IJMMS 26:7 (2001) 417–426
PII. S0161171201004379
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
MIXED PROBLEM WITH NONLOCAL BOUNDARY CONDITIONS
FOR A THIRD-ORDER PARTIAL DIFFERENTIAL EQUATION
OF MIXED TYPE
M. DENCHE and A. L. MARHOUNE
(Received 7 January 2000)
Abstract. We study a mixed problem with integral boundary conditions for a third-order
partial differential equation of mixed type. We prove the existence and uniqueness of the
solution. The proof is based on two-sided a priori estimates and on the density of the range
of the operator generated by the considered problem.
2000 Mathematics Subject Classification. 35B45, 35K20, 35M10.
1. Introduction. In the rectangle Ω = (0, ) × (0, T ), we consider the equation
ᏸu =
∂2u
∂2u
∂
a(x,
t)
= f (x, t),
−
∂t 2 ∂x
∂x∂t
(1.1)
where a(x, t) is bounded with 0 < a0 < a(x, t) ≤ a1 and has bounded partial derivatives such that 0 < a2 ≤ ∂a(x, t)/∂t ≤ a3 and 0 < a4 ≤ ∂a(x, t)/∂x ≤ a5 for (x, t) ∈ Ω.
To (1.1) we add the initial conditions
l1 u = u(x, 0) = ϕ(x),
l2 u =
∂u
(x, 0) = ψ(x),
∂t
x ∈ (0, ),
(1.2)
the Dirichlet condition
u(0, t) = 0,
t ∈ (0, T ),
(1.3)
and the integral condition
0
u(ξ, t) dξ = 0,
t ∈ (0, T ),
(1.4)
where ϕ and ψ are known functions which satisfy the compatibility conditions given
by (1.3) and (1.4), that is,
ϕ(0) = 0,
0
ϕ(x) dx = 0,
ψ(0) = 0,
0
ψ(x) dx = 0.
(1.5)
Boundary-value problems for parabolic equations with integral boundary conditions are investigated by Batten [1], Bouziani and Benouar [2], Cannon [3, 4], Perez
Esteva and van der Hoeck [5], Ionkin [8], Kamynin [9], Kartynnik [10], Shi [11], Yurchuk
[13], and many references therein. We remark that integral boundary conditions for
evolution problems have various applications in chemical engineering, thermoelasticity, underground water flow and population dynamics; see for example, [6, 7, 11, 12].
418
M. DENCHE AND A. L. MARHOUNE
The present paper is devoted to the study of a mixed problem with boundary integral conditions for a third-order partial differential equation of mixed type.
We associate to problem (1.1), (1.2), (1.3), and (1.4) the operator L = (ᏸ, l1 , l2 ), defined from E into F , where E is the Banach space of functions u ∈ L2 (Ω), satisfying
(1.3) and (1.4), with the finite norm
∂ 2 u 2 ∂ 3 u 2
+
u2E = ( − x)2 ∂t 2 ∂x 2 ∂t dx dt
Ω
2 2 ∂u 2
∂u 2
2 ∂ u 2
+ sup
( − x) +
dx + sup
∂t + |u| dx,
∂x∂t ∂x 0≤t≤T 0
0≤t≤T 0
(1.6)
and F is the Hilbert space of vector-valued functions Ᏺ = (f , ϕ, ψ) obtained by completion of the space L2 (Ω) × W22 (0, ) × W22 (0, ) with respect to the norm
2
Ᏺ2F = (f , ϕ, ψ)F
2 dψ 2
2
2
2 dϕ dx +
= ( − x) |f | dx dt+ ( − x) +
|ϕ|2 + |ψ|2 dx.
dx
dx
Ω
0
0
(1.7)
Using the energy inequalities method proposed in [13], we establish two-sided a priori estimates. Then, we prove that the operator L is a linear homeomorphism between
the spaces E and F .
2. Two-sided a priori estimates
Theorem 2.1. For any function u ∈ E, there is the a priori estimate
LuF ≤ cuE ,
(2.1)
where the constant c is independent of u.
Proof. Using (1.1) and the initial conditions (1.2), we obtain
2 2
3 2 2 2
2 ∂ u 2 ∂ u 2
2
2 ∂ u
(−x) |ᏸu| dx dt ≤ 3 (−x) 2 +a5
+a1 ∂x 2 ∂t dx dt,
∂t
∂x∂t
Ω
Ω
2 dϕ 2
2 2 ∂u 2
2 dψ 2 ∂ u dx, (2.2)
( − x) +
( − x) +
dx ≤ sup
dx dx ∂x∂t ∂x 0
0≤t≤T 0
∂u 2
2
2
2
|ψ| + |ϕ| dx ≤ sup
∂t + |u| dx.
0
0≤t≤T 0
Combining the inequalities (2.2), we obtain (2.1) for u ∈ E.
Theorem 2.2. For any function u ∈ E, there is the a priori estimate
uE ≤ αLuF ,
with the constant
α=
max 167/10, a1
,
min exp(−cT )/20, exp(−cT )a20 /15
(2.3)
(2.4)
MIXED PROBLEM WITH NONLOCAL BOUNDARY CONDITIONS . . .
419
and c is such that
ca0 − 1 ≥ a3 + 2a25 .
c ≥ 1,
(2.5)
Before proving this theorem, we first give the following two lemmas.
Lemma 2.3. For u ∈ E satisfying the first condition in (1.2),
1
2
2 2
τ
∂ u
∂u 2
2
dx dt+ c−1
(−x)2 exp(−ct)
(−x)
exp(−ct)
∂x∂t ∂x dx dt
2
0
0 0
τ
0
≥
1
2
2
2
∂u
dx − 1 ( − x)2 dϕ dx.
(x,
τ)
( − x)2 exp(−cτ)
∂x
2 0
dx 0
(2.6)
Proof. Starting from
τ
0
0
( − x)2 exp(−ct)
∂ ∂u ∂u
dx dt,
∂t ∂x ∂x
(2.7)
then integrating by parts and using elementary inequalities, we obtain (2.6).
Lemma 2.4. For u ∈ E satisfying the initial conditions (1.2),
0
2
exp(−cτ)u(x, τ) dx ≤
τ
0
0
∂u 2
2
exp(−ct)
∂t dx dt + |ϕ| dx,
0
(2.8)
with c ≥ 1.
Proof. Integrating by parts the expression
τ
0
0
exp(−ct)u
∂u
dx dt
∂t
(2.9)
and using elementary inequalities yield (2.8).
Remark 2.5. We note that Lemmas 2.3 and 2.4 hold for weaker conditions on u.
Proof of Theorem 2.2. First, define
D(L) = u ∈ E |
∂5u
∈ L2 (Ω) ,
2
3
∂x ∂t
where
Ju =
Mu = ( − x)2
x
0
∂2u
∂2u
+ 2( − x)J 2 ,
2
∂t
∂t
u(ξ, t) dξ.
(2.10)
(2.11)
We consider for u ∈ D(L) the quadratic formula
Re
τ
0
0
exp(−ct)ᏸuMu dx dt,
(2.12)
with the constant c satisfying (2.5), obtained by multiplying (1.1) by exp(−ct)Mu, by
420
M. DENCHE AND A. L. MARHOUNE
integrating over Ωτ , where Ωτ = (0, ) × (0, τ), with 0 ≤ τ ≤ T , and by taking the real
part. Integrating by parts (2.12) with the use of boundary conditions (1.3) and (1.4),
we obtain
τ
exp(−ct)ᏸuMu dx dt
Re
0
=
0
2 2
2 2
τ
∂ u
∂ u
J
dx dt+ 1
(−x)2 exp(−ct)
exp(−ct)
∂t 2 ∂t 2 dx dt
2 0 0
0 0
τ
2 ∂ u
∂2u ∂
(2.13)
dx dt
( − x)2 exp(−ct)a
+ Re
∂x∂t ∂t ∂x∂t
0 0
τ
∂u ∂ 2 u
a
+ 2 Re
exp(−ct)
dx dt
∂t ∂t 2
0 0
τ
∂a ∂u ∂ 2 u
J
exp(−ct)
dx dt.
+ 2 Re
∂x ∂t ∂t 2
0 0
τ
On the other hand, by using the elementary inequalities we get
Re
τ
0
0
exp(−ct)ᏸuMu dx dt
≥
2 2
∂ u
( − x)2 exp(−ct)
∂t 2 dx dt
0 0
τ
2 ∂2u ∂
∂ u
( − x)2 exp(−ct)a
+ Re
dx dt
∂x∂t ∂t ∂x∂t
0 0
τ
∂u ∂ 2 u
a
exp(−ct)
dx dt
+ 2 Re
∂t ∂t 2
0 0
τ
∂a 2 ∂u 2
exp(−ct)
− 2 Re
∂x ∂t dx dt.
0 0
τ
(2.14)
Again, integrating by parts the second and third terms of the right-hand side of the
inequality (2.14) and taking into account the initial conditions (1.2) give
dψ 2
a(x, 0)(−x)2 dx dx dt
0 0
0
0
2 2
τ
τ
∂a 2 ∂u 2
∂ u
dx dt − 2
exp(−ct)( − x)2 exp(−ct)
≥
∂x ∂t dx dt
∂t 2 0 0
0 0
2
2
∂ u
1 (x, τ)
+
a(x, τ) exp(−cτ)( − x)
dx
2 0
∂x∂t
2 2
∂ u ∂a
1 τ ( − x)2 exp(−ct)
−
∂x∂t dx dt
2 0 0
∂t
2
2 2
∂ u c τ dx dt+ exp(−cτ)a(x, τ) ∂u (x, τ) dx
exp(−ct)(−x)2 a
+
2 0 0
∂x∂t
∂t
0
2
2
τ
τ
∂u ∂a ∂u dx dt + c
−
exp(−ct)
exp(−ct)a
∂t dx dt.
∂t ∂t 0 0
0 0
(2.15)
Re
τ
exp(−ct)ᏸuMu dx dt+
a(x, 0)|ψ|2 dx+
1
2
MIXED PROBLEM WITH NONLOCAL BOUNDARY CONDITIONS . . .
421
By using the elementary inequalities on the first integral in the left-hand side of (2.15),
we obtain
2 2
∂ u
3 τ 33 τ exp(−ct)( − x)2 |ᏸu|2 dx dt +
exp(−ct)( − x)2 ∂t 2 dx dt
4 0 0
2 0 0
dψ 2
1 + a(x, 0)|ψ|2 dx +
a(x, 0)( − x)2 dx dx
2 0
0
2 2
τ
τ
∂a 2 ∂u 2
∂ u
dx dt − 2
≥
exp(−ct)( − x)2 exp(−ct)
∂x ∂t dx dt
∂t 2 0 0
0 0
2 2
2
τ
∂ u(x, τ)2
1 ∂a dx− 1
∂ u dx dt
exp(−cτ)(−x)2 exp(−ct)(−x)2
+
2 0
∂x∂t
2 0 0
∂t ∂x∂t
2
2 2
∂ u c τ dx dt + exp(−cτ)a(x, τ) ∂u(x, τ) dx
exp(−ct)( − x)2 a
+
2 0 0
∂x∂t
∂t
0
2
τ
τ
∂u 2
∂a ∂u dx dt + c
−
exp(−ct)
exp(−ct)a
∂t dx dt.
∂t ∂t 0 0
0 0
(2.16)
Now, from (1.1) we have
1 τ exp(−ct)( − x)2 |ᏸu|2 dx dt
5 0 0
∂a 2 ∂ 2 u 2
1 τ exp(−ct)( − x)2 +
∂x ∂x∂t dx dt
5 0 0
2 2
∂ u
1 τ exp(−ct)( − x)2 +
∂t 2 dx dt
5 0 0
3
∂ u 2
1 τ exp(−ct)( − x)2 a2 ≥
∂x 2 ∂t dx dt.
15 0 0
(2.17)
Combining inequalities (2.16), (2.17), and Lemmas 2.3 and 2.4, we get
167
10
dψ 2
( − x)2 dx dx
Ω
0
dϕ 2
1 dx + |ϕ|2 dx
|ψ|2 dx +
( − x)2 + a1
dx 2 0
0
0
2
2
2
τ 2 1 ∂ u
1
2 ∂ u
dx dt
(x,
τ)
( − x)2 dx
dt
+
(
−
x)
≥ exp(−cT )
2
20 0 0
∂t
2 0
∂x∂t
2
2
∂u
1 2 ∂u
+ |u(x, τ)|2 dx+a0
∂t (x, τ) dx+ 2 (−x) ∂x (x, τ) dx
0
0
0
2
3
a20 τ 2 ∂ u ( − x) 2 dx dt .
+
15 0 0
∂x ∂t
(2.18)
( − x)2 |ᏸu|2 dx dt +
a1
2
As the left-hand side of (2.18) is independent of τ, by replacing the right-hand side
by its upper bound with respect to τ in the interval [0, T ], we obtain the desired
inequality.
422
M. DENCHE AND A. L. MARHOUNE
3. Solvability of the problem. From estimates (2.1) and (2.3) it follows that the
operator L : E → F is continuous and its range is closed in F . Therefore, the inverse
operator L−1 exists and is continuous from the closed subspace R(L) onto E, which
means that L is a homomorphism from E onto R(L). To obtain the uniqueness of
solution, it remains to show that R(L) = F . The proof is based on the following lemma.
Lemma 3.1. Suppose that ∂ 3 a/∂x 2 ∂t is also bounded. Let D0 (L) = {u ∈ D(L) : l1 u = 0,
l2 u = 0}. If for u ∈ D0 (L) and some ω ∈ L2 (Ω),
( − x)ᏸu" dx dt = 0,
(3.1)
Ω
then ω = 0.
Proof. From (3.1) we have
∂2u
∂2u
∂
a
" dx dt.
( − x) 2 " dx dt = ( − x)
∂t
∂x
∂x∂t
Ω
Ω
(3.2)
If we introduce the smoothing operators with respect to t (see [13]) Jξ−1 = (I+ξ(∂/∂t))−1
and (Jξ−1 )∗ , then these operators provide the solutions of the respective problems
ξ
−ξ
dgξ (t)
+ gξ (t) = g(t),
dt
dgξ∗ (t)
dt
gξ (t)|t=0 = 0,
+ gξ∗ (t) = g(t),
gξ∗ (t)|t=T = 0,
(3.3)
(3.4)
and also have the following properties: for any g ∈ L2 (0, T ), the functions gξ = (Jξ−1 )g
and gξ∗ = (Jξ−1 )∗ g are in W21 (0, T ) such that gξ |t=0 = 0 and gξ∗ |t=T = 0. Moreover, Jξ−1
T
T
commutes with ∂/∂t, so 0 |gξ − g|2 dt → 0 and 0 |gξ∗ − g|2 dt → 0 for ξ → 0.
Now, for given ω(x, t), we introduce the function
x
ω(ξ, t)
dξ.
(3.5)
v(x, t) = ω(x, t) −
−ξ
0
Integrating by parts with respect to ξ, we obtain
x
0
ξ
x
∂
ω(η, t)
( − ξ)
dη dξ
ω(ξ, t) dξ +
−η
0
0 ∂ξ
0
= ( − x) ω(x, t) − v(x, t) ,
v(ξ, t) dξ =
x
(3.6)
which implies that
( − x)v + Jv = ( − x)w,
0
v(x, t) dx = 0.
Then, from equality (3.2) we obtain
2
∂u
∂ u
−
Nv dx dt =
A(t)
v dx dt,
2
∂t
Ω ∂t
Ω
where
Nv = ( − x)v + Jv,
A(t)u = −
∂
∂u
( − x)a(x, t)
.
∂x
∂x
(3.7)
(3.8)
(3.9)
MIXED PROBLEM WITH NONLOCAL BOUNDARY CONDITIONS . . .
423
Replace ∂u/∂t by the smoothed function Jξ−1 (∂u/∂t) in (3.8) and use the relation
A(t)Jξ−1 = Jξ−1 A(τ) + ξJξ−1
∂A(τ) −1
Jξ .
∂τ
(3.10)
Then, by taking the adjoint of the operator Jξ−1 , and by integrating by parts with
respect to t in the left-hand side, we obtain
∗
Ω
∂u ∂vξ
dx dt =
N
∂t
∂t
Ω
A(t)
∂u ∗
v dx dt + ξ
∂t ξ
∂A ∂u
v ∗ dx dt.
∂t ∂t ξ ξ
Ω
(3.11)
The operator A(t) has a continuous inverse on L2 (0, ) defined by the relation
A−1 (t)g = −
where
x
0
dξ
a(ξ, t)( − ξ)
ξ
0
g(η) dη + c
x
dx/a(x, t) 0 g(ξ) dξ
,
0 dx/a(x, t)
c=
0
0
x
0
dξ
,
a(ξ, t)( − ξ)
A−1 (t)g dx = 0.
(3.12)
(3.13)
Hence, the function (∂u/∂t)ξ can be represented in the form
∂u
∂t
ξ
= Jξ−1 A−1 (t)A(t)
∂u
.
∂t
(3.14)
Then, (∂A/∂t)(∂u/∂t)ξ = Aξ (t)A(t)(∂u/∂t), where
Aξ (t) =
x
∂ 2 a −1 ∂a −1 ∂a 1 1
∂a −1 1
J −
J
J
g,
g(η, t) dη − c +
∂x∂t ξ
∂t ξ ∂x a a
∂t ξ a
0
(3.15)
where the constant c is given by (3.13).
Consequently, equation (3.11) becomes
∗
Ω
∂u ∂vξ
dx dt =
N
∂t
∂t
Ω
A(t)
∂u ∗
∗
v + ξA∗
ξ vξ dx dt,
∂t ξ
(3.16)
in which the conjugate operator A∗
ξ (t) of Aξ (t) is defined by
∗
A∗
ξ vξ
∗
dξ/a(ξ, t)
1 −1 ∗ ∂a ∗ ∗ Jξ
vξ + Bvξ (x) − Bvξ (0) x =
,
a
∂τ
dξ/a(ξ, t)
(3.17)
0
where
Bvξ∗ (x) =
x
−1 ∗ ∂ 2 a
1
∂a −1 ∗ ∂a
1
Jξ
−
Jξ
vξ∗ (ξ, τ) dξ.
a(ξ, t)
∂ξ∂τ a(ξ, t) ∂ξ
∂τ
(3.18)
424
M. DENCHE AND A. L. MARHOUNE
The left-hand side of (3.16) is a continuous linear functional of ∂u/∂t. Hence, the
∗
function hξ = vξ∗ + ξA∗
ξ vξ has the derivatives ( − x)(∂hξ /∂x) ∈ L2 (Ω), (∂/∂x)(( −
x)(∂hξ /∂x)) ∈ L2 (Ω), and the following conditions are satisfied
hξ |x=0 = 0,
hξ |x= = 0,
( − x)
∂hξ = 0.
∂x x=
(3.19)
From (3.17) we have
( − x)
∗
∂hξ
1 −1 ∗ ∂a ∂vξ
= I +ξ
Jξ
,
∂x
a
∂τ ∂x
(3.20)
∂vξ∗
∂hξ
∂
1 −1 ∗ ∂a ∂
( − x)
= I +ξ
Jξ
( − x)
∂x
∂x
a
∂τ ∂x
∂x
+ξ −
∗
(∂a/∂x) Jξ−1 (∂a/∂τ)
a2
I +ξ
I +ξ
∂vξ∗
1 −1∗ ∂ 2 a
(−x)
,
+ Jξ
a
∂x∂τ
∂x
(3.21)
1 −1 ∗ ∂a
Jξ
vξ∗
= 0,
a
∂τ
x=0
(3.22)
1 −1 ∗ ∂a
Jξ
vξ∗
= 0,
a
∂τ
x=
(3.23)
∂vξ∗
1 −1 ∗ ∂a
J
( − x)
= 0.
I +ξ
a ξ
∂τ
∂x x=
(3.24)
Since ξ(1/a)(Jξ−1 )∗ (∂a/∂τ)L2 (Ω) < 1 for sufficiently small ξ, the operator I +
ξ(1/a)(Jξ−1 )∗ (∂a/∂τ) has a continuous inverse on L2 (Ω). In addition, the derivative
of the above operator with respect to x is a bounded operator in L2 (Ω). Therefore,
from (3.20) and (3.21), the function vξ∗ has derivatives ( − x)(∂vξ∗ /∂x) ∈ L2 (Ω) and
(∂/∂x)(( − x)(∂vξ∗ /∂x)) ∈ L2 (Ω).
In a similar way, we show that for each fixed x ∈ [0, ] and sufficiently small ξ, the
operator I +ξ(1/a)(Jξ−1 )∗ (∂a/∂τ) has a continuous inverse on L2 (0, T ); hence, (3.22),
and (3.23), and (3.24) imply that
vξ∗ x=0 = 0,
vξ∗ x= = 0,
( − x)
∂vξ∗ ∂x = 0.
(3.25)
x=
So, for ξ sufficiently small, the function vξ∗ has the same properties as hξ . In addition,
vξ∗ satisfies the integral condition in (3.7).
tτ
Putting u = 0 0 exp(cη)vξ∗ (η, τ) dη dτ in (3.8), where the constant c satisfies ca0 −
a3 − a23 /a0 ≥ 0, and using (3.4), we obtain
Ω
exp(ct)vξ∗ Nv dx dt = −
Ω
A(t)
∂2u
∂u
exp(−ct) 2 dx dt + ξ
∂t
∂t
∗
Ω
A(t)
∂u ∂vξ
dx dt.
∂t ∂t
(3.26)
425
MIXED PROBLEM WITH NONLOCAL BOUNDARY CONDITIONS . . .
Integrating by parts each term in the left-hand side of (3.26) and taking the real parts
yield
Re
Ω
A(t)
≥
∂2u
∂u
exp(−ct) 2 dx dt
∂t
∂t
2 2
∂ u ( − x)a(x, t) exp(−ct)
∂x∂t dx dt
Ω
2 2
∂ u 1
∂a
−
exp(−ct)
( − x)
∂x∂t dx dt,
2 Ω
∂t
c
2

Re  − ξ
(3.27)

2 2
∗
∂ u −ξa23
∂u ∂vξ
dx dt  ≥
A(t)
( − x) exp(−ct)
∂x∂t dx dt.
∂t
∂t
2a
Ω
Ω
0
Now, using (3.27) in (3.26) with the choice of c indicated above we have
2 Re exp(ct)vξ∗ Nv dx dt ≤ 0.
Ω
Then, for ξ → 0 we obtain 2 Re
2 Re
Since Re
Ω
Ω exp(ct)vNv dx dt
exp(ct)( − x)|v|2 dx dt + 2 Re
Ω exp(ct)vJv dx dt
(3.28)
≤ 0, that is,
Ω
exp(ct)vJv dx dt ≤ 0.
(3.29)
= 0, we conclude that v = 0; hence, ω = 0, which ends
the proof of the lemma.
Theorem 3.2. The range R(L) of L coincides with F .
Proof. Since F is a Hilbert space, we have R(L) = F if and only if the relation
dl2 u dψ
2 dl1 u dϕ
+
dx+
(−x) ᏸuf dx dt+
l1 uϕ+l2 uψ dx = 0,
(−x)
dx dx
dx dx
Ω
0
0
(3.30)
for arbitrary u ∈ E and (f , ϕ, ψ) ∈ F , implies that f = 0, ϕ = 0 and ψ = 0. Putting
2
u ∈ D0 (L) in (3.30), we conclude from Lemma 3.1 that ( − x)f = 0. Hence,
0
( − x)2
Setting
dl1 u dϕ dl2 u dψ
+
+ l1 uϕ + l2 uψ dx = 0
dx dx
dx dx
D0k (L) = u ∈ D(L) : u(k) t=0 = 0, k = 0, 1 ,
∀u ∈ D(L).
(3.31)
(3.32)
and taking u ∈ D01 (L) in (3.31) yield
0
( − x)2
dl1 u dϕ
+ l1 uϕ dx = 0.
dx dx
(3.33)
The range of the trace operator l1 is everywhere dense in Hilbert space with the norm
[ 0 (( − x)2 |dϕ/dx|2 + |ϕ|2 ) dx]1/2 ; hence, ϕ = 0. Likewise, for u ∈ D00 (L), we get
ψ = 0.
426
M. DENCHE AND A. L. MARHOUNE
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M. Denche: Institut de Mathématiques, Université Mentouri Constantine, Constantine 25000, Algeria
E-mail address: m_denche@hotmail.com
A. L. Marhoune: Institut de Mathématiques, Université Mentouri Constantine, Constantine 25000, Algeria