5. Initial value problem - Homogeneous medium
Deep water waves (ω2 = gk). Again method of stationary phase
+∞
Put η(x,t) = ∫ [C(k)ei (kx+ωt ) + D(k)ei (kx−ωt ) ]dk
−∞
superposition of waves going in opposite directions
We need to specify at t = o
+∞
η(x,o) = ∫ [C(k)+ D(k)]eikx dk
−∞
and
+∞
ηt (x,o) = ∫ iw[C(k)− D(k)]eikx dk
−∞
Assume for simplicity ηt(x,o) = 0
then C(k) = D(k) and C(k)+ D(k) = 2C(k) = η
˜o
Then
η(x,t) =
1 +∞
∫ η˜ o (k) e i(kx+ωt) + e i(kx−ωt) dk
2 −∞
[
]
Now ω = (sign k)(g|k|)1/2 to ensure that waves at frequency ω propagate in the same
direction (even though oppositely) regardless of the sign of k. Separate into left and
right-going wave contributions:
η(x,t) =
+
−
1 +∞
1 +∞
∫ η˜ o (k)e iθ t dk + ∫ η˜ o (k)e iθ t dk
2 −∞
2 −∞
where
θ± ≡
kx
± (sign k)(g | k |)1 / 2
t
Points of stationary phase (giving the major contribution to the integrals) are those where
∂ ±
θ =0
∂k
1
Let us consider the half-plane (x>o) for k>o |k| = R
θ+ (k) =
kx
+ (gk)1/ 2
t
left-going wave
∂θ+ x 1 g
= +
=0
∂k
t 2 k
gives
x=−
t g
< 0 always for increasing time
2 k
No stationary points in θ+ for x>o
θ− =
kx 1
− (gk)1 / 2 = 0
t
2
right going wave
∂θ−
x 1 g
|k=k o = − ( )1/ 2 = 0 at k = ko
∂k
t 2 k
x 1 g 1/2
− ( )
|= c g |k o
t 2 ko
group velocity of packet centered at ko
∂2θ−
1 g1/ 2
2x 3
|
=
(
)
=
k
∂k 2 o 4 k 3o / 2
gt 3
as
8x 3 1
=
t 3 g3 / 2
k 3o / 2
1
only right going wave
−
Then
+∞
1
ηk>o (x > o,t)~ η
˜ o (k o )eiθ (k o )t ∫
2
−∞
+∞
2
π
As ∫ e−αz dz = ( )1/ 2
α
−∞
and
(k−k o ) 2 θ'' (k o )t
2i
e
dk
z2 = (k-ko)2
α=
θ''(k o )t
2i
⎡ 2πi ⎤1/ 2
1
iθ− (k o )t
ηk>o (x > o,t)~ η
˜ o (k o )e
⎢
⎥
2
⎣ tθ"(k o )⎦
π
i ⎛ πgt 2 ⎞
2πi
⎟
= e 2 ⎜⎜
But
3 ⎟
tθ''(k o )
x
⎝
⎠
2
Hence
⎡ x
⎤
θ − (k o ) = ⎢ k o − g1 / 2k1o/ 2⎥ t
⎣ t
⎦
⎡
⎤
1
g1/ 2
1/ 2 1/ 2⎥
2
⎢
= ko
− g k o t = − g1/ 2k1/
o t
1/
2
2
⎢⎣ 2k o
⎥⎦
As k 3o / 2 =
g 3 / 2t 3
1/2
1/2 g t
⇒
k
=
o
2x
8x 3
And
1
g1/ 2t
1 gt 2
θ− (t o ) = − g1/ 2t
=−
2x
2
4 x
1/ 2
So
⎡ 2πi ⎤
−
e iθ (k o )t ⎢ ''
⎥
⎢⎣ tθ (k o )⎥⎦
1/ 2
1 gt 2 π ⎡
− i
i πgt 2 ⎤
=e 4 x e 4⎢ 3 ⎥
⎢
⎥
⎣ x ⎦
⎡ 2 ⎤
gt π
1/ 2
−i ⎢ − ⎥⎡
⎢⎣4x 4 ⎥⎦ πgt 2 ⎤
⎢ 3 ⎥
=e
⎢⎣ x ⎥⎦
Take the real part
⎡ πgt 2 ⎤1 / 2 ⎡ gt 2 π ⎤
1
ηk >o (x > o,t) ≅ η
− ⎥
˜ o (k o )⎢ 3 ⎥ cos⎢
2
⎢⎣ x ⎥⎦
⎢⎣ 4x 4 ⎦⎥
~
⎡ gt 2 π ⎤
1
t
η
− ⎥
˜ o (k o )(πg)1/ 2( 3/ 2 )cos⎢
2
⎢⎣ 4x 4 ⎥⎦
x
⎡ πgt 2 ⎤1 / 2
⎢
⎥ is the modulating amplitude
⎢⎣ x 3 ⎥⎦
Central wavelength
2π 8πx 2
=
ko
gt 2
x -> increases λ -> increases at fixed x
The final solution is:
3
⎡ gt 2 π ⎤
1
t
η(x > o,t)~ η
− ⎥
˜ o (k o )(πg)1/ 2( 3/ 2 )cos⎢
2
⎢⎣ 4x 4 ⎥⎦
x
modulating
wave
amplitude
part
Plot η(x,t) as a function of x at fixed t, i.e. a snapshot of the wave:
The central wavelength
2π
increases and the amplitude decreases with x going away
ko
from the initial position.
η
x=- 3
2
x
0
x = 0 Singular point
Figure by MIT OpenCourseWare.
If we put a wavestaff at fixed x and record η(x,t) as a function of time
The wavelength decreases and the amplitude
η
increases linearly
t = o amplitude = o
t
Figure by MIT OpenCourseWare.
4
2.0
N0(k)
1.5
1.0
Re exp(itθ(k))
0.5
0.0
-0.5
k = k0
-1.0
-3
-2
-1
0
1
2
3
The behavior of the exponential factor for a large t showing the interval of stationary phase
Figure by MIT OpenCourseWare.
5
MIT OpenCourseWare
http://ocw.mit.edu
12.802 Wave Motion in the Ocean and the Atmosphere
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.