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 ψ_i(x,y,t) means ψ(x,y,p_i,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:

(∂/∂t + U₁∂/∂x)(∂²ψ'₁/∂x²) + β(∂ψ₁/∂x) - (f₀/2Δp)ω'₂ = 0
(∂/∂t + U₃∂/∂x)(∂²ψ'₃/∂x²) + β(∂ψ₃/∂x) + (f₀/2Δp)ω'₂ = 0
 

Let

U₂ = [U₁+U₂]/2
and
U_Th = [U₁-U₂]/2
 

where the Th in U_Th stands for thermal.

With these definitions the momentum equations become

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

Or, upon rearrangement,

(∂/∂t + U₂∂/∂x)(∂²ψ'₁/∂x²) + U_Th∂/∂x)(∂²ψ'₁/∂x²)
+ β(∂ψ'₁/∂x) - (f₀/2Δp)ω'₂ = 0
(∂/∂t + U₂∂/∂x)(∂²ψ'₃/∂x²) - U_Th∂/∂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)
+ U_Th∂/∂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)
+ U_Th∂/∂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)
+ U_Th∂/∂x)(∂²ψ'_Th/∂x²) + β(∂ψ'₂/∂x) = 0
and
(∂/∂t + U₂∂/∂x)(∂²ψ'_Th/∂x²)
+ U_Th∂/∂x)∂²ψ'₂/∂x²) + β(∂ψ'_Th/∂x)
- (f₀/2Δp)ω'₂₂ = 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'₃
and 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,

(∂/∂t + U₂∂/∂x)(∂ψ'_Th/∂x) - U_Th(∂ψ'₂/∂x)
- (σΔp/f₀)ω'₂ = 0
 

The Existence of Nontrivial Solutions

If the variables are to have wave-like solutions

ψ₂ = Ae^[ik(x-ct)], ψ_th = 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 + ikU_Th(-k²)B = 0
ikU_Th(-k²)A + ik[(-c + U₂)(-k²)+β]B - (f₀/2Δp)C = 0
-ikU_ThA + 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

ikU_Th(-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 - U_Th(k²)B = 0
U_Th(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 + U_Th²k²(k²-Λ²)
 

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

d₂(c-U₂)² + d₁(c-U₂) + d₀ = 0
where
d₂ = k²(k²+Λ²)
d₁ = β(2k²+Λ²)
d₀ = β² + U_Th²k²(k²-Λ²)
 

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 = 4d₂d₀
which is
β²(2k²+Λ²)² = 4[k²(k²+Λ²)][β² + U_Th²k²(k²-Λ²)]
or
β²(2k²+Λ²)² - 4k²(k²+Λ²)β² = 4k²(k²+Λ²)U_Th²k²(k²-Λ²)
or
β²Λ⁴ = 4U_Th²k⁴(k²-Λ²)(k²+Λ²)
or
β²Λ⁴/(4U_Th²) = k⁴(k⁴-Λ⁴)
 

The latter equation can be put into the convenient form

β²/(4Λ⁴U_Th²) = (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/(4Λ⁴U_Th²))^1/2
 

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

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

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

1 - β²/(4Λ⁴U_Th²) = 0
that is to say, where
U_Th² = 4Λ⁴/β²
 

The value of (k/Λ)⁴ where that minimum of U_Th 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 U_th is 12.1 m/s, which occurs for a wave of length equal to 1181 km.

(To be continued.)