|
KARRAD
MOKHTAR, Département de maths, Universite Constantine.
REGHIOUA MOHAMED, Ecole Normale Supérieure Constantine |
Abstract
The aim of this work is to obtain one regularity of
solutions to Laplace
problems in a section from Green nucleus in order to apply
these results
to varying coefficient problems or to curvilinear
domainns.The results are
tackled under a general from by Kondratiev ( Transformation
of the domain
to bring it to the band ), Merigot ( Solution in Lp,limited
development
from Mellin transformation ), Moussaoui (C.
(..)solution,Dauge
(written soluton from a resolvent ), Grisvard ( Freedholm
alternative ).
Resumé :
Le but de ce travail est d’obtenir la régularité des
solutions du problème
de Laplace dans un secteur à partir du noyau de Green dans
le but
d’appliquer ces résultats aux problèmes à coefficients
variables ou aux
domaines curvilignes. Les résultats sont abordés sous la
forme générale:
Kondratiev (Transformation du domaine pour se ramener à une
bande ),
Merigot (Solution dans Lp, développement limité à partir de
la transformation
de Mellin ), Moussaoui (Solution C. (..)), Dauge (Solution
écrite
à partir d’une résolvante ),Grisvard (Alternative de
Freedholm )
mots clés: Noyau de Green, Elliptique,Convolution
multiplicative, Transformée
de Mellin
1
.
.
1 Noyau de Green du Laplacien
1.1 Introduction
Many work were realises concerning the problems in extreme
caseselliptic in
fields with " corners "(polygons in R2, polyhedres in
R3).Grisvard [1] show
that the formula of Green for the Laplacian known in the
traditional case is
also valid in a polygon, the alternative of Freedholm
enables him to find the
results traditionals.Kondratiev[2] ,to avoid them angular
points, was based
on the transformation . = e.t, what raméne the study with
the traditional
case in a band, what is has problem badly posed.Il
introduced spaces with
weight in the beginning that one remaque while writing dt =
..
. .Mérigot[7]
built a core of Poisson for the elliptic problems with
constant coefficients in
one sector . = {(., .) : 0< . . . < 2. , 0 < .} by using the
transform of
Mellin, which underlines convolutions multiplicative of the
type : (f . g) (.) =
. Z0
f ³.
t ´g (t)
dt
t
=
. Z0
f (t) g ³.
t ´dt
t
.In this work, we used the core of Green to
solve the Laplacian in a sector.This method is original and
its generalization
remains open.
That is to say the polygon .; the problem: ½ .u = f
u |.= 0
f . L2 (.) (1.1)
admits a solutionu, who in the vicinity of the top Si of
angle .i check:
If .i . . , u . H2 ...i¢ If .i > ., u = u0 + k.
.
. sin . .
. , u0 . H2 ...i¢ ..ibeing the intersection of . with the
disc of centerSi and of ray .0.
Let us consider the problem model: ½ .u = f on ..
u (., 0) = 0 and u (., .) = 0
(1.2)
It is supposed that the solution exists, is with compact
support and belongs
with H2 (..). The transform ofMellin: f (t) .. ef (.) = Z +.
..
f (t) t. dt
t
,equivalent
to the transform of Fourier for measurement dt
t , transform the problem(1.1) in
a problem with a variable ., depending on a parameter S on
the segment [ 0,]: ( ^[.2.u] (., .) = (2) .2eu (., .) + g.2u
.2.2 (., .) = hg.2f (., .)i sur [0, .]
eu (., 0) = eu (., .) = 0
(1.3)
The solutions of the homogeneous differential equation of
the 2^{éme} order
are goniometrical functions and the core of Green of the
problem (1.3) is not
other than the elementary solution of the problem and is
written:
2
eK (., ., .) = ..
..
.
.
sin . (. . .) sin..
. sin ..
, 0 . . . .
.
sin . (. . .) sin..
. sin ..
, . . . . .
(1.4)
Demonstration
While considering . like variable and . like parameter eK
check: .2 eK +
.2 eK
..2 = .0 who is the definition even of the solution
elementary.One can also
make a demonstration direct. eu (., .) =
. Z0
sin . (. . .) . sin ..
. sin ..
^ [.2f (., .)]d. +
. Z0
sin . (. . .) . sin ..
. sin ..
^ [.2f (., .)]d. (¢)
.eu (., .)
..
= ...
.
. Z0
cos . (. . .) . sin ..
sin ..
^ [.2f (., .)]d. +
sin . (. . .) . sin ..
. sin ..
^ [.2f (., .)]
+
. Z0
sin . (. . .) cos ..
sin ..
^ [.2f (., .)]d. .
sin . (. . .) . sin ..
. sin ..
^ [.2f (., .)]...
From where
.eu (., .)
..
= .
. Z0
cos . (. . .) . sin ..
sin ..
