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-!J. (gradp) = - grad (div (gradp» = -gradp (!J.p) = Agradp
Furthermore gradp verifies the boundary conditions of (Pl) which demonstrates that
(A.,gradp) is solution of (PI).
b) In the same way (P3) admits an infinity of (11, q) solutions belonging to H 2 (Q). If
(11, q) is solution of (P3) then:
-!J. (Curlq) = Curl (curl (Curlq» = -Curl (!J.q) = IlCurlq
which demonstrates that (J.1, Curlq) verifies the equation of (Pl). Furthermore:
curlu = curl (Curlq) = -!J.q = Ilq = 0 on y
and as Curlq.n is the tangential differential coefficient of q that is constant on y, Curlq. n = 0
on y which demonstrates that (Il,curlq) is solution of (PI).
c) If r is solution of (Pi) then Curlr verifies !J. (Curlr) = 0, Curlr. n = 0 on y (same
demonstration as the one in paragraph b) and curl (Curlr) = -!J.r = 0 on y. This demonstrates
that (0, Curlr) is solution of (PI).
d) This is a consequence of the previous lemma.
e) Let us suppose u belonging to V, u orthogonal to gradpi' Curlqj and to the
Curlrk,then using the Green's formulae (11.2,5) and (II.2,7) :
(u, gradpi) = - (Pi' divu) = 0 for each Pi
as the whole of the Pi constitutes a base of L 2 (Q)
then,
divu=O
(u, Curlrqj) = (curlu, qj) = 0 for each qj , and as the set of the qj form a base ofL 2(Q)
curlu = 0
Consequently u belongs to Ho (div 0, curl 0) and as by hypothesis u is orthogonal to all the
Curlrk that generate Ho(div 0, curl 0), u = O. The density of V (that contains H I O (Q)2) into
L 2(Q)2 allows to conclude that the whole of the linear combinations of the gradp, Curlq and
Curlr is dense into L 2( Q)2. The verification of the orthogonality is instantaneous.
-!J. (gradp) = - grad (div (gradp» = -gradp (!J.p) = Agradp
Furthermore gradp verifies the boundary conditions of (Pl) which demonstrates that
(A.,gradp) is solution of (PI).
b) In the same way (P3) admits an infinity of (11, q) solutions belonging to H 2 (Q). If
(11, q) is solution of (P3) then:
-!J. (Curlq) = Curl (curl (Curlq» = -Curl (!J.q) = IlCurlq
which demonstrates that (J.1, Curlq) verifies the equation of (Pl). Furthermore:
curlu = curl (Curlq) = -!J.q = Ilq = 0 on y
and as Curlq.n is the tangential differential coefficient of q that is constant on y, Curlq. n = 0
on y which demonstrates that (Il,curlq) is solution of (PI).
c) If r is solution of (Pi) then Curlr verifies !J. (Curlr) = 0, Curlr. n = 0 on y (same
demonstration as the one in paragraph b) and curl (Curlr) = -!J.r = 0 on y. This demonstrates
that (0, Curlr) is solution of (PI).
d) This is a consequence of the previous lemma.
e) Let us suppose u belonging to V, u orthogonal to gradpi' Curlqj and to the
Curlrk,then using the Green's formulae (11.2,5) and (II.2,7) :
(u, gradpi) = - (Pi' divu) = 0 for each Pi
as the whole of the Pi constitutes a base of L 2 (Q)
then,
divu=O
(u, Curlrqj) = (curlu, qj) = 0 for each qj , and as the set of the qj form a base ofL 2(Q)
curlu = 0
Consequently u belongs to Ho (div 0, curl 0) and as by hypothesis u is orthogonal to all the
Curlrk that generate Ho(div 0, curl 0), u = O. The density of V (that contains H I O (Q)2) into
L 2(Q)2 allows to conclude that the whole of the linear combinations of the gradp, Curlq and
Curlr is dense into L 2( Q)2. The verification of the orthogonality is instantaneous.