The transmission problem for quasi-linear elliptic second order equations in a conical domain
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.
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Інститут прикладної математики і механіки НАН України
2007
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| Цитувати: | The transmission problem for quasi-linear elliptic second order equations in a conical domain / M. Borsuk // Український математичний вісник. — 2007. — Т. 4, № 4. — С. 485-524. — Бібліогр.: 15 назв. — англ. |
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nasplib_isofts_kiev_ua-123456789-1245302025-02-09T13:50:35Z The transmission problem for quasi-linear elliptic second order equations in a conical domain Borsuk, M. 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. 2007 Article The transmission problem for quasi-linear elliptic second order equations in a conical domain / M. Borsuk // Український математичний вісник. — 2007. — Т. 4, № 4. — С. 485-524. — Бібліогр.: 15 назв. — англ. 1810-3200 2000 MSC. 35J65, 35J70, 35B05, 35B45, 35B65. https://nasplib.isofts.kiev.ua/handle/123456789/124530 en Український математичний вісник application/pdf Інститут прикладної математики і механіки НАН України |
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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. |
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Article |
| author |
Borsuk, M. |
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Borsuk, M. The transmission problem for quasi-linear elliptic second order equations in a conical domain Український математичний вісник |
| author_facet |
Borsuk, M. |
| author_sort |
Borsuk, M. |
| title |
The transmission problem for quasi-linear elliptic second order equations in a conical domain |
| title_short |
The transmission problem for quasi-linear elliptic second order equations in a conical domain |
| title_full |
The transmission problem for quasi-linear elliptic second order equations in a conical domain |
| title_fullStr |
The transmission problem for quasi-linear elliptic second order equations in a conical domain |
| title_full_unstemmed |
The transmission problem for quasi-linear elliptic second order equations in a conical domain |
| title_sort |
transmission problem for quasi-linear elliptic second order equations in a conical domain |
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Інститут прикладної математики і механіки НАН України |
| publishDate |
2007 |
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https://nasplib.isofts.kiev.ua/handle/123456789/124530 |
| citation_txt |
The transmission problem for quasi-linear elliptic second order equations in a conical domain / M. Borsuk // Український математичний вісник. — 2007. — Т. 4, № 4. — С. 485-524. — Бібліогр.: 15 назв. — англ. |
| series |
Український математичний вісник |
| work_keys_str_mv |
AT borsukm thetransmissionproblemforquasilinearellipticsecondorderequationsinaconicaldomain AT borsukm transmissionproblemforquasilinearellipticsecondorderequationsinaconicaldomain |
| first_indexed |
2025-11-26T11:45:31Z |
| last_indexed |
2025-11-26T11:45:31Z |
| _version_ |
1849853252631265280 |
| fulltext |
Український математичний вiсник
Том 4 (2007), № 4, 485 – 524
Dedicated to the memory of
Academician I. V. Skrypnik
The transmission problem for quasi-linear
elliptic second order equations in
a conical domain
Mikhail Borsuk
(Presented by A. E. Shihskov)
Abstract. We investigate the behavior of weak solutions to the trans-
mission problem for quasi-linear elliptic divergence second order equa-
tions in a neighborhood of the boundary conical point.
2000 MSC. 35J65, 35J70, 35B05, 35B45, 35B65.
Key words and phrases. Elliptic quasi-linear equations, interface
problem, conical points.
1. Introduction
The transmission problems often appear in different fields of physics
and technics. For instance, one of the importance problem of the elec-
trodynamics of solid media is the electromagnetic processes research in
ferromagnetic media with different dielectric constants. These problems
appear as well as in solid mechanics if a body consists of composite ma-
terials.
In this work we obtain estimates of weak solutions of the nonlin-
ear 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 trans-
mission problem was investigated only in smooth domains (see works of
M. V. Borsuk [1], V. Ja. Rivkind, N. N. Ural’tseva [15], N. Kutev, P. L. Li-
ons [11]). Later other mathematicians are studied transmission problems
Received 22.10.2007
ISSN 1810 – 3200. c© Iнститут математики НАН України
486 The transmission problem...
in non-smooth domains in some particular linear cases (see the refer-
ences cited in [5, 13, 14]). General linear interface problems in polygonal
and polyhedral domains was considered in [13, 14]. Regularity results in
terms of weighted Sobolev–Kondratiev spaces are obtained in [5] for two
and three dimensional transmission problems for the Laplace operator.
D. Kapanadze and B.-W. Schulze studied boundary-contact problems
with conical [8] singularities and edge [9] singularities at the interfaces
for general linear any order elliptic equations (as well as systems). They
constructed parametrix and showed regularity with asymptotics of solu-
tions in weighted Sobolev–Kondratiev spaces. We knew only one paper,
namely the D. Knees work [10], that concerns with the study of the reg-
ularity of weak solutions of special nonlinear transmission problem on
polyhedral domains. A principal new feature of our work is the consider-
ation of estimates of solutions for elliptic general divergence quasi-linear
second order equations in n-dimensional conic domains.
Let G ⊂ R
n, 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.
We consider the elliptic transmission problem
− d
dxi
ai(x, u,∇u) + b(x, u,∇u) = 0, x ∈ G \ Σ0;
[u]Σ0
= 0,
S[u] ≡
[
∂u
∂ν
]
Σ0
+ 1
|x|m−1σ
(
x
|x|
)
u · |u|q+m−2 = h(x, u), x ∈ Σ0;
B[u] ≡ ∂u
∂ν + 1
|x|m−1γ
(
x
|x|
)
u · |u|q+m−2 = g(x, u), x ∈ ∂G \ O
(QL)
(summation over repeated indices from 1 to n is understood); here:
•
u(x) =
{
u+(x), x ∈ G+,
u−(x), x ∈ G−;
ai(x, u, ux) =
{
a+
i (x, u+,∇u+), x ∈ G+,
a−i (x, u−,∇u−), x ∈ G−
etc.;
• [u]Σ0 = u+(x)
∣∣
Σ0
−u−(x)
∣∣
Σ0
, where u±(x)
∣∣
Σ0
= lim
G±∋y→x∈Σ0
u±(y);
M. Borsuk 487
• ∂u
∂ν = ai(x, u, ux) cos(−→n , xi), where −→n denotes the unite outward
with respect to G+ (or G) normal to Σ0 (respectively ∂G \ O);
•
[
∂u
∂ν
]
Σ0
denotes the saltus of the co-normal derivative of the function
u(x) on crossing Σ0, i.e.
[∂u
∂ν
]
Σ0
= a+
i (x, u+,∇u+) cos(−→n , xi)
∣∣
Σ0
− a−i (x, u−,∇u−) cos(−→n , xi)
∣∣
Σ0
.
We obtain estimates of weak solutions to problem (QL) near a conical
boundary point. Such estimates are sharp in the case of quasi-linear
equations with semi-linear principal part, namely for the transmission
problem
− d
dxi
(
|u|qaij(x)uxj
)
+ b(x, u,∇u) = 0, q ≥ 0, x ∈ G \ Σ0;
[u]Σ0
= 0, S[u] ≡
[
∂u
∂ν
]
Σ0
+ 1
|x|σ
(
x
|x|
)
u · |u|q = h(x, u), x ∈ Σ0;
B[u] ≡ ∂u
∂ν + 1
|x|γ
(
x
|x|
)
u · |u|q = g(x, u), x ∈ ∂G \ O
(WL)
(summation over repeated indices from 1 to n is understood).
Remark 1.1. The problem (WL) is the problem (QL) with ai(x, u,∇u)
= |u|qaij(x)uxj , i = 1, . . . n and m = 2.
We introduce the following notations:
• Sn−1 : a unit sphere in R
n centered at O;
• (r, ω), ω = (ω1, ω2, . . . , ωn−1) : the spherical coordinates of x ∈ R
n
with pole O;
• C : the rotational cone {x1 > r cos ω0
2 } with the vertex at O;
• ∂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 with the sphere Sn−1;
• Ω+ = Ω∩{xn > 0}, Ω− = Ω∩{xn < 0} =⇒ Ω = Ω+∪Ω−∪σ0; σ0 =
Σ0 ∩ Ω;
• ∂Ω = ∂C ∩ Sn−1, ∂±Ω = Ω± ∩ ∂C, ∂Ω± = ∂±Ω ∪ σ0;
488 The transmission problem...
• Gba = {(r, ω) | 0 ≤ a < r < b; ω ∈ Ω} ∩G : a layer in R
n;
• Γba = {(r, ω) | 0 ≤ a < r < b;ω ∈ ∂Ω} ∩ ∂G : the lateral surface of
layer Gba;
• Σb
a = Gba ∩ {xn = 0} ⊂ Σ0;
• Gd = G \Gd0, Γd = ∂G \ Γd0, Σd = Σ0 \ Σd
0, d > 0;
• Ωρ = Gd0 ∩ {|x| = ρ}; 0 < ρ < d.
0 0
O
1
x
0
0
Fig. 1
We use the standard function spaces: Ck(G±) with the norm |u±|k,G±
,
Lebesgue space Lm(G±), m ≥ 1 with the norm ‖u±‖m,G±
, the Sobolev
space W k,m(G±) with the norm ‖u±‖k,m;G±
, and introduce their di-
rect sums Ck(G) = Ck(G+) ∔ Ck(G−), Lm(G) = Lm(G+) ∔ Lm(G−),
M. Borsuk 489
Wk,m(G) = W k,m(G+) ∔ W k,m(G−). We define the weighted Sobolev
spaces: Vk
m,α(G) for integer k ≥ 0 and real α as the space of distribu-
tions u ∈ D′(G) with the finite norm
‖u‖Vk
m,α(G) =
( ∫
G+
k∑
|β|=0
rα+m(|β|−k)|Dβu+|
m dx
) 1
m
+
( ∫
G−
k∑
|β|=0
rα+m(|β|−k)|Dβu−|
m dx
) 1
m
and V
k− 1
m
m,α (∂G) as the space of functions ϕ, given on ∂G, with the norm
‖ϕ‖
V
k− 1
m
m,α (∂G)
= inf ‖Φ‖Vk
m,α(G), where the infimum is taken over all func-
tions Φ such that Φ|∂G = ϕ in the sense of traces. We denote
Wk(G) ≡ Wk,2(G),
◦
Wk
α(G) ≡ Vk
2,α(G),
◦
W
k− 1
2
α (∂G) ≡ V
k− 1
2
2,α (∂G).
Definition 1.1. The function u(x) is called a weak solution of the prob-
lem (QL) provided that u(x) ∈ C0(G)∩W1,m(G) and satisfies the integral
identity
∫
G
{ai(x, u, ux)ηxi + b(x, u, ux)η(x)} dx
+
∫
Σ0
σ(ω)
rm−1
u|u|q+m−2η(x) ds+
∫
∂G
γ(ω)
rm−1
u|u|q+m−2η(x) ds
=
∫
∂G
g(x, u)η(x) ds+
∫
Σ0
h(x, u)η(x) ds (II)
for all functions η(x) ∈ C0(G) ∩ W1,m(G).
Lemma 1.1. Let u(x) be a weak solution of (QL). For any function
η(x) ∈ C0(G) ∩ W1,m(G) the equality
∫
G̺
0
{
ai(x, u, ux)ηxi + b(x, u, ux)η(x)
}
dx
=
∫
Ω̺
ai(x, u, ux) cos(r, xi)η(x)dΩ̺
490 The transmission problem...
+
∫
Γ̺
0
(
g(x, u) −
γ(ω)
rm−1
u|u|q+m−2
)
η(x) ds
+
∫
Σ̺
0
(
h(x, u) −
σ(ω)
rm−1
u|u|q+m−2
)
η(x) ds (II)loc
holds for a.e. ̺ ∈ (0, d).
Proof. The proof is analogous to the proof of Lemma 5.2 [3, pp. 167–
170].
Now we formulate our assumptions and main results.
Problem (WL).
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:
a±ξ
2 ≤ aij±(x)ξiξj ≤ A±ξ
2, ∀x ∈ G±, ∀ ξ ∈ R
n; a±, A± = const > 0,
aij(0) = aδji , where δji is the Kronecker symbol;
a =
{
a+, x ∈ G+,
a−, 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
(∑n
i,j=1 |a
ij
±(x) − aij±(y)|2
) 1
2 ≤
A(|x − y|) holds for x, y ∈ G, where A(r) is a monotonically in-
creasing, 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;
(e) ∂h(x,u)
∂u ≤ 0, ∂g(x,u)
∂u ≤ 0;
(f) |b0(x)| ≤ f1|x|
β, |g(x, 0)| ≤ g1|x|
s−1, |h(x, 0)| ≤ h1|x|
s−1.
M. Borsuk 491
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
Γd0 =
{
(r, ω)
∣∣x2
1 = cot2
ω0
2
n∑
i=2
x2
i ; r ∈ (0, d), ω1 =
ω0
2
, ω0 ∈ (0, 2π)
}
.
(1.1)
Our main result about problem (WL) is the following statement. Let
λ =
2 − n+
√
(n− 2)2 + 4ϑ
2
, (1.2)
where ϑ is the smallest positive eigenvalue of the problem (NEV P ) with
m = 2 (see Subsection 2.1).
Theorem 1.1. Let u be a weak solution of the problem (WL), the as-
sumptions (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,measG, diamG and on the quantity
∫ 1
0
A(r)
r dr such that ∀x ∈ Gd0
|u(x)| ≤ C0
(
‖u‖2(q+1),G + f1 + g1 + h1
)
×
|x|
λ(1+q−µ)
(q+1)2 , if s > λ1+q−µ
1+q ,
|x|
λ(1+q−µ)
(q+1)2 ln3/2(q+1)
(
1
|x|
)
, if s = λ1+q−µ
1+q ,
|x|
s
q+1 , if s < λ1+q−µ
1+q .
(1.3)
Suppose, in addition, that coefficients of the problem (WL) are satisfied
such conditions, which guarantee the local a-priori estimate |∇u|0,G′ ≤
M1 for any smooth G′ ⊂⊂ G \ {O} (see for example [1, §4] or [15]).
Then for ∀x ∈ Gd0
|∇u(x)| ≤ C1 ·
|x|
λ(1+q−µ)
(q+1)2
−1
, if s > λ1+q−µ
1+q ,
|x|
λ(1+q−µ)
(q+1)2
−1
ln3/2(q+1)
(
1
|x|
)
, if s = λ1+q−µ
1+q ,
|x|
s
q+1
−1
, if s < λ1+q−µ
1+q
(1.4)
with C1 = c1
(
‖u‖2(q+1),G + f1 + g1 + h1
)
, where c1 depends on M0, M1
and C0 from above.
Problem (QL).
Regarding the equation we assume that the following conditions are
satisfied:
492 The transmission problem...
Let
a =
{
a+, x ∈ G+,
a−, x ∈ G−,
a± > 0;
a∗ = min{a+, a−} > 0, a∗ = max{a+, a−} > 0;
1 < m < n, mn > p > n > m, q ≥ 0, k1 ≥ 0, s > 1, 0 ≤ µ <
q +m− 1
m− 1
be given numbers; a0(x), α(x), b0(x) are nonnegative measurable func-
tions; ai(x, u, ξ), i = 1, . . . , n; b(x, u, ξ) are Caratheodory functions G×
R × R
n → R and h(x, u) is a continuously differentiable with respect to
the u variable function Σ0 × R → R but g(x, u) is a continuously differ-
entiable with respect to the u variable function ∂G × R → R possessing
the properties:
1) ai(x, u, ξ)ξi ≥ a|u|q|ξ|m − a0(x); a0(x) ∈ Lp/m(G);
2)
√√√√
n∑
i=1
a2
i (x, u, ξ) +
√√√√
n∑
i=1
∣∣∣∂ai(x, u, ξ)
∂xi
∣∣∣
2
≤ a|u|q|ξ|m−1 + α(x);
α(x) ∈ L p
m−1
(G);
3a) |b(x, u, ξ)| ≤ aµ|u|q−1|ξ|m + b0(x); b0(x) ∈ L p
m
(G);
3b) b(x, u, ξ) = β(x, u) + b̃(x, u, ξ), u · β(x, u) ≥ a|u|q+m;
|̃b(x, u, ξ)| ≤ aµ|u|q−1|ξ|m + b0(x), b0(x) ∈ L p
m
(G);
4)
∂h(x, u)
∂u
≤ 0,
∂g(x, u)
∂u
≤ 0;
5) σ(ω) ≥ ν0 ≥ 0 on σ0; γ(ω) ≥ ν0 ≥ 0 on ∂G.
In addition, suppose that the functions ai(x, u, ξ) are continuously differ-
entiable with respect to the u, ξ variables in Md,M0 = Gd0×[−M0,M0]×R
n
and satisfy in Md,M0
6) (m− 1)u∂ai(x,u,ξ)
∂u = q ∂ai(x,u,ξ)
∂ξj
ξj ; i = 1, . . . , n;
7)
∣∣∣ai(x, u, ux) cos(−→n , xi) − a|u|q|∇u|m−2 ∂u
∂−→n
∣∣∣
∣∣∣∣
Ωd
≤ k1|x|
s−1, |x| ≤ d.
Our main result about problem (QL) is the following statement.
M. Borsuk 493
Theorem 1.2. Let u be a weak solution of the problem (QL) and as-
sumptions 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 (NEV P )
(see Subsection 2.1). Suppose, in addition, that h(x, 0) ∈ L∞(Σ0),
g(x, 0) ∈ L∞(∂G) and there exist real numbers ks ≥ 0, K ≥ 0 such
that
ks =: sup
̺>0
̺1−n−s
{ ∫
G̺
0
r
q
q+m−1 |a0(x)|
m(q+m−1)
(m−1)(q+m) dx+
∫
G̺
0
r
1
m−1 |b0(x)|
m
m−1
+
∫
Σ̺
0
1
r
|h(x, 0)|
m
m−1 ds+
∫
Γ̺
0
1
r
|g(x, 0)|
m
m−1 ds
}
; (1.5)
K =: sup
̺>0
̺
n
m
−1
ψ(̺)
{
̺
m(1−n
p
) q+m−1
(m−1)(q+m) ‖a0‖
q+m−1
(m−1)(q+m)
p
m
, G̺
0
+ ̺
1−n
p ‖α(x)‖
1
m−1
p
m−1
, G̺
0
+ ̺
(1−n
p
) m
m−1 ‖b0(x)‖
1
m−1
p
m
, G̺
0
+ ̺
(
‖g(x, 0)‖
1
m−1
∞,Γ̺
0
+ ‖h(x, 0)‖
1
m−1
∞,Σ̺
0
)}
, (1.6)
where
ψ(̺) =
̺
ϑ
1
m (m)
C(m)
·
q+(m−1)(1−µ)
q+m−1 , s > ϑ
1
m (m)
C(m) · q+(m−1)(1−µ)
q+m−1 ;
̺
ϑ
1
m (m)
C(m)
·
q+(m−1)(1−µ)
q+m−1 ln d
̺ , s = ϑ
1
m (m)
C(m) · q+(m−1)(1−µ)
q+m−1 ;
̺s, s < ϑ
1
m (m)
C(m) · q+(m−1)(1−µ)
q+m−1 ,
(1.7)
C(m) = (m− 1)
m−1
m max{1; 2
m−2
2(m−1) }. (1.8)
Then there are d ∈ (0, 1) and a constant C0 > 0 independent of u such
that
|u(x)| ≤ C0
(
|x|1−
n
mψ(|x|)
) m−1
q+m−1 , ∀x ∈ Gd0. (1.9)
Moreover, if coefficients of the problem (QL) satisfy such conditions
which guarantee the local a-priori estimate |∇u|0,G′ ≤M1 for any smooth
G′ ⊂⊂ G \ {O} (see for example [1, §4] or [15]), then there is a constant
C1 > 0 independent of u such that
|∇u(x)| ≤ C1|x|
−
n(m−1)+qm
m(q+m−1) ψ
m−1
q+m−1 (|x|), ∀x ∈ Gd0. (1.10)
494 The transmission problem...
2. Preliminaries
2.1. The eigenvalue problem
Let Ω ⊂ Sn−1 with smooth boundary ∂Ω be the intersection of the
cone C with the unit sphere Sn−1. Let −→ν be the exterior normal to ∂C at
points of ∂Ω and −→τ be the exterior with respect 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 m-Laplace–Beltrami operator on
the unit sphere:
a
(
divω(|∇ωψ|
m−2∇ωψ) + ϑ|ψ|m−2ψ
)
= 0, ω ∈ Ω,
[ψ]σ0 = 0,
[
a|∇ωψ|
m−2 ∂ψ
∂−→τ
]
σ0
+ σ(ω)|ψ|m−2ψ
∣∣∣
σ0
= 0;
a|∇ωψ|
m−2 ∂ψ
∂−→ν
+ γ(ω)|ψ|m−2ψ
∣∣∣
∂Ω
= 0,
(NEV P )
which consists of the determination of all values ϑ (eigenvalues) for which
(NEV P ) has a non-zero weak solutions (eigenfunctions); here
a =
{
a+, x ∈ Ω+,
a−, x ∈ Ω−,
a± are positive constants.
Definition 2.1. Non-zero function ψ is called a weak eigenfunction of
the problem (NEV P ) provided that ψ ∈ C0(Ω) ∩ W1,m(Ω) and satisfies
the integral identity
∫
Ω
a
{
|∇ωψ|
m−2 1
qi
∂ψ
∂ωi
∂η
∂ωi
− ϑ|ψ|m−2ψη
}
dΩ
+
∫
σ0
σ(ω)|ψ|m−2ψη dσ +
∫
∂Ω
γ(ω)|ψ|m−2ψη dσ = 0
for all η(x) ∈ C0(Ω)∩W1,m(Ω); here: q1 = 1, qi = (sinω1 . . . sinωi−1)
2;
i ≥ 2.
Remark 2.1. We observe that ϑ = 0 is not an eigenvalue of (NEV P ).
In fact, setting η = ψ and ϑ = 0 we have
∫
Ω
a|∇ωψ|
m dΩ +
∫
σ0
σ(ω)|ψ|m dσ +
∫
∂Ω
γ(ω)|ψ|m dσ = 0 =⇒ ψ ≡ 0,
M. Borsuk 495
since a > 0, σ(ω) > 0, γ(ω) > 0.
We characterize the first eigenvalue ϑ(m) of the eigenvalue problem
for m-Laplacian by
ϑ(m) = inf
ψ∈W 1,m(Ω)
ψ 6=0
∫
Ω a|∇ωψ|
m dΩ +
∫
σ0
σ(ω)|ψ|m dσ +
∫
∂Ω γ(ω)|ψ|m dσ∫
Ω a|ψ|
m dΩ
.
(2.1)
Theorem 2.1. Let Ω ⊂ Sn−1 be a bounded 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 the eigenvalue
ϑ > 0 and the corresponding weak eigenfunction ψ.
Proof. The proof is similar to Theorem 8.20 [3].
Now from the variational principle we obtain
The Friedrichs–Wirtinger type inequality. Let ϑ be the smallest
positive eigenvalue of the problem (NEV P ) (it exists according to The-
orem 2.1). Let Ω ⊂ Sn−1. Let ψ ∈ W1,m(Ω) and satisfies the boundary
and conjunction conditions from (NEV P ) 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
∫
Ω
a|ψ|m dΩ ≤
1
ϑ
{ ∫
Ω
a|∇ωψ|
m dΩ +
∫
σ0
σ(ω)|ψ|m dσ +
∫
∂Ω
γ(ω)|ψ|m dσ
}
(W )m
with the sharp constant 1
ϑ .
Remark 2.2. In the case m = 2 we consider the value λ by (1.2); there-
fore the Friedrichs–Wirtinger inequality will be written in the following
form
λ(λ+ n− 2)
∫
Ω
aψ2(ω) dΩ ≤
∫
Ω
a|∇ωψ|
2 dΩ +
∫
σ0
σ(ω)ψ2(ω) dσ
+
∫
∂Ω
γ(ω)ψ2(ω) dσ, ∀ψ(ω) ∈ W1(Ω) (W )2
satisfying the boundary and conjunction conditions from (NEV P ) in the
weak sense; σ(ω) ≥ ν0 > 0, γ(ω) ≥ ν0 > 0.
496 The transmission problem...
Corollary 2.1. Let ϑ be the smallest positive eigenvalue of the problem
(NEV P ). Let v(x) ∈ W1,m(Gd0) and v(·, ω) satisfies the boundary and
conjunction conditions from (NEV P ) 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 for any ̺ ∈ (0, d) and ∀α
∫
G̺
0
arα|v|m dx ≤
1
ϑ
{ ∫
G̺
0
arα+m|∇v|m dx+
∫
Σ̺
0
rα+m σ(ω)
rm−1
|v|m ds
+
∫
Γ̺
0
rα+m γ(ω)
rm−1
|v|m ds
}
, (2.2)
provided that integrals on the right are finite. In particular,
∫
G̺
0
a|v|m dx ≤
̺m
ϑ(m)
{ ∫
G̺
0
a|∇v|m dx+
∫
Σ̺
0
σ(ω)
rm−1
|v|m ds
+
∫
Γ̺
0
γ(ω)
rm−1
|v|m ds
}
(H −W )m
as well
∫
G
a|v|m dx ≤ C(m,n, ϑ,G)
{∫
G
a|∇v|m dx+
∫
Σ0
σ(ω)
rm−1
|v|m ds
+
∫
∂G
γ(ω)
rm−1
|v|m ds
}
. (F −W )m
Proof. We consider the inequality (W )m for the function v(r, ω), multiply
it by rα+n−1 and integrate over r ∈ (0, ̺). Hence it follows the desired
(2.2). Setting α = 0 we get (H −W )m. The inequality (F −W )m is the
consequence of (H −W )m and the Poincaré inequality.
Corollary 2.2. Let v ∈ C0(G) ∩
◦
W1
α−2(G) and v(·, ω) satisfies the
boundary and conjunction conditions from (NEV P ) for m = 2 in the
weak sense. Let σ(ω), ω ∈ σ0; γ(ω), ω ∈ ∂Ω be positive bounded piece-
wise smooth functions with σ(ω) ≥ ν0 > 0, γ(ω) ≥ ν0 > 0. Then
∫
Gd
0
arα−4v2 dx ≤
1
λ(λ+ n− 2)
{ ∫
Gd
0
arα−2|∇v|2 dx
M. Borsuk 497
+
∫
Σd
0
rα−3σ(ω)v2(x) ds+
∫
Γd
0
rα−3γ(ω)v2(x) ds
}
, ∀α, (2.3)
where λ is defined by (1.2).
Proof. Multiplying (W )2 by rn−5+α and integrating over r ∈ (0, d) we
obtain the required (2.3).
2.2. One auxiliary integral inequality
Lemma 2.1. Let Gd0 be the conical domain, ∇v(̺, ·) ∈ W1,m(Ω) for
almost all ̺ ∈ (0, d) and
V (̺) =
∫
G̺
0
a|∇v|m dx+
∫
Σ̺
0
σ(ω)
rm−1
|v|m ds+
∫
Γ̺
0
γ(ω)
rm−1
|v|m ds <∞. (2.4)
Let ϑ(m) be the smallest positive eigenvalue of the problem (NEV P ) 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)
∫
Ω̺
av
∂v
∂r
|∇v|m−2dΩ̺ ≤ (m− 1)
m−1
m max{1; 2
m−2
2(m−1) } ·
̺
mϑ
1
m
V ′(̺). (2.5)
Proof. Writing V (̺) in spherical coordinates
V (̺) =
̺∫
0
rn−1
( ∫
Ω
a|∇v(r, ω)|m dΩ
)
dr
+
̺∫
0
rn−m−1
( ∫
∂Ω
γ(ω)|v(r, ω)|m dσ
)
dr
+
̺∫
0
rn−m−1
( ∫
σ0
σ(ω)|v(r, ω)|m dσ
)
dr
and differentiating with respect to ̺ we obtain
V ′(̺) = ̺n−1
∫
Ω
a|∇v(̺, ω)|m dΩ
+ ̺n−m−1
( ∫
∂Ω
γ(ω)|v(̺, ω)|m dσ +
∫
σ0
σ(ω)|v(̺, ω)|m dσ
)
.
498 The transmission problem...
Now using the Young inequality with p = m, p′ = m
m−1 we have
∫
Ω̺
av
∂v
∂r
|∇v|m−2 dΩ̺ = ̺n−1
∫
Ω
av
∂v
∂r
|∇v|m−2
∣∣∣∣
r=̺
dΩ
= ̺n
∫
Ω
a
(v
̺
)(∂v
∂r
|∇v|m−2
)∣∣∣∣
r=̺
dΩ ≤
≤ ̺n
∫
Ω
a
{ ε
m
(v
̺
)m
+
m− 1
m
ε−
1
m−1
(∂v
∂r
) m
m−1
|∇v|(m−2) m
m−1
}∣∣∣∣
r=̺
dΩ,
∀ ε > 0.
Further, applying the Friedrichs–Wirtinger type inequality (W )m we ob-
tain
∫
Ω̺
av
∂v
∂r
|∇v|m−2dΩ̺
≤
ε̺n
mϑ(m)
{ ∫
σ0
σ(ω)
∣∣∣v(̺, ω)
̺
∣∣∣
m
dσ +
∫
∂Ω
γ(ω)
∣∣∣v(̺, ω)
̺
∣∣∣
m
dσ
}
+
1
m
̺n
∫
Ω
a
{ ε
ϑ(m)
∣∣∣∇ωv
̺
∣∣∣
m
+(m−1)ε−
1
m−1
∣∣∣∂v
∂r
∣∣∣
m
m−1
|∇v|(m−2) m
m−1
}∣∣∣∣
r=̺
dΩ.
But, by |∇v|2 = v2
r + 1
r2
|∇ωv|
2 and
∣∣∇ωv
̺
∣∣ ≤ |∇v|, we get from above
∫
Ω̺
av
∂v
∂r
|∇v|m−2 dΩ̺
≤
ε̺n−m
mϑ(m)
{ ∫
σ0
σ(ω)|v(̺, ω)|m dσ +
∫
∂Ω
γ(ω)|v(̺, ω)|m dσ
}
+
1
m
̺n
∫
Ω
a|∇v|(m−2) m
m−1
{ ε
ϑ(m)
(∣∣∣∇ωv
̺
∣∣∣
2) m
2(m−1)
+ (m− 1)ε−
1
m−1 (v2
r )
m
2(m−1)
}∣∣∣∣
r=̺
dΩ.
Now we choose
ε = 〈(m− 1)ϑ(m)〉
m−1
m =⇒ (m− 1)ε−
1
m−1 =
ε
ϑ(m)
(2.6)
M. Borsuk 499
and therefore
∫
Ω̺
av
∂v
∂r
|∇v|m−2dΩ̺
≤
ε̺n−m
mϑ(m)
{ ∫
σ0
σ(ω)|v(̺, ω)|m dσ +
∫
∂Ω
γ(ω)|v(̺, ω)|m dσ
}
+
ε
mϑ(m)
̺n
∫
Ω
a|∇v|(m−2) m
m−1
{(∣∣∣∇ωv
̺
∣∣∣
2) m
2(m−1)
+ (v2
r )
m
2(m−1)
}∣∣∣∣
r=̺
dΩ.
Applying yet the Jensen inequality (see e.g. [7, Theorem 65]), by |∇v|2 =
v2
r + 1
r2
|∇ωv|
2, hence it follows
∫
Ω̺
av
∂v
∂r
|∇v|m−2dΩ̺
≤
ε̺n−m
mϑ(m)
{ ∫
σ0
σ(ω)|v(̺, ω)|m dσ +
∫
∂Ω
γ(ω)|v(̺, ω)|m dσ
}
+
ε
mϑ(m)
·
1
min{1; 2
m
2(m−1)
−1
}
̺n
∫
Ω
a|∇v|m
∣∣∣∣
r=̺
dΩ.
Substituting here ε from (2.6) we get the desired inequality (2.5).
In the case m = 2 we establish the more exact statement:
Lemma 2.2. Let Gd0 be the conical domain and ∇v(̺, ·) ∈ L2(Ω) a.e.
̺ ∈ (0, d). Assume that for a.e. ̺ ∈ (0, d)
V (ρ) =
∫
G̺
0
ar2−n|∇v|2 dx+
∫
Σ̺
0
r1−nσ(ω)v2(x) ds
+
∫
Γ̺
0
r1−nγ(ω)v2(x) ds <∞. (2.7)
Then ∫
Ω
a
(
̺v
∂v
∂r
+
n− 2
2
v2
)∣∣∣∣
r=̺
dΩ ≤
̺
2λ
V ′(̺), (2.8)
where λ is defined by (1.2).
Proof. The proof is similar to the [3, Lemma 2.35] proof, if we use in-
equality (W )2.
500 The transmission problem...
We need also the well known inequalities:
∫
Γ
v ds ≤ C
∫
G
(|v| + |∇v|) dx, ∀ v(x) ∈ W1,1(G), ∀Γ ⊆ ∂G, (2.9)
∫
∂G
v2 ds ≤
∫
G
(
δ|∇v|2 +
1
δ
c0v
2
)
dx, ∀ v(x) ∈ W1,2(G), ∀ δ > 0, (2.10)
where C, c0 depend only on n,G,Γ.
2.3. Quasi-distance rε(x)
Further, we define the function rε(x) as follows. We fix the point
Q = (−1, 0, . . . , 0) ∈ Sn−1 \ Ω̄ and consider the unit radius-vector ~l =
OQ = {−1, 0, . . . , 0}. We denote by ~r the radius-vector of the point
x ∈ Ḡ and introduce the vector ~rε = ~r − ε~l, ∀ ε > 0. Since ε~l /∈ Gd0 for
all ε ∈]0, d[, it follows that rε(x) = |~r − ε~l| 6= 0 for all x ∈ Ḡ. It is easy
to see that rε(x) has the following properties (see in detail [3, §1.4]):
1. ∃h > 0 such that: rε(x) ≥ hr and rε(x) ≥ hε, ∀x ∈ G, where
h =
{
1, if x1 ≥ 0 ,
sin ω0
2 , if x1 < 0.
2. If x ∈ Gd, then rε(x) ≥
d
2 for all ε ∈]0, d2 [.
3. limε→0+ rε(x) = r, for all x ∈ Ḡ.
4. |∇rε|
2 = 1, and △rε = n−1
rε
.
Lemma 2.3. Let v ∈ C0(G)∩W1(G) and v(·, ω) satisfies the boundary
and conjunction conditions from (EV P ) in the weak sense. Let σ(ω) ≥
σ0 > 0, γ(ω) ≥ γ0 > 0 and λ be as above in (1.2). Then for any ε > 0
∫
Gd
0
arα−2
ε r−2v2 dx ≤
1
λ(λ+ n− 2)
{ ∫
Gd
0
arα−2
ε |∇v|2 dx
+
∫
Σd
0
r−1rα−2
ε σ(ω)v2(x) ds+
∫
Γd
0
r−1rα−2
ε γ(ω)v2 ds
}
. (2.11)
Proof. Multiplying both sides of the Friedrichs–Wirtinger inequality
(W )2 by (̺+ ε)α−2rn−3 and integrating over r ∈ (̺2 , ̺) we obtain
M. Borsuk 501
∫
G̺
̺/2
a(̺+ ε)α−2r−2v2 dx
≤
1
λ(λ+ n− 2)
{ ∫
G̺
̺/2
a(̺+ ε)α−2|∇v|2 dx+
∫
Γ̺
̺/2
r−1(̺+ ε)α−2γ(ω)v2 ds
+
∫
Σ̺
̺/2
r−1(̺+ ε)α−2σ(ω)v2 ds
}
, ∀ ε > 0
or since ̺+ ε ∼ rε
∫
G̺
̺/2
arα−2
ε r−2v2 dx ≤
1
λ(λ+N − 2)
{ ∫
G̺
̺/2
arα−2
ε |∇v|2 dx
+
∫
Σ̺
̺/2
r−1rα−2
ε σ(ω)v2 ds+
∫
Γ̺
̺/2
r−1rα−2
ε γ(ω)v2 ds
}
, ∀ ε > 0.
Letting ρ = 2−kd, (k = 0, 1, 2, . . . ) and summing the obtained inequali-
ties over all k we get the desired inequality (2.11).
3. Local estimate at the boundary.
The maximum principle
Applying the Moser iteration method (see e.g. [6, §§ 8.6, 8.10] or [4])
we derive a result asserting the local boundedness (near the conical point)
of the weak solution of problem (QL).
Theorem 3.1. Let u(x) be a weak solution of the problem (QL). Let
assumptions 1), 2), 3a), 4), 5), 6), (1.6) be satisfied and, in addition,
h(x, 0) ∈ L∞(Σ0), g(x, 0) ∈ L∞(∂G). Then the inequality
sup
x∈Gκ̺
0
|u(x)| ≤
C
(1 − κ)nς/t
{
̺−nς/t‖u‖ t
ς
,G̺
0
+ ̺
mς(p−n)
p(m−1+ς) · ‖a0(x)‖
ς
m−1+ς
p
m
, G̺
0
+ ̺
ς(1−n
p
)
‖α(x)‖
ς
m−1
p
m−1
, G̺
0
+ ̺
ς(1−n
p
) m
m−1 ‖b0(x)‖
ς
m−1
p
m
, G̺
0
+ ̺ς
(
‖g(x, 0)‖
ς
m−1
∞,Γ̺
0
+ ‖h(x, 0)‖
ς
m−1
∞,Σ̺
0
)}
, p > n > m (3.1)
holds for any t > 0, κ ∈ (0, 1), ̺ ∈ (0, d) and ς = m−1
q+m−1 , where C =
C
(
n,m, p, t, q, a∗, a
∗,m∗,m
∗, d, ‖a0(x)‖ p
m
, G, ‖α(x)‖ p
m−1
, G, ‖b0(x)‖ p
m
, G
)
.
502 The transmission problem...
We consider also one in a possible case of the deriving L∞(G)−a
priori estimate of the weak solution to problem (QL).
Theorem 3.2. 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
(Σ0),
g(x, 0) ∈ L j
j−1
(∂G), 1 < j < n−1
m−1 . Then there exists the constant M0 >
0, depending only on measG, meas ∂G, meas Σ0, n, m, ‖b0(x)‖L p
m
(G),
‖h(x, 0)‖L j
j−1
(Σ0), ‖g(x, 0)‖L j
j−1
(∂G), µ, q, ‖a0(x)‖L p
m
(G), such that
‖u‖L∞(G) ≤M0.
Proof. The proof is similar to [3, Theorem 9.11].
4. Global integral estimates
Theorem 4.1. Let u(x) be a weak solution of the problem (QL). 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(Σ0), g(x, 0)∈ L1(∂G).
Then the inequality
∫
G
a|u|
qm
m−1 |∇u|m dx+
∫
Σ0
σ(ω)
rm−1
|u|
m
m−1
(q+m−1) ds
+
∫
∂G
γ(ω)
rm−1
|u|
m
m−1
(q+m−1) ds ≤ c(M0, a∗, ν0, q,m, µ, n,measG)
×
( ∫
G
(a0(x) + b0(x)) dx+
∫
Σ0
|h(x, 0)| ds+
∫
∂G
|g(x, 0)| ds
)
(4.1)
holds.
Proof. At first we make the function change
u = v|v|ς−1 with ς =
m− 1
q +m− 1
. (4.2)
By virtue of the assumption 6), the identity (II) takes the form:
∫
G
〈
Ai(x, vx)ηxi + B(x, v, vx)η
〉
dx
M. Borsuk 503
+
∫
∂G
γ(ω)
rm−1
v|v|m−2η(x) ds+
∫
Σ0
σ(ω)
rm−1
v|v|m−2η(x) ds
=
∫
∂G
G(x, v)η(x) ds+
∫
Σ0
H(x, v)η(x) ds (4.3)
and the identity (II)loc takes the form
∫
G̺
0
〈
Ai(x, vx)ηxi + B(x, v, vx)η
〉
dx+
∫
Γ̺
0
γ(ω)
rm−1
v|v|m−2η(x) ds
+
∫
Σ̺
0
σ(ω)
rm−1
v|v|m−2η(x) ds =
∫
Ω̺
Ai(x, vx) cos(r, xi)η(x) dΩ̺
+
∫
Γ̺
0
G(x, v)η(x) ds+
∫
Σ̺
0
H(x, v)η(x) ds (4.4)
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−1vx),
B(x, v, vx) ≡ b(x, v|v|ς−1, ς|v|ς−1vx),
G(x, v) ≡ g(x, v|v|ς−1), H(x, v) ≡ h(x, v|v|ς−1).
(4.5)
We verify that coefficients Ai, i = 1, . . . n do not depend on v explicit.
In fact, by the change (4.2) and the assumption 6), we calculate
∂Ai
∂v
=
∂ai(x, u, ξ)
∂u
·
∂
∂v
(
|v2|
ς−1
2 · v
)
+
∂ai(x, u, ξ)
∂ξj
· ςvxj
∂
∂v
(
|v2|
ς−1
2
)
= ς|v|ς−1∂ai
∂u
+ ς(ς − 1)vxjv|v|
ς−3 ·
∂ai
∂ξj
= ς ·
u
v
·
∂ai
∂u
+ (ς − 1) ·
ξj
v
·
∂ai
∂ξj
=
1
v
(
ςu·
∂ai
∂u
+(ς−1)·
m− 1
q
u
∂ai
∂u
)
=
u
v
·
∂ai
∂u
·
(
ς+(ς−1)·
m− 1
q
)
= 0,
because of (4.2).
Our assumptions regarding problem (QL) take the form:
1)′ Ai(x, vx)vxi ≥ aςm−1|∇v|m − 1
ς |v|
1−ςa0(x); a0(x) ∈ Lp/m(G);
2)′
√√√√
n∑
i=1
A2
i (x, vx) +
√√√√
n∑
i=1
∣∣∣∂Ai(x, vx)
∂xi
∣∣∣
2
≤ aςm−1|∇v|m−1 + α(x);
α(x) ∈ L p
m−1
(G);
504 The transmission problem...
3a)′ |B(x, v, vx)| ≤ aµςm|v|−1|∇v|m + b0(x); b0(x) ∈ L p
m
(G),
4)′
∂H(x, v)
∂v
≤ 0,
∂G(x, v)
∂v
≤ 0;
7)′
∣∣Ai(x, vx) cos(−→n , xi) − aςm−1|∇v|m−2 ∂v
∂−→n
∣∣
∣∣∣
Ωd
≤ k1|x|
s−1, |x| < d;
We consider integral identity (4.3) and we put η(x) = v(x). Then we
have
∫
G
〈
Ai(x, vx)vxi + B(x, v, vx)v
〉
dx
+
∫
∂G
γ(ω)
rm−1
|v|m ds+
∫
Σ0
σ(ω)
rm−1
|v|m ds
=
∫
∂G
G(x, v)v(x) ds+
∫
Σ0
H(x, v)v(x) ds.
With regard to assumptions 1)′, 3a)′, 5)′, since ςm−1(1−ςµ) < 1 by (4.2),
we obtain
ςm−1(1 − ςµ)
{ ∫
G
a|∇v|m dx+
∫
Σ0
σ(ω)
rm−1
|v|m ds+
∫
∂G
γ(ω)
rm−1
|v|m ds
}
≤
∫
G
|v|b0(x) dx+
1
ς
∫
G
|v|1−ςa0(x) dx
+
∫
Σ0
|h(x, 0)| · |v| ds+
∫
∂G
|g(x, 0)| · |v| ds.
From M0 = supG |u(x)|, by the change (4.2), it follows that |v(x)| ≤M
1
σ
0 .
Therefore we get
∫
G
a|∇v|m dx+
∫
Σ0
σ(ω)
rm−1
|v|m ds+
∫
∂G
γ(ω)
rm−1
|v|m ds
≤ c(M0,m, q, µ,measG) ·
( ∫
G
(a0(x) + b0(x)) dx+
∫
Σ0
|h(x, 0)| ds
+
∫
∂G
|g(x, 0)| ds
)
. (4.6)
M. Borsuk 505
Returning to the function u(x) by (4.2) we obtain the desired estimate
(4.1).
For the problem (WL) we obtain a global estimate for the weighted
Dirichlet integral.
Theorem 4.2. 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
b0(x) ∈
◦
W0
α(G),
∫
Σ0
rα−1h2(x, 0) ds <∞,
∫
∂G
rα−1g2(x, 0) ds <∞, 4 − n ≤ α ≤ 2.
Then |u(x)|q+1 ∈
◦
W1
α−2(G) and
∫
G
a
(
rα−2|u|2q|∇u|2 + rα−4|u|2(q+1)
)
dx
+
∫
Σ0
rα−3σ(ω)|u|2(q+1) ds+
∫
∂G
rα−3γ(ω)|u|2(q+1) ds
≤ C
{ ∫
G
(
|u|2(q+1) + (1 + rα)b20(x)
)
dx
+
∫
Σ0
rα−1h2(x, 0) ds+
∫
∂G
rα−1g2(x, 0) ds
}
, (4.7)
where the constant C > 0 depends only on a∗, α, λ, µ, q, n and the domain
G.
Proof. Making the function change (4.2) we consider the integral identity
for the function v(x) :
∫
G
〈
ςaij(x)vxjηxi + B(x, v, vx)η
〉
dx
+
∫
∂G
γ(ω)
r
vη(x) ds+
∫
Σ0
σ(ω)
r
vη(x) ds
=
∫
∂G
G(x, v)η(x) ds+
∫
Σ0
H(x, v)η(x) ds. (ĨI)
506 The transmission problem...
Setting in this identity η(x) = rα−2
ε v(x), with regard to
ηxi = rα−2
ε vxi + (α− 2)rα−3
ε
xi − εli
rε
v(x)
we obtain
ς
∫
G
arα−2
ε |∇v|2 dx+
∫
Σ0
r−1rα−2
ε σ(ω)v2(x) ds
+
∫
∂G
r−1rα−2
ε γ(ω)v2(x) ds = ς
2 − α
2
∫
G
arα−4
ε (xi − εli)(v
2)xi dx
+ ς(2 − α)
∫
G
(
aij(x) − aij(0)
)
rα−4
ε (xi − εli)vxjv(x) dx
− ς
∫
G
(
aij(x) − aij(0)
)
rα−2
ε vxivxj dx−
∫
G
B(x, v, vx)r
α−2
ε v(x) dx
+
∫
Σ0
rα−2
ε v(x)H(x, v) ds+
∫
∂G
rα−2
ε v(x)G(x, v) ds. (4.8)
We transform the first integral on the right by integrating by parts:
∫
G
arα−4
ε (xi − εli)
∂v2
∂xi
dx
=
∫
G+
a+r
α−4
ε (xi − εli)
∂v2
+
∂xi
dx+
∫
G−
a−r
α−4
ε (xi − εli)
∂v2
−
∂xi
dx
= −
∫
G
av2 ∂
∂xi
(
rα−4
ε (xi− εli)
)
dx+
∫
∂G+
a+v
2
+r
α−4
ε (xi− εli) cos(−→n , xi) ds
+
∫
∂G−
a−v
2
−r
α−4
ε (xi− εli) cos(−→n , xi) ds = −
∫
G
av2 ∂
∂xi
(
rα−4
ε (xi− εli)
)
dx
+
∫
∂G
av2rα−4
ε (xi − εli) cos(−→n , xi) ds
+ [a]Σ0
∫
Σ0
v2rα−4
ε (xi − εli) cos(−→n , xi) ds, (4.9)
because of [v]Σ0 = 0. By elementary calculation we have:
1) ∂
∂xi
(
rα−4
ε (xi − εli)
)
= nrα−4
ε + (α − 4)(xi − εli)r
α−5
ε
xi−εli
rε
=
(n+ α− 4)rα−4
ε ;
M. Borsuk 507
2) because of cos(−→n , xi)
∣∣
Σ0
= cos(xn, xi) = δni ,
(xi− εli) cos(−→n , xi)
∣∣
Σ0
= δni (xi− εli)
∣∣
Σ0
= (xn− εln)
∣∣
Σ0
= xn
∣∣
Σ0
= 0,
since Σ0 = {xn = 0} ∩G and ln = 0;
3) from the representation ∂G = Γd0 ∪Γd and (xi− εli) cos(−→n , xi)
∣∣
Γd
0
=
−ε sin ω0
2 =⇒
∫
∂G
av2rα−4
ε (xi − εli) cos(−→n , xi) ds = −ε sin
ω0
2
∫
Γd
0
av2rα−4
ε ds
+
∫
Γd
av2rα−4
ε (xi − εli) cos(−→n , xi) ds.
Hence and from (4.9) it follows
2 − α
2
∫
G
arα−4
ε (xi − εli)
∂v2
∂xi
dx
=
(2 − α)(4 − n− α)
2
∫
G
arα−4
ε v2 dx− ε
2 − α
2
sin
ω0
2
∫
Γd
0
av2rα−4
ε ds
+
2 − α
2
∫
Γd
av2rα−4
ε (xi − εli) cos(−→n , xi) ds (4.10)
From (4.8), (4.10) we obtain following equality:
ς
∫
G
arα−2
ε |∇v|2 dx+ ες
2 − α
2
sin
ω0
2
∫
Γd
0
av2rα−4
ε ds
+
∫
Σ0
r−1rα−2
ε σ(ω)v2(x) ds+
∫
∂G
r−1rα−2
ε γ(ω)v2(x) ds
=
2 − α
2
ς
∫
Γd
av2rα−4
ε (xi − εli) cos(−→n , xi) ds
+ ς
(2 − α)(4 − n− α)
2
∫
G
arα−4
ε v2 dx
+ ς(2 − α)
∫
G
(
aij(x) − aij(0)
)
rα−4
ε (xi − εli)vxjv(x) dx
508 The transmission problem...
− ς
∫
G
(
aij(x) − aij(0)
)
rα−2
ε vxivxj dx−
∫
G
B(x, v, vx)r
α−2
ε v(x) dx
+
∫
Σ0
rα−2
ε v(x)H(x, v) ds+
∫
∂G
rα−2
ε v(x)G(x, v) ds. (4.11)
Now we estimate the integral over Γd. Because on Γd : rε ≥ hr ≥ hd ⇒
(α− 3) ln rε ≤ (α− 3) ln(hd), since α ≤ 2, we have rα−3
ε |Γd
≤ (hd)α−3
and therefore:
2 − α
2
ς
∫
Γd
av2rα−4
ε (xi − εli) cos(−→n , xi) ds ≤
2 − α
2
ς
∫
Γd
arα−3
ε v2 ds
≤
2 − α
2
ς (hd)α−3
∫
Γd
av2 ds ≤ c
∫
Gd
(v2 + |∇v|2) dx, (4.12)
by (2.10). Now, in virtue of assumption (c) and the function change
(4.2), we have
|v · B(x, v, vx)| ≤ aµς2|∇v|2 + b0(x)|v|. (4.13)
Therefore using the Cauchy inequality we deduce the following
∫
G
B(x, v, vx)r
α−2
ε v(x) dx
≤ µς2
∫
G
arα−2
ε |∇v|2 dx+
∫
G
rα−2
ε |v|b0(x) dx
≤ µς2
∫
G
arα−2
ε |∇v|2 dx+
δ
2
∫
G
ar−2rα−2
ε v2 dx
+
1
2δa∗
∫
G
r2rα−2
ε b20(x) dx, ∀ δ > 0. (4.14)
Now we use the representation G = Gd0∪Gd. At first we estimate integrals
over Gd0. By assumption (b) and the Cauchy inequality, we obtain:
∫
Gd
0
(
aij(x) − aij(0)
)(
rα−2
ε vxivxj + rα−4
ε (xi − εli)v(x)vxj
)
dx
≤ A(d)
∫
Gd
0
a
(
rα−2
ε |∇v|2 + rα−3
ε |∇v| · |v(x)|
)
dx
M. Borsuk 509
≤
3
2
A(d)
∫
Gd
0
a
(
rα−2
ε |∇v|2 + rα−4
ε v2
)
dx. (4.15)
Now we estimate integrals over Gd. By assumptions (a), the Cauchy
inequality and taking into account that rε ≥ hd for r ≥ d, we obtain:
∫
Gd
(
aij(x) − aij(0)
)(
rα−2
ε vxivxj + rα−4
ε (xi − εli)v(x)vxj
)
dx
≤ A∗
∫
Gd
(3
2
rα−2
ε |∇v|2 + rα−4
ε |v|2
)
dx
≤ C(A∗, h, α, d)
∫
Gd
(
|∇v|2 + v2
)
dx. (4.16)
Next, in virtue of assumption (e),
vG(x, v) = vG(x, 0) + v2 ·
1∫
0
∂G(x, τv)
∂(τv)
dτ ≤ |g(x, 0)| · |v|. (4.17)
Further, by the Cauchy inequality and because of γ(ω) ≥ ν0 > 0,
|g(x, 0)| · |v| =
(
r
1
2
1√
γ(ω)
|g(x, 0)|
)(
r−
1
2
√
γ(ω)|v|
)
≤
δ
2
r−1γ(ω)v2 +
1
2δν0
rg2(x, 0), ∀ δ > 0;
taking into account that rε ≥ hr (see Subsection 2.3) we obtain
∫
∂G
rα−2
ε vG(x, v) ds ≤
δ
2
∫
∂G
rα−2
ε
1
r
γ(ω)v2 ds
+
1
2δν0
∫
∂G
rα−1g2(x, 0) ds, ∀ δ > 0. (4.18)
Similarly, because of σ(ω) ≥ ν0 > 0,
∫
Σ0
rα−2
ε vH(x, v) ds ≤
δ
2
∫
Σ0
rα−2
ε
1
r
σ(ω)v2 ds
+
1
2δν0
∫
Σ0
rα−1h2(x, 0) ds, ∀ δ > 0. (4.19)
510 The transmission problem...
Taking into account that ς ≤ 1 and 4 − n ≤ α ≤ 2 as a result from
(4.11)–(4.19) we obtain:
ς(1 − µς)
∫
G
arα−2
ε |∇v|2 dx+
∫
Σ0
1
r
rα−2
ε σ(ω)v2(x) ds
+
∫
∂G
1
r
rα−2
ε γ(ω)v2(x) ds ≤
3
2
A(d)
∫
Gd
0
a
(
rα−2
ε |∇v|2 + rα−4
ε v2
)
dx
+
δ
2
∫
G
ar−2rα−2
ε v2 dx+ C
∫
Gd
(
|∇v|2 + v2
)
dx+
1
2a∗δ
∫
G
rαb20(x) dx
+
1
2δν0
{ ∫
∂G
rα−1g2(x, 0) ds+
∫
Σ0
rα−1h2(x, 0) ds
}
+
δ
2
{ ∫
Σ0
rα−2
ε
1
r
σ(ω)v2ds+
∫
∂G
rα−2
ε
1
r
γ(ω)v2 ds
}
, ∀ δ > 0. (4.20)
In virtue of the inequality rε ≥ hr, we have rα−4
ε ≤ h−2r−2rα−2
ε . Hence,
by Lemma 2.3, from (4.20) it follows
ς(1 − µς)
{ ∫
G
arα−2
ε |∇v|2 dx+
∫
Σ0
1
r
rα−2
ε σ(ω)v2(x) ds
+
∫
∂G
1
r
rα−2
ε γ(ω)v2(x) ds
}
≤ c(λ, ω0) (δ + A(d))
{ ∫
G
arα−2
ε |∇v|2 dx
+
∫
Σ0
r−1rα−2
ε σ(ω)v2(x) ds+
∫
∂G
r−1rα−2
ε γ(ω)v2 ds
}
+ C
∫
G
(
|∇v|2 + v2
)
dx+
1
2a∗δ
∫
G
rαb20(x) dx
+
1
2δν0
∫
∂G
rα−1g2(x, 0) ds+
1
2δν0
∫
Σ0
rα−1h2(x, 0) ds,
∀ δ > 0, ∀ ε > 0. (4.21)
Because of 0 ≤ µ < 1+q, we can choose δ = 1
4c(λ,ω0)(1−µς) and next d >
0 such that, by the continuity of A(r) at zero, c(λ, ω0)A(d) ≤ 1
4(1 − µς)
Thus, from (4.21) we get
M. Borsuk 511
∫
G
arα−2
ε |∇v|2 dx+
∫
Σ0
1
r
rα−2
ε σ(ω)v2(x) ds+
∫
∂G
1
r
rα−2
ε γ(ω)v2(x) ds
≤ C(a∗, α, λ, µ, q, n, d)
{∫
G
(|∇v|2 + v2) dx+
∫
G
rαb20(x) dx
+
1
ν0
∫
∂G
rα−1g2(x, 0) ds+
1
ν0
∫
Σ0
rα−1h2(x, 0) ds
}
, ∀ ε > 0. (4.22)
We observe that the right hand side of (4.22) does not depend on ε.
Therefore we can perform the passage to the limit as ε → +0 by the
Fatou Theorem. Hence it follows that v(x) ∈
◦
W1
α−2(G) and
∫
G
arα−2|∇v|2 dx+
∫
Σ0
rα−3σ(ω)v2(x) ds+
∫
∂G
rα−3γ(ω)v2(x) ds
≤ C(a∗, α, λ, µ, q, n, d)
{∫
G
(|∇v|2 + v2) dx+
∫
G
rαb20(x) dx
+
1
ν0
∫
∂G
rα−1g2(x, 0) ds+
1
ν0
∫
Σ0
rα−1h2(x, 0) ds
}
. (4.23)
Further, returning to the integral identity (ĨI) and setting in it η(x) =
v(x), we get
∫
G
〈
ςaij(x)vxjvxi + B(x, v, vx)v
〉
dx+
∫
∂G
γ(ω)
r
v2 ds+
∫
Σ0
σ(ω)
r
v2 ds
=
∫
∂G
G(x, v)v ds+
∫
Σ0
H(x, v)v ds.
By the ellipticity condition (a), inequalities (4.13), (4.18)–(4.19) for α =
2, δ = 2, hence it follows
ς(1 − µς)
∫
G
a|∇v|2 dx
≤ c(a∗, ν0,diamG)
{ ∫
G
(
|v|2 + b20(x)
)
dx+
∫
∂G
rα−1g2(x, 0) ds
+
∫
Σ0
rα−1h2(x, 0) ds
}
. (4.24)
512 The transmission problem...
Now, using the inequality (2.3) and returning to the function u(x), by
means of the function change (4.2), from (4.23)–(4.24) we get the desired
estimate (4.7).
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 (QL) and
M0 = maxx∈G |u(x)| be known. Let ϑ(m) be the smallest positive eigen-
value of (NEV P ). Let assumptions of Theorem 4.1 and 7) be satisfied.
Suppose, in addition, that there exists real number ks ≥ 0 defined by
(1.5). Then there are d ∈ (0, 1) and a constant c > 0 independent of u
and depending only on m,n, s, q, d, ϑ(m), k1, ks,measΩ and M0 such that
for any ̺ ∈ (0, d)
∫
G̺
0
a|u|
qm
m−1 |∇u|m dx+
∫
Σ̺
0
σ(ω)
rm−1
|u|
m
m−1
(q+m−1) ds
+
∫
Γ̺
0
γ(ω)
rm−1
|u|
m
m−1
(q+m−1) ds ≤ cψm(̺) (5.1)
where ψ(̺) is defined by (1.7)–(1.8).
Proof. By virtue of Theorem 4.1 (see (4.6)), we have that
V (̺) =
∫
G̺
0
a|∇v|m dx+
∫
Σ̺
0
σ(ω)
rm−1
|v|m ds
+
∫
Γ̺
0
γ(ω)
rm−1
|v|m ds <∞, ̺ ∈ (0, d). (5.2)
Therefore we can set η(x) = v(x) in the identity (4.4) after the making
the change (4.2)):
∫
G̺
0
〈
Ai(x, vx)vxi + B(x, v, vx)v(x)
〉
dx+
∫
Σ̺
0
σ(ω)
rm−1
|v|m ds
+
∫
Γ̺
0
γ(ω)
rm−1
|v|m ds =
∫
Ω̺
Ai(x, vx) cos(r, xi) · v(x) dΩ̺
M. Borsuk 513
+
∫
Γ̺
0
G(x, v) · v(x) ds+
∫
Σ̺
0
H(x, v) · v(x) ds. (5.3)
By assumptions 1)′, 3a)′, 4)′, 5), 7)′ and since ςµ < 1, we obtain
(1 − ςµ)ςm−1V (̺) ≤
∫
G̺
0
|v|b0(x) dx+
1
ς
∫
G̺
0
|v|1−ςa0(x) dx
+ k1
∫
Ω̺
rs−1|v| dΩ̺ + ςm−1
∫
Ω̺
a|∇v|m−2 · v
∂v
∂r
dΩ̺
+
∫
Σ̺
0
|h(x, 0)| · |v| ds+
∫
Γ̺
0
|g(x, 0)| · |v| ds. (5.4)
Applying Lemma 2.1 from (5.4) it follows that
(1 − ςµ)ςm−1V (̺)
≤ ςm−1(m− 1)
m−1
m max{1; 2
m−2
2(m−1) } ·
̺
mϑ
1
m (m)
V ′(̺)
+
1
ς
∫
G̺
0
|v|1−ςa0(x) dx+
∫
G̺
0
|v|b0(x) dx+ k1̺
s−1
∫
Ω̺
|v| dΩ̺
+
∫
Σ̺
0
|h(x, 0)| · |v| ds+
∫
Γ̺
0
|g(x, 0)| · |v| ds. (5.5)
Further, by the Young inequality and inequality (2.2),
1
ς
∫
G̺
0
|v|1−ςa0(x) dx =
1
ς
∫
G̺
0
(
r
(1−ς)(1−ς−m)
m |v|1−ς
)
·
(
r
(1−ς)(m+ς−1)
m a0(x)
)
dx
≤
∫
G̺
0
(1 − ς
mς
r1−ς−m|v|m +
m+ ς − 1
mς
r1−ς |a0(x)|
m
m+ς−1
)
dx
≤
1 − ς
a∗mςϑ(m)
̺1−ςV (̺) +
m+ ς − 1
mς
∫
G̺
0
r1−ς |a0(x)|
m
m+ς−1 dx; (5.6)
∫
G̺
0
|v|b0(x) dx =
∫
G̺
0
(
r−
1
m |v|
)(
r
1
m b0(x)
)
dx
514 The transmission problem...
≤
∫
G̺
0
( 1
m
r−1|v|m +
m− 1
m
r
1
m−1 |b0(x)|
m
m−1
)
dx
≤
1
a∗mϑ(m)
̺m−1V (̺) +
m
m− 1
∫
G̺
0
r
1
m−1 |b0(x)|
m
m−1 dx
=
1
a∗mϑ(m)
̺m−1V (̺) +
m
m− 1
∫
G̺
0
r
1
m−1 |b0(x)|
m
m−1 dx. (5.7)
Now we estimate the integral over Ω̺ on the right in (5.5). By the Young
inequality with ∀ δ > 0 and inequality (H −W )m,
k1̺
s−1
∫
Ω̺
|v| dΩ̺ ≤ c̃k1̺
s−1
∫
G̺
0
(|v| + |∇v|) dx
≤ δ̺s−1
∫
G̺
0
(|v|m + |∇v|m) dx+ c(k1, c̃,m, δ) · meas Ω · ̺s+n−1
≤
δ
a∗
̺s−1
(
1 +
̺m
ϑ(m)
)
V (̺) + c(k1, c̃,m, δ) · meas Ω · ̺s−1+n
≤ ̺s−1V (̺) + c(k1, c̃,m, a∗, ϑ(m)meas Ω) · ̺s−1+n, (5.8)
if we choose δ > 0 in a reasonable way.
Further, by the Young inequality with ∀ δ > 0, we have
∫
Σ̺
0
|v| · |h(x, 0)| ds ≤
δ
m
∫
Σ̺
0
|v|m ds+
m− 1
m
δ
1
1−m
∫
Σ̺
0
|h(x, 0)|
m
m−1 ds.
Applying again inequality (H −W )m we obtain
∫
Σ̺
0
|v|m ds ≤ c̃
∫
G̺
0
(
|v|m +m|v|m−1|∇v|
)
dx
≤ c̃
∫
G̺
0
(m|v|m + |∇v|m) dx
=⇒
δ
m
∫
Σ̺
0
|v|m ds ≤ δc̃
∫
G̺
0
|v|m dx+
δ
m
c̃
∫
G̺
0
|∇v|m ds
≤
δ
a∗
c̃ ·
( ̺m
ϑ(m)
+
1
m
)
V (̺).
M. Borsuk 515
Therefore, if we choose δ = ̺m−1, hence it follows that ∀ ε > 0
∫
Σ̺
0
|h(x, 0)| · |v| ds ≤ c1̺
m−1V (̺) +
m− 1
m
∫
Σ̺
0
1
r
|h(x, 0)|
m
m−1 ds. (5.9)
where c1 = c̃
a∗
( 1
ϑ(m) + 1
m). In the same way ∀ ε > 0
∫
Γ̺
0
|g(x, 0)| · |v| ds ≤ c1̺
m−1V (̺) +
m− 1
m
∫
Γ̺
0
1
r
|g(x, 0)|
m
m−1 ds. (5.10)
Thus, from (5.5)–(5.10) with regard to (1.8) it follows that
(1 − ςµ) (1 − δ(̺))V (̺) ≤
̺C(m)
mϑ
1
m (m)
V ′(̺)
+
m+ ς − 1
mςm
∫
G̺
0
r1−ς |a0(x)|
m
m+ς−1 dx+ ς1−m
m
m− 1
∫
G̺
0
r
1
m−1 |b0(x)|
m
m−1 dx
+c∗̺s−1+n+ ς1−m
m− 1
m
( ∫
Σ̺
0
1
r
|h(x, 0)|
m
m−1 ds+
∫
Γ̺
0
1
r
|g(x, 0)|
m
m−1 ds
)
,
(5.11)
where δ(̺) = const(m, a∗, q, ϑ(m)) ·
(
̺ + ̺s−1 + ̺m−1 + q
q+m−1̺
q
q+m−1
)
,
s > 1, c∗ = const(k1, c̃,m, q, a∗, ϑ(m),meas Ω).We observe that
∫
0
δ(̺)
̺ d̺
<∞. Thus, from (5.11) in virtue of assumption (1.5) we have the Cauchy
problem for the differential inequality:
{
V ′(̺) − P(̺)V (̺) + Q(̺) ≥ 0, 0 < ̺ < d,
V (d) ≤ V0,
(CP)
where P(̺) = 1−δ(̺)
̺ ·mϑ
1
m (m)
C(m) (1−ςµ), Q(̺) = k̺s+n−2; k = const(ks,m,
q, ϑ(m)). We estimate now:
V (d) =
∫
Gd
0
a|∇v|m dx+
∫
Σd
0
σ(ω)
rm−1
|v|m ds+
∫
Γd
0
γ(ω)
rm−1
|v|m ds
≤
∫
G
a|∇v|m dx+
∫
Σ0
σ(ω)
rm−1
|v|m ds+
∫
∂G
γ(ω)
rm−1
|v|m ds
≤ c(M0,m, q, µ,measG) ·
( ∫
G
(a0(x) + b0(x)) dx+
∫
Σ0
|h(x, 0)| ds
516 The transmission problem...
+
∫
∂G
|g(x, 0)| ds+ 1
)
≡ V0 (5.12)
in virtue of (4.6) The solution of problem (CP ) is the following inequality
V (̺) ≤ V0 exp
(
−
d∫
̺
P(τ) dτ
)
+
d∫
̺
Q(τ) exp
(
−
τ∫
̺
P(ξ) dξ
)
dτ. (5.13)
(see [3, Theorem 1.57, §1.10]).
Direct calculations give:
−
τ∫
̺
P(ξ) dξ = −
mϑ
1
m (m)(1 − ςµ)
C(m)
τ∫
̺
1 − δ(ξ)
ξ
dξ
≤
mϑ
1
m (m)(1 − ςµ)
C(m)
(
ln
̺
τ
+
d∫
0
δ(ξ)
ξ
dξ
)
=⇒ exp
(
−
τ∫
̺
P(ξ) dξ
)
≤
(̺
τ
)mϑ
1
m (m)(1−ςµ)
C(m)
· exp
( d∫
0
δ(ξ)
ξ
dξ
)
;
d∫
̺
Q(τ) exp
(
−
τ∫
̺
P(ξ) dξ
)
dτ
≤ k exp
( d∫
0
δ(ξ)
ξ
)
̺
mϑ
1
m (m)(1−ςµ)
C(m)
d∫
̺
τms−1τ
−
mϑ
1
m (m)(1−ςµ)
C(m) dτ
= k exp
( d∫
0
δ(ξ)
ξ
)
̺
mϑ
1
m (m)(1−ςµ)
C(m)
×
d
m
(
s−
ϑ
1
m (m)(1−ςµ)
C(m)
)
−̺
m
(
s−
ϑ
1
m (m)(1−ςµ)
C(m)
)
m
(
s−
ϑ
1
m (m)(1−ςµ)
C(m)
) , s 6= ϑ
1
m (m)(1−ςµ)
C(m) ;
ln d
̺ , s = ϑ
1
m (m)(1−ςµ)
C(m) .
V0 · exp
(
−
d∫
̺
P(ξ) dξ
)
≤ V0
(̺
d
)mϑ
1
m (m)(1−ςµ)
C(m)
· exp
( d∫
0
δ(ξ)
ξ
dξ
)
M. Borsuk 517
=⇒
V (̺) ≤ c · exp
( d∫
0
δ(ξ)
ξ
dξ
)
· (V0 + k)
×
̺
mϑ
1
m (m)(1−ςµ)
C(m) , s > ϑ
1
m (m)(1−ςµ)
C(m) ;
̺
mϑ
1
m (m)(1−ςµ)
C(m) ln d
̺ , s = ϑ
1
m (m)(1−ςµ)
C(m) ;
̺ms, s < ϑ
1
m (m)(1−ςµ)
C(m) ,
(5.14)
where c = const(m, s, ϑ(m)). Thus we proved the statement of our theo-
rem.
For problem (WL) we establish the estimate with best possible expo-
nents.
Theorem 5.2. Let u(x) be a weak solution of the problem (WL) and λ
be as above in (1.2). Let assumptions (a)–(f), be satisfied with A(r) that
is Dini-continuous at zero.
Then |u(x)|q+1 ∈
◦
W1
2−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 on
∫ 1
0
A(r)
r dr such
that ∀ ̺ ∈ (0, d)
∫
G̺
0
a
(
r2−n|u|2q|∇u|2 + r−n|u|2(q+1)
)
dx
+
∫
Σ̺
0
r1−nσ(ω)|u|2(q+1) ds+
∫
Γ̺
0
r1−nγ(ω)|u|2(q+1) ds
≤ C
(∫
G
|u|2(q+1) dx+ f2
1 +
1
ν0
g2
1 +
1
ν0
h2
1
)
×
̺2λ(1−µς), if s > λ(1 − µς),
̺2λ(1−µς) ln3
(
1
̺
)
, if s = λ(1 − µς),
̺2s, if s < λ(1 − µς),
(5.15)
where ς = 1
1+q .
Proof. The proof is similar to above. We use sharp inequality (2.8) and
therefore we obtain the estimate with best possible exponents. See also
proofs of Theorems 4.18, 7.21, 10.39 [3].
518 The transmission problem...
6. The power modulus of continuity
at the conical point for weak solutions
Proof of Theorems 1.1, 1.2. We consider the function ψ(̺), 0 < ̺ < d
that is determined by (1.7). By Theorem 3.1 (about the local bound of
the weak solution modulus) with t = m for |v| = |u|
1
ς , we have
sup
x∈G
̺/2
0
|v(x)|m ≤ C
{
̺−n‖v‖mm,G̺
0
+Km(̺)
}
, (6.1)
K(̺) = ̺
m(p−n)
p(m−1+ς) · ‖a0(x)‖
1
m−1+ς
p
m
,G̺
0
+ ̺
(1−n
p
) m
m−1 ‖b0(x)‖
1
m−1
p
m
,G̺
0
+ ̺
1−n
p ‖α(x)‖
1
m−1
p
m−1
,G̺
0
+ ̺
(
‖g(x, 0)‖
1
m−1
∞,Γ̺
0
+ ‖h(x, 0)‖
1
m−1
∞,Σ̺
0
)
,
p > n > m. (6.2)
Hence, in virtue of inequality (H − W )m with regard to the notation
(2.4), we get
̺−n
∫
G̺
0
a|v(x)|m dx ≤
̺m−n
ϑ(m)
V (̺) ≤ C̺m−nψm(̺), (6.3)
by inequality (5.14). From (6.1)–(6.3) it follows
sup
x∈G
̺/2
0
|v(x)| ≤ C
{
̺1− n
mψ(̺) +K(̺)
}
, (6.4)
Now, in virtue of assumption (1.6), it follows K(̺) ≤ K̺1− n
mψ(̺).
Therefore hence and from (6.4) we get |v(x)| ≤ C0̺
1− n
mψ(̺), x ∈ G
̺/2
0 .
Putting |x| = 1
3 we obtain
|v(x)| ≤ C0|x|
1− n
mψ(|x|), x ∈ Gd0. (6.5)
Finally, because of (4.2), from (6.5) we establish the first desired estimate
(1.9).
Repeating verbatim the Theorem 10.35 [3] proof we obtain the in-
equality
|∇v(x)| ≤ C1|x|
− n
mψ(|x|), x ∈ Gd0. (6.6)
But since, by (4.2), |∇u(x)| ≤ |v|ς−1|∇v(x)| from (6.5)–(6.6) we establish
the second desired estimate (1.10).
The Theorem 1.1 proof is similar.
M. Borsuk 519
7. Example
Here we consider two dimensional transmission problem for the Lapla-
ce operator with absorbtion term in an angular domain and investigate
the corresponding eigenvalue problem. Suppose n = 2, the domain G lies
inside the corner
G0 =
{
(r, ω) |r > 0; −
ω0
2
< ω <
ω0
2
}
, ω0 ∈]0, 2π[;
O ∈ ∂G and in some neighborhood of O the boundary ∂G coincides with
the sides of the corner ω = −ω0
2 and ω = ω0
2 . We denote
Γ± =
{
(r, ω) | r > 0; ω = ±
ω0
2
}
, Σ0 = {(r, ω) | r > 0; ω = 0}
and we put σ(ω)|Σ0 = σ(0) = σ = const > 0, γ(ω)|ω=±
ω0
2
= γ± =
const > 0. We consider the following problem:
d
dxi
(|u|quxi) = a0r
−2u|u|q − µu|u|q−2|∇u|2, x ∈ G0 \ Σ0;
[u]Σ0
= 0,
[
a|u|q ∂u∂n
]
Σ0
+ 1
|x|σ(0)u|u|q = 0, x ∈ Σ0;
α±a±|u±|
q ∂u±
∂n + 1
|x|γ±u±|u±|
q = 0, x ∈ Γ± \ O
(AL)
where a0 ≥ 0, 0 ≤ µ < 1 + q, q ≥ 0; α± ∈ {0; 1}; a± > 0. We make the
function change (4.2) and consider our problem for the function v(x) :
△v + µςv−1|∇v|2 = a0(1 + q)r−2v; ς = 1
1+q , x ∈ G0 \ Σ0;
[v]Σ0
= 0,
[
a ∂v∂n
]
Σ0
+ (1 + q)σ(0)v(x)|x| = 0, x ∈ Σ0;
α±a±
∂v±
∂n + (1 + q)γ±
v±(x)
|x| = 0, x ∈ Γ± \ O.
We want find the exact solution of this problem in the form v(r, ω) =
rκψ(ω). For ψ(ω) we obtain the problem
ψ′′(ω) + µς
ψ(ω)ψ
′2(ω) +
{
(1 + µς)κ2 − a0(1 + q)
}
· ψ(ω) = 0,
ω ∈
(
−ω0
2 , 0
)
∪
(
0, ω0
2
)
;
[ψ]ω=0 = 0, [aψ′(0)] = (1 + q)σ(0)ψ(0);
±α±a±ψ
′
±
(
±ω0
2
)
+ (1 + q)γ±ψ±
(
±ω0
2
)
= 0.
520 The transmission problem...
We assume that κ
2 > a0
(1+q)2
1+q+µ and define the value
Υ =
√
κ2 − a0
(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(Υω) +B± sin(Υω), (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− a+Υ ·B+ + a−Υ ·B− = 0;{
(1 + q)γ+ cos
(
Υ ω0
2
)
− α+a+Υ sin
(
Υ ω0
2
) }
·A+
{
(1 + q)γ+ sin
(
Υ ω0
2
)
+ α+a+Υ cos
(
Υ ω0
2
) }
·B+ = 0;{
(1 + q)γ− cos
(
Υ ω0
2
)
− α−a−Υ sin
(
Υ ω0
2
) }
·A−
{
(1 + q)γ− sin
(
Υ ω0
2
)
+ α−a−Υ cos
(
Υ ω0
2
) }
·B− = 0.
The Dirichlet problem: α± = 0, γ± 6= 0.
Direct calculations will give
ψ±(ω) = cos(Υω) ∓ cot
(
Υ
ω0
2
)
· sin(Υω), Υ =
{
π
ω0
, if σ(0) = 0;
Υ ∗, if σ(0) 6= 0,
where Υ ∗ is the least positive root of the transcendence equation
Υ · cot
(
Υ
ω0
2
)
= −
1 + q
a+ + a−
σ(0)
and from the graphic solution we obtain π
ω0
< Υ ∗ < 2π
ω0
.
The corresponding eigenfunctions
ψ±(ω) =
{
cos
(
πω
ω0
)
, if σ(0) = 0;
cos(Υ ∗ω) ∓ cot
(
Υ ∗ ω0
2
)
· sin(Υ ∗ω), if σ(0) 6= 0.
The Neumann problem: α± = 1, γ± = 0.
Direct calculations will give:
Υ =
{
π
ω0
, if σ(0) = 0;
Υ ∗, if σ(0) 6= 0,
M. Borsuk 521
where Υ ∗ is the least positive root of the transcendence equation Υ ·
tan
(
Υ ω0
2
)
= 1+q
a++a−
σ(0) and from the graphic solution we obtain 0 <
Υ ∗ < π
ω0
. The corresponding eigenfunctions
ψ±(ω) =
{
a∓ sin
(
πω
ω0
)
, if σ(0) = 0;
cos(Υ ∗ω) ± tan
(
Υ ∗ ω0
2
)
· sin(Υ ∗ω), 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 posi-
tive root of the transcendence equation
a+ tan
(
Υ
ω0
2
)
− a− cot
(
Υ
ω0
2
)
=
1 + q
Υ
σ(0).
The corresponding eigenfunctions
ψ+(ω) = cos(Υ ∗ω) + tan
(
Υ ∗ω0
2
)
· sin(Υ ∗ω), ω ∈
[
0,
ω0
2
]
;
ψ−(ω) = cos(Υ ∗ω) + cot
(
Υ ∗ω0
2
)
· sin(Υ ∗ω), ω ∈
[
−
ω0
2
, 0
]
.
The Robin problem: α± = 1, γ± 6= 0.
Direct calculations of the above system will give:
1) γ+
γ−
= a+
a−
=⇒ ψ±(ω) = a∓ sin(Υ ∗ω), where Υ ∗ is the least positive
root of the transcendence equation
Υ · cot
(
Υ
ω0
2
)
= −(1 + q)
γ+
a+
and from the graphic solution we obtain π
ω0
< Υ ∗ < 2π
ω0
.
2) γ+
γ−
6= a+
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 : By setting y(ω) = ψ ′(ω)
ψ(ω) , we arrive at the problem for y(ω)
y′ + (1 + µς)y2(ω) + (1 + µς)κ2 − a0(1 + q) = 0,
ω ∈
(
−ω0
2 , 0
)
∪
(
0, ω0
2
)
;
a+y+(0) − a−y−(0) = (1 + q)σ(0);
±α±a±y±
(
±ω0
2
)
+ (1 + q)γ± = 0.
522 The transmission problem...
Integrating the equation of our problem we find
y±(ω) = Υ tan {Υ (C± − (1 + µς)ω)} , ∀C±. (7.3)
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 κ :
a+ ·
α+a+Υ tan
{
(1 + µς)Υ ω0
2
}
− (1 + q)γ+
α+a+Υ + (1 + q)γ+ tan
{
(1 + µς)Υ ω0
2
}
+ a− ·
α−a−Υ tan
{
(1 + µς)Υ ω0
2
}
− (1 + q)γ−
α−a−Υ + (1 + q)γ− tan
{
(1 + µς)Υ ω0
2
} =
1 + q
Υ
σ(0), (7.5)
where 1 + µς = 1+q+µ
1+q . Further, from (7.3) and (7.4) we obtain
y±(ω) = Υ tan
{
Υ
1 + q + µ
1 + q
(
±
ω0
2
− ω
)
∓ arctan
(1 + q)γ±
α±a±Υ
}
(7.6)
and, because of (lnψ(ω))′ = y(ω), hence it follows
ψ±(ω) = cos
1+q
1+q+µ
{
Υ
1 + q + µ
1 + q
(
±
ω0
2
− ω
)
∓ arctan
(1 + q)γ±
α±a±Υ
}
.
(7.7)
At last, returning to the function u, by (4.2), we establish a solution of
(AL)
u±(r, ω) = r
κ
1+q cos
1
1+q+µ
{
Υ
1 + q + µ
1 + q
(
±
ω0
2
− ω
)
∓ arctan
(1 + q)γ±
α±a±Υ
}
,
(7.8)
where Υ is defined by (7.1) and κ 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)
u(r, ω) = rλ̃ cos
1
1+q+µ
(
πω
ω0
)
; λ̃ =
√
(π/ω0)2 + a0(1 + q + µ)
1 + q + µ
.
It is well known result (see [2, Example 4.6, p. 374]).
Now we can verify that the derived exact solution satisfies the estimate
(1.3) of Theorem 1.1. In fact, in our case we have: the value λ for (1.2)
is equal ϑ = π
ω0
and therefore
|u(r, ω)| ≤ rλ̃ ≤ r
π
ω0
· 1
1+q+µ ≤ r
π
ω0
· 1+q−µ
(1+q)2 ,
M. Borsuk 523
since a0 ≥ 0 and 1
1+q+µ ≥ 1+q−µ
(1+q)2
.
References
[1] M. V. Borsuk, A priori estimates and solvability of second order quasilinear el-
liptic equations in a composite domain with nonlinear boundary conditions and
conjunction condition // Proc. Steklov Inst. of Math. 103 (1970), 13–51.
[2] M. Borsuk, The behavior of weak solutions to the boundary value problems for el-
liptic quasilinear equations with triple degeneration in a neighborhood of a bound-
ary edge // Nonlinear Analysis: Theory, Methods and Applications 56 (2004),
N 3, 347–384.
[3] M. Borsuk, V. Kondratiev, Elliptic Boundary Value Problems of Second Order
in Piecewise Smooth Domains. North-Holland Mathematical Library, 69, ELSE-
VIER, 2006, 531 p.
[4] Ya-Zhe Chen, Lan-Cheng Wu, Second order elliptic equations and elliptic systems.
Translated of Mathematical Monographs, 174, 1998. AMS, Providence, Rhode
Island, 246 p.
[5] W. Chikouche, D. Mercier and S. Nicaise, Regularity of the solution of some
unilateral boundary value problems in polygonal and polyhedral domains // Com-
munications in partial differential equations, 29 (2004), N 1&2, 43–70.
[6] D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of Sec-
ond Order. Springer-Verlag, Berlin/Heidelberg/New York, 1977. Revised Third
Printing, 1998.
[7] G. H. Hardy, J. E. Littlewood and G. Pólya, Inequalities. 1934; University Press,
Cambridge, 1952.
[8] D. Kapanadze, B.-W. Schulze, Boundary-contact problems for domains with con-
ical singularities // Journal of Differential Equations, 217 (2005), N 2, 456–500.
[9] D. Kapanadze, B.-W. Schulze, Boundary-contact problems for domains with edge
singularities // Journal of Differential Equations, 234 (2007), 26–53.
[10] D. Knees, On the regularity of weak solutions of nonlinear transmission problems
on polyhedral domains // Zeitschrift fur Analysis und ihre Anwendungen, 23
(2004), N 3, 509–546 (Preprint).
[11] N. Kutev, P. L. Lions, Nonlinear second-order elliptic equations with jump dis-
continuous coefficients. Part I: Quasilinear equations // Differential and Integral
Equations, 5 (1992), N 6, 1201–1217.
[12] Gary M. Lieberman, Boundary regularity for solutions of degenerate elliptic equa-
tions // Nonlineare Analysis. 12 (1988), N 11, 1203–1219.
[13] S. Nicaise, Polygonal Interface Problems. Peter Lang, 1993, 250 p. (Methoden
und Verfahren der mathematischen Physik; Bd. 39).
[14] S. Nicaise, A.-M. Sändig, General interface problems I, II // Math. Meth. Appl.
Sci, 17 (1994), N 6, 395–450.
[15] V. Ja. Rivkind, N. N. Ural’tseva, The classical solvability and linear schemes
of the approximate solution of diffraction problems for quasilinear equations of
parabolic and elliptic types // Problems of Math. Analysis, Leningrad. Univ., 3
(1972), 69–110.
524 The transmission problem...
Contact information
Mikhail Borsuk Department of Mathematics
and Informatics
University of Warmia and
Mazury in Olsztyn
10-957 Olsztyn-Kortowo,
Poland
E-Mail: borsuk@uwm.edu.pl
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