The transmission problem for quasi-linear elliptic second order equations in a conical domain. I, II

Nonlinear Analysis 71 (2009) 5032–5083

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The transmission problem for quasi-linear elliptic second order equations in a conical domain. I, II Mikhail Borsuk ∗ Department of Mathematics and Informatics, University of Warmia and Mazury in Olsztyn, 10-957 Olsztyn-Kortowo, Poland

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Article history: Received 25 November 2007 Accepted 25 March 2009 MSC: 35J65 35J70 35B05 35B45 35B65

abstract We investigate the behavior of weak solutions to the transmission problem for quasi-linear elliptic divergence second order equations in a neighborhood of the boundary conical point. In Part I we study the problem for quasi-linear equation with semi-linear principal part. In Part 2 we study the problem for general quasi-linear elliptic divergence second order equations. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Elliptic quasi-linear equations Interface problem Conical points

PART I 1. Introduction The transmission problems often appear in different fields of physics and technics. For instance, one of the important problems of the electrodynamics of solid media is the electromagnetic process research in ferromagnetic media with different dielectric constants. These problems appear as well as in solid mechanics if a body consists of composite materials. In this work we obtain the estimates of weak solutions of the nonlinear elliptic transmission problem near conical boundary point. Namely, for weak solutions of this problem we establish the possible exponent in the local bound of the solution modulus. Earlier the quasi-linear transmission problem was investigated only in smooth domains (see works of Borsuk [1], Rivkind–Ural’tseva [2], Kutev–Lions [3]). Later other mathematicians are studied transmission problems in nonsmooth domains in some particular linear cases (see the references cited in [4–6]). General linear interface problems in polygonal and polyhedral domains were considered in [4,5]. Regularity results in terms of weighted Sobolev–Kondratiev spaces are obtained in [6] for two- and three-dimensional transmission problems for the Laplace operator. D. Kapanadze and B.-W. Schulze studied boundary-contact problems with conical [7] singularities and edge [8] singularities at the interfaces for general linear any order elliptic equations (as well as systems). They constructed parametrix and showed regularity with asymptotics of solutions in weighted Sobolev–Kondratiev spaces. We knew only one paper, namely D. Knees’ work [9], that concerns with the study of the regularity of weak solutions of special nonlinear transmission problem on polyhedral

∗

Tel.: +48 89 524 60 14; fax: +48 89 524 60 07. E-mail address: [email protected].

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.03.090

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5033

Fig. 1.

domains. A principal new feature of our work is the consideration of estimates of solutions for general quasi-linear second order equations in n-dimensional conic domains. Let G ⊂ Rn , n ≥ 2 be a bounded domain with boundary ∂ G that is a smooth surface everywhere except at the origin O ∈ ∂ G and near the point O it is a conical surface with vertex at O . We assume that G = G+ ∪ G− ∪ Σ0 is divided into two subdomains G+ and G− by a Σ0 = G ∩ {xn = 0}, where O ∈ Σ0 . In Part I we consider the transmission problem for a quasi-linear equation with semi-linear principal part and establish the best possible exponent in the local bound of the solution modulus

 d
 |u|q aij (x)uxj + b(x, u, ∇ u) = 0, q ≥ 0, x ∈ G \ Σ0 ; −    dxi       1 x ∂u [u]Σ0 = 0, S [u] ≡ + σ u · |u|q = h(x, u), ∂ν | x | | x |     Σ0    B [u] ≡ ∂ u + 1 γ x u · |u|q = g (x, u), x ∈ ∂ G \ O ∂ν |x| | x|

x ∈ Σ0 ;

(WL)

(summation over repeated indices from 1 to n is understood); here (Fig. 1):

• u( x ) =



u+ (x), u− (x),

x ∈ G+ , x ∈ G− ;

aij (x) =



ij

a+ (x), ij a− (x),

x ∈ G+ , x ∈ G−

etc.;

• [u]Σ0 = u+ (x)|Σ0 − u− (x)|Σ0 , where u± (x)|Σ0 = limG± 3y→x∈Σ0 u± (y); → − → ∂ • ∂ν = aij (x) cos(− n , xi ) ∂∂x , where n denotes the unit outward with respect to G+ (or G) normal to Σ0 (respectively j ∂ G \ O ); ∂u denotes the saltus of the co-normal derivative of the function u(x) on crossing Σ0 , i.e. • ∂ν Σ0   ∂u ∂ u ∂ u ij → − → − a ( x ) cos ( n , x ) . = aij+ (x) cos(− n , xi ) j − ∂ν Σ0 ∂ xj Σ0 ∂ xj Σ0 We introduce the following notations:

• S n−1 : a unit sphere in Rn centered at O ; • (r , ω), ω = (ω1 , ω2 , . . . , ωn−1 ): the spherical coordinates of x ∈ Rn with pole O : x1 = r cos ω1 , x2 = r cos ω2 sin ω1 ,

.. .

xn−1 = r cos ωn−1 sin ωn−2 . . . sin ω1 , xn = r sin ωn−1 sin ωn−2 . . . sin ω1 .

• C : the convex rotational cone {x1 > r cos ω20 } with the vertex at O ; • ∂ C : the lateral surface of C : {x1 = r cos ω20 };

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M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

• Ω : a domain on the unit sphere S n−1 with smooth boundary ∂ Ω obtained by the intersection of the cone C with the sphere S n−1 ;

• • • • • • • •

Ω+ = Ω ∩ {xn > 0}, Ω− = Ω ∩ {xn < 0} =⇒ Ω = Ω+ ∪ Ω− ∪ σ0 ; σ0 = Σ0 ∩ Ω ; ∂ Ω = ∂ C ∩ S n−1 , ∂± Ω = Ω± ∩ ∂ C , ∂ Ω± = ∂± Ω ∪ σ0 ; Gba = {(r , ω) | 0 ≤ a < r < b; ω ∈ Ω } ∩ G: a layer in Rn ; Γab = {(r , ω) | 0 ≤ a < r < b; ω ∈ ∂ Ω } ∩ ∂ G: the lateral surface of layer Gba ; Σab = Gba ∩ {xn = 0} ⊂ Σ0 ; Gd = G \ Gd0 , Γd = ∂ G \ Γ0d , Σd = Σ0 \ Σ0d , d > 0; Ωρ = Gd0 ∩ {|x| = ρ}; 0 < ρ < d.

We use the standard function spaces: C k (G± ) with the norm |u± |k,G± , Lebesgue space Lm (G± ), m ≥ 1 with the norm ku± km,G± , the Sobolev space W k,m (G± ) with the norm ku± kk,m;G± , and introduce their direct sums Ck (G) = C k (G+ ) u C k (G− ) with the norm |u|k,G = |u+ |k,G+ + |u− |k,G− ; Lm (G) = Lm (G+ ) u Lm (G− ) with the norm

Z

|u+ |m dx

kukm,G =

 m1

Z

|u− |m dx

+

 m1 ;

G−

G+

Wk,m (G) = W k,m (G+ ) u W k,m (G− ) with the norm k X

Z kukm,k;G =

! m1 β

m

|D u+ | dx

+

G+ |β|=0

! m1

k X

Z

β

.

m

|D u− | dx

G− |β|=0

We define the weighted Sobolev spaces: Vkm,α (G) for integer k ≥ 0 and real α as the space of distributions u ∈ D 0 (G) with the finite norm

kukVkm,α (G) =

! m1

k X

Z

r

α+m(|β|−k)

β

m

|D u+ | dx

Z +

G+ |β|=0

k X

! m1 r

α+m(|β|−k)

β

m

|D u− | dx

G− |β|=0

k− 1

and Vm,αm (∂ G) as the space of functions ϕ , given on ∂ G, with the norm

kϕk

k− 1 Vm,αm (∂ G)

= inf kΦ kVkm,α (G) ,

where the infimum is taken over all functions Φ such that Φ |∂ G = ϕ in the sense of traces. We denote Wk (G) ≡ Wk,2 (G),

◦k

Wα (G) ≡ Vk2,α (G),

1 ◦ k− 2

Wα

k− 1

(∂ G) ≡ V2,α2 (∂ G).

Definition 1. The function u(x) is called a weak solution of the problem (WL) provided that u(x) ∈ C0 (G) ∩ W1 (G) and satisfies the integral identity

Z



|u|q aij (x)uxj ηxi + b(x, u, ux )η(x) dx +

G

γ (ω)

Z +

r

∂G

u|u|q η(x)ds =

Z ∂G

σ (ω)

Z

r

Σ0

g (x, u)η(x)ds +

u|u|q η(x)ds

Z Σ0

h(x, u)η(x)ds

(II)

for all functions η(x) ∈ C0 (G) ∩ W1 (G). Lemma 1.1. Let u(x) be a weak solution of (WL). For any function η(x) ∈ C0 (G) ∩ W1 (G) the equality

Z n % G0

Z o |u| a (x)uxj ηxi + b(x, u, ux )η(x) dx = q ij



Z +

%

Γ0

g ( x , u) −

holds for a.e. % ∈ (0, d).

γ (ω) r

u|u|q



η(x)ds +

Ω%

|u|q aij (x)uxj cos(r , xi )η(x)dΩ% 

Z %

Σ0

h(x, u) −

σ (ω) r

u|u|q



η(x)ds

(IIloc )

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Proof. The proof is analogous to the proof of Lemma 5.2 [10] (pp. 167–170).

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Regarding the equation we assume that the following conditions are satisfied: Let q ≥ 0, 0 ≤ µ < q + 1, s > 1, f1 ≥ 0, g1 ≥ 0, h1 ≥ 0, β ≥ s − 2 be given numbers; (a) the condition of the uniform ellipticity: ij

a± ξ 2 ≤ a± (x)ξi ξj ≤ A± ξ 2 , j

∀x ∈ G± ,

∀ξ ∈ Rn ; a± , A± = const > 0,

j

aij (0) = aδi , where δi is the Kronecker symbol; a+ , a− ,

a=

(

x ∈ G+ , x ∈ G− ;



we denote

a∗ = min{a+ , a− } > 0, a∗ = max{a+ , a− } > 0, A∗ = max(A− , A+ );

(b) aij (x) ∈ C0 (G) and the inequality

X n

|aij± (x) − aij± (y)|2

 12

≤ A(|x − y|)

i,j=1

holds for x, y ∈ G, where A(r ) is a monotonically increasing, nonnegative function, continuous at 0, A(0) = 0; (c) |b(x, u, ux )| ≤ aµ|u|q−1 |∇ u|2 + b0 (x); b0 (x) ∈ Lp/2 (G), n < p < 2n; (d) σ (ω) ≥ ν0 > 0 on σ0 ; γ (ω) ≥ ν0 > 0 on ∂ G;

≤ 0, ∂ g∂(xu,u) ≤ 0. (f) |b0 (x)| ≤ f1 |x|β , |g (x, 0)| ≤ g1 |x|s−1 , |h(x, 0)| ≤ h1 |x|s−1 .

(e)

∂ h(x,u) ∂u

We shall consider the function change u = v|v|ς−1

with ς =

1 q+1

.

(1.1)

Then identities (II) and (II)loc take the form

Z D Z E ς aij (x)vxj ηxi + B (x, v, vx )η dx + G

Z = ∂G

G(x, v)η(x)ds +

Z Σ0

γ (ω)

%

G0

ς a (x)vxj cos(r , xi )η(x)dΩ% + ij

= Ω%

σ (ω)

Z

r

Σ0

vη(x)ds

H (x, v)η(x)ds.

Z D Z E ς aij (x)vxj ηxi + B (x, v, vx )η dx + Z

r

∂G

vη(x)ds +

(e II)

γ (ω) %

r

Γ0

Z %

Γ0

vη(x)ds +

σ (ω)

Z

G(x, v)η(x)ds +

%

r

Σ0

Z %

Σ0

vη(x)ds

H (x, v)η(x)ds

((e II)loc )

for a.e. % ∈ (0, d), v(x) ∈ C0 (G) ∩ W1 (G) and any η(x) ∈ C0 (G) ∩ W1 (G), where

B (x, v, vx ) ≡ b(x, v|v|ς−1 , ς|v|ς −1 vx ),

G(x, v) ≡ g (x, v|v|ς −1 ),

H (x, v) ≡ h(x, v|v|ς −1 ).

(1.2)

We assume without loss of generality that there exists d > 0 such that Gd0 is a rotational cone with the vertex at O and the aperture ω0 , thus

Γ0d

  n ω0 X ω0 = (r , ω) x21 = cot2 x2i ; r ∈ (0, d), ω1 = , ω0 ∈ (0, 2π ) . 2 i=2 2

(1.3)

Our main result is the following statement. Let

λ=

2−n+

p

(n − 2)2 + 4ϑ 2

,

where ϑ is the smallest positive eigenvalue of the problem (EVP) (see Section 2.2).

(1.4)

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M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Theorem 1.2. Let u be a weak solution of the problem (WL), assumptions (a)–(f) are satisfied with A(r ) Dini-continuous at zero. Let us assume that M0 = maxx∈G |u(x)| is known. Then there are d ∈ (0, 1) and a constant C0 > 0 depending only on n, a∗ , A∗ , p, q, λ, µ, f1 , h1 , g1 , ν0 , s, M0 , meas G, diam G and on the quantity

 λ(1+q−µ)   |x| (q+1)2 ,      λ(1+q−µ)  1 1 |u(x)| ≤ C0 kuk2(q+1),G + f1 + g1 + h1 · |x| (q+1)2 ln q+1 ,  |x|   s   |x| q+1 ,

R1

A(r )

0

r

dr such that ∀x ∈ Gd0

if s > λ if s = λ if s < λ

1+q−µ 1+q 1+q−µ 1+q 1+q−µ 1+q

, ,

(1.5)

.

Suppose, in addition, that coefficients of the problem (WL) satisfy such conditions, which guarantee the local a priori estimate |∇ u|0,G0 ≤ M1 for any smooth G0 ⊂⊂ G \ {O } (see for example §4 [1] or [2]). Then for ∀x ∈ Gd0

 λ(1+q−µ) −1   , |x| (q+1)2     λ(1+q−µ) 1 1 −1 |∇ u(x)| ≤ C1 · |x| (q+1)2 ln q+1 ,  |x|   s   |x| q+1 −1 , 

if s > λ if s = λ if s < λ

1+q−µ 1+q 1+q−µ 1+q 1+q−µ

, ,

(1.6)

1+q



with C1 = c1 kuk2(q+1),G + f1 + g1 + h1 , where c1 depends on M0 , M1 and C0 from above.

2. Preliminaries 2.1. Auxiliary formulae Let us recall some well-known formulae related to spherical coordinates (r , ω1 , . . . , ωn−1 ) centered at the conical point

O

• • • • • • • • •

dx = r n−1 drdΩ , dΩρ = ρ n−1 dΩ , dΩ = J (ω)dω denotes the (n − 1)-dimensional area element of the unique sphere, J (ω) = sinn−2 ω1 sinn−3 ω2 . . . sin ωn−2 , dω = dω1 . . . dωn−1 , ds denotes the (n − 1)-dimensional area element on ∂ G; dσ denotes the (n − 2)-dimensional area element on ∂ Ω ; ds = r n−2 drdσ ;

|∇ u|2 =


∂u 2 ∂r

• |∇ω u|2 =

Pn−1

+

the point ω,

• ∆u =

1 r2

1 i =1 q i

|∇ω u|2 , where |∇ω u| is the projection of the vector ∇ u onto the tangent plane to the unit sphere at



∂u ∂ωi

2

, where q1 = 1, qi = (sin ω1 · · · sin ωi−1 )2 , i ≥ 2,

∂2u ∂r2

• ∆ω u =

+ n−r 1 ∂∂ur + r12 ∆ω u, Pn−1 ∂  J (ω) ∂ u  1

J (ω)

→ • divω − u =

1 J (ω)

i=1 ∂ωi

qi

Pn−1



∂

i=1 ∂ωi

∂ωi

J (ω) ui qi

, the Beltrami–Laplace operator,



.

C = C (. . .), c = c (. . .) denote the constants depending only on the quantities appearing in parentheses. In what follows, the same letters C , c will (generally) be used to denote different constants depending on the same set of arguments. By means of the direct calculation we obtain Lemma 2.1. xi cos(E n, xi )|Γ d = 0, 0

and

cos(E n, x1 )|Γ d = − sin 0

ω0 2

.

(2.1)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

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2.2. Auxiliary inequalities We need some statements and inequalities. The eigenvalue problem. − → Let Ω ⊂ S n−1 with a smooth boundary ∂ Ω be the intersection of the cone C with the unit sphere S n−1 . Let ν be the − → exterior normal to ∂ C at points of ∂ Ω and τ be the exterior with respect to Ω+ normal to Σ0 (lying in the tangent to Ω plane). Let γ (ω) be a positive bounded piecewise smooth function on ∂ Ω , σ (ω) be a positive continuous function on Σ . We consider the eigenvalue problem for the Laplace–Beltrami operator 4ω on the unit sphere 0f

 a± (4ω u± + ϑ u± ) = 0, a± are positive constants;       ∂u  [u]σ = 0, a − + σ (ω)u = 0; 0 → ∂ τ σ0 σ0    ∂ u±   a + γ (ω) u = 0 ,  ± − ± ± ∂→ ν

ω ∈ Ω± , (EVP)

∂± Ω

which consists of the determination of all values ϑ (eigenvalues) for which (EVP) has a non-zero weak solutions (eigenfunctions). Definition 2. A function ψ is called a weak solution of the problem (EVP) provided that ψ ∈ C0 (Ω ) ∩ W1 (Ω ) and satisfies the integral identity

Z  a Ω

1 ∂ψ ∂η

 Z Z − aϑψη dΩ + σ (ω)ψηdσ +

qi ∂ωi ∂ωi

σ0

∂Ω

γ (ω)ψηdσ = 0

for all η(x) ∈ C0 (Ω ) ∩ W1 (Ω ). Remark 1. We observe that ϑ = 0 is not an eigenvalue of (EVP). In fact, setting η = ψ and ϑ = 0 we have

Z Ω

a|∇ω ψ|2 dΩ +

Z σ0

σ (ω)|ψ|2 dσ +

Z ∂Ω

γ (ω)|ψ|2 dσ = 0

=⇒ ψ ≡ 0,

since a± > 0, σ (ω) > 0, γ (ω) > 0. Now, let us introduce the following functionals on C0 (Ω ) ∩ W1 (Ω )

Z

a|∇ω ψ| dΩ +

Z

Z

σ (ω)|ψ| dσ + γ (ω)|ψ| dσ , G[ψ] = ∂Ω Z D Z Z E H [ψ] = a |∇ω ψ|2 − ϑψ 2 dΩ + σ (ω)|ψ|2 dσ + γ (ω)|ψ|2 dσ 2

F [ψ] =

Ω

2

2

σ0

σ0

Ω

Z Ω

aψ 2 d Ω ,

∂Ω

and the corresponding bilinear forms

F (ψ, η) =

Z a Ω

1 ∂ψ ∂η qi ∂ωi ∂ωi

dΩ +

Z σ0

σ (ω)ψηdσ +

Z ∂Ω

γ (ω)ψηdσ ,

G(ψ, η) =

Z Ω

aψηdΩ .

We define yet the set K = ψ ∈ C0 (Ω ) ∩ W1 (Ω )| G[ψ] = 1 . Since K ⊂ C0 (Ω ) ∩ W1 (Ω ), F [ψ] is bounded from below for ψ ∈ K . The greatest lower bound of F [ψ] for this family we denote by ϑ = infψ∈K F [ψ]. We formulate the following statement:





Theorem 2.2. Let Ω ⊂ S n−1 be a bounded domain with smooth boundary ∂ Ω . Let γ (ω) be a positive bounded piecewise smooth function on ∂ Ω , σ (ω) be a positive continuous function on σ0 . There exist ϑ > 0 and a function ψ ∈ K such that

F (ψ, η) − ϑ G(ψ, η) = 0 for arbitrary η ∈ C0 (Ω ) ∩ W1 (Ω ). In particular F [ψ] = ϑ. Proof. The proof is analogous to the proof of Theorem 2.18 [10] (pp. 56–59).



Now from the variational principle we obtain the Friedrichs–Wirtinger type inequality (see e.g. the proof of Theorem 2.18 [10], pp. 59–60): Theorem 2.3. Let ϑ be the smallest positive eigenvalue of problem (EVP). (It exists according to Theorem 2.2.) Let Ω ⊂ S n−1 . Let ψ ∈ W1 (Ω ) and satisfies the boundary and conjunction conditions from (EVP) in the weak sense. Let γ (ω) be a positive bounded piecewise smooth function on ∂ Ω , σ (ω) be a positive continuous function on σ0 . Then

ϑ

Z Ω

aψ 2 (ω)dΩ ≤

Z Ω

a|∇ω ψ(ω)|2 dΩ +

Z σ0

σ (ω)ψ 2 (ω)dσ +

Z ∂Ω

γ (ω)ψ 2 (ω)dσ .

(2.2)

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M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Because of (1.4), the Friedrichs–Wirtinger inequality will be written in the following form

Z

λ(λ + n − 2)

Ω

aψ 2 (ω)dΩ ≤

Z

a|∇ω ψ|2 dΩ +

Ω

Z

σ (ω)ψ 2 (ω)dσ +

σ0

Z ∂Ω

γ (ω)ψ 2 (ω)dσ ,

(2.3)

∀ψ(ω) ∈ W1 (Ω ) satisfying the boundary and conjunction conditions from (EVP) in the weak sense; σ (ω) ≥ 0, γ (ω) ≥ 0. ◦1

Corollary 2.4. Let u ∈ C0 (G) ∩ Wα−2 (G), u(·, ω) satisfies the boundary and conjunction conditions from (EVP) in the weak sense and λ be as above in (1.4). Let σ (ω), ω ∈ σ0 ; γ (ω), ω ∈ ∂ Ω be nonnegative bounded piecewise smooth functions. Then

Z

ar α−4 u2 dx ≤ Gd0

nZ

1

λ(λ + n − 2)

ar α−2 |∇ u|2 dx +

Z Σ0d

Gd0

r α−3 σ (ω)u2 (x)ds +

Z

o

Γ0d

r α−3 γ (ω)u2 (x)ds ,

Proof. Multiplying (2.3) by r n−5+α and integrating over r ∈ (0, d) we obtain the required (2.4).

∀α.

(2.4)



Lemma 2.5. Let Gd0 be the conical domain, ∇v(%, ·) ∈ L2 (Ω ) a.e. % ∈ (0, d) and λ be as above in (1.4). Assume that for a.e. % ∈ (0, d) V (ρ) =

Z

ar 2−n |∇v|2 dx +

% G0

Z %

Σ0

r 1−n σ (ω)v 2 (x)ds +

Z %

Γ0

r 1−n γ (ω)v 2 (x)ds < ∞.

(2.5)

Then



Z a Ω

%v

 % 0 ∂v n−2 2 + v dΩ ≤ V (%). ∂r 2 2λ r =%

(2.6)

Proof. Writing V (%) in spherical coordinates V (%) =

%

Z

r 2−n

Z Ω

0

%

Z =

r Ω



r n−1 dr +

%

Z

r 1 −n

Z σ0

0



Z

0

a|∇v|2 dΩ

a vr2 +



1 r

|∇ω v|2 dΩ dr + 2

%

Z 0

1

Z

r

σ0

σ (ω)|v|2 dσ +

Z

σ (ω)|v|2 dσ +

Z

∂Ω

∂Ω

 γ (ω)|v|2 dσ r n−2 dr  γ (ω)|v|2 dσ dr

and differentiating with respect to % we obtain

∂v V (%) = a % ∂r Ω 

Z

0

2

! + |∇ω v| % 1

dΩ +

2

r =%

Z

1

%

σ0

σ (ω)v 2 (%, ω)dσ +

Z ∂Ω

 γ (ω)v 2 (%, ω)dσ .

