Zdunkowski W., Bott A. Dynamics of the atmosphere: A course in theoretical meteorology

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15.4 Barotropic instability 461

In order to obtain an analytic solution to the problem we must linearize (15.105).

This results in the following equation:

∂

∂t



∇





∂

∂x



∇





−

∂ψ



∂x

+ β

∂ψ



∂x

= 0(15.108)

which can be solved by assuming the solution



= F (y)exp[ik

(x − ct)] (15.109)

We now introduce the absolute vorticity of the basic current

η =−

+ f (15.110)

Differentiation of this equation with respect to y leads to the introduction of the

Rossby parameter β:

=−

+ β (15.111)

Hence, we speak of a representation in the β-plane. Substituting (15.109) and

(15.111) into (15.108) gives the desired differential equation for the amplitude

F (y) of the disturbance:

−



c − u



F = 0

(15.112)

This is the famous Orr–Sommerfeld equation for frictionless ﬂow, which cannot

be solved for arbitrary values of

u(y).

To make the solution of (15.112) more tractable we assume that we have a

channel ﬂow that is bounded by rigid walls at y =±a. We may think of the walls

of the channel as latitude circles. At the rigid walls the normal components of the

velocity must vanish:

v(y =±a) =

∂ψ

∂x



y=±a

∂ψ



∂x



y=±a

= 0since

∂ψ

∂x

= 0(15.113)

so the amplitude of the disturbance must vanish also, that is F (y =±a) = 0. In

order to have a very general solution we permit not only the phase velocity c but

also the amplitude F to be complex:

c = c

+ ic

,F= F

+ iF

(15.114)

462 The barotropic model

Substitution of (15.114) into (15.112) gives

+ i

−



− ic

− u

dη



+ iF

) = 0(15.115)

with λ

= (c

− u)

+ c

. By separating the real and the imaginary parts we ﬁnd

(a)

−



− u

dη



−

dη

= 0

(b)

−



− u

dη



dη

= 0

(15.116)

The next step is to multiply (15.116a) by F

and (15.116b) by −F

and then add

the resulting equations together. Application of the identity

− F



− F



(15.117)

and then integrating over the width of the channel yields



−a



− F



dy = c



−a

+ F

dη

dy (15.118)

Since F (y =±a) = 0, the left-hand side of (15.118) must vanish. Thus, the

remaining part



−a



+ F



dy = 0

(15.119)

determines whether the ﬂow is barotropically stable or unstable. If c

= 0then

the ﬂow is always stable. If, however, c

= 0 then unstable solutions exist only if

the integral is zero. This type of stability treatment originated with Rayleigh. For

comparison let us recall the result of the previous section, where we considered a

situation with a linear wind shear and a constant Coriolis parameter. In this case

η/dy = 0 and the instability was determined solely by c

Let us now discuss the stability integral (15.119). Since the expression in paren-

theses is positive deﬁnite, it is the product c

dη/dy that determines the stability.

The derivative of the absolute vorticity was given by (15.111), permitting three

possibilities:

=−

+ β

0(15.120)

(i) The ﬂow ﬁeld is always stable if dη/dy does not change sign in the region of integration

so that the integral cannot vanish. If d

η/dy > 0ordη/dy < 0 between −a ≤ y ≤ a

15.5 The mechanism of barotropic development 463

then c

must be zero so that (15.119) remains valid. This condition is sufﬁcient for

barotropic stability since (15.109) results in stable oscillations.

(ii) If c

differs from zero then dη/dy must have at least one zero in order to satisfy

(15.119). This condition is necessary but not sufﬁcient for barotropic instability to

occur. Thus the necessary condition for barotropic instability is that β −d

u/dy

= 0

somewhere in the range of integration. Suppose that d

η/dy has a zero to make the

integrand zero. The ﬂow ﬁeld is not necessarily unstable since c

could be zero also so

that (15.119) vanishes under all circumstances. According to the results of the previous

section, we may speculate that a sufﬁcient condition for instability would result from

the requirement that the wavelength of the disturbance is large in comparison with

the lateral extent of the shear

zone.

(iii) Suppose that c

differs from zero and that dη/dy = 0 everywhere. If the Coriolis

parameter is assumed to be ﬁxed, then d

η/dy =−d

u/dy

= 0. In this case we have

either a vorticity maximum or a vorticity minimum. Therefore, the wind proﬁle must

have an inﬂexion point. This is a particular example of inﬂexion-point instability.This

type of stability investigation goes back to Lord Rayleigh (1880) for a nonrotating

system. Kuo (1951) has extended the theory by including the β-term.

15.5 The mechanism of barotropic development

Let us return to the divergence-free vorticity equation (15.105). This nonlinear

equation is satisﬁed by the so-called Neamtan solution (Neamtan, 1946)

ψ =

ψ + ψ



with

ψ = constant − u

y + A cos(ky)



= B cos[k

(x − ct)] cos(k

(15.121)

The term

ψ represents the stream function of the basic current, k = 2π/L is the

wavenumber, and

L the wavelength of the basic current. The term ψ



is the stream

function of the disturbance which is superimposed on the basic ﬂow.

We will now investigate the Neamtan solution. Assuming that

k =



+ k

(15.122)

the disturbance moves with the constant phase velocity

c = u

− β/k

(15.123)

which is known as the phase velocity of the Rossby wave. The validity of (15.123)

will be shown later. Note also that, in this particular situation, the wave moves

without change of shape since c is independent of y. This situation is known as the

indifferent case.

464 The barotropic model

Fig. 15.6 A schematic view of the basic ﬂow ﬁeld u(y) for the Neamtan solution.

The basic current itself is obtained by differentiating the basic current part of the

stream function with respect to y; see Figure 15.6.

u =−

∂ψ

∂y

= u

+ kA sin(ky)(15.124)

Since the indifference of the ﬂow ﬁeld results from the equality (15.122), we may

expect that the formation of barotropic stability or instability in some way depends

on the inequality

k =



+ k

(15.125)

For this general case it is impossible to determine the phase velocity c.

Let us reconsider the vorticity equation (15.68) in the form

∂ζ

∂t

+ u

∂ζ

∂x

+ v

∂ζ

∂y

+ βv = 0(15.126)

We split the vorticity and the velocity components:

ζ =

ζ + ζ



,u= u + u



,v= v



since v =

∂ψ

∂x

= 0(15.127)

We substitute these into (15.126) but retain the nonlinear terms and ﬁnd

∂ζ



∂t

∂

∂t

=−



∂ζ



∂x





∂ζ



∂x

+ v



∂ζ



∂y



+ v



∂ζ

∂y

+ βv





(15.128)

Note that ∂

ζ/∂x = 0sinceψ = ψ(y). By averaging this equation and recalling

that the mean over the ﬂuctuation vanishes, i.e.

∂ζ



∂t

= 0,

∂ζ



∂x

= 0,



= 0(15.129)

15.5 The mechanism of barotropic development 465

we ﬁnd

∂

∂t

=−





∂ζ



∂x

+ v



∂ζ



∂y



=−



·∇



(15.130)

Thus, the change of

ζ with time results from the advection of the perturbation

vorticity due to the perturbation velocity.

Let us now consider the Neamtan formula at the initial time t = 0. For clarity

we have collected the basic relationships needed to evaluate (15.130) in



=−

∂ψ



∂y



∂ψ



∂x

,ζ



=∇



=−



+ k





∂ζ



∂x

=−



+ k





∂ζ



∂y



+ k





∂ζ

∂y

A sin(ky)

(15.131)

The reason why these expressions are valid at t = 0 is that in general the phase

velocity c depends on y. Performing the required differentiations and substituting

the resulting expressions into (15.130) results in



∂

∂t



t=0

= 0(15.132)

We conclude that, at t = 0, the nonlinearity of the vorticity equation (15.126) is

removed due to the Neamtan solution. Again using (15.131), we ﬁnd at t = 0 from

(15.128) for the change with time of the perturbation vorticity that



∂ζ



∂t



t=0

=−



u −

+ k

∂ζ

∂y

−

+ k



∂ζ



∂x

(15.133)

This formula has the form of an advection equation, so the expression within

the large parentheses is the momentary phase velocity c(y)

t=0

of the vorticity in

the x-direction. By substituting u according to (15.124) and replacing the partial

derivative of the mean vorticity, we obtain the ﬁnal form for the phase speed at

t = 0:

c(y)

t=0



−

+ k



+ kA



1 −

+ k



sin(ky)

(15.134)

There are three cases that need to be discussed.

466 The barotropic model

Fig. 15.7 Positive and negative areas of the perturbation ﬁeld cos(k

x)cos(k

y)att = 0

with k

x = k

y. u



= 0 at positions a and c; v



= 0 at positions b and d.

(i)

= k

+ k

The disturbance moves with a constant velocity given by (15.123), which is inde-

pendent of y. The phase velocity and the disturbance do not adapt to the proﬁle of

the basic current. In this case of indifference the disturbance moves without change

of shape.

Before we discuss the second case let us obtain a picture of the disturbance



of (15.121) for the special case that k

x = k

y; see Figure 15.7. Positive and

negative areas correspond to regions of high and low pressure, thus deﬁning the

sense of the circulation. On the boundaries of each region one of the two velocity

components, u



or v



, is zero so that the product u



is zero. Thus, the zonal

mean of the perturbation product is zero also. However, the product u



,which

is usually nonzero, can be interpreted as the meridional transport of the mean

zonal momentum along a latitudinal circle. The direction and the magnitude of

the transport strongly depend on the structure of the disturbance. As shown in the

appendix to this chapter, the kinetic energy of the ﬂow may be decomposed into a

zonal part

K plus a perturbation part K



. There it is demonstrated that the change

of K with time is given by

∂

∂t

=−



∂

∂y

(



) dy dx (15.135)

where W and L represent the width and the length of the channel. Since in the

present case



is zero, there is no meridional transport of zonal kinetic energy.

Hence, K does not change with time.

(ii) k

+ k

This means that the wavelengths L

and L

are small in comparison with the

wavelength of the basic current. According to (15.134) the phase velocity and thus

the shape of the perturbation ﬁeld adjust to the shape of the basic ﬂow proﬁle

15.5 The mechanism of barotropic development 467

Fig. 15.8 Meridional transport of momentum, ∂u



/∂y < 0.

(see Figure 15.6) since the factor multiplying sin(ky) remains positive. The largest

contribution occurs at ky = π/2. The situation is displayed schematically in

Figure 15.8. To illustrate, let us consider the ﬂow at points (a, c) and (b, d). In both

cases u



< 0, so the zonal mean of u



is also less than zero. The central part

of Figure 15.8 repeats the situation of case I, in which the meridional transport of

momentum vanishes. Consider the points (A, C) and (B, D) in the lower section of

Figure 15.8. In both cases we have u



> 0, so the zonal mean of u



is also larger

than zero. In summary, the situation is characterized by ∂



/∂y < 0. Thus, for the

conﬁguration of the perturbed part of the ﬂow ﬁeld shown in Figure 15.8 the zonal

kinetic energy increases with time due to the meridional transport of momentum.

This leads to the formation of a jet stream. Referring to Figure 15.4, this situation

expresses barotropic stability.

From energy balances it follows that the real (baroclinic) atmosphere in the

majority of cases is barotropically stable. Friction would cause the western ﬂow to

slow down if kinetic energy of the perturbations were not transferred to the kinetic

energy of the basic current. To state it differently, barotropic stability is an essential

requirement for the maintenance of the westward wind drift. All barotropic models

exhibit the tendency of zonalization.

(iii) k

+ k

In this case L

and L

are not small in comparison with the wavelength of the basic

current. Now the term multiplying sin(ky) in (15.134) is negative. The contributions

of the basic current and the perturbations to the phase velocity occur with opposite

signs, in contrast to the previous case. Therefore, the disturbance has the opposite

shape, as shown in Figure 15.9. According to (15.135) this results in a depletion of

the zonal kinetic energy and in an accumulation of K



. According to Figure 15.4

this corresponds to barotropic instability. This situation is relatively rare and occurs

only in connection with large wavelengths of the disturbances. It is known that this

type of process contributes to the formation and maintenance of blocking highs.

468 The barotropic model

Fig. 15.9 Meridional transport of momentum, ∂u



/∂y > 0.

15.6 Appendix

In this appendix we are going to brieﬂy derive a formula for the transformation of

zonal to perturbation kinetic energy and vice versa for the ﬁltered model. Since for

the moment we are interested in midlatitude ﬂow only, we limit the ﬂow to a channel

of width W and length L. We may think of this restricted atmospheric ﬂow as being

bounded by two latitude circles. Moreover, the northern and southern boundaries of

the channel are treated as rigid walls. We also assume that the Coriolis parameter

varies according to (15.104). The average kinetic energy of the horizontal ﬂow

within the channel is then given by

K =



dy dx =



(∇

ψ)

dy dx (15.136)

since the divergence-free wind is given by v

= i

×∇

ψ.Inviewof

∇

ψ ·∇

∂ψ

∂t

=∇



ψ ∇

∂ψ

∂t



− ψ ∇

∂ψ

∂t

(15.137)

the change with time of the kinetic energy is given by

∂K

∂t

=−



∂ζ

∂t

dy dx (15.138)

Application of the two-dimensional Gaussian divergence theorem (M6.34) shows

that the divergence term of (15.137) does not contribute to the change in energy. This

results from the assumption that we have rigid walls and cyclic boundary conditions

in the x-direction. Using (15.68) and recalling that the horizontal divergence of the

velocity ﬁeld is zero in this equation, we ﬁnd

∂K

∂t



∇

· (v

ψη) dy dx (15.139)

15.6 Appendix 469

since v

·∇

ψ = 0. By again applying the divergence theorem and the adopted

boundary conditions we ﬁnd that the change with time of the kinetic energy vanishes

completely:

∂K

∂t

= 0(15.140)

Next we split the horizontal velocity by introducing the zonal average

and the

perturbation velocity v



= v

+ v



, v



dx, v



= 0(15.141)

Analogously, we split the kinetic energy into the zonal and perturbation parts:

K =

K + K





)

+ (v



)

dy dx (15.142)

In view of (15.138) we may write for the change of

K with time

∂

∂t

=−



∂ζ

∂t

dy dx =



ψ v



·∇



dy dx (15.143)

since the zonal average of ∂

ζ/∂t has the form (15.130). On rewriting this expression

we ﬁnd

∂

∂t



∇

· (ψ v



) dy dx −





·∇

ψ dy dx

=−





∂ψ

∂y

dy dx

(15.144)

where the ﬁrst integral on the right-hand side must vanish due to the divergence

theorem of Gauss and due to the boundary conditions.

Equation (15.144) can be rewritten giving a formula that consists of three inte-

grals:

∂

∂t

=−



∂

∂y

(



) dy dx +



u v



∂v



∂x

dy dx



u u



∂v



∂y

dy dx

(15.145)

Only the ﬁrst of these differs from zero, so

∂

∂t

=−



∂

∂y

(



) dy dx (15.146)

470 The barotropic model

This is the desired equation for the change with time of the zonal kinetic energy.

The other two integrals in (15.145) are zero, as follows immediately from partial

integration and by application of the boundary conditions. Finally, due to (15.140),

we ﬁnd

∂

∂t

=−

∂K



∂t

(15.147)

showing the only possible way for transformation of kinetic energy to occur.

Several parts of this chapter are based on the synopsis The Barotropic Model

reported in Promet (1973), which is a publication of the German Weather Service. It

refers to articles by Reiser (1973), Edelmann (1973), Edelmann and Reiser (1973),

and Tiedtke (1973).

15.7 Problems

15.1: Show that the conditions (15.1)

and (15.2) guarantee that the horizontal

pressure gradient in the barotropic model atmosphere is independent of height.

Hint: Start with the relation ∇×[(1/ρ) ∇p] = 0. Prove it.

15.2: Show that any two of the three relations

∇ρ ×∇p = 0,

∂

∂z



∇



= 0,

∂p

∂z

= f (z)

imply the correctness of the third one.

Hint: First prove that ∇

ρ ×∇

p = 0.

15.3: Verify equation (15.62).

15.4: Verify equation (15.103).

15.5: An upper-level midlatitude jet proﬁle may be approximated by

u = u

sech

(y/y

)

Plot the wind proﬁle u/u

(x-axis) versus y/y

(a) Suppose that β = 0. Is the necessary condition for barotropic instability satis-

ﬁed?

(b) Suppose that β = 0. Find the necessary condition for barotropic instability.