ON BASIN OF ZERO-SOLUTIONS TO
A SEMILINEAR PARABOLIC EQUATION
WITH ORNSTEIN-UHLENBECK OPERATOR
YASUHIRO FUJITA
Received 27 April 2005; Accepted 10 July 2005
We consider the basin of the zero-solution to a semilinear parabolic equation on
R
N
with
the Ornstein-Uhlenbeck operator. Our aim is to show that the Ornstein-Uhlenbeck oper-
ator contributes to enlargement of the basin by using the logarithmic Sobolev inequality.
Copyright © 2006 Yasuhiro Fujita. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let α, β>0 be given constants. We consider the following semilinear parabolic problem:
u
t
=
1
2
Δu
− αx · Du + βulogu in (0, ∞) × R
N
,
u(0,
·) = ϕ in R
N
,
(1.1)
where the initial data ϕ satisfies

(1.1), the Ornstein-Uhlenbeck operator L contributes to enlargement of the basin of the
zero-solution.
Our aim of this paper is to clarify this contribution by using the relation between
the parameters α, β. Our result states that if α is sufficiently larger than β/2 then the
basin of the zero-solutions is large enough; on the other hand, if α is sufficiently smaller
than β/2 then it is small enough. Note that as α increases the attractive power to the
origin is stronger in the Ornstein-Uhlenbeck operator. Hence, the results above show that
enlargement of the basin arises from a contribution of the Ornstein-Uhlenbeck operator.
The contents of the paper are organized as follows: in Section 2, we state existence
and uniqueness of a classical solution to (1.1). In Section 3,wederiveL
q
-estimates of the
classical solution to (1.1). These estimates are based on the logarithmic Sobolev inequality
and the Jensen inequality. In Section 4, we state our main results and prove them.
2. A classical solution to (1.1)
In this section, we will show existence and uniqueness of a classical solution to (1.1). In
order to show existence and uniqueness of a classical solution to (1.1), we consider first
the following semilinear parabolic problem:
η
t
=
1
2
Δη
− αx · Dη +
1
2
e
βt
|Dη|

was shown in [6]. Our proof for (2.1) is almost same as that of [6]. So, we omit it. Let
Q
= (0,∞) × R
N
. (2.3)
Theorem 2.1 ([6]). Assume (1.2).Then, (2.1) admits at least one classical solution η such
that η
∈ C(Q)

C
1,2
(Q) with the property


D
x
η


∞,Q
< ∞. (2.4)
Now, we state existence and uniqueness of a classical solution to (1.1).
Theorem 2.2. Assume (1.2). Then (2.1) admits the unique classical solution u
∈ C(Q)

C
1,2
(Q) satisfying the follow ing: u(·) > 0 in Q, and for each T>0 there exists a constant
C
T


t
=
1
2
Δη
j
− αx · Dη
j
+
1
2
e
βt


Dη
j


2
in (0, ∞) × R
N
,
η
j
(0,·) = ψ(·)inR
N
.
(2.6)

Dη
1
+ Dη
2


·
D

η
1
− η
2

. (2.7)
Note that, for each T>0, there exists a constant K
T
> 0suchthat




−
αx +
1
2
e
βt

Dη

K
T

1+|x|

,(t,x) ∈ [0,T] × R
N
.
(2.8)
Hence, by the comparison theorem for parabolic equations (cf. [5, Theorem 9, page 43]),
we deduce that η
1
≡ η
2
on [0, T] × R
N
.SinceT>0 is arbitrarily, we conclude the theorem.
The proof is complete.

3. L
q
-estimates of the solution to (1.1)
In this section, we will give L
q
-estimates of the unique classical solution to (1.1). Let ν be
the Borel probability measure on
R
N
defined by
dν(y)

χ
q
logχ
q
dν ≤−
q
2
2α

q − 1


R
N
χ
q−1
Lχ dν + χ
q
L
q
(ν)
logχ
q
L
q
(ν)
. (3.3)
Next, we have the following lemma.
Lemma 3.2. .Foranyq>1 and 0 <χ
∈ C

n

dν = 0. (3.5)
Since
L

χ
q
n

=
qχ
q−1
n
Lχ +
1
2
q(q
− 1)χ
q−2
n
|Dχ|
2
, (3.6)
we obtain

R
N
χ
q−1

,(t,x) ∈ (0,T] × R
N
.
(3.8)
Now, we state the main results of this section.
Theorem 3.4. . Assume that (1.2)holdsand2α>β.Letu be the unique classical s olution
to (1.1)obtainedinTheorem 2.2.Then,foranyq
≥ 2α/(2α − β),


u(t,·)


L
q
(ν)
≤ exp

e
βt
logϕ
L
q
(ν)

, t ≥ 0. (3.9)
Proof. Let ρ
∈ C
∞
(R

u
q
n

t
= qu
q−1
n
ρ
n
(Lu + βulogu),
ρ
n
Lu = Lu
n
− uLρ
n
− Du· Dρ
n
,
u
q
n
logu = u
q
n
logu
n
− u
q

q−1
Lu
n
(t,·)dν + β

R
N
u
n
(t,·)
q
logu
n
(t,·)dν

−
q

R
N

u
n
(t,·)
q−1

u(t,·)Lρ
n
+ Du(t,·) · Dρ
n

n
(t,·)
q−1
dν + β


u
n
(t,·)


q
L
q
(ν)
log


u
n
(t,·)


q
L
q
(ν)
≤ β



J(t)


|
t ∈ [0,T]

−→
0asn −→ ∞ . (3.16)
Then the function f
n
(t)definedby
f
n
(t) =


u
n
(t,·)


q
L
q
(ν)
(3.17)
satisfies
d
dt
f

dt
log f
n
(t) ≤ βlog f
n
(t)+
θ
n
(T)
γ
T
,0<t<T. (3.20)
From this inequality, we have
e
−βt
log


u
n
(t,·)


q
L
q
(ν)
≤ log





q
L
q
(ν)
≤ logϕ
q
L
q
(ν)
,0≤ t ≤ T. (3.22)
Since T>0 is arbitrary, we obtain the desired result easily. The proof is complete.

Theorem 3.5. Assume that (1.2)holdsandα,β>0.Letu be the unique classical solution to
(1.1)obtainedinTheorem 2.2.Then,


u(t,·)


L
1
(ν)
≥ exp

e
βt
logϕ
L

N
u
n
(t,·)logu
n
(t,·)dν

−

R
N

u(t,·)Lρ
n
+ Du(t,·) · Dρ
n
+ βu
n
(t,·)logρ
n

dν
:
=

I(t) −

J(t).
(3.24)
By (3.2) and the Jensen inequality, we have

|

J(T)||t ∈ [0,T]

−→
0asn −→ ∞ . (3.26)
Then the function g
n
(t)definedby
g
n
(t) =


u
n
(t,·)


L
1
(ν)
(3.27)
satisfies
d
dt
g
n
(t) ≥ βg
n

T
,0<t<T. (3.30)
Yasuhiro Fujita 7
From this inequality, we have
e
−βt
log


u
n
(t,·)


L
1
(ν)
≥ log


u
n
(0,·)


L
1
(ν)
−



4. The main results
In this section, we will state our main results of this paper and prove them. For α,β>0,
we write (1.1)
α,β
for the parabolic problem (1.1) to emphasize the dependence on α,β>0.
We denote by u
ϕ,α,β
the unique solution of (1.1)
α,β
for ϕ with (1.2).
Definit ion 4.1. Let α,β>0andq>1. We define Γ
q
(α,β)by
Γ
q
(α,β) =

ϕ | ϕ(·) > 0, logϕ ∈ Lip

R
N

,lim
t→∞


u
ϕ,α,β
(t,·)

<δ

. (4.2)
Theorem 4.2. Let α,β>0 and q>1.Then,
Γ
q
(α,β) ⊂ B
1
(1). (4.3)
Proof. Let ϕ
∈ B
1
(1). Then, ϕ
L
1
(ν)
≥ 1. Since ν is the probability measure, it follows
from Theorem 3.5 that
liminf
t→∞


u
ϕ,α,β
(t,·)


L
q
(ν)

(α,β), α ≥ α
0
. (4.5)
(ii) For each 0 <δ
≤ 1, there exists a c onstant α
1
= α
1
(β,δ,q)(0<α
1
<β/2) such that
B
q
(δ) ⊂ Γ
q
(α,β), 0 <α≤ α
1
. (4.6)
8 On the basin of zero-solutions
By Theorem 4.3, we see that the Ornstein-Uhlenbeck operator L contributes to en-
largement of the basin. Indeed, if α
≥ α
0
, then the basin is large enough to include B
q
(1).
On the other hand, if 0 <α
≤ α
1
, the basin is small enough not to include B

(δ)suchthatϕ
1
∈ Γ
q
(α,β)for0<α≤ α
1
.For0<α≤ α
1
,weset
ρ(x)
= exp

−

β
2
− α

|
x|
2
−
N
β

, x ∈ R
N
. (4.9)
It is easy to see that
ρ

2β

2α
β

N/2q
≤ exp

N
2
−
αN
β

δe
−N/2
<δ. (4.11)
Now, choose C>0sothate
C
ρ
L
q
(ν)
<δ. This is possible by (4.11). We define the
function u
0
by
u
0
(t,x) = ρ(x)exp

(ν)
<δ,lim
t→∞
u
0
(t,x) = +∞

x ∈ R
N

. (4.14)
However, note that ϕ
0
does not fulfill (1.2). Hence, we need the following device.
First of all, let us choose R>0sothat
R>

2C
β − 2α
+
N
β
, ν

|
x| >R

<δ
q
−

2
−
N
β

. (4.16)
Then,itiseasytoseethat
ψ
0
(x) < 0, |x| >R. (4.17)
Next, we choose χ
∈ C
∞
(R
N
)suchthat0≤ χ(·) ≤ 1onR
N
and
χ(x)
=
⎧
⎨
⎩
1, |x|≤R,
0,
|x|≥R +1.
(4.18)
Then, we define the functions ψ
1
and ϕ

α,β,ϕ
1
which is ensured by Theorem 2.2.By(4.15)–(4.17), we see that


ϕ
1


q
L
q
(ν)
=

|x|≤R
exp

qψ
0

dν +

|x|>R
exp

qχψ
0

dν ≤

−βt
logu
j
(t,x). Since
η
j
satisfies

η
j

t
=
1
2
Δη
j
− αx · Dη
j
+
1
2
e
βt


Dη
j



− η
0

+

−
αx +
1
2
e
βt

Dη
1
+ Dη
0


·
D

η
1
− η
0

. (4.23)
By Theorem 2.2,weseethatforanyT>0 there exists a constant K
T
such that

η
1
(t,x) − η
0
(t,x) ≥−K
T

1+|x|

,(t,x) ∈ [0,T] × R
N
.
(4.24)
10 On the basin of zero-solutions
By (4.20) and the comparison theorem for parabolic equations (cf. [5,Theorem9,page
43]) we deduce that
η
1
(t,x) − η
0
(t,x) ≥ 0, (t,x) ∈ [0,∞) × R
N
. (4.25)
Hence, by (4.14), we see that
lim
t→∞
u
1
(t,x) = +∞, x ∈ R
N

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Ornstein-Uhlenbeck operator, to appear in Communications in PDE.
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1061–1083.
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Yasuhiro Fujita: Department of Mathematics, Toyama University, Toyama 930-8555, Japan
E-mail address: