Electronic Journal of Differential Equations, Vol. 2000(2000), No. 60, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
ftp ejde.math.swt.edu ftp ejde.math.unt.edu (login: ftp)
High-order mixed-type differential equations with
weighted integral boundary conditions ∗
M. Denche & A. L. Marhoune
Abstract
In this paper, we prove the existence and uniqueness of strong solutions for high-order mixed-type problems with weighted integral boundary
conditions. The proof uses energy inequalities and the density of the range
of the operator generated.
1
Introduction
Let α be a positive integer and Q be the set (0, 1) × (0, T ). We consider the
equation
∂ 2α+1 u
∂2u
(1)
Lu := 2 + (−1)α a(t) 2α = f (x, t),
∂t
∂x ∂t
where the function a(t) and its derivative are bounded on the interval [0, T ]:
0 < a0 ≤ a(t) ≤ a1 ,
da(t)
≤ a2 .
dt
(2)
(3)
To equation (1) we attach the initial conditions
l1 u = u(x, 0) = ϕ(x),
l2 u =
∂u
(x, 0) = ψ(x)
∂t
x ∈ (0, 1),
(4)
the boundary conditions
∂i
∂i
u(0,
t)
=
u(1, t) = 0,
∂xi
∂xi
for 0 ≤ i ≤ α − k − 1, t ∈ (0, T ),
(5)
and integral conditions
Z
0
1
xi u(x, t)dx = 0, for 0 ≤ i ≤ 2k − 1,
1 ≤ k ≤ α t ∈ (0, T ).
∗ Mathematics Subject Classifications: 35B45, 35G10, 35M10.
Key words: Integral boundary condition, energy inequalities, equation of mixed type.
c
2000
Southwest Texas State University.
Submitted March 27, 2000. Published September 21, 2000.
1
(6)
2
High-order mixed-type differential equations
EJDE–2000/60
where ϕ and ψ are known functions which satisfy the compatibility conditions
given in (5)-(6).
Various problems arising in heat conduction [3, 4, 8, 9], chemical engineering
[6], underground water flow [7], thermo-elasticity [14], and plasma physics [12]
can be reduced to the nonlocal problems with integral boundary conditions.
This type of boundary value problems has been investigated in [1, 2, 3, 4, 5, 6, 8,
9, 10, 13, 16] for parabolic equations and in [11, 15] for hyperbolic equations. The
basic tool in [2, 10, 16] is the energy inequality method which, of course, requires
appropriate multipliers and functional spaces. In this paper, we extend this
method to the study of a high-order mixed-type partial differential equations.
2
Preliminaries
In this paper, we prove existence and uniqueness of a strong solution of problem
(1)-(6). For this, we consider the problem (1)-(6) as a solution of the operator
equation
Lu = F ,
(7)
where L = (L, l1 , l2 ), with domain of definition D(L) consisting of functions
∂ i ∂ α−k+1 u
u ∈ W22α,2 (Q) such that ∂x
i ( ∂xα−k ∂t ) ∈ L2 (Q), i = 0, α + k − 1 and u satisfies
conditions (5)-(6); the operator L is considered from E to F , where E is the
Banach space consisting of functions u ∈ L2 (Q), satisfying (5)-(6), with the
finite norm
2
kukE
Z
=
X ∂ i+1 u 2
k ∂ 2 u 2 α−k
dx dt
J
+
i ∂t
∂t2
∂x
Q
i=0
Z 1 α−k
X ∂ i u 2 ∂ α−k+1 u 2
dx ,
+
+ sup
i
∂xα−k ∂t
0≤t≤T 0 i=0 ∂x
(8)
RxRξ
Rξ
where J k u = 0 0 1 . . . 0 k−1 u(ξ, t)dξ. Here F is the Hilbert space of vectorvalued functions F = (f, ϕ, ψ) obtained by completing of the space L2 (Q) ×
W22α (0, 1) × W22α (0, 1) with respect to the norm
2
kF kF =
Z
Q
2
f dx dt +
Z
1 α−k
X
0
i=0
i α−k ψ 2
∂ ϕ 2 + ∂
dx .
i
∂x
∂xα−k
(9)
Then we establish an energy inequality
kukE ≤ C1 kLukF ,
(10)
and we show that the operator L has the closure L.
Definition A solution of the operator equation Lu = F is called a strong
solution of the problem (1)-(6).
EJDE–2000/60
M. Denche & A. L. Marhoune
3
Inequality (10) can be extended to u ∈ D(L), i.e.,
kukE ≤ C1 LuF , ∀u ∈ D(L).
