theory of ship waves (wave-body interaction … of ship waves (wave-body interaction theory) quiz...

Theory of Ship Waves (Wave-Body Interaction Theory)

Quiz No. 2, April 20, 2016

Your Name:

Write the meaning of the following words in Japanese or in other ways in English.

(1) conservation of mass (2) conservation of momemtum

(3) viscous effects (4) shear stress

(5) normal pressure (6) prescribed volume

(7) pursue (8) differentiable

(9) bear in mind (10) be denoted by

(11) variation (12) adjacent surfaces

(13) be composed of (14) integrand

(15) be referred to as (16) partial differential equation

(17) hold for (18) incompressible fluid

(19) continuity equation (20) substantial derivative

Let us consider a general volume integral of the form

I(t) =

∫∫∫V (t)

F (x, t) dV .

The time derivative of I(t), which is known as the transport theorem, can be written as

dI

dt=

where Un denotes the normal velocity of the boundary surface S.

By using this transport theorem and Gauss’ theorem together with Un = ujnj (≡∑3

j=1 ujnj), the

principles of conservation of mass and momemtum expressed by

d

dt

∫∫∫V (t)

ρ dV = 0

d

dt

∫∫∫V (t)

ρui dV = −∫∫

S(t)

pni dS +

∫∫∫V (t)

ρg δi3 dV

can be transformed as follows:


Quiz No. 3, April 27 (Wednesday), 2016

Your Name:


(1) vorticity (2) scalar function

(3) identity (4) be recast

(5) atmospheric pressure (6) readily

(7) normal vector (8) imply

(9) kinematic condition (10) dynamic condition

(11) wave elevation (12) eliminate

(13) explicitly (14) expedient

(15) pragmatic (16) subsequently

Let us consider an ideal fluid with irrotational motion; that is, ω = ∇×u = 0 (where ω is the vorticity).

In the vector analysis, it is known that an identity of ∇×∇Φ ≡ 0 is always satisfied irrespective of the

kind of scalar function Φ. Then, write the relation in this case between u and Φ.

Write the continuity equation (derived from the conservation of mass) in terms of u and also of Φ.

Euler’s equations, which are to be obtained from the conservation of momentum, can be written in

the form∂u

∂t+ u · ∇u = −1

ρ∇p+ gk (1)

where ρ denotes the density of fluid, p the pressure, g the acceleration due to gravity, and k the unit

vector in the vertical z-axis. Then by using a transformation of the vector quantity

u · ∇u =1

2∇(u · u

)in the case of ∇× u = 0 ,

write the result of Eq. (1) in terms of Φ, which is known as Bernoulli’s pressure equation.

Write the substantial derivative of function F (x, y, z, t) = z − ζ(x, y, t) = 0 being equal to zero, in

terms of the velocity potential Φ instead of the velocity vector u.


Quiz No. 4, May 11, 2016

Your Name:

Let us consider a solution of the velocity potential for plane progressive waves, satisfying the following

Laplace equation and the free-surface and deep-water conditions:

[L]∂2ϕ

∂x2+∂2ϕ

∂y2= 0 for y ≥ 0 (1)

[F ]∂2ϕ

∂t2− g

∂ϕ

∂y= 0 on y = 0 (2)

[B]∂ϕ

∂y= 0 on y = h (3)

Skipping the details (I believe you understand), we can write a homogeneous solution satisfying (1) and

(3), in the following form (with D as unknown):

ϕ(x, y, t) = Dcosh k0(y − h)

cosh k0hsin(ωt− k0x) (4)

However, the free-surface condition (2) is not imposed yet. In fact, the so-called dispersion relation can

be obtained from Eq. (2). Write explicitly the relation to be obtained from Eq. (2).

(5)

Next, obtain asymptotic forms of Eqs. (4) and (5) in the limit of h → ∞. Then, from the dispersion

relation to be obtained from Eq. (5) for the case of h → ∞, obtain the relation between period (T ) and

wavelength (λ), and evaluate the wave period for λ = 100 m and the wavelength for T = 10 s. (You may

use√2π/g ≃ 0.8.)


(1) nevertheless (2) advection

(3) specify (4) physically relevant

(5) trivial (6) necessary and sufficient

(7) sinusoidal (8) eigen value

(9) homogeneous (10) spatial part

(11) time-dependent (12) likewise

(13) mutually (14) wave elevation

(15) envisage (16) monotonically

(17) estimation (18) schematically

(19) no longer (20) provided that


Quiz No. 5, May 18, 2016

Your Name:

(1) Prove (confirm) Eq. (2.16) by substituting Eq. (2.15) in Eq. (2.10).

(2) Determine two unknowns C and D in Eqs. (2.16) and (2.17) from the conditions of Eqs. (2.13) and

(2.14), and show that the result can be expressed in a unified form of Eq. (2.18).


Quiz No. 6, June 1, 2016

Your Name:

Let us consider the mass conservation in the 2D potential flow by using the s1s2 coordinate system.

