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Damped wave equation in the subcritical case
J. Differential Equations 207 (2004) 161 – 194 www.elsevier.com/locate/jde
Damped wave equation in the subcritical case Nakao Hayashia,∗ , Elena I. Kaikinab , Pavel I. Naumkinc a Department of Mathematics, Graduate School of Science, Osaka University, Osaka,
Toyonaka 560-0043, Japan b Departamento de Ciencias Básicas, Instituto Tecnológico de Morelia, Morelia CP 58120,
Michoacán, Mexico c Instituto de Matemáticas, Unam Campus Morelia, AP 61-3 (Xangari), Morelia CP 58089,
Michoacán, Mexico Received 22 July 2003; revised 31 March 2004 Available online 12 August 2004
Abstract We study large time asymptotics of small solutions to the Cauchy problem for the one dimensional nonlinear damped wave equation vtt + vt − vxx + v 1+ = 0, x ∈ R, t > 0, (1) v (0, x) = εv0 (x) , vt (0, x) = εv1 (x) −1 v1 ∈ in the sub critical case ∈ 2 − ε 3 , 2 . We assume that the initial data v0 , 1 + *x L∞ ∩ L1,a , a ∈ (0, 1) where
L1,a = ∈ L1 ; L1,a = · a L1 < ∞ , x = 1 + x 2 . Also we suppose that the mean value of initial data (v0 (x) + v1 (x)) dx > 0. R
∗ Corresponding author. E-mail addresses: [email protected] (N. Hayashi), [email protected] (P.I. Naumkin).
0022-0396/$ - see front matter © 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jde.2004.06.018
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
Then there exists a positive value ε such that the Cauchy problem (1) has a unique global ∞ 1,a , satisfying the following time decay estimate: solution v (t, x) ∈ C [0, ∞) ; L ∩ L v (t) L∞ Cε t −
1
for large t > 0, here 2 − ε 3 < < 2. © 2004 Elsevier Inc. All rights reserved. MSC: 35B33; 35B40 Keywords: Damped wave equation; Subcritical nonlinearity; Asymptotic expansion; Large time behavior
1. Introduction We study the one-dimensional nonlinear damped wave equation
vtt + vt − vxx + v 1+ = 0, x ∈ R, t > 0, v (0, x) = εv0 (x) , vt (0, x) = εv1 (x) , x ∈ R
(2)
in the sub critical case ∈ 2 − ε 3 , 2 , where ε > 0. Note that the case = 2 is critical for the asymptotic behavior of solutions to (2). Recently, much attention was drawn to nonlinear wave equations with dissipative terms. The blow-up results were proved in [28] for the case of nonlinearity − |v|1+ , with < 2, when the initial data are such that R v0 (x) dx > 0, R v1 (x) dx > 0. Blow-up results for the critical and sub critical cases 2 were obtained in [21,28,30]. In paper [28], it was proved global existence and large time decay estimates of solutions to the Cauchy problem for the damped wave equation (2) with nonlinearities ± |v|1+ or ± |v| v, for the super critical case > 2, if the initial data are sufficiently small and have a compact support. The large time asymptotic behavior of solutions to (2) for the super critical case > 2 was obtained in paper [15] in the framework of the usual L2 -theory. When the initial data are in the usual Sobolev space * u0 ∈ L1 ∩ L∞ , || 1, u1 ∈ L1 ∩ L∞ , problem (2) was considered in [24,27]. Via the energy-type estimates obtained in papers [22,19] it was proved in [17] that solutions of the nonlinear damped wave equation (2) in the super critical cases > 3 with arbitrary initial data u0 ∈ H1 ∩ L1 , u1 ∈ L2 ∩ L1 (i.e. without smallness assumption on the initial data) have the same large time asymptotics 2 as that for the linear heat equation *t − *x , that is u (t) − MG0 (t) Lp
1 1 − 1− p =o t 2 |x|2
as t → ∞, where 2 p ∞, and G0 (t, x) = (4t)− 2 e− 4t is the heat kernel, M is a constant. Recently the critical case = 2 was considered in paper [14], where it was 1
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
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proved that solutions of (2) have an additional large time decay v (t) L∞ C t − 2 (1 + log t )− 2 , 1
1
√ where t = 1 + t 2 . Note that similar behavior first was discovered for the nonlinear heat equation vt − vxx − v 1+ = 0 in the critical case = 2, comparing with the linear heat equation, (see [6,11]). For blow-up results we refer [5,8,20]. Large time behavior of solutions to the nonlinear heat equations in the sub critical cases ∈ (1, 2) was obtained in papers [3,4,9,18,29]. −1 −1 −1 Taking v = u1 and 1 + *x vt = u2 , with 1 + *x = F→x 1 + i Fx→ = x e−x −∞ dx ex , we rewrite Eq. (2) in the form of a system of nonlinear evolutionary equations ut + N (u) + Lu = 0
(3)
for the vector u (t, x) =
u1 (t, x) u2 (t, x)
.
The initial data are u (0, x) = u (x) ≡
εv0 (x) −1 ε 1 + *x v1 (x)
.
The linear part of system (3) is a pseudodifferential operator defined by the Fourier transformation as follows: Lu = F→x L Fx→ u, with a matrix-symbol L = Lj k j,k=1,2 =
− 1 + i
0 2
1
1+i
The nonlinearity is defined by
N (u) = 1 + *x
−1
0
u1+ 1
.
We define the direct Fourier transformation Fx→ by 1 uˆ ≡ Fx→ u = (2)− 2
R
e−i x u (x) dx,
.
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then the inverse Fourier transformation F→x is − 21
uˇ (x) ≡ F→x u = (2)
eix u d . R
Denote by 1 1 1 − 4 2 , 1 = + 2 2
1 1 2 = − 1 − 4 2 2 2
the eigenvalues of the matrix L . Note that the matrix
Q =
1 + i 1 + i Q11 Q12 = −1 −2 Q21 Q22
and Q−1 =
1 1 + i 1 − 2
−2 − 1 + i 1 1 + i
diagonalize the matrix L , i.e. Q
−1
L Q =
0 1 . 0 2
Consider the system of ordinary differential equations with constant coefficients depending on the parameter ∈ R d u t, = 0. u t, + L dt
(4)
Multiplying system (4) by Q−1 from the left and changing u t, = Q w t, we diagonalize system (4) d dt
0 w1 t, 1 w1 t, =− , w2 t, 0 2 w2 t,
whence integrating with respect to time t 0 we find
−t w1 t, 0 w1 0, e 1( ) = w2 t, w2 0, 0 e−t 2 ()
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165
Returning to the solution u t, we get w1 t, u1 (t, x) u t, = =Q u2 (t, x) w2 t, e−t 1 () u0 0 −1 =Q Q u1 0 e−t 2 () u0 = e−tL() , u1
where the fundamental Cauchy matrix has the form e
−tL()
0 e−t 1 () Q−1 −t ( ) 2 0 e − 2 − 1 + i 1 = e−t 1 () 2 1 2 1+i 1 − 4 1 1 + i 1 e−t 2 () . + 2 − − 1+i 2 2 1 − 4
=Q
We rewrite the Cauchy problem (3) in the form of the integral equation u (t) = G (t) u−
t
G (t − ) N (u) () d ,
(5)
0
ˆ . So by the solution of the where the Green operator G (t) = F→x e−tL() Cauchy problem (3) we always understand the solution u (t, x) of the corresponding integral equation (5), belonging to C0 ([0, ∞); X) with an appropriate choice of a functional space X. In the present paper we prove the following result. Denote
L1,a = ∈ L1 : L1,a = · a L1 < ∞ . 2 Theorem 1.1. We assume that the initial data u ∈ L∞ ∩ L1,a , a ∈ (0, 1), and the mean value
=ε
u1 (x) + u2 (x)) dx > 0. ( R
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Then there exists a positive ε such that the Cauchy problem for Eq. (3) has a unique 2 satisfying the following time decay mild solution u (t, x) ∈ C [0, ∞) ; L∞ ∩ L1,a estimate: u (t) L∞ Cε t − 1
for large t > 0 and any ∈ 2 − ε 3 , 2 . Furthermore, the asymptotic formula
− 1
u (t, x) = e1 (t )
V
x √
+O t
t
− 1 −
,
is valid for t → ∞ uniformly with respect to x ∈ R, where = 1 e1 = , V ∈ L1,a ∩ L∞ is the solution of the integral equation 0 V =
1
2
e− 4 − 1
(4) 2
1 1
(4) 2
dz
1
1
z (1 − z) 2
0
e−
√ 2 z) 4(1−z)
(−y
1 2
min a, 1 − 2 ,
F (y) dy,
(6)
R
where
=
1−
2
V 1+ (y) dy R
and F (y) = V 1+ (y) − V (y)
V 1+ d . R
Remark 1.1. As a consequence of Theorem 1.1 we have the following asymptotics for the damped wave equation (2): v (t, x) = (t )− V 1
x √
t
1 + O t − −
−1 v1 ∈ for t → ∞ uniformly with respect to x ∈ R if the initial data v0 , 1 + *x 2 ∞ 1,a , a ∈ (0, 1), are such that the mean value L ∩L R
−1 v0 (x) + 1 + *x v1 (x) dx > 0.
(7)
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We believe that condition (7) is necessary for the global existence of solutions for the −1 v1 (x) dx 0 solutions damped wave equation. In the case R v0 (x) + 1 + *x could blow up in a finite time. Remark 1.2. The restriction ∈ 2 − ε 3 , 2 with ε sufficiently small is only technical. It comes from the application of the contraction mapping principle in the proof of Lemma 3.1 below. The asymptotic behavior of solutions obtained in Theorem 1.1 is general whenever ∈ (0, 2) with some decay restrictions on the initial data, as it happens for the nonlinear heat equation (see [4]). Indeed in [26], a sharp time decay of solutions in Lp norm is obtained recently under the condition such that the data decay exponentially at infinity but do not have any restriction on the size, where 2 2 p n−2 for the space dimension n 3, 2 p < ∞ for n = 2 and 2 p ∞ for n = 1. For higher space dimension with super critical nonlinearities see [16,23–25] in which asymptotic expansion of small solutions in uniform norm was studied. We organize the rest of the paper as follows. In Section 2 we obtain some preliminary estimates of the Green operator solving the linearized Cauchy problem corresponding to (3). Section 3 is devoted to the proof of Theorem 1.1. 2. Preliminary lemmas We first collect some preliminary estimates for the Green operator G (t) = F→x e−tL() ˆ , in the weighted Lebesgue norms Lp and L1,a , where L1,a = · a L1 , a ∈ (0, 1), 1 p ∞. Also we show that G (t) asymptotically behaves as a Green function G0 (t) for the heat equation 2 G0 (t) = F→x e−t ˆ = G0 (t, x − y) (y) dy R
with a kernel 2 1 1 x2 G0 (t, x) = (2)− 2 F→x e−t = (4t)− 2 e− 4t . Denote a matrix A0 =
1 1 0 0
and a vector
ϑ=
(x) dx. R
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2 Lemma 2.1. Suppose that the vector-function ∈ L∞ ∩ L1,a , where a ∈ (0, 1). Then the estimates G (t) p C p , L L and 2 |·| G (t) − (4t)− 21 e− x4t A0 ϑ
Lp
Ct
− 21 1− p1 − a− 2
L1,a
are valid for all t > 0, 1 p ∞, 0 a. Proof. We represent the fundamental Cauchy matrix e−tL() in the form e−tL()
cos t = e− 2
t 2
−
√ sin 2t 42 −1 4 − 1 + √ 2 4 −1 √ 2 t 2
22
(1+i )
sin
2 √
4 −1
42 −1
√ sin 2t 42 −1 √ 2 2 1 + i 4 −1 √ sin 2t 42 −1 2 t cos 2 4 − 1 − √ 2
.
4 −1
We can see that the components of the matrix e−tL() are C∞ -functions. Define the ∞ 2 and = 0 for ∈ C = 1 for such that
cut off function
(R) 1 1 1 3, = 1 − . Denote 2 1 Gj (t) = F→x j e−tL() ˆ = Gj (t, x − y) (y) dy.
(8)
R
1 with matrix-kernels Gj (t, x) = (2)− 2 F→x j e−tL() . Note that the kernel 1 G1 (t, x) = (2)− 2 F→x 1 e−tL() in representation (8) is a smooth matrixfunction G1 (t, x) ∈ (C∞ (R × R))4 . In view of the estimates e
− 2t
m k 2 * cosh t 1 − 42 C t m−k 2 e −Ct 1 2
and
t
e− 2
sinh 2t 1 − 42 m k 2 C t m−k * 2 e −Ct 1 1 − 4 2
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169
for all ∈ R, t > 0, where m, k 0, we get 1 m m k x *x G1 (t, x) = (2)− 2 F→x * 1 k e−tL() m C * 1 k e−tL() 1
C t
m−k 2
−Ct 2 e
L
L1
(|| 3)
C t
m−1−k 2
for all x ∈ R, t > 0. Therefore G1 (t, x) have the following estimate: 1 sup x t − 2
x∈R
3 k *x G1 (t, x)
C t −
1+k 2
(9)
for all t > 0, where k 0. By virtue of (9) with k = 0 we find − 21
G1 (t) L1 C t
x t − 2
1
−3
dx C,
R
whence applying the Young inequality we obtain G1 (t) p C G1 (t) L1 p C p L L L for all t > 0, where 1 p ∞. Then applying the Lagrange finite differences Theorem, in view of (9) we obtain |x| |G1 (t, x − y) − G1 (t, x)| a |x| |y|a *x G1 (t, x ) (|G1 (t, x − y)| + |G1 (t, x)|)1−a 1 a− 1 −2 C t − 2 − 2 |y|a x t − 2 for all x, y ∈ R, if |y| |x| 2 or if
|x| 2
√ √ |y| t . When |y| max |x| t we get , 2
|x| |G1 (t, x − y) − G1 (t, x)| C |y| (|G1 (t, x − y)| + |G1 (t, x)|) −3 −3 − 21 − a− a − 21 − 21 2 |y| . C t + x t (x − y) t
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Thus we obtain |x| (|G1 (t, x − y) − G1 (t, x)|) 1 − 21 − a− a 2 |y| C t (x − y) t − 2
−2
+ x t
− 21
−2
for all x, y ∈ R, whence |x| (y) dy x − y) − G x)) (G (t, (t, 1 1 p R L −2 1 a− −1 · a 1 C t − 2 − 2 t 2 (·) L
Lp
− 21 1− p1 − a− 2
C t
a ·
L1
for all t > 0, 1 p ∞, ∈ [0, a]. Furthermore in view of the estimates m 2 2 * e− 2t cosh t 1 − 42 − 1 e−t C t m−1 2 e −Ct 1 2 2 and sinh 2t 1 − 42 m −t 2 2 m−1 1 −t * e 2 − e C t 2 e−Ct 1 2 1 − 4 2 for all ∈ R, t > 0, m 0, we estimate the difference 2 m − 21 − x4t x G t) e A x) − (t, (4 1 0 2 1 m = (2)− 2 F→x * 1 e−tL() − e−t A0 2 m C * 1 e−tL() − e−t A0 1
C t
m−1 2
−Ct 2 e
L1
L
m
Ct 2 −1
for all x ∈ R, t > 0, 0 m 2. Thus we obtain
sup xt
x∈R
− 21
3−
2 − 21 − x4t |x| G1 (t, x) − (4t) e A0 Ct 2 −1
(10)
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171
for all t > 0. By virtue of (10) we find 2 |x| G1 (t, x) − (4t)− 21 e− x4t A0 −2 − 21 −1 2 Ct (·) t
Lp
Ct
Lp − 21 1− p1 − 1− 2
.
Therefore the second estimate of the lemma follows 2 |x| G1 (t) − (4t)− 21 e− x4t A0 ϑ
Lp
2 − 21 − x4t |x| G t) e A (y) dy x − y) − = (t, (4 1 0 p R L |x| (G1 (t, x − y) − G1 (t, x)) (y) dy Lp
R
2 − 21 − x4t + |x| G1 (t, x) − (4t) e A0 ϑ
Ct
− 21
1− p1
− a− 2
a ·
L1
Lp
.
Next we estimate G2 (t) = F→x 2 e−tL() ˆ 1+ 2*t R F2 (t,x − y)1 (y) dy t +2 R F2 (t, x − y) 1 + *y 2 (y) dy, = e− 2 2 F2 (t, x − y) *y − 1 1 (y) dy − F3 (t, x − y) 1 (y) dy R R + 2*t − 1 R F2 (t, x − y) 2 (y) dy, with kernels
1 F2 (t, x) = (2)− 2 F→x 2
sinh
t 2
1 − 4
1 − 42
2
and sinh t 1 − 42 2 2 2 . F→x 1 + i 2 1 − 4
F3 (t, x) = (2)− 2
1
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
We use the asymptotic formula e
it 2
√
42 −1
=e
it ||
t eit || + O t 2 −2 , + 2i 1 + 2
(11)
whence
k *
sin
t 2
4 − 1
t cos t − sin t ! "2 − 2 2 8 1 + 2
2
4 2 − 1
−3 = O t 2+k
(12)
for → ∞, k = 0, 1, 2, therefore
1 F2 (t, x) = √ 2
eix
sin
t 2
R
4 − 1 2
4 2 − 1
2 d
= H1 (t, x) + H2 (t, x) + R1 (t, x) , where 1 H1 (t, x) = √ 2 2 H2 (t, x) =
t √
e R
8 2
ix
d sin t ! "2 ,
d eix cos t ! "2 R
and d e sin t 1 ! "2 R sin t 2 − 1 1 ix +√ e 2 R 2 − 1
1 R1 (t, x) = √ 2 2
ix
t cos t 2 d . − sin t ! "2 − 2 2 8 1+
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173
Computing the Fourier transform we get (see [1,7]) ∞ d 1 cos x sin t H1 (t, x) = √ 2 0 1 + 2 1 e−t cosh x, |x| < t, 2 2 = − 1 e−|x| sinh t, |x| > t, 2 2 H2 (t, x) =
t √
∞
cos x cos t
d
4 2 0 1 + 2 t e−t cosh x, |x| < t, 8 2 = t e−|x| cosh t, |x| > t. 8 2
In view of asymptotics (12) we have the estimate l k *t *x R1 (t, x) C t 2 x −2 for all t > 0, x ∈ R, where k + l = 0, 1. Therefore by the Young inequality we get F2 (t, x − y) 1 (y) dy
L∞
R
C t 2 1 L∞
and F2 (t, x − y) 1 (y) dy
L1,a
R
C t 4 1 L1,a .
In the same manner we estimate the kernel F3 (t, x) |F3 (t, x)| C t 2 x −2 for all t > 0, x ∈ R, hence F x − y) dy (t, (y) 3 1
L∞
R
C t 2 1 L∞
and F3 (t, x − y) 1 (y) dy R
L1,a
C t 4 1 L1,a .
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
We now estimate terms with derivatives *t and *y . Since *t F2 (t, x − y) 1 (y) dy R
=
#
4 1− H2 (t, x − y) 1 (y) dy 1 (x − t) − 1 (x + t) + 2 t R 1 − H1 (t, x − y) 1 (y) dy + *t R1 (t, x − y) 1 (y) dy, 2 R R 1 2
whence we get estimates *t x − y) dy F (t, (y) 2 1
L∞
R
C t 2 1 L∞
and * t F2 (t, x − y) 1 (y) dy
L1,a
R
C t 4 1 L1,a .
Similarly
1 F2 (t, x − y) *y 1 (y) dy = 2 R
#
1 (x − t) − 1 (x + t) 2
+
R
R
+ + R
*x H1 (t, x − y) 1 (y) dy *x H2 (t, x − y) 1 (y) dy *x R1 (t, x − y) 1 (y) dy,
whence the estimates x − y) dy F * (t, (y) 2 y 1
L∞
R
C t 2 1 L∞
and F2 (t, x − y) *y 1 (y) dy
L1,a
R
follow. Lemma 2.1 is proved.
C t 4 1 L1,a
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175
Next lemma will be used for estimates of the nonlinearity via contraction mapping principle. Lemma 2.2. Let the vector-function (t, x) be such that A0 norm
R
(t, x) dx = 0 and the
1 a sup t + 2 (t) L∞ + sup t − 2 · a (t) L1 ≡ F t>0
t>0
be finite, where a ∈ (0, 1), ∈ (0, 1). We also suppose that the function g (t) is such that g (t) t for all t > 0, where > 0 is such that + − a2 < 1. Then the following inequalities are valid: t g −1 () G (t − ) () d
L∞
0
1 C t 2 −− F
and a t −1 · g () G (t − ) () d
L1
0
a C t 1+ 2 −− F
for all t > 0.
Proof. Since A0
R
(t, x) dx = 0, we have by the estimates of Lemma 2.1 G (t − ) () ∞ C () ∞ , L L
(13)
G (t − ) () ∞ C (t − )− 1+a 2 · a () L L1
(14)
a · G (t − ) () 1 C · a () 1 L L
(15)
and
for all 0 < < t. Therefore by virtue of (13) and (14) we get t g −1 () G (t − ) () d
L∞
0
C
t 2
0
a 1 (t − )− 2 − 2 · a () L1 − d + C
t t 2
()
L∞
− d
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
C F
t 2
− a2 − 21
(t − )
a 2 −−
d +
t t 2
0
−−− 21
d
1 C t 2 −− F for all t > 0. In the same manner via (15) we find a t −1 · g G − d ( ) (t ) ( )
L1
0
C
t
− d C F L1
a · ()
0
1+ a2 −−
C t
for all t > 0. Lemma 2.2 is proved. Denote = that 1.
(4) 2
1+
1
(1− 2 )(1+) 2
t
a
2 −− d
0
F
. Since 2 − ε 3 < < 2 and Cε, we may suppose
2 with a ∈ (0, 1) and ≡ Lemma ∈ L1,a ∩ L∞ 2.3. We assume that w (x) dx 1 Cε > 0. Let the function (w (t, x))1 = w1 (t, x) satisfy the ε A0 R w estimates
w1 (t) L1+ Cε t − 2(1+) and
w1 (t) − (G (t) ε w )1 L1+ Cε 1+ t − 2(1+) for all t > 0. Then there exists ε > 0 such that the following inequality is valid: 1+
t 0
for all t > 0.
d R
w11+ (, x) dx
1 1− t 2 2
(16)
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177
Proof. By the Hölder inequality with r ∈ (1, 1 + a), a > 0, we obtain choosing 1 − 1 = · a a 1 1a > 0 L
1 L
L
a − a 2r 2r (x) dx = 2 + x 2 2 + x 2 R
1 1− 1 a − a r r r 2 2(r−1) 2 + x 2 dx 2 + x 2 (x) dx R
C
R
1− 1r
1 1 1− 1r 1r ∞ 1 + C 1− 1r − ar 1−∞r · a r 1 L L L L
r−1 1 − r−1 r−1 C Lr∞ · a Lar1 Lr 1 ar , whence
L1
1 a a 1+a · 1 . C L1+a ∞ L
Again applying the Hölder inequality we have 1− 1 1 p C ∞p(1+a) · a p(1+a) L L L1
(17)
for all 1 p ∞. Then via Lemma 2.1 we get (G (t) ε w )1 − G0 (t, x)
L1+
a
Cε t − 2(1+) − 2 ,
and via the assumption
w1 (t) − (G (t) ε w )1 L1+ Cε 1+ t − 2(1+) we have applying the Hölder inequality 1+ 1+ 1+ G − ε w (t) ( (t) ) w1 − 1+ G1+ w 0 1 1 L1 L1 1+ 1+ w )1+ x) + (G (t) ε − G (t, 1 1 0 L
w )1 L1+ w1 (t) − (G (t) ε w)1 L1+ C w1 L1+ + (G (t) ε +C (G (t) ε w )1 L1+ + G0 L1+ (G (t) ε w)1 − G0 (t, x) L1+
a
Cε2+ t − 2 + Cε1+ t − 2 − 2
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
for all t > 0. By a direct computation we obtain for the heat kernel G0 (t, x) = x2
(4t)− 2 e− 4t 1
(G0 (t, x))1+ dx = (1 + )− 2 (4t)− 2 . 1
R
Therefore we get 1+ − 2 t w11+ (t, x) dx − 1 R (4) 2 (1 + ) 2 = w11+ (t, x) dx − 1+ (G0 (t, x))1+ dx R R a 1+ 2+ t − 2 + Cε 1+ t − 2 − 2 C w1 − 1+ G1+ 0 1 Cε L
for all t > 0, whence t 1+ t 1− 2 1+ w1 (t, x) dx − d 1 0 2 2 R (4) 1 − 2 (1 + ) t d v (, x) dx − t 1− 2 = 0
Cε2+
R
t
− 2 d + Cε 1+
0
Cεt
1− 2
t
a
− 2 − 2 d
0
+ Cε 1+ t
1− 2 − a2
1 t 1− 2 2
for all t > 0. Thus we get
t 0
d R
w11+ (t, x) dx t 1− 2 .
Estimate (18) implies (16), since 1. Lemma 2.3 is proved.
(18)
Lemma 2.4. Let the function f (x) have the zero mean value fˆ(0) = 0 and the norm f L∞ + · a f L1
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
179
be finite, where a ∈ (0, 1), 1 p ∞. Then the following inequalities are valid: √ ! " 1 −y z 2 dz a − ( 4(1−z)) dye f (y) C · a f L1 + C f L∞ 1 p R 0 z (1 − z) 2 L and √ ! " 1 −y z 2 dz 1 t a − ( 4(1−z)) dye f (y) − 1 p tz z R 0 (1 − z) 2 L − a2 a · f L1 + f L∞ C t for all t > 0, where 1 p ∞.
Proof. By the Young inequality for convolutions we get √ −y z 2 ! "a − ( 4(1−z)) e f (y) dy p R L √ 2 √ C −y z 2 ! √ "a − C (−y z) − ( 1−z ) a 1−z |f (y)| dy + e y |f (y)| dy −y z e p Lp L 2 2 ! "a − C − C a 1−z · f p (1 − z) 21 · a f p 1−z e f Lp + e L L L1
for all z ∈
$1
L1
%
2, 1
, since
· − y √z f (y) dy
y = z (· − y) f √ dy p z R L 1 · Cz− 2p L1 f √z p C L1 f Lp . L 1 − 2p
Lp
Therefore ! " a
1
1 2
z (1 − z) 2
C
1
e
1
1 1 2
√
−y z 2 − ( 4(1−z))
R
(1 − z)− 2
1
f (y) dy dz
Lp
√ −y z 2 ! "a − ( 4(1−z)) e f dy (y) R
Lp
dz C · a f Lp
(19)
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
and ! " a
√ −y z)2 ( 1 t − 4(1−z) dye f − (y) 1 1 p z tz R 2 (1 − z) 2 L 1 √ 2 (−y z) 1 ! "a e− 4(1−z) f (y) dy C t −1 (1 − z)− 2 dz dz
1
1 2
C t −1 · a f
Lp
Lp
,
(20)
where 1 p ∞. Via the condition R f (y) dy = 0, we write √ −y z 2 ! "a − ( 4(1−z)) e f dy (y)
Lp
R
(−y √z)2 ! "a 2 − 4(1−z) − 4(1−z) e f (y) dy −e =
Lp
R
√ ! "a ! √ "a − (−y z)2 4(1−z) f (y) dy − −y z e
Lp
R
√ ! √ "a − (−y z)2 ! "a − 2 4(1−z) − e 4(1−z) f dy + − y z e (y) p R L √ 2 a e−C (−y z) y a |f (y)| dy Cz 2 p R L √ 2 a −C (−y z) −C 2 a |f y dy +Cz 2 + e e (y)| R
2 a Cz 2 · a f L1 e−C
Lp
a Cz 2 · a f L1
Lp
for all z ∈ 0, 21 , since √ · − y z f (y) dy p R L 1 y − 2p =z (· − y) f √ dy p z R L 1 1 1 · Cz 2 1− p p f L1 f Cz− 2p Lp √ L z L1 C Lp f L1 .
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181
Thus √ ! " 21 −y z 2 1 a − ( 4(1−z)) e f dy dz (y) 1 p 0 z (1 − z) 2 R L 1 √ 2 −y z ! "a 2 dz − ( 4(1−z)) e f (y) dy 1 p 0 z (1 − z) 2 R L 1 a 2 dz · f 1 C · a f 1 C 1 a L L 0 z1− 2 (1 − z) 2
(21)
and √ ! " 21 −y z 2 dz 1 t a − ( 4(1−z)) e f (y) dy − 1 p tz z R 0 (1 − z) 2 L 1 √ 2 −y z ! "a 2 1 t dz − ( 4(1−z)) f dy e − (y) 1 p z tz R 0 (1 − z) 2 L a 1 2 a 1 t z2 − C dz · a f L1 C t − 2 · a f L1 , 1 tz z 0 (1 − z) 2
(22)
where 1 p ∞. Collecting estimates (19)–(22), we get the results of the lemma. Lemma 2.4 is proved.
3. Proof of Theorem 1.1 Following the idea of papers [12,13], we make a change of the dependent variable u (t, x) = e−(t) w (t, x), then we get from (3) wt + Lw + e− N (w) − w = 0
(23)
or in the integral form − w (t) = G (t) w
t
G (t − ) e− N (w) − w d .
(24)
0
According to the large time asymptotic behavior described by Lemma 2.1 we need to demand that the real-valued function (t) satisfies the zero total
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mass condition
e− N (w) − w dx = 0,
A0 R
which is reduced to one equation e−(t)
1 + *x
R
= e−(t)
R
−1
w11+ (t, x) dx − (t)
w11+ (t, x) dx − (t)
R
R
(A0 w (t, x))1 dx
(A0 w (t, x))1 dx = 0.
We also can assume that (0) = 0. Since A0 L (0) = 0 via Eq. (23) we get d dt
R
(A0 w (t, x))1 dx = 0
that is
R
(A0 w (t, x))1 dx = ε =ε
R
u (x))1 dx (A0
R
−1 v0 (x) + 1 + *x v1 (x) dx
(v0 (x) + v1 (x)) dx ≡
=ε R
for all t > 0. Therefore we obtain the equation
(t) = e−(t)
1
R
w11+ (t, x) dx.
(25)
Multiplying Eq. (25) by the factor e(t) , then integrating with respect to time t > 0 and making a change of the dependent variable e(t) = g (t), we get
g (t) = 1 +
t 0
R
w11+ (, x) dx d .
(26)
Thus we need to solve the following integral equation: w = M (w) ,
(27)
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
183
where the operator
t
u− M(w) (t) = G (t) ε
g −1 () G (t − ) f () d
0
and 1
f (t) = N (w (t)) − w (t)
R
w11+ (t, x) dx.
We define w(0) = G (t) u and successive approximations w(k+1) = M w (k) for k = 0, 1, 2,... We prove that M is a contraction mapping in the set
w ∈ C [0, ∞); L1,a ∩ L∞ :
X =
& 1 a sup t 2 w (t) L∞ + t − 2 · a w (t) L1 2ε . t>0
We prove that the mapping M transforms the set X into itself, also we prove the estimates w(k) ∈ X; w (k) (t) − G (t) ε u
L1+
g (k) (t) A0 f
(k)
Cε1+ t − 2(1+) ,
1 1− t 2 , 2
(29)
A0 w (k) (t, x)
(t, x) dx = 0,
R
(28)
R
1
dx =
(30)
for all t > 0 and k 0, where f
(k)
(t) = N w
(k)
1+ 1 (k) (k) dx w1 (t, x) (t) − w (t) R
and g (k) (t) = 1 +
t 0
R
(k)
1+
w1 (, x)
dx d .
For k = 0 estimates (28) follow from Lemma 2.1. Then applying Lemma 2.3 we get estimate (29) with k = 0. Equalities (30) are true due to the definition of w (0) . Then by induction we suppose that (28)–(30) are valid for some k 0.
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
Applying the identity
1 + *x
−1
(x) = e
−x
x −∞
ey (y) dy
we see that −1 1 + *x
L∞
−1 1 + *x
C L∞ ,
L1,a
C L1,a .
Therefore as a consequence of (28) via interpolation inequality (17) we get estimates for the function f (k) (t) (k) f (t)
L∞
1 (k) (k) 1+ C w (t) ∞ 1 + w (t) 1 L L
Cε1+ t −
1+ 2
and a (k) · f (t)
L1
C w(k) (t) ∞ · a w(k) (t)
1+
L1
L
Cε1+ t
1 (k) w (t) 1 L
a− 2
which imply 1+ sup t 2 f (k) (t)
L∞
t>0
+ t
−a 2
a (k) · f (t) 1 Cε 1+ . L
(31)
By (29)–(31) we can apply Lemma 2.2 with = 1 − 2 , = 2 , to obtain (k+1) u (t) − G (t) ε w
L∞
C t − 2 sup t 1
t>0
1+ 2
C
t
1
0
g (k) ()
(k) f (t)
L∞
+ t
G (t − ) f (k) () d
L∞
−a 2
1 a (k) · f (t) 1 Cε 1+ t − 2 L
and a u · w (k+1) (t) − G (t) ε
L1
a · C
1+ a C t 2 sup t 2 f (k) (t) t>0
L∞
+ t
t 0
−a 2
1
G (t − ) f (k) () d
g (k) () L1 a a (k) · f (t) 1 Cε 1+ t 2 L
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
185
for all t > 0. In particular, we see that estimates w (k) (t) L1+ Cε t − 2(1+) and (k) w (t) − G (t) ε u L1+ Cε 1+ t − 2(1+) are true. Then application of Lemma 2.3 yields g (k+1) (t) = 1 +
t 0
(k+1)
R
w1
1+
(, x)
dx d
1 1− t 2 2
for all t > 0. Therefore estimates (28) and (29) are valid with k replaced by k + 1. Integrating (k+1)
wt
+ Lw (k+1) + e− N w (k) − w (k) = 0
with respect to x ∈ R we get d dt
w (k+1) dx +
e− N w (k) − w (k) dx = 0
R
R
In view of (30) A0
w
(k+1)
u (x) dx = e1 ,
(t, x) dx = A0
R
(32)
R
where 1 e1 = . 0 Therefore by (32)
f (k+1) (t, x) dx
A0 R
N w
= A0
R
= A0 R
(k+1)
(t, x) − w
(k+1)
1 (t)
R
1+ (k+1) w1 (t, x)
dx dx
1+ (k+1) w1 N w(k+1) (t, x) dx − dx = 0. (t, x) R
By induction we see that properties (28)–(30) are true for all k 0. Thus the mapping M transforms the set X into itself. In the same manner we can estimate the differences M(w(k) ) − M(w(k−1) ) to see that M is the contraction mapping in X. Therefore there exists a unique solution w of integral equation (27) in the set X.
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
Now let us compute the asymptotics of the solution. We first show the existence of solutions to the integral equation 2 1 V = (4)− 2 e− 4 −
1 1
(4) 2
dz
1
e
1
z (1 − z) 2
0
√
−y z 2 − ( 4(1−z))
F (y) dy,
(33)
R
where
= 1−
V 1+ d
2
R
and F (y) = V 1+ (y) − V (y)
V 1+ d . R
Lemma 3.1. Let ∈ 2 − ε 3 , 2 , where ε > 0 is sufficiently small. Then there exists an unique solution V ∈ L1,a ∩ L∞ to the integral equation (33). 2 1 Proof. Denote V (0) = (4)− 2 e− 4 and define the successive approximations 2 1 V (k+1) = (4)− 2 e− 4 −
1 1
(k) (4) 2
1 0
dz z (1 − z)
1 2
e−
√ 2 z) 4(1−z)
(−y
F (k) (y) dy
R
for k = 0, 1, 2, . . ., where (k)
= 1−
2
1+ d V (k)
R
and 1+ 1+ F (k) (y) = V (k) (y) − V (k) (y) d . V (k) R
First by the induction argument we prove the identities V R
(k)
F (k) (y) dy = 0
(y) dy = 1,
(34)
R
for all k 0. Since R V (0) (y) dy = 1 by the definition of F (k) (y), we see that (34) is true for k = 0. Now by induction we assume that equalities (34) hold for some k.
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
187
Then we have
V (k+1) d = 1 −
1
dz
1
1
1 2
0 z (1 − z) 2
(k) (4) √ (−y z)2 × e− 4(1−z) d F (k) (y) dy = 1,
R
R
R
hence it follows that
F (k+1) (y) dy = 0. R
Thus (34) are true for all k 0. Now let us prove the estimates
(k)
1 Ca ε 3
(35)
and sup
1p∞
a · V (k+1) − V (0)
Lp
Ca ε 2
(36)
for all k 0, where the constant Ca > 0 does not depend on k 0. For k = 0 we have
(0)
= 1+ 1 − 2 (4) 2 =
1−
2
e−
d
R
2
(1+)2 4
(4) (1 + )
1 2
1 , Ca ε 3
since ∈ 2 − ε 3 , 2 . For (36) with k = 0, by virtue of Lemma 2.4 we get sup · a V (1) − V (0) p
1p∞
L
√ ! " 1 −y z 2 dz C a − ( 4(1−z)) (0) = (0) sup dye F (y) 1 p
1p∞ R 0 z (1 − z) 2 L C (0) sup · a F (0) p L
1p∞ a (0) 1+ 1+ C (0) (0) · V V = (0) sup − V d (·) (·) p
1p∞ R L
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
CCa ε3
sup
1p∞
a (0) sup · V p
1p∞
L
(0) V
Lp
+
sup
1p∞
(0) V
+1
Lp
Ca ε 2 , since ε > 0 is sufficiently small. Now by induction we assume that inequalities (35) and (36) hold for some k. Then via (36) we have
(k+1) =
1− +
=
1+ d = V (k+1)
2
1 − 2
(0)
1−
1+ d V (0) 1− 2 R R 1+ (0) 1+ (k+1) V d − V
R
+ 1− 2
1+ (0) 1+ (k+1) V d − V R
2
1
1
− 4 V (0) ∞ + V (k+1)
L (4) 2 (1 + ) 2 −V (0) ∞ V (k+1) − V (0) 1 L L 1 1 2 − 4Ca ε , 1 1 − 2 (4) 2 (1 + ) 2 Ca ε 3
therefore (35) with k replaced by k + 1 is also true. Note that by (36) a (k+1) · V
Lp
· a V (k+1) − V (0)
Lp
+ · a V (0)
Lp
C.
