STUDY OF A PARABOLIC PROBLEM IN A CONICAL DOMAIN
Boubaker-Khaled Sadallah
Abstract. In this paper we consider the heat equation with Dirichlet boundary conditions in a conical domain. We look for a sufficient con- dition on the lateral surface of the cone in order to have the optimal regularity of the solution in an anisotropic Sobolev space when the right hand side of the equation is in a Lebesgue space.
1. Intoduction
Let ϕ be a locally Lipschitz function ϕ : (0, T ] → R + where T > 0, such that
ϕ(0) = 0,
ϕ (t) > 0, t ∈ (0, T ], and Ω ⊂ R 3 be the open subset of conical type :
Ω = n
(t, x, y) ∈ R 3 : 0 < t < T, 0 ≤ p
x 2 + y 2 < ϕ(t) o . Consider in Ω the parabolic problem
(1.1)
∂ t u − ∂ x 2 u − ∂ y 2 u = f ∈ L 2 (Ω), u |∂Ω\D(T,ϕ(T )) = 0,
where D(s, r) denotes the disc of radius r centred at (s, 0, 0), and L 2 (Ω) is the usual Lebesgue space on Ω. We look for the solution u in the anisotropic Sobolev space
H 1,2 (Ω) = u ∈ H 1 (Ω) : ∂ x 2 u ∈ L 2 (Ω), ∂ xy u ∈ L 2 (Ω), ∂ y 2 u ∈ L 2 (Ω) , here, H 1 (Ω) stands for the Sobolev space defined by
H 1 (Ω) = u ∈ L 2 (Ω) : ∂ t u ∈ L 2 (Ω), ∂ x u ∈ L 2 (Ω), ∂ y u ∈ L 2 (Ω) . The space H 1,2 (Ω) is equipped with the natural norm, that is
kuk H1,2 = q
kuk 2 H1+ k∂ x 2 u k 2 L2 +
∂ xy u
+ ∂ xy u
2 L
2+
∂ y 2 u
2 L
2.
Mathematics Subject Classification. 35K05, 35K20.
Key words and phrases. Key Words and Phrases. Heat equation, Parabolic equation, Nonregular domain, Cone.
157
The belonging of the solution u to H 1,2 (Ω) depends on the function ϕ. Our aim is to find a sufficient condition on ϕ, as weak as possible in order to obtain a solution u ∈ H 1,2 (Ω). Our main result is
Theorem 1. Assume that ϕ(0) = 0 and
t→0 lim ϕ(t)ϕ ′ (t) = 0 or
ϕ ′ (t) ≥ 0 a.e. in a neighborhood of 0.
Then, for all f ∈ L 2 (Ω), Problem (1) admits a (unique) solution u ∈ H 1,2 (Ω).
For example, if ϕ(t) = at α , Problem (1) admits a unique solution for any positive constants a and α. The same result holds true if ϕ(t) = t α ln t, with α > 1 2 .
Observe that the main difficulty is due to the condition ϕ(0) = 0 (instead of ϕ(0) > 0) which gives a conical point at (0, 0, 0) and then, it does not allow to transform the cone Ω into a usual cylindrical domain. This type of problems has been studied by Sadallah [14] in one space dimension for the parabolic operator ∂ t u + (−1) m ∂ x 2m u with m ≥ 1.
Alkhutov [1] has treated, in some weighted Sobolev L p -spaces, the heat equation in bounded and unbounded domains of paraboloid type. He has considered in [2] the case of the heat equation in a ball; this case corresponds here to ϕ(t) = pt(2R − t) in the neighborhood of 0 (for a ball of radius R centred at (R, 0, 0)). It is clear that this function satisfies the hypothesis of Theorem 1. In [15] Sadallah has obtained the optimal regularity of the solution for any disc Ω = n
(t, x) ∈ R 2 : 0 ≤ √
t 2 + x 2 < R o .
In the case of one space variable, the heat equation in non Hilbretian spaces has been considered in some works, for instance in Labbas et al. [10]
and [11]. Berroug et al. [3] have treated the same equation with a singular domain in H¨ older spaces, whereas Kheloufi et al. [7] have studied the case when the domain is cylindrical, not with respect to the time variable, but with respect to one of space variables.
Some authors have considered the singularities which appear in the solu- tions when the domain is non cylindrical as Kozlov and Maz’ya in [8] and [9].
We can find in Nazarov [13] some results about the solution of Neumann problem in a conical domains. In Degtyarev [5] a more general parabolic equation has been studied in domains with a conical point.
The method used here is to approach the domain Ω for T small enough by a sequence of subsomains (Ω n ) n∈N in which we can solve Problem (1).
More precisely, when T is any positive real number, we divide Ω into two
parts : Ω T′ , with T ′ small enough,
Ω T′ = (t, x, y) ∈ Ω : 0 < t < T ′ and
Ω T′T = (t, x, y) ∈ Ω : T ′ < t < T .
So, we obtain two solutions u 1 ∈ H 1,2 (Ω T′) in Ω T′ and u 2 ∈ H 1,2 (Ω T′T ) in Ω T
′T . Finally, we prove that the function u defined by
and u 2 ∈ H 1,2 (Ω T′T ) in Ω T
′T . Finally, we prove that the function u defined by
u :=
u 1 in Ω T′
u 2 in Ω T′T
is the solution of Problem (1) and has the optimal regularity, that is u ∈ H 1,2 (Ω).
It is not difficult to prove the uniqueness of the solution. Indeed, if we consider the inner product (∂ t u − ∂ x 2 u − ∂ y 2 u, u) L2 with ∂ t u − ∂ x 2 u − ∂ y 2 u = 0 and take into account the boundary conditions, we obtain
0 = (∂ t u − ∂ x 2 u − ∂ y 2 u, u) L2
= 1
2 Z
D(T,ϕ(T ))
u 2 dxdy + 1 2
Z
Ω |∇u| 2 dtdxdy
where |∇u| 2 = (∂ x u) 2 +(∂ y u) 2 . Then |∇u| 2 = 0. This leads to ∂ x u = ∂ y u = 0 and therefore, ∂ x 2 u = ∂ y 2 u = 0. From the equation ∂ t u − ∂ x 2 u − ∂ y 2 u = 0 we deduce that ∂ t u = 0. Accordingly, the solution u is constant, but it is null on one part of the boundary. So, u = 0. This proves the uniqueness of the solution of Problem (1). This is why we will be interested in the sequel only by the question of the existence of the solution when f ∈ L 2 (Ω).
Our work is motivated by the interest of researchers for many mathe- matical questions related to non-regular domains. In fact, some important applied problems reduce to the study of boundary-value problems for par- tial differential equations (Laplace equation, heat equation, Fokker-Planck equation, Chapman-Kolmogorov equation...) in domains with non-regular points on the boundary.
The plan of the paper is : In Section 2 we give a change of variables
which transforms a subdomain of Ω into a cylindrical domain, so we confine
ourselves to the neighborhood of the origin (0, 0, 0). Section 3 is devoted to
the proof of an estimate in order to get a subsequence of functions converging
to the solution of Problem (1) in a subdomain of Ω. Finally, in Section 4,
we complete the proof of Theorom 1.
2. Change of variables
Let ε > 0 be a real which we will choose small enough. The continuity of ϕ at 0 proves the existence of a real number T ′ < T such that :
(2.1) ∀t ∈ 0, T ′ , |ϕ(t)| ≤ ε.
Consider a decreasing sequence (t n ) n∈N of real numbers such that 0 <
t n < T ′ for all n ∈ N and lim
t→0
+t n = 0. We define the sequence of subdomains (Ω n ) n of Ω T′ by
Ω n = (t, x, y) ∈ Ω : t n < t < T ′ .
The solution u of Problem (1) will be approached by the solutions u n of the problems
(2.2)
∂ t u n − ∂ x 2 u n − ∂ y 2 u n = f n ∈ L 2 (Ω n ), u n|∂Ωn\D(T
′,ϕ(T
′)) = 0,
where f n = f |Ωn.
Then, we perform the change of variables : Ω n → (t n , T ′ ) × B(0, 1) (t, x, y) 7−→ (t, x
ϕ(t) , y
ϕ(t) ) = (t, x ′ , y ′ )
where B(0, 1) = (x, y) ∈ R 2 : x 2 + y 2 < 1 . Problem (3) is transformed into the problem
(2.3)
∂ t v n − ϕ21 (t) ∆v n − ϕ ϕ(t)
′(t) x ′ ∂ x
′v n + y ′ ∂ y′v n
v n
= g n ∈ L 2 ((t n , T ′ ) × B(0, 1)), v n|{tn}×B(0,1) = 0,
v n|(tn,T
′)×∂B(0,1) = 0,
here f n (t, x, y) = g n (t, x ′ , y ′ ), u n (t, x, y) = v n (t, x ′ , y ′ ) and ∆v n = ∂ x 2′v n +
∂ y 2′v n . It is easy to see that the operator ϕ ′ (t)
ϕ(t) x ′ ∂ x′ + y ′ ∂ y′ : H 1,2 ((t n , T ′ ) × B(0, 1)) → L 2 ((t n , T ′ ) × B(0, 1)) is compact and the problem
: H 1,2 ((t n , T ′ ) × B(0, 1)) → L 2 ((t n , T ′ ) × B(0, 1)) is compact and the problem
∂ t v n − ϕ21 (t) ∆v n = g n ∈ L 2 ((t n , T ′ ) × B(0, 1)), v n|{t
n}×B(0,1) = 0,
v n|(tn,T
′)×∂B(0,1) = 0,
admits a unique solution v n ∈ H 1,2 in the cylindrical domain (t n , T ′ ) ×
B(0, 1). This proves that Problem (4) has a unique solution
v n ∈ H 1,2 ((t n , T ′ ) × B(0, 1)). The previous change of variables shows that
Problem (3) has also a unique solution u n ∈ H 1,2 (Ω n ) for all n ∈ N. The goal
of the following section is to establish an estimate concerning the sequence (u n ) which allows us to extract a subsequence converging to the solution of Problem (1).
3. An estimate
Consider the sequence (u n ) and look for a constant C > 0 independent of n satisfying the estimate
(3.1) ku n k H1,2(Ω
n) ≤ C kf n k L
2(Ω
n) . For this purpose we will estimate
∂ t u n − ∂ x 2 u n − ∂ y 2 u n
2 L
2(Ω
n) .
Lemma 2. There exists a constant C > 0 independent of n such that for almost every t ∈ (0, T ′ )
ku n k L2(D(t,ϕ(t))) ≤ Cϕ 2 (t) k∆u n k L
2(D(t,ϕ(t))) , max
k∂ x u n k L2(D(t,ϕ(t))) , k∂ y u n k L
2(D(t,ϕ(t)))
≤ Cϕ(t) k∆u n k L2(D(t,ϕ(t))) . Proof. Let H 2 and H 0 1 be the usual Sobolev spaces defined, for instance, in Lions-Magenes [12]. We know that the Laplace operator ∆ : H 2 (D(0, 1)) ∩ H 0 1 (D(0, 1)) → L 2 (D(0, 1)) is an isomprphism. Then, there exists a constant C > 0 such that
(3.2) kvk H2(D(0,1)) ≤ C k∆vk L
2(D(0,1)) , ∀v ∈ H 2 (D(0, 1)).
Let t n < t < T ′ . The change of variables D(0, 1) → D(t, ϕ(t))
(x, y) 7−→ (ϕ(t)x, ϕ(t)y) = (x ′ , y ′ ) allows to write, thanks to (6) :
kvk L2(D(0,1)) = 1
ϕ(t) ku n k L2(D(t,ϕ(t))
≤ C k∆vk L2(D(0,1))
= Cϕ(t) k∆u n k L2(D(t,ϕ(t)) ,
with the notation v(x, y) = u n (x ′ , y ′ ). Notice that u n ∈ H 2 (D(t, ϕ(t)) for almost every t ∈ (t n , T ′ ). Hence
ku n k L2(D(t,ϕ(t)) ≤ Cϕ 2 (t) k∆u n k L
2(D(t,ϕ(t)) . On the other hand, we have
k∂ x′u n k L2(D(t,ϕ(t))) = k∂ x v k L
2(D(0,1))
(D(t,ϕ(t))) = k∂ x v k L
2(D(0,1))
≤ C k∆vk L2(D(0,1))
= Cϕ(t) k∆u n k L2(D(t,ϕ(t)) . The same result is true for
∂ y′u n
L2
(D(t,ϕ(t))) .
The following lemma (see Theorem 2.1, Lions-Magenes [12]) will justify some calculations in this section.
Lemma 3. The space
u ∈ H 2 ((0, T ) × D(0, 1)) : u |{0}×D(0,1) = 0, u |(0,T′)×∂D(0,1) = 0 is dense in the space
u ∈ H 1,2 ((0, T ) × D(0, 1)) : u |{0}×D(0,1) = 0, u |(0,T′)×∂D(0,1) = 0 . Now, consider the solution u n ∈ H 1,2 (Ω n ) of Problem (3) and let ∆u n =
∂ x 2 u n +∂ y 2 u n . Expanding the inner product kf n k 2 L2(Ω
n) = (∂ t u n −∆u n , ∂ t u n −
∆u n ) L2(Ω
n) we obtain
(3.3) kf n k 2 L2(Ω
n) = k∂ t u n k 2 L
2(Ω
n) + k∆u n k 2 L
2(Ω
n) − 2(∂ t u n , ∆u n ) L
2(Ω
n) . We deduce the following result from Lemma 2 and Grisvard-Ioss [6] (see Theorem 2.2).
Lemma 4. There exists a constant C > 0 independent of n such that (3.4)
∂ xy u n
2
L
2(Ω
n) +
∂ x 2 u n
2
L
2(Ω
n) +
∂ y 2 u n
2
L
2(Ω
n) ≤ C k∆u n k 2 L
2(Ω
n) . We have to estimate the term (∂ t u n , ∆u n ) L
2(Ω
n) . For this end, we need to prove the result
Proposition 5. One has
−2(∂ t u n , ∆u n ) L2(Ω
n)
= Z 2π
0
Z T′
t
n
|∇u n | 2 (t, ϕ(t) cos θ, ϕ(t) sin θ)ϕ(t)ϕ ′ (t)dtdθ +
Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy.
Proof. To establish this relationship we perform the following parametriza- tion of Γ 1 :
(t n , T ′ ) × (0, 2π) → Γ 1
(t, θ) 7→ (t, ϕ(t) cos θ, ϕ(t) sin θ) = (t, x, y) and we denote by ν = √ 1
1+ϕ
′2(t) (−ϕ ′ (t), cos θ, sin θ) = (ν t , ν x , ν y ) the out- ward unit normal vector to the part Γ 1 of ∂Ω n defined by
Γ 1 = n
(t, x, y) : p
x 2 + y 2 = ϕ(t) o
.
The other part of ∂Ω n will be denoted by Γ T′, i.e., Γ T′ = D(T ′ , ϕ(T ′ )).We have
= D(T ′ , ϕ(T ′ )).We have
(∂ t u n , ∆u n ) L2(Ω
n) = Z
Ω
n∂ t u n .∆u n dtdxdy
= Z
∂Ω
n∂ t u n .(ν x ∂ x u n + ν y ∂ y u n )dσ
− 1 2
Z
Ω
n∂ t (∂ x u n ) 2 + (∂ y u n ) 2 dtdxdy
= Z
Γ
1∂ t u n .(cos θ.∂ x u n + sin θ.∂ y u n ) 1
p1 + ϕ ′2 (t) dσ + 1
2 Z
Γ
1|∇u n | 2 ϕ ′ (t)
p1 + ϕ ′2 (t) dσ − 1 2
Z
Γ
T ′|∇u n | 2 dxdy.
The Dirichlet boundary condition on Γ 1 leads to
∂ t u n = −ϕ ′ (t) (cos θ.∂ x u n + sin θ.∂ y u n ) , sin θ.∂ x u n = cos θ.∂ y u n .
Taking into account these relationships we deduce (∂ t u n , ∆u n ) L2(Ω
n) = −
Z
Γ
1(cos θ.∂ x u n + sin θ.∂ y u n ) 2 ϕ ′ (t)
p1 + ϕ ′2 (t) dσ + 1
2 Z
Γ
1|∇u n | 2 ϕ ′ (t)
p1 + ϕ ′2 (t) dσ
− 1 2
Z
Γ
T ′|∇u n | 2 (T ′ , x, y)dxdy
= − Z 2π
0
Z T′
t
n
(cos θ.∂ x u n + sin θ.∂ y u n ) 2 ϕ(t)ϕ ′ (t)dtdθ + 1
2 Z 2π
0
Z T′
t
n
|∇u n | 2 ϕ(t)ϕ ′ (t)dtdθ
− 1 2
Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy
= − Z 2π
0
Z T′
t
n
|∇u n | 2 ϕ(t)ϕ ′ (t)dtdθ + 1
2 Z 2π
0
Z T′
t
n
|∇u n | 2 ϕ(t)ϕ ′ (t)dtdθ
− 1 2
Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy
= − 1 2
Z 2π 0
Z T′
t
n
|∇u n | 2 ϕ(t)ϕ ′ (t)dtdθ
− 1 2
Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy.
This ends the proof of the proposition.
Remark 6. Observe that the integral R
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy which appears in Proposition 5 is non negative.This is a good sign for our estimate because we can deduce immediately
kf n k 2 L2(Ω
n) ≥ k∂ t u n k 2 L
2(Ω
n) + k∆u n k 2 L
2(Ω
n) +
Z 2π 0
Z T′
t
n
|∇u n | 2 (t, ϕ(t) cos θ, ϕ(t) sin θ)ϕ(t)ϕ ′ (t)dtdθ.
So, in the sequel, we will attach no importance to this integral. On the other hand, if ϕ is an increasing function in the interval (t n , T ′ ), then
Z 2π 0
Z T′
t
n
|∇u n | 2 (t, ϕ(t) cos θ, ϕ(t) sin θ)ϕ(t)ϕ ′ (t)dtdθ ≥ 0.
Consequently,
(3.5) kf n k 2 L2(Ω
n) ≥ k∂ t u n k 2 L
2(Ω
n) + k∆u n k 2 L
2(Ω
n) . But, thanks to Lemma 2
Z T′
t
n
ku n k 2 L2(D(t,ϕ(t))) dt ≤ C 2 Z T
′
t
nϕ 4 (t) k∆u n k 2 L2(D(t,ϕ(t))) dt, Z T
′
t
nk∂ x u n k 2 L2(D(t,ϕ(t))) dt ≤ C 2 Z T
′
t
nϕ 2 (t) k∆u n k 2 L2(D(t,ϕ(t))) dt, Z T
′
t
nk∂ y u n k 2 L2(D(t,ϕ(t))) dt ≤ C 2 Z T
′
t
nϕ 2 (t) k∆u n k 2 L2(D(t,ϕ(t))) dt.
Since ϕ is bounded on (0, T ), there exists a constant C ′ > 0 such that max
ku n k 2 L2(Ω
n) , k∂ x u n k 2 L
2(Ω
n) , k∂ y u n k 2 L
2(Ω
n)
≤ C ′ k∆u n k 2 L2(Ω
n) . Taking into account (8) and (9), this proves (5). So, it remains to establish (5) under the hypothesis lim
t→0
+ϕ(t)ϕ ′ (t) = 0. For this purpose, we need the following proposition
Proposition 7. One has
−2(∂ t u n , ∆u n ) L2(Ω
n) = 2 Z
Ω
nϕ ′ (t)
ϕ(t) (x∂ x u n + y∂u n )∆u n dtdxdy
+ Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy.
Proof. For t n < t < T ′ , denote by D t the disc D(t, ϕ(t)) and consider the inner product (∆u n , x∂ x u n + y∂u n ) L2(D
t) . We have
(∆u n , x∂ x u n + y∂ y u n ) L2(D
t) = 1 2
Z
D
tx∂ x (∂ x u n ) 2 + y∂ y (∂ y u n ) 2 dxdy +
Z
∂D
t(xν y + yν x ) ∂ x u n ∂ y u n dσ
− Z
D
t(x∂ y u n ∂ xy u n + y∂ x u n ∂ xy u n ) dxdy.
Then
(∆u n , x∂ x u n + y∂ y u n ) L2(D
t) = 1 2
Z
∂D
txν x (∂ x u n ) 2 + yν y (∂ y u n ) 2 dσ
− 1 2
Z
D
t|∇u n | 2 dxdy +2
Z
∂D
t∂ x u n ∂ y u n cos θ sin θdσ
− 1 2
Z
D
tx∂ x (∂ y u n ) 2 + y∂ y (∂ x u n ) 2 dxdy.
The boundary condition u(t, ϕ(t) cos θ, ϕ(t) sin θ) = 0 leads to sin θ∂ x u n = cos θ∂ y u n . Consequently
(∆u n , x∂ x u n + y∂ y u n ) L2(D
t) = ϕ 2 (t) 2
Z 2π 0
(cos θ.∂ x u n ) 2 + (sin θ.∂ y u n ) 2 )dθ +ϕ 2 (t)
Z 2π 0
(sin θ.∂ x u n ) 2 + (cos θ.∂ y u n ) 2 )dθ
− 1 2
Z
D
t|∇u n | 2 dxdy
− ϕ 2 (t) 2
Z 2π 0
(sin θ.∂ x u n ) 2 + (cos θ.∂ y u n ) 2 )dθ + 1
2 Z
D
t|∇u n | 2 dxdy.
So
(∆u n , x∂ x u n + y∂ y u n ) L2(D
t) = ϕ 2 (t) 2
Z 2π
0 |∇u n | 2 dθ,
and 2
Z
Ω
nϕ ′ (t)
ϕ(t) (x∂ x u n + y∂ y u n )∆u n dtdxdy = Z T′
t
nZ 2π
0 |∇u n | 2 ϕ(t)ϕ ′ (t)dθdt.
Finally, in virtue of Proposition 5, it follows
−2(∂ t u n , ∆u n ) L2(Ω
n) = 2 Z
Ω
nϕ ′ (t)
ϕ(t) (x∂ x u n + y∂ y u n )∆u n dtdxdy +
Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy.
Theorem 8. Assume that lim
t→0 ϕ(t)ϕ ′ (t) = 0 or ϕ ′ (t) ≥ 0 in a neighborhood of 0.Then, there exists a constant C > 0 independent of n satisfying the estimate
ku n k H1,2(Ω
n) ≤ C kf n k L
2(Ω
n) .
Proof. The case when ϕ ′ (t) ≥ 0 in a neighborhood of 0 has been treated in Remark 6. Then, assume that lim
t→0 ϕ(t)ϕ ′ (t) = 0. We have
Z
Ω
nϕ ′ (t)
ϕ(t) (x∂ x u n + y∂ y u n )∆u n dtdxdy
≤ k∆u n k L2(Ω
n)
x.ϕ ′ (t) ϕ(t) ∂ x u n
L2
(Ω
n)
+ k∆u n k L2(Ω
n)
y.ϕ ′ (t) ϕ(t) ∂ y u n
L2(Ω
n)
(3.6) ,
but Lemma 2 yields
x.ϕ ′ (t) ϕ(t) ∂ x u n
2 L
2(Ω
n)
=
Z T′
t
n
ϕ ′2 (t) Z
D(t,ϕ(t))
x ϕ(t)
2
(∂ x u n ) 2 dtdxdy
≤
Z T′
t
n
ϕ ′2 (t) Z
D(t,ϕ(t))
(∂ x u n ) 2 dtdxdy
≤ C 2 Z T′
t
nϕ(t)ϕ ′ (t) 2 Z
D(t,ϕ(t))
(∆u n ) 2 dtdxdy.
The hypothesis lim
t→0 ϕ(t)ϕ ′ (t) = 0 implies the existence of 0 < α < 1 2 such that |ϕ(t)ϕ ′ (t)| < 2C α for all t ∈ (0, T ′ ) since T ′ is chosen small enough. So
x.ϕ ′ (t) ϕ(t) ∂ x u n
2 L
2(Ω
n)
≤ α 2
4 k∆u n k 2 L2(Ω
n) .
