IJMMS 2004:28, 1503–1507
PII. S0161171204303017
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
A THIRD-ORDER NONLOCAL PROBLEM
WITH NONLOCAL CONDITIONS
LAZHAR BOUGOFFA
Received 1 March 2003
We study an equation with dominated lower-order terms and nonlocal conditions. Using the
Riesz representation theorem and the Schauder fixed-point theorem, we prove the existence
and uniqueness of a generalized solution.
2000 Mathematics Subject Classification: 35K35, 35B30, 35D05.
1. Introduction. Various problems arising in heat conduction, chemical engineering,
plasma physics, thermoelasticity, and so forth, can be reduced to the nonlocal problems with integral conditions. This type of nonlocal boundary value problems has been
investigated in [1, 2, 3, 4, 5, 6, 8] for parabolic equations and in [7, 10] for hyperbolic
equations. However, some partial differential equations of higher order with dominated
low terms and nonlocal conditions are encountered when studying models for certain
natural and physical processes. An example of such type of equations is the equation of
longitudinal waves in a thin elastic stem taking into account the effects of transversal
inertia [9]:
∂4u
∂2u ∂2u
−
− 2 2 = 0.
2
2
∂t
∂x
∂t ∂x
(1.1)
Another example is the equation of moisture transfer:
∂u
∂
∂u
∂2u
=
D
+A
,
∂t
∂x
∂x
∂x∂t
(1.2)
where u is the concentration of moisture per unit, D is the coefficient of diffusivity,
and A > 0 is the varying coefficient of Hallaire. Motivated by this, we study the equation
lu =
∂2u
∂3u
∂u
∂u
+ a(t, x)
+ c(t, x)u = f (t, x)
−
+ b(t, x)
2
∂t∂x
∂t
∂x 2
∂x
(1.3)
in the rectangular domain Ω = (0, T ) × (0, 1).
To (1.3), we attach the nonlocal conditions
T
0
u(t, x)dt = 0,
u(t, 0) = 0,
∀x ∈ (0, 1),
1
0
(1.4)
u(t, x)dx = 0,
∀t ∈ (0, T ).
1504
LAZHAR BOUGOFFA
We assume that the coefficients of l are smooth and bounded on Ω:
σ (x)
0 < c(t, x) ≤ c0 ,
0 < b(t, x) ≤ b0 √ ,
2
∀x ∈ (0, 1), ∀t ∈ (0, T ), where σ (x) = 1 − x.
0 < a(t, x) ≤ a0 ,
(1.5)
2. Generalized solution. Define the operator l1 by
l1 u = −
∂2u
∂u
∂u
+ a(t, x)
+ c(t, x)u,
+ b(t, x)
∂t
∂x 2
∂x
(2.1)
and
F (t, x, u) = f (t, x) − l1 u.
(2.2)
Then (1.3) can be assumed to have the form
∂3u
= F (t, x, u).
∂t∂x 2
(2.3)
We introduce the function space
σ (x) ∂v
∂v
∂2v
√
∈ L2 (Ω),
∈ L2 (Ω),
∈ L2 (Ω),
∂t
∂x 2
2 ∂x
1
T
v(t, x)dt = 0, v(t, 0) =
v(t, x)dx = 0 .
V = v : v ∈ L2 (Ω),
0
(2.4)
0
The completion of this space, with respect to the norm
v21,2,σ
=
2 2 σ (x)2 ∂v 2
∂v 2
∂ v
+
+
v +
dt dx,
2
∂x
∂t
∂x 2
2
Ω
(2.5)
σ1,2 (Ω) is a Hilbert space with
σ1,2 (Ω). Notice that H
is denoted by H
(u, v)H 1,2 (Ω) =
σ
σ 2 (x) ∂u ∂v ∂ 2 u ∂ 2 v
+
uv +
dt dx.
2 ∂x ∂x ∂x 2 ∂x 2
Ω
(2.6)
σ1,2 (Ω), define the operator M by
For v ∈ H
tx
t 2
(x − 1)2 t
∂ v
Mv = (1 − x)
v(τ, ξ)dξ dτ −
(τ,
x)dτ
+
v(τ, x)dτ
2
2
0 0
0 ∂x
0
x
x
∂v
(t, ξ)dξ.
Jv(t, ξ)dξ, where Jv =
+
0
0 ∂t
(2.7)
σ1,2 (Ω) is called a generalized solution to problem
Definition 2.1. A function u ∈ H
(1.3)-(1.4) if
(u, v)H 1,2 (Ω) = F (t, x, u), Mv L2 (Ω)
σ
σ1,2 (Ω).
for every v ∈ H
(2.8)
A THIRD-ORDER NONLOCAL PROBLEM . . .
1505
3. Existence and uniqueness theorem. In this section, we prove the existence and
uniqueness of a generalized solution for the problem (1.3)-(1.4). For this, we first study
the subsidiary problem
∂3u
= F (t, x, 0)
∂t∂x 2
l0 u ≡
(3.1)
with integral conditions (1.4), where
F (t, x, 0) = f (t, x).
(3.2)
Theorem 3.1. Let F (t, x, 0) ∈ L2 (Ω). Then there exists one and only one generalized
solution u0 of the subsidiary problem
l0 u ≡
T
0
∂3u
= F (t, x, 0),
∂t∂x 2
u(t, x)dt = 0,
∀x ∈ (0, 1),
(3.3)
1
u(t, 0) = 0,
0
u(t, x)dx = 0,
∀t ∈ (0, T ),
such that
c1 u0 1,2,σ ≤ F L2 (Ω) ,
(3.4)
where c1 is a positive constant.
Proof. For F (t, x, 0) ∈ L2 (Ω), Ψ (v) = (F , Mv)L (Ω) is a bounded linear functional on
2
σ1,2 (Ω).
H
Indeed,
F , Mv)
L2 (Ω)
≤ F L
2 (Ω)
MvL2 (Ω) .
(3.5)
By substituting the expression of Mv in (3.5) and using the Poincaré estimates
Ω
v 2 (t, x)dt dx ≤ 4
2
t
Ω
0
v(τ, x)dτ
Ω
(1 − x)2
dt dx ≤ 4
∂v
∂x
Ω
2
dt dx,
v(t, 0) = 0,
(3.6)
(1 − x)2 v 2 (t, x)dt dx,
we find that |Ψ (v)| ≤ 4 max{2T 2 , 4}F L2 (Ω) v1,2,σ .
Consider the scalar product (l0 , Mv)L2 (Ω) = Ω l0 u·Mv dt dx; employing integration
σ1,2 (Ω), we obtain
by parts and taking account of v ∈ H
∂3u
, Mv
∂t∂x 2
L2 (Ω)
= (u, v)H 1,2 (Ω) .
(3.7)
σ
Thus, by the Riesz representation theorem, there exists a unique solution
σ1,2 (Ω):Ψ (v) = (F , Mv)
u0 ∈ H
L
2 (Ω)
= u0 , v H 1,2 (Ω) ,
σ
σ1,2 (Ω).
∀v ∈ H
(3.8)
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LAZHAR BOUGOFFA
Hence, (u, v)H 1,2 (Ω) = (u0 , v)H 1,2 (Ω) , that is, u0 is a generalized solution. Letting 1/c1 =
σ
σ
4 max{2T 2 , 4}, we obtain inequality (3.4).
σ1,2 (Ω) → L2 (Ω) is bounded, that is, there exists a
Lemma 3.2. The operator l1 : H
positive constant c2 such that l1 uL2 (Ω) ≤ c2 u1,2,σ .
Proof. By using conditions (1.5), we directly obtain
2
2 2
∂u 2
2 ∂ u 2 ∂u 2
2
+ a0 + b0 + c0 uL2 ,
L2 (Ω) ≤ 4 ∂t L2
∂x 2 L2
∂x L2,σ
l1 u2
where ∂u/∂x2L2,σ =
Hence,
l1 u2L2 (Ω)
(3.9)
2
2
Ω (σ (x)/2)(∂u/∂x) dt dx.
2
2
2
≤ c2 u1,2,σ , where c2 = 4 max{1, a20 , b02 , c02 }.
√
√
Since l1 is linear, then l1 ( 2µu) = 2µl1 (u) for an arbitrary µ.
√
Let l1,µ (w) = l1 ( 2µw) for µ > 1/c1 .
Now, consider the general case. The idea in the proof is to derive the results for the
equation lu = f with integral conditions (1.4).
√
Theorem 3.3. Let f (t, x) ∈ L2 (Ω) and |f (t, x)| ≤ λ/ 2, where λ is a constant. Then
σ1,2 (Ω) to problem (1.3)-(1.4). Furtherthere exists at least one generalized solution u0 ∈ H
more, the solution is uniquely determined if c2 < c1 .
Proof. Let W = {l1,µ w : l1,µ w ∈ L2 (Ω), l1 u2L2 (Ω) ≤ λ2 T /κ 2 } be a closed ball, where
κ = c12 − 1/µ 2 .
It is clear that
2
F (t, x, w) ≤ f (t, x) +
c12 − k2 l1,µ (w),
2
(3.10)
and we have F (t, x, w)2L2 (Ω) ≤ c12 λ2 T /κ 2 for all l1,µ w ∈ W .
From Theorem 3.1, there exists a unique generalized solution of the problem
∂3u
= F (t, x, w)
∂t∂x 2
(3.11)
with integral conditions (1.4), so that
(u, v)H 1,2 (Ω) = (F , Mv)L
σ
2 (Ω)
.
(3.12)
σ1,2 (Ω), S(W ) ⊂ W .
Define an operator S : l1 w ∈ W → u = Sl1 w ∈ H
Notice that S is completely continuous. To show this, let (l1 w)n , (l1 w)0 ∈ W and
(l1 w)n − (l1 w)0 2L2 (Ω) → 0, as n → ∞.
Then, for un = S(l1 w)n , u0 = S(l1 w)0 , we have
un − u0 , v H 1,2 (Ω) = F t, x, (w)n − F t, x, (w)0 , Mv
σ
σ1,2 (Ω).
= l1 w n − l1 w 0 , Mv
for every v ∈ H
L2 (Ω)
(3.13)
1507
A THIRD-ORDER NONLOCAL PROBLEM . . .
Now, from Theorem 3.1,
c1 un − u0 1,2,σ ≤ l1 w n − l1 w 0 L2 (Ω) → 0
as n → ∞.
(3.14)
Again, taking a sequence {(l1 w)n } ⊂ W , (l1 w)n 2L2 (Ω) ≤ λ2 T /κ 2 . For un = S(l1 w)n , we
σ1,2 (Ω); therefore there
have un 2L (Ω) ≤ λ2 T /κ 2 , so a sequence {un } is bounded in H
2
σ1,2 (Ω).
exists a subsequence weakly convergent in H
1,2
σ (Ω) is compact in L2 (Ω), then there exists a subseSince any bounded set in H
quence, which we also denote by {un }, strongly convergent in L2 (Ω) to u0 , as n → ∞.
As l1 is a bounded operator, S is completely continuous, and so Sl1 is completely
continuous. Thus, from Schauder’s fixed-point theorem, there exists at least one fixed
point u0 ∈ W such that u0 = Sl1 u0 and (u0 , v)H 1,2 (Ω) = (F (t, x, u0 ), Mv)L2 (Ω) for every
σ
σ (Ω).
v ∈H
Now, assume that u1 , u2 are distinct generalized solutions, then (u1 −u2 , v)H 1,2 (Ω) =
1,2
σ (Ω).
(F (t, x, u1 ) − F (t, x, u2 ), Mv)L2 (Ω) for all v ∈ H
From (3.4) and Lemma 3.2, we have
1,2
u1 − u 2 1,2,σ
≤
1
l 1 u 1 − l1 u 2 ≤ c 2 u 1 − u 2 1,2,σ .
c1
c1
σ
(3.15)
Thus, if c2 < c1 , then it gives a contradiction; therefore u1 = u2 .
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Lazhar Bougoffa: Department of Mathematics, Faculty of Science, King Khalid University, P. O.
Box 9004, Abha, Saudi Arabia
E-mail address: abogafah@kku.edu.sa