162
which demonstrates that:
Curl (curl (Curlq» = 0
that is to say :
curl (Curlq) = -8q = cte and q = 0 on y
hence we deduce that this constant is null because curlu = curl(Curlq) = 0 on yand that q = 0
which ends the demonstration.
Corolary. Let HO (div 0) = { vEL 2 (Q )2, divv = 0, v.n = O} and u solution of (PI), A-:f.
O. We write u = gradp + Curlq as in (11.2,14). Then:
( i ) Curl ( curl (Curlq) ) E HO (div 0)
( i i ) Curl ( curl ( Curlq) ) = A Curlq
( iii) grad ( div ( gradp) ) = Agradp.
We do not find a similar property in the case of the Stokes's problem:
- 8U = AU into Q ; u = 0 on y
11.3. Numerical results.
II. 3.1. Principle of resolution.
We approach v and ~ by a finite linear combinations solutions of (Pl) and (P2) :
v= 1: aj (t) Uj + 1: bj (t) Vj + 1: Ck (t) Wk ; ~ = 1: dl (t) PI
with
-8Uj = -8(gradpj) = A; Uj ; -8Vj = -8(curl (qj) = IIj Vj 8Wk = curlwk = diVWk = 0
-8PI = ~ PI ; -~ = I1j CJ.j
We change into (PI) u and h by their approached value and we scalary multiply into L 2(Q)2
the equation (i) consecutively by Uj, i = 1 to n ; Vj, j = 1 to m ; Wk> k = 1 to p, then the
equation (ii) by PI> I = 1 to q. We then get a system of non linear ordinary differential
equations n + m + p + q equations of the type :
6'a +1:Aj 6i + 1: Ai,j 6i 6j =fa
that we solve by the Adams-Moulton's method.
II. 3.2 The test problem
Real domains are of very important sizes (up to 1000 km x 1000 km in the case of the
occidental Mediterranean Sea), thus it is not realist to be willing to determine the eigen real
elements. So, we reduced the time and scale spaces by a 10 6 factor, the sequence of the
which demonstrates that:
Curl (curl (Curlq» = 0
that is to say :
curl (Curlq) = -8q = cte and q = 0 on y
hence we deduce that this constant is null because curlu = curl(Curlq) = 0 on yand that q = 0
which ends the demonstration.
Corolary. Let HO (div 0) = { vEL 2 (Q )2, divv = 0, v.n = O} and u solution of (PI), A-:f.
O. We write u = gradp + Curlq as in (11.2,14). Then:
( i ) Curl ( curl (Curlq) ) E HO (div 0)
( i i ) Curl ( curl ( Curlq) ) = A Curlq
( iii) grad ( div ( gradp) ) = Agradp.
We do not find a similar property in the case of the Stokes's problem:
- 8U = AU into Q ; u = 0 on y
11.3. Numerical results.
II. 3.1. Principle of resolution.
We approach v and ~ by a finite linear combinations solutions of (Pl) and (P2) :
v= 1: aj (t) Uj + 1: bj (t) Vj + 1: Ck (t) Wk ; ~ = 1: dl (t) PI
with
-8Uj = -8(gradpj) = A; Uj ; -8Vj = -8(curl (qj) = IIj Vj 8Wk = curlwk = diVWk = 0
-8PI = ~ PI ; -~ = I1j CJ.j
We change into (PI) u and h by their approached value and we scalary multiply into L 2(Q)2
the equation (i) consecutively by Uj, i = 1 to n ; Vj, j = 1 to m ; Wk> k = 1 to p, then the
equation (ii) by PI> I = 1 to q. We then get a system of non linear ordinary differential
equations n + m + p + q equations of the type :
6'a +1:Aj 6i + 1: Ai,j 6i 6j =fa
that we solve by the Adams-Moulton's method.
II. 3.2 The test problem
Real domains are of very important sizes (up to 1000 km x 1000 km in the case of the
occidental Mediterranean Sea), thus it is not realist to be willing to determine the eigen real
elements. So, we reduced the time and scale spaces by a 10 6 factor, the sequence of the