159
II. 2.4 Corollary
If 0 is simply connected then the Ho (div 0, curl 0) space is reduced to to} and we
have the orthogonal decomposition for the scalar product of L 2 ( 0) 2 and V :
(11.2,14)
u = gradp + Curlq'
where p is the solution of (2,12), and q' solution of
(11.2,15)
-~q' = curlu into 0; q' = 0 on Y
If 0 is multiplicatively connected then we have the orthogonal decomposition for the
scalar product ofL 2 (0)2 and V:
(11.2;16)
u = gradp + Curlq' + Curlr
where p is solution of (2,12), q' is solution of (2,15) and r is solution of:
(11.2,17)
-~ = 0 into 0; r = 0 on YO' r = ci on Yi
with Curlr E Ho(div 0, curl 0). Reciprocally the curl of the solutions of the n problems (Pi)
generate Ho (div 0, curiO).
Demonstration
For the first point, we just have to note that in the decomposition of the Lemma 1, if u
E Ho (div 0, curl 0) and if 0 is simply connected, we get p = cst, q = 0 and so u = O. The
decomposition (11.2,14) is obvious as the (11.2,13) and (II.2,lS) problems are equivalent if 0
is simply connected.
If 0 is not simply connected, we have q' + r = q (q is defined in (II.2,11) , q' in
(II.2,1S) and r in (II.2,17)), hence the decomposition (11.2,16) (in the simply connected case,
we have r = 0).
At last, Curir E Ho(div 0, curiO) as div (Curlr) = 0, curl (Curir) = -~ = 0 and Curlr. n
that is the tangential differential coefficient to Y is null as r is constant on each component of
y. These three properties demonstrate that Curlr belongs to Ho(div 0, curl 0).
In the same way if u belongs to Ho(div 0, curl 0) then the decomposition (II.2,16)
shows that gradp = Curlq' = 0 and so u = Curir with r solution of (II.2, 17).
II. 2.5 Demonstration of the theorem
a) The (P2) problem is classic and admits an infinity of solutions (A., p) belonging to
H 2 (0). If (A., p) is solution of (P2) then:
II. 2.4 Corollary
If 0 is simply connected then the Ho (div 0, curl 0) space is reduced to to} and we
have the orthogonal decomposition for the scalar product of L 2 ( 0) 2 and V :
(11.2,14)
u = gradp + Curlq'
where p is the solution of (2,12), and q' solution of
(11.2,15)
-~q' = curlu into 0; q' = 0 on Y
If 0 is multiplicatively connected then we have the orthogonal decomposition for the
scalar product ofL 2 (0)2 and V:
(11.2;16)
u = gradp + Curlq' + Curlr
where p is solution of (2,12), q' is solution of (2,15) and r is solution of:
(11.2,17)
-~ = 0 into 0; r = 0 on YO' r = ci on Yi
with Curlr E Ho(div 0, curl 0). Reciprocally the curl of the solutions of the n problems (Pi)
generate Ho (div 0, curiO).
Demonstration
For the first point, we just have to note that in the decomposition of the Lemma 1, if u
E Ho (div 0, curl 0) and if 0 is simply connected, we get p = cst, q = 0 and so u = O. The
decomposition (11.2,14) is obvious as the (11.2,13) and (II.2,lS) problems are equivalent if 0
is simply connected.
If 0 is not simply connected, we have q' + r = q (q is defined in (II.2,11) , q' in
(II.2,1S) and r in (II.2,17)), hence the decomposition (11.2,16) (in the simply connected case,
we have r = 0).
At last, Curir E Ho(div 0, curiO) as div (Curlr) = 0, curl (Curir) = -~ = 0 and Curlr. n
that is the tangential differential coefficient to Y is null as r is constant on each component of
y. These three properties demonstrate that Curlr belongs to Ho(div 0, curl 0).
In the same way if u belongs to Ho(div 0, curl 0) then the decomposition (II.2,16)
shows that gradp = Curlq' = 0 and so u = Curir with r solution of (II.2, 17).
II. 2.5 Demonstration of the theorem
a) The (P2) problem is classic and admits an infinity of solutions (A., p) belonging to
H 2 (0). If (A., p) is solution of (P2) then: