djibibe

Electronic Journal of Differential Equations, Vol. 2008(2008), No. 17, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
CONTINUOUS DEPENDENCE OF SOLUTIONS TO MIXED
BOUNDARY VALUE PROBLEMS FOR A PARABOLIC
EQUATION
MOUSSA ZAKARI DJIBIBE, KOKOU TCHARIE, NIKOLAY IOSSIFOVICH YURCHUK
Abstract. We prove the continuous dependence, upon the data, of solutions
to second-order parabolic equations. We study two boundary-value problems:
One has a nonlocal (integral) condition and the another has a local boundary
condition. The proofs are based on a priori estimate for the difference of
solutions.
1. Introduction
This paper is devoted to the proof of the continuous dependence, upon the
data, of generalized solutions of a second order parabolic equation. The boundary
conditions are of mixed type. This article contributes to the development of the
a priori estimates method for solving such problems. The questions related to
these problems are so miscellaneous that the elaboration of a general theory is still
premature. Therefore, the investigation of these problems requires at every time a
separate study.
The importance of problems with integral condition has been pointed out by
Samarskii [23]. Mathematical modelling by evolution problems with a nonlocal
R1
1
constraint of the form 1−α
u(x, t) dx = E(t) is encountered in heat transmission
α
theory, thermoelasticity, chemical engineering, underground water flow, and plasma
physic. See for instance Benouar-Yurchuk [1], Benouar-Bouziani [2]-[3], Bouziani
[4]-[7], Cannon et al [13]-[15], Ionkin [17]-[18], Kamynin [19] and Yurchuk [25][27]. Mixed problems with nonlocal boundary conditions or with nonlocal initial
conditions were studied in Bouziani [7]-[9], Byszewski et al [10]-[12], Gasymov [16],
Ionkin [17]-[18], Lazhar [21], Mouravey-Philipovski [22] and Said-Nadia [24]. The
results and the method used here are a further elaboration of those in [1]. We should
mention here that the presence of integral term in the boundary condition can
greatly complicate the application of standard functional and numerical techniques.
This work can be considered as a continuation of the results in [7], [25] and [27].
The remainder of the paper is divided into four section. In Section 2, we give
the statement of the problem. Then in Section 3, we establish a priori estimate.
2000 Mathematics Subject Classification. 35K20, 35K25, 35K30.
Key words and phrases. Priori estimate; mixed problem; continuous dependence;
boundary conditions.
c 2008 Texas State University - San Marcos.
Submitted October 16, 2007. Published February 5, 2008.
1
2
M. Z. DJIBIBE, K. TCHARIE, N. I. YURCHUK
EJDE-2008/17
Finally, in section 4, we show the continuous dependence of a solution upon the
data.
2. Statement of the problem
In the rectangle G = {(x, t) : 0 < x < 1, 0 < t < T }, we consider following two
mixed boundary value problems for parabolic equations:
The mixed problem with boundary integral condition
∂
∂uα ∂uα
−
a(x, t)
= f (x, t),
∂t
∂x
∂x
∂uα (0, t)
uα (x, 0) = ϕα (x),
=0
∂x
Z 1
1
uα (x, t) dx = h(t),
1−α α
(2.1)
(2.2)
(2.3)
where 0 ≤ α < 1.
The mixed problem with local boundary conditions
∂
∂u1
∂u1
−
(a(x, t)
) = f (x, t)
∂t
∂x
∂x
∂u1 (0, t)
u1 (x, 0) = ϕ1 (x),
=0
∂x
u1 (1, t) = h(t).
(2.4)
(2.5)
(2.6)
We assume the following conditions:
(A1) The coefficient a(x, t) in (2.1) and (2.4) is a continuous differentiable function, and 0 < a0 ≤ a(x, t) ≤ a1 .
(A2) f ∈ L2 (G), h ∈ W21 (0, T ), ϕα , ϕ1 ∈ W21 (0, T ), ϕ0α (0) = ϕ01 (0) = 0, ϕ1 (1) =
R1
1
h(0), 1−α
ϕ (x) dx = h(0).
α α
In [25], it is shown that if the conditions (A1)–(A2) are satisfied, then there
exists a unique solution to (2.1)–(2.3) and (2.4)–(2.6), and that the solution is
differentiable almost everywhere
R 1 on G. It is clear that for every function g ∈
1
C([0, 1]) : g(1) = limα→1 1−α
g(x) dx Therefore, the problem (2.4)–(2.6) is the
α
limit when α → 1 of the problem (2.1)–(2.3). In this paper, we establish a priori
estimate for the difference uα − u1 and use it to prove that if α → 1 and ϕα → ϕ1
then uα → u1 . In this way we prove a new important property; a continuous
dependence of solutions of mixed problems for parabolic equations of the form of
boundary conditions.
EJDE-2008/17
CONTINUOUS DEPENDENCE OF SOLUTIONS
3
3. A priori estimate
Theorem 3.1. Under assumptions (A1)–(A2), there exists a positive constant c
independent of uα , u1 and α such that
Z 1Z T
∂uα
∂ 2 u1 2
∂u1 2
∂ 2 uα
(1 − x)[|
−
| ] dx dt
−
| +|
2
∂t
∂t
∂x
∂x2
0
0
Z 1
∂uα
∂u1 2
+ sup
[|uα − u1 |2 + (1 − x)|
−
| ] dx
∂x
∂x
0≤t≤T 0
Z 1
nZ 1
1
2
u1 (x, 0) dx
≤c
|ϕ0α (x) − ϕ01 (x)|2 dx + h(0) −
1−α α
0
Z 1
Z 1
Z T
o
0
1
∂u1 (x, t)
1
2
2
h (t) −
u1 (x, t) dx dt
+
dx + h(t) −
1−α α
∂t
1−α α
0
(3.1)
Proof. Consider the problem (2.1)–(2.3), with
uα = vα +
3x2
h(t)
1 + α + α2
(3.2)
where vα is a solution of the problem
∂
∂vα
∂vα
−
(a(x, t)
) = gα (x, t)
∂t
∂x
∂x
Z 1
∂vα (0, t)
1
vα (x, 0) = ψα (x),
= 0,
vα (x, t) dx = 0 ,
∂x
1−α α
where
3x2
6h(t)
∂a(x, t)
gα (x, t) = f (x, t) −
h0 (t) +
(a(x, t) + x
)
1 + α + α2
1 + α + α2
∂x
3x2
ψα (x) = ϕα (x) −
h(0)
1 + α + α2
In (2.4)-(2.6) we also pose
Z 1
3x2
u1 (x, t) dx
(3.3)
u1 = v1 +
1 − α3 α
where v1 is a solution of the problem
∂v1
∂
∂v1 −
a(x, t)
= g1 (x, t)
∂t
∂x
∂x
∂v(0, t)
v1 (x, 0) = ψ1 (x),
= 0,
∂x
Z 1
3
v1 (1, t) = h(t) −
u1 (x, t) dx
1 − α3 α
where
g1 (x)
= f (x, t) −
3x2
1 − α3
Z
1
∂u1 (x, t)
6
∂a(x, t)
dx +
(a(x, t) + x
)
3
∂t
1−α
∂x
α
Z 1
3x2
u1 (x, 0) dx .
ψ( x) = ϕ1 (x) −
1 − α3 α
Z
1
u1 (x, t) dx,
α
4
M. Z. DJIBIBE, K. TCHARIE, N. I. YURCHUK
EJDE-2008/17
Then the function wα = v1 − vα is a solution of the problem
∂wα
∂
∂wα
−
(a(x, t)
) = Fα (x, t)
∂t
∂x
∂x
Z 1
∂wα (0, t)
1
wα (x, 0) = φα (x),
wα (x, t) dx = 0 ,
= 0,
∂x
1−α α
(3.4)
(3.5)
where
Z 1
6
∂a(x, t) 1
u1 (x, t) dx
a(x,
t)
+
x
h(t)
−
2
1+α+α
∂x
1−α α
(3.6)
Z 1
2
0
∂u1 (x, t) 3x
1
+
h (t) −
dx
1 + α + α2
1−α α
∂t
Z 1
3x2
1
φα (x) = ϕ1 (x) − ϕα (x) +
u1 (x, 0) dx .
(3.7)
h(0) −
2
1+α+α
1−α α
Fα (x, t) = −
For this purpose we need the following result.
Lemma 3.2. For a function wα satisfying (3.4))–(3.5), there exists a positive constant c independent of wα , φα and Fα such that
Z
Z 1
∂wα 2
∂ 2 wα 2 ∂wα 2
ψα (x) |
| +|
|
dx
dt
+
sup
ψα (x)|
| + |wα |2 dx
2
∂t
∂x
∂x
0≤t≤T 0
G
(3.8)
Z 1
Z
0
2
2
≤ c1
ψα (x)|φα (x)| dx +
|Fα (x, t)| dx dt
0
G
(
1
where ψα (x) =
1−x
1−α
if 0 ≤ x ≤ α
if α ≤ x ≤ 1 .
Suppose, for the moment, that Lemma 3.2 has been proved, and turn to the
proof of Theorem 3.1. In accordance with the equality
Z 1
3x2
1
u1 − uα = wα −
h(t)
−
u1 (x, t) dx ,
2
1+α+α
1−α α
and the estimates
Z
∂u1
∂uα 2
(1 − x)|
−
| dx dt
∂t
∂t
G
Z
Z 1
Z
3 T 0
1
∂u1 (x, t)
∂wα 2
≤
|h (t) −
dx|2 dt + 2 (1 − x)|
| dx dt ,
5 0
1−α α
∂t
∂t
G
Z
∂ 2 u1
∂ 2 uα 2
−
| dx dt
2
∂x
∂x2
Z 1
Z
∂ 2 wα 2
1
u1 (x, t) dx|2 dt + 2 (1 − x)|
|h(t) −
| dx dt ,
1−α α
∂x2
G
(1 − x)|
G
Z
T
≤ 36
0
Z
1
∂u1
∂uα 2
−
| dx
∂x
∂x
0≤t≤T 0
Z 1
Z 1
1
∂wα 2
u1 (x, t) dx|2 + 2 sup
(1 − x)|
≤ 6 sup |h(t) −
| dx ,
1−α α
∂x
0≤t≤T
0≤t≤T 0
sup
(1 − x)|
EJDE-2008/17
CONTINUOUS DEPENDENCE OF SOLUTIONS
Z
sup
0≤t≤T
5
1
|u1 − uα |2 dx
0
Z 1
Z 1
1
18
u1 (x, t) dx|2 + 2 sup
|wα |2 dx ,
sup |h(t) −
≤
5 0≤t≤T
1−α α
0≤t≤T 0
Z 1
Z 1
Z 1
1
0
0 2
0 2
|ϕ1 (x) − ϕα | dx + 24|h(0) −
|φα | dx ≤ 2
u1 (x, 0) dx|2 ,
1−α α
0
0
Z
Z
∂a(x, t) 2
|Fα (x, t)|2 dx dt ≤ 72
|a(x, t) + x
| dx dt
∂x
G
G
Z T
Z 1
1
|h(t) −
×
u1 (x, t) dx|2 dt
1
−
α
0
α
Z
Z 1
18 T 0
1
∂u1 (x, t)
+
|h (t) −
dx|2 dt ,
5 0
1−α α
∂t
Z 1
1
u1 (x, t) dx|2
1−α α
0≤t≤T
Z 1
Z T
Z 1
1
∂u1 (x, t)
1
2
0
u1 (x, 0) dx| + 2T
|h (t) −
dx|2 dt .
≤ 2|h(0) −
1−α α
1
−
α
∂t
0
α
sup |h(t) −
We have the estimate
Z 1
Z
|ϕ1 (x) − ϕα (x)|2 dx ≤ 2
0
1
|ϕ01 (x) − ϕ0α (x)|2 dx + 2|ϕ1 (0) − ϕα (0)|2 .
(3.9)
0
Then we obtain
Z
∂u1
∂uα 2
∂ 2 u1
∂ 2 uα 2 −
| +|
−
| dx dt
(1 − x) |
2
∂t
∂t
∂x
∂x2
G
Z 1
∂u1
∂uα 2
+ sup
(1 − x)|
−
| + |u1 − uα |2 dx dt
∂x
∂x
0≤t≤T 0
Z 1
Z 1
1
5c1 + 2
0
0
2
) h(0) −
u1 (x, 0) dx
≤ 4c1
|ϕ1 (x) − ϕα (x)| dx + 48(
5
1−α α
0
Z T
Z 1
12c1 + 1 + 32T
1
∂u1 (x, t)
2
+ 3(
)
dx dt
h0 (t) −
5
1−α α
∂t
0
Z 1
Z
∂a(x, t) 2 T
| dx
+ 36 + 144c1 max
|a(x, t) + x
h(t)
0≤t≤T 0
∂x
0
Z 1
1
2
u1 (x, t) dx dt .
−
1−α α
From the above inequality, it follows that (3.1) holds with
5c1 + 2
12c1 + 1 + 32T
c = max 4c1 , 48(
), max 3(
), N ,
5
5
where
Z 1
∂a(x, t) 2
N = 36 + 144c1 max
|a(x, t) + x
| dx .
0≤t≤T 0
∂x
To complete the proof of Theorem 3.1, it remains to prove the Lemma.
2
(3.10)
6
M. Z. DJIBIBE, K. TCHARIE, N. I. YURCHUK
EJDE-2008/17
Proof of Lemma 3.2. Let
(
Mα wα =
∂wα
∂t
1−x ∂wα
1−α ∂t
+
∂wα
1
1−α Iα ∂t
if 0 ≤ x ≤ α
if α ≤ x ≤ 1 ,
Rx
where Iα wα = α wα (ξ, t) dξ. Multiplying both sides of (3.4) by Mα wα and integrating the result with respect to x ∈ [0; 1], we obtain
Z 1
Z 1
Z 1
∂
∂wα
∂wα
Fα (x; t)Mα wα dx dt .
Mα wα dx −
(a(x, t)
)Mα wα dx =
∂t
∂x
0 ∂x
0
0
(3.11)
Integrating by parts the first two integrals on the left-hand of (3.11), and using the
boundary conditions (3.5), we obtain
Z 1
Z 1
∂wα
∂wα 2
(3.12)
Mα wα dx =
) dx;
ψα (x)(
∂t
∂t
0
0
and
1
Z
∂
∂wα a(x, t)
Mα wα dx
∂x
0 ∂x
Z
Z
1 ∂ 1
∂a(x, t) ∂wα 2
∂wα 2 1 1
=
| dx −
ψα (x)
|
| dx .
ψα (x)a(x; t)|
2 ∂t 0
∂x
2 0
∂t
∂x
−
(3.13)
Substituting (3.12) and (3.13) into (3.11), we obtain
Z
Z 1
1 ∂ 1
∂wα 2 ∂wα 2
| dx +
| dx
ψα (x)a(x; t)|
ψα (x)|
∂t
2 ∂t 0
∂x
0
Z
Z 1
1 1
∂a(x, t) ∂wα 2
=
|
| dx +
ψα (x)
Fα (x; t)Mα wα dx dt .
2 0
∂t
∂x
0
Multiplying both sides of the above equality by ec(τ −t) , and integrating the result
with respect t ∈ [0; τ ], we obtain
Z
Z τZ 1
1 1
∂wα 2
∂wα 2
| dx dt +
| dx |t=τ
a(x, t)ψα (x)|
ec(τ −t) ψα (x)|
∂t
2 0
∂x
0
0
Z τZ 1
Z Z
1 τ 1 c(τ −t) ∂a(x, t)
∂wα 2
=
ec(τ −t) Fα (x, t)Mα wα dx dt +
e
ψα (x)|
| dx dt
2
∂t
∂x
0
0
0
0
Z
Z Z
1 1 cτ
1 τ 1
∂wα 2
0
2
+
e a(x, 0)ψα (x)|φ (x)| dx −
ca(x, t)ec(τ −t) ψα (x)|
| dx dt .
2 0
2 0 0
∂x
(3.14)
Using the following estimates
Z 1
Fα (x, t)Mα wα dx
0
1
Z
|Fα (x, t)|2 dx +
≤
0
Z
1
ψα (x)|
0
Z
Fα (x, t)Iα
1
∂wα 2
1
| dx +
∂t
1−α
Z
1
Fα (x, t)Iα
α
∂wα
dx ,
∂t
Z 1
∂wα
1
∂wα 2
|Fα (x, t)|2 dx +
|Iα
dx ≤ 2
| dx ,
2
∂t
8(1 − α) α
∂t
α
α
Z 1
Z
Z
1
∂wα 2
1 1 (1 − x)2 ∂wα 2
1 1
∂wα 2
|Iα
| dx ≤
|
| dx ≤
ψα (x)|
| dx;
8(1 − α)2 α
∂t
2 α (1 − α)2 ∂t
2 0
∂t
1
1−α
Z
1
1
4
EJDE-2008/17
CONTINUOUS DEPENDENCE OF SOLUTIONS
7
and choosing a constant c in (3.14) such that ca0 ≥ | ∂a(x,t)
∂t |, we deduce the inequality
Z Z
Z 1
∂wα 2
∂wα 2
1 τ 1 c(τ −t)
1
e
ψα (x)|
| dx dt + a0
ψα (x)|
| dx |t=τ
4 0 0
∂t
2
∂x
0
Z τZ 1
Z 1
ec(τ −t) |Fα (x, t)|2 dx dt +
≤3
ecτ a(x, t)ψα (x)|φ0 (x)|2 dx
0
0
0
which can be written as
Z Z
Z 1
∂wα 2
∂wα 2
1 τ 1
ψα (x)|
| dx dt + a0
ψα (x)|
| |t=τ dx
4 0 0
∂t
∂t
0
Z
Z 1
ecT ψα (x)a(x, t)|φ0α (x)|2 dx .
≤3
ecT |Fα (x, t)|2 dx dt +
(3.15)
0
G
For the function wα (x, t), there hold the relation
Z 1
Z 1
Z 1
∂wα (x, t) 2
2
2 ∂wα (x, t) 2
|wα (x, t)| dx ≤ 4
(1 − x) |
| dx ≤ 4
| dx .
ψα (x)|
∂x
∂x
0
0
0
(3.16)
It follows from (3.15) and (3.16) that
Z Z
Z
Z
1 τ 1
a0 1
a0 1
∂wα 2
∂wα 2
| dx dt +
| |t=τ dx +
ψα (x)|
|wα |2 |t=τ dx
ψα (x)|
4 0 0
∂t
2 0
∂x
8 0
Z 1
Z
≤ ecT
a(x, t)ψα (x)|φ0α (x)|2 dx + 3ecT
|Fα (x, t)|2 dx dt .
0
G
(3.17)
Since the right-hand side of (3.17) does not depend on τ , we take the supremum of
the left-hand side of (3.17) with respect to 0 ≤ t ≤ T ; we obtain
Z 1
Z
∂wα 2
1
∂wα 2 1
1
ψα (x)|
| dx dt + a0 sup
[ψα (x)|
| + |wα |2 ] dx
4 G
∂t
2 0≤t≤T 0
∂x
4
(3.18)
Z 1
Z
cT
0
2
cT
2
≤e
a(x, t)ψα (x)|φα (x)| dx + 3e
|Fα (x, t)| dx dt .
0
G
It follows from (3.4) that
Z
∂w2
a0
ψα (x)| 2α |2 dx dt
∂x
G
Z
∂wα 2
∂a(x, t) 2
≤2
ψα (x)|
| dx dt + 2T max |
|
0≤t≤t
∂t
∂t
G
Z 1
Z
∂wα 2
× sup
ψα (x)|
| dx + 2
ψα (x)|Fα (x, t)|2 dx dt .
∂x
0≤t≤T 0
G
Combining this inequality and (3.18), we obtain (3.8), with
c1 =
2
48ecT max(a0 ,3) max(1,T | ∂a
∂t | )
+
min(a0 ,2)
∂a 2
min(a0 , 3) max(1, T | ∂t | )
3
.
(3.19)
Thus the proof is complete, which also proves Theorem 3.1.
8
M. Z. DJIBIBE, K. TCHARIE, N. I. YURCHUK
EJDE-2008/17
Corollary 3.3. Since 1 − x ≤ 1 − α < 1 for α ≤ x ≤ 1 it follows that 1 − x ≤
ψα (x) ≤ 1 and the inequality (3.8) yields
Z 1
Z
∂wα 2
∂wα 2
∂ 2 wα 2 [(1 − x)|
| dx dt + sup
(1 − x) |
| +|
| + |wα |2 ] dx
2
∂t
∂x
∂x
0≤t≤T 0
G
Z
Z 1
|φ0α (x)|2 dx +
|Fα (x, t)|2 dx dt} .
≤ c1 {
0
G
4. Continuous dependence of solutions
Using a priori estimate (3.1) for the difference (uα − u1 ) of solution uα of the
mixed problem (2.1)–(2.3) with integral boundary condition, and the solution u1 of
the the mixed problem (2.4)–(2.6) with the local condition, we come to the following
result. We remark that this important property has not been established prior to
this work.
Theorem 4.1. Assume that (A1)–(A2) hold. If
Z 1
lim
|ϕ01 (x) − ϕ0α (x)| dx = 0,
α→1
(4.1)
0
then
1
T
∂uα
∂u1 2
∂ 2 uα
∂u1 2 −
| +|
−
| dx dt
(1 − x) |
2
α→1
∂t
∂t
∂x
∂t
0
0
Z 1
o
∂uα
∂u1 2
+ sup
(1 − x)|
−
| + |uα − u1 |2 dx = 0 .
∂x
∂t
0≤t≤T 0
lim
nZ
Z
(4.2)
Proof. Since (A1) and (A2) are satisfied, the a priori estimate (3.1) holds for the
difference uα − u1 . Moreover, the equality u1 (1, t) = h(t) yields
Z 1
Z 1
1
1
2
lim
u1 (x, t) dx − h(t) = lim
u1 (x, t) dx − u1 (1, t) = 0
α→1 1 − α α
α→1 1 − α α
Z 1
Z 1
1
1
∂u1 (x, t)
∂u1 (x, t)
∂u1 (1, t) 2
2
0
− h (t) = lim
dx −
=0
lim
α→1 1 − α α
α→1 1 − α α
∂t
∂t
∂t
Z 1
Z 1
1
1
lim
u1 (x, 0) dx − h(0)|2 = lim
u1 (x, 0) dx − u1 (1, 0)|2 = 0 .
α→1 1 − α α
α→1 1 − α α
From these three equalities, (4.1), and (3.1), we obtain (4.3). Thus theorem is
proved.
To complete our investigation we show that for any function ϕ1 ∈ W21 (0, T )
satisfying the conditions ϕ01 (0) = 0 and ϕ1 (1) = h(0), there exist functions ϕα ∈
W21 (0, T ) such that
Z 1
1
ϕ0α (0) = 0,
ϕα (x) dx = h(0)
(4.3)
1−α α
and the equality (4.1) is valid. Set
3x2
1
ϕα (x) = ϕ1 (x) +
h(0) −
1 + α + α2
1−α
Z
1
α
ϕ1 (x) dx .
EJDE-2008/17
CONTINUOUS DEPENDENCE OF SOLUTIONS
9
Then ϕα ∈ W21 (0, T ), ϕ0α (0) = ϕ01 (0) = 0,
Z 1
Z 1
1
1
ϕα (x) dx =
ϕ1 (x) dx
1−α α
1−α α
Z 1
Z 1
3
1
2
+
x
ϕ1 (x) dx
dx
h(0)
−
3
1−α α
1−α α
Z 1
Z 1
1
1
=
ϕ1 (x) dx + h(0) −
ϕ1 (x) dx
1−α α
1−α α
= h(0)
and
Z
lim
α→1
1
|ϕ0α (x)
−
ϕ01 (x)| dx
0
1
lim
α→1 1 − α
This completes the proof.
Z
3
1
= lim
h(0) −
α→1 1 + α + α2
1−α
= |ϕ1 (1) − h(0)| = 0 ,
Z
1
ϕ1 (x)
α
1
|ϕ1 (x) dx = ϕ1 (1) = h(0) .
α
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M. Z. DJIBIBE, K. TCHARIE, N. I. YURCHUK
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Moussa Zakari Djibibe
University of Lomé - Togo, Department of Mathematics, PO Box 1515 Lomé, Togo
E-mail address: [email protected] [email protected] tel +228 924 45 21
Kokou Tcharie
University of Lomé - Togo, Department of Mathematics, PO Box 1515 Lomé, Togo
E-mail address: [email protected]
Nikolay Iossifovich Yurchuk
Dept. of Mechanics and Mathematics, Belarussian State University, 220050, Minsk,
Belarus
E-mail address: [email protected]