San José State University

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Thayer Watkins
Silicon Valley
USA

 The Analysis of the Baroclinic Two-Layer (Five-Level) Modelof the Atmosphere

The governing equations for the model, which are derived elsewhere, are

#### (∂/∂t + V1·∇)∇2ψ1 + β(∂ψ1/∂x) - (f0/2Δp)ω2 = 0 (∂/∂t + V3·∇)∇2ψ3 + β(∂ψ3/∂x) + (f0/2Δp)ω2 = 0 (∂/∂t + V2·∇)(ψ1-ψ3) - (2σΔp/f0)ω2 = 0

where V1, V2 and V3 are the wind velocity vectors at levels 1, 2 and 3, respectively, and are determined from the stream function ψ at those two levels. ω2 is the vertical pressure velocity at level 2. The relationships to the levels is shown below.

It is assumed that V2=(V1+V2)/2. Rather than replacing V2 by this relationship it is more convenient to express the 1 and 3 levels of the variables in terms of the level 2 variables and the difference between level 1 and level 3 values. But first it is necessary to obtain the linearized form of the equations.

## The Linearization of the Equations

It is assumed that the stream function ψ is of the form

#### ψ(x,y,p,t) = -U(p)y + ψ'(x,p,t) and ω(x,p,t) = ω'(x,p,t)

This notation will be later replaced by a notation in which ψi(x,y,t) means ψ(x,y,pi,t).

Under the assumption concerning the stream function then

#### ∇2ψ = ∂2ψ'/∂x2 V = (-∂ψ/∂y, ∂ψ/∂x) = (U(p)+∂ψ'/∂y, ∂ψ'/∂x) ∂ψ/∂x = ∂ψ'/∂x

In the linearization process the only parts of the wind velocity vectors that can survive in the momentum equations are the background velocities; i.e., (U(p), 0). Thus the linearized momentum equations are:

Let

#### U2 = [U1+U2]/2 and UTh = [U1-U2]/2

where the Th in UTh stands for thermal.

With these definitions the momentum equations become

#### (∂/∂t + [U2+UTh]∂/∂x)(∂2ψ'1/∂x2) + β(∂ψ'1/∂x) - (f0/2Δp)ω'2 = 0 (∂/∂t + [U2-UTh]∂/∂x)(∂2ψ'3/∂x2) + β(∂ψ'3/∂x) + (f0/2Δp)ω'2 = 0

Or, upon rearrangement,

#### (∂/∂t + U2∂/∂x)(∂2ψ'1/∂x2) + UTh∂/∂x)(∂2ψ'1/∂x2)+ β(∂ψ'1/∂x) - (f0/2Δp)ω'2 = 0 (∂/∂t + U2∂/∂x)(∂2ψ'3/∂x2) - UTh∂/∂x)(∂2ψ'3/∂x2)+ β(∂ψ'3/∂x) + (f0/2Δp)ω'2 = 0

Now if these two equations are added and the result is:

#### (∂/∂t + U2∂/∂x)(∂2(ψ'1+ψ'3)/∂x2) + UTh∂/∂x)(∂2(ψ'1-ψ'1)/∂x2) + β(∂(ψ'1+ψ'3)/∂x) = 0

Likewise if the second equation is subtracted from the first the result is:

#### (∂/∂t + U2∂/∂x)(∂2(ψ'1-ψ'3)/∂x2) + UTh∂/∂x)(∂2(ψ'1+ψ'1)/∂x2) + β(∂(ψ'1-ψ'3)/∂x) - (f0/Δp)ω'2<<< = 0

Using the defintions

#### ψ'2 = [ψ'1+ψ'3]/2 ψ'Th = [ψ'1-ψ'3]/2

the previous two equations can be expressed as

#### (∂/∂t + U2∂/∂x)(∂2ψ'2/∂x2) + UTh∂/∂x)(∂2ψ'Th/∂x2) + β(∂ψ'2/∂x) = 0 and (∂/∂t + U2∂/∂x)(∂2ψ'Th/∂x2) + UTh∂/∂x)∂2ψ'2/∂x2) + β(∂ψ'Th/∂x) - (f0/2Δp)ω'2222 = 0

The linearization of the equation which comes from the thermodynamic equation is only a bit more complex. First, we must go back to the original stream functions

#### ψ1 = -U1y + ψ'1 ψ3 = -U3y + ψ'3and hence V2 = (U1+U3, ∂(ψ'1+ψ'3)/∂x) ∇(ψ'1-ψ'3) = (∂(ψ'1-ψ'3)/∂x, -(U1-U3)) so V2·∇(ψ1-ψ3)) = (U1+U3)(∂(ψ'1-ψ'3) - (U1-U3)(∂(ψ'1+ψ3)/∂x)

When this last expression is substituted into the thermodynamic equation the result is, after division by 2,

## The Existence of Nontrivial Solutions

If the variables are to have wave-like solutions

#### ψ2 = Ae[ik(x-ct)], ψth = Be[ik(x-ct)], ω2 = Ce[ik(x-ct)]

then the complex constants A, B and C must be such that

#### ik[(-c + U2)(-k2)+β]A + ikUTh(-k2)B = 0 ikUTh(-k2)A + ik[(-c + U2)(-k2)+β]B - (f0/2Δp)C = 0 -ikUThA + ik(-c + U2)(ik)B - (σΔp/f0)C = 0

C may be eliminated by dividing the third equation by (σΔp/f0) and multiplying by (f0/2Δp) and subtracting from the second equation. This is equivalent to multiplying the coefficients of A and B in the third equation by [(f0/2Δp)]/[(σΔp/f0)] = f02/(4σ(Δp)2) and subtracting from the second equation. This factor of f02/(4σ(Δp)2) will be denoted as Λ2. The result is

#### ikUTh(-k2+Λ2)A + ik[(-c + U2)(-k2-Λ2)+β]B = 0

After elimination of common factors of ik the set of equations to be satisfied by A and B is

#### [(c - U2)k2+β]A - UTh(k2)B = 0 UTh(k2-Λ2)A + [(c - U2)(k2+Λ2)+β]B = 0

For a nontrivial solution the determinant of the coefficient matrix for these equations must be zero.

The value of the determinant is

#### (c-U2)2k2(k2+Λ2) +β(c - U2)(2k2+Λ2) + β2 + UTh2k2(k2-Λ2)

When this is set equal to zero the result is a quadratic equation in (c-U2); i.e.,

Its solution is

#### (c-U2) = -d1/2d2 ±D where D2 = [d1/2d2]2 - d0/d2

The dispersion relation for the wave solutions is

#### c = U2 -d1/2d2 ±D

The dividing line between stable and unstable waves is then

#### D2 = 0 d12 = 4d2d0which is β2(2k2+Λ2)2 = 4[k2(k2+Λ2)][β2 + UTh2k2(k2-Λ2)] or β2(2k2+Λ2)2 - 4k2(k2+Λ2)β2 = 4k2(k2+Λ2)UTh2k2(k2-Λ2) or β2Λ4 = 4UTh2k4(k2-Λ2)(k2+Λ2) or β2Λ4/(4UTh2) = k4(k4-Λ4)

The latter equation can be put into the convenient form

#### β2/(4Λ4UTh2) = (k/Λ)4((k/Λ)4-1)

which is a quadratic equation in (k/Λ)4 and the solution is

#### (k/Λ)4 = 1/2 ± (1/2)(1 - β2/(4Λ4UTh2))1/2

Thus as UTh → ∞ k goes asymptotically to 0 or 1.

The relationship of UTh to (k/Λ) is as is shown below:

The minimum value of UTh is where the two roots for (k/Λ) are equal; i.e.,

#### 1 - β2/(4Λ4UTh2) = 0 that is to say, where UTh2 = 4Λ4/β2

The value of (k/Λ)4 where that minimum of UTh is achieved is (1/2) so the value of k is (1/2)1/4Λ.

## Empirical Values

For the following values of the parameters, all in SI units,

• σ = 2.5×10-6
• f0 = 1.0×10-4
• Δp = 25,000 Pa
• β = 2.3×10-11

the value of Λ is 6.32×10-6 and the corresponding wavelength is 0.9935×106 m = 993.5 km. The minimal value of an unstable thermal wind Uth is 12.1 m/s, which occurs for a wave of length equal to 1181 km.

(To be continued.)