Amath-Atm. Sci. 505/Ocean 511
Autumn 2014
Homework 6 Solutions
1a) By definition, the trajectory equations are
dx p
∂φ
= u(x p , z p , t) =
(x p , z p , t) = aω cos(kx p − ω t)exp(kz p )
dt
∂x
dz p
∂φ
= w(x p , z p , t) =
(x p , z p , t) = aω sin(kx p − ω t)exp(kz p )
dt
∂z
b) To linearize these equations, we note that for small-amplitude oscillatory motion,
x p (t) = x0 + O(a), z p (t) = z0 + O(a)
Taking x0 = 0 and defining z ′p (t) = z p (t) − z0 = O(a) , and Taylor expanding the RHS in xp and
zp about x0 and z0:
dx p
= aω [ cos(kx0 − ω t) + O(a)][ exp(kz0 ) + O(a)]
dt
= aω cos(kx0 − ω t)exp(kz0 ) + O(a 2 )
By linearization, we mean to drop the quadratic term so the trajectory is a purely linear
function of the wave amplitude a:
dx p
≈ aω cos(kx0 − ω t)exp(kz0 ) = a exp(kz0 )cos(ω t) for x0 = 0
dt
Similarly
dz p
≈ aω sin(kx0 − ω t)exp(kz0 ) = −a exp(kz0 )sin(ω t) for x0 = 0
dt
We can integrate each of these equations with respect to time to get
x p (t) = a exp(kz0 )sin(ω t) for x0 = 0
z p (t) = z0 + a exp(kz0 )[ cos(ω t) −1]
This traces out a clockwise circle of radius a exp(kz0), with period 2π/ω, starting at (0, z0).
The radius of the circle decreases exponentially with depth in proportion with the size of the
velocity perturbations.
A fluid parcel starting on the free surface at x0 = t = 0 would have z0 = a, corresponding to
the free surface height η(x0, t = 0). Since a << 1 we can approximate exp(kz0) = exp(ka) ≈ 1.
Hence the linearized trajectory of this fluid parcel would be:
x p (t) = asin(ω t), z p (t) = a cos(ω t)
Thus, the parcel height zp(t) exactly follows the free-surface height η(x0, t) at the parcel’s
mean horizontal location; this is expected because the free surface is a material surface.
c) Substituting X = kx, Z = kz, and T = ωt into (a):
dX p k dx p k
=
= aω cos(kx p − ω t)exp(kz p ) = ak cos(X p − T )exp(Z p )
dT ω dt ω
dZ p k dz p k
=
= aω sin(kx p − ω t)exp(kz p ) = ak sin(X p − T )exp(Z p )
dT ω dt ω
d) The plot below was made by the Matlab script DeepWaterWave_trajplot.m on the
class web page. It shows that for each ak, the trajectory approximately goes clockwise
around a circle of radius ak in each period, like the linearized trajectory in (b) starting at the
water surface. However, there is also a Stokes drift to the right that rapidly increases with
wave amplitude. The lower right panel shows that this Stokes drift increases quadratically
with ak. There is no corresponding Stokes drift in Z. If one naively calculates the Stokes
drift from the beginning to the end of the oscillation, this will include a spurious negative
contribution due to the particle not executing a full oscillation by the final time. To avoid
this, the Stokes drift was computed around an integral number (4) of particle oscillations by
extracting the times and Xp’s at which the particle reaches the rightmost point of each
oscillation.
2. In linearizing the dynamic BC,
∂φ 1 2
+ (u + w 2 ) + gη = 0 at z = η (x, t)
∂t 2
we:
(i) Neglected u2 + w2
∂φ
∂φ
∂ 2φ
∂φ
∂ 2φ
(x, η, t) =
(x, 0, t) + η
…≈
(x, 0, t) (i.e. neglected η
(ii) Approximated
)
∂t
∂t
∂t∂z
∂t
∂t∂z
In linearizing the kinematic BC,
∂η
∂η
+u
= w at z = η (x, t)
∂t
∂x
we:
(iii) Neglected u
∂η
∂x
∂w
∂w
… ≈ w(x, 0, t) (i.e. neglected η
)
∂z
∂z
Thus, what we need to do in each of the two limits is to estimate how big the neglected terms
are compared to the retained linear terms. We use a {} to indicate ‘estimate of’
(iv) Approximated w(x, η, t) = w(x, 0, t) + η
a) In the shallow-water limit kH << 1, {x} = k-1, {z} = H, {t} = ω-1, and
ω = c0k,
η = a cos(kx - ωt) = O(a),
φ = ag/ω sin(kx - ωt) = O(ag/kc0),
u = agk/ω cos(kx - ωt) = O(ag/c0),
w = (1+z/H)∂η/∂t = aω sin(kx - ωt) = O(ac0k) = kH O(u).
For the dynamic BC:
⎧ ∂φ ⎫
Linear:
⎨ ⎬ = {gη} = ag
⎩ ∂t ⎭
⎧
⎫ ⎛ ag ⎞ 2
ag
a
2⎪
2
2
2⎪
u + w = {u} ⎨1 + {kH}
=
=
ag
⋅
=
ag
⋅
⎬
2
⎜ ⎟
!"#
c0
H
⎪⎩
small ⎪
⎭ ⎝ c0 ⎠
Nonlinear:
⎧ ∂ 2φ ⎫ ⎧ ∂w ⎫ ⎧ ∂ 2η ⎫
2 2
2 2 2
2 a
⎨η
⎬ = ⎨η
⎬ = ⎨η 2 ⎬ = a ω = a k c0 = ag ⋅(kH ) ⋅
H
⎩ ∂t∂z ⎭ ⎩ ∂t ⎭ ⎩ ∂t ⎭
Since kH << 1, the second nonlinear term will always be smaller than the first, and the first
nonlinear term will be negligible compared to the linear terms if a/H << 1.
{
}
For the kinematic BC:
⎧ ∂η ⎫
Linear:
⎨ ⎬ = {w} = ac0 k
⎩ ∂t ⎭
ag
a
⎧ ∂η ⎫ ag
⋅ ak = ac0 k ⋅ 2 = ac0 k ⋅
⎨u ⎬ =
c0
H
⎩ ∂x ⎭ c0
Nonlinear:
ac0 k
a
⎧ ∂w ⎫
= ac0 k ⋅
⎨η
⎬=a
H
H
⎩ ∂z ⎭
Again, the nonlinear terms will be negligible compared to the linear terms if a/H << 1.
Thus, overall, the linearization is justified if a/H << 1.
b) In the deep-water limit kH << 1, {x} = {z} = k-1, {t} = ω-1, and
ω = (gk)1/2,
η = a cos(kx - ωt) = O(a),
φ = ag/ω sin(kx - ωt) = O(ag/ω),
u = agk/ω cos(kx - ωt) = O(agk/ω),
w = agk/ω sin(kx - ωt) = O(agk/ω) = O(u).
For the dynamic BC:
Linear:
⎧ ∂φ ⎫
⎨ ⎬ = {gη} = ag
⎩ ∂t ⎭
{u
Nonlinear:
2
+w
2
}
2
agk
⎛ agk ⎞
=⎜
= ag ⋅ 2 = ag ⋅ ak
⎟
⎝ ω ⎠
ω
2
⎧ ∂ 2φ ⎫ ⎧ ∂w ⎫
⎨η
⎬ = ⎨η
⎬ = a ⋅ agk = ag ⋅ ak
⎩ ∂t∂z ⎭ ⎩ ∂t ⎭
For the kinematic BC:
agk
⎧ ∂η ⎫
Linear:
⎨ ⎬ = { w} =
ω
⎩ ∂t ⎭
⎧ ∂η ⎫ agk
⋅ ak
⎨u ⎬ =
⎩ ∂x ⎭ ω
Nonlinear:
agk
agk
⎧ ∂w ⎫
k=
⋅ ak
⎨η
⎬=a
ω
ω
⎩ ∂z ⎭
All the nonlinear terms will be negligible compared to the linear terms in the corresponding
equations, justifying the linearization, if ak << 1.