Boundary value problems for quasi-linear elliptic second order equations in unbounded cone-like domains

Cent. Eur. J. Math. • 10(6) • 2012 • 2051-2072DOI: 10.2478/s11533-012-0127-2

Central European Journal of Mathematics

Boundary value problems for quasi-linearelliptic second order equations in unboundedcone-like domains

Research Article

Mikhail V. Borsuk1∗, Damian Wiśniewski1†

1 Department of Mathematics and Informatics, University of Warmia and Mazury in Olsztyn, Słoneczna 54,10-957 Olsztyn-Kortowo, Poland

Received 15 February 2012; accepted 16 July 2012

Abstract: We study the behaviour of weak solutions (as well as their gradients) of boundary value problems for quasi-linearelliptic divergence equations in domains extending to infinity along a cone.

MSC: 35B05, 35B45, 35B65, 35D30, 35J66

Keywords: Elliptic divergence quasi-linear equations • Weak solutions • Unbounded domains© Versita Sp. z o.o.

1. Introduction

Let B1(O) be the unit ball in Rn, n ≥ 2, with the center at the origin O and G ⊂ Rn \B1(O) be an unbounded domain withthe smooth boundary ∂G. We assume that G = G0 ∪GR , where G0 is a bounded domain in Rn, GR = x = (r, ω) ∈ Rn :r ∈ (R,∞), ω ∈ Ω ⊂ Sn−1, n ≥ 2, R 1, Sn−1 is the unit sphere.We consider the following boundary value problem for the elliptic second order divergence quasi-linear equation:

− ddxi

ai(x, u,∇u) + b(x, u,∇u) = 0, x ∈ G,

α(x) ∂u∂ν + 1|x|m−1 γ

(x|x|

)u|u|q+m−2 = g(x, u), x ∈ ∂G,

lim|x|→∞

u(x) = 0,(QL)

∗ E-mail: [email protected]† E-mail: [email protected]

2051

Boundary value problems for quasi-linear elliptic second order equations in unbounded cone-like domains

where ai : G×R×Rn → R, b : G×R×Rn → R, m > 1, q ≥ 0,α(x) = 0 if x ∈ D,1 if x 6∈ D,

D ⊆ ∂G is the part of the boundary ∂G, where the Dirichlet boundary condition is posed, γ : (0,+∞) → (0,+∞) and∂u/∂ν = ai(x, u,∇u) cos (~n, xi), ~n denotes the unit outward with respect to G normal to ∂G. Thus, if D = ∂G then wehave the Dirichlet problem, if D = ∅ then α(x) = 1 and we have the Robin problem, if D ⊂ ∂G then we have the mixedproblem.Recently in [12] D. Wiśniewski investigated the behavior of weak solutions to the boundary value problems (Dirichlet,Neumann, Robin and mixed) for linear elliptic divergence second order equations in a neighborhood of the infinity foran unbounded cone-like domain. There was found the exact exponent of the solution decreasing rate under the minimalsmoothness assumptions on the problem coefficients.After creating the linear theory, a number of mathematicians [4, 6, 8–10] took the study of semi-linear elliptic problemsin unbounded domains. The problem of existence and nonexistence of positive solutions to a weak linear second-orderdivergence type elliptic equation in an unbounded cone-like domain was studied in [6].We introduce the following notation:

• (r, ω), ω = (ω1, ω2, . . . , ωn−1): spherical coordinates of x ∈ Rn,x1 = r cosω1, x2 = r cosω2 sinω1,

. . .xn−1 = r cosωn−1 sinωn−2 · · · sinω1, xn = r sinωn−1 sinωn−2 · · · sinω1,

• C: a rotational cone x1 > r cos (ω0/2),• ∂C: the lateral surface of C, x1 = r cos (ω0/2),• Ω: a domain on the unit sphere Sn−1 with smooth boundary ∂Ω obtained by the intersection of the cone C withthe sphere Sn−1, ∂Ω = ∂C ∩ Sn−1,• Gb

a = (r, ω) : a < r < b, ω ∈ Ω ∩ G, the layer in Rn,2052

M.V. Borsuk, D. Wiśniewski

• Γba = (r, ω) : a < r < b, ω ∈ ∂Ω ∩ ∂G, the lateral surface of layer Gba ,• GR = G∞R , ΓR = Γ∞R , ΩR = G ∩ |x| = R, R 1.

We use standard function spaces C k (G) with the norm |u|k,G , the Lebesgue space Lp(G), p ≥ 1, with the norm ‖u‖p,G ,the Sobolev space W k,p(G) with the norm ‖u‖p,k;G . We define the weighted Sobolev spaces V kp,α (G), for an integer k ≥ 0and a real α , as the spaces of distributions u ∈ D′(G) with the finite norm

‖u‖V kp,α (G) = ∫G

k∑|β|=0 r

α+p(|β|−k)|Dβu|pdx

1/p.

Definition 1.1.A function u(x) is said to be a weak solution of problem (QL) provided that u(x) ∈ C 0(G) ∩ V 1m,0(G) and satisfies theintegral identity∫

G

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

∂G

α(x) γ(ω)rm−1 u|u|q+m−2η(x)ds = ∫

∂G

α(x)g(x, u)η(x)ds (II)for all functions η(x) ∈ C 0(G) ∩ V 1

m,0(G) such that lim|x|→∞

η(x) = 0.Remark 1.2.In the case of the Dirichlet boundary condition we assume, without loss of generality, that g∂G∩D = 0 implies u∂G∩D = 0.Lemma 1.3.Let u(x) be a weak solution of (QL). For any function η(x) ∈ C 0(G) ∩ V 1

m,0(G) with lim|x|→∞

η(x) = 0 the equality

∫Gρ

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

Ωρai(x, u, ux ) cos(r, xi)η(x)dΩρ

+ ∫Γρα(x)(g(x, u)− γ(ω)

rm−1 u|u|q+m−2)η(x)ds (IIloc)

holds for a.e. ρ > R 1.

Proof. Let χρ(x) be the characteristic function of the set Gρ. We consider the integral identity (II), replacing η(x) byη(x)χρ(x). As a result we obtain∫

Gρ

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

Gρ

ai(x, u, ux )η(x) ∂χρ∂xidx + ∫

Γρα(x)(g(x, u)− γ(ω)

rm−1 u|u|q+m−2)η(x)ds.Let δ(r−ρ) be the Dirac distribution lumped on the sphere r = ρ. Using formula (7’) of [5, subsection 3, § 1, Chapter 3],

∂χρ∂xi

= xir δ(r−ρ),

we obtain, see [5, subsection, § 1, Chapter 3, Example 4],∫Gρ

ai(x, u, ux )η(x) ∂χρ∂xidx = ∫

Gρ

ai(x, u, ux )η(x) xir δ(r−ρ)dx = ∫Ωρai(x, u, ux )η(x) cos(r, xi)dΩρ.

The lemma is proved.2053


The following conditions will be needed throughout the paper: Let 1 < m < n < p, q ≥ 0, 0 ≤ µ < (q+m−1)/(m−1)be given numbers; a0(x), α1(x) and b0(x) be non-negative measurable functions; ai(x, u, ξ), i = 1, . . . , n, b(x, u, ξ)be Carathéodory functions continuously differentiable with respect to xi; g(x, u) be Carathéodory and continuouslydifferentiable with respect to u function ∂G×R→ R with the properties:1) ai(x, u, ξ)ξi ≥ |u|q|ξ|m − a0(x), a0(x) ∈ Lp/m(G),2) √√√√ n∑

i=1 a2i (x, u, ξ) +√√√√ n∑

i=1∣∣∣∣∂ai(x, u, ξ)∂xi

∣∣∣∣2 ≤ |u|q|ξ|m−1 + α1(x), α1(x) ∈ Lp/(m−1)(G) ∩ Lm/(m−1)(G),3a) |b(x, u, ξ)| ≤ µ|u|q−1|ξ|m + b0(x), b0(x) ∈ Lp/m(G) ∩ L1(G),3b) b(x, u, ξ) = β(x, u) + b(x, u, ξ), u · β(x, u) ≥ |u|q+m, |b(x, u, ξ)| ≤ µ|u|q−1|ξ|m + b0(x),4) ∂g(x, u)/∂u ≤ 0,5) γ(ω) ≥ γ0 > 0 on ∂G.

In addition, suppose that the functions ai(x, u, ξ) are continuously differentiable with respect to u, ξ variables in Mρ,M0 =Gρ× [−M0,M0]×Rn and satisfy in Mρ,M0 the following conditions:

6) (m−1)u∂ai(x, u, ξ)/∂u = q(∂ai(x, u, ξ)/∂ξj )ξj , i = 1, . . . , n,7) √√√√ n∑

i=1∣∣ai(x, u, ux )− |u|q|∇u|m−2uxi ∣∣2 ≤ A

( 1|x|

)|u|q|∇u|m−1, x ∈ Gρ,

where A(t) is a function Dini-continuous at zero.We shall consider the substitutionu = v|v|ς−1 with ς = m− 1

q+m− 1 . (1)By virtue of assumption 6), the identity (IIloc) takes the form

∫Gρ

Ai(x, vx )ηxi + B(x, v, vx )η(x)dx + ∫

Γρα(x) γ(ω)

rm−1 v|v|m−2η(x)ds= −∫

ΩρAi(x, vx ) cos(r, xi)η(x)dΩρ + ∫

ΓρG(x, v)α(x)η(x)ds (2)

for a.e. ρ > R 1, v(x) ∈ C 0(G) ∩ V 1m,0(G) and any η(x) ∈ C 0(G) ∩ V 1

m,0(G), with lim|x|→∞

η(x) = 0, whereAi(x, vx ) ≡ ai(x, v|v|ς−1, ς|v|ς−1vx), B

(x, v, vx ) ≡ b(x, v|v|ς−1, ς|v|ς−1vx), G(x, v) ≡ g(x, v|v|ς−1). (3)

It is easy to check that coefficients Ai, i = 1, . . . n, do not depend on v explicitly.Our assumptions regarding problem (QL) take the following form:1’) Ai(x, vx )vxi ≥ ςm−1|∇v|m − |v|1−ςa0(x)/ς, a0(x) ∈ Lp/m(G),2’) √√√√ n∑

i=1 A2i (x, vx ) +√√√√ n∑

i=1∣∣∣∣∂Ai(x, vx )

∂xi

∣∣∣∣2 ≤ ςm−1|∇v|m−1 + α1(x), α1(x) ∈ Lp/(m−1)(G) ∩ Lm/(m−1)(G),3a’) |B(x, v, vx )| ≤ µςm|v|−1|∇v|m + b0(x), b0(x) ∈ Lp/m(G) ∩ L1(G),

2054


4’) ∂G(x, v)/∂v ≤ 0,7’) √√√√ n∑

i=1∣∣Ai(x, vx )− ςm−1|∇v|m−2vxi ∣∣2 ≤ ςm−1A

( 1|x|

)|∇u|m−1, x ∈ Gρ.

Our main result is the following statement.Theorem 1.4.Let u be a weak solution of the problem (QL), assumptions 1)–7) hold and θ be the smallest positive eigenvalue of theproblem (NEVP), see Section 2. Suppose, in addition, that g(x, 0) ∈ Lj/(j−1)(∂G), 1 ≤ j < (n−1)/(m−1), and there arefinite numbers ks ≥ 0, K ≥ 0 such that

ks = supρ>R

ρms

∫Gρ

rq(m+1)/((q+m)(m−1)) am(q+m−1)/((m−1)(q+m))0 (x)dx + ∫Gρ

r(m+1)/(m−1)bm/(m−1)0 (x)dx+ ∫

Γρα(x)r(m−2)/(m−1)|g(x, 0)|m/(m−1)ds

, s > 0,(4)

K = supρ>R

ρn/m−1ψ(ρ)

ρm(1−n/p)(q+m−1)/((m−1)(q+m))‖a0(x)‖(q+m−1)/((m−1)(q+m))

p/m,G2ρρ

+ ρ(1−n/p)m/(m−1)‖b0(x)‖1/(m−1)p/m,G2ρ

ρ

+ ρ1−n/p‖α1(x)‖1/(m−1)p/(m−1),G2ρ

ρ+ ρ‖g(x, 0)‖1/(m−1)

∞,Γ2ρρ

,

(5)where

ψ(ρ) =

ρ−(θ1/m(m)/Ξ(m))·(q+(m−1)(1−µ))/(q+m−1), s > θ1/m(m)Ξ(m) · q+ (m−1)(1−µ)q+m− 1 ,

ρ−(θ1/m(m)/Ξ(m))·(q+(m−1)(1−µ))/(q+m−1) ln1/m ρR , s = θ1/m(m)Ξ(m) · q+ (m−1)(1−µ)

q+m− 1 ,

ρ−s, 0 < s < θ1/m(m)Ξ(m) · q+ (m−1)(1−µ)q+m− 1 ,

(6)

and

Ξ(m) =m

√m2 ·

( 2m+ 2

)(m+2)/(2m), m ≥ 2,

(m−1)(m−1)/m · 2(2−m)/2, 1 < m ≤ 2. (7)Then there exist R0 > R 1 and a constant C0 > 0 independent of u such that

|u(x)| ≤ C0(|x|1−n/mψ(|x|))(m−1)/(q+m−1) for all x ∈ GR0 . (8)Furthermore, if coefficients of the problem (QL) satisfy such conditions which guarantee the local a priori estimate|∇u|G′ ≤ M1 for any smooth G′ ⊂ G, see for example [7, 11], then there exists a constant C1 > 0 independent of u suchthat

|∇u(x)| ≤ C1|x|−(n(m−1)+qm)/(m(q+m−1))ψ(m−1)/(q+m−1)(|x|) for all x ∈ GR0 . (9)2. Preliminaries

We shall use well-known formulae related to the spherical coordinates (r, ω), see either [3, § 1.3] or [2, § 1.2]. C = C (· · · ),c = c(· · · ) denote the constants depending only on the quantities appearing in parentheses. In what follows, the sameletters C, c will (generally) be used to denote different constants depending on the same set of arguments.We need some auxiliary statements and inequalities.

2055


The eigenvalue problem for the m-Laplacian in a bounded domain on the unit sphere.

Let −→ν be the exterior normal to ∂C at points of ∂Ω. Let γ(ω), ω ∈ ∂Ω, be a positive bounded piecewise smooth functionon ∂Ω. We consider the eigenvalue problem for the m-Laplace–Beltrami operator, m > 1, on the unit sphere,

divω(|∇ωψ|m−2∇ωψ) + θ|ψ|m−2ψ = 0, ω ∈ Ω,α(ω)|∇ωψ|m−2 ∂ψ

∂−→ν+ γ(ω)|ψ|m−2ψ(ω) = 0, ω ∈ ∂Ω, (NEVP)

which consists of determination of all values θ (eigenvalues) for which (NEVP) has weak solutions ψ(ω) 6= 0 (eigen-functions); hereα(ω) = 0 if ω ∈ ∂DΩ,1 if ω ∈ ∂Ω \ ∂DΩ,

where ∂DΩ ⊆ ∂Ω is the part of the boundary ∂Ω for which we consider the Dirichlet boundary condition.Definition 2.1.A function ψ is said to be a weak solution of problem (NEVP) provided that ψ ∈ W 1,m(Ω) ∩ C 0(Ω) and satisfies theintegral identity ∫

Ω|∇ωψ|m−2 1

qi∂ψ∂ωi

∂η∂ωi− θ|ψ|m−2ψη

dΩ + ∫

∂Ωα(ω)γ(ω)|ψ|m−2ψηdσ = 0

for all η(ω) ∈ W 1,m(Ω) ∩ C 0(Ω).Remark 2.2.θ = 0 is not an eigenvalue of (NEVP). In fact, setting η = ψ and θ = 0 we have

∫Ω|∇ωψ|mdΩ + ∫

∂Ωα(ω)γ(ω)|ψ|mdσ = 0 =⇒ ψ ≡ 0,

since γ(ω) > 0 if α(ω) = 1, and ψ∂Ω = 0 if α(ω) = 0.We characterize the first eigenvalue θ(m) of the eigenvalue problem for the m-Laplacian by

θ(m) = inf06=ψ∈W 1,m(Ω)∫

Ω|∇ωψ|mdΩ + ∫

∂Ωα(ω)γ(ω)|ψ|mdσ/∫

Ω|ψ|mdΩ.

The proof of existence of θ > 0 and associated eigenfunction ψ(ω) can be found in [3, Chapter 8] or in [2, § 2.1]. Next,from the variational principle we have the Friedrichs–Wirtinger type inequality:Theorem 2.3.Let θ be the smallest positive eigenvalue of problem (NEVP), Ω ⊂ Sn−1 and ψ ∈ W 1,m(Ω) satisfy the boundary conditionfrom (NEVP) in the weak sense. Let γ(ω) be a positive bounded piecewise smooth function on ∂Ω. Then

∫Ω|ψ|mdΩ ≤ 1

θ(m)∫Ω |∇ωψ|mdΩ + ∫

∂Ωα(ω)γ(ω)|ψ|mdσ

(Wm)with the sharp constant 1/θ(m).

2056


For the proof see [2, § 2.2].Corollary 2.4.Let θ be the smallest positive eigenvalue of problem (NEVP), v(x) ∈ W 1,m(GR ) and v( · , ω) satisfy the boundary conditionfrom (NEVP) in the weak sense. Let γ(ω) be a positive bounded piecewise smooth function on ∂Ω. Then for any ρ > Rand for all α ∫

Gρ

rα |v|mdx ≤ 1θ(m)

∫Gρ

rα+m|∇v|mdx + ∫Γρα(x)rα+m γ(ω)

rm−1 |v|mds , (10)

provided that integrals on the right are finite.

Proof. Consider the inequality (Wm) for the function v(r, ω). Multiplying it by rα+n−1 and integrating over r ∈ (ρ,∞),we obtain the desired inequality (10).Lemma 2.5.Let GR be a conical domain, v(ρ, · ) ∈ W 1,m(Ω) for almost all ρ > R 1 and

V (ρ) = ∫Gρ

|∇v|mdx + ∫Γρα(x) γ(ω)

rm−1 |v|mds <∞. (11)Let θ(m) be the smallest positive eigenvalue of problem (NEVP) and γ(ω) be a positive bounded piecewise smoothfunction on ∂Ω. Then for almost all ρ > R 1,

∫Ωρv ∂v∂r |∇v|

m−2dΩρ ≥ Ξ(m) · ρmθ1/m V ′(ρ), (12)

where Ξ(m) is determined by (7).Proof. Writing the function V (ρ) in spherical coordinates yields

V (ρ) = ∞∫ρ

rn−1∫

Ω|∇v(r, ω)|mdΩdr + ∞∫

ρ

rn−m−1 ∫

∂Ωα(ω)γ(ω)|v(r, ω)|mdσdr.

Differentiating it with respect to ρ we obtainV ′(ρ) = −ρn−1 ∫

Ω|∇v(ρ, ω)|mdΩ− ρn−m−1 ∫

∂Ωα(ω)γ(ω)|v(ρ, ω)|mdσ. (13)

Case m ≥ 2. Using the Cauchy inequality and next the Young inequality with p = m/2, p′ = m/(m−2) we obtain∫Ωρv ∂v∂r |∇v|

m−2dΩρ = ρn∫Ωvρ ·

∂v∂r |∇v|

m−2∣∣∣∣r=ρdΩ ≥ −ρn∫

Ωε2∣∣∣∣ vρ∣∣∣∣2 + 12ε

∣∣∣∣∂v∂r∣∣∣∣2|∇v|m−2dΩ

≥ −ρn∫Ωεδm

∣∣∣∣ vρ∣∣∣∣m + m− 2

m · ε2 δ−2/(m−2)|∇v|m + 12ε∣∣∣∣∂v∂r

∣∣∣∣2|∇v|m−2∣∣∣∣r=ρdΩ for all ε, δ > 0.

2057


Applying the Friedrichs–Wirtinger type inequality (Wm) we see that∫Ωρv ∂v∂r |∇v|

m−2dΩρ ≥ −εδρnmθ(m)

∫∂Ω

α(ω)γ(ω)∣∣∣∣ v(ρ, ω)ρ

∣∣∣∣mdσ− ρn

∫Ω

εδmθ(m)

∣∣∣∣∇ωvρ

∣∣∣∣m+ (m−2) ε2m δ−2/(m−2)|∇v|m + 12ε∣∣∣∣∂v∂r

∣∣∣∣2|∇v|m−2∣∣∣∣r=ρdΩ for all ε, δ > 0. (14)

Because |∇ωv/ρ| ≤ |∇v| and |∇v|2 = v2r + |∇ωv|2/r2, we get

εδmθ(m)

∣∣∣∣∇ωvρ

∣∣∣∣m+ 12ε∣∣∣∣∂v∂r

∣∣∣∣2|∇v|m−2 ≤

εδmθ(m)

∣∣∣∣∇ωvρ

∣∣∣∣2+ 12ε∣∣∣∣∂v∂r

∣∣∣∣2|∇v|m−2 = εδ

mθ(m) |∇v|mif we choose

ε =√mθ(m)2δ . (15)Therefore from (14) it follows∫

Ωρv ∂v∂r |∇v|

m−2dΩρ ≥−εδ

mθ(m) ρn−m∫∂Ωα(ω)γ(ω)|v(ρ, ω)|mdσ

− εm

(m− 22 δ−2/(m−2) + δ

θ(m))ρn∫Ω|∇v(ρ, ω)|mdΩ for all δ > 0. (16)

Let us put δ = βθ (m−2)/m(m), β > 0. (15) now becomes ε = √m/2β−1/2θ1/m(m), β > 0, and for this reasonε(m− 22 δ−2/(m−2) + δ

θ(m)) =√m2 θ−1/m(m)φ(β), φ(β) = β1/2 + m− 22 β−(m+2)/(2(m−2)) for all β > 0.

