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Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 145, pp. 1–23.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

FRACTIONAL ELLIPTIC EQUATIONS WITH SIGN-CHANGING AND SINGULAR NONLINEARITY

Abstract. In this article, we study the fractional Laplacian equation with singular nonlinearity

(−∆)su=a(x)u−q+λb(x)up in Ω, u >0 in Ω, u= 0 in∂Ω,

where Ω is a bounded domain inRn with smooth boundary∂Ω,n >2s,s (0,1),λ >0. Using variational methods, we show existence and multiplicity of positive solutions.

1. Introduction

Let Ω⊂Rnbe a bounded domain with smooth boundary,n >2sands∈(0,1).

We consider the fractional elliptic problem with singular nonlinearity (−∆)su=a(x)u−q+λb(x)up in Ω,

u >0 in Ω, u= 0 in ∂Ω. (1.1) We use the following assumptions onaandb:

(A1) a: Ω⊂Rn→Rsuch that 0< a∈L

2 s 2

s−1+q(Ω).

(A2) b : Ω⊂Rn →Ris a sign-changing function such thatb+ 6≡0 andb(x)∈ L

2 s 2

s−1−p(Ω).

Here λ > 0 is a parameter, 0 < q < 1 < p < 2s−1, with 2s = n−2s2n , known as fractional critical Sobolev exponent and where (−∆)s is the fractional Laplacian operator in Ω with zero Dirichlet boundary values on∂Ω.

To define the fractional Laplacian operator (−∆)s in Ω, let {λk, φk} be the eigenvalues and the corresponding eigenfunctions of −∆ in Ω with zero Dirichlet boundary values on∂Ω

(−∆)sφkkφk in Ω, φk = 0 on∂Ω.

normalized bykφkkL2(Ω)= 1. Then one can define the fractional Laplacian (−∆)s fors∈(0,1) by

(−∆)su=

X

k=1

λskckφk,

2010Mathematics Subject Classification. 35A15, 35J75, 35R11.

Key words and phrases. Non-local operator; singular nonlinearity; Nehari manifold;

sign-changing weight function.

c

2016 Texas State University.

Submitted January, 21, 2016. Published June 14, 2016.

1

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which clearly maps H0s(Ω) :=

u=

X

k=1

ckφk ∈L2(Ω) :kukHs0(Ω)=X

k=1

λskc2k1/2

<∞ intoL2(Ω). MoreoverH0s(Ω) is a Hilbert space endowed with an inner product

X

k=1

ckφk,

X

k=1

dkφk

Hs0(Ω)=

X

k=1

λskckdkφk, if

X

k=1

ckφk,

X

k=1

dkφk ∈H0s(Ω).

Definition 1.1. A function u∈ H0s(Ω) such that u(x)> 0 in Ω is a solution of (1.1) such that for every functionv∈H0s(Ω), it holds

Z

(−∆)s/2u(−∆)s/2vdx= Z

a(x)u−qvdx+λ Z

b(x)upvdx.

Associated with (1.1), we consider the energy functional for u∈H0s(Ω),u > 0 in Ω such that

Iλ(u) = Z

|(−∆)s/2u|2dx− 1 1−q

Z

a(x)|u|1−qdx− λ p+ 1

Z

b(x)|u|p+1dx.

The fractional power of Laplacian is the infinitesimal generator of L´evy stable diffusion process and arise in anomalous diffusions in plasma, population dynamics, geophysical fluid dynamics, flames propagation, chemical reactions in liquids and American options in finance. For more details, we refer to [3, 14] and reference therein.

Recently the study of existence, multiplicity of solutions for fractional elliptic equations attracted a lot of interest by many researchers. Among the works dealing with fractional elliptic equations we cite [6, 9, 21, 22, 23, 24, 25, 26] and references therein, with no attempt to provide a complete list. Caffarelli and Silvestre [8]

gave a new formulation of fractional Laplacian through Dirichlet-Neumann maps.

This formulation transforms problems involving the fractional Laplacian into a local problem which allows one to use the variational methods.

On the other hand, there are some works where multiplicity results are shown using the structure of associated Nehari manifold. In [15, 16] authors studied subcritical problems and in [28] the authors obtained the existence of multiplicity for critical growth nonlinearity. In the case of the square root of Laplacian, the multiplicity results for sublinear and superlinear type of nonlinearity with sign- changing weight functions are studied in [7, 27].

In the local setting, s= 1, the paper by Crandall, Robinowitz and Tartar [12]

is the starting point on semilinear problem with singular nonlinearity. There is a large body of literature on singular problems, see [1, 2, 11, 12, 17, 18, 19, 20] and reference therein. Recently, Chen and Chen in [10] studied the following problem with singular nonlinearity

−∆u− λ

|x|2 =h(x)u−q+µW(x)up in Ω\ {0}, u >0 in Ω\ {0}, u= 0 on∂Ω, where 0 ∈ Ω⊂ Rn(n ≥ 3) is a bounded smooth domain with smooth boundary, 0 < λ < (n−2)4 2 and 0 < q < 1 < p < n+2n−2. Also h, W both are continuous functions in Ω with h >0 and W is sign-changing. By variational methods, they showed that there exists Tλ such that forµ ∈(0, Tλ) the above problem has two positive solutions.

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In case of the fractional Laplacian, Fang [13] proved the existence of a solution of the singular problem

(−∆)sw=w−p, u >0 in Ω, u= 0 inRn\Ω,

with 0< p < 1, using the method of sub and super solution. Recently, Barrios, Peral and et al [4] extend the result of [13]. They studied the existence result for the singular problem

(−∆)su=λf(x)

uγ +M up, u >0 in Ω, u= 0 inRn\Ω,

where Ω is a bounded smooth domain of Rn, n > 2s, 0 < s < 1,γ > 0, λ > 0, p >1 andf ∈Lm(Ω), m≥1 is a nonnegative function. For M = 0, they proved the existence of solution for every γ >0 and λ > 0. ForM = 1 andf ≡1, they showed that there exist Λ such that it has a solution for every 0 < λ < Λ, and have no solution forλ >Λ. Here the authors first studied the uniform estimates of solutions{un}of the regularized problems

(−∆)su=λ f(x)

(u+1n)γ +up, u >0 in Ω, u= 0 inRn\Ω. (1.2) Then they obtained the solutions by taking limit in the regularized problem (1.2).

