Poincaré Problem for Nonlinear Elliptic Equations of
Second Order in Unbounded Domains
Guochun Wen
LMAM, School of Mathematical Sciences, Peking University, Beijing, China Email: wengc@math.pku.edu.cn
Received September 27, 2012; revised November 2, 2012; accepted November 10,2012
ABSTRACT
In [1], I. N. Vekua propose the Poincaré problem for some second order elliptic equations, but it can not be solved. In [2], the authors discussed the boundary value problem for nonlinear elliptic equations of second order in some bounded domains. In this article, the Poincaré boundary value problem for general nonlinear elliptic equations of second order in unbounded multiply connected domains have been completely investigated. We first provide the formulation of the above boundary value problem and corresponding modified well posed-ness. Next we obtain the representation theorem and a priori estimates of solutions for the modified problem. Finally by the above estimates of solutions and the Schauder fixed-point theorem, the solvability results of the above Poincaré problem for the nonlinear elliptic equations of second order can be obtained. The above problem possesses many applications in mechanics and physics and so on.
Keywords: Poincaré Boundary Value Problem; Nonlinear Elliptic Equations; Unbounded Domains
1. Formulation of the Poincaré Boundary
Value Problem
Let D be an
N1
-connected domain including the0
N
infinite point with the boundary
j j
2 0
C
in ,
where . Without loss of generality, we
assume that D is a circular domain in
1
1
z , where the boundary consists of N1 circles 0 N1
z 1
,
z z r
j j j , 1, , N and . In this
article, the notations are as the same in References [1-8]. We consider the second order equation in the complex form
j z D
, , , Re
, , ,
zz z zz
zz z
u F z u u u
F Qu A u
G G z u u Q
1 2 3
, , ,
,
, , , ,
, , , 1, 2,3,
z
z z zz
j j z
G z u u A u A Q z u u u
A A z u u j
w j
1, 2,3,
z D U 0,
, 2,3
(1.1)
satisfying the following conditions.
Condition C. 1) ,
are continuous in for almost every point
and Q A for z D.
, , ,
Q z u w U A z uj , ,
,
u w
,
j 0 j 1
z D
2) The above functions are measurable in for all continuous functions u z
,w z
in D, and satisfy
,2 0
1
, , , , 1, 2,
, ,
p j
L A z u w D k j w D k
0, 2 < 0 ,
p p p p k k0, 1
2 ,
U
,2
p
L A3 z u, , (1.2)
in which are non-negative
constants.
3) The Equation (1.1) satisfies the uniform ellipticity condition, namely for any number u and w, U1,
the inequality
F z u w U, , , 1 F z u w U, , , 2 q U0 1U2,
z D holds, where q0
1for almost every point is
a non-negative constant.
4) For any function u z C D
1
,2
p
w z W D , ,
, ,
G z u w
satisfies the condition
1 2
, , z z , 0 , ,
G z u u B u B u
, ,
1, 2
j j z
B B z u u j satisfy the condition in which
,2 , 0 , 1, 2,
p j
L B D k j k
(1.3)
with a non-negative constant 0.
Now, we formulate the Poincaré boundary value problem as follows.
Problem P. In the domain D, find a solution u z
of Equation (1.1), which is continuously differentiable inD, and satisfies the boundary condition
1 2
1 2
1 ,
2
. . Re z , ,
u c z u c z
i e z u c z u c z z
(1.4)
D
in which is any unit vector at every point on ,
z cos ,
x icos ,
y
,
known functions satisfying the conditions
, 0,
1,
0,
C k C c k C c 2, k2, (1.5)
where
0 ,
1 2 1
k0 k2cos c 0
, , are non-negative
constants.
If and 1 on , where n is the
outward normal vector on
,n 0
, then Problem P is the Dirichlet boundary value problem (Problem D). If
and a1 on
,ncos 1 0 , then Problem P is the
Neumann boundary value problem (Problem N), and if , and 1 on , then Problem P is the regular oblique derivative problem, i.e. the third bound-ary value problem (Problem III or O). Now the direc-tional derivative may be arbitrary, hence the boundary condition is very general.
,ncos 0 c 0
The integer
arg π 1 2
K z
0, K is called the index of Problem P. When the index
Problem P may not be solvable, and when the
solution of Problem P is not necessarily unique. Hence we consider the well-posedness of Problem P with modi-fied boundary conditions.
0,
K
Problem Q. Find a continuous solution w z u z,
of the complex equation
1
2 3, , , , , ,
Re ,
z z
z
w F z u w w G z u w
1 2 ,
F Qw A w A u A G B w B u
(1.6) satisfying the boundary condition
1
2
Re z w z c z u c z h z z
, , (1.7)and the relation
d 0,j j
j z
z b
z
1, ,
d j N
D
,
0 ,
0,
z K N
K N
K
2 1
1
2 Re z N
j id w z
u z
z z z
(1.8)where j are appropriate real constants
such that the function determined by the integral in (1.8) is single-valued in , and the undetermined function
is as stated in
h z
1
0 0
1
0, ,
, , 1, , ,
0, , 1, , 1
, , 1, , ,
Re ,
j j
j
j j K
m m m m
h z j N K
h z
z j N K N
h z j N
h h ih z z
in which h j0,1, , N
h m
1, , K 1,K0
0
K
j , m
are unknown real constants to be determined appropri- ately. In addition, for the solution w z
is as- sumed to satisfy the point conditions
Im ,
1, , 2 1, ,
1, , 1, 0 ,
j j j
a w a b
K N K N
j J
N K N K N
(1.9)
where
0
1, , ,
1, , 2 1,
j j
j
a j N
a j N K N K N
are distinct points, and b j Jj 0
are all real con- stants satisfying the conditions
3, 0 ,
j
b k j J
3
k
(1.10)
for a non-negative constant .
2. Estimates of Solutions for the Poincaré
Boundary Value Problem
First of all, we give a prior estimate of solutions of Problem Q for (1.6).
Theorem 2.1. Suppose that Condition C holds and ε = 0 in (1.6) and (1.7). Then any solution w z u z,
of Problem Q for (1.6) satisfies the estimates
,
, 1 ,Cw z DC u z D M k (2.1)
0,2 , 2 ,
p z z
L w w DM k (2.2) in which
0
min ,1 2 p ,
0, 0, , , ,0
, 1, 2, j jM M q p k K D j
1 2 3 0 , , .
k k k k k C w D C u D
, w z u z
Proof. Noting that the solution of Prob-
lem Q satisfies the equation and boundary conditions
1
3
Re , , in ,
z z
w Qw A w A G z u w D
(2.3)
2Re z w c z h z on ,
(2.4)
0Im a w aj j b j J uj, , 1 b , (2.5) according to the method in the proof of Theorem 4.3, Chapter II, [2] or Theorem 2.2.1, [5], we can derive that the solution w z
satisfies the estimates
, 3 ,Cw z D M k (2.6)
0,2 , 4 ,
p z z
L w w DM k (2.7)
K D j,
, 3, 4
,w z u z
where
0, 0, , ,0
j j
M M q p k
and
1 2
k k k k3L G Dp , . From (1.8), it follows that
, 5 w z D
, k3,C u z D M C (2.8)
0,2 5
p z z
L u u D, M Cw z D, k3,
,
(2.9)
in which M5 M5 p D0 is a non-negative constant.
Moreover, it is easy to see that
,
, ,
, , .
w D
u D
C u D
,2 ,2 1 ,2 2
0
, ,
p p
p
L G D L B D C
L B D C
k C w D
(2.10)
Combining (2.6)-(2.10), the estimates (2.1) and (2.2) are obtained.
Theorem 2.2. Let the Equation (1.6) satisfy Condition
C and in (1.6)-(1.7) be small enough. Then any solu-
tion of Problem Q for (1.6) satisfies the estimates
,
, 6 ,Cw z DC u z D M k (2.11)
0,2 , 0,2 , 7 ,
p z z p z
L w w DL u DM k
0
,p k,
(2.12)
are as stated in Theorem 2.1,
here
0, 0, , , ,0
, 6,7. j jM M q p k K D j
,w z u z
Proof. It is easy to see that satisfies
the equation and boundary conditions
1 2 3Re , ,
z z
w Qw A wA u A G z D
(2.13)
1 2
Re z w z c u c z h z z, ,
(2.14)
0Im a w aj j b j J uj, , 1 b.
