A NNALES DE L ’I. H. P., SECTION C
A. K OSHELEV
Regularity of solutions for some problems of mathematical physics
Annales de l’I. H. P., section C, tome 12, n
o4 (1995), p. 355-413
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Regularity of solutions for
someproblems of mathematical physics
A. KOSHELEV
Univ. Stuttgart Mathematisches Institut A 801140 Stuttgart, D-70511 Germany
and
St. Petersburg Univ. Fac. of Math.
and Mech. 1 Ulianovskayast.
198 904 St. Petersburg-Petrodvoretz Russia.
Vol. 12, n° 4, 1995, p. 355-413. Analyse non linéaire
ABSTRACT. -
Coercivity
estimates of solutions for someproblems
ofmathematical
physics including parabolic
and Stokes systems are obtained.Key words: Elliptic parabolic systems.
1. PREFACE
This paper is devoted to some
boundary
valueproblems
for systems ofpartial
differentialequations.
Inparticular
we consider Stokes system andquasilinear elliptic degenerate
systems ofdivergent
type with bounded nonlinearities. The author would like to express hisdeep gratitude
to Prof.S. Hildebrandt for fruitfull discussions and
general
support.It was shown in
[3] (see
also[4]
and[9])
that thequestion
ofregularity
of weak solutions for
quasilinear elliptic
andparabolic
systems isclosely
attached to the
dispersion
of the spectrum of the matrix which defines theellipticity (parabolicity)
of the system. The upper bound for thisdispersion
isdetermined
by
some coercive constants forelementary elliptic
orparabolic
operators. The
explicit
form of these constants leads to some conditions which are easy to check in order to obtain theregularity
of weak solutions.This
approach
can beapplied
forexample
to suchimportant
systems as the Stokes system. We divide the paper in three sections.Annales de l’lnstitut Henri Poincaré - Analyse non linéaire - 0294-1449 Vol. 12/95/04/$ 4.00/© Gauthier-Villars
356 A. KOSHELEV
The first one is devoted to some constants
concerning
theoperators A
and
A,
where e is anarbitrary positive
constant.Let B be a unit ball in
Rm (m 2 2)
with the center at theorigin
and leta =
2-m- 2, (0
~y1).
Ifu(x)
isequal
zero on aB then theinequality
holds true. Here and
IDul2
are,respectively,
the sums of allsquared
derivatives of u of the second and the first order. An
analogous
result wasat first obtained
by
H. O. Cordes[1].
This estimate could be also obtained with the
help
of the result of E. Stein[4] concerning
the boundness of thesingular integral operators
in theweighted
spaces But this method doesn’tgive
theexplicit
constant in front of theright-hand
sideintegral containing Au.
The
nonstationary
case is also considered in this section. LetQ
=(0, T)
xB, u
= 0 for t =0, and (
is a cut-off function. Then theinequality
holds for m > 3, and the constant C doesn’t
depend
on ~ > 0.Section 2 is devoted to some
coercivity
estimates. In section 3 we consider the Stokes system both forstationary
andnonstationary
cases. Consider forexample
hereonly
thestationary
systemin a bounded domain Q C R"2 with a smooth
boundary
and with u = 0Then the estimates for the weak solution u, p
and
hold true and C doesn’t
depend
on xo. The results of thisparagraph
wereobtained in
cooperation
with A.Wagner (Cologne).
The third
paragraph
contains some results about theelliptic
systemUnder natural
analytic
conditions on the coefficientsa2 (x, p)
we assumethat the
eigenvalues Aj
of thesymmetric
matrixsatisfy
thefollowing inequalities
with
A,A
= const. > 0 and 0 s 1.It is
proved,
forexample,
that if theinequality
holds then the "small" weak solution of the system satisfies the Holder condition in H.
Vol. 12, nO 4-1995.
358 A. KOSHELEV
2. SOME COERCIVITY
INEQUALITIES
WITH EXPLICIT CONSTANTS
Consider in R"’
(m
>2)
a ballBR(xo)
with the center xo and radiusR. The ball
Bi(0)
will be denoted B. In this ball anequation
with a
boundary
conditionis
given.
Suppose
thatf
EL2,~(B),
where is the space ofsquared integrable
functions with a
weight Ixla. Throughout
this paper we assume thata = 2 - m -
2~y (0
~y1), and Ixl
denotes the distance from theorigin.
The norm inL2,a(B)
as usual is determinedby
Set
where Ui are the derivatives with respect to xi.
By
we shall denote those functions in the Sobolev spaceW22~ (B) whose
second derivatives are square summable with theweight
As norm in this space we could take for
example
theexpression
One of the aims of this section is to prove for the solution of the
problem (2.1), (2.2)
theinequality
where
Ca
has form. For a’LEMMA 2.1. -
If
~c EW22~ (B),
then theinequalities
and
hold.
Here r~ is as usual an
arbitrary positive
constant andwhere
|S|
is thesurface of
the unitsphere
in R"‘ and 03BB is the smallest absolute valueof
theeigenvalues for operator 0
in B with thecondition
(2.2).
Proof. - Evidently
Square
both sides of thisequality
andintegrate
over the ballB8(0)
=jB5
with 8 1. We get
The first term on the
right
hand side we can write in theequivalent
form and get
Applying
the Holderinequality
to the innerintegral,
we obtain theestimate
Vol. I 2, n ° 4-1995.
360 A. KOSHELEV
Putting 8
instead of the upper bound of the innerintegral
we get thefollowing :
Dividing by
8m andtaking
into account thatm(rrc
+2~y)-1 l,
weobtain the
inequality
Using
the notation(2.6)
we obtain theinequalities (2.4)
and(2.5).
0COROLLARY 2.1. - Let 03BB be the smallest absolute value
of
theeigenvalues for
theoperator 0
with condition(2.2).
Then theinequalities
and
take
place if
usatisfies (2.2).
Proof. -
In fact both of the second terms on theright
hand side of(2.4)
and
(2.5)
can beeasily
estimatedby
theintegral
ofUsing
the condition(2.2)
andintegrating by
parts we haveThen
and from the
previous inequality
we haveLEMMA 2.2. - For u E
satisfying (2.2),
theequality
holds,
where uT is the derivative with respect to r.Proof - Integrating
twiceby
parts we haveTherefore
With the same kind of calculations
(see
forexample [3]
p. 142etc.)
wecome to the
identity
Vol. 12, nO 4-1995.
362 A. KOSHELEV
After
applying (2.10)
we arrive atUnder condition
(2.2)
we have from(2.10)
thatand we come to
(2.9).
Consider a function
which
evidently
satisfies the conditionsTake a
complete
orthonormal set ofspherical
functionsand consider the
expansion
The derivatives of with respect to r we denote
by va,l (r).
LEMMA 2.3. - For any u E
W22~ (B), satisfying (2.2),
theidentity
holds true
(by ~
we understand the summation in the same limits asj,l in
(2.12)).
Proof -
We can write theidentity (2.9)
in the formUsing (2.11)
we canintegrate by
parts in the last term on theright
hand side
So
Integrating by
parts we getVol. 12, n ° 4-1995.
364 A. KOSHELEV
Finally
Substituting
in(2.14)
we come toLet us transform the last three terms on the
right
hand side of(2.15)
withthe
help
of theexpansion (2.12).
ThenSo
Now
Vol. 12, n ° 4-1995.
366 A. KOSHELEV
Then
Combining (2.15), (2.16)
and(2.17)
we get(2.13)
0LEMMA 2.4. -
If
v Esatisfies (2.2)
and vanishes with allfirst
derivatives at the center
of
B then theidentity
Proof. - Using
theexpansion (2.12)
weget
Continuing
this process we come top
Vol. 12, n ° 4-1995.
368 A. KOSHELEV
In the same way we get
After
simple
calculations we come to(2.18).
DLEMMA 2.5. - For any u E
satisfying (2.2)
theinequality
holds,
whereProof. -
From(2.18)
we haveAccording
to[3] (p.
51 and p.54)
for a =2-rrc-27(0
~y1)
we haveThe fact that in this case v and Vv can differ from zero on aB
plays
no role.
Vol. 12, nO 4-1995.
370 A. KOSHELEV
Then from
(2.13)
and(2.18)
we haveSince u vanishes on aB
and, according
to(2.11 ),
v is a linear functionon ~B we have that l = 0 on aB
for j
> 1. ThereforeSo
In the same way we come to the
inequality
With the
help
of(2.21)
and(2.22)
weget
Let us estimate now the
right-hand
side of(2.23). Evidently
from(2.11)
we have
Taking
into account that on~B |~u|2
=u;
and Ui = urcos(r, xj ) ,
aftercancelling
some terms we come to theequality
After easy calculations we have
Vol. 12, n ° 4-1995.
372 A. KOSHELEV
Applying
theinequality
to the middle term in
quadratic
brackets on theright-hand
side of(2.23)
and
taking
into account that a - 3= -(m
+ 1 +2~y),
we obtainSince
we come to
(2.19).
DLEMMA 2.6. -
If
u Esatisfies (2.2),
thenfor
any r~ > 0 theinequality
takes
place.
Here a = 2 - m -2q (0
~y1),
andMy
aredetermined
by (2.6)
and(2.20).
The value~S~ (the
areaof
the unitsphere
in
Rm)
is determinedby
theformula
Proof -
From theidentity
follows the
inequality
Evidently
Since u satisfies the condition
(2.2)
we haveFrom the
Hardy inequality
followsSo, taking
into account thatr
1, we come to theinequality
Applying (2.26)
and(2.27)
we getNow
combining (2.4), (2.5), (2.19)
and(2.28)
we come to theinequality (2.25).
DVol. 12, nO 4-1995.
374 A. KOSHELEV
COROLLARY 2.2. - We can also
apply
theinequalities (2.4)
and(2.5 ).
Thenwe arrive at the relation
COROLLARY 2.3. -
Taking
into account that r l, we can writeand
from (2.29)
thenfollows
It is easy to see, that if
is
nonnegative,
then theexpression
is
nonpositive.
Afterrescaling
in x we come to thefollowing
THEOREM 2.1. - Let u E
satisfies
the condition(2.2).
Let alsothe
inequality
holds. Then the
following
estimatesVol. 12, n ° 4-1995.
376 A. KOSHELEV
and
hold true.
Let us recall that a = 2 - m -
2q (0
~y1~, My
and aredefined by (2.20)
and(2.6) respectively, ~
is the least absolute valueof
theeigenvalues for
theoperator ~
in B with condition(2.2)
andis the area
of
the unitsphere
in R"°.Consider now the
cylinder QR
=(0, T)
xBR (Q1
=Q)
withboundary
conditions
for a function
u ( t, x ) given
inQ.
Denote/3
= -a and omit for a while the index R.For m > 2 the
inequality
LEMMA 2.7. - Let m = 2 and
therefore ,Q
=2q (0
~y1). Suppose
that u
satisfies (2.36)
and u ELZ { (O, T );
Then theinequality
holds,
where ~ is anarbitrary nonnegative
constant.Proof. -
Denotemultiply
thisequality by
Au .r~
andintegrate by parts
on the left-hand side. Thenaccording
to lemma 2 in our paper([2], (2.38))
weget
After
using
foru(x, t)
theexpansion, analogous
to(2.12), according
tothe same lemma in
[2] ((2.39)),
we come to theinequality
Now we have to estimate the
right-hand
side term of(2.40).
As in[2]
we
multiply (2.39) by 1). Integrating by
parts we come to theinequality ([2], (3.30) etc.)
Vol. 12, nO 4-1995.
378 A. KOSHELEV It is clear that
Thus we have
After
applying
the Holder’sinequality
we getWith the
help
of(2.40)
we come to theinequality
Applying
well knowninequalities
we getAccording
toexpansion (2.12)
and we can write
So,
from(2.41)
we come to2-/3
Taking ~ = 201420142014
andapplying
theinequality
we get
(2.38).
Set
THEOREM 2.2. -
Suppose
u E andsatisfies
the
boundary
conditions(2.36),
a E(-m,
-2 -m)
U(3 - m, 0)
andq =
(2 -
m -a) /2.
Then thefollowing
estimatesVol. 12, n ° 4-1995.
380 A. KOSHELEV
and
hold where ~ and T are
arbitrary positive
values, and all other constants aredefined
at the endof
theformulation of
theorem(2.1 )
andby (2.42) (C
does notdepend
on ~ andR).
The
function (
is a smooth monotonecut-off function, defined by
therelation
Proof -
We can assume at first that u is as smooth as we wish. Letw(t, x)
be a solution inQ
of thefollowing boundary
valueproblem
Multiply equation (2.47) by Au(
andintegrate
onceby parts
withrespect
to t and twice with
respect
to x. Then we getwhere the nonwritten terms are those
containing
the derivatives of(.
Applying
the Holderinequality,
we getIt is trivial that w also satisfies the
inequalities (2.37)
and(2.38) (one only
has toexchange t by
T -t).
ThereforeNow from
(2.48)
afterrescaling
we get the results of thetheorem,
ifwe take into account that
u
Let us return now to
inequalities (2.4)
and(2.5)
of lemma(2.1 ).
Since thepower of the
integrals
on theright-hand
side of theseinequalities
isequal
to one
they belong
to the so-called class of the linearinequalities.
HoweverVol. 12, n° 4-1995.
382 A. KOSHELEV
in some
problems
it isimportant
to have the so-calledmultiplicative inequalities.
We shall obtain them in thefollowing.
LEMMA 2.8. -
If u
E and(2. 2)
takesplace,
then theinequalities
hold.
Proof. - Evidently
it isenough
to proveonly
the first of theinequalities (2.49). Substituting
in(2.4)
theexpression (2.6),
we getTake now
and we come to
(2.49) (if IDul
= 0 then t6 = 0 and(2.49)
istrivial.)
DRemark 2.1. - Under the
assumption
of the lemma theinequality
holds.
In fact
From the well known
Hardy inequality
and from(2.5)
it follows thatThen from
(2.2)
andinequalities r
1 and a 0 we getApplying (2.49)
and(2.50)
to(2.19)
we come toTHEOREM 2.3. - Let u E and
satisfy (2.2).
Then theinequality
holds true.
Proof. -
In fact we omit on theright-hand
side of(2.19)
thenegative
term and
apply (2.49)
and(2.50).
Then we getAfter
using
the relationwe come to the result. D
Suppose
now that condition(2.2)
is notsatisfied;
how will estimate(2.51) change
in this case?Vol. 12, n° 4-1995.
384 A. KOSHELEV
THEOREM 2.4. - Let u E Then the
inequality
holds true,
where (
isdefined by (2.46)
and r~ is anarbitrary
smallpositive
number.
The result follows
immediately by substituting
the functionu(
for uin
(2.51 ).
0THEOREM 2.5. -
Suppose
that u EL2 ~ (o, T ); satisfies only
the second
of
the conditions(2.36),
u = 0, when t = 0. Then the estimateholds, where C does not
depend
on ~.Proof. -
Take a function w which satisfies theequation (2.47)
with thesame conditions and
multiply
both sides of the differentialequation by
Au .
(.
Afterintegration
overBR
we come toAfter two
integrations by parts
on the left-hand side withrespect
to x,we get
Integrating
on the left-hand sideby
parts once with respect to t and on theright-hand
side in the secondintegral
once with respect to x, we getTherefore
Let us now estimate the
integrals
on theright-hand
side. Afterapplying
an
elementary inequality
we come to thefollowing
relations:Vol. 12, nO 4-1995.
386 A.
KOSHELEV
Then
by (2.37)
we getTaking 1J
we obtain theinequality
3. COERCIVITY ESTIMATES FOR THE STOKES SYSTEM IN WEIGHTED SPACES Consider now at first the
stationary
Stokessystem
with the
boundary
conditionWe assume that the mean value
of p
isequal
to zero.Using
theinequalities (2.52)
and(2.53)
we shall derive some estimates withexplicit
constantsfor the solution of the
problem (3.1), (3.2).
The Stokes system was veryextensively
discussed in many books and papers we refer hereonly
to thepaper of V. Solonnikov
[6]
andmonograph
of O.Ladyzhenskaya [10].
From the results of these paper and
monograph
inparticular
follows that iff
E >1)
then the second derivatives of u and the first derivatives of p alsobelong
to this space. Theanalogous
result for thenonstationary
system is also included there.Suppose
that Q c R"° is a bounded domain and 8Q issufficiently
smooth.THEOREM 3.1. -
If f
E with a = 2 - m -2y(0
~y1)
then the weak solutions
of
system(3.1)
with theboundary
condition(3.2) satisfy
theinequalities
Vol. 12, nO 4-1995.
388 A. KOSHELEV
where
xo is
anarbitrary point
insideSZ,
Rdist (xo , a52),
~7 = const > 0 isarbitrary, M~,
isdefined by (2.20)
and 1 -2-1 a ( a
+ m -2 ) (m - 1 ) -1
> 0.The
proof
of this theorem isanalogous
to theproof
which wasgiven
in
[3]
for the solution of the Poissonequation.
Let us sketch thisproof.
According
to the above mentioned results of V. Solonnikov we can at firstassume that both
f
and the solution u, p are as smooth as we wish. Takea
point xo E S2
and consider a ballBR(xo)
with Rdist(xo, c7S2).
Afterrescaling
we can consideronly
the ballBi(0)
= B.Let E
S)
be acomplete
orthonormal set ofspherical
functionsand let
Construct the function
where
Take the function
where the cut-off function
((r)
is determinedby (2.46).
Multiplying
the Stokes system(3.1) by
Vw andtaking
into account thatwe come to the
equality
Integrating by
parts andsubstituting
theexpansions (3.5), (3.6)
for p and w we havewhere the unwritten terms contain
only integrals
withoutsingularity.
From this
immediately
followsFinally
On the other hand
Comparing
the last two relations we come to theinequality
Vol. 12, nO 4-1995.
390 A. KOSHELEV
In our book
([3],
p. 120 see also[9]) by
the same method it was shownthat
Therefore one of the statements of the theorem is
proved.
Take now in the Stokes system
(3.1) Vp
to theright
hand side andapply
theinequality (2.52).
After small calculations you come to theinequality (3.4).
DConsider now the
nonstationary
Stokes systemwith
boundary
conditionsAt first we consider the inner estimates.
Suppose f
ELZ ~ (o, T ); L2,~ (S2) ~
andQR
=(0, T)
xBR,
whereR
dist(xo,
It is trivial that estimate(3.3)
holds if wechange BR
forQ R .
Then, dividing
the firstequation
of(3.8) by v
andapplying (2.53),
wecome to
THEOREM 3.2. - Let the conditions
for
a in theorem 3.1 hold. The solutionof
theproblem (3.8), (3.9) satisfies
theinequalities
Here C does not
depend
on v,M~,
andAa,m
arerelatively
definedby (2.20)
and(2.42).
For
small,
> 0 we haveTHEOREM 3.3. - Let the conditions
of
theorem 3.1. besatisfied,
andassume
that 03B3
> 0 is small. Then thefollowing
estimatesfor
the solutionsof
system(3 .1 )
are true.
THEOREM 3.4. - Let the conditions
of
theorem 3.2 besatisfied,
and letq > 0 be small. Then the inner
following
estimatesfor
the solutionsof
theVol. 12, nO 4-1995.
392 A. KOSHELEV
system
(3.8)
hold true, where C doesn’t
depend
on v.It is necessary now for the solution of the
problem (3.1) (3.2),
to get the estimates in theneighbourhood
of theboundary
an. For this purpose suppose that apiece
of theboundary
is flat and has theequation
xm = 0.Thus in the
neighbourhood
the domain n lies in the half space xm 0. TakeConsider also a
parallelelepiped
n-Suppose
for a moment that = 0(k
=1,...,
m -1)
and R = x. Theprincipal
part of the estimates doesn’tdepend
on theseassumptions.
Wecan suppose at first that all the functions
f, p
and u are smooth.Expand f (x)
in II in thefollowing
Fourier serieswhere n =
(ni ,
...,n~.,.t_ 1, )
and all nk arenonnegative integers.
TakeLet
Then
We see that the functions
(k m) satisfy
also thefollowing boundary
conditionsVol. 12, nO 4-1995.
394 A. KOSHELEV
The functions
u(x), p(x) satisfy
the Stokes system(3.1)
in II- if thefollowing equalities
where
and dots over denote derivatives with respect to Xm.
Multiply
the firstm - 1
equations
of(3.23) respectively by
nk. After summation andusing
the last
equation (3.23)
we haveIf we differentiate the second
equation
of(3.23)
with respect to xm and subtract the relation(3.24)
from the result we shall haveThe bounded solution of this
equation
for xm 0 isgiven by
Here is a function which coincides with on xm > -x and is
continuously expanded
on xm -~r. We suppose also that all the functions areabsolutely
summable on(-oo, 0].
Let us also consider the
equation (3.25)
in xm > 0 with such suitableright-hand
sideFn (xm)
that is continuous andabsolutely
summableon the whole
strip
-oo +00. The solution for xm > 0 isSubstitute xm = 0 in
(3.26)
and(3.27). Suppose
that C- =C+
andThen
We see that
F;;
should be extendedsymmetrically
on xm > 0.According
to
(3.25)
the functions and =1, ... , m-1)
should beextended
respectively
in anantisymmetric
andsymmetric
ways.Integrating
once
by parts
we come toThe
analogous
formulafollowing
from(3.27)
holds true for xm > 0.Denote
by
the Fourier transform of the functions(k
=1, ... , ~n)
on -oo xm -f-oo.We have
For
example
belongs
to(II)
and therefore itsboundary
values are defined forMoreover
they
can be estimatedby
the norm of inVol. 12, n° 4-1995.
396 A. KOSHELEV
Then
according
to(2.52) (theorem 2.4)
Differentiating p(x)
with respect toxk(k
=1, ... ,
m -1)
we get thesame estimates. Then
Denote
As far as xo E
BRo
then for x EBRo
where x is
symmetric
to x with respect to xm = 0. Fora = 2 - rrz - 2~y 0
we have
and from the
monotonicity of (
followsSince
f
and p areexpanded
onBRo
insymmetric
andantisymmetric
ways we have
The same estimate is true for the
integral fB Ro Taking
intoaccount that
we come with the
help
of(3.29)
toTHEOREM 3.5. -
If
the conditionsof
theorem 3.1 aresatisfied
then thesolution
of
theboundary
valueproblem (3.1 ), (3.2) satisfies
theinequalities
Vol. 12, n ° 4-1995.
398 A. KOSHELEV
and
where
SZR
= 52 nBR(xo), R sufficiently small,
r = ESZ, C
doesn’tdepend
on xoand ~
is anarbitrary
smallpositive
number.Proof. -
Theinequality (3.32)
followsby comparing
the estimates(3.4)
and
(3.31).
In fact it isenough
to compare the coefficients in front of~
and to take the greatest one. To get(3.33),
the system(3.1)
should be written in the form
and the
boundary
conditions(3.2)
are to be used.In the interior the
inequality
follows from the estimate(3.4).
In theboundary strip
the solution should be continued in anantisymmetric
way and estimated with thehelp
of(3.30).
Theproof
of the theorem can nowbe
completed by
somesimple
calculations. DFor 03B3 ~
0 we haveTHEOREM 3.6. -
If the
conditionshold.
Consider now the
nonstationary
system(3.8)
with condition(3.9).
THEOREM 3.7. -
If f
EL2 ~ (0, T ); L2,a (SZ) ~
with asatisfying
theconditions
of theorem
3.1 then the solutionof system (3.8)
with theboundary
condition
(3.9) satisfies
the estimatesVol. 12, nO 4-1995.
400 A. KOSHELEV
and
Here r
= ~ x - xo
andQ R
=(0, T)
xBR (xo )
n SZ withsufficiently
smallR. The constant C doesn’t
depend
on xo and v.The
proof
iscompletely analogous
to that of the theorem 3.5. Theonly
difference is that one has to refer to estimate
(2.53).
For
small,
> 0 the last two theorems can be formulated in a moreexplicit
way.THEOREM 3.8. -
If the
conditionsof
theorem 3.5 aresatisfied,
thenfor
thesolution
of problem (3.1 ), (3.2)
thefollowing
estimateshold.
THEOREM 3.9. -
If
the conditionsof
theorem 3.5 aresatisfied,
then thesolution
of
theproblem (3.8), (3.9) satisfies
thefollowing inequalities:
Vol. 12, nO 4-1995.
402 A. KOSHELEV
4.
REGULARITY
OFSOLUTIONS
FORDEGENERATED
ELLIPTICSYSTEMS
In a bounded domain n C
2)
we consider a systemwhere u and
ai (x, p) (i
=0,1,..., m)
are N-dimensional vector functions with componentsu~~~ (x),
=1, ... , N),
(now,
different from§,
we include u inDu).
About the functionsai (x, p)
we assume, that
they satisfy
thefollowing
conditions:(1)
Allsatisfy
theCaratheodory
conditions and are differentiable with respect to variables p;(2)
The(m
+1)N
x(m
+1)N
matrixis
symmetric
and theeigenvalues
of this matrixsatisfy
theinequalities
where
A,A
= const > 0 and 0 s1;
(3)
Forarbitrary
u EW~1~(S2)(q
>1)
the result of the substitutionai(x,
=0,..., m)
willbelong
to(4)
Theinequality
holds,
where b is asufficiently
smallnonnegative value;
(5)
For all u E the result of substitution inL(u) belongs
toLQ(st).
Consider
In
[3] (chapter 1, § 4)
it wasproved
that the universal iterational processconverges in to the weak solution u of
(4.1), (4.5)
if u E(S2).
Consider also the process
(4.6)
with apenalty
term,with the same condition
(4.5).
In
[3]
it was also shown that asubsequence
of the iterations ofprocess (4.7)
convergesweakly
to the solution.So,
if we want to show that the solution has Holder continuous first derivatives it isenough
to show thatthe iterations
of(4.6)
or(4.7) satisfy
theinequality
where
52R
= xo ES2,
a =2 -rn- 2q(0
71),
r = ,and C doesn’t
depend
on xo and n. It is also assumed that R issufficiently
small and fixed.
o (1) LEMMA 4.1. -
If
the conditions1)-3)
aresatisfied
and EWZ (S2)
then
holds,
whereProof. - Multiply
both sides of the system(4.6) by
Un+1 andintegrate
once
by
parts. ThenVol. 12, n° 4-1995.
404 A. KOSHELEV
where summation as
always
runs overrepeated
indices.Adding
andsubtracting ai (x, 0)
under the square brackets on theright
hand
side,
we getApplying
the mean value theorem we come towhere
A
denotes the matrix A with intermediate values of variables.The Holder
inequality gives
It can be
easily proved (see
forexample [3]
p.58, (2.29)),
thatUsing
theright
side of theinequalities (4.3)
weget
a
Suppose
that the cut-off function((r)(2.46)
satisfies in addition theinequality
to the same space. The iterations can be extented outside the domain 0
to a
sufficiently
narrowstrip preserving
the class. This can be made withthe
help
of the well-knownprocedure
which we have used in theprevious paragraph.
First one considers aplane piece
of theboundary
andexpands
all of the Un in an
antisymmetric
way. Thisgives
one the same classof U
S2R)
for ballsBR(xo)
which don’tcompletely
lie in H. Aswe have shown in
[3] (chapter 4, § 3)
all the conditions1) - 5)
don’tchange,
and thevalues, s,
A and A will be the same. Thisgives
also us thepossibility
to consideronly
the case whenS2R
nBR(xo)
=BR(xo)
and this
gives
the fixed smallRo.
LEMMA 4.2. -
If the
conditions1)-5)
aresatisfied
and uo E n2)
then the iterations(4.6)
or(4.7) satisfy
theinequality
where C doesn’t
depend
on xo ESZ,
n and in the caseof (4.7)
on 6.Proof. - According
to ourprevious
consideration we can suppose that52R
=BR(xo). Multiply (4.6) (or 4.7) by
andintegrate by parts
as in the
proof
of lemma 1.2 or lemma 1.5 for a = 0. In[3] (theorem 1)
it is shown that
if (
satisfies(4.13)
thenFrom this and from
(4.12) immediately
follows(4.14).
Let
wk(x) satisfy
theequation
and the
boundary
conditionVol. 12, n ° 4-1995.
406 A. KOSHELEV
where M is a
positive integer
and ak is monotone andsatisfy
thefollowing
relations:
According
to results of E. M. Stein[5]
and V. A.Kondratjev [8]
theinequality
holds,
if -maBR.
Multiply (4.6)
or(4.7) by Awk(
andintegrate
twiceby
parts. It is obvious that(2
also satisfies(2.46)
and(4.13).
Then we get(the
unwritten terms containonly
the first derivatives of un and Letus estimate at first the
integral 7i.
It is easy to see thatFurthermore,
According
to theinequality
of S. Chelkak([9],
p.28,
Lemma1.2),
wehave
Then,
from(4.16)
and the fact thatD( =
0 forr R/2
it follows thatIf ak satisfies
(4.17)
we canapply (4.18)
and come to theinequality
Vol. 12, n° 4-1995.
408 A. KOSHELEV
Carrying
out the same considerations forI2
and13, (4.12), (4.19)
and(4.20) yield
the relationInequality (2.52) (th. 2.4) gives
for k = MFor k M
according
to[2] (p. 51,
lemma2.2) (see
also[8])
THEOREM 4.1. -
Suppose
the conditions1)-5)
aresatisfied
and theinequalities
hold true
(~,
a, b = const > 0 aresuficiently
smallnumbers).
If
the relationis
satisfied,
then the solutionof
theproblem (4.1 ), (4.5) belongs
towith ~y = -(cx
+ m -2)/2
and thesubsequences of
iterationsof (4.6)
and(4.7)
converge to this solution.Proof -
Consider at first the case m > 4. As we have mentioned before it is sufficient to proveinequality (4.8). Suppose
that u EW(2)q (SZ)
withq >
m(m
+Then all un are in
Wq 2~ (SZ) .
From this follows thatdun
EW {2~ SZ .
If we write
(2.49)
for the functionsu~ ~ ,
we get2,cx
( )
Vol. 12, nO 4-1995.
410 A. KOSHELEV
where =
Diuj
Infact,
from(2.49)
and u =Uj(
we obtain after somecalculations that
Now
(4.25)
follows from(4.23) for j
= 0.Applying (4.9)
and theinequality (a
+b)Z 2(a2
+b2),
we haveAfter
using (4.21)
and(4.22),
theinequalitiy labl
+ theestimates
give
the relationTherefore
From
(4.25)
follows thatfor j = 0,1
(4.21) gives
Set
Inequality (4.26)
now turns towhich can be written in the form
Let the condition
holds. Then there exists such a qo E
(0,1)
thatqo)
> 1. Let Hbe so small that H
(1 - qo) -
1 -Hms 2(m+203B3]2(m+203B3) ms ~
Vol. 12, nO 4-1995.
412 A. KOSHELEV
P(l - qo ) .
Then after small calculations we get from theinequality Xo
Pthe relation
Let us return now to
(4.22).
From(4.25)
andwe get
With the
help
of(4.27)
and(4.29)
we have for k MAll ak are
negative
anddecreasing.
Then from the lastinequality
weobtain
From