Electronic Journal of Differential Equations, Vol. 2007(2007), No. 109, pp. 1–25.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
INITIAL-BOUNDARY VALUE PROBLEMS FOR NONLINEAR
PSEUDOPARABOLIC EQUATIONS IN A CRITICAL CASE
ELENA I. KAIKINA
Abstract. We study nonlinear pseudoparabolic equations, on the half-line in
a critical case,
x ∈ R+ , t > 0,
∂t (u − uxx ) − αuxx = λ|u|u,
u(0, x) = u0 (x),
x ∈ R+ ,
u(t, 0) = 0,
where α > 0, λ ∈ R. The aim of this paper is to prove the existence of global
solutions to the initial-boundary value problem and to find the main term of
the asymptotic representation of solutions.
1. Introduction
We study the following nonlinear pseudoparabolic equation on the half line, with
Dirichlet boundary condition,
∂t (u − uxx ) − αuxx = λ|u|u,
u(0, x) = u0 (x),
x ∈ R+ , t > 0,
x ∈ R+ ,
(1.1)
u(t, 0) = 0,
where α > 0 and λ ∈ R.
The Cauchy problem for nonlinear pseudoparabolic type equations was studied
in many papers (for example, see [11, 12, 13, 14, 15, 26, 30, 31]). The large time
asymptotics of solutions to the Cauchy problem was obtained in papers [1]-[4],
[18, 22, 23, 27, 28, 33].
In this paper we study the initial boundary-value problem (1.1) in a critical
case, when the nonlinear term of equation (1.1) has the same time decay rate as
the linear terms. Recently much attention was drown to the study of the global
existence and large time asymptotic behavior of solutions to the Cauchy problems
for nonlinear equations in the critical cases (see papers [6]-[10, 20, 21, 24, 25] and
literature cited therein). A general theory of nonlinear nonlocal equations on a
half-line was developed in the book [16], however the case of nonanalytic symbols
K(p) in the right-hand complex plane was not studied previously. In the present
paper we fill this gap, considering as example the pseudoparabolic type equation
2000 Mathematics Subject Classification. 35Q35.
Key words and phrases. Dissipative nonlinear evolution equation; Sobolev equation;
large time asymptotic behavior.
c
2007
Texas State University - San Marcos.
Submitted March 22, 2007. Published August 7, 2007.
1
2
E. I. KAIKINA
EJDE-2007/109
(1.1) with a symbol K(p) = αp2 /(1 − p2 ). We construct the Green operator for
problem (1.1) posing some necessary condition at the singularity point p = 1 of
the symbol K(p). Another difficulty which we overcome in the present paper is in
evaluating the contribution of the boundary data into the large time asymptotic
behavior of solutions. In this paper we will prove that in the case of the initialboundary value problem due to boundary data the solution obtains an additional
decay comparing with the case of Cauchy problem. As a result the main term of the
asymptotic expansion does not depend on the mean value of the solution instead
it is determined by the first moment of the solution. Thus we have to estimate
the evolution of the first moments to obtain an optimal time decay estimate of the
solution. Also we are interested in the case of large initial data. Using the energy
type a-priori estimates for the first moment of the solution we are able to remove
the smallness condition for the initial data. The asymptotic behavior of solutions
is founded by the standard way developed in the book [19].
Below φ̂ is the Laplace transform of φ defined by
Z +∞
φ̂(ξ) =
e−xξ φ(x)dx,
0
Here
−1
L
−1
Z
i∞
exξ φ̂(ξ)dξ
φ̂(ξ) = (2πi)
−i∞
is the inverse Laplace transform of φ. By C(I; B) we denote the space of continuous
functions from a time interval I to the Banach space B. The usual Lebesgue space
is denote by Lp (R+ ), 1 ≤ p ≤ ∞, the weighted Lebesgue space Lp,a (R+ ) is defined
by
Lp,a (R+ ) = φ ∈ Lp (R+ ); kφkLp,a = khxia φkLp < ∞ ,
p
where hxi = 1 + |x|2 , a ≥ 0. Weighted Sobolev spaces we define as follows
k
n
o
X
Wpk,a (R+ ) = φ ∈ Lp (R+ ); kφkWpk,a =
k∂ j φkLp,a < ∞ ,
j=0
where k ≥ 0, a ≥ 0, 1 ≤ p ≤ ∞.Also we denote by Hk,a (R+ ) = W2k,a (R+ ). Define
Z +∞
|λ|t
x(G0 (t, x))2 dx = η,
0
where the heat kernel
G0 (t, x) = (4παt)−1/2
x − x2
e 4αt .
αt
Denote
Z
g(t) = 1 + |θ|η log(1 + t), θ =
+∞
xu0 (x)dx.
0
Now we state the results of this paper.
Theorem 1.1. Assume that λθ ≤ −Cε < 0. Let the initial data u0 ∈ L∞ (R+ ) ∩
L1,1+a (R+ ), a ∈ (0, 1] are small ku0 kL∞ +ku0 kL1,1+a ≤ ε. Then the initial-boundary
value problem (1.1) has a unique global solution
u ∈ C([0, ∞); L∞ (R+ ) ∩ L1,1+a (R+ ))
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
3
satisfying the time decay estimate
ku(t) − θG0 (t)g −1 (t)kL∞ ≤ Chti−1 g −2 (t).
(1.2)
Using the method of papers [17, 18] we can remove the smallness condition on
the initial data u0 (x).
2,0
Theorem 1.2. Assume the initial data u0 ∈ W∞
(R+ ) ∩ W12,0 (R+ ) ∩ L1,1+a (R+ ),
0 < a ≤ 1, are such that λθ < 0. Then the initial-boundary value problem (1.1) has
a unique global solution
u(t, x) ∈ C([0, ∞); L∞ (R+ ) ∩ L1,1+a (R+ ))
satisfying the time decay estimate
ku(t) − θG0 (t)g −1 (t)kL∞ ≤ Chti−1 g −2 (t) log log g(t).
(1.3)
We start with a section where we obtain preliminary estimates for the linearized
initial boundary-value problem corresponding to (1.1). Also we prove a local existence theorem. In the next two sections we prove Theorems 1.1 and 1.2 respectively.
2. Preliminaries
2.1. Green operator. Consider the linear initial boundary-value problem on halfline
∂t (u − uxx ) − αuxx = f (t, x), x ∈ R+ , t > 0,
u(0, x) = u0 (x),
x ∈ R+
(2.1)
u(t, 0) = 0.
Taking Laplace transform with respect to space variable we get
2
2
X
X
∂xj−1 u(t, 0) ∂xj−1 u(t, 0) 2
−
αp
= fb(t, p),
∂t u
u
b
−
b − p2 u
b−
j
p
pj
j=1
j=1
(2.2)
u
b(0, p) = u
b0 (p),
u(t, 0) = 0.
We rewrite
(1 − p2 )∂t u
b − αp2 u
b = f1 (t, p),
(2.3)
where
f1 (t, p) = fb(t, p) − p2
2
X
∂ j−1 ut (0, t)
x
pj
j=1
− αp2
Integrating (2.3) with respect to time, we obtain
Z t
−K(p)t
u
b(t, p) = e
u
b0 (p) +
e−K(p)(t−τ )
0
where
2
X
∂ j−1 u(t, 0)
x
j=1
pj
1
f1 (τ, p)dτ,
1 − p2
.
(2.4)
αp2
.
1 − p2
Note that symbol K(p) is not analytic in right-half complex plane. Therefore we
have for solution u(t, x)
Z 1+ε+i∞
1
u(t, x) =
epx u
b(t, p)dp.
2πi 1+ε−i∞
K(p) = −
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E. I. KAIKINA
EJDE-2007/109
To satisfy zero condition for x → +∞ we need to have
res(b
u(t, p), 1) = 0.
(2.5)
We have by definition
Z
res(b
u(t, p), 1) = lim
ρ→0
Cρ,1
u
b(t, p)dp
Z
Z
= lim
ρ→0
Cρ,1 ,p∈D +
u
b(t, p)dp + lim
ρ→0
Cρ,1 ,p∈D −
u
b(t, p)dp,
where
Cρ,1 = p ∈ C : p = 1 + ρeiφ , ρ > 0, φ ∈ [0, 2π) ,
and by D+ , D− we denote domains where Re K(p) > 0 and Re K(p) < 0. Since for
p ∈ Cρ,1
α(1 + ρeiφ )2
K(p) = − iφ
,
ρe (2 + ρeiφ )
it is easy to see that
Z
lim
ρ→0
Cρ,1, p∈D +
u
b(t, p)dp = 0.
We rewrite formula (2.4) in a domain, where Re K(p) < 0 in the form
Z +∞
1
u
b(t, p) = e−K(p)t (b
u0 (p) +
e−K(p)(t−τ )
f1 (τ, p)dτ )
1
−
p2
0
Z +∞
1
−
e−K(p)(t−τ )
f1 (τ, p)dτ.
1
−
p2
t
We have
Z
Z
lim
ρ→0
+∞
e−K(p)(t−τ )
dp
Cρ,1,
p∈D −
t
1
f1 (τ, p)dτ = 0.
1 − p2
To satisfy (2.5) we have to put the following condition for all p ∈ D− , Re p > 0
Z +∞
1
u
b0 (p) +
eK(p)τ
f1 (τ, p)dτ = 0.
(2.6)
1
−
p2
0
Note that function f1 includes two unknown boundary data u(t, 0) and ux (0, t).
Also we have two roots φj (ξ) of equation ξ = −K(p). By direct calculation we
obtain
s
s
ξ
ξ
φ1 =
and φ2 = −
.
ξ+α
ξ+α
Since we are interested in Re ξ > 0 and Re p > 0 making the change of variable
ξ = −K(p) we rewrite condition (2.6) as one equation with two unknown boundary
data u(t, 0) and ux (t, 0)
Z +∞
1
u
b0 (φ1 ) +
e−ξτ f1 (τ, ξ)dτ = 0,
(2.7)
1 − φ21 0
where
f1 (τ, ξ) = fb(τ, φ1 ) − φ21
2
X
∂ j−1 uτ (τ, 0)
x
j=1
φj1
− αφ21
2
X
∂ j−1 u(τ, 0)
x
j=1
φj1
.
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
5
So we have to put only one boundary conditions in the problem (2.1) and rest
boundary data we will find from equation (2.7). Putting u(t, 0) = 0 in equation
(2.7) we get
b
L(b
utx (t, 0) + αux (t, 0)) = (1 − φ21 )b
u0 (φ1 ) + fb(ξ, φ1 ).
So that
u
btx (t, 0) + αux (t, 0) =
1
2πi
=
1
2πi
1
=
2πi
Z
i∞
b
eξt ((1 − φ21 )b
u0 (φ1 ) + fb(ξ, φ1 ))dξ
−i∞
Z i∞
eξt (1 −
−i∞
i∞
Z
eξt (
−i∞
ξ
b
)(b
u0 (φ1 ) + fb(ξ, φ1 ))dξ
ξ+α
αb
u0 (φ1 ) bb
+ f (ξ, φ1 ))dξ.
ξ+α
Substituting this representation into (2.4) we obtain
u
b(t, p) = I1 (t, p) + I2 (t, p),
(2.8)
where
I1 (t, p) = e−K(p)t u
b0 (p) −
α
1
1 − p2 2πi
Z
t
dτ e−K(p)(t−τ )
Z
i∞
−i∞
0
eξτ
u
b0 (φ1 )
dξ
ξ+α
(2.9)
= J1 + J2
and
Z
1 t −K(p)(t−τ )
I2 (t, p) =
e
f (τ, p)dτ
1 − p2 0
Z t
Z i∞
1
b
−
dτ e−K(p)(t−τ )
eξτ fb(ξ, φ1 )dξ .
2πi 0
−i∞
(2.10)
Now we consider I1 in the representation (2.8). Changing the order of integration,
J2 =
1
α
2
1 − p 2πi
= e−K(p)t
t
Z
dτ e−K(p)(t−τ )
0
Z
α
1
1 − p2 2πi
i∞
−i∞
Z
i∞
eξτ
u
b0 (φ1 )
dξ
ξ+α
−i∞
(ξ+K(p))t
u
b0 (φ1 ) e
−1
dξ.
ξ + α K(p) + ξ
Since φ1 is analytic in right-half complex plane and Re φ1 > 0, Re K(p) > 0 for all
Re p = 0, Re ξ > 0 via Cauchy Theorem we obtain
Z
i∞
−i∞
u
b0 (φ1 )
1
dξ = 0.
ξ + α K(p) + ξ
Therefore,
J2 =
α
1
1 − p2 2πi
Z
i∞
−i∞
u
b0 (φ1 )
eξt
dξ.
ξ + α K(p) + ξ
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E. I. KAIKINA
EJDE-2007/109
Taking inverse Laplace transform with respect to space variable we obtain
Z i∞
Z i∞
α
u
b0 (φ1 )
1
eξt
px
dpe
L−1
(J
)
=
dξ
2
x
(2πi)2 −i∞
1 − p2 −i∞ ξ + α K(p) + ξ
Z i∞
Z
α
b0 (φ1 ) i∞
1
ξt u
dξe
dpepx
=
(2πi)2 −i∞
ξ + α −i∞
(K(p) + ξ)(1 − p2 )
Z i∞
Z
α
b0 (φ1 ) i∞
1
ξt u
dξe
dpepx
.
=
(2πi)2 −i∞
ξ + α −i∞
−αp2 + ξ(1 − p2 )
Since
Z
i∞
dpepx
−i∞
Z
i∞
1
(p − φ1 )(p − φ2 )
−i∞
1
1
= −2πieφ2 x
,
α + ξ φ2 − φ1
1
1
=−
2
−(α + ξ)p + ξ
α+ξ
dpepx
using φ1 + φ2 = 0, we obtain
L−1
x (J2 ) = −
α
2πi
Z
i∞
−i∞
dξeξt+φ2 x
1
u
b0 (−φ2 ).
2(ξ + α)2 φ2
We have
1
ξ 0
α
1
(
) =
.
2φ2 ξ + α
2φ2 (ξ + α)2
Making the change of the variables p = φ2 we obtain
Z i∞
1
dξeξt+φ2 x φ02 u
b0 (−φ2 )
L−1
(J
)
=
−
2
x
2πi −i∞
Z i∞
1
=
dpe−K(p)t+px u
b0 (−p)
2πi −i∞
αt 2
= e−αt L−1
b0 (−p) .
e 1+|p| u
x
φ02 =
Substituting this representation into (2.9) we get
L−1
x (I1 ) = G(t)u0 ,
where the Green operator G(t) is given by
αt 2
e 1+|p| (φ̂(p) − φ̂(−p)) .
G(t)φ = e−αt L−1
x
Now we consider I2 in the formula (2.10). In the same way we get
Z t Z +∞
−K(p)(t−τ ) 1
L−1
(I
)
=
dτ
dyf (τ, y)L−1
e
(e−py − epy ) .
2
x
x
2
1−p
0
0
We have
−K(p)(t−τ ) 1
e
L−1
(e−py − epy )
x
1 − p2
Z +∞
n
o
−K(p)(t−τ )
=
dzL−1
L−1
z
x−z e
−∞
1
(e−py − epy ) .
