Solution of the Wave Theory exercises
J. Rabault
5th February 2016
1 Solution: plot the wave velocity field
1. In the limit of deep water waves, kh >> 1, we have:
cosh[k(z + h)] =
ek(z+h) + e−k(z+h)
ek(z+h)
≈
2
2
ekh − e−kh
ekh
≈
2
2
So that the velocity potential becomes, neglecting infinitesimal terms:
sinh(kh) =
φ=
aω kz
e sin(kx − ωt)
k
(1)
(2)
(3)
2. By definition of the gradient, vx = ∂φ/∂x and vz = ∂φ/∂z, so that:
vx = aωekz cos(kx − ωt)
(4)
vz = aωekz sin(kx − ωt)
(5)
3. This can be plotted in Matlab, in Fig. 1 is an example of snapshot for vx , vz and the
velocity as a vector field for waves of period 2 seconds and amplitude 0.1 meter.
2 Solution: Dispersion relation for gravity waves
1. limx→∞ tanh(x) = 1, therefore ω 2 = gk for deep water waves. By plotting the curve of
the tanh function, you can see that tanh ≈ 1 already for about 0.5.
2. It is easy to compute that:
ω
cp (k) = =
k
cg (k) =
g
g
=
k
ω
(6)
∂ω
1
= cp (k)
∂k
2
(7)
r
3. See Fig. 2.
4. See Fig. 3.
1
Solution Wave exercices
Figure 1: vx (top, left), vz (top, right) and velocity vectors (bottom) for deep water waves of period
2 seconds and amplitude 0.1 meter. With the convention chosen, i.e. θ = kx − ωt, waves
are propagating to the right as time goes.
5. According to figures 2 and 3, there are significant deviations from the deep water dispersion
relation for such waves. We cannot consider those waves as deep water gravity waves.
We obtain from the dispersion relation that λ = 4.6 m. The group velocity describes
the velocity at which energy is transmitted through the medium. Therefore, the group
velocity describes the length of the wave packet generated. Calculation in the case of the
waves generated leads to cg = 2.1 m/s. The phase velocity describes how fast wave crests
are advected, calculations lead to cp = 2.3 m/s. Therefore, if we take a snapshot, we will
observe 2.1 ∗ 20/4.6 waves i.e. about 9 waves. I we stand at a point and count the crests
that go by, we will observe 20/2 crests i.e. 10 crests.
3 Solution: Stokes drift
1. By symmetry of the cos function, we only need to find the smallest positive Tm such that
η(0, Tm ) = z. This is Tm = arccos(z/a)/ω
2. We should now compute:
1 Z arccos(z/a)/ω
Vm (z) =
aωekz cos(ωt)dt
T −arccos(z/a)/ω
(8)
Any formal computation tool (example: Maple, SymPy) will be able to give you the
following result (or, if y ou like integrals by substitution of variable, you can do it by
hand):
2aekz
Vm (z) =
T
s
1−
2
z
a
(9)
J. Rabault
2
Solution Wave exercices
10
8
Gravity waves
Deep water gravity waves
Gravity waves
Deep water gravity waves
8
6
6 (m)
k (rad/m)
6
4
4
2
0
2
0
0
2
4
6
8
0.5
1
! (rad/s)
1.5
2
2.5
T (s)
Figure 2: Dispersion relation for Gravity Waves taking into account finite depth effects, and Deep
Water Gravity Waves. The water depth for the finite depth case is h = 0.7 m.
4
4
c p Gravity waves
c g Deep Gravity waves
2
1
0
c p Deep Gravity waves
3
c g Gravity waves
c (m/s)
c (m/s)
3
c p Gravity waves
c p Deep Gravity waves
c g Gravity waves
c g Deep Gravity waves
2
1
0.5
1
1.5
2
2.5
0
T (s)
0.5
1
1.5
2
2.5
f (Hz)
Figure 3: Phase and group velocities for Gravity and Deep Water Gravity waves. The water depth
for the finite depth case is h = 0.7 m.
J. Rabault
3
Solution Wave exercices
0
z (m)
-0.5
-1
Stokes drift
Eulerian mean current
-1.5
0
0.02
0.04
0.06
0.08
0.1
velocity (m/s)
Figure 4: Comparison of the Stokes drift obtained in the Eulerian and Lagrangian approaches.
3. The results are shown in figure 4.
4. We simply perform the calculations that are required. In the computation of the inteagrated
mean Eulerian drift, we must limit ourselves to second order terms in = ka << 1:
Mm 1 =
Mm 2 =
1
ωka2 e2kz dz = ωa2
2
−∞
Z a
Z a
−a
2aekz
T
s
2
z
1−
a
(10)
dz ≈
Z a
−a
s
2
2a
z
1−
T
a
1
dz = ωa2
2
(11)
4 Solution: third order Stokes waves
1. The crest is obtained at for example θ = 0, the trough at for example θ = π. One then
finds that:
3
H = 2a 1 + (ka)2
8
!
(12)
2. Since by definition λk = 2π, H/λ ≈ 0.1412 can be written as:
!
1
3
Hk
= (ka) 1 + (ka)2 ≈ 0.1412
2π
π
8
(13)
Which can be solved into max ≈ 0.42
3. The results are presented in Fig. 5. You can observe that Stokes waves are higher, have
higher and sharper crests and less deep and smoother troughs. If corrections to all orders
would be included in the solution expansion, we would obtain a 120 degrees angle at waves
crests.
J. Rabault
4
0.1
0.1
0
0
2 (m)
2 (m)
Solution Wave exercices
-0.1
Linear
Stokes second order
Stokes third order
-0.2
0
0.5
1
1.5
-0.1
Linear
Stokes second order
Stokes third order
-0.2
2
x (m)
0
0.5
1
1.5
2
x (m)
Figure 5: Elevation of linear, second order and third order Stokes waves. Wave length λ = 1 m, wave
steepness ka = max /2 (left), ka = max (right).
J. Rabault
5