The total net flux through the boundary surface may be given as

dQ1 + dQ2 =∂

∂s1

(∂ϕ

∂s1δs2

)δs1 +

∂

∂s2

(∂ϕ

∂s2δs1

)δs2 = 0 (1)

In the polar coordinate system, the differential element can be given by δs1 = δr, δs2 = rδθ. Thus by

substituting these relations into Eq. (1) and dividing the result by rδrδθ, obtain the Laplace equation

expressed in the 2D polar coordinate system.

Let us consider the following integral:

F (x, y) = −∫ ∞

0

1

ke−ky cos kx dk = −Re

∫ ∞

0

1

keik(x+iy) dk ≡ −Re

∫ ∞

0

1

keikz dk, z ≡ x+ iy (2)

First consider differentiation of Eq. (2) with respect to z and then perform integration of the result

(dF/dz) with respect to k. From the result to be obtained, prove the following expression:

F (x, y) = log r + c

where r =√x2 + y2 and c is an arbitrary constant.


(1) complementary (2) modulation

(3) slowly varying (4) envelope

(5) encompass (6) persist

(7) preceding (8) retain

(9) discard (10) crest

(11) trough (12) transportation

(13) decay (14) subsequent

(15) simplicity (16) be replaced with



Your Name:


(1) decay (2) diverge

(3) noteworthy (4) discard

(5) envisage (6) quadrant

(7) transient (8) plausible

(9) residue theorem (10) without recourse to

With assumption of x > 0 and y > 0, let us consider the

following complex integral in the complex plane

J ≡∮C

e−ζy+iζx

ζ −Kdζ

where C denotes a certain round integration path.

When considering the integration path shown on the right

side, we note that J = 0 because there are no singularities inside

of the round integration path. With this fact and by taking only

the real part of the result, please show the following:

k = K

i

for x>0

kO

1

1

∫ ∞

0

Ce−ky cos kx

k −Kdk =

∫ ∞

0

k cos ky −K sin ky

k2 +K2e−kx dk − π e−Ky sinKx



Your Name:


(1) reciprocity (2) artificial

(3) argument (4) be imposed

(5) advantageous (6) in return for

(7) explicitly (8) that is to say

(9) homogeneous condition (10) be regarded as

(11) asymptotic form (12) wave elevation

(13) characteristics (14) body geometry

With Green’s theorem, we can obtain the following expression for the velocity potential at an arbitrary

point P = (x, y) in the fluid region:

ϕ(P) =

∫SH

{∂ϕ(Q)

∂nQ− ϕ(Q)

∂

∂nQ

}G(P;Q) ds(Q). (1)

where SH denotes the wetted surface of a floating body.

Using this expression, let us consider the asymptotic form of ϕ(P) for the case of |x| → ∞. Since

P(x, y) is included only in the Green function G(P;Q), we have the following expression for |x| → ∞:

G(P;Q) ∼ i e−K(y+η)∓iK(x−ξ) = i e−Kη±iKξ e−Ky∓iKx as x→ ±∞. (2)

In terms of this result, please show the asymptotic form of the velocity potential valid for |x| → ∞.

Furthermore, show that the wave at a distance from the body (|x| → ∞) is the outgoing progressive

wave, and obtain the complex amplitude of the wave elevation from the relation ζ(x) = iωg ϕ(x, 0).



Your Name:


(1) characteristics (2) distinction

(3) wave absorption (4) perfect reflection

(5) superposition (6) incident wave

(7) calm water (8) schematic

(9) displacement (10) time invariable

The body boundary condition is given by

∂Φ

∂n= α(t) · n (1)

where Φ(x, t) = Re[ϕ(x) eiωt

], x = (x, y), n = (n1, n2)

α(t) = iξ1(t) + jξ2(t) + kξ3(t)× x, ξj(t) = Re[Xj e

iωt]

Without the time-dependent part eiωt, rewrite Eq. (1) in a form of summation of three modes (sway,

heave, and roll) of body motion; that is, confirm the following result:

∂ϕ

∂n=

3∑j=1

iωXj nj (2)

If the velocity potential is written in the form

ϕ(x) = ϕ0(x) + ϕ4(x) +

3∑j=1

ϕj(x) =ga

iω

{φ0(x) + φ4(x)

}+

3∑j=1

iωXj φj(x) (3)

what form of the boundary condition does each velocity potential φj(x) (j = 1, 2, 3, 4) have to satisfy?


Quiz No. 11, July 06, 2016

Your Name:


(1) specifically (2) implicitly

(3) wetted surface (4) discard

(5) reference value (6) be associated with

(7) transpose (8) transfer function

(9) be referred to as (10) advent

(11) variance (12) couple of vertical forces

When the velocity potential is given in the following form

ϕ(x, y) =ga

iω

(φ0 + φ4

)+

3∑j=1

iωXjφj , φ0 = e−Ky+iKx (incident wave)

and the asymptotic form of the velocity potential due to body disturbance can be written as

φj(x, y) ≃ iH±j (K) e−Ky e∓iKx as x→ ±∞ ,

write the wave elevation on the free surface ζ(x) valid at a distance from the body and the pressure in

the fluid p(x, y) by using the following relation:

ζ(x) =iω

gϕ(x, 0) , p(x, y) = −ρiωϕ(x, y)

Then write the hydrodynamic force acting on the body in the i-th direction Fi


Quiz No. 12, July 06, 2016

Your Name:


(1) restoring moment (2) buoyancy

(3) reflection (4) transmission

(5) deformation (6) subsquently

(7) asymmetric (8) indicate

(9) in contrast to (10) nevertheless

(11) take into account (12) likewise

The wave elevation at a distance from a body can be expressed as

ζ(x) = a eiKx + iaH±4 (K) e∓iKx −K

3∑j=1

Xj iH±j (K) e∓iKx as x→ ±∞

where we note that the incident wave comes from the positive x-axis.