(2.15)
Moreover from (2.6) and (2.7), we have
0
3 0
,2 4 0
, , ,
, , ,
p z z
C w z D M k k C u D
L w w D M k k C u D
(2.16)
and from (2.8)-(2.10), it follows that
3 0 5 3
,2 4 0 5 3
, , ,
, , .
p z z
C w z D M k k M C w z D k
L w w D M k k M C w z D k
0
(2.17)
If the positive constant is small enough such that Combining (2.8) and (2.18), we obtain
0 3 5
1k M M 1 2, then the first inequality in (2.17)
implies that
0 3 5
0 3 8
1 ,
2 1 .
k M
C w z k
M
k M k M k
0 3
1 D
k M
(2.18)
5
8 6, ,
1 1 ,
C w z D C u z D M M k M k
(2.19)
which is the estimate (2.11). As for (2.12), it is easily derived from (2.9) and the second inequality in (2.17),
i.e.
0,2 0,2
4 0 5 3 5 3
4 0 5 8 0 4 7
, ,
, ,
1 1 1 .
p z z p z
L w w D L u D
M k k M C w z D k M C w z D k
M k M M k M k M k
(2.20)
3. Solvability Results of the Poincaré
Boundary Value Problem
ed in Condition C, then the nonlinear mapping G:
We first prove a lemma.
Lemma 3.1. If G z u w
, ,
satisfies the condition stat-
p,2
C D C D L D
, ,
z u z w z
G G
defined by is continuous and
bounded
p B D C u D2,
, ,,2 , , , ,2 1, , ,2
p p
where p p02.
f. In order t ove that e mapping G:
Proo o pr th
p,2
C D C D L D
Defined by z u z w z,
, is continuous, we
G G
choose any sequence of functions w z u zn
, n
w z u zn , n C D n, 0,1, 2,
such that0, u D0, 0
as .n Similarly to Lemma 2.2.1, [5], we n prove
0 0
, ,
n G z u w
possesses the property
n n
C w w DC u
ca that
, ,
n n
C G z u w
,2 ,
p n
L C D 0 asn . (3.2)
And the inequality (3.1) is obviously true.
1.1) satisfy C
Theorem 3.2. Let the complex Equation (
ondition C, and the positive constant in (1.6) and (1.7) is small enough.
1) When 0, 1, Problem Q for (1.6) has a solution w
z u, z , where w z
,
10
,2
p
u z W D ,
0 p
is a co state .en
0 2
p p nstant as d before
2) Wh min
,
1, Problem Q for (1.6) has a solution w
z u z, , where
10
,2 ,
p
w z W D pro-
vided that
9 ,2 3 2
0
, ,
p j
j J
M L A D C c b
(3.3)is sufficiently small.
3) If F z u w w
, , , z
, , ,
i.e
G z u w satisfy the conditions,
ition C and for any funct
. Cond ions w zj
,
1, 2
j
u z C D j and
p0,2
D ere are
1 1 2 2 1 1 2 2 1 2
1 1 2 2 1 1 2 2 1 2
, , , , , , Re ,
, , , , Re ,
V z L , th
F z u w V F z u w V A w w A u u
G z u w G z u w B w w B u u
(3.4)
where
,2A D L , , p,2B Dj, k0 , j1, 2,
p j
L
is a sufficiently small positive constant, th the
above solution of Problem Q is unique.
ation for t is as en
Proof. 1) In this case, the algebraic equ follows
6 7
,2 3 ,2 1 ,2 2 2{0}
, , , , ,
p p p j
j J
M M L A D L B D t L B D t L a b t
(3.5)stated in (2.11) and (2.12).
here M6, M7 are constants as
w
Because 0, 1, the Equation (3.5) has a unique
solution 10 Now we introduce a bounded,
closed and convex subset t M 0. B* of the Banach space C D
,C D whose elements are of the form w z
,u z s g the condition
atisfyin
10(3.6 se a pair of functions w z u z
, B, ,
w z u z C D Cw z D, C u z D , M .
We choo
)
ositions of
and substitute it into the appropriate p
, , , z
F z u w w , G z u w
, ,
in (1.6) and the boundary con- dition (1.7), and obtain
, , , , zw F z u w u w w , z
G z u w
, ,
, (3.7)
1
2
Re z w z c z u c z h z z,
where
,
(3.8)
1
2 3
, , , Re , , , , ,
, , , , .
z z z
, ,
F z u w u w w Q z u w w w A z u w w
A z u w u A z u w
In accordance with the method in the proof of Theo-re
m 1.2.5, [5], we can prove that the boundary value problem (3.7), (3.8) and (1.6) has a unique solution
, w z u z
. Denote by
w u, T w z u z
, themapping from w z u z
, otingthat
to w z u z
, . N
,2 2
p A u D, M k C10 0, c u1 , M k10 0.
provided that the positive number L
is sufficiently small, and noting that the coefficients of complex Equa-tion (3.7) satisfy the same condiEqua-tions as in CondiEqua-tion C, from Theorem 2.2, we can obtain
0,2 0,2
6 7 9 ,2 1 10 ,2 2 1
, , , ,
,
,
, ,
p z z p z
p
p p
C w z D L w w D C u z D L u D
D
M M M L B D M L B D M
6 7 ,2 3 2 ,2
0
6 7 9 ,2 1 ,2 2
, ,
, , ,
p j
j J
p p
M M L A D C c b L G D
M M M L B D C w D L B D C u
0
M10.(3.9)
This shows that T maps B* onto a compact subset in B*.
Next, we verify that T in B* is a continuous operator. In
fact, we arbitrarily select a sequence
w z u zn
,n
inB*, such that
n 0,
n 0,
C w w D C u u D 0 as n . (3.10) By Lemma 3.1, we can see that
,2 0 0
1, 2,3 as .
p j n n j
j n
(3.11)
Moreover, from
, , ,
L A z u w A z u ,w ,D0
w un, n
T w u n, n
,
w u0 0
0 0
, T w u , ,
it is clear that n 0 n 0
lution of Problemequation ,
w w u u is a so
Q for the following
0
0
, , , , ,
, , , , ,
, , in ,
n z n n n n nz
z
n n
w w F z u w u w w
z u w u w w
G z u w G z D
(3.12)
0
0 0 0 0
0
, ,u w F
0
Re z wnw
1 n 0 on ,
c z u u h z
(3.13)
0
Im
, 1 1
j n j
n
a w a
j J u u
0
0,0.
j
w a
(
In accordance with the method in proof of Theorem 2.2, we can obtain the estimate
3.14)
0
0
0 ,2 0
0 ,2 0
11 ,2 2 2 0 0
0 0
1 0
, ,
, ,
, , , ,
, ,
, ,
n p n z n z
n p n z
p n n n
n
C w w D L w w w
C u u D L u u D
M L A z u w u A z u w
w D
C c z u u
(3.15)
0, , ,
.0
w D
0,
u D
,2 3 3 0 0
,2
, , , , ,
, , ,
p n n
p n n
L A z u w A z u w D
L G z u w G z u
in which M11M11 q p0, 0,
(3.11) and the above estimate
k K D From (3.10),
, we obtain
0, u D0, 0 as n .
int theorem, there
ex
n n
C w w DC u
On the basis of the Schauder fixed-po
ists a function
u z C D
such , and from Th, ,
w z u z w z
at w z u z
, T w z u z
, th
sy to see t
eorem
2.2, it is ea hat w z
,
10
,2
p
W D , and
, w z u z
is a solution of Problem Q
u z
for the Equa-tion (1.6) and the relaEqua-tion (1.8) with the condition 0,
1
.
In addition, if G z u w
, ,
ReB w B u1 2
in D, where 0 1,Lp,2B Dj, k0 , j1, 2, then the above solvability result still
similar method.
2) Secondly, we discuss the case:
hold by using the above
min , 1. In this case, (3.5) has the solution tM10 provided that
M9 in (3.3) is small enough. Now we consider a closed
and convex subset B in the Banach space
,C D C D i.e.
, ,
.B w z u z C D C w D , C u D , M10 (3.16) Applying a method similar as before, we can verify that there exists a solution
, p0,2
p0,2
w z u z W D W D
) with the condition
1 1
,
.of Problem Q for (1.6 min
over, if
1 More
1 2
, , Re
G z u w B w B u in D, where
1 , Lp,2B Dj, k0 , j = 1, 2. Under the same
condition, we can derive the above solvability result by the similar method.
3) When G z u w
, ,
satisfies the condition (3.4), we can verify the uniqu ess of solutions in this ten heorem. In fact, if w z1
,u z1 , w z u z2
, 2 are two solu-tions of Problem Q for the Equation (1.6), then
, 1
2
, 1 2
w z u z w z w z u z u z
sa
tisfies the equation and boundary conditions
1 1
Re
z z
w Qw A B w
2 2 , ,
A B u z D
(3.17)
Re z w z c u h z z1 , , (3.18)
0, .j j
a w a j J
in which
Im (3.19)
0 1
Q q . Similarly to Theorem 2.2, we can
derive the following estimates of the solution
w z u z,
r cofo mplex Equation (3.17):
,
, 12 ,Cw z DC u z D M k
(3.20)
0,2 , 13 ,
p z z
L w w DM k
(3.21)
where
0
min ,1 2 p ,
0, 0, , , ,0
, 12,13
j j
M M q p k K D j
are two non-negative constants, k2k C u0
,over the estimate
.
D More-
, 1 2 0 13Cw z D k M C u z ,D0
can be derived. Provided that the positive constant
(3.22)
small enough such that
1 2 k M0 13
0, from (3.22) itfollows u z
u z1
u z2
0, i.e. u z1
D. Th e proof of the theorem.
bove theorem, the next result can
2
u z in
be de-
itions as in
for (1.1) has N
of Problem P
stants. is co
From rived.
rem 3.2, t
depends o 2) Whe
mpletes th the a
Theorem 3.3. Under th
he following statem en the index K > N
y conditions, and n 2K N 2 arb
n 0 ,
e same cond ents hold. , Problem P
the solution itrary real con
Theo-1) Wh solvabilit
K N
Problem P for (1.1) is solvable,
if 2N K solvability conditions are satisfied, and the solution of Problem P depends on K2 arbitrary real co
is solvable under 2
nstants.
3) When K < 0, Problem P for (1.1)
2 1
N K conditions, and the solution of Problem P
depends on 1 arbitrary real constant.
Moreover, we can write down the solvability
condi-tio lem P for a
L
dary
tion (1.8). If the function h z
0,z , i.e.1, ,
0,1, , , 0,
j m
N
N h m
ns of Prob ll other cases.
Proof. et the solution w z u z
, of Problem Q for (1.6) be substituted into the boun condition (1.7) and the rela0, 0, j
h j
h j
, K
1, , K 1, ifif 0KK0,N,
and dj0, j1, , N, then we have w z
uz Dand the function w z
of Problem Pfor (1.1). Hence the t r of above equalities is
just the number of solvability conditions as s this heorem. Also note that the real constants b0
in n
tated in in (1.8) and This shows that .1) includes the stated in the
theo-Classes of P
y atics Monograph American Mathematical Society, Providence, 1992. [5] G. C. Wen, D. C. Chen and Z. L. Xu, “Nonlinear
Com-plex Analysis and Its Applications,” Mathematics Mono-graph Series 12, Science Press, Beijing, 2008.
proximate Methods and Numerical Anal-
is otal num
in (1.9) are arbitrarily ution of Proble of arbitrary real consta
REF
.
just a solutio be
chosen. m P for (1
nts as t
bj
j J
the general sol number
m.
[2] G. C
re
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