As far as we know, there is no work related to fractional Laplacian for singular nonlinearity and sign-changing weight functions. So, in this paper, we study the multiplicity results for problem (1.1) for 0< q <1< p <2s−1 andλ >0. This work is motivated by the work of Chen and Chen in [10]. Due to the singularity of problem, it is not easy to deal the problem (1.1) as the associated functional is not differentiable even in sense of Gˆateaux and the strong maximum principle is not applicable to show the positivity of solutions. Moreover one can not directly extend all the results from Laplacian case to fractional Laplacian, due to the non- local behavior of the operator and the bounded support of the test function is not preserved. To overcome these difficulties, we first use the Cafferelli and Silvestre [9] approach to convert the problem (1.1) into the local problem. Then we use the variational technique to study the local problem as in [10]. In this paper, the proofs of some Lemmas follow the similar lines as in [10] but for completeness, we give the details.

The article is organized as follows: In section 2 we present some preliminaries on extension problem and necessary weighted trace inequalities required for variational settings. We also state our main results. In section 3, we study the decomposition of Nehari manifold and local charts using the fibering maps. In Section 4, we show the existence of a nontrivial solutions and show how these solutions arise out of nature of Nehari manifold.

We will use the following notation throughout this paper: The same symbol k · kdenotes the norms in the three spaces L

2 s 2

s−1−q(Ω),L

2 s 2

s−1−p(Ω), andH0,Ls (C) defined by (2.1). AlsoS:=ksS(s, n) whereS(s, n) is the best constant of Sobolev embedding (see (2.2)).

2. Preliminaries and main results

In this section we give some definitions and functional settings. At the end of this section, we state our main results. To state our main result, we introduce some

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notation and basic preliminaries results. Denote the upper half-space inRn+1 by Rn+1+ :={z= (x, y) = (x1, x2, . . . , xn, y)∈Rn+1|y >0},

the half cylinder standing on a bounded smooth domain Ω ⊂ Rn by C := Ω× (0,∞)⊂Rn+1+ and its lateral boundary is denoted by∂LC=∂Ω×[0,∞). Define the function spaceH0,Ls (C) as the completion ofC0,L(C) ={w∈C(C) :w= 0 on∂LC}under the norm

kwkHs

0,L(C)= ks

Z

C

y1−2s|∇w(x, y)|2dx dy1/2

, (2.1)

where ks := 21−2sΓ(s)Γ(1−s) is a normalization constant. Then it is a Hilbert space endowed with the inner product

hw, viHs

0,L(C)=ks

Z

Ω×{0}

y1−2s∇w∇v dx dy.

If Ω is a smooth bounded domain then it is verified that (see [8, Proposition 2.1], [6, Proposition 2.1], [26, section 2])

H0s(Ω) :={u= tr|Ω×{0}w:w∈H0,Ls (C)}

and there exists a constantC >0 such that

kw(·,0)kH0s(Ω)≤CkwkH0,Ls (C) for allw∈H0,Ls (C).

Now we define the extension operator and fractional Laplacian for functions in H0s(Ω).

Definition 2.1. Given a function u∈H0s(Ω), we define its s-harmonic extension w=Es(u) to the cylinderC as a solution of the problem

div(y1−2s∇w) = 0 in C, w= 0 on∂LC, w=u on Ω× {0}.

Definition 2.2. For any regular function u ∈ H0s(Ω), the fractional Laplacian (−∆)sacting onuis defined by

(−∆)su(x) =−ks lim

y→0+y1−2s∂w

∂y(x, y) for all (x, y)∈ C.

From [5] and [9], the mapEs(·) is an isometry between H0s(Ω) andH0,Ls (C).

Furthermore, we have (1)

k(−∆)sukH−s(Ω)=kukHs

0(Ω)=kEs(u)kHs

0,L(C), whereH−s(Ω) denotes the dual space ofH0s(Ω);

(2) For any w ∈ H0,Ls (C), there exists a constant C independent of w such that

ktrwkLr(Ω)≤CkwkHs

0,L(C)

holds for everyr∈[2,n−2s2n ]. Moreover, H0,Ls (C) is compactly embedded intoLr(Ω) forr∈[2,n−2s2n ).

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Lemma 2.3. For every 1≤r≤ n−2s2n and every w∈H0,Ls (C), it holds Z

Ω×{0}

|w(x,0)|rdx2/r

≤Cks Z

C

y1−2s|∇w(x, y)|2dx dy, where the constant C depends onr,s,nand|Ω|.

Lemma 2.4. For every w∈Hs(Rn+1+ ), it holds S(s, n)Z

Rn

|u(x)|n−2s2n dxn−2sn

≤ Z

Rn+1+

y1−2s|∇w(x, y)|2dx dy, (2.2) whereu= trw. The constant S(s, n)is known as the best constant and takes the value

S(s, n) =2πsΓ(2−2s2 )Γ(n+2s2 )(Γ(n2))2sn Γ(s)Γ(n−2s2 )(Γ(n))2sn .

Now we can transform the nonlocal problem (1.1) into the local problem

−div(y1−2s∇w) = 0 inC:= Ω×(0,∞), w= 0 on∂LC, w >0 on Ω× {0},

∂w

∂v2s =a(x)w−q+λb(x)wp on Ω× {0},

(2.3)

where ∂v∂w2s :=−kslimy→0+y1−2s ∂w∂y(x, y), for allx∈Ω.

Definition 2.5. A weak solution of (2.3) is a function w ∈H0,Ls (C), w >0 in Ω× {0}such that for every v∈H0,Ls (C),

ks

Z

C

y1−2s∇w∇v dx dy

= Z

Ω×{0}

a(x)(w−qv)(x,0)dx+λ Z

Ω×{0}

b(x)(wpv)(x,0)dx.

Ifwsatisfies (2.3), thenu= trw=w(x,0)∈H0s(Ω) is a weak solution to problem (1.1).

Letc= 1−2s, then the associated functionalJλ:H0,Ls (C)→Rto the problem (2.3) is

Jλ(w) =ks

2 Z

C

yc|∇w|2dx dy− 1 1−q

Z

Ω×{0}

a(x)|w|1−qdx

− λ p+ 1

Z

Ω×{0}

b(x)|w(x,0)|p+1dx.

Now forw∈H0,Ls (C), we define the fiber mapφw:R+ →Ras φw(t) =Jλ(tw) =t2

2kwk2− 1 1−q

Z

Ω×{0}

a(x)|tw|1−qdx

−λtp+1 p+ 1

Z

Ω×{0}

b(x)|w(x,0)|p+1dx.

Then

φ0w(t) =tkwk2−t−q Z

Ω×{0}

a(x)|w(x,0)|1−qdx−λtp Z

Ω×{0}

b(x)|w(x,0)|p+1dx,

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φ00w(t) =kwk2+qt−q−1 Z

Ω×{0}

a(x)|w(x,0)|1−qdx

−pλtp−1 Z

Ω×{0}

b(x)|w(x,0)|p+1dx.

It is easy to see that the energy functionalJλ is not bounded below on the space H0,Ls (C). But we will show that it is bounded below on an appropriate subset of H0,Ls (C) and a minimizer on subsets of this set gives rise to solutions of (2.3). In order to obtain the existence results, we define

Nλ: ={w∈H0,Ls (C) :hJλ0(w), wi= 0}

=n

w∈H0,Ls (C) :kwk2= Z

Ω×{0}

a(x)|w(x,0)|1−qdx +λ

Z

Ω×{0}

b(x)|w(x,0)|p+1dxo .

Note thatw∈ Nλ ifwis a solution of (2.3). Also one can easily see thattw∈ Nλ

if and only if φ0w(t) = 0 and in particular, w ∈ Nλ if and only if φ0w(1) = 0. In order to obtain our result, we decomposeNλ withNλ±,Nλ0as follows:

Nλ±={w∈ Nλ00w(1)≷0}

=

w∈ Nλ: (1 +q)kwk2≷λ(p+q) Z

Ω×{0}

b(x)|w(x,0)|p+1dx , Nλ0={w∈ Nλ00w(1) = 0}

=

w∈ Nλ: (1 +q)kwk2=λ(p+q) Z

Ω×{0}

b(x)|w(x,0)|p+1dx . Inspired by [9] and [10], we show that how variational methods can be used to established some existence and multiplicity results. Our results are as follows:

Theorem 2.6. Suppose that λ∈(0,Λ), where Λ := 1 +q

p+q p−1

p+q p−11+q 1

kbk Sp+q

kakp−1

1/(1+q)

.

Then problem (2.3) has at least two solutions w0 ∈ Nλ+, W0 ∈ Nλ with kW0k>

kw0k. Moreover, u0(x) = w0(·,0) ∈ H0s(Ω) and U0(x) = W0(·,0) ∈ H0s(Ω) are positive solutions of the problem (1.1).

Next, we obtain the blow up behavior of the solutionW∈ Nλ of problem (2.3) withp= 1 +as →0+.

Theorem 2.7. letW∈ Nλ be the solution of problem (2.3)withp= 1 +, where λ∈(0,Λ), then

kWk> C

Λ λ

1/

, where

C=

1 + 1 +q

1/(1+q)

kak1/(q+1) 1

√ S

1−q1+q

→ ∞ as→0+. Namely,W blows up faster than exponentially with respect to .

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We remark that ifwis a positive solution of the problem (−∆)su=a(x)u−q+λb(x)up in Ω

u >0 in Ω, u= 0 on∂Ω.

Then one can easily see thatv=λ1/(p−1)uis a positive solution of the problem (−∆)sv=λ1+qp−1a(x)v−q+b(x)vp in Ω

v >0 in Ω, v= 0 in∂Ω. (2.4) That is, the problem (2.4) has two positive solution forλ∈(0,Λp−1).

3. Fibering map analysis

In this section, we show thatNλ± is nonempty andNλ0 ={0}. Moreover,Jλ is bounded below and coercive.

Lemma 3.1. Let λ∈(0,Λ). Then for eachw∈H0,Ls (C)with Z

Ω×{0}

a(x)|w(x,0)|1−qdx >0, we have the following:

(i) R

Ω×{0}b(x)|w(x,0)|p+1dx ≤ 0, then there exists a unique t1 < tmax such that t1w∈ Nλ+ andJλ(t1w) = inft>0Jλ(tw);

(ii) R

Ω×{0}b(x)|w(x,0)|p+1dx > 0, then there exists a unique t1 and t2 with 0 < t1 < tmax < t2 such that t1w ∈ Nλ+, t2w ∈ Nλ and Jλ(t1w) = inf0≤t≤tmaxJλ(tw),Jλ(t2w) = supt≥t

1Jλ(tw).

Proof. Fort >0, we define ψw(t) =t1−pkwk2−t−p−q

Z

Ω×{0}

a(x)|w(x,0)|1−qdx−λ Z

Ω×{0}

b(x)|w(x,0)|p+1dx.

One can easily see thatψw(t)→ −∞ast→0+. Now ψw0 (t) = (1−p)t−pkwk2+ (p+q)t−p−q−1

Z

Ω×{0}

a(x)|w(x,0)|1−qdx.

ψw00(t) =−p(1−p)t−p−1kwk2

−(p+q)(p+q+ 1)t−p−q−2 Z

Ω×{0}

a(x)|w(x,0)|1−qdx.

Then ψw0 (t) = 0 if and only if t = tmax := [(p+q)R (p−1)kwk2

Ω×{0}a(x)|w(x,0)|1−qdx]−1/(1+q). Also

ψw00(tmax) =p(p−1)h (p−1)kwk2 (p+q)R

Ω×{0}a(x)|w(x,0)|1−qdx ip+1q+1

kwk2

−(p+q)(p+q+ 1)h (p−1)kwk2 (p+q)R

Ω×{0}a(x)|w(x,0)|1−qdx ip+q+2q+1

× Z

Ω×{0}

a(x)|w(x,0)|1−qdx

=−kwk2(p−1)(1 +q)h (p−1)kwk2 (p+q)R

Ω×{0}a(x)|w(x,0)|1−qdx ip+1q+1

<0.

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Thusψwachieves its maximum att=tmax. Now using the H¨older’s inequality and Sobolev inequality (2.2), we obtain

Z

Ω×{0}

a(x)|w(x,0)|1−qdx

≤hZ

Ω×{0}

|a(x)|

2 s 2

s−1+qi

2 s+q−1

2 s hZ

Ω×{0}

|w(x,0)|2sdxi1−q2 s

≤ kakkwk

√ S

1−q

.

(3.1)

Z

Ω×{0}

b(x)|w(x,0)|p+1dx

≤hZ

Ω×{0}

|b(x)|

2 s 2

s−1−pi

2 s−p−1

2 s hZ

Ω×{0}

|w(x,0)|2sdxip+12 s

≤ kbkkwk

√S p+1

.

(3.2)

Using (3.1) and (3.2), we obtain ψw(tmax)

=1 +q p+q

p−1 p+q

p−11+q kwk2(p+q)1+q [R

Ω×{0}a(x)|w(x,0)|1−qdx]p−11+q

−λ Z

Ω×{0}

b(x)|w(x,0)|p+1dx

≥h1 +q p+q

p−1 p+q

p−11+q(√ S)(1−q)

kak

p−11+q

−λkbk 1

√S p+1i

kwkp+1

≡Eλkwkp+1,

(3.3) where

Eλ:=h1 +q p+q

p−1 p+q

p−11+q(√ S)(1−q)

kak

p−11+q

−λkbk 1

√S p+1i

. Then we see thatEλ= 0 if and only ifλ= Λ, where

Λ := 1 +q p+q

p−1 p+q

p−11+q 1 kbk

Sp+q kakp−1

1/(1+q)

.

Thus for λ ∈ (0,Λ), we have Eλ > 0, and therefore it follows from (3.3) that ψw(tmax)>0.

(i) IfR

Ω×{0}b(x)|w(x,0)|p+1dx≥0, then ψw(t)→ −λ

Z

Ω×{0}

b(x)|w(x,0)|p+1dx <0

as t → ∞. Consequently, ψw(t) has exactly two points 0 < t1 < tmax < t2 such that

ψw(t1) = 0 =ψw(t2) andψ0w(t1)>0> ψw0 (t2).

From the definition ofNλ±, we gett1w∈ Nλ+andt2w∈ Nλ. Nowφ0w(t) =tpψw(t).

Thus φ0w(t)< 0 in (0, t1), φ0w(t) >0 in (t1, t2) and φ0w(t) <0 in (t2,∞). Hence Jλ(t1w) = inf0≤t≤tmaxJλ(tw),Jλ(t2w) = supt≥t

1Jλ(tw).

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(ii) IfR

Ω×{0}b(x)|w(x,0)|p+1dx <0 and ψw(t)→ −λ

Z

Ω×{0}

b(x)|w(x,0)|p+1dx >0

ast→ ∞. Consequently,ψw(t) has exactly one point 0< t1< tmax such that ψw(t1) = 0 and ψ0w(t1)>0.

Now φ0w(t) = tpψw(t). Then φ0w(t) < 0 in (0, t1), φ0w(t) > 0 in (t1,∞). Thus Jλ(t1w) = inft≥0Jλ(tw). Hence, it follows thatt1w∈ Nλ+. Corollary 3.2. Suppose thatλ∈(0,Λ), then Nλ±6=∅.

Proof. From (A1) and (A2), we can choosew∈H0,Ls (C)\ {0}such that Z

Ω×{0}

a(x)|w(x,0)|1−qdx >0 and Z

Ω×{0}

b(x)|w(x,0)|p+1dx >0.

Then by (ii) of Lemma 3.1, there exists a unique t1 and t2 such thatt1w ∈ Nλ+,

t2w∈ Nλ. In conclusion,Nλ± 6=∅.

Lemma 3.3. Forλ∈(0,Λ), we have Nλ0={0}.

Proof. We prove this by contradiction. Assume that there exists 0 6≡ w ∈ Nλ0. Then it follows fromw∈ Nλ0 that

(1 +q)kwk2=λ(p+q) Z

Ω×{0}

b(x)|w(x,0)|p+1dx and consequently

0 =kwk2− Z

Ω×{0}

a(x)|w(x,0)|1−qdx−λ Z

Ω×{0}

b(x)|w(x,0)|p+1dx

= p−1 p+qkwk2

Z

Ω×{0}

a(x)|w(x,0)|1−qdx.

