A THREE-POINT BOUNDARY VALUE PROBLEM WITH AN INTEGRAL CONDITION FOR A
THIRD-ORDER PARTIAL DIFFERENTIAL EQUATION
C. LATROUS AND A. MEMOU Received 9 February 2004
We prove the existence and uniqueness of a strong solution for a linear third-order equa- tion with integral boundary conditions. The proof uses energy inequalities and the den- sity of the range of the operator generated.
1. Introduction
In the rectangleΩ=(0, 1)×(0,T), we consider the equation f(x,t)=∂3u
∂t3 + ∂
∂x
a(x,t)∂u
∂x
(1.1) with the initial conditions
u(x, 0)=0, ∂u
∂t(x, 0)=0, x∈(0, 1), (1.2) the final condition
∂2u
∂t2(x,T)=0, x∈(0, 1), (1.3)
the Dirichlet condition
u(0,t)=0 ∀t∈(0,T), (1.4)
and the integral condition 1
l u(x,t)dx=0, 0≤l <1,t∈(0,T). (1.5)
Copyright©2005 Hindawi Publishing Corporation Abstract and Applied Analysis 2005:1 (2005) 33–43 DOI:10.1155/AAA.2005.33
In addition, we assume that the functiona(x,t) and its derivatives satisfy the conditions 0< a0< a(x,t)< a1 ∀x,t∈Ω,
∂a
∂x
≤b ∀x,t∈Ω,
ck<∂ku
∂tk(x,t)< ck ∀x,t∈Ω,k=1, 3, withc1>0.
(1.6)
Over the last few years, many physical phenomena were formulated into nonlocal mathe- matical models with integral boundary conditions [1,9,10,11]. The reader should refer to [13,14] and the references therein. The importance of these kinds of problems has also been pointed out by Samarskii [22]. This type of boundary value problems has been investigated in [2,3,4,6,7,8,12,18,19,20,23,25] for parabolic equations, in [21,24]
for hyperbolic equations, and in [15,16,17] for mixed-type equations. The basic tool in [5,15,16,17,20,25] is the energy inequality method which, of course, requires appro- priate multipliers and functional spaces. In this paper, we extend this method to the study of a linear third-order partial differential equation.
2. Preliminairies
In this paper, we prove the existence and uniqueness of a strong solution of the problem (1.1)–(1.5). For this, we consider the solution of problem (1.1)–(1.5) as a solution of the operator equation
Lu=Ᏺ, (2.1)
where the operatorLhas domain of definitionD(L) consisting of functionsu∈L2(Ω) such that (∂k+1u/∂tk∂x)(x,t)∈L2(Ω),k=1, 3 and satisfing the conditions (1.4)-(1.5).
The operatorLis considered fromEtoF, whereEis the Banach space consisting of functionu∈L2(Ω), with the finite norm
u2E=
ΩΘ(x)∂3u
∂t3
2+∂2u
∂x2 2
dx dt
+
ΩΘ(x)∂u
∂x
2+∂2u
∂t∂x 2
dx dt
+
ΩΦ(x)∂u
∂t
2+|u|2
dx dt.
(2.2)
Fis the Hilbert space of functionsᏲ=(f, 0, 0, 0), f ∈L2(Ω), with the finite norm Ᏺ2F=
ΩΘ(x)f(x,t)2dx dt, (2.3)
where
Θ(x)=
(1−l)2, 0< x≤l, (1−x)2, l≤x <1, Φ(x)=
0, 0< x < l, 1, l≤x <1.
(2.4)
3. An energy inequality and its application
Theorem3.1. For any functionu∈D(L), the a priori estimate
uE≤kLuF foru∈D(L), (3.1)
where k2=40 exp(cT)/k1 with k1=inf{1/4, (c3−3cc1+ 3c2c1−c3a1−b2)/2, a20/2, (3/
2)(ca0−c1)}. The constantcsatisfies
sup
(x,t)∈Ω
1 a
∂a
∂t
< c < inf
(x,t)∈Ω
1 a
∂a
∂t + 1
, c3−3cc1+ 3c2c1−c3a1−b2>0,
c2−2cc1+c2a21+ca0−c1>0.
(3.2)
Proof. Let
Mu=
(1−l)2∂3u
∂t3, 0< x < l, (1−x)2∂3u
∂t3 + 2(1−x)Jx∂3u
∂t3, l < x <1,
(3.3)
whereJxu=x
l u(x,t)dx.
We consider the quadratic form obtained by multiplying (1.1) by exp(−ct)Mu, with the constantc satisfying (3.2), integrating overΩ=(0, 1)×(0,T), and taking the real part:
Φ(u,u)=Re
Ωexp(−ct)f(x,t)Mudx dt. (3.4)
By substituting the expression ofMuin (3.4), integrating with respect tox, and using the Dirichlet and integral conditions, we obtain
Re
Ωexp(−ct)f(x,t)Mudx dt
= T
0
1
0Θ(x) exp(−ct)∂3u
∂t3 2dx dt
−3 2
T
0
1
0Θ(x) exp(−ct) ∂a
∂t −ca∂2u
∂x∂t 2dx dt +
T
0
1
0
Θ(x)
2 exp(−ct) ∂3a
∂t3 −3c∂2a
∂t2 + 3c∂a
∂t −c3a∂u
∂x 2dx dt +
T
0
1
l exp(−ct)Jx∂3u
∂t3 2dx dt
−2 Re T
0
1
l exp(−ct)a(x,t)u∂3u
∂t3dx dt +
1
0Θ(x) exp(−ct)a(x,t)∂2u
∂x∂t
2dx|t=T
− 1
0Θ(x) exp(−ct) ∂a
∂t−ca ∂u
∂x
∂2u
∂x∂tdx|t=T
− 1
0
Θ(x)
2 exp(−ct) ∂2a
∂t2 −2c∂a
∂t +c2a∂u
∂x
2dx|t=T
−2 Re T
0
1
l exp(−ct)∂a
∂xuJx∂3u
∂t3dx dt.
(3.5)
Integrating by parts−2 Re0Tl1exp(−ct)a(x,t)u(∂3u/∂t3)dx dtwith respect tot, and us- ing the initial conditions, the final conditions, and the elementary inequalities, we obtain
T
0
1
0
Θ(x)
2 exp(−ct)∂3u
∂t3 2dx dt
−3 2
T
0
1
0Θ(x) exp(−ct) ∂a
∂t −ca∂2u
∂x∂t 2dx dt +
T
0
1 0
Θ(x)
2 exp(−ct) ∂3a
∂t3 −3c∂2a
∂t2 + 3c∂a
∂t −c3a∂u
∂x 2dx dt +
T
0
1
l exp(−ct)Jx∂3u
∂t3 2dx dt +
T
0
1
l exp(−ct) ∂3a
∂t3 −3c∂2a
∂t2 + 3c∂a
∂t −c3a
|u|2dx dt
−3 2
T
0
1
l exp(−ct) ∂a
∂t−ca∂u
∂t 2dx dt +
1 0
Θ(x)
2 exp(−ct)
a− ∂a
∂t −ca∂2u
∂x∂t
2dx|t=T
− 1
0
Θ(x)
2 exp(−ct) ∂2a
∂t2 −2c∂a
∂t +c2a+∂a
∂t −ca∂u
∂x
2dx|t=T
+ 1
0Φ(x) exp(−ct)
a− ∂a
∂t −ca∂u
∂t
2dx|t=T
− 1
0Φ(x) exp(−ct) ∂2a
∂t2 −2c∂a
∂t +c2a+∂a
∂t −ca
|u|2dx|t=T
≤17 T
0
1
l Θ(x) exp(−ct)|f|2dx dt.
(3.6) From (1.1), we get
ΩΘ(x)a2∂2u
∂x2 2dx dt
≤2
ΩΘ(x)∂3u
∂t3
2dx dt+ 2
ΩΘ(x)∂a
∂x 2
∂u
∂x 2dx dt + 4
ΩΘ(x)|f|2dx dt.