Then, write the equations for the nondimensional (in terms of the amplitude of incident wave a )

complex amplitude of reflection and transmission waves:

Transform the following integral by using the 2D Gauss theorem and write the result in terms of the

centers of gravity (G) and buoyancy (B), and the metacenter (M):

SG3 = −ρgX3

∫SH

x{n2x− n1(y −OG)

}dℓ

where n1 = nx and n2 = ny.


Quiz No. 13, July 13, 2016

Your Name:


(1) in what follows (2) interpret

(3) asymptotic (4) be imparted to

(5) be exerted by (6) superficially

(7) remarkable (8) extension

(9) specifically (10) owing to

(11) irrespective of (12) be associated with

(13) consequence (14) hold for

For two different velocity potentials, ϕ and ψ, satisfying the same boundary conditions on SF and SB

but not necessarily the same on SH and S±∞, the Green’s theorem gives the following equation:∫SH

(ϕ∂ψ

∂n− ψ

∂ϕ

∂n

)dℓ =

1

2K

[(ϕ∂ψ

∂x− ψ

∂ϕ

∂x

)y=0

]x=+∞

x=−∞(1)

Then let us consider a combination of ϕ = φD (diffraction) and ψ = φj (radiation in the j-th mode).

The boundary conditions satisfied by these on SH can be written as

∂φD

∂n= 0 ,

∂φj

∂n= nj

and at x→ ±∞, the following relations hold: φD ∼ eiKx +Re−iKx , φj ∼ iH+j e−iKx , at x = +∞

φD ∼ T eiKx , φj ∼ iH−j eiKx , at x = −∞

In this case, obtain the relation to be derived from Eq.(1):


Quiz No. 14, July 20, 2016

Your Name:


(1) namely (2) asymmetric

(3) antisymmetric (4) take account of

(5) restoring force (6) variation

(7) inversely (8) resonance

(9) moment of inertia (10) preceding

(11) acquire (12) noteworthy

For a symmetric body, the relation between the scattered wave H±4 in the diffraction problem and the

radiated wave H+j (j = 1 ∼ 3) in the radiation problem can be shown to be expressed as

H±4 =

Im(H+

2

)H

+

2

∓ iRe

(H+

j

)H

+

j

(j = 1 or 3) (1)

From (3.18) in the lecture note, the radiation Kochin functionH+j can be written with the wave amplitude

ratio Aj and the phase difference εj in the form

H+j =

i

KAj e

iεj (j = 1, 2, 3), where K =ω2

g(2)

Then, rewrite Eq. (1) in terms of εj (in fact, Aj will not appear in the result).

The definition of the radiation Kochin function in heave is:

H+2 =

∫SH

(∂φ2

∂n− φ2

∂

∂n

)e−Ky+iKx dℓ (3)

From this, obtain analytically the expression of H+2 in the limit of K → 0 (ω → 0).


Quiz No. 2, April 20, 2016

Your Name:


(1) conservation of mass 質量保存 (2) conservation of momemtum 運動量保存

(3) viscous effects 粘性影響 (4) shear stress 剪断応力

(5) normal pressure 法線圧力 (6) prescribed volume 規定された体積

(7) pursue 追跡する (8) differentiable 微分可能な

(9) bear in mind 心に留める (10) be denoted by ～で表される

(11) variation 変化 (12) adjacent surfaces 隣り合った表面

(13) be composed of ～で構成される (14) integrand 被積分関数

(15) be referred to as ～と称せられる (16) partial differential equation 偏微分方程式

(17) hold for ～に対して成り立つ (18) incompressible fluid 非圧縮性流体

(19) continuity equation 連続方程式 (20) substantial derivative 実質微分

Let us consider a general volume integral of the form

I(t) =

∫∫∫V (t)

F (x, t) dV .

The time derivative of I(t), which is known as the transport theorem, can be written as

dI

dt=

∫∫∫V

∂F

∂tdV +

∫∫S

F Un dS

where Un denotes the normal velocity of the boundary surface S.