(2.7)


2 Moreover, by the Cauchy inequality, we have ρv ∂v ≤ 2ε v 2 + 21ε ρ 2 ∂v for all ε > 0. Then ∂r ∂r 

Z a Ω

%v

 Z Z  2 ∂v n−2 2 ε+n−2 %2 ∂v + v dΩ ≤ av 2 d Ω + a dΩ . ∂r 2 2 2 ε ∂r Ω Ω r =%

Thus choosing ε = λ we obtain, by the Friedrichs–Wirtinger inequality (2.3),

Z

Z Z  2  ∂v n − 2 2  ε+n−2 %2 ∂v + v dΩ ≤ a|∇ω v|2 dΩ + a dΩ a %v ∂r 2 2λ(λ + n − 2) Ω 2ε Ω ∂r Ω r =% Z  Z ε+n−2 % 0 + σ (ω)v 2 (%, ω)dσ + γ (ω)v 2 (%, ω)dσ = V (%).  2λ(λ + n − 2) 2λ σ0 ∂Ω

We need also the well-known inequalities:

Z Γ

v ds ≤ C

Z ∂G

Z

(|v| + |∇v|)dx,

∀v(x) ∈ W1,1 (G), ∀Γ ⊆ ∂ G,

(2.8)

G

v 2 ds ≤

 Z  1 δ|∇v|2 + c0 v 2 dx, δ G

where C , c0 depend only on n, G, Γ .

∀v(x) ∈ W1,2 (G), ∀δ > 0,

(2.9)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5039

2.3. Quasi-distance rε (x)

¯ and consider the Further, we define the function rε (x) as follows. We fix the point Q = (−1, 0, . . . , 0) ∈ S n−1 \ Ω ¯ and introduce the vector unit radius-vector El = O Q = {−1, 0, . . . , 0}. We denote by E r the radius-vector of the point x ∈ G ¯ It is easy to verify that Erε = Er − εEl, ∀ε > 0. Since εEl 6∈ Gd0 for all ε ∈]0, d[, it follows that rε (x) = |Er − εEl| 6= 0 for all x ∈ G. rε (x) has the following properties (see in detail §1.4 [10]): 1. ∃h > 0 such that: rε (x) ≥ hr and rε (x) ≥ hε, ∀x ∈ G, where 1,

( h=

sin

ω0 2

if x1 ≥ 0 ,

,

if x1 < 0.

2. If x ∈ Gd , then rε (x) ≥

d 2

for all ε ∈]0, 2d [.

¯ 3. limε→0+ rε (x) = r, for all x ∈ G. n−1 2 4. |∇ rε | = 1, and 4rε = r . ε Lemma 2.6. Let v ∈ C0 (G)∩ W1 (G) and v(·, ω) satisfies the boundary and conjunction conditions from (EVP) in the weak sense. Let σ (ω) ≥ 0, γ (ω) ≥ 0 and λ be as above in (1.4). Then for any ε > 0

Z Gd0

arεα−2 r −2 v 2 dx

≤

Z

1

λ(λ + n − 2)

Gd0

arεα−2 |∇v|2 dx

Z + Σ0d

r −1 rεα−2 σ (ω)v 2 (x)ds

Z + Γ0d



r −1 rεα−2 γ (ω)v 2 ds

. (2.10) %

Proof. Multiplying both sides of the Friedrichs–Wirtinger inequality (2.3) by (% + ε)α−2 r n−3 and integrating over r ∈ ( 2 , %) we obtain

Z % G%/2

Z

1

a(% + ε)α−2 r −2 v 2 dx ≤

a(% + ε)α−2 |∇v|2 dx λ(λ + n − 2) G%%/2  Z Z −1 α−2 2 −1 α−2 2 + r (% + ε) γ (ω)v ds + r (% + ε) σ (ω)v ds , %

%

Γ%/2

Σ%/2

∀ε > 0

or since % + ε ∼ rε

Z % G%/2

arεα−2 r −2 v 2 dx

≤

Z

1

arεα−2 |∇v|2 dx

λ(λ + n − 2) G%%/2  Z + r −1 rεα−2 γ (ω)v 2 ds ,

Z +

%

Σ%/2

r −1 rεα−2 σ (ω)v 2 ds

∀ε > 0.

%

Γ%/2

Letting ρ = 2−k d, (k = 0, 1, 2, . . .) and summing the obtained inequalities over all k we get the desired inequality (2.10).  2.4. The Cauchy problem for differential inequality Theorem 2.7 (For the Proof See Section 1.10 (Theorem 1.57) [10]). Let V (%) be a monotonically increasing, nonnegative differentiable function defined on [0, 2d] and satisfy the problem



V 0 (ρ) − P (%)V (%) + N (ρ)V (2ρ) + Q(ρ) ≥ 0, V (d) ≤ V0 ,

0 < ρ < d,

(CP)

where P (%), N (%), Q(%) are nonnegative continuous functions defined on [0, 2d] and V0 is a constant. Then d

Z

V (%) ≤ exp

%

B (τ )dτ 2%

Z

B (%) = N (%) exp

%



d

 Z

V0 exp −



P (σ )dσ .

%

P (τ )dτ



d

Z + %

 Z

Q(τ ) exp −

%

τ





P (σ )dσ dτ ,

(2.11)

5040

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

3. Local estimate at the boundary We derive here a result asserting the local boundedness (near the conical point) of the weak solution of problem (WL). Theorem 3.1. Let u(x) be a weak solution of the problem (WL). Let assumptions (a), (c)–(e) be satisfied. Suppose, in addition, that h(x, 0) ∈ L∞ (Σ0 ), g (x, 0) ∈ L∞ (∂ G). Then the inequality sup |u(x)| ≤ ~% G0

C

n

1

2

%−n/t (q+1) kukt (q+1),G% + % q+1 (1−n/p) kb0 kpq/+21,G%

0 (1 − ~)n/t (q+1)  o 1 1 1 q+1 q+1 + % q+1 kg (x, 0)k∞, % + kh(x, 0)k % Γ ∞,Σ 0

0

(3.1)

0

holds for any t > 0, ~ ∈ (0, 1) and % ∈ (0, d), where C = const (n, a∗ , A∗ , t , p, µ, G) and d ∈ (0, 1). Proof. We can assume without loss of generality that n ≥ 3 (with respect to the general case see §8.6 [11] or §1 chapter 4 [12]). We apply the Moser iteration method. At first, we introduce the change of function (1.1) and consider the integral identity (e II). We make the coordinate transformation x = %x0 . Let G0 be the image of G, ∂ G0 be the image of ∂ G, Σ00 be the image of Σ0 . We have dx = %n dx0 , ds = %n−1 ds0 . In addition, we denote z (x0 ) = v(%x0 ),

F (x0 ) = %2 b0 (%x0 ).

(3.2)

Then (e II) means: for all η(x0 ) ∈ C0 (G0 ) ∩ W1 (G0 )

Z D Z E ς aij (%x0 )zx0j ηx0i + %2 B (%x0 , z , %−1 zx0 )η(x0 ) dx0 + G0

=%

Z ∂ G0

G(%x0 , z )η(x0 )ds0 + %

Z Σ00

∂ G0

γ (ω) z η(x0 )ds0 + |x0 |

Z Σ00

σ (ω) z η(x0 )ds0 |x0 |

H (%x0 , z )η(x0 )ds0 .

((e II)0 )

Now we define the quantity k by



k ≡ k(%) = (a∗ ς )−1 kF kp/2,G1 + %kG(%x0 , 0)k∞,Γ 1 + %kH (%x0 , 0)k∞,Σ 1 0 0 0



(3.3)

and we set z (x0 ) = |z (x0 )| + k.

(3.4)

Now we observe that

|F |z = |H |z ≤

1 k 1 k

|F | · kz =

1 k

|H | · z 2 ;

|F |(z − |z |) · z = |G|z ≤

1 k

1 k

1

1

k

k

|F | · z 2 − |F | · |z |z ≤

|F | · z 2 ; (3.5)

|G| · z 2

in the same way. First we assume that t ≥ 2. As the test function in the integral identity (e II)0 we choose

η(x0 ) =

1

ς

z (x0 )z

t −2

(x0 )ζ 2 (|x0 |),

where ζ (|x0 |) ∈ C∞ 0 ([0, 1]) is nonnegative function to be further specified. By the chain and product rules, η is a valid test function in (e II)0 and also

ηx0i =

1 

ς

1 + (t − 2)

z z

· z t −2 zx0i ζ 2 (|x0 |) +

2

ς

ζ ζx0i z (x0 )z t −2 (x0 ),

so that by substitution into (e II)0 with regard to 0 ≤ z ≤ z , ς ≤ 1, in virtue of assumptions (a), (c), (d) we obtain

Z az G10

t −2

|∇ 0 z |2 ζ 2 dx0 ≤

 |F (x0 )| t −1 2 z ζ dx0 ς G10 Z Z % % t −2 t −2 + z (x0 )G(%x0 , z )z (x0 )ζ 2 (|x0 |)ds0 + z (x0 )H (%x0 , z )z (x0 )ζ 2 (|x0 |)ds0 . ς Γ01 ς Σ01

Z 

2Az

t −1

|∇ 0 z |ζ |∇ 0 ζ | + µς az t −2 |∇ 0 z |2 ζ 2 +

We estimate every term by the Cauchy inequality:

(3.6)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

• z t −1 |∇ 0 z |ζ |∇ 0 ζ | ≤ •

1

ς

|F (x0 )|z t −1 ζ 2 ≤

t −2 t ε 1 0 2 2 az z az 2a∗ 2a∗ ε 0 |F (x )| t 2 az , in virtue of z ka∗ ς

|∇ | ζ +

ζ

5041

|∇ 0 ζ |2 , ∀ε > 0;

≥ k.

For estimating the integrals over the boundaries we have: 1

Z

z G(%x0 , z ) = z G(%x0 , 0) + z

d dτ

0

because of

% ς

∂ G(%x0 ,τ z )

z G(%x , z )z 0

Γ01

∂ G(%x0 , τ z ) dτ ≤ z¯ G(%x0 , 0), ∂(τ z )

1

Z 0

≤ 0. Therefore

∂(τ z )

Z

G(%x0 , τ z )dτ = z G(%x0 , 0) + z 2 ·

t −2

(x )ζ ds ≤ 0

0

2

%kG(%x0 , 0)k∞,Γ 1 Z ka∗ ς

Z

az ζ ds ≤ t

0

Γ01

0

2

az ζ 2 ds0 , t

Γ01

in virtue of z ≥ k and the definition of k by (3.3). In the same way we have

% ς

Z Σ01

z H (%x0 , z )z

t −2

Z

ζ 2 ds0 ≤

az ζ 2 ds0 . t

Σ01

For estimating the integrals over the boundaries we apply the inequality (2.9):

Z

Z

az ζ 2 ds0 ≤ t

Γ01 ∪Σ01

n

G10

a δ|∇ 0 (z

t /2

c0

ζ )|2 +

δ

o

z ζ 2 dx0 , t

∀δ > 0.

Thus we get

% ς

(Z

z G(%x , z )z 0

Γ01

t −2

(x )ζ ds + 0

0

2

)

Z

z H (%x , z )z 0

Σ01

t −2

ζ ds 2

Z

0

n

≤ G10

a δ|∇ 0 (z

t /2

ζ )|2 +

c0

δ

o

z ζ 2 dx0 , t

∀δ > 0. (3.7)

But from relations


|∇ 0 (ζ z t /2 )|2 ≤ 2 ζ 2 |∇ 0 (z t /2 )|2 + z t |∇ 0 ζ |2 ,

|∇ 0 (z t /2 )|2 =

t2 4

z

t −2

|∇ 0 z |2

(3.8)

it follows the inequality

|∇ 0 (ζ z t /2 )|2 ≤

t2 2

z

t −2

|∇ 0 z |2 ζ 2 + 2z t |∇ 0 ζ |2 .

(3.9)

Now, by (3.6) and (3.9), we find that

(1 − µς )

Z az

t −2

G10

2

0

2

A∗ ε



|∇ z | ζ dx ≤ 0

+

a∗

δt 2

Z az

2

G10

A∗

Z  +

2+

(1 − µς )

(1−µς) 2t 2

Z az

2

t −2

G10

,ε =

(1−µς)a∗ 4A∗

|∇ 0 z |2 ζ 2 dx0 ≤



a∗ ε

G10

We choose now δ =

t −2

|∇ z | ζ dx + 0

2

2

az |∇ 0 ζ |2 dx0 , t

0

Z  G10

 |F (x0 )| t + az ζ 2 dx0 δ ka∗ ς

c0

∀ε > 0, ∀δ > 0, % ∈ (0, d).

(3.10)

. Then the last inequality (3.10) rewrite in the such way 2c0 t 2

Z 

1 − µς

G10

+

  Z  |F (x0 )| 4A∗ t t az ζ 2 dx0 + 2+ 2 az |∇ 0 ζ |2 dx0 . ka∗ ς a∗ (1 − µς ) G10

But, by (3.8), the last means

Z G10

a|∇ 0 (z

t /2

)|2 ζ 2 dx0 ≤

c0 t 4

Z 

(1 − µς )2

G10

t2

Z +

1 − µς

G10

+

+

 |F (x0 )| t az ζ 2 dx0 2(1 − µς ) ka∗ ς ! t2

·

2A∗ 2 t 2

a2∗ (1 − µς )2

az |∇ 0 ζ |2 dx0 . t

Further, the above inequality can be rewritten as follows

Z

a|∇ (z 0

G10

t /2

)| ζ dx ≤ C1 t 2

2

0

4

Z

a |∇ ζ | + ζ 0

G10

2

2




t

0

z dx + C2 t

2

Z G10

|F (x0 )| · az t ζ 2 dx0 , ka∗ ς

(3.11)

5042

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

where C1 > 0, C2 > 0 depend only on c0 , a∗ , A∗ , µ, q and are independent of t. Setting

w(x0 ) =

√ a·z

t /2

(x0 )

(3.12)

from (3.11) we obtain

Z

|∇ w| ζ (|x |)dx ≤ C1 t 0

G10

2

2

0

0

4

Z

|∇ ζ | + ζ (|x |) w (x )dx + C2 t 0

G10

2

0

2

2




0

0

2

Z G10

|F (x0 )| · w2 (x0 )ζ 2 (|x0 |)dx0 . ka∗ ς

(3.13)

Now the desired iteration process can be developed from (3.13). By the Sobolev imbedding theorem, we have

kζ wk22n

,G 1 n−2 0

Z

≤ C∗



|∇ 0 ζ |2 + ζ 2 w 2 (x0 ) + ζ 2 (|x0 |)|∇ 0 w|2 dx0 ,

G10

n≥3

(3.14)

where C ∗ depends only on n and the domain G. Using the Hölder inequality for integrals

Z G10

|F (x0 )| · w2 (x0 )ζ 2 (x0 )dx0 ≤ kF kp/2,G1 · kwζ k22p 0

p−2

,G10

,

p > 2,

(3.15)

we get from (3.13)–(3.15)

kζ wk

2 2n ,G 1 n−2 0

≤ C3 t

4

Z

0


2 0 0 2 |F (x )| |∇ ζ | + ζ (|x |) w (x )dx + C4 t · kwζ k22p 1 , ,G ka∗ ς p/2,G1 p−2 0 0

G10

2

2

0

p > n.

(3.16)

0

By the interpolation inequality for Lp -norms

kζ wk

2p ,G 1 p−2 0

≤ εkζ wk

n

+ ε n−p kζ wk2,G1 ,

2n ,G1 n−2 0

p > n, ∀ε > 0,

0

(3.17)

from (3.16) and (3.17) it follows that

kζ wk

2n ,G 1 n−2 0

p |F (x0 )| 1/2 p

≤ t 2 C3 k(ζ + |∇ 0 ζ |)wk2,G1 + t C4

ka ς 0 ∗ p/2,G10   n × εkwζ k 2n ,G1 + ε n−p kζ wk2,G1 , p > n, ∀ε > 0 n−2

0

0

(3.18)

√ (here we take into account that

t < 1 + t , ∀t ≥ 0). Choosing

1

|F (x0 )| − 2 1

ε= √ ka∗ ς p/2,G1 2t C4 0 we obtain

kζ wk

p

2n ,G 1 n−2 0

≤ Ct p−n k(ζ + |∇ 0 ζ |)wk2,G1 , 0

2n ≥ p > n ≥ 3,

or recalling the definition of w by (3.12) we establish finally the inequality

kζ · z t /2 k

p

2n ,G 1 n−2 0

≤ Ct p−n k(ζ + |∇ 0 ζ |) · z t /2 k2,G1 , 0

2n ≥ p > n ≥ 3,

(3.19)

where C depends only on c0 , n, a∗ , A∗ , µ, p, diam G and is independent of t. This inequality can now be iterated to yield the desired estimate. For ∀~ ∈ (0, 1) we define sets ~+(1−~)2−j

G0(j) ≡ G0

,

j = 0, 1, 2, . . .

It is easy to verify that G~0 ≡ G0(∞) ⊂ · · · ⊂ G0(j+1) ⊂ G0(j) ⊂ · · · ⊂ G0(0) ≡ G10 . Now we consider the sequence of cut-off functions ζj (x0 ) ∈ C∞ (G0(j) ) such that 0 ≤ ζj (x0 ) ≤ 1 in G0(j)

|∇ 0 ζj | ≤

2j+1 1−~

and ζj (x0 ) ≡ 1 in G0(j+1) ,

ζj (x0 ) ≡ 0 for |x0 | > ~ + 2−j (1 − ~);

for ~ + 2−j−1 (1 − ~) < |x0 | < ~ + 2−j (1 − ~).

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5043

We define as yet the number sequence

 tj = t

j

n n−2

,

j = 0, 1, 2, . . .

Now we rewrite the inequality (3.19) replacing ζ (|x0 |) by ζj (x0 ) and t by tj ; then taking tj th root, we obtain

 kz ktj+1 ,G0(j+1) ≤

2/tj

C 1−~

2p

j

· 4 tj · (tj ) p−n

· t1 j

kz ktj ,G0(j) .

After iteration, we find that

( kz ktj+1 ,G0(j+1) ≤

Ct

p p−n



1−~

Notice that the series

P∞

j j = 0 tj

·

 p−p n )2

n

∞ P j=0

1 tj

∞ P

· 4j=0

n−2

j tj

· k z k t ,G 1 .

(3.20)

0

is convergent by the d’Alembert ratio test, but the series

P∞

1 j = 0 tj

=

n 2t

as a geometric series.

Therefore from (3.20) we get

kz ktj+1 ,G0(j+1) ≤

C

(1 − ~)n/t

kz kt ,G1 . 0

Consequently, letting j → ∞, we have sup z (x0 ) ≤

x0 ∈G~ 0

C

(1 − ~)n/t

k z k t ,G 1 . 0

Hence, because of the definitions of the function z (x0 ) by (3.4) and of the number k by (3.3), we obtain C

sup |z (x0 )| ≤



(1 − ~)n/t

x0 ∈G~ 0

 kz kt ,G1 + kF kp/2,G1 + %kG(%x0 , 0)k∞,Γ 1 + %kH (%x0 , 0)k∞,Σ 1 . 0

0

0

0

Returning to the variables x, v , by (3.2), we obtain in the case t ≥ 2 the estimate: sup |v(x)| ≤ ~% G0

K (%) = %

C

n

(1 − ~)n/t

2(1−n/p)

o %−n/t kvkt ,G% + K (%) , 0

(3.21)

kb0 kp/2,G% 0

  + % kg (x, 0)k∞,Γ % + kh(x, 0)k∞,Σ % . 0

0

Let now 0 < t < 2. We consider (3.21) with t = 2: sup |v(x)| ≤ ~% G0

C

n

(1 − ~)n/2

o %−n/2 kvk2,G% + K (%) .

Now, using the Young inequality with p = C

[(1 − ~)%]n/2

kvk2,G% = 0

(3.22)

0

2 t

2 and p0 = 2− we can write t

!1/2

Z

C

% [(1 − ~)%]n/2 G0 !1−t /2

sup |v(x)|

≤

%

v ·v t

·

G0

2−t

C

t /2

[(1 − ~)%]n/2

kvkt ,G% ≤ 0

1 2

sup |v(x)| + %

G0

C1

[(1 − ~)%]n/t

kvkt ,G% . 0

(3.23)

Let us define the function ψ(s) = supx∈Gs |v(x)|. Then from (3.22) and (3.23) it follows that 0

ψ(~%) ≤

1 2

ψ(%) +

C1

n

(1 − ~)n/t

o n %− t kvkt ,G% + K (%) , 0

~ ∈ (0, 1).

(3.24)

Further we apply the following statement: Proposition 3.2 (See Lemma 4.1 in Chapter 2 [12]). Let ψ(s) be a bounded nonnegative function defined on the interval [0, %]. Suppose that for any 0 ≤ σ < s ≤ % the function ψ(s) satisfy

ψ(σ ) ≤ δψ(s) +

A

(s − σ )α

+ B,

5044

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

where δ ∈ (0, 1), A, B and α are nonnegative constants. Then



ψ(r ) ≤ C

A

(R − r )α

 +B ,

0 ≤ r < R ≤ %,

(3.25)

where C depends only on α, δ. C1 , α = nt , A = C1 kvkt ,G% , B = (1−~) n/t K (%) from (3.24) we obtain the 0 validity of required estimate (3.21) in the case 0 < t < 2. Finally, returning to the function u(x) by the change of function (1.1) we complete the proof of Theorem 3.1. 

By this Proposition 3.2, letting r = ~%, R = %, δ =

1 2

4. Global integral estimate At first we obtain a global estimate for the weighted Dirichlet integral. Theorem 4.1. Let u(x) be a weak solution of the problem (WL). Let assumptions (a)–(e) be satisfied with a function A(r ) that is continuous at zero. Suppose, in addition, that ◦0

b0 (x) ∈ Wα (G),

Z Σ0

r α−1 h2 (x, 0)ds < ∞,

Z ∂G

r α−1 g 2 (x, 0)ds < ∞,

4 − n ≤ α ≤ 2.

◦1

Then |u(x)|q+1 ∈ Wα−2 (G) and

Z

a r α−2 |u|2q |∇ u|2 + r α−4 |u|2(q+1) dx +




Z Σ0

G

≤C

nZ


|u|2(q+1) + (1 + r α )b20 (x) dx +

Z Σ0

G

r α−3 σ (ω)|u|2(q+1) ds + r α−1 h2 (x, 0)ds +

Z

Z ∂G

r α−3 γ (ω)|u|2(q+1) ds

o

∂G

r α−1 g 2 (x, 0)ds ,

(4.1)

where the constant C > 0 depends only on a∗ , α, µ, q, n, λ from (1.4) and the domain G. Proof. Making the function change (1.1) we consider the integral identity (e II) for the function v(x). Setting in this identity η(x) = rεα−2 v(x), with regard to

ηxi = rεα−2 vxi + (α − 2)rεα−3

xi − ε li rε

v(x)

we obtain

ς

Z G

arεα−2 |∇v|2 dx +

=ς

2−α

Z

2

Z

−ς

G

Z + Σ0

G

Z Σ0

r −1 rεα−2 σ (ω)v 2 (x)ds +

Z Z

arεα−4 (xi − ε li )(v 2 )xi dx + ς (2 − α)


a (x) − a (0) rεα−2 vxi vxj dx − ij

ij

rεα−2 v(x)H (x, v)ds

Z + ∂G

Z G

r −1 rεα−2 γ (ω)v 2 (x)ds

∂G

G


aij (x) − aij (0) rεα−4 (xi − ε li )vxj v(x)dx

B (x, v, vx )rεα−2 v(x)dx

rεα−2 v(x)G(x, v)ds.

(4.2)

We transform the first integral on the right by integrating by parts:

Z

Z Z 2 2 ∂v+ ∂v− ∂v 2 dx = a+ rεα−4 (xi − ε li ) dx + a− rεα−4 (xi − ε li ) dx ∂ xi ∂ xi ∂ xi G G+ G−   Z Z ∂ − → 2 α−4 = − av 2 rεα−4 (xi − ε li ) dx + a+ v + rε (xi − ε li ) cos( n , xi )ds ∂ xi G ∂ G+   Z Z ∂ − → 2 α−4 + a− v − rε (xi − ε li ) cos( n , xi )ds = − av 2 rεα−4 (xi − ε li ) dx ∂ xi G ∂ G− Z Z − → → + av 2 rεα−4 (xi − ε li ) cos( n , xi )ds + [a]Σ0 v 2 rεα−4 (xi − ε li ) cos(− n , xi )ds, arεα−4 (xi − ε li )

∂G

Σ0

because of [v]Σ0 = 0. By elementary calculation we have:

(4.3)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5045

(1)

  ∂ xi − ε li rεα−4 (xi − ε li ) = nrεα−4 + (α − 4)(xi − ε li )rεα−5 = (n + α − 4)rεα−4 ; ∂ xi rε − → (2) because of cos( n , xi )|Σ0 = cos(xn , xi ) = δin , → (x − ε l ) cos(− n , x )| = δ n (x − ε l )| = (x − ε l )| = x | = 0, i

i

Σ0

i

i

i

i

Σ0

n

n Σ0

Σ0

n

since Σ0 = {xn = 0} ∩ G and ln = 0; (3) from the representation ∂ G = Γ0d ∪ Γd and by (2.1),

ω0 → (xi − ε li ) cos(− n , xi )|Γ d = −ε sin =⇒ 0 2 Z Z ω0 − → av 2 rεα−4 (xi − ε li ) cos( n , xi )ds = −ε sin 2

∂G

av 2 rεα−4 ds +

Γ0d

Z

− →

Γd

av 2 rεα−4 (xi − ε li ) cos( n , xi )ds.

Hence and from (4.3) it follows 2−α

Z

2

G

−ε

Z (2 − α)(4 − n − α) ∂v 2 dx = arεα−4 v 2 dx ∂ xi 2 G Z 2−α − → av 2 rεα−4 (xi − ε li ) cos( n , xi )ds. av 2 rεα−4 ds +

arεα−4 (xi − ε li )

2−α 2

sin

ω0

Z

2

Γ0d

2

(4.4)

Γd

From (4.2) and (4.4) we obtain the following equality:

ς

Z G

arεα−2 |∇v|2 dx

+ ες

2−α 2

ω0

sin

Z

2

2 α−4

Γ0d

av r ε

Z ds + Σ0

r −1 rεα−2 σ (ω)v 2 (x)ds

Z + ∂G

r −1 rεα−2 γ (ω)v 2 (x)ds

Z (2 − α)(4 − n − α) − → av rε (xi − ε li ) cos( n , xi )ds + ς ς arεα−4 v 2 dx = 2 2 Γd G Z Z

+ ς (2 − α) aij (x) − aij (0) rεα−4 (xi − ε li )vxj v(x)dx − ς aij (x) − aij (0) rεα−2 vxi vxj dx 2−α

Z

2 α−4

G

Z − G

G

B (x, v, vx )rεα−2 v(x)dx

Z + Σ0

rεα−2 v(x)H (x, v)ds

Z + ∂G

rεα−2 v(x)G(x, v)ds.

(4.5)

Now we estimate the integral over Γd . Because of Γd : rε ≥ hr ≥ hd ⇒ (α − 3) ln rε ≤ (α − 3) ln(hd), since α ≤ 2, we have rεα−3 |Γd ≤ (hd)α−3 and therefore: 2−α 2

ς

Z

2 α−4

Γd

av r ε

2−α → (xi − ε li ) cos(− n , xi )ds ≤ ς

Z

2

≤

2−α 2

Γd

arεα−3 v 2 ds

ς (hd)α−3

Z Γd

av 2 ds ≤ c

Z

(v 2 + |∇v|2 )dx,

(4.6)

Gd

by (2.9). Now, in virtue of assumption (c) and the function change (1.1), we have

|v · B (x, v, vx )| ≤ aµς 2 |∇v|2 + b0 (x)|v|.

(4.7)

Therefore using the Cauchy inequality we deduce the following

Z G

B (x, v, vx )rεα−2 v(x)dx ≤ µς 2

≤ µς 2

Z G

arεα−2 |∇v|2 dx +

δ 2

Z

Z G

G

arεα−2 |∇v|2 dx +

ar −2 rεα−2 v 2 dx +

Z G

rεα−2 |v|b0 (x)dx

Z

1 2δ a∗

G

r 2 rεα−2 b20 (x)dx,

∀δ > 0.

(4.8)

Now we use the representation G = Gd0 ∪ Gd . At first we estimate the integrals over Gd0 . By assumption (b) and the Cauchy inequality, we obtain:

Z n Gd0

aij (x) − aij (0)

≤ A(d)




Z a Gd0



rεα−2 vxi vxj + rεα−4 (xi − ε li )v(x)vxj dx

rεα−2 |∇v|2




+

rεα−3 |∇v|

o

Z 
3 · |v(x)| dx ≤ A(d) a rεα−2 |∇v|2 + rεα−4 v 2 dx. 2

Gd0

(4.9)

5046

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Now we estimate the integrals over Gd . By assumption (a), the Cauchy inequality and taking into account rε ≥ hd for r ≥ d, we obtain:

Z n

rεα−2 vxi vxj + rεα−4 (xi − ε li )v(x)vxj

aij (x) − aij (0)




Gd

≤ A∗

Z  Gd

3 α−2 rε |∇v|2 + rεα−4 |v|2 2




o

dx

Z

dx ≤ C (A∗ , h, α, d)


|∇v|2 + v 2 dx.

(4.10)

Gd

Next, in virtue of assumption (e),

v G(x, v) = v G(x, 0) + v 2 ·

∂ G(x, τ v) dτ ≤ |g (x, 0)| · |v|. ∂(τ v)

1

Z 0

(4.11)

Further, by the Cauchy inequality and because of γ (ω) ≥ ν0 > 0,

 1 p  1  δ 1 1 |g (x, 0)| r − 2 γ (ω)|v| ≤ r −1 γ (ω)v 2 + rg 2 (x, 0), |g (x, 0)| · |v| = r 2 √ 2 2δν0 γ (ω)

∀δ > 0;

taking into account rε ≥ hr (see Section 2.3) we obtain

Z ∂G

rεα−2 v G(x, v)ds

≤

δ

Z

1 rεα−2 γ (ω)v 2 ds

2 ∂G

r

+

Z

1 2δν0

∂G

r α−1 g 2 (x, 0)ds,

∀δ > 0.

(4.12)

Similarly, because of σ (ω) ≥ ν0 > 0,

Z Σ0

rεα−2 v H (x, v)ds ≤

δ 2

Z

1 1 rεα−2 σ (ω)v 2 ds + r 2δν0

Σ0

Z

r α−1 h2 (x, 0)ds,

Σ0

∀δ > 0.

(4.13)

Taking into account ς µ ≤ 1 and 4 − n ≤ α ≤ 2 as a result from (4.5)–(4.13) we obtain:

Z

ς (1 − µς )

arεα−2 |∇v|2 dx +

G

≤

3 2

+ +

A(d)

Z a Gd0

Z

1 2a∗ δ

δ

r α b20 (x)dx + G

Z

2

rεα−2 |∇v|2

α−2 1

Σ0

rε

r

Z

Z

1 α−2 1 α−2 rε σ (ω)v 2 (x)ds + rε γ (ω)v 2 (x)ds r r ∂G

Σ0


δ + rεα−4 v 2 dx +

2

Z

1 2δν0

∂G

Z

σ (ω)v 2 ds +

G

ar −2 rεα−2 v 2 dx + C

r α−1 g 2 (x, 0)ds + α−2 1

rε

∂G

Z

r

Z Σ0



γ (ω)v 2 ds ,

Z


|∇v|2 + v 2 dx Gd

r α−1 h2 (x, 0)ds



∀δ > 0.

(4.14)

In virtue of the inequality rε ≥ hr, we have rεα−4 ≤ h−2 r −2 rεα−2 . Hence, by Lemma 2.6, from (4.14) it follows

ς (1 − µς )

Z G

arεα−2 |∇v|2 dx +

Z

G

+

Z

2δν0

Σ0

Z

1 α−2 1 α−2 rε σ (ω)v 2 (x)ds + rε γ (ω)v 2 (x)ds r ∂G r

Σ0

arεα−2 |∇v|2 dx

≤ c (λ, ω0 ) (δ + A(d)) G Z
+C |∇v|2 + v 2 dx + 1

Z

Z

1 2a∗ δ

r α−1 h2 (x, 0)ds,

Z + Σ0

r −1 rεα−2 σ (ω)v 2 (x)ds

r α b20 (x)dx +

G

1 2δν0

Z ∂G

+ ∂G

r −1 rεα−2 γ (ω)v 2 ds



r α−1 g 2 (x, 0)ds

∀δ > 0, ∀ε > 0.

Because of 0 ≤ µ < 1 + q, we can choose δ =

Z



(4.15)

ς

(1 − µς ) and next d > 0 such that, by the continuity of A(r ) at zero, ( 1 − µς ) . Thus, from (4.15) we get 4 Z Z Z 1 α−2 1 α−2 α−2 2 2 arε |∇v| dx + rε σ (ω)v (x)ds + rε γ (ω)v 2 (x)ds

c (λ, ω0 )A(d) ≤

ς

Σ0

G

4c (λ,ω0 )

r

∂G

r

Z Z 2 2 ≤ C (a∗ , α, λ, µ, q, n, d) (|∇v| + v )dx + r α b20 (x)dx G

+

1

ν0

Z r ∂G

α−1 2

g (x, 0)ds +

G

1

ν0

Z r Σ0

α−1 2



h (x, 0)ds ,

∀ε > 0.

(4.16)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5047

We observe that the right-hand side of (4.16) does not depend on ε . Therefore we can perform the passage to the limit as ◦1

ε → +0 by the Fatou Theorem. Hence it follows that v(x) ∈ Wα−2 (G) and Z Z Z α−3 2 α−2 2 r σ (ω)v (x)ds + r α−3 γ (ω)v 2 (x)ds ar |∇v| dx + ∂G

Σ0

G

Z Z 2 2 (|∇v| + v )dx + r α b20 (x)dx ≤ C (a∗ , α, λ, µ, q, n, d) G

G

+

Z

1

r

ν0

α−1 2

g (x, 0)ds +

∂G

1

ν0

Z r



α−1 2

h (x, 0)ds .

Σ0

(4.17)

Further, returning to the integral identity (e II) and setting in it η(x) = v(x), we get

Z Z D E ij ς a (x)vxj vxi + B (x, v, vx )v dx + G

γ (ω) r

∂G

v ds + 2

σ (ω)

Z Σ0

r

v ds = 2

Z ∂G

G(x, v)v ds +

Z Σ0

H (x, v)v ds.

By the ellipticity condition (a), inequalities (4.7), (4.12) and (4.13) for α = 2, δ = 2, hence it follows

ς (1 − µς )

Z

nZ a|∇v| dx ≤ c (a∗ , ν0 , diam G) 2

G

2

|v| +

b20


(x) dx +

G

Z r

α−1 2

g (x, 0)ds +

∂G

Z

o

Σ0

r α−1 h2 (x, 0)ds . (4.18)

Now, using the inequality (2.4) and returning to the function u(x), by means of the function change (1.1), from (4.17) and (4.18) we get the desired estimate (4.1). 

5. Local integral weighted estimates Now we will obtain a local estimate for the weighted Dirichlet integral. Theorem 5.1. Let u(x) be a weak solution of the problem (WL) and λ be as above in (1.4). Let assumptions (a)–(f), be satisfied with A(r ) that is Dini-continuous at zero. ◦1

Then |u(x)|q+1 ∈ W2−n (G) and there are d ∈ (0, 1) and a constant C > 0 depending only on n, s, λ, q, µ, ν0 , a∗ , G, Σ0 and R 1 A(r ) on 0 r dr such that ∀% ∈ (0, d)

Z

a r 2−n |u|2q |∇ u|2 + r −n |u|2(q+1) dx +




% G0

≤C

Z %

Σ0

Z

r 1−n σ (ω)|u|2(q+1) ds +

  %2λ(1−µς ) ,   1  1 1 , |u|2(q+1) dx + f12 + g12 + h21 · %2λ(1−µς ) ln2  % ν0 ν0 G   2s % ,

Z %

Γ0

r 1−n γ (ω)|u|2(q+1) ds

if s > λ(1 − µς ), if s = λ(1 − µς ),

(5.1)

if s < λ(1 − µς ),

1 where ς = 1+ . q

Proof. Making the function change (1.1) we consider the integral identity (e II)loc for the function v(x). Therefore from ◦1

Theorem 4.1 it follows that v(x) belongs to W2−n (G), so it is enough to prove the estimate (5.1). Using the function V (%) that is defined by (2.5) and setting η(x) = r 2−n v(x) in the integral identity (e II)loc we obtain

Z
∂v ς V (%) ≤ ς % av(x) r 2−n v(x) aij (x) − aij (0) vxj cos(r , xi )dΩ% dΩ + ς ∂ r Ω Ω% r =% Z Z Z 2−n 2−n + r v(x)G(x, v)ds + r v(x)H (x, v)ds + ς (n − 2) ar −n xi vvxi dx Z

%

%

Γ0

Z +

Σ0

 %

% G0



−ς r 2−n aij (x) − aij (0) vxi vxj + ς (n − 2)r −n v(x) aij (x) − aij (0) xi vxj

G0

−r

2 −n

v(x)B (x, v, vx ) dx.

(5.2)

5048

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Now we transform some integrals on the right. By the Gauss–Ostrogradskiy divergence theorem

(n − 2)

Z % G0

ar

−n

v(x)xi vxi dx =

n−2

axi ∂v 2

Z % G0

2

Z + [a]Σ0

dx =

r n ∂ xi %

r

n − 2n 1

%n

2

Z Ω%

v (x)xi cos(n, xi )ds +

−n 2

Σ0

av 2 (x)xi cos(r , xi )dΩ%

Z

o

%

Γ0

ar −n v 2 (x)xi cos(n, xi )ds .

Since xi cos(n, xi )|Γ % = 0,

xi cos(n, xi )|Σ0 = xi cos(xn , xi )|Σ0 = xn |Σ0 = 0,

xi cos(r , xi )|Ω% = %;

0

we have from above

(n − 2)

Z %

ar −n v(x)xi vxi dx =

n−2

Z

2

G0

av 2 (x)dΩ .

Ω

(5.3)

Because of Lemma 2.5, from (5.2) and (5.3) it follows that

Z Z
% 0 1 ij ij V (%) ≤ v(x) a (x) − a (0) vxj cos(r , xi )dΩ + r 2−n v(x)G(x, v)ds V (%) + % 2λ ς Γ0% Ω Z Z 
1 + r 2−n v(x)H (x, v)ds + −r 2−n aij (x) − aij (0) vxi vxj % % ς Σ0 G0 
1 2 −n −n ij ij + (n − 2)r v(x) a (x) − a (0) xi vxj − r v(x)B (x, v, vx ) dx. ς

(5.4)

Hence, in virtue of assumptions (b)–(d) together with inequalities (4.7) and (4.11), it follows that

(1 − µς )V (%) ≤ + +

1

ς 1

ς

Z %

Σ0

Z % G0

% 0 V (%) + %A(%) 2λ

Z Ω

a|v||∇v|dΩ +

r 2−n |v(x)||h(x, 0)|ds + c1 (n)A(%)

Z

1

ς

Z %

Γ0

r 2−n |v(x)||g (x, 0)|ds

a r 2−n |∇v|2 + r 1−n |v||∇v| dx




% G0

r 2−n |v(x)||b0 (x)|dx.

(5.5)

We shall obtain an upper bound for each integral on the right. At first, applying the Cauchy and Friedrichs–Wirtinger inequalities we have (see (2.3) and (2.7))

Z Ω

a%|v||∇v|dΩ ≤

1

Z

2

a %2 |∇v|2 + |v|2 dΩ ≤ c2 (λ)%V 0 (%)




Ω

(5.6)

as well as, in virtue of the inequality (2.4) with α = 4 − n,

Z % G0

ar 1−n |v||∇v|dx ≤

Z

a r 2−n |∇v|2 + r −n |v|2 dx ≤ c3 (λ)V (%).




% G0

(5.7)

Further, by the Cauchy inequality with ∀δ > 0,

   3−n 1 γ (ω)|v| r 2 √ |g (x, 0)| ds % % γ (ω) Γ0 Γ0 Z Z δ 1 1 −n 2 ≤ r γ (ω)|v| ds + r 3−n |g (x, 0)|2 ds; 2 Γ% 2δν0 Γ % 0 0  Z Z    3−n 1 1−n p r 2−n |v||h(x, 0)|ds = r 2 σ (ω)|v| r 2 √ |h(x, 0)| ds % % σ (ω) Σ0 Σ0 Z Z δ 1 1−n 2 ≤ r σ (ω)|v| ds + r 3−n |h(x, 0)|2 ds; 2 Σ% 2δν0 Σ % 0 0 Z Z Z δ 1 r 2−n |v(x)||b0 (x)|dx ≤ ar −n |v|2 dx + r 4−n |b0 |2 dx % 2a∗ G% 2δ G% G0 0 0 Z δ 1 4−n ≤ c4 (λ)V (%) + r |b0 |2 dx 2a∗ 2δ G% 0 Z

r 2−n |v||g (x, 0)|ds =

Z



r

1−n 2

p

(5.8)

(5.9)

(5.10)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5049

in virtue of the inequality (2.4) with α = 4 − n. Thus, from (5.5)-(5.10) we get

% {(1 − µς ) − c5 (n, λ, q, a∗ )(δ + A(%))} V (%) ≤ (1 + c6 (λ)A(%)) V 0 (%) 2λ (Z Z Z +

1

2δ

% G0

r 4−n |b0 |2 dx +

1

ν0

%

Γ0

r 3−n |g (x, 0)|2 ds +

1

ν0

%

Σ0

)

r 3−n |h(x, 0)|2 ds ,

∀δ > 0.

(5.11)

But, by condition (f),

Z % G0

r 4−n |b0 |2 dx +

Z

1

ν0

%

Γ0

r 3−n |g (x, 0)|2 ds +

1

ν0

Z %

Σ0

r 3−n |h(x, 0)|2 ds ≤



1 2s

c0 (G) f12 +

1

ν0

g12 +

1

ν0

h21



· %2s .

Thus, from (5.11) we have the differential inequality (CP) Section 2.4 with

P (%) =

2λ

· {(1 − µς ) − c5 (n, λ, a∗ )(δ + A(%))}, ∀δ > 0; N (%) ≡ 0;   1 1 Q(%) = c0 (G) f12 + g12 + h21 · δ −1 %2s−1 , ∀δ > 0 s ν0 ν0 Z  Z Z
2 4−n 2 3−n 2 3−n 2 V0 = C v + (1 + r )b0 (x) dx + r h (x, 0)ds + r g (x, 0)ds , % λ

∂G

Σ0

G

(5.12)

by (4.1) with α = 4 − n. (1) s > λ(1 − µς ) Choosing δ = %ε , ∀ε > 0, 2λ

· {(1 − µς ) − c5 (n, λ, a∗ )(%ε + A(%))};   1 2 1 2 2 Q(%) = c0 (G) f1 + g1 + h1 · %2s−1−ε . s ν0 ν0 P (%) =

% λ

Since

P (%) =

2λ(1 − µς )

%

−

K (%)

%

,

where K (%) satisfies the Dini condition at zero we have

  Z τ  % 2λ(1−µ) Z d K (r ) τ K (s) + ds ≤ ln + dr =⇒ % s τ r % % 0 Z d   Z d     % 2λ(1−µς ) K (τ ) % 2λ(1−µς) exp d τ = K0 ; exp − P (τ )dτ ≤ d τ d 0 %  Z τ    Z d   % 2λ(1−µς ) % 2λ(1−µς) K (τ ) exp − P (τ )dτ ≤ exp d τ = K0 . τ τ τ % 0 Z

τ

−

P (s)ds = −2λ(1 − µς ) ln

We have as well: d

Z %

τ

 Z

Q(τ ) exp −

%



P (σ )dσ dτ ≤

≤

λc0 K0 s

λc0 K0 s

f12 + g12 + h21 %2λ(1−µς ς )




f12 + g12 + h21 ·




Z

d

% s−λ(1−µς )

τ 2s−2λ(1−µς )−ε−1 dτ

d

s − λ(1 − µς )

%2λ(1−µς ) ,

since s > λ(1 − µς ) and we can choose ε = s − λ(1 − µς ). Now we apply Theorem 2.7: then from (2.11), by virtue of the deduced inequalities and with regard to (2.4) for α = 4 − n, we get

Z

a r 2−n |∇v|2 + r −n v 2 dx +




% G0

Z +

Z %

Σ0

r 1−n σ (ω)v 2 (x)ds

r 1−n γ (ω)v 2 (x)ds ≤ C kvk22,G + f12 + g12 + h21 %2λ(1−µς ) ,




%

Γ0

(5.13)

5050

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

where C = const(n, s, q, λ, µ, ν0 , G). Returning to the function u(x) by means of the function change (1.1) hence we obtain the statement of (5.1) for s > λ(1 − µ). (2) s = λ(1 − µς ). Taking in (5.12), any function δ(%) > 0 instead of δ > 0, we obtain the problem (CP) with 2λ{(1 − µς ) − c5 δ(%)}

A(%) − c5 ; N (%) = 0; % %   1 1 Q(%) = c0 (G) f12 + g12 + h21 · δ −1 (%)%2λ(1−µ)−1 . ν0 ν0 P (%) =

We choose 1

δ(%) =

 ,

0 < % < d,

2λc5 ln ed %

where e is the Euler number. Then we obtain τ

Z − %

 % 2λ(1−µς)

P (σ )dσ ≤ ln

= ln

 % 2λ(1−µς) τ

τ

τ

Z

dσ

%

σ ln

   ln ed % ed

ln τ

τ

 Z ⇒ exp − %

P (σ )dσ



ed

σ

A(τ )

d

Z

+ ln     + c5

d

Z

  + c5

+

A(τ )

τ

0

dτ

dτ

τ  

0

 Z d   % 2λ(1−µς) ln ed % A(τ ) ≤ dτ , ·   exp c5 τ τ 0 ln ed τ

d

 Z exp − %

P (τ )dτ



≤

 % 2λ(1−µς) d

 ed 

ln

%

d

Z



exp C5 0

A(τ )

τ



dτ .

In this case we also have d

Z %

τ

 Z

Q(τ ) exp −

%

 ed  Z d dτ   · ν0 ν0 % % τ δ(τ ) ln ed τ    ed  1 1 ≤ 2λc5 c6 f12 + g12 + h21 · %2λ(1−µς ) ln2 . ν0 ν0 %





1

P (σ )dσ dτ ≤ c6 f12 +

g12 +

1

h21



%2λ(1−µς ) ln

Now we apply Theorem 2.7: then from (2.11), by virtue of the deduced inequalities and with regard to (2.4) for α = 4 − n, we get

Z

a r 2−n |∇v|2 + r −n v 2 dx +




% G0

Z

Z %

Σ0

r 1−n σ (ω)v 2 (x)ds

r 1−n γ (ω)v 2 (x)ds ≤ C kvk22,G + f12 + g12 + h21 %2λ(1−µς ) ln2




+

%

Γ0

 ed  %

,

(5.14)

where C = const(n, s, q, λ, µ, ν0 , G). Returning to the function u(x) by means of the function change (1.1) we obtain the statement of (5.1) for s = λ(1 − µς ). (3) 0 < s < λ(1 − µς ). Now similar to case (1) with regard to (5.12) we have

 Z exp − %

d

P (τ )dτ

 ≤

 % 2λ(1−µς−δ) d

d

 Z exp c5 0

A(τ )

τ

dτ

 = c8

 % 2λ(1−µς−δ) d

.

In this case we also have d

Z %

 Z Q(τ ) exp −

≤

c0 (ν0 , G)

δ

f12

τ %

+



P (σ )dσ dτ g12

+

h21




%

2λ(1−µς −δ)

d

Z %

τ

2s−2λ(1−µς−δ)−1

dτ ≤ c9



f12

+

1

ν0

g12

+

1

ν0

h21



· %2s ,

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5051

if we choose δ ∈ (0, 1 − µς − λs ). Now we apply Theorem 2.7: then from (2.11), by virtue of the deduced inequalities and with regard to (2.4) for α = 4 − n, we get

Z

a r 2−n |∇v|2 + r −n v 2 dx +




% G0

Z %

Σ0

r 1−n σ (ω)v 2 (x)ds +

Z %

Γ0

r 1−n γ (ω)v 2 (x)ds


≤ C kvk22,G + f12 + g12 + h21 %2s ,

(5.15)

where C = const(n, s, q, λ, µ, ν0 , G). Returning to the function u(x) by means of the function change (1.1) we obtain the statement of (5.1) for s < λ(1 − µς ).  6. The power modulus of continuity at the conical point for weak solutions Proof of Theorem 1.2. We define the function

 λ(1−µς) % ,   1 λ(1−µς) ψ(%) = % ln ,  %  s %,

if s > λ(1 − µς ), if s = λ(1 − µς ),

(6.1)

if s < λ(1 − µς )

for 0 < % < d. We make the function change (1.1) and consider the function v(x). For it, by Theorem 3.1 about the local bound of the weak solution modulus, we have (see (3.22)–(3.21))

n



sup |v(x)| ≤ C %−n/2 kvk2,G% + %2(1−n/p) kb0 kp/2,G% + % kg (x, 0)k∞,Γ % + kh(x, 0)k∞,Σ % 0

%/2 G0

0

0

0

o ,

% ∈ (0, d),

(6.2)

where C = C (n, a∗ , A∗ , p, µ, G) and n < p < 2n. Now, by Theorem 5.1 (see (5.13)-(5.15)), we have

%−n/2 kvk2,G% ≤

Z % G0

0

!1/2 r −n v 2 (x)dx

  ≤ C kvk2,G + f1 + g1 + h1 ψ(%).

(6.3)

Further, by assumption (f), we obtain

  %2(1−n/p) kb0 kp/2,G% + % kg (x, 0)k∞,Γ % + kh(x, 0)k∞,Σ % ≤ c (f1 + g1 + h1 ) ψ(%). 0

0

(6.4)

0

From (6.2)–(6.4) it follows that





sup |v(x)| ≤ C kvk2,G + f1 + g1 + h1 ψ(%). %/2 G%/4

Putting now |x| = 31 % and returning to the function u(x) by (1.1) we obtain finally the desired estimate (1.5). Now we will estimate the gradient modulus of the problem (WL) solution near a conical point. We consider two sets 2% % 2% G%/4 and G%/2 ⊂ G%/4 , % > 0. We perform the change of variables x = %x0 and v(%x0 ) = ψ(%)z (x0 ). Then the function z (x0 ) satisfies in the weak sense the problem

    %2 d  ij 0 0 0 ψ(%)   0 0 = 0, −ς a (% x ) z + B % x , ψ(%) z ( x ), z x  xj  dx0i ψ(%) %      ∂z σ (ω) % z (x0 ) Σ 1 = 0, ς + 0 z ( x0 ) = H (%x0 , ψ(%)z (x0 )), 0 1/2 ∂ν | x | ψ(%) 1  Σ1/2      ς ∂ z + γ (ω) z (x0 ) = % G(%x0 , ψ(%)z (x0 )), ∂ν 0 | x0 | ψ(%)

x ∈ G11/2 , x ∈ Σ11/2 ,

((WL)0 )

x ∈ Γ11/2 .

Now we apply our assumption about a priori estimate of the gradient modulus of the problem (WL)0 solution as an estimate inside the domain and near a smooth boundary portion max |∇ 0 z (x0 )| ≤ M10

(6.5)

x0 ∈G11/2

(see §4 Theorem 4.4 [1] or §§3, 5 [2] as well as [13]). Returning to the variable x and the function v(x) we obtain from (6.5)

|∇v(x)| ≤ M10 %−1 ψ(%), Putting now |x| =

2 3

%

x ∈ G%/2 , 0 < % < d.

% and returning to the function u(x) by (1.1), we obtain the desired estimate (1.6).



5052

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

7. Example Here we consider two-dimensional transmission problem for the Laplace operator with absorption term in an angular domain and investigate the corresponding eigenvalue problem. Suppose n = 2, the domain G lies inside the corner

n

G0 = (r , ω) |r > 0; −

ω0 2

<ω<

ω0 o 2

,

ω0 ∈]0, 2π [;

O ∈ ∂ G and in some neighborhood of O the boundary ∂ G coincides with the sides of the corner ω = − ω20 and ω = denote

n ω0 o , Γ± = (r , ω) | r > 0; ω = ± 2

ω0 2

. We

Σ0 = {(r , ω) | r > 0; ω = 0}

and we put σ (ω)|Σ0 = σ (0) = σ = const > 0, γ (ω)|ω=± ω0 = γ± = const > 0. We consider the following problem: 2

 d
 |u|q uxi = a0 r −2 u|u|q − µu|u|q−2 |∇ u|2 ,    dx     i ∂u 1 [u]Σ0 = 0, a|u|q + σ (0)u|u|q = 0, ∂ n Σ0 |x|     ∂ u± 1  q α± a± |u± | + γ± u± |u± |q = 0, ∂n |x|

x ∈ G0 \ Σ0 ; x ∈ Σ0 ;

(AL)

x ∈ Γ± \ O

where a0 ≥ 0, 0 ≤ µ < 1 + q, q ≥ 0; α± ∈ {0; 1}; a± > 0. We make the function change (1.1) and consider our problem for the function v(x):

 1  , 4v + µς v −1 |∇v|2 = a0 (1 + q)r −2 v; ς =    1 + q     v(x) ∂v [v ]Σ0 = 0, + (1 + q)σ (0) = 0, a ∂ n Σ0 |x|      α± a± ∂v± + (1 + q)γ± v± (x) = 0, ∂n | x|

x ∈ G0 \ Σ0 ; x ∈ Σ0 ; x ∈ Γ± \ O .

v(r , ω) = r ~ ψ(ω). For ψ(ω) we obtain the problem   ω   ω   µς 0 2 0 0  ψ (ω) + (1 + µς )~ 2 − a0 (1 + q) · ψ(ω) = 0, ω ∈ − , 0 ∪ 0, ; ψ 00 (ω) +    ψ(ω)  2 2  [ψ ]ω=0 = 0, aψ 0 (0) = (1 + q)σ (0)ψ(0);   ω   ω   0 0   ±α± a± ψ±0 ± + (1 + q)γ± ψ± ± = 0.

We want to find the exact solution of this problem in the form

2

We assume that ~ > 2

s Υ =

~ 2 − a0

2

(1+q)2 a0 1+q+µ

and define the value

(1 + q)2 . 1+q+µ

(7.1)

We consider separately two cases: µ = 0 and µ 6= 0.

µ = 0. In this case we get

ψ± (ω) = A cos(Υ ω) + B± sin(Υ ω),

(7.2)

where constants A, B± it should be determined from conjunction and boundary conditions; namely, they satisfy the system

(1 + q)σ (0) · A − + · B− =0 + Υ · B+ n  ω ao a− Υ  ω o ω0 ω0  0 0 (1 + q)γ+ cos Υ − α+ a+ Υ sin Υ · A + (1 + q)γ+ sin Υ + α+ a+ Υ cos Υ · B+ = 0 n  2   2 o n  2   2 o    (1 + q)γ cos Υ ω0 − α a Υ sin Υ ω0 · A − (1 + q)γ sin Υ ω0 + α a Υ cos Υ ω0 · B = 0. − − − − − − −   n  

2

2

2

The Dirichlet problem: α± = 0, γ± 6= 0. Direct calculations will give

 ω  0 ψ± (ω) = cos(Υ ω) ∓ cot Υ · sin(Υ ω), 2

(π , Υ = ω0 ∗ Υ ,

if σ (0) = 0; if σ (0) 6= 0,

2

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5053

where Υ ∗ is the least positive root of the transcendence equation

 ω  1+q 0 =− σ (0) Υ · cot Υ 2 a+ + a− and from the graphic solution we obtain ωπ < Υ ∗ < 2ωπ . 0 0 The corresponding eigenfunctions

    cos π ω , ω0 ψ± (ω) =    cos(Υ ∗ ω) ∓ cot Υ ∗ ω0 · sin(Υ ∗ ω), 2

if σ (0) = 0; if σ (0) 6= 0.

The Neumann problem: α± = 1, γ± = 0. Direct calculations will give:

(π , Υ = ω0 ∗ Υ ,

if σ (0) = 0; if σ (0) 6= 0,

where Υ is the least positive root of the transcendence equation ∗

 ω  0 Υ · tan Υ = 2

1+q a+ + a−

σ (0)

and from the graphic solution we obtain 0 < Υ ∗ < ωπ . 0 The corresponding eigenfunctions

    a∓ sin π ω , ω0 ψ± (ω) =    cos(Υ ∗ ω) ± tan Υ ∗ ω0 · sin(Υ ∗ ω), 2

if σ (0) = 0; if σ (0) 6= 0.

Mixed problem: α+ = 1, α− = 0; γ+ = 0, γ− = 1. Direct calculations will give: Υ = Υ ∗ , where Υ ∗ is the least positive root of the transcendence equation

 ω  0

a+ tan Υ

2

 ω  1+q 0 σ (0). − a− cot Υ = 2 Υ

The corresponding eigenfunctions

h ω i  ω  0 0 · sin(Υ ∗ ω), ω ∈ 0, ; ψ+ (ω) = cos(Υ ∗ ω) + tan Υ ∗ 2 2  ω  h ω i 0 0 ψ− (ω) = cos(Υ ∗ ω) + cot Υ ∗ · sin(Υ ∗ ω), ω ∈ − , 0 . 2

2

The Robin problem: α± = 1, γ± 6= 0. Direct calculations of the above system will give: γ a (1) γ+ = a+ . =⇒ ψ± (ω) = a∓ sin(Υ ∗ ω), −

−

where Υ ∗ is the least positive root of the transcendence equation

 ω  γ+ 0 Υ · cot Υ = −(1 + q) 2

a+

and from the graphic solution we obtain ωπ < Υ ∗ < 2ωπ . 0 0 γ a (2) γ+ 6= a+ . =⇒ A 6= 0 and from (7.2) it follows that ψ± (0) 6= 0; further see below the general case µ 6= 0. −

−

µ 6= 0. ψ 0 (ω)

It is obvious that in this case ψ(0) 6= 0. By setting y(ω) = ψ(ω) , we arrive at the problem for y(ω)

 0 2 2   y + (1 + µς )y (ω) + (1 + µς )~ − a0 (1 + q) = 0, a+ y+ (0) −a− y− (0) = (1 + q)σ (0);  ω  ±α± a± y± ± 0 + (1 + q)γ± = 0.

 ω   ω  0 0 ω ∈ − , 0 ∪ 0, ; 2

2

2

Integrating the equation of our problem we find y± (ω) = Υ tan {Υ (C± − (1 + µς )ω)} ,

∀C± .

(7.3)

5054

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

From boundary conditions we have C± = ±(1 + µς)

ω0

∓

2

1

Υ

arctan

(1 + q)γ± . α± a± Υ

(7.4)

Finally, in virtue of the conjunction condition, we get the equation for required ~ :

 α− a− Υ tan (1 + µς )Υ ω20 − (1 + q)γ− − (1 + q)γ+ + a− ·   a+ · α+ a+ Υ + (1 + q)γ+ tan (1 + µς )Υ ω20 α− a− Υ + (1 + q)γ− tan (1 + µς )Υ ω20  α+ a+ Υ tan (1 + µς )Υ

=

1+q

Υ

σ (0),

ω0 2

where 1 + µς =

1+q+µ 1+q

.

(7.5)

Further, from (7.3) and (7.4) we obtain



y± (ω) = Υ tan Υ

1+q+µ  1+q

ω0

(1 + q)γ± ± − ω ∓ arctan 2 α± a± Υ 

 (7.6)

and, because of (ln ψ(ω))0 = y(ω), hence it follows

   1+q (1 + q)γ± 1 + q + µ  ω0 ψ± (ω) = cos 1+q+µ Υ ± − ω ∓ arctan . 1+q 2 α± a± Υ

(7.7)

At last, returning to the function u, by (1.1), we establish a solution of (AL) u± (r , ω) = r

~

1+q

cos

1 1+q+µ



Υ

 (1 + q)γ± − ω ∓ arctan , ± 2 α± a± Υ

1+q+µ  1+q

ω0



(7.8)

where Υ is defined by (7.1) and ~ is the smallest positive root of the transcendence equation (7.5). If we consider the Dirichlet problem without the interface: α± = 0, a± = 1, σ (0) = 0, then we can calculate from (7.5) and (7.8) e λ

u(r , ω) = r cos

1 1+q+µ



 πω ; ω0

p e λ=

(π /ω0 )2 + a0 (1 + q + µ) . 1+q+µ

It is a well-known result (see Example 4.6, p. 374 [14]). Now we can verify that the derived exact solution satisfies the estimate (1.5) of Theorem 1.2. In fact, in our case we have: the value λ for (1.4) is equal to ϑ = ωπ and therefore 0

e λ

|u(r , ω)| ≤ r ≤ r since a0 ≥ 0 and

1 1+q+µ

π · 1+q−µ (1+q)2

1 π ω0 · 1+q+µ

≥

1+q−µ (1+q)2

≤ r ω0

,

.

PART II 8. Introduction This paper is the second of our work concerning concerning the transmission problem for quasi-linear equations. In Part 2 we study the problem for general elliptic divergence quasi-linear equations in n-dimensional domain with a conical point at the boundary. Let G ⊂ Rn , n ≥ 2 be a bounded domain with boundary ∂ G that is a smooth surface everywhere except at the origin O ∈ ∂ G and near the point O it is a conical surface with vertex at O . We assume that G = G+ ∪ G− ∪ 60 is divided into two subdomains G+ and G− by a 60 = G ∩ {xn = 0}, where O ∈ 60 . We consider the elliptic transmission problem

 d  − ai (x, u, ∇ u) + b(x, u, ∇ u) = 0,    dxi        ∂u 1 x [u]60 = 0, S [u] ≡ + m−1 σ u · |u|q+m−2 = h(x, u), ∂ν | x | | x |     60   ∂ u 1 x  B [u] ≡ + m−1 γ u · |u|q+m−2 = g (x, u), ∂ν | x| |x| (summation over repeated indices from 1 to n is understood); here:

x ∈ G \ 60 ; x ∈ 60 ; x ∈ ∂G \ O

(QL)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5055

• u( x ) =



u+ (x), u− (x),

x ∈ G+ , x ∈ G− ;

ai (x, u, ux ) =



a+ i (x, u+ , ∇ u+ ), a− i (x, u− , ∇ u− ),

x ∈ G+ , etc.; x ∈ G−

• [u]60 = u+ (x) −u− (x) , 60

60

− →



where u± (x)

60

=

lim

G± 3y→x∈60

u± (y);

− →

∂u = ai (x, u, ux ) cos( n , xi ), where n denotes the unit outward vector with respect to G+ (or G) normal to 60 ∂ν (respectively ∂ G \ O );  ∂u  • ∂ν 6 denotes the saltus of the co-normal derivative of the function u(x) on crossing 60 , i.e.

•

0



∂u ∂ν

 60

− → − → = a+ −a − . i (x, u+ , ∇ u+ ) cos( n , xi ) i (x, u− , ∇ u− ) cos( n , xi ) 60

60

We obtain estimates of weak solutions to problem (QL) near a conical boundary point. We use notations from Part I. Definition 3. The function u(x) is called a weak solution of the problem (QL) provided that u ∈ C0 (G)∩ W1,m (G) and satisfies the integral identity

Z

ai (x, u, ux )ηxi + b(x, u, ux )η(x) dx +





G

γ (ω)

Z + ∂G

σ (ω)

Z 60

r m−1

u|u|q+m−2 η(x)ds

Z

u|u|q+m−2 η(x)ds = g (x, u)η(x)ds + r m−1 ∂G

Z 60

h(x, u)η(x)ds

(II)

for all functions η ∈ C0 (G) ∩ W1,m (G). Lemma 8.1. Let u(x) be a weak solution of (QL). For any function η ∈ C0 (G) ∩ W1,m (G) the equality

Z n % G0

o

ai (x, u, ux )ηxi + b(x, u, ux )η(x) dx =



Z +

%

Γ0

g (x, u) −

γ (ω)

u|u|q+m−2 r m−1



Z Ω%

ai (x, u, ux ) cos(r , xi )η(x)dΩ%

η(x)ds +



Z %

60

h(x, u) −

σ (ω)

u|u|q+m−2 r m−1



η(x)ds

((II )loc )

holds for a.e. % ∈ (0, d). Proof. The proof is analogous to the proof of Lemma 5.2 [10, pp. 167–170].



We assume without loss of generality that there exists d > 0 such that Gd0 is a rotational cone with the vertex at O and the aperture ω0 ∈ (0, 2π ), thus

Γ0d

  n X ω0 2 2 2 ω0 , ω0 ∈ (0, 2π ) . xi ; r ∈ (0, d), ω1 = = (r , ω) x1 = cot 2

i=2

(8.1)

2

Regarding the n equation we assume that the following conditions are satisfied: Let a =

a+ , a− ,

x ∈ G+ , x ∈ G− ,

a± > 0, a∗ = min{a+ , a− } > 0, a∗ = max{a+ , a− } > 0;

1 < m < n, mn > p > n > m, q ≥ 0, ν0 > 0, 0 ≤ µ < m−1 be given numbers; a0 (x), α(x), b0 (x) are non-negative measurable functions; ai (x, u, ξ ), i = 1, . . . , n; b(x, u, ξ ) are the Caratheodory functions G × R × Rn → R and continuously differentiable with respect to xi , h(x, u) is continuously differentiable with respect to u variable function 60 × R → R but g (x, u) is continuously differentiable with respect to u variable function ∂ G × R → R possessing the properties: q+m−1

(1) ai (x, u, ξ )ξi ≥ a|u|q |ξ |m − a0 (x); a0 (x) ∈ Lp/m (G); v v u u n n uX uX ∂ ai (x, u, ξ ) 2 ≤ a|u|q |ξ |m−1 + α(x); α(x) ∈ L p (G); (2) t a2i (x, u, ξ ) + t m−1 ∂ x i i=1 i =1 (3a) |b(x, u, ξ )| ≤ aµ|u|q−1 |ξ |m + b0 (x); (3b) b(x, u, ξ ) = β(x, u) + e b(x, u, ξ ), u · β(x, u) ≥ a|u| |e b(x, u, ξ )| ≤ aµ|u|q−1 |ξ |m + b0 (x), ∂ h(x, u) ∂ g ( x , u) (4) ≤ 0, ≤ 0; ∂u ∂u (5) σ (ω) ≥ ν0 > 0 on σ0 ; γ (ω) ≥ ν0 > 0 on ∂ G.

b0 (x) ∈ L p (G); m

q+m

; b0 (x) ∈ L p (G); m

5056

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

In addition, suppose that the functions ai (x, u, ξ ) are continuously differentiable with respect to u, ξ variables in Md,M0 = Gd0

× [−M0 , M0 ] × Rn and satisfy in Md,M0 ∂ ai ( x , u, ξ ) ∂ ai ( x , u, ξ ) (6) (m − 1)u =q ξj ; i = 1, . . . , n; ∂u ∂ξj v u n uX ai (x, u, ux ) − a|u|q |∇ u|m−2 ux 2 ≤ aA(|x|)|u|q |∇ u|m−1 , (7) t i

x ∈ Gd0 ,

i=1

where A(r ) is Dini-continuous at zero function. We shall consider the function change u = v|v|ς −1

with ς =

m−1 q+m−1

.

(8.2)

By virtue of the assumption (6), the identity ((II )loc ) takes the form

Z D % G0

E

Ai (x, vx )ηxi + B (x, v, vx )η dx +

Z = Ω%

γ (ω)

Z

Ai (x, vx ) cos(r , xi )η(x)dΩ% +

%

r

Γ0

Z %

Γ0

v|v|m−2 η(x)ds + m−1

G(x, v)η(x)ds +

Z %

60

σ (ω)

Z %

60

r m−1

v|v|m−2 η(x)ds

H (x, v)η(x)ds

(8.3)

for a.e. % ∈ (0, d), v(x) ∈ C0 (G) ∩ W1,m (G) and any η(x) ∈ C0 (G) ∩ W1,m (G), where

Ai (x, vx ) ≡ ai (x, v|v|ς−1 , ς|v|ς −1 vx ), G(x, v) ≡ g (x, v|v|ς −1 ),

B (x, v, vx ) ≡ b(x, v|v|ς −1 , ς|v|ς −1 vx ),

H (x, v) ≡ h(x, v|v|ς −1 ).

(8.4)

We verify that coefficients Ai , i = 1, . . . , n do not depend on v explicit. In fact, by the change (8.2) and the assumption (6), we calculate

∂ Ai ∂ ai (x, u, ξ ) ∂  2 ς −1  ∂ ai (x, u, ξ ) ∂  2 ς −1  = · |v | 2 · v + · ς vxj |v | 2 ∂v ∂u ∂v ∂ξj ∂v ∂ ai ∂ ai u ∂ ai ξj ∂ ai = ς |v|ς−1 + ς (ς − 1)vxj v|v|ς−3 · =ς· · + (ς − 1) · · ∂u ∂ξj v ∂u v ∂ξj     ∂ ai m − 1 ∂ ai u ∂ ai m−1 1 ςu · = · = 0, + (ς − 1) · u · ς + (ς − 1) · = v ∂u q ∂u v ∂u q because of (8.2). Our assumptions regarding problem (QL) take the form: 1

(1)0 Ai (x, vx )vxi ≥ aς m−1 |∇v|m − |v|1−ς a0 (x); a0 (x) ∈ Lp/m (G); ς v v u n u n uX ∂ Ai (x, vx ) 2 uX ≤ aς m−1 |∇v|m−1 + α(x); (2)0 t A2i (x, vx ) + t ∂ x i i =1 i =1 (3a)0 |B (x, v, vx )| ≤ aµς m |v|−1 |∇v|m + b0 (x); (4)0 (7)0

α(x) ∈ L

p m−1

(G);

b0 (x) ∈ L p (G), m

∂ H (x, v) ∂ G(x, v) ≤ 0, ≤ 0; ∂v ∂v v u n uX t Ai (x, vx ) − aς m−1 |∇v|m−2 vx 2 ≤ aς m−1 A(|x|)|∇v|m−1 , i

x ∈ Gd0 .

i =1

Our main result is the following statement. Theorem 8.2. Let u be a weak solution of the problem (QL) and assumptions (1)–(7) are satisfied. Let us assume that M0 = maxx∈G |u(x)| is known. Let ϑ be the smallest positive eigenvalue of the problem (NEVP ) (see Section 2.2). Suppose, in addition,

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5057

that h(x, 0) ∈ L∞ (60 ), g (x, 0) ∈ L∞ (∂ G) and there exist real numbers ks ≥ 0, K ≥ 0 such that ks =: sup %−ms

Z %

%>0

m(q+m−1)

q

r q+m−1 |a0 (x)| (m−1)(q+m) dx +

G0

Z %

60

1

%

m

r m−1 |b0 (x)| m−1

G0

Z

m

+

Z

|h(x, 0)| m−1 ds +

%

Γ0

 m |g (x, 0)| m−1 ds ;

(8.5)

 n q+m−1 1 n % m −1 m(1− np ) (mq−+1m)(−q+1m) K =: sup ka0 k (pm,−G1%)(q+m) + %1− p kα(x)k m−p 1 , G% % m 0 0 m−1 %>0 ψ(%)   1 1 1 (1− np ) mm m−1 m−1 m−1 − 1 +% kb0 (x)k p , G% + % kg (x, 0)k∞, Γ % + kh(x, 0)k∞, 6 % , m

0

0

(8.6)

0

where

ψ(%) =

 1 ϑ m (m) q+(m−1)(1−µ)  · q+m−1  % Ξ (m) ,       1 ϑ m (m) q+(m−1)(1−µ) · q+m−1

% Ξ (m)        % s ,

d

%

ϑ m (m) q + (m − 1)(1 − µ) · ; Ξ (m) q+m−1

s=

ϑ m (m) q + (m − 1)(1 − µ) · ; Ξ (m) q+m−1

s<

ϑ m (m) q + (m − 1)(1 − µ) · , Ξ (m) q+m−1

1

,

(8.7)

1

( Ξ (m) = (m − 1)

ln

1

s>

m−1 m

·

m−2

2 2(m−1) ,

2

2−m 2

,

m ≥ 2; 1 < m ≤ 2.

(8.8)

Then there are d ∈ (0, 1) and a constant C0 > 0 independent of u such that

 q+mm−−1 1  n |u(x)| ≤ C0 |x|1− m ψ(|x|) ,

∀x ∈ Gd0 .

(8.9)

Moreover, if coefficients of the problem (QL) satisfy such conditions which guarantee the local a priori estimate |∇ u|0,G0 ≤ M1 for any smooth G0 ⊂⊂ G \ {O } (see for example Section 4 [1] or [2]), then there is a constant C1 > 0 independent of u such that

|∇ u(x)| ≤ C1 |x|

−1)+qm − nm(m (q+m−1)

m−1

ψ q+m−1 (|x|),

∀x ∈ Gd0 .

(8.10)

9. Preliminaries 9.1. Auxiliary inequalities We need some statements and inequalities. First of all, we will use the well-known primary inequalities: The Jensen inequality ([15, Theorem 65]). Let ai , i = 1, . . . , n, be any non-negative real numbers and m > 0. Then

θ

n X i=1

am i

≤

n X i=1

!m ai

≤Θ

n X

am i ,

(9.1)

i=1

where θ = min(1, nm−1 ) and Θ = max(1, nm−1 ). Proposition 9.1 ([15, Theorem 41]). Let a, b be non-negative real numbers and m ≥ 1. Then mam−1 (a − b) ≥ am − bm ≥ mbm−1 (a − b).

(9.2)

Proposition 9.2. Let a, b ∈ R and m > 1. Then the inequality

|b|m ≥ |a|m + m|a|m−2 a(b − a) is valid.

(9.3)

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M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Proof. By the Young inequality with ε = 1, p = m, q =

m , m−1

we have

m|a|m−2 ab ≤ m|b| · |a|m−1 ≤ |b|m + (m − 1)|a|m =⇒ (9.3).  We need also the well-known inequalities:

Z

Z |v|ds ≤ e c

Γ

Z ∂G

v 2 ds ≤

(|v| + |∇v|)dx,

∀v(x) ∈ W1,1 (G), ∀Γ ⊆ ∂ G,

(9.4)

G

 Z  1 δ|∇v|2 + c0 v 2 dx, δ G

∀v(x) ∈ W1,2 (G), ∀δ > 0,

(9.5)

where e c , c0 depend only on G. We shall need the following statements in dealing with integral estimates:

• The Sobolev imbedding Theorem. Let G be a lipschitzian domain. Let 1 < m < n and m# = and the inequality

mn . n −m

Then W1,m (G) ,→ Lm# (G)

kukLm# (G) ≤ e c kukW1,m (G)

(9.6)

holds;

• The Sobolev boundary trace imbedding Theorem. Let G be a lipschitzian domain and let Γ ⊆ G \ O be a piecewise C 1 -smooth (n − 1)-dimensional manifold. Let 1 < m < n and 1 ≤ j ≤ mn(n−−m1) . Then W1,m (G) ,→ Lj (Γ ) and the inequality kukLj (Γ ) ≤ e c kukW1,m (G)

(9.7)

holds;

• The Stampacchia Lemma. Let ϕ : [k0 , ∞) → R be a non-negative and non-increasing function which satisfies ϕ(h) ≤

C

(h − k)α

[ϕ(k)]β for h > k > k0 ,

(9.8)

where C , α, β are positive constants with β > 1. Then

ϕ(k0 + d) = 0, where dα = C |ϕ(k0 )|β−1 2αβ/(β−1) . 9.2. Eigenvalue problem for m-Laplacian in a bounded domain on the unit sphere

− →

Let Ω ⊂ S n−1 with a smooth boundary ∂ Ω be the intersection of the cone C with the unit sphere S n−1 . Let ν be the − → exterior normal to ∂ C at points of ∂ Ω and τ be the exterior with respect to Ω+ normal to 60 (lying in the tangent to Ω plane). Let γ (ω) be a positive bounded piecewise smooth function on ∂ Ω , σ (ω) be a positive continuous function on 6 . We consider the eigenvalue problem for the m-Laplace–Beltrami operator on the unit sphere:


 m−2 a divω (|∇ω ψ|m−2 ∇ω ψ) + ϑ|ψ| ψ = 0,     ∂ψ  m−2 a|∇ω ψ|m−2 − + σ (ω)|ψ| ψ [ψ]σ0 = 0, = 0; σ0 ∂→ τ σ0    ∂ψ a|∇ ψ|m−2 + γ (ω)|ψ|m−2 ψ = 0, ω → ∂Ω ∂− ν

ω ∈ Ω, (NEVP)

which consists of the determination of all values ϑ (eigenvalues) for which (NEVP ) has a non-zero weak solutions n a+ , a− ,

(eigenfunctions); here a =

x ∈ Ω+ , x ∈ Ω− ,

a± are positive constants.

Definition 4. A function ψ is called a weak eigenfunction of the problem (NEVP ) provided that ψ ∈ C0 (Ω ) ∩ W1,m (Ω ) and satisfies the integral identity



Z Ω

a |∇ω ψ|

m−2

1 ∂ψ ∂η qi ∂ωi ∂ωi

− ϑ|ψ|

m−2



ψη dΩ +

Z

σ (ω)|ψ|

m−2

σ0

ψηdσ +

Z ∂Ω

γ (ω)|ψ|m−2 ψηdσ = 0

for all η ∈ C0 (Ω ) ∩ W1,m (Ω ). Remark 2. We observe that ϑ = 0 is not an eigenvalue of (NEVP ). In fact, setting η = ψ and ϑ = 0 we have

Z Ω

a|∇ω ψ|m dΩ +

Z σ0

σ (ω)|ψ|m dσ +

since a > 0, σ (ω) > 0, γ (ω) > 0.

Z ∂Ω

γ (ω)|ψ|m dσ = 0

=⇒ ψ ≡ 0,

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5059

We characterize the first eigenvalue ϑ(m) of the eigenvalue problem for the m-Laplacian by

R ϑ(m) =

inf

Ω

ψ∈W 1,m (Ω ) ψ6=0

a|∇ω ψ|m dΩ + σ σ (ω)|ψ|m dσ + ∂ Ω γ (ω)|ψ|m dσ 0 R . a|ψ|m dΩ Ω

R

R

(9.9)

Let us introduce the following functionals on C0 (Ω ) ∩ W1,m (Ω )

Z

a|∇ω ψ| dΩ +

Z

Z

σ (ω)|ψ| dσ + γ (ω)|ψ| dσ , G[ψ] = ∂Ω Z D Z Z E H [ψ] = a |∇ω ψ|m − ϑ|ψ|m dΩ + σ (ω)|ψ|m dσ + γ (ω)|ψ|m dσ m

F [ψ] =

Ω

m

m

σ0

σ0

Ω

Z Ω

a|ψ|m dΩ ,

∂Ω

and the corresponding forms

F (ψ, η) =

Z

G(ψ, η) =

Z

Ω

Ω

a|∇ω ψ|m−2

1 ∂ψ ∂η qi ∂ωi ∂ωi

dΩ +

Z σ0

σ (ω)|ψ|m−2 ψηdσ +

Z ∂Ω

γ (ω)|ψ|m−2 ψηdσ ,

|ψ|m−2 ψηdΩ .

Consider now the set



K = ψ ∈ C0 (Ω ) ∩ W1,m (Ω ) G[ψ] = 1 .





Observing that the functional F [ψ] is bounded from below for ψ ∈ K , we let infψ∈K F [ψ] = ϑ . Now our aim is to establish the following statement: Theorem 9.3. Let Ω ⊂ S n−1 be a domain with smooth boundary ∂ Ω . Let γ (ω), ω ∈ ∂ Ω be a positive bounded piecewise smooth function, σ (ω) be a positive continuous function on σ0 . There exist ϑ > 0 and a function ψ ∈ K such that

F (ψ, η) − ϑ G(ψ, η) = 0

for arbitrary η ∈ C0 (Ω ) ∩ W1,m (Ω ).

In particular F [ψ] = ϑ . Proof. Consider a sequence {vk } ⊂ K such that limk→∞ F [vk ] = ϑ (such a sequence exists by the definition of the infimum). Since K ⊂ W1,m (Ω ) it follows that {vk } is bounded in W1,m (Ω ) and therefore compact in Lm (Ω ). Choosing a subsequence if needed we can assume that {vk } is converging in Lm (Ω ). As a consequence we obtain the following property of the functional G : given any ε > 0 we can find N (ε) such that G[vk − vl ] ≤ a∗ · kvk − vl km < ε, Lm (Ω )

∀ε > 0

(9.10)

for all k, l > N (ε). Now we use the inequality (9.3):

vk + vl m m m m−2 2 ≥ |vk | + 2 |vk | vk (vl − vk ),

m > 1.

By integrating this inequality over Ω

Z Ω

Z Z vk + vl m m m a dΩ ≥ a|vk | dΩ + a|vk |m−2 vk (vl − vk )dΩ 2 2 Ω

Ω

and applying the Young inequality with p =

m m−1

, q = m, we obtain

m m m−1 m 1 δ m−1 |vk |m + m |vl − vk |m , vk |vk |m−2 (vl − vk ) ≤ |vk |m−1 |vl − vk | ≤ 2 2 2 2δ

∀δ > 0

and consequently

Z

 Z Z vk + vl m 1 m dΩ ≥ 1 − m − 1 δ mm−1 a a |v | d Ω − a|vl − vk |m dΩ , k 2 2 2δ m Ω Ω Ω

This implies that

 G

vk + vl



2

  m−1 m 1 ≥ 1− δ m−1 G[vk ] − m G[vl − vk ], 2 2δ

By using G[vk ] = G[vl ] = 1 and G[vl − vk ] < ε1 we obtain

 G

vk + vl 2



>1−

m−1 2

m

δ m−1 −

ε1 , 2δ m

∀δ, ε1 > 0

∀δ > 0.

∀δ > 0.

5060

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083 m−1 m

for large k, l. Now by choosing δ m = ε1

 G

vk + vl



2

and setting ε =

mϑ 2

1

ε1m we get

ε ϑ

>1−

(9.11)

for large k, l. The functionals F [v] and G[v] are homogeneous functionals and therefore their ratio GF [v] does not change under [v] the passage from v to c v (c = const 6= 0) and hence F [v]

inf

v∈W1,m (Ω )

G[v]

= inf F [v] = ϑ. v∈K

Therefore F [v] ≥ ϑ G[v] for all v ∈ W1,m (Ω ). Since

 F

vk + vl



2

≥ ϑG



vk + vl



2

vk +vl 2

∈ W1,m (Ω ) together with vk , vl ∈ K , then

 ε = ϑ − ε, >ϑ 1− ϑ

k, l > N (ε).

Let us take k and l large enough so that F [vk ] < ϑ +ε and F [vl ] < ϑ +ε . We apply the Clarkson inequalities (see, e.g., Section 3.2, Chapter I [16]):

• (1) m ≥ 2   vl − vk F

2

1

≤

2

F [vl ] +

1 2

 F [vk ] − F

vl + vk



2

< ϑ + ε − (ϑ − ε) = 2ε • (2) 1 < m ≤ 2 1

F m−1



vl − vk



 ≤

2

1 2

F [vk ] +

1 2

 m−1 1

1

− F m−1

F [vl ]

1

1

< (ϑ + ε) m−1 − (ϑ − ε) m−1 <



vk + vl



2 2ε

m−1

2−m

(θ + ε) m−1 ,

by inequality (9.2). Consequently, F [vk − vl ] → 0,

as k, l → ∞.

(9.12)

From (9.10), (9.12) it follows that kvk − vl kW1,m (Ω ) → 0, as k, l → ∞. Thus {vk } is a Cauchy sequence in W 1 ,m

by the completeness of W

(Ω ) there exists a function ψ ∈ W

kvk − ψkW1,m (Ω ) → 0,

1,m

1,m

(Ω ) and hence

(Ω ) such that

as k → ∞.

In addition,

Z F [vk ] − F [ψ] =

Ω

a |∇ω vk | − |∇ω ψ| m

m




dΩ +

Z

σ (ω) |vk | − |ψ| m

σ0

m




dσ +

Z ∂Ω


γ (ω) |vk |m − |ψ|m dσ .

Now, by (9.2) and the Hölder inequality we have

Z Z
a |∇ω vk |m − |∇ω ψ|m dΩ ≤ m a|∇vk |m−1 |∇ω (vk − ψ)|dΩ Ω Ω Z 1/m Z (m−1)/m m m ≤m a|∇ω (vk − ψ)| dΩ a|∇ω vk | dΩ → 0, Ω

Ω

since vk ∈ W1,m . Furthermore, by using (9.3), the Hölder inequality for integrals and (9.4), we obtain

Z σ0


σ (ω) |vk |m − |ψ|m dσ ≤ m

Z σ0

σ (ω)|vk |m−1 · |vk − ψ|dσ

≤ m max |σ (ω)| · kvk − ψkLm (σ0 ) · kvk kLm (σ0 ) σ0

≤ c kvk − ψkW1,m (Ω ) · kvk kW1,m (Ω ) → 0,

as k → ∞,

as k → ∞,

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5061

and also

Z ∂Ω


γ (ω) |vk |m − |ψ|m dσ ≤ m

Z ∂Ω

γ (ω)|vk |m−1 · |vk − ψ|dσ

≤ m max |γ (ω)| · kvk − ψkLm (∂ Ω ) · kvk kLm (∂ Ω ) ∂Ω

≤ c kvk − ψkW1,m (Ω ) · kvk kW1,m (Ω ) → 0,

as k → ∞.

Therefore we get F [ψ] = lim F [vk ] = ϑ. k→∞

Analogously it can be seen that G[ψ] = 1. F [ψ+εη] Suppose now that η is a given function from C0 (Ω ) ∩ W1,m (Ω ). Consider the ratio G[ψ+εη] . It is a continuously differentiable function of ε on some interval around the point ε = 0. This ratio has a minimum at ε = 0 equal to ϑ and therefore, by the Fermat Theorem, we have



F [ψ + εη]

0

G[ψ + εη] ε=0

=m

F (ψ, η)G[ψ] − F [ψ]G(ψ, η) G2 [ψ]

= 0,

which, because F [ψ] = ϑ, G[ψ] = 1, gives

F (ψ, η) − ϑ G(ψ, η) = 0,

∀η ∈ C0 (Ω ) ∩ W1,m (Ω ). 

Now from the variational principle we obtain The Friedrichs–Wirtinger type inequality. Let ϑ be the smallest positive eigenvalue of the problem (NEVP ) (it exists according to Theorem 9.3). Let Ω ⊂ S n−1 and ψ ∈ W1,m (Ω ) satisfy the boundary and conjunction conditions from (NEVP ) in the weak sense. Let γ (ω) be a positive bounded piecewise smooth function on ∂ Ω and σ (ω) be a positive continuous function on σ0 . Then

Z Ω

a|ψ|m dΩ ≤

1

Z

ϑ(m)

Ω

a|∇ω ψ|m dΩ +

Z σ0

σ (ω)|ψ|m dσ +

Z ∂Ω

γ (ω)|ψ|m dσ



((W )m )

with the sharp constant ϑ(1m) . Proof. By approximation arguments, it is clearly sufficient to consider the above described functionals F [ψ], G[ψ], H [ψ] on C0 (Ω ) ∩ W2,m (Ω ). We will find the minimum of the functional F [ψ] on the set K . For this we investigate the minimization of the functional H [ψ] on all functions ψ(ω), for which the integrals exist and which satisfy the boundary conditions from (NEVP ). We use formally the Lagrange multipliers and get the Euler equation from the condition δ H [ψ] = 0. By the calculation of the first variation δ H, we have

δ H [ψ] = δ

Z

*   2 + m2  Z Z N −1  X  1 ∂ψ m m m a σ (ω)(ψ 2 ) 2 dσ + − ϑ(ψ 2 ) 2 dΩ + γ (ω)(ψ 2 ) 2 dσ  ∂ωi Ω  i=1 qi σ0 ∂Ω

  Z Z N −1 X ∂ J (ω) ∂ψ ∂ψ · |∇ω ψ|m−2 · δψ dω − mϑ a|ψ|m−2 ψ · δψ dΩ + m a|ψ|m−2 − · δψ dσ ∂ωi qi ∂ωi ∂→ ν Ω Ω ∂Ω i=1  Z  Z Z ∂ψ m−2 +m a|ψ|m−2 − · δψ d σ + m σ (ω)|ψ| ψ · δψ d σ + m γ (ω)|ψ|m−2 ψ · δψ dσ ∂→ τ σ0 σ0 ∂Ω   Z Z   ∂ψ m−2 = −m a divω (|∇ω ψ|m−2 ∇ω ψ) + ϑ|ψ|m−2 ψ · δψ dΩ + m a|ψ|m−2 − + σ (ω)|ψ| ψ · δψ dσ ∂→ τ Ω σ0  Z  ∂ψ m−2 + γ (ω)|ψ| ψ · δψ dσ . +m a|ψ|m−2 − ∂→ ν ∂Ω Z

= −m

a

Hence, because of δ H [ψ] = 0, ∀δψ ∈ C0 (Ω ) ∩ W1,m (Ω ), it follows the eigenvalue problem (NEVP ) for m-Laplacian. Conversely, let ϑ, ψ(ω) be a weak solution of the eigenvalue problem for m-Laplacian. From the definition of the weak eigenfunction under η = ψ(ω) it follows 0 = F [ψ] − ϑ G[ψ] = F [ψ] − ϑ ⇒ ϑ = F [u], (by K )

consequently, the required minimum is the least eigenvalue of the eigenvalue problem for m-Laplacian. The existence of a function ψ ∈ K such that F [ψ] ≤ F [v] for all v ∈ K has been proved above. 

5062

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Corollary 9.4. Let ϑ be the smallest positive eigenvalue of the problem (NEVP ) and v(x) ∈ W1,m (Gd0 ) satisfies the boundary and conjunction conditions from (NEVP ) in the weak sense. Let γ (ω) be a positive bounded piecewise smooth function on ∂ Ω and σ (ω) be a positive continuous function on σ0 . Then for any % ∈ (0, d) and ∀α

Z % G0

ar α |v|m dx ≤

Z

1

ϑ(m)

Z

ar α+m |∇v|m dx +

% G0

r α+m

%

60

σ (ω) r

|v|m ds + m−1

Z %

r α+m

Γ0

γ (ω) r

 m |v| ds , m−1

(9.13)

provided that integrals on the right are finite. In particular,

Z

%m a|v| dx ≤ % ϑ(m) G0 m

(Z % G0

σ (ω)

Z

m

a|∇v| dx +

%

r m−1

60

γ (ω)

Z

m

|v| ds +

%

Γ0

r m−1

) |v| ds . m

((H − W )m )

Proof. Consider the inequality ((W )m ) for the function v(r , ω), by multiplying it by r α+n−1 and integrating over r ∈ (0, %), we obtain the desired inequality (9.13). Setting in it α = 0 we get ((H − W )m ).  9.3. One auxiliary integral inequality Lemma 9.5. Let Gd0 be the conical domain, ∇v(%, ·) ∈ W1,m (Ω ) for almost all % ∈ (0, d) and V (%) =

Z %

σ (ω)

Z

m

a|∇v| dx +

%

60

G0

r m−1

γ (ω)

Z

m

|v| ds +

%

r m−1

Γ0

|v|m ds < ∞.

(9.14)

Let ϑ(m) be the smallest positive eigenvalue of the problem (NEVP ) and γ (ω) be a positive bounded piecewise smooth function on ∂ Ω , σ (ω) be a positive continuous function on σ0 . Then for almost all % ∈ (0, d)

Z Ω%

av

∂v % V 0 (%), |∇v|m−2 dΩ% ≤ Ξ (m) · 1 ∂r m mϑ

(9.15)

where Ξ (m) is determined by (8.8). Proof. Writing the function V (%) in spherical coordinates V (%) =

%

Z

r n −1

Z Ω

0

%

Z +

r

n−m−1

a|∇v(r , ω)|m dΩ



%

Z dr +

r n−m−1

∂Ω

0

Z

 γ (ω)|v(r , ω)|m dσ dr



σ (ω)|v(r , ω)| dσ dr m

σ0

0

Z

and differentiating it with respect to % we obtain V 0 (%) = %n−1

Z Ω

a|∇v(%, ω)|m dΩ + %n−m−1

Now using the Young inequality with p = m, p0 =

Z

Z

m m−1

∂Ω

γ (ω)|v(%, ω)|m dσ +

Z σ0

 σ (ω)|v(%, ω)|m dσ .

(9.16)

we have

 Z    ∂v ∂v ∂v v m−2 n−1 m−2 n m−2 av |∇v| dΩ% = % av |∇v| dΩ = % a · |∇v| dΩ ∂ r ∂ r % ∂ r Ω% Ω Ω r =% r =%   mm−1 Z n  m o m m−1 − 1 ∂v ε v ≤ %n a + ε m−1 |∇v|(m−2) m−1 dΩ , ∀ε > 0. m % m ∂r Ω r =% Z

Next, applying the Friedrichs–Wirtinger type inequality ((W )m ) we obtain

Z

∂v ε%n av |∇v|m−2 dΩ% ≤ ∂r mϑ(m) Ω%

 Z v(%, ω) m v(%, ω) m dσ σ (ω) d σ + γ (ω) % % σ0 ∂Ω ( ) m Z ∂v m−1 1 1 n ε ∇ω v m − m− (m−2) mm dΩ . 1 − 1 + % a + (m − 1)ε |∇v| m ϑ(m) % ∂r Ω r =% Z

Now we choose

ε = h(m − 1)ϑ(m)i

m−1 m

1

=⇒ (m − 1)ε− m−1 =

ε ϑ(m)

(9.17)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5063

and therefore the above gives

Z v(%, ω) m v(%, ω) m dσ + σ (ω) γ (ω) % dσ % Ω% σ0 ∂Ω ! m  ∇ω v m ∂v m−1 m (m−2) m−1 + · |∇v| % dΩ . ∂r Ω r =% But, because |∇v|2 = vr2 + r12 |∇ω v|2 and ∇%ω v ≤ |∇v|, we get: Z

av

∂v ε%n |∇v|m−2 dΩ% ≤ ∂r mϑ(m) Z + a

Z

m ∇ω v m ∂v m−1 m + · |∇v|(m−2) m−1 % ∂r

   ! m  ∇ v 2 2(m−1)  m 
m ( m − 2 ) ω  2 2(m−1)  m−1 · + v = |∇v|  r   %     m−2 ≤ 2 2(m−1) |∇v|m , m ≥ 2;    !m   ∇ω v 2 2 m 2−m   + (vr2 ) 2 ≤ 2 2 |∇v|m , 1 < m ≤ 2 ≤ % (m−2) mm −1

because of the Jensen inequality (9.1); in the last case we use |∇v| ≥ |vr | =⇒ |∇v| from (9.18) it follows that

Z Ω%

av

(9.18)

m

≤ |vr |(m−2) m−1 . Hence and

 Z Z Z v(%, ω) m v(%, ω) m ε%n ∂v m dσ + dσ + d Ω |∇v|m−2 dΩ% ≤ σ (ω) γ (ω) a |∇v| % ∂r mϑ(m) % σ0 ∂Ω Ω r =% ( m−2 2 2(m−1) , m ≥ 2 × 2−m 2 2 , 1 < m ≤ 2.

Substituting here ε from (9.17) and recalling the definition (8.8) we get the desired inequality (9.15).



10. Maximum principle In this section we consider one in a possible case of the deriving L∞ (G) a priori estimate of the weak solution to problem

(QL).

Theorem 10.1. Let u(x) be a weak solution of (QL) and assumptions (1), (3b), (4), (5) hold. Suppose, in addition, that h(x, 0) ∈ L

j j−1

(60 ),

g ( x, 0 ) ∈ L

j j−1

(∂ G),

1
n−1 m−1

.

Then there exists the constant M0 > 0, depending only on meas G, meas ∂ G, meas 60 , n, m, kh(x, 0)kL

µ, q, ka0 (x)kL p (G) , kb0 (x)kL p (G) , such that m

j j−1

(60 ) ,

kg (x, 0)kL

j j−1

(∂ G) ,

m

kukL∞ (G) ≤ M0 . Proof. For the proof we apply the following Lemma 10.2 (See Lemma 1.60 [10]). Let us consider the function

η(x) =

e~ x − 1, −e−~ x + 1,

x ≥ 0, x ≤ 0,



where ~ > 0. Let a, b be positive constants, m > 1. If ~ > (2b/a) + m, then we have aη0 (x) − b|η(x)| ≥

a ~x e ,

2

∀x ≥ 0

(10.1)

and

h  x im η(x) ≥ η , m

∀x ≥ 0.

(10.2)

5064

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Moreover, there exist a d ≥ 0 and an M > 0 such that

h  x im h  x im η(x) ≤ M η and η0 (x) ≤ M η , m

|η(x)| ≥ x,

∀x ≥ d;

m

(10.3)

∀x ∈ R.

(10.4)

Let us introduce the set A(k) = {x ∈ G, |u(x)| > k} and let χA(k) be a characteristic function of the set A(k). We note that A(k + d) ⊆ A(k) ∀d > 0. By setting η((|u| − k)+ )χA(k) · sign u as test function in (II ), where η is defined by Lemma 10.2 and k ≥ k0 (without loss of generality we can assume k0 ≥ 1), on the strength of the assumptions (1), (3b), (5) we get the inequality

Z

 A(k)

a|u|q |∇ u|m η0 ((|u| − k)+ ) + a|u|q+m−1 η((|u| − k)+ ) dx



σ (ω)

Z + Z

60 ∩A(k)

r

|u|q+m−1 η((|u| − k)+ )ds + m−1

γ (ω)

Z

|u|q+m−1 η((|u| − k)+ )ds

r m−1

∂ G∩A(k)

aµ|u|q−1 |∇ u|m η((|u| − k)+ ) + a0 (x)η0 ((|u| − k)+ ) + b0 (x)η((|u| − k)+ ) dx



≤ A(k)

Z + 60 ∩A(k)



h(x, u)sign u · η((|u| − k)+ )ds +

Z ∂ G∩A(k)

g (x, u)sign u · η((|u| − k)+ )ds.

(10.5)

Because of g (x, u) − g (x, 0) =

1

Z 0

d dτ

g (x, τ u)dτ = u ·

1

Z 0

∂ g (x, τ u) dτ ∂(τ u)

and the assumption (4),

Z ∂ G∩A(k)

g (x, u) · sign u · η((|u| − k)+ )ds =

Z + ∂ G∩A(k)

Z ∂ G∩A(k)

g (x, 0) · sign u · η((|u| − k)+ )ds ≤

|u(x)|

1

Z 0

Z ∂ G∩A(k)

 ∂ g (x, τ u) dτ η((|u| − k)+ )ds ∂(τ u)

|g (x, 0)| · η((|u| − k)+ )ds.

In the same way

Z 60 ∩A(k)

h(x, u) · sign u · η((|u| − k)+ )ds ≤

Z 60 ∩A(k)

|h(x, 0)| · η((|u| − k)+ )ds.

Therefore from (10.5), by assumption (5), it follows

Z

 A(k)

a|u|q |∇ u|m η0 ((|u| − k)+ ) + a|u|q+m−1 η((|u| − k)+ ) dx



Z

1 q m 0 aµ k − 0 |u| |∇ u| η((|u| − k)+ ) + a0 (x)η ((|u| − k)+ ) + b0 (x)η((|u| − k)+ ) dx



≤ A(k)

Z + 60 ∩A(k)



|h(x, 0)|η((|u| − k)+ )ds +

Now we define the function wk (x) := η

Z ∂ G∩A(k)



(|u|−k)+

Z ∂ G∩A(k)



m

|g (x, 0)|η((|u| − k)+ )ds ≤ M ·

|g (x, 0)|η((|u| − k)+ )ds.

(10.6)

. By (10.3) from Lemma 10.2, we have

Z

|g (x, 0)||wk | ds + e m

∂ G∩A(k+d)

~d

Z · ∂ G∩{A(k)\A(k+d)}

|g (x, 0)|ds.

(10.7)

In virtue of the Hölder inequality

Z ∂ G∩A(k+d)

|g (x, 0)| · |wk |m ds ≤ k|wk |m kLj (∂ G∩A(k)) · kg (x, 0)kL

j j−1

(∂ G)

= kwk km Lmj (∂ G∩A(k)) · kg (x, 0)kL

and the Sobolev boundary trace imbedding theorem (9.7) we get

Z ∂ G∩A(k+d)

|g (x, 0)| · |wk |m ds ≤ C kg (x, 0)kL

Z j j−1

(∂ G)

· A(k)


|∇wk |m + |wk |m dx,

1
n−1 n−m

.

j j−1

(∂ G) ,

∀j > 1

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5065

In the same way

Z

Z

|h(x, 0)| · |wk |m ds ≤ C kh(x, 0)kL

60 ∩A(k+d)

(60 ) ·

j


|∇wk |m + |wk |m dx,

A(k)

j−1

n−1

1
n−m

.

Therefore from (10.6) and (10.7) it follows that

Z

1 q+m−1 a|u|q |∇ u|m η0 ((|u| − k)+ ) − µk− η((|u| − k)+ ) dx 0 η((|u| − k)+ ) + a|u|

 A(k)



Z



≤ A(k)





a0 (x)η0 ((|u| − k)+ ) + b0 (x)η((|u| − k)+ ) dx



 + CM kh(x, 0)kL

(60 )

j

+ kg (x, 0)kL

j−1

j

 Z ·

60 ∩{A(k)\A(k+d)}

|h(x, 0)|ds +


|∇wk |m + |wk |m dx

A(k)

j−1

Z

+ e ~d

(∂ G)

Z ∂ G∩{A(k)\A(k+d)}

 |g (x, 0)|ds ,

1
n−1 n−m

.

(10.8)

By the definition of η(x) (see Lemma 10.2) and wk (x), e ~(|u|−k)+ |∇ u|m =

 m m

|∇wk |m ,

~

Therefore, by the choice of ~ > m + q

k0

 m m Z ~

2µ k0

a|∇wk |m dx + k0

 1 + Ma− c kh(x, 0)kL 2 ∗

j

Z A(k)

(60 )

(10.9)

according to Lemma 10.2, using (10.1)–(10.3), from (10.8), we obtain

q+m−1

A(k)

~ > 0.

+ kg (x, 0)kL

j−1

+ c3 e ~ d

Z {A(k)\A(k+d)}

Z

a|wk |m dx ≤ c1 M (∂ G)

j

 Z ·

B0 (x)dx +

60 ∩{A(k)\A(k+d)}

a|∇wk |m + a|wk |m dx




A(k)

j−1

Z

B0 (x)|wk |m dx

A(k+d)

|h(x, 0)|ds +

Z ∂ G∩{A(k)\A(k+d)}

 |g (x, 0)|ds ,

(10.10)

−1 and where 1 < j < nn− m

B0 (x) = a0 (x) + b0 (x).

(10.11)

By assumptions (1), (3b) we get that B0 (x) ∈ Ls (G), where s > where

1 s

+

1 s0

> 1. By the Hölder inequality with exponents s and s0 ,

= 1:

Z

Z

B0 (x)|wk | dx ≤ kB0 (x)kLs (G) · m

A(k+d)

From the inequality

Z

ms0

A(k)

n m

|wk |

<

1 s

m n

 10 s

dx

Z

s

θ Z |wk | dx ·

≤ A(k)

A(k)

1 ms0

=

B0 (x)|wk | dx ≤ m

1

θ −1 1 θ kB0 (x)k θ a− ∗ θε Ls (G)

m

|wk | dx θ m

+

1 with exponents θ1 and (1−θ) from (10.12) we obtain

A(k+d)

(10.12)

m  (1−θ) #

m#

m

with θ ∈ (0, 1), which is defined by the equality

Z

.

it follows that ms0 < m# = nmn and then the interpolation inequality gives −m

 10 dx

A(k)

|wk |

ms0

Z

1−θ m#

=⇒ θ = 1 −

a|wk | dx + (1 − θ )ε · m

A(k)

n . ms

Therefore by using the Young inequality

Z

m#

A(k)

|wk | dx



m m#

,

∀ε > 0.

(10.13)

Now, it follows from (10.10), (10.13) that



q

k0

 m m ~ Z

≤ c6

− c4

A(k)

Z



A(k)

B0 (x)dx +

+ c1 M (1 − θ )ε ·

q+m−1

a|∇wk |m dx + k0

Z 60 ∩A(k)

Z A(k)

|h(x, 0)|ds + m#

|wk | dx



m m#

,

− c4 − c5 ε

θ −1 θ

Z A(k)

a|wk |m dx



Z ∂ G∩A(k)

|g (x, 0)|ds

∀ε > 0, ∀k ≥ k0 ,

(10.14)

5066

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

where 1 c4 = Ma− ∗ c2



kh(x, 0)kL

j j−1

(60 ) + kg (x, 0)kL

1

1 θ Ma− ∗ c1 θ kB0 (x)kLs (G) ,

c5 =

~d

c6 = c3 e

 j j−1

(∂ G)

, (10.15)

.

We use yet the Sobolev imbedding theorem (9.6); as a result we obtain from (10.14) the inequality



q

k0

 m m

− c4 − c7 ε

~ Z

≤ c6

A(k)

Z

B0 (x)dx +



A(k)

q+m−1

a|∇wk |m dx + k0

Z

|h(x, 0)|ds +

60 ∩A(k)

− c4 − c5 ε

θ −1 θ

− c7 ε



Z ∂ G∩A(k)

|g (x, 0)|ds ,

Z A(k)

a|wk |m dx

∀ε > 0, ∀k ≥ k0 ,

(10.16)

1 where c7 = e c m a− ∗ c1 M (1 − θ ). Let us choose

c5 ε

θ −1 θ

(

q

= c7 ε ⇒ ε =



c5

θ (10.17)

c7

and

 m m

k0

~ q+m−1

≥ 2(c4 + c7 ε) = 2(c4 + c71−θ c5θ );

≥ 2(c4 + 2c7 ε) = 2c4 + 4c71−θ c5θ    ~  mq
1
1 1 1−θ θ q 1−θ θ q+m−1 q · c4 + c7 c5 ; 2c4 + 4c7 c5 . =⇒ k0 ≥ max 1; 2 k0

(10.18)

m

Thus, from (10.14)–(10.18) we obtain the inequality

Z

a|∇wk |m + a|wk |m dx ≤




A(k)

Z

c6 c4 + c71−θ c5θ

A(k)

B0 (x)dx +

Z 60 ∩A(k)

|h(x, 0)|ds +

Z ∂ G∩A(k)

|g (x, 0)|ds



for all k ≥ k0 . By virtue of the Sobolev imbedding theorem (9.6)–(9.7), hence we obtain

Z

m#

A(k)



|wk | dx

Z 60 ∩A(k)

c4 + c71−θ c5θ

A(k)

 m∗

|wk | ds

B0 (x)dx +

Z

j

j∗

+

Z

e cc6

≤

m m#

j∗

+ ∂ G∩A(k)

Z 60 ∩A(k)

|h(x, 0)|ds +

 m∗

Z

j

|wk | ds

≤e c

∂ G∩A(k)




A(k)



Z

a|∇wk |m + a|wk |m dx

|g (x, 0)|ds ,

j∗ =

m(n − 1) n−m

, ∀k ≥ k0 .

(10.19)

At last, by the Hölder inequality, we get

Z

1

A(k)

B0 (x)dx ≤ kB0 (x)kLs (G) meas 1− s A(k);

Z 60 ∩A(k)

Z ∂ G∩A(k)

|h(x, 0)|ds ≤ kh(x, 0)kL

j

(60 )

 1 · meas(60 ∩ A(k)) j ;

(∂ G)

 1 · meas(∂ G ∩ A(k)) j ,

j−1

|g (x, 0)|ds ≤ kg (x, 0)kL

j

n−m n−1

j−1

<

1 j

< 1.

(10.20)

Therefore from (10.19) it follows that

Z

m#

A(k)

≤



|wk | dx e cc6

c4 +

m m#

j∗

+ 60 ∩A(k)

n

c71−θ c5θ

+ kg (x, 0)kL

Z

 m∗

|wk | ds 1− 1s

kB0 (x)kLs (G) meas

Z

j

j∗

+ ∂ G∩A(k)

A(k) + kh(x, 0)kL

 m∗ j

|wk | ds j

(60 )

 1 · meas(60 ∩ A(k)) j

j−1

j j−1

(∂ G)

· meas(∂ G ∩ A(k)) 

 1j o

,

n−m n−1

<

1 j

< 1 , j∗ =

m(n − 1) n−m

, ∀k ≥ k0 .

(10.21)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5067

Let now l > k > k0 . By (10.4) and the definition of the function wk (x) we have |wk | ≥

Z

#

A(l)

|wk |m dx ≥

Z





Z

|wk | ds ≥ ∗

∂ G∩A(l)

|wk |j ds ≥

j∗

l−k m



l−k

(|u| − k)+ and therefore

· measA(l);

m

j∗

60 ∩A(l)

m#

l−k

1 m

 j∗

m

· meas (60 ∩ A(l)); · meas (∂ G ∩ A(l)).

From (10.21) it now follows that



 ≤

m l−k

m# Z · A(k)

 m∗#

  m# + meas(∂ G ∩ A(l)) j∗  m∗# Z Z j j∗ m# |wk | ds + |wk | dx +

meas A(l) + meas(60 ∩ A(l))

j

60 ∩A(k)

j∗

∂ G∩A(k)

 m∗#  j

|wk | ds

m#   mm# m ≤ c8 · kB0 (x)kLs (G) + kh(x, 0)kL j (60 ) + kg (x, 0)kL j (∂ G) l−k j−1 j−1     m# 1 m# 1   m# · 1   · 1− s × meas m A(k) + meas(60 ∩ A(k)) m j + meas(∂ G ∩ A(k)) m j , 

n−m

<

n−1

1 j

m(n − 1)

< 1, j∗ =

n−m

, ∀l > k ≥ k0 .

(10.22)

Now we set

  m#   m# ψ(k) = meas A(k) + meas(60 ∩ A(k)) j∗ + meas(∂ G ∩ A(k)) j∗ . Then from (10.22) it follows that

ψ(l) ≤ c9 m# =



m l−k

mn n−m

m#     m# j∗ 1 1− 1s · [ψ(k)] m + [ψ(k)] m · j ,

,

n

s>

m

n−m

> 1,

n−1

<

1 j

< 1,

∀l > k ≥ k0 ; m(n − 1)

j∗ =

n−m

(10.23)

.

(10.24)

In virtue of (10.24), we note that



γ = min

m# m

 1−

1



s

,

j∗ m

·

1 j



> 1.

From (10.23)–(10.24) we then get

ψ(l) ≤

c19

(l − k)m#

ψ γ (k),

γ > 1; ∀l > k ≥ k0

and therefore we have, because of the Stampacchia Lemma, that ψ(k0 + δ) = 0 with δ depending only on quantities in the formulation of Theorem 10.1. This fact means that |u(x)| < k0 + δ for almost all x ∈ G. Theorem 10.1 is proved.  11. Local estimate at the boundary We derive here a result asserting the local boundedness (near the conical point) of the weak solution of problem (QL). Theorem 11.1. Let u(x) be a weak solution of the problem (QL). Let assumptions (1), (2), (3a), (4), (5), (6), (8.6) be satisfied and, in addition, h(x, 0) ∈ L∞ (60 ), g (x, 0) ∈ L∞ (∂ G). Then the inequality sup |u(x)| ≤

~% x∈G0



C

(1 − ~) +%

nς/t

ς

mς(p−n)

%−nς/t kuk t ,G% + % p(m−1+ς) · ka0 (x)k mp −, 1G+ς % +%

  ς 1− np mm −1

ς

0

ς

m



ς

m−1 kb0 (x)k mp −, 1G% + %ς kg (x, 0)k∞, % Γ m

0

0

  ς 1− np

0

ς

kα(x)k m−p 1 , G% m−1

 ς m−1 + kh(x, 0)k∞, 6 % , 0

0

p>n>m

(11.1)

5068

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

m−1 holds for any t > 0, ~ ∈ (0, 1), % ∈ (0, d) and ς = q+ , where m−1



C = C n, m, p, t , q, a∗ , a∗ , m∗ , m∗ , d, ka0 (x)k p , G , kα(x)k m

p m−1

 p , k b ( x )k 0 ,G ,G . m

Proof. We apply the Moser iteration method. At first, we introduce the change of function (8.2). By virtue of the assumption (6), the identity (II ) takes the form (see Section 8):

Z D

E

Ai (x, vx )ηxi + B (x, v, vx )η dx +

G

Z = ∂G

G(x, v)η(x)ds +

Z 60

γ (ω)

Z

r

∂G

v|v|m−2 η(x)ds + m−1

σ (ω)

Z 60

r m−1

v|v|m−2 η(x)ds

H (x, v)η(x)ds

(11.2)

with coefficients that determine by (8.4). Now first we assume that t ≥ m. We consider the integral identity (11.2) and make the coordinate transformation x = %x0 . Let G0 be the image of G, ∂ G0 be the image of ∂ G, 600 be the image of 60 , and z (x0 ) = v(%x0 ). We have dx = %n dx0 , ds = %n−1 ds0 . Then (11.2) means

Z n G0

Ai (%x , % 0

+

1

%m−1

−1

Z ∂ G0

zx0

)η i + %B (%x , z , % 0

x0

−1

zx0

γ (ω) z |z |m−2 η(x0 )ds0 = |x0 |m−1

σ (ω) z |z |m−2 η(x0 )ds0 | x0 |m−1 0 Z G(%x0 , z )η(x0 )ds0 + H (%x0 , z )η(x0 )ds0

o )η(x ) dx0 + 0

Z ∂ G0

1

%m−1

Z

60

600

((II )0 )

for all η(x0 ) ∈ C0 (G0 ) ∩ W1,m (G0 ). Now we define the quantity k by

  m−m1+ς   1 % % · ka0 (%x0 )k mp −,G11+ς + ς ς m 0    m−1 1  1 1 1 m−1 m−1 0 m−1 0 0 0 p × kα(%x )k p ,G1 + %kb0 (%x )k ,G1 + kG(%x , 0)k∞,Γ 1 + kH (%x , 0)k∞,6 1

k = k(%) =

m−1

m

0

0

0

(11.3)

0

and we set z (x0 ) = |z (x0 )| + k.

(11.4)

As the test function in the integral identity ((II )0 ) we choose

η(x0 ) =

 m−1 % · z (x0 )z t −m (x0 )ζ m (|x0 |), ς

where ζ (|x0 |) ∈ C∞ 0 ([0, 1]) is non-negative function to be further specified. By the chain and product rules, η is a valid test function in ((II )0 ) and also

ηx0i =

 m−1   m−1 % z % · 1 + (t − m) z t −m zx0i ζ m (|x0 |) + m · z (x0 )z t −m (x0 )ζ m−1 ζx0i , ς z ς

so that by substitution into ((II )0 ) with regard to that 0 ≤ z ≤ z and t ≥ m, in virtue of assumptions (1)0 , (2)0 , (3a)0 , (4)0 , (5) we obtain

Z az G10

t −m

|∇ 0 z |m ζ m (|x0 |)dx0 ≤

+ aµς z

t −m

Z  amz G10

t −m+1

|∇ 0 z |m−1 ζ m−1 (|x0 |)|∇ 0 ζ |

 m % |∇ z | ζ + (t − m + 1) · a0 (%x0 )z t −m+1−ς (x0 )ζ m (|x0 |) ς 0

m

m

 m−1  % %m t −m+1 m−1 0 0 0 0 t −m+1 m 0 +m · |α(%x )| · |∇ ζ |z ζ (|x |) + m−1 b0 (%x )z ζ (|x |) dx0 ς ς  m−1 Z  m−1 Z % % t −m 0 t −m 0 0 0 m 0 0 + · z (x )G(%x , z )z (x )ζ (|x |)ds + · z (x0 )H (%x0 , z )z (x )ζ m (|x0 |)ds0 . 1 1 ς ς Γ0 60

(11.5)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5069

We estimate every term by the Young inequality and in virtue of z ≥ k:

   t  m−1 • amz t −m+1 |∇ 0 z |m−1 ζ m−1 |∇ 0 ζ | = am z (t −m) m |∇ 0 z |m−1 ζ m−1 | · z m |∇ 0 ζ | ≤ ε(m − 1)az t −m |∇ 0 z |m ζ m + ε 1−m az t |∇ 0 ζ |m , ∀ε > 0;  m  m % 1 % • · a0 (%x0 )z t −m+1−ς ζ m ≤ m−1+ς · a0 (%x0 )z t ζ m ; ς k ς  m−1   t % • m · |α(%x0 )| · |∇ 0 ζ |z t −m+1 ζ m−1 = m · z m |∇ 0 ζ | ς !  m−1  m −1 m % % (t −m) mm 0 m−1 × · |α(%x )|z ζ ≤ (m − 1) · |α(%x0 )| m−1 · z t −m ζ m ς ς  m m % + z t |∇ 0 ζ |m ≤ z t |∇ 0 ζ |m + (m − 1) · |α(%x0 )| m−1 · z t ζ m ; kς
b0 (%x0 ) %m %m %m t −m+1 m t b0 (%x0 )z ≤ b0 (%x0 ) · z ζ m . • ζ = m−1 z t ζ m m−1 m − 1 ς ς (kς )m−1 z For the estimating integrals over the boundaries we have: z · G(%x0 , z ) = z · G(%x0 , 0) + z ·

1

Z

d dτ

0

= z · G(%x0 , 0) + z 2 ·

∂ G(%x0 , τ z ) dτ ≤ z · |G(%x0 , 0)|, ∂(τ z )

1

Z 0

because of

Z

∂ G(%x0 ,τ z ) ∂(τ z )

G(%x0 , τ z )dτ

≤ 0. Therefore just as above

( )  m−1  m−1 Z m−1 t % % t −m m ( t − m ) 0 m 0 0 m z · G(%x , z )z ζ · z m · |G(%x , 0)| ·z ζ ds ≤ ds0 ς ς Γ01 Γ01   m  1   m Z  Z  m−1 0 m % · kG(%x , 0)k∞   t m 0 % t t −m 1+ ≤ z + |G(%x0 , 0)| m−1 z ζ m ds0 ≤ z ζ ds .  ς kς Γ01 Γ01 

In the same way we have

Z

  m  1  m−1 Z   m−1 0 % % · k H (% x , 0 )k ∞ t −m 0 t  z · H (%x0 , z )z (x )ζ m (|x0 |)ds0 ≤ z ζ m ds0 . 1+  ς kς 601 601 

Now from (11.5) it follows that for all ε > 0

(1 − ς µ)

Z az G10

t −m

|∇ 0 z |m ζ m dx0 ≤

Z  ε(m − 1)az t −m |∇ 0 z |m ζ m + ε 1−m az t |∇ 0 ζ |m G10

 m t % %m t t 0 m b0 (%x0 ) · z ζ m + z |∇ ζ | + m−1+ς · a0 (%x0 )z t ζ m + %m z t ζ m + k ς (kς )m−1  m  % t m 0 mm − 1 + (m − 1) · |α(%x )| · z ζ dx0 kς   m   Z m m % t m−1 m−1 0 + 1+ · kG(%x0 , 0)k∞, + k H (% x , 0 )k z ζ m ds0 . Γ01 ∞,601 kς Γ01 ∪601 For the estimating integrals over the boundaries we apply the inequality (9.4):

Z

z ζ ds ≤ e c t

Γ01 ∪601

m

0

Z

z ζ m + |∇ 0 (z ζ m )| dx0 . t

G10

t




(11.6)

5070

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Now, by the Young inequality, we have

|∇ 0 (z t ζ m )| ≤ tz t −1 |∇ 0 z |ζ m + mz t ζ m−1 |∇ 0 ζ |  t −m   m−1  = t z m |∇ 0 ζ | · z t m ζ m + mz t ζ m−1 |∇ 0 ζ | tδ

≤

m

t −m

z

|∇ 0 z |m ζ m + t

m−1 m


1 δ 1−m z t ζ m + |∇ 0 ζ |m + (m − 1)ζ m z t ,

∀δ > 0.

Therefore from above we get for any δ > 0

Z

z ζ ds ≤ t

Γ01 ∪601

0

m

Z 

tδ

G10

m

z

t −m

|∇ z | ζ + t 0

m

m

m−1 m

δ

1 1−m

z ζ t

+ |∇ ζ | + mζ 0

m

m

m




z

t



dx0 .

(11.7) 1−ς µ

Thus, from (11.6)–(11.7) with regard to the definition (11.3) of the number k and choosing ε = 4(m−1) , δ = get 1 − ςµ

Z az

2

t −m

G10



|∇ z | ζ dx ≤ c1 1 + t 0

m

m

0

m m−1

Z

ζ + |∇ ζ | 0

m

G10

m




t

Z

0

z dx + c2 t

ma∗ (1−ς µ) , 8e ct

F (x0 )z ζ m dx0 , t

G10

we

(11.8)

where c1 = const (m, µ, ς , a∗ ,e c ), c2 = const (m) and F (x0 ) =

a0 (%x0 ) km−1+ς

 m  m m % %m % + |α(%x0 )| m−1 · + b0 (%x0 ) · . ς kς (kς )m−1

(11.9)

Let us introduce the function 1

w(x0 ) = a m

m t

t t z m =⇒ z = a−1

 m t

m

w m , |∇ 0 w|m = az t −m |∇ 0 z |m .

(11.10)

Then, in virtue of 1 − ς µ > 0, (11.8) takes the form

Z G10

Z  m |∇ 0 w|m ζ m dx0 ≤ C1 t m 1 + t m−1

G10


ζ m + |∇ 0 ζ |m wm dx0 + C2 t m+1

Z G10

F (x0 )w m ζ m dx0 .

(11.11)

The desired iteration process can now be developed from the (11.11). For this we apply the Hölder inequality for integrals

Z G10

|F (x0 )| · wm (x0 )ζ m (x0 )dx0 ≤ kF kp/m,G1 · kwζ kmmp

1 p−m ,G0

0

,

p > m,

(11.12)

then the interpolation inequality for Lp -norms

kζ wk

mp 1 p−m ,G0

≤ εkζ wk

n

mn 1 n−m ,G0

+ ε n−p kζ wkm,G1 , 0

p > n > m, ∀ε > 0,

(11.13)

and the Sobolev imbedding inequality

kζ wkmmn

1 n−m ,G0

≤ C∗

Z

 G10


|∇ 0 ζ |m + ζ m |w|m + ζ m |∇ 0 w|m dx0 ,

n > m,

(11.14)

where C ∗ depends only on n, m and the domain G. Then we get from (11.11)–(11.14)

kζ wk

1

m+1

1/m

≤ c3 t (1 + t m−1 ) · k(ζ + |∇ 0 ζ |)wkm,G1 + c4 t m kF kp/m,G1 0 0   n n − p × εkwζ k mn ,G1 + ε kζ wkm,G1 , p > n > m, ∀ε > 0. mn 1 n−m ,G0

n−m

0

0

We note, in virtue of definitions of k by (11.3) and F (x0 ) by (11.9), that kF k we obtain

kζ wk

mn 1 n−m ,G0

≤ C (1 + t )

p m+1 m · p−n

k(ζ + |∇ 0 ζ |)wkm,G1 , 0

where C depends only on m, µ, ς , a∗ ,e c , n, p , k F k

1/m p/m,G10

1/m p/m,G10

≤ c (p, m). Choosing ε =

mn > p > n > m,

+1 1 − mm t 2c4

−1

m kF kp/m ,G 1 0

(11.15)

and it is independent of t. Recalling the definition of w by (11.10)

and taking into account that t ≥ m > 1 from (11.15) we establish finally the inequality

kζ · z t /m k

mn 1 n−m ,G0

≤ Ct

p m+1 m · p−n

k(ζ + |∇ 0 ζ |) · z t /m km,G1 , 0

mn > p > n > m.

This inequality can now be iterated to yield the desired estimate.

(11.16)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5071

For ∀~ ∈ (0, 1) we define sets ~+(1−~)2−j

G0(j) ≡ G0

,

j = 0, 1, 2, . . . .

It is easy to verify that G~0 ≡ G0(∞) ⊂ · · · ⊂ G0(j+1) ⊂ G0(j) ⊂ · · · ⊂ G0(0) ≡ G10 . Now we consider the sequence of cut-off functions ζj (x0 ) ∈ C∞ (G0(j) ) such that 0 ≤ ζj (x0 ) ≤ 1 in G0(j)

|∇ 0 ζj | ≤

2j+1

ζj (x0 ) ≡ 1 in G0(j+1) ,

and

ζj (x0 ) ≡ 0 for |x0 | > ~ + 2−j (1 − ~);

for ~ + 2−j−1 (1 − ~) < |x0 | < ~ + 2−j (1 − ~).

1−~

We define as yet the number sequence

 tj = t

j

n

,

n−m

j = 0, 1, 2, . . . .

Now we rewrite the inequality (11.16) replacing ζ (|x0 |) by ζj (x0 ) and t by tj ; then taking tj th root, we obtain

 kz ktj+1 ,G0(j+1) ≤

C

m/tj

mj

· 2 tj · (tj )

1−~

(m+1)p 1 p−n · tj

kz ktj ,G0(j) .

After iteration, we find that

 kz ktj+1 ,G0(j+1) ≤

t j=0 j

=

1

∞ m P

j=0

1 tj

m

·2

1−~

Notice that the series ∞ X 1

C

P∞

j j = 0 tj

∞  X

j=0

j tj

 ·

 (mp+−1n)p

nt

∞ P j=0

n−m

1 tj

· k z k t ,G 1 . 0

is convergent by the d’Alembert ratio test, but the series

n−m

t j =0

∞ P

j

n

=

n

mt

as a geometric series. Therefore we get

kz ktj+1 ,G0(j+1) ≤

C

(1 − ~)n/t

kz kt ,G1 . 0

Consequently, letting j → ∞, we have sup z (x0 ) ≤

x0 ∈G~ 0

C

(1 − ~)n/t

k z k t ,G 1 . 0

Hence, because of the definitions of the function z (x0 ) by (11.4) and of the number k by (11.3), we obtain sup |z (x0 )| ≤

x0 ∈G~ 0

   m−m1+ς   D 1 1 % % 0 m−1+ς k z k + · k a (% x )k + · kα(%x0 )k m−p 1 ,G1 1 p 0 1 t ,G0 n / t , G (1 − ~) ς ς m 0 m−1 0    m−1 1 E 1 1 m−1 m−1 + %kb0 (%x0 )k p ,G1 + kG(%x0 , 0)k∞, + kH (%x0 , 0)k∞, . Γ1 61 C

m

0

0

0

Returning to the variables x, v , we obtain in the case t ≥ m the estimate sup |v(x)| ≤

~% x∈G0

C

n

(1 − ~)n/t

o %−n/t kvkt ,G% + K (%) , t ≥ m,

(11.17)

0

where 1 m−1+ς

m(p−n)

n

1

m

+ %(1− p ) m−1 kb0 (x)k mp −,G1% m ,G 0 m 0   1 1 1 n m−1 m−1 + %1− p kα(x)k m−p 1 ,G% + % kg (x, 0)k∞, + k h ( x , 0 )k , % % Γ ∞,6

K (%) = % p(m−1+ς) · ka0 (x)k p

m−1

0

and the constant C depends on m, q.

%

0

0

p>n>m

(11.18)

5072

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Let now 0 < t < m. We consider (11.17) with t = m: sup |v(x)| ≤

~% x∈G0

C

n

(1 − ~)n/m

o %−n/m kvkm,G% + K (%) . m t

Using the Young inequality with s =

m m−t

and s0 =

kvkm,G% = 0 [(1 − ~)%]n/m [(1 − ~)%]n/m !1−t /m ·

% G0

we can write

!1/m

Z

C

C

≤ sup |v(x)|

(11.19)

0

C

n/ m

[(1 − ~)%]

m−t

t

% G0

t /m

|v| · |v|

kvkt ,G% ≤

m−t

0

sup |v(x)| +

m

% G0

C1

[(1 − ~)%]n/t

kvkt ,G% .

(11.20)

0

Let us define the function ψ(s) = supx∈Gs |v(x)|. Then from (11.19)–(11.20) it follows that 0

ψ(~%) ≤

m−t m

ψ(%) +

C1

kvkt ,G% +

[(1 − ~)%]n/t

0

Further we apply Proposition 3.2. By this proposition, letting r = ~%, R = %, δ = 1 − 

1 n 1t − m

(1 − ~)

C

· K (%),

(1 − ~)n/m t , m

~ ∈ (0, 1).

α = nt , A = C1 kvkt ,G% , B =

C

(1−~)n/m

0



(11.21)

· K (%) and observing that

< 1 for m > t, from (11.21) we obtain the validity of estimate (11.17) in the case 0 < t < m. Returning to the variable u by (8.2), from (11.17)–(11.18) we get the estimate (11.1). The proof of Theorem 11.1 is complete.  12. Integral estimates At first we will obtain a global estimate for the Dirichlet integral. Theorem 12.1. Let u(x) be a weak solution of the problem (QL) and let us assume that M0 = maxx∈G |u(x)| is known. Let assumptions (1), (3a), (4), (5) with ν0 > 0 be satisfied. Suppose, in addition, that a0 (x) ∈ L1 (G),

b0 (x) ∈ L1 (G),

h(x, 0) ∈ L1 (60 ),

g (x, 0) ∈ L1 (∂ G).

Then the inequality

Z

qm

a|u| m−1 |∇ u|m dx +

σ (ω)

Z

r

60

G

m

|u| m−1 (q+m−1) ds + m−1

≤ c (M0 , a∗ , ν0 , q, m, µ, n, meas G) ·

Z

γ (ω)

Z

r m −1

∂G

m

|u| m−1 (q+m−1) ds

(a0 (x) + b0 (x)) dx +

Z 60

G

|h(x, 0)|ds +

Z ∂G

|g (x, 0)|ds



(12.1)

holds. Proof. At first we make the function change (8.2) and consider the integral identity (11.2) for v(x). Putting η(x) = v(x) we have

Z D

E

Ai (x, vx )vxi + B (x, v, vx )v dx +

γ (ω)

Z ∂G

G

r

|v|m ds + m−1

σ (ω)

Z 60

r

|v|m ds = m−1

Z ∂G

G(x, v)v(x)ds +

Z 60

H (x, v)v(x)ds.

With regard to assumptions (1)0 , (3a)0 , (4)0 , since ς m−1 (1 − ς µ) < 1 by (8.2), we obtain

ς

Z

Z

σ (ω)

Z

γ (ω)



(1 − ς µ) a|∇v| dx + |v| ds + |v| ds ≤ m−1 m−1 G 60 r ∂G r Z Z Z 1 + |v|1−ς a0 (x)dx + |h(x, 0)| · |v|ds + |g (x, 0)| · |v|ds. ς G 60 ∂G

m−1

m

m

m

Z

|v|b0 (x)dx G

1

ς

From M0 = supG |u(x)|, by the change (8.2), it follows that |v(x)| ≤ M0 . Therefore we get

Z

m

σ (ω)

Z

a|∇v| dx + G

60

r m−1

γ (ω)

Z

m

|v| ds +

≤ c (M0 , m, q, µ, meas G) ·

∂G

Z

r m−1

|v|m ds

(a0 (x) + b0 (x)) dx + G

Z 60

|h(x, 0)|ds +

Z ∂G

 |g (x, 0)|ds .

(12.2)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Returning to the function u(x) by (8.2) we obtain the desired estimate (12.1).

5073



Now we establish a local integral weighted estimate. Theorem 12.2. Let u(x) be a weak solution of the problem (QL) and let us assume that M0 = maxx∈G |u(x)| is known. Let ϑ(m) be the smallest positive eigenvalue of (NEVP ). Let assumptions of Theorem 12.1 and (7) be satisfied. Suppose, in addition, that there exists real number ks ≥ 0 defined by (8.5). Then there are d ∈ (0, 1) and a constant c > 0 independent of u such that for any % ∈ (0, d)

Z

qm

%

a|u| m−1 |∇ u|m dx +

σ (ω)

Z %

r

60

G0

m

|u| m−1 (q+m−1) ds + m−1

γ (ω)

Z %

Γ0

r m−1

m

|u| m−1 (q+m−1) ds ≤ c ψ m (%)

(12.3)

where ψ(%) is defined by (8.7)–(8.8). Proof. We make the change (8.2). By virtue of Theorem 12.1, we have that V (%) =

Z %

a|∇v|m dx +

σ (ω)

Z %

r

60

G0

|v|m ds + m−1

γ (ω)

Z %

r m−1

Γ0

|v|m ds < ∞,

% ∈ (0, d).

(12.4)

Therefore we can set η(x) = v(x) in the identity (8.3):

Z D %

σ (ω)

Ai (x, vx )vxi + B (x, v, vx )v(x) dx +

Z

Z

Z

E

G0

= Ω%

Ai (x, vx ) cos(r , xi ) · v(x)dΩ% +

%

r m−1

60

γ (ω)

Z

m

|v| ds +

%

G(x, v) · v(x)ds +

%

Γ0

r m−1

Γ0

Z %

60

|v|m ds

H (x, v) · v(x)ds.

(12.5)

By assumptions (1)0 , (3a)0 , (4)0 , (5), (7)0 and since µς < 1, we obtain

(1 − ς µ)ς

m−1

Z

V (%) ≤

% G0

|v|b0 (x)dx +

+ ς m−1

Z Ω%

1

Z

ς

% G0

|v|

a|∇v|m−2 · v

1−ς

a0 (x)dx + ς

∂v d Ω% + ∂r

Z %

60

m−1

A(%)

Z Ω%

a|v| · |∇v|m−1 dΩ%

|h(x, 0)| · |v|ds +

Z %

Γ0

|g (x, 0)| · |v|ds.

(12.6)

Applying Lemma 9.5 from (12.6) it follows that

(1 − ς µ)V (%) ≤

Ξ (m) 1

mϑ m (m)

+ ς 1 −m

%V (%) + A(%) 0

Z

Z a|v| · |∇v| Ω%

|v|b0 (x)dx + ς 1−m

% G0

m−1

Z %

60

d Ω% +

1

ςm

Z % G0

|h(x, 0)| · |v|ds + ς 1−m

|v|1−ς a0 (x)dx Z %

Γ0

|g (x, 0)| · |v|ds.

(12.7)

Now we estimate the integral over Ω% on the right in (12.7). At first, by the Hölder inequality for integrals

Z

m −1

a|v| · |∇v| Ω%

! m1

Z

d Ω% ≤

a|v| dΩ% m

Ω%

! mm−1

Z

a|∇v| dΩ% m

· Ω%

.

(12.8)

Next, in virtue of the inequality |∇ω v| ≤ %|∇v|, the inequality ((W )m ) and formula (9.16), we have

Z

% n −1 a|v| dΩ% ≤ ϑ Ω% m

Z

a|∇ω v| dΩ + m

Ω

Z

σ (ω)|v| dσ + m

σ0

Z

γ (ω)|v| dσ m

∂Ω

 ≤

%m 0 V (%). ϑ

(12.9)

Now from (12.8)–(12.9) it follows that

Z Ω%

a|v| · |∇v|m−1 dΩ% ≤

% 1

ϑm

V 0 (%).

(12.10)

5074

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Further, by the Young inequality and inequality (9.13),

Z

1

ς

%

Z

1

|v|1−ς a0 (x)dx =



ς G0 Z 

%

G0

≤

r

(1−ς)(1−ς −m) m

1−ς mς

% G0

1−ς

≤

m

|v| +

%1−ς V (%) +

m+ς −1 mς

m+ς −1

r

1−ς

Z

|a0 (x)|



m m+ς −1

dx

m

r 1−ς |a0 (x)| m+ς−1 dx;

mς G0    1 1 r − m |v| r m b0 (x) dx |v|b0 (x)dx = % % G0  ZG0  m 1 −1 m m−1 1 ≤ r |v| + r m−1 |b0 (x)| m−1 dx % m m G0 Z 1 m 1 m m−1 r m−1 |b0 (x)| m−1 dx ≤ % V (%) + a∗ mϑ(m) m − 1 G% Z0 1 m m 1 m−1 % V (%) + r m−1 |b0 (x)| m−1 dx; = a∗ mϑ(m) m − 1 G% 0 ! Z Z Z Z Z 1 m−1 m m |h(x, 0)| · |v|ds + |v| ds + |g (x, 0)| · |v|ds ≤ |v| ds +

Z

Z

a∗ mς ϑ(m)

r

1−ς −m

  (1−ς)(m+ς −1)  m |v|1−ς · r a0 (x) dx



%

%

60

+

Γ0

Z +

!

m

%

%

Γ0

|g (x, 0)| m−1 ds ≤ |g (x, 0)|

m m−1

%

m

Γ0

Z

%

!

%

m−1

σ (ω)

Z

mν0

%

60

r m−1

60

m

Γ0

|v|m ds +

γ (ω)

Z %

Γ0

m−1 %m−1 V (%) + ≤ mν0 m

ds

%

Z %

60

r m−1



|v|m ds +

|h(x, 0)|

m m−1

m−1 m

Z ds +

%

Γ0

(12.11) m

%

60

Z

|h(x, 0)| m−1 ds m

%

60

|g (x, 0)|

|h(x, 0)| m−1 ds

m m−1

! ds ,

(12.12)

by assumption (5) and (12.4). Thus, from (12.7)–(12.12) it follows that

Ξ (m)

Z
0 m 1 mς 1−m e (1 − ς µ) (1 − δ(%)) V (%) ≤ 1 + A(%) %V (%) + r m−1 |b0 (x)| m−1 dx 1 % m − 1 G0 mϑ m (m) ! Z Z Z m m m m+ς −1 m−1 1−ς |g (x, 0)| m−1 ds , + r |a0 (x)| m+ς −1 dx + |h(x, 0)| m−1 ds + % % % mς m mς m−1 Γ0 G0 60   q q e(%) = m A(%). We observe that where δ(%) = const (m, a∗ , µ, q, ν0 , ϑ(m)) · %m−1 + q+m−1 % q+m−1 ; A Ξ (m) Z 0

δ(%) d% < ∞, %

Z e A(%) d% < ∞. % 0

(12.13)

(12.14)

Thus, from (12.13) in virtue of assumption (8.5) we have the Cauchy problem for the differential inequality: V 0 (%) − P (%)V (%) + Q(%) ≥ 0, V (d) ≤ V0 ,



0 < % < d,

(CP)

where

P (%) =

1

1 − δ(%)

·

e(%) % 1+A

1

·

mϑ m (m)

Ξ (m)

(1 − ς µ),

Q(%) = k%ms−1 ;

k = const (ks , m, q, ϑ(m)).

(12.15)

We estimate now: V (d) =

Z

a|∇v|m dx + Gd0

Z

γ (ω)

σ (ω)

Z 60d

r

|v|m ds + m−1

γ (ω)

Z Γ0d

r

|v|m ds ≤ m −1

|v|m ds ≤ c (M0 , m, q, µ, meas G)V0 , ∂G Z Z Z V0 ≡ |h(x, 0)|ds + |g (x, 0)|ds, (a0 (x) + b0 (x)) dx + +

G

r m−1

60

∂G

Z

a|∇v|m dx + G

σ (ω)

Z 60

r m−1

|v|m ds

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5075

in virtue of (12.2). The solution of problem (CP) is the following inequality

 Z V (%) ≤ V0 exp −

d

P (τ )dτ

%



d

Z +

 Z Q(τ ) exp −

τ



P (ξ )dξ dτ .

%

%

(12.16)

(see Theorem 1.57 Section 1.10 [10]). Direct calculations give: τ

Z − %

1

P (ξ )dξ = −

mϑ m (m)(1 − ς µ)

τ

Z

Ξ (m)

1

ξ

%

1

mϑ m (m)(1 − ς µ)

τ

1 − δ(ξ )

·

e(ξ ) 1+A

dξ

  e(ξ ) δ(ξ ) + A dξ · 1− e(ξ ) Ξ (m) 1+A % ξ   Z d 1 e(ξ ) δ(ξ ) + A mϑ m (m)(1 − ς µ) % ≤ ln + dξ Ξ (m) τ ξ 0 !    mϑ m1 (m)(1−ςµ)  Z τ Z 1 e(ξ ) % mϑ m (m)(1 − ς µ) d δ(ξ ) + A Ξ (m) P (ξ )dξ ≤ =⇒ exp − · exp dξ ; τ Ξ (m) ξ % 0  Z τ  1 1 Z d Z d mϑ m (m)(1−ςµ) mϑ m (m)(1−ςµ) ms−1 − Ξ ( m ) Ξ (m) Q(τ ) exp − P (ξ )dξ dτ ≤ kC1 % τ τ dτ = −

%

Z

1

%

= kC1 %

where C1 = exp

1 mϑ m (m)(1−ςµ) Ξ (m)

%

·

 m     d   

1

ϑ m (m)(1−ςµ) s− Ξ (m)



Ξ (m)

ξ

0

−%

ϑ m (m)(1−ςµ) m s− Ξ (m)

1

ϑ m (m)(1−ς µ) Ξ (m)

m s−

      ln d , %  1 R e(ξ ) d δ(ξ )+A mϑ m (m)(1−ςµ)

1

!

! 1

,



s 6=

ϑ m (m)(1 − ς µ) ; Ξ (m)

s=

ϑ m (m)(1 − ς µ) , Ξ (m)

1



dξ . 1

 Z V0 · exp −

d

%

P (ξ )dξ

 ≤ V0 C1

 %  mϑ m (m)(1−ςµ) Ξ (m)

d

 1 mϑ m (m)(1−ςµ)  Ξ (m)  , %      mϑ m1 (m)(1−ςµ) d Ξ (m) =⇒ V (%) ≤ cC1 (V0 + k) · % ln ,  %      %ms ,

1

ϑ m (m)(1 − ς µ) ; Ξ (m) 1 ϑ m (m)(1 − ς µ) s= ; Ξ (m) 1 ϑ m (m)(1 − ς µ) , s< Ξ (m) s>

where c = const (m, s, ϑ(m)). Thus we proved the statement of our Theorem.

(12.17)



13. The power modulus of continuity at the conical point for weak solutions Proof of Theorem 1.2. We consider the function ψ(%), 0 < % < d that is determined by (8.7). By Theorem 11.1 about the local bound of the weak solution modulus we have (see (11.18)–(11.19))

n

o

m sup |v(x)|m ≤ C %−n kvkm % + K (%) . m,G

%/2 x∈G0

(13.1)

0

Hence, in virtue of inequality ((H − W )m ) with regard to the notation (9.14), we get

%

−n

Z % G0

a|v(x)|m dx ≤

%m−n V (%) ≤ C %m−n ψ m (%), ϑ(m)

(13.2)

by inequality (12.17). From (13.1)–(13.2) it follows

n

n

o

sup |v(x)| ≤ C %1− m ψ(%) + K (%) .

%/2 x∈G0

(13.3)

5076

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Now from (11.18), in virtue of assumption (8.5), it follows n

K (%) ≤ K %1− m ψ(%). Therefore hence and from (13.3) we get n

|v(x)| ≤ C0 %1− m ψ(%), Putting |x| =

% 3

%/2

x ∈ G%/4 .

we obtain n

|v(x)| ≤ C0 |x|1− m ψ(|x|),

x ∈ Gd0 .

(13.4)

Finally, because of (8.2), from (13.4) we establish the first desired estimate (8.9). Repeating verbatim second part of Theorem 14.2 proof we obtain the inequality n

|∇v(x)| ≤ C1 |x|− m ψ(|x|),

x ∈ Gd0 .

(13.5)

But since, by (8.2),

|∇ u(x)| ≤ |v|ς−1 · |∇v(x)| from (13.4)–(13.5) we establish the second desired estimate (8.10).



Acknowledgment This article was supported by the Polish Ministry of Science and Higher Education through the grant Nr N201 381834. Appendix If a neighborhood of the conical point is a convex conical domain then we can construct the barrier function and apply the comparison principle for the estimating of weak solutions of the transmission problem. We will assume that

ν0 >

a∗ 2 m cos

ω0
m−1

max 1; 2m−2




(14.1)

2

as well that the functions ai (x, u, ξ ), b(x, u, ξ ) are continuously differentiable with respect to the x, u, ξ variables in Md,M0 = Gd0 × [−M0 , M0 ] × Rn and satisfy in Md,M0

∂ ai ( x , u, ξ ) pi pj ≥ a|u|q |ξ |m−2 p2 , ∀p ∈ Rn \ {0}; ∂ξj ∂ ai (x, u, ξ ) (9) − b(x, u, ξ ) ≤ B (r )|u|q−1 |ξ |m + b1 r β , ∂ xi v u n uX ∂ b(x, u, ξ ) 2 t ≤ aµ|u|q−1 |ξ |m−1 + b0 (x); (10) ∂ξi i=1 (8)

β > −1;

∂ b(x, u, ξ ) ≥ a|u|q−2 |ξ |m ; ∂u (12) α(x) + b0 (x) + |h(x, 0)| + |g (x, 0)| ≤ k1 |x|s−1 , (11)

where B (r ) is a non-negative monotonically increasing function that is continuous at zero with B (0) = 0. After the function change (8.2) these assumptions take the form:

∂ Ai (x, vx ) pi pj ≥ aς m−1 |∇v|m−2 p2 , ∀p ∈ Rn \ {0}; ∂vxj 0 ∂ Ai (x, η) (9) − B (x, v, η) ≤ B (r )v −1 |η|m + b1 r β ; ∂ xi v u n uX ∂ B (x, v, η) 2 ≤ aµς m−1 |v|−1 · |η|m−1 + b0 (x); (10)0 t ∂η i i=1 (8)0

(11)0

∂ B (x, v, η) ≥ aς m−1 |v|−2 |η|m . ∂v

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5077

A.1. The comparison principle We consider the second order quasi-linear degenerate operator Q of the form Q (v, η) ≡

Z D Gd0

E

σ (ω)

Z + 60d

r

v|v|m−2 η(x)ds − m−1

Z Ωd

γ (ω)

Z

v|v|m−2 η(x)ds − m−1

Z

Ai (x, vx ) cos(r , xi )η(x)dΩd −

Z

Ai (x, vx )ηxi + B (x, v, vx )η dx +

Γ0d

r

60d

Γ0d

H (x, v)η(x)ds G(x, v)η(x)ds

(14.2)

for v(x) ∈ C0 (G) ∩ W1,m (G) and for all non-negative η belonging to C0 (G) ∩ W1,m (G) under the following assumptions: the functions Ai (x, ξ ), B (x, v, ξ ), G(x, v), H (x, v) are Caratheodory, continuously differentiable with respect to the v, ξ variables in M = Ω × R × RN and satisfy in M the following inequalities:

∂ Ai (x, ξ ) pi pj ≥ aγm |ξ |m−2 p2 , ∀p ∈ Rn \ {0}; ∂ξj v u N uX ∂ B (x, v, ξ ) 2 ≤ a|v|−1 |ξ |m−1 ; ∂ B (x, v, ξ ) ≥ a|v|−2 |ξ |m ; (ii) t ∂ξi ∂v i=1 (i)

∂ H (x, v) ∂ G(x, v) ≤ 0, ≤ 0, ∂v ∂v Here: m > 1, γm > 0 and a > 0.

γ (ω) ≥ 0,

(iii)

σ (ω) ≥ 0.

Proposition 14.1. Let operator Q satisfy assumptions (i)–(iii) and d ≪ 1. Let the functions v, w ∈ C0 (Gd0 ) ∩ W1,m (Gd0 ) satisfy the inequality Q (v, η) ≤ Q (w, η)

(14.3)

for all non-negative η ∈ C0 (Gd0 ) ∩ W1,m (Gd0 ) and also the inequality

v(x) ≤ w(x) on Ωd holds. Then v(x) ≤ w(x) in

(14.4) Gd0 .

Proof. Let us define z =v−w

and

v τ = τ v + (1 − τ )w,

τ ∈ [0, 1].

Then we have

 Z 1 Z 1 ∂ Ai (x, vxτ ) ∂ B (x, v τ , vxτ ) ∂ B (x, v τ , vxτ ) d τ + η z d τ + η z d τ dx xi ∂vxτj ∂vxτi ∂v τ 0 Gd0 0 0 Z 1  Z 1  Z Z γ (ω) ∂(v τ |v τ |m−2 ) σ (ω) ∂(v τ |v τ |m−2 ) + dτ z (x)η(x)ds + dτ z (x)η(x)ds m−1 m−1 ∂v τ ∂v τ Γ0d r 0 60d r 0 !  Z Z 1 Z Z 1 ∂ Ai (x, vxτ ) ∂ G(x, v τ ) − d τ cos ( r , x ) · z η( x ) d Ω − d τ z (x)η(x)ds i xj d ∂vxτj ∂v τ 0 0 Ωd Γ0d  Z Z 1 ∂ H (x, v τ ) − d τ z (x)η(x)ds (14.5) ∂v τ 60d 0

0 ≥ Q (v, η) − Q (w, η) =

Z Z  ηxi zxj

1

for all non-negative η ∈ C0 (Gd0 ) ∩ W1,m (Gd0 ). Now we define the sets

(Gd0 )+ := {x ∈ Gd0 | v(x) > w(x)} ⊂ Gd0 , (60d )+ := {x ∈ 60d | v(x) > w(x)} ⊂ 60d , (Γ0d )+ := {x ∈ Γ0d | v(x) > w(x)} ⊂ Γ0d and assume that (Gd0 )+ 6= ∅. Let k ≥ 1 be any odd number. As the test function in the integral inequality (14.5) we choose

η = max{(v − w)k , 0}.

5078

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

We calculate the integral 1

Z 0

∂(v τ |v τ |m−2 ) dτ = (m − 1) ∂v τ

1

Z

|v τ |m−2 dτ = (m − 1)

Ωd

 (Gd0 )+

kγm az

k−1

1

Z

τ m−2

|∇v |



dτ |∇ z | dx + az 2

k+1

≤

az

k

1

Z

(Gd0 )+

v m−1 − wm−1

>0

= 0, we obtain from (14.5)

1

Z

z

τ −2

τ m

 

|v | |∇v | dτ dx 0

0

Z

|τ z + w|m−2 dτ = 0

0

on (60d )+ ∪ (Γ0d )+ . Then, by assumptions (i)–(iii) and since η

Z

1

Z

 |v τ |−1 |∇v τ |m−1 dτ |∇ z |dx.

(14.6)

0

Now we use the Cauchy inequality z k |∇ z ||v τ |−1 |∇v τ |m−1 =

 ε

≤

2

|v τ |−1 z

k+1 2

   k−1 |∇v τ |m/2 · z 2 |∇ z ||∇v τ |m/2−1

|v τ |−2 z k+1 |∇v τ |m +

1 2ε

z k−1 |∇ z |2 |∇v τ |m−2 ,

∀ε > 0.

Hence, taking ε = 2, we obtain from (14.6) the inequality



Z (Gd0 )+

a kγm −

1



4

z k−1 |∇ z |2

Z

1

 |∇v τ |m−2 dτ dx ≤ 0.

(14.7)

0





Now choosing the odd number k ≥ max 1; 2γ1 , in view of z (x) ≡ 0 on ∂(Gd0 )+ , we get from (14.7) z (x) ≡ 0 in (Gd0 )+ . We m have finished with the contradiction to our definition of the set (Gd0 )+ . By this fact, the proposition is proved.



A.2. The barrier function. The preliminary estimate of the solution modulus Now we can estimate |u(x)| and |∇ u(x)| for (QL) in the neighborhood of the conical point. Theorem 14.2. Let u be a weak solution of the problem (QL). Let us assume that M0 = maxx∈G |u(x)| is known. Let assumptions (1)–(12) are satisfied. Then there are ~ ∈ (0, 1), d ∈ (0, 1) and a constant C0 > 0 independent of u such that

|u(x)| ≤ C0 |x|

(m−1)(1+~) q+m−1

,

∀x ∈ Gd0 .

(14.8)

Moreover, if coefficients of the problem (QL) satisfy such conditions which guarantee the local a priori estimate |∇ u| ≤ M1 for any smooth G0 ⊂⊂ G \ {O } (see for example Section 4 [1] or [2]), then there is a constant C1 > 0 independent of u such that 0,G0

|∇ u(x)| ≤ C1 |x|

(m−1)(1+~) −1 q+m−1

,

∀x ∈ Gd0 .

(14.9)

Proof. At first we make the function change (8.2) and will consider the function v(x). For the proof we construct the barrier function w(x) and apply the comparison principle to v(x) and w(x). We will show that Q (Aw, η) ≥ 0 for all non-negative

η ∈ C0 (Gd0 ) ∩ W1,m (Gd0 ) and some A > 0. From the definition (14.2), integrating by parts in the first integral, we have: Z D Z D E dAi (x, A∇w) − → Q (Aw, η) ≡ − + B (x, Aw, A∇w) η(x)dx + Ai (x, A∇w) cos( n , xi ) dxi

Gd0

Γ0d

Z D E γ (ω) → [Ai (x, A∇w) cos(− n , xi )] + Am−1 m−1 w|w|m−2 − G(x, Aw) η(x)ds + r

+ Am−1

60d

σ (ω) r

E m−2 w|w| − H ( x , A w) η(x)ds. m−1

(14.10)

We recall that functions Ai , B , G, H were determined by (8.4) with ς from (8.2). Let (x, y, x0 ) ∈ Rn , where x = x1 , y = x2 , x0 = (x3 , . . . , xn ). In {x1 ≥ 0} we consider the cone K with the vertex in O such that K ⊃ Gd0 (we recall that Gd0 ⊂ {x1 ≥ 0}). Let ∂ K be the lateral surface of K and let the equation of ∂ K ∩ yO x = Γ± be ω x = ±hy, where h = cot 20 , 0 < ω0 < π such that in the interior of K holds the inequality x > h|y|. We shall consider the barrier function:

w(x; y, x0 ) ≡ x~−1 (x2 − h2 y2 ) + Bx~+1 with some ~ ∈ (0; 1), B ≥ 1.

(14.11)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5079

Step 1. At first, we show that in Gd0

−

dAi (x, A∇w) dxi

+ B (x, Aw, A∇w) ≥ 0

(14.12)

for some A > 0, B > 1, 0 < d  1. By the direct calculation we obtain:

wx = (1 + ~)(1 + B)x~ − (~ − 1)h2 y2 x~−2 , wy = −2h2 yx~−1 , wxx = ~(1 + ~)(1 + B)x~−1 − (1 − ~)(2 − ~)h2 y2 x~−3 ,

(14.13)

wxy = 2h2 (1 − ~)yx~−2 , wyy = −2h2 x~−1 and therefore we get for (x, y) ∈ Gd0

 B

√

~+1

h

1 + h2

 √

~

h

1 + h2

· r ~+1 ≤ w(x, y) ≤ (1 + B)r ~+1 ;

· r ~ ≤ |∇w| ≤ 2(1 + h + B)r ~ ,

(14.14)

(14.15)

as well as

  ∂ Ai (x, ∇w) ∂ A1 ∂ A2 2 2 ∂ A2 ~−1 − wxi xj = 2h x − 2h (1 − ~) + yx~−2 ∂wxj ∂wy ∂wy ∂wx

 ∂ A1 + (1 − ~)(2 − ~)h2 y2 x~−3 − ~(1 + ~)(1 + B)x~−1 ∂wx ≥ x~−1 · φ(~),

(14.16)

where

  ∂ A1 ∂ A1 ∂ A2 ∂ A2 φ(~) = −B~ − (B + 4)~ + 2 − 2h(1 − ~) + + 2h2 . ∂wx ∂wy ∂wx ∂wy 



2

Because of the ellipticity condition (8)0 ,

φ(0) = 2

  ∂ A1 ∂ A2 ∂ A2 ∂ A1 − 2h + + 2h2 ≥ 2aς m−1 |∇w|m−2 (1 + h2 ). ∂wx ∂wy ∂wx ∂wy

(14.17)

Since φ(~) is the square functions with the negative leading coefficient and with regard to (14.17), there exists the number ~0 > 0 such that φ(~) ≥ aς m−1 |∇w|m−2 (1 + h2 ) for ~ ∈ [0; ~0 ]. Therefore from (14.16) it follows that

−

∂ Ai (x, A∇w) Awxi xj ≥ a∗ (Aς )m−1 |∇w|m−2 r ~−1 , ∂(Awxj )

(14.18)

in virtue of ~ < 1. Hence, by the assumption (9)0 , we obtain in Gd0

−

dAi (x, A∇w)

+ B (x, Aw, A∇w) ≥ a∗ (Aς )m−1 |∇w|m−2 r ~−1 − Am−1 w −1 |∇w|m B (r ) − b1 r β

 = Am−1 |∇w|m−2 a∗ ς m−1 r ~−1 − B (r )w −1 |∇w|2 − b1 r β * + 4(1 + h + B)2 m−1 m−2 ~−1 m−1 β ≥ A |∇w| r a∗ ς − B (d)  ~+1 − b1 r . B √h dxi

1+h2

Because of the continuity at zero of the function B (r ) and B (0) = 0, we can choose d > 0 so small that

 B (d) ≤

1 2

B a∗ ς m−1

√h

~+1

1+h2

4(1 + h + B)2

.

(14.19)

5080

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

Further, from (14.15) we obtain

|∇w|m−2 ≥ m1 r ~(m−2) ,  ~(m−2) h  , √ m1 = 1 + h2  h2(1 + h + B)im−2 ,

if m ≥ 2;

(14.20)

if 1 < m < 2

and therefore, in virtue of (14.19), we get

−

dAi (x, A∇w) dxi

+ B (x, Aw, A∇w) ≥

1 2

a∗ (Aς )m−1 m1 r ~(m−1)−1 − b1 r β .

Hence it follows the desired inequality (14.12), if we choose

~≤

1+β

and A ≥

m−1



1

ς

 m−1 1

2b1 a∗ m1

.

(14.21)

Step 2. Now, we go over to the estimating on Γ0d and we shall show that

− →

Ai (x, A∇w) cos( n , xi ) + Am−1

γ (ω) r m−1

w|w|m−2 − G(x, Aw) d ≥ 0.

(14.22)

Γ0

By (14.11) and because of x = ±hy on Γ± , we have

 ~ p  ~ h h ~ ~ 2 2 |∇w| = r {(1 + ~)B + 2} + 4h ≤ 2(B + 1 + h)r , √ √ Γ± 1 + h2 1 + h2 √ since ~ ≤ 1 and a2 + b2 ≤ |a| + |b|. Next, in virtue of assumptions (4)0 and (8.4), Z 1 ∂ G(x, τ w) G(x, w) = G(x, 0) + w · dτ ≤ |g (x, 0)|. ∂(τ w) 0

(14.23)

Further, since ς ≤ 1, a ≤ a∗ , because of assumptions (5), (2)0 , (12) and (14.23), on Γ0d :

− →

Ai (x, A∇w) cos( n , xi ) + Am−1

≥ ν0 (AB)

m−1 ~(m−1)

r

s−1

−a 2

w|w|m−2 − G(x, Aw)

(~+1)(m−1)

h

√

m−1



(~+1)(m−1)

h

√

1 + h2



≥A

∗ m−1

r m−1

1 + h2

≥ ν0 (AB)m−1 r ~(m−1) − k1 r



γ (ω)

√

~(m−1) 

hr

1 + h2

max{1, 2

m−2

}(1 + h)

m−1



− a∗ Am−1 |∇w|m−1 − α(x) − |g (x, 0)| − a∗ (2A)m−1

ν0

 √

h

 √

hr

1 + h2

(m−1)

1 + h2

~(m−1)

(B + 1 + h)m−1 !

− a∗ 2m−1 max{1, 2m−2 } Bm−1

− k1 r s−1 .

(14.24)

Now, in virtue of the assumption (14.1), we obtain from (14.24)

− →

Ai (x, A∇w) cos( n , xi ) + Am−1

≥ Am−1

 √

hr

~(m−1)

1 + h2

γ (ω) r m−1

w|w|m−2 − G(x, Aw)

a∗ 2m−1 max{1, 2m−2 }(Bm−1 − (1 + h)m−1 ) − k1 r s−1 .

If we take at first 1

B ≥ 2 m−1 (1 + h),

(14.25)

then ~ > 0 may be yet smaller

~≤

s−1 m−1

(14.26)

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

5081

and at last we choose

 A≥

k1

√

 m−1 1 ·

a∗

1 + h2

!~ ·

h

1

(14.27)

m−2

2(1 + h) max{1, 2 m−1 }

we get the required (14.22). Step 3. By analogy as above we shall derive that

h

− →

Ai (x, A∇w) cos( n , xi )

  σ (ω) + Am−1 m−1 w|w|m−2 − H (x, Aw) d ≥ 0.

i 60d

60

r

(14.28)

At first, on 60 we have:

w = (1 + B)r 1+~ ; cos(E n, x) = 0, cos(E n, y) = ±1; 6 0 wx = (1 + ~)(1 + B)r ~ ; wy = 0 =⇒ |∇w| = (1 + ~)(1 + B)r ~ . 60

60

60

Next, in virtue of assumptions (4) , (12), and (8.4), 0

H (x, w) = H (x, 0) + w ·

∂ H (x, τ w) dτ ≤ |h(x, 0)|. ∂(τ w)

1

Z 0

Therefore, since ς ≤ 1, a ≤ a , because of assumptions (5), (7)0 , on 60d we obtain ∗

h

i

− →

Ai (x, A∇w) cos( n , xi ) + Am−1

≥ ν0 Am−1

1 r m−1

σ (ω) r m−1

w|w|m−2 − H (x, Aw)

w|w|m−2 − a∗ Am−1 |∇w|m−1 − α(x) − |h(x, 0)|

≥ ν0 Am−1 (1 + B)m−1 r ~(m−1) − a∗ Am−1 (1 + ~)m−1 (1 + B)m−1 r ~(m−1) − k1 r s−1
= Am−1 (1 + B)m−1 r ~(m−1) ν0 − a∗ (1 + ~)m−1 − k1 r s−1
≥ Am−1 r ~(m−1) ν0 − a∗ (1 + ~)m−1 − k1 r s−1 ≥ 0, in virtue of (14.1), if we choose ~ > 0 as in (14.26) as well as 0<~ ≤

 ν  m−1 1 0

−1

2a∗

(14.29)

and

 A≥

2k1

 m−1 1

ν0

.

(14.30)

Thus (14.28) is proved. Step 4. Let us take the barrier function w(x) defined by (14.11) and the function v(x) from (8.2) that satisfy the integral identity (8.3). For them we shall verify Proposition 14.1. From the above proved estimates we obtain Q (w, η) ≥ 0 = Q (v, η)

(14.31)

for all non-negative η ∈ C0 (Gd0 ) ∩ W1,m (Gd0 ).

Now we compare v(x) and w(x) on Ωd . Since x2 ≥ h2 y2 in K from (14.11) we have

Aw(x)

≥ Ad



1+~

√

r =d

~+1

h 1 + h2

.

On the other hand

v(x)

Ωd

1 ≤ |u(x) | ς

q+m−1

Ωd

≤ M0 m−1

and therefore hence it follows:



Aw(x)

Ωd

≥ Ad~+1

 √

h

1 + h2

~+1

q+m−1 ≥ M0 m−1 ≥ v , Ωd

5082

M. Borsuk / Nonlinear Analysis 71 (2009) 5032–5083

if we choose q+m−1

M0 m−1 · A≥

√

1+h2 h

~+1 .

d~+1

(14.32)

Thus, if we choose at first a large B ≥ 1 according to (14.25), then a small d > 0 according to (14.19) and a value m1 > 0 according to (14.20), ~ ∈ (0, ~0 ] according to (14.21), (14.26), (14.29) and finally a large A > 0 according to (14.21), (14.27), (14.30) and (14.32), we provide the validity of Proposition 14.1. Therefore, by the comparison principle (Proposition 14.1), we have:

v(x) ≤ Aw(x),

x ∈ Gd0 .

(14.33)

Similarly, we derive the estimate

v(x) ≥ −Aw(x), if we replace v(x) with −v(x). Returning to the function u(x) by (8.2) thus we establish the first desired estimate (14.8). Now we will estimate the gradient modulus of the problem (QL) solution near a conical point. Let us consider problem % (QL) for the function v(x) after the change (8.2) in the set G%/2 ⊂ G, 0 < ρ < d. We make the transformation x = %x0 ; z (x0 ) = %−~−1 v(%x0 ). The function z (x0 ) satisfies the problem

 dAi (%x0 , %~ zx0 )  + %B (%x0 , %~+1 z , %~ zx0 ) = 0, x0 ∈ G11/2 ,  − dx0i      0   z (x ) 6 1 = 0, 1/2   σ (ω) − → 0  Ai (%x , %~ zx0 ) cos( n , x0i ) 6 1 + 0 m−1 %~(m−1) z |z |m−2 = H (%x0 , %~+1 z ), x0 ∈ 611/2 ,   1/2  |x |    γ (ω) ~(m−1) → A (%x0 , %~ z 0 ) cos(− 0 z |z |m−2 = G(%x0 , %~+1 z ), x0 ∈ Γ11/2 . n , xi ) + 0 m−1 % i x |x |

((QL)0 )

Now we apply our assumption about a priori estimate of the gradient modulus of the problem ((QL)0 ) solution as an estimate inside the domain and near a smooth boundary portion max |∇ 0 z (x0 )| ≤ M10

(14.34)

x0 ∈G11/2

(see Section 4 Theorem 4.4 [1] or Sections 3, 5 [2]). Returning to the variable x and the function v(x) we obtain from (14.34)

|∇v(x)| ≤ M10 %~ , Putting now |x| =

2 3

%

x ∈ G%/2 , 0 < % < d.

% we obtain the estimate

|∇v(x)| ≤ M10 |x|~ ,

x ∈ Gd0 .

Returning to the function u(x) by (8.2), we obtain desired estimate (14.9).



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