ¿
T 4
M
¿ 2h
4
2
1 a σρ
¿ …………………………………(vii)
2 sinh2 kh

Again,
η

∫ ρg(η−z )dz
0

T η
1
¿ ∫∫ ρg ( η−z ) dz dt
T 0 0
T 2 η
ρg
¿
T 0
∫ [ηz− z2 ] dt
0

T 2
ρg η
¿ ∫ (η − ¿)dt ¿
2
T 0 2
T 2
ρg η
¿ ∫ dt
T 0 2
 T 2
ρg H
¿ ∫
2T 0 4
2
cos ( kx−σt ) dt

2 T
ρg H 1
¿
8T 2 0
∫ 2
2 cos ( kx −σt ) dt

2 T
ρg H
¿
16 T
∫ ¿¿ ¿
0

T
ρg H 2 sin ( 2 kx−2 σt )
¿ [t− ]
16 T 2σ 0

¿
ρg H 2
16 T
T−
[
sin ( 2 kx −2 σT ) sin ( 2 kx )
2σ
+
2σ ]
¿
ρg H 2
16 T
T−
2σ [
sin ( 2 kx ) sin (2 kx )
+
2σ ]
2
ρg H
¿ T
16 T
2
ρg H
¿
16
E
¿
2
Hence from (vi) we get,
2
1 a σρ E
S xx = 2
+
2 sinh kh 2
2 2
1 H a σρ E H
¿ + [a= ]
2
2 4 sinh kh 2 2
2
ρh H gk tanh kh E
¿ 2
+
8 sin kh 2
2
ρh H kh E
¿ +
8 sinh kh cosh kh 2
E .2 kh E
¿ +
sinh 2 kh 2
2 kh 1 1 2
¿ E( + ) where E= ρg H
sinh 2 kh 2 8
 E 1 2
For deep water (ie kh>>1). The ratio tends to zero and so S xx = = ρg H
2 16
3 3 2
And, for shallow water (i.e kh<<1) the ratio tends to 1 and so , S xx = E= ρg H
2 16

Question: A wave in water 100m deep has a period of 10s a height of 2m. Determine the
wave celerity /velocity, length and steepness.

Solution:
Given that,
Water depth n=100m
Wave period T=10s
Wave height H=2m
We know,
2 2
gT 9.81 ×10
L= = =156.13 m
2 π 2× 3.1416
Again we know for deep water wave.
L 156.13
Wave celerity/velocity c= = =15.6 m/ s
T 10
H 2
Wave sharpness = = =0.0128
L 156.14

Question: A wave with a period 10s is propagated shore word over a uniformly sloping shell
from a depth h=200m to depth h=3m.

Solution:
Here, T=10s, h=200m
2 2
gT 9.81 ×10
L= = =156
2 π 2× 3.1416
h 200 1
Then, = =1.2821>
L 156 2
Therefore this is a deep wave,
L=156m
 L 156
And c= = =15.6 m/s
T 10
For h=3m
h 3
= =0.0192
L 156
By trial and error solution it is found that
d
=0.05641
L
∴ L−53.2 m
1 d 1
Transitional depth, since < <
25 L 2
L 53.2 −1
c= = =5.32 m s
T 2
An approximate value of L can be found by

L=
g T2
2π √tanh
2 πh
L
=54.1 m

L 54.1 −1
c= = =5.41 m s
T 10

Question: A wave period T=8 sec in a water depth d=15m and height H=5.5m. Find the
horizontal and vertical velocities and acceleration a x , az at an elevation Z= -5m (below SWL)
π
and kx −σt = (6o degree).
3

Solution:
Given that,
T=8 sec, H= 5.5 m, d=15m
o
kx −σt =θ=60
2
gT 9.81× 64
We know that L0= = =99.8 m
2 π 2× 3.1416
d
=0.1503
L0
 d
By trial and error solution for =0.1503
L0

d
We find =0.1835
L
15
⇒ L= =81.7
0.1835
Horizontal velocity
−δϕ
u=
δx
Hgk cosh k (h+ z)
¿ cos kx−σt
2σ cos kh
HgT cosh k (h+ z )
¿ cos θ
2L cos kh
−1
¿ 0.99 m s
−δϕ
w=
δz
HgT sinh k (h+ z)
¿ sin θ
2L sin kh
−1
¿ 1.21 m s
δu
And acceleration a x =
δt
δ HgT cosh k ( h+ z )
¿ [ cos kx−σt ]
δt 2 L cos kh
Hgk cosh k ( h+ z ) sin kx−σt
¿ σ
2σ cos kh
Hgπ cosh k ( h+ z ) sin θ
¿
L cos kh
−2
¿ 1.35 m s
δ
az= ¿
δt
Hgπ sin k (h+ z)
¿− cos θ
L cos kh
−2
¿−0.50 m s
Question: A wave traveling in water depth h=3m, with a period T= 15 Seconds and height
H= 0.1m. Using cnoidal wave theory find the wave length and compare this length with the
length determined using Airy Theory.

Solution: Given, h=3 m, T =15 Sec , H=1 m

H 1
Calculate = =0.33
h 3

And T
√ √ g
h
=15
9.8
3
=27.11

H
h
and T

2
g
h √
to determine the square of the modulus of the compute Elliptical integral k 2

−5
X =10−10
With the values of k 2gives
2
L H
3
=290
h

L=
√ 290 h3
H √ 3
= 290× =88.5 m
1
∴ Lecroidal=88.5 m
According to airy / Linearly wave theory
2
gT 2 πh
L= tan h( )
2π L
2
gT 2 πh
¿ ×
2π L
¿ T √ gh=15 √ 9.8 ×3=81.33 m
Lairy = 81.33
To Check whether the wave condition are in the range for which cordial wave theory is valid.
h 3 1
= =0.0339<
L 88.5 3
2
L H
And 3
=290>27
h
Therefore, Cordial theory is Applicable.
Wave Celerity
L 88.5 −1
Ccroidal ¿ = =5.9 m s
T 15
L 87.33 −1
Cairy¿ = =5.37 m s
T 15
If it is assumed that wave period same then,
Ccroidal L croidal
= ≈1
C airy Lairy

Question: What is wave breaking? Discuss the different types of waves breaking with
approximate graphs and surf similarity parameter with the breaking conditions.

Solution:
Wave breaking: Waves break when their height reaches a certain limiting value relative to their
length or the water depth. That is, the wave breaking will typically take place when the wave
height is about 0.8 times the local water depth.
Type of waves breaking: Breaking waves exhibit different forms, principally depending on the
incident wave height, period and on the beach slope. Commonly breaking waves have been
classified into four different types based on the physical changes of the surface profile during
the breaking process.
They are,
i. Spilling breakers
ii. Plunging breakers
iii. Surging breakers
iv. Collapsing
breakers.
1. Spilling breakers: Spilling breakers are observed when waves of large steepness are incident
to a gently sloping beach. Turbulence and foam first appear at the wave crest and spread down
the front face of the wave as the wave propagates forward. It appears as if the wave is
“plowing” the foam as it moves forward. The turbulence is uniformly dissipating wave energy,
resulting in a continual decrease in the wave height as the wave propagates forward.

(a). Spilling breakers.

2. Plunging breakers: Plunging breakers are observed on relatively steep beach slopes. Waves
become asymmetric about the crest, which exhibits a sharpening of the wave crest with an
almost vertical front face. Then the waves over turn, and the overturned crest plunges into the
water ahead. This process is schematically illustrated in fig-(b). The deformation of the wave
prior to breaking in a plunging breaker is more rapid than that in a spilling breaker. The water
particle motion in a plunging breaker is also violent and large sand clouds typically appear at
the plung point. The impact force on a structure produced by a plunging breaker is also large.

(b). Plunging breakers.

3. Surging breakers: Surging breakers are observed when waves of small steepness propagate
into a steep sloping beach. The waves break as though they were tripped over by the return
flow of the preceding waves, as shown by the figure-©.
 ©. Surging breakers.

4. Collapsing breakers: As the front face of the wave steepness at incident breaking, the lower
portion of the face plungs forward and the wave collapses. The collapsing breaker is an
intermediate form between the plunging and surging form and is not as clearly defined as
other.