(3.7)
Combining this last inequality with (3.6) and using the conditions (3.2) yield
ΩΘ(x)∂3u
∂t3
2+∂2u
∂x2 2
dx dt +
ΩΘ(x)∂u
∂x
2+∂2u
∂t∂x 2
dx dt+
ΩΦ(x)∂u
∂t
2+|u|2
dx dt
≤k
ΩΘ(x)f(x,t)2dx dt,
(3.8)
which is the desired inequality.
It can be proved in a standard way that the operatorL:E→Fis closable. LetLbe the closure of this operator, with the domain of definitionD(L).
Definition 3.2. A solution of the operator equationLu=Ᏺis called a strong solution of problem (1.1)–(1.5).
The a priori estimate (3.1) can be extended to strong solutions, that is, we have the estimate
uE≤cLuF ∀u∈D(L). (3.9)
This last inequality implies the following corollaries.
Corollary3.3. A strong solution of (1.1)–(1.5) is unique and depends continuously onᏲ.
Corollary3.4. The rangeR(L)ofLis closed inFandR(L)=R(L).
Corollary 3.4shows that to prove that problem (1.1)–(1.5) has a strong solution for arbitraryᏲ, it suffices to prove that setR(L) is dense inF.
4. Solvability of problem (1.1)–(1.5)
To prove the solvability of problem (1.1)–(1.5) it is sufficient to show thatR(L) is dense inF. The proof is based on the following lemma.
Lemma4.1. Suppose that the functiona(x,t)and its derivatives are bounded. Letu∈D0(L)
= {u∈D(L), u(x, 0)=0, (∂u/∂t)(x, 0)=0, (∂2u/∂t2)(x,T)=0}. If for u∈D0(L)and some functionsw(x,t)∈L2(Ω),
Ωh(x)f wdx dt=0, (4.1)
where
h(x)=
1−l, 0< x < l,
1−x, l < x <1, (4.2)
holds, for arbitraryu∈D0(L), and thenw=0.
Proof. The equality (4.1) can be written as follows:
Ωh(x)∂3u
∂t3wdx dt=
ΩA(t)uvdx dt, (4.3)
for a givenw(x,t), where
v=
(1−l)w, 0< x < l, w−
x
l
w
1−ζdζ, l < x <1, A(t)u= ∂
∂x
h(x)a(x,t)∂u
∂x
, Nv=
(1−l)v, 0< x < l, (1−x)v+Jxv, l < x <1.
(4.4)
Forv=w−x
l (w/(1−ζ))dζ,l < x <1 we deducelxv(ζ,t)dζ=(1−x)lx(w/(1−ζ))dζ, thenl1v(ζ,t)dζ=0.
Following [25], we introduce the smoothing operators with respect tot, (J−1)=(I− (∂3/∂t3))−1, and (J−1)∗=(I+(∂3/∂t3))−1which provide the solution of the respective problems:
u−∂3u
∂t3 =u, u(x, 0)=0, ∂u
∂t (x, 0)=0, ∂2u
∂t2 (x,T)=0, v∗ +∂3v∗
∂t3 =v, v∗(x, 0)=0, ∂v∗
∂t (x,T)=0, ∂2v∗
∂t2 (x,T)=0.
(4.5)
And also, we have the following properties: for anyu∈L2(0,T), the functionJ−1u∈ W23(0,T), (J−1)∗u∈W23(0,T). Ifu∈D(L),J−1u∈D(L).
lim→0
J−1u−uL2(0,T)=0, lim
→0
J−1∗u−uL2(0,T)=0. (4.6)
Substituting the functionuin (4.3) by the smoothing functionuand using the relation A(t)u=J−1A(t)u+J−1B(t)u, where B(t)=(3∂/∂t)((∂A(t)/∂t)(∂u/∂t)) + (∂3A(t)/
∂t3)u, we obtain
ΩuN∂3v∗
∂t3 dx dt=
ΩA(t)uv∗dx dt−
ΩB(t)uv∗dx dt. (4.7) The operatorA(t) has a continuous inverse inL2(0, 1) defined by
A−1(t)g=
− 1 1−l
x
0
dζ a(ζ,t)
ζ
0g(η)dη+C1(t) 1−l
x
0
dζ
a(ζ,t), 0< x < l, x
l
−dζ (1−ζ)a(ζ,t)
ζ
lg(η)dη+C2(t) x
l
dζ
(1−ζ)a(ζ,t)+u(l), l < x <1, (4.8)
where
C1(t)=(1−l)u(l) +0ldζ/a(ζ,t)0ζg(η)dη l
0
dζ/a(ζ,t) ,
C2(t)=−(1−l)u(l) +l1dζ/a(ζ,t)lζg(η)dη 1
l
dζ/a(ζ,t) .
(4.9)
Then we havel1A−1(t)u=0, hence, the functionJ−1u=uε can be represented in the form
uε=J−1A−1(t)A(t)u. (4.10)
The adjoint ofB(t) has the form B∗(t)v=1
a
J−1∗∂3a
∂t3v+3 a
J−1∗∂
∂t ∂a
∂t
∂v
∂t
−G(v)(x) +
x
0
dζ/a(ζ,t) 1
0
dζ/a(ζ,t)G(v)(1),
(4.11)
where
G(v)(x)= x
0
3 a
J−1∗∂
∂t ∂2a
∂t∂ζ
∂v
∂t
− 3 a2
∂a
∂ζ
J−1∗∂
∂t ∂a
∂t
∂v
∂t
+1 a
J−1∗∂
∂t ∂4a
∂t3∂ζv
− 1 a2
∂a
∂ζ J−1∗
∂3a
∂t3v
dζ.
(4.12)
Consequently, equality (4.7) becomes
ΩuN∂3v∗
∂t3 dx dt=
ΩA(t)uhdx dt, (4.13)
whereh=v∗−B∗(t)v∗.
The left-hand side of (4.13) is a continuous linear functional ofu, hence the function hhas the derivatives∂h/∂x, (1−x)(∂h/∂x)∈L2(Ω), and the conditionh(0,t)=0 is satisfied.
From the equality (1−x)∂h
∂x =
I−1 a
J−1∗ ∂3a
∂t3
(1−x)∂v∗
∂x −31 a
J−1∗∂
∂t ∂a
∂t
∂
∂t(1−x)∂v∗
∂x
, (4.14) and since the operator (J−1)∗ is bounded in L2(Ω), for sufficiently small , we have (1/a)(J−1)∗(∂3a/∂t3)<1. Hence, the operatorI−(1/a)(J−1)∗(∂3a/∂t3) has a bounded inverse inL2(Ω). We conclude that (1−x)(∂v∗/∂x)∈L2(Ω). Similarly, we conclude that (∂/∂x)((1−x)(∂v∗/∂x)) exists and belongs toL2(Ω), and the condition v∗(0,t)=0 is satisfied.
Puttingu=t
0
ζ
0
T
η exp(cτ)v∗dτ dη dζin (4.3), where the constantcsatisfies (3.2) and using the proprieties of smoothing operator, we obtain
Ωexp(ct)v∗εNv dx dt= −
ΩA(t)uv∗ε dx dt−ε
ΩA(t)u∂3v∗
∂t3 dx dt, (4.15) and from
−ε
ΩA(t)u∂3v∗
∂t3 dx dt
=3
Ωh(x) exp(−ct)∂2a
∂t2 ∂3u
∂t2∂x 2dx dt
−3
Ωh(x) exp(−ct) ∂3a
∂t3 −c∂2a
∂t2 ∂3u
∂t2∂x
∂2u
∂t∂xdx dt + 3
1
0
h(x)
2 exp(−ct)∂a
∂t ∂3u
∂t2∂x
2dx|t=T
+ 3 1
0
h(x)
2 exp(−ct) ∂2a
∂t2 −c∂a
∂t ∂2u
∂t∂x
2dx|t=T
−
Ωh(x) exp(−ct)a∂3v∗
∂t3
2dx dt
−
Ωh(x) exp(−ct)∂3a
∂t3
∂u
∂x
∂3u
∂t2∂xdx dt,
(4.16)
we have
−εRe
ΩA(t)u∂3v∗
∂t3 dx dt
≤ε
3
Ωh(x) exp(−ct) ∂2a
∂t2 +1 2
∂3a
∂t3 −c∂2a
∂t2
∂3u
∂t2∂x 2dx dt +3
2
Ωh(x) exp(−ct) ∂2a
∂t2 −c∂a
∂t +∂3a
∂t3 −c∂2a
∂t2
∂2u
∂t∂x 2dx dt
−
Ωh(x) exp(−ct)a∂3v∗
∂t3
2dx dt +3
2
Ωh(x) exp(−ct)∂3a
∂t3
∂u
∂x 2dx dt +1
2
Ωh(x) exp(−ct)∂3a
∂t3
∂4u
∂t3∂x 2dx dt +1
2
Ωh(x) exp(−ct)∂a
∂t ∂3u
∂t2∂x 2dx dt
.
(4.17)
Integrating the first term on the right-hand side by parts in (4.15), we obtain
−εRe
ΩA(t)uvε∗dx dt
=3 2
Ωh(x) exp(−ct) ∂a
∂t −ca∂2u
∂t∂x 2dx dt
−
Ωh(x) exp(−ct) ∂3a
∂t3 −3c∂2a
∂t2 + 3c2∂a
∂t −c3a∂u
∂x 2dx dt
− 1
0
1
2h(x) exp(−ct)a∂2u
∂t∂x
2dx|t=T
+ 1
0
1
2h(x) exp(−ct) ∂2a
∂t2 −2c∂a
∂t +c2a∂u
∂x
2dx|t=T
− 1
0h(x) exp(−ct) ∂a
∂t −ca ∂u
∂x
∂2u
∂t∂xdx|t=T.
(4.18)
This last equality gives
−εRe
ΩA(t)uvε∗dx dt
≤ − 1
0h(x) exp(−ct)∂a
∂t +a−ca∂2u
∂x∂t
2dx|t=T
+ 1
0
1
2h(x) exp(−ct) ∂2a
∂t2 −2c∂a
∂t +c2a+ca−∂a
∂t ∂u
∂x
2dx|t=T.
(4.19)
By using the conditions (3.2), inequalities (4.17) and (4.19), we obtain Re
Ωexp(ct)vε∗Nvdx dt≤0 as−→0. (4.20)
This implies ReΩexp(ct)(vε∗−v)Nvdx dt+ ReΩexp(ct)vNvdx dt≤0, that is, T
0
l
0exp(−ct)(1−l)|v|2dx dt +
T
0
1 l
l
0exp(−ct)(1−x)|v|2dx dt+ T
0
1
l exp(−ct)Jxv2dx dt +
T
0
l
0
1−l
2l exp(−ct)Jxv2dx dt≤0.
(4.21)
Thenv=0.
Finally from (4.4), we concludew=0.
Theorem4.2. The rangeR(L)ofLcoincides withF.
Proof. SinceFis Hilbert space, thenR(L)=Fif and only if the relation
ΩΘ(x)f g dx dt=0 (4.22)
holds.
Arbitraryu∈D0(L) andᏲ=(f, 0, 0, 0)∈Fimpliesf =0. Taking in (4.22),u∈D0(L), and usingLemma 4.1, we obtain
w=
(1−l)g, 0< x < l,
(1−x)g, l < x <1, (4.23)
theng=0.
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C. Latrous: Laboratoire Equations Differentielles, D´epartement de Mathematiques, Universit´e Mentouri Constantine, 25000 Constantine, Algeria
E-mail address:[email protected]
A. Memou: Laboratoire Equations Differentielles, D´epartement de Mathematiques, Universit´e Mentouri Constantine, 25000 Constantine, Algeria
E-mail address:[email protected]
