Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients
The mixed boundary-value problem is considered for linear second-order nondivergent parabolic equations with discontinuous coefficients satisfying the Cordes conditions. The one-valued strong (almost everywhere) solvability of this problem is proved in the space $Ŵ_p^{2,1}$, where $p$ belongs to the...
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| author | Guliyev, A. F. Ismayilova, S. H. Гулієв, А. Ф. Ісмаїлова, С. Х. |
| author_facet | Guliyev, A. F. Ismayilova, S. H. Гулієв, А. Ф. Ісмаїлова, С. Х. |
| author_sort | Guliyev, A. F. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T10:27:00Z |
| description | The mixed boundary-value problem is considered for linear second-order nondivergent parabolic equations with discontinuous coefficients satisfying the Cordes conditions. The one-valued strong (almost everywhere) solvability of this problem is proved in the space $Ŵ_p^{2,1}$, where $p$ belongs to the same segment containing point 2. |
| first_indexed | 2026-03-24T02:21:17Z |
| format | Article |
| fulltext |
UDC 517.9
A. F. Guliyev, S. H. Ismayilova (Inst. Math. and Mech. Nat. Acad. Sci. Azerbaijan, Baku)
THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR
SECOND ORDER NONDIVERGENT PARABOLIC EQUATIONS
WITH DISCONTINUOUS COEFFICIENTS
МIШАНА КРАЙОВА ЗАДАЧА ДЛЯ ЛIНIЙНИХ
БЕЗДИВЕРГЕНТНИХ ПАРАБОЛIЧНИХ РIВНЯНЬ ДРУГОГО ПОРЯДКУ
З РОЗРИВНИМИ КОЕФIЦIЄНТАМИ
The mixed boundary-value problem is considered for linear second order nondivergent parabolic equations with disconti-
nuous coefficients satisfying the Cordes conditions. The one-valued strong (almost everywhere) solvability of this problem
is proved in the space Ŵ 2,1
p , where p belongs to the same segment containing point 2.
Розглядається мiшана крайова задача для лiнiйних бездивергентних параболiчних рiвнянь другого порядку з роз-
ривними коефiцiєнтами, що задовольняють умови Корде. Однозначну сильну (майже скрiзь) розв’язнiсть цiєї задачi
доведено у просторi Ŵ 2,1
p , де p належить тому ж вiдрiзку, що мiстить точку 2.
1. Introduction. Let En and Rn+1 be n and (n + 1)-dimensional Euclidean spaces of points
x = (x1, x2, . . . , xn) and (t, x) = (t, x1, x2, . . . , xn), respectively, let Ω ⊂ En be a bounded domain
with boundary ∂Ω ∈ C2, let Bx0
R be an n-dimensional open sphere of radius R with center at
the point x0 = (x0
1, x
0
2, . . . , x
0
n), Qx
0
R × (0, T ) ≡ QTR, QT = {(t, x)|0 < t < T < ∞, x ∈ Ω},
ST = {(t, x)|0 < t < T < ∞, x ∈ ∂Ω}, and let A(QTR) be the set of all functions u(t, x) from
C∞(Q̄TR) with support in Bx0
ρ × [0, T ], ρ < R, for which u(0, x) = 0.
In the domain QT , we consider a mixed boundary-value problem for linear parabolic equations
of the form
Lu =
n∑
i,j=1
aij(t, x)
∂2u
∂xi∂xj
+
n∑
i=1
bi(t, x)
∂u
∂xi
+ c(t, x)u− ∂u
∂t
= f(t, x), (1)
u|t=0 = 0,
∂u
∂n
∣∣∣∣
ST
= 0, (2)
under the assumptions that ‖aij(t, x)‖ is a real symmetric matrix. Moreover, for all (t, x) ∈ QT and
ξ ∈ En, the conditions
γ|ξ|2 ≤
n∑
i,j=1
aij(t, x)ξiξj ≤ γ−1|ξ|2, γ ∈ (0, 1]− const, (3)
are satisfied.
In addition, we suppose that all coefficients of the operator L are real and measurable functions
in QT .
c© A. F. GULIYEV, S. H. ISMAYILOVA, 2014
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11 1443
1444 A. F. GULIYEV, S. H. ISMAYILOVA
The aim of the present paper is to find the conditions on thecoefficients of equations (1) under
which the mixed boundary-value problem (1), (2) is identically strongly (almost everywhere) solvable
in the space Ŵ 2,1
p for any f(t, x) ∈ Lp(QT ), p ∈ [p1, p2], where p1 ∈ (1, 2), p2 ∈ (2,∞).
In case where the leading coefficients of the linear operator are uniformly continuous in the
cylindrical domain and the minor coefficients are the elements of the corresponding Lebesgue spaces,
the uniform strong (almost everywhere) solvability of the Dirichlet and mixed problems for the
parabolic and elliptic equations in the space Sobolev was proved in [1, 2]. An example indicating the
exactness of the Cordes conditions is presented in [3]. In [4, 5] the indicated fact is taken to the class
of nonlinear parabolic equations of the second order under the stronger condition than the Cordes
condition. Note that the Dirichlet problem for linear and quasilinear second-order parabolic and
elliptic equations with nondivergent structure and discontinuous coefficients was studied in [6 – 12].
1. Some auxiliary assertions. First, we present some necessary notation and definitions. We
denote by ut, ui and uij the derivatives
∂u
∂t
,
∂u
∂xi
, and
∂2u
∂xi∂xj
, i, j = 1, . . . , n, respectively. Let
W 1,0
p (QT ) and W 2,1
p (QT ) be Banach spaces of measurable functions u(t, x) given on QT with
bounded norms
‖u‖
W 1,0
p (QT )
=
∫
QT
(
|u|p +
n∑
i=1
|ui|p
)
dtdx
1/p
and
‖u‖
W 2,1
p (QT )
=
∫
QT
|u|p +
n∑
i=1
|ui|p +
n∑
i,j=1
|uij |p + |ut|p
dtdx
1/p
,
respectively. By Ŵ 2,1
p (QT ),we denote the subspaceW 2,1
p (QT ) in which the dense set is the collection
of all functions from C∞(QT ) vanishing at t = 0 and
∂u
∂n
∣∣∣∣
ST
= 0.
Definition. A function u(t, x) ∈ Ŵ 2,1
p (QT ) is called a strong solution of the mixed boundary-
value problem (1), (2) if it satisfies equation (1) almost everywhere in QT .
Further, throughout the paper, the notation C(. . .) means that the positive constant C depends
only on the content of the parentheses.
Lemma 1. If u(t, x) ∈ A(QTR), then∫
QTR
n∑
i,j=1
|uij |2 + |ut|2
dtdx ≤
∫
QTR
(M0u)2dtdx,
whereM0 = 4− ∂
∂t
.