(d). Collapsing breakers.

Mathematical Definition:
The types wave breaking can be identified based on mathematical definition.
Battjes (1974) defined surf similarly parameter I as
tanβ tanβ
I 0= ∨, I b =

√ H0
L0 √ H0
L0

For spilling breaker,

I 0 <0.5∨, I b <0.4
For plunging breaker,
0.4 < I 0< 3.3∨, 0.4 < I b< 2.0

For surging breaker.

3.3< I 0∨,2.0< I b

Question: Define capillary wave. Show that the wave celerity for capillary waves can be

√
'
found as c= g + σ k .
k ρ

Solution:
Capillary wave: A capillary wave is a wave travelling along the phase boundary of a fluid, whose
dynamics and phase velocity are dominated by the effects of surface tension. Unlike larger
waves, capillary waves create short-wavelength ripples when a disturbance (small objects,
raindrops or wind) hits a calm body of water.
The wave length of capillary wave on water is typically less than few cm, with a phase speed in
excess of 0.2-0.3 m/s.
2nd part: Our assumption made when we arrived the small amplitude theory as the surface
tension is negligible. This assumption is reasonable if the wave height exceeds about 3 cm.
We consider a velocity potential of the form,
∅ = A cosh k (h+ z)sin(kx−σt ) ------------------------(1)

Which satisfies the Laplace equation.

The surface displacement associated with equation (1) will be
H
η= cos(kx−σt ) --------------(2)
2
Including surface tension σ ', substituting (1) and (2) into the linearized form of

−∂ ϕ Pa
[( ) ( ) ]
2 2 2
' ∂ η 1 ∂ϕ ∂ϕ
+ −σ 2
+ + + gz=c (t)
∂t ρ ∂x 2 ∂x ∂z

We get,
2
−∂ ϕ ' ∂ η
−σ 2
+ gη=0 at z=η
∂t ∂x
 2
∂ϕ ' ∂ η
⇒ +σ 2
−gη=0 at z=η ----------------------(3)
∂t ∂x
The bottom and kinematic condition doesn’t change, Solving the boundary value yields,

( )
'
kσ
−H g−
ρ cosh k ( h+ z ) ----------------(4)
∅= sin(kx−σt )
2σ cosh ( kh )
Which is the velocity potential for capillary wave.
Now for dispersion relation,

( )
'
g σk
2
C= + tanh( kh)
k ρ

(( ) )
' 1
g σk 2
⇒ C= + tanh ( kh ) ---------------(5)
k ρ

It can be seen that the effect of surface tension is to increase celerity for all wave frequency
that have significant surface tension effects will be short.
For the deep water,
From the equation (5) will represent wave celerity of capillary wave,

√( )
'
g σ k since, tanh(kh)≈ 1.
C= +
k ρ

Question: What is wave shoaling. Find the wave height before breaking and hence find the
shoaling co-efficient.
OR,

Show that, H=H 0

√ 1 1
2n tanh( kh)
and
H
H0
=K s=
1
√
1
2 n tanh(kh)
.

OR,

Show that, the wave shoaling can be expressed as H=H 0

√ C0
2Cg
.

Solution:
Wave Shoaling: If the waves are incident normal to the beach with straight and parallel bottom
contours change in the wave profile is called solely by the change in water depth. This
transformation is called wave shoaling.
2nd part: wave height before breaking:
We know the energy flux of small amplitude wave is/ can be expressed as,
2
ρg H
E f =E cn= cn -------------------(1)
8

Where, n=
1
2(1+
2 kh
sin 2 kh )
------------------(2)
H
2 kh
For deep water, kh ≫ 1 then →0
sin 2 kh
Then equation (1) becomes,
h
2
ρg H 0 c0
Ef = --------------(3)
0
16
Where H 0 and c 0 are the deep water wave Height and wave velocity.