1 − p2
Taking into account
L−1
z
1
1
(e−py − epy ) = (e−|z−y| − e−|z+y| ),
1 − p2
2
(2.11)
(2.12)
(2.13)
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
7
we get
−K(p)(t−τ ) 1
L−1
e
(e−py − epy )
x
2
1−p
Z +∞
o
n
−K(p)(t−τ ) 1 −|z−y|
=
dzL−1
(e
− e−|z+y| )
x−z e
2
−∞
Z +∞
−K(p)(t−τ ) −K(p)(t−τ ) 1 −|z−y|
=
(L−1
− e−|z+y| )dz.
− L−1
) (e
x−z e
x−z e
2
0
Substituting into (2.13) we obtain
L−1
x (I2 )
Z
t
G(t − τ )Bf (τ )dτ,
=
(2.14)
0
where operator G was defined into (2.12) and operator B is defined as
Z
1 +∞ −|x−y|
(e
− e−|x+y| )φ(y)dy.
Bφ =
2 0
Also by direct calculation we can obtain another representation of operator B
B = (1 − ∂x2 )−1 .
Indeed, we put Bw = u. Taking Laplace transform with respect to space variable
and using u(t, 0) = 0 we get
u
b(t, ξ) =
1
(w(t,
b ξ) − ux (0, t)).
1 − ξ2
(2.15)
To satisfy limx→∞ u(t, x) = 0 we need to put the following condition
resξ=1 eξx (w(t,
b ξ) − ux (0, t)) = 0.
Therefore we obtain
ux (0, t) = w(t,
b 1).
(2.16)
Substituting (2.16) into (2.15) we find
Z +∞
Bw =
B(x, y)w(y)dy,
0
where
1 −|x−y|
(e
− e−|x+y| ).
2
Note that B(x, y) ≥ 0 for any x ≥ 0, y ≥ 0. From (2.8),(2.11) and (2.14) we obtain
integral formula for solution of (2.1)
Z t
u(t, x) = G(t)u0 +
G(t − τ )Bf (τ )dτ.
(2.17)
B(x, y) =
0
We can easily see that
kBφkLr ≤ CkφkLr
(2.18)
kBφkL1,b ≤ CkφkL1,b
(2.19)
for all 1 ≤ r ≤ ∞ and
for any b ≥ 0.
We first collect some preliminary estimates of the Green operator G(t) in the
norms kφkLp and kφkL1,1+w , where w ∈ (0, 1), 1 ≤ r ≤ ∞.
8
E. I. KAIKINA
EJDE-2007/109
2.2. Preliminary lemmas. We introduce the operator G0 (t) given by
Z +∞
G1 (t, x, y)φ(y)dy,
G0 (t)φ =
0
where the kernel is
(x−y)2
(x+y)2
G1 (t, x, y) = (4παt)−1/2 e− 4αt − e− 4αt .
We first prepare some preliminary estimates of the operator G0 (t) in the Lebesgue
norms kφkLq and kφkL1,a = kh·ia φkL1 , where a ≥ 0, 1 ≤ q ≤ ∞. Denote
G0 (t, x) = ∂y G1 (t, x, y)|y=0 = (4παt)−1/2
x − x2
e 4αt .
αt
Lemma 2.1. Let φ ∈ Lr (R+ ). Then
1
1
1
kG0 (t)φkLq ≤ Chti 2 ( q − r ) kφkLr ,
for all t > 0, 1 ≤ q ≤ ∞, 1 ≤ r ≤ ∞. Furthermore we assume that φ ∈ L1,1+a ,
then the estimate
1
k(·)b (G0 (t)φ − ϑG0 (t))kLq ≤ Ct−1+ 2q +
b−a
2
kφkL1,1+a
is valid for all t > 0, where 1 ≤ q ≤ ∞, b ∈ [0, 1 + a] and
Z +∞
ϑ=
xφ(x)dx.
0
Proof. Since
C
|G1 (t, x, y)| ≤ Ct−1/2 e− t |x−y|
2
for all x, y ∈ R+ , by the Young inequality we have for p = qr+r−q
qr
Z +∞
2
C
kG0 (t)φkLq ≤ Ct−1/2 k
e− t |x−y| φ(y)dykLq
≤ Ct
−1/2
0
2
−C
t |x|
ke
1
1
1
kLp kφkLr ≤ Chti 2 ( q − r ) kφkLr
for all t > 0, where 1 ≤ q ≤ ∞, Hence the first estimate of the lemma follows. For
the second estimate we write
Z +∞
xb (G0 (t)φ − ϑG0 (t, x)) =
xb (G1 (t, x, y) − G0 (t, x)y)φ(y)dy
0
for any b ∈ [0, 1 + a]. Applying Taylor expansion, we obtain
2
C
a
C
2
|G1 (t, x, y) − G0 (t, x)y| ≤ Ct−1− 2 y 1+a (e− t |x−y| + e− t |x| )
for all x, y ∈ R+ . Hence in the domain y ≤
x
2
a
C
2
C
2
xb |G1 (t, x, y) − G0 (t, x)y| ≤ Ct−1− 2 y a+1 xb e− t |x|
≤ Ct−1+
b−a
2
y a+1 e− t |x| .
By the Lagrange finite differences Theorem we have
|G1 (t, x, y)| ≤ Ct−
1+ν
2
C
y ν e− t |x−y|
2
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
9
for all x, y ∈ R+ , where ν ∈ [0, 1]. Taking ν = 1 + a − b, in the case b ∈ [1, a + 1]
we get for y ≥ x2
xb |G1 (t, x, y) − G0 (t, x)y| ≤ xb (|G1 (t, x, y)| + |G0 (t, x)y|)
≤ Ct−1+
b−a
2
xb y a+1−b e− t |x−y| + Ct− 2 xb+1 ye− t |x|
≤ Ct−1+
b−a
2
y a+1 (e− t |x−y| + e− t |x| ).
2
C
2
C
C
3
C
2
2
In the case b ∈ [0, 1], we write
xb |G1 (t, x, y) − G0 (t, x)y|
≤ xb (|G1 (t, x, y)| + |G0 (t, x)y|)b |G1 (t, x, y) − G0 (t, x)y|1−b
2
C
a
C
≤ Ct−b |y|(1+a)b t−(1+ 2 )(1−b) |y|(a+1)(1−b) e− t |x−y| + e− t |x|
≤ Ct−1+
b−a
2
C
2
C
2
2
y 1+a (e− t |x−y| + e− t |x| ),
for all x, y ∈ R+ , y ≥ x2 . Thus we obtain the estimate
C
b−a
2
2
C
xb |G1 (t, x, y) − G0 (t, x)y| ≤ Ct−1+ 2 |y|a+1 e− t |x−y| + e− t |x|
for all x, y ∈ R+ , and for any b ∈ [0, 1 + a]. Applying the above estimate with
Young inequality we find
k(·)b (G0 (t)φ − ϑG0 (t))kLp
Z +∞
=k
xb (G1 (t, x, y) − G0 (t, x)y)φ(y)dykLqx
0
Z +∞
2
2
C
C
−1+ b−a
2
≤ Ct
k
(e− t |x−y| + e− t |x| )y 1+a |φ(y)|dykLqx
0
≤ Ct
1
+ b−a
−1+ 2q
2
kφkL1,1+a .
Thus the second estimate follows and the lemma is proved.
Denote by G(t) the expression
αt 2
G(t)φ = e−αt L−1
e 1+|p| (φ̂(p) − φ̂(−p)) .
x
Lemma 2.2. Suppose that the function φ ∈ L∞ (R+ )∩L1,1+a (R+ ), where a ∈ (0, 1).
Then the estimates
1
1
1
kG(t)φkLr ≤ e−at kφkLr + Chti− 2 ( r1 − r ) kφkLr1 ,
a
kG(t)φ − ϑG0 (t)kL∞ ≤ Ct−1− 2 kφkL1,1+a + e−αt kφkL∞ ,
1
k(·)b (G(t)φ − ϑG0 )kL1 ≤ Ct− 2 +
b−a
2
kφkL1,1+a + e−αt k(·)b φkL1
are valid for all t > 0, where 1 ≤ r ≤ ∞.
Proof. Note that the Green operator G(t) can be represented as
G(t)φ = G0 (t)φ + e−αt φ(x) + R(t)φ,
where the remainder is
Z
R(t)φ =
+∞
(R(t, x − y) − R(t, x + y))φ(y)dy
0
(2.20)
10
E. I. KAIKINA
EJDE-2007/109
with kernel
1
R(t, x) =
2πi
αtp2
Z
i∞
b p)dp,
epx R(t,
−i∞
2
b p) = e 1−p2 − eαtp − e−αt . From Lemma 2.1 the operator G0 (t) satisfies
where R(t,
the estimates of the lemma . Also it is easily to see that the term e−αt φ(x) satisfies
the estimate of the lemma.
Now we estimate the remainder R(t). We represent
p4 αtp2
b p) = e 1−p2 1 − e−αt 1−p2 − e−αt
R(t,
for all |p| ≤ 1, and
αt
b p) = −eαtp2 + e−αt e 1−p2 − 1
R(t,
for all |p| ≥ 1. Then we see that
j
α
2
b p)| ≤ Chti 2 −1 e 2 tp + Chti2 e−αt (1 − p2 )−3
|∂pj R(t,
for all Re p = 0, t > 0, 0 ≤ j ≤ 4. Therefore,
1
|R(t, x)| ≤ Chxhti−1/2 i−4 hti− 2 −1 + Chxi−4 hti2 e−αt
1
1
≤ Chxhti− 2 i−4 hti− 2 −1
for all x ∈ R, t > 0. Applying this estimate by the Young inequality we find
1
1
1
kR(t)φkLr ≤ Chti− 2 ( q − r )−1 kφkLq
for all 1 ≤ q ≤ r ≤ ∞ and
w
kR(t)φkL1,w ≤ Chti−1 (hti 2 kφkL1 + kφkL1,w )
for all t > 0. Now by representation (2.20) the estimates of the lemma follow.
Lemma 2.2 is proved.
In the next lemma we estimate the Green operator in our basic norm
kφkX = sup(htikφ(t)kL∞ + hti−a/2 kφ(t)kL1,1+a ),
t>0
where a ∈ (0, 1). Note that the L1 -norm is estimated by the norm in X,
Z hti
Z +∞
1
hti 2 kφ(t)kL1 =
|φ(t, x)|dx +
|1 + x|−1−α |x|1+α |φ(t, x)|dx
0
hti
a
≤ Chtikφ(t)kL∞ + Chti− 2 kφ(t)kL1,a ≤ CkφkX .
We define
g(t) = 1 + κ loghti
with κ > 0.
R +∞
Lemma 2.3. Let the function f (t, x) have a zero first moment 0 xf (t, x)dx = 0.
Then the following inequality
Z t
kg(t)
g −1 (τ )G(t − τ )Bf (τ )dτ kX ≤ Ckhtif kX
0
is valid, provided that the right-hand side is finite.
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
11
Proof. Since g −1 (t) < C in view of Lemma 2.2 we get
Z t
Z t
−1
g −1 (τ )G(t − τ )f (τ )dτ kL1,1+a
g (τ )G(t − τ )f (τ )dτ kL∞ + k
k
0
0
≤ Ckhtif kX
for all 0 ≤ t ≤ 4. We now consider t > 4. From definition of function g(t) we have
Cg −1 (t) ≥ hti−a/4 ,
κ
sup
g −1 (τ ) ≤ C(1 + loghti) ≤ Cg −1 (t).
√
2
τ ∈[ t,t]
Therefore by Lemma 2.2 and (2.18)-(2.19) we obtain
Z t
G(t − τ )Bf (τ )dτ kL∞
k
0
√
Z
t
a
(t − τ )−1− 2 (kBf (τ )kL∞ + kBf (τ )kL1,1+a )dτ
≤C
0
+ Cg −1 (t)
Z
+ Cg −1 (t)
Z
t/2
a
(t − τ )−1− 2 (kBf (τ )kL∞ + kBf (τ )kL1,1+a )dτ
√
t
t
kBf (τ )kL∞ dτ,
t
2
hence using the definition of the norm in X, we get
Z t
k
g −1 (τ )G(t − τ )Bf (τ )dτ kL∞
0
√
Z
t
a
a
(t − τ )−1− 2 hτ i 2 −1 dτ
≤ Ckhtif kX
0
+ Ckhtif kX g −1 (t)
Z
t/2
√
a
a
(t − τ )−1− 2 hτ i 2 −1 dτ + Ckhtif kX g −1 (t)
Z
t
hτ i−2 dτ
t
2
t
a
≤ C(t−1− 4 + t−1 g −1 (t))khtif kX ≤ Ct−1 g −1 (t)khtif kX
and similarly
Z
t
g −1 (τ )G(t − τ )f (τ )dτ kL1,1+a
Z t
≤C
g −1 (τ )kf (τ )kL1,1+a dτ
k
0
0
√
Z
≤ Ckhtif kX
t
τ
a
2 −1
dτ + Ckhtif kX g
0
a
−1
Z
(t)
t
√
a
τ 2 −1 dτ
t
a
≤ Cε(t 4 + g −1 ta/2 )khtif kX ≤ Ct 2 g −1 khtif kX
for all t > 4. Hence the result of the lemma follows. Lemma 2.3 is proved.
We now prove the local existence of weak solutions to the initial boundary-value
problem (1.1).
Proposition 2.4. Let u0 ∈ L1,1+a (R+ ) ∩ L∞ (R+ ), a ≥ 0. Then for some T > 0,
there exists a unique solution u ∈ C([0, T ]; L1,1+a (R+ ) ∩ L∞ (R+ )) to the problem
(1.1).
12
E. I. KAIKINA
EJDE-2007/109
Proof. We apply the contraction mapping principle. We choose a functional space
Z = φ ∈ C([0, T ]; L1,1+a (R+ ) ∩ L∞ (R+ )) : ||u||Z < ∞ ,
with the norm
kukZ = sup (ku(t)kL1,1+a + ku(t)kL∞ ).
t∈[0,T ]
For v ∈ Z we define the mapping M(v) by
Z t
M(v) = G(t)u0 + λ
dτ G(t − τ )B|v|σ v(τ ).
(2.21)
0
Suppose that the norm kvkZ < δ. We first prove that
kM(v)kZ < δ.
Applying the Lemma2.2 we have
Z
t
kM(v)kL∞ ≤ kG(t)u0 kL∞ + |λ|
kG(t − τ )B|v|v(τ )kL∞ dτ
0
Z t
≤ Cku0 kL∞ + C
kB|v|v(τ )kL∞ dτ.
(2.22)
0
Similarly we obtain the estimate
kM(v)kL1,1+a ≤ kG(t)u0 − ϑ1 G0 kL1,1+a + kG0 kL1,1+a ku0 kL1 + e−αt ku0 kL1,1+a
Z t
+ |λ|
kG(t − τ )B|v|v(τ ) − ϑ2 G0 kL1,1+a dτ
0
Z t
+ |λ|
kF (t − τ )kL1,1+a kB|v|v(τ )kL1 dτ
0
Z t
+ |λ|
e−α(t−τ ) kB|v|v(τ )kL1,1+a dτ
0
Z t
≤ Cku0 kL1,1+a + C
kB|v|v(τ )kL1,1+a dτ.
0
(2.23)
Using the estimate
kB|v|v(τ )kZ ≤ kv(τ )kL∞ kv(τ )kZ ≤ δ 2 ,
From (2.22) and (2.23), we obtain
kM(v)kZ ≤ Cku0 kZ + CT δ 2 ≤ δ
if we choose δ ≥ 2Cku0 kZ and T > 0 sufficiently small. Therefore, the mapping M
transforms a ball of a radius δ > 0 into itself in the space Z. We now estimate the
difference
1
kM(v1 ) − M(v2 )kZ ≤ kv1 − v2 kZ .
2
We have
Z T
||B(|v1 |v1 (τ ) − |v2 |v2 (τ ))||Z ht − τ ia/2 dτ.
kM(v1 ) − M(v2 )kZ ≤ C
0
Since
||B(|v1 |v1 (τ ) − |v2 |v2 (τ ))||Z ≤ Cδ||v1 − v2 ||Z ,
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
13
it follows that
kM(v1 ) − M(v2 )kZ ≤ CT δkv1 − v2 kZ ≤
1
kv1 − v2 kZ
2
if we choose T > 0 sufficiently small. Thus M is a contraction mapping, therefore
there exists a unique solution u(t, x) ∈ C([0, T ]; L1,1+a (R+ ) ∩ L∞ (R+ )) to the
problem (1.1). Proposition 2.4 is proved.
3. Proof of Theorem 1.1 (Small data)
As in reference [20] by making a change of the dependent variable u(t, x) =
v(t, x)e−ϕ(t) , for the new function v(t, x) we get the equation
∂t (v − vxx ) − αvxx − λe−ϕ |v|v − ϕ0 v = 0.
We assume that ϕ(t) is such that ϕ(0) = 1 and
Z +∞
x(λe−ϕ |v|v + ϕ0 v)dx = 0.
0
Since by construction
Z
+∞
u(x)dx = 0
0
the first moment of new function v(t, x) satisfies a conservation law:
Z
d +∞
xv(t, x)dx = 0,
dt 0
R +∞
R +∞
hence 0 xv(t, x)dx = 0 xu0 (x)dx for all t > 0. Thus we consider the initialboundary value problem for the new dependent variables (v(t, x), ϕ(t)),
Z
1 +∞
x|v|vdx v,
∂t (v − ∆v) − α∆v = λe−ϕ |v| −
θ 0
Z +∞
(3.1)
λ
∂t ϕ(t) = − e−ϕ
x|v|vdx,
θ
0
v(0, x) = u0 (x), v(t, 0) = 0, ϕ(0) = 0.
We denote h(t) = eϕ(t) and write (3.1) as
∂t (v − vxx ) − αvxx = f (v, h)
v(0, x) = u0 (x), v(t, 0) = 0
Z
λ +∞
∂t h = −
x|v|vdx, h(0) = 1,
θ 0
where
f (v, h) = λh
−1
1
|v| −
θ
Z
+∞
x|v|vdx v.
0
We note that the first moment of the nonlinearity is
Z +∞
xf (v, h)(t, x)dx = 0
0
(3.2)
14
E. I. KAIKINA
EJDE-2007/109
for all t > 0. We now prove the existence of the solution (v(t, x), h(t)) for the
problem (3.2) by the successive approximations (vm (t, x), hm (t)), m = 1, 2, . . . ,
defined as follows
∂t (vm − ∂x2 vm ) − α∂x2 vm = f (vm−1 , hm−1 ),
Z
λ +∞
|vm−1 |vm−1 dx,
∂t hm = −
θ 0
vm (0, x) = u0 (x), vm (t, 0) = 0, hm (0) = 1,
(3.3)
for all m ≥ 2, where v1 = G(t)u0 , h1 = g(t), g(t) = 1 + |θ|η loghti.
We now prove by induction the following estimates
kvm kX ≤ Cε,
kvm (t) − G(t)u0 kL1,1 ≤ Cε2 g −1 (t),
(3.4)
|hm (t) − g(t)| ≤ Cε(1 + log g(t))
for all m ≥ 1, the norm k · kX is defined as above by
kφkX = sup(htikφ(t)kL∞ + hti−a/2 kφ(t)kL1,1+a ).
t>0
By virtue of Lemma 2.2 we have
kG(t)u0 kL∞ ≤ Cεhti−1 ,
kG(t)u0 kL1,1 ≤ Cε,
a
k| · | (G(t)u0 − θG0 (t, x))kL1,1 ≤ Cε,
k| · |a G0 (t, x)kL1,1 ≤ Cta/2 .
Therefore, estimates (3.4) are valid for m = 1. We assume that estimates (3.4) are
true with m replaced by m − 1. The integral equations associated with (3.3) are
written as
Z t
vm (t) = G(t)u0 +
G(t − τ )Bf (vm−1 (τ ), hm−1 (τ ))dτ,
0
Z
Z +∞
λ t
hm (t) = 1 −
dτ
|vm−1 |vm−1 dx.
θ 0
0
We have
f (v, h) = λh
−1
1
|v| −
θ
Z
+∞
x|v|vdx v.
0
1
1+1
1,1
kf (vm−1 (t), hm−1 (t))kL∞ ≤ Ch−1
(t)kv
(t)k
1
+
kv
(t)k
∞
m−1
m−1
L
m−1
L
|θ|
(3.5)
2
−2 −1
≤ Cε hti g (t)
and
kf (vm−1 (t), hm−1 (t))kL1,1+a
1
≤ Ch−1
kvm−1 (t)kL1,1
m−1 (t)kvm−1 (t)kL∞ kvm−1 (t)kL1,1+a 1 +
|θ|
−1+ a
2
≤ Cε2 hti
g −1 (t)
(3.6)
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
15
for all t > 0, provided that (vm−1 (t), hm−1 (t)) satisfies (3.4). This yields the
estimate
khtig(t)f (vm−1 (t), hm−1 (t))kX ≤ Cε2 .
Since f (vm−1 (τ ), hm−1 (τ )) have the zero first moment we get via Lemma 2.3
Z t
g −1 (τ )G(t − τ )Bf (vm−1 (τ ), hm−1 (τ ))dτ kX ≤ Cε2
kg(t)
0
hence it follows that
kvm kX ≤ Cε, kvm (t) − G(t)u0 kL1,1 ≤ Cε2 g −1 (t).
(3.7)
To prove the third estimate in (3.4) we need the following lemma, where we
evaluate the large time behavior of the first moment
R +∞of the nonlinearity in equation
(1.1) in the critical case. As above we take θ = 0 xu0 (x)dx.
Lemma 3.1. Assume that u0 ∈ L∞ (R+ )∩L1,1+a (R+ ), that ku0 kL∞ +ku0 kL1,1+a =
ε is sufficiently small, and that θλ ≤ −Cε < 0. Let a function v(t, x) satisfy the
estimates
htikvkL∞ + kvkL1,1 ≤ Cε,
kv(t) − G(t)u0 kL1,1 ≤ Cε2 g −1 (t)
for all t > 0. Then the inequality
Z t Z +∞
1 − λ
dτ
x|v|v(τ, x)dx − g(t) ≤ Cε(1 + log g(t))
θ 0
0
(3.8)
is valid for all t > 0.
Proof. In view of the condition kvkL∞ +kvkL1,1 ≤ Cε we get
Z
Z +∞
λ t
|
dτ
x|v|v(τ, x)dx| ≤ Cεt,
θ 0
0
hence estimate (3.8) is true for all 0 < t < 1.
We now consider the case t ≥ 1. By the last estimate of Lemma 2.2 we get
kx(G(t)u0 − θG0 (t, x))kL1 ≤ Cεt−a/2 .
Hence we find
kx(|v|v − |θ|θ(G0 (t, x))1+1 )kL1
≤ C(kx(v(t) − G(t)u0 )kL1 + kx(G(t)u0 − θG0 (t, x))kL1 )
× (kvkL∞ + kG(t)u0 kL∞ + |θ|kG0 (t)kL∞ )
a
≤ Cε2 t−1 (εg −1 (t) + t− 2 )
for all t ≥ 1. Since
Z
t
0
+∞
x(G0 (t, x))2 dx =
η
,
|λ|
it follows that
Z
+∞
η x|v|v(t, x)dx − |θ|θt−1 ≤ Ckx(|v|v − |θ|θ(G0 (t, x))2 )kL1
|λ|
0
a
≤ Cε2 t−1 (εg −1 (t) + t− 2 )
16
E. I. KAIKINA
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for all t ≥ 1. Therefore,
Z
Z +∞
|λ| t
|
dτ
x|v|v(τ, x)dx − |θ|η log t|
θ 1
0
Z t
Z t
a
Cε2 dτ
≤
+ Cε
τ −1− 2 dτ
1 τ (1 + |θ|η log(1 + τ ))
1
≤ Cε(1 + log g(t))
for all t ≥ 1. Thus we obtain (3.8) and complete the proof.
By virtue of (3.7) and applying Lemma 3.1 we find that
|hm (t) − g(t)| ≤ Cε(1 + log g(t))
for all t > 0. Thus by induction we see that estimates (3.4) are valid for all m ≥ 1.
In the same way by induction we can prove that
1
kvm − vm−1 kX ≤ kvm−1 − vm−2 kX ,
4
sup g −1 (t)|hm (t) − hm−1 (t)|
t>0
≤
1
1
kvm−1 − vm−2 kX + sup g −1 (t)|hm−1 (t) − hm−2 (t)|
4
4 t>0
for all m > 2. Therefore taking the limit m → ∞, we obtain a unique solution
limm→∞ vm (t, x) = v(t, x) ∈ X, limm→∞ hm (t) = h(t) = eϕ(t) ∈ C(0, ∞) satisfying
the equalities
Z t
v(t) = G(t)u0 +
G(t − τ )Bf (v(τ ), h(τ ))dτ,
0
(3.9)
Z
Z +∞
λ t
h(t) = 1 −
dτ
x|v|vdx,
θ 0
0
and the estimates
kv(t) − G(t)u0 kL1,1 ≤ Cε2 g −1 (t),
|hm (t) − g(t)| ≤ Cε(1 + log g(t)).
(3.10)
Applying (3.5) and (3.6) to (3.9), we have
kv(t) − G(t)u0 kL∞ ≤ Cε2 hti−1 g −1 (t).
(3.11)
Then via formulas u(t, x) = e−ϕ(t) v(t, x) = h−1 (t)v(t, x) we find the estimates
ku(t) − θG0 (t, x)e−ϕ(t) kL∞
≤ ku(t) − (G(t)u0 )e−ϕ(t) kL∞ + k(G(t)u0 − θG0 (t, x))e−ϕ(t) kL∞
2
−1 −1
≤ Cε hti
g
(3.12)
(t),
where we used the estimate
a
k(G(t)u0 − θG0 (t, x))e−ϕ(t) kL∞ ≤ Ct−1− 2 ku0 kL1,1+a
and (3.11). By (3.10), we have
kθG0 (t)h−1 (t) − θG0 (t, x)g −1 (t)kL∞ ≤ Cεt−1 g −2 (t)|h(t) − g(t)|,
hence via (3.12) it follows that
ku(t) − θG0 (t, x)g −1 (t)kL∞ ≤ Cε2 hti−1 g −2 (t).
(3.13)
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
17
This completes the proof of Theorem 1.1.
4. Proof of Theorem 1.2 (Large data)
Before proving Theorem 1.2 we need a lemma, where we compare the solutions
of the following two problems
ut − uxx + |u|u = f,
u(0, x) = u0 (x),
and
x ∈ R+ , t > 0,
x ∈ R+ , t > 0
u(t, 0) = 0,
vt − vxx + v 2 = |f |,
v(0, x) = |u0 (x)|,
x ∈ R+ , t > 0,
x ∈ R+ , t > 0
v(t, 0) = 0,
(4.1)
(4.2)
Lemma 4.1. Suppose that u0 ∈ L∞ (R+ )∩C0 (R+ ), and 0 ≤ ≤ 1. Then |u(t, x)| ≤
v(t, x) for all t ≥ 0, x ∈ R+ .
Proof. Define r = v − u. Then we obtain
rt − rxx + v 2 − |u|u = |f | − f,
r(0, x) = |u0 (x)| − u0 (x),
x ∈ R+ , t > 0,
r(t, 0) = 0.
(4.3)
We need to prove that r ≥ 0 for all t ≥ 0, x ∈ R+ . Define R(t) ≡ inf x∈R+ r(t, x).
On the contrary, suppose that there exists a time T > 0 such that R(T ) < 0. By
the continuity we can find an interval [T1 , T ] such that R(t) ≤ 0 for all t ∈ [T1 , T ]
and R(T1 ) = 0. By [5, Theorem 2.1] there exists a point ζ(t) ∈ R+ such that
d
R(t) = r(t, ζ(t)), moreover R0 (t) = dt
r(t, ζ(t)) almost everywhere on t ∈ [T1 , T ].
We have
|u|u − v 2 = (v − R)2 − v 2 ≥ 0
for all t ∈ [T1 , T ]. For the Laplacian ∂x2 at the point of maximum ζ(t) we have
−∂x2 r(t, ζ(t)) ≤ 0.
Therefore by equation (4.3) we get R0 (t) ≥ 0 for all t ∈ [T1 , T ]. Integration with
respect to time yields R(t) ≥ 0. This gives a contradiction, hence u(t, x) ≤ |v(t, x)|
for all x ∈ R+ and t > T1 . In the same manner we prove that v + u ≥ 0 for all
x ∈ Rn and t > T1 . Lemma 4.1 is proved.
2
Lemma 4.2. Let u0 ∈ W∞
(R+ ) ∩ W12 (R+ ). Then we have the following estimate
for solution of initial-boundary value problem (1.1):
kukL2 ≤ Chti−3/4 ,
kxukL2 ≤ Chti−1/4
Proof. Multiplying equation (1.1) by 2u and integrating with respect to x ∈ R+ we
get
d
(ku(t)k2L2 + kux (t)k2L2 ) + 2αkux (t)k2L2 = 2λku(t)kσ+2
Lσ+2 ,
dt
hence integrating we see that
Z t
Z t
ku(τ )kσ+2
ku(t)k2L2 + kux (t)k2L2 + 2α
kux (τ )k2L2 dτ − 2λ
Lσ+2 dτ
0
≤ ku0 k2L2 + ku0x k2L2
= ku0 k2H1
0
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E. I. KAIKINA
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for all t ≥ 0. In particular we have
kuk∞,2 ≡ sup ku(t)kL2 ≤ ku0 kH1 ,
(4.4)
t≥0
kuk3,3 ≡ kku(t, x)kL3x kL3t (0,∞) ≤ Cku0 kH1
(4.5)
From equation (1.1) we have
uxx (t, 0) = 0.
Differentiating (1.1), multiplying by 2ux and integrating with respect to x ∈ R+
we get
Z +∞
d
(kux (t)k2L2 + kuxx (t)k2L2 ) + 2αkuxx (t)k2L2 = 2λ
|u|u2x dx,
dt
0
hence integrating we see that
ku(t)k2L2 + kux (t)k2L2 + 2α
Z
t
kux (τ )k2L2 dτ − 2λ
0
Z
+∞
|u|u2x dx
0
≤ ku0x k2L2 + ku0xx k2L2 = ku0 k2H2
for all t ≥ 0. In particular we obtain that the solution u(t, x) ∈ C([0, +∞), C1 (R+ )).
Now we prove the estimates
ku(t)kL2 ≤ Chti−3/4
(4.6)
for all t > 0. Denote Θ(x) = 1 for all x > 0 and Θ(x) = −1 for all x < 0; Θ(0) = 0.
We multiply equation (1.1) by xΘ(u(t, x)) and integrate with respect to x over R+
to get
Z +∞
Z +∞
∂t
xu(t, x)Θ(u(t, x))dx −
xuxx (t, x)Θ(u(t, x))dx
0
0
Z +∞
(4.7)
−α
xuxx (t, x)Θ(u(t, x))dx
0
Z +∞
=λ
x|u|2 dx.
0
Since u(t, x) ∈ C([0, +∞), C1 (R+ )) we get
Z +∞
xuxx (t, x)Θ(u(t, x))dx = 0.
(4.8)
0
Also we have
Z
0
+∞
d
∂t xu(t, x)Θ(u(t, x))dx = kxu(t)kL1 ,
dt
Z +∞
λ
x|u|2 dx ≤ 0.
(4.9)
0
Therefore by (4.7), (4.9) and (4.8) we find
d
kxukL1 ≤ 0.
dt
Integration of inequality (4.10) yields
kxukL1 ≤ C.
(4.10)
(4.11)
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
19
Now we can prove the estimate (4.6) for all t > 0. Indeed, using (4.11) in particular,
we find
sup |b
uξ (t, ξ)| ≤ Ckxu(t)kL1 ≤ C.
(4.12)
ξ∈R
We now multiply equation (1.1) by 2u, then integrating with respect to x ∈ R+ we
get
d
(4.13)
ku(t)k2L2 + kux (t)k2L2 = −2αkux (t)k2L2 + λku(t)k3L3 .
dt
By construction we have
Z +∞
u
b(t, 0) =
u(t, x)dx = 0
0
By the Plansherel theorem using the Fourier splitting method due to [29], we have
kux (t)k2L2 = kξb
u(t)k2L2
Z
Z
≥
|b
u(t, ξ)|2 |ξ|2 dξ ≥ δ 2
=δ
|ξ|≥δ
Z +∞
2
2
|b
u(t, ξ)| dξ − δ
2
−∞
= δ2
Z
|b
u(t, ξ)|2 dξ
|ξ|≥δ
Z
|ξ|<δ
+∞
|b
u(t, ξ)|2 dξ − δ 2
|b
u(t, ξ) − u
b(t, 0)|2 dξ
Z
Z
|
|ξ|<δ
0
0
ξ
u
bξ1 (t, ξ1 )dξ1 |2 dξ
≥ δ 2 ku(t)k2L2 − 2δ 5 sup |b
uξ (t, ξ)|2 ,
|ξ|≤δ
where δ > 0. Thus from (4.13) we have the inequality
d
ku(t)k2H1 ≤ −αδ 2 ku(t)k2H1 + 4αδ 5 sup |b
uξ (t, ξ)|2 .
dt
|ξ|≤δ
(4.14)
We choose αδ 2 = 2(1 + t)−1 and change ku(t)k2H1 = (1 + t)−2 W (t). Then via (4.13)
we get from (4.14)
d
W (t) ≤ C(1 + t)−1/2 .
(4.15)
dt
Integration of (4.15) with respect to time yields
1
W (t) ≤ ku0 k2H1 + C (1 + t)− 2 +1 − 1 .
Therefore we obtain a time decay estimate in the L2 -norm,
ku(t)kL2 ≤ C(1 + t)−3/4
(4.16)
for all t > 0. We now multiply equation (1.1) by x2 u, then integrating with respect
to x ∈ R+ we get
d
kxu(t)k2L2 + kxux (t)k2L2 − 2ku(t)k2L2
dt
Z +∞
2
2
x2 |u|3 dx.
= −2αkxux (t)kL2 + 2αku(t)kL2 + λ
0
(4.17)
20
E. I. KAIKINA
EJDE-2007/109
By the Plansherel theorem using the Fourier splitting method due to [29], we have
kxux (t)k2L2
=
k∂ξ ξb
u(t)k2L2
Z
Z
|b
u + ξb
uξ |2 |dξ|
−∞
∞
∞
Z
2
≥
∞
=
|b
u|2 |dξ|
|ξb
uξ | |dξ| −
Z−∞
≥
=δ
−∞
2
|ξb
uξ | dξ − ku(t)k2L2
|ξ|≥δ
Z +∞
2
Z
|b
uξ (t, ξ)|2 dξ − δ 2
−∞
|ξ|<δ
|b
uξ (t, ξ)|2 dξ − ku(t)k2L2
uξ (t, ξ)|2 − ku(t)k2L2
≥ δ 2 kxu(t)k2L2 − 2δ 3 sup |b
|ξ|≤δ
where δ > 0. Thus from (4.17) we have the inequality
d
kxu(t)k2H1 − 2ku(t)k2L2
dt
≤ −αδ 2 kxu(t)k2H1 + 2αδ 3 sup |b
uξ (t, ξ)|2 + 2(α + 1)ku(t)k2L2
(4.18)
|ξ|≤δ
We choose αδ 2 = 2(1 + t)−1 and change ku(t)k2H1 = (1 + t)−2 W (t). Then via
(4.12),(4.16) we get from (4.18)
1
d
W (t) ≤ C(1 + t) 2 .
dt
(4.19)
Integration of (4.19) with respect to time yields
1
W (t) ≤ kxu0 k2H1 + C (1 + t) 2 +1 − 1 .
Therefore, we obtain a time decay estimate of the L2 - norm
kxu(t)kL2 ≤ C(1 + t)−1/4
Lemma 4.2 is proved
By Lemmas 2.2 and 4.2, we have
Z
t
ku(t)kL∞ ≤ kG(t)u0 kL∞ + C
t
2
Z
1
ht − τ i− 2 ku(τ )k2L2 dτ
t/2
+C
ht − τ i−1 kx|u|u(τ )kL1 dτ
0
−1
≤ Chti
Z
+C
0
≤ Chti−1 loghti
t
2
ht − τ i−1 hτ i−1 dτ
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
21
for all t > 0. In the same manner
k∂x2 u(t)kL∞
≤ Chti−2 +
t
Z
k∂x G(t − τ )B|u|u(τ )kL∞ dτ
0
≤ Chti−2 + C
Z
t
2
ht − τ i−2 ku(τ )k2L2 dτ + C
Z
t
2
0
−2
≤ Chti
Z
t
2
−2
ht − τ i
+C
− 23
hτ i
Z
≤ Chti
ht − τ i−1 ku(τ )k2L∞ dτ
t
ht − τ i−1 hτ i−2 log2 hτ idτ
dτ + C
t
2
0
−2
t
3
log hti
for all t > 0. Denote f (t, x) = uxxt . Then by Lemmas 4.2 and 2.2 we have the
estimates
kf (t)kL∞
= k∂x2 ut (t)kL∞
≤ k∂x2 ∂t G(t)u0 kL∞ + k∂x2 B|u|u(t)kL∞
Z
Z t
ht − τ i−1 k∂x2 B|u|u(τ )kL∞ dτ + C
+C
t
2
t/2
ht − τ i−3 kx|u|u(τ )kL1 dτ
0
≤ Chti−3 log4 hti + C
Z
t
ht − τ i−1 hτ i−3 log4 hτ idτ + C
t
2
Z
t/2
ht − τ i−3 hτ i−1 dτ
0
≤ Chti−3 log4 hti
for all t > 0. We a sufficiently small ε > 0 and consider the following two auxiliary
problems
Ut − Uxx + U 2 = ε|f |,
U (0, x) = ε|u0 (x)|,
U (x, 0) = 0,
x ∈ R+ , t > 0,
x ∈ R+ ,
(4.20)
t>0
and
Vt − Vxx + εV 2 = |f |,
V (0, x) = |u0 (x)|,
V (x, 0) = 0,
x ∈ R+ , t > 0,
x ∈ R+ ,
(4.21)
t > 0.
Note that problem (4.21) can be reduced to problem (4.20) by the change of variable
V = ε−1 U . Also note that (4.20) has a sufficiently small initial data and a small
force ε|uxxt |. Applying results of paper we obtain an almost optimal time decay
estimate
kU (t)kL∞ ≤ Chti−1 ,
hence by Lemma 4.1 we get ku(t)kL∞ ≤ Chti−1 .
22
E. I. KAIKINA
EJDE-2007/109
Now we estimate the L1,1+a -norm of the solution
ku(t)kL1,1+a ≤ kG(t)u0 kL1,1+a
Z t
+C
(kB|u|u(τ )kL1,1+a + ht − τ ia/2 kB|u|u(τ )kL1,1 )dτ
0
Z t
Z t
a
a/2
−1
≤ Chti
+C
hτ i ku(τ )kL1,a dτ + C
ht − τ i 2 hτ i−1 dτ
0
0
Z t
hτ i−1 ku(τ )kL1,1+a dτ.
≤ Chtia/2 + C
0
Hence by Granwall’s inequality we obtain
ku(t)kL1,1+a ≤ Chtia/2
(4.22)
for all t > 0. In the same manner we estimate the L1,1+a - norm of f . By Lemma
2.2 we have
kf (t)kL1,1+a = kuxxt (t)kL1,1+a
≤ k∂x2 ∂t G(t)u0 kL1,1+a + k∂x2 B|u|u(t)kL1,1+a
Z t
+C
ht − τ i−1 (k|u|u(τ )kL1,1+a + ht − τ ia/2 k|u|u(τ )kL1,1 )dτ
0
Z t
Z t
a
a
a
≤ Chti−1+ 2 + C
ht − τ i−1 hτ i 2 −1 dτ + C
ht − τ i 2 −1 hτ i−1 dτ
0
−1+ a
2
≤ Chti
0
.
Thus we can apply the results of paper [21] to get the estimate of the functions
U (t, x) and V (t, x). Then by Lemma 4.1 we get an optimal time decay estimate for
the solution
ku(t)kL∞ ≤ Cε−1 hti−1 (log(2 + t))−1
(4.23)
for all t > 0.
We make a change of the dependent variable u(t, x) = v(t, x)e−ϕ(t) as in the proof
of Theorem 1.1. Then we obtain problem (3.1) for new functions (v(t, x), ϕ(t)). Now
we prove the following estimate
kv(t)kL1,1+a ≤ Chtia/2
for all t > 0. From (3.1) we obtain the integral formula
Z t
Z
v(τ )
v(t) = G(t)u0 + λ
dτ G(t − τ )B |u(τ )|v(τ ) −
|u(τ )|v(τ )dx .
θ
0
Rn
Using (4.23) and Lemma 2.2 we have
Z
Z
t
v(τ ) +∞
dτ G(t − τ )B(|u(τ )|v(τ ) −
|u(τ )|v(τ )dx)L1,1+a
θ
0
0
Z t
Z +∞
v(τ )
≤C
k|u(τ )|v(τ ) −
|u(τ )|v(τ )dxkL1,1+a dτ
θ
0
0
Z t
≤C
hτ i−1 (log(2 + τ ))−1 kv(τ )kL1,1+a dτ
0
(4.24)
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NONLINEAR PSEUDOPARABOLIC EQUATIONS
23
for all t > 0. Therefore, by (4.22) we find
Z t
hτ i−1 (log(2 + τ ))−1 kv(τ )kL1,1+a dτ
kv(t)kL1,1+a ≤ kG(t)v(0)kL1,1+a + C
0
Z t
a/2
hτ i−1 (log(2 + τ ))−1 kv(τ )kL1,1+a dτ
≤ Chti
+C
0
for all t > 0. Hence by Granwall’s inequality we obtain
kv(t)kL1,1+a ≤ Chtia/2
for all t > 0. In the same way from (4.23) and Lemma 2.2 we get
kv(t)kLp ≤ kG(t)v(0)kLp
Z
Z
t
v(τ ) +∞
|u(τ )|v(τ )dx)Lp
dτ G(t − τ )B(|u(τ )|v(τ ) −
+
θ
0
0
1
1
1
1
1
≤ Chti− 2 − 2 (1− p )
Z t/2
1
1
1
a
a
+C
(t − τ )− 2 − 2 (1− p )− 2 hτ i 2 −1 (log(2 + τ ))−1 dτ
0
Z t
hτ i−1 (log(2 + τ ))−1 kv(τ )kLp dτ
+C
t
2
1
≤ Chti− 2 − 2 (1− p ) +
t
Z
hτ i−1 kv(τ )kLp dτ
t
2
for all t > 0. So by Granwall’s inequality we have
1
1
1
kv(t)kLp ≤ Chti− 2 − 2 (1− p )
for all t > 0. Now from (4.24) and Lemma 2.2 we get
Z t/2
Z
a
v(τ )
kv(t) − G(t)u0 kL1,1 ≤
|u(τ )|v(τ )dxL1,1+a dτ
(t − τ )− 2 |u(τ )|v(τ ) −
θ
Rn
0
Z t
Z +∞
|u(τ )|v(τ ) − v(τ )
+
|u(τ )|v(τ )dxL1,1 dτ
t
θ
0
2
Z t/2
a
a
≤C
(t − τ )− 2 hτ i 2 −1 (log(2 + τ ))−1 dτ
0
Z t
+C
hτ i−1 (log(2 + τ ))−1 dτ
t
2
≤ C log(2 + t)−1
(4.25)
for all t > 0. Therefore using Lemma3.1 we find for h(t) = eϕ(t) ,
|h(t) − g(t)| ≤ C log g(t),
for all t > 0. Then from u(t, x) = e−ϕ(t) v(t, x) = h−1 (t)v(t, x), we have
ku(t) − θe−ϕ(t) G0 (t)kL∞
≤ ku(t) − e−ϕ(t) G(t)u0 kL∞ + kG(t)u0 − θe−ϕ(t) G0 (t)kL∞
a
≤ Ct−1 g −2 (t) + Ct−1− 2 ku0 kL1,1+a
≤ Ct−1 g −2 (t)
(4.26)
24
E. I. KAIKINA
EJDE-2007/109
for all t > 0. Also we have
kθG0 (t)(h−1 (t) − g −1 (t))kL∞ ≤ Ct−1 g −2 (t)|h(t) − g(t)|,
and so by (4.26),
ku(t) − θG0 (t)g −1 (t)kL∞ ≤ C(1 + t)−1 g −2 (t) log g(t).
Theorem 1.2 is proved.
Acknowledgements. The author would like to thank the anonymous referee for
many useful suggestions and comments. This work is partially supported by CONACyT.
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Elena I. Kaikina
Instituto de Matemáticas, Universidad Nacional Autónoma de México, Campus Morelia,
AP 61-3 (Xangari), Morelia CP 58089, Michoacán, Mexico
E-mail address: ekaikina@matmor.unam.mx