Theory of Ship Waves (Wave-Body Interaction …...Theory of Ship Waves (Wave-Body Interaction...

Theory of Ship Waves (Wave-Body Interaction Theory)

Quiz No. 1, April 17, 2019Your Name:

1) In the vector calculus, there are three important quantities:

1) 勾配 (Gradient) ∇ϕ =∂ϕ

∂xi≡ ∂iϕ ： V ector

2) 発散 (Divergence) ∇ ·A =

3∑j=1

∂Aj

∂xj≡ ∂jAj ：Scalar

3) 回転 (Rotation) ∇×A = εijk∂Ak

∂xj≡ εijk∂jAk ： V ector

In the calculations related to the above quantities, there exist two indentities, which has nothing to

do with the form of scalar ϕ or vector A. Namely, the following two relations are always satisfied:

(1) ∇×∇ϕ = 0 (2) ∇ · (∇×A) = 0

Please confirm (prove) these two identities. In the above, εijk is called alternating tensor and

defined as

εijk =

1 (i, j, k) が偶置換 (even permutation)

−1 (i, j, k) が奇置換 (odd permutation)

0 otherwise

Regarding (1)

∇×∇ϕ = εijk∂j∂kϕ = εijk∂

∂xj

∂ϕ

∂xk

= i( ∂2ϕ

∂y∂z− ∂2ϕ

∂z∂y

)+ j

( ∂2ϕ

∂z∂x− ∂2ϕ

∂x∂z

)+ k

( ∂2ϕ

∂x∂y− ∂2ϕ

∂y∂x

)= 0

Reagrding (2)

∇ · (∇×A) = ∂jεjkℓ∂kAℓ =

3∑j=1

∂

∂xjεjkℓ

∂Aℓ

∂xk

=∂

∂x

( ∂Az

∂y− ∂Ay

∂z

)+

∂

∂y

( ∂Ax

∂z− ∂Az

∂x

)+

∂

∂z

( ∂Ay

∂x− ∂Ax

∂y

)=

( ∂2

∂y∂z− ∂2

∂z∂y

)Ax +

( ∂2

∂z∂x− ∂2

∂x∂z

)Ay +

( ∂2

∂x∂y− ∂2

∂y∂x

)Az = 0

2) u · ∇u = −u × (∇ × u) +1

2∇(u · u) will be used in the transformation for the advection

equation. Please confirm (prove) this transformation.

First we have the following:

∇× u = i(∂w∂y− ∂v

∂z

)+ j

(∂u∂z− ∂w

∂x

)+ j

(∂v∂x− ∂u

∂y

)−u×

(∇× u

)= −εijkuj

(∇× u

)k

= −i{v(∂v∂x− ∂u

∂y

)− w

(∂u∂z− ∂w

∂x

)}− j

{w(∂w∂y− ∂v

∂z

)− u

(∂v∂x− ∂u

∂y

)}−k

{u(∂u∂z− ∂w

∂x

)− v

(∂w∂y− ∂v

∂z

)}1

2∇(u · u

)=

1

2∇(u2 + v2 + w2

)= i

(u∂u

∂x+ v

∂v

∂x+ w

∂w

∂x

)+ j

(u∂u

∂y+ v

∂v

∂y+ w

∂w

∂y

)+ k

(u∂u

∂z+ v

∂v

∂z+ w

∂w

∂z

)Thus −u×

(∇× u

)+

1

2∇(u · u

)=

(u∂

∂x+ v

∂

∂y+ w

∂

∂z

)(iu+ jv + kw

)= u · ∇u

Using the alternating tensor, we can write in a compact form:

u · ∇u = uj∂jui = uj(∂jui − ∂iuj) + uj∂iuj = ujεkji(∇× u)k +1

2∂i(ujuj)

= −εijkuj(∇× u)k +1

2∂i(ujuj) = −u× (∇× u) +

1

2∇(u · u) ←− εkji = −εijk


Quiz No. 2, April 24, 2019

Your Name:

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

(1) viscous effects 粘性影響 (2) shear stress 剪断応力

(3) normal pressure 法線圧力 (4) pursue 追跡する

(5) bear in mind 心に留める (6) be denoted by ～で表される

(7) variation 変化 (8) adjacent surfaces 隣り合った表面

(9) be composed of ～で構成される (10) integrand 被積分関数

(11) be referred to as ～と称せられる (12) partial differential equation 偏微分方程式

(13) hold for ～に対して成り立つ (14) incompressible fluid 非圧縮性流体

(15) continuity equation 連続方程式 (16) 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

(=

∫∫∫V

[∂F

∂t+

∂

∂xj

(Fuj

)]dV

)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: (Write the transformation and the result.)

∫∫∫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, May 8 (Wednesday), 2019

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 for incompressible fluid) in terms

of u and also in terms 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 22, 2019

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 29, 2019

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 (1)

C

{e|k|η + e−|k|η − 2|k|

|k| −Ke−|k|η

}+De−|k|η = − 1

|k|(2)

These simultaneous equations for C and D may be solved with Cramer’s formula.

Or by adding (1) and (2), we have

2C e|k|η = − 1

|k|−→ C = − 1

2|k|e−|k|η (3)

Substituting this in (1), we can obtain the solution for D as follows:

De−|k|η = − 1

2|k|e−|k|η

{e|k|η +

|k|+K

|k| −Ke−|k|η

}Thus −→ D = − 1

2|k|

{e|k|η +

|k|+K

|k| −Ke−|k|η

}(4)

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)


Quiz No. 7, June 5, 2019

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

LC ≡∫ ∞

0

Ce−ky cos kx

k −Kdk =

∫ ∞

0

k cos ky −K sin ky

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:

LC ≡∫ ∞

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.

By using this result, we can obtain the following result:

I1 = limµ→0

∫ ∞

0

e−ky cos kx

k − (K − iµ)dk =

∫ ∞

0

Ce−ky cos kx

k −Kdk − πi e−Ky cosKx

= LC − πi e−Ky cosKx =

∫ ∞

0


k2 +K2e−kx dk − πi e−Ky−iKx


Quiz No. 8, June 26, 2019

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)


Quiz No. 9, July 3, 2019

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)



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

]



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. 12, July 17, 2019

Your Name:

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

In Section 4.1, we will study that the reflection-wave and transmission-wave coefficients in the diffrac-

tion problem can be expressed as

R = iH+4 = − cos(ε2 − εj) ei(ε2+εj) (3)

T = 1 + iH−4 = −i sin(ε2 − εj) ei(ε2+εj) (4)

Prove these relations.

In terms of the result obtained above i.e. Eq. (), we can transform as follows:

iH+4 = − eiε2 cos ε2 − i eiεj sin εj

= − ei(ε2+εj){ e−iεj cos ε2 + i e−iε2 sin εj }

= − ei(ε2+εj){cos ε2 cos εj − i cos ε2 sin εj + i cos ε2 sin εj + sin ε2 sin εj }

= − cos(ε2 − εj) ei(ε2+εj) (5)

1 + iH−4 = 1− eiε2 cos ε2 + i eiεj sin εj

= − ei(ε2+εj){− e−i(ε2+εj) + e−iεj cos ε2 − i e−iε2 sin εj }

= − ei(ε2+εj){− cos(ε2 + εj) + i sin(ε2 + εj)

+ cos ε2 cos εj − i cos ε2 sin εj − i cos ε2 sin εj − sin ε2 sin εj }

= − ei(ε2+εj){i (sin ε2 cos εj + cos ε2 sin εj − cos ε2 sin εj − cos ε2 sin εj )

}= −i sin(ε2 − εj) ei(ε2+εj) (6)


Quiz No. 13, July 17, 2019

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 17, 2019

Your Name:


(1) decompose 分解する (2) namely すなわち

(3) asymmetric 非対称の (4) antisymmetric 反対称の

(5) split equally 等分割する (6) independently 独立に

(7) variation 変動，変化分 (8) inversely 逆に

(9) resonance 同調 (10) moment of inertia 慣性モーメント

(11) preceding すぐ前の，先行する (12) noteworthy 注目すべき

(13) step by step 一歩づつ (14) general-shaped 任意形状の

The equation of harmonic heave motion with circular frequency ω can be written in the form[C22 − ω2

(m+A22

)+ iω B22

]X2 = E2 (1)

where A22, B22, and C22 are the added mass, damping coefficient, restoring-force coefficient, respectively,

and E2 denotes the (complex) wave-exciting force.

By introducing the following relations

C22 − ω2(m+A22

)≡ ρω2E2, B22 = ρω

∣∣H+2

∣∣2, E2 = ρgaH+2

(H+

2 =i

KA2 e

iε2

)prove that the complex amplitude of heave motion X2/a will be given analytically in the following form

X2

a=| cosαH |iKH

+

2

eiαH =| cosαH |A2

ei(αH+ε2) (2)

where, tanαH = E2/∣∣H+

2

∣∣2. (Note that at the resonance, E = 0 and thus αH = 0.)

(ρω2E2 + iωρω

∣∣H+2

∣∣2)X2 = ρgaH+2

Thus X2

a=

1

K

H+2

(E2 + i|H+2 |2)

=H+

2

iK|H+2 |2

(1− iE2/|H+

2 |2)

where

1− i E2

|H+2 |2

= 1− i tanαH =√

1 + tan2 αH e−iαH =e−iαH

| cosαH |,

∣∣H+2

∣∣2 = H+2 H

+

2

ThereforeX2

a=

1

iKH+

2

∣∣ cosαH

∣∣ eiαH =

∣∣ cosαH

∣∣A2

ei(αH+ε2) ←− H+

2 =1

iKA2 e

−iε2

Theory of Ship Waves (Wave-Body Interaction …...Theory of Ship Waves (Wave-Body Interaction...

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