Continuing from the previous post, suppose we want to look at solutions where the (s) box is close to the short-time-constant (τ-) solution, and the (o) box then close to the long-time-constant (τ+) solution. That suggests setting the corresponding inverse time constants perturbatively close to τ- and τ+, respectively. Define dimensionless small numbers εs and εo as follows:
Eq. 27: αs = (1 - εs)/τ-
Eq. 28: αo = (1 - εo)/τ+
Then from the definitions of ν+ and ν- (unnumbered equation between 22 and 23 in the previous post) we find:
Eq. 29: ν+ = (1 - εs - εo + εsεo)/(τ+τ-) - (1 - εs)/(τ+τ-) - (1 - εo)/τ+2 + 1/τ+^2
= -εo(1/τ- - 1/τ+)/τ+ + εsεo/(τ+τ-)
Eq. 30: ν- = εs(1/τ- - 1/τ+)/τ- + εsεo/(τ+τ-)
Setting δ-+ = 1/τ- - 1/τ+ (a positive value since τ- < τ+) these can be simplified to:
ν+ = -εo ( δ-+ - εs/τ-)/τ+
ν- = εs ( δ-+ + εo/τ+)/τ-
Eq. 23 from the previous post then gives us λ (the heat capacity ratio
Co/Cs Cs/Co) - dropping terms of second or higher order in the ε's we get:
Eq. 31: λ ≈ - (ν+(1/τ+ - 1/τ-))/(-ν-(1/τ- - 1/τ+))
≈ - ν+/ν- ≈ εoτ-/(εsτ+)
so, to first order in the perturbations, the ratio of the (o) and (s) perturbations relative to their respective time constants must be equal to the corresponding heat capacity ratio λ =
Eq. 22 from the previous post then gives γs:
Eq. 32: γs = ν+/((1 + λ)/τ+ - (αo + λαs))
≈ -εo δ-+/(τ+ λ (-δ-+)) = εs/τ-
And similarly γo = λ γs = εo/τ+. Since the γ's are proportional to the heat transfer coefficient β from the original two-box problem, it's not surprising that they are of order epsilon (i.e. small) in this perturbative example (the pure eigenstates if they represented real systems would be thermally isolated). Note also the "second law" requirement that β be positive implies that the ε's must also be positive (thanks to the choice of perturbation direction in Eq. 27 and 28).
Eq. 18 from the previous post then gives r+:
Eq. 33: r+ = 1 + ((1 - εs)/τ- - 1/τ+)/γs ≈ δ-+τ-/εs
which is very large, being proportional to the inverse of the perturbation. This just means that most of the weight for the (o) solution is in the (+) term, and similarly most of the weight for the (s) solution would be in the (-) term. In fact, Eq. 19 gives r- ≈ 0 with the approximations we have so far, and it would need to be obtained to first order in the perturbations ε in some other fashion.
Using the result that r+ >> 1 >> r-, Eq. 26 from the previous post resolves to:
Eq. 34 through 36 have been corrected given corrections in Eq. 26 of the previous post:
Eq. 34: -r+ y^2 +( r+ - b2 r+/λ + b3r+) y - b3r+ ≈ 0
with 2 real solutions
Eq. 35: y = (1/2)(1 + b3 - b2/λ ± sqrt((1 + b3 - b2/λ)^2 - 4 b3))
as long as the term inside the square root is positive. If a3 (and thus b3) is negative, the sqrt term becomes greater than the first piece and the (-) solution for y is negative; however the (+) solution may be less than 1 if 1 + b3 - b2/λ < 1. Since the b's are defined as products of the a's with Cs (and divided by the associated time constant, one can think of varying Cs while holding the a's and the ratio λ = Cs/Co fixed. For very small Cs and given parameters, the square root term because close to 1, and so definitely positive. For very large values of Cs the squared term is of order Cs squared, and so will trend larger than the subtracted term, and again give a positive square root. However, the square root term does become negative for intermediate values of Cs, between the (-) and (+) solutions of:
Cs(l,r) = (a3/τ- + a2/(λτ+) ± 2 sqrt(a2a3/(λτ+τ-))/(a3/τ- - a2/(λτ+))^2
and for these values of of the "fast" heat capacity Cs there will be no allowed values of y, at least for small values of γs for which the perturbative approximation applies. I.e. for given fitting parameters and λ the "fast" heat capacity must be below the left-hand or above the right-hand end of the interval:
Eq. 36A: Cs < (sqrt(a3/τ-) - sqrt(a2/(λτ+)))^2/(a3/τ- - a2/(λτ+))^2
Eq. 36B: Cs > (sqrt(a3/τ-) + sqrt(a2/(λτ+)))^2/(a3/τ- - a2/(λτ+))^2
If a3/τ- is close to a2/(λτ+) then this excluded interval can span more than an order of magnitude. For the Tamino parameters with 30 years, and 1 year as the time constants, where the fit gives a2 = 0.739 and a3 = 0.038 in SI units, if the areal heat capacities are equal (λ = 1) the excluded range for Cs is 2.5x10^8 to 2.2x10^10 J/Km^2. For λ = 10, the excluded range is about 6.6x10^7 to 3.5x10^8 J/Km^2, while for λ = 100 the range is a much smaller 1.0x10^7 to 1.7x10^7 J/Km^2.
Note: Obsolete inequality from earlier erroneous analysis:
Eq. 36: 0 <= a2 <= λ τ+/(4 Cs)
From Eq. 24 and 25 of the previous post we can then find expressions for the w's:
Eq. 37: w+s ≈ a2εs/((1-y)δ-+τ-)
Eq. 38: w-s ≈ a3/y
Note these terms have the same sign as the fitted coefficients, given the previous constraints that require ε and δ-+ to be positive and y to be between 0 and 1.
Finally from Eq. 11 of the first post we get the relative weighting x for the forcing:
Eq. 39: x ≈ Csa3/(y τ-)
and requiring x > 0 is equivalent to requiring a3 > 0 - so the physical constraints of the underlying model do require that both a2 and a3 be positive. The constraint x <= 1 is strictest if we choose the smaller (-) solution for y - either way, given y the constraints on a3 become:
Eq. 40: 0 <= a3 <= y τ-/Cs
Since Cs is a free parameter here we can always pick a value that makes physical sense (as long as it allows the fitted coefficients to satisfy inequalities 36 and 40); the similar physical constraint on Co is manifested in the value of λ (the heat capacity ratio) which fixes the ratio of the (s) and (o) perturbations, so constraining Cs and Co leaves us with two one-parameter infinities of possible two-box solutions (one for each of the choices for y and the value of εs is free), as long as constraints 36 and 40 are met.
Thinking back it might have been more natural to let the three free parameters be Cs, Co and γs, and look at perturbative solutions with γs (i.e. the heat transfer coefficient β) small. The choice here was essentially equivalent to that set given Eq. 31 and 32.
UPDATE: I had the wrong expression for the meaning of λ - corrected now, above.
UPDATE 2 8 Sep 2009: Eq. 34 through 36 have been updated given the error in Eq. 26 in the previous post.
Next time: let's plug in some examples and see how they do...