Nonlinear Analysis 41 (2000) 501 – 521
Linear and semilinear eigenvalue problems
in exterior domains
Ru-Ying Xue ∗ , Yu-Chun Qin
Department of Mathematics, Hangzhou University, Hangzhou 310028, Zhejiang,
People’s Republic of China
Received 1 December 1997; accepted 3 May 1998
Keywords: Positive solution; Eigenvalue problem; Compact operator; Harnack inequality; Kelvin
transform
1. Introduction
In this paper, we will investigate the existence and asymptotic properties of positive
eigenfunctions of the problem
−u = f(x)u;
u=0
x ∈ ;
for x ∈ @
;
u(x) → 0
as |x| → ∞;
(1.1)
where ¿0 and the corresponding semilinear problem
−u = f(x)u + f(x)u ;
u=0
for x ∈ @
;
x ∈ ;
u(x) → 0
as |x| → ∞;
(1.2)
where is an exterior domain in Rn (n ≥ 3) with smooth boundary @
, 0 ∈ Rn − ,
(
) for some 0¡¡1, and satises
¿0
and 0¡¡1, f : → R is positive and Cloc
R
2−n
|x|
f(x)
dx¡+
∞.
More
precisely,
we
assume
that there exists a positive, radially
symmetric function P(|x|) dened in Rn , which satises
Z
|x| 2−n P(|x|) dx¡+∞:
(1.3)
f(x) ≤ P(|x|) for all x ∈ ;
∗
Projects supported by the Natural Science Foundation of China.
Corresponding author.
0362-546X/00/$ - see front matter ? 2000 Elsevier Science Ltd. All rights reserved.
PII: S 0 3 6 2 - 5 4 6 X ( 9 8 ) 0 0 2 9 3 - 4
502
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
Problems (1:1) and (1:2) have been considered by several authors (see [1–5, 8, 10]).
When = Rn , Edelson and Rumbos [5, 10] have shown that Eq. (1.1) has a positive,
principal eigenvalue and a corresponding positive, principal eigenfunction which
satises the asymptotic law
lim |x|n−2 (x) = C¿0
|x|→+∞
if P(|x|) satises (1.3), and that Eq. (1.2) has a positive solution, u, satisfying
lim |x|n−2 u(x) = C¿0
|x|→+∞
if P(|x|) satises
Z
|x|(2−n) P(|x|) dx¡+ ∞
Rn
R∞
and satises 0¡¡(n − 2)=( 0 sP(s) dx). When is an exterior domain, Edelson
[3] has shown that problem (1:2), without the condition u|@
= 0, has a positive solution satisfying the asymptotic law lim|x|→+∞ |x|n−2 u(x) = C¿0 if 0¡f(x) ≤ P(|x|) and
R
|x|(2−n) P(|x|) dx¡+ ∞. Those results are based on the classical integral operator
Rn
equation in Rn and the Green function dened in a ball (see [3–5, 8, 10]).
By a minimal solution of Eq. (1.1) or Eq. (1.2) we mean a solution which satises
|u(x)|
¡+ ∞:
(1
+
|x|) 2−n
x∈
sup
(1.4)
The main results of this paper show the existence of positive minimal solutions of
Eq. (1.1) or Eq. (1.2) under appropriate conditions. Our method is of interest, because it
does not depend on the fundamental solution of the Laplacian and the classical integral
operator equation in Rn , so it allows us to consider more general elliptic boundary value
problems.
Our approach is via the Harnack inequality and the barrier method. In Section 2, we
dene a norm for functions satisfying (1.4), state an operator associated with Eqs. (1.1)
and (1.2), and prove that the operator is compact. The existence of minimal solutions
of Eq. (1.1) is obtained in Section 3. In Section 4 we prove the existence of minimal
solutions of Eq. (1.2).
In the rest of this paper, C; C0 ; C1 ; : : : denote (possibly dierent) positive numbers.
2. Preliminaries
Throughout this section we assume P(|x|) satises (1.3). We begin by dening a
weighted L 2 space and proving some embedding results. Denote by L 2 (
; P) the class
of real-valued Lebesgue measurable functions, u, satisfying
Z
P(|x|)|u(x)| 2 dx¡∞:
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
503
Let D01; 2 (
) be the completion of the space C0∞ (
) with respect to the norm
Z
1=2
|∇u| 2 dx
:
kuk0 =
Obviously, L 2 (
; P) and D01; 2 (
) are two Hilbert spaces.
Lemma 2.1. If P(|x|) satisÿes (1.3), n ≥ 3, then the embedding D01; 2 (
) ,→L 2 (
; P)
is completely continuous.
The proof of Lemma 2.1 is analogous to those of Lemmas 1.1–1.3 in [5], we
omit it.
Denote by E the class of real-valued continuous functions, u(x), satisfying
kuk = sup (1 + |x|)n−2 |u(x)|¡+ ∞:
x∈
The set E, together with the norm k : k, constitutes a Banach space.
(
) for some 0¡¡1. We also
Lemma 2.2. Assume f : → R is positive and Cloc
assume that there is a radially symmetric function, P(|x|), satisfying (1.3) and f(x) ≤
P(|x|) for all x ∈ . Then, for h(x) ∈ E, there exists a unique solution, u(x), of the
problem
−u = f(x)h(x);
u|@
= 0
x ∈ ;
u(x) → 0 as |x| → ∞:
(2.1)
1+
(
) for some ∈ (0; 1) and kuk ≤ Ckhk, where
Moreover, u(x) ∈ E ∩ D01; 2 (
) ∩ Cloc
C is a positive constant depending solely on and P(|x|).
Proof. (i). we shall prove that Eq. (2.1) has a solution u(x) ∈ D01; 2 (
). Without loss
of generality, we assume h(x) 6≡ 0. Consider
Z
Z
2
|∇u| dx f(x)h(x)u dx = 1; u ∈ D01; 2 (
) :
J = inf
1; 2
Obviously, 0¡J ¡+ ∞. Let {uk }∞
k=1 be a sequence in D0 (
) satisfying
Z
Z
|∇uk | 2 dx → J and
f(x)h(x)uk dx = 1 as k → ∞:
(2.2)
Since D01; 2 (
) is a real Hilbert space, there are a subsequence of {uk }∞
k=1 (still denote
1; 2
)
and
an
element
u
∈
D
(
)
such
that
by {uk }∞
0
k=1
0
uk → u0
weakly in
uk → u0
in L 2 (
; P)
D01; 2 (
)
as k → ∞;
thus,
as k → ∞
(2.3)
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R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
by Lemma 2:1. It follows from (2.3) that
Z
f(x)h(x)(uk − u0 ) dx
Z
≤ khk
P(|x|)(1 + |x|)
2−n
1=2 Z
dx
2
1=2
P(|x|)|uk − u0 | dx
≤ Ckuk − u0 kL 2 (
; P) → 0
as k → ∞. Hence,
Z
|∇u0 | 2 dx = J
Z
and
f(x)h(x)u0 dx = 1:
Thus, u0 minimizes J , the Euler–Lagrange principle implies u0 is a D01; 2 (
)-solution
of Eq. (2.1).
(ii). We prove that the D01; 2 (
)-solution, u0 , obtained in (i) belongs to E. Set
Z
1
khk(1 + |y|) 2−n |x − y| 2−n P(|y|) dy;
1 (x) =
(n − 2)nwn Rn
where wn is the volume of the unit ball in Rn . It is easy to prove that 1 (x) ∈ D01; 2 (Rn ),
1 ¿0 in Rn , k1 k ≤ Ckhk, and 1 (x) satises
−1 (x) = khk(1 + |x|) 2−n P(|x|);
x ∈ Rn ;
sup 1 (x)(1 + |x|)n−2 ¡+ ∞;
(2.5)
Rn
(see [3]). Set w = u0 − 1 when x ∈ and w = −1 (x) when x ∈ Rn − , w = w+ − w−
where w+ = max{w; 0} and w− = max{−w; 0}. We get w ∈ D01; 2 (Rn ) and W + ∈ D01; 2 (
)
because of w+ = 0 for x ∈ Rn − . Multiply Eqs. (2.5) and (2.1) by w+ , integrate over
and apply integration by parts, thus obtain
Z
Z
+
∇1 ∇w dx = khk(1 + |x|) 2−n P(|x|)w+ dx;
Z
∇u0 ∇w+ dx =
Z
f(x)h(x)w+ dx:
Subtracting the rst equation from the second one, we get
Z
∇(u0 − 1 )∇w+ dx
Z
=
[f(x)h(x) − khk(1 + |x|) 2−n P(|x|)]w+ dx ≤ 0:
(2.6)
Now set = + ∪ − , where + = {x ∈ | u0 ¿1 } and − = {x ∈ | u0 ≤ 1 }. Note
that w+ = u0 − 1 and ∇w+ = ∇(u0 − 1 ) in + , w+ = 0 and ∇w+ = 0 in − . We can
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
505
rewrite (2.6) as follows:
Z
|∇(u0 − 1 )| 2 dx
0≤
Z
=
+
+
[f(x)h(x) − khk(1 + |x|) 2−n P(|x|)](u0 − 1) dx ≤ 0;
which implies meas(
+) = 0, that is, u0 ≤ 1 a.e. in . Similarly, we can prove that
u0 ≥ − 1 a.e. in , hence khk ≤ k1 k ≤ Ckhk: Elliptic regularity arguments can then
1+
(
) for some ∈ (0; 1).
be used to show u0 ∈ Cloc
(iii) We prove that problem Eq. (2.1) has a unique (D01; 2 (
) ∩ E)-solution. Suppose
u and v are two (D01; 2 (
) ∩ E)-solutions of Eq. (2.1). Then u − v satises
−(u − v) = 0;
(u − v)|@
= 0;
x ∈ ∩ {|x| ≤ R};
(u − v)||x|=R = u||x|=R − v||x|=R ;
(2.7)
where R is a positive constant large enough. It follows from the weak maximum
principle for Eq. (2.7) that
|u − v| ≤ max |u − v| ≤ (1 + R) 2−n (kuk + kvk)
|x|=R
(2.8)
for all x ∈ ∩ {|x| ≤ R}. Taking R → ∞ in (2.8) we get u ≡ v. We complete the
proof.
For h(x) ∈ E, we dene the mapping T : h → u0 = T [h], where u0 is a solution ob1+
(
)
tained in Lemma 2.2. It is obvious that by Lemma 2.2, T : E → E ∩ D01; 2 (
) ∩ Cloc
is a linear operator satisfying kT [h]k ≤ Ckhk, where C is a positive constant depending
on and P(|x|).
Remark 2.1. We can prove that T is a monotone mapping, i.e. h ≤ h0 in E implies
T [h] ≤ T [h0 ]. In fact, T [h] − T [h0 ] is a solution of the problem
−u = f(x)(h0 − h) ≥ 0;
u|@
= 0;
x∈
u||x|=R = (T [h0 ] − T [h])||x|=R :
The weak maximum principle implies
0
0
T [h ] − T [h] ≥ min 0; inf (T [h ] − T [h])
|x|=R
≥ −Ckh0 − hk(1 + R) 2−n
for all x ∈ ∩ {|x| ≤ R}. Let R → ∞, we get T [h0 ] − T [h] ≥ 0 for all x ∈ .
Let {hk }∞
k=1 be a bounded sequence in E. Set M = sup1≤k¡∞ {khk k}. For j¿0,
choose Rj so large that
Rj ≥ sup{|x| + 1 | x ∈ }
506
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
and
1
n(n − 2)wn
Z
|x|≥Rj
P(|x|)(1 + |x|) 2−n dx ≤
1
:
8Mj
Choose a function j (x) ∈ C 1 (Rn ) satisfying 0 ≤
|x| ≤ Rj , j (x) = 0 for |x| ≥ Rj + 1.
