Ecole doctorale de Physique de la région Parisienne (ED107)

The divergence-free approximation
B.2 Solving Euler or Navier-Stokes equations 185
The spirit of this approximation consists of replacing the mass-conservation Equation
(B.19) by the condition
div W = 0 (B.21)
or equivalently Ψ = 0. The problem is then to find the enthalpy h in a such a way that
the condition (B.21) is satisfied. This can be done easily by solving the Poisson Equation
∆h − divF j+1/2 = 0 (B.22)
reached by taking the divergence of the EE. As the gradient of a harmonic function is a
divergence-free vector, such a function can then be added to a particular solution of the
Equation (B.22) in order to satisfy the boundary conditions. Once h is obtained, the EE
has to be solved with ˆ F = F − ∇h. If viscous terms are present, NSE case, a vectorial
type equation must be solved.
Anelastic approximation
As was already explained, the anelastic approximation consists of neglecting the term
∂th in the Equation (B.19). Once again, the game consists of finding the enthalpy h in a
such a way that the following equation is satisfied :
ΓH0 div W + Wr∂rH0 = 0. (B.23)
To find h, it is sufficient to derive with respect to time the Equation (B.23), and to
replace ∂tdiv ˜ W by its value reached from the EE. It gives
Γ H0 ∆h + ∂rH0∂rh = Γ H0 div F j+1/2 + F j+1/2
r ∂rH0 (B.24)
where F contains all the force terms (Coriolis force included).
The solution of the above equation is achieved with a semi-implicit scheme very similar
to the one used to solve the Equation (B.1). The problem is then reduced to solving the
following Equation (for simplicity, we shall consider only the case of a polytropic EOS) :
(γ − 1)(a + br 2 )∆h + 2br∂rh = ¯ H0∆h + ∂r ¯ H0∂rh + Γ H0 div ˜ F j+1/2 + ∂rH0F j+1/2
r
(B.25)
where a and b are two constants defined as was done in solving the Equation (B.1). The
LHS operator can be easily inverted and the Equation (B.25) can be solved by iteration [see
Gourgoulhon et al. (2001)]. Since the iteration at each time step is quite time consuming,
a more convenient strategy (although less accurate) consists of replacing h in the right
hand side of the Equation (B.25) by h j+1/2 .