Two-box continued: solution space

This is essentially a continuation of the math in the previous post. The same warnings apply!

The previous analysis indicated we have 3 free variables to play with. Let them be αs, αo and Cs (the initial inverse time constants of the two boxes, and the heat capacity of the "surface" box).

Equations 9 and 12 of the previous post show a relationship between w+s and w+o depending on the α's, γ's and τ+, and equations 10 and 13 show the same relationship for w-s and w-o with τ- instead of τ+.

Define ratios r+ and r- by:

w+o = r+w+s

(and similarly for '-') - this gives r(+,-) the same meanings as in Lucia's analysis.

The value of w+s then factors out of Eq 9 and 12, and these become:

Eq. 16: 1/τ+ = αs + γs - γs r+
Eq. 17: r+ (1/τ+ - αo - γo) + γo = 0

Solving Eq. 16 for r+ gives:

Eq. 18: r+ = 1 + (αs - 1/τ+)/γs

Similarly for r-:

Eq. 19: r- = 1 + (αs - 1/τ-)/γs

Define the heat capacity ratio Cs/Co as a new parameter λ we have replacements:

Eq. 20: Co = Cs

and

Eq. 21: γo = λ γs

(the latter from Eq. 15).

Then combining Eq. 17, 18, (and their '-' versions) and using the replacement from Eq. 21 we get two equations for γs and λ in terms of the known values and the two free parameters αs and αo:

0 = (1 + (αs - 1/τ+)/γs) (1/τ+ - αo - γo) + γo
= 1/τ+ - αo + αs/(τ+γs) - αsαos - αsλ - 1/(τ+2γs) + αo/(τ+γs) + λ/τ+

or multiplying through by γs and rearranging:

Eq. 22: γs = (αsαo - (αs + αo)/τ+ + 1/τ+2)/((1 + λ)/τ+ - αo - λαs)

together with a second version of the same equation with τ+ replaced by τ-. Set

ν(+,-) = αsαo - (αs + αo)/τ(+,-) + 1/τ(+,-)2

Note these are defined in terms of our free parameters αs and αo and the input τ values.

Then solving for λ gives:

Eq. 23: λ = - (αo+ - ν-) - (ν+- - ν-+)) / (αs+ - ν-) - (ν+- - ν-+))

From which Eq. 22 then gives us γs, Eq. 21 gives us γo, Eq. 20 gives Co, and Eq. 18 and 19 give us r+ and r-.

The remaining variables to find are x, y, w+s and w-s. From Eq. 6 and 7 we can find the latter two in terms of y and the fitted coefficients:

Eq. 24: w+s = a2/(y + (1-y) r+)
Eq. 25: w-s = a3/(y + (1-y) r-)

Adding equation 11 and 14 together and substituting Eq. 24 and 25, we eliminate x and find one equation to solve for y in terms of known values:

1 = a2 Cs(1 + r+/λ)/(τ+ (y + r+(1 - y))) + a3 Cs(1 + r-/λ)/(τ-(y + r-(1 - y)))

which can be rearranged to a quadratic equation:

Eq. 26: (1 - r+)(1 - r-) y2 + (r+ + r- - 2 r+r-) y + r+r- - a2 Cs(1 + r+/λ)/τ+ - a3 Cs(1 + r-/λ)/τ- = 0

CORRECTION (see UPDATE 2 below):
Define b2 = a2Cs+ and b3 = a3Cs-:

Eq. 26: (1 - r+)(1 - r-) y2 + (r+ + r- - 2 r+r- - b2(1 + r+/λ)(1 - r-) - b3(1 + r-/λ)(1 - r+)) y + r+r- - b2r-(1 + r+/λ) - b3r+(1 + r-/λ) = 0
END OF CORRECTION

which we can solve with the usual assignment of a, b, c to the coefficients, and which gives two real solutions as long as (b2 - 4 a c) is positive.

Picking one or the other value for y (there may be just one that meets the constraint 0 <= y <= 1) we then can find w+s and w-s from Eq. 24-25, and finally x from Eq. 11.

And that gives us all the parameters of the two-box model in terms of the three free variables αs, αo and Cs, the chosen time constants, and fitted temperature series. The next thing to do is probably to substitute in some real values and see what the resulting box models look like... later!

UPDATE: Eq. 26 and the preceding unnumbered equation have been corrected for missing factors of τ+ and τ- in the coefficients (deriving from updates to Eq. 11 and 14 in the previous post).

UPDATE 2 (Sep 8, 2009): Eq. 26 was missing important factors involving the products of the fitted coefficients a2 and a3 with y (and also r+ and r- factors in the last two terms). Never trust algebra where they say "it can be shown" near the end of the discussion when the author was tired... Anyway, this has an important impact on subsequent posts, which have not been corrected at the time of this writing, but should be soon.

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Equations 16 & 17 are the

Equations 16 & 17 are the eigenvectors. So, I guess I won't get to use those as extra.

Is equation 26 dimensionally homogeneous? I think in SI it's got units (K-m*2 /Watt) (Joule/K m*2) so the time constant crept in from elsewhere.

Yup, same problem as in the

Yup, same problem as in the other post - if you divide a2 by τ+ and a3 by τ- then it's right. Thanks, I'll get this fixed later.

Ok-- Since other than the

Ok--
Since other than the time constant that propagated through, we seem to agree, I'll be organizing this the order I want to discuss. It's a bit different from your way mostly because I like to think of the dividing the planet into different size boxes instead of thinking of the temperature as a linear combination of box temperatures. (It works out the same-- I just think of it differently.)

I'll probably post graphs using what I consider somewhat reasonable ranges for dividing the planet earth, but probably ... wed? (Life can get in the way of blogging. Imagine that?)