^ [.2f (., .)]d.+
. Z0
sin . (. . .) . cos ..
sin ..
^ [.2f (., .)]d.
In the same way
.2eu (., .)
..2 = ...
.
. Z0
sin . (. . .) . sin ..
sin ..
^ [.2f (., .)]d. .
cos . (. . .) . sin ..
sin ..
^ [.2f (., .)]
+
. Z0
. sin . (. . .) . sin ..
sin ..
^ [.2f (., .)]d. .
sin . (. . .) . cos ..
sin ..
^ [.2f (., .)]...
.2eu (., .)
..2 = ..2eu . ^ [.2f (., .)].It is necessary to take ³.
eK´so that calculation
falls right, in addition,in ghost wi (¢) eu (., 0) = eu (.,
.) = 0,
The function eK is meromorphe compared to .,poles being
points µk.
. ¶, k . N..
Fork = 0, there is on the right a limit and a limit on the
left.It is known that
the solution of (1.3) is written:
eu (., .) = Z .
0 eK (., ., .)g.2f (., .) d.
The function eK (., ., .), regarded as function of the
variable ., is with
exponential decay for any couple (., .) 6= 0 when Im . tends
towards the infinite
one.One can thus define a function reversesK (., ., .).
Indeed, we have the
theorem according to:
3
Théorème 1 either f (z) an analytical function in the angle
.. < . < . .z = rei.¢ such as:
... > 0, f (z) = ...
0 ³|z|.a..´ when |z| . 0
0 ³|z|.b+.´ when |z| . +.
(with 0 < a < b < .) uniformly in the preceding angle, then
:
.. > 0, ef (.) = ½ 0 £e.(...).¤ when . . +.
0 £e(...).¤ when . . ..
(. = . + i.) ,uniformly in
the band Ba,b = {., a < Re . < b} and reciprocally [ 9 ]
This function depends on the value which one gives with Re
..We will pose
K0 (., ., .) the function reverses of eK défine for 0 < Re .
< .
. :
K0 (., ., .) =
1
2.i Z c+i.
c.i. eK (., ., .) ...d.
Pose Kj (., ., .) = 1
2.i Rcj+i.
cj.i. eK (., ., .) ...d.
with .j .
. < cj < . (j . 1) .
. ; j . 1
1.2 Relation between Kj.1 and Kj
( .)
Cj-1 -(j-1)( ./ .)
-(j-2)( ./ .)
Cj -j( ./ .)
On the interior of ., the function eK (., ., .) ...is
holomorphic except at the
point . (j . 1) .
. . Moreover, on both horizontal segments, the function
tends
towards zero (|Im.| .. +..)
4
The theorem of the residues makes it possible to write :
Kj.1 (., ., .) . Kj (., ., .) = Res.(j.1) .
. eK (., ., .) ....
In the example which interests us, in³..
.´ :
K2 (., ., .) = K1 (., ., .) . Res. ..
eK (., ., .) ....
If R1 is the residue, one checks easily that: R1 (., ., .) =
.2
.2 .
.
. sin ..
. sin ..
. .
The function R1 is singular if .
. / . N, regular if .
. is whole, but as it thereafter
will be seen, it then appears a problem of convergence of
integral. We consider
the function:
u (., .) =
+. Z0
..
. Z0
K0 ³.
t
, ., .´t2f (t, .) d...
dt
t
= Z .
0 £K0 . .t2f (t, .)¢¤d.
where. indicate the multiplicative convolution compared to
measurement dt
t .
We know that: if . < ., u . H2 (..); if . < . < 2. then u =
u0 + ..
..
sin ..
.
with u0 . H2 (..).
We will find these results directly.
Théorème 2 ( Mikhlin ) [9]
If f . LP (R+) and if M (t) check
(1) fM ³1
p + i.´ defined ..
(2) sup
. ¯¯¯
fM ³1
p + i.´¯¯¯
< +. and sup
. ¯¯¯
. fM ³1
p + i.´¯¯¯
< +., then
h (.) = R+.
0 f (t) M ..
t ¢dt
t belongs with LP (R+) and there exists c > 0 such
as
khkLP . c kfkLP .
u (., .) =
. R0
K (., ., .) . £.2f (., .)¤d.. One is interested primarily if
f .
L2 (..), with compact support.
Théorème 3 Under the preceding hypothéses, there are the
following properties:
a)
u
.. . L2 (..) ;
.u
..
and
1
.
.u
.. . L2 (..)
b) if . . . ,
.2u
..2 ,
1
.
.u
..
;
1
.2
.2u
..2 . L2 (..)
if . < . < 2. , u = u0+..
.
. sin ..
. , u0 being related to H2 (..) in the vicinity
of the top
5
Demonstration
a)
u
.. . L2 (..).
Indeed :°°°
u ..°°°
2
L2(..)
= ZZ ..
u2
.
(., .) d.d. . while using the theorem ofMikhlin,
it comes: Z +.
0
u2
.
(., .) .d. . Z .
0
c (.) °°
t2f (t, .)°°
2
L2(R+) d. ; while integrating compared
to . , the result is obtained.
.u
..
= ZZ ..
.K0
.. ³.
t
, ., .´tf (t, .)
dt
t
d..
..
.u
..
(., .) = Z .
0 Z +.
0
..
.t
.K0
.. ³.
t
, ., .´t
3
2 f (t, .)
dt
t
d.
= Z +.
0 Z .
0
G1 ³.
t
, ., .´t
3
2 f (t, .)
dt
t
d..
The same demonstration as for the function
u
..
us results in studying the
function fG1 (., ., .) = . .. . 1
2 ¢fK0 .. . 1
2 , ., .¢. The theorem of Mikhlin still
applies. It should be noted that each
derivation shifts the core of a unit.
1
.
.u
..
=
1
.
+. Z0
..
. Z0
.
..
K0 ³.
t
, ., .´t2f (t, .) d...dt
t
and
..
.
.u
..
=
+. Z0
..
. Z0
rt
.
.
..
K0 ³.
t
, ., .´t
3
2 f (t, .) d...
dt
t
=
+. Z0
. Z0
G2 ³.
t
, ., .´t
3
2 f (t, .) d.
dt
t
fG2 (., ., .) = .
.. fK0 .. . 1
2 , ., .¢. The function fG2 is to be studied on the
same line thangG1. The sup are with to take on the line . =
0.
If . < . ,the first pole is lower with(.1) and the
demonstration remains unchanged.
If . = . , the problem is in fact posed in a half-plane and
the result is
well-known. It is in spite of very interesting to study it
using the core of Green
for it underlines whole poles.
.u
..
= ZZ ..
.K0
.. ³.
t
, ., .´tf (t, .)
dt
t
d. .
By using the relation between K0 and K1, it comes:
6
.u
..
= ZZ ..
.K1
.. ³.
t
, ., .´tf (t, .)
dt
t
d. + sin. Z .
0 Z +.
0
f (t) sin.dtd. . (1)
The second integral converges like product of two functions
of L2 .If one calculates
.2u
..2 , the increase of the core is with to evaluate on the
line . = .1,what
is impossible. Two solutions are to be considered:
• to insulate the pole using a function which one connait
the opposite transform
of Mellin.
• to make a direct calculation.
It is the second step which we will choose. According to
(1.3) the same
demonstration than a) show that:
.. .2u
..2 . L2 (..) , what shows that
.2u
..2 . L2 (..) if . . r0 and according to
(1.4) ,one has: ... .2u
..2 . L2 (.. . B0 (0, r0)) , what gives the result by
density.
.2u
..2 (., .) = ZZ ..
.2K0
..2 ³.
t
, ., .´f (t, .)
dt
t
d. = ZZ ..
.2K0
..2 (t, ., .) f ³.
t
, .´dt
t
d. .
lim
..0
.2u
..2 (., .) has a direction only if f (0, .) = 0 for
1
t
.2K0
..2 (t, ., .) does not
belong with L1 (.. . B0 (0, r0)). In this case:
.2u
..2 (., .) = ZZ ..
.2K1
..2 (t, ., .) f ³.
t
, .´dt
t
d. . 2 sin
2.
. ZZ ..
f (t, .)
dt
t
d.,
and as in the case b) the derivative
.3u
..3 (., .) . L2 (..) .The demonstration is
identical for the others derived from order three which "
shift " the cores under
the same conditions.The relation (1) is written:
.u
..
= ZZ ..
.K1
.. ³.
t
, ., .´tf (t, .)
dt
t
d. +
.
.
.
.
. .1 sin
..
. Z +.
0
f (t)
1
t
.
. .1 dt .
The second integral made apparaitre a singular part which is
in fact a clean function
of the problem. While posing
.u1
..
= ZZ ..
.K1
.. ³.
t
, ., .´tf (t, .)
dt
t
d. , one
will have :
.2u1
..2 (., .) . L2 (..) as in a) and u (., .) = u1 (., .) + .2.
.
. sin ..
. .
7
The others derived which intervene in the expression of
.2
.x2 and
.2
.y2 treat
themselves in the same way.
1
.
.u
..
= ZZ ..
t
.
.K0
.. ³.
t
, ., .´f (t, .)
dt
t
d.,
.2u
..2 = f (., .) .
1
.
.u
.. .
.2u
..2 ,
1
.
.2u
....
= ZZ ..
t
.
.2K0
.... ³.
t
, ., .´f (t, .)
dt
t
d.. These derivative make apparaitre
a shift of the core of two units, for
example:µ1
.
.K0
.. ¶e= (. . 2) eK0 (. . 2, ., .)
1.3 Case general
So now f . Hs (..), several cases are to be considered:
Case: s + 1 <
.
.
.2u
..2 (., .) = ZZ ..
.2K0
..2 (t, ., .) f ³.
t
, .´dt
t
d..One supposes s . N and the
other cases are obtained by interpolation.One makes, for the
derivative of a
higher nature, to carry derivations on f , it then appears a
factor
1
t
for each
order of derivation, factor which shifts the transforms of
Mellin of a unit, but calculation
in its principle is unchanged and u belongs with Hs+2 (.. .
B0 (0, r0)) .
Case: s + 1 >
.
.
The study is identical to that of the derived second if
.
. . ¤1
2 , 1£. Each pole
nonwhole contents in the band .1 . s < Re. < .1 , fact
apparaitre a singular
function in the limited development of u in the vicinity of
the origin.The study
of a whole pole is somewhat different. The study of the case
. =
.
2
allows to
analyze the situation well. Let us suppose that u that is to
say defined in the
quadrant (x . 0, y . 0) and is solution of the problem :
...
.u = f (x, y) ,
u (x, 0) = g1 (x) ,
u (0, y) = g2 (y) .
When one makes tend x and y towards zero in the equation, it
comes:
.2g1
.x2 (0) +
.2g2
.y2 (0) = f (0, 0) who is a condition of compatibility. In
the
same way, by combining the operator . with operators of
order two one will
make apparaitre the conditions relating to the pole .2.
.
= .4. For facilities
of writing, one will limit oneself to the case . =
.
2
,but it is quite obvious
that the demonstration applies to all the poles entireties.
The first pole which
8
intervenes is . = .2. The difficulties thus appear for the
poles whole. For good
to include/understand the step, one supposes . =
.
2
and f . H2 (..) because
one can hope that the solution u belongswith H4 (..) . The
first pole of the
transform of Mellin is . = .2, it makes apparaitre
difficulties for the derivative
of orderit makes apparaitre difficulties for the derivative
of order three. Let us
interest, first of all with
.2u
..2 .If u . H4 (..), one can to define the value into
zero.
.2u
..2 = ZZ ..
.2K
..2 (t, ., .) f ³.
t
, .´dt
t
d.. When . tends towards zero, the
limit does not have feel that if f (0, .) = 0. One finds
then the conditions of
compatibility referred to above .
One can then write:
.2u
..2 = ZZ ..
.2K1
..2 (t, ., .) f ³.
t
, .´dt
t
d.+Residue and then:
.4u
..4 = ZZ ..
1
t2
.2K1
..2 (t, ., .)
.2
..2 f ³.
t
, .´dt
t
d..
This new integral belongs with L2 (..), from where the
theorem:
Théorème 4 In the vicinity of a top of the polygon, the
solution of:
...
.u = f
u (., 0) = 0
u (., .) = 0
f . Hs (..) , s . N
a limited development of the form admits:
u (., .) = u0 (., .)+Xk
ak .
k.
. sin k..
. , u0 . Hs+2 (..) ,
.
.
< k . (s + 1)
.
.
, if
k.
. / . N.
If
k.
. . N ,there is no term in the singular development. If it
is wanted
that u . Hs+2 (..), with each pole in the interval ..
.
, (s+ 1)
.
.¸ a condition
corresponds.
Remarque 1 : if
.2f
..2 (0, .) 6= 0 then u . H32
....
9
References
[1] P.Grisvard : Alternative de Fredholm relative au
probléme de Dirichlet dans
un polygone ou un pôlyèdre BolletinoUMI 5(1972)
[2] V.A.Kondratiev : Problèmes aux limites pour les
équations elliptiques dans
un domaine avec point conique ou anguleux.
[3] H. Bresis ”Analyse fonctionnelle, théorie et
application”, Masson (1983).
[4] M.Dauge: “Problème de Dirichlet sur un polyédre de R3
pour un opérateur
fortement elliptique” Séminaire d’équations aux dérivées
partielles
(1982/1983) Nantes
[5] C. Goudjo ”Problémes aux limites dans les espaces de
Sobolev avec poids”
(Thèse de spécialité, Nice 1970).
[6] J.L. Lions ”Problémes aux limites dans les équations aux
dérivées partielles”,
Dunod (1968).
[7] M. Merigot ”Solutions en normes LP des problémes
elliptiques dans des
polygones plans” (These de doctorat d’etat, Nice 1974).
[8] L. Schwartz ”Théorie des distributions ”, Hermann
(Nouvelle edition1973).
[9] E.C. Titchmarsh ”Introduction to the theory of Fourier
integrals ” Oxford
(1937).
[10] K.Yosida ”Fonctional analysis”, Springer (1965).
10 |
|
|
|
|
|
|
|
|
|
|
|
|