(2.9)
j (x) ≤ 1
for all x ∈ ,
j (x) = 1
for
Lemma 2.3. There is R1j such that
|T [hk (1 −
j )](x)| ≤
1
(1 + |x|) 2−n
4j
(2.10)
for all |x| ≥ R1j ; k = 1; 2; : : : :
Proof. Set
2 (x) =
1
n(n − 2)wn
Z
Rn
M (1 −
j (y))|x
− y| 2−n P(|y|)(1 + |y|) 2−n dy:
It is easy to prove that 2 (x) satises (see [3] or [5])
−2 = (1 −
j )P(|x|)M (1
+ |x|) 2−n ;
x ∈ Rn ;
(2.11)
sup (1 + |x|)n−2 2 (x)¡+ ∞
x∈Rn
and
(1 + |x|)n−2 2 (x) →
1
n(n − 2)wn
Z
Rn
M (1 −
j (x))(1
+ |x|) 2−n P(|x|) dx
(2.12)
as |x| → ∞. By (2.9) and (2.12), we can choose R1j so large that
Z
1
1
+
M (1 − j (x))(1 + |x|) 2−n P(|x|) dx
0 ≤ (1 + |x|)n−2 2 (x) ≤
8j n(n − 2)wn Rn
¡
1
1
1
+
=
8j 8j 4j
(2.13)
for all |x| ≥ R1j . The inequality f(x)|hk |(1 − j ) ≤ MP(|x|)(1 + |x|) 2−n (1 −
that (use the same proof as that used in Lemma 2.2(ii)),
|T [hk (1 −
j )](x)| ≤ 2 (x)
for all x ∈ :
Hence,
|T [hk (1 −
j )](x)| ≤ |2 (x)|¡
1
(1 + |x|) 2−n
4j
for all |x| ≥ R1j :
j)
implies
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
507
Lemma 2.4. There exist a constant, C1 , depending on P(|x|); Rj ; and M , and another constant, C0 , C0 ≥ 1, depending solely on n, such that
0≤
max
(1 + |x|)n−2 T [hk j ] −
|x|≥2l (Rj +1)
≤ C1
C0 − 1
C0 + 1
inf
(1 + |x|)n−2 T [hk j ]
|x|≥2l (Rj +1)
l
;
(2.14)
l = 1; 2; : : : ; k = 1; 2; : : : .
Proof. Let vk (x) = T [hk j ](x) and vbk (x) = |x| 2−n vk (x=|x| 2 ), the Kelvin transform of
vk (x), for all x ∈ , set
Z
1
M j (y)|x − y| 2−n P(|y|)(1 + |y|) 2−n dy:
3 (x) =
n(n − 2)wn Rn
The same argument as that used in Lemma 2.2(ii) shows
|vk (x)| ≤ 3 (x) ≤ C(1 + |x|) 2−n :
(2.15)
Obviuosly, vbk (x) satises
−vbk (x) = 0;
|vbk (x)| ≤ C
for all 0¡|x| ≤ 1=(Rj +1). By Theorem 2.69 in [6], we know that vbk (x) has a removable
singularity at 0, that is, vbk (x) is harmonic in |x| ≤ 1=(Rj + 1). Let
Ml = max{vbk (x)| |x| ≤ 2−l (1 + Rj )−1 };
ml = inf {vbk (x)| |x| ≤ 2−l (1 + Rj )−1 }
for l = 1; 2; : : : . Note that Ml−1 − vbk (x) and vbk (x) − ml−1 are harmonic and nonnegative
in |x| ≤ 2−l (1 + Rj )−1 . The Harnack inequality (see Theorem 8.20 in [7]) implies
Ml−1 − ml = max{Ml−1 − vbk (x)| |x| ≤ 2−l (1 + Rj )−1 }
≤ C0 inf {Ml−1 − vbk (x)| |x| ≤ 2−l (1 + Rj )−1 }
≤ C0 (Ml−1 − Ml )
and
Ml − ml−1 = max{vbk (x) − ml−1 | |x| ≤ 2−l (1 + Rj )−1 }
≤ C0 inf {vbk (x) − ml−1 ||x| ≤ 2−l (1 + Rj )−1 }
≤ C0 (ml − ml−1 );
508
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
where C0 is a positive constant depending solely on n. Adding the second inquality to
the rst one, we arrive at
Ml − ml ≤
C0 − 1
(Ml−1 − ml−1 );
C0 + 1
l = 1; 2; : : : :
Thus,
C0 − 1
(Ml−1 − ml−1 ) ≤ · · ·
C0 + 1
l
l
C0 − 1
C0 − 1
(M0 − m0 ) ≤ C1
;
≤
C0 + 1
C0 + 1
Ml − ml ≤
hence (2.14) holds.
Let be a positive constant, and let be a C ∞ (
)-function satisfying 0 ≤ ≤ 1
for x ∈ , = 0 for x ∈ {x ∈ | dist(x; @
) ≤ } and = 1 for x ∈ {x ∈ | dist(x; @
) ≥
2}. Denote by
Z
Z
f(x) j (x)hk (x) dx −
∇T [hk j ](x)∇ dx:
A; k; j =
∞
Lemma 2.5. There is a subsequence of {hk }∞
k=1 (still denote by {hk }k=1 ) satisfying
1
|A; k; j − A; m; j |¡ ;
j
k; m = 1; 2; : : :
and
1
|T [ j hk ](x) − T [ j hm ](x)| ≤ (1 + |x|) 2−n
j
for |x| ≥ 2l0 (Rj + 1), k; m = 1; 2; : : : , where l0 is an integer satisfying
C1
C0 − 1
C0 + 1
l0
≤
1
:
4j
Proof. Let vk (r) denote the spherical mean of vk (x). We write (r; w) for polar coordinates in R n and denote by @r the radial derivative, that is, the normal derivative on
sphere about the origin,
@r u(x) = |x|−1
n
X
xi @i u;
i=1
or equivalently, if x = rw where w ∈ S n−1 , the unit sphere in R n ,
d
u(w) :
@r u(x) =
d
=r
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
509
Denote by vk (x) = T [hk j ](x). For ¿0 small enough, r ≥ Rj + 1, we dene (x) = 1
for |x| ≤ r, (x) = 0 for |x| ≥ r + , and (x) = ( + r − |x|)= for r ¡ |x| ¡ r + . Note
that j (x) (x) ≡ j (x) for x ∈ . Then, ∈ D01; 2 (
) and
Z
Z
∇vk ∇( ) =
f(x) j (x)hk (x) dx:
(2.17)
Taking → 0 in Eq. (2.17), we get, for r ≥ Rj + 1,
Z
Z
Z
1 r+ n−1
@r vk (rw) dw = lim −
d
@r vk (w) ds
− r n−1
→0
r
|w|=1
|w|=1
Z
1
@r vk (x) dx
= lim −
→0
r≤|x|≤r+
Z
∇vk ∇ dx = A; k; j :
= lim
→0
(2.18)
On the other hand, the fact that |vk (x)| ≤ C(1 + |x|) 2−n for x ∈ implies
Z ∞
@r vk (t w) dt for x = rw; w ∈ S n−1 :
vk (x) = vk (rw) = −
r
Integrating over the unit sphere S n−1 in R n , using Eq. (2.18), we get
Z
Z ∞ Z
1
1
vk (rw) dw = −
dt
@r vk (tw) dw
vk (r) =
nwn |w|=1
nwn r
|w|=1
Z
A; k; j
A; k; j ∞ 1−n
t
dt =
r 2−n for r ≥ Rj + 1:
=
nwn r
n(n − 2)wn
(2.19)
Inequality (2:14) implies
max vk (x) − vk (rw) ≤ max vk (x) − inf vk (x) ≤ C1
|x|=r
|x|=r
|x|=r
and
vk (rw) − inf vk (x) ≤ max vk (x) − inf vk (x) ≤ C1
|x|=r
|x|=r
|x|=r
C0 − 1
C0 + 1
C0 − 1
C0 + 1
l
l
(1 + r) 2−n
(1 + r) 2−n
for x = rw; w ∈ S n−1 ; r = |x| ≥ 2l (Rj + 1); k = 1; 2; : : : : Integrating on both sides over
the unit sphere S n−1 we obtain
l
C0 − 1
(1 + r) 2−n
max vk (x) − vk (r) ≤ C
C0 + 1
|x|=r
and
vk (r) − inf vk (x) ≤ C
|x|=r
C0 − 1
C0 + 1
l
(1 + r) 2−n
510
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
for r = |x| ≥ 2l (Rj + 1), k = 1; 2; : : : . Thus,
l
C0 − 1
(1 + r) 2−n ≤ inf vk (x) ≤ vk (x)
vk (r) − C
C0 + 1
|x|=r
C0 − 1
≤ max vk (x) ≤ vk (r) + C
C0 + 1
|x|=r
l
(1 + r) 2−n :
(2.20)
It is obvious that {A; k; j }∞
k=1 is a bounded sequence. There exists a subsequence of
∞
{hk }∞
k=1 (still denote by {hk }k=1 ) such that
|A; k; j − A; m; j |¡
1
2j
(2.21)
for k; m = 1; 2; : : : . (2.19)–(2.21) imply
|vk (x) − vm (x)|
≤ |vk (x) − vk (r)| + |vk (r) − vm (r)| + |vm (r) − vm (x)|
≤
1
1
(1 + |x|) 2−n +
|A; k; j − A; m; j |(1 + |x|) 2−n
2j
n(n − 2)wn
1
≤ (1 + |x|) 2−n
j
for |x| = r ≥ R2 = 2l0 (Rj + 1). We complete the proof.
By the uniqueness of the solution of problem (2:1) we have
T [hk ](x) = T [(1 −
j )hk ](x)
+ T [ j hk ](x):
Lemma 2.6. Denote by
Z
Z
f(x)hk (x) dx −
∇T [hk ]∇ dx:
B; k =
Then, we get
lim (1 + |x|) n−2 T [hk ](x) =
|x|→∞
1
B; k ;
n(n − 2)wn
k = 1; 2; : : : :
Proof. For any integer j, we have
Z
1
khk k
P(|x|)(1 + |x|) 2−n dx ≤
n(n − 2)wn |x|≥Rj
j
(2.22)
by (2.9),
|(1 + |x|) n−2 T [(1 −
j )hk ](x)| ≤
1
;
j
|x| ≥ R1j ;
k = 1; 2; : : :
(2.23)
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
by Lemma 2.3, and
T [ j hk ](x) −
511
l
1
C0 − 1
2−n A; k; j (1 + |x|) ≤ C
(1 + |x|) 2−n
n(n − 2)wn
C0 + 1
(2.24)
for |x| ≥ 2l (Rj + 1); k = 1; 2 : : : , by (2.19) and (2.20), where Rj ; R1j are numbers depending solely on P(|x|), supk khk k; j and , C is a constant depending solely on
P(|x|), supk khk k, j; Rj ; R1j and , j is the function dened in (2.10).
