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Boundary Value Problems
Volume 2007, Article ID 14731, 25 pages
doi:10.1155/2007/14731
Research Article
Eigenvalue Problems and Bifurcation of Nonhomogeneous
Semilinear Elliptic Equations in Exterior Strip Domains
Tsing-San Hsu
Received 19 July 2006; Revised 10 October 2006; Accepted 20 October 2006
Recommended by Patrick J. Rabier
We consider the following eigenvalue problems:
−Δu + u = λ( f (u)+h(x)) in Ω, u>0
in Ω, u
∈ H
1
0
(Ω), where λ>0, N = m + n ≥ 2, n ≥ 1, 0 ∈ ω ⊆ R
m
is a smooth bounded
domain,
S
=
ω × R
n
, D is a smooth bounded domain in R
N
such that D ⊂⊂ S, Ω =
S\
––
D
. Under some suitable conditions on f and h, we show that there exists a positive

constant λ
∗
such that the above-mentioned problems have at least two solutions if λ ∈
(0,λ
∗
), a unique positive solution if λ = λ
∗
, and no solution if λ>λ
∗
.Wealsoobtain
some bifurcation results of the solutions at λ
= λ
∗
.
Copyright © 2007 Tsing-San Hsu. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Throughout this article, let N
= m +n ≥ 2, n ≥ 1, 2
∗
= 2N/(N − 2) for N ≥ 3, 2
∗
=∞for
N
= 2, x = (y,z) be the generic point of R
N
with y ∈ R
m
, z ∈ R

n
.
In this article, we are concerned with the following eigenvalue problems:
−Δu +u = λ

f (u)+h(x)

in Ω, u in H
1
0
(Ω), u>0inΩ, N ≥ 2, (1.1)
λ
where λ>0, 0 ∈ ω ⊆ R
m
is a smooth bounded domain, S
=
ω × R
n
, D is a smooth
bounded domain in
R
N
such that D ⊂⊂ S, Ω =
S \
D is an exterior strip domain in R
N
,
h(x)
∈ L
2

(Ω) ∩ L
q
0
(Ω)forsomeq
0
>N/2ifN ≥ 4, q
0
= 2ifN = 2, 3, h(x) ≥ 0, h(x) ≡ 0
and f satisfies the following conditions:
(f1) f
∈ C
1
([0,+∞),R
+
), f (0) = 0, and f (t) ≡ 0ift<0;
(f2) there is a positive constant C such that


f (t)


≤
C

|
t| + |t|
p

for some 1 <p<2
∗

− 1; (1.1)
2 Boundary Value Problems
(f3) lim
t→0
t
−1
f (t) = 0;
(f4) there is a number θ
∈ (0,1) such that
θt f

(t) ≥ f (t) > 0fort>0; (1.2)
(f5) f
∈ C
2
(0,+∞)and f

(t) ≥ 0fort>0;
(f5)
∗
f ∈ C
2
(0,+∞)and f

(t) > 0fort>0;
(f6) lim
t→0
+
t
1−q

1
f

(t) ≤ C where C is some constant, 0 <q
1
< 4/(N − 2) if N ≥ 3,
q
1
> 0ifN = 2.
If Ω
=
R
N
or Ω =
R
N
\ D (m = 0 in our case), then the homogeneous case of problem
(1.1)
λ
(i.e., the case h(x) ≡ 0) has been studied by many authors (see Cao [4] and the
references therein). For the nonhomogeneous case (h(x)
≡ 0), Zhu [18] has studied the
special problem
−Δu + u = u
p
+ h(x)inR
N
,
u in H
1


R
N

, u>0inR
N
, N ≥ 2.
(1.3)
They have proved that (1.3) has at least two positive solutions for
h
L
2
sufficiently small
and h exponentially decaying.
Cao and Zhou [5] have considered the following general problems:
−Δu + u = f (x,u)+h(x)inR
N
,
u in H
1

R
N

, u>0inR
N
, N ≥ 2,
(1.4)
where h
∈ H

−1
(R
N
), 0 ≤ f (x,u) ≤ c
1
u
p
+ c
2
u with c
1
> 0, c
2
∈ [0,1) being some con-
stants. They also have shown that (1.4) has at least two positive solutions for
h
H
−1
<
C
p
S
(p+1)/2(p−1)
and h ≥ 0, h ≡ 0inR
N
,whereS is the best Sobolev constant and C
p
=
c
−1/(p−1)

1
(p − 1)[(1 − c
2
)/p]
p/(p−1)
.
Zhu and Zhou [19] have investigated the existence and multiplicity of positive solu-
tions of (1.1)
λ
in R
N
\ D for N ≥ 3. They have shown that there exists λ
∗
> 0suchthat
(1.1)
λ
admits at least two positive solutions if λ ∈ (0,λ
∗
)and(1.1)
λ
has no positive solu-
tions if λ>λ
∗
under the conditions that h(x) ≥ 0, h(x) ≡ 0, h(x) ∈ L
2
(Ω) ∩ L
(N+γ)/2
(Ω)
(γ>0ifN
≥ 4andγ = 0ifN = 3), and f satisfies conditions (f1)–(f5). However, their

method cannot know whether λ
∗
is bounded or infinite.
In the present paper, motivated by [19], we extend and improve the paper by Zhu and
Zhou [19]. First, we deal with the more general domains instead of the exterior domains,
and second, we prove that λ
∗
is finite, and third, we also obtain the behavior of the two
solutions on (0, λ
∗
) and some bifurcation results of the solutions at λ = λ
∗
.Now,westate
our main results.
Theorem 1.1. Le t Ω
=
S \
D or Ω =
R
N
\ D.Supposeh(x) ≥ 0, h(x) ≡ 0, h(x) ∈ L
2
(Ω) ∩
L
q
0
(Ω) for some q
0
>N/2 if N ≥ 4, q
0

= 2 if N = 2,3,and f (t) satisfies (f1)–(f5). Then there
exists λ
∗
> 0, 0 <λ
∗
< ∞ such that
(i) equation (1.1)
λ
has at least two positive solutions u
λ
, U
λ
,andu
λ
<U
λ
if λ ∈ (0,λ
∗
),
where u
λ
is the minimal solution of (1.1)
λ
and U
λ
is the second solution of (1.1)
λ
constructed in Section 5;
Tsing-San Hsu 3
(ii) equation (1.1)

λ
has at least one minimal positive solution u
λ
∗
;
(iii) equation (1.1)
λ
has no positive solutions if λ>λ
∗
.
Moreover, assume that condition (f5)
∗
holds, then (1.1)
λ
∗
has a unique positive solution u
λ
∗
.
Theorem 1.2. Suppose the assumptions of Theorem 1.1 and condition (f5)
∗
hold, then
(i) u
λ
is strictly increasing with respect to λ, u
λ
is uniformly bounded in L
∞
(Ω) ∩ H
1

0
(Ω)
for all λ
∈ (0,λ
∗
],and
u
λ
−→ 0 in L
∞
(Ω) ∩ H
1
0
(Ω) as λ −→ 0
+
, (1.5)
(ii) U
λ
is unbounded in L
∞
(Ω) ∩ H
1
0
(Ω) for λ ∈ (0,λ
∗
),thatis,
lim
λ→0
+



U
λ


=
lim
λ→0
+


U
λ


∞
=∞, (1.6)
(iii) moreover, assume that condition (f6) holds and h(x) is in C
α
(Ω) ∩ L
2
(Ω),thenall
solutions of (1.1)
λ
are in C
2,α
(Ω) ∩ H
2
(Ω),and(λ
∗

,u
λ
∗
) is a bifurcation point for
(1.1)
λ
and
u
λ
−→ u
λ
∗
in C
2,α
(Ω) ∩ H
2
(Ω) as λ −→ λ
∗
,
U
λ
−→ u
λ
∗
in C
2,α
(Ω) ∩ H
2
(Ω) as λ −→ λ
∗

.
(1.7)
2. Preliminaries
In this paper, we denote by C and C
i
(i = 1,2, ) the universal constants, unless otherwise
specified. Now, we will establish some analyt ic tools and auxiliar y results which will be
used later. We set
F(u)
=

u
0
f (s)ds,
u=


Ω

|∇
u|
2
+ u
2

dx

1/2
,
u

p
=


Ω
|u|
q
dx

1/q
,1≤ q<∞,
u
∞
= sup
x∈Ω


u(x)


.
(2.1)
First, we give some properties of f (t). The proof can be found in Zhu and Zhou [19].
Lemma 2.1. Under conditions (f1), (f4), and (f5),
(i) let ν
= 1+θ
−1
> 2, one has that tf(t) ≥ νF(t) for t>0;
(ii) t
−1/θ

f (t) is monotone nondecreasing for t>0 and t
−1
f (t) is st rictly monotone in-
creasing if t>0;
(iii) for any t
1
,t
2
∈ (0,+∞), one has
f

t
1
+ t
2

≥
f

t
1

+ f

t
2

, f

t

1
+ t
2

≡
f

t
1

+ f

t
2

. (2.2)
4 Boundary Value Problems
In order to get the existence of positive solutions of (1.1)
λ
, consider the energy functional
I : H
1
0
(Ω) → R defined by
I(u)
=
1
2

Ω


|∇
u|
2
+ u
2

dx − λ

Ω
F

u
+

dx − λ

Ω
hudx. (2.3)
By the strong maximum principle, it is easy to show that the critical points of I are the positive
solutions of (1.1)
λ
.
Now, introduce the following elliptic equation on
S:
−Δu + u = λf(u) in S, u ∈ H
1
0
(S), N ≥ 2, (2.4)
λ

and its associated energ y functional I
∞
defined by
I
∞
(u) =
1
2

S

|∇
u|
2
+ u
2

dx − λ

S
F

u
+

dx, u ∈ H
1
0
(S). (2.4)
If (f1)–(f4) hold, using results of Esteban [8]andLions[15, 16], one knows that (2.4)

λ
has a
ground state w(x) > 0 in
S such that
S
∞
= I
∞
(w) = sup
t≥0
I
∞
(tw). (2.5)
Now, establish the following decomposition lemma for later use.
Proposition 2.2. Let conditions (f1), (f2), and (f4) be satisfied and suppos e that
{u
k
} is a
(PS)
α
-sequence of I in H
1
0
(Ω),thatis,I(u
k
) = α + o(1) and I

(u
k
) = o(1) strong in H

−1
(Ω).
Then there exist an integer l
≥ 0,sequence{x
i
k
}⊆R
N
of the form (0,z
i
k
) ∈ S,asolutionu of
(1.1)
λ
,andsolutionsu
i
of (2.4)
λ
, 1 ≤ i ≤ l, such that for some subsequence {u
k
}, one has
u
k
u
weakly in H
1
0
(Ω),
I


u
k

−→
I(u)+
l

i=1
I
∞

u
i

,
u
k
−

u +
m

i=1
u
i

x − x
i
k



−→
0 strong in H
1
0
(Ω),


x
i
k


−→ ∞
,


x
i
k
− x
j
k


−→ ∞
,1≤ i = j ≤ l,
(2.6)
where one agrees that in the case l
= 0, the above hold without u

i
, x
i
k
.
Proof. This result can be derived from the arguments in [3] (see also [15–17]). Here we
omit it.

3. Asymptotic behavior of solutions
In this section, we establish the decay estimate for solutions of (1.1)
λ
and (2.4)
λ
.Inorder
to get the asymptotic behavior of solutions of (1.1)
λ
, we need the following lemmas. First,
we quote regularity Lemma 1 (see Hsu [12] for the proof). Now, let
X be a C
1,1
domain
in
R
N
.
Tsing-San Hsu 5
Lemma 3.1 ( regular ity Lemma 1). Let g :
X × R → R be a Carath
´
eodory function such that

for almost every x
∈ X, there holds


f (x,u)


≤
C

|
u| + |u|
p

uniformly in x ∈ X, (3.1)
where 1 <p<2
∗
− 1.
Also, let u
∈ H
1
0
(X) be a weak solution of equation −Δu = f (x,u)+h(x) in X,where
h
∈ L
N/2
(X) ∩ L
2
(X). Then u ∈ L
q

(X) for q ∈ [2,∞).
Now, we quote Regularity Lemmas 2–4, (see Gilbarg and Trudinger [ 9,Theorems8.8,
9.11, and 9.16] for the proof).
Lemma 3.2 (regularity Lemma 2). Let
X ⊂ R
N
be a domain, g ∈ L
2
(X),andu ∈ H
1
(X)
a weak solution of the equation
−Δu + u = g in X. Then for any subdomain X

⊂⊂ X with
d

= dist(X

,∂X) > 0, u ∈ H
2
(X

) and
u
H
2
(X

)

≤ C


u
H
1
(X)
+ g
L
2
(X)

(3.2)
for some C
= C(N,d

).Furthermore,u satisfies the equation −Δu + u = g almost everywhere
in
X.
Lemma 3.3 (regularity Lemma 3). Let g
∈ L
2
(X) and let u ∈ H
1
0
(X) beaweaksolutionof
the equation
−Δu + u = g. Then u ∈ H
2
0

(X) satisfies
u
H
2
(X)
≤ Cg
L
2
(X)
, (3.3)
where C
= C(N,∂X).
Lemma 3.4 (regularity Lemma 4). Let g
∈ L
2
(X) ∩ L
q
(X) for some q ∈ [2,∞) and let u ∈
H
1
0
(X) be a weak solution of the equation −Δu +u = g in X. Then u ∈ W
2,q
(X) satisfies
u
W
2,q
(X)
≤ C



u
L
q
(X)
+ g
L
q
(X)

, (3.4)
where C
= C(N,q,∂X).
By Lemmas 3.1 and 3.4, we obtain the first asymptotic behavior of solution of (1.1)
λ
.
Lemma 3.5 (asymptotic Lemma 1). Let condition (f2) hold and let u be a weak solution of
(1.1)
λ
, then u(y,z) → 0 as |z|→∞uniformly for y ∈ ω.Moreover,ifh(x) is bounded, then
u
∈ C
1,α
(Ω) for any 0 <α<1.
Proof. Suppose that u is a solution of (1.1)
λ
,then−Δu + u = λ( f (u)+h(x)) in Ω.Since
f satisfies condition (f2) and h
∈ L
2

(Ω) ∩ L
q
0
(Ω)forsomeq
0
>N/2ifN ≥ 4, q
0
= 2if
N
= 2,3, this implies that h ∈ L
2
(Ω) ∩ L
N/2
(Ω)forN ≥ 4andh ∈ L
2
(Ω)forN = 2,3. By
Lemma 3.1,weconcludethat
u
∈ L
q
(Ω)forq ∈ [2,∞). (3.5)
Hence, λ( f (u)+h(x))
∈ L
2
(Ω) ∩ L
q
0
(Ω)andbyLemma 3.4,wehave
u
∈ W

2,2
(Ω) ∩ W
2,q
0
(Ω), q
0
>
N
2
if N
≥ 4, q
0
= 2ifN = 2,3. (3.6)
6 Boundary Value Problems
Now, by the Sobolev embedding theorem, we obtain that u
∈ C
b
(Ω). It is well known that
the Sobolev embedding constants are independent of domains (see [1]). Thus there exists
aconstantC such that, for R>0,
u
L
∞
(Ω\B
R
)
≤ Cu
W
2,q
0

(Ω\B
R
)
for N ≥ 2, (3.7)
where B
R
={x = (y,z) ∈ Ω ||z|≤R}. From this, we conclude that u(y,z) → 0as|z|→∞
uniformly for y ∈ ω.ByLemma 3.4 and condition (f2), we also have that
u
∞
≤u
W
2,q
0
(Ω)
≤ C


u
q
0
+


λf(u)+λh(x)


q
0


≤
C
1
u
q
0
+ λC
2


u
p
pq
0
+ h
q
0

,
(3.8)
where C
1
, C
2
are constants independent of λ.
Moreover , if h(x) is bounded, then we have u
∈ W
2,q
(Ω)forq ∈ [2,∞). Hence, by the
Sobolev embedding theorem, we obtain that u

∈ C
1,α
(Ω)forα ∈ (0,1). 
We use Lemma 3.5, and modify the proof in Hsu [11]. We obtain the following precise
asymptotic behavior of solutions of (1.1)
λ
and (2.4)
λ
at infinit y.
Lemma 3.6 (asymptotic Lemma 2). Let w be a positive s olution of (2.4)
λ
,letu be a positive
solution of (1.1)
λ
,andletϕ be the first positive eigenfunction of the Dirichlet problem −Δϕ =
λ
1
ϕ in ω,thenforanyε>0 with 0 <ε<1+λ
1
, there exist constants C,C
ε
> 0 such that
w(y,z)
≤ C
ε
ϕ(y)exp

−

1+λ

1
− ε|z|

,
w(y,z)
≥ Cϕ(y)exp

−

1+λ
1
|z|

|z|
−(n−1)/2
as |z|−→∞, y ∈ ,
u(y,z)
≥ Cϕ(y)exp

−

1+λ
1
|z|

|z|
−(n−1)/2
.
(3.9)
Proof. (i) First, we claim that for any ε>0with0<ε<1+λ

1
, there exists C
ε
> 0suchthat
w(y,z)
≤ C
ε
ϕ(y)exp

−

1+λ
1
− ε|z|

as |z|−→∞, y ∈ . (3.10)
Without loss of generality, we may assume ε<1. Now given ε>0, by condition (f3) and
Lemma 3.5,wemaychooseR
0
large enough such that
λf

w(y,z)

≤
εw(y,z)for|z|≥R
0
. (3.11)
Let q
= (q

y
,q
z
), q
y
∈ ∂ω, |q
z
|=R
0
,andB a small ball in Ω such that q ∈ ∂B.Sinceϕ(y) >
0forx
= (y,z) ∈ B, ϕ(q
y
) = 0, w(x) > 0forx ∈ B, w(q) = 0, by the strong maximum
principle (∂ϕ/∂y)(q
y
) < 0, (∂w/∂x)(q) < 0. Thus
lim
x→q
|z|=R
0
w(x)
ϕ(y)
=
(∂w/∂x)(q)
(∂ϕ/∂y)

q
y


> 0. (3.12)
Note that w(x)ϕ
−1
(y) > 0forx = (y,z), y ∈ ω, |z|=R
0
.Thusw(x)ϕ
−1
(y) > 0forx =
(y,z), y ∈ , |z|=R
0
.Sinceϕ(y)exp(−

