The sign-changing solutions for nonlinear elliptic problem with Carrier type

Journal Pre-proof The sign-changing solutions for nonlinear elliptic problem with Carrier type

Fengfei Jin, Baoqiang Yan

PII:

S0022-247X(20)30164-5

DOI:

https://doi.org/10.1016/j.jmaa.2020.124002

Reference:

YJMAA 124002

To appear in:

Journal of Mathematical Analysis and Applications

Received date:

29 August 2019

Please cite this article as: F. Jin, B. Yan, The sign-changing solutions for nonlinear elliptic problem with Carrier type, J. Math. Anal. Appl. (2020), 124002, doi: https://doi.org/10.1016/j.jmaa.2020.124002.

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THE SIGN-CHANGING SOLUTIONS FOR NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE FENGFEI JIN AND BAOQIANG YAN

Abstract. In this paper, using the fixed-point index method, we are concerned with the existence of sign-changing solutions of the following problem ⎧    ⎨ |u(x)|γ dx Δu = f (u), x in Ω, − a+b (0.1) Ω ⎩ u(x) = 0, x on ∂Ω, where a > 0, b > 0 and Ω ⊆ RN is a bounded open set with smooth boundary and γ ≥ 1. Keywords: fixed-point index, sign-changing solutions, elliptic problem with Carrier type. MSC2010: 35J60, 35J75, 47H10.

1. Introduction In this paper, we consider the following nonlocal elliptic problem ⎧ ⎨ ⎩

   − a+b |u(x)|γ dx Δu = f (u), x in Ω, Ω

(1.1)

u(x) = 0, x on ∂Ω,

where a > 0, b > 0 and Ω ⊆ RN is a bounded open set with smooth boundary and γ ≥ 1. This problem is related to the stationary analogue of the problem   2  α−2 π ∂ u ∂2u 2 − 1 + |u(x, t)| dx = 0, in (0, π) × (0, +∞), ∂t2 2π 0 ∂x2

(1.2)

proposed by Carrier [7]. Problem (1.2) describes the vibration of the elastic string when the change of the tension is not very little. It is worth mentioning that problem (1.1) is similar to the following Kirchhoff equation

⎧ ⎨

   − a+b |∇u(x)|2 dx Δu = f (x, u), x in Ω,

(1.3) Ω u(x) = 0, x on ∂Ω. Problem (1.3) or its generalizations received much attention after Lions [31] pro⎩

posed an abstract framework to the problem, see [2], [5], [10], [14], [22], [23], [26], [27], [29], [30], [35], [36], [37], [38], [41], [42], [43], [44], [49], [50], [51] and references therein. For examples, Alves el. [2] used, for the first time, variational arguments 1

2

F. JIN, B. YAN

to obtain the existence of positive solutions for problem (1.3); in [38], Perera and Zhang got the nontrivial solutions of problem (1.3) via the Yang index; by variational method and quantitative deformation lemmas, some interesting results on the the existence of sign-changing solutions for Kirchhoff equations have been obtained in [26], [27], [36], [41], [42], [43], [49]; the works on the uniqueness of Kirchhoff equations can be found in [5], [30], [44]. In contrast to the much research achievement on problem  (1.3), workson problem (1.1) or its generalizations are very scarce because − a +

|u(x)|γ dx Δu is lack Ω

of variational structure and it is difficult to study problem (1.1) via variational method. Some authors focus on the existence of positive solutions for problem (1.1) or some generalized cases via the theory of topological theory, the method of lower and upper solutions and pseudomontone operators theory, see [1], [3], [4], [11, 12, 13], [15, 16, 17, 18, 19], [21], [46, 47, 48]. Up to now, to the best of our knowledge, there is no result on the existence of sign-changing solutions for problem (1.1). In this paper, using the fixed-point index method (see [24], [25]), based on the existence of at least one positive solution and at least one negative solutions, we obtain the existence of at least one sign-changing solution for problem (1.1). This paper is organized as follows. In Section 2, firstly, we list some lemmas which point out the existence of at least one positive solution and one negative solution for problem (1.1); secondly, we define some operators and construct some sets and present some properties of the operators and the sets; finally, we list an important lemma (Lemma 2.16) which denotes the formula of index of the Fr´echet differentiable operators on a neighborhood of isolated x0 . Section 3 presents the existence of at least one sign-changing solution for the problem (1.1) via the formula of the fixed-point index on some special open sets we construct. Some ideas of our proof come from [6], [9], [20], [32], [33], [45].

2. Preliminaries Let R = (−∞, +∞), R+ = [0, +∞). First, we introduce the sub-supersolution method for nonlocal problem: ⎧  ⎨−a uγ γ Δu = F (x, u), L ⎩

u = 0,

x on ∂Ω,

x in Ω, (2.1)

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

where Ω ⊆ RN is a smooth bounded domain, γ ∈ [1, +∞), uLγ =



|u(x)|γ dx

3

γ1

Ω

and a : R+ → (0, +∞) is continuous function with a0 =: a(0) = inf+ a(t) > 0. t∈R

Let C(Ω) = {u : Ω → R|u be a continuous function on Ω} with the norm u0 = maxx∈Ω |u(x)|. For the case that F (x, u) ≥ 0 for all (x, u) ∈ Ω × R, we have the following definition and lemma in [48]. Definition 2.1. The pair functions α and β with α, β ∈ C(Ω) ∩ C 2 (Ω) are subsolution and supersolution of (2.1) if α(x) ≤ β(x) for x ∈ Ω and ⎧ 1 ⎨ −Δα(x) ≤ F (x, α(x)), x in Ω, b0

⎩ α ≤ 0 ∂Ω and

⎧ ⎨

1 −Δβ(x) ≥ F (x, β(x)), a ⎩ β

≥ 0, 0

x in Ω,

∂Ω

where a0 = a(0) and b0 =

sup

 t∈[0, Ω max{|α(x)|,|β(x)|}γ dx]

a(t).