The relative wave height can be found by (1) and (2), then
E f =E f 0

(∵ n= 12 )
2 2
ρg H ρg H 0 c 0
⇒ cn=
8 16

H
2
c0
⇒ 2
=
H 0 2 nc

⇒
H
H0
=
2n c√
1 c0
---------------------(4)

We know,
L
c=
T
2
g T tanh(kh)
¿
2 πT
But for deep water wave c=c 0 and tanh(kh)→1
2
gT
c 0=
2 πT
 ¿ ¿
2π
Now,
¿
c0 2π
=
c tanh ( kh )
> ¿¿
2π
¿ ∗2 π
2π
¿
¿ tanh( kh)
1
¿ -----------------(5)
tanh( kh)

From (4) and (5) we get,

H
H0
=
1
√ 1
2 n tanh (kh)
H
Since, =K s then
H0

H
H0
=K s=
1
√
1
2 n tanh(kh)

¿
√1 c0
2n c

¿
√ c0
2 cg
[ ∵ cn=cg ]

∴ H=H 0
√ c0
2 cg
. (Showed.)

Question: Define wave refraction and derive Snell’s law for water wave.
Solution:
Wave Refraction: If the wave period is constant, the celerity of wave mainly depends on the
local water depth while wave height has a minor effect, the celerity distribution is also distorted
if local currents exist in the field.
A gradient in the wave celerity along the crest line results in a modification of wave directions
such a kinematic transformation of wave is referred as wave refraction.

^
Snell’s law: The wave number vector K̅ represents in the x-direction, K= K i+0 ĵ and |K|=K .

^ K y ĵ and |K|=√ K 2 + K 2 .
Now for x-y space it becomes, K= K x i+ x y

If θ is the angle between the beach normal and the wave direction then,
K x =K cos θ

K y =K sin θ … … … … (1)

Consider irrotational flow, then we have ∇ × K=0

| |
i^ ^j k^
∂ ∂ ∂
⇒ =0
∂x ∂y ∂z
Kx Ky 0

⇒ ( ∂∂Kx − ∂∂Ky ) k=0

y ^ x

⇒(
∂y )
∂K ∂K y x
− =0
∂x
∂ Ky ∂Kx
⇒ = … … … … (2)
∂x ∂y
∂ K sin θ ∂ K cos θ
Using (2) & (3) we have, = … … … …(3)
∂x ∂y
For a shore line where the long shore variation in the y-direction of all variables are zero.
d K sin θ
If the wave varies only in X-direction then, equation (3) becomes =0
dx
⇒ K sin θ=constant
K sin θ
⇒ =constant
σ
sin θ
⇒ =constant
σ /k
sin θ
⇒ =constant
c
For deep water, c=c ˳ and θ=θ ˳then,
 sin θ ˳
⇒ =constant
c˳
sin θ sin θ ˳
Therefore, ⇒ =
c c˳
Which is the Snell’s law.

Question: Define dimensional analysis. Derive the dimensionless form of equation of

continuity and dimensionless of equation of motion.

Solution:
Dimensional analysis: Dimensional analysis is that each quantity is made up of the fundamental
dimensionless of mass, length and time that any relationship in their fundamental dimensions.
Since mass, length and time are considered fundamental dimensions, the units attached those
quantities are called fundamental units.
xi ¿ ui ¿ u ˳ p
2nd part: The dimensionless variables are defined as, x ¿i = ¿
, ui = , t = t , p = ,
L˳ u˳ L˳ p˳
¿ h ¿ ρ
h= ∧ρ =
h˳ ρ˳
Where L˳, u˳, p˳ and ρ ˳ are dimensional constant of reference and characteristic in the flow
system.
Under the above transformation the equation of continuity becomes,
∂ρ ∂
+
∂t ∂ x i
( ρ ui )=0 becomes

u˳ ρ ˳ ∂ ρ ¿ u ˳ ρ ˳ ∂
¿ ( ρ ui ) =0
¿ ¿
¿+
L˳ ∂ t L˳ ∂ xi

⇒
{
u˳ ρ ˳ ∂ ρ ¿ ∂
+
L˳ ∂ t ¿ ∂ x ¿i
¿ ¿
}
( ρ ui ) =0
¿
∂ρ ∂ ¿ ¿
⇒ ¿ + ¿ ( ρ u i )=0
∂ t ∂ xi

Which is the dimensionless form of the equation of continuity.

Now we know the equation of motion,
 2
D ui ∂h ∂ p μ ∂ ∂uj ∂ ui
ρ =−g ρ − + +μ
Dt ∂ x i ∂ xi 3 ∂ x i ∂ x j ∂ xi ∂ x j

And the equation of motion becomes,

2 ¿ ¿ ¿ ¿ 2 ¿
u˳ ρ˳ ¿ D ui ¿ ∂h ρ ˳ ∂ ρ 1 μ ˳ ∂ ∂ u j μu ˳ ∂ ui
ρ =−ρ ˳ g ρ − + μ + … … … …(1)
L˳ D t¿ ∂ x ¿i L ˳ ∂ x¿i 3 L ˳2 ∂ x ¿i ∂ x j¿ L ˳2 ∂ x¿i ∂ x j¿
2
u˳ ρ ˳
Now dividing (1) by , we get,
L˳
¿ ¿ 2 ¿
¿D ui −ρ¿ ∂ h ¿ ¿
∂ p 1 1 ∂ ∂u j ∂ ui
ρ ¿ = ¿ −2 c ¿+ ( ¿+ )
Dt 2
F ∂ xi
p
∂ xi R 3 ∂ x i ∂ x j ∂ x¿i ∂ x j¿
¿

L˳u˳ ρ ˳
Where, R= is the Reynolds number,
μ
p˳
c p= 2 is the pressure coefficient,
2 ρ ˳u˳
u˳
And F= is the Froude number.
√ g̅ L ˳
Which is the dimensionless form of equation of motion.

Question: Define wave run-up.

Solution: Wave run-up is defined as the maximum vertical elevation above the still water level
to which the water rises on the beach.

Mathematically, it can be obtained (Hunts ,1959),

 Ru tanθ
=

√
H0 H 0 , where H 0 and L0 are the deep water wave height and wave length.
L0

Question: Define solitary wave and show the solitary wave in the form

η=a sech 2
(√ )
3a
4 h3
x.

Solution:
Solitary wave: A solitary wave is a wave having a surface displacement that is completely above
the still water lines. Solitary waves have only wave crest, no wave through.

Wave Crest

Still Water Level

2nd part: We know the steady state form of Korteweg-de Vries equation in non-dimensional
form:

()
3π 2
1 d 3 α dπ dπ ga 1 x η h h
3
+ π + ( 2
− )=0 …….(1) ; where X = ; π = ; α = ; β=
3 dx β dx dx βα c β 2 2 2 L

Now integrating (1) With respect to x we get ,

( )
2π
1 d 3 α 2 dπ ga 1
2
+ π +π 2
− + D=0……..(2)
3 d x 2 β dx βα c β
dπ
multiplying (2) by again integrating with respect to x we get ,
dx

( )
2
1 d π dπ 3 α 2 dπ ga 1 dπ dπ
2
+ π +π 2
− + D =0
3 d x dx 2 β dx βα c β dx dx

( ) ( )
2 2
1 dπ α 3 π ga 1
or, + π + 2
− + Dπ + E=0 … . ( 3 )
6 dx 2β 2 βα c β

where D∧E areintregation constants .

dπ
we consider a single wave which was no influence at infinity then =π=0 at x=∞
dx
from equation (2) and (3) we get D=0 & E=0 .the equation (3) cam be written as ,
 ( ) ( )
2 2
1 dπ α 3 π ga 1
or, + π + − =0
6 dx 2β 2 βα c 2 β

( )
2 2
dπ 3π ga
or, = (1− 2 −απ ) … .(4)
dx 2β c α
dπ
for the above from to be the symmetric about x-axis =0 at π =1
dx
then equation (4) becomes,

( )
2
3.1 ga
or,0= 1− 2
−α
β α .c

(
0r, 1−
ga
α .c
2
−α =0
)
a −1 a
c= √ gh ( 1−α ) ( since α = h ,∨, h= α ¿
2

α
( )
or,c= √ gh 1+ (neglecting higher order)
2

then from equation (4) becomes,

( )
2 2
dπ 3π
= α (1−π )
dx β

or,
dπ
dx
=
3α
β √
π √ (1−π )

or,∫
1. dπ
π √ 1−π
=∫
3α
β
dx ……..(5)
√
let π=z 2 , dπ=2 zdz

so (5) becomes ∫ 2 zdz

z √ 1−z
2 2
=
3α
β
x
√
or,∫
zdz
z √ 1−z
2 2
=
3α
4β
x
√
−1
¿ , sech ¿ ¿)=
√ 3α
4β
x

or,z=sech
√ 3α
4β
x
1
or,√ π =sech ( √ 34 αβ x )
 (√ 34 αβ x)
or, π=sec h2

or, =sec h (
η
a
2

√ 4 β x)
3α

(√ )
a
3
h x
or,η=a sec h 2 2
h L
4 2
L

or,η=a sech
2
(√ )
3a
4 h3
x Figure: dimensionless free surface profile of a solitary wave

Question: Define cnoidal wave theory and show that a cnoidal wave can be expressed as

η=asech 2 x
(√ 3a
4 h3 k 2).

Solution:
Cnoidal wave: The wave profile in developed of Jacobian elliptic integral, c n (a ) they are called
the cnoidal to be consistent with the sinusoidal or airy theory.
2nd part: We know the steady state form of Korteweg-De Vries equation in non-dimensional
form

()
3π 2
1 d 3 α dπ dπ ga 1 x η h h
3
+ π + ( 2
− )=0 …….(1) ; where X = ; π = ; α = ; β=
3 dx β dx dx βα c β 2 2 2 L

Now integrating (1) With respect to x we get ,

( )
2π
1 d 3 α 2 dπ ga 1
2
+ π +π 2
− + D=0……..(2)
3 d x 2 β dx βα c β
dπ
multiplying (2) by again integrating with respect to x we get
dx

( ) ( )
2 2
1 dπ α 3 π ga 1
+ π + 2
− + Dπ + E=0 … . ( 3 )
6 dx 2β 2 βα c β

For cnoidal wave, we force π=0 and z=1 define as the wave trough. Since the wave is periodic
dπ
then =0.
dx
Then from equation, (3) we get, E=0
1
¿ ) + Dπ ¿ 0 …………(4)
6
Again, we consider π =1 at the wave crest
dπ
Then =0,then from equation (4)
dx
α 1 ga −¿ 1
 + ( )+D=0
2 β 2 β c2 α β
α 1
+ ¿ 1) + D=0
2β 2β
α 1 1 −ga
 − F+ D=0 , where F= (1 2 )
2β 2 β c α
1
 D= ¿ )
2
Putting the value of D in equation (4)
1
¿ ) π=0
6
¿) π
2 3α 3 α
=3 π F− π −3 π ( F− )
β β
¿
3α β 2 3 β
¿ [ F π −π −F π +π ¿
β α α
3α β 2
¿ [ F π ( π −1 )−π (π −1)¿
β α

¿
3α
β [β
π ( π−1 ) F −π −1
α ]
¿−
3α
β [β
π ( π −1 ) 1−F + π
α ]
β
¿, Where s=(1−F )
α
Now substituting π=cos 2 x
 π
The values of X are seen to be 0 and
2
π
Since when π=0 then X =
2
When π=1 then X=0
Now, π=cos 2 x
dπ d 2
= (cos x)
dx dx
dx
¿−2 cosx sinx
dX
Now from equation (5)

( )
2
dx 3α
cos x ( 1−cos x )( cos x+ 5 )
2 2 2
−2 cosx sinx =
dX β

2 cosx sinx
dx
dX
=
√
3α
β
cos2 x sin 2 x ( cos2 x+5 )


2 dX=
3α( 2
β √
cos x+ s ) cosx sinx
cosx sinx
dX

1 dx
dX =
 3α
4β √ √1−sin2 x + s

1 dx
¿

√ 3α
4β
√1+ s−sin2 x

1 dx
¿

√ √
2
3α 1
(1+ s ) {1− sin x }
4β ( 1+ s )
1 dx
dX = 1

√ √1−k 2 sin2 x
2
 3α where k =
(1+s ) 1+ s
4β
Now integrating
 1 x ds
X= ∫
√ 3α
4β
(1+s )
√1−k 2 sin2 s …………(6)

[changing variables]
We know the incomplete Jacobi elliptic integral of first kind
x dx
V =F ( x , k )=∫
√1−k 2 sin 2 s
Then from (6)
1
x= F (x , k )

√ 3α
β
( 1+s )

 F ( x , k )=x
√ 3α
4β
(1+ s)

¿x
√ 3α
4 β k2

Now for the Jacobi elliptic integral of first kind the elliptic cosine cnu is given by
C n .U =cosx

Cn . F ( x , k )=cosx

 Cn x {√ } 3α
4 β k2
=cosx

{√
 cosx =Cn x
3α
4 β k2 }
−1
 x=cos [Cn x{√ 3α
4 β k2
]
}
cosx =Cn x {√ 3α
4 β k2 }
{√ }
2
2 3α 2
cos x=Cn x 2 ——(7)
4 βk

We know the complete Jacobi elliptic function of first kind

 π
2
ds ——(8)
k ( m )=∫
0 √ 1−m 2 2
sin s
π
Since x has values, 0 and
2
Then from equation (5) we can write
π

√ 3α
2
ds 1
where m=
2 ∫
X = 1+s
4 β k 0 √ 1−msin2 s

X
√ 3α
4 β k2
=k ,[ from (5)]

Then mod ( k )=X

√ 3α
4 β k2
Then equation (7) becomes

{√ }
cos 2 x=cn2 x
3α
4 β k2
mod ( k )

 π=cn2 x
{√ } 3α
4 β k2
❑ 2
 a =cn ¿

2
¿ acn x {√ } 3a
4 h3 k 2

Taking the modulus of Jacobian elliptic function, mod(k) approximately 1

2
¿ acn F ( k , x )
Which is the dimensional form.

Question: Derive the water surface profile and horizontal velocity component for
geostrophic effects.
OR,
Geostrophic effects on long waves.

Solution:
The earth’s rotation plays an important role in long wave motion when the Coriolis acceleration
becomes significant.

The frictionless equation of motion for long waves, on a rotating surface, are modified by the
introduction of two terms as follows:

∂u ∂u ∂u ∂η
+u +v −f c v =−g … … … … (1)
∂t ∂x ∂y ∂x

And

∂v ∂v ∂y ∂η
+u +v −f u=−g … … … …(2)
∂t ∂x ∂y c ∂y

Where shear stresses have been neglected. The continuity equation is:

∂ η ∂ u( η+ h) ∂ v (η+ h)
+ + =0 … … … …(3)
∂t ∂x ∂y

Consider the propagation of long progressive waves in an infinitely long straight equal in the x-
direction with a flat bottom. The transverse velocity v is considered negligible. The equation of
motion in the x-direction, therefore is not affected by the presence of the Coriolis force.

∂η
In the y-direction the equation reduces to f c u=−g … … … …( 4)
∂y
If we linearize the equation of motion in the x-direction i.e., equation (1), a solution can be
assumed as:
η=η^ ( y )cos( kx−σt )

η c c η^ ( y )cos (kx −σt )

U= =
h h
∂η ∂ η^
Therefore, =cos(kx−σt )
∂y ∂y
Then equation (4) becomes,
 c η^ ( y )cos (kx −σt ) ∂ η^
fc =−g cos (kx−σt )
h ∂y

⇒−f c
c η^ ( y ) ∂ η^
=
hg ∂y

⇒
∂ η^ c
=−f c ∂ y
η gh

⇒log η̂ =−f c
c H
y + log , where c 2= gh and log H is a constant of integration.
c
2
2 2

⇒ η^ =
y
H −f c
c
e
2
The total water surface profile and horizontal water profile motions are now:
y
H −f c
η^ = e c cos(kx−σt )
2
And
y
Hc −f c c
U= e cos (kx−σt )
2h

At the wave crest, the wave amplitudes and velocity decrease across the channel (y increasing),
while at the wave trough the amplitude increases. This wave is called Kelvin wave.