9.8.4 Old biharmonic algorithm

The following algorithm summarizes the steps used to compute the biharmonic friction vector in MOM prior to Summer 1999. This method for non-constant viscosities was incorrect as it distributed the viscosity only on the second part of the calculation, rather than evenly as necessitated by the constraints of kinetic energy dissipation (Section 9.5.2). Hence, this approach may lead to problems with non-constant viscosities, although none have been found, and it is thought that the differences will be small.

• Second order calculation: The physical components (F^x,F^y) of the second order friction vector are summarized in Section 9.4.8. The same calculation is made for the biharmonic scheme, only using here a unit viscosity. Define the physical components (F^u,F^v), which equal (F^x,F^y) sans the vertical derivative term $(\kappa \, {\bf u}_{,z})_{,z}$
  

  $$g_{22}^{-1} \, (g_{22} \, D_{T})_{,x} + g_{11}^{-1} \, (g_{11} \, D_{S})_{,y}$$ F^u = (9.196)

  $$g_{11}^{-1} \, (g_{11} \, D_{T})_{,y} - g_{22}^{-1} \, (g_{22} \, D_{S})_{,x}$$ F^v = (9.197)

  
  
  where the horizontal tension and strain are given by the usual forms
  

  $$\sqrt{g_{22}} \, (u / \sqrt{g_{22}} \, )_{,x} - \sqrt{g_{11}} \, (v / \sqrt{g_{11}} \, )_{,y}$$ D_T = (9.198)

  $$\sqrt{g_{11}} \, (u/\sqrt{g_{11}} \, )_{,y} + \sqrt{g_{22}} \, (v/\sqrt{g_{22}} \, )_{,x}.$$ D_S = (9.199)

  
  

• Fourth order calculation: Use again the results from the second order calculation, only now with the second order friction vector taking the place of the velocity vector and the viscosity A set to the biharmonic viscosity. That is, compute
  
    

  $$g_{22}^{-1} \, (g_{22} \, A \, D^{B}_{T})_{,x} + g_{11}^{-1} \, (g_{11} \, A \, D^{B}_{S})_{,y}$$ F^u_B = (9.200)

  $$g_{11}^{-1} \, (g_{11} \, A \, D^{B}_{T})_{,y} - g_{22}^{-1} \, (g_{22} \, A \, D^{B}_{S})_{,x}$$ F^v_B = (9.201)

  
  
  where the new horizontal tension and strain are given by
  

  $$\sqrt{g_{22}} \, (F^{u} / \sqrt{g_{22}} \, )_{,x} - \sqrt{g_{11}} \, (F^{v} / \sqrt{g_{11}} \, )_{,y}$$ D^B_T = (9.202)

  $$\sqrt{g_{11}} \, (F^{u}/\sqrt{g_{11}} \, )_{,y} + \sqrt{g_{22}} \, (F^{v}/\sqrt{g_{22}} \, )_{,x}$$ D^B_S = (9.203)

  
  

For spherical coordinates in MOM, the following steps are taken:

• Second order calculation: The Laplacian friction vector is computed as in Section 9.8.2, only with a unit viscosity:
  

  $$\nabla_{h} \cdot \nabla_{h} u + old\_metric^{u}$$ F^u = (9.204)

  $$\nabla_{h} \cdot \nabla_{h} v + old\_metric^{v},$$ F^v = (9.205)

  
  
  where
  

  $$old\_metric^{u} \frac{u \, (1-\tan^{2}\phi)}{a^{2}} - \frac{2 \, v_{,\lambda} \, \sin\phi}{a^{2} \cos^{2}\phi},$$ = (9.206)

  $$old\_metric^{v} \frac{v \, (1-\tan^{2}\phi)}{a^{2}} + \frac{2 \, u_{, \lambda} \, \sin\phi}{a^{2} \cos^{2}\phi}.$$ = (9.207)

  
  

• Fourth order calculation: Now use the full biharmonic viscosity to compute the biharmonic friction
  

  $$\nabla_{h} \cdot (A \, \nabla_{h} F^{u}) + old\_metric^{u} + new\_metric^{u}$$ F^u_B = (9.208)

  $$\nabla_{h} \cdot (A \, \nabla_{h} F^{v}) + old\_metric^{v} + new\_metric^{v}.$$ F^v_B = (9.209)

  
  
  The metric terms are given by
  

  $$old\_metric^{u} \frac{A \, F^{u} \, (1-\tan^{2}\phi)}{a^{2}} - \frac{2 \, A \, F^{v}_{,\lambda} \, \sin\phi}{a^{2} \cos^{2}\phi}$$ = (9.210)

  $$old\_metric^{v} \frac{A \, F^{v} \, (1-\tan^{2}\phi)}{a^{2}} + \frac{2 \, A \, F^{u}_{, \lambda} \, \sin\phi}{a^{2} \cos^{2}\phi}$$ = (9.211)

  $$new\_metric^{u} - \frac{ A_{,\lambda} }{a^{2} \cos\phi} (F^{v}_{,\phi} + F^{v} \,... ...) + \frac{ A_{, \phi} }{a^{2} \cos\phi} ( F^{v}_{,\lambda} + F^{u} \, \sin\phi)$$ = (9.212)

  $$new\_metric^{v} \frac{ A_{,\lambda} }{a^{2} \cos\phi} ( F^{u}_{, \phi} + F^{u} \,... ...+ \frac{ A_{,\phi} }{a^{2} \cos\phi} (- F^{u}_{, \lambda} + F^{v} \, \sin\phi).$$ = (9.213)