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Int. Journal of Math. Analysis, Vol. 3, 2009, no. 18, 871 - 877

A Class of Third Order Parabolic Equations

with Integral Conditions

M. Bouzit and N. Teyar

Université Mentouri ConstantineInstitut de Mathématiques25000 Constantine, Algeria

bouzit med@yahoo.fr, teyarnadir@gmail.com

Abstract

In this paper, we study a mixed problem for a third order parabolicequation with non classical boundary condition. We prove the existence

and uniqueness of the solution. The proof of the uniqueness is based on

a priori estimate and the existence is established by Fourier’s method.

Mathematics Subject Classification:  35B45, 35K35

Keywords:   Integral Boundary Condition, Energy Inequalities, Parabolicequation of mixed type

1. INTRODUCTION

In the set Ω = (0, T ) × (0, 1), we consider the equation∂ 2u∂t2 −   1

xr

  ∂ ∂x

xr+1   ∂ 

2u∂x∂t

+ k ∂u

∂t  = F (t, x), r 0 and   k ≥ 0, (1.1)

To equation (1.1) we attach the initial conditionu(0, x) =  ϕ(x)   x ∈ (0, 1) (1.2)

∂u∂t

(0, x) =  ψ(x)   x ∈ (0, 1) (1.3)and the integral condition 1

0  u(t, x)dx = 0,

 10

  xru(t, x)dx = 0 for   t ∈ (0, T ) (1.4)Where   ϕ(x), ψ(x)  ∈   L2(0, 1) are known functions which satisfy the com-

patibility conditions given in (1.4).The boundary value problems with integrals conditions are mainly moti-

vated by the work of Samarskii [3]. Regular case of this problem for secondorder equations is studied in [4]. The problem where the equation of mixedtype contains an operator of the form   a(t) ∂ 

2α+1u∂x2α∂t

  is treated in [17], the oper-ator of the form   ∂ 

∂x

a(t, x) ∂u

∂x

 and   ∂ 

α

∂xα

a(t, x) ∂ 

αu∂xα

 is treated in [4] and [14].

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872   M. Bouzit and N. Teyar 

Two-point boundary value problems for parabolic equations, with an integralcondition, are investigated using the energy inequalities method in [8, 9, 10, 11]and the Fourier’s method [12]. Three-point boundary value problem with anintegral condition for parabolic equations with the Bessel operator is studied in[12]. And recently parabolic and hyperbolic equations with integral boundary

condition are treated by Fourier’s method in [1, 5].The presence of nonlocal conditions raises complications in applying stan-

dard methods to solve (1.1)-(1.4). Therefore to over come this difficulty wewill transfer this problem to another which we can handle more effectly. Forthat, we have the following lemma.

Lemma 1. Problem (1.1)-(1.4) is equivalent to the following problem

(Pr)1

∂ 2u∂t2 −   1

xr

  ∂ ∂x

xr+1   ∂ 

2u∂x∂t

+ k ∂u

∂t  = F (t, x)

u(0, x) =  ϕ(x)∂u∂t

(0, x) = ψ(x)∂u

∂t

(t, 1) −   ∂u∂t

(t, 0) =   1r 

1

0

  xrF (t, x)dx −  1

0

  F (t, x)dx∂ 2u

∂x∂t(t, 1) = −

 10

  xrF (t, x)dx

Proof. Let  u(t, x) be a solution of (1.1)-(1.4). Integrating equation (1.1)with respect to  x over (0, 1), and taking into account of (1.4), we obtain

−

x  ∂ 2u

∂x∂t

10− r

 10

∂ 2u∂x∂t

dx = 10

  F (t, x)dx

And so∂ 2u

∂x∂t(t, 1) + r( ∂u

∂t(t, 1) −   ∂u

∂t(t, 0)) = −

 10

  F (t, x)dx

To eliminate the second nonlocal condition 10

  xru(t, ξ )dξ  = 0 multiplying

both sides of (1.1) by   xr

and integrating the resulting over (0,1), and takingin account of (1.4), we obtain

∂ 2u∂x∂t

(t, 1) = − 10

  xrF (t, x)dxThese may also be written

∂u∂t

(t, 1)−   ∂u∂t

(t, 0) =   1r

 10

  xrF (t, x)dx− 10

  F (t, x)dx

And∂ 2u

∂x∂t(t, 1) = −

 10

  xrF (t, x)dxLet u(t, x) now be a solution of (Pr)1, it remains to prove that 1

0  u(t, x)dx = 0,

And 10   x

r

u(t, x)dx = 0We integrate Eq.(1.1) with respect to  x  , we obtain

d2

dt2

 10   u(t, x)dx + k

  ddt

 10   u(t, x)dx = 0, t ∈ (0, T )

And it also follows thatd2

dt2

 10

  xru(t, x)dx + k   ddt

 10

  xru(t, x)dx = 0, t ∈ (0, T )Introduce now the new function  v(t, x) = u(t, x) − u0(t, x),  where

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Third order parabolic equations    873

u0(t, x) =  α(x) t0

 m1(τ )dτ  + β (x) t0

 m2(τ )dτ,α(x) = −x + x2

, β (x) = 2x − x2, m1(t) = − 10

  xrF (t, x)dx,

m2(t) =  1

r

−m1(t)−

 10

  F (t, x)dx

Then (Pr)1  is transformed into the following problem

(Pr)2

v ≡   ∂ 2v∂t2  −  1xr

  ∂ ∂x

xr+1   ∂ 2v

∂x∂t

+ k ∂v∂t   = f (t, x)

lv  =   v(0, x) = ϕ(x)qv  =   ∂v

∂t(0, x) = Ψ(x)

∂v∂t

(t, 1) =   ∂v∂t

(t, 0)∂ 2v

∂x∂t(t, 1) = 0

Wheref (t, x) = F (t, x) + (α(x)− β (x)

r  ) 10

  xrF t(t, x)dx + β (x)

r

 10

  F t(t, x)dx + γ (t, x),γ (t, x) = (−(r  + 1) + 2(r  + 2)x)m1(t) + (2(r  + 1)  −  2(r  + 2)x)m2(t)  −

kα(x)m1(t) − kβ (x)m2(t)

Ψ(x) = ψ(x) + α(x)  1

0  xrF (0, x)dx +  β (x)

r  (−  

1

0  xrF (0, x)dx +  

1

0  F (0, x)dx)

2. A PRIORI ESTIMATE

we consider (P r)2  as a solution of the operator equationLv = F ,

where   L   = (,l ,q  ) ,  F   = (f,ϕ, Ψ). The operator   L   is acting from theBanach space  D(L) = E    to  F    where

E  =

v :  xr+1

2   v, xr+1

2  ∂v

∂t, x

r+1

2  ∂ 2v

∂x∂t ∈ L2(0, 1) and

xr+1

2  ∂v

∂t, x

r+1

2  ∂ 2v

∂x∂t, x

r+1

2  ∂ 2v

∂t2, xr+1   ∂ 

3v∂x2∂t

, xr   ∂ 2v

∂x∂t ∈ L2(Ωr)

With respect to the normx

2E  =

 Ωr

 xr+1

( ∂v∂t

)2 + (   ∂ 2v

∂x∂t)2 + ( ∂ 

2v∂t2

)2

+ Ωr

  ∂ ∂x

xr+1   ∂ 

2v∂x∂t

2+ sup

0≤t≤T 

 10

  xr+1

v2 + ( ∂v∂t

)2 + (   ∂ 2v

∂x∂t)2

Here  F   is the Hilbert space with the normF2F   = ((f,ϕ, Ψ)

2F   =

 Ωr

 f   2 + 10 (Ψ2 + (Ψ)2 + ϕ2)

Lemma 2. For any function  u ∈ E , we haveexp(−cT )

8

 10   x

r(v(τ, x))2 ≤   18 10   x

rϕ2 +  18 10

 τ 0   x

r( ∂v∂t

)2 (2.1)with the constant c satisfying  c ≥ 1 .Proof. Integrating by parts (exp(−ct)xrv,   ∂v

∂t) and using elementary in-

equalities yields (2.1).Theorem 1. For (P r)2  We have

vE  ≤ C LvF  ,Where C   0 is independent on  v   .Proof. Let

Mv = 2xr ∂ v∂t

 + xr ∂ 2v

∂t2

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874   M. Bouzit and N. Teyar 

Consider the scalar product (v,Mv),  and integrating overΩτ  = (0, τ ) × (0, 1),  we get

(v,Mv)L2(Ωτ )  = (1+k2

) 10

  xr( ∂v∂t

(τ, x))2−(1+ k2

) 10

  xrΨ2+2k Ωτ 

 xr( ∂v∂t

)2+

2 Ωτ  x

r+1(   ∂ 2v

∂x∂t)2 +

 Ωτ  x

r( ∂ 2v

∂t2)2 +   12

 10   x

r+1(   ∂ 2v

∂x∂t(τ, x))2 −   12

 10   x

r+1(Ψ)2 ≥

(1 +  k

2) 10   x

r

(∂v

∂t (τ, x))2

− (1 +  k

2) 10   x

r

Ψ2

+ 2k Ωτ  x

r

(∂v

∂t )2

+2 Ωτ 

 xr+1(   ∂ 2v

∂x∂t)2+

 Ωτ 

 xr( ∂ 2v

∂t2 )2+ 12

 10

  xr+1(   ∂ 2v

∂x∂t(τ, x))2− 12

 10

  xr(Ψ)2 (2.2)

We now apply an  ε-inequality to the term (v, 2xr ∂v∂t

 + xr ∂ 2v

∂t2 ) we obtain

(v, 2xr ∂v∂t

 + xr ∂ 2v

∂t2 ) =   1

ε1

 Ωτ 

 xrf   2 + ε1 Ωτ 

 xr( ∂v∂t

)2 +   12ε2

 Ωτ 

 xrf   2+22

 Ωτ  x

r( ∂ 2v

∂t2)2 (2.3)

From equation  v ≡   ∂ 2v

∂t2 −   1

xr

  ∂ ∂x

xr+1   ∂ 

2v∂x∂t

+ k ∂v

∂t  = f (t, x) we have

18

 Ωτ 

  ∂ ∂x

xr+1   ∂ 

2v∂x∂t

≤   14

 Ωτ 

 xrf   2+14

 Ωτ 

 xr( ∂ 2v

∂t2)2 +1

4k Ωτ 

 xr( ∂ 2v

∂t2 )2 (2.4)

Combining inequalities (2.1), (2.2), (2.3) and (2.4) and since (x  ≤  1) weobtain

42+21+12412

 Ωτ  f   2 + (1 +   k2 )

 10  Ψ2 +   12

 10 (Ψ)2 +   18

 10   ϕ2 ≥

(1 +   k2) 10

  xr+1( ∂v∂t

(τ, x))2 + ( 7k4   − 1 −  18) Ωτ 

 xr( ∂v∂t

)2 + 2 Ωτ 

 xr+1(   ∂ 2v

∂x∂t)2

+(34 −   2

2 ) Ωτ 

 xr( ∂ 2v

∂t2)2 +   1

2

 10

  xr+1(   ∂ 2v

∂x∂t(τ, x))2 +   1

8

 Ωτ 

  ∂ ∂x

xr+1   ∂ 

2v∂x∂t

+

exp(−cT )8

 10   x

r+1(v(τ, x))2 (2.5)Next choosing as and  i, i = 1, 2 as

  7k4  −1−

18  = k1   0 and

  34−

22   = k2  0.

The left-hand side of (2.5) is independent of  τ  , hence replacing the right-hand side by its upper bound with respect to  τ , in the interval [0, 1], we obtainthe desired inequality.

This completes the proof.

3. EXISTENCE AND UNIQUENESS OF SOLUTION

We shall establish the existence of solution of (P r)2. For this we make useof the Fourier’s method.

Consider the function  vn(t, x) = T n(t)X n(x) where  X n(x) is an eigenfunc-tion of the BVP  

1

xr

  ddx

xr+1 dX n

dx

− kX n =  λnX n

X n(1) = X n(0)d X n

dx (1) = 0

λn, n = 1, 2.......  is called the eigenvalue corresponding to the eigenfunction

X n(1),  and  T n(t) is satisfying the initial problem

d2T ndt2  − λn

dT ndt

  = f n(t)T n(0) = ϕn

d T ndt

 (0) = Ψn

ϕ(x) =∞

n=1

ϕnX n(x)

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Third order parabolic equations    875

Ψ(x) =∞

n=1

ΨnX n(x)

Ψ(x) =∞

n=1Ψ∗nX n(x)

f (t, x) =∞

n=1

f n(t)X n(x)

And by the Parseval-Steklov equality

ϕ2L2(0,1)  =∞

n=1

ϕ2n,

Ψ2L2(0,1)  =∞

n=1

Ψ2n,

Ψ2L2(0,1)  =∞

n=1

(Ψ∗n )2,

And 10

  f (t, x)dx =∞

n=1

f 2n(t).

Hence  Ω

f   2(t, x)dxdt =∞

n=1

 T 0

  f   2

n

 (t).

Then direct computation yieldsT n(t) = ϕn +

 t0

 Ψn exp(λnτ )dτ  + t0

 s0

  f n(τ ) exp(λns − λnτ )dτds 10

  xrX n(x)X m(x)dx = 0, n = mAnd

ϕn =  1

0  xrϕ(x)X n(x)dx

  1

0  xrX 

2n (x)dx

Ψn =  1

0  xrΨ(x)X 

n(x)dx

  1

0  xrX 

2n (x)dx

By principle of superposition, the solution of (P r)2  is given by the series

v(t, x) =∞

n=1T n(t)X n(x).   (3.1)

Then we haveTheorem 2. Let  f, ϕ ∈ L2(Ω),  and Ψ ∈ H 

1(0, 1).  Then the solution  v(t, x)of (P r)2  exists and is represented by series (3.1) wich converges in  E .

Proof. Consider the partial sum   S N (t, x) =N 

n=1T n(t)X n(x) of the series

(3.1) then by theorem 1N 

n=1

T n(t)X n(x)

2

E 

≤ C 1N 

n=1

 T 0

  f   2n   (t)dt + ϕ 2n   + Ψ

 2n   + (Ψ

n)

2

  (3.2)

The seriesN 

n=1

 T 0

  f   2n   (t)dt,N 

n=1

ϕ  2n   ,N 

n=1

Ψ  2n   ,andN 

n=1

(Ψn)2 converge. There-

fore it follows from (3.2) that the series (3.1) converges in E and accordinglyits sum  v ∈ E.

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876   M. Bouzit and N. Teyar 

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Received: November, 2008