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Next: 9.6 Comments on frictional Up: 9.5 Biharmonic friction Previous: 9.5.1 General formulation

   
9.5.2 Effects on kinetic energy

The manipulations necessary to show that the biharmonic friction dissipates kinetic energy are analogous to those used for second order friction in Section 9.4.8. As with that discussion, the relevant contribution from horizontal biharmonic friction is given by $dV \, u_{m} \, \Theta^{mn}_{; n}$. Assuming either no-slip or no-normal $\Theta$ stress at the boundaries brings this expression to the form

$\displaystyle \int dV \, u_{m} \, \Theta^{mn}_{; n}$ = $\displaystyle -\int dV \, \Theta^{mn} \, e_{mn}$  
  = $\displaystyle \int dV \, \rho \, B \, [ 2 \, E^{mn} \, e_{mn} - e^{n}_{n} \, E^{m}_{m}].$ (9.139)

For the product of traces, one has
$\displaystyle \int dV \, \rho \, B \, e^{n}_{n} \, E^{m}_{m}$ = $\displaystyle \int dV \, \rho \, B \, e^{n}_{n} \, F^{m}_{\, ; m}$  
  = $\displaystyle \int dV \,
[ (\rho \, B \, e^{n}_{n} \, F^{m})_{; m} - (\rho \, B \, e^{n}_{n})_{; m} \, F^{m}].$ (9.140)

The first term reduces to a boundary contribution, which will be assumed to vanish. For the contraction of the two strain tensors, one has
$\displaystyle 2 \, \int dV \, \rho \, B \, e^{mn} \, E_{mn}$ = $\displaystyle 2 \, \int dV \, \rho \, B \, e^{mn} \, F_{m \, ; n}$  
  = $\displaystyle -2 \, \int dV \, F_{m} \, (\rho \, B \, e^{mn})_{\, ; n},$ (9.141)

where the boundary term $(\rho \, B \, e^{mn} \, F_{m})_{; n}$ was assumed to vanish. Combining the two contributions leads to
$\displaystyle \int dV \, u_{m} \, \Theta^{mn}_{; n}$ = $\displaystyle -\int dV \, F_{m} \, [ 2 \, B \, \rho \, e^{mn} - g^{mn} \, B \, \rho \, e_{k}^{k} ]_{\, ; n}$  
  = $\displaystyle -\int dV \, \rho \, F_{m} \, F^{m},$ (9.142)

which is non-positive. If the viscosity B is distributed non-symmetrically, then the effects on kinetics energy are guaranteed to be dissipative only for the special case of constant viscosity. That is, in cartesian coordinates, the operator $\nabla_{h} \cdot B \, \nabla_{h} \, (
\nabla_{h} \cdot B \nabla_{h} \psi)$ is dissipative for all B > 0, whereas $\nabla_{h} \cdot B \, \nabla_{h} \, ( \nabla_{h}^{2} \psi)$ or $\nabla_{h}^{2} \, ( \nabla_{h} \cdot B \nabla_{h} \psi)$ can be proven to be dissipative only for constant B. Until May 1999, this point was not recognized when implementing the biharmonic friction with non-constant viscosities in MOM.


next up previous contents
Next: 9.6 Comments on frictional Up: 9.5 Biharmonic friction Previous: 9.5.1 General formulation
RC Pacanowski and SM Griffies, GFDL, Jan 2000