ON THE WEAK SOLUTION OF A THREE-POINT
BOUNDARY VALUE PROBLEM FOR A CLASS
OF PARABOLIC EQUATIONS WITH ENERGY
SPECIFICATION
ABDELFATAH BOUZIANI
Received 3 September 2002
This paper deals with weak solution in weighted Sobolev spaces, of three-point
boundary value problems which combine Dirichlet and integral conditions, for
linear and quasilinear parabolic equations in a domain with curved lateral
boundaries. We, firstly, prove the existence, uniqueness, and continuous dependence of the solution for the linear equation. Next, analogous results are established for the quasilinear problem, using an iterative process based on results
obtained for the linear problem.
1. Introduction
Many physical phenomena are modeled by one-dimensional second-order parabolic equation which involves nonlocal boundary condition of the form
b
0
θ(x,t)dx = E(t),
(1.1)
the so-called energy specification, where b ∈]0,1] is a constant, θ(x,t) is an unknown function, and E(t) is a given function.
For heat conduction theory, condition (1.1) represents the internal energy
content of a portion of the conductor [3, 9, 11, 16, 17]. For diffusion processes,
the condition is equivalent to the specification of mass in a portion of the domain of diffusion [3, 10, 12]. We note that such problems have other important
applications, for instance, in thermoelasticity [15, 18], in electrochemistry [14],
and in medical science [13].
In this paper, we prove in weighted Sobolev spaces the existence, uniqueness,
and continuous dependence of the weak solution of three-point boundary value
problem for a class of linear and quasilinear parabolic equations with nonlocal
condition in a domain with curved boundaries varying with respect to the time.
Copyright © 2003 Hindawi Publishing Corporation
Abstract and Applied Analysis 2003:10 (2003) 573–589
2000 Mathematics Subject Classification: 35K20, 35K55, 35B45, 35D05, 35B30
URL: http://dx.doi.org/10.1155/S1085337503210010
574
Parabolic equations with energy specification
We will first investigate the linear case. Particular cases of it have been treated by
several authors; most of the works were directed to strongly generalized solution
for two-point boundary value problems [1, 2, 3, 4, 5, 6, 7, 19] and to the classical
solutions of the heat equation [8, 9, 10, 11, 12, 16, 17]. In contrast to previous
papers, we consider a weak solution by using a functional analysis method based
on a priori estimates. Then, we investigate the quasilinear problem by combining
an iterative process with results established for the linear case.
The outline of the paper is as follows. In Section 2, we study the linear problem. In Section 2.1, we give the statement of the problem, the basic assumptions,
and some function spaces needed in the remainder of the work. Section 2.2 is devoted to establishing the existence of the solution. The uniqueness and continuous dependence with respect to the data are proved in Section 2.3. In Section 3,
analogous investigation for the quasilinear problem is considered.
2. The linear problem
2.1. Statement of the problem, hypothesis, and notations. Let Γ p (τ), where
τ ∈ I = (0,T) and p = 1,2,3, be nonintersecting curves, varying with time, in
the plane (ξ,τ), such that Γ1 (τ) < Γ3 (τ) < Γ2 (τ). In Q = (Γ1 (τ),Γ2 (τ)) × I, we
consider the following problem: given a sufficiently smooth data Γ p (p = 1,2,3),
h, θ 0 , Ψ, and E, find the solution θ(ξ,τ) of
∂θ
∂θ
∂θ
∂
−
p1 (ξ,τ)
+ p2 (ξ,τ) + p3 (ξ,τ)θ = h(ξ,τ),
∂τ ∂ξ
∂ξ
∂ξ
Γ1 (0) < ξ < Γ2 (0),
θ(ξ,0) = θ 0 (ξ),
θ Γ2 (τ),τ = Ψ(τ),
Γ3 (τ)
Γ1 (τ)
(ξ,τ) ∈ Q,
(2.1)
0 < τ < T,
θ(ξ,τ)dξ = E(τ),
0 < τ < T,
with
Γ3 (0)
θ 0 Γ2 (0) = Ψ(0),
Γ1 (0)
θ 0 (ξ)dξ = E(0).
(2.2)
Problem (2.1) can be reduced to a problem with homogeneous boundary
conditions by setting
θ(ξ,τ) = Θ(ξ,τ) + Ψ(τ)
2 Γ2 (τ) − ξ Ψ(τ) Γ3 (τ) − Γ1 (τ) − E(τ)
,
− 2Γ2 (τ) − Γ1 (τ) − Γ3 (τ) Γ3 (τ) − Γ1 (τ)
(2.3)
Abdelfatah Bouziani 575
then the original problem is equivalent to the problem, for Θ(ξ,τ) in Q:
∂Θ ∂
∂Θ
∂Θ
−
p1 (ξ,τ)
+ p2 (ξ,τ)
+ p3 (ξ,τ)Θ = g(ξ,τ),
∂τ ∂ξ
∂ξ
∂ξ
(ξ,τ) ∈ Q,
(2.4)
Θ(ξ,0) = Θ0 (ξ),
Γ3 (τ)
Γ1 (τ)
Γ1 (0) < ξ < Γ2 (0),
Θ Γ2 (τ),τ = 0,
Θ(ξ,τ)dξ = 0,
(2.5)
0 < τ < T,
0 < τ < T,
(2.6)
Θ0 (ξ)dξ = 0.
(2.7)
with
Θ0 Γ2 (0) = 0,
Γ3 (0)
Γ1 (0)
We now introduce new variables x and t connected with ξ and τ by the relations
 b ξ − Γ1 (τ)



,


 Γ3 (τ) − Γ1 (τ)
Γ1 (τ) < ξ < Γ3 (τ),


(1 − b)ξ + bΓ2 (τ) − Γ3 (τ)


,

Γ2 (τ) − Γ3 (τ)
Γ3 (τ) < ξ < Γ2 (τ),
x=
(2.8)
where t = τ. Under transformation (2.8), the region Q becomes the rectangle
Ω = {(x,t) : 0 < x < 1, 0 < t < T }. In the new variable, problem (2.4), (2.5), and
(2.6) assumes the form
∂u
∂u
∂u ∂
−
q1 (x,t)
+ q2 (x,t) + q3 (x,t)u = f (x,t),
∂t ∂x
∂x
∂x
u(x,0) = u0 (x), 0 < x < 1,
b
0
(x,t) ∈ Ω,
(2.9)
(2.10)
u(1,t) = 0,
0 < t < T,
u(x,t)dx = 0,
0 < t < T,
(2.11)
u0 (x)dx = 0.
(2.12)
with
u0 (1) = 0,
b
0
Thus, problem (2.1) for θ(ξ,τ) in the region Q has been reduced to problem
(2.9), (2.10), and (2.11) for u(x,t) in the rectangle Ω.
576
Parabolic equations with energy specification
Assumption 2.1. For all (x,t) ∈ Ω, we assume that
0 < c0 ≤ q1 ≤ c1 ,
∂q1 ∂t ≤ c2 ,
∂q1 ∂x ≤ c3 ,
q2 ≤ c4 ,
q3 ≤ c5 .
(2.13)
Here and in the rest of the paper ci are positive constants.
We now introduce some function spaces which are related to the study. By
L2 (0,1) we represent the usual space of Lebesgue square integrable functions
on (0,1) whose scalar product and norm will be denoted by (·, ·)L2 (0,1) and
· L2 (0,1) , respectively. By C0 (0,b) we denote the space of continuous functions with compact support in (0,b). Let H be a Hilbert space with the norm
uH and let u : I → H be an abstract function. By u(·,t)H we denote the
norm of the element u(·,t) ∈ H at a fixed t. The space C(I,H) is the set of all
continuous functions u : I → H with uC(I,H) = maxt∈[I] u(·,t)H < ∞, and
L2 (I,H) is the set of the measurable abstract functions u such that uL2 (I,H) =
( I u(·,t)2H dt)1/2 < ∞.
Definition 2.2. Denote by L2ρ (0,1) the weighted L2 -space with the scalar product
(u,w)L2ρ (0,1) = (ρu,w)L2 (0,1)
(2.14)
uL2ρ (0,1) = ρuL2 (0,1) ,
(2.15)
and the associated norm
where ρ(x) is a continuous function defined by

x2 ,
0 ≤ x ≤ b,
ρ(x) =  2
b , b ≤ x ≤ 1.
(2.16)
Definition 2.3. Denote by Hρ1 (0,1) the space equipped with the scalar product
(u,w)Hρ1 (0,1) := (u,w)L2 (0,1) +
du dw
,
dx dx
(2.17)
L2ρ (0,1)
and the associated norm
uHρ1 (0,1) :=
2
du u2L2 (0,1) + dx
1/2
.
(2.18)
L2ρ (0,1)
Definition 2.4. Denote by B21,∗ (0,b) a completion of the space C0 (0,b) for the
scalar product
(u,w)B21,∗ (0,b) :=
b
0
∗
∗
x ux w dx
(2.19)
Abdelfatah Bouziani 577
and the associated norm
uB21,∗ (0,b) := ∗
x u L2 (0,b) ,
where ∗x u =
b
x
(2.20)
u(η, ·)dη.
As L2 (0,1) and B21,∗ (0,b) are Hilbert spaces, so are the spaces L2 (I,L2 (0,1))
and L2 (I,B21,∗ (0,b)).
It is easy to see that
uL2ρ (0,1) ≤ uL2 (0,1) ,
b
2
1
2
uB21,∗ (0,b) ≤ √ uL2 (0,b) ≤ √ uL2 (0,b) ,
(2.21)
uB21,∗ (0,b) ≤ 2xuL2 (0,b) .
2.2. Existence of the solution. First, we make precisely the concept of the solution of problem (2.9), (2.10), and (2.11) we are considering in this paper. For
this, we take a function v(x,t) ∈ V , the space of functions belonging to C 1 (Ω),
which satisfies the following conditions:
v(x,T) = 0,
b
v(1,t) = 0,
0
v(x,t)dx = 0.
(2.22)
It is easy to observe that ∗0 v = ∗b v = 0 and −(d/dx)∗x v = v, for all t ∈ I.
We now consider the inner product in L2 (I,L2 (0,1)) of (2.9) and the operator

x2 v + 2x∗ v,
x
Mv =  2
b v,
0 ≤ x ≤ b,
b ≤ x ≤ 1,
(2.23)
it follows that
∂u
,Mv
∂t
L2 (I,L2 (0,1))
+ q2
−
∂u
,Mv
∂x
∂
∂u
q1
,Mv
∂x
∂x
L2 (I,L2 (0,1))
L2 (I,L2 (0,1))
+ q3 u,Mv
(2.24)
L2 (I,L2 (0,1))
= ( f ,Mv)L2 (I,L2 (0,1)) .
We integrate by parts the first two terms on the left-hand side of (2.24). To
do this, we assume that u,v ∈ C 1 (Ω), u(x,0) = 0, v(x,T) = 0, v(1,t) = 0, and
578
b
0
Parabolic equations with energy specification
u(x,t)dx =
b
0
v(x,t)dx = 0. In light of the above assumptions, we have
∂u
,Mv
∂t
L2 (I,L2 (0,1))
=
∂u
,v
∂t
∂u
=
,v
∂t
+2
L2 (I,L2ρ (0,1))
L2 (I,L2ρ (0,1))
∂u
,x∗x v
∂t
∂
∂u
−
q1
,Mv
∂x
∂x
L2 (I,L2 (0,b))
b
2
L (I) 0
∂u
+ 2 ∗x ,x∗x v
∂t
∂u
+ 2 ∗x ,xv
∂t
L2 (I,L2 (0,b))
−2
b
∂u = − q1 , x2 v + 2x∗
xv
2
2
∂x
L (I,L (0,1))
∂u
,v
∂t
L2 (I,B21,∗ (0,b))
,
L2 (I) 0
1
∂u 2
− q1 ,b v
2
∂x
L (I) b
b
+ 2 q1 u, x v L2 (I) L2 (I,L2ρ (0,1))
0
∂q1
∗
+ 2 q1 u,v L2 (I,L2 (0,b)) − 2
u, x v
∂u ∂v
+ q1 ,
∂x ∂x
∗
∂x
.
L2 (I,L2 (0,b))
(2.25)
Substituting (2.25) into (2.24), we obtain
∂u
,v
∂t
L2 (I,L2ρ (0,1))
+ q1
∂u
,xv
∂t
+ 2 ∗x
∂u ∂v
,
∂x ∂x
+ 2 q2
L2 (I,L2 (0,b))
L2 (I,L2ρ (0,1))
−2
+ 2 q1 u,v
∂u
,x∗x v
∂x
+ 2 q3 u,x∗x v
∂q1
−2
u, ∗x v
∂x
L2 (I,B21,∗ (0,b))
L2 (I,L2 (0,b))
∂u
+ q2 ,v
2
2
∂x
L (I,L (0,b))
L2 (I,L2 (0,b))
∂u
,v
∂t
+ q3 u,v
(2.26)
L2 (I,L2ρ (0,1))
L2 (I,L2ρ (0,1))
L2 (I,L2 (0,b))
0
= ( f ,v)L2 (I,L2ρ (0,1)) + 2 f ,x∗
x v L2 (I,L2 (0,b)) + u ,Mv(·,0) L2 (0,1) .
This equality may be written as
A(u,v) = ( f ,v)L2 (I,L2ρ (0,1)) + 2 f ,x∗x v
L2 (I,L2 (0,b)) +
u0 ,Mv(·,0)
L2 (0,1) ,
(2.27)
where A(u,v) is the left-hand side of (2.26).
Definition 2.5. We say that u ∈ L2 (I,Hρ1 (0,1)) is a weak solution of problem
(2.9), (2.10), and (2.11) if identity (2.27) holds for all v ∈ V , and u verifies conditions (2.10).
Abdelfatah Bouziani 579
We now give an approximation of problem (2.9), (2.10), and (2.11). To this
end, we suppose that there are sequences of functions fn ∈ C(Ω) and u0n ∈C 1 [0,1],
n = 1,2,..., such that
in L2ρ (0,1), ∀t ∈ I,
fn −→ f
u0n −→ u0
(2.28a)
in Hρ1 (0,1),
(2.28b)
∂un
∂
∂u
∂u
−
q1 n + q2 n + q3 un = fn (x,t),
∂t
∂x
∂x
∂x
un (x,0) = u0n (x),
b
un (1,t) = 0,
0
(2.29)
un (x,t)dx = 0.
Problem (2.29) possesses a unique classical solution un = un (x,t). For the proof,
we refer the reader to [17].
Taking the difference of (2.29) for n = i and n = j, we get
∂ui j
∂ui j
∂ui j
∂
−
q1
+ q2
+ q3 ui j = fi j (x,t),
∂t
∂x
∂x
∂x
ui j (x,0) = u0i j (x),
b
ui j (1,t) = 0,
0
(2.30)
ui j (x,t)dx = 0,
with ui j (x,t) = ui (x,t) − u j (x,t), fi j (x,t) = fi (x,t) − f j (x,t), and u0i j (x) = u0i (x) −
u0j (x).
We must derive some a priori estimates for ui j . To this end, we first compute
t
the integral 0 ( f (·,τ),M(∂ui j (·,τ)/∂t))L2 (0,1) dτ:
t
0
t
0
∂ui j (·,τ) ∂ui j (·,τ)
,M
∂t
∂t
dτ
L2 (0,1)
t ∂ui j (·,τ) 2
=
2
∂t
0
Lρ (0,1)
∂ui j (·,τ)
∂ui j (·,τ)
∂
q1
,M
∂x
∂x
∂t
=
b
0
dτ,
B2 (0,b)
dτ
L2 (0,1)
q1 u2i j (x,t)dx
1
+
2
−
∂ui j (·,τ) 2
+
1,∗
∂t
1
2
1
0
1
0
∂ui j (x,t)
q1 (x,t)ρ(x)
∂t
q1 (x,0)ρ(x)
tb
du0i j (x)
∂x
2
dx −
2
dx −
b
0
q1 (x,0)
tb
0
0
1
2
2
u0i j (x) dx
∂q1 2
u dx dτ
∂t i j
∂ui j 2
∂q1
dx dτ
ρ(x)
∂x
0 0 ∂t
t
∂ui j (·,τ)
∂q1
∗
−2
dτ.
ui j (·,τ), x
∂x
∂t
0
L2 (0,b)
−
(2.31)
580
Parabolic equations with energy specification
From identities (2.31), we get
t ∂ui j (·,τ) 2
2
∂t
0
+
1
2
Lρ (0,1)
1
∂ui j (·,τ) 2
+
1,∗
∂t
dτ
B2 (0,b)
b
∂ui j (x,t) 2
dx + q1 u2i j dx
∂t
0
0
t
b
2
∂ui j (·,τ)
=
f (·,τ),M
dτ + q1 (x,0) u0i j (x) dx
2
∂t
0
0
L (0,1)
q1 (x,t)ρ(x)
1
+
2
1
0
du0i j (x)
q1 (x,0)ρ(x)
dx
tb
2
tb
dx +
0
0
∂q1 2
u dx dτ
∂t i j
∂ui j 2
∂q1
1
dx dτ
+
ρ(x)
2 0 0 ∂t
∂x
t
∂ui j (·,τ)
∂q1
+2
dτ
ui j (·,τ), ∗x
∂x
∂t
0
L2 (0,b)
t ∂ui j (·,τ)
∂ui j (·,τ)
q2
dτ
+2
+ q3 ui j (·,τ) ,
∂x
∂t
0
L2ρ (0,1)
t +2
0
x q2
∂ui j (·,τ)
∂ui j (·,τ)
+ q3 ui j (·,τ) , ∗x
∂x
∂t
dτ.
L2 (0,b)
(2.32)
According to the Cauchy inequality, the first and the last three integrals on the
right-hand side of (2.32) are controlled from above as follows:
t
0
f (·,τ),M
≤
ε1
+ ε2
2
1
ε2
+
t
2
0
t 0
t
0
dτ
L2 (0,1)
fi j (·,τ)2 2
Lρ (0,1) dτ +
t
∂ui j (·,τ) 2
1,∗
∂t
0
tb
0
0
∂q1
∂x
t1
0
0
dτ
Lρ (0,1)
2
dτ
L2 (0,b)
u2i j dx dτ +
1
ε3
t
∂ui j (·,τ) 2
1,∗
∂t
0
0
q22 ρ
∂ui j
∂x
2
+ q32 u2i j
dτ,
B2 (0,b)
∂ui j (·,τ)
∂ui j (·,τ)
+ q3 ui j (·,τ) ,
∂x
∂t
≤ 2ε4
t
∂ui j (·,τ) 2
2
∂t
dτ,
q2
1
2ε1
B2 (0,b)
∂ui j (·,τ)
∂q1
ui j (·,τ), ∗x
∂x
∂t
≤ ε3
2
∂ui j (·,τ)
∂t
dτ
L2ρ (0,1)
dx dτ +
1
ε4
t
∂ui j (·,τ) 2
2
∂t
0
Lρ (0,1)
dτ,
Abdelfatah Bouziani 581
t x q2
2
0
∂ui j (·,τ)
∂ui j (·,τ)
+ q3 ui j (·,τ) , ∗x
∂x
∂t
≤ 2ε5
t1
0
0
q22 ρ(x)
∂ui j
∂x
dτ
L2 (0,b)
2
+ q32 u2i j dx dτ +
1
ε5
t
∂ui j (·,τ) 2
1,∗
∂t
0
dτ.
B2 (0,b)
(2.33)
Substituting (2.33) into (2.32) and taking into account Assumption 2.1, we get
by choosing ε1 = 2, ε2 = ε3 = 3,ε4 = 4, and ε5 = 3
1
2
t
∂ui j (·,τ) 2
2
∂t
0
Lρ (0,1)
≤4
t
0
+
fi j (·,τ)2 2
Lρ (0,1)
c2 + 3c32 + 14c52
+
c2
+ 14c42
2
2
dτ + c0 ui j (·,t)L2ρ (0,b) +
t
0
2
c1 dui j L (0,b) +
2 dx L2 (0,1)
2
dτ + c1 u0 2
ij
2
c0 ∂ui j (·,t) 2
∂x L2ρ (0,1)
0
(2.34)
ρ
ui j (·,τ)2 2
L (0,b) dτ
t ∂ui j (·,τ) 2
∂x 2
0
dτ.
Lρ (0,1)
In light of the elementary inequality
b2 ui j (·,t)2 2
L (b,1)
4
2
2
2
∂ui j (·,τ) b2 t b2 t b 0 2
≤
u 2
+
ui j (·,τ) L2 (b,1) dτ +
dt
2
4 i j L (b,1) 4 0
4 0 ∂t
L (b,1)
2
b2 t 1 t
b2 2
ui j (·,τ)2 2
∂ui j (·,τ) ≤ u0i j L2 (b,1) +
dτ
+
dt,
2
L (b,1)
4
4 0
4 0
∂t
Lρ (0,1)
(2.35)
the last estimate becomes
t
∂ui j (·,τ) 2
2
∂t
0
≤ c6
t
0
Lρ (0,1)
2
dτ + ui j (·,t)Hρ1 (0,1)
fi j (·,τ)2 2
2
0
Lρ (0,1) dτ + ui j Hρ1 (0,1) + c7
t
0
ui j (·,τ)2 1
Hρ (0,1) dτ,
(2.36)
where
max 4,c1 ,b2 /4
,
min c0 /2,b2 /4
max c2 + 3c32 + 14c52 ,b2 /4,c2 /2 + 14c42
c7 =
.
min c0 /2,b2 /4
c6 =
(2.37)
582
Parabolic equations with energy specification
Therefore, from Gronwall’s lemma, we come to the conclusion that
t
∂ui j (·,τ) 2
2
∂t
0
Lρ (0,1)
2
dτ + ui j (·,t)Hρ1 (0,1)
2
2
≤ c6 exp c7 T fi j L2 (I,L2ρ (0,1)) + u0i j Hρ1 (0,1) .
(2.38)
If we omit the first term on the left-hand side of (2.38) and integrate the result
over I, it yields
2
u i j 2
L (I,Hρ1 (0,1))
2
2
≤ c6 T exp c7 T fi j L2 (I,L2ρ (0,1)) + u0i j Hρ1 (0,1) .
(2.39)
On the other hand, if in the left-hand side of (2.38) we take the upper bound
with respect to t, from 0 to T, since the right-hand side of the inequality does
not depend on t, we obtain
∂ui j 2
∂t 2
L
(I,L2ρ (0,1))
2
+ ui j C(I,Hρ1 (0,1))
2
2
≤ c6 exp c7 T fi j L2 (I,L2ρ (0,1)) + u0i j Hρ1 (0,1) .
(2.40)
Thus we have proved the following theorem.
Theorem 2.6. Suppose that Assumption 2.1 holds. Then, the solution of problem
(2.30) satisfies the following estimates, for T > 0:
2
0 2
u 1
fi j 2 2
≤
c
+
8
i
j
L
L (I,Lρ (0,1))
Hρ (0,1) ,
2
2
0 2
u i j fi j 2 2
C(I,Hρ1 (0,1)) ≤ c9
L (I,Lρ (0,1)) + ui j Hρ1 (0,1) ,
2
∂ui j 2
2
≤ c9 fi j L2 (I,L2ρ (0,1)) + u0i j Hρ1 (0,1) ,
∂t 2 2
2
u i j 2
(I,Hρ1 (0,1))
(2.41)
L (I,Lρ (0,1))
where c8 = c6 T exp(c7 T) and c9 = c6 exp(c7 T).
From estimates (2.41), it follows immediately that {un }n is a Cauchy sequence. Thus we have the following corollary.
Corollary 2.7. Under the assumptions of Theorem 2.6, the sequence {un }n converges to u in the following sense:
un −→ u
un −→ u
∂u
∂un
−→
∂t
∂t
in L2 I,Hρ1 (0,1) ,
in C
I,Hρ1 (0,1)
(2.42)
,
(2.43)
in L2 I,L2ρ (0,1) .
(2.44)
We must prove that the limit function u(x,t) is a solution of problem (2.9),
(2.10), and (2.11) in the sense of Definition 2.5. For this purpose, we consider
Abdelfatah Bouziani 583
the weak formulation of problem (2.29)
A un ,v = fn ,v
L2 (I,L2ρ (0,1)) + 2
fn ,x∗x v
L2 (I,L2 (0,b)) +
u0n ,Mv(·,0)
L2 (0,1) .
(2.45)
However, un = (un − u) + u, u0n = (u0n − u0 ) + u0 , and fn = ( fn − f ) + f , then it
follows from the last identity that
A un − u,v + A(u,v) = fn − f ,v
L2 (I,L2ρ (0,1)) + ( f ,v)L2 (I,L2ρ (0,1))
+ 2 fn − f ,x∗x v
+
L2 (I,L2 (0,b)) + 2
u0n − u0 ,Mv(·,0) L2 (0,1) +
f ,x∗x v
L2 (I,L2 (0,b))
0
u ,Mv(·,0)
L2 (0,1) .
(2.46)
In light of the Schwarz inequality and inequalities (2.21), we get the following
estimates:
∂ un − u
,v
∂t
A un − u,v =
L2 (I,L2ρ (0,1))
∂ un − u
,v
∂t
−2
+ q2
∂ un − u
,v
∂x
+ 2 ∗x
L2 (I,B21,∗ (0,b))
+ q1
L2 (I,L2ρ (0,1))
∂ un − u
,xv
∂t
L2 (I,L2 (0,b))
∂ un − u ∂v
,
∂x
∂x
+ 2 q2
L2 (I,L2ρ (0,1))
∂ un − u
,x∗x v
∂x
L2 (I,L2 (0,b))
∂q1 un − u , ∗x v
∂x
L2 (I,L2 (0,b))
∗
+ q3 un − u ,v L2 (I,L2ρ (0,1)) + 2 q3 un − u ,xx v L2 (I,L2 (0,b))
+ 2 q1 un − u ,v
L2 (I,L2 (0,b)) − 2
√
√ √
√ ≤ max 5 + 2,c1 + 1 + 2 c4 ,2c1 + 2c3 + 1 + 2 c5
∂ un − u
+ u − un L2 (I,Hρ1 (0,1)) vL2 (I,Hρ1 (0,1)) ,
× 2 2
∂t
L (I,Lρ (0,1))
u0n − u0 ,Mv(·,0)
L2 (0,1)
√ ≤ 1 + 2 u0n − u0 Hρ1 (0,1) · v(·,0)L2 (0,1) ,
fn − f ,v
L2 (I,L2ρ (0,1))
+ 2 fn − f ,x∗x v
L2 (I,L2 (0,b))
√ ≤ 1 + 2 fn − f L2 (I,L2ρ (0,1)) v L2 (I,L2ρ (0,1)) .
(2.47)
Therefore, if we pass to the limit in equality (2.46), by taking into account the
limit relations (2.42) and (2.44), and estimates (2.47), we conclude that u satisfies equality (2.27). On the other hand, it follows from Corollary 2.7 that
t
(x,t)dx is in C(Ω); from which we deduce that u
(1,t) = 0 almost everywhere.
0u
584
Parabolic equations with energy specification
It remains to prove that u satisfies the initial condition. Let
u
(·,0) − u0 Hρ1 (0,1)
≤ u
(·,0) − un (·,0)Hρ1 (0,1) + u0n − u0 Hρ1 (0,1)
≤ u
− un + u 0 − u 0 1 .
1
n
C(I,Hρ (0,1))
(2.48)
Hρ (0,1)
Passing to the limit as n → +∞ in the above inequality, by taking into account
(2.43) and (2.28b), we get u(x,0) = u0 (x).
We have thus proved the following theorem.
Theorem 2.8. If u0 belongs to Hρ1 (0,1) and f to L2 (I,L2ρ (0,1)), then there exists a weak solution u of problem (2.9), (2.10), and (2.11) possessing the following
properties:
A(u,v) = ( f ,v)L2 (I,L2ρ (0,1)) + 2 f ,x∗x v
0
+ u ,Mv(·,0)
L2 (0,1) ,
L2 (I,L2 (0,b))
u ∈ L2 I,Hρ1 (0,1) .
(2.49)
Moreover,
u ∈ C I,Hρ1 (0,1) ,
∂u
∈ L2 I,L2ρ (0,1) .
∂t
(2.50)
2.3. Continuous dependence and uniqueness. We first establish the continuous dependence of the solution with respect to the data. The uniqueness then
follows directly.
Theorem 2.9. Suppose that Assumption 2.1 holds. Let ( f ,u0 ), ( f ∗ ,u∗0 ) ∈ L2 (I,
Hρ1 (0,1)) × Hρ1 (0,1) and let u, u∗ be two solutions of problem (2.9), (2.10), and
(2.11) corresponding, respectively, to the above data. Then
2
2
≤ c8 f − f ∗ L2 (I,L2ρ (0,1)) + u0 − u∗0 Hρ1 (0,1) ,
0
∗0 2
u − u ∗ 2
f − f ∗ 2 2 2
C(I,Hρ1 (0,1)) ≤ c9
L (I,Lρ (0,1)) + u − u
Hρ1 (0,1) ,
2
2
∂u ∂u∗ 2
−
≤ c9 f − f ∗ L2 (I,L2ρ (0,1)) + u0 − u∗0 Hρ1 (0,1) .
u − u ∗ 2 2
L (I,Hρ1 (0,1))
∂t
∂t
(2.51)
L2 (I,L2ρ (0,1))
Proof. Let {un (x,t)}n be a sequence of classical solutions of problem (2.29),
converging to the weak solution. Therefore, we have the analogue of estimates
(2.41), with ui j replaced by un . If we pass to the limit, in the resulting inequalities, as n → ∞, by taking into account (2.28), we obtain
2
u2L2 (I,Hρ1 (0,1)) ≤ c8 f 2L2 (I,L2ρ (0,1)) + u0 Hρ1 (0,1) ,
2
u2C(I,Hρ1 (0,1)) ≤ c9 f 2L2 (I,L2ρ (0,1)) + u0 Hρ1 (0,1) ,
2
2
∂u ≤ c9 f 2L2 (I,L2ρ (0,1)) + u0 Hρ1 (0,1) .
∂t
L2 (I,L2ρ (0,1))
(2.52)
Abdelfatah Bouziani 585
Set u = u − u∗ , u0 = u0 − u0∗ , and f = f − f ∗ . From (2.52), with u replaced
by u, u0 by u0 , and f by f, we get inequalities (2.51).
Corollary 2.10. Under the assumptions of Theorem 2.8, the weak solution of
problem (2.9), (2.10), and (2.11) is unique.
3. The quasilinear problem
3.1. Statement of the problem. In this section, we consider a boundary value
problem for quasilinear parabolic equation which combines Dirichlet and integral conditions in a domain with curved boundaries. Precisely, we state the
problem as follows. Find a function θ satisfying
∂θ
∂θ
∂θ
∂
∂θ
−
p1 (ξ,τ)
+ p2 (ξ,τ) + p3 (ξ,τ)θ = h ξ,τ,θ,
,
∂τ ∂ξ
∂ξ
∂ξ
∂ξ
θ(ξ,0) = θ 0 (ξ),
Γ1 (0) < ξ < Γ2 (0),
θ Γ2 (τ),τ = Ψ(τ),
Γ3 (τ)
Γ1 (τ)
(ξ,τ) ∈ Q,
0 < τ < T,
θ(ξ,t)dξ = E(t),
0 < τ < T.
(3.1)
Here, we conserve the notations given in Section 2.
As in Section 2.1, we reduce problem (3.1) to the following:
∂u
∂u
∂u
∂u ∂
−
q1 (x,t)
+ q2 (x,t) + q3 (x,t)u = f x,t,u,
,
∂t ∂x
∂x
∂x
∂x
u(x,0) = u0 (x),
b
0
(x,t) ∈ Ω,
0 < x < 1,
u(1,t) = 0,
0 < t < T,
u(x,t)dx = 0,
0 < t < T.
(3.2)
Consider now the following auxiliary problem:
∂ζ
∂ζ
∂
∂ζ
−
q1 (x,t)
+ q2 (x,t) + q3 (x,t)ζ = 0,
∂t ∂x
∂x
∂x
ζ(x,0) = u0 (x),
ζ(1,t) = 0,
b
0
0 < x < 1,
0 < t < T,
ζ(x,t)dx = 0,
0 < t < T.
(3.3)
586
Parabolic equations with energy specification
Let ω = u − ζ, where u is the solution of problem (3.2). Then it follows from
(3.2) and (3.3) that ω verifies
∂ω
∂ω
∂ω ∂ζ
∂
∂ω
−
q1 (x,t)
+ q2 (x,t)
,
+ q3 (x,t)ω = f x,t,ω + ζ,
+
∂t ∂x
∂x
∂x
∂x ∂x
ω(x,0) = 0,
ω(1,t) = 0,
b
0
ω(x,t)dx = 0.
(3.4)
Section 2 implies that problem (3.3) admits a unique weak solution that depends
continuously with respect to the initial condition. Thus it remains to establish
the proof for problem (3.4). For this purpose, we employ the following iteration
procedure.
Let ω(0) = 0 and let {ω(n) }n be defined as follows: if ω(n−1) is given, then solve
the following problem:
∂ω(n−1) ∂ζ
∂
∂ω(n)
∂ω(n)
∂ω(n)
−
q1
+ q2
,
+ q3 ω(n) = f x,t,ω(n−1) + ζ,
+
∂t
∂x
∂x
∂x
∂x
∂x
ω(n) (x,0) = 0,
b
ω(n) (1,t) = 0,
0
ω(n) (x,t)dx = 0.
(3.5)
Theorem 2.8 implies, for fixed n, that each of problems (3.5) possesses a unique
solution ω(n) (x,t). Taking its difference for n = k and n = k + 1, it follows that
∂z(k)
∂
∂z(k)
∂z(k)
−
q1
+ q2
+ q3 z(k) = f(k−1) (x,t),
∂t
∂x
∂x
∂x
z(k) (x,0) = 0,
b
z(k) (1,t) = 0,
0
(3.6)
z(k) (x,t)dx = 0,
where
z(k) (x,t) = ω(k+1) (x,t) − ω(k) (x,t),
f(k−1) (x,t) = f x,t,ω(k) + ζ,
∂ω(k) ∂ζ
∂ω(k−1) ∂ζ
.
+
+
− f x,t,ω(k−1) + ζ,
∂x
∂x
∂x
∂x
(3.7)
3.2. A priori estimates
Theorem 3.1. Let Assumption 2.1 be fulfilled, let f (x,t,0,0) ∈ L2 (I,L2ρ (0,1)), and
let f (x,t,r, y) satisfy the Lipschitz condition, that is,
∃L > 0 : f x,t,r1 , y1 − f x,t,r2 , y2 ≤ L r1 − r2 + y1 − y2 .
(3.8)
Abdelfatah Bouziani 587
Then
(k) z ≤ c12 z(k−1) ,
B
B
k = 1,2,...,
(3.9)
where c12 = L 2Tc11 .
t
Proof. Integrating by parts 0 ( f(k−1) (·,τ),Mz(k) (·,τ))L2 (0,1) dτ and proceeding as
in the establishment of Theorem 2.6, we get
t (k)
∂z (·,τ) 2
2
∂t
0
Lρ (0,1)
2
dτ + z(k) (·,t)Hρ1 (0,1) ≤ c11
t
0
(k−1)
2
f
(·,τ) 2
Lρ (0,1) dτ,
(3.10)
where c11 = 4exp(c7 T)/ min(c0 /2,b2 /4).
Applying Lipschitz inequality to the right-hand side of (3.10), we obtain
t (k)
∂z (·,τ) 2
2
∂t
0
Lρ (0,1)
≤ 2c11 L
2
2
dτ + z(k) (·,t)Hρ1 (0,1)
(k−1) 2
z
2
L
2
≤ 2c11 L2 z(k−1) 2
(k−1) 2
∂z
(I,L2ρ (0,1)) + ∂x 2
L (I,L2ρ (0,1))
L (I,Hρ1 (0,1))
(3.11)
2
≤ 2c11 TL2 z(k−1) C(I,Hρ1 (0,1))
2
≤ 2c11 TL2 z(k−1) .
B
The right-hand side of the above inequality being independent of t, we take the
supremum with respect to t in the left-hand side. It then follows that
(k) 2
z ≤ 2c11 TL2 z(k−1) 2 ,
B
B
(3.12)
from which we have estimate (3.9).
3.3. Existence of the solution
Theorem 3.2. Let the hypotheses of Theorem 3.1 hold. If
L< 1
,
2Tc11
(3.13)
then problem (3.4) admits a weak solution in the space B.
Proof. It is easy to see that if c12 < 1, that is, L < 1/ 2Tc11 , then the series
(k)
(n)
k≥1 z (x,t) converges. Consequently, the sequence {ω }n∈N defined by
ω(n) (x,t) =
n
−1
k =0
ω(k+1) (x,t) − ω(k) (x,t) + ω(0) (x,t),
n = 1,2,...,
(3.14)
588
Parabolic equations with energy specification
To prove that this function is a weak
converges, in the norm of space B, to ω(x,t).
solution of problem (3.4), we proceed exactly as in the proof of Theorem 2.8.
3.4. Uniqueness of the solution. In this section, we first establish an a priori
estimate. The uniqueness of the solution is then a direct corollary of it.
Theorem 3.3. Under the hypotheses of Theorem 3.1, assume that ω1 and ω2 are
two solutions in B of problem (3.4). Then
ω1 − ω2 ≤ c12 ω1 − ω2 .
B
B
(3.15)
Proof. Let (x,t) = ω1 (x,t) − ω2 (x,t), (x,t) ∈ Ω. Then (x,t) satisfies
∂
∂
∂
∂
−
q1 (x,t)
+ q2 (x,t)
+ q3 (x,t)
= f(x,t),
∂t ∂x
∂x
∂x
(x,0) = 0,
(1,t) = 0,
b
0
(3.16)
(x,t)dx = 0,
where
∂ω1 ∂ζ
∂ω2 ∂ζ
− f x,t,ω2 + ζ,
.
+
+
∂x ∂x
∂x ∂x
f(x,t) = f x,t,ω1 + ζ,
(3.17)
Performing similar calculations to that for Theorem 3.1, we obtain
B ≤ c12 B .
(3.18)
From estimate (3.15), we conclude that
1 − c12 ω1 − ω2 B ≤ 0.
(3.19)
Since c12 < 1, then ω1 − ω2 B = 0, that is, ω1 (x,t) ≡ ω2 (x,t), for all (x,t) ∈ Ω,
from which we have the following corollary.
Corollary 3.4. Suppose that the hypotheses of Theorem 3.1 are satisfied, then the
solution of problem (3.4) is unique.
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Abdelfatah Bouziani: Department of Mathematics, The Larbi Ben M’hidi University
Centre, Oum El Bouagui 04000, Algeria
Current address: The Abdus Salam International Centre for Theoretical Physics, Strada
Costiera 11, 34100 Trieste, Italy