By using this transport theorem and Gauss’ theorem together with Un = ujnj (≡∑3

j=1 ujnj), the

principles of conservation of mass and momemtum expressed by

d

dt

∫∫∫V (t)

ρ dV = 0

d

dt

∫∫∫V (t)

ρui dV = −∫∫

S(t)

pni dS +

∫∫∫V (t)

ρg δi3 dV

can be transformed as follows:∫∫∫V

∂ρ

∂tdV +

∫∫S

ρujnj dS =

∫∫∫V

[∂ρ

∂t+

∂

∂xj

(ρuj

)]dV = 0

−→ ∂ρ

∂t+

∂

∂xj

(ρuj

)= 0 −→ ∂uj

∂xj= 0∫∫∫

V

[∂

∂t

(ρui

)+

∂

∂xj

(ρuiuj

)]=

∫∫∫V

[− ∂p

∂xi+ ρg δi3

]dV

−→ ∂

∂t

(ρui

)+

∂

∂xj

(ρuiuj

)= − ∂p

∂xi+ ρg δi3

−→ ui

{∂ρ

∂t+

∂

∂xj

(ρuj

)}+ ρ

{∂ui∂t

+ uj∂ui∂xj

}= − ∂p

∂xi+ ρg δi3

−→ ∂ui∂t

+ uj∂ui∂xj

= −1

ρ

∂p

∂xi+ g δi3


Quiz No. 3, April 27 (Wednesday), 2016

Your Name:


(1) vorticity 渦度 (2) scalar function スカラー関数

(3) identity 恒等式 (4) be recast 書き換えられる

(5) atmospheric pressure 大気圧 (6) readily 直ぐに，容易に

(7) normal vector 法線ベクトル (8) imply 意味する

(9) kinematic condition 運動学的条件 (10) dynamic condition 力学的条件

(11) wave elevation 波の変位 (12) eliminate 消去する

(13) explicitly 陽（な形）に (14) expedient 便利な，都合の良い

(15) pragmatic 実用的な (16) subsequently 後ほど

Let us consider an ideal fluid with irrotational motion; that is, ω = ∇×u = 0 (where ω is the vorticity).

In the vector analysis, it is known that an identity of ∇×∇Φ ≡ 0 is always satisfied irrespective of the

kind of scalar function Φ. Then, write the relation in this case between u and Φ.

u = ∇Φ

Write the continuity equation (derived from the conservation of mass) in terms of u and also of Φ.

∇ · u = 0 −→ ∇ · ∇Φ = ∇2Φ = 0

Euler’s equations, which are to be obtained from the conservation of momentum, can be written in

the form∂u

∂t+ u · ∇u = −1

ρ∇p+ gk (1)

where ρ denotes the density of fluid, p the pressure, g the acceleration due to gravity, and k the unit

vector in the vertical z-axis. Then by using a transformation of the vector quantity

u · ∇u =1

2∇(u · u

)in the case of ∇× u = 0 ,

write the result of Eq. (1) in terms of Φ, which is known as Bernoulli’s pressure equation.

∇[∂Φ

∂t+

1

2∇Φ · ∇Φ+

p

ρ− gz

]= 0

−→ − 1

ρ

(p− pa

)=∂Φ

∂t+

1

2∇Φ · ∇Φ− gz

Write the substantial derivative of function F (x, y, z, t) = z − ζ(x, y, t) = 0 being equal to zero, in

terms of the velocity potential Φ instead of the velocity vector u.

DF

Dt=∂F

∂t+∇Φ · ∇F

= −∂ζ∂t

− ∂Φ

∂x

∂ζ

∂x− ∂Φ

∂y

∂ζ

∂y+∂Φ

∂z= 0

−→ ∂Φ

∂z=∂ζ

∂t+∂Φ

∂x

∂ζ

∂x+∂Φ

∂y

∂ζ

∂y


Quiz No. 4, May 11, 2016

Your Name:

Let us consider a solution of the velocity potential for plane progressive waves, satisfying the following

Laplace equation and the free-surface and deep-water conditions:

[L]∂2ϕ

∂x2+∂2ϕ

∂y2= 0 for y ≥ 0 (1)

[F ]∂2ϕ

∂t2− g

∂ϕ

∂y= 0 on y = 0 (2)

[B]∂ϕ

∂y= 0 on y = h (3)

Skipping the details (I believe you understand), we can write a homogeneous solution satisfying (1) and

(3), in the following form (with D as unknown):

ϕ(x, y, t) = Dcosh k0(y − h)

cosh k0hsin(ωt− k0x) (4)

However, the free-surface condition (2) is not imposed yet. In fact, the so-called dispersion relation can

be obtained from Eq. (2). Write explicitly the relation to be obtained from Eq. (2).

− ω2 − g k0(− tanh k0h

)= 0

→ k0 tanh k0h =ω2

g

(≡ K

)(5)

Next, obtain asymptotic forms of Eqs. (4) and (5) in the limit of h → ∞. Then, from the dispersion

relation to be obtained from Eq. (5) for the case of h → ∞, obtain the relation between period (T ) and

wavelength (λ), and evaluate the wave period for λ = 100 m and the wavelength for T = 10 s. (You may

use√2π/g ≃ 0.8.)

(4) : ϕ→ De−k0(y−h)

ek0hsin(ωt− k0x) = De−k0y sin(ωt− k0x)

(5) : tanh k0h→ 1, Thus k0 =ω2

g(6)

2π

λ=

1

g

(2πT

)2

→ λ =g

2πT 2 ≃ 1.56T 2 → T =

√2π

gλ ≃ 0.8

√λ

From these formulae, λ = 100 m → T = 8 s , T = 10 s → λ = 156 m


(1) nevertheless にもかかわらず (2) advection 移流

(3) specify 明示（明記）する (4) physically relevant 物理的に関連した

(5) trivial 取るに足らない (6) necessary and sufficient 必要（で）十分な

(7) sinusoidal 正弦関数的な (8) eigen value 固有値

(9) homogeneous 同次の (10) spatial part 空間（座標に関する）部分

(11) time-dependent 時間に依存した (12) likewise 同様に

(13) mutually お互いに (14) wave elevation 波の（垂直）変位

(15) envisage 心に描く (16) monotonically 単調に

(17) estimation 推定，算定 (18) schematically 図式的に

(19) no longer もはや～でない (20) provided that もし～であれば


Quiz No. 5, May 18, 2016

Your Name:

(1) Prove (confirm) Eq. (2.16) by substituting Eq. (2.15) in Eq. (2.10).

G∗1(k; y) = C1 e

|k|y + C2 e−|k|y

−→ dG∗1

dy+KG∗

1 = C1

(|k|+K

)+ C2

(− |k|+K

)= 0 on y = 0

Thus we can have the following:

C2 = C1|k|+K

|k| −K= C1

{−1 +

2|k||k| −K

}

G∗1(k; y) = C1

{e|k|y − e−|k|y +

2|k||k| −K

e−|k|y}

(2) Determine two unknowns C and D in Eqs. (2.16) and (2.17) from the conditions of Eqs. (2.13) and

(2.14), and show that the result can be expressed in a unified form of Eq. (2.18).

From the conditions of Eqs. (2.13) and (2.14), we have the followings:

C

{e|k|η − e−|k|η +

2|k||k| −K

e−|k|η}−De−|k|η = 0

C

{e|k|η + e−|k|η − 2|k|

|k| −Ke−|k|η

}+De−|k|η = − 1

|k|These can be written in a matrix form:∣∣∣∣∣∣∣∣

e|k|η +|k|+K

|k| −Ke−|k|η − e−|k|η

e|k|η − |k|+K

|k| −Ke−|k|η + e−|k|η

∣∣∣∣∣∣∣∣

C

D

=

0

− 1

|k|

Its determinant is

∆ = 1 +|k|+K

|k| −Ke−2|k|η + 1− |k|+K

|k| −Ke−2|k|η = 2

Thus we obtain the following results:

C = − 1

2|k|e−|k|η

D = − 1

2|k|

{e|k|η +

|k|+K

|k| −Ke−|k|η

}Substituting these in the original, we have

G∗1 = − 1

2|k|

{e|k|(y−η) +

|k|+K

|k| −Ke−|k|(y+η)

}for y − η < 0

G∗2 = − 1

2|k|

{e−|k|(y−η) +

|k|+K

|k| −Ke−|k|(y+η)

}for y − η > 0

Therefore we may write these in a unified form as follows:

G∗ = − 1

2|k|

{e−|k||y−η| − e−|k|(y+η)

}− 1

|k| −Ke−|k|(y+η)



Your Name:

Let us consider the mass conservation in the 2D potential flow by using the s1s2 coordinate system.

The total net flux through the boundary surface may be given as

dQ1 + dQ2 =∂

∂s1

(∂ϕ

∂s1δs2

)δs1 +

∂

∂s2

(∂ϕ

∂s2δs1

)δs2 = 0 (1)

In the polar coordinate system, the differential element can be given by δs1 = δr, δs2 = rδθ. Thus by

substituting these relations into Eq. (1) and dividing the result by rδrδθ, obtain the Laplace equation

expressed in the 2D polar coordinate system.

∂

∂r

(∂ϕ

∂rrδθ

)δr +

∂

r∂θ

(∂ϕ

r∂θδr

)rδθ = 0

−→ ∂

r∂r

(r∂ϕ

∂r

)+

1

r2∂2ϕ

∂θ2= 0

Let us consider the following integral:

F (x, y) = −∫ ∞

0

1

ke−ky cos kx dk = −Re

∫ ∞

0

1

keik(x+iy) dk ≡ −Re

∫ ∞

0

1

keikz dk, z ≡ x+ iy (2)

First consider differentiation of Eq. (1) with respect to z and then perform integration of the result

(dF/dz) with respect to k. From the result to be obtained, prove the following expression:

F (x, y) = log r + c

where r =√x2 + y2 and c is an arbitrary constant.

dF

dz= −Re

∫ ∞

0

i eikz dk = −Re

[1

zeikz

]∞0

= Re

[1

z

]

−→ F = Re[log z

]+ c = Re

[log(x+ iy)

]+ c

= Re[log r + iθ

]+ c = log r + c


(1) complementary 補助的な (2) modulation 変調

(3) slowly varying ゆっくり変化する (4) envelope 包絡線

(5) encompass 含む (6) persist 続く

(7) preceding すぐ前の，先行する (8) retain 保持する

(9) discard 捨てる (10) crest （波の）山

(11) trough （波の）谷 (12) transportation 輸送

(13) decay 減衰する (14) subsequent 次の，後の

(15) simplicity 簡単化 (16) be replaced with ～で置き換える



Your Name:


(1) decay 減衰する (2) diverge 発散する

(3) noteworthy 注目すべき（価値のある） (4) discard 捨てる

(5) envisage 心に描く (6) quadrant 象限

(7) transient 過渡的な (8) plausible 尤もらしい，妥当な

(9) residue theorem 留数定理 (10) without recourse to ～の助けなしで

With assumption of x > 0 and y > 0, let us consider the

following complex integral in the complex plane

J ≡∮C

e−ζy+iζx

ζ −Kdζ

where C denotes a certain round integration path.

When considering the integration path shown on the right

side, we note that J = 0 because there are no singularities inside

of the round integration path. With this fact and by taking only

the real part of the result, please show the following:

k = K

i

for x>0

kO

1

1

∫ ∞

0

Ce−ky cos kx

k −Kdk =

∫ ∞

0


k2 +K2e−kx dk − π e−Ky sinKx

J =

∫ ∞

0

Ce−ky+ikx

k −Kdk − πi e−Ky+iKx +

∫ 0

∞

e−iky−kx

ik −Kidk = 0

Here the first integral along the real axis (k) should be understood as Cauchy’s principla-value integral

excluding the neighborhood of the singular point (k = K) from the integration range. We can write this

result in the following form:∫ ∞

0

Ce−ky+ikx

k −Kdk = πi e−Ky+iKx +

∫ ∞

0

(k − iK) e−iky

k2 +K2e−kx dk

Therefore, taking only the real part of both sides, we can obtain the following result:∫ ∞

0

Ce−ky cos kx

k −Kdk = −π e−Ky sinKx+

∫ ∞

0


k2 +K2e−kx dk

This is the result to be shown, and needless to say, both sides are of real quantity.



Your Name:


(1) reciprocity 相反（関係） (2) artificial 人工的な

(3) argument 議論 (4) be imposed 課される

(5) advantageous 都合の良い、有利な (6) in return for ～の見返りに

(7) explicitly 陽（な形）に (8) that is to say すなわち

(9) homogeneous condition 同次条件 (10) be regarded as ～と見なされる

(11) asymptotic form 漸近形 (12) wave elevation 波の変位

(13) characteristics 特性、特色 (14) body geometry 物体形状

With Green’s theorem, we can obtain the following expression for the velocity potential at an arbitrary

point P = (x, y) in the fluid region:

ϕ(P) =

∫SH

{∂ϕ(Q)

∂nQ− ϕ(Q)

∂

∂nQ

}G(P;Q) ds(Q). (1)

where SH denotes the wetted surface of a floating body.

Using this expression, let us consider the asymptotic form of ϕ(P) for the case of |x| → ∞. Since

P(x, y) is included only in the Green function G(P;Q), we have the following expression for |x| → ∞:

G(P;Q) ∼ i e−K(y+η)∓iK(x−ξ) = i e−Kη±iKξ e−Ky∓iKx as x→ ±∞. (2)

In terms of this result, please show the asymptotic form of the velocity potential valid for |x| → ∞.

Furthermore, show that the wave at a distance from the body (|x| → ∞) is the outgoing progressive

wave, and obtain the complex amplitude of the wave elevation from the relation ζ(x) = iωg ϕ(x, 0).

Substituting (2) in (1), we can obtain the following:

ϕ(x, y) ∼ i

∫SH

{∂ϕ(Q)

∂nQ− ϕ(Q)

∂

∂nQ

}e−Kη±iKξ ds(Q)

[e−Ky∓iKx

]as x→ ±∞

Thus we may write this result as follows:

ϕ(x, y) ∼ iH±(K) e−Ky∓iKx as x→ ±∞, (3)

whereH±(K) =

∫SH

{∂ϕ(Q)

∂nQ− ϕ(Q)

∂

∂nQ

}e−Kη±iKξ ds(Q) (4)

The wave elevation can be written as

ζ(x) =iω

gϕ(x, 0) ∼ −ω

gH±(K) e∓iKx as x→ ±∞ (5)

Thus the complex amplitude of this outgoing progressive wave is

A± = −ωgH±(K) as x→ ±∞ (6)



Your Name:


(1) characteristics 特性 (2) distinction 区別

(3) wave absorption 波吸収 (4) perfect reflection 完全反射

(5) superposition 重ね合わせ (6) incident wave 入射波

(7) calm water （攪乱のない）静水 (8) schematic 図式的な

(9) displacement （直線）変位 (10) time invariable 時間的に不変

The body boundary condition is given by

∂Φ

∂n= α(t) · n (1)

where Φ(x, t) = Re[ϕ(x) eiωt

], x = (x, y), n = (n1, n2)

α(t) = iξ1(t) + jξ2(t) + kξ3(t)× x, ξj(t) = Re[Xj e

iωt]

Without the time-dependent part eiωt, rewrite Eq. (1) in a form of summation of three modes (sway,

heave, and roll) of body motion; that is, confirm the following result:

∂ϕ

∂n=

3∑j=1

iωXj nj (2)

∂ϕ

∂n= iω

[iX1 + jX2 + i(−X3 y) + j(X3 x)

]·[in1 + jn2

]= iω

[X1n1 +X2n2 + (−X3 y)n1 + (X3 x)n2

]= iω

[X1n1 +X2n2 +X3(xn2 − n1 y)

]−→ ∂ϕ

∂n=

3∑j=1

iωXj nj n3 ≡ xn2 − y n1 = (x× n)3 (3)

If the velocity potential is written in the form

ϕ(x) = ϕ0(x) + ϕ4(x) +

3∑j=1

ϕj(x) =ga

iω

{φ0(x) + φ4(x)

}+

3∑j=1

iωXj φj(x) (4)

what form of the boundary condition does each velocity potential φj(x) (j = 1, 2, 3, 4) have to satisfy?

By comparison of Eq. (4) with Eq. (2), it is obvious that the following relations should hold

∂

∂n

(φ0 + φ4

)= 0

∂φj

∂n= nj (j = 1, 2, 3)


Quiz No. 11, July 06, 2016

Your Name:


(1) specifically はっきりと (2) implicitly 暗に

(3) wetted surface 没水表面（濡れ面） (4) discard 捨てる

(5) reference value 参照値 (6) be associated with ～に関連した

(7) transpose 移項する (8) transfer function 伝達関数

(9) be referred to as ～と称せられる (10) advent 出現

(11) variance 変化分，食い違い (12) couple of vertical forces 垂直力の偶力

When the velocity potential is given in the following form

ϕ(x, y) =ga

iω

(φ0 + φ4

)+

3∑j=1

iωXjφj , φ0 = e−Ky+iKx (incident wave)

and the asymptotic form of the velocity potential due to body disturbance can be written as

φj(x, y) ≃ iH±j (K) e−Ky e∓iKx as x→ ±∞ ,

write the wave elevation on the free surface ζ(x) valid at a distance from the body and the pressure in

the fluid p(x, y) by using the following relation:

ζ(x) =iω

gϕ(x, 0) , p(x, y) = −ρiωϕ(x, y)

ζ(x) = a{φ0(x, 0) + φ4(x, 0)

}−K

3∑j=1

Xjφj(x, 0)

= a eiKx + iaH±4 (K) e∓iKx︸︷︷︸

Scattered wave

−K3∑

j=1

Xj iH±j (K) e∓iKx

︸︷︷︸Radiation wave

as x→ ±∞

p(x, y) = −ρga{φ0(x, y) + φ4(x, y)︸︷︷︸

φD

}− ρ(iω)2

3∑j=1

Xj φj(x, y)

Then write the hydrodynamic force acting on the body in the i-th direction Fi

Fi = −∫SH

pni ds = ρga

∫SH

φDni ds︸︷︷︸Ei

+

3∑j=1

ρ(iω)2Xj

∫SH

φjni ds︸︷︷︸fij

where Ei is the wave-exciting force and fij denotes the radiation force acting in the i-th direction due to

the j-th mode of motion, which can be written further as

fij = −[(iω)2Xj Re

{− ρ

∫SH

φjni ds}

︸︷︷︸−ρ

∫SH

Re{φj

}ni ds

+iωXj Re{− ρiω

∫SH

φjni ds}

︸︷︷︸ρω

∫SH

Im{φj

}ni ds

]


Quiz No. 12, July 06, 2016

Your Name:


(1) restoring moment 復原モーメント (2) buoyancy 浮力

(3) reflection 反射 (4) transmission 透過

(5) deformation 変形 (6) subsquently 後で

(7) asymmetric 非対称の (8) indicate 示す

(9) in contrast to ～とは違って (10) nevertheless にもかかわらず

(11) take into account ～を考慮する (12) likewise 同様に

The wave elevation at a distance from a body can be expressed as

ζ(x) = a eiKx + iaH±4 (K) e∓iKx −K

3∑j=1

Xj iH±j (K) e∓iKx as x→ ±∞

where we note that the incident wave comes from the positive x-axis.

Then, write the equations for the nondimensional (in terms of the amplitude of incident wave a )

complex amplitude of reflection and transmission waves:

ζ(x) = a{iH+

4 (K)− iK3∑

j=1

Xj

aH+

j (K)}

︸︷︷︸ζR=aCR

e−iKx + a{1 + iH−

4 (K)− iK3∑

j=1

Xj

aH−

j (K)}

︸︷︷︸ζT=aCT

e+iKx

Therefore

CR =ζRa

= iH+4 (K)︸︷︷︸≡R

−iK3∑

j=1

Xj

aH+

j (K), R ≡ iH+4 (K)

CT =ζTa

= 1 + iH−4 (K)︸︷︷︸

≡T

−iK3∑

j=1

Xj

aH−

j (K), T ≡ 1 + iH−4 (K)

Transform the following integral by using the 2D Gauss theorem and write the result in terms of the

centers of gravity (G) and buoyancy (B), and the metacenter (M):

SG3 = −ρgX3

∫SH

x{n2x− n1(y −OG)

}dℓ

where n1 = nx and n2 = ny.

From Gauss’ theorem in the 2D case, we have the following formula:∫SH

Ani dℓ =

∫SF

Aδi2 dx+

∫∫V

∂A

∂xidS

because n1 = 0, n2 = −1 on the interior free surface SF .

Applying this formula, we have the following:

SG3 = −ρgX3

∫ B/2

−B/2

x2 dx+ ρgX3

∫∫V

(y −OG) dS

= −ρgX3

[V BM − V

(OB −OG

) ]= −ρgV

{BM −OB +OG

}X3 = −ρgV GMX3


Quiz No. 13, July 13, 2016

Your Name:


(1) in what follows 以下に於いては (2) interpret 解釈する，理解する

(3) asymptotic 漸近形の (4) be imparted to ～に与えられる（加えられる）

(5) be exerted by ～によって誘起される (6) superficially 表面的には

(7) remarkable 注目すべき，特筆すべき (8) extension 拡張

(9) specifically 明確に，具体的に (10) owing to ～のために

(11) irrespective of ～に関係なく (12) be associated with ～に関連した

(13) consequence 結果 (14) hold for ～に対して成り立つ

For two different velocity potentials, ϕ and ψ, satisfying the same boundary conditions on SF and SB

but not necessarily the same on SH and S±∞, the Green’s theorem gives the following equation:∫SH

(ϕ∂ψ

∂n− ψ

∂ϕ

∂n

)dℓ =

1

2K

[(ϕ∂ψ

∂x− ψ

∂ϕ

∂x

)y=0

]x=+∞

x=−∞(1)

Then let us consider a combination of ϕ = φD (diffraction) and ψ = φj (radiation in the j-th mode).

The boundary conditions satisfied by these on SH can be written as

∂φD

∂n= 0 ,

∂φj

∂n= nj

and at x→ ±∞, the following relations hold: φD ∼ eiKx +Re−iKx , φj ∼ iH+j e−iKx , at x = +∞

φD ∼ T eiKx , φj ∼ iH−j eiKx , at x = −∞

In this case, obtain the relation to be derived from Eq.(1):

The left-hand side of Eq.(1), denoted as L, can be written as

L =

∫SH

φD nj dℓ =Ej

ρga

where Ej denotes the wave-exciting force acting in the j-th direction, ρ is the density of fluid, g is the

gravitational acceleration, and a is the incident-wave amplitude.

On the other hand, the right-hand side of Eq.(1), denoted as R, can be calculated as follows:

R =1

2K

[KH+

j

(1 +Re−i2Kx

)+KH+

j

(1−Re−i2Kx

)+KH−

j T ei2Kx −KH−

j T ei2Kx

]= H+

j

Therefore, equating L and R gives the following relation:

Ej = ρgaH+j

This relation is known as Haskind-Newman’s relation.


Quiz No. 14, July 20, 2016

Your Name:


(1) namely すなわち (2) asymmetric 非対称の

(3) antisymmetric 反対称の (4) take account of ～を考慮する

(5) restoring force 復原力 (6) variation 変化，変化量

(7) inversely 逆に，逆比例して (8) resonance 同調

(9) moment of inertia 慣性モーメント (10) preceding すぐ前の

(11) acquire 習得する，学ぶ (12) noteworthy 注目すべき，顕著な

For a symmetric body, the relation between the scattered wave H±4 in the diffraction problem and the

radiated wave H+j (j = 1 ∼ 3) in the radiation problem can be shown to be expressed as

H±4 =

Im(H+

2

)H

+

2

∓ iRe

(H+

j

)H

+

j

(j = 1 or 3) (1)

From (3.18) in the lecture note, the radiation Kochin functionH+j can be written with the wave amplitude

ratio Aj and the phase difference εj in the form

H+j =

i

KAj e

iεj (j = 1, 2, 3), where K =ω2

g(2)

Then, rewrite Eq. (1) in terms of εj (in fact, Aj will not appear in the result).

Im(H+

2

)=A2

Kcos ε2, Re

(H+

j

)= − Aj

Ksin εj

Thus

H±4 =

(A2

Kcos ε2

)(K

−iA2eiε2

)∓ i

(− Aj

Ksin εj

)(K

−iAjeiεj

)= i eiε2 cos ε2 ∓ eiεj sin εj (3)

The definition of the radiation Kochin function in heave is:

H+2 =

∫SH

(∂φ2

∂n− φ2

∂

∂n

)e−Ky+iKx dℓ (4)

From this, obtain analytically the expression of H+2 in the limit of K → 0 (ω → 0).

e−Ky+iKx → 1 as K → 0 ,∂φ2

∂n= n2

ThusH+

2 ≃∫SH

n2 dℓ =

∫ B/2

−B/2

dx = B as K → 0

From Haskind-Newman’s relation, we have immediately the following result:

E2 = ρgaH+2 ≃ ρgaB as ω → 0 (λ→ ∞)

This is the same as the heave restoring force when X2 = a.

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