The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges
In this paper we deals with the mixed boundary value problem for secondorder elliptic equations in a polyhedral domain. We obtain exact estimates for solutions of the problem in a neighbourhood of an vertex. A special section is dedicated to the examples.
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| Zitieren: | The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges / M. Plesha // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 230-244. — Бібліогр.: 8 назв. — англ. |
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irk-123456789-1242642017-09-24T03:03:09Z The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges Plesha, M. In this paper we deals with the mixed boundary value problem for secondorder elliptic equations in a polyhedral domain. We obtain exact estimates for solutions of the problem in a neighbourhood of an vertex. A special section is dedicated to the examples. 2008 Article The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges / M. Plesha // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 230-244. — Бібліогр.: 8 назв. — англ. 0236-0497 MSC (2000): 35J25, 35J55, 35C15, 35Q72 http://dspace.nbuv.gov.ua/handle/123456789/124264 en Інститут прикладної математики і механіки НАН України |
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In this paper we deals with the mixed boundary value problem for secondorder elliptic equations in a polyhedral domain. We obtain exact estimates for solutions of the problem in a neighbourhood of an vertex. A special section is dedicated to the examples. |
| format |
Article |
| author |
Plesha, M. |
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Plesha, M. The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
| author_facet |
Plesha, M. |
| author_sort |
Plesha, M. |
| title |
The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
| title_short |
The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
| title_full |
The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
| title_fullStr |
The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
| title_full_unstemmed |
The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
| title_sort |
behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges |
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Інститут прикладної математики і механіки НАН України |
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2008 |
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http://dspace.nbuv.gov.ua/handle/123456789/124264 |
| citation_txt |
The behavior of solutions of the mixed boundary value problem for a linear second-order elliptic equation in a neighbourhood of intersecting edges / M. Plesha // Нелинейные граничные задачи. — 2008. — Т. 18. — С. 230-244. — Бібліогр.: 8 назв. — англ. |
| work_keys_str_mv |
AT plesham thebehaviorofsolutionsofthemixedboundaryvalueproblemforalinearsecondorderellipticequationinaneighbourhoodofintersectingedges AT plesham behaviorofsolutionsofthemixedboundaryvalueproblemforalinearsecondorderellipticequationinaneighbourhoodofintersectingedges |
| first_indexed |
2025-07-09T01:09:12Z |
| last_indexed |
2025-07-09T01:09:12Z |
| _version_ |
1837129641313173504 |
| fulltext |
230 Нелинейные граничные задачи 18, 230-244 (2008)
c©2008. M. Plesha
THE BEHAVIOR OF SOLUTIONS OF THE MIXED
BOUNDARY VALUE PROBLEM FOR A LINEAR
SECOND-ORDER ELLIPTIC EQUATION
IN A NEIGHBOURHOOD OF INTERSECTING EDGES
In this paper we deals with the mixed boundary value problem for second-
order elliptic equations in a polyhedral domain. We obtain exact estimates for
solutions of the problem in a neighbourhood of an vertex. A special section is
dedicated to the examples.
Keywords and phrases: second-order elliptic equations, mixed boundary
value problem, nonsmooth domains
MSC (2000): 35J25, 35J55, 35C15, 35Q72
1. Introduction.
Let G ⊂ R
3 be a bounded domain. We consider the problem
with mixed boundary conditions
Lu := aij(x)uxixj
+ ai(x)uxi
+ a(x)u = f(x), x ∈ G,
∂u
∂~n
= 0, x ∈ Γ1,
u(x) = 0, x ∈ ∂G \ Γ1.
(1.1)
(1.2)
(1.3)
Here and throughout summation from 1 to 3 over repeated indices
is assumed. The problem with boundary conditions (1.2)–(1.3), also
known as the Zaremba problem. The main purpose of this paper is to
analyze the behavior of solutions, in the case when the boundary of G
contain singular points having the type of the vertex of a polyhedron.
The assumptions on the coefficients of the equation are essential
for obtaining sharp estimates of the modulus of a solution. Elliptic
boundary problems in nonsmooth domains have been studied in many
works. In particular, the exact solution estimates for boundary value
problems in domains with angular and conical points at the boundary
were obtained in [1]. Mixed boundary problem with conormal deri-
vative was studied in [6] and in [7]. There the authors have obtained
Schauder and weighted Lp estimates of solutions for equations with
constant coefficients in polyhedral domains. The vast bibliography of
elliptic boundary problems in nonsmooth domains was compiled by
authors of [1].
The behavior of solutions in a neighbourhood of intersecting edges 231
2. Basic symbols, definitions and assumptions.
Let us introduce the following notations: x = (x1, x2, x3): an
element of R
3; (r, ω) = (r, ω1, ω2): spherical coordinates in R
3, defined
by: x1 = r cosω1 sinω2, x2 = r sinω1 sinω2, x3 = r cosω2; ~n: exterior
unit normal vector on ∂G; δj
i : Kronecker’s delta; the quasi-distance
function rε(x) :=
(
3
∑
i=1
(xi + δ3
i ε)
2
)1/2
; ux :=
(
3
∑
i=1
u2
xi
)1/2
; uxx :=
(
3
∑
i,j=1
u2
xixj
)1/2
. On the unit sphere S2 we consider the domain Ω =
{ω : ω1 ∈ [0, $1], ω2 ∈ [0, $2]}, where $1 ∈ (0, 2π), $2 ∈ (0, π). Our
domain G coincides in some neighbourhood of the boundary point
O1 (point of origin) with the domain {(r, ω) : ω ∈ Ω}. ∂G =
3
∪
i=1
Γi,
Γ1 coincides in some neighbourhood of O1 with the set {(r, ω) : ω ∈
∂Ω, ω1 = 0}, Γ2 with the set {(r, ω) : ω ∈ ∂Ω, ω1 = $1} and Γ3
with the set {(r, ω) : ∈ ∂Ω ω2 = $2}. `i are edges of boundary,
`1 := Γ1∩Γ2, `2 := Γ2∩Γ3, `3 := Γ2∩Γ3, ` :=
3
∪
i=1
`i. Edges `i intersect
at the vertices O :=
3
∩
i=1
`i of the polyhedron. We suppose that O =
{O1,O2} and analyze the behavior in a neighbourhood of O1. Γi are
smooth surfaces everywhere except at O. We introduce the notation
Gb
a := {x ∈ G : 0 ≤ a < r < b}. We denote the following spaces.
Ck(G): the Banach space of functions having all the derivatives of
order at most k (k ∈ N) continuous in G; Ck
0 (G): the set of functions
in Ck(G) with compact support in G; Lp(G): the space of functions
whose absolute value raised to the p-th power (p ≥ 1) has a finite
Lebesgue integral; W k,p(G): Sobolev space, is defined to be the subset
of Lp(G) such that function and its weak derivatives up to some order
k (k ∈ N) have a finite Lp norm, for given p ≥ 1; W k,p
0 (G): is the
closure of C∞
0 (G) with respect to the norm ||·||W k,p(G)||; W k,p
loc (G): the
space of functions that belong to W k,p(G′), for all G′ ⊂ G; V 2
2,α(G):
weighted Sobolev space, is defined as the closure of C∞
0 (G \O1) with
respect to the norm
||u||V 2
2,α(G) = ||rα/2uxx||L2(G) + ||rα/2−1ux||L2(G) + ||rα/2−2u||L2(G),
where α ∈ R.
Definition 2.1. A (strong) solution of the problem (1.1)–(1.3) in
232 M. Plesha
domain G is a function u ∈ W 2,2
loc (G \O)∩C0(G), which satisfies the
equations (1.1) for almost x ∈ G, boundary condition (1.2) in the
sense of traces and the boundary condition (1.3) for all x ∈ Γ2 ∪ Γ3.
In the following we assume that the coefficients aij(x), ai(x) and
a(x) satisfy the following conditions.
Assumption 2.2. The uniform ellipticity condition
ν|ξ|2 ≤ aij(x)ξiξj ≤ µ|ξ|2, ∀ξ ∈ R
3, x ∈ G.
with some ν, µ > 0.
Assumption 2.3. aij(0) = δj
i .
Assumption 2.4. aij ∈ C0(G), ai ∈ Lp(G) and a ∈ Lp/2(G), where
p > 3.
Assumption 2.5. There exists a monotonically increasing nonnega-
tive function A such that
(
n
∑
i,j=1
|aij(x) − aij(y)|2
)1/2
≤ A(|x− y|),
r
(
3
∑
i=1
a2
i (x)
)1/2
+ r2|a(x)| ≤ A(r),
for x, y ∈ G.
3. Some auxiliary assertions.
We denote by G̃ := {x : ω1 ∈ [0, 2π], ω2 ∈ [0, $0]}}, $0 ∈ (0, π).
Obviously, there exists an $0, such that G ⊂ G̃. It is easy to see
(see lemma 1.11 [1], lemma 1.4.1 [2]) that rε(x) has the following
properties.
Lemma 3.1. There exists an h > 0 such that
rε(x) ≥ h · r, rε(x) ≥ h · ε
for all x ∈ G, where h = 1 if $0 ∈ (0, π/2] and h = sin($0) if
$0 ∈ (π/2, π).
The behavior of solutions in a neighbourhood of intersecting edges 233
We consider the problem of the eigenvalues for the Laplace-Beltrami
operator ∆ω on the unit sphere.
∆ωv + ϑv = 0, ω ∈ Ω,
∂v
∂ω1
= 0, ω1 = 0,
v($1, ω2) = v(ω1, 0) = v(ω1, $2) = 0.
(3.1)
(3.2)
(3.3)
According to the variational principle of eigenvalues we have the
Wirtinger inequality (see 2.3.1, 2.4.6 [1], 2.2.1 [2]).
Theorem 3.2. (the Wirtinger inequality). The following inequality
is valid for all v ∈ W 2,2(Ω), that satisfies (3.2)–(3.3) in the sense of
traces
∫
Ω
v2(ω)dΩ ≤ 1
ϑ0
∫
Ω
(
(
∂v
∂ω1
)2
+
1
sin2 ω1
(
∂v
∂ω2
)2
)
dΩ,
where ϑ0 is the smallest positive eigenvalue of the problem (3.1)–
(3.3).
Let us define the value
λ =
−1 +
√
1 + 4ϑ0
2
, (3.4)
where ϑ0 is the smallest positive eigenvalue of the problem (3.1)–
(3.3). Next theorem follows from the Wirtinger inequality (see 2.5.2–
2.5.9, corollary 2.29 [1], 2.3.2–2.3.9, corollary 2.3.6 [2]).
Theorem 3.3. Let v ∈ W 2,2(Gd
0) satisfy the boundary value condition
(1.2)–(1.3) in the sense of traces. Then following estimates are held
∫
Gd
ε
rα−4v2dx ≤ 1
λ(λ+ 1)
∫
Gd
ε
rα−2v2
xdx, α ∈ R (3.5)
∫
Gd
ε
rα−4v2dx ≤ H(λ, α)
∫
Gd
ε
rα−2v2
xdx, α ≤ 1, (3.6)
where H(λ, α) = ((1 − α)2/4 + λ(λ+ 1))
−1
, ε ∈ [0, d] i
∫
Gd
0
rα−2
ε r−2v2dx ≤
(
3
h
)2−α
1
λ(λ+ 1)
∫
Gd
0
rα−2
ε v2
xdx, α ∈ R, (3.7)
234 M. Plesha
where h is a number from the lemma 0.0, also, if V (ρ) :=
∫
Gρ
0
r−1v2
xdx <
∞, then
∫
Ω
(
ρv
∂v
∂r
+
v2
2
)
∣
∣
∣
∣
r=ρ
dΩ ≤ ρ
2λ
V ′(ρ), ρ ∈ (0, d). (3.8)
4. Integral estimates.
At first, we will obtain a local integral estimate in the neighbour-
hood of an edge.
Lemma 4.1. Let u(x) be a solution of (1.1)–(1.3). Suppose that
lim
r→+0
A(r) = 0 and that f ∈ L2(G). Then there are d > 0 and constant
c > 0 depends only on ν, µ, α, λ, max
x∈G
A(|x|) and G, such that
||uxx||L2(G2d
d ) ≤ c
(
||ux||L2(G3d
d/2
) + ||u||L2(G3d
d/2
) + ||f ||L2(G3d
d/2
)
)
. (4.1)
Proof. Let us introduce the function v(x) = u(x)η(x), where η(x) ∈
C2(G2d
0 ) is a cutoff function such that: η(x) ≡ 1 if r(x) ∈ [d, 2d],
0 ≤ η(x) ≤ 1 if r(x) ∈ (d/2, d) ∪ (2d, 3d) and η(x) ≡ 0 when r(x) ∈
[0, d/2] ∪ [3d,∞). Then the function v satisfies the equation
aij(x)vxixj
+ ai(x)vxi
+ a(x)v = f1(x), (4.2)
where f1 = fη + aij(2uxi
ηxj
+ uηxixj
) + aiuηxi
. Since aij(0) = δj
i , we
have
∆v = f1(x)− (aij(x)−aij(0))vxixj
−ai(x)vxi
−a(x)v := f2(x). (4.3)
For the equation (4.3) we use (7.19) [5] (f2 ∈ V 0
2,0(G
3d
d/2)), applying it
for the domains G3d
d/2 with edges on the boundary
||vxx||L2(G3d
d/2
) ≤ c1||∆v||L2(G3d
d/2
).
Using the assumption (0.0) we obtain
||vxx||L2(G3d
d/2
) ≤ c2 · A2(3d)||vxx||L2(G3d
d/2
)+
+c3
(
||vx||L2(G3d
d/2
) + ||v||L2(G3d
d/2
) + ||f1||L2(G3d
d/2
)
)
.
The behavior of solutions in a neighbourhood of intersecting edges 235
Now, let d > 0 chosen according to the inequality c2 · A2(3d) < 1,
then from properties of the cutoff function we obtain (4.1).
Theorem 4.2. Let u(x) be a solution of (1.1)–(1.3) and λ be defined
by (3.4). Suppose that lim
r→+0
A(r) = 0, f ∈ V 0
2,α(G), where α ∈ (1 −
2λ, 2]. Then u ∈ V 2
2,α(Gd
0) and
||u||V 2
2,α(Gd
0
) ≤ c
(
||u||V 0
2,α(Gd
0
) + ||ux||V 0
2,α(Gd
0
) + ||f ||V 0
2,α(Gd
0
)
)
, (4.4)
where c > 0 depends only on ν, µ, α, λ, max
x∈G
A(|x|) i G.
Proof. Let us introduce the function v(x) = u(x)η(x), where η(x) ∈
C2(G2d
0 ) is a cutoff function such that: η(x) ≡ 1 if r(x) ∈ [0, d],
0 ≤ η(x) ≤ 1 if r(x) ∈ (d, 2d) and η(x) ≡ 0 when r(x) ∈ [2d,∞).
Case I: 1 ≤ α ≤ 2. We multiply both parts of the (4.3) by rα−2v(x)
and integrate over the domain G2d
ε . Twice integrating by parts we
obtain the analog of (4.3.6) [1] (see also (4.2.6) [2])
εα−2
∫
Ωε
v
∂v
∂r
dΩε +
∫
G2d
ε
rα−2v2
xdx+
(2 − α)εα−3
2
∫
Ωε
v2dΩε+
+
(2 − α)(α− 1)
2
∫
G2d
ε
rα−4v2dx =
=
2 − α
2
∫
Γ1∩∂G2d
ε
rα−4v2xi cos(~n, xi)dσ+
(4.5)
+
∫
G2d
ε
rα−2v(−f1(x) + (aij(x) − aij(0))vxixj
+ ai(x)vxi
+ a(x)v)dx.
where dσ area element of Γ1. Since xi cos(~n, xi) = x2 cos(~n, x2) =
−x2 = 0, ∀x ∈ Γ1 ∩ ∂G2d
ε , therefore
∫
Γ1∩∂G2d
ε
rα−4v2xi cos(~n, xi)dσ = 0. (4.6)
Let us estimate in the above equation the integrals over Ωε. We
consider the set G2ε
ε and we have Ωε ⊂ ∂G2ε
ε . Now we use the
inequality (6.23) [4]
∫
Ωε
|w|dΩε ≤ c1
∫
G2ε
ε
(|w| + |wx|)dx. (4.7)
236 M. Plesha
Setting w = v
∂v
∂r
we find (see (4.3.8) [1], (4.2.8) [2])
∫
Ωε
∣
∣
∣
∣
v
∂v
∂r
∣
∣
∣
∣
dΩε ≤ c2
∫
G2ε
ε
(
r2v2
xx + v2
x + r−2v2
)
dx. (4.8)
Twice using (3.5) we obtain
∫
G2ε
ε
(v2
x + r−2v2dx) dx ≤ c3
∫
G2ε
ε
v2
xdx ≤
≤ 4c3ε
2
∫
G2ε
ε
r−2v2
xdx ≤ c4ε
2
∫
G2ε
ε
v2
xxdx ≤ c5
∫
G2ε
ε
r2v2
xxdx,
therefore from (4.8) we get
∫
Ωε
∣
∣
∣
∣
v
∂v
∂r
∣
∣
∣
∣
dΩε ≤ c6
∫
G2ε
ε
r2v2
xxdx.
Applying the local integral estimate (4.1) we obtain
∫
Ωε
∣
∣
∣
∣
v
∂v
∂r
∣
∣
∣
∣
dΩε ≤
≤ c6
∫
G2ε
ε
r2vxxdx ≤ c7
∫
G3ε
ε/2
r2(v2
x + v2 + f 2
1 )dx ≤
≤ c8ε
2−α
∫
G3ε
ε/2
rα(v2
x + v2 + f 2
1 )dx ≤
≤ c8ε
2−α
∫
G2d
ε/2
rα(v2
x + v2 + f 2
1 )dx.
(4.9)
The behavior of solutions in a neighbourhood of intersecting edges 237
Let us apply again (4.7), in analogy to (4.9) we have
∫
Ωε
v2dΩε ≤
≤ c9
∫
G2ε
ε
(v2 + |v||vx|)dx ≤ c10
∫
G2ε
ε
(rv2
x + r−1v2)dx ≤
≤ c11
∫
G2ε
ε
r3v2
xxdx ≤ c12ε
3−α
∫
G3ε
ε/2
rα(v2
x + v2 + f 2
1 )dx ≤
≤ c12ε
3−α
∫
G2d
ε/2
rα(v2
x + v2 + f 2
1 )dx.
(4.10)
Writing the inequality (4.1) for the ρ ∈ (0, d) and taking into account
that ρ ∼ r in G2ρ
ρ , we obtain
∫
G2ρ
ρ
rαv2
xxdx ≤ c13
∫
G3ρ
ρ/2
rα(v2
x + v2 + f 2
1 )dx.
We replace ρ by 2−kd. Summing up this inequalities for k = 0, 1, . . . ,
[log2(d/ε)] + 1, we get
∫
G2d
ε
rαv2
xxdx ≤ c14
∫
G2d
ε/4
rα(v2
x + v2 + f 2
1 )dx. (4.11)
Applying assumption (0.0) together with the Hölder and the Cauchy
inequality
rα−2
ε v
(
(aij(x) − δj
i )vxixj
+ ai(x)vxi
+ a(x)v
)
≤
≤ c15A(r)
(
rα−2
ε r2v2
xx + rα−2
ε v2
x + rα−2
ε r−2v2
)
rα−2
ε vf1 ≤
δ
2
rα−2
ε r−2v2 + c16r
αf 2
1 ,
(4.12)
for all δ > 0, ε ≥ 0. Let ε = 0. From (4.5), (4.6) and (4.9)–(4.12)
238 M. Plesha
follows that
∫
G2d
ε
rαv2
xxdx +
∫
G2d
ε
rα−2v2
xdx +
(2 − α)(α− 1)
2
∫
G2d
ε
rα−4v2dx ≤
≤ c17
∫
G2d
ε
(
A(2d)(rα−2v2
x + rα−4v2) + δrα−4v2
)
dx+
+c18
∫
G2d
ε/4
rα(v2
x + v2 + f 2
1 )dx,
for all δ > 0 and 0 < ε < d. Furthermore, if (2− α)(α− 1) = 0, then
we apply the inequality(3.5). Now, let δ > 0, d > 0 are small enough.
Then we obtain
∫
G2d
ε
(rαv2
xx + rα−2v2
x + rα−4v2)dx ≤ c19
∫
G2d
ε/4
rα(v2
x + v2 + f 2
1 )dx,
where the constants c19 do not depend on ε. Letting ε → +0 we
obtain the assertion of our theorem in the case I.
Case II: 1 − 2λ < α < 1, α ≥ 0. From the inequality (4.1) we have
∫
G2ρ
ρ
ρ2(ρ+ ε)α−2v2
xxdx ≤ c20
∫
G3ρ
ρ/2
ρ2(ρ + ε)α−2(v2
x + v2 + f 2
1 )dx.
Since rε ≤ r + ε ≤ 2rε/h in G, we obtain
∫
G2ρ
ρ
r2rα−2
ε v2
xxdx ≤ c21
∫
G3ρ
ρ/2
r2rα−2
ε (v2
x + v2 + f 2
1 )dx.
Let ρ = 2−kd. Summing up this inequalities for k = 0, 1, . . . , we
finally obtain
∫
G2d
0
r2rα−2
ε v2
xxdx ≤ c22
∫
G2d
0
r2rα−2
ε (v2
x + v2 + f 2
1 )dx. (4.13)
Multiplying both sides of (4.3) by rα−2
ε v(x) and integrating by parts
The behavior of solutions in a neighbourhood of intersecting edges 239
twice we obtain (compare with case I)
∫
G2d
0
rα−2
ε v2
xdx =
(2 − α)(1 − α)
2
∫
G2d
0
rα−4
ε v2dx+
+
2 − α
2
∫
Γ1∩∂G2d
ε
rα−4
ε v2(xi + εδ3
i ) cos(~n, xi)dσ+
+
∫
G2d
0
rα−2
ε v(−f1(x) + (aij(x) − aij(0))vxixj
+ ai(x)vxi
+ a(x)v)dx,
where dσ area element of Γ1. The second integral on the right is equal
to zero (see (4.6)). Therefore from (4.12) for ε = ε we get
∫
G2d
0
rα−2
ε v2
xdx ≤ (2 − α)(1 − α)
2
∫
G2d
0
rα−4
ε v2dx+
+c23
∫
G2d
0
(
rα−2
ε
(
δr−2v2 + A(2d)(r2v2
xx + v2
x + r−2v2)
)
+ rαf 2
1
)
dx.
Since by case I u ∈ V 2
2,2(G
d
0) and f1 ∈ V 0
2,α(Gd
0) (α ≥ 0) the integral
from the right side is finite. Therefore from (3.6), (3.7) and (4.13) we
have
C(λ, α)
∫
G2d
0
rα−2
ε v2
xdx ≤
≤ c24
∫
G2d
0
(
rα−2
ε
(
(A(2d) + δ)v2
x + r2(v2
x + v2 + f1)
)
+ rαf 2
1
)
dx,
where C(λ, α) = 1 − 1
2
(2 − α)(1 − α)H(λ, α) > 0. Choosing δ > 0
and d > 0 small enough and passing to the limits as ε → 0, by the
Fatou Theorem we obtain the assertion, if we recall (3.6) and (4.13).
Case III: 1− 2λ < α < 1, α < 0. We take any α0 ∈ [max(−2, α), 0].
Then we have u, ux, f1 ∈ V 0
2,α0+2(G
d
0). Now, we can repeat verbatim
the proof of case II. We get u ∈ V 2
2,α0
(Gd
0) and (4.4). Repeating the
stated process k times we obtain u ∈ V 2
2,αk
(Gd
0), where αk = αk−1−2.
Obviously, we can find such an integer k that αk+1 ≤ α ≤ αk. Finally,
repeating the proof of case II once again, we obtain the assertion.
Corollary 4.3. Let u(x) be a solution of (1.1)–(1.3). Suppose that
240 M. Plesha
lim
r→+0
A(r) = 0 and f ∈ L2(G2d
0 ). Then
||u||V 2
2,0(G
d
0
) ≤ c
(
||u||L2(G2d
0
) + ||f ||L2(G2d
0
)
)
. (4.14)
Proof. Let us fix d > 0, such that the inequality (4.4) would be
fulfilled. We take ρ ∈ (0, d/2) and ς ∈ (0, 1). Let us introduce the
cutoff function η ∈ C2(G2ρ
0 ), such that η(x) ≡ 1 if r(x) ∈ [0, ςρ],
0 ≤ η(x) ≤ 1 if r(x) ∈ (ςρ, ς ′ρ), η(x) ≡ 0 when r(x) ∈ [ς ′ρ,∞),
|ηx| ≤ 4/((1 − ς)ε), |ηxx| ≤ 16/((1 − ς)2ε2), where ς ′ = (1 + ς)/2.
Now, if v = ηu we apply the estimate (4.4) to the solution v of the
(4.2) with α = 0
||uxx||L2(Gςρ
0
) ≤ c1
(
||u||
L2(Gς′ρ
0
)
+ ||ux||L2(Gς′ρ
0
)
+ ||f1||L2(Gς′ρ
0
)
)
=
= c1
(
||u||
L2(Gς′ρ
0
)
+ ||ux||L2(Gς′ρ
0
)
+
+||aij(2uxi
ηxj
+ uηxixj
) + aiuηxi
+ fη||
L2(Gς′ρ
0
)
)
≤
≤ c2
(
||f ||
L2(Gς′ρ
0
)
+
1
(1 − ς)ρ
||ux + r−1u||
L2(Gς′ρ
0
)
+
+
1
(1 − ς)2ρ2
||u||
L2(Gς′ρ
0
)
)
.
Rewriting this inequality in the form
sup
0<ς<1
(1 − ς)2ρ2||uxx||L2(Gςρ
0
) ≤
≤ c3
(
ρ2||f ||L2(Gρ
0
) + sup
0<ς<1
(1 − ς)ρ||ux||L2(Gς′ρ
0
)
+ sup
0<ς<1
||u||
L2(Gς′ρ
0
)
)
=
= c3
(
ρ2||f ||L2(Gρ
0
)+
+2 sup
1/2<ς′<1
(1 − ς ′)ρ||ux||L2(Gς′ρ
0
)
+ 2 sup
1/2<ς′<1
||u||
L2(Gς′ρ
0
)
)
≤
≤ c4
(
ρ2||f ||L2(Gρ
0
) + sup
0<ς<1
(1 − ς)ρ||ux||L2(Gςρ
0
) + sup
0<ς<1
||u||L2(Gςρ
0
)
)
≤
from the interpolation inequality (see (7.61), example 7.19 [3])
≤ c5
(
ρ2||f ||L2(Gρ
0
) + sup
0<ς<1
(1 − ς)ρ
(
ε(1 − ς)ρ||ux||L2(Gςρ
0
)+
+ε−1(1 − ς)−1ρ−1||u||L2(Gςρ
0
)
)
+ sup
0<ς<1
||u||L2(Gςρ
0
)
)
.
The behavior of solutions in a neighbourhood of intersecting edges 241
Hence, choosing ε > 0 sufficiently small, we can write
||uxx||L2(Gςρ
0
) ≤
c6
(1 − ς)2ρ2
(
||f ||L2(Gρ
0
) + ||u||L2(Gςρ
0
)
)
.
Taking ς = 1/2 and using (3.5), we arrive to the sought estimate
(4.14).
Theorem 4.4. Let u(x) be a strong solution of problem (1.1)–(1.3)
and assumptions (2.2)–(0.0) are satisfied with A(r) Dini continuous
at zero. Suppose, in addition f ∈ V 0
2,1(G) and there exist real numbers
s > 0, ks ≥ 0 such that ks = sup
ρ>0
ρ−s||f ||V 0
2,1(G
ρ
0
). Then there are d > 0
and a constant c > 0 depends only on ν, µ, A(d), s, λ, G and on the
quantity
d
∫
0
t−1A(t)dt such that ∀ρ ∈ (0, d)
||u||V 2
2,1(G
ρ
0
) ≤ c
(
||u||L2(G) + ||f ||V 0
2,1(G) + ks
)
·
ρλ, s > λ,
ρλ ln3/2(1/ρ), s = λ,
ρs, s < λ.
(4.15)
Proof. We consider the equation (4.3) with η ≡ 1 (v ≡ u). Let us
now multiply both parts of the (4.3) by r−1u and integrate over Gρ
0;
twice having applied the formula of integration by parts. As a result
we have
∫
Ω
(
ρu∂u
∂r
+ u2
2
)
dΩ −
∫
Gρ
0
r−1u2
xdx =
=
∫
Gρ
0
r−1u
(
(aij(x) − aij(0))uxixj
+ ai(x)uxi
+ a(x)u)
)
dx.
Let U(ρ) :=
∫
Gρ
0
r−1u2
xdx. From the assumption (0.0), estimates (3.6),
(3.8) and the Cauchy inequality we obtain for ∀δ > 0
U(ρ) ≤ ρ
2λ
V ′(ρ) + c1A(ρ)
∫
Gρ
0
ru2
xxdx+
+c2A(ρ)U(ρ) + δ
2
U(ρ) + 1
2δ
||f ||2
V 0
2,1(G
ρ
0
)
If we take into account (4.4) and condition on the function f , we get
U(ρ) ≤ ρ
2λ
U ′(ρ) + c3A(ρ)U(2ρ)+
+c4(A(ρ) + δ)U(ρ) + c5
1
δ
k2
sρ
2s, ∀δ > 0, ρ ∈ (0, d).
(4.16)
242 M. Plesha
Moreover, because of (4.14) in virtue of the obvious embedding
V 0
2,0(G
ρ
0) ⊂
→
V 0
2,1(G
ρ
0), we have the initial condition U(d) ≡ U0 < ∞.
The estimate (4.15) follow from (4.16), in the same way as (4.3.43)
[1] from (4.3.47) [1] (see also (4.2.43) and (4.2.47) [2]).
5. The estimate of the solution modulus.
Theorem 5.1. Let u(x) be a strong solution of problem (1.1)–(1.3)
and let the assumptions of theorem 4.4 be satisfied. Then there are
d > 0 and a constant c > 0, depends on the same values as constant
c in the theorem 4.4, such that for ∀x ∈ Gd
0
|u(x)|≤c
(
||u||L2(G) + ||f ||V 0
2,1(G) + ks
)
·
rλ, s > λ,
rλ ln3/2(1/r), s = λ,
rs, s < λ.
(5.1)
Proof. Let us introduce the function
ψ(ρ) =
ρλ, s > λ,
ρλ ln3/2(1/ρ), s = λ,
ρs, s < λ,
for ρ ∈ (0, d). We make the transformation x = ρx′, u(x) = v(ρx′) =
ψ(ρ)w(x′). By the Sobolev Imbedding theorems (see (7.30) [3])
W 2,2(G1
1/2) ⊂→C0(G1
1/2) and we have
sup
G1
1/2
|w(x′)| ≤ c1||w||W 2,2(G1
1/2
)
Returning to the variables x, u considering the inequality (4.15), we
have for ∀ρ ∈ (0, d)
sup
Gρ
ρ/2
ψ−1(ρ)|u(x)| ≤ c2ψ
−2(ρ)||u||V 2
2,1(G
ρ
ρ/2
) ≤ c3(||u||L2(G)+||f ||V 0
0,1(G)+ks)
Putting now r = 2ρ/3, we obtain finally the desired estimate.
6. Remarks and examples.
Remark 6.1. The solution of (1.1)–(1.3) can be taken as a function
from W 2,2
loc (G \ `) ∩ C0(G). Then from [8] we obtain W 2,2
loc (G \ O) ∩
C0(G).
The behavior of solutions in a neighbourhood of intersecting edges 243
Remark 6.2. The number λ that is defined by (3.4) cannot in general
be expressed as explicit functions of $1 and $2. There are a few
examples (see below), where λ can be calculated directly. They shows
that the exponent λ in (5.1) cannot be increased.
Example 6.3. Let Ω = {ω : ω1 ∈ [0, $1], ω2 ∈ [0, $2]}, where $1 =
π cos2$2
1 − 3 cos2$2
, $2 =
5π
12
is domain on the unit sphere S2. Then ϑ0 =
4(γ + 2)(γ + 3) is the smallest positive eigenvalue of the eigenvalue
problem (3.1)–(3.3) (γ =
π
2$1
) and λ = γ + 2. Let us consider the
function
u(x) = rγ+2 cos γω1 sinγ ω2(cos2 ω2 − cos2$2)
in G = {(r, ω) : 0 < r < ∞, ω ∈ Ω}. It is the solution of (1.1)–(1.3)
for Laplacian.
Example 6.4. Let Ω = {ω : ω1 ∈ [0, $1], ω2 = π/2}, where $1 ∈
(0, 2π). Then ϑ0 = 4(γ+ 1)(γ + 2) is the smallest positive eigenvalue
of the eigenvalue problem (3.1)–(3.3) (γ =
π
2$1
) and λ = γ + 1. Let
us consider the function
u(x) = rγ+1 cos γω1 sinγ ω2 cosω2
in domain G = {(r, ω) : 0 < r < ∞, ω ∈ Ω}. It is the solution of
(1.1)–(1.3) for Laplacian.
Example 6.5. Let γ and domain G be defined as in the example 6.4
and let
u(x) = rγ+1 ln(1/r) cos γω1 sinγ ω2 cosω2.
The function u satisfies in the domain Gd
0 following equations
aij(x)uxixj
:=
=
(
δj
i −
2γ + 3
γ(γ + 1) ln(1/r)
(
2γ + 1
2γ + 3
· δj
i −
xixj
r2
))
uxixj
= 0, (6.1)
∆u = −ai(x)uxi
:= − (2γ + 3)xi
r2((γ + 1) ln(1/r) − 1)
uxi
, (6.2)
∆u = −a(x)u := − 2γ + 3
r2 ln(1/r)
u, (6.3)
∆u = f(x) := −(2γ + 3)rγ−1 cos γω1 sinγ ω2 cosω2. (6.4)
244 M. Plesha
If d < e(6γ+7)/(γ2+γ), then the equation (6.1) is uniformly elliptic with
ellipticity constants ν=1− 6γ + 7
γ(γ + 1) ln(1/d)
, µ=1+
4γ + 8
γ(γ + 1) ln(1/d)
.
Furthermore, A(r) =
6γ + 9
γ(γ + 1) ln(1/r)
. If d < e−1, then for the
equation (6.2) we have A(r) =
6γ + 9
γ ln(1/r)
and for the equation (6.3) we
have A(r) =
2γ + 3
ln(1/r)
. In all these cases
d
∫
0
r−1A(r)dr = +∞, that is
the leading coefficients of the (6.1) and the lower order coefficients of
(6.2), (6.3) are continuous but not Dini continuous at zero. From the
explicit form of the solution u(x) we have |u(x)| ≤ crγ+1−ε = crλ−ε,
for all ε > 0, x ∈ Gd
0. Thus the assumptions about the coefficients
are essential. In the case of (6.4) all assumptions on the coefficients
are satisfied, but ||f ||V 0
2,1(G
ρ
0
) ≤ cρs with s = λ. This verifies the
importance of conditions of our theorems.
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LAC, Samchuka 9,
79011, Lviv, Ukraine
milan@lac.lviv.ua
Received 6.11.2007
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