Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 116, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
RESOLUTIONS OF PARABOLIC EQUATIONS IN
NON-SYMMETRIC CONICAL DOMAINS
AREZKI KHELOUFI
Abstract. This article is devoted to the analysis of a two-space dimensional
linear parabolic equation, subject to Cauchy-Dirichlet boundary conditions.
The problem is set in a conical type domain and the right hand side term of
the equation is taken in a Lebesgue space. One of the main issues of this work
is that the domain can possibly be non regular. This work is an extension of
the symmetric case studied in Sadallah [13].
1. Introduction
Let Q be an open set of R3 defined by
Q = {(t, x1 , x2 ) ∈ R3 : (x1 , x2 ) ∈ Ωt , 0 < t < T }
where T is a finite positive number and for a fixed t in the interval ]0, T [, Ωt is a
bounded domain of R2 defined by
Ωt = {(x1 , x2 ) ∈ R2 : 0 ≤
x2
x21
+ 2 22
< 1}.
2
ϕ (t) h (t)ϕ (t)
Here, ϕ is a continuous real-valued function defined on [0, T ], Lipschitz continuous
on [0, T ] and such that
ϕ(0) = 0, ϕ(t) > 0
for every t ∈ ]0, T ]. h is a Lipschitz continuous real-valued function defined on
[0, T ], such that
0 < δ ≤ h(t) ≤ β
(1.1)
for every t ∈ [0, T ], where δ and β are positive constants.
In Q, we consider the boundary-value problem
∂t u − ∂x21 u − ∂x22 u = f ∈ L2 (Q),
u
= 0,
(1.2)
∂Q−ΓT
where L2 (Q) is the usual Lebesgue space on Q, ∂Q is the boundary of Q and ΓT is
the part of the boundary of Q where t = T .
The difficulty related to this kind of problems comes from this singular situation
for evolution problems; i.e., ϕ is allowed to vanish for t = 0, which prevents the
domain Q from being transformed into a regular domain without the appearance of
2000 Mathematics Subject Classification. 35K05, 35K20.
Key words and phrases. Parabolic equation; conical domain; anisotropic Sobolev space.
c
2012
Texas State University - San Marcos.
Submitted May 31, 2012. Published July 9, 2012.
1
2
A. KHELOUFI
EJDE-2012/116
some degenerate terms in the parabolic equation, see for example Sadallah [12]. In
order to overcome this difficulty, we impose a sufficient condition on the function
ϕ; that is,
ϕ0 (t)ϕ(t) → 0 as t → 0,
(1.3)
and we obtain existence and regularity results for Problem (1.2) by using the domain
decomposition method. More precisely, we will prove that Problem (1.2) has a
solution with optimal regularity, that is a solution u belonging to the anisotropic
Sobolev space
H01,2 (Q) := {u ∈ H 1,2 (Q) : u∂Q−Γ = 0},
T
with
H 1,2 (Q) = {u ∈ L2 (Q) : ∂t u, ∂xj 1 u, ∂xj 2 u, ∂x1 ∂x2 u ∈ L2 (Q), j = 1, 2}.
In Sadallah [13] the same problem has been studied in the case of a symmetric
conical domain; i.e., in the case where h = 1. Further references on the analysis of
parabolic problems in non-cylindrical domains are: Alkhutov [1, 2], Degtyarev [4],
Labbas, Medeghri and Sadallah [8, 9], Sadallah [12]. There are many other works
concerning boundary-value problems in non-smooth domains (see, for example,
Grisvard [6] and the references therein).
The organization of this article is as follows. In Section 2, first we prove an
uniqueness result for Problem (1.2), then we derive some technical lemmas which
will allow us to prove an uniform estimate (in a sense to be defined later). In
Section 3, there are two main steps. First, we prove that Problem (1.2) admits a
(unique) solution in the case of a domain which can be transformed into a cylinder.
Secondly, for T small enough, we prove that the result holds true in the case of
a conical domain under the above mentioned assumptions on functions ϕ and h.
The method used here is based on the approximation of the conical domain by
a sequence of subdomains (Qn )n which can be transformed into regular domains
(cylinders). We establish an uniform estimate of the type
kun kH 1,2 (Qn ) ≤ Kkf kL2 (Qn ) ,
where un is the solution of Problem (1.2) in Qn and K is a constant independent of
n. This allows us to pass to the limit. Finally, in Section 4 we complete the proof
of our main result (Theorem 4.4).
2. Preliminaries
Proposition 2.1. Problem (1.2) is uniquely solvable.
Proof. Let us consider u ∈ H01,2 (Ω) a solution of Problem (1.2) with a null righthand side term. So,
∂t u − ∂x21 u − ∂x22 u = 0 in Q.
In addition u fulfils the boundary conditions
u
= 0.
∂Q−ΓT
Using Green formula, we have
Z
Z
(∂t u − ∂x21 u − ∂x22 u)u dt dx1 dx2 =
Q
1
( |u|2 νt − ∂x1 u.uνx1 − ∂x2 u.uνx2 )dσ
2
∂Q
Z
+ (|∂x1 u|2 + |∂x2 u|2 )dt dx1 dx2
Q
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RESOLUTIONS OF PARABOLIC EQUATIONS
3
where νt , νx1 , νx2 are the components of the unit outward normal vector at ∂Q.
Taking Rinto account the boundary conditions, all the boundary integrals vanish
except ∂Q |u|2 νt dσ. We have
Z
Z
|u|2 νt dσ =
|u|2 dx1 dx2 .
∂Q
ΓT
Then
Z
(∂t u − ∂x21 u − ∂x22 u)u dt dx1 dx2
Z
Z
1 2
|u| dx1 dx2 + (|∂x1 u|2 + |∂x2 u|2 )dt dx1 dx2 .
=
Q
ΓT 2
Q
Consequently,
Z
Q
(∂t u − ∂x21 u − ∂x22 u)u dt dx1 dx2 = 0
yields
Z
(|∂x1 u|2 + |∂x2 u|2 )dt dx1 dx2 = 0,
Q
because
1
2
Z
|u|2 dx1 dx2 ≥ 0.
ΓT
This implies |∂x1 u|2 + |∂x2 u|2 = 0 and consequently ∂x21 u = ∂x22 u = 0. Then, the
hypothesis ∂t u − ∂x21 u − ∂x22 u = 0 gives ∂t u = 0. Thus, u is constant. The boundary
conditions imply that u = 0 in Q. This proves the uniqueness of the solution of
Problem (1.2).
Remark 2.2. In the sequel, we will be interested only by the question of the
existence of the solution of Problem (1.2).
The following result is well known (see, for example, [11])
Lemma 2.3. Let D(0, 1) be the unit disc of R2 . Then, the Laplace operator ∆ :
H 2 (D(0, 1))∩H01 (D(0, 1)) → L2 (D(0, 1)) is an isomorphism. Moreover, there exists
a constant C > 0 such that
kvkH 2 (D(0,1)) ≤ Ck∆vkL2 (D(0,1)) , ∀v ∈ H 2 (D(0, 1)).
In the above lemma, H 2 and H01 are the usual Sobolev spaces defined, for instance, in Lions-Magenes [11]. In section 3, we will need the following result.
Lemma 2.4. Let t ∈]αn , T [, where (αn )n is a decreasing sequence to zero. Then,
there exists a constant C > 0 independent of n such that for each un ∈ H 2 (Ωt ), we
have
(a) k∂x1 un k2L2 (Ωt ) ≤ Cϕ2 (t)k∆un k2L2 (Ωt ) ,
(b) k∂x2 un k2L2 (Ωt ) ≤ Cϕ2 (t)k∆un k2L2 (Ωt ) .
Proof. It is a direct consequence of Lemma 2.3. Indeed, let t ∈]αn , T [ and define
the following change of variables
D(0, 1) →
(x1 , x2 ) 7→
Ωt
(ϕ(t)x1 , h(t)ϕ(t)x2 ) = (x01 , x02 ).
4
A. KHELOUFI
EJDE-2012/116
Set
v(x1 , x2 ) = un (ϕ(t)x1 , h(t)ϕ(t)x2 ),
2
then if v ∈ H (D(0, 1)), un belongs to H 2 (Ωt ).
(a) We have
Z
2
k∂x1 vkL2 (D(0,1)) =
(∂x1 v)2 (x1 , x2 ) dx1 dx2
D(0,1)
Z
1
dx0 dx0
=
(∂x01 un )2 (x01 , x02 )ϕ2 (t)
h(t)ϕ2 (t) 1 2
Ωt
Z
1
=
(∂x0 un )2 (x01 , x02 ) dx01 dx02
h(t) Ωt 1
1
k∂x01 un k2L2 (Ωt ) .
=
h(t)
On the other hand,
k∆vk2L2 (D(0,1)) =
Z
D(0,1)
Z
=
Ωt
[(∂x21 v + ∂x22 v)(x1 , x2 )]2 dx1 dx2
(ϕ2 (t)∂x201 un + (hϕ)2 (t)∂x202 un )2 (x01 , x02 )
dx01 dx02
(hϕ2 )(t)
Z
ϕ2 (t)
=
(∂ 20 un + h2 (t)∂x202 un )2 (x01 , x02 ) dx01 dx02
h(t) Ωt x1
1
≤ ϕ2 (t)k∆un k2L2 (Ωt ) ,
δ
where δ is the constant which appears in (1.1). Using Lemma 2.3 and the condition
(1.1), we obtain the desired inequality.
(b) We have
Z
k∂x2 vk2L2 (D(0,1)) =
(∂x2 v)2 (x1 , x2 ) dx1 dx2
D(0,1)
Z
1
dx01 dx02
=
(∂x02 un )2 (x01 , x02 )h2 (t)ϕ2 (t)
2 (t)
h(t)ϕ
Ωt
Z
(∂x02 un )2 (x01 , x02 ) dx01 dx02
= h(t)
Ωt
= h(t)k∂x02 un k2L2 (Ωt ) .
On the other hand,
k∆vk2L2 (D(0,1)) ≤
1 2
ϕ (t)k∆un k2L2 (Ωt ) .
δ
Using the inequality
k∂x2 vk2L2 (D(0,1)) ≤ Ck∆vk2L2 (D(0,1))
of Lemma 2.3 and condition (1.1), we obtain the desired inequality
k∂x02 un k2L2 (Ωt ) ≤ Cϕ2 (t)k∆un k2L2 (Ωt ) .
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RESOLUTIONS OF PARABOLIC EQUATIONS
5
3. Local in time result
3.1. Case of a truncated domain Qα . In this subsection, we replace Q by Qα
Qα = {(t, x1 , x2 ) ∈ R3 :
1
x2
x2
< t < T, 0 ≤ 2 1 + 2 2 2
< 1}
α
ϕ (t) h (t)ϕ (t)
with α > 0.
Theorem 3.1. The problem
∂t u − ∂x21 u − ∂x22 u = f ∈ L2 (Qα ),
u
= 0,
(3.1)
∂Qα −ΓT
admits a unique solution u ∈ H 1,2 (Qα ).
Proof. The change of variables
(t, x1 , x2 ) 7→ (t, y1 , y2 ) = (t,
x2
x1
,
)
ϕ(t) h(t)ϕ(t)
transforms Qα into the cylinder Pα =] α1 , T [×D( α1 , 1), where D( α1 , 1) is the unit disk
centered on ( α1 , 0, 0). Putting u(t, x1 , x2 ) = v(t, y1 , y2 ) and f (t, x1 , x2 ) = g(t, y1 , y2 ),
then Problem (3.1) is transformed, in Pα into the variable-coefficient parabolic
problem
1
ϕ0 (t)y1
(hϕ)0 (t)y2
1
∂y22 v −
∂y1 v −
∂y v = g
∂t v − 2 ∂y21 v − 2
2
ϕ (t)
h (t)ϕ (t)
ϕ(t)
h(t)ϕ(t) 2
v
= 0.
∂Pα −ΓT
This change of variables conserves the spaces H 1,2 and L2 . In other words
f ∈ L2 (Qα ) ⇒ g ∈ L2 (Pα )
u ∈ H 1,2 (Qα ) ⇒ v ∈ H 1,2 (Pα ).
Proposition 3.2. The operator
ϕ0 (t)y1
(hϕ)0 (t)y2
−
∂y1 +
∂y2 : H01,2 (Pα ) → L2 (Pα )
ϕ(t)
h(t)ϕ(t)
is compact.
Proof. Pα has the horn property of Besov (see [3]). So, for j = 1, 2
∂yj
H01,2 (Pα ) →
v
7→
1
H 2 ,1 (Pα )
∂yj v,
1
is continuous. Since Pα is bounded, the canonical injection is compact from H 2 ,1 (Pα )
into L2 (Pα ) (see for instance [3]), where
1
1
1
1
, T ; H 1 D( , 1) ∩ H 1/2
, T ; L2 D( , 1) .
H 1/2,1 (Pα ) = L2
α
α
α
α
For the complete definitions of the H r,s Hilbertian Sobolev spaces see for instance
[11].
Consider the composition
∂yj : H01,2 (Pα ) →
v
7→
1
H 2 ,1 (Pα ) →
∂yj v
7→
L2 (Pα )
∂yj v,
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A. KHELOUFI
EJDE-2012/116
0
0
(hϕ) (t)
(t)
, − h(t)ϕ(t)
then ∂yj is a compact operator from H01,2 (Pα ) into L2 (Pα ). Since − ϕϕ(t)
0
0
(t)y1
(t)y2
are bounded functions, the operators − ϕϕ(t)
∂y1 , − (hϕ)
h(t)ϕ(t) ∂y2 are also compact
from H01,2 (Pα ) into L2 (Pα ). Consequently,
ϕ0 (t)y1
(hϕ)0 (t)y2
∂y1 +
∂y2
−
ϕ(t)
h(t)ϕ(t)
is compact from H01,2 (Pα ) to L2 (Pα ).
So, to complete the proof of Theorem 3.1, it is sufficient to show that the operator
1
1
∂t − 2 ∂y21 − 2
∂2
ϕ (t)
h (t)ϕ2 (t) y2
is an isomorphism from H01,2 (Pα ) into L2 (Pα ).
Lemma 3.3. The operator
∂t −
1
1
∂2 −
∂2
ϕ2 (t) y1 h2 (t)ϕ2 (t) y2
is an isomorphism from H01,2 (Pα ) to L2 (Pα ).
1
Proof. Since the coefficients ϕ21(t) and h2 (t)ϕ
2 (t) are bounded in Pα , the optimal
regularity is given by Ladyzhenskaya-Solonnikov-Ural’tseva [10].
We shall need the following result to justify the calculus of this section.
Lemma 3.4. The space
{u ∈ H 4 (Pα ) : u∂
p Pα
= 0}
is dense in the space
{u ∈ H 1,2 (Pα ) : u∂
p Pα
= 0}.
Here, ∂p Pα is the parabolic boundary of Pα and H 4 stands for the usual Sobolev
space defined, for instance, in Lions-Magenes [11].
The proof of the above lemma can be found in [7].
Remark 3.5. In Lemma 3.4, we can replace Pα by Qα with the help of the change
of variables defined above.
3.2. Case of a conical type domain. In this case, we define Q by
Q = {(t, x1 , x2 ) ∈ R3 : 0 < t < T, 0 ≤
x22
x21
+
< 1}
ϕ2 (t) h2 (t)ϕ2 (t)
with
ϕ(0) = 0,
ϕ(t) > 0,
t ∈]0, T ].
(3.2)
We assume that the functions h and ϕ satisfy conditions (1.1) and (1.3). For each
n ∈ N∗ , we define Qn by
Qn = {(t, x1 , x2 ) ∈ R3 :
x2
x2
1
< t < T, 0 ≤ 2 1 + 2 2 2
< 1}
n
ϕ (t) h (t)ϕ (t)
and we denote fn = f/Qn and un ∈ H 1,2 (Qn ) the solution of Problem (1.2) in Qn .
Such a solution exists by Theorem 3.1.
EJDE-2012/116
RESOLUTIONS OF PARABOLIC EQUATIONS
7
Proposition 3.6. There exists a constant K1 independent of n such that
kun kH 1,2 (Qn ) ≤ K1 kfn kL2 (Qn ) ≤ K1 kf kL2 (Q) ,
where kun kH 1,2 (Qn ) = kun k2H 1 (Qn ) +
2
P
i,j=1
k∂xi ∂xj un k2L2 (Qn )
1/2
.
To prove Proposition 3.6, we need the following result which is a consequence of
Lemma 2.4 and Grisvard-Looss [5] (see Theorem 2.2).
Lemma 3.7. There exists a constant C > 0 independent of n such that
k∂x21 un k2L2 (Qn ) + k∂x22 un k2L2 (Qn ) + k∂x21 x2 un k2L2 (Qn ) ≤ Ck∆un k2L2 (Qn ) .
Proof of Proposition 3.6. Let us denote the inner product in L2 (Qn ) by h·, ·i,
then we have
kfn k2L2 (Qn ) = h∂t un − ∆un , ∂t un − ∆un i
= k∂t un k2L2 (Qn ) + k∆un k2L2 (Qn ) − 2h∂t un , ∆un i
Estimation of −2h∂t un , ∆un i: We have
1
∂t un .∆un = ∂x1 (∂t un ∂x1 un ) + ∂x2 (∂t un ∂x2 un ) − ∂t [(∂x1 un )2 + (∂x2 un )2 ].
2
Then
Z
−2h∂t un , ∆un i = −2
∂t un .∆un dt dx1 dx2
Qn
Z
= −2
[∂x1 (∂t un ∂x1 un ) + ∂x2 (∂t un ∂x2 un )]dt dx1 dx2
Qn
Z
+
∂t [(∂x1 un )2 + (∂x2 un )2 ]dt dx1 dx2
Qn
Z
=
[|∇un |2 νt − 2∂t un (∂x1 un νx1 + ∂x2 un νx2 )]dσ
∂Qn
where νt , νx1 , νx2 are the components of the unit outward normal vector at ∂Qn .
We shall rewrite the boundary integral making use of the boundary conditions. On
the part of the boundary of Qn where t = n1 , we have un = 0 and consequently
∂x1 un = ∂x2 un = 0. The corresponding boundary integral vanishes. On the part
of the boundary where t = T , we have νx1 = 0, νx2 = 0 and νt = 1. Accordingly
the corresponding boundary integral
Z
A=
|∇un |2 dx1 dx2
ΓT
is nonnegative. On the part of the boundary where
νx1 = q
νx2 = q
νt = q
x21
ϕ2 (t)
+
x22
h2 (t)ϕ2 (t)
= 1, we have
h(t) cos θ
,
(ϕ0 (t)h(t) cos2 θ + (hϕ)0 (t) sin2 θ)2 + (h(t) cos θ)2 + sin2 θ
sin θ
(ϕ0 (t)h(t) cos2 θ + (hϕ)0 (t) sin2 θ)2 + (h(t) cos θ)2 + sin2 θ
−(ϕ0 (t)h(t) cos2 θ + (hϕ)0 (t) sin2 θ)
(ϕ0 (t)h(t) cos2 θ + (hϕ)0 (t) sin2 θ)2 + (h(t) cos θ)2 + sin2 θ
,
8
A. KHELOUFI
EJDE-2012/116
and un (t, ϕ(t) cos θ, h(t)ϕ(t) sin θ) = 0. Differentiating with respect to t then with
respect to θ we obtain
∂t un = −ϕ0 (t) cos θ.∂x1 un − (hϕ)0 (t) sin θ.∂x2 un ,
sin θ.∂x1 un = h(t) cos θ.∂x2 un .
Consequently the corresponding boundary integral is
Z 2π Z T
Jn = −2
∂t un .(hϕ cos θ.∂x1 un + hϕ sin θ.∂x2 un ) dt dθ
0
1/n
2π
Z
T
Z
|∇un |2 ((hϕ)0 ϕ sin2 θ + ϕ0 (hϕ) cos2 θ) dt dθ
−
0
1/n
2π
Z
T
Z
{(ϕ0 cos θ.∂x1 un + (hϕ)0 sin θ.∂x2 un )
=2
0
1/n
× (hϕ cos θ.∂x1 un + hϕ sin θ.∂x2 un )} dt dθ
Z 2π Z T
|∇un |2 ((hϕ)0 ϕ sin2 θ + ϕ0 hϕ cos2 θ) dt dθ
−
0
1/n
2π
Z
T
Z
|∇un |2 ((hϕ)0 ϕ sin2 θ + ϕ0 hϕ cos2 θ) dt dθ
=2
0
Z
1/n
2π
T
Z
|∇un |2 ((hϕ)0 ϕ sin2 θ + ϕ0 hϕ cos2 θ) dt dθ
−
0
Z
2π
1/n
Z
T
|∇un |2 ((hϕ)0 ϕ sin2 θ + ϕ0 hϕ cos2 θ) dt dθ.
=
0
1/n
Finally,
2π
Z
Z
T
|∇un |2 ((hϕ)0 ϕ sin2 θ + ϕ0 hϕ cos2 θ) dt dθ
−2h∂t un , ∆un i =
0
1/n
Z
(3.3)
2
|∇un | (T, x1 , x2 ) dx1 dx2 .
+
ΓT
Lemma 3.8. One has
Z
−2h∂t un , ∆un i = 2
(
Qn
Z
(hϕ)0
ϕ0
x1 ∂x1 un +
x2 ∂x2 un )∆un dt dx1 dx2
ϕ
hϕ
|∇un |2 (T, x1 , x2 ) dx1 dx2 .
+
ΓT
Proof. For
1
n
< t < T , consider the following parametrization of the domain Ωt
(0, 2π) →
θ
→
Ωt
(ϕ(t) cos θ, h(t)ϕ(t) sin θ) = (x1 , x2 ).
Let us denote the inner product in L2 (Ωt ) by h·, ·i, and set
(hϕ)0
ϕ0
x2 ∂x2 un
In = ∆un , x1 ∂x1 un +
ϕ
hϕ
then we have
Z
ϕ0
(hϕ)0
In =
(∂x21 un + ∂x22 un )( x1 ∂x1 un +
x2 ∂x2 un ) dx1 dx2
ϕ
hϕ
Ωt
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RESOLUTIONS OF PARABOLIC EQUATIONS
9
(hϕ)0
ϕ0
x1 ∂x21 un ∂x1 un +
x2 ∂x22 un ∂x2 un ) dx1 dx2
hϕ
Ωt ϕ
Z
(hϕ)0
ϕ0
x2 ∂x21 un ∂x2 un ) dx1 dx2 .
+
( x1 ∂x22 un ∂x1 un +
hϕ
Ωt ϕ
Z
=
(
Using Green formula, we obtain
Z
1
ϕ0
(hϕ)0
In =
x2 ∂x2 (∂x2 un )2 ) dx1 dx2
( x1 ∂x1 (∂x1 un )2 +
2 Ωt ϕ
hϕ
Z
ϕ0
(hϕ)0
x2 ∂x1 (∂x1 un )∂x2 un ) dx1 dx2
+
( x1 ∂x2 (∂x2 un )∂x1 un +
hϕ
Ωt ϕ
Z
ϕ0
(hϕ)0
1
x2 νx2 (∂x2 un )2 )dσ
( x1 νx1 (∂x1 un )2 +
=
2 ∂Ωt ϕ
hϕ
Z
1
ϕ0
(hϕ)0
−
( (∂x1 un )2 +
(∂x2 un )2 ) dx1 dx2
2 Ωt ϕ
hϕ
Z
ϕ0
(hϕ)0
+
( x1 νx2 +
x2 νx1 )∂x1 un ∂x2 un dσ
hϕ
∂Ω ϕ
Z t 0
ϕ
(hϕ)0
−
( x1 ∂x2 un ∂x21 x2 un +
x2 ∂x1 un ∂x21 x2 un ) dx1 dx2
hϕ
Ωt ϕ
where νx1 , νx2 are the components of the unit outward normal vector at ∂Ωt . Then
Z
(hϕ)0
ϕ0
1
x2 νx2 (∂x2 un )2 )dσ
( x1 νx1 (∂x1 un )2 +
In =
2 ∂Ωt ϕ
hϕ
Z
ϕ0
1
(hϕ)0
( (∂x1 un )2 +
−
(∂x2 un )2 ) dx1 dx2
2 Ωt ϕ
hϕ
Z
ϕ0
(hϕ)0
+
( x1 νx2 +
x2 νx1 )∂x1 un ∂x2 un dσ
hϕ
∂Ωt ϕ
Z
ϕ0
1
(hϕ)0
( x1 ∂x1 (∂x2 un )2 +
−
x2 ∂x2 (∂x1 un )2 ) dx1 dx2 .
2 Ωt ϕ
hϕ
Thus,
In =
ϕ0
(hϕ)0
x2 νx2 (∂x2 un )2 dσ
ϕ
hϕ
∂Ωt
Z 0
0
1
ϕ
(hϕ)
−
(∂x1 un )2 +
(∂x2 un )2 dx1 dx2
2 Ω ϕ
hϕ
Z t 0
0
ϕ
(hϕ)
+
x1 νx2 +
x2 νx1 ∂x1 un ∂x2 un dσ
ϕ
hϕ
∂Ωt
Z
0
1
ϕ
(hϕ)0
( x1 νx1 (∂x2 un )2 +
−
x2 νx2 (∂x1 un )2 ) dx1 dx2
2 ∂Ωt ϕ
hϕ
Z 0
1
ϕ
(hϕ)0
+
(∂x1 un )2 +
(∂x2 un )2 dx1 dx2
2 Ωt ϕ
hϕ
1
2
Z
x1 νx1 (∂x1 un )2 +
and then
In =
1
2
Z
∂Ωt
ϕ0
ϕ
x1 νx1 (∂x1 un )2 +
(hϕ)0
x2 νx2 (∂x2 un )2 dσ
hϕ
10
A. KHELOUFI
EJDE-2012/116
(hϕ)0
ϕ0
x1 νx2 +
x2 νx1 )∂x1 un ∂x2 un dσ
hϕ
∂Ωt ϕ
Z
ϕ0
(hϕ)0
1
x1 νx1 (∂x2 un )2 +
x2 νx2 (∂x1 un )2 dx1 dx2 .
−
2 ∂Ωt ϕ
hϕ
Z
+
(
Consequently,
In =
1
2
2π
Z
ϕ0
ϕ
0
ϕhϕ(cos θ.∂x1 un )2 +
(hϕ)0
ϕhϕ(sin θ.∂x2 un )2 dθ
hϕ
2π
ϕ0 2 (hϕ)0
ϕ +
(hϕ)2 ) sin θ cos θ.∂x1 un ∂x2 un dθ
ϕ
hϕ
0
Z
(hϕ)0
1 2π ϕ0
ϕhϕ(cos θ.∂x2 un )2 +
ϕhϕ(sin θ.∂x1 un )2 dθ
−
2 0
ϕ
hϕ
Z 2π 2
1
ϕ0 hϕ cos θ.∂x1 un + ϕ(hϕ)0 (sin θ.∂x2 un )2 dθ
=
2 0
Z 2π +
ϕ0 ϕ + (hϕ)0 hϕ sin θ cos θ.∂x1 un ∂x2 un dθ
Z
+
(
0
1
−
2
2π
Z
ϕ0 hϕ(cos θ.∂x2 un )2 + ϕ(hϕ)0 (sin θ.∂x1 un )2 dθ.
0
The boundary condition un (t, ϕ(t) cos θ, h(t)ϕ(t) sin θ) = 0 leads to
sin θ.∂x1 un = h(t) cos θ.∂x2 un ;
then
sin θ cos θ.∂x1 un ∂x2 un = h(t)(cos θ.∂x2 un )2
and
h(t) sin θ cos θ.∂x1 un ∂x2 un = (sin θ.∂x1 un )2 .
Consequently,
In =
1
2
Z
2π
ϕ0 hϕ(cos θ.∂x1 un )2 + ϕ(hϕ)0 (sin θ.∂x2 un )2 dθ
0
Z
+
2π
ϕ0 hϕ(cos θ.∂x2 un )2 + ϕ(hϕ)0 (sin θ.∂x1 un )2 dθ
0
Z
1 2π 0
−
(ϕ hϕ(cos θ.∂x2 un )2 + ϕ(hϕ)0 (sin θ.∂x1 un )2 )dθ
2 0
Z
1 2π 0
=
ϕ hϕ(cos θ.∂x1 un )2 + ϕ(hϕ)0 (sin θ.∂x2 un )2 dθ
2 0
Z
1 2π 0
+
ϕ hϕ(cos θ.∂x2 un )2 + ϕ(hϕ)0 (sin θ.∂x1 un )2 dθ
2 0
Z
1 2π n 0
ϕ hϕ(cos θ.∂x1 un )2 + ϕ(hϕ)0 (sin θ.∂x2 un )2
=
2 0
o
+ ϕ0 hϕ(cos θ.∂x2 un )2 + ϕ(hϕ)0 (sin θ.∂x1 un )2 dθ
Z
1 2π
=
[(∂x1 un )2 + (∂x2 un )2 ](ϕ(hϕ)0 sin2 θ + ϕ0 hϕ cos2 θ)dθ.
2 0
EJDE-2012/116
RESOLUTIONS OF PARABOLIC EQUATIONS
So
1
2
In =
Z
11
2π
|∇un |2 (ϕ(hϕ)0 sin2 θ + ϕ0 hϕ cos2 θ)dθ
0
and
Z
T
2π
Z
1/n
|∇un |2 (ϕ(hϕ)0 sin2 θ + ϕ0 hϕ cos2 θ) dt dθ
0
Z
=2
(
Qn
ϕ0
(hϕ)0
x1 ∂x1 un +
x2 ∂x2 un )∆un dt dx1 dx2 .
ϕ
hϕ
Finally, by (3.3), it follows that
Z 0
(hϕ)0
ϕ
x1 ∂x1 un +
x2 ∂x2 un ∆un dt dx1 dx2
−2h∂t un , ∆un i = 2
ϕ
hϕ
Q
Z n
+
|∇un |2 (T, x1 , x2 ) dx1 dx2 .
ΓT
Now, we continue the proof of Proposition 3.6. We have
Z 0
(hϕ)0
ϕ
x1 ∂x1 un +
x2 ∂x2 un ∆un dt dx1 dx2 |
|
ϕ
hϕ
Qn
≤ k∆un kL2 (Qn ) k
(hϕ)0
ϕ0
x1 ∂x1 un kL2 (Qn ) + k∆un kL2 (Qn ) k
x2 ∂x2 un kL2 (Qn ) ,
ϕ
hϕ
but Lemma 2.4 yields
ϕ0
k x1 ∂x1 un k2L2 (Qn ) =
ϕ
Z
T
02
Z
ϕ (t)
1/n
Z
Ωt
T
ϕ02 (t)
≤
1/n
≤ C2
Z
(
Z
x1 2
) (∂x1 un )2 dt dx1 dx2
ϕ(t)
(∂x1 un )2 dt dx1 dx2
Ωt
T
(ϕ(t)ϕ0 (t))2
1/n
2 2
Z
(∆un )2 dt dx1 dx2
Ωt
≤ C k∆un k2L2 (Qn ) ,
since (ϕ(t)ϕ0 (t)) ≤ . Similarly, we have
k
(hϕ)0
x2 ∂x2 un k2L2 (Qn ) ≤ C 2 2 k∆un k2L2 (Qn ) .
hϕ
Then
Z
|
Qn
ϕ0
ϕ
x1 ∂x1 un +
(hϕ)0
x2 ∂x2 un ∆un dt dx1 dx2 | ≤ 2Ck∆un k2L2 (Qn ) .
hϕ
Therefore, Lemma 3.8 shows that
Z 0
ϕ
(hϕ)0
x1 ∂x1 un +
x2 ∂x2 un ∆un dt dx1 dx2 |2h∂t un , ∆un i| ≥ −2
ϕ
hϕ
Qn
Z
+
|∇un |2 (T, x1 , x2 ) dx1 dx2
ΓT
≥ −4Ck∆un k2L2 (Qn ) .
12
A. KHELOUFI
EJDE-2012/116
Hence
kfn k2L2 (Qn ) = k∂t un k2L2 (Qn ) + k∆un k2L2 (Qn ) − 2h∂t un , ∆un i
≥ k∂t un k2L2 (Qn ) + (1 − 4C)k∆un k2L2 (Qn ) .
Then, it is sufficient to choose such that 1 − 4C > 0 to get a constant K0 > 0
independent of n such that
kfn kL2 (Qn ) ≥ K0 kun kH 1,2 (Qn ) ,
and since
kfn kL2 (Qn ) ≤ kf kL2 (Qn ) ,
there exists a constant K1 > 0, independent of n satisfying
kun kH 1,2 (Qn ) ≤ K1 kfn kL2 (Qn ) ≤ K1 kf kL2 (Q) .
This completes the proof of Proposition 3.6.
Passage to the limit. We are now in position to prove the main result of this
work.
Theorem 3.9. Assume that the functions h and ϕ verify the conditions (1.1),
(1.3) and (3.2). Then, for T small enough, Problem (1.2) admits a unique solution
u ∈ H 1,2 (Q).
Proof. Choose a sequence Qn n = 1, 2, . . . , of truncated conical domains (see subsection 3.2) such that Qn ⊆ Q. Then we have Qn → Q, as n → ∞.
Consider the solution un ∈ H 1,2 (Qn ) of the Cauchy-Dirichlet problem
∂t un − ∂x21 un − ∂x22 un = f
un = 0,
in Qn
∂Qn −ΓT
where ΓT is the part of the boundary of Qn where t = T . Such a solution un exists
by Theorem 3.1. Let u
en the 0-extension of un to Q. By Proposition 3.6, we know
that there exists a constant C such that
2
X
ke
un kL2 (Q) + k∂t u
en kL2 (Q) +
k∂xj 1 ∂xj 2 u
en kL2 (Q) ≤ Ckf kL2 (Q) .
i,j=0, 1≤i+j≤2
This means that u
en , ∂t u
en , ∂xj 1 ∂xj 2 u
en for 1 ≤ i + j ≤ 2 are bounded functions in
2
L (Q). So for a suitable increasing sequence of integers nk , k = 1, 2, . . . , there exist
functions u, v, vi,j , 1 ≤ i + j ≤ 2 in L2 (Q) such that
u
enk * u
∂t u
enk * v
∂xj 1 ∂xj 2 u
enk * vi,j
weakly in L2 (Q) as k → ∞
weakly in L2 (Q) as k → ∞
weakly in L2 (Q) as k → ∞, 1 ≤ i + j ≤ 2.
Clearly,
v = ∂t u, vi,j = ∂xi 1 ∂xj 2 u, 1 ≤ i + j ≤ 2
in the sense of distributions in Q and so in L2 (Q). So, u ∈ H 1,2 (Q) and
∂t u − ∂x21 u − ∂x22 u = f
in Q.
On the other hand, the solution u satisfies the boundary conditions u∂Q−Γ = 0
T
since uQn = un for all n ∈ N∗ . This proves the existence of a solution to Problem
(1.2).
EJDE-2012/116
RESOLUTIONS OF PARABOLIC EQUATIONS
13
4. Global in time result
Assume that Q satisfies (3.2). In the case where T is not in the neighborhood of
zero, we set Q = D1 ∪ D2 ∪ ΓT1 where
x21
x22
+
< 1}
ϕ2 (t) (hϕ)2 (t)
x22
x2
< 1}
D2 = {(t, x1 , x2 ) ∈ R3 : T1 < t < T, 0 ≤ 2 1 +
ϕ (t) (hϕ)2 (t)
x2
x22
ΓT1 = {(T1 , x1 , x2 ) ∈ R3 : 0 ≤ 2 1 +
< 1}
ϕ (T1 ) (hϕ)2 (T1 )
D1 = {(t, x1 , x2 ) ∈ R3 : 0 < t < T1 , 0 ≤
with T1 small enough.
In the sequel, f stands for an arbitrary fixed element of L2 (Q) and fi = f D ,
i
i = 1, 2.
Theorem 3.9 applied to the conical domain D1 , shows that there exists a unique
solution u1 ∈ H 1,2 (D1 ) of the problem
∂t u1 − ∂x21 u1 − ∂x22 u1 = f1 , f1 ∈ L2 (D1 )
u1 ∂D1 −Γ = 0.
(4.1)
T1
Hereafter, we denote the trace u1/ΓT1 by ψ which is in the Sobolev space H 1 (ΓT1 )
because u1 ∈ H 1,2 (D1 ) (see [11]).
Now, consider the following problem in D2 ,
∂t u2 − ∂x21 u2 − ∂x22 u2 = f2
f2 ∈ L2 (D2 )
u2/ΓT1 = ψ
u2 ∂D2 −(ΓT1 ∪ΓT )
(4.2)
=0
We use the following result, which is a consequence of [11, Theorem 4.3, Vol. 2],
to solve Problem (4.2).
Proposition 4.1. Let Q be the cylinder ]0, T [×D(0, 1), f ∈ L2 (Q) and ψ ∈ H 1 (γ0 ).
Then, the problem
∂t u − ∂x21 u − ∂x22 u = f in Q
u γ0 = ψ
uγ0 ∪γ1 = 0
where γ0 = {0} × D(0, 1), γ1 =]0, T [×∂D(0, 1), admits a (unique) solution u ∈
H 1,2 (Q).
Remark 4.2. In the application of [11, Theorem 4.3, Vol.2], we can observe that
there are no compatibility conditions to satisfy because ∂x ψ is only in L2 (γ0 ).
Thanks to the transformation
(t, x1 , x2 ) 7→ (t, y1 , y2 ) = (t, ϕ(t)x1 , (hϕ)(t)x2 ),
we deduce the following result.
Proposition 4.3. Problem (4.2) admits a (unique) solution u2 ∈ H 1,2 (D2 ).
14
A. KHELOUFI
EJDE-2012/116
So, the function u defined by
(
u1
u=
u2
in D1
in D2
is the (unique) solution of Problem (1.2) for an arbitrary T . Our second main result
is as follows.
Theorem 4.4. Assume that the functions h and ϕ verify conditions (1.1), (1.3)
and (3.2). Then, Problem (1.2) admits a unique solution u ∈ H 1,2 (Q).
Acknowledgments. The author is thankful to Professor B. K. Sadallah (Ecole
Normale Supérieure de Kouba, Algeria) for his help during the preparation of this
work, and to the anonymous referees for their careful reading of a previous version
of the manuscript, which led to a substantial improvement of this manuscript.
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Arezki Kheloufi
Department of Technology, Faculty of Technology, Béjaia University, 6000 Béjaia,
Algeria
E-mail address: arezkinet2000@yahoo.fr