I am trying to work through the following paper specifically trying to get from equation (1) to equation (6).

Equation 1 states that

$$ Q = \beta S e^{-\beta z} + 2B \delta(z) $$

where $\delta z$ is essentially equal to $\frac{1}{2}$.


$$ Q = \beta S e^{-\beta z} + 2B \frac{1}{2} $$ $$ Q = \beta S e^{-\beta z} + B $$

Equation 3 of the paper than specifies that

$$ \frac{dT}{dt} + \frac{\partial}{\partial z} \left(\overline{W'T'} \right) = Q $$

Thus equation 3 can be written as:

$$ \frac{dT}{dt} + \frac{\partial}{\partial z} \left(\overline{W'T'} \right) = \beta S e^{-\beta z} + B $$

The paper then describes that equation 4 is the integration of 3, which should produce:

$$ \frac{dT}{dt} + \left(\overline{W'T'} \right) = S + B - Se^{-\beta z}$$

How do they get this outcome?

Furthermore, how do we get to equation 6 in the paper:

$$ h\frac{dT_{s}}{dt} + \Lambda \left( T_{s} - T_{h} \right) \frac{dh}{dt} = S + B-Se^{-\beta h} \approx S+B $$


First, note that $\delta(z)\neq\frac12$ but $$ \int_0^h\delta(z)\,dz=\frac12 $$ which is different than your assertion that $\delta(z)\simeq\frac12$.

If you insert Equation (1) into Equation (3), you get $$ \frac{dT}{dt}+\frac{\partial}{\partial z}\left(\overline{W'T'}\right)=\beta Se^{-\beta z}+2B\delta(z) $$ Now integrate this over $z$ (using $z'$ to denote that it's different than the $z$ in the end): $$ \int_0^z\frac{dT}{dt}dz'+\int_0^z\frac{\partial}{\partial z'}\left(\overline{W'T'}\right)dz'=\int_0^z\beta Se^{-\beta z'}dz'+\int_0^z2B\delta(z')dz' $$ By assuming that $T$ is constant over the integration range, your first term pops out as $dT_s/dt$ where $T_s$ is the surface layer temperature, and the other terms are integrated straight forward to get $$ \frac{dT_s}{dt}+\left(\overline{W'T'}\right)_z=S+B-Se^{-\beta z} $$

The application of Equation (5) into the above is straight forward. In Equation 6, they let $z=h$ because they are looking at the particular depth $h$, the depth of the layer, rather than the arbitrary depth $z$. Since $\beta h>1$, then $\exp(-\beta h)\ll1$ and you can assert that $$ S+B-Se^{-\beta h}\approx S+B $$ to get Equation 6.

  • $\begingroup$ How do you get three terms on the rhs after integrating two terms (if that makes sense)? I get -Se^(-beta z) + B, I can;t see where the additional S comes from? Is the extra S the 'constant' term? $\endgroup$ – KatyB Jun 10 '14 at 14:34
  • $\begingroup$ If you integrate from $0$ to $z'$, then the sole $S$ comes from the $z=0$ case. I'll make it a little more clear in my answer. $\endgroup$ – Kyle Kanos Jun 10 '14 at 14:51

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