UNIVERSITATIS IAGELLONICAE ACTA MATHEMATICA, FASCICULUS XL
2002
THE DIRICHLET PROBLEM FOR A PARABOLIC
SEMILINEAR DIFFERENTIAL EQUATION IN AN
UNBOUNDED DOMAIN
by Katarzyna Urbańska
Abstract. The aim of the paper is to give theorems about the uniqueness and existence of a weak solution for a parabolic semilinear differential
equation in an unbounded domain. The uniqueness and existence do not
depend on the behaviour of the solution for |x| → +∞.
It is generally known that the uniqueness of the solutions of the Dirichlet
problem for parabolic semilinear differential equations in unbounded domains
take place in the class of the functions increasing for |x| → +∞ not faster than
2
eλ|x| , where λ depends on the coefficients of the equation.
It turned out ([1], [2]) that for some semilinear parabolic equations the
uniqueness and existence of the solution of the Dirichlet problem did not depend on the behaviour of the solution for |x| → +∞.
In this paper the analogous result was obtained for a parabolic equation
with a nonlinear derivative with respect to time.
Let Ω ⊂ Rn be unbounded domain. We will denote by QT = Ω × (0, T )
for T > 0 and ST = ∂Ω × (0, T ). We consider the following equation in QT
n n
X
X
(1) |ut |p−2 ut −
aij (x)uxi
+
bi (x, t)uxi + c(x, t)u = f (x, t), p > 2
xj
i,j=1
i=1
with the boundary condition
(2)
u|ST = 0
and with the initial condition
(3)
u(x, 0) = u0 (x) for x ∈ Ω.
The following assumptions will be needed throughout the paper:
70
(i) Ω ∩ BR is a domain for every R > 0, where BR ⊂ Rn is the open ball of
radius R and the centre at the origin of coordinates, ∂Ω∩BR = ΓR ∈ C 1 ;
(ii) aij ∈ L∞ (Ω); bi , c, ct ∈ L∞ (QT ); functions aij satisfy inequality
n
X
aij (x)ξi ξj ≥ ν
i,j=1
n
X
ξi2 ∀ξ ∈ Rn , ν = const > 0, x ∈ Ω,
i=1
aij (x) = aji (x) for almost every x ∈ Ω,
and
c(x, t) ≥ c0 > 0
q
L ((0, T ); Lqloc (Ω̄)); u0
for almost every (x, t) ∈ QT ;
1
(iii) f ∈
∈ H0,loc
(Ω̄), where
n
o
Lrloc (Ω̄) = u : u ∈ Lr (Ω ∩ BR ) for every R > 0 , 1 ≤ r < +∞,
n
o
1
(Ω̄) = u : u ∈ H01 (Ω ∩ BR ) for every R > 0 ,
H0,loc
and
1
p
+
1
q
= 1.
1
(Ω̄)) such that ut ∈ Lploc (Q̄T )
Definition. A function u ∈ L∞ ((0, T ); H0,loc
is said to be a weak solution of the equation (1), if u satisfies the integral
equality
Z n
n
X
X
|ut |p−2 ut v +
aij (x)uxi vxj +
bi (x, t)uxi v
QT
i,j=1
i=1
+c(x, t)uv − f (x, t)v dxdt = 0
for every v ∈ C ∞ ([0, T ]; C0∞ (Ω)) and u satisfies the initial condition (3).
We can now formulate our result.
Theorem 1. If assumptions (i)–(iii) hold and p satisfies inequalities
2n
2 < p < n−2
, for n > 2,
p > 2,
for n = 1, 2,
then problem (1)–(3) has no more than one weak solution.
Proof. We prove that if problem (1)–(3) has a weak solution, then this
solution is unique. On the contrary, suppose that there are two weak solutions
u1 , u2 of problem (1)–(3).
Let u = u1 − u2 . By the definition of weak solution we have
Z n
n
X
X
p−2
|ukt | ukt v +
aij (x)ukxi vxj +
bi (x, t)ukxi v
QT
i,j=1
i=1
+c(x, t)uk v − f (x, t)v dxdt = 0
71
for k = 1, 2. Subtracting above the equation for k = 1 from the equation for
k = 2 we obtain
Z n
X
p−2
p−2
(|u1t | u1t − |u2t | u2t )v +
aij (x)uxi vxj +
QT
(4)
+
n
X
i,j=1
bi (x, t)uxi v + c(x, t)uv dxdt = 0.
i=1
Let v be a function defined by the formula
λt
λt
v = (θm ut ϕβ e− 2 ) ∗ ρl ∗ ρl θm e− 2
Z +∞ Z +∞
λτ
λt
(θm (τ )ut (x, τ )ϕβ (x)e− 2 )ρl (s − τ )dτ × ρl (t − s)dsθm (t)e− 2 ,
=
−∞
−∞
where
ϕ(x) =
1
2
R (R
0,
− |x|2 ), |x| ≤ R,
|x| > R;
β ≥ 1, λ > 0, R > 0.
For fixed τ0 , τ ∈ (0, T ) such that τ0 < τ let θm be the function defined in [0, T ]
as follows:
2
2
θm (t) = 1, if τ0 + m
<t<τ−m
,
1
1
,
θm (t) = 0, if t > τ − m or t < τ0 + m
2
2
1
1
, τ0 + m
) and (τ − m
,τ − m
);
θm is linear and continuous on (τ0 + m
R +∞
ρl satisfy the condition ρl (t) = ρl (−t), −∞ ρl (t)dt = 1 and
supp ρl ⊂ [− 1l , 1l ], l ∈ N.
Using the properties of operation ∗ and letting l → +∞ we can rewrite (4) as
follows
Z 2
|u1t |p−2 u1t − |u2t |p−2 u2t θm
ut ϕβ e−λt dxdt
QT
−
+
+
+
n Z Z
X
+∞
0
aij (x)θm (t)θm
(t)ϕβ (x)e−λt uxi (x, t)uxj (x, t)dtdx
Ω
−∞
i,j=1
n Z Z +∞
λ X
2
aij (x)ϕβ (x)θm
(t)uxi (x, t)uxj (x, t)e−λt dtdx
2
i,j=1 Ω −∞
n Z Z +∞
X
2
β
aij (x)ϕβ−1 ϕxj (x)θm
(t)uxi (x, t)ut (x, t)e−λt dtdx
Ω
−∞
i,j=1
Z X
Z
n
2
2
bi (x, t)uxi θm
ut ϕβ e−λt dxdt +
c(x, t)uθm
ut ϕβ e−λt dxdt
QT i=1
QT
= 0.
72
Letting m → +∞ we have
(5)
Z
|u1t |p−2 u1t − |u2t |p−2 u2t ut ϕβ e−λt dxdt
Qτ0 ,τ
n Z
X
1
aij (x)ϕβ (x) uxi (x, τ )uxj (x, τ )e−λτ −uxi (x, τ0 )uxj (x, τ0 )e−λτ0 dx
2
i,j=1 Ω
n Z Z τ
λ X
+
aij (x)ϕβ (x)uxi (x, t)uxj (x, t)e−λt dtdx
2
Ω
τ
0
i,j=1
n Z Z τ
X
+ β
aij (x)ϕβ−1 ϕxj (x)uxi (x, t)ut (x, t)e−λt dtdx
+
Z
+
i,j=1 Ω τ0
n
X
β −λt
bi (x, t)uxi ut ϕ e
Z
c(x, t)uut ϕβ e−λt dxdt = 0.
dxdt +
Qτ0 ,τ i=1
Qτ0 ,τ
Prolong the functions u and f by zero for τ < 0,while bi and c by bi (x, 0),
c(x, 0) for τ < 0. We claim that (5) is true for almost every τ0 , τ0 < τ, τ ∈
(0, T ), τ0 ∈ (−T, 0). From (5) we thus get the equality
Z p−2
p−2
|u1t | u1t − |u2t | u2t ut ϕβ e−λt dxdt
Qτ
n Z
1 X
−
aij (x)ϕβ (x)uxi (x, τ )uxj (x, τ )e−λτ dx
2
i,j=1 Ωτ
n Z
λ X
+
aij (x)ϕβ (x)uxi (x, t)uxj (x, t)e−λt dxdt
2
i,j=1 Qτ
n Z
X
+β
aij (x)ϕβ−1 ϕxj (x)uxi (x, t)uxj (x, t)e−λt dxdt
Z
+
i,j=1 Qτ
n
X
β −λt
bi (x, t)uxi ut ϕ e
Qτ i=1
Z
dxdt +
c(x, t)uut ϕβ e−λt dxdt = 0.
Qτ
Now we can obtain the following estimates:
Z I1 =
|u1t |p−2 u1t − |u2t |p−2 u2t u1t − u2t ϕβ e−λt dxdt
QT
Z
2−p
≥2
|ut |p ϕβ e−λt dxdt;
QT
73
−I22 ≥
−
ν
2
n
X
Z
ΩT i=1
n
X
Z
i,j=1 Ω
I23
I24
1
≥ λν
2
n
X
Z
u2xi ϕβ e−λT dx
1
aij (x)ϕβ (x)uxi (x, s0 )uxj (x, s0 )e−λs0 dx;
2
n
X
QT i=1
u2xi ϕβ e−λt dxdt;
Z
n
X
1
2
u2xi ϕβ e−λt dxdt
aij (x)uxi ut ϕ
ϕxj e dxdt ≤ βnδµ
2
QT i,j=1
QT i,j=1
Z
Z
n
X β− 2p
2p
1
β
+ βn2 δ
|ut |p ϕβ e−λt dxdt +
ϕ p−2 |ϕxj | p−2 e−λt dxdt,
p
%(δ) QT
QT
Z
=β
β−1
−λt
i,j=1
where κ =
(p−2)β
2p
− 1, |aij (x)| ≤ µ, δ > 0;
Z X
n
1
bi (x, t)uxi ut ϕβ e−λt dxdt ≤ µ22 δ1
u2xi ϕβ e−λt dxdt
2
QT i=1
QT i=1
Z
Z
n
nδ1
|ut |p ϕβ e−λt dxdt +
ϕβ e−λt dxdt,
+
p QT
ζ(δ1 ) QT
Z
I3 =
n
X
where |bi (x, t)| ≤ µ2 for i = 1, . . . , n and δ1 > 0;
Z
Z
1
β −λt
c(x, t)uut ϕ e dxdt ≥ c0
I4 =
u2 ϕβ e−λT dx
2
QT
ΩT
Z
Z
1
1
2 β −λt
−
ct (x, t)u ϕ e dxdt + λc0
u2 ϕβ e−λt dxdt.
2 QT
2
QT
Summarizing, we have
Z
βn2 δ nδ1 2−p
2
−
−
|ut |p ϕβ e−λt dxdt
p
p
QT
n
1
Z X
1
1
+ λν − nβδµ2 − µ22 δ1
u2xi ϕβ e−λt dxdt
2
2
2
QT i=1
Z
Z
n
X β− 2p
2p
n
β
(6)
−
ϕ p−2 |ϕxj | p−2 e−λt dxdt −
ϕβ e−λt dxdt
%(δ) QT
ζ(δ1 ) QT
i,j=1
Z
1
1 + λc0 − c1
u2 ϕβ e−λt dxdt ≤ 0.
2
2
QT
74
We can choose the constants δ, δ1 so that the expression 22−p − p1 βn2 δ − p1 nδ1
is positive. Then we choose the constant λ so that the expression 21 λν −
1
2 − 1 µ2 δ
nβδµ
2
2 2 1 is greater than 0. Addtionally, we note that
Z
Z
Z
2p
2p
2p
2p
2p
β− p−2
β− p−2
β− 2p
p−2
p−2
p−2
ϕ
|ϕxj | dx ≤ 2
ϕ
dx ≤ 2
(R + |x|) p−2 dx
Ω
Ω
ΩR
Z
2p
2p
2p
β−
β−
+n
≤ 2 p−2 (2R) p−2
dx = 2β R p−2 .
BR
Then from (6) it follows that
Z
β− 2p +n
u2 ϕβ e−λt dxdt ≤ CR p−2 .
QR
T
It is easily seen that
Z
Z
2 β −λt
u ϕ e dxdt =
QR
T
2 β −λt
u ϕ e
R
Z
dxdt +
QT 0
Z
≥
R−R0
u2 ϕβ e−λt dxdt
QT
R
u2 ϕβ e−λt dxdt
QT 0
and
Z
R
QT 0
u2 e−λt (R − R0 )β dxdt ≤ CR
where R0 < R. Therefore
Z
2 −λt
u
e
dxdt
≤
C
R
QT 0
2p
+n
β− p−2
,
2p
R β n− p−2
R
.
R − R0
Let ε be any small positive number. Then for R large enough and n <
obtain
Z
u2 e−λt dxdt < ε.
R
2p
p−2
we
QT 0
Thus
Z
R
QT 0
0
u2 e−λt dxdt = 0 and u = 0 in QR
T .
As R0 is an arbitrary number, u = 0 in QT . This completes the proof.
Theorem 2. Let assumptions (i)–(iii) hold; futhermore we assume that
the functions bi ≡ 0 for i = 1, . . . , n. Then there exists a weak solution of
problem (1)–(3).
75
Proof. We apply the Galerkin method. Set ΩR = Ω ∩ BR .
Let

x ∈ ΩR−1 ,
 u0 (x),
R
s(x)u0 (x), x ∈ ΩR \ ΩR−1 ,
u0 (x) =

0,
x ∈ Ω \ ΩR ,
where the function s ∈ C 1 (Rn ), 0 ≤ s(x) ≤ 1 in BR \ BR−1 ,
s(x) =
1, x ∈ BR−1 ,
0, x ∈ Rn \ BR ;
while
R
f (x, t) =
f (x, t), (x, t) ∈ QR
T,
0,
(x, t) ∈ QT \ QR
T.
Let a function uN (x, t) be of the form
N
u (x, t) =
N
X
k
cN
k (t)ϕ (x),
N = 1, 2, . . .
k=1
where c1k (t), . . . , cN
k (t) are the solutions of the following Cauchy problem
(7)
Z n
X
l
N p−2 N l
εuN
ut ϕ +
tt ϕ + |ut |
ΩR
l
N l
R l
aij (x)uN
ϕ
+
c(x,
t)u
ϕ
−
f
ϕ
dx = 0
xi xj
i,j=1
cN
k (0)
(8)
and uN,R
=
0
=
N
X
uN
0,k
, cN
kt (0) = 0, l, k = 1, . . . , N
k
uN
0,k ϕ (x),
1
→ uR
uN,R
0 in H0 (ΩR ); ε > 0.
0
k=1
Multiplying equation (7) by cN
kt (t), summing over l, for l = 1, . . . , N and integrating over t ∈ (0, τ ), we obtain
Z
(9)
QR
τ
N
N p−2 N N
εuN
ut ut +
tt ut + |ut |
n
X
N
aij (x)uN
xi uxj t
i,j=1
R N
+ c(x, t)uN uN
−
f
u
t
t dxdt = 0.
76
It is easy to get the following estimates
Z
Z
1
N N
2
εutt ut dxdt = ε
J1 =
(uN
t ) dx;
2
R
Q
Ωτ,R
Z
Z τ
p
N p−2 N 2
|uN
|ut | (ut ) dxdt =
J2 =
t | dxdt;
QR
τ
QR
τ
n
X
Z
J3 =
QR
τ
=
1
2
Z
Z
N
aij (x)uN
xi uxj t dxdt
i,j=1
n
X
Ωτ,R i,j=1
N
aij (x)uN
xi uxj dx −
1
2
Z
n
X
Ω0,R i,j=1
2
aij (x)(uN
xi (x, 0)) dx,
Z
1
(uN )2 dx
c(x, t)uN uN
t dxdt ≥ c0
2
R
Q
Zτ
ZΩτ,R
1
1
N,R 2
−
c(x, 0)(u0 ) dx −
ct (x, t)(uN )2 dxdt,
2 Ω0,R
2 QR
τ
Z
Z
Z
δ
1
1
R
N
N p
J5 =
f (x, t)ut dxdt ≤
|ut | dxdt + p
|f R |q dxdt,
p
R
R
R
q
Qτ
Qτ
qδ1 Qτ
J4 =
where δ1 > 0. So we have
Z
Z
Z
Z
δ1 1
1
1
N 2
N 2
N p
ε
(u ) dx + 1 −
|ux | dx + c0
(uN )2 dx
|ut | dxdt + a0
2 Ωτ,R t
p QR
2
2
Ωτ,R
Ωτ,R
τ
Z
1
≤ c1
|uN |2 dxdt + c2 ku0 kW 1,2 (Ω0,R ) + c3 kf kLq (QR
τ )
2
R
Qτ
and taking into account the Gronwall–Bellman lemma we obtain
Z
Z τ Z
Z
1
1
N 2
N 2
(u ) dxdt = σ + c1
(uN )2 dx dt.
(u ) dx ≤ σ + c1
2
2
QR
0
ΩR
ΩR
τ
It follows that
Z
2
|uN |2 + |uN
x | dx ≤ µ,
Ωτ,R
Z
(10)
QR
τ
p
|uN
t | dxdt ≤ µ,
Z
ε
2
(uN
t ) dx ≤ µ,
Ωτ,R
where µ is a constant independent of N .
77
1
k.
Let εk =
Then εk ukt → 0 ∗-weakly in L∞ (0, T ); H01 (ΩR ) . From the
sequence {uN } we can select a subsequence {uk } such that
uk −→ uR ∗ -weakly in L∞ ((0, T ); H01 (ΩR )),
p
R
ukt −→ uR
t weakly in L (QT ),
|ukt |p−2 ukt −→ z R weakly in Lq (QR
T ),
as k → +∞.
p−2 uR . Consider the sequence {X }, where
We show that z R = |uR
k
t |
t
Z
0 ≤ Xk =
QR
T
|ukt |p−2 ukt − |v|p−2 v
Z
−
(11)
QR
T
Z
ukt − v dxdt =
QR
T
Z
|ukt |p−2 ukt vdxdt −
QR
T
|ukt |p−2 (ukt )2 dxdt
|v|p−2 v(ukt − v)dxdt,
for every v ∈ Lp (QR
T ). From (9) we have
Z
QR
T
|ukt |p dxdt
Z
QR
T
+
+
1
2
1
2
Z
=
QR
T
f R ukt dxdt
=
Z
Z
n
X
Ω0,R i,j=1
f R ukt
−
1
− εk
2
Z
εk uktt ukt
n
X
−
aij (x)ukxi ukxj t − c(x, t)uk ukt dxdt
i,j=1
(ukt )2 dx
ΩT,R
aij (x)ukxi ukxj dx −
c(x, t)(uk )2 dx +
Ω0,R
1
2
Z
QR
T
1
2
1
−
2
Z
Z
n
X
ΩT,R i,j=1
aij (x)ukxi ukxj dx
c(x, t)(uk )2 dx
ΩT,R
ct (x, t)(uk )2 dxdt.
Hence
Z
0 ≤ Xk =
QR
T
f R ukt
−
|ukt |p−2 ukt v
p−2
− |v|
v(ukt
− v) dxdt
Z
Z
Z
n
X
1
1
1
k 2
k k
|ut | dx −
− εk
aij (x)uxi uxj dx −
c(x, T )(uk )2 dx
2
2
2
Ω0,R
ΩT,R i,j=1
ΩT,R
Z
Z
Z
n
X
1
1
1
k k
k 2
+
aij (x)uxi uxj dx +
c(x, 0)(u ) dx +
ct (x, t)(uk )2 dxdt.
2 Ω0,R
2 Ω0,R
2 QR
T
i,j=1
78
Thus ([4])
Z
0 ≤ sup limXk ≤
ΩT,R i,j=1
Z
n
X
(12)
+
+
QR
T
n
X
Z
1
−
2
1
2
Ω0,R i,j=1
Z
1
2
QR
T
R
p−2
R
f R uR
−
z
v
−
|v|
v(u
−
v)
dxdt
t
t
R
aij (x)uR
xi uxj dx
1
−
2
R
aij (x)uR
0,xi u0,xj dx +
Z
c(x, T )(uR )2 dx
ΩT,R
1
2
Z
2
c(x, 0)(uR
0 ) dx
Ω0,R
ct (x, t)(uR )2 dxdt.
We have (from (7), (8))
Z n
X
R
R
zRv +
aij uR
v
+
cu
v
−
f
v
dxdt = 0
x
xi j
QR
T
i,j=1
2 (0, T ); H 1 (Ω ) . Analogously as in the proof of the
for every v ∈ Lp (QR
)
∩
L
R
0
T
uniqueness we can receive v = ut . Hence
Z
Z
n
X
1
1
R R
0≤
c(x, T )(uR )2 dx
aij (x)uxi uxj dx +
2 ΩT,R
2 ΩT,R
i,j=1
Z
Z
n
X
1
1
2
R
(13)
−
c(x, 0)(uR
aij (x)uR
u
dx
−
0 ) dx
0,xi 0,xj
2 Ω0,R
2 Ω0,R
i,j=1
Z 1
R 2
R R
+
z R uR
−
c
(x,
t)(u
)
−
f
u
t
t
t dxdt
2
QR
T
Adding (12) and (13) we obtain
Z R R
− z R v − |v|p−2 v(uR
−
v
)
+
z
u
t
t
t dxdt ≥ 0
QR
T
and hence
Z
z R − |v|p−2 v
QR
T
uR
t − v dxdt ≥ 0.
p
R
Let v = uR
t − λωt , λ > 0, ω ∈ L (QT ) . Therefore
Z z R − |v|p−2 v ωt dxdt ≥ 0.
QR
T
79
Letting λ → 0
Z
QR
T
p−2 R
z R − |uR
|
u
t
t ωt dxdt = 0.
p−2 uR .
This gives z R = |uR
t |
t
From (10) uk (·, 0) → uR (·, 0) weakly in L2 (Ω0,R ). On the other hand
1
R,εk (x, 0) = uR (x).
uk (·, 0) → uR
0 in H0 (ΩR ). The result is u
0
If R receives values 1, 2, 3, . . . than we have a sequence {um (x, t)}. Prolong
every function um by zero beyond the domain Qm
T . Then for the elements of
the sequence {um } we have the equality (for fixed R):
Z n
X
m p−2 m
m
(14)
|ut | ut v +
aij (x)um
xi vxj + c(x, t)u v − f (x, t)v dxdt = 0
QR
τ
i,j=1
for every v ∈ L2 ((0, T ); H01 (ΩR )) ∩ Lp (QR
T ). Let
uk − um = uk,m for k, m > R.
We have
Z QR
τ
|ukt |p−2 ukt v
+
n
X
aij (x)ukxi vxj
k
+ c(x, t)u v − f (x, t)v dxdt = 0
i,j=1
and hence
Z n
X
p−2 m
k,m
k,m
|ukt |p−2 ukt − |um
|
u
v
+
a
(x)u
v
+
c(x,
t)u
v
dxdt = 0.
ij
xj
t
t
xi
QR
τ
i,j=1
Analogously as in the proof of Theorem 1, we can prove that the above
β
equation is satisfied by v = uk,m
t ϕ .
We estimate respective expressions
Z k,m β
p−2 m
I1 =
|ukt |p−2 ukt − |um
|
u
t
t ut ϕ dxdt
Qτ
Z
2−p
p β
≥2
|uk,m
t | ϕ dxdt;
Qτ
Z
I2 =
Z
+
n
X
Qτ i,j=1
n
X
Qτ i,j=1
aij (x)uk,m
xi vxj dxdt
Z
=
n
X
Qτ i,j=1
k,m β
aij (x)uk,m
xi utxj ϕ dxdt
k,m
β−1
aij (x)uk,m
ϕxj dxdt = I21 + I22 ,
xi ut βϕ
80
where
I21
n
X
Z
=
=
k,m β
aij (x)uk,m
xi utxj ϕ dxdt
Qτ i,j=1
Z X
n
1
2
Ωτ i,j=1
k,m β
aij (x)uk,m
xi uxj ϕ dx
Z X
n
1
k,m
β
aij (x)uk,m
−
xi (x, 0)uxj (x, 0)ϕ dx,
2 Ω0
i,j=1
Z X
n
k,m β−1
I22 = β
aij (x)uk,m
ϕxj dxdt
xi u t ϕ
Qτ i,j=1
1
≤ βnδ1 µ2
2
+
where κ =
β
%(δ1 )
p−2
2p β
n X
Z
Qτ i,j=1
Z X
n
uk,m
xi
ϕ
2
2p
β− p−2
1
ϕ dxdt + βn2 δ1
p
β
Z
Qτ
p β
|uk,m
t | ϕ dxdt
2p
|ϕxj | p−2 dxdt,
Qτ i,j=1
− 1;
Z
Z
β
c(x, t)uk,m uk,m
t ϕ dxdt
Qτ
Qτ
Z
Z
2
2
1
1
=
c(x, τ ) uk,m ϕβ dx −
c(x, 0) uk,m
ϕβ dx
0
2 Ωτ
2 Ω0
Z
2
1
−
ct (x, t) uk,m ϕβ dxdt.
2 Qτ
I3 =
c(x, t)u
k,m
vdxdt =
Summarizing we obtain
Z
Z X
n 2
βn2 δ1
1
k,m p β
2
−
uk,m
ϕβ dx
|ut | ϕ dxdt + ν
xi
p
2 Ωτ
Qτ
i=1
Z Z X
n 2
2
1
1
+ c0
uk,m ϕβ dx ≤ βnδ1 µ2
uk,m
ϕβ dxdt
xi
2
2
Ωτ
Qτ i=1
Z Z X
n
2
2p
2p
1
β
β− p−2
p−2
|ϕxj | dxdt + c1
+
ϕ
uk,m ϕβ dxdt.
2
%(δ1 ) Qτ
Qτ
2−p
i,j=1
81
Let ε > 0. Then, analogously as in the proof of the uniqueness, from the
above inequality, we may obtain the following estimates
Z
n 2 X
k,m p
k,m
|ut | +
u xi
dxdt < ε,
R
Qτ 0
Z
i=1
|uk,m (x, t)|2 dx < ε, τ ∈ [0, T ],
Ωτ,R0
where R0 is an arbitrary positive number.
We have proved that the sequence {uk } satisfies the Cauchy condition. It
follows that this sequence is convergent.
Observe that
p
k
k
2
1
ut → ut in Lloc (QT ), u → u in L (0, T ); H0,loc
(Ω) ,
uk → u in C [0, T ]; L2loc (Ω) .
1
(Ω). Thus u satisfies the initial conOn the other hand uk0 → u0 in H0,loc
dition. Letting m → +∞ in (14) and by above convergences we conclude
that u is a weak solution of problem (1)–(3). The proof of the theorem is
complete.
References
1. Brezis H., Semilinear equations in Rn without condition at infinity, Appl. Math. Optim.
1984, Vol.12, 271–282.
2. Bokalo M.M., On the existence and uniqueness of a solution of the problem without
conditions at infinity for semilinear parabolic equations in unbounded domain, Sibirsky
Mathematichesky Zhurnal, 34 (1993), N4 (in Russian).
, Boundary value problem for semilinear parabolc equations without conditions at
3.
infinity in unbounded domain, Sibirsky Mathematichesky Zhurnal, 37 (1996), N5, 977–985
(in Russian).
4. Lions J.L., Quelques methodes de résolution des problémes aux limites non linéaires,
Dunod Gautier – Villars, Paris, 1969.
Received
March 23, 2002