# integration for upper ocean mixed layer equation

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 $\frac{1}{2}$.

Thus

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

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.
• 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. – Kyle Kanos Jun 10 '14 at 14:51