It is clear that we have also
y.ϕ ′ (t) ϕ(t) ∂ y u n
2 L
2(Ω
n)
≤ α 2
4 k∆u n k 2 L2(Ω
n) . Accordingly, Relationship (10) leads to
2 Z
Ω
nϕ ′ (t)
ϕ(t) (x∂ x u n + y∂ y u n )∆u n dtdxdy
≤ 2α k∆u n k 2 L2(Ω
n) . Therefore, Proposition 7 shows that
−2(∂ t u n , ∆u n ) L2(Ω
n) ≥ −2 Z
Ω
nϕ ′ (t)
ϕ(t) (x∂ x u n + y∂ y u n )∆u n dtdxdy +
Z
D(T
′,ϕ(T
′)) |∇u n | 2 (T ′ , x, y)dxdy
≥ −2α k∆u n k 2 L2(Ω
n) . Hence
kf n k 2 L2(Ω
n) = k∂ t u n k 2 L
2(Ω
n) + k∆u n k 2 L
2(Ω
n) − 2(∂ t u n , ∆u n ) L
2(Ω
n)
≥ k∂ t u n k 2 L2(Ω
n) + (1 − 2α) k∆u n k 2 L
2(Ω
n) .
Since 2α < 1, this inequality with Lemma 4 completes the proof. 4. Proof of Theorom 1
From the previous theorem we deduce that ku n k H1,2(Ω
n) ≤ C kf k L
2(Ω) for all n ∈ N because kf n k L
2(Ω
n) ≤ kf k L
2(Ω) . For any function v defined in Ω n , denote by ˜ v the function
˜ v =
v : in Ω n
0 : in Ω T′\Ω n . It is then possible to extract a subsequence ˜ u nj
j from (˜ u n ) ∈ L 2 (Ω T
′) which converges weakly in H 1,2 (Ω T′). Denote by u 1 the limit of ˜ u nj
j . Therefore, passing to the limit for n j → +∞, we get
(4.1)
∂ t u 1 − ∂ x 2 u 1 − ∂ y 2 u 1 = f |ΩT ′ ∈ L 2 (Ω T′), u 1|∂ΩT ′\D(T,ϕ(T )) = 0.
), u 1|∂ΩT ′\D(T,ϕ(T )) = 0.
This proves Theorem 1 in the subdomain Ω T′.
Let T be any positive real number, f ∈ L 2 (Ω), and T ′ < T small enough in order to Problem 11 admits a solution u 1 ∈ H 1,2 (Ω T′)
We have to solve Problem (1) in Ω. We know (see Section 2) that the
Cauchy-Dirichlet problem
(4.2)
∂ t v − ∂ x 2 v − ∂ y 2 v = f |ΩT ′ T ∈ L 2 (Ω T′T ), v |∂Ω
T ′T\D(T,ϕ(T )) = 0.
T ), v |∂Ω
T ′T\D(T,ϕ(T )) = 0.
has a unique solution v ∈ H 1,2 (Ω T′T ).
We need the following trace result (see Lions-Magenes [12])
Lemma 9. For v ∈ H 1,2 (Ω T′T ), one has v |D(T
′,ϕ(T
′)) ∈ H 1 (D(T ′ , ϕ(T ′ ))).
Moreover, for each ψ ∈ H 0 1 (D(T ′ , ϕ(T ′ ))), there exists w ∈ H 1,2 (Ω T′T ) such that w |D(T
′,ϕ(T
′)) = ψ and w |∂Ω
T ′ T\D(T
′,ϕ(T
′)) = 0 .
Since u 1 ∈ H 1,2 (Ω T′), from this lemma and the homogeneous Problem (12), we obtain
Proposition 10. The problem (4.3)
∂ t u 2 − ∂ x 2 u 2 − ∂ y 2 u 2 = f |ΩT ′ T ∈ L 2 (Ω T′T ), u 2|D(T
′,ϕ(T
′)) = u 1|D(T
′,ϕ(T
′)) ,
T ), u 2|D(T
′,ϕ(T
′)) = u 1|D(T
′,ϕ(T
′)) ,
u 2|∂Ω
T ′ T