From this inequality we obtain the uniqueness of a strong solution if it exists,
and the equality of sets R(L) and R(L). Thus, to prove the existence of a strong
solution of the problem (1)-(6) for any F ∈ F , it remains to prove that the set
R(L) is dense in F .
Lemma 2.1 For any function u ∈ E, we have
Z 1
Z 1
2k ∂ 2 u 2
k ∂ 2 u 2
J
dx ≤ 4k
J
dx .
2
∂t
∂t2
0
0
(11)
R1
2
2
Proof Integrating − 0 xJ k ∂∂t2u J k+1 ∂∂tu2 dx by parts, and using elementary inequalities we obtain (11).
♦
Lemma 2.2 For u ∈ E and 0 ≤ i ≤ α − k, we have
Z 1 i+1
Z 1
∂ u 2
∂ α k ∂u 2
dx ≤ 4(α−k)−i
) dx .
(J
i
∂x ∂t
∂xα
∂t
0
0
(12)
R 1 ∂ α−i k ∂u ∂ α−i−1 k ∂u
Proof Integrating by parts − 0 x ∂x
α−i (J
∂t ) ∂xα−i−1 (J ∂t ) dx and using elementary inequalities yield (12).
♦
Lemma 2.3 For u ∈ E satisfying the condition (4) we have
α−k
XZ 1
∂ i u(x, τ ) 2
dx
exp(−cτ )
∂xi
i=0 0
α−k
XZ 1
∂ i ϕ 2
(1 − x) i dx
≤
∂x
i=0 0
Z τZ 1
∂α
1
∂u 2
+ (4α−k+1 − 1)
exp(−ct) α (J k ) dx dt ,
3
∂x
∂t
0
0
(13)
where c ≥ 1 and 0 ≤ τ ≤ T .
Rτ
i+1
i
Proof Integrating by parts 0 exp(−ct) ∂∂xi ∂tu ∂∂xui dt for i = 0, α − k − 1, using
elementary inequalities, and lemma 2.2, we obtain (13).
3
An energy inequality and its applications
Theorem 3.1 For any function u ∈ D(L), we have
kukE ≤ C1 kLukF ,
where C1 =
exp(cT ) max(8.4k , a21 )/ min( a20 , 78 )
c≥1
and
(14)
, with the constant c satisfying
3(ca0 − a2 ) ≥ 2(4α−k−1 − 1).
(15)
4
High-order mixed-type differential equations
EJDE–2000/60
Proof Let
∂2u
.
∂t2
For a constant c satisfying (15), we consider the quadratic form
M u = (−1)k J 2k
Z
τ
Z
1
exp(−ct)LuM u dx dt,
Re
0
(16)
0
which is obtained by multiplying (1) by exp(−ct)M u, then integrating over Qτ ,
with Qτ = (0, 1) × (0, τ ), 0 ≤ τ ≤ T , and then taking the real part. Integrating
by parts in (16) with the use of boundary conditions (5) and (6), we obtain
Z
τ
Z
1
Re
exp(−ct)LuM u dx dt
0
=
Z
0
τ
Z
∂ 2 u 2
exp(−ct)J k 2 dx dt
∂t
0
0
Z τZ 1
∂ α−k+1 u ∂ α−k+2 u
exp(−ct)a(t) α−k
dx dt .
+ Re
∂x
∂t ∂xα−k ∂t2
0
0
1
(17)
By substituting the expression of M u in (16), using elementary inequalities and
Lemma 2.1 we obtain
Z τZ 1
Z τZ 1
2
k
Re
exp(−ct)LuM u dx dt ≤ 8.4
exp(−ct)Lu dx dt (18)
0
0
1
+
8
Z
0
τ
0
Z
0
1
0
∂ 2 u 2
exp(−ct)J k 2 dx dt .
∂t
By integrating the last term on the right-hand side of (17) and combining the
obtained results with the inequalities (15), (18) and lemmas 2.2, 2.3 we obtain
Z
α−k
X Z 1 ∂ i ϕ 2
2
∂ α−k ψ 2
1 1
dx
exp(−ct)Lu dx dt +
a(0) α−k dx +
2 0
∂x
∂xi
0
0
i=0 0
Z Z
α−k
XZ 1
∂ 2 u 2
∂ i u(x, τ ) 2
7 τ 1
dx
≥
exp(−ct)J k 2 dx dt +
exp(−cτ )
8 0 0
∂t
∂xi
i=0 0
α−k
XZ τZ 1
∂ i+1 u 2
+
exp(−ct) i dx dt
(19)
∂x ∂t
0
i=0 0
Z
∂ α−k+1 u(x, τ ) 2
1 1
dx
exp(−cτ )a(τ )
+
2 0
∂xα−k ∂t
8.4k
Z
τ
Z
1
Using elementary inequalities and (2) we obtain
8.4
k
Z
2
Lu dx dt + a1
2
Q
Z
0
1
α−k
X
∂ α−k ψ 2
dx +
∂xα−k
i=0
Z
1
0
∂ i ϕ 2
dx
∂xi
EJDE–2000/60
M. Denche & A. L. Marhoune
5
Z Z
α−k
X Z 1 ∂ i u(x, τ ) 2
7 τ 1 k ∂ 2 u 2
J
dx dt +
dx (20)
≥ exp(−cT )
8 0 0
∂t2
∂xi
i=0 0
Z 1 α−k+1
α−k
X Z τ Z 1 ∂ i+1 u 2
∂
u(x, τ ) 2 dx dt + a0
+
dx
i
α−k
∂x ∂t
2 0
∂x
∂t
0
i=0 0
As the left hand side of (20) 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.
♦
Lemma 3.2 The operator L from E to F admits a closure.
Proof Suppose that (un ) ∈ D(L) is a sequence such that
un → 0 in E
(21)
and
Lun → F
in F .
(22)
We need to show that F = (f, ϕ1 , ϕ2 ) = 0. The fact that ϕi = 0; i = 1, 2;
results directly from the continuity of the trace operators li .
Introduce the operator
L0 v =
∂ α ((1 − x)α+k J k v)
∂ 2 ((1 − x)α+k J k v)
∂ α+1
+ (−1)α+1 α (a(t)
),
2
∂t
∂x ∂t
∂xα
defined on the domain D(L0 ) of functions v ∈ W22α,2 (Q) satisfying
v|t=T = 0
∂iv
∂ iv
|x=0 =
|x=1 = 0 ,
i
∂x
∂xi
∂v
|t=T = 0
∂t
i = 0, α − 1 .
we note that D(L0 ) is dense in the Hilbert space obtained by completing L2 (Q)
with respect to the norm
Z
2
2
(1 − x)2(α+k) J k v dx dt.
kvk =
Q
Since
Z
Z
Q
f (1 − x)α+k J k v dx dt
=
=
lim
n→∞
Z
Q
lim
n→∞
Q
Lun (1 − x)α+k J k v dx dt
un L0 v dx dt = 0 ,
holds for every function v ∈ D(L0 ), it follows that f = 0.
Theorem 3.1 is valid for strong solutions, i.e., we have the inequality
kukE ≤ C1 LuF , ∀ u ∈ D(L),
hence we obtain the following.
♦
6
High-order mixed-type differential equations
EJDE–2000/60
Corollary 3.3 A strong solution of (1)-(6) is unique if it exists, and depends
continuously on F = (f, ϕ, ψ) ∈ F .
Corollary 3.4 The range R(L) of the operator L is closed in F , and R(L) =
R(L).
4
Solvability of Problem (1)-(6)
To proof solvability of (1)-(6), it is sufficient to show that R(L) is dense in F .
The proof is based on the following lemma
Lemma 4.1 Let D0 (L) = {u ∈ D(L) : l1 u = 0, l2 u = 0}. If for u ∈ D0 (L) and
some ω ∈ L2 (Q), we have
Z
(1 − x)2k Lu$ dx dt = 0 ,
(23)
Q
then ω = 0.
Proof The equality (23) can be written as follows
Z
Z
∂2u
∂ α ∂ α+1 u
− (1 − x)2k 2 $ dx dt = (−1)α (1 − x)2k α (a α )$ dx dt.
∂t
∂x
∂x ∂t
Q
Q
(24)
For ω(x, t) given, we introduce the function v(x, t) = (−1)k ∂ 2k ((1−x)2k ω)/∂x2k ,
R1
then we have 0 xi v(x, t)dx = 0 for i = 0, 2k − 1. Then from equality (24) we
have
Z
Z
∂ 2 u 2k
∂ α ∂ α+1 u 2k
α
J
v
dx
dt
=
(−1)
(a α )J v dx dt.
(25)
−
2
α
∂x ∂t
Q ∂t
Q ∂x
Integrating by parts the right hand side of (25) 2k times, we get
Z
Z
∂u
∂ 2 u 2k
−
J v dx dt =
A(t) v dx dt,
2
∂t
∂t
Q
Q
α−k
(26)
α−k
∂
u
∂
where A(t)u = (−1)α ∂x
α−k (a ∂xα−k ).
∂ −1
) and (Jε−1 )∗
When we introduce the smoothing operators Jε−1 = (I + ε ∂t
with respect to t [16], then these operators provide solutions of the problems
ε
dgε (t)
+ gε (t) =
dt
gε (t)|t=0 =
g(t),
(27)
0,
and
−ε
dgε∗ (t)
+ gε∗ (t)
dt
gε∗ (t)|t=T
=
g(t),
=
0.
(28)
EJDE–2000/60
M. Denche & A. L. Marhoune
7
The solutions have the following properties: for any g ∈ L2 (0, T ), the functions
gε = (Jε−1 )g and gε∗ = (Jε−1 )∗ g are in W21 (0, T ) such that gε |t=0 = 0 and
2
RT ∂
gε∗ |t=T = 0. Moreover, Jε−1 commutes with ∂t
, so 0 gε − g dt → 0 and
2
RT ∗
gε − g dt → 0, for ε → 0.
0
−1 ∂u
Replacing in (26), ∂u
∂t by the smoothed function Jε ∂t , using the relation
−1
−1
−1 ∂A(τ ) −1
A(t)Jε = Jε A(τ ) + εJε
∂τ Jε , and using properties of the smoothing
operators we obtain
Z
Z
Z
∂u
∂u 2k ∂vε∗
∂A ∂u ∗
) dx dt =
( )ε vε dx dt.
J (
A(t) vε∗ dx dt + ε
(29)
∂t
∂t
Q ∂t
Q
Q ∂t ∂t
Passing to the limit,(29) is satisfied for all functions satisfying the conditions
∂i
∂ α+1 u
(4)-(6) such that ∂x
i (a ∂xα ∂t ) ∈ L2 (Q) for 0 ≤ i ≤ α.
The operator A(t) has a continuous inverse on L2 (0, 1) defined by
A−1 (t)g
=
(−1)α
Z
x Z ηα−k−1
+
η1
...
0
0
α−k
X
Z
0
i−1
Ci (t)
i=1
Z
η
Z
(30)
Z
ξα−k−1
ξ1
...
0
0
0
1
g(ξ)dξdξ1 . . . dξα−k−1
a
η
dηdη1 . . . dηα−k−1 .
(i − 1)!
Then we have
Z
1
A−1 (t)g dx = 0 .
(31)
0
∂u
−1 −1 ∂u
A ∂t .
Hence the function ( ∂u
∂t )ε can be represented in the form ( ∂t )ε = Jε A
∂A ∂u
∂u
Then ∂t ( ∂t )ε = Aε (t)A(t) ∂t , where
α−k
g X ∂ α−k xi−1 −1 J Ci .
Aε (t)g = (−1)α a0 (t)Jε−1 +
a i=1 ∂xα−k (i − 1)! ε
Consequently, (29) becomes
Z
Z
∂u
∂u 2k ∂vε∗
) dx dt =
J (
A(t) (vε∗ + εA∗ε vε∗ ) dx dt,
∂t
∂t
∂t
Q
Q
(32)
(33)
in which A∗ε (t) is the adjoint of the operator Aε (t). The left-hand side of (33)
∗
∗ ∗
is a continuous linear functional of ∂u
∂t . Hence the function hε = vε + εAε vε
i
has the derivatives ∂∂xhiε ∈ L2 (Q),
following conditions are satisfied
∂ α−k hε
∂i
∂xi (a ∂xα−k )
∈ L2 (Q), i = 0, α − k, and the
∂ i hε ∂ i hε =
= 0 , i = 0, α − k − 1.
∂xi x=0
∂xi x=1
(34)
The operators A∗ε (t) are bounded in L2 (Q), for ε sufficiently small we have
kεA∗ε (t)kL2 (Q) < 1; hence the operator I + εA∗ε (t) has a bounded inverse in
8
High-order mixed-type differential equations
L2 (Q). In addition, the operators
From the equality
∂ i A∗
ε (t)
∂xi ,
EJDE–2000/60
i = 0, α − k are bounded in L2 (Q).
i
X ∂ k A∗ (t) ∂ i−k v ∗
∂ i hε
∂ i v∗
ε
ε
= (I + εA∗ε (t)) iε + ε
Cik
,
i
∂x
∂x
∂xk ∂xi−k
i = 0, α − k − 1
(35)
k=1
∂ i v∗
we conclude that vε∗ has derivatives ∂xiε in L2 (Q), i = 0, α − k − 1. Taking into
account (34) and (35), for i = 0, α − k − 1, we have
i
X
∂ i v∗
∂ k A∗ε (t) ∂ i−k vε∗ (I + εA∗ε (t)) iε + ε
Cik
∂x
∂xk ∂xi−k x=0
= 0,
(36)
i
X
∂ i v∗
∂ k A∗ε (t) ∂ i−k vε∗ (I + εA∗ε (t)) iε + ε
Cik
∂x
∂xk ∂xi−k x=1
= 0.
(37)
k=1
k=1
∂ i A∗ (t)
Similarly, for ε sufficiently small, and each fixed x ∈ [0, 1] the operators ∂xεi ,
i = 0, α − k are bounded in L2 (Q) and the operator I + εA∗ε (t) is continuously
invertible in L2 (Q). From (36) and (37) result that vε∗ satisfies the conditions
∂ i vε∗ ∂ i vε∗ =
= 0, i = 0, α − k − 1,
∂xi x=0
∂xi x=1
So, for ε sufficiently small, the function vε∗ has the same properties as hε . In
addition vε∗ satisfies the integral conditions (6).
RtRτ
Putting u = 0 0 exp(cη)vε∗ (η, τ )dηdτ in (26), with the constant c satisfying
a2
ca0 − a2 − a20 ≥ 0, and using (28), we obtain
Z
Z
∂2u
∂u
k
∗ 2k
exp(−ct) 2 dx dt
(−1) exp(ct)vε J v dx dt = − (−1)k A(t)
∂t
∂t
Q
Q
Z
∗
∂u
∂v
ε
dx dt .
(38)
+ε (−1)k A(t)
∂t ∂t
Q
Integrating by parts each term in the right-hand side of (38), we have
Z
∂2u
∂u
Re (−1)k A(t)
exp(−ct) 2 dx dt
(39)
∂t
∂t
Q
Z
Z
α−k+1 u 2
c
1
∂a −ct ∂ α−k+1 u 2
−ct ∂
e
≥
a(t)e
dx
dt
−
dx dt .
2 Q
∂xα−k ∂t
2 Q ∂t
∂xα−k ∂t
Z
Z
∂ α−k+1 u 2
−εa22
∂u ∂vε∗
k
dx dt) ≥
exp(−ct) α−k dx dt. (40)
Re(−ε (−1) A(t)
∂t ∂t
2a0 Q
∂x
∂t
Q
Now, using (39) and (40) in (38), with the choice of c indicated above, we
R
R
have 2 Re Q exp(ct)vε∗ j 2k v dx dt ≤ 0, then 2 Re Q exp(ct)vJ 2k v dx dt ≤ 0 as ε
2
R
approaches zero. Since Re Q exp(ct)J k v dx dt = 0, we conclude that J 2k v =
0, hence ω = 0, which completes the present proof.
♦
Theorem 4.2 The range R(L) of L coincides with F.
EJDE–2000/60
M. Denche & A. L. Marhoune
9
Proof Since F is a Hilbert space, we have R(L) = F if and only if the following
implication is satisfied:
Z
Q
Z
Luf dx dt +
1 α−k
X
(
0
i=0
∂ i l1 u ∂ i ϕ ∂ α−k l2 u ∂ α−k ψ
+
)dx = 0,
∂xi ∂xi
∂xα−k ∂xα−k
(41)
for arbitrary u ∈ E and F = (f, ϕ, ψ) ∈ F , implies
that f , ϕ, and ψ are zero.
R
Putting u ∈ D0 (L) in (41), we obtain Q Luf dx dt = 0. Taking ω =
f /(1 − x)2k , and using lemma 4.1 we obtain that f /(1 − x)2k = 0, then f = 0.
Consequently, ∀u ∈ D(L) we have
Z
0
1 α−k
X
i=0
∂ i l1 u ∂ i ϕ ∂ α−k l2 u ∂ α−k ψ
+
dx = 0 .
∂xi ∂xi
∂xα−k ∂xα−k
(42)
The range of the trace operator (l1, l2 ) is everywhere dense in a Hilbert space
with norm
Z X i
∂ ϕ 2 ∂ α−k ψ 2 1/2
1 α−k
+
dx
.
i
∂xα−k
0 i=0 ∂x
Therefore, (ϕ, ψ) = (0, 0) and the present proof is complete.
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Mohamed Denche (e-mail: m denche@hotmail.com)
A. L. Marhoune
Institut de Mathematiques
Université Mentouri Constantine
25000 Constantine, Algeria