1+λ
1
− ε|z|)andw(x)areC
1
(ω × ∂B
R
0
(0)), if
Tsing-San Hsu 7
set
C
ε
= sup
y∈,|z|=R
0

w(x)ϕ
−1

(y)exp


1+λ
1
− εR
0

, (3.13)
then 0 <C
ε
< +∞ and
C
ε
ϕ(y)exp

−

1+λ
1
− εR
0

≥
w(x)fory ∈ , |z|=R
0
. (3.14)
Let Φ
1
(x) = C

ε
ϕ(y)exp(−

1+λ
1
− ε|z|), for x ∈ Ω.Then,for|z|≥R
0
,wehave
Δ

w − Φ
1

(x) −

w − Φ
1

(x) =−λf

w(x)

+

ε +

1+λ
1
− ε(n − 1)
|z|


Φ
1
(x)
≥−εw(x)+εΦ
1
(x) = ε

Φ
1
− w

(x) .
(3.15)
Hence Δ(w
− Φ
1
)(x) − (1 − ε)(w − Φ
1
)(x) ≥ 0, for |z|≥R
0
.
The strong maximum pr inciple implies that w(x)
− Φ
1
(x) ≤ 0forx = (y,z), y ∈ ,
|z|≥R
0
, and therefore we get this claim.
(ii) Let

Ψ(y,z)
=

1+
1

|z|

ϕ(y)exp

−

1+λ
1
|z|

|z|
−(n−1)/2
for (y, z) ∈ Ω. (3.16)
It is very easy to show that
−ΔΨ + Ψ ≤ 0fory ∈ , |z| large. (3.17)
Therefore, by means of the maximum principle, there exists a constant C>0suchthat
w(y,z)
≥ Cϕ(y)exp

−

1+λ
1
|z|


|z|
−(n−1)/2
u(y,z) ≥ Cϕ(y)exp

−

1+λ
1
|z|

|z|
−(n−1)/2
as |z|−→∞, y ∈ . (3.18)
This completes the proof of Lemma 3.6.

4. Existence of minimal solution
In this section, by the barrier method, we prove that there exists some λ
∗
> 0suchthat
for λ
∈ (0,λ
∗
), (1.1)
λ
has a minimal positive solution u
λ
(i.e., for any positive solution u
of (1.1)
λ

,thenu ≥ u
λ
).
Lemma 4.1. If conditions (f1) and (f2) hold, then for any given ρ>0,thereexistsλ
0
> 0 such
that for λ
∈ (0,λ
0
), one has I(u) > 0 for all u ∈ S
ρ
={u ∈ H
1
0
(Ω) |u=ρ}.
For the proof, see Zhu and Zhou [19].
Remark 4.2. For any ε>0, there exists δ>0(δ
≤ ρ)suchthatI(u) ≥−ε for all u ∈{u ∈
H
1
0
(Ω) | ρ − δ ≤u≤ρ} and for λ ∈ (0,λ
0
)ifλ
0
is small enough (see Zhu and Zhou
[19]).
8 Boundary Value Problems
For the number ρ>0giveninLemma 4.1, we denote
B

ρ
=

u ∈ H
1
0
(Ω) |u <ρ

. (4.1)
Thus we have the following local minimum result.
Lemma 4.3. Under conditions (f1), (f2), and (f4), if λ
0
is chosen as in Remark 4.2 and
λ
∈ (0,λ
0
), then there is a u
0
∈ B
ρ
such that I(u
0
) = min{I(u) | u ∈ B
ρ
} < 0 and u
0
is a
positive solution of (1.1)
λ
.

Proof. Si nce h
≡ 0andh ≥ 0, we can choose a function ϕ ∈ H
1
0
(Ω)suchthat

Ω
hϕ > 0.
For t
∈ (0,+∞), then
I(tϕ)
=
t
2
2

Ω

|∇
ϕ|
2
+ ϕ
2

−
λ

R
N
+

F

tϕ
+

−
λt

Ω
hϕ
≤
t
2
2
ϕ
2
+ λCt
2

Ω

|
ϕ|
2
+ t
p−1
|ϕ|
p+1

−

λt

Ω
hϕ.
(4.2)
Then for t small enough, I(tϕ) < 0. So α
= inf{I(u) | u ∈ B
ρ
}.Clearly,α>−∞.ByRemark
4.2, there is ρ

such that 0 <ρ

<ρand α = inf{I(u) | u ∈ B
ρ

}. By Ekeland variational
principle [7], there exists a (PS)
α
-sequence {u
k
}⊂B
ρ

.ByProposition 2.2, there exists a
subsequence
{u
k
},anintegerl ≥ 0, a solution u
i

of (2.4)
λ
,1≤ i ≤ l, and a solution u
0
in
B
ρ

of (1.1)
λ
such that u
k
 u
0
weakly in H
1
0
(Ω)andα = I(u
0
)+

l
i
=1
I
∞
(u
i
). Note that
I

∞
(u
i
) ≥ S
∞
> 0fori = 1,2, ,m.Sinceu
0
∈ B
ρ
,wehaveI(u
0
) ≥ α.Weconcludethat
l
= 0, I(u
0
) = α,andI

(u
0
) = 0. 
By the standard barrier method, we prove the following lemma.
Lemma 4.4. Let conditions (f1), (f2), and (f4) be satisfied, then there exists λ
∗
> 0 such that
(i) for any λ
∈ (0,λ
∗
), (1.1)
λ
has a minimal positive solution u

λ
and u
λ
is strictly increas-
ing in λ;
(ii) if λ>λ
∗
, (1.1)
λ
has no positive solution.
Proof. Set Q
λ
={0 <λ<+∞|(1.1)
λ
is solvable},byLemma 4.3,wehaveQ
λ
is nonempty.
Denoting λ
∗
= supQ
λ
> 0, we claim that (1.1)
λ
has at least one solution for all λ ∈ (0, λ
∗
).
In fact, for any λ
∈ (0,λ
∗
), by the definition of λ

∗
, we know that there exists λ

> 0and
0 <λ<λ

<λ
∗
such that (1.3)
λ

has a solution u
λ

> 0, that is,
−Δu
λ

+ u
λ

= λ


f

u
λ



+ h

≥
λ

f

u
λ


+ h

. (4.3)
Then u
λ

is a supersolution of (1.1)
λ
.Fromh ≥ 0andh ≡ 0, it is easy to see that 0 is a
subsolution of (1.1)
λ
. By the standard barrier method, there exists a solution u
λ
> 0of
(1.1)
λ
such that 0 ≤ u
λ
≤ u

λ

. Since 0 is not a solution of (1.1)
λ
and λ

>λ,themaximum
principle implies that 0 <u
λ
<u
λ

. Using the result of Graham-Eagle [10], we can choose
a minimal positive solution u
λ
of (1.1)
λ
. 
Tsing-San Hsu 9
Let u
λ
be the minimal positive solution of (1.1)
λ
for λ ∈ (0,λ
∗
), we study the following
eigenvalue problem
−Δv + v = μ
λ
f



u
λ

v in Ω,
v
∈ H
1
0
(Ω), v>0inΩ,
(4.4)
then we have the following lemma.
Lemma 4.5. Under conditions (f1)–(f5), the first eigenvalue μ
λ
of (4.4)isdefinedby
μ
λ
= inf


Ω

|∇
v|
2
+ v
2

dx | v ∈ H

1
0
(Ω),

Ω
f


u
λ

v
2
dx = 1

. (4.5)
Then
(i) μ
λ
is achieved;
(ii) μ
λ
>λand is strictly decreasing in λ, λ ∈ (0,λ
∗
);
(iii) λ
∗
< +∞ and (1.1)
λ
∗

has a minimal positive solution u
λ
∗
.
Proof. (i) Indeed, by the definition of μ
λ
,weknowthat0<μ
λ
< +∞.Let{v
k
}⊂H
1
0
(Ω)
be a minimizing sequence of μ
λ
, that is,

Ω
f


u
λ

v
2
k
dx = 1,


Ω



∇
v
k


2
+ v
2
k

dx −→ μ
λ
as k −→ ∞ . (4.6)
This implies that
{v
k
} is bounded in H
1
0
(Ω), then there is a subsequence, still denoted by
{v
k
} and some v
0
∈ H
1

0
(Ω)suchthat
v
k
v
0
weakly in H
1
0
(Ω),
v
k
−→ v
0
a.e. in Ω.
(4.7)
Thus,

Ω



∇
v
0


2
+ v
2

0

dx ≤ liminf

Ω



∇
v
k


2
+ v
2
k

dx = μ
λ
. (4.8)
By Lemma 3.5 and the conditions (f1), (f3), we have f

(u
λ
) → 0as|x|→∞, it follows that
there exists a constant C>0suchthat


f



u
λ



≤
C ∀x ∈ Ω. (4.9)
Furthermore, for any ε>0, there exists R>0suchthatforx
∈ Ω and |x|≥R, f

(u
λ
) <ε.
Then





Ω
f


u
λ




v
k
− v
0


2
dx




≤

B
R
∩Ω
f


u
λ



v
k
− v
0



2
dx +

Ω\B
R
f


u
λ



v
k
− v
0


2
dx
≤ C

B
R
∩Ω


v

k
− v
0


2
dx + ε

Ω\B
R


v
k
− v
0


2
dx.
(4.10)
10 Boundary Value Problems
It follows from the Sobolev embedding theorem that there exists k
1
,suchthatfork ≥ k
1
,

B
R

∩Ω


v
k
− v
0


2
dx < ε. (4.11)
Since
{v
k
} is bounded in H
1
0
(Ω), this implies that there exists a constant C
1
> 0suchthat

Ω\B
R


v
k
− v
0



2
dx ≤ C
1
. (4.12)
Therefore, we conclude that for k
≥ k
1
,





Ω
f


u
λ



v
k
− v
0


2

dx




≤
Cε + C
1
ε. (4.13)
Take i ng ε
→ 0, we obtain that

Ω
f


u
λ

v
2
0
dx = 1. (4.14)
Hence

Ω



∇

v
0


2
+ v
2
0

dx ≥ μ
λ
. (4.15)
This implies that v
0
achieves μ.Clearly,|v
0
| also achieves μ
λ
.By(4.17) and the maximum
principle, we may assume v
0
> 0inΩ.
(ii) We now prove μ
λ
>λ. Setting λ

>λ>0andλ

∈ (0,λ
∗

), by Lemma 4.4, (1.1)
λ

has
a positive solution u
λ

.Sinceu
λ
is the minimal positive solution of (1.1)
λ
,thenu
λ

>u
λ
as
λ

>λ.Byvirtueof(1.1)
λ

and (1.1)
λ
,weseethat
−Δ

u
λ


− u
λ

+

u
λ

− u
λ

=
λ

f

u
λ


−
λf

u
λ

+(λ

− λ)h. (4.16)
Applying the Taylor expansion and noting that λ


>λ, h(x) ≥ 0and f

(t) ≥ 0, f (t) > 0
for all t>0, we get
−Δ

u
λ

− u
λ

+

u
λ

− u
λ

≥
(λ

− λ) f

u
λ

+ λ


f


u
λ

u
λ

− u
λ

>λf


u
λ

u
λ

− u
λ

.
(4.17)
Let v
0
∈ H

1
0
(Ω)andv
0
> 0solve(4.4). Multiplying (4.17)byv
0
and noting (4.4), then we
get
μ
λ

Ω
f


u
λ

u
λ

− u
λ

v
0
dx > λ

Ω
f



u
λ

u
λ

− u
λ

v
0
dx, (4.18)
hence μ
λ
>λ.Nowletv
λ
be a minimizer of μ
λ
,then

Ω
f


u
λ



v
2
λ
dx >

Ω
f


u
λ

v
2
λ
dx = 1, (4.19)
Tsing-San Hsu 11
and there is t,with0<t<1suchthat

Ω
f


u
λ


tv
λ


2
dx = 1. (4.20)
Therefore,
μ
λ

≤ t
2


v
λ


2
<


v
λ


2
= μ
λ
, (4.21)
showing that μ
λ
is strictly decreasing in λ,forλ ∈ (0,λ
∗

).
(iii) We show next that λ
∗
< +∞.Letλ
0
∈ (0,λ
∗
)befixed.Foranyλ ≥ λ
0
,wehave
μ
λ
>λand by (4.21), then
μ
λ
0
≥ μ
λ
>λ (4.22)
for all λ
∈ [λ
0
,λ
∗
). Thus λ
∗
< +∞.
By (4.4)andμ
λ
>λ,wehave


Ω



∇
u
λ


2
+


u
λ


2

dx >

Ω
λf


u
λ

u

2
λ
dx, (4.23)
and also we have

Ω



∇
u
λ


2
+


u
λ


2

dx −

Ω
λf

u

λ

u
λ
dx −

Ω
λh(x)u
λ
dx = 0. (4.24)
By condition (f4) and (4.23), we have that

Ω



∇
u
λ


2
+


u
λ


2


dx =

Ω
λf

u
λ

u
λ
dx +

Ω
λh(x)u
λ
dx ≤ θ

Ω
λf


u
λ

u
2
λ
dx
+ λ

h
2


u
λ


≤
θ


u
λ


2
+ λh
2


u
λ


.
(4.25)
This implies that



u
λ


≤
λ
1 − θ
h
2
(4.26)
for all λ
∈ (0,λ
∗
). Since λ
∗
< +∞,by(4.26) we can obtain that u
λ
≤C<+∞ for all
λ
∈ (0,λ
∗
). Thus, there exists u
λ
∗
∈ H
1
0
(Ω)suchthat
u
λ

u
λ
∗
weakly in H
1
0

Ω

,
u
λ
−→ u
λ
∗
strongly in L
q
loc
(Ω)for2≤ q<
2N
N − 2
,asλ
−→ λ
∗
,
u
λ
−→ u
λ
∗

almost everywhere in Ω.
(4.27)
For ϕ
∈ H
1
0
(Ω), by condition (f2), we obtain that

Ω

∇
u
λ
·∇ϕ + u
λ
ϕ

dx −→

Ω

∇
u
λ
∗
·∇ϕ + u
λ
∗
ϕ


dx
λ

Ω

f

u
λ

+ h

ϕdx −→ λ
∗

Ω

f

u
λ
∗

+ h

ϕdx
as λ
−→ λ
∗
. (4.28)

12 Boundary Value Problems
From
I

λ
(u
λ
),ϕ=0andletλ → λ
∗
,wededuceI

λ
∗
(u
λ
∗
) = 0inH
−1
(Ω). Hence, u
λ
∗
is a
positive solution of (1.1)
λ
∗
.
Let u be any positive solution of (1.1)
λ
∗
. By adopting the argument as in Lemma 4.4,

we have u
≥ u
λ
in Ω for λ ∈ (0,λ
∗
). Let λ → λ
∗
,wededucethatu ≥ u
λ
∗
in Ω. This implies
that u
λ
∗
is a minimal solution of (1.1)
λ
∗
. 
5. Existence of second solution
When λ
∈ (0,λ
∗
), we have known that (1.1)
λ
has a minimal positive solution u
λ
by Lemma
4.4,thenweneedonlytoprovethat(1.1)
λ
has another positive solution in the form of

U
λ
= u
λ
+ v,wherev is a solution of the following equation:
−Δv + v = λ

f

u
λ
+ v

−
f

u
λ

in Ω,
v>0inΩ, v
∈ H
1
0
(Ω).
(5.1)
For (5.1), we define the energy functional J : H
1
0
(Ω) → R as follows:

J(v)
=
1
2

Ω

|∇
v|
2
+ v
2

dx − λ

Ω

F

u
λ
+ v
+

−
F

u
λ


−
f

u
λ

v
+

dx. (5.2)
Using the monotonicity of f and the maximum principle, we know that the nontrivial
critical points of energy functional J are the positive solutions of (5.1).
First, we give an inequality about f and u
λ
.
Lemma 5.1. Under conditions (f1), (f2), and (f5), then for any ε>0,thereexistsC
ε
> 0 such
that
f

u
λ
+ s

−
f

u
λ


−
f


u
λ

s ≤ εs + C
ε
s
p
, s ≥ 0, (5.3)
where 1 <p<2
∗
− 1 and u
λ
is the minimal solution of (1.1)
λ
.
For the proof, see Zhu and Zhou [19].
Lemma 5.2. Under conditions (f1), (f2), (f4), and (f5), there exist ρ>0 and α>0 such that
J(v)
|
S
ρ
≥ α>0, (5.4)
where S
ρ
={u ∈ H

1
0
(Ω) |u=ρ}.
Proof. By Lemma 4.5, it is easy to see that, for all v
∈ H
1
0
(Ω),

Ω

|∇
v|
2
+ v
2

dx ≥ μ
λ

Ω
f


u
λ

v
2
dx. (5.5)

Tsing-San Hsu 13
Again, by Lemma 5.1 and Sobolev embedding, we obtain that
J(v)
=
1
2

Ω

|∇
v|
2
+ v
2

dx − λ

Ω

F

u
λ
+ v
+

−
F

u

λ

−
f

u
λ

v
+

dx
=
1
2
v
2
−
λ
2

Ω
f


u
λ




v
+


2
dx − λ

Ω

v
+
0

f

u
λ
+ s

−
f

u
λ

−
f


u

λ

s

dsdx
≥
1
2
v
2
−
λ
2

Ω
f


u
λ



v
+


2
dx −
1

2
λε

Ω


v
+


2
dx −
λ
p +1

Ω
C
ε


v
+


p+1
dx
≥
1
2
v

2
−
λ
2
μ
−1
v
2
−
1
2
λε
v
2
− C
ε
v
p+1
=
1
2
μ
−1
λ

μ
λ
− λ − λμ
λ
ε



v
2
− C
ε
v
p+1
.
(5.6)
Since μ
λ
>λ,wemaychooseε>0 small enough such that μ
λ
− λ − λμ
λ
ε>0. If we take
ε
= (μ
λ
− λ)/2λμ
λ
,then
J(v)
≥
1
4
μ
−1
λ


μ
λ
− λ


v
2
− Cv
p+1
. (5.7)
Hence, there exist ρ>0andα>0suchthatJ(v)
|
S
ρ
≥ α>0. 
Similar to Proposition 2.2, for the energy functional J, we also have the following re-
sult.
Proposition 5.3. Under conditions (f1), (f2), and (f4), let
{v
k
} be a (PS)
c
-sequence of J.
Then there ex ists a subsequence (still denoted by
{v
k
}) for which the follow ing holds: there
exist an integer l
≥ 0,asequence{x

i
k
}⊆R
N
of the form (0,z
i
k
) ∈ S,asolutionv of (5.1), and
solutions u
i
of (2.4)
λ
, 1 ≤ i ≤ l, such that for some subsequence {v
k
},ask →∞, one has
v
k
v
weakly in H
1
0
(Ω),
J

v
k

−→
J(v)+
l


i=1
I
∞

u
i

,
v
k
−

v +
l

i=1
u
i

x − x
i
k


−→
0 strongly in H
1
0
(Ω),



x
i
k


−→ ∞
,


x
i
k
− x
j
k


−→ ∞
,1≤ i = j ≤ l,
(5.8)
where one agrees that in the case l
= 0, the above hold without u
i
, x
i
k
.
Now, let δ be small enoug h, D

δ
a δ-tubular neighborhood of D such that D
δ
⊂⊂ S.
Let η(x):
S → [0,1] be a C
∞
cut-off function such that 0 ≤ η ≤ 1and
η(x)
=
⎧
⎨
⎩
0ifx ∈ D;
1ifx
∈ S \ D
δ
.
(5.9)
14 Boundary Value Problems
Let e
N
= (0,0, ,0,1) ∈ R
N
, denote
τ
0
= 2sup
x∈D
δ

|x| +1,
w
τ
(x) = w

x − τe
N

, τ ∈ [0,∞),
(5.10)
where w is a ground state solution of (2.4)
λ
.
Lemma 5.4. Let conditions (f1)–(f5) be satisfied. Then
(i) there exists t
0
> 0 such that J(tηw
τ
) < 0 for t ≥ t
0
, τ ≥ τ
0
,
(ii) there exists τ
∗
> 0 such that the follow ing inequality holds for τ ≥ τ
∗
:
0 < sup
t≥0

J

tηw
τ

<I
∞
(w) = S
∞
. (5.11)
Proof. (i) By the definition of η and Lemma 2.1(iii), we have
J

tηw
τ

=
1
2

Ω



∇

tηw
τ




2
+

tηw
τ

2

dx − λ

Ω

tηw
τ
0

f

u
λ
+ s

−
f

u
λ

dsdx

≤
t
2
2

Ω



∇

ηw
τ



2
+

ηw
τ

2

dx − λ

S\D
δ
F


tw
τ

dx.
(5.12)
Noting part (ii) of Lemma 2.1,weseethatF(u)/(ν
−1
u
ν
) is monotone nondecreasing for
u>0, where ν
= 1+θ
−1
> 2. Thus, for any given constant C>0, there is u
0
≥ 0suchthat
F(u)
≥ Cu
ν
∀u ≥ u
0
. (5.13)
Let r
0
be a positive constant such that B
m
(0;r
0
) ={y ||y|≤r
0

}⊂⊂ω, B
n
(0;1) ={z |
|
z|≤1}, Ω
1
= B
m
(0;r
0
) × B
n
(0;1), and Ω
1τ
= B
m
(0;r
0
) ×{z + τe
N
||z|≤1}.Bythedefi-
nition of τ
0
,wehavethatΩ
1τ
⊂⊂ Ω \ D
δ
for all τ ≥ τ
0
. T his also implies that there exists

t
0
≥ 0, as t ≥ t
0
,wehave
F

tw
τ

≥
Ct
ν
w
ν
τ
∀τ ≥ τ
0
, ∀x ∈ Ω
1τ
. (5.14)
Therefore, as t>t
0
and τ ≥ τ
0
,
J

tηw
τ


≤
t
2
2

Ω



∇

ηw
τ



2
+

ηw
τ

2

dx − λCt
ν

Ω
1τ

w
ν
τ
dx
≤
t
2
2


ηw
τ


2
− λCt
ν

Ω
1
w
ν
dx.
(5.15)
Since ν > 2, we can choose t
0
> 0 large enough such that (i) holds.
(ii) By (i), J is continuous on H
1
0

(Ω), J(0) = 0, and Lemma 5.2, we know that there
exists t
1
with 0 <t
1
<t
0
such that
sup
t≥0
J

tηw
τ

=
sup
t
1
≤t≤t
0
J

tηw
τ

∀
τ ≥ τ
0
. (5.16)

Tsing-San Hsu 15
For τ
≥ τ
0
, t
1
≤ t ≤ t
0
, by condition (f2), (2.5), Lemmas 2.1 and 3.6,wehave
J

tηw
τ

=
t
2
2

Ω



∇

ηw
τ




2
+

ηw
τ

2

dx − λ

Ω
F

tηw
τ

dx
− λ

Ω

tηw
τ
0

f

u
λ
+ s


−
f

u
λ

−
f (s)

dsdx
≤
t
2
2

S
(−Δw + w)

η
2
τ
w

dx +
t
2
2

S



∇
η
τ


2
|w|
2
dx − λ

S
F

tw
τ

dx
+ λ

S

tw
τ
tηw
τ
f (s)dsdx − λ

Ω


tηw
τ
0

f

u
λ
+ s

−
f

u
λ

−
f (s)

dsdx
≤ S
∞
+
t
2
0
2

D

δ
\D
|∇η|
2



τ


2
dx + λ

D
δ

tw
τ
0
f (s)dsdx
− λ

Ω

tηw
τ
0

f


u
λ
+ s

−
f

u
λ

−
f (s)

dsdx
≤ S
∞
+ C
ε
exp

−
2

1+λ
1
− ετ

+ λC

D

δ


tw
τ

2
2
+

tw
τ

p+1
p +1

dx
− λ

Ω

tηw
τ
0

f

u
λ
+ s


−
f

u
λ

−
f (s)

dsdx
≤ S
∞
+ C
ε
exp

−
2

1+λ
1
− ετ

−
λ

Ω

tηw

τ
0

f

u
λ
+ s

−
f

u
λ

−
f (s)

dsdx,
(5.17)
where 0 <ε<1+λ
1
and C
ε
is independent of τ.
It follows from the Taylor’s expansion that
f

u
λ

+ s

=
f (s)+ f

(s)u
λ
+
1
2
f

(ξ)u
2
λ
, ξ ∈

s,u
λ
+ s

. (5.18)
From (f5) and the above formula, for t
1
≤ t ≤ t
0
,weobtainthat

tηw
τ

0

f

u
λ
+ s

−
f

u
λ

−
f (s)

ds
≥

t
1
ηw
τ
0

f

(s)u
λ

− f

u
λ

ds =

t
1
w
τ

−1
f

t
1
ηw
τ

−
ηu
−1
λ
f

u
λ

t

1
w
τ
u
λ
.
(5.19)
Since w
τ
> 0inS, there exists γ
1
> 0suchthat
w
τ
≥ γ
1
in Ω
1τ
. (5.20)
By the definition of w
τ
and u
λ
(x) → 0as|x|→∞,weseethatforτ large enough,
t
1
w
τ
≥ u
λ

in Ω
1τ
, (5.21)
then part (ii) of Lemma 2.1 implies that there exist γ
2
> 0andτ
1
> 0suchthat,forτ ≥ τ
1
,

t
1
w
τ

−1
f

t
1
w
τ

−
u
−1
λ
f


u
λ

>γ
2
in Ω
1τ
. (5.22)
16 Boundary Value Problems
Now by Lemma 3.6,forτ
≥ max(τ
0
,τ
1
)andt
1
≤ t ≤ t
0
,weobtainthat

Ω
1τ

tηw
τ
0

f

u

λ
+ s

−
f

u
λ

−
f (s)

dsdx
≥

Ω
1τ

t
1
w
τ

−1
f

t
1
w
τ


−
u
−1
λ
f

u
λ

t
1
w
τ
u
λ
dx
≥ γ
1
γ
2

Ω
1τ
t
1
u
λ
dx ≥ C
2

exp

−

1+λ
1
τ

,
(5.23)
where C
2
is independent of τ.
Therefore, we obtain that
J

tηw
τ

≤
S
∞
+ λC
ε
exp

−
2

1+λ

1
− ετ

−
λC
2
exp

−

1+λ
1
τ

, (5.24)
for t
∈ [t
1
,t
0
]andτ ≥ max(τ
0
,τ
1
).
Now, let ε
= (1 + λ
1
)/2, then we can find some τ
∗

large enough such that
λC
ε
exp

−

2

1+λ
1

τ

−
λC
2
exp

−

1+λ
1
τ

< 0, (5.25)
for all τ
≥ τ
∗
and we complete the proof. 

Theorem 5.5. Let c onditions (f1)–(f5) be satisfied. Then (5.1) has a positive solution v if
λ
∈ (0,λ
∗
).
Proof. Now , set
Γ
=

p ∈ C

[0,1],H
1
0
(Ω)

|
p(0) = 0, p(1) = t
0
ηw
τ
∗

,
c
= inf
p∈Γ
max
s∈[0,1]
J


p(s)

.
(5.26)
By Lemmas 5.2 and 5.4,wehave
0 <α
≤ c<S
∞
. (5.27)
Applying the mountain pass theorem of Ambrosetti and Rabinowitz [2], there exists a
(PS)
c
-sequence {v
k
} such that
J

v
k

−→
c,
J


v
k

−→

0stronglyinH
−1
(Ω).
(5.28)
By Proposition 5.3, there exists a sequence (still denoted by
{v
k
}), an integer l ≥ 0, a se-
quence
{x
i
k
} in Ω,1≤ i ≤ l,asolutionv of (5.1), and solutions u
i
of (2.4)
λ
such that
c
= J(v)+
l

i=0
I
∞

u
i

. (5.29)
Tsing-San Hsu 17

By the strong maximum principle, to complete the proof, we only need to prove
v ≡ 0in
Ω.Infact,wehave
c
= J(v) ≥ α>0ifl = 0, S
∞
>c≥ J(v)+S
∞
if l ≥ 1. (5.30)
This implies
v ≡ 0inΩ. 
6. Properties and bifurcation of solutions
Denote by A
={(λ,u) | u solves problem (1.1)
λ
} the set of solutions of (1.1)
λ
, λ ∈ (0, λ
∗
].
For each (λ,u)
∈ A,letμ
λ
(u) denote the number defined by
μ
λ
(u) = inf


Ω


|∇
v|
2
+ v
2

dx | v ∈ H
1
0
(Ω),

Ω
f

(u)v
2
dx = 1

, (6.1)
which is the smallest eigenvalue of the following problem:
−Δv + v = μ
λ
(u) f

(u)v in Ω,
v>0, v
∈ H
1
0

(Ω).
(6.2)
In this section, we always assume that conditions (f1)–(f4), (f5)
∗
, and (f6) hold. With
the same arguments used in the proof of part (i) of Lemma 4.5, we can show that μ
λ
(u)is
achieved for all (λ,u)
∈ A.ByLemma 3.5,wehaveA ⊂ R × L
∞
(R
N
) ∩ H
1
0
(Ω). Moreover,
if we assume that h(x)
∈ C
α
(Ω) ∩ L
2
(Ω), then by elliptic regular theory (see [9]), we can
deduce that A
⊂ R × C
2,α
(Ω) ∩ H
2
(Ω).
Lemma 6.1. Let u be a solution and let u

λ
be the minimal solution of (1.1)
λ
for λ ∈ (0,λ
∗
).
Then
(i) μ
λ
(u) >λif and only if u = u
λ
;
(ii) μ
λ
(U
λ
) <λ,whereU
λ
is the second solution of (1.1)
λ
constructed in Section 5.
Proof. Now , let ψ
≥ 0andψ ∈ H
1
0
(Ω). Since u and u
λ
are the solution of (1.1)
λ
,then


Ω
∇ψ ·∇

u
λ
− u

dx +

Ω
ψ

u
λ
− u

dx
= λ

Ω

f

u
λ

−
f (u)


ψdx= λ

Ω


u
λ
u
f

(t)dt

ψdx≥ λ

Ω
f

(u)

u
λ
− u

ψdx.
(6.3)
Let ψ
= (u − u
λ
)
+

≥ 0andψ ∈ H
1
0
(Ω). If ψ ≡ 0, then (6.3) implies
−

Ω

|∇
ψ|
2
+ ψ
2

dx ≥−λ

Ω
f

(u)ψ
2
dx (6.4)
and, therefore, the definition of μ
λ
(u) implies

Ω

|∇
ψ|

2
+ ψ
2

dx ≤ λ

Ω
f

(u)ψ
2
dx < μ
λ
(u)

Ω
f

(u)ψ
2
dx ≤

Ω

|∇
ψ|
2
+ ψ
2


dx,
(6.5)
18 Boundary Value Problems
which is impossible. Hence ψ
≡ 0, and u = u
λ
in Ω. On the other hand, by Lemma 4.5,
we also have that μ
λ
(u
λ
) >λ. This completes the proof of (i).
By (i), we get that μ
λ
(U
λ
) ≤ λ for λ ∈ (0,λ
∗
). We claim that μ
λ
(U
λ
) = λ cannot occur.
We proceed by contradiction. Set w
= U
λ
− u
λ
;wehave
−Δw + w = λ


f

U
λ

−
f

U
λ
− w

, w>0inΩ. (6.6)
By μ
λ
(U
λ
) = λ, we have that the problem
−Δφ + φ = λf


U
λ

φ, φ ∈ H
1
0
(Ω) (6.7)
possesses a positive solution φ

1
.
Multiplying (6.6)byφ
1
and (6.7)byw, integrating and subtracting, we deduce that
0
=

Ω
λ

f

U
λ

−
f

U
λ
− w

−
f


U
λ


w

φ
1
dx =−
1
2

Ω
λf


ξ
λ

w
2
φ
1
dx, (6.8)
where ξ
λ
∈ (u
λ
,U
λ
). By condition (f5)
∗
,weobtainthatw ≡ 0, that is, U
λ

= u
λ
for λ ∈
(0,λ
∗
). This is a contradiction. Hence, we have that μ
λ
(U
λ
) <λfor λ ∈ (0,λ
∗
). 
Lemma 6.2. Let u
λ
be the minimal solution of (1.1)
λ
for λ ∈ [0,λ
∗
] and μ
λ
(u
λ
) >λ. Then
for any g(x)
∈ H
−1
(Ω),problem
− Δw + w = λf



u
λ

w + g(x), w ∈ H
1
0
(Ω), (6.9)
λ
has a solution.
Proof. Consider the functional
Φ(w)
=
1
2

Ω

|∇
w|
2
+ w
2

dx −
1
2
λ

Ω
f



u
λ

w
2
dx −

Ω
g(x)wdx, (6.9)
where w
∈ H
1
0
(Ω). From H
¨
older inequality and Young’s inequality, we have, for any  > 0,
that
Φ(w)
≥
1
2

1 − λμ
λ

u
λ


−1


w
2
−
1
2


w
2
−
C

2
g
2
H
−1
(Ω)
≥−Cg
2
H
−1
(Ω)
(6.10)
if we choose
 small.
Now, let

{w
k
}⊂H
1
0
(Ω) be the minimizing sequence of variational problem
d
= inf

Φ(w) | w ∈ H
1
0
(Ω)

. (6.11)
From (6.10)andμ
λ
(u
λ
) >λ, we can also deduce that {w
k
} is bounded in H
1
0
(Ω), if we
choose
 small. So we may suppose that
w
k
w

weakly in H
1
0
(Ω)ask −→ ∞ ,
w
k
−→ w a.e. in Ω as k −→ ∞ .
(6.12)
Tsing-San Hsu 19
By Fatou’s lemma,
w
2
≤ liminf


w
k


2
. (6.13)
By Lemma 3.5,wehavethatu
λ
(x) → 0as|x|→∞, conditions (f1)–(f5) and the weak
convergence imply

Ω
gw
k
dx −→


Ω
gwdx,

Ω
f


u
λ

w
2
k
dx −→

Ω
f


u
λ

w
2
dx as k −→ ∞ . (6.14)
Therefore
Φ(w)
≤ lim
k→∞

Φ

w
k

=
d, (6.15)
and hence Φ(w)
= d which gives that w is a solution of (6.9)
λ
. 
Remark 6.3. Fro m Lemma 6.2,weknowthat(6.9)
λ
has a solution w ∈ H
1
0
(Ω). Now, we
also assume that h(x)andg(x)areinC
α
(Ω) ∩ L
2
(Ω), then by Lemmas 3.1, 3.3, conditions
(f1)–(f5), and the e lliptic regular theory (see [9]), we can deduce that w
∈ C
2,α
(Ω) ∩
H
2
(Ω).
Lemma 6.4. Suppose u

λ
∗
is a solution of (1.1)
λ
∗
, then μ
λ
∗
(u
λ
∗
) = λ
∗
and the solution u
λ
∗
is
unique.
Proof. Define F :
R × H
1
0
(Ω) → H
−1
(Ω)by
F(λ,u)
= Δu − u + λ

f (u)+h(x)


. (6.16)
Let g(λ)
= μ
λ
(u
λ
) = inf

Ω
f

(u
λ
)v
2
dx=1
v
2
for λ ∈ (0,λ
∗
], then it is easy to see that g is con-
tinuous on (0,λ
∗
]. Since μ
λ
(u
λ
) >λfor λ ∈ (0,λ
∗
), so μ

λ
∗
(u
λ
∗
) ≥ λ
∗
.Ifμ
λ
∗
(u
λ
∗
) >λ
∗
,
the equation F
u
(λ
∗
,u
λ
∗
)φ = 0 has no nontrivial solution. From Lemma 6.2, F
u
maps
R × H
1
0
(Ω)ontoH

−1
(Ω). Applying the implicit function theorem to F, we can find
a neighborhood (λ
∗
− δ, λ
∗
+ δ)ofλ
∗
such that (1.1)
λ
possesses a solution u
λ
if λ ∈
(λ
∗
− δ, λ
∗
+ δ). This is contradictory to the definition of λ
∗
.Hence,weobtainthat
μ
λ
∗
(u
λ
∗
) = λ
∗
.
Next, we are going to prove that u

λ
∗
is unique. In fact, suppose (1.1)
λ
∗
has another
solution U
λ
∗
≥ u
λ
∗
.Setw = U
λ
∗
− u
λ
∗
;wehave
−Δw + w = λ
∗

f

w + u
λ
∗

−
f


u
λ
∗

, w>0inΩ. (6.17)
By μ
λ
∗
(u
λ
∗
) = λ
∗
, we have that the problem
−Δφ + φ = λ
∗
f


u
λ
∗

φ, φ ∈ H
1
0
(Ω) (6.18)
possesses a positive solution φ
1

.
Multiplying (6.17)byφ
1
and (6.18)byw, integrating and subtracting, we deduce that
0
=

Ω
λ
∗

f

w + u
λ
∗

−
f

u
λ
∗

−
f


u
λ

∗

w

φ
1
dx =
1
2

Ω
λ
∗
f


ξ
λ
∗

w
2
φ
1
dx, (6.19)
20 Boundary Value Problems
where ξ
λ
∗
∈ (u

λ
∗
,u
λ
∗
+ w). By condition (f5)
∗
,weobtainthatw ≡ 0. Thus, u
λ
∗
is
unique.

Proposition 6.5. Le t u
λ
be the minimal solution of (1.1)
λ
. Then u
λ
is uniformly bounded
in L
∞
(Ω) ∩ H
1
0
(Ω) for all λ ∈ (0,λ
∗
],and
u
λ

−→ 0 in L
∞
(Ω) ∩ H
1
0
(Ω) as λ −→ 0
+
. (6.20)
Proof. By (4.26), we have that


u
λ


≤
λ
1 − θ
h
2
(6.21)
for λ
∈ (0,λ
∗
), and u
λ
is strictly increasing with respect to λ, we can easily deduce that u
λ
is uniformly bounded in L
∞

(Ω) ∩ H
1
0
(Ω)forλ ∈ (0,λ
∗
]andu
λ
→ 0inH
1
0
(Ω)asλ → 0
+
.
By (3.8), (4.26), and u
λ
is uniformly bounded in L
∞
(Ω) ∩ H
1
0
(Ω), we have that


u
λ


∞
≤ C
1



u
λ


q
0
+ λC
2



u
λ


p
pq
0
+ h
q
0

≤
C
1


u

λ


(q
0
−2)/q
0
∞


u
λ


2/q
0
2
+C
3
λ ≤ C

λ
2/q
0
+ λ

,
(6.22)
where C is independent of λ,andλ
∈ (0,λ

∗
]. Hence, we obtain that u
λ
→ 0inL
∞
(Ω)as
λ
→ 0
+
. 
Proposition 6.6. For λ ∈ (0,λ
∗
),letU
λ
be the second solution of (1.1)
λ
constructed in
Section 5. Then U
λ
is unbounded in L
∞
(Ω) ∩ H
1
0
(Ω),and
lim
λ→0
+



U
λ


=
lim
λ→0
+


U
λ


∞
=∞. (6.23)
Proof. First, we show that {U
λ
: λ ∈ (0,λ
0
)} is unbounded in L
∞
(Ω)foranyλ
0
∈ (0,λ
∗
).
We proceed by contradiction. Assume to the contrary that there exists c
0
> 0suchthat



U
λ


∞
≤ c
0
< ∞∀λ ∈

0,λ
0

. (6.24)
Now, let ϕ
λ
be a minimizer of μ
λ
(U
λ
)forλ ∈ (0,λ
0
), that is,

Ω
f


U

λ

ϕ
2
λ
= 1,


ϕ
λ


2
= μ
λ

U
λ

. (6.25)
By condition (f1) and (6.24), there exists a constant M independent of λ,suchthat
f

(U
λ
(x)) ≤ M for all λ ∈ (0, λ
0
)andx ∈ Ω.Hence,by(6.25)andμ
λ
(U

λ
) <λfor all
λ
∈ (0,λ
0
), we obtain that
1
=

Ω
f


U
λ

ϕ
2
λ
≤ M


ϕ
λ


2
= Mμ
λ


U
λ

<Mλ. (6.26)
This is a contradiction for all λ<1/M.Hence,foranyλ
0
∈ (0,λ
∗
), we have that {U
λ
: λ ∈
(0,λ
∗
)} is unbounded in L
∞
(Ω). From this result, it is to be seen that lim
λ→0
+
U
λ

∞
=∞.
Now, we show that
{U
λ
: λ ∈ (0,λ
0
)} is unbounded in H
1

0
(Ω)foranyλ
0
∈ (0,λ
∗
). If
not, then there exists a constant M independent of λ such that


U
λ


≤
M ∀λ ∈

0,λ
0

. (6.27)
Tsing-San Hsu 21
Since U
λ
is a solution of (1.1)
λ
, and by condition (f2) and (6.27), we have that


U
λ



2
=

Ω
λf

U
λ

U
λ
dx +

Ω
λhU
λ
dx ≤ λC


Ω
U
2
λ
dx +

Ω
U
p+1

λ
dx

+ λh
2


U
λ


2
≤ λC



U
λ


2
+


U
λ


p+1


+ λh
2


U
λ


2
≤ λC
1
,
(6.28)
where C
1
is independent of λ. Without loss of generality, we may assume that q
0
= 2if
N
= 2,3 and N/2 <q
0
< 2
∗
/(p − 1) if N ≥ 4. By (3.8), (6.27), and the Sobolev embedding
theorem, we obtain that


U
λ



∞
≤ C
1


U
λ


q
0
+ λC
2



U
λ


p
pq
0
+ h
q
0

≤
C

1


U
λ


1−2/q
0
∞


U
λ


2/q
0
2
+ λC
2


U
λ


p−2
∗
/q

0
∞


U
λ


2
∗
/q
0
2
∗
+ λC
2
h
q
0
≤ C
3


U
λ


1−2/q
0
∞

+ λC
4


U
λ


1−(2
∗
−q
0
(p−1))/q
0
∞
+ λC
2
h
q
0
.
(6.29)
This implies that
1
≤ C
3


U
λ



−2/q
0
∞
+ λC
4


U
λ


−(2
∗
−q
0
(p−1))/q
0
∞
+ λC
2


h


q
0



U
λ


−1
∞
, (6.30)
where C
2
, C
3
,andC
4
are constants independent of λ.Now,letλ → 0
+
and by
lim
λ→0
+
U
λ

∞
= +∞, then we obtain a contradiction. Hence, {U
λ
: λ ∈ (0,λ
∗
)} is un-
bounded in H

1
0
(Ω)andlim
λ→0
+
U
λ
=+∞. This completes the proof of Proposition 6.6.

In order to get bifurcation results, we need the following bifurcation theorem which
can be found in Crandall and Rabinow itz [6].
Theorem B. Let X, Y be Banach space. Let (
λ,x) ∈ R × X and let F be a continuously
differentiable mapping of an open neighborhood of (
λ,x) into Y.LetthenullspaceN(F
x
(λ,
x)) = span{x
0
} be one-dimensional and codim R(F
x
(λ,x)) = 1.LetF
λ
(λ,x) ∈ R(F
x
(λ,x)).
If Z is the complement of span
{x
0
} in X, then the solutions of F(λ, x) = F(λ,x) near (λ,x)

form a curve (λ(s),x(s))
= (λ + τ(s),x + sx
0
+ z(s)),wheres → (τ(s), z(s)) ∈ R × Z is a con-
tinuously differentiable function near s
= 0 and τ(0) = τ

(0) = 0, z(0) = z

(0) = 0.
Proof of Theorems 1.1 and 1.2. First, we consider the case Ω
=
S \
D. Theorem 1.1 now
follows from Lemmas 4.4, 4.5, 6.4,andTheorem 5.5. The conclusions (i) and (ii) of
Theorem 1.2 follow immediately from Lemma 4.5, and Propositions 6.5, 6.6.Nowwe
are going to prove that (λ
∗
,u
λ
∗
)isabifurcationpointinC
2,α
(Ω) ∩ H
2
(Ω) by using an
idea in [13]. We also assume that h(x)isinC
α
(Ω) ∩ L
2

(Ω)anddefine
F :
R
1
× C
2,α
(Ω) ∩ H
2
(Ω) −→ C
α
(Ω) ∩ L
2
(Ω) (6.31)
by
F(λ,u)
= Δu − u + λf

u
+

+ λh, (6.32)
22 Boundary Value Problems
where C
2,α
(Ω) ∩ H
2
(Ω)andC
α
(Ω) ∩ L
2

(Ω) are endowed with the natural norm; then
they become Banach spaces. It can be easily verified that F( λ, u)isdifferentiable. From
Lemma 6.2 and Remark 6.3,weknowthat
F
u

λ,u
λ

w = Δw − w + λf


u
λ

w (6.33)
is an isomorphism of
R
1
× C
2,α
(Ω) ∩ H
2
(Ω)ontoC
α
(Ω) ∩ L
2
(Ω). It follows from implicit
function theorem that the solutions of F(λ,u)
= 0 near (λ,u

λ
) are given by a continuous
curve.
Now we are going to prove that (λ
∗
,u
λ
∗
) is a bifurcation point of F.Weshowfirst
that at the critical point (λ
∗
,u
λ
∗
), Theorem B applies. Indeed, from Lemma 6.4,problem
(6.18)hasasolutionφ
1
> 0inΩ. By the standard elliptic regular theory, we have that φ
1
∈
C
2,α
(Ω) ∩ H
2
(Ω)ifh ∈ C
α
(Ω) ∩ L
2
(Ω). Thus F
u

(λ
∗
,u
λ
∗
)φ = 0, φ ∈ C
2,α
(Ω) ∩ H
2
(Ω)
has a solution φ
1
> 0. This implies that N(F
u
(λ
∗
,u
λ
∗
)) = span{φ
1
}=1 is one dimen-
sional and codim R(F
u
(λ
∗
,u
λ
∗
)) = 1 by the Fredholm alternative. It remains to check that

F
λ
(λ
∗
,u
λ
∗
) ∈ R(F
u
(λ
∗
,u
λ
∗
)).
Assuming the contrary would imply existence of v
≡ 0suchthat
Δv
− v + λ
∗
f


u
λ
∗

v = f

u

λ
∗

+ h, v ∈ H
1
0
(Ω). (6.34)
From F
u
(λ
∗
,u
λ
∗
)φ
1
= 0, we conclude that

Ω
( f (u
λ
∗
(x)) + h(x))φ
1
(x) dx = 0. This is im-
possible because f (t) > 0, for t>0, u
λ
∗
(x) > 0, h(x) ≥ 0, h(x) ≡ 0andφ
1

(x) > 0forx ∈ Ω.
ApplyingTheoremB,weconcludethat(λ
∗
,u
λ
∗
) is a bifurcation point near which the
solution of (1.1)
λ
forms a curve (λ
∗
+ τ(s),u
λ
∗
+ sφ
1
+ z(s)) with s near s = 0andτ(0) =
τ

(0) = 0, z(0) = z

(0) = 0. We claim that τ

(0) < 0 which implies that the bifurcation
curve turns strictly to the left in (λ,u) plane. In order to obtain that τ

(0) < 0, we need
the following lemma.

Lemma 6.7. For R>0,letΩ

R
={x = (y,z) ∈ Ω : |z| <R}=(ω × B
R
) \ D,whereB
R
={z ∈
R
n
: |z| <R}. Suppose conditions (f1)–(f6) hold, then

Ω
f


u
λ
∗

φ
3
1
dx < +∞. (6.35)
Proof. Si nce u
λ
∗
(x) → 0as|x|→∞, and by conditions (f1) and (f3), we have that there is
R
1
> 0suchthat
0

= Δφ
1
− φ
1
+ λ
∗
f


u
λ
∗

φ
1
≤ Δφ
1
−
1
4
φ
1
for y ∈ ω, |z|≥R
1
. (6.36)
It is well known that the Dirichlet equation Δw
− (1/4)w =−w
p
in S has a positive
ground-state solution, denoted by

w (see [14] and the references there). We can mod-
ify the proof in Hsu [11] and obtain that for any ε>0with0<ε<1/4+λ
1
, there exist
constants C
ε
> 0andR
2
> 0suchthat
w(y,z) ≤ C
ε
ϕ(y)exp

−

1
4
+ λ
1
− ε|z|

for y ∈ ω, |z|≥R
2
, (6.37)
Tsing-San Hsu 23
where ϕ is the first positive eigenfunction of the Dirichlet problem
−Δϕ = λ
1
ϕ in ω.Now,
let ε

= (1/2)λ
1
.SinceΔw − (1/4)w =−w
p
≤ 0inS, hence by the maximum principle we
obtain that there exist constants C
1
> 0andR
3
> 0suchthat
φ
1
(x) ≤ C
1
ϕ(y)exp

−
1
2

1+2λ
1
|z|

for y ∈ ω, |z|≥R
3
. (6.38)
By condition (f6), (3.9), (6.38), and u
λ
∗

(x) → 0as|x|→∞, we have that there exist con-
stants C
2
> 0andR
0
≥ R
1
+ R
2
+ R
3
such that D ⊂⊂ ω × B
R
0
and
f


u
λ
∗

≤
C
2
u
q
1
−1
λ

∗
u
−1
λ
∗
(x) φ
2
1
(x) ≤ C
2
for x = (y,z) ∈ Ω \ Ω
R
0
, (6.39)
where 0 <q
1
< 4/(N − 2) if N ≥ 3, q
1
> 0ifN = 2.
By the strong maximum principle and modifying the proof in Lemma 3.6(i), we have
that u
−1
λ
∗
φ
1
∈ C
1
(Ω)andu
−1

λ
∗
φ
1
> 0onΩ. Therefore, there exists C
3
> 0suchthat
u
−1
λ
∗
(x) φ
1
(x) ≤ C
3
for x ∈ Ω
R
0
. (6.40)
Since u
λ
∗
≡ 0onU
=
∂D

(∂ω × B
R
0
)andu

λ
∗
is uniformly continuous on Ω
R
0
,andby
conditions (f5) and (f6), there exist δ>0andC
4
such that
f


u
λ
∗

≤
C
4
u
q
1
−1
λ
∗
for x ∈ U
δ
,
f



u
λ
∗

≤
C
4
for x ∈ Ω
R
0
\ U
δ
,
(6.41)
where
U
δ
=
U
δ

Ω
R
0
, U
δ
is a δ-tubular neighborhood of U,0<q
1
< 4/(N − 2) if N ≥ 3,

q
1
> 0ifN = 2.
From (6.38)–(6.41) and the H
¨
older inequality, we derive that

Ω
f


u
λ
∗

φ
3
1
dx =

U
δ
f


u
λ
∗

φ

3
1
dx +

Ω
R
0
\U
δ
f


u
λ
∗

φ
3
1
dx +

Ω\Ω
R
0
f


u
λ
∗


φ
3
1
dx
≤

U
δ
C
4
u
q
1
−1
λ
∗
φ
3
1
dx +

Ω
R
0
\U
δ
C
4
φ

3
1
dx +

Ω\Ω
R
0
C
2
u
q
1
−1
λ
∗
φ
3
1
dx
≤ C
3
C
4

U
δ
u
q
1
λ

∗
φ
2
1
dx + C
5
+ C
2
2

Ω\Ω
R
0
u
q
1
λ
∗
φ
1
dx
≤ C
6
+ C
2
2


u
λ

∗


q
1
q
1
+2


φ
1


(q
1
+2)/2
≤ C.
(6.42)

Since λ = λ
∗
+ τ(s), u = u
λ
∗
+ sφ
1
+ z(s)in
−Δu + u− λf(u) − λh = 0, u>0, u ∈ C
2,α

(Ω) ∩ H
2
(Ω). (6.43)
Differentiating (6.43)ins twice, we have
−Δu
ss
+ u
ss
− λf

(u)u
ss
− 2λ
s
f

(u)u
s
− λf

(u)

u
s

2
− λ
ss

f (u)+h


=
0. (6.44)
24 Boundary Value Problems
Setting here s
= 0 and using the facts that τ

(0) = 0, u
s
= φ
1
(x)andu = u
λ
∗
as s = 0, we
obtain
−Δu
ss
+ u
ss
− λ
∗
f


u
λ
∗

u

ss
− λ
∗
f


u
λ
∗

φ
2
1
− τ

(0)

f

u
λ
∗

+ h

=
0. (6.45)
Multiplying F
u
(λ

∗
,u
λ
∗
)φ
1
= 0byu
ss
and (6.45)byφ
1
, integrating and subtr acting the
result, and by (6.35), we obtain

Ω
λ
∗
f


u
λ
∗

φ
3
1
dx + τ

(0)


Ω

f

u
λ
∗

+ h

φ
1
dx = 0, (6.46)
which immediately gives τ

(0) < 0. Thus
u
λ
−→ u
λ
∗
in C
2,α
(Ω) ∩ H
2
(Ω)asλ −→ λ
∗
,
U
λ

−→ u
λ
∗
in C
2,α
(Ω) ∩ H
2
(Ω)asλ −→ λ
∗
,
(6.47)
and we complete the proof of Theorem 1.2 for Ω
=
S \
D.
With the same argument, we also have that Theorems 1.1 and 1.2 hold for Ω
=
R
N
\ D.
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Tsing-San Hsu: Center of General Education, Chang Gung University, Kwei-San,
Tao-Yuan 333, Taiwan
Email address: