INTEGRAL CONDITIONS FOR A LINEAR THIRD-ORDER EQUATION

INTEGRAL CONDITIONS FOR A LINEAR THIRD-ORDER EQUATION

M. DENCHE AND A. MEMOU

Received 6 March 2003 and in revised form 29 July 2003

We prove the existence and uniqueness of a strong solution for a linear third-order equation with integral boundary conditions. The proof uses energy inequalities and the density of the range of the generated opera- tor.

1. Introduction

In the rectangleΩ = [0,1]×[0, T], we consider the equation

£u=∂³u

∂t³ + ∂

∂x

a(x, t)∂u

∂x

=f(x, t), (1.1a)

with the initial conditions

u(x,0) =0, ∂u

∂t(x,0) =0, x∈(0,1), (1.1b) the final condition

∂²u

∂t²(x, T) =0, x∈(0,1), (1.1c) the Dirichlet condition

u(0, t) =0, ∀t∈(0, T), (1.1d)

Copyright^c2003 Hindawi Publishing Corporation Journal of Applied Mathematics 2003:11(2003)553–567 2000 Mathematics Subject Classification: 35B45, 35G10 URL:http://dx.doi.org/10.1155/S1110757X03303031

and the integral condition ₁

0

u(x, t)dx=0, ∀t∈(0, T). (1.1e) In addition, we assume that the functiona(x, t)is bounded with

0< a0≤a(x, t)≤a1, (1.2) and has bounded partial derivatives such that

c_k≤∂^ka

∂t^k(x, t)≤ck, ∀x∈(0,1), t∈(0, T), k=1,3,withc₁≥0, ∂a

∂x(x, t)

≤b1, for(x, t)∈Ω.

(1.3)

Various problems arising in heat conduction[4,6,14,15], chemical en- gineering[9], underground water flow[13], thermoelasticity[21], and plasmaphysics [19] can be reduced to the nonlocal problems with in- tegral boundary conditions. This type of boundary value problems has been investigated in [1, 2, 3,5,6,7,9, 14,15,16, 20, 23] for parabolic equations, in [18, 22] for hyperbolic equations, and in [10,11,12]for mixed-type equations. The basic tool in[4,10,11,12,16,23]is the en- ergy inequality method which, of course, requires appropriate multipli- ers and functional spaces. In this paper, we extend this method to the study of a linear third-order partial differential equation. This type of problems is encountered in the study of thermal conductivity[17]and microscale heat transfer[8].

2. Preliminaries

In this paper, we prove the existence and uniqueness of a strong so- lution of problem (1.1). For this, we consider the solution of problem (1.1)as a solution of the operator equationLu=F,whereLis the oper- ator with domain of definitionD(L)consisting of functionsu∈E such that√

1−x(∂^k+1u/∂t^k∂x)(x, t)∈L²(Ω),k=0,3 andusatisfies conditions (1.1d)and(1.1e). The operatorLis considered fromEtoF, whereEis the Banach space of the functionsu,u∈L²(Ω), with the finite norm

u²_E=

Ω

(1−x)² 2

∂³u

∂t^{3 2}+

∂²u

∂x^{2 2}

dx dt

+

Ω

(1−x)² 2

∂u

∂x

²+|u|²

dx dt,

(2.1)

andFis the Hilbert space of the functionsF= (f,0,0,0),f∈L²(Ω), with the finite norm

F²_F=

Ω(1−x)²|f|²dx dt. (2.2) Then we establish an energy inequality

uE≤kLuF, ∀u∈D(L), (2.3)

and we show that the operatorLhas the closureL.

Definition 2.1. A solution of the operator equation Lu=F is called a strong solution of problem(1.1).

Inequality(2.3)can be extended tou∈D(L), that is, uE≤k Lu _F, ∀u∈D

L

. (2.4)

From this inequality, we obtain the uniqueness of a strong solution, if it exists, and the equality of the setsR(L)andR(L). Thus, to prove the existence of a strong solution of problem(1.1)for anyF ∈F, it remains to prove that the setR(L)is dense inF.

3. An energy inequality and its applications

Theorem3.1. For any functionu∈D(L), there exists the a priori estimate

uE≤kLuF, (3.1)

where

k²=17 exp(ct) 5+4

b12

/

c₃−3cc2+3c²c₁−c³a1−b₁² +1 min

1, a²₀, c₃−3cc2+3c²c₁−c³a1−b₁² , (3.2) with the constantcsatisfying

sup

(x,t)∈Ω

1 a

∂a

∂t

≤c < inf

(x,t)∈Ω

1 a

∂a

∂t +1

,

c₃−3cc2+3c²c₁−c³a1− b1

2

>0, c2−2cc₁+c²a1−c₁+ca1<0.

(3.3)

Proof. Let

Mu= (1−x)²∂³u

∂t³ +2(1−x)Jx∂³u

∂t³, (3.4)

where

Jxu= _x

0

u(ζ, t)dζ. (3.5)

We consider the quadratic form Φ(u, u) =Re

Ωexp(−ct)£uMu dx dt, (3.6) with the constantc satisfying(3.3), obtained by multiplying(1.1a) by exp(−ct)Mu, integrating overΩ, and taking the real part. Substituting the expression ofMuin(3.6), we obtain

Re

Ωexp(−ct)£uMu dx dt

=Re

Ωexp(−ct)(1−x)² ∂³u

∂t³

²dx dt +2 Re

Ωexp(−ct)(1−x)∂³u

∂t³Jx∂³u

∂t³dx dt +Re

Ωexp(−ct) ∂

∂x

a(x, t)∂u

∂x

Mu dx dt.

(3.7)

Integrating the last two terms on the right-hand side by parts with re- spect toxin(3.7)and using the Dirichlet condition(1.1d), we obtain

2 Re ₁

0(1−x)exp(−ct)∂³u

∂t³Jx

∂³u

∂t³dx= ₁

0exp(−ct) Jx

∂³u

∂t³

²dx, (3.8) Re

Ωexp(−ct) ∂

∂x

a∂u

∂x

Mu dx dt

=−Re

Ωexp(−ct)(1−x)²a∂u

∂x

∂⁴u

∂t³∂xdx dt

−2 Re

Ωexp(−ct)∂a

∂xuJx∂³u

∂t³dx dt

−2 Re

Ωexp(−ct)au∂³u

∂t³dx dt.

(3.9)

Integrating each term by parts in(3.9) with respect tot and using the initial and final conditions(1.1b)and(1.1c), we get

Re

Ωexp(−ct) ∂

∂x

a∂u

∂x

Mu dx dt

=−2 Re

Ωexp(−ct)∂a

∂xuJx

∂³u

∂t³dx dt +

Ωexp(−ct) ∂³a

∂t³ −3c∂²a

∂t² +3c²∂a

∂t −c³a

×

(1−x)² 2

∂u

∂x

²+|u|²

dx dt

−3

Ωexp(−ct) ∂a

∂t −ca

(1−x)² 2

∂²u

∂t∂x ²+

∂u

∂t ²

dx dt

+ ₁

0

exp(−ct)a

(1−x)² 2

∂²u

∂t∂x ²+

∂u

∂t ²

dx

T=t

− ₁

0

exp(−ct)

∂²a

∂t² −2c∂a

∂t +c²a

(1−x)² 2

∂u

∂x ²+|u|²

dx

t=T

+Re 1

0

exp(−ct)

∂a

∂t −ca

(1−x)² ∂²u

∂t∂x

∂u

∂x+2u∂u

∂t

T=tdx.

(3.10) Substituting(3.8) and(3.10) in (3.7) and using conditions (1.2), (1.3), and(3.3), we obtain

Ωexp(−ct)(1−x)² ∂³u

∂t³

²dx dt +

Ωexp(−ct)

c₃−3cc2+3c²c₁−c³a1−b₁²

×

(1−x)² 2

∂u

∂x ²+|u|²

dx dt

≤Re

Ωexp(−ct)£uM u dx dt.

(3.11)

Again, substituting the expression ofMuin(3.11)and using elementary inequality, we get

Ωexp(−ct)(1−x)² 2

∂³u

∂t³

²dx dt +

Ωexp(−ct)

c₃−3cc2+3c²c₁−c³a1−b₁²

×

(1−x)² 2

∂u

∂x ²+|u|²

dx dt

≤17

Ωexp(−ct)(1−x)²|f|²dx dt.

(3.12)

By virtue of(1.1a), we have

Ωa0

∂²u

∂x²

²(1−x)² 2 dx dt

≤

Ω(1−x)²|f|²dx dt+

Ω2(1−x)² ∂³u

∂t³

²dx dt +4

Ωb²₁

(1−x)² 2

∂u

∂x

²+|u|²

dx dt.

(3.13)

This last inequality combined with(3.12)yields

Ω

(1−x)² 2

∂³u

∂t³

²dx dt +

Ω

c₃−3cc2+3c²c₁−c³a1−b²₁ (1−x)²

2 ∂u

∂x

²+|u|²

dx dt

+

Ωa²₀(1−x)² 2

∂²u

∂x^{2 2}dx dt

≤

17 exp(cT)

5+ 4b₁²

c₃−3cc2+3c²c₁−c³a1−b₁²

+1

×

Ω(1−x)²|f|²dx dt.

(3.14) Thus, this inequality implies

Ω

(1−x)² 2

∂³u

∂t^{3 2}+

∂²u

∂x^{2 2}

dx dt+

Ω

(1−x)² 2

∂u

∂x

²+|u|²dx dt

≤k²

Ω(1−x)²|f|²dx dt,

(3.15)

where

k²=17 exp(cT)

5+4b²₁/

c₃−3cc2+3c²c₁−c³a1−b²₁ +1 min

1, a²₀, c₃−3cc2+3c²c₁−c³a1−b²₁ . (3.16) Then,

uE≤kLuF, ∀u∈D(L). (3.17)

Thus, we obtain the desired inequality.

Lemma3.2. The operatorLfromEtoFadmits a closure.

Proof. Suppose that(un)∈D(L)is a sequence such that

un−→0 inE, Lun−→ F inF. (3.18) We need to show thatF=0. We introduce the operator

£0v=−(1−x)²∂³v

∂t³ + ∂

∂x

a(x, t) ∂

∂x

(1−x)²v

, (3.19)

with domainD(£0)consisting of functionsv∈W₂^2,3(Ω)satisfying

v|t=0=0, ∂v

∂t

_t=0=0, ∂²v

∂t²

_t=0=0, v|x=0=0, ∂v

∂x

_x=0=0.

(3.20) We note thatD(£0)is dense in the Hilbert space obtained by completing L²(Ω)with respect to the norm

Ω(1−x)²|v|²dx dt=v². (3.21) Since

Ω(1−x)²fv dx dt= lim

n→+∞

Ω(1−x)²£unv dx dt

= lim

n→+∞

Ωun£0v dx dt=0,

(3.22)

for any functionv∈D(£0), it follows thatf=0.

Theorem 3.1is valid for a strong solution, then we have the inequality uE≤k Lu _F, ∀u∈D

L

. (3.23)

Hence we obtain the following corollary.

Corollary3.3. A strong solution of problem (1.1) is unique if it exists, and depends continuously onF.

Corollary3.4. The rangeR(L)of the operatorLis closed inF, andR(L) = R(L).

4. Solvability of problem(1.1)

To prove the solvability of problem(1.1), it is sufficient to show thatR(L) is dense inF. The proof is based on the following lemma.

Lemma4.1.Suppose thata(x, t)and its derivatives∂⁴a/∂t³∂xand∂²a/∂t∂x are bounded. LetD0(L) ={u∈D(L):u(x,0) =0,(∂u/∂t)(x,0) =0,(∂²u/

∂t²)(x, T) =0}. If, foru∈D0(L)and for some functionsw∈L²(Ω),

Ω(1−x)£uw dx dt=0, (4.1)

thenw=0.

Proof. Equality(4.1)can be written as follows:

Ω(1−x)w∂³u

∂t³dx dt=−

Ω

∂

∂x

a(1−x)∂u

∂x

w− _x

0

w 1−ζdζ

dx dt.

(4.2)

For a givenw(x, t), we introduce the functionv(x, t)such that v(x, t) =w(x, t)−

_x

0

w(ζ, t)

1−ζ dζ. (4.3)

From(4.3), we conclude that₁

0v(x, t)dx=0, and thus, we have

Ω

∂³u

∂t³Nv dx dt=−

ΩA(t)uv dx dt, (4.4) whereA(t)u= (∂/∂x)(a(1−x)(∂u/∂x))andNv= (1−x)v+Jv.

Following[23], we introduce the smoothing operators

J_ε⁻¹= I−

∂³

∂t³ −1

,

J_ε⁻¹_∗

= I+

∂³

∂t³ −1

, (4.5)

with respect tot, which provide the solutions of the respective problems

g−∂³gε

∂t³ =g, g(0) =0, ∂g

∂t (0) =0, ∂²g

∂t² (T) =0, g^∗+∂³g^∗

∂t³ =g, g^∗(0) =0, ∂g^∗

∂t (T) =0, ∂²g^∗

∂t² (T) =0.

(4.6)

We also have the following properties: for anyg∈L²(0, T), the functions J⁻¹(g),(J⁻¹)^∗g∈W₂³(0, T). Ifg∈D(L), thenJ⁻¹(g)∈D(L)and we have

lim J⁻¹_∗ g−g

L²[0,T]=0 for−→0, lim J⁻¹

g−g

L²[0,T]=0 for−→0. (4.7) Substituting the functionuin(4.4)by the smoothing functionuεand using the relation

A(t)uε=J_ε⁻¹Au−J_ε⁻¹β(t)uε, (4.8)

where

β(t)uε=3∂²A(t)

∂t²

∂uε

∂t +3∂A(t)

∂t

∂²uε

∂²t +∂³A(t)

∂t³ uε, (4.9)

we obtain

−

ΩuN∂³vε∗

∂³t dx dt=

ΩA(t)uvε^∗dx dt−

Ωβ(t)uεv^∗_εdx dt. (4.10)

Passing to the limit, the equality in the relation(4.10)remains true for all functionsu∈L²(Ω)such that(1−x)(∂u/∂x),(∂/∂x)((1−x)(∂u/∂x))∈ L²(Ω), and satisfying condition(1.1d).

The operatorA(t)has a continuous inverse inL²(0,1)defined by

A⁻¹(t)g=− _x

0

1 1−ζ

1 a(ζ, t)

_ζ

0

g(η, t)dη dζ

+C(t) _x

0

1 1−ζ

1 a(ζ, t)dζ,

(4.11)

where

C(t) = ₁

0

dζ/a(ζ, t)ζ

0g(η, t)dη ₁

0

dζ/a(ζ, t) . (4.12)

Then, we have₁

0A⁻¹(t)g dx=0, hence the functionuε= (Jε)⁻¹ucan be represented in the form

uε= Jε−1

A⁻¹(t)A(t)u. (4.13)

Then

Bε(t)g= ∂⁴a

∂t³∂xJ_ε⁻¹ 1

a(x, t) x

0

g(η, t)dη−C(t)

+∂³a

∂t³J_ε⁻¹ g

a− ax

a²(x, t) x

0

g(η, t)dη−C(t)

+3∂

∂t

∂²a

∂t²∂x

∂

∂tJ_ε⁻¹ 1 a(x, t)

x 0

g(η, t)dη−C(t)

+∂a

∂t

∂

∂tJ_ε⁻¹g a− ax

a²(x, t) x

0

g(η, t)dη−C(t) .

(4.14)

The adjoint ofBε(t)has the form

B_ε^∗(t) =1 a

J_ε⁻¹∗

∂³a

∂t³h

+3 a

J_ε⁻¹∗∂

∂t ∂a

∂t

∂h

∂t

+ Gεh

(x)− _x

0

1/a(η, t) ₁ dη

0

1/a(x, t) dx

Gεh (1),

(4.15)

where Gh

(x) = _x

0

− 3 a(ζ, t)

J⁻¹∗∂

∂t ∂²

∂t∂ζ

∂h

∂t

+3∂a

∂ζ 1 a²(ζ, t)

J⁻¹∗∂

∂t ∂a

∂t

∂h

∂t

− 1 a(ζ, t)

J_ε⁻¹∗

∂⁴a

∂t³∂ζh

+∂a

∂ζ 1 a²(ζ, t)

J_ε⁻¹∗

∂³a

∂t³h

dζ.

(4.16) Consequently, equality(4.10)becomes

−

ΩuN∂³v^∗ε

∂t³ dx dt=

ΩA(t)uhεdx dt, (4.17) wherehε=v^∗_ε−εB_ε^∗v_ε^∗.

The left-hand side of (4.17) is a continuous linear functional of u.

Hence the functionhεhas the derivatives (1−x)(∂hε/∂x), (∂/∂x)((1− x)(∂hε/∂x))∈L²(Ω)and the following conditions are satisfied:hε|x=0= 0,hε|x=1=0, and(1−x)(∂hε/∂x)|x=1=0.

From the equality (1−x)∂hε

∂x =

I−ε1 a

J_ε⁻¹∗∂³a

∂t³

(1−x)∂v_ε^∗

∂x

−3ε1 a

J_ε⁻¹∗ ∂

∂t ∂a

∂t

∂

∂t(1−x)∂v^∗_ε

∂x

,

(4.18)

and since the operator(Jε⁻¹)^∗is bounded inL²(Ω), for sufficiently smallε, we haveε(1/a)(Jε⁻¹)^∗(∂³a/∂t³)<1. Hence the operatorI−ε(1/a)(Jε⁻¹)^∗ (∂³a/∂t³)has a bounded inverse inL²(Ω). We conclude that(1−x)(∂vε^∗/

∂x)∈L²(Ω).

Similarly, we conclude that (∂/∂x)((1−x)(∂v^∗ε/∂x)) exists and be- longs toL²(Ω), and the following conditions are satisfied:

v^∗_ε|x=0=0, v_ε^∗|x=1=0, (1−x)∂v^∗_ε

∂x

x=1=0. (4.19) Substitutingu=t

0

η 0

T

ζ exp(cτ)v^∗_ε(τ)dτ dζ dηin(4.4), where the constant csatisfies(3.3), we obtain

Ωexp(ct)v^∗_εNv dx dt=−

ΩA(t)uv dx dt. (4.20)

Using the properties of smoothing operators, we have

Ωexp(ct)v^∗εNv dx dt=−

ΩA(t)uv^∗εdx dt−ε

ΩA(t)u∂³v^∗ε

∂t³ dx dt, (4.21) and from

εRe

ΩA(t)u∂³vε^∗

∂t³ dx dt=ε

Ω(1−x)a∂u

∂x

∂

∂x

∂³v^∗ε

∂t³ dx dt

=−εRe

Ω(1−x)∂a

∂t

∂u

∂x

∂²

∂t²

∂vε^∗

∂xdx dt +εRe

Ω(1−x)∂a

∂t

∂²u

∂t∂x

∂

∂t

∂v^∗ε

∂xdx dt +ε

Ωaexp(−ct)(1−x)

∂v^∗_ε

∂x

2

dx dt

+εRe

Ω(1−x)∂a

∂t

∂²u

∂t∂x

∂v^∗ε

∂xdx dt,

(4.22)

we have εRe

ΩA(t)u∂³vε^∗

∂t³ dx dt

≥ε

Ωaexp(+ct)(1−x) ∂vε^∗

∂x

2

dx dt

−ε

Ω(1−x) 1 4a

∂a

∂t 2

exp(−ct) ∂³u

∂t²∂x ²dx dt

−ε

Ωaexp(+ct)(1−x)

∂v^∗ε

∂x

2

dx dt

−ε

Ω

1−x 2

∂a

∂t 2

exp(−ct) ∂u

∂x ²dx dt

−ε

Ωexp(+ct)1−x 2

∂³v^∗_ε

∂t²∂x

2

dx dt

−ε

Ωexp(+ct)1 2 ∂²v^∗ε

∂t∂x

2

dx dt

−ε

Ω

1−x 2

∂a

∂t 2

exp(−ct) ∂²u

∂x∂t

²dx dt.

(4.23)

Integrating the first term on the right-hand side by parts in (4.21), we obtain

Re

ΩA(t)uv^∗εdx dt

≥ −3 2

Ω(1−x)exp(−ct)

∂a

∂t −ca ∂²u

∂t∂x ²dx dt +1

2 1

0

(1−x)exp(−ct) a−

∂a

∂t −ca

∂²u

∂t∂x ²dx

_t=T

−1 2

₁

0

(1−x)exp(−ct) ∂²a

∂t² −2c∂a

∂t+c²a+

∂a

∂t−ca

∂u

∂x ²

_t=Tdx +1

2

Ω(1−x)exp(−ct)

∂³a

∂t³ −3c∂²a

∂t² +3c²∂a

∂t −c³a ∂u

∂x

²dx dt.

(4.24) Combining(4.23)and(4.24), we get

Re

Ωexp(ct)v^∗εNv dx dt

≤3 2

Ω(1−x)exp(−ct)

c1−ca0∂²u

∂t∂x ²dx dt

−1 2

₁

0

(1−x)exp(−ct)

a0−c₁−ca1∂²u

∂t∂x ²dx

_t=T +1

2 ₁

0

(1−x)exp(−ct)

c2−2c₁c−c²a1−c₁+ca1∂u

∂x ²

_t=Tdx

−1 2

Ω(1−x)exp(−ct)

c₃−3c2c+3c²c₁−c³a1∂u

∂x ²dx dt +ε

Ω(1−x)exp(−ct) c²₁ 4a0

∂³u

∂t²∂x ²dx dt +

Ω(1−x)exp(−ct)c²₁ 2

∂u

∂x ²dx dt +

Ω

1−x

2 exp(ct) ∂³v^∗_ε

∂t²∂x ²dx dt +

Ω(1−x)exp(−ct)c²₁ 2

∂²u

∂t∂x ²dx dt +

Ω

1−x

2 exp(ct) ∂²v^∗_ε

∂t∂x ²dx dt

.

(4.25)

Using conditions(3.3)and inequalities(4.23)and(4.24), we obtain

Re

Ωexp(ct)vNv dx dt≤0, asε−→0. (4.26) Since Re

Ωexp(ct)vJxv dx dt=0, thenv=0 a.e.

Finally, from the equality(1−x)v+Jxv= (1−x)w, we concludew=0.

Theorem4.2. The rangeR(L)ofLcoincides withF.

Proof. SinceFis Hilbert space, thenR(L) =Fif and only if the relation

Ω(1−x)²£uf dx dt=0, (4.27) for arbitraryu∈D0(L)andF ∈F, implies thatf=0.

Takingu∈D0(L)in(4.27)and usingLemma 4.1, we obtain thatw=

(1−x)f=0, thenf=0.

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M. Denche: Laboratoire Equations Différentielles, Département de Mathéma- tiques, Faculté des Sciences, Université Mentouri, 25000 Constantine, Algeria

E-mail address:[email protected]

A. Memou: Laboratoire Equations Différentielles, Département de Mathéma- tiques, Faculté des Sciences, Université Mentouri, 25000 Constantine, Algeria

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