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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 [email protected], [email protected]
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