Lemma 2.2. (see [48]) Let Ω ⊆ RN (N ≥ 1) be a smooth bounded domain. Suppose that F : Ω × R → R is a continuous nonnegative function. Assume α and β are the subsolution and supersolution of (2.1) respectively. Then problem (2.1) has at least one solution u such that, for all x ∈ Ω, α(x) ≤ u(x) ≤ β(x). When F (x, u) ≤ 0 for all (x, u) ∈ Ω × R, we have the similar definition and lemma. Definition 2.3. The pair functions α and β with α, β ∈ C(Ω) ∩ C 2 (Ω) are supersolution and subsolution of (2.1) if α(x) ≥ β(x) for x ∈ Ω and ⎧ 1 ⎨ −Δα(x) ≥ F (x, α(x)), x in Ω, b0 ⎩ α

≥ 0 ∂Ω and

⎧ ⎨

1 −Δβ(x) ≤ F (x, β(x)), a ⎩ β

≤ 0, 0 ∂Ω

x in Ω,

4

F. JIN, B. YAN

where a0 = a(0) and b0 =

sup

 t∈[0, Ω max{|α(x)|,|β(x)|}γ dx]

a(t).

Lemma 2.4. Let Ω ⊆ RN (N ≥ 1) be a smooth bounded domain. Suppose that F : Ω × R → R is a continuous nonpositive function. Assume α and β are the supersolution and subsolution of (2.1) respectively. Then problem (2.1) has at least one solution u such that, for all x ∈ Ω, α(x) ≥ u(x) ≥ β(x). Proof. Let

⎧ ⎪ ⎨F (x, α(x)), if u > α(x); F (x, u) = F (x, u), if α(x) ≥ u ≥ β(x); ⎪ ⎩ F (x, β(x)), if u < β(x).

We introduce the following modified problem of (2.1) for λ > 0 ⎧ F (x, u) ⎨ −Δu + λu = + λχ(x, u), x ∈ Ω, a( Ω |χ(x, u(x))|γ dx) ⎩ u|∂Ω = 0,

(2.2)

where χ(x, u) = β(x) + (u − β(x))+ − (u − α(x))+ . In following we show that every solution u of (2.2) satisfies: α(x) ≥ u(x) ≥ β(x), x ∈ Ω. First, we prove that α(x) ≥ u(x) on Ω. Obviously, |χ(x, u(x))| ≤ max{|α(x)|, |β(x)|}, and so





a0 ≤ a

|χ(x, u(x))| dx γ

≤ b0 .

Ω

By contradiction, assume that minx∈Ω (α(x) − u(x)) = −m < 0. Note that α(x) − u(x) ≡ −m on Ω (α(x)−u(x) ≥ 0, x ∈ ∂Ω). Let x0 ∈ Ω be such that α(x0 )−u(x0 ) = −m. We have 0 ≥ −Δ(α(x0 ) − u(x0 )) 1 1 ≥ F (x0 , α(x0 )) − F (x0 , u(x0 )) − λχ(x0 , u(x0 )) + λu(x0 ) b0 a( Ω |χ(x, u(x))|γ dx) ≥ −λ(α(x0 ) − u(x0 )) > 0. This is a contradiction. Second, we prove that β(x) ≤ u(x) on Ω. By contradiction, assume maxx∈Ω (β(x)− u(x)) = M > 0. Note that β(x) − u(x) ≡ M on Ω (β(x) − u(x) ≤ 0, x ∈ ∂Ω). Let

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

5

x0 ∈ Ω be such that β(x0 ) − u(x0 ) = M . We have 0 ≤ −Δ(β(x0 ) − u(x0 )) 1 1 F (x0 , u(x0 )) − λχ(x0 , u(x0 )) + λu(x0 ) ≤ F (x0 , β(x0 )) − a0 a( Ω |χ(x, u(x))|γ dx) ≤ λ(u(x0 ) − β(x0 )) < 0. This is a contradiction. Consequently, α(x) ≥ u(x) ≥ β(x), x ∈ Ω. Other proof is the same as that of Theorem 2.4 in [48] and we omit it. 

The proof is completed.

Next, in order to study the poblem (1.1), we give some new symbols, definitions and lemmas. Let C 1 (Ω) = {u : Ω → R|u(x) is continuously differentiable on Ω} with the norm u = max{u0 , u1 }, where u0 = maxx∈Ω |u(x)|, u1 = maxx∈Ω |∇u(x)|. Obviously, C 1 (Ω) is a Banach space. Let P = {u ∈ C 1 (Ω)|u(x) ≥ 0, ∀x ∈ Ω}. Obviously, P is a cone in C 1 (Ω). Definition 2.5. Let u, v ∈ C 1 (Ω). We claim that u ≺ v if u(x) < v(x) on Ω and for x ∈ ∂Ω, either u(x) < v(x) or u(x) = v(x) and Define three operators

∂u ∂n

>

∂v ∂n .

 G(x, y)u(y)dy, x ∈ Ω, u ∈ C 1 (Ω),

(Ku)(x) :=

(2.3)

Ω

(F u)(x) :=

1 f (u(x)), x ∈ Ω, u ∈ C 1 (Ω), a + buγLγ

(2.4)

and (T u)(x) := (K(F u))(x), x ∈ Ω, u ∈ C 1 (Ω), where G(x, y) is the Green’s function of following problem

−u = h, x ∈ Ω, u|∂Ω = 0.

(2.5)

(2.6)

Suppose e(x) is the unique positive solution of the following problem

−u = 1, x ∈ Ω, u|∂Ω = 0. Obviously, from (2.6), we have

 G(x, y)dy.

e(x) = Ω

(2.7)

6

F. JIN, B. YAN

Define

 e0 := max |e(x)|, d := max x∈Ω

x∈Ω

|∇Gx (x, y)|dy.

(2.8)

Ω

Obviously, the fixed points of the operator T in C 1 (Ω) are the solutions of problem (1.1). Let λ1 < λ2 < λ3 < · · · < λk < · · · denote all the distinct eigenvalues of the following elliptic problem

−u = λu, x ∈ Ω, u|∂Ω = 0

(2.9)

and the corresponding eigenfunction ψk to λk (k = 1, 2, · · · ). Obviously, 1 ψk = Kψk , k = 1, 2, · · · . λk In the following, some conditions are listed for convenience: (H1 ) f : R → R is continuous and strictly increasing, f (0) = 0; (H2 ) there exist positive integer n0 and positive constant k > 0 such that λ2n0 <

k < λ2n0 +1 , a

where lim

|t|→0

f (t) = k > 0; t

(H3 ) lim

|t|→+∞

a f (t) < , t c

where c = max{1, e0 , d} and e0 , d are defined in (2.8). Now we list following lemmas. Lemma 2.6. Assume that (H1 ) and (H3 ) hold. Then there exists an L > 0 such that T u < L, ∀ u ∈ {u ∈ C 1 (Ω) : −L ≤ u(x) ≤ L, ∀ x ∈ Ω}, where T is defined by (2.5). Proof. Since (H3 ) holds, suppose that l :=

a f (t) < , c |t|→+∞ t lim

there is an L1 > 0 such that |f (t)| <

l + ac |t|, ∀ |t| > L1 . 2

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

Let M = maxt∈[−L1 ,

L1 ]

