9.7.1 Continuum formulation

As shown in Section 9.4.7, the effects on kinetic energy dissipation from horizontal deformations in the fluid takes the form $-\rho_{o} \, \int dV \, A \, (D_{T}^{2} + D_{S}^{2})$, where a Boussinesq fluid has been assumed. One is therefore led to consider the functional

 

$${\mathcal S} = - \rho_{o} \int \, \sqrt{\mathcal{G}} \, d\xi^{1} \, d\xi^{2} \, dz \, A \, (D_{T}^{2} + D_{S}^{2}).$$ (9.164)

As shown in this section, the functional derivative $\delta {\mathcal S} / \delta u^{a}$ is proportional to the friction $g_{ab} \, F^{b}$. The connection between a functional and the friction is afforded through the self-adjointness of the friction operator. This result will then lead to a numerical discretization of the friction which is ensured to dissipate kinetic energy on the discrete lattice. To make this method work, two assumptions are needed: (1) The viscosity is functionally independent of the velocity field. (2) The flow satisfies ``natural boundary conditions'', of which no-slip is one. The Smagorinsky viscosity does not satisfy the first assumption. Nonetheless, the functional approach will lead to a discretization inside of which one can employ the Smagorinsky viscosity. A similar assumption was used to discretize the isoneutral diffusion operator when diffusing active tracers. Writing the functional as ${\mathcal S} = \int \, \sqrt{\mathcal{G}} \, d\xi^{1} \, d\xi^{2} \, dz \, \mathcal{L}$ leads to the variation

$$\delta \, \mathcal{S} = \int \, \sqrt{\mathcal{G}} \, d\xi^{1} \, d\xi^{2} \, dz \; \delta \mathcal{L}.$$ (9.165)

Note that the metric components are held fixed, since the only variation considered here is that of the velocity field $u^{a} \rightarrow u^{a} + \delta u^{a}$, not the underlying space-time geometry. Since $\mathcal{L}$ is a function of the velocity and its derivative, $\mathcal{L}[u^{a}, u^{a}_{,b}]$, its variation leads to

$$\delta \, \mathcal{S} \int \, \sqrt{\mathcal{G}} \, d\xi^{1} \, d\xi^{2} \, dz \, \left... ... + \frac{\delta \, \mathcal{L}}{\delta u^{a}_{,b}} \, \delta u^{a}_{,b} \right]$$ =

$$\int \, d\xi^{1} \, d\xi^{2} \, dz \, \left[ \sqrt{\mathcal{G}} \... ...frac{\delta \, \mathcal{L}}{\delta u^{a}_{,b}} \right) \, \delta u^{a} \right],$$ = (9.166)

where an integration by parts has been performed. The total derivative reduces to a surface term, which vanishes when either $\delta u^{a} = 0$ on all boundaries, or $\hat{n}_{b} \, (\delta \, \mathcal{L}/\delta u^{a}_{,b}) \, \delta u^{a} = 0$, where $\hat{n}_{b}$ are components to the outward normal at the boundaries. These two conditions define the ``natural boundary conditions'' mentioned above. If the velocity, and hence its variation, satisfy the no-slip condition, then $\delta u^{a} = 0$ on all boundaries and the total derivative can be dropped. More general boundary conditions can be derived from the second type of natural boundary condition, yet they are not considered here since MOM employs no-slip on the side boundaries. With natural boundary conditions, the variation of the functional takes the form

$$\delta \, \mathcal{S} \int \, \sqrt{\mathcal{G}} \, d\xi^{1} \, d\xi^{2} \, dz \, \left... ...rac{\delta \, \mathcal{L}}{\delta u^{a}_{,b}} \right) \right] \, \delta u^{a} ,$$ = (9.167)

which then leads to the functional derivative

$$\frac{ \delta \, \mathcal{S}}{\delta u^{a}} \frac{\delta \, \mathcal{L}}{\delta u^{a}} - \mathcal{G}^{-1/2} \... ... \sqrt{\mathcal{G}} \, \frac{\delta \, \mathcal{L}}{\delta u^{a}_{,b}} \right).$$ = (9.168)

To reach this result, it was necessary to use the identity

$$\frac{\delta u^{a}(\vec{x})}{\delta u{^b}(\vec{y})} = \delta^{a}_{b} \, \delta(\vec{x} - \vec{y}),$$ (9.169)

where $\delta(\vec{x} - \vec{y})$ is the Dirac delta-function. The delta-function has physical dimensions of inverse volume L^-3. Hence,

$$\int \, \sqrt{\mathcal{G}} \, d\xi^{1} \, d\xi^{2} \, dz \, \delta(\vec{x} - \vec{y}) = 1,$$ (9.170)

so long as the integration is over a domain containing the singular point $\vec{x} = \vec{y}$; otherwise, the integral vanishes. Now that the general functional derivative of $\mathcal{S}$ has been computed, it remains to prove the connection to the friction vector. For this purpose, recall equation (9.103), for which is was shown that the horizontal tension can be written

$$D_{T} = u^{1}_{\; , 1} - u^{2}_{\; , 2} + u^{m} \, \partial_{m} \, \ln (h_{1}/h_{2}),$$ (9.171)

and the horizontal shearing strain can be written

$$D_{S} = \frac{h_{1}}{h_{2}} \, u^{1}_{\; , 2} +\frac{h_{2}}{h_{1}} \, u^{2}_{\; , 1}.$$ (9.172)

Hence, the horizontal tension is functionally dependent on both the velocity and its partial derivatives, whereas the shearing strain is dependent only on the velocity partial derivatives. These results lead to the functional derivatives

$$-\frac{\delta \mathcal{L}}{\delta u^{a}} = 2 \, \rho_{o} \, ... ... + \delta^{2}_{a} \, \partial_{2} \, \ln(h_{1}/h_{2}) \, ],$$ (9.173)

and

$$-\frac{\delta \mathcal{L}}{\delta u^{a}_{,b}} 2 \, \rho_{o} \, A \, \left( D_{T} \, \frac{\delta D_{T}}{\delta u^{a}_{,b}} + D_{S} \, \frac{\delta D_{S}}{\delta u^{a}_{,b}} \right)$$ =

$$2 \, \rho_{o} \, A \, \left( D_{T} \, (\delta^{1}_{a} \, \delta^{... ...elta^{b}_{2} +\frac{h_{2}}{h_{1}} \, \delta^{2}_{a} \, \delta^{b}_{1}) \right).$$ = (9.174)

For a=1, these results lead to

$$\frac{1}{2 \, \rho_{o}} \, \frac{ \delta \, \mathcal{S}}{\delta u^{1}} -A \, D_{T} \, \partial_{1} \, \ln(h_{1}/h_{2}) + A \, D_{T} \, \... ... (A \, D_{T})_{,1} + \frac{1}{h_{1} \, h_{2}} \, (h_{1}^{2} \, A \, D_{S})_{,2}$$ =

$$h_{2}^{-2} \, (h_{2}^{2} \, A \, D_{T} )_{,1} + \frac{1}{h_{1} \, h_{2}} \, (h_{1}^{2} \, A \, D_{S} )_{,2}$$ =

$$h_{1}^{2} \, F^{1}.$$ = (9.175)

Similar manipulations with a=2 lead to

$$\frac{1}{2 \, \rho_{o}} \, \frac{\delta \, \mathcal{S}}{\delta u^{a}} = g_{ab} \, F^{b},$$ (9.176)

which is the desired general result.