A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
A BDERRAHMAN M AGHNOUJI S ERGE N ICAISE
On a coupled problem between the plate equation and the membrane equation on polygons
Annales de la faculté des sciences de Toulouse 6
esérie, tome 1, n
o2 (1992), p. 187-209
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187
On
acoupled problem between the plate equation
and the membrane equation
onpolygons
ABDERRAHMAN MAGHNOUJI and SERGE
NICAISE(1)
Annales de la Faculte des Sciences de Toulouse Vol. 2, n° 2, 1992
RÉSUME. - Nous étudions un problème d’interface sur un domaine
polygonal du plan, ou les operateurs laplacien et bilaplacien sont con-
sidérés sur chacune des faces respectives. Nous précisons les conditions nécessaires et suffisantes pour que l’opérateur associé soit de Fredholm dans les espaces de Hilbert appropries. Lorsque c’est le cas, nous don-
nons une decomposition de la solution variationnelle en parties regulieres
et singulière.
ABSTRACT. - We study an interface problem on polygonal domains of
the plane, where on one face, we consider the biharmonic operator and
on the other one, the Laplace operator. We instigate if the associated operator on appropriate Hilbert spaces is a Fredholm operator or not. If it is, we give an expansion of the variational solution into regular and singular parts.
1. Introduction
We introduce a new kind of interface
problems
onpolygonal
domains ofthe
plane.
Thenovelty
is that the order of thepartial
differentialoperators
is different on each face. Weonly study
a modelproblem corresponding
tothe mechanical
example
of acoupling
between aplate
and a membrane. Weexpect
that the methods wedeveloped
could be extended to moregeneral problems.
In classical interface
problems (see ~3, 10]
and the references citedthere),
the variational solution has
singularities
at the common vertices betweenthe interface and the
boundary. Therefore,
we canexpect
the sametype
(1) Université des Sciences et Techniques de Lille-Flandre-Artois, U.F.R. de Mathe-
matiques Pures et Appliquées, U.R.A. C.N.R.S. D 751, F-59655 Villeneuve-d’Ascq
Cedex
(France)
of results for our
problem. Indeed,
for interior data inL~,
we cangive
thedecomposition
of the variational solution of ourproblem
into aregular part
with theoptimal regularity
and asingular
one. The main idea is to usea two
steps argument by splitting
up two of the interfaceconditions,
anduse
successively
thedecomposition
results for aninhomogeneous boundary
value
problem
on each facerespectively
associated with theLaplace operator
and the biharmonic one.
For more
regular data,
we could argueiteratively
asbefore,
but thisinduces too much
geometrical
conditions(on
theangles
of thedomains).
Therefore we
prefer
acompact perturbation argument
as forboundary
value
problems
withnon-homogeneous partial
differentialoperators [7, 2].
Indeed,
we shall see that the difference of order of theoperators
on the faces will induce interface conditions withnon-homogeneous operators (i.e.
it issum of
operators
of differentorder).
Thisargument only
holds under someconditions on the Sobolev
exponents.
One of them is also necessary sincewe shall show that if this condition fails then the induced
operator
is not Fredholm.We finish this paper
by solving,
in section6,
a differentialequation
withoperator
coefficientswhere A is a closed
operator
defined on a Hilbert space~~, ~
E~,
q E lNU~O~
and
fq E X.
As shown in[11] (see
also the references citedthere), solving (1.1)
allows us tu solveexplicitly
someboundary
valueproblems
in a finitecone of ~n with a
right-hand
side which is a linear combination of functions oftype
,
where
(r, 8)
are thespherical coordinates,
a EC,
q E IN U{0}
and 03C6q isregular enough.
This result agrees with those of[6].
.2. Formulation of the
problem
Let
S~2
be two boundedsimply
connectedpolygonal
domains of theplane
such that their boundaries have a common side denotedby
r. Wedenote
by Fi (resp. F2)
theboundary
ofn1 (resp. n2) except I‘,
i.e.rj BYB for j = 1,
2.For j
=1, 2, vj
will denoteunitary
outer normal vector on theboundary
~03A9j
of03A9j
and Tj theunitary tangent
vectoralong ~03A9j
so thatj)
isa direct orthonormal basis.
Along
the common sider,
we omit the indexby setting (v, r)
=(v2, r2 ~.
We shall denoteSjk,
for k E~1, ... , N~ ~,
the vertices of numbered
according
to thetrigonometric
orientation forHi
and numbered clockwise for will be the interiorangle
at5jk. Moreover,
forconvenience,
we assume that511
=S21
andS12
=522 belong
to F and denote them51 and 52 respectively.
We also denoteby 17jk,
a cut-off functionequal
to 1 in aneighbourhood
of5jk
andequal
to0 in a
neighbourhood
of the other vertices. Aspreviously,
we may suppose that: ~11 = 1721 =: 171 and 7712 = 1722 =: ~2.Finally, rj
will denote the traceoperator
on theboundary ~03A9j
of03A9j; 03B3j0393
will be the restriction of03B3j
to F.For E > 0 and (j
E ] 0, 1 [ (respectively
theYoung
modulus and thePoisson coefficient of the constitutive material of the
plate S~2),
we setp =
.~/( 1 _
and we introduce theboundary operator
definedonly
on FWe recall that we use here classical Sobolev spaces, i.e. if H is a bounded open set of
R2
and s anon-negativite integer,
thenits norm
being
denoted For otherdefinitions,
we follow Gris- vard’s book[4].
We consider the
following
interfaceproblem: given f 1
Ef2 E Hs2 2(~2)~ h1
Eh2
E~s2 3~2(r)
UHsl 1~2(I‘~,
for si,s2 E IN with
s2 > 2,
find ui E u2 EHs2+2(S~2),
solutions of(2.1)-(2.7)
below:We first
give
the variational formulation of thisproblem.
We setIt is a Hibert space
equipped
with the innerproduct
inducedby
xH2 (SZ2 )
with the normWe define the
sesquilinear
form a on V as follows:where we take
LEMMA 2.1. - For all
f 1
Ef 2
EL2 (S22 ) ~ ~1 t ~2
EL2 (r),
thereexists a
unique
solutionu
~ Vof
Proof.
- In order toapply
theLax-Milgram lemma,
we need to showthe
continuity
and the coerciveness of the form a on V.The
continuity
is a direct consequence of thecontinuity
of the form ajon and of the
Cauchy-Schwarz inequality.
Owing
toinequality (2.15)
of[12],
we deduce thatwhere
~u~~ ~,~
. denotes the semi-norm of But fori E V,
theboundary
conditions(2.3)
and(2.4) respectively
fulfilledby
U1 and u2imply
that the norms and the semi-norms are
equivalent (see
e.g. Theorem I.1.9 of[8]).
Therefore theprevious
estimate leads to the coerciveness of the form a on V. DLet us now show that a solution
of (2.8)
is a weak solution of(2.1)-(2.7).
We follow the
arguments
of section 1.5.3 of[4]:
we introduce the spacesthese are Banach spaces for the norms
Lemma 1.5.3.9 of
[4]
proves that is dense inE(A, L2 (S~1 ~~ ; analogous
arguments
lead to thedensity
intoL2(5~2~~.
.LEMMA 2.2. - The
mapping
which is
defined
on xD(SZ2),
has aunique continuous extension
as anoperator from E (0 , L2 (SZ1 )~
xL2 (S~2 ~~
intoH1/2 (I‘)~
xH3/2 (r)~
(identifying
F with a realinterval,
we recall that u EHs (r) iff u,
theeztension
of
uby
0 outsideF,
remains inH$(R)~.
Proof.
-Owing
to theorem 1.5.2.8 andcorollary
1.4.4.10 of~4~, given w2~
EH1/2(I‘~
x~j3/2(r~~
there existsv
=(vl, v2~ E ~ satisfying
and
where the constant
Ci
isindependent
of ~w2 .For a
fixed u
=u2~
E x let ~a setBy integration by parts,
weget (see
lemma 2.3 of[12]
for the biharmonicoperator) :
Therefore
using
thecontinuity
of the form a on V and the estimate(2.10),
there exists a constant
C2 independant
of wl , w2 such thatBy density,
we deduce that l is a continuous linear form onH1~2(I‘)
xH3~2 (~).
0Let us notice that the Green formula
(2.11)
still holds for every v E Vfulfilling (2.9)
and every ui EL2(S~1)~,
u2 E.E(~2, L2(~2)~,
wherethe left-hand side has to be understood as a
duality
bracket.LEMMA 2.3.- Let
11
E V be theunique
solutionof ~,~.8~.
Thenu fulfils (2.1) to (2. 7).
Proof. - Applying (2.8)
with(vl, v2)
E xD(SZ2),
we see that ul(resp, u2 )
fulfils(2.1) (resp. (2.2))
in the distributional sense. This also shows thatU1 , u2 E
.l~’ (02, ~2)) .
.Therefore,
forarbitrary (z,vl, u~2)
E.H1~2(I‘)
xH3~2(I‘), comparing (2.8)
with
(2.11),
when v E V fulfils(2.9),
we deduce thatThis
obviously implies
thatu
satisfies(2.6)
and(2.7).
03.
Regularity
for interior data inL2
In this
paragraph,
we look for conditions on/i
Ef2
EL2 (SZ2~,
h1
EI~3/2 (I‘), h2
EH1/2 (r),
which ensure that ul and u2 have theoptimal
regularity,
i.e. ui E u2 E~4(SZ2~;
indeed we shall prove thatU1 and u2 admit a
decomposition
into aregular part
with theoptimal regularity
and a finite sum ofsingular
functions. We shall see that thisdecomposition
result will be determinedby analogous decomposition
resultsof two
decoupled boundary
valueproblems
set inHi
and Moreprecisely,
the first one is the Dirichlet
problem
inHi
withnon-homogeneous
Dirichletboundary
conditions onr,
i.e.The second one is the
following
mixedboundary
valueproblem
for thebiharmonic
operator
in~22 :
The
regularity
of the solution ofproblem (3.1)
wasgiven
in theorem 5.1.3.5 of[4],
whileproblem (3.2)
was studied in theorem 5.2 of[12] (see
also[1]).
In order to recall these
results,
let us define thesingular exponents
andsingular
functions ofproblems (3.1)
and(3.2).
For
problem (3.1),
we setFor A E the assosiated
singular
function iswhere
(r, 8)
arepolar
coordinates withorigin S1k (such
that the half-lines 8 = 0 and 8 = contain theedges containing Slk).
For
problem (3.2),
in order to avoid toocomplicated notations,
weonly
recall that the
singular exponents
are the roots of thefollowing
characteristicequation :
We
only
say that there exists a setA2k
of roots of theequation (3.4)
fork 2 and of
(3.5) for k > 3, repeated according
to theirmultiplicity;
toeach A E
A2k corresponds
asingular
function denotedby ~~~ (see [12]
formore
details).
ForI~ = 1, 2,
thepolynomial
resolution(cf. §
3.C of[12]
and(2.9)
of[1]) implies
that A = 2 induces asingular
functiongiven by
where
(r, 8)
arepolar
coordinates withorigin Sk
and p~ is apolynomial
of
degree
2(in
the cartesiancoordinates).
Remark that EH2 (S~2 ~
.Therefore,
forconvenience,
we shall add .~ = 2 toA2k,
for k =1,
2 and still denoted itby A2k.
THEOREM 3.1. Let
fl
E g EHsl ~1/2(I‘) fulfilling
=
g(S‘2 ~
=0,
with sl E IN . .Suppose
thatthen there exists a
unique
solution u1 Eof problem ~3.1~
whichadmits the
following expansion:
where u10 E
c1k03BB ~
Cdepend continuously
onfi
and g, and1k(s1)
=1k ~ ] 0 , s1].
THEOREM 3.2.- Let
,f2
EHs2 2~5~2~, hi
EHs2 1~2~I‘~, h2
EHsZ
-3~2 (r~,
with s2 EIN,
s2 > 2. Assume thatthen there exists a
unique
solution u2 E~2 (~2 ~ of problem ~3.2~
such thatwhere u20 E
Hs2+2(03A92),
c2k03BB ~depend continuously
onf 2, hl , h2.
Herewe denote
2k(s2) = {03BB
EA2k 1
R03BB s2 +1}.
Let us now go back to our
boundary
valueproblem (2.1)-(2.7).
Farfrom the interface
F,
we see that itcorresponds
to(3.1)
or(3.2) ; therefore,
the
regularity
of ul and u2 isgiven by
theprevious
theorems.Analogous arguments
as thosedeveloped
in thesequel
show that ul and u2 have theoptimal regularity
in aneighbourhood
of apoint
of r.Therefore,
weonly
have to
study
the behaviour of ul and u2 in aneighbourhood
of the commonvertices of
S~1
and5~2.
THEOREM 3.3.- Let
f 1
E,f 2
EL2 (~2 ~
~1~1
EH~/~(C),
h2
EH1/2(r)
andu
=(ul, u2)
E V be the solutionof ~,~.1~-~2.7~.
Fork E
~1, 2~,
we have:a) if
03C91k > 03C0, then ul admits thefollowing decomposition
in aneigh-
bourhood
Vk of S~
:where ulp E , ~k E
C;
b) if 03C91k ~
03C0, then ui EH2(Vk
r1Proo,f .
We may look ui EH1 (~1 )
as the solution ofSince u2 E
H2 ~SZ2 ~
and fulfils(2.4),
theorem 1.6.1.5 of~4~ implies
thatTherefore, applying
theorem 3.1 with sl = 1 toproblem (3. II),
weget
theresult
except
if = x. In this last case,(3.12)
and theorem 1.5.1.2 of[4]
allow to reduce(3.11)
in theneighbourhood Vk
DHi
to ahomogeneous
Dirichlet
problem
with interior data inL2.
Classicalregularity
results onsmooth domains leads to the conclusion. 0
To
study
theregularity of u2,
we now use theequations (2.2), (2.4), (2.6)
and
(2.7),
i.e. u2 is seen as a solution ofin a
neighbourhood Vk
ofSk.
~r, then and we may
directly apply
theorem 3.2 with s2 = 2 to
(3.13). But,
if 03C91k > 03C0,only
yl has theadequate regularity H1/2,
while the normal derivative of thesingular
function 03C3k03C0/03C91klap has not. The idea is to
compute explicitly
the contribution of thissingular
function.By
theorem 6.1hereafter,
there exists a solutionT2
EH2~~2
nVk)
ofTherefore,
the function u21 definedby
belongs
toH2(03A92
nVk)
and is a solution ofproblem (3.2)
inSZZ
nVk
withdata
f2
EL2~S~2
nVk), hl - 0, h2
= E~1~2(r
nVk).
Therefore, applying
theorem 3.2 with s2 = 2 to u21, we obtain thefollowing
theorem.
THEOREM 3.4. - Let
u
=u2)
~ V be the weak solutionof (2.1)- (2.7)
with dataf1
~ L2(03A91),f2
EL2(03A92), h1
EH3/2(0393), h2
EH1/2(0393).
For k = 1 or
2,
let us suppose that~aE~~~a=3~n~12~=~,
then u2 admits the
following decomposition
in aneighbourhood of S k:
:where u2o E
H4(S~2),
c~ E ~C and the last termof
theright-hand
sideof (3.14) is
zero x.4. More
regular
dataIn theorems 3.3 and
3.4,
if we increase theregularity
of thedata,
weexpect
to increase in the same way theregularity
of theregular parts.
Onemethod is to use the same iterative
procedure
as in section3; unfortunately,
it
imposes
too much conditions and iscomplicated. Therefore,
weprefer
touse a
compact perturbation argument.
We need to introduce the Hilbert spaces
The
operator induced by
theboundary
valueproblem (2.1)-(2.7)
isclearly
thefollowing:
We look for conditions on sI, s2 which ensure that is a Fredholm
operator.
To do that wesplit
upinto
itsprincipal part and
a remainder as follows:
Obviously,
we haveTHEOREM 4.1. -
If
slE ( s2 - 1 ,
s2 +1 ~ ]
and the Fredholm conditions(3.6)
and~3. 8~ hold,
then~
is a Fredholmoperator
andBut this is
equivalent
to the fact that u2 is a solution of(3.2)
and ui isa solution of
(3.1) with g
= 12ru2 on r.Therefore, applying
theorem 3.2to u2, we deduce that there exists a
unique
solution r2 E of(3.1),
which admits the
decomposition (3.9).
As in theorem3.4, looking
for ui weuse this
decomposition (3.9)
of u2. For all k E{1, 2},
A EA2k,
theorem 6.1below
gives
theexplicit
solution E of(4.6)
in aneighbourhood v~
ofs~ :
Furthermore,
theorem 3.1 proves the existence of aunique
solution v1 E ofproblem (3.1)
with datafi, g
= ;2ru20 EH~z+3/2 (r)
Hsl
+1/2 (~),
since sl s2 +1,
which admits thedecomposition (3.7).
Letus notice that in that
decomposition (3.7),
the coefficients c1k03BBdepend
continuous on
f 1
and 03B320393u20; and thereforecontinuously
on, f 1, f 2 , h1, h2 . Setting
we have proven that there exists a
unique
solutionu
=(ul , u2 )
E V of(4.5),
which admits thedecomposition
where
Wo E
E Cdepend continuously
onfl, f 2, hl, h2
andwe have set
with the
agreement
that =0,
3.This establishes that if
( fi f 2 , hl , h2 ) E B~ sl ’s2 ~
is such thatthen it
belongs
to the range ofReciprocally,
if such a datumbelongs
to the range, then there exists asolution of
(4.5);
then u2 is a solution of(3.2)
and ulof
(3.1) with g
= Due to theorem 3.2 and after theorem3.1,
thisimplies
that it fulfils(4.8).
So we have proven that the range of is closed and that(4.4)
holds sinceis clearly injective. []
THEOREM 4.2. -
If
slE s2 - 1 , s2
+1 ~ ]
andif ~3.6~
and~3.8~ hold,
then J is a Fredholm
operator
andProof.
- theassumption
sl > s2 - 1implies
that is acompact operator,
becauseH81-1/2(r)
iscompactly
imbedded intoHs2-3/2(I‘).
Using
a classicalperturbation
theorem(see
theorem IV.5.26 of[5],
forinstance),
we deduce the theorem. 0Since we want to
give
theasymptotic
behaviour of the solution of ourproblem (2.1)-(2.7),
we need thesingularities
of thisproblem,
i.e. thesingularities
of . As M.Dauge
in~2J
fornon-homogeneous operators,
we
compute
themby
recurrencestarting
from thesingularities
of theprincipal part
In view of(4.7),
we see that thesingularities
ofare
the So weproceed
as follows: for k = 1 or2,
we setand for p E
IN,
is a solution of(4.11 ~
hereafter in aneighbourhood Vk
of
S k
:Splitting up ~ ~ ~~
into itscomponents,
problem (4.11)
isequivalent
to(4.12)
and(4.13)
hereafter solved in that orderusing
theorem 6.1.The associated
singularity
ofL~~1’$2 ~
is definedby (compare with §
5.C of[2])
Let us recall that
T ~~~
is called asingularity
ofL~s1 ~s2~
because itbelongs
to V and not to whileL~~1 ~s2 ~ T ~~~‘ belongs
toB{$1 ~s2 ).
Let us check this last
assumption.
From(4.11)
and(4.14),
it is clear thatwhere pmax is such that
Therefore,
E} iff
In view of the form of and theorem
6.1, ~ir(~/~)~~ i
behaveslike in a
neighbourhood
of So(4.15)
leads to theadequate regularity.
Let us
finally
notice that the aboveprocedure only
concerns thesingu-
larities induced
by 5i
and52 (i.e.
the for k = 1 or2). Indeed,
for~ ~ 3, ~(~’~) ~ ~~~~~
in aneighbourhood
so thesingularities
ofL~’’~
are those ofZ(~’~B
i.e.=~~’~~~, V~~3.
We now recall lemma B.I of
[2] concerning
therelationship
between the index and asingularities
space.LEMMA 4.3
(M. Dauge [2]).2014
Let~i
CAo
and~i
CBo be
twopairs of Hilbert
spaces such thatj4.i
is dense inAo
andB1 is
dense inBo.
LetMo be
a Fredholmoperator from Ao
intoBo,
which may be restricted to a semi-Fredholmoperator, denoted by M1, from Ai
intoB1
.We suppose that there exists a
finite
dimensionalspace E having
thefollowing properties:
Then the
following
conditions areequivalent:
Ml
is a Fredholmoperator
and dim E = indMo -
indM1
,(4.20) for
any u EAo
such that Mu EB1 (4.21)
there exists v E
Al
and w E E such that u = v + w .We are
ready
to prove the theorem 4.4.THEOREM 4.4. - Under the
assumptions of
theoreml~ .2, given ( f 1, f 2, hl, h2)
E there exists aunique
solutionu
E Vof problem (2.1)- (2.7),
which admits thefollowing decomposition
where
p
E ~ and E C.Proof.
- Weapply
theprevious
lemma withwhere A is the natural
isomorphism
between V andVI
definedby
Actually,
is identified with asubspace
ofBo by
thefciowing
continuous
injection:
forF
=( fl, f 2, ~1, h2)
E we setThe restriction of
Mo
toA1
isclearly Mi
because Green’s formula(2.11) implies
thatThe space
is
dense in V because we can prove that USZ 2 )
isdense in V. Since A is a
isomorphism,
we deduce thatA(s1 ,s2)
is dense inV.
Thisimplies
thedensity
sinceFinally,
the space E is the vector spacespanned by
the forj
E{1, 2}, A;
E{1,...,
A E We havepreviously
checked that it fulfils theassumptions (4.1?)-(4.19). ~
To finish this
section,
let us show that it ispossible
to hit the limit casesl = s2 - 1.
THEOREM 4.5. - Let sl = s2
- 1,
assume that ~~e conditions~3.6~
and~3.8~
hold and moreover that the Fredholm condition~3.8~
holdsfor
s2 - 1too. Then the conclusion
of
theoreml~ .l~
still holds.Proof.
- Wefirstly apply
theorem 4.4 with the same s1, but with s2replaced by
s2 - 1.Therefore,
the variational solutionu
of(2.1)-(2.7)
admits the
decomposition (4.22)
withs2 -1
instead of s2 . So theregular part
o
has theoptimal regularity
inHi
but not inS~t2.
The secondcomponent
of thisregular part
isactually
solution of aboundary
valueproblem (3.2)
with data which are the sum of an
optimal regularity part
and a contribution of thesingularities.
As in theorem3.4,
wecompute explicitly
the solution of thisboundary
valueproblem
with asingular right-hand
sideusing
theorem6.1. The
regular right-hand
side induces adecomposition
into a newregular part
inHs2+2 (~2~
andsingularities
of theboundary
valueproblem (3.2)
for ~a
E ~ [ 1,
s2 + 1[,
due to theorem 3.2.This allows to show that
~
is a Fredholmoperator
of indexgiven by (4.9).
At thisstep,
we follow thearguments
of theorem 4.4. ~5. The non Fredholm
property
The aim of this section is to show that in theorems 4.4 and
4.5,
the condition siE ~ s2 - 1,
s2 +1] ]
isoptimal.
In otherwords,
we shall prove that if this condition fails then theoperator L~’~) }
is never a Fredholmoperator.
Theproof
of this result isagain
based on acompact perturbation argument.
In the
sequel,
we shall need thefollowing
technical result.LEMMA 5.1.2014 Let
X,
, Y be two Hilbert spaces and A a linearoperator from
X into Y. .Suppose
that there exists afinite
dimensionalsubspace
Eof
Y such thatR(A)
D~ .
.(5.1)
Then the range
of A, R(A),
is closed and its codimension isfinite.
Proof.
- Due to(5.1 ), R(A)
admits thefollowing orthogonal decompo-
sition
R(A)
=(R(A)
nE)
~E1
.Since
R(A)
n E is a finite dimensional linear manifold ofY,
it is closed.Therefore,
theprevious decomposition implies
thatR(A)
is closed. 0In the
following,
we supposethat 81
> s2+ 1;
the case si s2 - 1being
treated
analogously.
We need to introduce a variant of the
operator L~S1 ~s2 ~,
which take into account thenon-homogeneous
interface condition(2.5).
We setLet us notice that theorem 1.6.1.5 of
[4]
shows that is well defined.LEMMA 5.2. -
Suppose
that theangles at
the endso, f I‘
aredifferent from
?r, then Fredholm
operator iff
is a Fredholmoperator.
Proof.
-Clearly, and
areinjective ;
therefore the asser-tion
only
concerns their ranges..
Suppose
that(s1 ,s2) is a
Fredholmoperator.
Then there exists a finite dimensionalsubspace
E of such thatBut this
implies
thatwhere P is the
projection
in~
onB{sl ~s~ ~.
Since PE has a finitedimension,
lemma 5.1 allows us to conclude that is a Fredholmoperator.
. Let us now assume that
L{$1 ~s2 is
Fredholm. Aspreviously,
there existsa finite dimensional
subspace El
ofB{~1 ~s2 such
thatFurthermore, using
theorem 1.6.1.5 of[4]
and theassumptions
of thelemma,
we can prove that the
operator
is onto. So it admits a continuous
right inverse,
denoted itby
R.Let us now fix
( F , h~
Esatisfying
Then there exists
u
Esuch
thatThis means
( F , h) belongs
to the range of(s1 ,s2 because v
definedby
belongs
to~s~ and
fulfilsAgain,
lemma 5.1implies
that~{si ~s2 ~
is Fredholm since(5.2)
isequivalent
to
THEOREM 5.3. -
Suppose
that theangles
at the endsof
F aredifferent from
~,if
sl > s2 + 1, ; then is not a Fredholmoperator.
Proo, f
- We introduce theoperator
Setting
we remark that
This
implies
that K is acompact operator
becauseHsl +1~2 (I‘~ (resp.
H$1+1~2(I‘)) is compactly
imbedded intoHs2+3~2(I‘~ (resp. Hs2+3~2(I‘)).
Let us suppose that
L~s1 °$2 is
Fredholm. Thenis also Fredholm. This leads to a contradiction since the kernel of -~- K is not
finite-dimensional (indeed
U1 does notsatisfy
any condition on theinterface).
0Remark 5.4. - The
amplitude
2 in the condition slE ~ s2 - 1,
s2 +1 ~ ]
is
exactly
the difference between the order of the biharmonicoperator
and theLaplace operator.
This means that if we considerelliptic operators
ofrespective
order2m1
and2m2,
then theamplitude
wouldbe 2m1 - 2m2 I.
6.
Logarithmico-polynomial
resolutionTheorem 1.3 of
[6] gives
the existence of a solution to theboundary
valueproblems (4.12)
and(4.13). Here, following [11],
wegive
anotherproof
ofthis
result,
based on the use the Jordan chains.Indeed,
it was shown in[11]
how to reduce each of theseboundary
valueproblems
into an abstractdifferential
equation
in a Hilbert spaceusing
thechange
of variable r =e~
tand
reducing
the order. With theparticular right-hand
side of(4.12)
or(4.13),
theequivalent
differentialequation
weget
iswhere A is a closed
operator
defined in anappropriate
Hilbert spaceX,
forsome A E
~, Q
E IN U~0~
andfq E X,
for all q E~0,
...In order to solve
(4.12)
or(4.13),
it is thereforeequivalent
to solve theircorresponding problems (6.1).
The caseQ
= 0 was solved inparagraph
4 of[11]
in an abstractsetting (it
was called thepolynomial
resolution because for A EIN,
itcorresponds
to the resolution ofproblems (4.12)
or(4.13)
withpolynomial data).
We shall extend thistechnique
to thegeneral
caseQ ~
0.Let us recall the abstract
setting
of[11] :
X is a Hilbert space, A a closedoperator
from X into X such that its domainD(A)
is also a Hilbert space with its owntopology.
We assume that there exists a closedsubspace
Z ofX such that
D(A)
is dense in Z and iscompactly
imbedded into Z.Finally,
the resolvent set of A is assumed to be
nonempty.
The first idea to solve
(6.1)
is to look for a solution u in the same form than theright-hand side,
i.e.where pq E
D(A)
are the new unknowns. In that case,problem (6.1)
isequivalent
toIf A is not an
eigenvalue
ofA,
then forarbitrary f q
EX (6.3)
hasunique
solutions
given by
If A is an
eigenvalue
ofA,
theprevious technique
fails ingeneral.
Asin
(11],
we shall use the associated Jordan basis{{03C603BB, ,k} 03BA(03BB, )-1K=0 }M(03BB) =1
and the dual Jordan basis
( {03C803BB, ,k }03BA(03BB, )-1 =0}M(03BB) =1
>= . Let us recall thatthey
,=i fulfil
(see
lemma 2.3 of[11] ):
for
every k
=0, ... , J~C~a, ~c~ - 1,
,k’
=0, ... , J~C~a, ~’~ - 1,
, =1,
...,M(À)
and the conventions~p~‘~~‘~-1
= 0 and~~.~‘~~~~~~‘~
= 0.For all q E IN U
~0~,
let us denoteUsing (6.5),
we check thatNow,
we look for a solution u of(6.1)
in the formwhere pq E
D(A)
and c;~q are unknown. In view of(6.9), problem (6.1)
isthus
equivalent
towith the convention
~p~+1
= 0. Since the range of A - A is theorthogonal
of
ker ((A - A)*)
=Sp ~~~~~‘~~~~‘~~‘~-1 ~Mt~~ ,
, thisproblem (6.11)
hassolutions q =
0,
...,Q
iffUsing
theorthogonal
conditions(6.7), (6.12)
isequivalent
toThis means that we solve
problem (6.11) by
recurrencestarting
with thevalue q =
Q. Indeed,
for each q,assuming
that~ ~ exists,
thentaking given by (6.13),
we deduce the existence of at least one solution ~pq of(6.11).
Since
~p~
exists(recall
that~0~+1
=0),
we have proven the theorem 6.1.THEOREM 6.1. - For all .1 E
C, Q
E IN U~0~,
pqE X
q E{0,..., Q},
there exists a solution
u(t) of problem (6.1 ) in
the formwhere the sum over ~u
disappears if
~l is not aneigenvalue of A; otherwise,
the are
given by ~6.13~
and are solutionsof ~6.11~.
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