UNIVERSITATIS IAGELLONICAE ACTA MATHEMATICA,
FASCICULUS XLIV
2006
EXISTENCE OF A SOLUTION FOR THE INITIAL
BOUNDARY VALUE PROBLEM FOR A HIGH ORDER
PARABOLIC EQUATION IN AN UNBOUNDED DOMAIN
by Lech Zar¦ba
In this paper we consider the initial boundary value problem
for the equation utt +A1 u+A2 ut +g(ut ) = f (x, t) in an unbounded domain,
where A1 is a linear elliptic operator of the fourth order and A2 is a linear
elliptic operator of the second order. We obtain the conditions of the
existence of a weak solution for this problem.
Abstract.
Many authors have considered initial boundary value problems in unbounded domains for parabolic equations of a high order with the rst derivative with respect to time. There are much fewer papers concerning the
problems for parabolic equations with the second derivative with respect to
time. In particular, in [123123123] and [5656], the authors have considered the Cauchy problem and initial boundary value problem for a parabolic
equation of a high order and the properties of its solutions. In [7], the authors obtained some conditions for the uniqueness of the solution of the initial
boundary value problem for the general linear parabolic systems in unbounded
domains. By introducing a parameter, they have shown the uniqueness of a
weak solution
in the class of functions which do not grow faster than the funcα
tion ea|x| for |x| → ∞. In [11], the author proved with this method that the
solution of problem (1){(3) is unique. The main goal of this paper is to obtain
some conditions for the existence of a weak solution for a parabolic equation
of the fourth order with the second derivative with respect to time and with
Dirichlet boundary conditions on some weighted Sobolev space.
Let Ω ⊂ Rn be an unbounded domain and ∂Ω ∈ C 1 , Ω ∩ BR = ΩR
be a regular domain for all R > 0, where BR = {x ∈ Rn , |x| < R} and
R
QT = Ω × (0, T ), QR
T = Ω × (0, T ), Ωτ = Qτ ∩ {t = τ }, Qτ0 ,τ1 = Ω × (τ0 , τ1 ).
2000 Mathematics Subject Classication. 35K55, 35K60.
98
We shall consider the equation of the form
A(u) ≡ utt (x, t) +
(1)
−
n
X
n
X
(akl
ij (x, t)uxi xj (x, t))xk xl −
n
X
(aij (x, t)uxi (x, t))xj
i,j=1
i,j,k,l=1
(bij (x, t)utxi (x, t))xj + a(x, t)u(x) + g(x, ut ) = f (x, t)
i,j=1
in the domain QT .
For this equation, we put the following boundary and initial conditions
(2)
∂u
|S = 0,
∂ν T
= u0 (x), ut |t=0 = u1 (x),
u|ST = 0,
(3)
u|t=0
where ST = ∂Ω × (0, T ) and ν is the normal vector for ST .
We dene the function Ψ ∈ C 2 (R), Ψ : [0, +∞) → (0, +∞) such that Ψ > 0
in [0, +∞), Ψ0 (ξ) ≤ 0 or Ψ0 (ξ) ≥ 0 in [0, +∞) and
(4)
Ψxi (|x|) Ψ(|x|) ≤ Ψ0 ,
Ψxi xj (|x|) Ψ(|x|) ≤ Ψ0 ,
i, j = 1, . . . , n, Ψ0 < ∞.
Let us introduce the weighted spaces
LpΨ (Ω)
=
Z
u :
p
|u(x)| Ψ(|x|)dx < ∞ , p ∈ (1, ∞),
Ω
Z X
∂ k−1 u
0,k
α
2
HΨ (Ω) = u :
|D u(x)| Ψ(|x|)dx < ∞, u|∂Ω = 0,
|∂Ω = 0 ,
∂ν k−1
Ω |α|≤k
where k = 1, 2; Dα =
∂xα1 1
∂ |α|
, |α| = α1 + · · · + αn , and ν is the normal
. . . ∂xαnn
vector for ∂Ω.
For equation (1), we adopt the following system of assumptions:
ij
kl
kl
∞
kl
(A) akl
ij , (aij )xk xl , (aij )tt ∈ L (QT ); aij (x, t) = akl (x, t), i, j, k, l = 1, . . . , n
for almost all (x, t) ∈ QT ;
n
P
i,j,k,l=1
n
P
i,j,k,l=1
akl
ij (x, t)ξij ξkl ≥ a2
(akl
ij (x, t))t ξij ξkl ≤
n
P
2,
ξij
i,j=1
n
P
2
a12
ξij
i,j=1
i, j, k, l = 1, . . . , n,
for almost all (x, t) ∈ QT and for
all ξ ∈ Rn(n−1)/2 , where a2 > 0 is a constant;
aij , (aij )xj , (aij )t ∈ L∞ (QT ),
i, j = 1, . . . , n;
a, at ∈ L∞ (QT ),
99
(B) bij , (bij )xj , (bij )t ∈ L∞ (QT ),
n
P
bij (x, t)ξi ξj ≥ b0
i=1
ij=1
(G)
n
P
i, j = 1, . . . , n;
ξi2 for almost all (x, t) ∈ QT and for all ξ ∈ Rn ,
where b0 > 0 is a constant;
The functions x → g(x, ξ), x → gξ (x, ξ) are continuous for every ξ ∈ R
and the functions ξ → g(x, ξ), ξ → gξ (x, ξ) are measurable for almost
all x ∈ Ω and satisfy the following inequalities:
(g(x, ξ) − g(x, µ))(ξ − µ) ≥ g0 |ξ − µ|q
for almost all x ∈ Ω and for all ξ, µ ∈ R, g0 = const > 0;
|g(x, ξ)| ≤ g1 |ξ|q−1 for almost all x ∈ Ω and for all ξ ∈ R, q ∈ (2, +∞).