7

|f (t)| and L > max{L1 , c(M + 1)/a}. For t ∈ [−L, L], if

t ∈ [−L1 , L1 ], then |f (t)| < M + 1 <

a L c

(2.10)

and if t ∈ [−L, L] \ [−L1 , L1 ], then |f (t)| < Thus

and

l + ac a L < L. 2 c

T u0

= max |(T u)(x)| x∈Ω





1

= max

G(x, y)f (u(y))dy

γ x∈Ω a+ buLγ Ω 1 a < max G(x, y)dy L a x∈Ω Ω c ≤ L, ∀ u ∈ {u ∈ C 1 (Ω) | − L ≤ u(x) ≤ L}

T u1

= max |(∇T u)(x)| x∈Ω





1

= max

∇Gx (x, y)f (u(y))dy

γ x∈Ω a+ buLγ Ω 1 a < max |∇Gx (x, y)|dy L a x∈Ω Ω c ≤ L, ∀u ∈ {u ∈ C 1 (Ω) | − L ≤ u(x) ≤ L},

(2.11)

which implies that T u < L, ∀u ∈ {u ∈ C 1 (Ω) : −L ≤ u(x) ≤ L, ∀ x ∈ Ω}. 

The proof is completed. Let D = {u ∈ C 1 (Ω) |u0 ≤ L}.

(2.12)

(F0 u)(x) := f (u(x)), x ∈ Ω, u ∈ D

(2.13)

Define

and 1 (KF0 u)(x), x ∈ Ω, u ∈ D, a + bLγ |Ω| where K is defined by (2.3). (TL u)(x) :=

(2.14)

Lemma 2.7. (see [39]) Operator K : C(Ω) → C 1 (Ω) defined by (2.3) is continuous and compact. Lemma 2.8. Assume that (H1 ) holds. Then operators T : C(Ω) → C 1 (Ω) defined by (2.5) and TL : D → C 1 (Ω) defined by (2.14) are continuous and compact. Moreover, TL : D → C 1 (Ω) is strictly increasing.

8

F. JIN, B. YAN

Proof. From (H1 ), operators F , F0 : C(Ω) → C(Ω) are continuous and for any bounded set D1 ⊆ C(Ω), F D1 , F0 D1 are bounded. From Lemma 2.7, we have that T : C(Ω) → C 1 (Ω) and TL : D → C 1 (Ω) are continuous and compact. Moreover, for v(x) ≤ u(x) with v(x) ≡ u(x), we have ⎧ 1 ⎪ (f (u(x)) − f (v(x))) ⎨ −(TL u − TL v)(x) = a + bLγ |Ω| ≥ 0, x ∈ Ω, ⎪ ⎩ (TL u − TL v)|∂Ω = 0. The strong maximum principle guarantees that TL v ≺ TL u. 

The proof is completed.

Remark: Suppose that D1 is an open set of C 1 (Ω) satisfying there is an M > 0 such that u0 < M for all u ∈ D1 . Obviously, D1 is perhaps unbounded in C 1 (Ω). But by Lemma 2.8, we can define the degree of T on D1 (see [24]). Lemma 2.9. Assume that (H1 ) and (H3 ) hold. For any ε > 0, there exists a δ > 0 such that max |(F u)(x) − (F v)(x)| < ε, max |(F0 u)(x) − (F0 v)(x)| < ε x∈Ω

x∈Ω

for any u, v ∈ D with u − v < δ, where F and F0 are defined by (2.4) and (2.13). Proof. The continuity of f (t) implies that there exists a k0 > a + bLγ |Ω| such that |f (t)| ≤ k0 , ∀ t ∈ [−L, L], and for given ε > 0, there is a δ  > 0 such that |f (t1 ) − f (t2 )| <

1 a2 ε 2 k0

(2.15)

for all |t1 − t2 | < δ  , t1 , t2 ∈ [−L, L]. For u, v ∈ D, let C1 (u(x), v(x)) := f (u(x)) − f (v(x)). It deduces from (2.15) that if u − v < δ  , u, v ∈ D, then |C1 (u(x), v(x))| <

1 a2 ε, ∀ x ∈ Ω. 2 k0

Let C2 (u, v) := (a +

(2.16)

 bvγLγ )

− (a +

buγLγ )

 (|v(x)| − |u(x)| )dx . γ

=b Ω

γ

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

Since

9

 



γ γ

(|v(x)| − |u(x)| )dx



 Ω ≤ ||v(x)|γ − |u(x)|γ |dx Ω |γ(|u(x)| + θ(x)(|v(x)| − |u(x)|))γ−1 (|v(x)| − |u(x)|)|dx = Ω γLγ−1 |v(x) − u(x)|dx ≤ ≤

Ω

|Ω|γLγ−1 v − u,

one has |C2 (u, v)| <

1 a2 a2 ε. ε, if u − v < 2 k0 2k0 b(|Ω|γLγ−1 )

(2.17)

2

a Let δ1 = min{δ  , 2k0 b(|Ω|γL γ−1 ) ε}, from (2.16) and (2.17), we have



f (u(x)) f (v(x))



|F (u(x)) − F (v(x))| =

γ − a + bvγLγ



a + buLγ

f (u(x))(a + bvγLγ ) − f (v(x))(a + buγLγ )



=

(a + buγLγ )(a + vγLγ )



C1 (u(x), v(x))(a + bvγLγ ) + f (v(x))C2 (u, v)



<

a2 |C1 (u(x), v(x))|k0 + k0 |C2 (u, v)| < a2 ε ε < + 2 2 = ε, ∀ x ∈ Ω, if u − v < δ1 .

Hence, one has max |(F u)(x) − (F v)(x)| < ε x∈Ω

for all u − v < δ1 , u, v ∈ D. The same argument shows that there is a δ2 > 0 such that max |(F0 u)(x) − (F0 v)(x)| < ε x∈Ω

for all u − v < δ2 , u, v ∈ D. Let δ = min{δ1 , δ2 }. Obviously, max |(F u)(x) − (F v)(x)| < ε, x∈Ω

max |(F0 u)(x) − (F0 v)(x)| < ε x∈Ω

for any u, v ∈ D with u − v < δ. The proof is completed.



Lemma 2.10. Suppose that u ≥ 0, u ≡ 0. Then there exist two positive numbers δ(u) > 0 and M (u) > 0 such that M e(x) ≥ Ku(x) ≥ δe(x), x ∈ Ω and −M e(x) ≤ K(−u(x)) ≤ −δe(x), x ∈ Ω.

10

F. JIN, B. YAN

Proof. Since

−(Ku)(x) = u(x), x ∈ Ω, Ku|∂Ω = 0, by the strong maximum principle, we have Ku(x) > 0, ∀x ∈ Ω and ∂Ku < 0, ∀ x ∈ ∂Ω. ∂n By the properties of e, there is a δ > 0 such that Ku(x) ≥ δe(x), ∀ x ∈ Ω. On the other hand, choose M > maxx∈Ω |u(x)| + 1. Then

−(M e − Ku)(x) = M − u(x) > 0, x ∈ Ω, (M e − Ku)|∂Ω = 0. The strong maximum principle guarantees that M e(x) − Ku(x) ≥ 0, ∀ x ∈ Ω. 

The proof is completed.

Lemma 2.11. Suppose that (H1 ) and (H3 ) hold. For any ε > 0, there exists δ > 0 such that

−εe(x) ≤ (T u)(x) − (T v)(x) ≤ εe(x), t ∈ [0, 1], −εe(x) ≤ (TL u)(x) − (TL v)(x) ≤ εe(x), t ∈ [0, 1] for any u, v ∈ D with u − v < δ. Proof. For u, v ∈ D, from (2.7), one has |(T u)(x) − (T v)(x)|

= |(KF u)(x) − (KF v)(x)| ≤

G(x, y)dyF u − F v Ω

= F u − F ve(x), x ∈ Ω, which together with Lemma 2.9 implies that, for any ε > 0, there exists a δ1 > 0 such that −εe(x) ≤ (T u)(x) − (T v)(x) ≤ εe(x), x ∈ Ω for any u, v ∈ D with u − v < δ1 . Repeating above argument for TL , there is a δ2 > 0 such that −εe(x) ≤ (TL u)(x) − (TL v)(x) ≤ εe(x), x ∈ Ω for any u, v ∈ D with u − v < δ2 . Let δ = min{δ1 , δ2 }. Obviously

−εe(x) ≤ (T u)(x) − (T v)(x) ≤ εe(x), x ∈ Ω, −εe(x) ≤ (TL u)(x) − (TL v)(x) ≤ εe(x), x ∈ Ω

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

11

for any u, v ∈ D with u − v < δ. The proof is completed.



Lemma 2.12. Suppose that (H1 ) and (H3 ) hold. Then the operator T is Fr´echet differentiable at θ, and T  (θ) = ka K. Proof. Since T u − T θ − ka Ku0 u u→0 f (u(y)) maxx∈Ω | Ω G(x, y)( a+bu − ka u(y))dy| γ Lγ = lim u u→0



af (u(y)) − (a + buγLγ )ku(y)



= lim max

dy

G(x, y) a(a + buγLγ )u u→0 x∈Ω Ω γ |G(x, y)||o(|u(y)|) − uLγ bku(y)|dy ≤ lim max Ω a(a + buγLγ )u u→0 x∈Ω =0 lim

and T u − T θ − ka Ku1 u u→0 f (u(y)) maxx∈Ω | Ω ∇Gx (x, y)( a+bu − ka u(y))dy| γ Lγ = lim u u→0 |∇G (x, y)||o(|u(y)|) − uγLγ bku(y)|dy x ≤ lim max Ω γ a(a + buLγ )u u→0 x∈Ω = 0, lim

one has T u − T θ − ka Ku = 0, u u→0 lim

which means that the operator T is Fr´echet differentiable at θ and T  (θ) =

k a K.



The proof is completed.

Lemma 2.13. Assume that (H1 ), (H2 ) and (H3 ) hold. Then there exists a b1 > 0 such that problem (1.1) has at least one positive solution and one negative solution for all b ∈ (0, b1 ]. Proof. Let β1 (x) = Le(x)/e0 , β2 (x) = −Le(x)/e0 , where L is defined in Lemma 2.6, e is defined by (2.7) and e0 is defined in (2.8). From (2.10) and (2.11), we have |f (Le(x)/e0 )| < a

L L ≤a , ∀ x ∈ Ω, c e0

which implies that − β1 (x) ≥

1 f (β1 (x)), x ∈ Ω, a

(2.18)

12

F. JIN, B. YAN

and 1 f (β2 (x)), x ∈ Ω. (2.19) a From (H2 ), there exist δ0 > 0 and ε0 > 0 such that λ1 (a + δ0 ) < k, ε0 ψ1 (x) < − β2 (x) ≤

Le(x)/e0 for all x ∈ Ω and |f (±ε0 ψ1 (x))| ≥

k + λ1 (a + δ0 ) ε0 ψ1 (x), x ∈ Ω. 2

Let α2 (x) = −ε0 ψ1 (x), α1 (x) = ε0 ψ1 (x). Choose b1 > 0 such that b1

(2.20) Ω

|Le(x)/e0 |γ dx ≤

δ0 . From (2.20), we have − α1 (x) ≤

1 f (α1 (x)), x ∈ Ω, a + b Ω |Le(x)/e0 |γ dx

(2.21)

− α2 (x) ≥

1 f (α2 (x)), x ∈ Ω a + b Ω |Le(x)/e0 |γ dx

(2.22)



and

for all b ∈ (0, b1 ]. Combined (2.18), (2.19), (2.21) and (2.22), we have ⎧ 1 ⎨ −Δα1 (x) ≤ f (α1 (x)), x in Ω, b0

⎩ α = 0, 1  ⎧ ⎨

∂Ω

1 −Δβ1 (x) ≥ f (β1 (x)),

a β1 = 0,

x in Ω,

∂Ω

1 −Δα2 (x) ≥ f (α2 (x)), b0

⎩ α =0 2

x in Ω,

∂Ω

and



1 −Δβ2 (x) ≤ f (β2 (x)),

a β2 = 0,

x in Ω,

∂Ω

where

 |Le(x)/e0 |γ dx.

b0 = a + b Ω

Consequently, Lemma 2.2 and Lemma 2.4 guarantee that problem (1.1) has at least one positive solution w1 and at least one negative solution w2 with α1 (x) ≤ w1 (x) ≤ β1 (x), β2 (x) ≤ w2 (x) ≤ α2 (x), ∀ x ∈ Ω. The proof is completed.



Lemma 2.14. Assume that (H1 ), (H2 ) and (H3 ) hold. Let X = [−L, L] be an order interval of C 1 (Ω), where L is defined Lemma 2.6. Let S1 = {u|u ∈ X ∩ (−(P − {θ})), u = T u}, S2 = {u|u ∈ X ∩ (P − {θ}), u = T u}.

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

13

Then there exists a δ0 > 0 such that u(x) ≤ −δ0 e(x), x ∈ Ω, ∀ u ∈ S1 and u(x) ≥ δ0 e(x), x ∈ Ω, ∀ u ∈ S2 . where e = e(x) is defined by (2.7). Proof. Now Lemma 2.13 guarantees that S1 and S2 are not empty sets. Since f is strictly increasing, we get that (F u)(x) =

f (u(x)) f (0) < = 0, ∀ x ∈ Ω, u ∈ S1 , a + buγLγ a + buγLγ

which implies that (F u)(x) < 0, x ∈ Ω for any u ∈ S1 . It follows from Lemma 2.10 that for any u ∈ S1 , there exist two positive numbers δu > 0 and Mu > 0 such that δu e(x) ≤ (K(−F u))(x) ≤ Mu e(x), i.e. δu e(x) ≤ −u(x) ≤ Mu e(x), x ∈ Ω.

(2.23)

For fixed u ∈ S1 and given δu > 0, Lemma 2.11 guarantees that there exists a δu > 0 such that −

δu δu e(x) ≤ (T u)(x) − (T v)(x) ≤ e(x), x ∈ Ω, 2 2

i.e. δu δu e(x) ≤ u(x) − v(x) ≤ e(x), x ∈ Ω (2.24) 2 2 for any v ∈ N (u, δu ) ∩ S1 , where N (u, δu ) = {v ∈ C 1 (Ω) : u − v < δu }. −

Combining (2.23) and (2.24), we obtain that for any v ∈ N (u, δu ) ∩ S1 , v(x) ≤ u(x) +

δu δu e(x) ≤ − e(x), x ∈ Ω. 2 2

(2.25)

Obviously, ∪u∈S1 N (u, δu ) is an open cover of the set S1 . Lemma 2.8 guarantees that T is a continuous and compact operator and since S1 = T S1 , we have S1 is a relatively compact set. Consequently, there exists a finite open cover of S1 . Without loss of generality, we may assume that {N (u1 , δ1 ), N (u2 , δ2 ), · · · , N (un , δn )} is a finite open cover of S1 , that is S1 ⊆ ∪ni=1 N (ui , δi ). Let δ = min{

δu1 2

,

δu2 2

, ··· ,

δun 2

}. It follows from (2.25) that

v(x) ≤ −δ  e(x), x ∈ Ω, ∀v ∈ S1 .

(2.26)

14

F. JIN, B. YAN

Repeating the same argument for S2 , it is easy to see that there exists a δ  > 0 such that v(x) ≥ δ  e(x), t ∈ Ω, ∀v ∈ S2 . 

(2.27)



Let δ0 = min{δ , δ }. Now (2.26) and (2.27) guarantees that u(x) ≤ −δ0 e(x), x ∈ Ω, ∀u ∈ S1 and u(x) ≥ δ0 e(x), x ∈ Ω, ∀u ∈ S2 . 

The proof is completed.

Let a and k be the positive numbers in (H2 ). It follows from (2.9) that the eigenvalues of the linear operator

k aK

are

k k , , aλ1 aλ2

··· ,

k , ··· aλn

and the algebraic multiplicity of each positive eigenvalue k aK

k aλn

of the linear operator

is equal to 1.

Lemma 2.15. Suppose that (H1 ), (H2 ) and (H3 ) hold, ψ1 is an eigenfunction corresponding to the eigenvalue

k aλ1

of the linear operator

k a K.

Then there exist

two positive constants b2 > 0 and τ0 > 0 such that

T (−τ ψ1 ) ≺ −τ ψ1 , τ ψ1 ≺ T (τ ψ1 ), TL (−τ ψ1 ) ≺ −τ ψ1 , τ ψ1 ≺ TL (τ ψ1 ) for any τ ∈ (0, τ0 ] and b ∈ (0, b2 ]. Proof. From (H2 ), we have lim

τ →0+,b→0+

f (τ ψ1 (x)) = k, τ →0+ τ ψ1 (x)

(a + bτ ψ1 γLγ ) = a and lim

and so k 1 f (τ ψ1 (x)) = > λ1 . a + bτ ψ1 γLγ τ ψ1 (x) a Therefore, there exist τ1 > 0 and b1 > 0 such that 1 f (τ ψ1 (x)) − τ λ1 ψ1 (x) > 0, ∀ x ∈ Ω, ∀ τ ∈ (0, τ1 ], b ∈ (0, b1 ]. a + bτ ψ1 γLγ Hence ⎧ 1 ⎪ ⎪ −(KF (τ ψ1 )(x) − τ ψ1 (x)) = f (τ ψ1 (x)) − τ λ1 ψ1 (x) ⎨ a + bτ ψ1 γLγ > 0, ∀ x ∈ Ω, ∀ τ ∈ (0, τ1 ], b ∈ (0, b1 ], ⎪ ⎪ ⎩ (KF (τ ψ ) − τ ψ )| 1 1 x∈∂Ω = 0. lim

τ →0+,b→0+

The strong maximum principle guarantees that τ ψ1 ≺ T (τ ψ1 ), ∀ τ ∈ (0, τ1 ], b ∈ (0, b1 ].

(2.28)

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

15

Similarly, there exist τ2 > 0 and b2 > 0 such that T (−τ ψ1 ) ≺ −τ ψ1

(2.29)

for all τ ∈ (0, τ2 ], b ∈ (0, b2 ]. From (H2 ), we have lim (a + bLγ |Ω|) = a and lim

τ →0+

b→0+

f (τ ψ1 (x)) = k, τ ψ1 (x)

and so f (τ ψ1 (x)) k 1 = > λ1 . a + bLγ |Ω| τ ψ1 (x) a Therefore, there exist b3 > 0 and τ3 > 0 such that 1 f (τ ψ1 (x)) − τ λ1 ψ1 (x) > 0, ∀ x ∈ Ω, ∀ τ ∈ (0, τ3 ], b ∈ (0, b3 ]. a + bLγ |Ω| lim

τ →0+,b→0+

Hence ⎧ ⎪ ⎨ −(TL (τ ψ1 )(x) − τ ψ1 (x)) ⎪ ⎩

(TL (τ ψ1 ) − τ ψ1 )|x∈∂Ω = 0.

1 f (τ ψ1 (x)) − τ λ1 ψ1 (x) a + bLγ |Ω| > 0, ∀ x ∈ Ω, ∀ τ ∈ (0, τ3 ], b ∈ (0, b3 ],

=

The strong maximum principle guarantees that τ ψ1 ≺ TL (τ ψ1 ), ∀ τ ∈ (0, τ3 ], b ∈ (0, b3 ].

(2.30)

Similarly, there exist τ4 > 0 and b4 > 0 such that TL (−τ ψ1 ) ≺ −τ ψ1 , ∀ τ ∈ (0, τ4 ],

b ∈ (0, b4 ].

(2.31)

Choose b2 = min{b1 , b2 , b3 , b4 } and τ0 = min{τ1 , τ2 , τ3 , τ4 }. From (2.28), (2.29), (2.30) and (2.31), one has

T (−τ ψ1 ) ≺ −τ ψ1 , τ ψ1 ≺ T (τ ψ1 ), TL (−τ ψ1 ) ≺ −τ ψ1 , τ ψ1 ≺ TL (τ ψ1 ) for any τ ∈ (0, τ0 ] and b ∈ (0, b2 ]. 

The proof is completed.

Lemma 2.16. (see [26]) Let E be a Banach space, H : E → E be a completely continuous operator, let x0 ∈ E be a fixed point of H and assume that H is defined in a neighborhood of x0 and Fr´echet differentiable at x0 . If 1 is not an eigenvalue of the linear operator H  (x0 ), then x0 is an isolated singular point of the completely continuous vector field Φ = I − H and ind(I − H, x0 ) = (−1)k , where k is the sum of the algebraic multiplicities of the real eigenvalues of H in (1, +∞).

16

F. JIN, B. YAN

3. Main Result In this section, using fixed point index method, we have the following theorem. Theorem 3.1. Suppose that (H1 )−(H3 ) hold. Then there exists a positive constant b > 0 such that the problem (1.1) has at least one sign-changing solution, one positive solution and one negative solution for all b ∈ (0, b]. Proof. Step 1. We establish a subspace X of C 1 (Ω) and there exists a positive constant b1 > 0 such that the problem (1.1) has at least one positive solution and one negative solution in X for all b ∈ (0, b1 ]. From condition (H1 ), Lemma 2.8 guarantees that operator T : C 1 (Ω) → C 1 (Ω) is continuous and compact. Obviously, u0 (x) ≡ 0 (for x ∈ Ω) is a zero solution of problem (1.1). And moreover, since (H1 ) and (H3 ) hold, Lemma 2.6 is true, i.e. there exists an L > 0 such that T u < L, ∀ − L ≤ u(x) ≤ L, u ∈ C 1 (Ω).

(3.1)

u1 = −L, v2 = L and X = [u1 , v2 ].

(3.2)

Let

Let d(u, v) = u − v. It is easy to see that (X, d) is a complete metric subspace of C 1 (Ω) and Lemma 2.6 guarantees that T (X) ⊂ X. By the definition of D in (2.12), we have X ⊆ D. Now Lemma 2.8 implies that TL : X → C 1 (Ω) is continuous, compact and strictly increasing. Furthermore, Lemma 2.13 guarantees that there is a b1 > 0 such that for all b ∈ (0, b1 ], problem (1.1) has at least one negative solution w2 ∈ [−L, 0) and at least one positive solution w1 ∈ (0, L] for all b ∈ (0, b1 ]. Let S1 = {u|u ∈ X ∩ (−(P − {θ})), u = T u}, S2 = {u|u ∈ X ∩ (P − {θ}), u = T u}. Obviously, S1 and S2 are not empty for all b ∈ (0, b1 ]. According to Lemma 2.14, there exists a δ0 > 0 such that u(x) ≤ −δ0 e(x), x ∈ Ω, ∀ u ∈ S1 ; u(x) ≥ δ0 e(x), x ∈ Ω, ∀ u ∈ S2 , where e = e(x) is defined by (2.7). Since ψ1 satisfies 1 ψ1 = Kψ1 , λ1

(3.3)

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

17

Lemma 2.10 implies that there exist two positive constants ε > 0 and M > 0 such that εe(x) ≤ (Kψ1 )(x) ≤ M e(x), x ∈ Ω and so ελ1 e(x) ≤ ψ1 (x) ≤ M λ1 e(x), x ∈ Ω. From Lemma 2.15,

lim

τ →0,b→0+

T (τ ψ1 )0 = 0 and

lim

τ →0,b→0+

TL (τ ψ1 )0 = 0(note

f (0) = 0 and f (t) is continuous on t = 0), there exist b2 > 0 and τ1 > 0 with δ0 L τ1 < min{ cψ , 4ψ } such that 1 1 ⎧ τ1 ψ1 ≺ δ0 e, −δ0 e ≺ −τ1 ψ1 , ⎪ ⎪ ⎨ L L − ≺ T (−τ1 ψ1 ) ≺ −τ1 ψ1 , τ1 ψ1 ≺ T (τ1 ψ1 ) ≺ , 2 2 ⎪ ⎪ L L ⎩ − ≺ TL (−τ1 ψ1 ) ≺ −τ1 ψ1 , τ1 ψ1 ≺ TL (τ1 ψ1 ) ≺ , ∀b ∈ (0, b2 ], 2 2 where δ0 comes from (3.3) and L is defined in (3.1).

(3.4)

Step 2. We claim that there is a b > 0 with b ≤ b1 such that problem (1.1) has at least one sign-changing solution in X for all b ∈ (0, b]. Choose b = min{b1 , b2 }. Let v1 = −τ1 ψ1 , u2 = τ1 ψ1 . From (3.2), (3.3) and (3.4), we have u1 ≺ v1 ≺ θ ≺ u2 ≺ v2 ,

(3.5)

TL v1 ≺ v1 , u2 ≺ TL u2 ,

(3.6)

u2 ≺ δ0 e ≺ u ≺ v2 , ∀ u ∈ S2 ; u1 ≺ u ≺ −δ0 e ≺ v1 , ∀ u ∈ S1 .

(3.7)

and

(1) We construct two disjoint open sets in X: Ω1 = {u ∈ X : there exists λ > 0 such that TL u ≤ TL v1 − λe} ∩ {u ∈ X : x ≺ θ} and Ω2 = {u ∈ X : there exists λ > 0 such that TL u ≥ TL u2 + λe} ∩ {u ∈ X : θ ≺ u}. First we show that Ω1 = ∅. From (3.5), we have u1 ≺ v1 , which togther with the monotonicity of f (t) means that f (u1 (x)) < f (v1 (x)), for all x ∈ Ω. Now Lemma 2.11 guarantees that there is a δ1 > 0 (TL v1 )(x) − (TL u1 )(x)

1 K(F0 v1 − F0 u1 )(x) a + bLγ |Ω| ≤ δ1 e(x),

=

and so (TL u1 )(x) ≤ (TL v1 )(x) − δ1 e(x),

18

F. JIN, B. YAN

which implies that u1 ∈ Ω1 . Hence, Ω1 = ∅. Next, we show that Ω2 = ∅. From (3.5), we have u2 ≺ v2 , which togther with the monotonicity of f (t) means that f (u2 (x)) < f (v2 (x)), for all x ∈ Ω. Now Lemma 2.11 implies that there is δ2 > 0 such that (TL v2 )(x) − (TL u2 )(x)

1 K(F0 v2 − F0 u2 ))(x) a + bLγ  ≥ δ2 e(x),

=

and so (TL v2 )(x) ≥ (TL u2 )(x) + δ2 e(x), which means that v2 ∈ Ω2 . Hence, Ω2 = ∅. Obviously, Ω1 ∩ Ω2 = ∅. Finally, we show that Ω1 and Ω2 are relatively open to X. Let u0 ∈ Ω1 . The definition of Ω1 guarantees that there exists a λ0 > 0 such that (TL u0 )(x) ≤ (TL v1 )(x) − λ0 e(x), x ∈ Ω.

(3.8)

From Lemma 2.11, for λ0 > 0, there exists a δ1 > 0 such that −

λ0 λ0 e(x) ≤ (TL u)(x) − (TL u0 )(x) ≤ e(x), x ∈ Ω, ∀ u − u0  < δ1 . 2 2

(3.9)

Then, for u ∈ B(u0 , δ1 ) ∩ X, from (3.8) and (3.9), one has (TL u)(x) ≤ (TL u0 )(x) +

λ0 e(x) 2

≤ (TL v1 )(x) − λ0 e(x) + = (TL v1 )(x) −

λ0 e(x) 2

λ0 e(x), x ∈ Ω, 2

i.e. (B(u0 , δ1 ) ∩ X) ⊂ Ω1 , which implies that Ω1 is relatively open to X. Let v0 ∈ Ω2 . The definition of Ω2 infers that there exists a λ1 > 0 such that (TL v0 )(x) ≥ (TL u2 )(x) + λ1 e(x), x ∈ Ω.

(3.10)

From Lemma 2.11, for λ1 > 0, there exists a δ2 > 0 such that −

λ1 λ1 e(x) ≤ (TL u)(x) − (TL v0 )(x) ≤ e(x), x ∈ Ω, ∀ u − v0  < δ2 . 2 2

(3.11)

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

19

Then, for u ∈ B(v0 , δ2 ) ∩ X, from (3.10) and (3.11), one has (TL u)(x) ≥ (TL v0 )(x) −

λ1 e(x) 2

≥ (TL u2 )(x) + λ1 e(x) − = (TL u2 )(x) +

λ1 e(x) 2

λ1 e(x), x ∈ Ω, 2

i.e. (B(v0 , δ2 ) ∩ X) ⊂ Ω2 , which implies that Ω2 is relatively open to X. (2) We show that i(T, Ω1 , X) = 1, i(T, Ω2 , X) = 1. First, we prove that u = μT u + (1 − μ)u1 , ∀ u ∈ ∂X Ω1 , μ ∈ [0, 1],

(3.12)

where ∂X Ω1 denotes the boundary of Ω1 relative to X. In fact, suppose that there exists a u0 ∈ ∂X Ω1 and μ0 ∈ [0, 1] such that u0 = μ0 T u0 + (1 − μ0 )u1 . There are two cases for μ0 ∈ [0, 1]. (a) μ0 = 0. Then u0 = u1 , which contradicts to u1 ∈ Ω1 . (b) μ0 ∈ (0, 1]. Since u0 ∈ ∂X Ω1 , one has u0 ≤ θ, which yields that T u0 ≤ TL u0 ,

(3.13)

and there exists a sequence {zn } ⊆ Ω1 with zn − u0  → 0 as n → +∞. Since zn ∈ Ω1 , there is a λn > 0 such that TL u1 ≤ TL zn ≤ TL v1 − λn e.

(3.14)

Obviously {λn } is bounded (otherwise TL u1  = +∞). Without loss of generality, we assume that λn → λ0 ≥ 0,

as n → +∞.

Letting n → +∞ in (3.14), we have TL u0 ≤ TL v1 − λ0 e ≤ TL v1 ,

20

F. JIN, B. YAN

which together with (3.6) and (3.13) implies that u0

= μ0 T u0 + (1 − μ0 )u1 ≤ μ0 TL u0 + (1 − μ0 )u1 ≤ μ0 TL v1 + (1 − μ0 )u1 < μ0 v1 + (1 − μ0 )v1 = v1 .

The monotonicity of F0 means (F0 v1 )(x) − (F0 u0 )(x) > 0, ∀ x ∈ Ω. Now Lemma 2.11 implies that there exists a δ1 > 0 such that (TL v1 )(x) − (TL u0 )(x) = ≥

1 (KF0 v1 − KF0 u0 )(x) a + bLγ  δ1 e(x),

i.e. TL u0 ≤ TL v1 − δ1 e, which implies that u0 ∈ Ω1 . This contradicts to u0 ∈ ∂X Ω1 . Hence, (3.12) holds. Second, we prove that v = μT v + (1 − μ)v2 , ∀ v ∈ ∂X Ω2 , μ ∈ [0, 1],

(3.15)

where ∂X Ω2 denotes the boundary of Ω2 relative to X. In fact, suppose that there exist v0 ∈ ∂X Ω2 and μ1 ∈ [0, 1] such that v0 = μ1 T v0 + (1 − μ1 )v2 . There are two cases for μ1 ∈ [0, 1]. (a) μ1 = 0. Then v0 = v2 , which contradicts to v2 ∈ Ω2 . (b) μ1 ∈ (0, 1]. Since v0 ∈ ∂X Ω2 , one has v0 ≥ θ, and so T v0 ≥ TL v0 ,

(3.16)

and there is a sequence {zn } ⊆ Ω2 with zn − v0  → 0 as n → +∞. Since zn ∈ Ω2 , there is a λn > 0 such that TL v2 ≥ TL zn ≥ TL u2 + λn e.

(3.17)

Obviously {λn } is bounded (otherwise TL v2  = +∞). Without loss of generality, we assume that λn → λ0 ,

as n → +∞.

Letting n → +∞ in (3.17), we have TL v0 ≥ TL u2 + λ0 e ≥ TL u2 ,

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

21

which together with (3.6), (3.16) implies that v0

= μ1 T v0 + (1 − μ1 )v2 ≥ μ1 TL v0 + (1 − μ1 )v2 ≥ μ1 TL u2 + (1 − μ1 )v2 > μ1 u2 + (1 − t0 )u2 = u2 .

The the monotonicity of F0 means (F0 v0 )(x) − (F0 u2 )(x) > 0, ∀x ∈ Ω. Now Lemma 2.11 implies that there is a δ2 > 0 such that 1 (TL v0 − TL u2 )(x) = (KF0 v0 − KF0 u2 )(x) a + bLγ  ≥ δ2 e(x), i.e. TL v0 ≥ TL u2 + δ2 e, which implies that v0 ∈ Ω2 . This contradicts to v0 ∈ ∂X Ω2 . Hence (3.15) holds. Finaly, from (3.12) and (3.15), we have i(T, Ω1 , X) = i(u1 , Ω1 , X) = 1

(3.18)

i(T, Ω2 , X) = i(v2 , Ω2 , X) = 1.

(3.19)

and

(3) We show that there is a δ > 0 small enough such that i(T, X − (Ω1 ∪ Ω2 ∪ B(θ, δ)), X) = −2, which guarantees that T has at least one fixed point w3 in X − (Ω1 ∪ Ω2 ∪ B(θ, δ)). By (H1 ) and Lemma 2.14, θ is a isolated fixed point of T . It follows from Lemma 2.16 that ind(I − T, θ) = (−1)k , where k is the sum of the algebraic multiplicities of the real eigenvalues of T  (θ) in (1, +∞). On the other hand, by Lemma 2.12, the real eigenvalues of T  (θ) = ka K in (1, +∞) are k k k , , ··· , . aλ1 aλ2 aλ2n0 By Lemma 2.16 and (H2 ), we get that ind(I − T, θ) = (−1)2n0 = 1. Therefore, there exists a small enough neighborhood B(θ, δ) ⊆ C 1 (Ω) such that θ is the only fixed point of T in B(θ, δ) and deg(I − T, B(θ, δ), θ) = ind(I − T, θ) = 1.

(3.20)

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F. JIN, B. YAN

By the definition of fixed point index in [24], we have i(T, B(θ, δ), X) = deg(I − T · r, DR ∩ r−1 (B(θ, δ)), θ),

(3.21)

where r : C 1 (Ω) → X is an arbitrary retraction and R > 0 is big enough such that B(θ, δ) ⊆ DR = {u ∈ C 1 (Ω) : u < R}. We claim that θ is the only fixed point of T · r in DR ∩ r−1 (B(θ, δ)). Assume that u ∈ DR ∩ r−1 (B(θ, δ)) such that u = T ·r(u). Since r(C 1 (Ω)) ⊆ X and T (X) ⊆ X, then u ∈ X. Since r : C 1 (Ω) ⊆ X is a retraction, we have u = r(u) ∈ B(θ, δ), and u = T u. On the other hand, θ is the only fixed point of T in B(θ, δ). Then we get that u = θ. By the excision property of the Leray-Schauder degree, we have deg(I − T · r, DR ∩ r−1 (B(θ, δ)), θ) = deg(I − T, B(θ, δ), θ).

(3.22)

It follows from (3.20), (3.21) and (3.22) that i(T, B(θ, δ), X) = 1.

(3.23)

Obviously, from (3.1), one has i(T, X, X) = 1.

(3.24)

It deduces from (3.18), (3.19), (3.23) and (3.24) that i(T, X − (Ω1 ∪ Ω2 ∪ B(θ, δ)), X) = 1 − 1 − 1 − 1 = −2, which implies that T has at least one fixed point w3 ∈ X − (Ω1 ∪ Ω2 ∪ B(θ, δ)). Consequently, problem (1.1) has a solution w3 in X − (Ω1 ∪ Ω2 ∪ B(θ, δ)) also. In the following we prove that w3 is sign-changing. For any u ∈ S1 , from (3.7), we have u ≺ v1 ≺ θ, which together with the strict monotonicity of f implies that f (u(x)) < f (v1 (x)) < 0, ∀ x ∈ Ω. Now Lemma 2.11 guarantees that there is a δ1 > 0 1 K(F0 v1 − F0 u)(x) (TL v1 )(x) − (TL u)(x) = a + bLγ |Ω|  ≤ δ1 e(x), and so (TL u)(x) ≤ (TL v1 )(x) − δ1 e(x), which implies that u ∈ Ω1 , i.e. S1 ⊆ Ω1 . Similarly, we can prove S2 ⊆ Ω2 .

NONLINEAR ELLIPTIC PROBLEM WITH CARRIER TYPE

23

Hence, w3 ∈ X − (Ω1 ∪ Ω2 ∪ B(θ, δ)) implies that w3 ∈ X − (P ∪ (−P )). Thus, w3 is a sign-changing solution of problem (1.1).



Acknowledgement: The authors thank the referee for the suggestions. The work is supported by the National Natural Science Foundation of China(61603226) and the Fund of Natural Science of Shandong Province (ZR2018MA022).

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F. JIN, B. YAN

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