Geography Reference
In-Depth Information
Neglecting products of the perturbation quantities and noting that the basic state
fields are constants, we obtain the linear perturbation equations
1
∂
∂t
+
u
+
∂p
∂x
=
∂
∂x
1
ρ
u
0
(7.12)
∂
∂t
+
p
+
γ p
∂u
∂
∂x
u
∂x
=
0
(7.13)
Eliminating u
by operating on (7.13) with
∂
∂t
u∂
∂x
and substituting from
+
(7.12), we get
2
∂
∂t
+
2
∂
2
p
∂x
2
∂
∂x
γ p
ρ
p
−
u
=
0
(7.14)
which is a form of the standard
wave equation
familiar from electromagnetic
theory. A simple solution representing a plane sinusoidal wave propagating in x is
p
=
A exp [ik (x
−
ct)]
(7.15)
where for brevity we omit the Re
notation, but it is to be understood that only
the real part of (7.15) has physical significance. Substituting the assumed solution
(7.15) into (7.14), we find that the phase speed c must satisfy
{}
−
γ p
ρ
(ik)
2
iku)
2
(
−
ikc
+
=
0
where we have canceled out the factor A exp [ik (x
−
ct)], which is common to
both terms. Solving for c gives
±
γRT
1/2
(γ p/ρ)
1/2
c
=
u
±
=
u
(7.16)
Therefore (7.15) is a solution of (7.14), provided that the phase speed satisfies
(7.16). According to (7.16) the speed of
wa
ve propagation relative to the zonal
current is c
≡
γRT
1
/
2
−
u
=±
c
s
, where c
s
is called the
adiabatic speed of
sound
.
1
It is not necessary that the perturbation velocity be small compared to the mean velocity for
linearization to be valid. It is only required that quadratic terms in the perturbation variables be small
compared to the dominant linear terms in (7.12) and (7.13).
2
Note that the squared differential operator in the first term expands in the usual way as
∂
∂t
+
u
2
∂
2
∂t
2
∂t∂x
+
u
2
∂
2
∂
2
∂
∂x
=
+
2u
∂x
2