Obviously, T [(1 − j )hk ] is a solution of the problem
−u = f(x)hk (x)(1 −
u = 0;
j (x));
x∈
(2.25)
x ∈ @
:
By Lemma 2.2 and (2.14), T [(1 −
j )hk ](x)| ≤ C(1
|T [(1 −
j )hk ] ∈ E
∩ D01; 2 (
) and
+ |x|) 2−n :
(2.26)
Multiplying Eq. (2.25) by T [(1− j )hk ](x), integrating over and using Lemma 2.1
and (2.2), we get
Z
|∇T [(1 − j )hk ](x)| 2 dx
Z
=
fhk (1 −
j )T [(1
Z
≤
≤
1
2
≤
1
2
P(|x|)(T [(1 −
Z
Z
−
j )hk ](x) dx
2
j )hk ](x)) dx + C()
|∇T [hk (1 −
2
j )]| dx + C()
|∇T [hk (1 −
j )]|
2
dx + C()
Z
|x|≥Rj
Z
P(|x|)hk2 (1 −
j)
2
dx
P(|x|)khk k(1 + |x|) 2−n dx
1
;
4j
where is a positive constant small enough, C() is a positive constant depending
solely on . Hence,
Z
1
(2.27)
|∇T [hk (1 − j )]| 2 dx ≤ C() :
2j
A combination of (2.22)–(2.24) with (2.27) yields
B; k
(1 + |x|) n−2 T [hk ] −
n(n − 2)wn A; k; j 1
n−2
+
≤ (1 + |x|) T [ j hk ] −
n(n − 2)wn n(n − 2)wn
Z
1
|∇T [hk (1 − j )](x)| |∇ | dx
+
n(n − 2)wn Z
fk (1 −
j ) dx 512
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
+ |(1 + |x|) n−2 T [(1 − j )hk ](x)|
l
Z
C0 − 1
+ Ckhk k
P(|x|)(1 + |x|) 2−n dx
≤C
C0 + 1
|x|≥Rj
1=2
C()
1
+
2j
j
l
1=2
1=2 Z
khk k
1
C()
C0 − 1
2
+C
|∇ | dx
+
+
≤C
C0 + 1
j
2j
j
Z
+
|∇ | 2 dx
1=2 (2.28)
for |x| ≥ max{R1j ; 2l (Rj + 1)}, k = 1; 2; : : : . Taking j → ∞ and |x| → ∞ in (2.28), we
obtain
B; k
= 0;
lim (1 + |x|) n−2 T [hk ] −
n(n − 2)wn |x|→∞
that is,
lim (1 + |x|) n−2 T [hk ] =
|x|→∞
B; k
:
n(n − 2)wn
Remark 2.2. For ÿxed h ∈ E, we denote by
Z
Z
f(x)h (x) dx −
∇T [h]∇ dx:
C =
Choosing hk ≡ h for k = 1; 2; : : : ; from Lemma 2.6 we obtain
lim (1 + |x|) 2−n T [h](x) = C :
|x|→∞
Lemma 2.7. For any integer j, there are a subsequence of {hk }∞
k=1 (still denote by
{hk }∞
k=1 ) and a positive constant R2j , which dependes solely on ; P(|x|); j and
supk khk k, such that
1
|T [hk ](x) − T [hm ](x)| ≤ (1 + |x|) 2−n
j
for all |x| ≥ R2j , k; m = 1; 2; : : : .
Proof. By Lemma 2.3 there is a positive constant R1j , depending on j; P(|x|); and
supk khk k, such that
|T [hk (1 −
j )]|¡
1
(1 + |x|) 2−n
4j
for |x| ≥ R1j ; k = 1; 2; : : : :
(2.29)
∞
From Lemma 2.5 we can choose a subsequence of {hk }∞
k=1 (still denote by {hk }k=1 )
and a positive constant R3j , which depends solely on ; P(|x|); j; R1j and supk khk k,
such that
1
(2.30)
|T [ j hk ](x) − T [ j hm ](x)| ≤ (1 + |x|) 2−n
3j
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
513
for all |x| ≥ R3j , k; m = 1; 2; : : : . Set R2j = max{R1j ; R3j }. Lemma 2.7 follows from
(2.29) and (2.30).
Theorem 2.8. Assume f : → R is positive and C for some ∈ (0; 1). We also
assume that there is a radially symmetric function P(|x|) satisfying (1.3) and
f(x) ≤ P(|x|) for all x ∈ . Then the mapping T : E → E is relatively compact.
Proof. Assume {hk }∞
k=1 is a bounded sequence in E. Denote by M = supk khk k; we
shall prove that there is a convergent subsequence of {T [hk ]}∞
k=1 in E.
∞
For j = 1, Lemma 2.7 implies that we can choose a subsequence, {h(1)
k }k=1 of
∞
{hk }k=1 and a positive constant R1 , such that
(1)
2−n
|T [h(1)
k ](x) − T [hm ](x)| ≤ (1 + |x|)
for |x| ≥ R1 ; k; m = 1; 2; : : : :
(2.31)
The analogous proof to that in Lemma 2.2(ii) shows
2−n
|T [h(1)
k ](x)| ≤ C(1 + |x|)
for x ∈ ;
(2.32)
where C is a positive constant depending solely on ; P(|x|); M and n. Let B1 =
{|x| ≤ R1 } denote the ball about the origin of radius R1 . Obviously, T [h(1)
k ](x), when
restricted to ∩ B1 , is a W 1; 2 (
∩ B1 ) – solution of the problem
−u = f(x)h(1)
k ;
x ∈ ∩ B1 :
Using Theorems 8.22 and 8.27 in [7], we get a constant , 0¡¡1, such that T [h(1)
k ]∈
C 1+ (
∩ B1 ) with the Hőlder norm
|T [h(1)
B1 ≤ C;
k ](x)|1+; ∩
(2.33)
where and C depend on R1 ; ; P(|x|); M and n. By the Ascoli–Arzela theorem (see
∞
Theorem 1.30 in [1]), we may choose a subsequence of {h(1)
k }k=1 (still denote by
(1) ∞
(1)
∞
1 ), where
{hk }k=1 ), such that {T [hk ]| ∩
B1 }k=1 is a Cauchy sequence in C( ∩ B
(1)
(1)
1 . Furthermore, we
T [hk ]| ∩
B1 denotes the restriction of the function T [hk ] on ∩ B
can assume
(1)
2−n
|T [h(1)
k ](x) − T [hm ](x)| ≤ (1 + R1 )
(2.34)
for x ∈ ∩ B1 , k; m = 1; 2; : : : . Hence, it follows from (2.31) and (2.34) that
(1)
2−n
|T [h(1)
k ](x) − T [hm ](x)| ≤ (1 + |x|)
for x ∈ , k; m = 1; 2; : : : .
(2) ∞
∞
Similarly, we can choose a subsequence of {h(1)
k }k=1 , denoted by {hk }k=1 , and a
positive constant R2 , such that
(2)
2−n
1
|T [h(2)
k ](x) − T [hm ](x)| ≤ 2 (1 + |x|)
for |x| ≥ R2 , k; m = 1; 2; : : : ; and
(2)
2−n
1
|T [h(2)
k ](x) − T [hm ](x)| ≤ 2 (1 + R2 )
514
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
for x ∈ ∩ {|x| ≤ R2 }, k; m = 1; 2; : : : . Hence,
(2)
2−n
1
|T [h(2)
k ](x) − T [hm ](x)| ≤ 2 (1 + |x|)
for x ∈ ; k; m = 1; 2; : : : :
(2.35)
Coutinuing in this fashion, for any integer j, we can obtain a subsequence, {hk( j) }∞
k=1 ,
( j−1) ∞
}k=1 , such that
of {hk
1
( j)
](x)| ≤ (1 + |x|) 2−n
|T [hk( j) ](x) − T [hm
j
(2.36)
for x ∈ , k; m = 1; 2; : : : . Set wk (x) = hk(k) (x). Thus {wk (x)}∞
k=1 is a subsequence of
and
satises
{hk }∞
k=1
1
|T [wk ](x) − T [wm ](x)| ≤ (1 + |x|) 2−n
j
for x ∈ ; m ≥ j:
(2.37)
(2.37) implies that {T [wk ](x)}∞
k=1 is a Cauchy sequence in E, the result now follows
from the completeness of E.
3. The linear eigenvalue problem
(
); 0¡f(x) ≤ P(|x|) for x ∈ and P(|x|) satIn this section, we assume f ∈ Cloc
ises (1.3). Let K = {u ∈ E| u ≥ 0; x ∈ } be a cone in E. The operator T is compact
and maps K − {0} into K − {0} since T is a monotone mapping. But T does not map
K − {0} into the interior of K, so we cannot apply the Krein–Rutman theorem to get
the existence of eigenvalues and eigenfunctions of problem (1.1).
Lemma 3.1. For h ∈ K − {0}, T [h](x) satisÿes
@T [h]
¡0 for x ∈ @
;
@
Z
Z
n−2
lim (1 + |x|) T [h](x) =
f(x)h (x) dx −
∇T [h]∇ dx¿0;
T [h](x)¿0 for x ∈ ;
|x|→∞
(3.1)
where denotes the exterior unit normal to @
.
Proof. It is obvious that T [h](x), when restricted to ∩ {|x| ≤ R}, is a solution of the
problem
−u = f(x)h(x);
u|@
= 0;
x ∈ ∩ {|x| ≤ R};
(3.2)
u||x|=R = T [h](x)||x|=R :
The weak maximum principle (Theorem 8.1 in [7]) then implies
min 0; min T [h](x) ≤ T [h](x) ≤ max 0; max T [h]
|x|=R
|x|=R
(3.3)
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
515
for x ∈ ∩ {|x| ≤ R}. Taking R → ∞ in (3.3) we get T [h](x) ≥ 0 for x ∈ since we
have proved in Lemma 2.2 that T [h](x) ∈ E. Obviously, T [h](x) 6≡ 0 in . Hence
T [h] ∈ K − {0}. Applying the strong maximum and boundary point principles (Theorem 5 in [11]) to problem (3:2) yields
T [h](x)¿0
for x ∈ ∩ {|x| ≤ R};
@T [h]
¡0
@
for x ∈ @
:
Taking R → ∞ we have
T [h](x)¿0
@T [h]
¡0
@
for x ∈ ;
for x ∈ @
:
Let (x) be a C0∞ (
)-function satisfying (x)h(x) 6≡ 0, 0 ≤ ≤ 1 for x ∈ . Clearly,
if lim|x|→∞ (1 + |x|) n−2 T [h ](x)¿0, we can deduce from Remarks 2.1 and 2.2 that
(3.1) holds. So it remains only to prove lim|x|→∞ (1 + |x|) n−2 T [h ](x)¿0.
Denote by bv(x) = |x| 2−n T [h ](x=|x| 2 ) for x ∈ {x ∈ R n | x=|x| 2 ∈ }, the Kelvin transform of T [h ], and assume that supp (x) ⊂ {|x| ≤ R0 } for some R0 . It is obvious that
bv(x) satises
−bv(x) = 0;
bv(x) ≥ 0
|bv(x)| ≤ C;
for |x| =
bv(x) 6≡ 0
for 0¡|x|¡
1
:
R0
1
;
R0
Theorem (2.69) in [6] implies that bv(x) has a removable singularity at 0, and satises
−bv(x) = 0;
bv(x) ≥ 0
|bv(x)| ≤ C;
for |x| =
bv(x) 6≡ 0
1
;
R0
for |x|¡
1
;
R0
(3.4)
here we dene bv(0) = lim|x|→∞ bv(x). An application of the strong maximum principle
to Eq. (3.4) yields bv(0)¿0. Hence,
lim (1 + |x|) n−2 T [ h](x) = lim bv(x) = bv(0)¿0:
|x|→∞
|x|→0
(3.5)
(
), 0¡¡1, satisfy 0¡f(x) ≤ P(|x|) for x ∈ , where
Theorem 3.2. Let f ∈ Cloc
P(|x|) satisÿes (1.3). There is a unique eigenvalue 1 for which Eq. (1.1) has a
minimal positive eigenfunction h(x). Furthermore, 1 is simple and positive,
Z
Z
1; 2
2
2
|∇u| dx f(x) |u| dx = 1; u ∈ D0 (
)
(3.6)
1 = inf
and
lim (1 + |x|) n−2 h(x) = C;
|x|→∞
for some positive constant C.
516
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
Proof. Let b
h(x) = (1+|x|)2−n for x ∈ . Denote by (x) = T [T [b
h]](x). From Lemma 3.1
we deduce
T [b
h](x)¿0
for x ∈ ;
T [b
h](x) = 0
for x ∈ @
;
@T [b
h]
¡0
@
for x ∈ @
;
h](x)¿0
lim (1 + |x|) n−2 T [b
|x|→∞
and
(x)¿0
for x ∈ ;
(x) = 0
for x ∈ @
;
@
¡0
@
for x ∈ @
;
lim (1 + |x|) n−2 (x) = C¿0:
|x|→∞
Thus, there exists a positive constant C satisfying C(x) ≥ T [b
h](x) for x ∈ . Furthermore, Remark 2.1 implies
T [](x) ≥
1
1
T [T [b
h]](x) = (x):
C
C
(3.7)
Denote by T [h] = T [h] + for ¿0 small enough. The mapping T : K → K is continuous and relatively compact, and satises
inf {kT [h]k | h ∈ K; khk = 1} ≥ kk¿0:
By Theorem 5.4.33 in [2] there is a pair ( ; h ) with ¿0, h ∈ K and kh k = 1
satisfying
h = T [h ] = T [h ] + :
(3.8)
Note that Eq. (3.8) implies
1
:
h ≥
Thus, there is a largest number t ≥ = satisfying
h ≥ t :
(3.9)
Combining (3.7) and (3.9) with Remark 2.1, we arrive at
T [h ] ≥
t
:
C
Thus,
h ≥
t
C
by (3.8) and (3.10), and
t ≥
t
C
(3.10)
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
517
by the maximality of t . Hence,
≥
1
¿0:
C
(3.11)
Since T is a compact operator on K and h satises kh k = 1, there are a subsequence
of {h } (still denote by {h }), an element h 0 ∈ K and a number 0 satisfying
T [h ] → h 0
in K
and
→ 0
as → 0:
Furthermore, 0 ¿0 by (3.11). Taking → 0 in Eq. (3.8) we deduce
h →
h0
0
in K
as → 0;
and
h = 1 T [h];
where we denote by h = h 0 = 0 ; 1 = 1= 0 . Hence 1 is a eigenvalue of Eq. (1.1) with
a nonnegative, nontrivial minimal eigenfunction h.
Now, we consider the constrained variational problem
Z
Z
b
|∇u|2 dx f(x)|u|2 dx = 1; u ∈ D01; 2 (
) :
(3.12)
= inf
b
¿0 is obvious by Lemma 2.1. Lemma 2.1 also implies that there exists some u0 ∈
D01; 2 (
) for which inmum (3:12) is attained, then u0 is a D01; 2 (
)-solution of Eq. (1.1)
by the Euler–Lagrange principle. If u0 minimizes (3.12), so does |u0 |. Hence, it can be
assumed that u0 ≥ 0 and u0 ¿0 in by the strong maximum principle. Thus, there is
a positive D01; 2 (
)-eigenfunction corresponding to b
. Using the same argument as that
b
in [9] we can show that is simple and that there are no other eigenvalues having
nonnegative eigenfunctions in D01; 2 (
). Note that T [h] ∈ E ∩ D01; 2 (
) is a nonnegative
= 1 and 1 is positive and simple and
eigenfunction associated with 1 . Hence, b
h = du0 for some constant d. That h satises lim|x|→∞ (1 + |x|) n−2 h(x) = C for some
constant C follows from Lemma 3.1.
4. The semilinear eigenvalue problem
In this section, we consider the existence of positive minimal solutions of the semilinear problem (1:2). We assume that P(|x|) is a radially symmetric positive function
satisfying
Z
|x|(2−n) P(|x|) dx¡+∞;
(4.1)
(
) satises 0¡f(x) ≤ P(|x|) for all x ∈ .
f(x) ∈ Cloc
Let T be the mapping dened in Lemma 2.2 with the special functions f(x) and
= f(x)(1 + |x|)(1−)(n−2) ; P(|x|) = P(|x|)(1 + |x|)(1−)(n−2) . Then
P(|x|). Denote by f(x)
518
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
P(|x|) satises (1.3). Let T be the mapping dened in Lemma 2.2 with the special
functions f(x)
and P(|x|). Then T : E → E ∩ D01; 2 (
) by Lemma 2.2 and T [h ] =
(1−)(2−n)
T [h (1 + |x|)
] for h ∈ K. Obviously,
kh (1 + |x|)(1−)(2−n) k ≤ khk ¡+∞:
Hence, T [h ] is well dened and satises kT [h ]k ≤ Ckhk for h ∈ K. The problem of
nding minimal solutions of Eq. (1.2) is equivalent to that of nding positive solutions
of the functional equation
u = T [u] + T [u ]
in K:
(4.2)
For 0 ≤ ¡1 , where 1 is the rst eigenvalue of Eq. (1.1), the Riesz–Schauder
theory shows (I − T )−1 : E → E is bounded. Thus, functional Eq. (4:2) is equivalent to
u = (I − T )−1 T [u ]
in K:
(4.3)
We denote by G(u) = (I − T )−1 T [u ] in K.
Lemma 4.1. The mapping G : K → E is a monotone operator in K.
Proof. Consider any pair h1 ≤ h2 in K and set u1 = G(h1 ) and u2 = G(h2 ). We shall
prove that u1 ≤ u2 . By our denition of G, u1 and u2 satisfy
u1 = T [u1 ] + T [h1 ];
u2 = T [u2 ] + T [h2 ]:
Thus, u1 ; u2 ∈ E ∩ D01; 2 (
), u2 − u1 satises
− (u2 − u1 ) = f(x)(u2 − u1 ) + f(x)(h2 − h1 ):
(4.4)
def :
def :
Set w = u2 − u1 and w = w+ − w− where w+ = max{w; 0} and w− = max{−w; 0}.
Multiplying Eq. (4.4) by w− , integrating over , we obtain
Z
Z
Z
−
−
∇w∇w dx = f(x)ww dx +
f(x)(h2 − h1 )w− dx:
def :
(4.5)
def :
Now set = + ∪ − where − = {x ∈ | u2 ¡u1 } and + = {x ∈ | u2 ≥ u1 }. Note
that w− ¿0 and ∇w− = −∇w in − , and w− = 0 and ∇w− = 0 in + . We can rewrite
Eq. (4.5) as follows
Z
Z
Z
− 2
− 2
|∇w | dx = − f(x)|w | dx +
f(x)(h2 − h1 )w− dx:
(4.6)
−
−
−
A combination of Eq. (4.6) with the fact that
Z
Z
− 2
|∇w | dx ≥ 1
f(x)|w− |2 dx
−
−
−
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
yields
Z
0 ≥ ( − 1 )
−
f(x)|w− |2 dx ≥
Z
−
519
f(x)(h2 − h1 )w− dx ≥ 0:
Thus, mes(
− ) = 0, i.e., u2 ≥ u2 in .
Lemma 4.2. There exist functions u;
u ∈ K, with u ≥ u for x ∈ , satisfying
u ≤ G(u):
u ≥ G( u);
Proof. Let (x) ∈ E ∩ C ∞ (
) be a given positive function satisfying (x) = 0 for
x ∈ @
; (1 + |x|) n−2 (x) ≥ C¿0 for |x| large enough, @ =@ ≤ −C¡0 for x ∈ @
,
where denotes the exterior unit normal to @
. Denote by 2 (x) = G( (x)). Then
2 (x) ≥ 0 by Lemma 4.1. Because 2 satises
2 (x) = T [ 2
+
];
2
+
¿0
for x ∈ , by Lemma 3.1 we deduce that 2 (x)¿0 for x ∈ , 2 (x) = 0 for x ∈ @
,
(1+|x|) n−2 ≥ C¿0 for |x| large enough, and @ 2 =@ ≤ −C¡0 for x ∈ @
. Hence, there
is a positive constant, C0 , such that
1
(x) ≤
C0
2 (x) ≤ C0
(x)
for x ∈ :
Choosing M so large that M 1= C0−1 ≥ M , we obtain
M
2 (x) = G(M
1=
(x)) ≥ G(M 1= C0−1
2 (x)) ≥ G(M 2 (x)):
Let 1 ; 1 (x) be the rst eigenvalue, the rst eigenfunction, respectively, of problem
(1:1). Choosing ¿0 so small that
1− 1−
(1 − ) max 1
≤ 1;
1 (x) ≤ M 2 (x) for x ∈ ;
we obtain
(I − T )(1 ) = (1 − )T (1 ) ≤ (1 − ) max 1
1−
Let u = M
2 (x)
u ≥ G( u);
T [(1 ) ] ≤ T [(1 ) ]:
and u(x) = 1 (x). Then u and u satisfy
u ≤ G(u):
Theorem 4.3. Assume f(x) and P(|x|) satisfy (4:1), 0 ≤ ≤ 1 . Then Eq: (1:2) has a
positive minimal solution satisfying (1+|x|) n−2 u(x) → C as |x| → ∞ for some positive
constant C.
Proof. Let u0 (x) = u(x) and let
uj (x) = G(uj−1 (x));
j = 1; 2; : : : :
(4.7)
520
R.-Y. Xue, Y.-C. Qin / Nonlinear Analysis 41 (2000) 501 – 521
It is obvious that Eq. (4.7) is well dened. Furthermore, we deduce from Lemmas 4.1
and 4.2 that
≤ u(x):
u0 (x) ≤ G(u0 (x)) = u1 (x) ≤ G( u)
Similarly, we have
u0 (x) ≤ G(uj−1 (x)) = uj (x) ≤ u(x):
(4.8)
(1−)(2−n) ∞
}j=1 . Hence,
Thus {uj }∞
j = 1 is a bounded sequence in E, so does {uj (1 + |x|)
∞
there are a subsequence (we still denote it by {uj }j=1 ) and an element v1 such that
T [uj ] = T [uj (1 + |x|)(1−)(2−n) ] → v1
as j → ∞:
(4.9)
It follows from Eqs. (4.9) and (4.7) that
uj (x) = G(uj−1 ) = (I − T )−1 T [uj ] → (I − T )−1 v1
in K as j → ∞. Denote by v0 = (I − T )−1 v1 . Obviously,
u ≤ v0 ≤ u
for x ∈ :
(4.10)
We claim that G(uj−1 ) → G(v0 ) as j → ∞ in K. In fact, from (4.8) and (4.10), we get
k(uj − v0 )(1 + |x|)(1−)(2−n) k = sup |uj − v0 |(1 + |x|)(2−n) ;
x∈
sup |uj − v0 |(1 + |x|)(2−n) |u|−1
x∈
≤ C sup |uj − v0 | (1 + |x|)(2−n) = Ckuj − v0 k ;
x∈
which implies
G(uj ) − G(v0 ) = (I − T )−1 T F[(uj − v0 )(1 + |x|)(1−)(2−n) ] → 0
as j → ∞. Taking j → ∞ in Eq. (4.7), we arrive at v0 = G(v0 ) with 0¡u ≤ v0 ≤ u
for x ∈ , and v0 is a positive minimal solution of Eq. (1.2). That v0 satises (1 +
|x|) n−2 v0 (x) → C as |x| → ∞ for some positive constant C follows from Lemma 3.1.
We complete the proof of Theorem 4.3.
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