Proof. We have ∫
QTR
(M0u)2dtdx =
∫
QTR
(
(4u)2 − 24uut + u2
t
)
dtdx =
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1445
=
∫
QTR
n∑
i,j=1
uiiujj − 2
n∑
i=1
uiiut + u2
t
dtdx =
= −
∫
QTR
n∑
i,j=1
uiujjidtdx+ 2
∫
QTR
n∑
i=1
uiutidtdx+
∫
QTR
u2
tdtdx =
=
∫
QTR
n∑
i,j=1
u2
ij + u2
t
dtdx+
∫
QTR
n∑
i=1
(u2
i )tdtdx =
=
∫
QTR
n∑
i,j=1
u2
ij + u2
t
dtdx+
∫
Bx
0
R
n∑
i=1
(u2
i (T, x)− u2
i (0, x))dx.
Since ui(0, x) = 0, we get the required inequality.
Lemma 2. If u(t, x) ∈ A(QTR) and p ∈ (1,∞), then∫
QTR
n∑
i,j=1
|uij |p + |ut|p
dtdx ≤ C1(p, n)
∫
QTR
|M0u|pdtdx.
Proof. Let
F (t, x) = 4u(t, x)− ut(t, x),
G(t, x) =
a0t
−n/2 exp
(
−|x|
2
4t
)
, at t > 0,
0 at t ≤ 0, (except for t = |x| = 0),
where a0 = 2−nπ−n/2. Then
u(t, x) =
∫
QTR
G(t− τ, x− y)F (τ, y)dτdy.
For i = 1, . . . , n we have
ui(t, x) =
∫
QTR
Gi(t− τ, x− y)F (τ, y)dτdy =
∫
QTR
Gi(t− τ, y − x)F (τ, y)dτdy =
=
∫
Rn+1
Gi(t− τ, v)F (τ, v + x)dτdv.
Further, acting as in the differentiation of integrals with weak singularity [12], we obtain
uij(t, x) =
∫
Rn+1
Gi(t− τ, v)F (τ, v + x)dτdv =
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1446 A. F. GULIYEV, S. H. ISMAYILOVA
= lim
ρ→0
−
∫
B
(t,x)
0,1/ρ
Gi(t− τ, x− y)F (τ, y)dτdy +
+
∫
∂B
(t,x)
0,1/ρ
Gi(t− τ, x− y)F (τ, y) cos (n, yj)dsτ,y
=
= lim
ρ→0
−
∫
∂B
(t,x)
0,1/ρ
Gi(t− τ, x− y)F (τ, y)dτdy
+
+ lim
ρ→0
F (t, x)
∫
∂B
(t,x)
0,1/ρ
Gi(t− τ, x− y) cos (n, yj)dsτ,y +
+
∫
∂B
(t,x)
0,1/ρ
[F (τ, y)− F (t, x)]Gi(t− τ, x− y) cos (n, yj)dsτ,y
=
= Gij ∗ F + F (t, x) lim
ρ→0
∫
∂B
(t,x)
0,1/ρ
Gi(t− τ, x− y) cos (n, yj)dsτ,y+
+ lim
ρ→0
∫
∂B
(t,x)
0,1/ρ
Kij(ρ)Gi(t− τ, x− y) ∗ cos (n, yj)dsτ,y,
where
Gij ∗ F = lim
ρ→0
∫
B
(t,x)
0,1/ρ
Gi(t− τ, x− y)F (τ, x)dτdy,
Kij(ρ) = F (τ, y)− F (t, x),
B
(t,x)
0,1/ρ =
{
(τ, y) : 0 <
G(t− τ, x− y)
t− τ
<
1
ρ
}
,
and ∂B(t,x)
0,1/ρ is its boundary.
We now find
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1447
Jij(ρ) =
∫
∂B
(x,t)
0,1/ρ
Gi(t− τ, x− y) cos (n, yj)dsτ,y =
=
∫
∂B
(0,0)
0,1/ρ
Gi(t− τ, x− y) cos (n, yj)dsτ,y =
1
ρ
∫
∂B
(0,0)
0,1/ρ
yi
2
cos (n, yj)dsτ,y.
If i 6= j, then Jij = 0. Now let i = j. Consider, e.g., the case i = j = n, because, in all remaining
cases, the proof is similar. Denote by Sρ the part of ∂B(x,t)
0,1/ρ in which yn > 0. By Πρ we denote the
projection of Sρ onto the hyperline yn = 0. Then
Jnn(ρ) =
2
ρ
∫
Sρ
yn
2
cos (n, yn)dsτ,y =
=
2
ρ
∫
Πρ
yn
2
cos (n, yn)
1
cos (n, yn)
dτdy1 . . . dyn−1 =
=
2
ρ
∫
Πρ
yn
2
dτdy1 . . . dyn−1 =
2
ρ
∫
Πρ
n∑
i=1
y2
i
4
dτdy1 . . . dyn−1 =
=
2
ρ
∫
Πρ
√√√√n+ 2
2
(−τ) ln
(a0ρ)
2
n+2
−τ
−
n−1∑
i=1
y2
i
4
dτdy1 . . . dyn−1.
We now perform the change of variables u = −τ(a0ρ)
− 2
n+2 , vi = yi(a0ρ)
− 1
n+2 , i = 1, 2, . . . , n−
− 1. Let Π+ be the image of transformation Πρ. We get
Jnn(ρ) = 2a0
∫
Π+
√√√√n+ 2
2
u ln
1
u
−
n−1∑
i=1
v2
i
4
dudv1 . . . dvn−1 =
=
2n+1
n+ 2
1∫
0
√
ln
1
r
dr
∫
En
exp
[
−
n−1∑
i=1
ξ2
i
]
dξ1 . . . dξn−1,
where r = exp
[∑n−1
i=1
v2
i
4u
− n+ 2
2
u ln
1
u
]
, ξi =
vi
2
√
u
, i = 1, 2, . . . , n− 1.
It is easy to see that the last integral is equal to
1
n+ 2
.
Kij(ρ) =
∫
∂B
(t,x)
0,1/ρ
[F (τ, y)− F (t, x)]Gi(t− τ, x− y) cos (n, yj)dsτ,y =
=
∫
∂B
(0,0)
0,1/ρ
[F (τ, y)− F (0, 0)]Gi(−τ,−y) cos (n, yj)dsτ,y =
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1448 A. F. GULIYEV, S. H. ISMAYILOVA
=
1
ρ
∫
∂B
(0,0)
0,1/ρ
[F (τ, y)− F (0, 0)]
yi
2
cos (n, yj)dsτ,y =
=
2
ρ
∫
Sρ
[F (τ, y)− F (0, 0)]
yi
2
cos (n, yj)dsτ,y =
=
2
ρ
∫
Πρ
[F (τ, y)− F (0, 0)]
yi
2
dτdy1 . . . dyidyi+1 . . . dyn.
Since
|F (τ, y)− F (0, 0)| =
C(n, F )
(
|x|1/2 + |y|
)
, for (τ, y) ∈ B(0,0)
1,∞ ,
2 max |F |, for (τ, y) ∈ B(0,0)
0,∞ ,
then, for i 6= j, we get
Kij(ρ) ≤ C(n, F )ρ
1
n+2
2
ρ
∫
Πρ
yi
2
dτdy1 . . . dyidyi+1 . . . dyn = C(n, F )ρ
1
n+2Jij(ρ) = 0
and, for i = j, we find Kii(ρ) ≤ C(n, F )ρ
1
n+2 , ρ
1
n+2 → 0 as ρ→ 0, where C1(n, F ) =
C(n, F )
n+ 2
.
As a result of these calculations for uij , we have
uij(t, x) = −Gij ∗ F +
δij
n+ 2
F (t, x), i, j = 1, . . . , n, (4)
where δij is the Kronecker symbol and Gij ∗F is a parabolic singular integral with the kernel in Gij .
By the Jones theorem [13], for p ∈ (1,∞) and i, j = 1, . . . , n, we conclude that
‖Gij ∗ F‖Lp(QTR) ≤ Cij(p, n)‖F‖Lp(QTR).
By using this inequality in (4), we obtain
n∑
i,j=1
‖uij‖Lp(QTR) ≤ C1(p, n)‖F‖Lp(QTR). (5)
We now show that ‖ut‖Lp(QTR) ≤ C2(p, n)‖F‖Lp(QTR). Indeed, from the relations ut = 4u − F
and (5), we get
‖ut‖Lp(QTR) ≤ ‖4u‖Lp(QTR) + ‖F‖Lp(QTR) ≤
n∑
i=1
‖uii‖Lp(QTR)+
+‖F‖Lp(QTR) ≤ C2(p, n)‖F‖Lp(QTR).
Then ∫
QTR
n∑
i,j=1
|uij |p + |ut|p
dtdx
1/p
≤
∫
QTR
n∑
i,j=1
|uij |pdtdx
1/p
+
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1449
+
∫
QTR
|ut|pdtdx
1/p
≤
n∑
i,j=1
‖uij‖Lp(QTR) + ‖ut‖Lp(QTR) ≤
≤ C3(p, n)
∫
QTR
‖M0u|pdtdx
1/p
.
Lemma 2 is proved.
By
◦
W 2,1
p (QTR) and
◦
V 2,1
p (QTR), we denote the of closures A(QTR) in the norms
‖ut‖ ◦
W 2,1
p (QTR)
=
∫
QTR
n∑
i,j=1
|uij |p + |ut|p
dtdx
1/p
and
‖ut‖ ◦
V 2,1
p (QTR)
=
∫
QTR
‖M0u|pdtdx
1/p
,
respectively, p ∈ (1,∞). According to the Friedrichs-type inequality and Lemma 2, the functionals
determined above are indeed norms. Denote by T (p) the operator associating each function u(t, x) ∈
∈
◦
V 2,1
p (QTR) with it-self as an element of the space
◦
W 2,1
p (QTR). By Lemma 2, the operator T (p) is
bounded. Denote by K(ρ) its norm. By Lemma 1, K(2) ≤ 1. Let p0 be an arbitrary number from
the interval (1, 2). According to the Riesz – Thorin theorem on convexity [14], for any p ∈ [p0, 2],
K(p) ≤ (K(p0))1−θ(K(2))θ ≤ (K(p0))1−θ,
where θ =
2(p− p0)
p(2− p0)
.
Thus,
K(p) ≤ K(p0)
2p0(p−p0)
p(2−p0) .
We now fix p0 =
5
3
and denote a = max
{(
5
3
)3
,
((
5
3
))3
}
. Since, for p ∈
[
5
3
, 2
]
,
p0(p− p0)
p(2− p0)
≤ 2− p
2− p0
= 3(2− p), we finally obtain
K(p) ≤ a2−p.
Thus, we have proved the following assertions:
Lemma 3. If u(t, x) ∈ Ŵ 2,1
p (QTR), then, for any p ∈
[
5
3
, 2
]
,
‖ut‖ ◦
W 2,1
p (QTR)
≤ a2−p‖ut‖ ◦
V 2,1
p (QTR)
.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1450 A. F. GULIYEV, S. H. ISMAYILOVA
Note that, in this case, the constant a > 1 depends only on n. For p ∈
[
5
3
, 2
]
, we denote
sup
QTR
n∑
i,j=1
|aij(t, x)− δij |
p
p−1
p−1
p
by δp (for the sake of brevity, we write sup instead of ess sup). Also let δ2 = δ, h = max
{
1− γ2
γ
, 1
}
.
Lemma 4. For p ∈
[
5
3
, 2
]
, the following inequality is true:
δp ≤ h
2−p
p δ
2(p−1)
p .
Proof. It follows from condition (3) that, for i = 1, . . . , n,
γ − 1 ≤ aij(t, x)− 1 ≤ γ−1 − 1,
and, since γ − 1 ≥ 1− γ−1, that
|aij(t, x)− 1| ≤ 1− γ
γ
. (6)
If i 6= j, then
2γ ≤ aii(t, x) + ajj(t, x) + 2aij(t, x) ≤ 2γ−1.
Therefore,
|aij(t, x)| ≤ 1− γ2
γ
. (7)
From (6) and (7), we conclude that, for i, j = 1, . . . , n,
|aij(t, x)− δij | ≤ h. (8)
On the other hand, in view of (8), we obtain
δp = sup
QTR
n∑
i,j=1
(aij(t, x)− δij)2|aij(t, x)− δij |
2−p
p−1
p−1
p
≤ h
2−p
p δ
2(p−1)
p .
Lemma 4 is proved.
Lemma 5. Let δ < 1. Then there exists p1(γ, δ, n) ∈
[
5
3
, 2
]
, such that for all p ∈ [p1, 2]
a2−pδp ≤ δ1/3.
Proof. According to the previous lemma,
a2−pδp ≤ a2−ph
2−p
p δ
2(p−1)
p .
But h1/p ≤ h
3
5 = h1,
p− 1
p
≥ 1
3
. Therefore,
a2−pδp ≤ (ah1)2−pδ
2
3 . (9)
Now let p1 = max
{
5
3
, 2− ln (1/δ)
3 ln (ah1)
}
. Then, for p ∈ [p1, 2] , we have (ah1)2−p ≤ δ−1/3 and the
assertions of the lemma follow from (9).
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1451
2. Internal priory estimation. Consider an operator
L0 =
n∑
i,j=1
aij(t, x)
∂2
∂xi∂xj
− ∂
∂t
,
together with the operator L.
Lemma 6. If condition (3) and inequality δ < 1 are satisfied for the coefficients of the operator
L0, then, for all p ∈ [p1, 2] and any function u(t, x) ∈
◦
W 2,1
p (QTR) the estimation
‖u‖ ◦
W 2,1
p (QTR)
≤ C4(γ, δ, n)‖L0u‖Lp(QTR)
is true.
Proof. According to Lemma 3,
‖u‖ ◦
W 2,1
p (QTR)
≤ a2−p‖M0u‖Lp(QTR) ≤
≤ a2−p‖L0u‖Lp(QTR) + a2−p
∥∥∥∥∥∥
n∑
i,j=1
(aij(t, x)− δij)uij
∥∥∥∥∥∥
Lp(QTR)
≤
≤ a2/5‖L0u‖Lp(QTR) + a2−p
∥∥∥∥∥∥
n∑
i,j=1
(aij(t, x)− δij)uij
∥∥∥∥∥∥
Lp(QTR)
. (10)
But, on the other hand, ∥∥∥∥∥∥
n∑
i,j=1
(aij(t, x)− δij)uij
∥∥∥∥∥∥
Lp(QTR)
≤
≤
∫
QTR
n∑
i,j=1
|uij |p
n∑
i,j=1
(aij(t, x)− δij)
p
p−1
p−1
dtdx
1/p
≤ δp‖u‖ ◦
W 2,1
p (QTR)
.
Therefore, from (10) and Lemma 5, we conclude that
‖u‖ ◦
W 2,1
p (QTR)
≤ a2/5‖L0u‖Lp(QTR) + a2−pδp‖u‖ ◦
W 2,1
p (QTR)
≤
≤ a2/5‖L0u‖Lp(QTR) + δ1/3‖u‖ ◦
W 2,1
p (QTR)
and the assertion of the lemma is proved.
In what follows, we everywhere assume that the radius R of the sphere Bx0
R (Bx0
R is the foundation
of the cylinder QTR) does not exceed 1.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1452 A. F. GULIYEV, S. H. ISMAYILOVA
Lemma 7. If the condition of previous lemma is true, then, for all p ∈ [p1, 2] and any function
u(t, x) ∈ A(QTR), the inequality
‖u‖
W 2,1
p (QTR)
≤ C5(γ, δ, n)‖L0u‖Lp(QTR)
is true.
To prove this, it suffices to apply the Friedrichs inequality and Lemma 6.
We now assume that the following Cordes condition for the leading coefficients of the operator
L is true:
σ =
supQT
∑n
i,j=1
a2
ij(t, x)[
infQT
∑n
i=1
aii(t, x)
]2 ≤
1
n− 1
. (11)
In this case, we suppose that condition (11) is satisfied to within a nonsingular linear transforma-
tion, i.e., we can cover the domain QT with finite number of subdomains Q1, . . . , Qm and, hence, in
every Qi, there exists a nonsingular linear transformation under which the image of the operator L
satisfies condition (11) in the image of subdomain Qi, i = 1, . . . ,m.
Lemma 8. To within a nonsingular linear transformation, the condition δ < 1 coincides with
condition (11).
Proof. We now perform the transformation τ = k2t, yi = kxi, i = 1, . . . , n, where
k =
supQT
∑n
i,j=1
a2
ij(t, x)
infQT
∑n
i=1
aii(t, x)
−1/2
.
Thus, if ‖Aij(τ, y)‖ is the matrix of leading part of the image of the operator L then Aij(τ, y) =
= k2aij(t, x), i, j = 1, . . . , n. In the new variables, the condition δ < 1 takes the form
sup
Q̃T
n∑
i,j=1
A2
ij(τ, y)− 2 inf
Q̃T
n∑
i=1
Aii(τ, y) + n < 1, (12)
where Q̃T is the image of the domain QT . It is clear, that it coincides with the conditions
supQT
∑n
i,j=1
a2
ij(t, x)[
infQT
∑n
i=1
aii(t, x)
]2 ≤
1
n− 1
.
Lemma 9. Let conditions (3) and (11) be satisfied for the coefficients of the operator L0. Then
there exists a constant C6(γ, σ, n) such that, for any function u(t, x) ∈ C∞(Q̄TR), u|t=0 = 0 for
every p ∈ [p1, 2] , and R1 ∈ (0, R), the estimate
‖u‖
W 2,1
p (QTR1
)
≤ C5‖L0u‖Lp(QTR) +
C6
(R−R1)2
‖u‖Lp(QTR) +
C6
R−R1
‖u‖
W 1,0
p (QTR)
is true.
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THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1453
Proof. Let the functions η(x) ∈ C∞0 (Bx0
R ) be such that η(x) = 1 in Bx0
R1
, 0 ≤ η(x) ≤ 1.
Moreover
|η1| ≤
C7
R−R1
, |ηij | ≤
C7
(R−R1)2
, i, j = 1, . . . , n, (13)
where C7 = C7(n). Applying Lemma 7 to the functions uη, we obtain
‖u‖
W 2,1
p (QTR1
)
≤ C5‖L0(uη)‖Lp(QTR). (14)
But, on the other hand,
|L0(uη)| ≤ |L0u|+ |u|
∣∣∣∣∣
n∑
i=1
aij(t, x)ηij
∣∣∣∣∣+ 2
∣∣∣∣∣∣
n∑
i,j=1
aij(t, x)uiηi
∣∣∣∣∣∣ . (15)
Further, in view of (13), we get ∣∣∣∣∣
n∑
i=1
aij(t, x)ηij
∣∣∣∣∣ ≤ C8(γ, n)
(R−R1)2
,
2
∣∣∣∣∣∣
n∑
i,j=1
aij(t, x)uiηi
∣∣∣∣∣∣ ≤ 2
n∑
i,j=1
aij(t, x)uiuj
1/2 n∑
i,j=1
aij(t, x)ηiηj
1/2
≤
≤ 2γ−1
(
n∑
i=1
u2
i
)1/2( n∑
i=1
η2
i
)1/2
≤ 2γ−1
n∑
i=1
|ui|
n∑
i=1
|ηi| ≤
2nγ−1C7
R−R1
n∑
i=1
|ui|.
Thus, from (15), we conclude
‖L0(uη)‖Lp(QTR) ≤ ‖L0u‖Lp(QTR) +
C8
(R−R1)2
‖u‖Lp(QTR)+
+
C9(γ, n)
R−R1
n∑
i=1
‖ui‖Lp(QTR) ≤ ‖L0u‖Lp(QTR) +
C8
(R−R1)2
‖u‖Lp(QTR)+
+
C9(γ, n)
R−R1
‖u‖
W 1,0
p (QTR)
. (16)
In view of (16) and (14), we denote max {C5C8, C5C9} by C10 and arrive at the required esti-
mate (13).
Lemma 10. Let the conditions of the previous lemma be satisfied for the coefficients of the
operator L0. Then there exists a constant C11(γ, σ, n) such that, for any function u(t, x) ∈ C∞(Q̄TR),
u|t=0 = 0 for any ε > 0, and p ∈ [p1, 2] , the estimate
‖u‖
W 2,1
p (QTR
2
)
≤ C5‖L0u‖Lp(QTR) + ε‖u‖
W 2,1
p (QTR)
+
C11
εR2
‖u‖Lp(QTR)
is true.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1454 A. F. GULIYEV, S. H. ISMAYILOVA
Proof. We use the following interpolation inequality [1]: let p ∈ (0,∞). Then, for any functions
u(t, x) ∈W 2,1
p (QTR), any ε > 0, and p ∈ [p1, 2] the following estimate is true:
‖u‖
W 1,0
p (QTR)
≤ ε‖u‖
W 2,1
p (QTR)
+
C12
ε
‖u‖Lp(QTR). (17)
We now fix an arbitrary number ε > 0 and let ε1 > 0 be a number which will be chosen later.
According to Lemma 9 and inequality (17), we have
‖u‖
W 2,1
p (QT
R/2
)
≤ C5‖L0u‖Lp(QTR) +
4C6
R2
‖u‖Lp(QTR) +
2C6
R2
‖u‖
W 1,0
p (QTR)
≤
≤ C5‖L0u‖Lp(QTR) +
4C6
R2
‖u‖Lp(QTR) +
2C6ε1
R
‖u‖
W 2,1
p (QTR)
+
2C6C13
Rε1
‖u‖Lp(QTR),
where C13 = supp∈[p1,2]C12(p, n). It is now sufficient to choose ε1 =
εR
2C6
.
Lemma 10 is proved.
Remark 1. If the minor coefficients of the operator L are bounded, then there exists
R0(γ, σ0, n,B, c) such that, for R ≤ R0, the assertion of Lemma 10 is also true for the opera-
tor L. Here, B = (b1(t, x), . . . , bn(t, x)). For ρ > 0, the set {x : x ∈ Ω, dist (x, ∂Ω) > ρ} is denoted
by Ωρ.
Lemma 11. Let conditions (3) and (11) be satisfied for the coefficients of the operator L. Then,
for any function u(t, x) ∈ C∞(Q̄TR), u|t=0 = 0 for any ε > 0, ρ > 0, and p ∈ [p1, 2] , the estimate
‖u‖
W 2,1
p (Qρ×(0,T ))
≤ C14(γ, σ, n, ρ,Ω)‖L0u‖Lp(QTR)+
+ε‖u‖
W 2,1
p (QTR)
+
C15(γ, σ, n, ρ,Ω)
ε
‖u‖Lp(QTR)
is true.
Proof. We now fix arbitrary ε > 0 and ρ > 0. Let ε2 > 0 be a number chosen in what follows.
Consider a covering Ω̄ρ by a system of spheres {Bxi
ρ/2} and choose a finite subcovering B1, . . . , BN
from this covering. It is evident that the number N depends only on ρ, n, and diam Ω. Applying, for
every i = 1, . . . , N, Lemma 10, we obtain
‖u‖
W 2,1
p (Bi×(0,T ))
≤ 3p−1
(
Cp5‖L0u‖pLp(QT ) + εp2‖u‖
p
W 2,1
p (QT )
+
Cp11
εp2ρ
2p
‖u‖pLp(QT )
)
.
Finding the sum of these inequalities over i from 1 to N, we conclude that
‖u‖
W 2,1
p (Ωρ×(0,T ))
≤ 3p−1N
(
Cp5‖L0u‖pLp(QT ) + εp2‖u‖
p
W 2,1
p (QT )
+
Cp11
εp2ρ
2p
‖u‖pLp(QT )
)
.
It is now sufficient to choose ε2 =
ε
3N
and the lemma is proved.
3. Basic coercive estimation. The assertion of Lemma 11 is true without any requirements
imposed on the domain ∂Ω. All subsequent assertions of the present paper hold under the condition
∂Ω ∈ C2, and we always assume that this condition is satisfied.
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THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1455
Lemma 12. Let conditions (3) and (11) be satisfied for the coefficients of the operator L0. Then
there exists positive constants p1, C16, and C17 depending on γ, σ0, and n and a domain Ω such
that, for any function u(t, x) ∈ Ŵ 2,1
p (QT ), any ε > 0, and p ∈ [p1, 2] the estimate
‖u‖
W 2,1
p ((Ω\Ωρ1×(0,T ))
≤ C16‖L0u‖Lp(QT ) + ε‖u‖
W 2,1
p (QT )
+
C17
ε
‖u‖Lp(QT )
is true.
Proof. It is sufficient to prove the lemma for the functions u(t, x) ∈ C∞(Q̄T ), u|Γ(QT ) = 0.
Moreover, without loss of generality, we can suppose that the coefficients of the operator L0 are
infinitely differentiable Q̄T . We now fix an arbitrary number ε > 0 and a point x0 ∈ ∂Ω. We
perform an orthogonal transformation of the coordinate x → y such that the tangent hyperline to
∂Ω̃ at the point y0 is perpendicular to the axis Oyn. Here, Ω̃ and y0 are images of the domain
Ω and the point x0, under this transformation, respectively. Denote by ũ(t, x) the image of the
function u(t, x). For simplicity, we suppose that the domain ∂Ω̃ at the intersection of ∂Ω̃ with
some neighborhood Oh of the point y0 is given by the equation yn = ϕ(y1, . . . , yn−1) with twice
continuously differentiable function ϕ and the part Ω̃ adjacent to ∂Ω̃ ∩ Oh belongs to the set {y :
yn > ϕ(y1, . . . , yn−1). Let A(t, x) = ‖aij(t, x)‖ be a matrix of coefficients of the operator L0,
Ã(t, x) = ‖ãij(t, y)‖, where ãij(t, y) are leading coefficients of the image L̃0 of the operator L0
under our transformation; i, j = 1, . . . , n. We now show that the eigenvalues of the matrices A and
à coincide. Indeed, we fix an arbitrary point (t, x) ∈ QT ; λ is an arbitrary eigenvalue of the matrix
A and xλ corresponds to its eigenvector. By virtue of the orthogonality of our transformation, there
exists a nondegenerate matrix T such that à = T−1AT. Denote T−1xλ. We get
Ãyλ = T−1Axλ = λyλ.
On the other hand, we can write condition (11) in the following form:
σ = sup
QT
∑n
i=1
λ2
i (t, x)[∑n
i=1
λi(t, x)
]2 ≤
1
n− 1
,
where λi(t, x) are eigenvalues of the matrix A(t, x), i = 1, . . . , n. Thus, condition (11) is also
satisfied for the operator L̃0 and, moreover, with the same constant σ. Analogously, it can be shown
that conditions (3) are satisfied for the operator L̃0 (with the same constant γ). We now perform one
more transformation: zi = yi, i = 1, . . . , n − 1, zn = yn − ϕ(y1, . . . , yn−1). Let L′0, Ω′, and z0 be
the images of the operator L̃0, domain Ω̃, and point y0, respectively, under our transformation, and
let a′ij(t, z) be the leading coefficients of the operator L′0; i, j = 1, . . . , n. It is easy to see that
a′ij(t, z) =
n∑
k,l=1
ãkl(t, y)
∂zl
∂yk
∂zj
∂yl
, i, j = 1, . . . , n.
Therefore,
a′ij(t, z) = ãij(t, y) if 1 ≤ i, j ≤ n− 1,
a′nj(t, z) = −
n−1∑
k=1
ãkj(t, y)
∂ϕ
∂yk
+ ãnj(t, y) if 1 ≤ i, j ≤ n− 1,
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1456 A. F. GULIYEV, S. H. ISMAYILOVA
a′nn(t, z) =
n∑
k,l=1
ãkl(t, y)
∂ϕ
∂yk
∂ϕ
∂yl
− 2
n−1∑
k=1
ãnk(t, y)
∂ϕ
∂yk
+ ãnn(t, y).
Since
∂ϕ
∂yi
(y0) = 0 for i = 1, . . . , n − 1, there exists h1(y0, ϕ) such that the condition (11)(
with the same constant σ′ =
σ + 1/(n− 1)
2
)
is satisfied for h ≤ h1 in the intersection Ω′ ∩
∩
(
Bz0
h × (0, T )
)
. Moreover, conditions (3) are satisfied
(
with the constant
γ
2
)
for the operator
L′0 in the indicated intersection. Assume that r = r(z0) = h1(y0, ϕ). Let u′(t, z) be the image of
the function ũ(t, y) under our transformation. It is clear that, in the variables z, the intersection
Ω′∩Bz0
r represents a hemisphere B+
r = {z : |z− z0| < r, zn > 0}. We continue the function u′(t, z)
and the coefficients of the operator L′0 by evenness relative to the hyperplane zn = 0 in Bz0
r \B+
r
and denote by u′(t, z) and L′0, respectively, the function and operator obtained in this case. Since
u′(t, z) ∈W 2,1
p
(
Bz0
h × (0, T )
)
, according to Lemma 10, we find
‖u′‖
W 2,1
p
(
Bz
0
r
2
×(0,T )
) ≤ C5‖L′0u′‖Lp(Bz0r ×(0,T )) + ε3‖u′‖W 2,1
p (Bz0r ×(0,T ))+
+
C11
ε3r2
‖u′‖
Lp(Bz0r ×(0,T )), (18)
where ε3 > 0 is chosen in what follows. However, on the other hand, each norm on the right-hand
side of (18) represents the corresponding norm taken for a semicylinder Q+
r = B+
r × (0, T ) and
multiplied by 21/p. Therefore, from (18), we conclude
‖u′‖
W 2,1
p
(
Q+
r
2
) ≤ C5‖L′0u′‖Lp(Q+
r ) + ε3‖u′‖W 2,1
p (Q+
r ) +
C11
ε3r2
‖u′‖Lp(Q+
r ). (19)
We cover ∂Ω′ by a system of spheres {Bzi
r
2
} and choose from this covering a finite subcovering
B1, . . . , BM . In this case, the number M is determined only by the quantities γ, σ0, and h and
the domain Ω. We write an inequality of the form (19) for every semicylinder B+
r (zi) × (0, T ),
i = 1, . . . ,M, raise both sides of the obtained inequalities to the power p, and find the sum of these
inequalities over i from 1 to M. This yields
‖u′‖p
W 2,1
p (B×(0,T ))
≤ 3p−1M
(
C5‖L′0u′‖
p
Lp(Ω′×(0,T )) + εp3‖u
′‖
W 2,1
p (Ω′×(0,T ))
+
+
Cp11
εp3r
2p
0
‖u′‖pLp(Ω′×(0,T ))
)
,
where B =
⋃M
i=1
B+
r
2
(zi), and r0 = min{r(z1), . . . , r(zM )}. We return to the variables x and note
that the preimage B contains the set Ω\Ωρ1 with some ρ1(γ, σ, n,Ω). This enables us to conclude
that
‖u‖
W 2,1
p ((Ω\Ωρ1×(0,T ))
≤ C18‖L0u‖Lp(QT ) + C19ε3‖u‖W 2,1
p (QT )
+
C20
ε3
‖u‖Lp(QT ),
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THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1457
where the constants C18, C19, and C20 depend only on γ, σ, and n and the domain Ω. It is now
sufficient to choose ε3 =
ε
C19
, and the lemma is proved.
Lemmas 11 and 12 now imply the following assertion:
Lemma 13. Let conditions (3) and (11) be satisfied for coefficients of the operator L0. Then,
for any function u(t, x) ∈ Ŵ 2,1
p (QT ) and any p ∈ [p1, 2] , the estimate
‖u‖
W 2,1
p (QT )
≤ C21(γ, σ, n,Ω)
(
‖L0u‖Lp(QT ) + ‖u‖Lp(QT )
)
is true.
We now impose the following conditions on the minor coefficient of the operator L. For p ∈
∈ [p1, 2],
bi(t, x) ∈ Ln+2(QT ), i = 1, . . . , n, (20)
c(t, x) ∈ Ll(QT ), l =
max
(
p,
n+ 2
2
)
, for p 6= n+ 2
2
,
2 + ν, for n = p = 2,
(21)
where ν is a positive constant. Let ψ(t, x) ∈ Lp(QT ), 1 < p <∞. The quantity
ωψ;p(δ) = sup
e∈QT ,mes e≤δ
∫
e
|ψ|pdtdx
1/p
is called the AC modulus of the function ψ(t, x). Denote max1≤i≤n{ωbi;p(δ)} by ωB;p(δ).
Let K =
∑n
i=1
‖bi‖Ln+2(QT ) + ‖c‖Lm(QT ).
Everywhere in what follows, the symbol C(L) means that the positive constant C depends only
on γ, σ, K, and ν.
Lemma 14. Let conditions (3), (11), and (20) be satisfied for the coefficients of the operator
L. Then there exist positive constants C22(L, n,Ω) and T0(L, n) such that if T ≤ T0, then, for any
function u(t, x) ∈ Ŵ 2,1
p (QT ), and any p ∈ [p1, 2] the estimate
‖u‖
W 2,1
p (QT )
≤ C22‖Lu‖Lp(QT ),
is true.
Proof. We use the following embedding theorems [1]: For any function u(t, x) ∈ Ŵ 2,1
q (QT ),
the following inequalities are true:
‖ui‖L q(n+2)
n+2−q
(QT ) ≤ C23(q, n)‖u‖
W 2,1
q (QT )
for 1 ≤ q < n+ 2, (22)
‖u‖L q(n+2)
n+2−2q
(QT ) ≤ C24(q, n)‖u‖
W 2,1
q (QT )
for 1 ≤ q < n+ 2
2
. (23)
According to Lemma 13,
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1458 A. F. GULIYEV, S. H. ISMAYILOVA
‖u‖
W 2,1
p (QT )
≤ C21‖Lu‖Lp(QT ) + C21‖(L − L0)u‖Lp(QT ) + C21‖u‖Lp(QT ) ≤
≤ C21‖Lu‖Lp(QT ) + C21
n∑
i=1
‖biui‖Lp(QT ) + C21‖cu‖Lp(QT ) + C21‖u‖Lp(QT ). (24)
We now fix an arbitrary i, 1 ≤ i ≤ n and assume that q = p in (21). We find
‖biui‖Lp(QT ) ≤ ‖bi‖Ln+2(QT )‖ui‖L p(n+2)
n+2−p
(QT ) ≤ C23‖bi‖Ln+2(QT )‖u‖W 2,1
p (QT )
.
Thus,
n∑
i=1
‖biui‖Lp(QT ) ≤ C23
n∑
i=1
‖bi‖Ln+2(QT )‖u‖W 2,1
p (QT )
≤
≤ C25(n)ωB;n+2(δ)‖u‖
W 2,1
p (QT )
, (25)
where δ = Tmes Ω and C25 = supp∈[p1,2]C23(p, n).
Similarly, by virtue of (23), for n ≤ 3, we get
‖cu‖Lp(QT ) ≤ ‖c‖Ln+2
2
(QT )‖u‖L p(n+2)
n+2−2p
(QT ) ≤ C24‖c‖Ln+2
2
(QT )‖u‖W 2,1
p (QT )
≤
≤ C26(n)ω
c;
n+2
2
(δ)‖u‖
W 2,1
p (QT )
,
where C26 = supp∈[p1,2]C24(p, n).
It is easy to see that an analogous estimate holds for n = 2 and p 6= 2. Now let n = p = 2. Thus,
according to embedding theorem [1], for any function u(t, x) ∈W 2,1
p (QT ) and every q ∈ [1,∞] , the
following estimate is true:
‖u‖Lq(QT ) ≤ C27(q, n)‖u‖
W 2,1
p (QT )
.
Therefore, if c(t, x) ∈ L2+ν1(QT ), then
‖cu‖L2(QT ) ≤ ‖c‖L2+ν1 (QT )‖u‖L 2(2+ν1)
ν1
(QT ) ≤ C28(ν)ωc;2+ν1(δ)‖u‖
W 2,1
2 (QT )
.
Finally, let n = 1. Then, according to the embedding theorem [1], for any function u(t, x) ∈
∈W 2,1
p (QT ), the estimate
sup
QT
|u| ≤ C28‖u‖W 2,1
p (QT )
is true. Therefore,
‖cu‖Lp(QT ) ≤ sup
QT
|u|‖c‖Lp(QT ) ≤ C28ωc;p(δ)‖u‖W 2,1
p (QT )
.
Thus, in any case,we get the inequality
‖cu‖Lp(QT ) ≤ C29(n)ωc;l(δ)‖u‖W 2,1
p (QT )
. (26)
Now let t ∈ (0, T ). We use the following inequality:
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THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1459
‖u‖Lp(QT ) ≤ T‖ut‖Lp(QT ). (27)
In view of (25), (26), (27), and (24) we arrive at the inequality
‖u‖
W 2,1
p (QT )
≤ C21‖Lu‖Lp(QT ) + C21(C25ωB;n+2(δ) + C29ωc;l(δ) + T )‖u‖
W 2,1
p (QT )
.
Then there exists a constant T0(L, n) such that, for T ≤ T0,
C25ωB;n+2(δ) + C29ω
c;
n+2
n
(δ) + T <
1
2C21
.
Lemma 14 is proved.
4. Case p > 2. Let p ∈
[
2,
7
3
]
and let K(p) have the same meaning as in Lemma 3. By the
Riesz – Theorin theorem, for any p ∈
[
2,
7
3
]
,
K(p) ≤ (K(2))1−θ
(
K
(
7
3
))θ
≤
(
K
(
7
3
))θ
,
where θ =
2(p− 2)
p
(
7
3
− 2
) . Denote max
{(
7
3
)3
,K
((
7
3
))3
}
by a1(n). We obtain
K(p) ≤ ap−2
1 .
Thus, the following analog of Lemma 3 is true:
Lemma 15. If u(t, x) ∈ Ŵ 2,1
p (QT ), then, for any p ∈
[
2,
7
3
]
, the inequality
‖u‖ ◦
W
2,1
p (QT )
≤ ap−2
1 ‖u‖ ◦
V 2,1
p (QTR)
is true.
The analogs of Lemmas 4 and 5 are proved in an absolutely similar way:
Lemma 16. For p ∈
[
2,
7
3
]
, the following inequality is true:
δp ≤ h
p−2
p δ.
Lemma 17. Let δ < 1. Then there exists p2(γ, δ, n) ∈
(
2,
7
3
]
such that, for all p ∈ [2, p2] ,
ap−2
1 δp ≤ δ1/3.
We now impose the following restrictions on the minor coefficients of the operator L for p ∈
∈ (2, p2] :
bi(t, x) ∈ Ln+2(QT ), i = 1, . . . , n, (28)
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1460 A. F. GULIYEV, S. H. ISMAYILOVA
c(t, x) ∈ Ll′(QT ), l′ = max
(
p,
n+ 2
2
)
. (29)
By using the scheme realized in Lemmas 6 – 12 and applied to Lemmas 14 – 16, we conclude
that Lemma 1 is true for p ∈ (2, p2] and u(t, x) ∈ Ŵ 2,1
p (QT ) if conditions (3), (11), (18), and (29)
are satisfied only for the coefficients of the operator L. We combine conditions (21) and (29) by
assuming that p ∈ [p1, p2] , i.e., we suppose that
c(t, x) ∈ Lm(QT ), where m =
l, for p ∈ [p1, 2],
l′, for p ∈ (2, p2].
(30)
Theorem 1. Let conditions (3), (11), (18), and (29) be satisfied for the coefficients of the operator
L. Then there exists positive constants T0(L, n) and C30(γ, σ,K, n,Ω) such that, for any functions
u(t, x) ∈ Ŵ 2,1
p (QT ) with T ≤ T0 and any p ∈ [p1, 2] , the estimate
‖u‖
W 2,1
p (QT )
≤ C30‖Lu‖Lp(QT ),
is true.
5. Solvability of the mixed boundary-value problem. We now consider the mixed boundary-
value problem (1), (2).
Theorem 2. Let conditions (3), (11), (28), and (30) be satisfied for the coefficients of the operator
L given in the domain QT . If T ≤ T0 and ∂Ω ∈ C2, then the mixed boundary-value problem is
identically strongly solvable in the space Ŵ 2,1
p (QT ) for every f(t, x) ∈ Lp(QT ), p ∈ [p1, 2] . In this
case, for the solution u(t, x) ∈ Ŵ 2,1
p (QT ) the estimate
‖u‖
W 2,1
p (QT )
≤ C30‖f‖Lp(QT ), (31)
is true.
Proof. We now prove the theorem by the method of continuation in the parameter. We introduce,
for s ∈ [0, 1], the family of operators Ls = sL+ (1− s)M0.
It is easy to see that conditions (3) and (11) are satisfied for the operator Ls with constants γ
and σ, respectively. We show this on the example of condition (11). According to Lemma 8, the
indicated condition coincides, to within a nonsingular linear transformation, with the condition δ < 1.
Let asij(t, x) be the leading coefficients of the operator Ls, i, j = 1, . . . , n, and let
δs = sup
QT
n∑
i,j=1
(asij(t, x)− δij)2
1/2
.
We have
δs = sup
QT
n∑
i,j=1
(sasij(t, x) + (1− s)δij − δij)2
1/2
= s sup
QT
n∑
i,j=1
(asij(t, x)− δij)2
1/2
= sδ ≤ δ.
In addition, if bsi (t, x), i, j = 1, . . . , n, and cs(t, x) are minor coefficients of the operator Ls,
then the quantity
∑n
i=1
‖bsi (t, x)‖Ln+2(QT ) +‖cs(t, x)‖Lm(QT ) is majorized by a constant depending
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
THE MIXED BOUNDARY-VALUE PROBLEM FOR LINEAR SECOND ORDER NONDIVERGENT . . . 1461
only on
∑n
i=1
‖bi‖Ln+2(QT ) + ‖cs(t, x)‖Lm(QT ). Hence, the assertion of Theorem 1 is true for the
operator Ls with constant C ′30 independent of s. Denote by E = {s : s ∈ [0, 1]} the set in which the
problem has a solution. Note that, by virtue of Theorem 2, this solution is unique. We now show
that the set E is nonempty and open and closed simultaneously relative to [0, 1]. Then
Ls0u = f(t, x), (t, x) ∈ QT , u(t, x) ∈ Ŵ 2,1
p (QT ), (32)
coincides with the segment [0, 1] and, in particular, problem (32) is identically solvable for s = 1
when L1 = L. In this case, estimate (31) follows from the fact that problem (32) is solvable for s = 0
(see [1]).
We now show that the set E is open relative to [0, 1]. Let s0 ∈ E, s ∈ [0, 1] be such that
|s− s0| < α, where α > 0 will be chosen later. We represent problem (32) as
Ls0u = f(t, x) + (Ls0 − Ls)u, (t, x) ∈ QT , u(t, x) ∈ Ŵ 2,1
p (QT ). (33)
It is easy to see that Ls0 − Ls = (s0 − s)(L −M0). Consider an auxiliary problem
Ls0u = f(t, x) + (s0 − s)(L −M0)v, (t, x) ∈ QT , u(t, x) ∈ Ŵ 2,1
p (QT ), (34)
where v(t, x) ∈ Ŵ 2,1
p (QT ). Acting as in Theorem 1, we can show that
‖(L −M0)v‖Lp(QT ) ≤ C31(L, n)‖v‖
W 2,1
p (QT )
.
Thus, the operator M associating every function v(t, x) ∈ Ŵ 2,1
p (QT ) with a solution u(t, x) of
the problem (34) is determined, i.e., u = Mv. We now show that if α is chosen in a certain way,
then the operatorM becomes contractive. Let u1 =Mv1 and u2 =Mv2. We have
Ls0(u1 − u2) = (s0 − s)(L −M0)(v1 − v2), u1 − u2 ∈ Ŵ 2,1
p (QT ).
Thus, according to Theorem 1
‖u1 − u2‖
W 2,1
p (QT )
≤ C30αC31‖v1 − v2‖
W 2,1
p (QT )
,
and it is sufficient to choose α =
1
2C30C31
. Then the operator M has a fixed point u = Mv.
However, for v = u, problem (34) coincides with problem (33), i.e., with (32). The openness of the
set E is proved. We now prove that it is closed. Let sm ∈ E, m = 1, 2, . . . , s0 = limm→∞ s
m. We
show that s0 ∈ E. Denote by um(t, x) the solution of the boundary-value problem
Lsmum = f(t, x), (t, x) ∈ QT , um ∈ Ŵ 2,1
p (QT ).
According to Theorem 1,
‖um‖
W 2,1
p (QT )
≤ C30‖f‖Lp(QT ).
Thus, the sequence {um(t, x)} is bounded in the norm W 2,1
p (QT ). Hence, it is weakly compact,
i.e., there exists a subsequence mk → ∞ as k → ∞ and a function u(t, x) ∈ Ŵ 2,1
p (QT ) such that
u(t, x) is the weak limit of the subsequence {umk(t, x)} as k → ∞ in Ŵ 2,1
p (QT ). Therefore, in
particular, we conclude that, for any function Ŵ 2,1
p (QT ),
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1462 A. F. GULIYEV, S. H. ISMAYILOVA
〈Ls0umk , ϕ〉 → 〈Ls0 , ϕ〉, k →∞,
where 〈u, v〉 =
∫
QT
uvdtdx. But
〈Ls0umk , ϕ〉 = 〈(Ls0 − Lsmk )umk , ϕ〉+ 〈Lsmkumk , ϕ〉 = i1 + i2.
We have
|i1| ≤ |s0 − smk | |〈(L −M0)umk , ϕ〉| ≤ |s0 − smk |C32(ϕ, p)C31‖umk‖W 2,1
p (QT )
≤
≤ C30C32C31|s0 − smk |‖f‖Lp(QT ).
Thus, i1 → 0 as k → ∞. On the other hand, i2 = 〈f, ϕ〉. Hence, for any function ϕ(t, x) ∈
∈
◦
W 2,1
p (Q̄T ),
〈Ls0u, ϕ〉 = 〈f, ϕ〉.
This means that Ls0u = f(t, x) almost everywhere in QT , i.e., s0 ∈ E.
The theorem is proved.
Remark 2. For p = 2 and the operator L, Theorem 2 is correct without the assumption T ≤ T0
(see [11]).
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Received 10.04.12,
after revision — 21.09.13
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
|
| id | umjimathkievua-article-2237 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:21:17Z |
| publishDate | 2014 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/5b/afd8d51f9b2d2bc95302bc7387a1285b.pdf |
| spelling | umjimathkievua-article-22372019-12-05T10:27:00Z Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients Мішана крайова задача для лінійних бездивергентних параболічних рівнянь другого порядку з розривними коефіцієнтами Guliyev, A. F. Ismayilova, S. H. Гулієв, А. Ф. Ісмаїлова, С. Х. The mixed boundary-value problem is considered for linear second-order nondivergent parabolic equations with discontinuous coefficients satisfying the Cordes conditions. The one-valued strong (almost everywhere) solvability of this problem is proved in the space $Ŵ_p^{2,1}$, where $p$ belongs to the same segment containing point 2. Розглядається мішана крайова задача для лінійних бездивергентних параболiчних рівнянь другого порядку з розривними коефiцiєнтами, що задовольняють умови Корде. Однозначну сильну (майже скрізь) розв'язність цієї задачі доведено у просторі $Ŵ_p^{2,1}$ , де $p$ належить тому ж відрізку, що містить точку 2. Institute of Mathematics, NAS of Ukraine 2014-11-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2237 Ukrains’kyi Matematychnyi Zhurnal; Vol. 66 No. 11 (2014); 1443-1462 Український математичний журнал; Том 66 № 11 (2014); 1443-1462 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2237/1475 https://umj.imath.kiev.ua/index.php/umj/article/view/2237/1476 Copyright (c) 2014 Guliyev A. F.; Ismayilova S. H. |
| spellingShingle | Guliyev, A. F. Ismayilova, S. H. Гулієв, А. Ф. Ісмаїлова, С. Х. Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients |
| title | Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients |
| title_alt | Мішана крайова задача для лінійних бездивергентних параболічних рівнянь другого порядку з розривними коефіцієнтами |
| title_full | Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients |
| title_fullStr | Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients |
| title_full_unstemmed | Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients |
| title_short | Mixed Boundary-Value Problem for Linear Second-Order Nondivergent Parabolic Equations with Discontinuous Coefficients |
| title_sort | mixed boundary-value problem for linear second-order nondivergent parabolic equations with discontinuous coefficients |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2237 |
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