In this way, (16) takes the form∫Ωρv ∂v∂r |∇v|

m−2dΩρ ≥√m2 φ(β) · ρ

mθ1/m V ′(ρ) for all β > 0.Now, by a direct calculation, we find that there exists β0 ≡ ((m + 2)/2)(m−2)/m such that φ′(β0) = 0 and φ′′(β0) > 0.Moreover, lim

β→+0φ(β) = limβ→+∞φ(β) = +∞. Thus, we get

infβ∈(0,+∞)φ(β) = φ(β0) = (m+ 22

)(m−2)/(2m)+ m− 22( 2m+ 2

)(m+2)/(2m) = m( 2m+ 2

)(m+2)/(2m).

Hence and from above we derive the required (12) for m ≥ 2.Case 1 < m ≤ 2. We conclude from |∇v| ≥ |vr | that −|∇v|m−2 ≥ −|vr |m−2. Therefore using the Young inequality withp = m, p′ = m/(m−1), we have∫


m−2dΩρ = ρn−1∫Ωv ∂v∂r |∇v|

m−2∣∣∣∣r=ρdΩ ≥ −ρn−1∫

Ω|v|∣∣∣∣∂v∂r

∣∣∣∣|∇v|m−2∣∣∣∣r=ρdΩ ≥

− ρn∫Ω∣∣∣∣ vρ∣∣∣∣ |vr |m−1∣∣∣∣

r=ρdΩ ≥ −ρn∫Ωεm

∣∣∣∣ vρ∣∣∣∣m + m− 1

m ε−1/(m−1)∣∣∣∣∂v∂r∣∣∣∣m∣∣∣∣

r=ρdΩ for all ε > 0.2058


Next, applying the Friedrichs–Wirtinger type inequality (Wm) we obtain∫Ωρv ∂v∂r |∇v|

m−2dΩρ ≥−ερn

mθ(m)∫∂Ω

α(ω)γ(ω) ∣∣∣∣ v(ρ, ω)ρ

∣∣∣∣mdσ− ρnm

∫Ω

εθ(m)

∣∣∣∣∇ωvρ

∣∣∣∣m + (m−1)ε−1/(m−1)∣∣∣∣∂v∂r∣∣∣∣m∣∣∣∣

r=ρdΩ for all ε > 0.Let us choose ε = (m−1)θ(m)(m−1)/m. Hence the inequality above gives∫


m−2dΩρ ≥ −ερn

mθ(m) ∫

∂Ωα(ω)γ(ω) ∣∣∣∣ vρ

∣∣∣∣mdσ + ∫Ω(∣∣∣∣∇ωv

ρ

∣∣∣∣m+ ∣∣∣∣∂v∂r∣∣∣∣m)∣∣∣∣

r=ρdΩ . (17)

But, because of |∇v|2 = v2r + |∇ωv|2/r2 and the Jensen inequality, we can conclude that∣∣∣∣∇ωvρ

∣∣∣∣m+ ∣∣∣∣∂v∂r∣∣∣∣m = (∣∣∣∣∇ωv

ρ

∣∣∣∣2)m/2+ (v2r )m/2 ≤ 2(2−m)/2|∇v|m, 1 < m ≤ 2.

From this and (17) it follows that∫Ωρv ∂v∂r |∇v|

m−2dΩρ ≥ −2(2−m)/2εmθ(m) ρn

∫∂Ω

α(ω)γ(ω) ∣∣∣∣ vρ∣∣∣∣mdσ + ∫

Ω|∇v|mdΩ

.

Substituting here ε chosen above and recalling (7), (13) we get the desired inequality (12).The Cauchy problem for a differential inequality

Theorem 2.6.Suppose that V (ρ) is monotonically decreasing, nonnegative differentiable function defined on [R,∞), R 1, satisfying

V ′(ρ) + P(ρ)V (ρ)− Q(ρ) ≤ 0, ρ > R,V (R) ≤ V0, (CP)

where P(ρ),Q(ρ) are nonnegative continuous functions defined on [R,∞) and V0 is a constant. Then

V (ρ) ≤ V0 exp− ρ∫R

P(s)ds+ ρ∫R

Q(t) exp− ρ∫t

P(s)dsdt. (18)For the proof we refer to [3, Theorem 1.57] and [12, Theorem 2.8].3. Maximum principle

In this section we consider one of the possible cases of deriving a priori L∞(G) estimate of the weak solution toproblem (QL). We first observe that there exists R∗ > 1 such that |u(x)| < 1 for all x ∈ GR∗ . We denote G∗ ≡ G \ GR∗and introduce the set A(k) = x ∈ G∗ : |u(x)| > k.Theorem 3.1.Let u(x) be a weak solution of (QL) and let assumptions 1), 3b), 4), 5) hold. Suppose, in addition, that g(x, 0) ∈ Lj/(j−1)(∂G),1 ≤ j < (n−1)/(m−1). Then there exists a constant M0 > 0, depending only on measA(k), meas (∂G∗ ∩A(k)), n,m, µ, q,‖g(x, 0)‖Lj/(j−1)(∂G), ‖a0(x)‖Lp/m(G), ‖b0(x)‖Lp/m(G), such that ‖u‖L∞(G) ≤ M0.

For the proof we refer to [2, Theorem 6.5, § 6.3].2059


4. Local estimate near the infinity

The weak solution of problem (QL) is locally bounded at the infinity. More precisely, we haveTheorem 4.1.Let u(x) be a weak solution of problem (QL). Let assumptions 1), 2), 3a), 4), 5), 6) be satisfied and, in addition,g(x, 0) ∈ L∞(∂G). Then the inequality

supx∈G2ρ

ρκ

|u(x)| ≤ C(κ−1)nς/tρ−nς/t‖u‖t/ς,G2ρ

ρ+ ρmς(p−n)/(p(m−1+ς)) · ‖a0(x)‖ς/(m−1+ς)

p/m,G2ρρ

+ ρς(1−n/p)‖α1(x)‖ς/(m−1)p/(m−1),G2ρ

ρ

+ ρς(1−n/p)m/(m−1)‖b0(x)‖ς/(m−1)p/m,G2ρ

ρ+ ρς‖g(x, 0)‖ς/(m−1)

∞,Γ2ρρ

, p > n > m,

holds for any t > 0, κ ∈ (1, 2), ρ > R 1 and ς = (m−1)/(q+m−1), where C > 0 depends only on n,m, p, t, q, d,‖a0(x)‖p/m,G , ‖α1(x)‖p/(m−1),G , ‖b0(x)‖p/m,G .

We refer to the proof of [2, Theorem 6.6, § 6.4].5. Integral estimates

At first we shall obtain the global estimate for the Dirichlet integral.Theorem 5.1.Assume u(x) is a weak solution of the problem (QL). Suppose that M0 = maxx∈G |u(x)| is known. Let assumptions1), 3a), 4), 5) with γ0 > 0 be fulfilled and, in addition, g(x, 0) ∈ L1(∂G). Then

∫G

|u|qm/(m−1)|∇u|mdx + ∫∂G

α(x) γ(ω)rm−1 |u|m(q+m−1)/(m−1)ds

≤ c(M0, q,m, µ, n) ·∫G

(a0(x)+b0(x))dx + ∫∂G

α(x)|g(x, 0)|ds .(19)

Proof. We first make the substitution (1). By the assumption 6), identity (II) takes the form∫G

⟨Ai(x, vx )ηxi + B(x, v, vx )η(x)⟩dx + ∫

∂G

α(x) γ(ω)rm−1 v|v|m−2η(x)ds = ∫

∂G

α(x)G(x, v)η(x)ds,where coefficients A,B,G are defined by (3). Putting η(x) = v(x) we obtain

∫G

⟨Ai(x, vx )vxi + B(x, v, vx )v⟩dx + ∫

∂G

α(x) γ(ω)rm−1 |v|mds = ∫

∂G

α(x)G(x, v)v(x)ds.We observe that

G(x, v)− G(x, 0) = 1∫0

ddτ G(x, τv)dτ = v ·

1∫0∂G(x, τv)∂(τv) dτ.

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Now, according to assumption 4’) we have vG(x, v) ≤ vg(x, 0). Next, by 1’), 3a’), since ςm−1(1−ςµ) < 1 (by (1)), weobtainςm−1(1−ςµ)

∫G

|∇v|mdx + ∫∂G

α(x) γ(ω)rm−1 |v|mds

≤ ∫G

|v|b0(x)dx + 1ς

∫G

|v|1−ςa0(x)dx + ∫∂G

α(x)|g(x, 0)| |v|ds.By the substitution (1), from M0 = sup

G|u(x)|, it follows that |v(x)| ≤ M1/ς0 . Therefore we get

∫G

|∇v|mdx + ∫∂G

α(x) γ(ω)rm−1 |v|mds ≤ c(M0, m, q, µ) ·

∫G

(a0(x)+b0(x))dx + ∫∂G

α(x)|g(x, 0)|ds . (20)Returning to the function u(x) we obtain the desired estimate (19).Now we establish the local integral weighted estimate.Theorem 5.2.Let u(x) be a weak solution of problem (QL), θ(m) be the smallest positive eigenvalue of (NEVP), M0 = maxx∈G |u(x)|be known and assumptions of Theorem 5.1 and 7) be fulfilled. Suppose, in addition, that there exists a real num-ber ks ≥ 0 defined by (4). Then there exist R 1 and a constant c > 0 independent of u and depending onlyon m,n, s, q, θ(m), k1, ks,measΩ and M0 such that for any ρ > R

∫Gρ

|u|qm/(m−1)|∇u|mdx + ∫Γρα(x) γ(ω)

rm−1 |u|m(q+m−1)/(m−1)ds ≤ cψm(ρ),where ψ(ρ) is defined by (6) with (7).Proof. We first make the substitution (1). By Theorem 5.1, we have

V (ρ) = ∫Gρ

|∇v|mdx + ∫Γρα(x) γ(ω)

rm−1 |v|mds <∞, ρ > R. (21)The substitute η(x) = v(x) in the identity (2) yields

∫Gρ

⟨Ai(x, vx )vxi + B(x, v, vx )v(x)⟩dx + ∫

Γρα(x) γ(ω)

rm−1 |v|mds

= − ∫Ωρ

Ai(x, vx ) cos(r, xi) · v(x)dΩρ + ∫Γρα(x)G(x, v) · v(x)ds.

By virtue of assumptions 1’), 3a’), 4’), 5), 7’) and since µς < 1, we obtain(1−ςµ)ςm−1V (ρ) ≤ ∫

Gρ

|v|b0(x)dx + 1ς

∫Gρ

|v|1−ςa0(x)dx + ςm−1A(1ρ

)∫Ωρ|v| · |∇v|m−1dΩρ

− ςm−1 ∫Ωρ|∇v|m−2 · v ∂v∂r dΩρ + ∫

Γρα(x)|g(x, 0)| · |v|ds.

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Applying Lemma 2.5 we can rewrite the latter relation in the form(1− ςµ)V (ρ) ≤− Ξ(m)

mθ1/m(m) ρV ′(ρ) + A

(1ρ

)∫Ωρ|v| · |∇v|m−1dΩρ + 1

ςm

∫Gρ

|v|1−ςa0(x)dx+ ς1−m∫

Gρ

|v|b0(x)dx + ς1−m∫Γρα(x)|g(x, 0)| · |v|ds, ρ > R 1. (22)

Now we estimate the integral over Ωρ on the right hand side in (22). The Hölder inequality for integrals yields∫Ωρ|v| · |∇v|m−1dΩρ ≤

∫Ωρ|v|mdΩρ

1/m∫

Ωρ|∇v|mdΩρ

(m−1)/m

. (23)Next, from the inequality (Wm), the inequality |∇ωv| ≤ ρ|∇v|, and formula (13) we conclude that

∫Ωρ|v|mdΩρ ≤

ρn−1θ(m)

∫Ω |∇ωv|mdΩ + ∫∂Ω

α(ω)γ(ω)|v|mdσ ≤ −ρmθ V ′(ρ). (24)

Now, according to (23)–(24), we have∫Ωρ|v| · |∇v|m−1dΩρ ≤

[−V ′(ρ)]1/mρθ1/m(m) ·

∫Ωρ|∇v|mdΩρ

(m−1)/m

= [−V ′(ρ)]1/mρθ1/m(m) ·

ρn−1∫Ω|∇v|mdΩ(m−1)/m

≤ [−V ′(ρ)]1/m ρθ1/m(m) · [−V ′(ρ)](m−1)/m ≤ − ρ

θ1/m(m)V ′(ρ).(25)

Next, in virtue of the Young inequality and inequality (10) with α = −m we get∫Gρ

|v|1−ςa0(x)dx = ∫Gρ

(r−(1−ς)(m+1)/m|v|1−ς) · (r(1−ς)(m+1)/ma0(x))dx

≤∫Gρ

(1− ςm r−1−m|v|m + m+ ς − 1

m r(1−ς)(m+1)/(m+ς−1)am/(m+ς−1)0 (x))dx≤ 1− ςmθ(m) ρ−1V (ρ) + m+ ς − 1

m

∫Gρ

r(1−ς)(m+1)/(m+ς−1)am/(m+ς−1)0 (x)dx,∫Gρ

|v|b0(x)dx = ∫GR

(r(−1−m)/m|v|)(r(1+m)/m · b0(x))dx

≤∫Gρ

( 1m r−1−m|v|m + m− 1

m r(m+1)/(m−1)bm/(m−1)0 (x))dx≤ 1mθ(m) ρ−1V (ρ) + m

m− 1∫Gρ

r(m+1)/(m−1)bm/(m−1)0 (x)dx,

(26)

∫Γρα(x)|g(x, 0)| · |v|dx = ∫

Γρα(x)(γ1/m(ω)

r(m−2)/m |v|)(

r(m−2)/mγ1/m(ω) |g(x, 0)|)dx

≤ 1m

∫Γρα(x) γ(ω)

rm−2 |v|mdx + m− 1m

∫Γρα(x) r(m−2)/(m−1)

γ1/(m−1)(ω) |g(x, 0)|m/(m−1)ds

≤ 1mρ V (ρ) + c(m, γ0) ∫Γρ α(x)r(m−2)/(m−1)|g(x, 0)|m/(m−1)dx

(27)

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by assumption 5) and (21). Now, (22)–(27) yield(1−ςµ)(1−δ(ρ))V (ρ) ≤− Ξ(m)

mθ1/m(m)(1 + A

(1ρ

))ρV ′(ρ) + mς1−m

m− 1∫Gρ

r(m+1)/(m−1)bm/(m−1)0 (x)dx+ m+ ς − 1

mςm

∫Gρ

r(1−ς)(m+1)/(m+ς−1)am/(m+ς−1)0 (x)dx+ c(m, γ0)

ςm−1∫Γρα(x)r(m−2)/(m−1) · |g(x, 0)|m/(m−1)dx,

(28)

where δ(ρ) = const (m, µ, q, θ(m)) · ρ−1, A(1/ρ) = mA(1/ρ)/Ξ(m). We observe that∞∫ δ(ρ)

ρ dρ <∞,∞∫A(1/ρ)ρ dρ <∞.

In this way, from (28), in virtue of assumption (4), we have the Cauchy problem (CP) from Section 2 withP(ρ) = 1

ρ ·1− δ(ρ)1 + A(1/ρ) · mθ1/m(m)Ξ(m) (1−ςµ), Q(ρ) = c0ksρ−ms−1, c0 = const (m,q, θ(m), γ0).

On account of (20) we haveV (R) ≤ ∫

G

|∇v|mdx + ∫∂G

α(x) γ(ω)rm−1 |v|mdx ≤ c(M0, m, q, µ) ·

∫G

(a0(x)+b0(x))dx + ∫∂G

α(x)|g(x, 0)|dx ≡ V0.

The solution of (CP) is determined by (18) from Theorem 2.6. Direct calculations give−

ρ∫R

P(s)ds = −mθ1/m(m)(1−ςµ)Ξ(m)ρ∫

R

1s ·

1−δ(s)1 + A(1/s) ds= −mθ1/m(m)(1−ςµ)Ξ(m)

ρ∫R

1s ·(1− δ(s) + A(1/s)1 + A(1/s)

)ds ≤ mθ1/m(m)(1−ςµ)Ξ(m)

ln Rρ + ∞∫R

δ(s) + A(1/s)s ds

.

Thusexp− ρ∫

R

P(s)ds ≤ (Rρ)mθ1/m(m)(1−ςµ)/Ξ(m) expmθ1/m(m)(1−ςµ)Ξ(m)

∞∫R

δ(s) + A(1/s)s ds

,

ρ∫R

Q(t) exp− ρ∫t

P(s)dsdt ≤ ksC1ρ−mθ1/m(m)(1−ςµ)/Ξ(m) ρ∫R

t−ms−1tmθ1/m(m)(1−ςµ)/Ξ(m)dt

= ksC1ρ−mθ1/m(m)(1−ςµ)/Ξ(m) ·ρm(θ1/m(m)(1−ςµ)/Ξ(m)−s) − Rm(θ1/m(m)(1− ςµ)/Ξ(m)−s)

m(θ1/m(m)(1−ςµ)/Ξ(m)− s) , s 6= θ1/m(m)(1−ςµ)Ξ(m) ,

ln ρR , s = θ1/m(m)(1−ςµ)Ξ(m) ,

whereC1 = c0 expmθ1/m(m)(1−ςµ)Ξ(m)

∞∫R

δ(s) + A(1/s)s ds

.

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The inequalityV0 exp− ρ∫

R

P(s)ds ≤ V0C1(Rρ

)mθ1/m(m)(1−ςµ)/Ξ(m)

impliesV (ρ) ≤ cC1(V0 + ks) ·

ρ−mθ1/m(m)(1−ςµ)/Ξ(m), s > θ1/m(m)(1−ςµ)Ξ(m) ,

ρ−mθ1/m(m)(1−ςµ)/Ξ(m) ln ρR , s = θ1/m(m)(1−ςµ)Ξ(m) ,

ρ−ms, s < θ1/m(m)(1−ςµ)Ξ(m) ,

(29)

where c = const (m, s, θ(m)). Thus, the theorem is proved.6. The power modulus of continuity near the infinity for weak solutions

Proof of Theorem 1.4. We consider the function ψ(ρ), ρ > R 1, determined by (6). By Theorem 4.1 with t = m,sup

x∈G2ρ3ρ/2|v(x)|m ≤ C ρ−n‖v‖m

m,G2ρρ

+ Km(ρ), (30)K (ρ) = ρm(p−n)/(p(m−1+ς))‖a0(x)‖1/(m−1+ς)

p/m,G2ρρ

+ ρ(1−n/p)m/(m−1)‖b0(x)‖1/(m−1)p/m,G2ρ

ρ+ ρ1−n/p‖α1(x)‖1/(m−1)p/(m−1),G2ρ

ρ+ ρ‖g(x, 0)‖1/(m−1)

∞,Γ2ρρ, p > n > m,

(31)where v is given by (1). Hence, in virtue of inequality (Wm) regarding to the notation (11), we get

ρ−n∫G2ρρ

|v(x)|mdx ≤ 2n∫G2ρρ

r−n|v|mdx ≤ 2nρm−nθ(m) V (ρ) ≤ Cρm−nψm(ρ), (32)

by inequality (29). Next, (30)–(32) yieldsup

x∈G2ρ3ρ/2|v(x)| ≤ Cρ1−n/mψ(ρ) + K (ρ). (33)

Now from (31), by assumption (5), it follows K (ρ) ≤ Kρ1−n/mψ(ρ). From this and by (33) we have|v(x)| ≤ C0ρ1−n/mψ(ρ), x ∈ G2ρ3ρ/2.

Putting |x| = 7ρ/4 we obtain|v(x)| ≤ C0|x|1−n/mψ(|x|), x ∈ GR . (34)

Eventually, in virtue of (1), from (34) we establish the first desired estimate (8).Now we consider problem (QL) for the function v(x) after the substitution (1) in the set Gρρ/2 ⊂ G, ρ > R 1. Let uschange variables: x = ρx ′, z(x ′) = ρn/m−1ψ−1(ρ)v(ρx ′). Then the function z(x ′) satisfies

− ddx ′i

Ai(ρx ′, ρ−n/mψ(ρ)zx′)+ ρB

(ρx ′, zρ1−n/mψ(ρ), zx′ρ−n/mψ(ρ)) = 0, x ′ ∈ G11/2,

α(ρx ′)Ai(ρx ′, ρ−n/mψ(ρ)zx′) cos(~n, x ′i) + γ(ω)

|x ′|m−1 ρ−n(m−1)/mψm−1z|z|m−2 = G(ρx ′, zρ1−n/mψ), x ′ ∈ Γ11/2.(QL’)

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Now we apply our assumption about a priori estimate of the gradient modulus of the problem (QL’) solution as anestimate inside the domain and near a smooth boundary portionmaxx′∈G11/2|∇

′z(x ′)| ≤ M ′1, (35)see [1, § 4. Theorem 4.4]. Returning to the variable x and function v(x), from (35) it follows that

|∇v(x)| ≤ M ′1ρ−n/mψ(ρ), x ∈ Gρρ/2, R < ρ <∞.

Putting now |x| = 2ρ/3 we obtain the estimate|∇v(x)| ≤ C1|x|−n/mψ(|x|), x ∈ GR . (36)

But by (1),|∇u(x)| ≤ |v|ς−1|∇v(x)|, x ∈ GR .From (34) and (36) we establish the second desired estimate (9).

7. Examples

7.1. Example I (n = 2)

Let GR = (r, ω) : r > R, −ω0/2 < ω < ω0/2, ω0 ∈ (0, π) and Γ±R = (r, ω) : r > R, ω = ±ω0/2. We putγ(ω)∣∣ω=±ω0/2 = γ± = const ≥ 0 and α(x)∣∣Γ±R = α± ∈ 0, 1.

Let q ≥ 0, a0 ≥ 0, 0 ≤ µ < 1 + q, m = 2. We consider the following problem:

ddxi

(|u|quxi ) = a0|x|−2u|u|q − µu|u|q−2|∇u|2, x ∈ GR ,(α±|u|q

∂u∂~n + 1

|x| γ±u|u|q)∣∣∣∣Γ±R = 0,

lim|x|→∞

u(x) = 0.We make the substitution (1), denote µ = µς, a0 = a0ς−1 and consider our problem for the function v(x),

∆v + µv−1|∇v|2 = a0r−2v(x), x ∈ GR ,(α±

∂v∂~n + (1+q) 1

|x| γ±v)∣∣∣Γ±R = 0,

lim|x|→∞

v(x) = 0.Our goal is to find the exact solution of this problem in the form v(r, ω) = rλ−ψ(ω), where λ− < 0. For ψ(ω) we obtainthe problem

ψ′′(ω) + µψ(ω) ψ′2(ω) + (1+µ)λ2

− − a0ψ(ω) = 0, ω ∈(−ω02 , ω02

),

±α±ψ′(± ω02

)+ (1+q)γ±ψ(± ω02) = 0.

We assume that λ2− > a0/(1+µ) and define the value

Λ =√λ2− −

a01 + µ ⇐⇒ λ− = − √Λ2 + a0 (1+q)21 + q+ µ . (37)We consider separately two cases: µ = 0 and µ 6= 0.

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Case µ = 0.

The solution of our equation has the form ψ(ω) = A cos(ωΛ) + B sin(ωΛ). In order to find A,B, from the boundaryconditions we obtain the system((1+q)γ+ cos ω0Λ2 − Λα+ sin ω0Λ2

)A+ ((1+q)γ+ sin ω0Λ2 + Λα+ cos ω0Λ2

)B = 0,((1+q)γ− cos ω0Λ2 − Λα− sin ω0Λ2

)A−

((1+q)γ− sin ω0Λ2 + Λα− cos ω0Λ2)B = 0.

Since A2 + B2 6= 0, the system determinant must be equal to zero; this means that Λ is defined via the transcendentequation (Λ2α−α+ − (1+q)2γ+γ−) · sin(ω0Λ)− (1+q)Λ(α+γ−+α−γ+) · cos(ω0Λ) = 0.Hence

tan(ω0Λ) = (1+q)Λ(α+γ−+α−γ+)Λ2α+α− − (1+q)2γ+γ− . (38)Then we find the eigenfunction

ψ(ω) = (1+q)γ+ sin[Λ(ω02 − ω)]+ α+Λ cos[Λ(ω02 − ω

)].

Now we investigate some particular cases of the boundary conditions.Dirichlet problem: α± = 0, γ± = 1. Equation (38) becomes tan(ω0Λ) = 0. Hence,

Λ = πω0 , λ− = −√π2

ω20 + a0(1+q)and the corresponding eigenfunction ψ(ω) = cos(πω/ω0).Neumann problem: γ± = 0, α± = 1. Equation (38) becomes tan(ω0Λ) = 0. Hence,

Λ = πω0 , λ− = −√π2

ω20 + a0(1+q)and the corresponding eigenfunction ψ(ω) = sin(πω/ω0).Mixed problem: α+ = 1, α− = 0, γ+ = 0, γ− = 1. Equation (38) becomes cos(ω0Λ) = 0. Hence,

Λ = π2ω0 , λ− = −√ π24ω20 + a0(1+q)and the corresponding eigenfunction ψ(ω) = cos(πω/(2ω0)−π/4).Robin problem: α± = 1, γ± 6= 0. In this case we obtain the least positive eigenvalue as the least positive root of thetranscendent equation tan(ω0Λ) = (1+q)Λ(γ−+γ+)/(Λ2 − (1+q)2γ+γ−) and the corresponding eigenfunction

ψ(ω) = (1+q)γ+ sin[Λ(ω02 − ω)]+ Λ cos[Λ(ω02 − ω

)].

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In particular, if γ+ = γ− = γ, we get either

• ψ(ω) = cos(Λ∗ω), tan(ω0Λ∗/2) = (1+q)γ/Λ∗ and from the graphic solution, see Figure 1,0 < Λ∗ < π

ω0 =⇒ −

√(πω0)2+ a0(1+q) < λ− < 0.

Figure 1.

• or ψ(ω) = sin(Λ∗ω), tan(ω0Λ∗/2) = −Λ∗/((1+q)γ) and from the graphic solution, see Figure 2,πω0 < Λ∗ < 2π

ω0 =⇒ −

√(2πω0)2+ a0(1+q) < λ− < −

√(πω0)2+ a0(1+q) .

Figure 2.

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Case µ 6= 0.

By setting y(ω) = ψ′(ω)/ψ(ω), we arrive at a problem for y(ω),y′(ω) + (1+µ)y2 + (1+µ)λ2

− − a0 = 0, ω ∈(−ω02 , ω02

),

±α±y(±ω02

)+ (1+q)γ± = 0.By integrating the equation of our problem we find

y(ω) = Λ · tanΛ(C − (1 + µ)ω) for all C ∈ R. (39)Now, from the boundary conditions we obtain

C = (1+µ) ω02 − 1Λ arctan (1+q)γ+α+Λ (40)

and equation for the needed λ−:Λω0(1+µ) = arctan (1+q)γ−

α−Λ + arctan (1+q)vγ+α+Λ . (41)

Thus tan(1 + q+ µ1 + q ω0Λ) = Q+ +Q−Λ2 −Q+Q− Λ (see Figure 3), where Q± = (1+q) γ±α± .

Figure 3.

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Further, from (39) and (40) we obtainy(ω) = Λ tanΛ(1+µ)(ω02 − ω

)− arctan (1+q)γ+

α+Λ

and, because (ln |ψ(ω)|)′ = y(ω), it follows thatψ(ω) = ∣∣∣∣cos1/(1+µ)Λ(1+µ)(ω02 − ω


α+Λ∣∣∣∣.

At last, returning to the function u, we establish a solution in the formu(r, ω) = rλ−/(1+q)∣∣∣∣cos1/(1+q+µ)Λ∗ 1 + q+ µ1 + q

(ω02 − ω


α+Λ∗∣∣∣∣,

where Λ∗ is the smallest positive root of the transcendent equation (41) and λ− is defined by (37) with Λ = Λ∗.Now we investigate some particular cases of the boundary conditions.Dirichlet problem: α± = 0, γ± 6= 0. Direct calculations give

Λ∗ = πω0 ·

1 + q1 + q+ µ λ− = − (1+q)√(π/ω0)2 + a0(1+q+µ)1 + q+ µ .

The corresponding function isu(r, ω) = r−

√(π/ω0)2+a0(1+q+µ)/(1+q+µ) · cos1/(1+q+µ) πωω0 .

The Neumann problem: α± = 1, γ± = 0. Direct calculations giveΛ∗ = 2π

ω0 ·1 + q1 + q+ µ , λ− = − (1+q)√(2π/ω0)2+ a0(1+q+µ)1 + q+ µ .


√(2π/ω0)2+a0(1+q+µ)/(1+q+µ) · cos1/(1+q+µ) 2πωω0 .

Mixed problem: α+ = 1, α− = 0, γ+ = 0, γ− = 1. Direct calculations giveΛ∗ = π2ω0 ·

1 + q1 + q+ µ , λ− = − (1+q)√(π/(2ω0))2 + a0(1+q+µ)1 + q+ µ .


√(π/(2ω0))2+a0(1 +q+ µ)/(1 +q+ µ) · cos1/(1+q+µ)(π4 − πω2ω0).

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7.2. Example II (n = 3)

Let GR = (r, ω, φ) : r > R, 0 < ω < ω0/2, ω0 ∈ (π, 2π), φ ∈ [0, 2π), q ≥ 0, 0 ≤ µ < q + 1, a0 ≥ 0, m = 2.We consider the following problem:

ddxi

(|u|quxi ) = a0|x|−2u|u|q − µu|u|q−2|∇u|2, x ∈ GR ,(|u|q ∂u∂n + 1

|x| γu|u|q)∣∣∣∣

ω=ω0/2 = 0, γ > 0,lim|x|→∞

u(x) = 0.We make the substitution (1) and denote µ = µς, a0 = a0ς−1, ς = 1/(1+q). Let us consider our problem for thefunction v(x),

∆v + µv−1|∇v|2 = a0r−2v(x), x ∈ GR ,(∂v∂n + γ

ς |x|−1v)∣∣∣∣

ω=ω0/2 = 0,lim|x|→∞

v(x) = 0.We want to find the exact solution of this problem in the form v(r, ω, φ) = rλ− sinβ ω · w(φ), where λ− < 0, β > 0.In virtue of ∂v/∂n = (1/r)(∂v/∂ω), we obtain the problem

w ′′(φ)w(φ) + µ (w ′(φ))2

w2(φ) + λ−(λ−+1) + µλ2− − β − a0 sin2 ω+ β2(1+µ) cos2 ω = 0, ω ∈

(0, ω02),

β = −γς · tan ω02 > 0, ω0 ∈ (π, 2π),w(φ) = w(φ+2π), φ ∈ [0, 2π).

Now, if we choose λ−(λ−+1) + µλ2− − β − a0 = β2(1+µ), then we have the equation for w(φ),

w ′′(φ)w(φ) + µ (w ′(φ))2

w2(φ) + β2(1+µ) = 0 (42)with

β = −1 +√1 + 4(1 + µ)(λ−(λ−+1) + µλ2− − a0)2(1 + µ) > 0, λ− <

−1−√1 + 4a0(1+µ)2(1 + µ) = λ∗−. (43)By setting y(φ) = w ′(φ)/w(φ), we arrive at an equation for y(φ),

y′(φ) + (1+µ)y2(φ) + (1+µ)β2 = 0, φ ∈ [0, 2π).By integrating the equation above, we find

y(φ) = β · tan(C−φ · β(1+µ)) for all C ∈ R.

Because y(φ) = y(φ+2π), we obtain β = 1/(2(1+µ)) > 0. Hence and from (43) it follows thatλ− = −1− 2√1 + a0(1+µ)2(1+µ) < λ∗−.

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Now we observe that solutions of (42) are determined uniquely up to a scalar multiple. Further, by the definition of y(φ)we get (ln |w(φ)|)′ = y(φ). Hence, because β = 1/(2(1+µ)), it follows thatw(φ) = ∣∣∣∣cos2β(C − φ2

)∣∣∣∣ for all C ∈ R.

As a result we obtain the solution v(r, ω, φ) in the formv(r, ω, φ) = r(−1−2√1+a0(1+µ))/(2(1+µ)) · sin1/(2(1+µ)) ω ·

∣∣∣∣cos1/(1+µ)(C − φ2)∣∣∣∣, r > R 1, 0 < ω < ω02 , φ ∈ [0, 2π).

Because µ = µς = µ/(q+1), a0 = a0ς−1 = a0(q+1), q ≥ 0, we can rewrite v(r, ω, ψ) as follows:v(r, ω, φ) = r−(q+1)(1+2√1+a0(q+1+µ))/(2(q+1+µ)) · sin(q+1)/(2(q+1+µ)) ω ·

∣∣∣∣cos(q+1)/(q+1+µ)(C − φ2)∣∣∣∣ for all C ∈ R.

At last, we establish a solution u(r, ω, ψ) of our problem:u(r, ω, φ) = r−(1+2√1+a0(q+1+µ))/(2(q+1+µ)) · sin1/(2(q+1+µ)) ω ·

∣∣∣∣cos1/(q+1+µ)(C − φ2)∣∣∣∣

for all C ∈ R, r > R 1, 0 < ω < ω0/2, φ ∈ [0, 2π).Acknowledgements

This is part of the second author’s Ph.D. thesis, written under supervision of the first author at the University of Warmiaand Mazury in Olsztyn.

References

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[10] Pao C.V., Nonlinear elliptic boundary value problems in unbounded domains, Nonlinear Anal., 1992, 18(8), 759–774[11] Rivkind V.Ja., Ural’ceva N.N., Classical solvability and linear schemes for the approximate solution of diffractionproblems for quasilinear equations of parabolic and of elliptic type, In: Problems of Mathematical Analysis III,Integral and Differential Equations, Leningrad University, Leningrad, 1972, 69–111 (in Russian)[12] Wiśniewski D., Boundary value problems for a second-order elliptic equation in unbounded domains, Ann. Univ.Paedagog. Crac. Stud. Math., 2010, 9, 87–122

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Boundary value problems for quasi-linear elliptic second order equations in unbounded cone-like domains