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Electronic Journal of Differential Equations, Vol. 2005(2005), No. 109, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
DIRICHLET PROBLEMS FOR SEMILINEAR ELLIPTIC
EQUATIONS WITH A FAST GROWTH COEFFICIENT ON
UNBOUNDED DOMAINS
ZHIREN JIN
Abstract. When an unbounded domain is inside a slab, existence of a posi-
tive solution is proved for the Dirichlet problem of a class of semilinear elliptic
equations that are similar either to the singular Emden-Fowler equation or
a sublinear elliptic equation. The result obtained can be applied to equa-
tions with coefficients of the nonlinear term growing exponentially. The proof
is based on the super and sub-solution method. A super solution itself is
constructed by solving a quasilinear elliptic equation via a modified Perron’s
method.
1. Introduction and Main Results
Let Ω be an unbounded domain in Rn(n≥3) with C2,α (0 < α < 1) boundary.
We assume that Ω is inside a slab of width 2M:
Ω⊂SM={(x, y)∈Rn:|y|< M }
where x= (x1, x2, . . . , xn−1) and throughout the paper, ywill be identified with
xn. We consider the existence of positive solutions for the Dirichlet problem
−
n
X
i,j=1
aij (x, y)Dij u=p(x, y)uγin Ω;
u= 0 on ∂Ω
(1.1)
where (aij ) is a positive definite matrix in which each entry is a local H¨older con-
tinuous function on Ω, p(x, y) is also local H¨older continuous on Ω, γ < 1 is a
constant. We note here that (aij) is not required to be uniformly elliptic on Ω.
When the principal part in (1.1) is the Laplace operator, γ < 0, (1.1) becomes a
boundary-value problem for the singular Emden-Fowler equation
−∆u=p(x, y)uγin Ω;
u= 0 on ∂Ω.(1.2)
The singular Emden-Fowler is related to the theory of heat conduction in electrical
conduction materials and in the studies of boundary layer phenomena for viscous
2000 Mathematics Subject Classification. 35J25, 35J60, 35J65.
Key words and phrases. Elliptic boundary-value problems; positive solutions;
semilinear equations; unbounded domains; Perron’s method; super solutions.
c
2005 Texas State University - San Marcos.
Submitted February 11, 2005. Published October 10, 2005.
1
2 Z. JIN EJDE-2005/109
fluids ([1], [15]) . The existence of positive solutions of the equation on exterior
domains (including Rn) has been widely considered (for example, see [3], [4], [7],
[10], [11], [14], and references therein). The main approach used to prove existence
results is to construct super and sub solutions. A super solution is usually found
in the class of radial symmetric functions. If Ω is an exterior domain (not inside a
slab), γ < 0 and there is Csuch that p(x, y)≥C
(1+|x|2+y2)for |x|2+y2large, then
(1.2) has no positive solutions ([10]). On the other hand, if there are constants
σ > 1 and C, such that 0 ≤p(x, y)≤C
(1+|x|2+y2)σfor |x|2+y2large, (1.2) has a
positive solution ([7]). When Ω is an unbounded domain inside a slab, the situation
is quite different since now one cannot construct a super solution which is a radial
symmetric function. In addition, the generality of the coefficient matrix (aij ) in the
equation in (1.1) also makes finding a radial symmetric super solution impossible.
However a super solution still can be constructed when Ω is a domain inside a slab.
In [8], the author combined an idea from [12] and a family of auxiliary functions
constructed in [9] to construct a super solution which is then used to prove the
following existence result.
Theorem 1.1. Assume
(1) p(x0, y0)>0for some (x0, y0)∈Ω;
(2) there is a positive constant Csuch that
0≤p(x, y)≤C(|x|+ 1)−γfor (x,y) ∈Ω; (1.3)
(3) T race(aij ) = 1 and there is a constant c1>0, such that
ann(x, y)≥c1on Ω.(1.4)
Then for γ < 0,(1.1) has a positive solution u∈C2(Ω) ∩C0(Ω).
Comparing Theorem 1 with the known results when the domain is an exterior
domain, we see a new phenomena appearing. That is, when the domain is inside
a slab, (1.1) has a positive solution even if the coefficient p(x, y) of the nonlinear
term is unbounded, while if the domain is an exterior domain, to assure that (1.1)
has a positive solution, the coefficient p(x, y) of the nonlinear term must go to zero
no slower than some functions ([10]). In this paper, we improve Theorem 1 in two
aspects. One is to allow the exponent γin (1.1) to be any number less than 1. The
other is to allow the coefficient p(x, y) of the nonlinear term to grow exponentially!
Here is the statement of the main result of the paper.
Theorem 1.2. Assume γ < 1,Trace(aij) = 1 and there is a constant c1>0, such
that
ann(x, y)≥c1in Ω.(1.5)
Then there is a positive constant α0depending only on c1,Mand n, such that for
any positive constant C,(1.1) has a positive solution u∈C2(Ω) ∩C0(Ω) for all
p(x, y)that is not identical to zero and satisfies
0≤p(x, y)≤Ce(1−γ)α0|x|for (x, y)∈Ω.(1.6)
Furthermore, there are constants c6and c7depending only on n,Mand c1such
that
u(x, y)≤c6ec7|x|in Ω.(1.7)
EJDE-2005/109 DIRICHLET PROBLEMS 3
The idea of the proof of this theorem is as follows: Consider the boundary-value
problem
−
n
X
i,j=1
aij (x, y)Dij u0=p(x, y)uγ
0in Ω; u0= 1 on ∂Ω.(1.8)
For any positive constant k0>0, that u0>0 satisfies (1.8) is equivalent to that
v0=1
k0ln u0(i.e. u0=ek0v0) satisfies
−
n
X
i,j=1
aij (x, y)Dij v0=k0|∇v0|2+1
k0
p(x, y)e(γ−1)k0v0in Ω
v0= 0 on ∂Ω,
(1.9)
where |∇v0|2=Pn
i,j=1 aij (x, y)Div0Djv0.
We will show that for an appropriately chosen k0, (1.9) has a positive solution
v0. Then (1.8) has a positive solution u0≥1 that will be a supersolution to (1.1).
From there the existence of a positive solution of (1.1) follows from a standard
procedure that approximates the solution by solutions on a sequence of bounded
domains.
The proof of the existence of a positive solution of (1.9) is very similar to that of
the existence of the supersolution in [8]. The main difference is that in [8], a super
solution is constructed directly for (1.1) while in this paper, a super solution is
constructed through a solution of (1.9). Therefore, the proofs here will be parallel
to that in [8] except we need to make some necessary changes to deal with different
equations. We give full details of the proofs here so that the paper is self contained
and convenient for readers to follow the argument. However for some technical
constructions we still refer readers to the paper [8].
2. A Family of Auxiliary Functions
In this section, we use a family of auxiliary functions constructed in [9] to con-
struct families of sub-domains Ωx0of SM, constants Tx0and functions zx0(see
definitions below) such that Tx0+zx0satisfies
−
n
X
i,j=1
aij (x, y)Dij (Tx0+zx0)
≥k0|∇(Tx0+zx0)|2+1
k0
p(x, y)e(γ−1)k0(Tx0+zx0)in Ωx0∩Ω
(2.1)
and the graphs of the functions Tx0+zx0have special relative positions (see below).
We first extend aij (1 ≤i, j ≤n) to be continuous functions on SMin such a
way that we still have T race(aij) = 1 and
ann(x, y)≥c1in SM.(2.2)
It was proved in [9] (the construction of the functions was inspired by [13] and
the details of the construction were also repeated in the appendix in [8]) that there
are positive decreasing functions χ(t), ha(t) and a positive increasing function A(t)
(χ(t) depending on c1only, ha(t) and A(t) depending on c1and Monly), such that
4 Z. JIN EJDE-2005/109
for any number K, there is a number H0, depending only on K,Mand c1, such
that for H≥H0, we have (for 0 < t < 2M)
A(H)≤h−1
a(t)≤A(H)eχ(H),22MH ≤c1A(H)eχ(H)≤66MH, (2.3)
8K≤A(H)eχ(H),0< χ(H)<1,(2.4)
and the non-negative function
z=zx0=A(H)eχ(H)− {(h−1
a(y+M))2− |x−x0|2}1/2(2.5)
satisfies
n
X
i,j=1
aij (x, y)Dij z≤−3c1
22eMH in Ωx0,H,K ,(2.6)
z≥Kon ∂Ωx0,H,K ∩ {|y|< M}, z(x0, y )≤2M
Hfor |y| ≤ M, (2.7)
|Dxz(x, y)| ≤ 2( c1K
M)1/21
√H,|Dyz(x, y)| ≤ 2
Hon Ωx0,H,K ,(2.8)
where
Ωx0,H,K =(x, y) : |y|< M, |x−x0|<s2K
A(H)eχ(H)h−1
a(y+M).(2.9)
Now we set
K= 100, H =H0+ 4M, c2=3c1
22eMH ,Ωx0= Ωx0,H,K .(2.10)
Then from (2.7), we have
z≥100 on ∂Ωx0∩ {|y|< M}, z(x0, y )≤1 for |y| ≤ M. (2.11)
For two points x0and x1in Rn−1, when Ωx1either covers the whole segment of
the set {(x0, y)||y| ≤ M}or does not intersect with the set, from (2.3) and (2.9),
we have either
|x1−x0| ≤ q200A(H)e−χ(H)or |x1−x0| ≥ q200A(H)eχ(H).(2.12)
When Ωx1covers part of the set {(x0, y) : |y| ≤ M}, we have
q195A(H)e−χ(H)≤ |x1−x0| ≤ q205A(H)eχ(H).(2.13)
Let x1and x0satisfy (2.13) and δ0be a small positive number such that 2δ0<
p195A(H)e−χ(H). If (x, y)∈Ωx1for some yand |x−x0| ≤ δ0, by (2.3), (2.5) and
(2.13), we have
zx1(x, y)
≥A(H)eχ(H)− {A(H)2e2χ(H)− |x−x1|}1/2
≥A(H)eχ(H)− {A(H)2e2χ(H)−(q195A(H)e−χ(H)−δ0)2}1/2
≥A(H)eχ(H)− {A(H)2e2χ(H)−195A(H)e−χ(H)+ 2δ0q195A(H)e−χ(H)}1/2
≥A(H)eχ(H)(1 −(1 −195
A(H)e3χ(H)+2δ0p195A(H)e−χ(H)
A(H)2e2χ(H))1/2)
EJDE-2005/109 DIRICHLET PROBLEMS 5
(by the inequality √1−t≤1−1
2tfor 0 < t < 1 and (2.4))
≥A(H)eχ(H)(195
2A(H)e3χ(H)−2δ0p195A(H)e−χ(H)
2A(H)2e2χ(H))
=195
2e2χ(H)−δ0p195A(H)e−χ(H)
A(H)eχ(H)>10 −δ0p195A(H)e−χ(H)
A(H)eχ(H).
Thus there is a δ0small such that for all |x−x0| ≤ δ0with (x, y)∈Ωx1, if x1and
x0satisfy (2.13), we have
zx1(x, y)≥8.(2.14)
From (2.9) and (2.11), we can choose a number δ2(x0)>0 such that for all x∈Rn−1
with |x0−x| ≤ δ2(x0), we have (x, y)∈Ωx0for all |y|< M , and
zx0(x, y)≤2.(2.15)
Now if we set δx0= min{δ0, δ2(x0)}, from (2.14) and (2.15), we have
zx0(x, y)≤2<8≤zx1(x, y) (2.16)
for all x0and x1satisfying (2.13), |x0−x| ≤ δx0and (x, y)∈Ωx1.
Since Trace(aij ) = 1, we have Paij ξiξj≤1 for any unit vector ξ= (ξ1, . . . , ξn).
Then from (2.8), we have
|∇z|2=Xaij zizj≤ |Dxz|2+|Dyz|2
≤400c1
MH +4
H2
≤400c1
MH +1
HM =400c1+ 1
MH =c3.
(2.17)
Here we have used that H > 4M. If we set k0=c2
2c3, then the function zx0satisfies
that on Ωx0
k0|∇z|2≤c2
2c3
c3=1
2c2.(2.18)
Now we set c4=p205A(H)eχ(H),c5=p200A(H)eχ(H),α0=1
c4k0and assume
that for some constant C,p(x, y) satisfies (1.6). Furthermore we set
Tx0=1
c4
(|x0|+A)
where A=c5−c4
(1−γ)k0ln(min{c2k0
2C,1}). Since on Ωx0,|x| ≤ |x0|+c5, from (1.6),
we have that on Ωx0∩Ω,
1
k0
p(x, y)e(γ−1)k0(Tx0+z)≤C
k0
e
1
c4k0(1−γ)|x|e(γ−1)k0Tx0
≤C
k0
e
1
c4k0(1−γ)(|x0|+c5)e(γ−1)k0
1
c4(|x0|+A)
=C
k0
e
1
c4k0(1−γ)c5e(γ−1)k0
1
c4A≤1
2c2
(2.19)
by the definition of A. Combining (2.6), (2.18) and (2.19), we have that on Ωx0∩Ω,
Tx0+zsatisfies
k0|∇(Tx0+z)|2+1
k0
p(x, y)e(γ−1)k0(Tx0+z)≤c2≤ −
n
X
i,j=1
aij (x, y)Dij (Tx0+z).
6 Z. JIN EJDE-2005/109
That is, Tx0+zx0satisfies (2.1). As x0varies, we get a family of such functions.
Now for all x0and x1satisfying (2.13),
Tx0=Tx1+Tx0−Tx1
=Tx1+1
c4
(|x0|−|x1|)
≤Tx1+1
c4|x0−x1| ≤ Tx1+ 1.
Then for all x0and x1satisfying (2.13), |x0−x| ≤ δx0and (x, y)∈Ωx1, from
(2.16), we have
Tx0+zx0(x, y)≤Tx1+1+2< Tx1+zx0(x, y).(2.20)
Finally we define a family of open subsets of Ω that will be needed in next section.
For each point (x0, y0)∈Ω, we define an open set O(x0, y0) as follows:
1) If (x0, y0)∈Ω, we choose a ball Bwith center (x0, y0) and a radius less than
δx0so that B⊂Ω. We then set O(x0, y0) = B;
2) If (x0, y0)∈∂Ω, since Ω has C2,α boundary, there is a ball Bwith center (x0, y0)
and a radius less than δx0, such that there is a C2,α diffeomorphism Φ that satisfies
Φ(B∩Ω) ⊂Rn
+,Φ(B∩∂Ω) ⊂∂Rn
+; Φ(x0, y0) = 0.
Now we choose a domain Jwith C3boundary with following properties: (a)
J⊂Φ(B∩Ω); (b) ∂J ∩∂Rn
+is a neighborhood of 0in ∂Rn
+. Certainly there are
many different J’s having those properties. One example on how to construct Jis
given in the Appendix II [8].
Now we set O(x0, y0)=Φ−1(J). It is easy to see that O(x0, y0)⊂B∩Ω,
O(x0, y0) has a C2,α boundary and ∂O(x0, y0)∩∂Ω is a neighborhood of (x0, y0)
in ∂Ω.
Let Π be the collection of all such open sets O(x0, y0) defined in 1) and 2).
3. A Solution of (1.9)
In [12], a modified version of the Perron’s method has been used to prove the
existence of solutions. In the modified version of the Perron’s method, one uses a
family of local upper barriers to replace the role played by the supersolution in the
normal version of the Perron’s method (the local upper barriers in [12] were inspired
by [5]). The modified version has been used in [8] to prove the existence of a super
solution by using a family of auxiliary functions similar to the one constructed in
section 2. In this section, we will follow the same part in [8] with the necessary
modifications. We will show that there is a positive function v0∈C2(Ω) ∩C0(Ω),
satisfies
−
n
X
i,j=1
aij (x, y)Dij v0=k0|∇v0|2+1
k0
p(x, y)e(γ−1)k0v0in Ω,(3.1)
v0= 0 on ∂Ω.(3.2)
Let v≥0 be a continuous function on Ω, for a point (x0, y0)∈Ω, we define a
new function M(x0,y0)(v), called the lift of vover O(x0, y0) as follows:
M(x0,y0)(v)(x, y) = (v(x, y) if (x, y)∈Ω\O(x0, y0)
w(x, y) if (x, y)∈O(x0, y0)
EJDE-2005/109 DIRICHLET PROBLEMS 7
where w(x, y) is the non-negative solution of the boundary-value problem
−
n
X
i,j=1
aij (x, y)Dij w=k0|∇w|2+1
k0
p(x, y)e(γ−1)k0win O(x0, y0),(3.3)
w=von ∂O(x0, y0).(3.4)
We claim that this system has a solution in C2(O(x0, y0)) ∩C0(O(x0, y0)), which
is positive and unique. Indeed the uniqueness and positivity of a solution easily
follow from a standard maximum principle. For existence, we notice that m1=
min{v(x, y)|(x, y)∈∂ O(x0, y0)}is a sub-solution, m2+Tx0+zx0is a super solution
by (2.1), where m2= max{v(x, y)|(x, y)∈∂O(x0, y0)}. We set a change of variable
u=ek0w.
Then wsatisfies (3.3)-(3.4) if and only if usatisfies
−
n
X
i,j=1
aij (x, y)Dij u=p(x, y)uγin O(x0, y0),(3.5)
u=ek0won ∂O(x0, y0).(3.6)
Then ek0m1and ek0(m2+Tx0+zx0)are sub- and super- solutions of (3.5)-(3.6). It is
then well known that (3.5)-(3.6) has a positive solution u(see [2]). Then w=1
k0ln u
solves (3.3)-(3.4).
We define a class Ξ of functions as follows: A function vis in Ξ if
(1) v∈C0(Ω), v≥0 on Ω and v≤0 on ∂Ω;
(2) For any (x0, y0)∈Ω, v≤M(x0,y0)(v);
(3) v≤Tx0+zx0on Ωx0∩Ω for any (x0, y0)∈Ω.
An application of a maximum principle implies that the function v= 0 is in Ξ.
Thus Ξ is not empty. Now we set
v0(x, y) = sup
v∈Ξ
v(x, y),(x, y)∈Ω.
We will show that v0is positive on Ω, in C2(Ω) ∩C0(Ω) and satisfies (3.1)-(3.2).
First we present some lemmas.
Lemma 3.1. Let Dbe a bounded domain, If w1,w2are in C2(D)∩C0(D),w1≤w2
on ∂D, and
−
n
X
i,j=1
aij (x, y)Dij w1≤k0|∇w1|2+1
k0
p(x, y)e(γ−1)k0w1in D,
−
n
X
i,j=1
aij (x, y)Dij w2≥k0|∇w2|2+1
k0
p(x, y)e(γ−1)k0w2in D
then w1≤w2on D.
Since 1
k0p(x, y)e(γ−1)k0tis decreasing on t, a straightforward application of a
maximum principle to w1−w2gives the proof of the above lemma.
Lemma 3.2. If 0< v1≤v2, then M(x0,y0)(v1)≤M(x0,y0)(v2)for any (x0, y0)∈Ω.
8 Z. JIN EJDE-2005/109
Proof. Let w1,w2be the positive solutions for the following problems, respectively,
−
n
X
i,j=1
aij (x, y)Dij wq=k0|∇wq|2+1
k0
p(x, y)e(γ−1)k0wqin O(x0, y0),
wq=vqon ∂O(x0, y0), q = 1,2.
Since w1=v1≤v2=w2on ∂O(x0, y0), from lemma 1, we see w1≤w2on
O(x0, y0). However, on Ω \O(x0, y0), M(x0,y0)(v1) = v1,M(x0,y0)(v2) = v2. Thus
M(x0,y0)(v1)≤M(x0,y0)(v2).
Lemma 3.3. If v1∈Ξ,v2∈Ξ, then max{v1, v2} ∈ Ξ.
Proof. If v1∈Ξ, v2∈Ξ, it is clear that max{v1, v2} ∈ C0(Ω), max{v1, v2} ≥ 0 on
Ω and max{v1, v2} ≤ 0 on ∂Ω. It is also clear that max{v1, v2} ≤ Tx0+zx0on
Ωx0∩Ω for any (x0, y0)∈Ω. Since
v1≤max{v1, v2}, v2≤max{v1, v2},
by lemma 2 we have that for any (x0, y0)∈Ω,
M(x0,y0)(v1)≤M(x0,y0)(max{v1, v2}),
M(x0,y0)(v2)≤M(x0,y0)(max{v1, v2}).
Since v1∈Ξ and v2∈Ξ imply
v1≤M(x0,y0)(v1),and v2≤M(x0,y0)(v2),
we have
max{v1, v2} ≤ M(x0,y0)(max{v1, v2}).
Thus max{v1, v2} ∈ Ξ.
Lemma 3.4. If v∈Ξ, then M(x0,y0)(v)∈Ξfor any (x0, y0)∈Ω.
Proof. By the definition of M(x0,y0)(v), it is clear that M(x0,y0)(v)≥0 on Ω,
M(x0,y0)(v)∈C0(Ω) and M(x0,y0)(v)≤0 on ∂Ω. For any (x∗, y∗)∈Ω, we first
show that
M(x0,y0)(v)(x, y)≤M(x∗,y∗)(M(x0,y0)(v))(x, y ).(3.7)
We only need to prove (3.7) for (x, y)∈O(x∗, y∗).
Since v≤M(x0,y0)(v), by lemma 2 we have
M(x∗,y∗)(v)≤M(x∗,y∗)(M(x0,y0)(v)).
Then from v≤M(x∗,y∗)(v), we have
v≤M(x∗,y∗)(M(x0,y0)(v)).
Thus for (x, y)∈O(x∗, y∗)\O(x0, y0),
M(x0,y0)(v)(x, y) = v(x, y)≤M(x∗,y ∗)(M(x0,y0)(v))(x, y).(3.8)
That is, (3.7) is true on O(x∗, y∗)\O(x0, y0), Now for Ω1=O(x∗, y∗)∩O(x0, y0),
if we set
M(x0,y0)(v) = w1, M(x∗,y∗)(M(x0,y0)(v)) = w2
EJDE-2005/109 DIRICHLET PROBLEMS 9
then
−
n
X
i,j=1
aij (x, y)Dij w1=k0|∇w1|2+1
k0
p(x, y)e(γ−1)k0w1on Ω1,
−
n
X
i,j=1
aij (x, y)Dij w2=k0|∇w2|2+1
k0
p(x, y)e(γ−1)k0w2on Ω1.
On ∂Ω1,w1≤w2on O(x∗, y∗)∩∂O(x0, y0) by (3.8) and w1≤w2on ∂O(x∗, y∗)∩
O(x0, y0) since (3.7) is true on Ω \O(x∗, y∗). Then lemma 1 implies w1≤w2on
Ω1. Thus (3.7) is true on O(x∗, y∗)∩O(x0, y0) and on O(x∗, y ∗).
Now we prove that M(x0,y0)(v)≤Tx1+zx1on Ωx1∩Ω for all (x1, y1)∈Ω. By
the definition of M(x0,y0)(v), we only need to consider the graph of the function
M(x0,y0)(v) over O(x0, y0). If O(x0, y0) is covered completely by Ωx1, since v≤
Tx1+zx1and Tx1+zx1satisfies (2.1), Tx1+zx1is a super solution of (3.3) on
O(x0, y0). Then Lemma 1 implies M(x0,y0)(v)≤Tx1+zx1on O(x0, y0). In the
case that O(x0, y0) does not intersect with Ωx1, the conclusion is trivial. Now we
consider the case that O(x0, y0) is partially covered by Ωx1. First by what we have
just proved, we always have
M(x0,y0)(v)≤Tx0+zx0on O(x0, y0).(3.9)
By the choice of δx0and O(x0, y0), the graph of Tx0+zx0over O(x0, y0)∩Ωx1is
under the graph of Tx1+zx1. Thus the conclusion follows from (3.9).
Now we are ready to prove that v0has the desired properties. Let (x0, y0)∈Ω.
By the definition of v0(x0, y0), there is a sequence of functions vkin Ξ such that
v0(x0, y0) = lim
k→∞ vk(x0, y0).
By the definition of Ξ, vk≥0 on Ω. We replace vkby M(x0,y0)(vk). Then we have
a sequence of functions wksuch that
v0(x0, y0) = lim
k→∞ wk(x0, y0),
−
n
X
i,j=1
aij (x, y)Dij wk=k0|∇wk|2+1
k0
p(x, y)e(γ−1)k0wkon O(x0, y0),
wk=vkon ∂O(x0, y0).
Then uk=ek0wksatisfies
−
n
X
i,j=1
aij (x, y)Dij uk=p(x, y)uγ
kon O(x0, y0),
uk=ek0vkon ∂O(x0, y0).
Further, from the fact that for all k,
0≤vk≤wk≤Tx0+zx0on O(x0, y0),
we have
1≤uk≤ek0(Tx0+zx0)on O(x0, y0).
By [6, Theorem 9.11] and an approximation of the boundary value by smooth
functions, we see that there is a subsequence of uk, for convenience still denoted by
10 Z. JIN EJDE-2005/109
uk, converges to a C2(O(x0, y0)) ∩C0(O(x0, y0)) function u∗(x) in C2(O(x0, y0)) ∩
C0(O(x0, y0)). Thus u∗(x) satisfies
−
n
X
i,j=1
aij (x, y)Dij u∗=p(x, y)uγ
∗on O(x0, y0)
Then w=1
k0ln u∗satisfies
−
n
X
i,j=1
aij (x, y)Dij w=k0|∇w|2+1
k0
p(x, y)e(γ−1)k0won O(x0, y0)
and v0(x0, y0) = w(x0, y0). We claim that v0=won O(x0, y0). Indeed, if there
is another point (x2, y2)∈O(x0, y0) such that v0(x2, y2) is not equal to w(x2, y2),
then v0(x2, y2)> w(x2, y2). Then there is a function v∗∈Ξ, such that
w(x2, y2)< v∗(x2, y2)≤v0(x2, y2).
Now the sequence max{v∗, M(x0,y0)(vk)}satisfies
vk≤max{v∗, M(x0,y0)(vk)} ≤ v0.
In a similar way, M(x0,y0)(max{v∗, M(x0,y0)(vk)}) will produce a function w1such
that
−
n
X
i,j=1
aij (x, y)Dij w1=k0|∇w1|2+1
k0
p(x, y)e(γ−1)k0w1on O(x0, y0),
w≤w1on O(x0, y0),
w(x2, y2)< v∗(x2, y2)≤w1(x2, y2),
w(x0, y0) = w1(x0, y0) = v0(x0, y0).
That is, w1(x, y)−w(x, y) is non-negative, not identically zero on O(x0, y0) and
achieves its minimum value zero inside O(x0, y0). However, from the equations
satisfied by wand w1, we have that on O(x0, y0),
−
n
X
i,j=1
ai,j (x, y)Dij (w1−w)−2k0(∇w+θ1∇(w1−w)) · ∇(w1−w)
−(γ−1)p(x, y)e(γ−1)k0(w+θ2(w−w1))(w−w1) = 0
for some continuous functions θ1and θ2. Then by the standard maximum principle
(see [6, Theorem 3.5]), we get a contradiction. Thus v0=won O(x0, y0). Therefore
v0∈C2(Ω) and
−
n
X
i,j=1
ai,j (x, y)Dij v0=k0|∇v0|2+1
k0
p(x, y)e(γ−1)k0v0.
When (x0, y0)∈∂Ω, ∂O(x0, y0)∩∂Ω is a neighborhood of (x0, y0) in ∂Ω. Since
max{0, vk}= 0 on ∂Ω, v0= 0 on ∂Ω and w= 0 on ∂O(x0, y0)∩∂Ω. Since wis
continuous up to the boundary of O(x0, y0), v0is continuous on ∂O(x0, y0)∩∂Ω
from inside O(x0, y0). Thus v0∈C0(Ω) and v0= 0 on ∂Ω. Now from the definition
of Tx0and v≤Tx0+zx0on Ωx0for all v∈Ξ, we have
v0(x0, y)≤Tx0+zx0(x0, y)≤1
c4|x0|+A+ 3
EJDE-2005/109 DIRICHLET PROBLEMS 11
for all (x0, y)∈Ω. If we let x0vary, we get
v0(x, y)≤1
c4|x|+A+ 3 on Ω.(3.10)
Then u0=ek0v0satisfies (1.8), u0≥1 and u0≤c6ec7|x|for some constants c6and
c7depending only on n,Mand c1.
4. The Proof of Theorem 1
Since Ω is an unbounded domain with C2,α boundary, we can choose a sequence
of subdomains in Ω, denoted by Ωm,m= 1,2,3, . . . , such that
(1) Ωm⊂Ωm+1 ⊂Ω for all m;
(2) ∪Ωm= Ω;
(3) Each Ωmis a bounded domain with C2,α boundary;
(4) dist(0, ∂Ω\∂Ωm)→ ∞ as m→ ∞.
We can find a number µ, such that the eigenvalue problem
−
n
X
i,j=1
aij (x, y)Dij φ=λ(µp(x, y))φon Ωm
φ= 0 on ∂Ωm
has a first eigenvalue λ1(m)<1 with eigenfunction φm. We assume max φm= 1.
Let δbe a number such that 0 < δ ≤1/2 and t≤tγfor 0 < t < δ. Let u0≥1 be
a positive solution to (1.8) with p(x, y) replaced by µp(x, y). Then
−
n
X
i,j=1
aij (x, y)Dij w=µp(x, y)wγon Ωm
w= 0 on ∂Ωm
(4.1)
has a pair of super and sub solutions u0(x, y), δφm, and u0(x, y)≥1≥δφm. Thus
(4.1) has a positive solution wm. Since wmis also a super solution of the equation
in (4.1) on Ωsfor all m>s,δφsis a subsolution of the equation in (4.1) on Ωs,
δφs= 0 ≤wmon ∂Ωs, we have
δφs≤wmon Ωs(4.2)
for all m>s. We also have wm≤u0on Ωs. Therefore a subsequence of wm
will converge to a positive C2(Ω) function uthat satisfies the equation in (1.1).
Furthermore usatisfies (1.7) since u0satisfies (1.7). Now we still need to prove
that u∈C0(Ω) and u= 0 on ∂Ω.
If (x0, y0)∈∂Ω, we let O(x0, y0)∈Π be the C2,α domain chosen in section 2.
Then ∂O(x0, y0)∩∂Ω is a neighborhood of (x0, y0) in ∂Ω. Now we choose a C1
function ψon ∂O(x0, y0) so that ψ≥0, ψ= 0 in a neighborhood of (x0, y0)∈
∂O(x0, y0) and ψ≥u0on ∂O(x0, y0)∩Ω. Let w0∈C0(O(x0, y0)) ∩C2(O(x0, y0))
be the solution of the problem
−
n
X
i,j=1
aij (x, y)Dij w0=µp(x, y)wγ
0on O(x0, y0),
w0=ψon ∂O(x0, y0).
(4.3)
Then for all mwith O(x0, y0)⊂Ωm, from wm≤u0≤ψon ∂O(x0, y0), we have
wm≤w0on O(x0, y0).
12 Z. JIN EJDE-2005/109
Thus combining this with (4.2), we have that for large fixed s,
δφs≤u≤w0on O(x0, y0).
Since w0and δψsare in C0(O(x0, y0)) and w0(x0, y0) = δψs(x0, y0) = 0, uis
continuous near (x0, y0) and u(x0, y0) = 0. Since (x0, y0)∈∂Ω can be arbitrary,
we get u∈C0(Ω) and u= 0 on ∂Ω.
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Zhiren Jin
Department of Mathematics and Statistics, Wichita State University, Wichita, Kansas,
67260-0033, USA
E-mail address:zhiren@math.wichita.edu
