Electronic Journal of Differential Equations, Vol. 2006(2006), No. 89, pp. 1–16.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS ON A
HALF-LINE WITH LARGE INITIAL DATA
ROSA E. CARDIEL, ELENA I. KAIKINA
Abstract. We study the initial-boundary value problem for nonlinear pseudodifferential equations, on a half-line,
ut + λ|u|σ u + Lu = 0,
(x, t) ∈ R+ × R+ ,
u(x, 0) = u0 (x),
x ∈ R+ ,
where λ > 0 and pseudodifferential operator L is defined by the inverse Laplace
transform. The aim of this paper is to prove the global existence of solutions
and to find the main term of the asymptotic representation in the case of the
large initial data.
1. Introduction
We consider the initial-boundary value problem for a general class of the nonlinear nonlocal equations, on a half-line,
ut + λ|u|σ u + Lu = 0,
u(x, 0) = u0 (x),
(x, t) ∈ R+ × R+ ,
x ∈ R+ ,
(1.1)
where λ > 0, σ > 0. The linear operator L is a pseudodifferential operator defined
by the inverse Laplace transform as follows
Z i∞
u(0, t) 1
epx Cα pα û(p, t) −
dp,
(1.2)
Lu =
2πi −i∞
p
where α ∈ (1, 2) and
Z
+∞
e−px u(x)dx
û(p) =
0
denotes the Laplace transform of u. We assume that the symbol K(p) = Cα pα is
dissipative, i.e. <(K(p)) > 0 for <(p) = 0. Here and below pα is the main branch
of the complex analytic function in the half-complex plane <p ≥ 0, so that 1α = 1
(we make a cut along the negative real axis (−∞, 0)).
The initial-boundary value problem (1.1) is of great interest from the physical
point of view , since it describes many physical phenomena , such as the focusing of
laser beams , waves on water (some other applications can be found [28]) . A great
number of publications have dealt with asymptotic representations of solutions to
2000 Mathematics Subject Classification. 35Q35, 35B40.
Key words and phrases. Pseudodifferential operator; large data; asymptotic behavior.
c 2006 Texas State University - San Marcos.
Submitted February 10, 2006. Published August 9, 2006.
1
2
R. E. CARDIEL, E. I. KAIKINA
EJDE-2006/89
the Cauchy problem for nonlinear evolution equations in the last twenty years.
While not attempting to provide a complete review of this publications, we do
list some known results [3]-[10],[12] ,[13],[15] -[18], [21], [22], [26]-[35], where there
were obtained optimal time decay estimates and asymptotic formulas of solutions
to different nonlinear local and nonlocal dissipative equations. The asymptotic
theory of the initial-boundary value problems for the nonlinear pseudodifferential
equations is relatively new and traditional questions of a general theory are far
from their conclusion . A description of the large time asymptotic behavior of
solutions for the initial-boundary value problems requires new approaches and the
reorientation of the points of view compared to the Cauchy problem.
The initial-boundary value homogeneous problems for nonlinear pseudodifferential equations were studied in the book [20]. In the present paper we continue the
study of pseudodifferential equations on a half-line, considering the case of a large
initial data .
The aim of this paper is to prove a global existence of solutions to the initialboundary value problem (1.1), and to find the main term of the asymptotic representation of solutions. We will obtain the a priory optimal time decay estimates of
solutions in the usual Lebesgue spaces Lr for 1 ≤ r ≤ ∞. These type of estimates
enable us to consider the critical and sub critical cases in future works.
To state our results we give some notations. The weighted Sobolev space is
r
Hk,s
= khi∂x ik hxis f kLr < ∞},
r = {f ∈ L : kf kHk,s
r
√
where hxi = 1 + x2 .
We introduce the function Λ0 (s) ∈ L∞ ,
Λ0 (s) = (−1)
{α}
α
Cα
A1
π{α}
)
sin(
2π 2 i π − A1
α
Z
i∞
sz {α}
−q
dze z
−i∞
{α}
∞
Z
dqe
0
q− α
,
K(z) + q
(1.3)
where
A1 = (−1){α}+1 (−Cα )−
[α]
α
Γ(1 − {α})Γ({α}) sin π{α}.
We prove the following theorem.
Theorem 1.1. Let α ∈ (1, 2), σ > α ,λ > 0 and real valued function u0 ∈ H1,0
2 ∩
0,0
0,µ
H∞ ∩ H1 , where µ ∈ [0, 1]. Then there exists a unique real valued solution of
(1.1) such that
0,0
0,µ
u(x, t) ∈ C([0, ∞); H1,0
2 ∩ H∞ ∩ H1 ).
Moreover there exists a constant A such that
u(x, t) = t−1/α AΛ0 (xt−1/α ) + O(t−
as t → ∞ uniformly with respect to x > 0, where
Z +∞
Z +∞ Z
A=
u0 dy + λ
dt
0
0
1+µ
α
)
(1.4)
∞
|u|σ u dy.
0
Remark 1.2. We canR guarantee that the coefficient A 6= 0 in the asymptotic
+∞
representation (1.4) if 0 u0 dy 6= 0 and |u|σ u is small. It can occur that A = 0,
R∞
for instance, for convective equations 0 |u|σ u dy ≡ 0 if the initial data have zero
R +∞
mean value 0 u0 dy = 0. In the last case formula (1.4) gives us only some time
decay estimate for the solutions.
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
3
2. Preliminaries
We introduce the function κ(ξ) = K −1 (−ξ), such that Re κ(ξ) > 0 for all Re ξ ≥
0. We denote
Z
+∞
G(t1 )φ(t2 ) =
G(x, y, t1 )φ(y, t2 ) dy,
0
where Green function G(x, y, t) is defined by
Z i∞
1
G(x, y, t) =
e−K(p)t+p(x−y) dp
2πi −i∞
Z i∞
Z i∞
1
epx K(p)
ξt
−1 −κ(ξ)y
+ 2
e κ(ξ)ξ e
dξ dp.
4π −i∞
−i∞ p(K(p) + ξ)
(2.1)
The solution u of the problem (1.1) can be represented as follows (see [20], pp.
23-24)
Z t
dτ G(t − τ )N (u)(τ ),
(2.2)
u(x, t) = G(t)u0 +
0
where N (u) = λ|u|σ u. We introduce complete metric space
1,0
X = {φ(x, t) ∈ C([0, ∞); H0,µ
r ) ∩ C((0, ∞); Hr )|kφkX < +∞, },
1
1
µ
1
1
kφkX = suphti α (hti− αr − α kxµ φkLr + t α − αr kφx kLr ),
t>0
where µ ∈ [0, 1], 1 ≤ r ≤ ∞. Let a continuous linear functional f (φ) : L1 → R is
defined as
Z +∞
f (φ) =
φ(x)dx.
0
Now we obtain some estimates for the Green operator G in the space X.
Lemma 2.1. The following two estimates are valid
kGφkX ≤ C(kφkL1 + kφkL∞ ),
sup t
1
α (1+µ)
kGφ − t
1
−α
Λ0 (xt
−1/α
)f (φ)kL∞ ≤ CkφkL1,µ
(2.3)
(2.4)
t>1
for all t > 0, provided that the right-hand sides are bounded.
Proof. We rewrite the Green function as
G(x, y, t) = F1 (x − y, t) + F2 (x, y, t),
where
Z i∞
1
e−K(p)t+px dp
F1 (x, t) =
2πi −i∞
Z i∞
Z i∞
epx K(p)
1
ξt−k(ξ)y
−1
e
k(ξ)ξ
dξ dp.
F2 (x, y, t) =
4π 2 −i∞
−i∞ p(K(p) + ξ)
Firstly we prove the estimate
µ
1
1
1
t α − αr kF1 (q, t)kLr + t− α kq µ F1 (q, t)kLr + t α k∂q F1 (q, t)kLr ≤ C
(2.5)
(2.6)
(2.7)
for all t > 0, where µ > 0 and 1 ≤ r ≤ ∞. Changing variables pα t = z α , and
qe = qt−1/α we get
Z i∞
Z i∞
α
1
|F1 (q, t)| =
eqp e−K(p)t dp ≤ Ct−1/α
eqez e−Cα z dz .
2πi −i∞
−i∞
4
R. E. CARDIEL, E. I. KAIKINA
EJDE-2006/89
Therefore,
1
1
kF1 (q, t)kLr ≤ Ct− α + αr .
After the change pα t = z α , we have
Z
µ
−1+
|q µ F1 (q, t)| ≤ Ct α α qeµ
|∂q F1 (q, t)| ≤ Ct
2
−α
Z
i∞
α
eqez e−Cα z dz ≤ Ct
−i∞
i∞
µ
−1+
α
α
α
he
q iµ−1−γ ,
2
zeqez e−Cα z dz ≤ Ct− α he
q i−1−γ ,
−i∞
where 0 < γ < α. Therefore,
µ
kq F1 (q, t)kLr ≤ Ct
µ
−1+ + 1
α
α
αr
Z
((
1
+
0
≤ Ct
∞
Z
1
)de
q (1 + qe2 )
(µ−1−γ)r
2
)1/r
(2.8)
µ
−1+ + 1
α
α
αr
and
2
1
k∂q F1 (q, t)kLr ≤ Ct− α + αr .
(2.9)
Estimate (2.7) is then proved. Now we prove
1
1
sup t α − rα +µ1 +γ hti−2γ y −µ1 α (kF2 (x, y, t)kLr
t>0,y>0
(2.10)
µ
1
+ t− α kxµ F2 (x, y, t)kLr + t α k∂x F2 (x, y, t)kLr ) ≤ C,
where µ ≥ 0, γ > 0, 0 ≤ µ1 < α1 (1 − 1r − γα) and 1 ≤ r ≤ ∞.
1
We have, by definition, κ(q) = C1 |q| α . Changing variables pα t = z α and ξt = q,
we obtain the estimate
1
Z i∞
Z i∞
eq−C1 q α ye
−1/α
x
ez
−1
−1
|F2 (x, y, t)| ≤ Ct
dze K(z)z
dq q κ(q)
. (2.11)
K(z) + q
−i∞
−i∞
We move the contours of integration with respect to z as follows
π
C1 = {z = ρe±iβ1 , ρ ≥ 0, β1 = + 1 }
2
and with respect to q by
π π
π
C2 = {q = eiφ2 φ2 ∈ [− , ]} ∪ {q = ρe±iφ2 ρ ≥ 1, φ2 = + 2 },
2 2
2
where 1 , 2 > 0 are small enough. It is apparent that
|eq | ≤ C|q|−γ ,
(2.12)
(2.13)
|K(z) + q|−1 ≤ C|z|−να |q|ν−1 ,
|ezx̃ | ≤ C|z x̃|−µ2 ,
1
1
|e−C1j q α ye| ≤ C|q α ye|−αµ1
(2.14)
for all q ∈ C2 and z ∈ C1 , where ν ∈ [0, 1], γ ≥ 0, and µ1 ≥ 0, µ2 ≥ 0. Taking into
account (2.14) we get
Z i∞
Z i∞
1
−1/α −µ1 α −µ2
α−1−να−µ2
|F2 (x, y, t)| ≤ Ct
y
x̃
|dz||z|
|dq| |q| α −2+ν−µ1 −γ .
−i∞
−i∞
(2.15)
To guarantee the convergence of this integrals we need to satisfy the following
conditions
1
− 2 + ν − µ1 > −1
(2.16)
α
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
and
α − 1 − να − µ2 > −1,
5
(2.17)
α − [α] − να − µ2 < −1.
respectively. Under the condition αµ1 + µ2 < 2 − [α] there exists some ν > 0 such
that the estimates (2.17) are valid. Therefore, we obtain
|F2 (p, y, t)| ≤ Ct−
1−µ2
α
+µ1 −µ2 −αµ1
x
y
.
(2.18)
From (2.18) it follows that
1
1
1
1
∞
Z
kF2 (x, y, t)kLr ≤ Ct− α + rα +µ1 ±γ y −αµ1
x−µ2 r dx
1/r
(2.19)
0
≤ Ct− α + rα +µ1 ±γ y −αµ1 ,
where 0 ≤ µ1 <
1
α (1
− 1r ). In the same manner we obtain
1
|xµ F2 (x, y, t)| ≤ Ct− α +
|∂x F2 (p, y, t)| ≤ Ct
µ−µ2
α
2−µ
− α2
+µ1 µ−µ2 −αµ1
y
x
+µ1 −µ1 α −µ2
y
x
,
.
Thus we obtain
kxµ F2 (x, y, t)kLr ≤ Ct−
1+µ
1
α + rα +µ1 ±γ
k∂x F2 (x, y, t)kLr ≤ Ct
y −µ1 α ,
2
1
−α
+ rα
+µ1 −µ1 α
y
(2.20)
,
(2.21)
and therefore estimate (2.10) is proved. From estimates (2.7) and (2.10) we easily
get result (2.3) of Lemma 2.1. To obtain (2.4) firstly we prove the following estimate
G(x, y, t) = t−1/α sin
1+µ
π{α}
Λ0 (xt−1/α ) + y µ O(t− α )
α
(2.22)
for t → ∞, where x, y > 0.
We write the representation (2.5) for the function F1 (x, t) as
F1 (x − y, t) = F1 (x, t) + [F1 (x − y, t) − F1 (x, t)].
After the change of variables pα t = z α , we easily find that
Z i∞
Z i∞
1
1
F1 (x, t) =
exp−K(p)t dp = t−1/α
ezs−K(z) dz.
2πi −i∞
2πi −∞
(2.23)
Using the estimate |e−py −1| ≤ C|py|µ for y > 0 and p ∈ (−i∞, i∞) with µ1 ∈ [0, 1]
we have
Z i∞
α
|F1 (x − y, t) − F1 (x, t)| ≤
exp−Cα p t (e−py − 1) dp
−i∞
Z
i∞
µ − 1+µ
α
1
2πi
µ − 1+µ
α
= y O t−
≤ Cy t
≤ Cy t
−1
α
|dz| |z|µ ext
z−Cα z α
(2.24)
−i∞
µ
1+µ
α
.
Therefore, from (2.23), we obtain
F1 (x − y, t) = t−1/α
1
2πi
Z
i∞
ezs−K(z) dz + y µ O(t−
1+µ
α
).
(2.25)
−∞
Now we write the representation (2.6) of the function F2 (x, y, t) as
F2 (x, y, t) = M (x, y, t) + R(x, y, t),
(2.26)
6
R. E. CARDIEL, E. I. KAIKINA
EJDE-2006/89
where M (x, t) = F2 (x, 0, t) and R = [F2 (x, y, t) − F2 (x, 0, t)]. Considering that
1/α
|e−Cξ y − 1| ≤ C|ξ 1/α y|µ and changing variables ξt = q and pα t = z α , we get
|F2 (x, y, t) − F2 (x, 0, t)|
Z i∞
Z
1
µ1
1
≤ Ct− α − α y µ1
|dz| |ext α z | |z|α−1
−i∞
µ
= y O(t
− 1+µ
α
i∞
µ
1
|dq||q| α −1+ α
−i∞
|eq |
|K(z) + q|
).
Therefore,
F2 (x, y, t − τ ) = M (x, t) + y µ O(t−
1+µ
α
),
(2.27)
where
M (x, t) = t
−1/α
1
4π 2
Z
i∞
sz
dze K(z)z
−1
Z
i∞
−i∞
−i∞
dqeq
q −1 κ(q)
,
K(z) + q
with s = xt−1/α . Applying the Cauchy Theorem, we obtain
Z i∞
Z i∞
q −1 κ(q)
dqeq
dzesz K(z)z −1
K(z) + q
−i∞
−i∞
Z i∞
Z i∞
Z
q −1 κ(q)
sz −K(z)
sz
−1
= 2πi
dze e
+
dze K(z)z
dq eq
K(z) + q
−i∞
−i∞
Γ
= I1 + I2 ,
where Γ = {z ∈ (−∞e−iπ , 0e−iπ ) ∪ (0eiπ , −∞eiπ )}. Using
Z i∞
Z
1
dzesz
dqeq
= 0,
K(z)
+q
−i∞
Γ
we get
{α}
Z i∞
α
A1
−1/α 1
sz −Cα z α
−1/α Cα (−1)
M (x, t) = t
dze e
+t
2πi −i∞
2π 2 i
π − A1
{α}
Z
Z +∞
− α
π{α} i∞
sz {α}
−q q
dze z
dqe
,
× sin
α
K(z) + q
−i∞
0
where
(2.28)
(2.29)
(2.30)
[α]
A1 = (−1){α}+1 (−Cα )− α Γ(1 − {α})Γ({α}) sin π{α}.
From (2.25), (2.27), (2.30), we obtain (2.22) and then estimate (2.4). This
completes the proof of Lemma 2.1.
We defined the space
W = {φ(x, t) ∈ C((0, ∞); L1 ∩ L∞,µ ) : kφkW < ∞}
where kφkW = (kφkL1 + kφkL∞,µ ).
Lemma 2.2. The following two estimates are valid
Z t
µ
1
1
kxµ
G(t − τ )φ(τ )dτ kLr ≤ Chti− α + αr + α khti1+γ φkW ,
0
Z t
1
1
1
k∂x
G(t − τ )φ(τ )dτ kLr ≤ Ct1− α + αr hti− α khti1+γ φkW ,
0
for all t > 0, µ ∈ (0, 1), r ≥ 1 and small enough γ > 0, provided that the right-hand
sides are bounded.
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
7
Proof. Let t ∈ [0, 1]. By estimate (2.21) of Lemma 2.1, we get
Z ∞
2
1
1
1
−α
+ αr
kk∂x F2 (x, y, t)kLrx kL1y ≤ Ct
he
y i−µ1 α dy ≤ Ct− α + αr ,
(2.31)
0
for all t > 0. Therefore, using (2.8) of Lemma 2.1 we obtain
Z t
G(t − τ )φ(τ )dτ kLr
kxµ
0
Z t
Z t
dτ kφkL∞ kxµ F2 kLrx L1y
dτ (kφkLr kxµ F1 kL1 + kxµ φkLr kF1 kL1 ) +
≤
(2.32)
0
0
1
1
1
µ
1
< C(t− α + rα +1 + t− α + rα + α +1 )kφkW < CkφkW
for all t ∈ [0, 1]. Using (2.31) and (2.9), we have
Z t
k∂x
G(t − τ )φ(τ )dτ kLr
0
Z t
≤
dτ (kφkLr k∂x F1 kL1 + kφkL∞ k∂x F2 kLrx L1y )
(2.33)
0
1
1
1
1
< Ct− α + αr +1 kφkW ≤ t− α + αr kφkW
for all t ∈ [0, 1].
We consider now the case t > 1,
Z t
Z
µ
kx
G(t − τ )φ(τ )dτ kLr ≤
0
t/2
dτ (kφkL1 kxµ F1 kLr + kxµ φkL1 kF1 kLr )
0
t
Z
dτ (kφkLr kxµ F1 kL1 + kxµ φkLr kF1 kL1 )
+
t/2
t/2
Z
dτ kφkL1 kxµ F2 kLrx L∞
y
+
0
Z
t
+
t/2
dτ kφkL∞ kxµ F2 kLrx L1y .
So in view of estimates (2.20), (2.21) , (2.8) and (2.9), we attain
Z t
µ
1
1
µ
kx
G(t − τ )φ(τ )dτ kLr ≤ Ct− α + rα + α khti1+γ φkW .
(2.34)
0
In the same way we estimate k∂x GφkLr for t > 1,
Z
k∂x
t
G(t − τ )φ(τ )dτ k
0
t/2
Z
Lr
≤
dτ (kφkL1 k∂x F1 kLr + kφkL1 k∂x F2 kLr L∞ )
0
Z
t
+
t/2
2
dτ (kφkLr k∂x F1 kL1 + kφkL∞ k∂x F2 kLrx L1y )
1
≤ Ct− α + rα khti1+γ φkW .
Then, by (2.32)-(2.35) Lemma 2.2 is proved.
(2.35)
8
R. E. CARDIEL, E. I. KAIKINA
EJDE-2006/89
Denote by
0,0
0,µ
Z = {φ(x) ∈ H1,0
2 ∩ H∞ ∩ H1 ; kφkz < +∞},
kφkZ = (khxiµ φkL∞ + kφkL1 + kφkH12 ),
where µ ∈ (0, 1]. Now we prove local existence theorem. Note that the existence
time T > 0 could be sufficiently small.
Theorem 2.3. Let initial data u0 ∈ Z. Then for some time interval T > 0 there
exists a unique solution u ∈ C([0, T ]; X) to problem (1.1). Moreover the existence
time T can be chosen as follows
−σ−1 µ11
1
1
, µ1 = 1 − .
T = ku0 kZ 1 +
ku0 kZ
2C
α
Proof. We apply the contraction mapping principle in a ball of a radius ρ > 0 in a
complete metric space XT ,
XT, ρ = {φ ∈ C([0, T ]; X) : supt∈[0,T ] kukX = kukXT ≤ ρ},
where ρ =
1
2C ku0 kZ .
For v ∈ XT, ρ we define the mapping M(v) by
Z t
M(v) = G(t)u0 −
G(t − τ )N (v(τ ))dτ,
(2.36)
0
where N (v(τ )) = λ|v|σ v. We first prove that
kM(v)kXT ≤ ρ,
when v ∈ XT, ρ . We have by Lemmas 2.1 and 2.2 for µ1 = 1 − α1 ,
Z t
kM(v)kXT ≤ kGu0 kXT + k
G(t − τ )N (v(τ ))dτ kXT
0
(2.37)
≤ Cku0 kZ + CT µ1 (1 + kvkXT )σ+1
ρ
≤ + CT µ1 (1 + ρ)σ+1 ≤ ρ,
2
if T > 0 is small enough. Therefore, the mapping M transforms a ball of a radius
ρ > 0 into itself in the space XT . As in the proof of (2.37), we have for w, v ∈ XT
Z t
kM(w) − M(v)kXT ≤ k
G(t − τ )(N (w(τ )) − N (v(τ )))dτ kXT
0
1
kw − vkXT ,
2
since T > 0 is small enough. Thus M is a contraction mapping in XT, ρ ; therefore,
there exists a unique solution u ∈ XT to the problem (1.1). Theorem 2.3 is proved.
≤ CT µ1 kw − vkXT (1 + kwkXT + kvkXT )σ ≤
Now we define the space X[T1 , T2 ] = C([T1 , T2 ]; Z) with the norm
kψkX[T1 ,T2 ] =
sup
kψ(t)kZ .
t∈[T1 ,T2 ]
Theorem 2.4. Let the initial data u0 ∈ Z and the following a priory estimate be
valid
kukX[0,T ) ≤ C(T )ku0 kZ ,
(2.38)
provided that there exists a solution u ∈ X[0, T ) for some T > 0. Then there exists
a unique global solution u ∈ X[0, ∞) to the problem (1.1).
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
9
Proof. From Lemma 2.1, we have that G(t−T1 ) : Z → X[T1 , T2 ] for any T2 > T1 ≥ 0
and
kG(t − T1 )φkX[T1 ,T2 ] ≤ CkφkZ .
Rt
Also using Lemma 2.2 we obtain that T1 G(t − τ )N (v(τ ))dτ ∈ X[T1 , T2 ] for any
v ∈ X[T1 , T2 ], T2 > T1 ≥ 0, and
Z t
k
G(t − τ )(N (w(τ )) − N (v(τ )))dτ kX[T1 ,T2 ]
T1
≤ Ckw − vkX[T1 ,T2 ] (1 + kwkX[T1 ,T2 ] + kvkX[T1 ,T2 ] )σ ,
for all v, w ∈ X[T1 , T2 ], where σ > 0. Using a priory estimates (2.38) we can
prolongate the local solution given by Theorem 2.3 for all times t > 0. Indeed,
by the contrary we can suppose that there exists a maximal existence time T > 0
such that u ∈ X[0, T ). If we choose a new initial time T1 ∈ [0, T ) and consider the
problem (1.1) with initial data u(T1 ), then via a priory estimate (2.38) the norm
ku(T1 )kZ is bounded uniformly with respect to T1 ∈ [0, T ). Then the existence
time given by the local existence Theorem 2.3 is bounded from below uniformly
with respect to T1 ∈ [0, T ). Therefore if a new initial time T1 > 0 is chosen to
be sufficiently close to the maximal time T , then by virtue of the local existence
Theorem 2.3 we can guarantee that there exists a unique solution u ∈ X[0, T ].
Now putting u(T ) as a new initial data at time T we can apply the local existence
Theorem 2.3 and prolongate the solution u(t) on some bigger time interval [0, T +
T2 ]. This contradicts to the fact that T is a maximal existence time. Hence there
exists a unique solution u ∈ X[0, ∞) to the problem (1.1). Theorem 2.4 is proved.
3. Large initial data (proof of Theorem 1.1)
To prove of Theorem 1.1 we first apply the so-called energy method to estimate
the L2 (R) norm: i.e. we multiply equation (1.1) by u and integrate with respect to
x ∈ R+ , to get
Z +∞
Z +∞
d
2
σ+1
ku(t)kL2 + 2λ
|u|
dx = −2
uLudx.
(3.1)
dt
0
0
By the Plancherel Theorem we have
Z +∞
Z i∞
1
u(0)
2
uLudx = 2 Re
u
b(p)Cα pα (b
u(p) −
) dp
2πi
p
0
−i∞
Z i∞
1
u(0) 2
= Re(V P
| dp
Cα pα |b
u(p) −
πi −i∞
p
Z i∞
u(0)
1
Cα pα−1 (b
u(p) −
) dp).
− Re u(0)V P
πi −i∞
p
By the Cauchy Theorem using the analyticity of u
b(p) and K(p) = Cα pα in the
right-half complex plane we attain
Z i∞
u(0)
VP
Cα pα−1 (b
u(p) −
) dp
p
−i∞
Z i∞
1
u(0)
u(0)
= res p=0 (Cα pα−1 (b
u(p) −
)) +
Cα pα−1 (b
u(p) −
) dp = 0.
2
p
p
−i∞
10
R. E. CARDIEL, E. I. KAIKINA
EJDE-2006/89
Thus from dissipative condition Re K(p) > 0 for Re p = 0 we get
Z +∞
Z ∞
1
u(0) 2
2
uLudx = V P
Re K(ip)|b
u(ip) −
| dp > 0.
π
ip
0
−∞
Also since λ > 0 we have
+∞
Z
|u|σ+1 dx > 0.
λ
0
Therefore, from equation (3.1) we obtain
d
kuk2L2 ≤ 0.
dt
Hence integrating with respect to time we see that
(3.2)
sup ku(t)kL2 ≤ ku0 kL2 .
(3.3)
t≥0
Now we prove that
sup kux (t)kL2 ≤ ku0x kL2 .
(3.4)
t≥0
We differentiate (1.1) with respect to the space variable, multiply equation (1.1) by
ux and integrate with respect to x ∈ R+ , to get
Z +∞
Z +∞
d
2
σ−1
2
kux (t)kL2 + 2σλ
|u|
|u|x dx = −2
ux ∂x Ludx.
(3.5)
dt
0
0
Since
Z +∞
Z ∞
1
u(0) 2
2
ux ∂x Ludx = V P
Re K(ip)|p|2 |b
u(ip) −
| dp > 0
π
ip
0
−∞
and
Z
+∞
σλ
|u|σ−1 |u|2x dx > 0
0
integrating (3.5) with respect to time we easily get (3.4).
Note that time decay estimates (3.3) and (3.4) are not optimal. To get an optimal
time decay estimates we need to show that the L1 - norm of the solution does not
grow with time. Using the idea of papers [2], [5] we multiply equation (1.1) by
u
and integrate with respect to x over R+ to get
S ≡ sign u ≡ |u|
Z
Z
Z
ut (x, t)S(x, t)dx +
N (u)(x, t)S(x, t)dx = −
S(x, t)Ludx,
(3.6)
R+
R+
R+
σ
where N (u) = λ|u| u. We have
Z
Z
∂
d
ut (x, t)S(x, t)dx =
|u(x, t)|dx = ku(t)kL1 ,
∂t
dt
+
+
R
R
Z
Z
N (u)(x, t)S(x, t)dx = λ
|u|σ dx ≥ 0.
R+
R+
Representing the operator Lu via the Riesz potential (see [29]) let us show that
Z
S(x, t)Ludx ≥ 0.
(3.7)
R+
By [1], we have
1
2πi
Z
i∞
−i∞
p−ν−1 epx dp =
1
xν .
Γ(ν + 1)
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
11
Thus for α ∈ (1, 2), we get
Z
Lu = −C∂x
x
(x − y)1−α ∂y u(y) dy, C > 0.
0
Denote S(x, t) = sign(u(x, t)) and represent u(x, t) = S(x, t)|u(x, t)|. We make a
regularization
(
∂x2 x1−α , for x ≥ ε
00
Kε (x) =
0,
for 0 ≤ x < ε,
such that Kε0 (x) ≤ 0 and Kε00 (x) ≥ 0 for all x > 0. We can easily see that
Z x
Z x
1−α
∂x
(x − y)
uy (y, t) dy = lim ∂x
Kε (x − y)uy (y, t) dy.
ε→0
0
0
(To justify our calculation we note that the linear operator L in equation (1.1) is
strongly dissipative, therefore by smoothing effect the solution obtain regularity
u ∈ C1 (R+ ) (see Theorem 2.3 and [28])).We have
Z
Z x
dxS(x, t)∂x
Kε (x − y)∂y u(y, t) dy
R+
0
Z
Z x
=
dxS(x, t)∂x
dyKε (x − y)S(y, t)∂y |u(y, t)|
R+
0
Z
= Kε (0)
dxS 2 (x, t)∂x |u(x, t)|
R+
Z
Z +∞
+
dy∂y |u(y, t)|
dxS(y, t)S(x, t)∂x Kε (x − y).
R+
y
Then via the identity S(y, t)S(x, t) = 1 − 21 (S(x, t) − S(y, t))2 ,
Z
Z
x
dxS(x, t)∂x
Kε (x − y)∂y u(y, t) dy
R+
0
Z
Z
Z
2
= Kε (0)
dxS (x, t)∂x |u(x, t)| +
dy∂y |u(y, t)|
R+
−
1
2
R+
Z
Z
dx∂x Kε (x − y)
y
+∞
dy∂y |u(y, t)|
R+
+∞
dx(S(x, t) − S(y, t))2 ∂x Kε (x − y).
y
Since
Z
Z
dy∂y |u(y, t)|
R+
+∞
Z
dx∂x Kε (x − y) = −Kε (0)
y
dy∂y |u(y, t)|
R+
we obtain
Z
Z x
dxS(x, t)∂x
Kε (x − y)∂y u(y, t) dy
R+
0
Z
Z +∞
1
=−
dy∂y |u(y, t)|
dx(S(x, t) − S(y, t))2 ∂x Kε (x − y).
2 R+
y
(3.8)
12
R. E. CARDIEL, E. I. KAIKINA
Integrating by parts, we have
Z
Z
dy∂y |u(y, t)|
R+
EJDE-2006/89
+∞
dx(S(x, t) − S(y, t))2 ∂x Kε (x − y)
y
Z +∞
= −|u(0, t)|
dx(S(x, t) − S(0, t))2 ∂x Kε (x)
0
Z
Z +∞
dxKε00 (x − y)(S(x, t) − S(y, t))2 .
+
dy|u(y, t)|
R+
y
Therefore, from (3.8) using ∂x Kε (x) < 0, we gain
Z
Z x
dxS(x, t)∂x
Kε (x − y)∂y u(y, t) dy
R+
0
Z
Z +∞
1
≤−
dy|u(y, t)|
dxKε00 (x − y)(S(x, t) − S(y, t))2
2 R+
y
and therefore since ∂x2 K(x) > 0 for all x > 0 we get
Z
Z x
dxS(x, t)∂x
Kε (x − y)∂y u(y, t) dy ≤ 0.
R+
0
Hence we have
Z
S(x, t)Ludx
R+
Z
Z
C
dy|u(y, t)|
dxKε00 (x − y)(S(x, t) − S(y, t))2 ≥ 0.
≥ − lim
2 ε→0 R+
R+
d
Thus (3.7) is true and from (3.6) we find dt
ku(t)kL1 ≤ 0. From this inequality, we
see that the norm ku(t)kL1 is bounded for all t ≥ 0.
We now prove that the norm kux (t)kL2 → 0 as t → ∞. Taking % ∈ (0, 1) by the
Plancherel theorem,
Z +∞
Z ∞
1
u(0) 2
| dp
ux ∂x Ludx = V P
Re K(ip)|p|2 |b
u(ip, t) −
π
ip
0
−∞
Z
≥
Re Cα pα |b
ux (p, t)|2 dp
|p|≥%
α
≥ C% kux (t)k2L2 − C%α+3 sup |b
u(p, t)|2
|p|<%
− Cρ
α+2
|u(0)| sup |b
u(p, t)| − C%α+1 |u(0)|2
|p|<%
≥ C%α ku(t)k2L2 − C%α+3 ku(t)k2L1
− C%α+2 ku(t)kL1 kukL∞ − C%α+1 kuk2L∞
≥ C%α ku(t)k2L2 − C%α+3 ku(t)k2L1
1
1
− C%α+2 ku(t)kL1 kukL2 2 kux kL2 2 − C%α+1 kukL2 kux kL2 ,
where we have used that
kux (t)k2L∞ ≤ Cku(t)kL2 kux (t)kL2 .
(3.9)
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
13
Since the norms ku(t)kL1 , ku(t)kL2 and kux (t)kL2 are bounded, choosing %(t) =
C −1/α (1 + t)−1/α , we obtain
1
d
kux (t)k2L2 ≤ −(1 + t)−1 kux (t)k2L2 + C(1 + t)−1− α .
dt
We substitute kux (t)k2L2 = 12 h(t)(1 + t)−2 , then for h(t) we have
1
h0 (t) ≤ C(1 + t)1− α ,
1
hence integration with respect to time yields h(t) ≤ C(1 + t)2− α . Therefore we get
the time decay estimate
kux (t)kL2 ≤ C(1 + t)−1/(2α) .
and therefore
1
1
1
kux (t)kL∞ ≤ Cku(t)kL2 2 kux (t)kL2 2 ≤ C(1 + t)− 4α .
We substitute this estimate in (3.9) to obtain
1
1
d
kux (t)k2L2 ≤ −(1 + t)−1 kux (t)k2L2 + C(1 + t)−1− α − 2α .
dt
Again after the change kux (t)k2L2 = 12 h(t)(1 + t)−2 we get
1
1
kux (t)kL2 ≤ C(1 + t)− 2α − 4α .
We can repeat this consideration to get the optimal time decay estimate
3
kux (t)kL2 ≤ C(1 + t)− 2α
(3.10)
for all t > 0.
We now prove that the norm ku(t)kL2 → 0 as t → ∞. Taking % ∈ (0, 1) by the
Plancherel theorem we get
Z +∞
uLudx
0
=
1
VP
π
Z
≥
≥%
≥%
|p|≥%
α
Z
i∞
u(0)
)b
u dp
ip
−i∞
Z i∞
α
2
b dp
Re Cα p |b
u(p, t)| dp − u(0)
Re Cα pα−1 u
ku(t)k2L2
α
ku(t)k2L2
Re K(ip)(b
u(ip, t) −
−i∞
α+1
−%
2
sup |b
u(p, t)| − |u(0)|%α sup |b
u(p, t)| − %α−1 |u(0)|2
|p|<%
α+1
−%
ku(t)k2L1
|p|<%
α
− |u(0)|% ku(t)kL1 − %α−1 |u(0)|2 .
Since the norms ku(t)kL1 , ku(t)kL2 are bounded and due to (3.10)
1
1
3
|u(0)| ≤ ku(t)kL∞ ≤ Cku(t)kL2 2 kux (t)kL2 2 ≤ C(1 + t)− 4α ,
choosing %(t) = C −1/α (1 + t)−1/α , we obtain
1
d
ku(t)k2L2 ≤ −(1 + t)−1 ku(t)k2L2 + C(1 + t)−1− 2α .
dt
We substitute ku(t)k2L2 = h(t)(1 + t)−2 , then for h(t) we have
1
h0 (t) ≤ C(1 + t)1− 2α ,
14
R. E. CARDIEL, E. I. KAIKINA
EJDE-2006/89
hence integration with respect to time yields
1
h(t) ≤ C(1 + t)2− 2α .
Therefore, we have the time decay estimates
1
ku(t)kL2 ≤ C(1 + t)− 4α ,
3
1
|u(0)| ≤ C(1 + t)− 4α − 8α .
We can repeat this consideration to get
1
d
ku(t)k2L2 ≤ −(1 + t)−1 ku(t)k2L2 + C(1 + t)−1− α .
dt
Therefore, we obtain the optimal estimate
1
ku(t)kL2 ≤ C(1 + t)− 2α
(3.11)
for all t > 0. Also from (3.10) and (3.11) we get the optimal time decay of the
L∞ (R+ )-norm of the solutions
ku(t)kL∞ ≤ C(1 + t)−1/α
(3.12)
for all t > 0.
Now we can estimate the norm L∞,µ (R+ ). By the integral formula (2.2) we
have
ku(t)kL∞,µ ≤ ku0 kL1 kG(t)kL∞,µ + ku0 kL∞,µ kG(t)kL1
Z t
+C
k|u(τ )|σ+1 kL1 kG(t − τ )kL∞,µ dτ
0
Z t
+C
k|u(τ )|σ+1 kL∞,µ kG(t − τ )kL1 dτ
0
Hence using Lemmas 2.1 and 2.2,
ku(t)kL∞,α+1
≤ Chti−
1−µ
α
t
Z
ku(τ )kσL∞ ht − τ i−
+C
1−µ
α
Z
dτ + C
0
− 1−µ
α
≤ Chti
σ
−α
ku(τ )kL∞,µ hτ i
+C
ku(τ )kσL∞ ku(τ )kL∞,µ dτ
0
t
Z
t
dτ.
0
Hence for the function h(t) = sup0≤τ ≤t ku(τ )kL∞,µ , we get the inequality
Z t
1−µ
σ
hτ i− α h(τ )dτ
h(t) ≤ Chti− α + C
0
and since σ > α by the Gronwall’s lemma it follows that
ku(t)kL∞,µ ≤ Chti−
1−µ
α
∀t > 0.
(3.13)
From a priory estimates, due to Theorem 2.4 there exists a unique solution
0,0
0,µ
u(x, t) ∈ C([0, ∞); H1,0
2 ∩ H∞ ∩ H1 )
to the problem (1.1), such that
ku(t)kL∞ ≤ Chti−1/α , ku(t)kL∞,µ ≤ Chti−
1−µ
α
1
and ku(t)kL2 ≤ Chti− 2α .
(3.14)
EJDE-2006/89
NONLINEAR PSEUDODIFFERENTIAL EQUATIONS
15
By Lemma 2.2 we have
1
1
sup t α (1+µ) kGφ − t− α Λ0 (xt−1/α )f (φ)kL∞ ≤ CkφkL1,µ ,
t>1
where µ ∈ [0, 1]. Substituting this formula into (2.2) we obtain
x
u(x, t) = t−1/α AΛ0 ( 1 ) + R(x, t),
tα
where by (3.14),
Z +∞
Z +∞ Z ∞
A=
u0 dy +
dτ
N (u) dy < ∞,
0
0
(3.15)
0
and
R(x, t) = O(t
Z
+
− 1+µ
α
Z
)
+∞
Z
µ
τ
2µ
α
− 1+µ
α
O (t − τ )
Z
0
+∞
+∞
y µ N (u) dy
0
N (u) dy
dτ
0
= O(t−
Z
dτ
y u0 (y) dy +
0
+∞
+∞
0
1+µ
α
).
Thus the asymptotic (1.4) is valid. Theorem 1.1 is then proved.
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Rosa E. Cardiel
Instituto de Matemáticas UNAM (Campus Cuernavaca), Av. Universidad s/n, col. Lomas
de Chamilpa, Cuernavaca, Morelos, Mexico
E-mail address: rosy@matcuer.unam.mx
Elena I. Kaikina
Instituto de Matemáticas, UNAM Campus Morelia, AP 61-3 (Xangari), Morelia CP
58089, Michoacán, Mexico
E-mail address: ekaikina@matmor.unam.mx