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The Robin problem for

singular p ( x ) -Laplacian equation in a cone

Mikhail Borsuk

B

Faculty of Mathematics and Computer Science, University of Warmia and Mazury in Olsztyn, 10-957 Olsztyn-Kortowo, Poland

Received 7 August 2018, appeared 1 December 2018 Communicated by Maria Alessandra Ragusa

Abstract. We study the behavior near the boundary angular or conical point of weak solutions to the Robin problem for an elliptic quasi-linear second-order equation with the variable p(x)-Laplacian.

Keywords: p(x)-Laplacian, angular and conical points.

2010 Mathematics Subject Classification: 35J20, 35J25, 35J70.

1 Introduction

The aim of our article is the investigation of the behavior of the weak solutions to the Robin problem for quasi-linear elliptic second-order equations with the variable p(x)-Laplacian in a neighborhood of an angular or a conical boundary point of the bounded cone. Boundary value problems for elliptic second order equations with a non-standard growth in function spaces with variable exponents have been an active investigations in recent years. We refer to [7] for an overview and the recent papers [1,9,10] and reference therein. Differential equa- tions with variable exponents-growth conditions arise from the nonlinear elasticity theory, electrorheological fluids, etc. There are many essential differences between the variable expo- nent problems and the constant exponent problems. In the variable exponent problems, many singular phenomena occurred and many special questions were raised. V. Zhikov [11,12]

has gave examples of the Lavrentiev phenomenon for the variational problems with variable exponent.

Most of the works devoted to the quasi-linear elliptic second-order equations with the variable p(x)-Laplacian refers to the Dirichlet problem in smooth bounded domains (see [7]).

Concerning the Robin problem for such equations we know only a few articles [2,5,6,8], but in these works a domain is smooth and lower order terms depend only on (x,u) and do not depend on |∇u|. Our article [3] is deduced to the Robin problemin a cone for such equations with a singular p(x)-power gradient lower order term. The present article is the continuation of [3]. Here we describe qualitatively the behavior of the weak solution near

BEmail: borsuk@uwm.edu.pl

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a conical point, namely we derive the sharp estimate of the type |u(x)| = O(|x|κ) (cf. §3.1) for the weak solution modulus (for the solution decrease rate) of our problem near a conical boundary point. As well as, we establish the comparison principle for weak solutions.

The Robin boundary conditions appear in the solving Sturm–Liouville problems which are used in many contexts of science and engineering: for example, in electromagnetic problems, in heat transfer problems and for convection-diffusion equations (Fick’s law of diffusion). The Robin problem plays a major role in the study of reflected shocks in transonic flow. Important applications of this problem is the capillary problem.

LetCbe an open cone inRn, n≥2, with the vertex at the origin Oand letBrbe an open ball with radiusr centered atO. We use the following standard notations:

• Sn1: a unit sphere inRn centered atO;

• (r,ω),ω = (ω1,ω2, . . . ,ωn1): the spherical coordinates ofx ∈Rnwith pole O: x1 =rcosω1,

x2 =rcosω2sinω1, ...

xn1 =rcosωn1sinωn2. . . sinω1, xn=rsinωn1sinωn2. . . sinω1;

• Ω : a domain on the unit sphere Sn1 with the smooth boundary ∂Ω obtained by the intersection of the coneCwith the sphereSn1;

∂Ω=∂C∩Sn1;

• G0dC∩Bd ={(r,ω)|0<r <d; ω};

• Γd0C∩Bd= {(r,ω)|0<r <d; ωΩ};

• Ωd= G0d∩ {|x|=d}.

We investigate the behavior in a neighborhood of the origin O of solutions to the Robin problem with the boundary condition on the lateral surface of the cone:

(−4p(x)u+b(u,∇u) = f(x), x ∈Gd00,

|∇u|p(x)−2 ∂u

n + γ

|x|p(x)−1u|u|p(x)−2=0, xΓd00, (RQL) where 0<d01 (d0is fixed) and

4p(x)u≡div

|∇u|p(x)−2∇u

. (1.1)

We will work under the following assumptions:

(i) 1< p≤ p(x)≤ p+ = p(0)<n, ∀x ∈G0d0;

(ii) the Lipschitz condition: p(x)∈C0,1(G0d0) =⇒ 0≤ p+−p(x)≤ L|x|, ∀x ∈Gd00; where Lis the Lipschitz constant for p(x).

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(iii) |f(x)| ≤ f0|x|β(x), f0 ≥ 0, β(x) > pp+1

+1+µ(p(x)−1)λ − p(x); ∀x ∈ Gd00; γ = const ≥ 1, 0 ≤ µ < 1 andλ is the least positive eigenvalue of problem (NEVP) (see below);

(iv) the functionb(u,ξ)is differentiable with respect to theu,ξvariables inM=R×Rnand satisfy inMthe following inequalities:

(iv)a |b(u,ξ)| ≤δ|u|1|ξ|p(x)+b0|u|p(x)−1, 0≤δ< µ; ifµ>0;

(iv)b b(u,ξ)≥ν|u|1|ξ|p(x)−b0|u|p(x)−1, ν>0; ifµ=0;

(iv)c s n

i=1

∂b

(u,ξ)

∂ξi

2 ≤b1|u|1|ξ|p(x)−1; ∂b(∂uu,ξ) ≥b2|u|2|ξ|p(x); b0≥0, b1≥0, b2≥0;

(iiv) the spherical region Ω⊂Sn1 is invariant with respect to rotations inSn2. We consider the functions class

N1,p1,∞(x)(G0d0) = (

u

u(x)∈ L(G0d0)and Z

G0d0

D|x|p(x)|u|p(x)+|u|1|∇u|p(x)Edx< )

which was introduced in [4]. It is obvious that N1,p1,∞(x)(G0d0)⊂W1,p(x)(G0d0).

Definition 1.1. The functionuis called a weak bounded solution of problem (RQL) provided that u(x)∈N1,p1,∞(x)(G0d0)and satisfies the integral identity

Q(u,η):≡

Z

G0d0

D|∇u|p(x)−2uxiηxi +b(u,∇u)η E

dx+γ Z

Γ0d0 r1p(x)u|u|p(x)−2ηdS

Z

d0

|∇u|p(x)−2∂u

∂rηdΩd=

Z

Gd00 f(x)η(x)dx (II)

for all η(x)∈N1,p1,∞(x)(G0d0).

Remark 1.2. It is easy to verify that the above assumptions(i), (iii), (iv)ensure the existence of integrals over G0d andΓd0. Therefore, Definition1.1is correct.

Main result is the following statement.

Theorem 1.3. Let u be a weak bounded solution of problem(RQL), M0 = sup

xGd00|u(x)|(see [3]) and let λ be the least positive eigenvalue of problem (NEVP) (see Section 2). Suppose that(i)–(iiv) hold. Then there exist de∈ (0,d0)and a constant C0 > 0depending only on λ,d0,M0,p+,p,L,n, (µδ),ν,b0,f0and such that

|u(x)| ≤C0|x|κ, κ= p+−1

p+−1+µλ; ∀x ∈Gd0e. (1.2)

2 Nonlinear eigenvalue problem

To prove the main result we shall consider the nonlinear eigenvalue problem for ψ(ω)∈C2()∩C1():









−divω

(λ2ψ2+|∇ωψ|2)(p+2)/2ωψ

=λ(λ(p+−1) +n−p+) (λ2ψ2+|∇ωψ|2)(p+2)/2ψ, ωΩ, (λ2ψ2+|∇ωψ|2)(p+2)/2∂ψ

ν +γ

p+1+µ p+1

p+1

·ψ|ψ|p+2 =0, ω∂Ω,

(NEVP)

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where|∇ωψ|denotes the projection of the vector∇ψonto the tangent plane to the unit sphere at the pointω:

ωψ= 1

√q1

∂ψ

∂ω1, . . . , 1

√qn1

∂ψ

∂ωn1

,

|∇ωψ|2=

n1 i

=1

1 qi

∂ψ

∂ωi 2

, q1=1, qi = (sinω1· · ·sinωi1)2, i≥2 and−→

ν denotes the exterior normal toCat points ofΩ.

If we rename ω = ω1, ω0 = (ω2, . . . ,ωn1), then, by assumption (iiv), we can assume that ψ(ω1,ω0) does not depend on ω0. Therefore, our problem (NEVP) is equivalent to the following













λ2ψ2+ (p+−1)ψ02

ψ00(ω) + (n−2)cotω λ2ψ2+ψ02 ψ0(ω) +λ(λ(2p+−3) +n−p+)ψ02ψ(ω)

+λ3(λ(p+−1) +n−p+)ψ3(ω) =0, ω ∈ −ω20,ω20 ,

±(λ2ψ2+ψ02)(p+2)/2ψ0(ω) +γ p

+1+µ p+1

p+1

·ψ(ω)|ψ(ω)|p+2

ωω20=0.

(OEVP)

2.1 Properties of the(OEVP)eigenvalue and corresponding eigenfunction

First of all, we note that any two eigenfunctions are scalar multiples of each other if they solve problem for the sameλ. Therefore, without loss of generality we can assumeψ ω20

=1.

Next, we observe that the following two cases are possible: either ψ(−ω) = −ψ(ω) or ψ(−ω) = ψ(ω). In Section 2 [4], it was shown that we obtain the least positive eigenvalueλ ifψ(−ω) =ψ(ω); thenψ0(−ω) =−ψ0(ω) =⇒ ψ0(0) =0, ψ(0)6=0, as well as the following inequalities for eigenvalue and for the corresponding eigenfunction:

λ(λ(p+−1) +n−p+)>0; 0< λ < π

ω0; (2.1)

1≤ψ(ω)≤ψ0 =const(n,p,λ,ω0) (2.2) hold.

Now we define the function y(ω) = ψ0(ω)

ψ(ω), ψ(0) 6= 0 and let y0 = y ω20

. From(OEVP) we obtain the Cauchy problem





(p+−1)y2+λ2

y0(ω) + (p+−1)y4+ (n−2)cotω y2+λ2 y(ω)

+λ(2λ(p+−1) +n−p+)y2+λ3(λ(p+−1) +n−p+) =0, ω∈ 0,ω20 , y(0) =0.

(CP)

and the following equation forλ: y0

λ2+y20

p+2

2 =−γ

p+−1+µ p+1

p+1

. (λ)

Since

(p+−1)y4+λ(2λ(p+−1) +n−p+)y2+λ3(λ(p+−1) +n−p+)

= (p+−1)(y2+λ2)

y2+λ2+ n−p+

p+−1λ

,

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the (CP) equation can be rewritten as follows y0(ω)

y2+λ2

=− (n−2)cotω (p+−1)y2+λ2

·y(ω)−(p+−1)(y2+λ2) + (n−p+)λ

(p+−1)y2+λ2 , ω0,ω0 2

. (2.3) By Lemma 2.2 [4], we have

y(ω)≤0, |y(ω)| ≤z0 =const(n,λ,ω0,p+), ∀ω∈ [0,ω0/2]. (2.4) Proposition 2.1. If assumption(i)is satisfied andγ≥1(see assumption(iii)), then

κ λ

q

λ2+y20

p(x)−p(0)

≤1, ∀x∈Γd0, (2.5)

whereκis defined by(1.2).

Proof. We rewrite (λ) with regard to (1.2):

|y0|= γλ

κ , if p+=2;

q

λ2+y20 =|yγ

0|

p 1

+2

· λ

κ

p+1

p+2 , if p+6=2. (2.6) Case p+ =2.

The inequality (2.5) is true if p(x)≡2. Now, let 1 < p(x)≤ p+ = 2, ∀x ∈ Γd0. From (λ) we have

|y0|= γ(1+µ) = γλ

κ =⇒ κ λ

q

λ2+y20 ≥ κ

λ|y0|=γ≥1 (p(x)−p+)ln

κ λ

q

λ2+y20

≤0 =⇒ (2.5)is true.

Case p+ >2.

From (λ) and (2.6) it follows that

|y0| ≤ λ κ ·γ

1

p+1 and

q

λ2+y20λ κ ·γ

1

p+1 =⇒

(p(x)−p+)ln κ

λ q

λ2+y20

p(x)−p+

p+−1 lnγ≤0 =⇒ (2.5)is true.

Case p+ <2.

From (λ) and (2.6) we obtain that

|y0|=γ λ

κ

p+1q

λ2+y20 2p+

γ λ

κ p+1

|y0|2p+ |y0| ≥ λ κ ·γ

p+11 =⇒

and q

λ2+y20= γ

1 p+2

λ κ

p+

1 p+2

|y0|21p+λ κ ·γ

1

p+1 =⇒

(p(x)−p+)ln κ

λ q

λ2+y20

p(x)−p+

p+−1 lnγ≤0 =⇒ (2.5)is true.

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3 Comparison principle

InG0d ⊂Gwe consider the second order quasi-linear degenerate operator Qof the form Q(u,η)≡

Z

Gd0

DAi(x,ux)ηxi +b(x,u,ux)η(x)Edx+

Z

Γd0

γ(ω)

rp(x)−1u|u|p(x)−2η(x)ds

Z

d

Ai(x,ux)cos(r,xi)η(x)dΩd; γ(ω)≥γ0 >0 (3.1)

foru(x) ∈ N1,p1,∞(x)(G0d) and for all non-negative η(x)belonging to N1,p1,∞(x)(G0d)under the fol- lowing assumptions:

functions Ai(x,ξ),b(x,u,ξ) are Carathéodory, continuously differentiable with respect to the u,ξ variables inM=G×R×Rnand satisfy inMthe following inequalities:

(i) A∂ξi(x,ξ)

j ζiζj ≥κp|ξ|p(x)−2ζ2, ∀ζRn\ {0}; κp >0;

(ii) s

n i=1

∂b(x,u,ξ)

∂ξi

2

≤ b1|u|1|ξ|p(x)−1; ∂b(x,u,ξ∂u ) ≥b2|u|2|ξ|p(x); b1 ≥0, b2≥0;

(iii) p(x)≥ p >1.

Proposition 3.1. Let Q satisfy assumptions (i)–(iii) and functions u,w ∈ N1,p1,∞(x)(Gd0) satisfy the inequality

Q(u,η)≤Q(w,η) (3.2)

for all non-negativeη∈N1,p1,∞(x)(Gd0).Assume also that the inequality

u(x)≤w(x)onΩd (3.3)

holds. Then u(x)≤ w(x)in G0d.

Proof. Let us define z= u−wanduτ =τu+ (1−τ)w, τ∈[0, 1]. Then we have 0≥Q(u,η)−Q(w,η)

=

Z

G0d

* ηxizxj

Z 1

0

Ai(x,uτx)

∂uτxj dτ+ηzxi Z 1

0

∂b(x,uτ,uτx)

∂uτxi dτ+ηz Z 1

0

∂b(x,uτ,uτx)

∂uτ dτ +

dx

Z

d

Z 1

0

Ai(x,uτx)

∂uτxj

!

cos(r,xi)·zxjη(x)dΩd

+

Z

Γd0

γ(ω) rp(x)−1

Z 1

0

(uτ|uτ|p(x)−2)

∂uτ

!

z(x)η(x)ds (3.4)

for all non-negativeη∈N1,p1,∞(x)(G0d).

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Now, we introduce the sets

(G0d)+:={x∈ G0d | u(x)>w(x)} ⊂G0d, (Γd0)+:={x∈ Γd0 | u(x)>w(x)} ⊂Γd0

and assume that (G0d)+ 6= ∅, (Γd0)+ 6= ∅. Let k ≥ 1 be any odd number. We choose η=max{(u−w)k, 0} as a test function in the integral inequality (3.4). We have

Z 1

0

(uτ|uτ|p(x)−2)

∂uτ dτ= (p(x)−1)

Z 1

0

|uτ|p(x)−2dτ>0.

Then, by assumptions(i)–(iii)andη

d=0, we obtain from (3.4) that Z

(G0d)+

pzk1 Z 1

0

|∇uτ|p(x)−2

|∇z|2+b2zk+1 Z 1

0

|uτ|2|∇uτ|p(x)

dx

≤ b1·

Z

(G0d)+

zk Z 1

0

|uτ|1|∇uτ|p(x)−1

|∇z|dx. (3.5)

By the Cauchy inequality, b1zk|∇z||uτ|1|∇uτ|p(x)−1=

|uτ|1zk+21|∇uτ|p(2x)

·

b1zk21|∇z||∇uτ|p(2x)1

ε

2|uτ|2zk+1|∇uτ|p(x)+ b

21

2εzk1|∇z|2|∇uτ|p(x)−2, ∀ε>0.

Hence, takingε=2b2, we obtain from (3.5) the inequality

pb

2 1

4b2 Z

(G0d)+

zk1|∇z|2 Z 1

0

|∇uτ|p(x)−2

dx ≤0. (3.6)

Choosing the odd number k ≥ max 1;2bb21

2κp

, in view of z(x) ≡ 0 on (G0d)+, we get from (3.6) that z(x) ≡ 0 in (G0d)+. We got a contradiction to our definition of the set (Gd0)+, this completes the proof.

Remark 3.2. For the p(x)-Laplacian assumption(i)is satisfied with

κp=

(1, if p(x)≥2;

p−1, if 1< p≤ p(x)<2.

3.1 Barrier function and eigenvalue problem(OEVP)

We shall study thebarrierfunctionw(r,ω)6≡0 as a solution of the auxiliary problem:





−4p+w=µw1|∇w|p+, x∈ G0d,

|∇w|p+2∂w

n + γ

|x|p+1w|w|p+2 =0, x∈ Γd0, 0<d≤d0.

(BFP)

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By direct calculations, we derive a solution of this problem in the form w=w(r,ω) =rκψκ(ω), κ= p+−1

p+−1+µλ, (BF)

where(λ,ψ(ω))is the solution of the eigenvalue problem (OEVP). For this function we cal- culate with regard toy(ω) = ψ0(ω)

ψ(ω):

∂w

∂rrκ−1ψκ(ω); ∂w

∂ω = κ

λrκψκλ1(ω)ψ0(ω);

|∇w|= κ

λrκ1ψκλ1(ω) q

λ2ψ2(ω) +ψ02(ω) = κ

λrκ1ψκλ(ω) q

λ2+y2(ω).

(3.7)

Proposition 3.3. w∈N1,p1,∞(x)(G0d).

Proof. From (BF) and (2.2) it follows thatw∈L(G0d). Next, Z

G0d

rp(x)wp(x)dx =

Z

G0d

r(κ−1)p(x)ψκλp(x)(ω)dx.

Byr≤d1 and assumption(i), we have

r(κ1)p(x) ≤r(κ1)p, ifκ ≥1;

r(κ−1)p(x) ≤r(κ−1)p+, ifκ ≤1;

ψκλp(x)(ω)≤ψ0p+ in virtue of (2.2) andκ ≤λ.

Hence it follows that Z

Gd0

rp(x)wp(x)dx≤ψ0p+measΩ·

d(κ−1)p+n

(κ1)p+n, ifκ≥1;

d(κ−1)p+ +n

(κ1)p++n, ifκ≤1.

(3.8) From (3.7) with regard to (2.4) we obtain that

Z

Gd0

w1|∇w|p(x)dx=

Z

Gd0

κ λ

p(x)

r(p(x)−1)κp(x)ψ(p(x)−1)κλp(x)(ω)λ2ψ2(ω) +ψ02(ω)p(x)/2dx

ψ0p+1 Z

G0d

r(p(x)−1)κp(x) λ2+y2(ω)p(x)/2dx

ψ0p+1 Z

G0d

r(p(x)−1)κ−p(x) λ2+z20p(x)/2

dx.

Since q

λ2+z20=const(n,λ,ω0,p+)andp(x)∈ [p,p+], we have λ2+y2(ω)p(x)/2 ≤C1=const(n,λ,ω0,p+,p). From the above inequality we obtain that

Z

Gd0

w1|∇w|p(x)dx≤C1ψ0p+1 Z

Gd0

r(p(x)−1)κ−p(x)dx

=C1ψ0p+1 Z

Gd0

r(κ−1)(p(x)−p+)·r(κ−1)p+−κdx. (3.9)

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Now, by assumptions(i)–(ii)andr1, we derive

r(κ1)(p(x)−p+)

(r(1κ)Lr, ifκ >1, 1, ifκ ≤1.

Using the well known inequality

rα|lnr| ≤ 1

αe, ∀α>0, 0<r<1, (3.10) wheree is the Euler number, we establish forκ>1 the inequality

r(1κ)Lr ≤eL(κ−e 1), 0<r<1.

Thus, from (3.9) it follows that Z

G0d

w1|∇w|p(x)dx≤ C1ψ0p+1eL(κ−e 1)measΩ· dκ

(p+1)+np+

κ(p+−1) +n−p+.

4 The proof of the main Theorem 1.3.

Let A > 1, and let w(r,ω) be the barrier function defined above. By the definition of the operatorQin(I I), we have

Q(Aw,η)≡

Z

G0d

D

Ap(x)−1|∇w|p(x)−2wxiηxi +b(Aw,A∇w)η(x)Edx

+γ Z

Γd0

Ap(x)−1r1p(x)wp(x)−1η(x)dS−

Z

d

Ap(x)−1|∇w|p(x)−2∂w

∂r η(x)dΩd

(4.1)

for all d∈(0,d0)and all non-negativeη∈N1,p1,(x)(Gd0). Integrating by parts, we obtain that Z

Gd0

Ap(x)−1|∇w|p(x)−2wxiηxidx

= −

Z

Gd0

d dxi

D

Ap(x)−1|∇w|p(x)−2wxi

E

η(x)dx+

Z

Γd0

Ap(x)−1|∇w|p(x)−2dw

dnη(x)dS +

Z

d

Ap(x)−1|∇w|p(x)−2∂w

∂rη(x)dΩd.

Hence and (4.1), with regard to problem (BFP), it follows that Q(Aw,η) =JGd

0 +JΓd

0, (4.2)

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where

JGd

0

Z

Gd0

*

µAp(x)−1w1|∇w|p(x)−Ap(x)−1|∇w|p+2wxid|∇w|p(x)−p+ dxi

∂A

p(x)−1

∂xi wxi|∇w|p(x)−2+b(Aw,A∇w) +

η(x)dx;

JΓd

0γ

Z

Γd0

Aw r

p(x)−1* 1−

r|∇w| w

p(x)−p++

η(x)dS.

(4.3)

At first, we assert that JΓd

0 ≥0. Indeed, by (3.7), r|∇w|

w

Γd0

= κ λ

q

λ2+y20

and desired inequality follows from Proposition2.1. Thus, from (4.2) it follows that Q(Aw,η)≥ JGd

0. (4.4)

Further, we proceed to the estimating of integralJGd

0. SettingW(x) =|∇w|p(x)−p+, we calculate lnW(x) = (p(x)−p+)ln|∇w|, =⇒ 1

W(x)·∂W

∂xi = ∂p

∂xi ln|∇w|+ p(x)−p+

|∇w| ·d|∇w|

dxi =⇒ d

dxi

|∇w|p(x)−p+= |∇w|p(x)−p+ p

∂xiln|∇w|+ p(x)−p+

|∇w| ·d|∇w| dxi

. Similarly,

d dxi

Ap(x)−1

= Ap(x)−1∂p

∂xi lnA.

By (4.3), we obtain that JGd

0

Z

G0d

(

Ap(x)−1|∇w|p(x)−2

*

µw1|∇w|2−(∇p· ∇w)(lnA+ln|∇w|)

p(x)−p+

|∇w| ·wxid|∇w| dxi

+

+b(Aw,A∇w) )

η(x)dx.

(4.5)

Passing to polar coordinates, we calculate wxid|∇w|

dxi = ∂w

∂r ·|∇w|

∂r + 1 r2

∂w

∂ω ·|∇w|

∂ω . Now, by (4.4)–(4.5) with regard to assumption(iv), we obtain that

Q(Aw,η)≥

Z

Gd0

Ap(x)−1 (

|∇w|p(x)−2

*

σw1|∇w|2−(∇p· ∇w)(lnA+ln|∇w|)

p(x)−p+

|∇w| · ∂w

∂r ·|∇w|

∂r + 1 r2

∂w

∂ω ·|∇w|

∂ω +

−b0wp(x)−1 )

η(x)dx,

(4.6)

(11)

with

σ= (

µδ, ifµ>0;

ν, ifµ=0.

Taking into account (3.7), the Lipschitz condition ofp(x)(ii)and ψψ0((ωω)) =y(ω), we directly calculate:

1)

|(∇p· ∇w)(lnA+ln|∇w|)| ≤ |∇p| · |∇w| ·(lnA+|ln|∇w||)

≤ L|∇w| ·(lnA+|ln|∇w||); in virtue of (3.7), (2.2), (2.4), we derive:

|ln|∇w|| ≤lnκ λ

+|κ−1| · |lnr|+κ

λψκλ1lnψ+ 1

2|ln(λ2+y2(ω))|

≤ln λ

κ +lnψ0+1

2|ln(λ2+z20)|+|κ−1| · |lnr|

=lnC1(n,p+,λ,ω0) +|κ−1| · |lnr|; (note that C1 ≥ 1 : indeed, by virtue of (BF) and (2.2) λ

κψ0 ≥ 1; from (13) [4] and (2.5) it follows that

q

λ2+z20qλ2+y20λ

κ ≥ 1 =⇒ C1 = λ

κψ0 q

λ2+z20 ≥ 1); using inequality (3.10) with α= 12, we get that

|∇w| · |ln|∇w|| ≤ κ

λrκ−1ψκλ(ω) q

λ2+y2(ω)·(lnC1+|κ−1| ·lnr)

= κ

λrκ2ψκλ(ω) q

λ2+y2(ω)· rlnC1+ (|κ−1|√ r)·√

rlnr

≤ κ

λrκ−2ψκλ(ω) q

λ2+y2(ω)·(rlnC1+|κ−1|√ r); from which we obtain that

|(∇p· ∇w)(lnA+ln|∇w|)| ≤ Lκ

λrκ2ψκλ(ω) q

λ2+y2(ω)·(rln(AC1) +|κ −1|√ r). 2)

|∇w|

∂r = κ−1 r |∇w|;

|∇w|

∂ω = |∇w| κ

λ + y

0(ω) λ2+y2(ω)

·y(ω) =|∇w|

p+−1

p+−1+µ+ y

0(ω) λ2+y2(ω)

·y(ω), and by (BF) it follows that

p+−p(x)

|∇w| · ∂w

∂r ·|∇w|

∂r + 1 r2

∂w

∂ω·|∇w|

∂ω

=κ(p+−p(x))w r2

κ−1+ y

2

λ

p+−1

p+−1+µ+ y

0(ω) λ2+y2(ω)

; using (2.3)–(2.4), we establish that

p+−1

p+−1+µ+ y

0(ω)

λ2+y2(ω) ≥ −λ (n−p+)

(p+−1)y2+λ2µ p+−1

p+−1+µ· y

2+λ2 (p+−1)y2+λ2

≥ −λ (n−p+)

(p+−1)y2+λ2µ y2+λ2 (p+−1)y2+λ2;

Hivatkozások

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