Under these assumptions, we will obtain the existence of a weak solution
of problem (1){(3).
Definition 1. We call a function u a weak solution of problem (1){(3) if
0,2
0,1
u ∈ L∞ (0, T ); HΨ
(Ω) ∩ C [0, T ]; HΨ
(Ω) ,
0,1
(Ω) ∩ Lq (0, T ); LqΨ (Ω) ∩ C [0, T ]; L2Ψ (Ω)
ut ∈ L2 (0, T ); HΨ
and u satises the following integral equality
Z
Z n
X
akl
ut (x, T )w(x, T )dx +
−ut wt +
ij (x, t)uxi xj wxk xl
ΩT
(5)
+
i,j,k,l=1
n
X
QT
n
X
aij (x, t)uxi wxj + a(x, t)uw +
bij (x, t)uxi t wxj
i,j=1
Z
Z
+ g(x, ut )w dxdt =
f (x, t)wdxdt + u1 (x)w(x, 0)dx
i,j=1
QT
∀w ∈
C 1 ([0, T ]; C0∞ (Ω))
Ω
and u(x, 0) = u0 (x).
We consider the following equation
(6)
A(u) = f R (x, t)
R
in the domain QR
T = Ω × (0, T ), R > 1, where
for (x, t) ∈ QR
T,
for (x, t) ∈ QT \ QR
T , R = 2, 3, 4, . . .
For this equation, we put the following boundary and initial conditions
R
(7)
u|t=0 = uR
0 (x), ut |t=0 = u1 (x),
(
f (x, t),
f R (x, t) =
0,
(8)
u|∂ΩR =
∂u
| R = 0,
∂ν ∂Ω
100
R
R
R
R
where uR
0 (x) = u0 (x)(· ζ (x), u1 (x) = u1 (x)ζ (x) for 0 ≤ ζ (x) ≤ 1, ζ ∈
1, for |x| ≤ R − 1,
C 2 (Rn ) and ζ R (x) =
0, for |x| ≥ R.
0,2
R
Let H (Ω ) = u : u ∈ H 2 (ΩR ); u|∂ΩR = 0, ∂u
∂ν |∂ΩR = 0 .
Definition
2. We call a function uR a weak solution of problem (6){(8) if
uR ∈ L∞ (0, T ); H 0,2 (ΩR ) ,
2
uR
t ∈L
2
2
R
uR
tt ∈ L (0, T ); L (Ω ) ,
(0, T ); H 0,1 (ΩR ) ∩ Lq (QR
T)
and uR satises the following integral equality
Z n
n
X
X
R
kl
R
R
utt w +
aij (x, t)uxi xj wxk xl +
aij (x, t)uR
xi wxj + a(x, t)u w
i,j=1
i,j,k,l=1
QR
τ
(9)
+
n
X
Z
R
bij (x, t)uR
w
+
g(x,
u
)w
dxdt
=
f R (x, t)wdxdt
xi t xj
i,j=1
QR
τ
∀τ ∈ (0, T ], ∀w ∈ L2 (0, T ); H 0,2 (ΩR ) ∩Lq (QR
T ) and the initial conditions (7).
Theorem
1. Suppose that conditions
(A),
(B),
(G)
hold and
u0 ∈ H 0,2 (ΩR ) ∩ H 4 (ΩR ), u1 ∈ H 0,1 (ΩR ) ∩ L2q−2 (ΩR ), f R , ftR ∈ L2 (QR
T ).
Then problem (4){(6) has a weak solution.
Proof.
Consider the space H 0,2 (ΩR ) ∩ H 4 (ΩR ) ∩ Lq (ΩR ) and the basis of
this space {Φk (x)}.
Next, consider the sequence of functions of the form
u
R,N
(x, t) =
N
X
CsN (t)Φs (x)
s=1
for N = 1, 2, . . . , where the functions C1N , . . . , CNN are the solution of the
following Cauchy problem
Z n
X
R,N
R,N
utt (x, t)Φs (x) +
akl
ij (x, t)uxi xj (x, t)Φsxk xl (x)
i,j,k,l=1
ΩR
(10)
+
n
X
aij (x, t)uR,N
xi (x, t)Φsxj (x)
i,j=1
+ a(x, t)u
+
n
X
bij (x, t)uR,N
txi (x, t)Φsxi (x)
i,j=1
R,N
(x, t)Φs (x) +
g(x, uR,N
)Φs (x)
t
− f (x, t)Φs (x) dx = 0,
R
t ∈ [0, T ]
101
with the conditions
R,N
N
(11)
CsN (0) = uR,N
0,s , Cst (0) = u1,s , s = 1, . . . , N,
R,N
and kuR,N
− uR
− uR
0 kH 0,2 (ΩR )∩H 4 (ΩR ) → 0, ku1
1 kH 0,1 (ΩR )∩L2q−2 (ΩR ) → 0,
0
where uR,N
(x) =
0
N
P
s=1
R,N
uR,N
(x) =
0,s Φs (x) and u1
N
P
s=1
uR,N
1,s Φs (x).
From the Caratheodory theorem, there exists the solution of problem (10),
(11), where CsN are continuous and CstN , s = 1, . . . , N are absolutely continuous
on the interval [0, h0 ]. Taking into account the estimations obtained below, we
can state that h0 = T .
Multiplying (10) by the functions CstN (t)e−ηt , η > 0, respectively, then
summing over s from 1 to N and integrating with respect to t from 0 to τ ,
τ ∈ (0, T ], we obtain
Z n
X
R,N
R,N R,N
akl
uR,N
u
+
ij (x, t)uxi xj utxk xl
tt
t
i,j,k,l=1
QR
τ
(12)
+
n
X
R,N
aij (x, t)uR,N
xi utxj
+
i,j=1
+
n
X
R,N
bij (x, t)uR,N
txi utxj
i,j=1
a(x, t)uR,N uR,N
t
+
g(x, uR,N
)uR,N
t
t
−f
R
(x, t)uR,N
t
e−ηt dxdt = 0.
If we consider the respective components of the last equality we will obtain
Z
R,N −ηt
uR,N
e dxdt
tt ut
1
=
2
Z
|uR,N
|2 e−ητ dx
t
ΩR
τ
QR
τ
1
−
2
Z
|2 dx
|uR,N
1
η
+
2
ΩR
0
Z
|uR,N
|2 e−ηt dxdt,
t
QR
τ
R
where ΩR
τ = QT ∩ {t = τ }. Next from (A) there follows:
Z
n
X
QR
τ
i,j,k,l=1
R,N R,N −ηt
akl
dxdt
ij (x, t)uxi xj utxk xl e
a2
≥
2
Z X
n
ΩR
τ
2 −ητ
|uR,N
dx
xi xj | e
i,j=1
1
+ (ηa2 − a12 )
2
Z X
n
QR
τ
i,j=1
1
−
2
Z
n
X
ΩR
0
i,j,k,l=1
2 −ηt
|uR,N
dxdt.
xi xj | e
R,N
R,N
akl
ij (x, 0)u0xi xj u0xk xl dx
102
It is easy to prove that
(13)
Z
(u
R,N
Z
2
(x, t)) dxdt ≤ 2T
QR
τ
(u
R,N
2
(x, 0)) dx + 2T
2
(uR,N
(x, t))2 dxdt,
t
QR
τ
ΩR
Z
Z
2 −ηt
(uR,N
dxdt
xi (x, t)) e
QR
τ
(14)
Z
2
(uR,N
xi (x, 0)) dx
≤ 2T
+ 2T
Z
2
2 −ηt
(uR,N
dxdt,
xi t (x, t)) e
QR
τ
ΩR
and
Z X
n
QR
τ
1
≤
2
2 −ηt
(uR,N
dxdt
xi ) e
i=1
Z X
n
QR
τ
2 −ηt
(uR,N
dxdt
xi xj ) e
i,j=1
M
+
2
Z
(uR,N )2 e−ηt dxdt,
QR
τ
where M does not depend on N .
Hence from assumption (A), (13) and the initial condition, we obtain
Z X
n
QR
τ
R,N −ηt
aij (x, t)uR,N
dxdt
xi utxj e
i,j=1
Z X
n
δ
≤
2
QR
τ
+
2 −ηt
dxdt
(uR,N
txi ) e
i=1
A1 nT 2
2δ
Z
Z X
n
A1
+
4δ
QR
τ
i,j=1
A1 nT
2δ
(uR,N
)2 e−ηt dxdt +
t
QR
τ
2 −ηt
(uR,N
dxdt
xi xj ) e
Z
)2 dx,
(uR,N
0
QR
τ
and
Z
a(x, t)uR,N uR,N
e−ηt dxdt
t
A0
≤
2
QR
τ
Z R,N 2 −ηt
R,N 2
(u
) + (ut ) e dxdt
QR
τ
Z
≤ A0 T
(uR,N
)2 dx
0
Z
2
+ (T + 1)A0
ΩR
where A1 = ess sup
n
P
QT i,j=1
(uR,N
)2 e−ηt dxdt,
t
QR
τ
a2ij (x, t), A0 = ess sup |a(x, t)| and δ > 0.
QT
Next, from the assumption (B) there follows:
Z X
n
QR
τ
i,j=1
R,N −ηt
bij (x, t)uR,N
dxdt
txj utxi e
≥ b0
Z X
n
QR
τ
i,j=1
2 −ηt
|uR,N
dxdt.
txi | e
103
Moreover, by virtue of condition (G) we get
Z
(g(x, uR,N
), uR,N
)e−ηt dxdt
t
t
Z
≥ g0
|uR,N
|q e−ηt dxdt
t
QR
τ
QR
τ
and it is evident that
Z
(f R , uR,N
)e−ηt dxdt ≤
t
QR
τ
1
2
Z
|f R |2 e−ηt dxdt +
1
2
QR
τ
Z
|uR,N
|2 e−ηt dxdt.
t
QR
τ
Taking into account the above estimates, from (9) we obtain the following
inequality
(15)
Z n
X
R,N 2
R,N 2 −ητ
(ut ) + a2
(uxi xj ) e dx
i,j=1
ΩR
τ
Z A1 nT 2
2
(uR,N
)2 + g0 |uR,N
|q
+
η − 2A0 (T + 1) − 1 −
t
t
δ
QR
τ
A1
+ ηa2 −
− a12
2δ
X
n
2
(uR,N
xi xj ) + (b0 − δ)
i,j=1
n
X
2 −ηt
dxdt
(uR,N
txi ) ]e
i=1
Z n
X
A1 nT
R,N 2
R,N
R,N
2
kl
u
+
2A
T
+
)
+
a
(x)u
≤
(uR,N
(u
)
dx
0
ij
0xi xj 0xk xl
1
0
δ
i,j,k,l=1
ΩR
0
Z
+
(f R (x, t))2 e−ηt dxdt ≤ µ1 .
QR
τ
Then from (15), choosing
b0
δ= ,
2
A1 + b0 a12
2A1 nT 2
η ≥ max
; 2 + 2A0 (T 2 + 1) +
a2 b0
b0
we get the estimate
Z n
X
2
R,N 2 −ητ
(uR,N
)
+
(u
)
e dx
xi xj
t
(16)
ΩR
τ
ij=1
Z n
X
R,N q
R,N 2 −ηt
+
(ut ) +
(utxi ) e dxdt ≤ µ2 ,
QR
T
i=1
104
τ ∈ [0, T ] and the constant µ2 does not depend on N. Moreover, assumption
(G)
implies the estimate
Z
(17)
0
|g(x, t, uR,N )|q dxdt ≤ µ3 ,
q0 =
q
.
q−1
QR
T
Due to the assumptions of Theorem 1, we can dierentiate system (7) with
respect to t :
(18)
Z n
X
R,N
R,N
uttt (x, t)φs (x) +
akl
ij (x, t)utxi xj (x, t)φsxk xl (x)
i,j,k,l=1
ΩR
+
n
X
aij (x, t)uR,N
txi (x, t)φsxj (x) +
n
X
bij (x, t)uR,N
ttxi (x, t)φsxj (x)
i,j=1
+
+
+
i,j=1
R,N
R,N R,N
a(x, t)ut (x, t)φs (x) + gξ (x, ut )utt (x, t)φs (x) − ft (x, t)φs (x)
n
n
X
X
R,N
akl
(x,
t)u
(x,
t)φ
(x)
+
aijt (x, t)uR,N
sxk xl
ijt
xi xj
xi (x, t)φsxj (x)
i,j=1
i,j,k,l=1
n
X
R,N
R,N
at (x, t)u
(x, t)φs (x) +
bijt (x, t)utxi (x, t)φsxi (x) dx = 0.
i,j=1
N (t)e−αt , α > 0, respectively, then
Hence, multiplying (18) by the functions Cstt
summing over s from 1 to N and integrating with respect to t from 0 to τ ,
τ ∈ (0, T ], we obtain
Z n
n
X
X
R,N
R,N R,N
N
kl
u
+
aij (x, t)uR,N
a
(x,
t)u
uR,N
u
+
tt
ij
ttt
txi xj ttxk xl
txi uttxj
i,j=1
i,j,k,l=1
QR
τ
+
n
X
R,N
R,N R,N
bij (x, t)uR,N
utt
ttxi uttxi + a(x, t)ut
i,j=1
(19)
+
+
2
gξ (x, uR,N
)(uR,N
t
tt )
Z X
n
QR
τ
−
ft (x, t)uR,N
tt
R,N R,N
akl
ijt (x, t)uxi xj uttxk xl +
+
i,j=1
R,N
aijt (x, t)uR,N
xi uttxj
i,j=1
i,j,k,l=1
n
X
n
X
e−αt dxdt
R,N
bijt (x, t)uR,N
txi uttxi
+
at (x, t)uR,N uR,N
tt
e−αt dxdt = 0.
105
where ξ :=: uR,N
. By analogy with (12), the following estimate for the rst
t
integral can be obtained
Z n
n
X
X
R,N N
R,N R,N
R,N
kl
uttt utt +
aij (x, t)utxi xj uttxk xl +
aij (x, t)uR,N
txi uttxj
i,j=1
i,j,k,l=1
QR
τ
+
n
X
R,N
R,N R,N
2
bij (x, t)uR,N
utt + gξ (x, uR,N
)(uR,N
ttxi uttxi + a(x, t)ut
t
tt )
i,j=1
−
1
≥
(20) 2
e−αt dxdt
ft (x, t)uR,N
tt
Z Z n
X
1
R,N 2
R,N 2 −ατ
2
(utt ) + a2
(α − A0 − 1)(uR,N
(utxi xj ) e
dx +
tt )
2
i,j=1
ΩR
τ
+ (αa2 −
a12 )
n
X
QR
τ
2
(uR,N
txi xj )
+ (2b0 − δ1 )
i,j=1
n
X
2
(uR,N
ttxi )
e−αt dxdt
i=1
Z n
n
X
A1 X R,N 2
1
2
2
(uR,N
)
+
(u
)
+
(2b
−
δ
)
(uR,N
−
0
1
t
txi
ttxi )
2
δ1
i=1
QR
τ
+
(ftR (x, t)))2
i=1
Z n
X
p
1
R,N 2
R,N
−αt
2
e dxdt −
(utt ) + A3
(u1,xi xj ) dx,
2
i,j=1
ΩR
0
where A3 = ess sup
n
P
Ω i,j,k,l=1
2
(akl
ij (x, 0)) , because gξ (x, ξ) ≥ 0.
Taking into account (A), we get
Z
n
X
QR
τ
i,j,k,l=1
1
≥−
2
R,N R,N
−αt
akl
dxdt
ijt (x, t)uxi xj uttxk xl e
Z n
n
X
A2 X R,N 2
R,N 2 −ατ
(uxi xj ) + δ1
(utxi xj ) e
dx
δ1
i,j=1
ΩR
τ
1
−
2
Z p
ΩR
0
1
−
2
Z QR
τ
A2
n
X
i,j=1
2
(uR,N
0,xi xj )
i,j=1
+
p
A2
n
X
2
(uR,N
1xi xj )
dx
i,j=1
n
n
X
α2 A2 + A4 X R,N 2
R,N 2 −αt
1
(uxi xj ) + (δ1 + 2a2 )
(utxi xj ) e dxdt,
δ1
i,j=1
i,j=1
106
Z X
n
R,N
aijt (x, t)uR,N
xi uttxj +
i,j=1
QR
τ
n
X
R,N
bijt (x, t)uR,N
txi uttxj
i,j=1
e−αt dxdt
+ at (x, t)uR,N uR,N
tt
1
≥−
2
Z n
n
n
X
A5 X R,N 2
B1 X R,N 2
R,N 2
(uxi ) + 2δ1
(uttxi ) +
(utxi )
δ1
δ1
i=1
QR
τ
i=1
i=1
R,N 2
2 R,N 2 −αt
+ (utt ) + A0 (u
) e dxdt,
n
P
where A2 = ess sup
QT i,j,k,l=1
ess sup
n
P
QT i,j=1
2
(akl
ijt (x, t)) , A4 = ess sup
n
P
QT i,j,k,l=1
(aijt (x, t))2 , B1 = ess sup
n
P
2
(akl
ijtt (x, t)) , A5 =
(bijt (x, t))2 .
QT i,j=1
Hence
Z n
X
R,N 2
R,N 2 −ατ
(utt ) + (a2 − δ1 )
dx
(utxi xj ) e
i,j=1
ΩR
τ
1
+
2
Z n
X
R,N 2
2
1
(uR,N
(α − A0 − 2)(utt ) + (αa2 − 3a2 − δ1 )
txi xj )
i,j=1
QR
τ
+ (2b0 − 3δ1 )
n
X
2 −αt
)
e dxdt
(uR,N
ttxi
i=1
Z n
n
A1 + B1 X R,N 2 α2 A2 + A4 X R,N 2
R,N 2
(ut ) +
(utxi ) +
(uxi xj )
(21) ≤
δ1
δ1
i=1
QR
τ
+
i,j=1
n
A5 X R,N 2
(uxi ) + A20 (uR,N )2 + (ftR (x, t)))2 e−αt dxdt
δ1
i=1
Z n
n
X
X
p
p
p
R,N
R,N
2
2
2
A
(u
A
+
A
)
(u
+
(uR,N
)
+
)
+
(
)
dx
2
2
3
tt
0,xi xj
1,xi xj
i,j=1
ΩR
0
A2
+
δ1
Z X
n
ΩR
τ
i,j=1
2 −ατ
(uR,N
dx.
xi xj ) e
i,j=1
107
Observe that for t = 0, multiplying (18) by Cstt (0), by analogy with (19),
we obtain
Z n
X
2
kl
R,N
(uR,N
(x,
0))
+
a
(x,
0)u
(x,
0)
ij
xi xj
tt
i,j,k,l=1
ΩR
0
(22)
−
−
n
X
uR,N
tt (x, 0)
R,N
aij (x, 0)uR,N
xi (x, 0) x utt (x, 0)
j
i,j=1
n X
bi,j (x, 0)uR,N
txi
xi
i,j=1
+
xk xl
R,N
uR,N
(x, 0)uR,N
tt (x, 0) + a(x, 0)u
tt (x, 0)
R,N
g(x, uR,N
(x, 0))uR,N
t
tt (x, 0)utt (x, 0)
−
dx = 0.
f (x, 0)uR,N
tt (x, 0)
By conditions of Theorem 1,

2 X
Z X
2
n
n R,N
R,N
kl
+
aij u0xi xj 
aij u0xi
xj i,j=1
i,j,k,l=1
xk xl
ΩR
0
n 2
X
R,N
R,N
2
N
2
+ |au
+ bi (x)u1xi
| + |g(x, u1 )| dx ≤ K1 .
0
xi
i=1
Then from (22) it follows that
(23)
Z
2
|uN
tt | dx ≤ M1 ,
ΩR
0
where M1 does not depend on N .
Choosing
a2 b0
δ1 = min
;
,
2 3
3a12 + δ1
α = min A0 + 2 ;
a2
and using (16), (17), we obtain
kL2 ((0,T );H 0,2 (ΩR )∩ Lq (QR ) ≤ M2 ,
(24) kuR,N kL2 ((0,T );H 0,2 (ΩR )) ≤ M2 , kuR,N
t
T
(25) kuR,N
kg(·, uR,N
kLq0 (QR ) ≤ M2 ,
tt kL2 ((0,T );H 0,1 (ΩR )) ≤ M2 ,
t
T
where M2 does not depend on N .
Then due to (24), (25) and [4, p. 70] there exists such subsequence {uR,Nk} ⊂
k
{uR,N } that uR,Nk → uR weakly in L2 ((0, T ); H 0,2 (ΩR )), uR,N
→ uR
t weakly
t
R,N
2
0,2
R
q
R
R
2
k
in L ((0, T ); H (Ω )) ∩ L (QT ), utt → utt weakly in L ((0, T ); H 0,1 (ΩR )),
0
R,Nk
2
R
k
g(·, uR,N
) → χR weakly in Lq (QR
→ uR
t strongly in L (QT ) when
t
T ), ut
Nk → ∞. Hence χR = g(·, uR
t ).
108
Taking into account (7), (8) it is easy to show that the function uR is a
weak solution of problem (4), (6):
Z n
X
R
uR
w
+
akl
tt
ij (x, t)uxi xj wxk xl
QR
T
(26)
i,j,k,l=1
n
X
n
X
R
aij (x, t)uR
w
+
a(x,
t)u
v
+
bij (x, t)uR
xi xj
xi t wxj
i,j=1
i,j=1
Z
R
+ g(x, u )w dxdt =
f R (x, t)wdxdt
+
QR
T
∀w ∈ L2 (0, T ); H 0,2 (ΩR ) ∩ Lq (QR
T ) and initial conditions (7). Moreover,
0,1
R
R
ut ∈ C([0, T ]; H (Ω )).
2. Suppose that
conditions (A)
, (B) and (G) hold and u0 ∈
∈ L2Ψ (Ω); f ∈ L2 (0, T ); L2Ψ (Ω) . Then there exists a weak solu-
Theorem
0,2
HΨ
(Ω), u1
tion of problem (1){(2), and this solution satises the following inequality
Z n
n
X
X
2
2
2
2
|u| + |ut | +
|uxi t | +
|uxi xj | Ψ(|x|)dxdt < µ,
(27)
i=1
QT
i,j=1
where µ is the constant which depends on f, u0 , u1 and coecients of equation
(1).
We consider the sequence of domains QR for R = 2, 3, . . . Observe
that the assumptions of Theorem 1 guarantee the existence of a weak solution
us of problem (1){(3) in the domain Qs , where s = 2, 3, . . . , in the sense of
Denition 2. Every function us satises Denition 1 and equality (5). We
extend all of these solutions by 0 on QT . We obtain the sequence {us }. For
this sequence, by analogy to (9), we can obtain the following equality
Proof.
Z n
X
s
s
ustt ust Ψ(|x|) +
akl
ij (x, t)uxi xj (ut Ψ(|x|))xk xl
i,j,k,l=1
QT
+
(28)
n
X
(bij (x, t)ustxi (ust Ψ(|x|))xj
i,j=1
+
n
X
aij (x, t)usxi (ust Ψ(|x|))xj
i,j=1
+ a(x, t)us ust Ψ(|x|) + g(x, ust )ust Ψ(|x|) e−βt dxdt
Z
=
QT
f s (x, t)ust Ψ(|x|)e−βt dxdt, β > 0.
109
Now we transform and estimate every term of (28). Obviously,
Z
ustt ust Ψ(|x|)e−βt dxdt =
I1 :=
QT
Z
1
+
2
(ust )2 Ψ(|x|)e−βt dx
Z
β
2
QZT
1
−
2
ΩT
(ust )2 Ψ(|x|)e−βt dxdt
(us1 (x))2 Ψ(|x|)dx.
Ω0
From assumption (A) there follows
n
X
Z
I2 :=
−βt
s
s
dxdt ≥ I21 + I22 + I23 ,
akl
ij (x, t)uxi xj (ut Ψ(|x|))xk xl e
QT i,j,k,l=1
where
I21
a2
≥
2
−
Z X
n
|usxi xj |2 Ψ(|x|)e−βt dx
ΩT i,j=1
Z X
n
1
2
s
s
akl
ij (x, 0)u0,xi xj u0xk xl Ψ(|x|)dx
Ω0 i,j,k,l=1
1
+ (βa2 − a12 )
2
Z X
n
|usxi xj |2 Ψ(|x|)e−βt dxdt,
QT i,j=1
Z X
n
A3 nΨ0
2
(usxi xj )2 Ψ(|x|)e−βt dxdt
I2 ≤
δ3
QT i,j=1
Z X
n
+δ3 Ψ0
(ustxk )2 Ψ(|x|)e−βt dxdt,
QT k=1
and
I23
A3 n2 Ψ0
≤
2
Z X
n
(usxi xj )2 Ψ(|x|)e−βt dxdt
QT i,j=1
Ψ0
+
2
Z
(ust )2 Ψ(|x|)e−βt dxdt.
QT
Next from assumption (B) we obtain:
I3 :=
Z X
n
bij (x, t)ustxi (ust Ψ(|x|))xj e−βt dxdt ≥ I31 + I32 ,
QT i,j=1
where
I31
:=
Z X
n
QT i,j=1
bij (x, t)ustxi ustxj Ψ(|x|)e−βt dxdt
≥ b0
Z X
n
QT i=1
(ustxi )2 Ψ(|x|)e−βt dxdt,
110
I32 ≤
B0 nΨ0 δ3
2
Z X
n
QT i=1
n
P
where B0 = ess sup
QT i,j=1
Z X
n
I4 :=
(ustxi )2 Ψ(|x|)e−βt dxdt +
Z
Ψ0
2δ3
(ust )2 Ψ(|x|)e−βt dxdt,
QT
b2ij (x, t). Next,
aij (x, t)usxi (ust Ψ(|x|))xj
+
e−βt dxdt ≥ I41 + I42 ,
a(x, t)us ust Ψ(|x|)
i,j=1
QT
where
I41
A1
≤
2δ3
Z X
n
QT i=1
A1 Ψ0
≤
2
I42
(usxi )2 Ψ(|x|)e−βt dxdt
Z X
n
δ3
+
2
Z X
n
QT i=1
(usxi )2 Ψ(|x|)e−βt dxdt
QT i=1
Z
n + A0
+
2
(ust )2 Ψ(|x|)e−βt dxdt +
A0
2
QT
I5 :=
Z
(us )2 Ψ(|x|)e−βt dxdt.
Qτ
From (G), there follows
Z
(usxi t )2 Ψ(|x|)e−βt dxdt,
g(x, ust )ust Ψ(|x|)e−βt dxdt
Z
≥ g0
QT
|ust |q Ψ(|x|)e−βt dxdt.
QT
It is evident that
Z
I6 :=
f
s
(x, t)ust Ψ(|x|)e−βt dxdt
1
≤
2
QT
Z s 2
s 2
(f ) + (ut ) Ψ(|x|)e−βt dxdt.
QT
Using the property of Ψ, it is easy to prove the following inequality:
Z X
n
(usxi )2 Ψ(|x|)dx
≤
Ω i=1
(29)
Z X
n
(usxi xj )2 Ψ(|x|)dx
Ω i,j=1
+(1 +
Ψ20 )
Z
(us )2 Ψ(|x|)dx.
Ω
Moreover,
Z
(30)
Z
(us (x, t))2 Ψ(|x|)e−βt dxdt ≤ 2T
QT
(us (0, t))2 Ψ(|x|)dx
Ω0
+2T
2
Z
QT
(ust )2 Ψ(|x|)e−βt dxdt.
111
Using the estimates for the integers I1 {I6 , from (29), (30), we get
Z n
X
(ust )2 + a2
(usxi xj )2 Ψ(|x|)e−βt dx
i,j=1
ΩT
Z n
X
(2b0 − 2Ψ0 δ3 − Ψ0 nB0 δ3 − δ3 )
+
(ustxi )2
i=1
QT
s q
+ 2g0 |ut | Ψ(|x|)e−βt dxdt
Z Ψ0
− n − A0 − 1 − 2T 2 (A0 + Ψ20 + 1))|ust |2
(β − Ψ1 −
+
δ3
(31) QT
X
n
2Ψ0 nA3
A1
1
2
s
2
+ βa2 − a2 −
− Ψ1 n A3 −
− A1 Ψ0
|uxi xj |
δ3
δ3
i,j=1
−βt
· Ψ(|x|)e
dxdt
Z ≤
(2T A0 + Ψ20 + 2T )(us0 )2 + |us1 |2
Ω0
n
X
+
s
s
akl
ij (x, 0)u0,xi xj u0,xk xl
Z
Ψ(|x|)dx + |f s |2 e−βt Ψ(|x|)dxdt.
i,j,k,l=1
QT
Choosing β , δ3 from the inequalities
2b0 − 2Ψ0 δ3 − Ψ0 nB0 δ3 − δ3 ≥ 1,
β − Ψ1 −
Ψ0
− n − A0 − 1 − 2T 2 (A0 + Ψ20 + 1) ≥ 1,
δ3
2Ψ0 nA3
A1
− Ψ1 n2 A3 −
− A1 Ψ0 ≥ 1,
δ3
δ3
we get the estimates kust kL∞ ((0,T );L2Ψ (Ω)) ≤ µ1 , kus kL2 ((0,T );H 0,2 (Ω)) ≤ µ1 ,
Ψ
R
0
kust kL2 ((0,T );H 0,1 (Ω))∩Lq ((0,T );Lq (Ω)) ≤ µ1 ,
|g(x, ust |q Ψe−βt dxdt ≤ µ1 , where
βa2 − a12 −
Ψ
µ1 does not depend on s.
Ψ
QT
s
Hence we can choose a subsequence {usk } of the sequence
{u } such that
0,2
ust k (·, T ) → w weakly in L2Ψ (Ω), usk → u weakly in L2 (0, T ); HΨ
(Ω) ,
0,1
ust k → ut weakly in L2 (0, T ); HΨ
(Ω) ∩ Lq (0, T ); LqΨ (Ω) , g(·, ust k ) → χ
0
q0
q
weakly in L (0, T ); LΨ (Ω) .
112
It is easy to prove similarly to [4, p. 24] that
Z
Z n
X
ut (x, T )w(x, T )dx +
−ut wt +
akl
ij (x, t)uxi xj wxk xl
ΩT
(32)
i,j,k,l=1
QT
+
n
X
n
X
aij (x, t)uxi wxj + a(x, t)uw +
i,j=1
bij (x, t)uxi t wxj
i,j=1
Z
Z
+ χw dxdt =
f (x, t)wdxdt + u1 (x)w(x, 0)dx
QT
Ω
∀w ∈ C 1 ([0, T ]; C0∞ (Ω)) and u(x, 0) = u0 (x). Moreover, equality (32) is true
for w(x) = ut e−βt Ψ(|x|).
For the function g we obtain
(33)Z 0≤
g(x, ust )
− g(x, v) (ust − v)Ψ(|x|)e−βt dxdt
QT
Z s
s
g(x, ut )v + g(x, v)(ut − v) Ψ(|x|)e−βt dxdt
Z
g(x, ust )ust Ψ(|x|)e−βt dxdt−
=
QT
QT
Z n
X
s
s
=
f s ust Ψ(|x|) − ustt ust Ψ(|x|) −
akl
ij uxi xj (ut Ψ(|x|))xk xl
i,j,k,l=1
n
X
s
s
s
s
s s
−
(bij utxi (ut Ψ(|x|))xj −
aij uxi (ut Ψ(|x|))xj −au ut Ψ(|x|) e−βt dxdt
i,j=1
i,j=1
QT
n
X
−
Z g(x, ust )v + g(x, v)(ust − v) Ψ(|x|)e−βt dxdt
QT
Z
≤−
ut (x, T )ut (x, T )e−βt Ψ(|x|)dx
ΩT
Z n
X
+
f ut Ψ(|x|) − βut ut Ψ(|x|) −
akl
ij uxi xj (ut Ψ(|x|))xk xl
QT
n
X
−
i,j=1
i,j,k,l=1
(bij utxi (ut Ψ(|x|))xj −
n
X
aij uxi (ut Ψ(|x|))xj
− auut Ψ(|x|) e−βt dxdt
i,j=1
Z Z
−βt
−
χv + g(x, v)(ut − v) Ψ(|x|)e dxdt + u1 (x)u1 (x)Ψ(|x|)dx,
QT
Ω
113
when s → +∞. Hence, summing (32) for w(x) = ut e−βt Ψ(|x|) and (33), we
obtain
Z χ − g(x, v) (ut − v)Ψ(|x|)e−βt dxdt ≥ 0.
QT
Let v = ut − λw ,λ > 0, w ∈ L2 (0, T ); HΨ0,2 (Ω) . Then
Z
(χ − g(x, ut ))wΨ(|x|)e−βt dxdt = 0
QT
for every w, which means that χ = g(x, ut ).
From (31) there follows (27). By virtue of Denition 1, the function u is a
weak solution of problem (1){(3).
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Received February 20, 2006
University of Rzeszów
Institute of Mathematics
Rejtana 16A
35-959 Rzeszów
Poland
e-mail : lzareba@univ.rzeszow.pl