Therefore, asλ∈(0,Λ) andw6≡0, we use similar arguments as those in (3.3) to obain

0< Eλkwkp+1

≤ 1 +q p+q

p−1 p+q

p−11+q kwk2(p+q)1+q R

Ω×{0}a(x)|w(x,0)|1−qdxp−11+q

−λ Z

Ω×{0}

b(x)|w(x,0)|p+1dx

= 1 +q p+q

p−1 p+q

p−11+q kwk2(p+q)1+q

p−1

p+qkwk2p−11+q

−1 +q

p+qkwk2= 0,

We note that Λ is also related to a gap structure inNλ:

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Lemma 3.4. Suppose that λ∈(0,Λ), then there exist a gap structure inNλ: kWk> Aλ> A0>kwk for allw∈ Nλ+, W ∈ Nλ,

where

Aλ=h 1 +q λ(p+q)kbk(√

S)p+1i1/(p−1)

and A0=hp+q p−1kak 1

√S

1−qi1/(q+1)

. Proof. Ifw∈ Nλ+⊂ Nλ, then

0<(1 +q)kwk2−λ(p+q) Z

Ω×{0}

b(x)|w(x,0)|p+1dx

= (1 +q)kwk2−(p+q)h kwk2

Z

Ω×{0}

a(x)|w(x,0)|1−qdxi

= (1−p)kwk2+ (p+q) Z

Ω×{0}

a(x)|w(x,0)|1−qdx.

Hence it follows from (3.1) that (p−1)kwk2<(p+q)

Z

Ω×{0}

a(x)|w(x,0)|1−qdx≤(p+q)kakkwk

√S 1−q

which yields

kwk<hp+q

p−1kak 1

√ S

1−qi1/(q+1)

≡A0. IfW ∈ Nλ, then it follows from (3.2) that

(1 +q)kWk2< λ(p+q) Z

Ω×{0}

b(x)|W(x,0)|p+1dx≤λ(p+q)kbkkWk

√ S

p+1

which yields

kWk>h 1 +q λ(p+q)kbk(√

S)p+1i1/(p−1)

≡Aλ. Now we show thatAλ=A0if and only if λ= Λ.

λ= Λ = 1 +q p+q

p−1 p+q

p−11+q 1 kbk

Sp+q kakp−1

1/(1+q)

if and only if Aλ−1/(p−1)1 +q

p+q

1/(p−1) 1 kbk

1/(p−1)√ Sp+1p−1

=p+q p−1

1/(1+q)

kak1/(q+1)

S(1+q)(p−1)2(p+q) +p+1p−1

=h (p+q)kak (p−1)(√

S)1−q

i1/(q+1)

≡A0. Thus for allλ∈(0,Λ), we can conclude that

kWk> Aλ> A0>kwk for allw∈ Nλ+, W ∈ Nλ.

This completes the proof.

Lemma 3.5. Suppose thatλ∈(0,Λ), thenNλis a closed set inH0,Ls (C)-topology.

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Proof. Let{Wk} be a sequence inNλ withWk→W0inH0,Ls (C). Then we have kW0k2= lim

k→∞kWkk2

= lim

k→∞

hZ

Ω×{0}

a(x)|Wk(x,0)|1−qdx+λ Z

Ω×{0}

b(x)|Wk(x,0)|p+1dxi

= Z

Ω×{0}

a(x)|W0(x,0)|1−qdx+λ Z

Ω×{0}

b(x)|W0(x,0)|p+1dx and

(1 +q)kW0k2−λ(p+q) Z

Ω×{0}

b(x)|W0(x,0)|p+1dx

= lim

k→∞

h

(1 +q)kWkk2−λ(p+q) Z

Ω×{0}

b(x)|Wk(x,0)|p+1dxi

≤0.

That is,W0∈ Nλ∩ Nλ0. Since{Wk} ⊂ Nλ, from Lemma 3.4 we have kW0k= lim

k→∞kWkk ≥A0>0,

which imply, W0 6= 0. It follows from Lemma 3.1, that W0 6∈ Nλ0 for any λ ∈ (0,Λ). Thus W0 ∈ Nλ. Hence, Nλ is a closed set inH0,Ls (C)-topology for any

λ∈(0,Λ).

Lemma 3.6. Let w ∈ Nλ±. Then for any φ ∈ C0,L (C), there exists a number >0 and a continuous functionf :B(0) :={v∈H0,Ls (C) :kvk< } →R+ such that

f(v)>0, f(0) = 1, f(v)(w+vφ)∈ Nλ± for allv∈B(0).

Proof. We give the proof only for the casew∈ Nλ+, the caseNλmay be preceded exactly. For anyφ∈C0,L(C), we defineF :H0,Ls (C)×R+→Ras follows:

F(v, r) =r1+qkw+vφk2− Z

Ω×{0}

a(x)|(w+vφ)(x,0)|1−q

−λrp+q Z

Ω×{0}

b(x)|(w+vφ)(x,0)|p+1. Sincew∈ Nλ+(⊂ Nλ), we have

F(0,1) =kwk2− Z

Ω×{0}

a(x)|w(x,0)|1−qdx−λ Z

Ω×{0}

b(x)|w(x,0)|p+1dx= 0, and

∂F

∂r(0,1) = (1 +q)kwk2−λ(p+q) Z

Ω×{0}

b(x)|w(x,0)|p+1dx >0.

Applying the implicit function theorem at (0,1), we have that there exists ¯ > 0 such that for kvk < ¯, v ∈ H0,Ls (C), the equation F(v, r) = 0 has a unique continuous solution r=f(v)>0. It follows from F(0,1) = 0 that f(0) = 1 and fromF(v, f(v)) = 0 forkvk<¯,v∈H0,Ls (C) that

0 =f1+q(v)kw+vφk2− Z

Ω×{0}

a(x)|(w+vφ)(x,0)|1−q

−λfp+q(v) Z

Ω×{0}

b(x)|(w+vφ)(x,0)|p+1

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=

kf(v)(w+vφ)k2− Z

Ω×{0}

a(x)|f(v)(w+vφ)(x,0)|1−q

−λ Z

Ω×{0}

b(x)|f(v)(w+vφ)(x,0)|p+1

/f1−q(v);

that is,

f(v)(w+vφ)∈ Nλ for allv∈H0,Ls (C),kvk<˜. Since ∂F∂r(0,1)>0 and

∂F

∂r(v, f(v))

= (1 +q)fq(v)kw+vφk2−λ(p+q)fp+q−1(v) Z

Ω×{0}

b(x)|(w+vφ)(x,0)|p+1dx

=

(1 +q)kf(v)(w+vφ)k2−λ(p+q)R

Ω×{0}b(x)|f(v)(w+vφ)(x,0)|p+1dx

f2−q(v) ,

we can take >0 possibly smaller ( <¯) such that for anyv∈H0,Ls (C),kvk< , (1 +q)kf(v)(w+vφ)k2−λ(p+q)

Z

Ω×{0}

b(x)|f(v)(w+vφ)(x,0)|p+1dx >0;

that is,

f(v)(w+vφ)∈ Nλ+ for allv∈B(0).

This completes the proof.

Lemma 3.7. Jλ is bounded below and coercive onNλ. Proof. Forw∈ Nλ, from (3.1), we obtain

Jλ(w) =1 2 − 1

p+ 1

kwk2− 1

1−q − 1 p+ 1

Z

Ω×{0}

a(x)|w(x,0)|1−qdx

≥1 2 − 1

p+ 1

kwk2− 1

1−q − 1 p+ 1

kakkwk

√ S

1−q

.

(3.4)

Now consider the functionρ:R+→Ras ρ(t) =αt2−βt1−q, where α,β are both positive constants. One can easily show that ρ is convex(ρ00(t)>0 for all t >0) with ρ(t) → 0 as t → 0 and ρ(t) → ∞ as t → ∞. ρ achieves its minimum at tmin= [β(1−q) ]1/(1+q) and

ρ(tmin) =αβ(1−q) 2α

1+q2

−ββ(1−q) 2α

1−q1+q

=−1 +q

2 β1+q2 1−q 2α

1−q1+q . Applying ρ(t) with α = 12p+11

, β = 1−q1p+11 kak 1

S

1−q

and t = kwk, w∈ Nλ, we obtain from (3.4) that

kwk→∞lim Jλ(w)≥ lim

t→∞ρ(t) =∞.

ThusJλ is coercive onNλ. Moreover, it follows from (3.4) that

Jλ(w)≥ρ(t)≥ρ(tmin) (a constant), (3.5) i.e.,

Jλ(w)≥ −1 +q

2 β1+q2 1−q 2α

1−q1+q

=− 1 +q (1−q)(p+ 1)

(p+q)kak 2(√

S)1−q

1+q2 1 p−1

1−q1+q .

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ThusJλ is bounded below onNλ. 4. Existence of solutions in Nλ±

Now from Lemma 3.5,Nλ+∪Nλ0andNλare two closed sets inH0,Ls (C) provided λ∈(0,Λ). Consequently, the Ekeland variational principle can be applied to the problem of finding the infimum of Jλ on bothNλ+∪ Nλ0 andNλ. First, consider {wk} ⊂ Nλ+∪ Nλ0 with the following properties:

Jλ(wk)< inf

w∈Nλ+∪Nλ0

Jλ(w) +1

k (4.1)

Jλ(w)≥Jλ(wk)−1

kkw−wkk, ∀w∈ Nλ+∪ Nλ0. (4.2) FromJλ(|w|) =Jλ(w), we may assume that wk ≥0 onC.

Lemma 4.1. Show that the sequence {wk} is bounded in Nλ. Moreover, there exists06=w0∈H0,Ls (C)such that wk * w0 weakly inH0,Ls (C).

Proof. By equations (3.5) and (4.1), we have at2−bt1−q =ρ(t)≤Jλ(w)< inf

w∈Nλ+∪Nλ0

Jλ(w) +1 k ≤C5,

for sufficiently large k and a suitable positive constant. Hence putting t =wk in the above equation, we obtain{wk} is bounded.

Let{wk} is bounded in H0,Ls (C). Then, there exists a subsequence of{wk}k, still denoted by{wk}k andw0∈H0,Ls (C) such thatwk * w0weakly inH0,Ls (C), wk(·,0)→w0(·,0) strongly inLp(Ω) for 1≤p <2s andwk(·,0)→w0(·,0) a.e. in Ω.

For anyw∈ Nλ+, we have from 0< q <1< pthat Jλ(w) =1

2 − 1 1−q

kwk2+ 1

1−q − 1 p+ 1

λ

Z

Ω×{0}

b(x)|w(x,0)|p+1dx

<1 2 − 1

1−q

kwk2+ 1

1−q − 1 p+ 1

1 +q p+qkwk2

= 1 p+ 1−1

2 1 +q

1−qkwk2<0, which means that infN+

λ Jλ <0. Now for λ ∈ (0,Λ), we know from Lemma 3.1, thatNλ0={0}. Together, these imply thatwk∈ Nλ+ forklarge and

inf

w∈Nλ+∪Nλ0

Jλ(w)≤ inf

w∈Nλ+

Jλ(w)<0.

Therefore, by weak lower semi-continuity of norm, Jλ(w0)≤lim inf

k→∞ Jλ(wk) = inf

Nλ+∪Nλ0

Jλ<0,

that is,w0≥0,w06≡0.

Lemma 4.2. Supposewk∈ Nλ+ such thatwk* w0 weakly inH0,Ls (C). Then for λ∈(0,Λ),

(1 +q) Z

Ω×{0}

a(x)w1−q0 (x,0)dx−λ(p−1) Z

Ω×{0}

b(x)w0p+1(x,0)dx >0. (4.3)

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Moreover, there exists a constantC2>0 such that (1 +q)kwkk2−λ(p+q)

Z

Ω×{0}

b(x)wkp+1(x,0)dx≥C2>0. (4.4) Proof. For{wk} ⊂ Nλ+(⊂ Nλ), since

(1 +q) Z

Ω×{0}

a(x)w01−q(x,0)dx−λ(p−1) Z

Ω×{0}

b(x)w0p+1(x,0)dx

= lim

k→∞

h (1 +q)

Z

Ω×{0}

a(x)w1−qk (x,0)dx−λ(p−1) Z

Ω×{0}

b(x)wp+1k (x,0)dxi

= lim

k→∞

h

(1 +q)kwkk2−λ(p+q) Z

Ω×{0}

b(x)wp+1k (x,0)dxi

≥0, we can argue by a contradiction and assume that

(1 +q) Z

Ω×{0}

a(x)w1−q0 (x,0)dx−λ(p−1) Z

Ω×{0}

b(x)w0p+1(x,0)dx= 0. (4.5) Sincewk∈ Nλ, from the weak lower semi continuity of norm and (4.5) we have

0 = lim

k→∞

hkwkk2− Z

Ω×{0}

a(x)w1−qk (x,0)dx−λ Z

Ω×{0}

b(x)wp+1k (x,0)dxi

≥ kw0k2− Z

Ω×{0}

a(x)w1−q0 (x,0)dx−λ Z

Ω×{0}

b(x)wp+10 (x,0)dx

=

(kw0k2−λp+q1+qR

Ω×{0}b(x)wp+10 (x,0)dx kw0k2p+qp−1R

Ω×{0}a(x)w1−q0 (x,0)dx.

Thus for anyλ∈(0,Λ) andw06≡0, by similar arguments as those in (3.3) we have that

0< Eλkw0kp+1

≤ 1 +q p+q

p−1 p+q

p−11+q kw0k2(p+q)1+q R

Ω×{0}a(x)w1−q0 (x,0)dxp−11+q

−λ Z

Ω×{0}

b(x)w0p+1(x,0)dx

= 1 +q p+q

p−1 p+q

p−11+q kw0k2(p+q)1+q

p−1

p+qkw0k2p−11+q

−1 +q

p+qkw0k2= 0, which is clearly impossible. Now by (4.3), we have that

(1 +q) Z

Ω×{0}

a(x)w1−qk (x,0)dx−λ(p−1) Z

Ω×{0}

b(x)wp+1k (x,0)dx≥C2 for sufficiently largekand a suitable positive constantC2. This, together with the

fact thatwk∈ Nλ we obtain equation (4.4).

Fixφ ∈C0,L(C) withφ≥0. Then we apply Lemma 3.6 with w=wk ∈ Nλ+ (k large enough such that (1−q)Ck 1 < C2), we obtain a sequence of functions fk : Bk(0)→Rsuch thatfk(0) = 1 andfk(w)(wk+wφ)∈ Nλ+ for allw∈Bk(0). It follows fromwk ∈ Nλ andfk(w)(wk+wφ)∈ Nλ that

kwkk2− Z

Ω×{0}

a(x)wk1−q(x,0)dx−λ Z

Ω×{0}

b(x)wkp+1(x,0)dx= 0 (4.6)

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and

fk2(w)kwk+wφk2−fk1−q(w) Z

Ω×{0}

a(x)(wk+wφ)1−q(x,0)dx

−λfkp+1(w) Z

Ω×{0}

b(x)(wk+wφ)p+1(x,0)dx= 0.

(4.7)

Choose 0 < ρ < k and w = ρv with kvk < 1. Then we find fk(w) such that fk(0) = 1 andfk(w)(wk+wφ)∈ Nλ+ for allw∈Bρ(0).

Lemma 4.3. Forλ∈(0,Λ)we have|hfk0(0), vi|is finite for every0≤v∈H0,Ls (C) withkvk ≤1.

Proof. By (4.6) and (4.7) we have

0 = [fk2(w)−1]kwk+wφk2+kwk+wφk2− kwkk2

−[fk1−q(w)−1]

Z

Ω×{0}

a(x)(wk+wφ)1−q(x,0)dx

− Z

Ω×{0}

a(x)[((wk+wφ)1−q−w1−qk )(x,0)]dx

−λ[fkp+1(w)−1]

Z

Ω×{0}

b(x)(wk+wφ)p+1(x,0)dx

−λ Z

Ω×{0}

b(x)[((wk+wφ)p+1−wp+1k )(x,0)]dx

≤[fk2(ρv)−1]kwk+ρvφk2+kwk+ρvφk2− kwkk2

−[fk1−q(ρv)−1]

Z

Ω×{0}

a(x)(wk+ρvφ)1−q(x,0)

−λ[fkp+1(ρv)−1]

Z

Ω×{0}

b(x)(wk+ρvφ)p+1(x,0)

−λ Z

Ω×{0}

b(x)[((wk+ρvφ)p+1−wkp+1)(x,0)]

Dividing byρ >0 and passing to the limitρ→0, we derive that 0≤2hfk0(0), vikwkk2+ 2ks

Z

C

yc∇wk∇(vφ)dx dy

−(1−q)hfk0(0), vi Z

Ω×{0}

a(x)wk1−q(x,0)dx

−λ(p+ 1)hfk0(0), vi Z

Ω×{0}

b(x)wp+1k (x,0)dx

−λ(p+ 1) Z

Ω×{0}

b(x)(wpkvφ)(x,0)dx

=hfk0(0), vih

2kwkk2−(1−q) Z

Ω×{0}

a(x)wk1−q(x,0)dx

−λ(p+ 1) Z

Ω×{0}

b(x)wp+1k (x,0)dxi + 2ks

Z

C

yc∇wk∇(vφ)dx dy−λ(p+ 1) Z

Ω×{0}

b(x)(wkpvφ)(x,0)dx

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=hfk0(0), vih

(1 +q)kwkk2

−λ(p+q) Z

Ω×{0}

b(x)wp+1k (x,0)dxi + 2ks

Z

C

yc∇wk∇(vφ)dx dy−λ(p+ 1) Z

Ω×{0}

b(x)(wkpvφ)(x,0)dx. (4.8) From this inequality and (4.4) we know thathfk0(0), vi 6=−∞. Now we show that hfk0(0), vi 6= +∞. Arguing by contradiction, we assume thathfk0(0), vi= +∞. Now we note that

|fk(ρv)−1|kwkk+fk(ρv)kρvφk ≥ k[fk(ρv)−1]wk+ρvfk(ρv)φk

=kfk(ρv)(wk+ρvφ)−wkk (4.9) andfk(ρv)> fk(0) = 1 for sufficiently largek.

From the definition of derivative hfk0(0), vi, applying equation (4.2) with w = fk(ρv)(wk+ρvφ)∈ Nλ+, we clearly have

[fk(ρv)−1]kwkk

k +fk(ρv)kρvφk k

≥ 1

kkfk(ρv)(wk+ρvφ)−wkk

≥Jλ(wk)−Jλ(fk(ρv)(wk+ρvφ))

=1 2 − 1

1−q

kwkk2+λ 1

1−q − 1 p+ 1

Z

Ω×{0}

b(x)wp+1k (x,0)dx + 1

1−q −1 2

fk2(ρv)kwk+ρvφk2

− λ(p+q)

(1−q)(p+ 1)fkp+1(ρv) Z

Ω×{0}

b(x)(wk+ρvφ)p+1(x,0)dx

= 1 1−q−1

2

(kwk+ρvφk2− kwkk2) + 1 1−q−1

2

[fk2(ρv)−1]kwk+ρvφk2

−λ 1

1−q− 1 p+ 1

fkp+1(ρv) Z

Ω×{0}

b(x)[((wk+ρvφ)p+1−wkp+1)(x,0)]dx

−λ 1

1−q− 1 p+ 1

[fkp+1(ρv)−1]

Z

Ω×{0}

b(x)wp+1k (x,0)dx.

Dividing byρ >0 and passing to the limit asρ→0, we can obtain hfk0(0), vikwkk

k +kvφk k

≥1 +q 1−q

ks

Z

C

yc∇wk∇(vφ)dx dy+1 +q 1−q

hfk0(0), vikwkk2

−λp+q 1−q

hfk0(0), vi Z

Ω×{0}

b(x)wp+1k (x,0)dx

−λp+q 1−q

Z

Ω×{0}

b(x)(wpkvφ)(x,0)dx

= hfk0(0), vi 1−q

h

(1 +q)kwkk2−λ(p+q) Z

Ω×{0}

b(x)wp+1k (x,0)dxi

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+1 +q 1−q

ks

Z

C

yc∇wk∇(vφ)dx dy−λp+q 1−q

Z

Ω×{0}

b(x)(wkpvφ)(x,0)dx.

That is, kvφk

k ≥ hfk0(0), vi 1−q

h

(1 +q)kwkk2−λ(p+q) Z

Ω×{0}

b(x)wp+1k (x,0)dx

−(1−q)kwkk k

i

+1 +q 1−q

ks

Z

C

yc∇wk∇(vφ)dx dy

−λp+q 1−q

Z

Ω×{0}

b(x)(wpkvφ)(x,0)dx,

(4.10)

which is impossible becausehfk0(0), vi= +∞and (1+q)kwkk2−λ(p+q)

Z

Ω×{0}

b(x)wp+1k (x,0)dx−(1−q)kwkk

k ≥C2−(1−q)C1 k >0.

In conclusion, |hfk0(0), vi| < +∞. Furthermore, (4.4) with kwkk ≤ C1 and two inequalities (4.8) and (4.10) also imply that

|hfk0(0), vi| ≤C3

forksufficiently large and a suitable constantC3. Lemma 4.4. For each 0 ≤φ ∈ C0,L (C) and for every 0 ≤ v ∈ H0,Ls (C) with kvk ≤1, we have a(x)w0−qvφ∈L1(Ω)and

ks Z

C

yc∇w0∇(vφ)dx dy− Z

Ω×{0}

a(x)(w−q0 vφ)(x,0)dx

−λ Z

Ω×{0}

b(x)(w0pvφ)(x,0)dx≥0.

(4.11)

Proof. Applying (4.9) and (4.2) again, we obtain [fk(ρv)−1]kwkk

k +fk(ρv)kρvφk k

≥Jλ(wk)−Jλ(fk(ρv)(wk+ρvφ))

=−fk2(ρv)−1

2 kwkk2−fk2(ρv)

2 (kwk+ρvφk2− kwkk2) +fk1−q(ρv)−1

1−q Z

Ω×{0}

a(x)(wk+ρvφ)1−q(x,0)

+ 1

1−q Z

Ω×{0}

a(x)[((wk+ρvφ)1−q−wk1−q)(x,0)]

+λfkp+1(ρv)−1 p+ 1

Z

Ω×{0}

b(x)(wk+ρvφ)p+1(x,0)

+ λ

p+ 1 Z

Ω×{0}

b(x)[((wk+ρvφ)p+1−wp+1k )(x,0)].

Dividing byρ >0 and passing to the limitρ→0+, we obtain

|hfk0(0), vi|kwkk

k +kvφk k

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Further, we develop a full symbolic calculus for pseudo- differential operators acting on algebras of Colombeau generalized functions.. As an application, we formulate a

[30] T. Guerin; Existence of nonnegative solutions to singular elliptic problems, a variational approach, Discrete Contin. Guerin; Multiplicity of weak solutions to subcritical

If there is an almost resolvable space X which is almost-ω- irresolvable, then there is a resolvable Baire open subspace U of X which is hereditarily almost-ω-irresolvable.. Proof: