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The previous expressions allow us to obtain estimations and to demonstrate the
existence of solutions of this problem.
111.4 Special basis
We present here a special basis that allows to solve the equations transformed by sigma
and the rigid lid equations when a constant depth is considered.
In both cases we will note 0 the domain filled by the fluid, with 0 =Ox x ]0, 1[.
IlI.4.} Theorem
To solve by the Galerkin's method, we use a special basis associated with the eigenvalue
problem:
[
-.1u = Au
(P}) u.n = 0
Cur[u/ln = 0
dans DJ
sur r
sur r
As in dimension two, the solutions of the problem (PI) can be obtained from the
solutions of the following scalar problems (Orenga [1992]):
(P2) [-~ = Ap
dans Ox]
gradp.n = 0 sur ' Yx
(P3)[ -t1q=J.Ul
q=O
dans Ox]
sur ' Yx
[
-..1r= 0
(Pi)
r=}
r= 0
dans Ox]
sur }j
sur Y,\}j
i = 0,2, ... n)
where p, q and r are functions of Ox into R.
We get the following result:
Theorem: Let {Pi; iEN*} a set of solutions of (P2) forming an orthogonal base of L2(Ox),
{qj ; jEN*} a set of solutions of (P3) forming an orthogonal base of L2(Ox), and (r1,r2,'" rn)
the n solutions of the problems (Pi). Then:
i)
the reunion of :