We now use Lemma 2.4 to obtain sup
1p∞
a · V (k+2) − V (0)
Lp
√ ! " 1 −y z 2 dz a − ( 4(1−z)) (k+1) = (k+1) sup dye F (y) 1
1p∞ p R 0 z (1 − z) 2 L C (k+1) sup · a F (k+1) p L
1p∞ a (k+1) 1+ 1+ C (k+1) (k+1) · V V (k+1) sup −V d (·) (·) p
R 1p∞ L C
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
CCa ε
3
sup
1p∞
+
sup
1p∞
a (k+1) · V
sup
Lp
(k+1) V
1p∞
+1
(k+1) V
189
Lp
Lp
Ca ε 2 , since ε > 0 is sufficiently small. Therefore (35) and (36) are true for any k. In the same manner we obtain the estimates sup
1p∞
(k+1) − V (k) V
Lp
1 sup V (k) − V (k−1) p L 2 1p∞
(37)
for any k 1. Thus the successive approximations V (k) converge to an unique solution V ∈ L1,a ∩ L∞ of integral equation (33). Lemma 3.1 is proved. We are now in the position to prove asymptotics of solutions. By induction let us prove that for all k 0 t
− b2
b · w (k) (t) − e1 t − 21 V (k) √· t
Lp
Cε t − 2 + 2p − 1
1
(38)
for all t > 0, 1 p ∞, b ∈ [0, a], where = 21 min a, 1 − 2 . Estimate (38) is true for k = 0 since by Lemma 2.1 we have · − b2 b − 21 (0) (0) t · w (t) − t e1 V √ p t L b −1+ 1 −a u − e1 G0 (t) p Cε t 2 2p 2 . = t − 2 · b G (t) (39) L
We assume that (38) is valid for some k. Then from (38) it follows that 1+ t (k) (k) dx d w1 (, x) g (t) − (k) t 1− 2 = 1 + 0 R 1+ (k) −t 1− 2 d V 1 − 2 R 1+ 1+ t (k) 1+ − 2 (k) = 1 + dx d − d d w1 (, x) V 0 R R 1+ x 1+ C t (k) 1+ − 2 − 21 (k) V 1 + − √ dx d w1 (, x) 0
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
1 C t (k) 1 + w ∞ + − 2 V (k) ∞ L L 0 (k) · − 1 (k) × d √ w (, ·) − e1 2 V L1 Cε 1+ t − − 2 d 1 + C (k) t 1− 2 − 1+ 0
(40)
for t > 0, where 1 p ∞. Changing variables such that = zt and √ = y we have 1
(k) =
=
t
2 −1
− 2 − 21
G0 (t − )
0
1 1
(k) (4) 2
1
(k) (4t) 2
= t− 2
1
d
−1− 21
· √
− 21
V (0)
d
(t − )
d e
2
− (x−) 4(t−)
F
(k)
R
0
1
t
F
(k)
dz
1
1 2
√
e−
(xt −1/2 −y z)2 4(1−z)
z (1 − z) x x (k+1) , √ −V √ t t 0
√
F (k) (y) dy
where the Green function G0 (t) for the heat equation
G0 (t) =
G0 (t, x − y) (y) dy R x2
have a kernel G0 (t, x) = (4t)− 2 e− 4t . Therefore we get 1
b · b − 21 (k+1) (k+1) t − 2 · e t V − w √ (t) 1 p t L t b 1 · b − 21 (k+1) (k) · u + − f d = t − 2 e t V G G − √ (t) (t ) ( ) 1 p (k) () t 0 g L b · b − 1 (0) C t − 2 √ − w (0) (t) · e1 t 2 V p t L t − 1 − b2 b −1 (k) 2 − +C t · G − f d (t ) ( ) p g (k) ()
(k) 0 L b t 2 d C t − 2 + (k) · b (G (t − ) − G0 (t − ) A0 ) f (k) () p
0 L
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194 b
C t − 2 + (k)
t 1+ (k) · 2 d b (k) G0 (t − ) A0 f () − e1 +1 F √ · 0 2
Lp
b t · 1 1 −1 (k) · G0 (t − ) F 2 d − √ p 0 L
b
+
191
C t − 2
(k)
≡ I1 + I2 + I3 + I4 + I5 . Via (39) we have I1 Cεt
− 21 1− p1
a
t − 2 .
By (31), (40) and Lemma 2.2 with = 2 , = 1 − 2 + , where =
1 2
min a, 1 − 2
t − (k) 1 b −1 1− (k) 2 2 − (k) g () (k) I2 t · G (t − ) f () d p g ()
0 L t b 1 − 1 1− p1 − b − (k) G (t − ) f (k) () d Cεt 2 . C t − 2 · g () 0 Lp − b2
We have by the estimates of Lemma 2.1 (G (t − ) − G0 (t − ) A0 ) () ∞ C () ∞ , L L (G (t − ) − G0 (t − ) A0 ) () ∞ C (t − )− 1+a 2 · a () , L L1 a a · (G (t − ) − G0 (t − ) A0 ) () 1 C · () 1 L L for all 0 < < t, hence as in the proof of Lemma 2.2 with = 2 , = 1 − 2 we have b
C t − 2 I3 = (k)
t 2 d b (k) (G (t − ) − G0 (t − ) A0 ) f () · 0
a
Cεt − 2 + 2p − 2 . 1
1
Lp
In the same manner via Lemma 2.2 with = 2 , = 1 − 2 + we get b
C t − 2 I4 = (k)
b t · 1+ (k) − +1 (k) · 2 G0 (t − ) A0 f () − e1 F √ 0 1− p1 −
1 b C − 21 1+− p1 + t − 2 sup sup t 2 t (k)
t>0 1 p ∞ b · 1+ (k) − 1+ (k) 2 F × · A0 f (t) − e1 t √ p t L
d 1− 2 Lp
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
C −1 (k) t 2
× sup
Cεt
1− p1 −
1
sup
t>0 1 p ∞ − 21 1− p1 −
t2
sup t w (k) (t)
t>0
1 2
∞
1− p1 +
b
t − 2
(k) + V ∞
b · w (k) (t) − t − 21 e1 V (k) √· t
Lp
.
Finally changing independent variables = zt and √ = y and applying Lemma 2.4 we obtain b t · 1 C t − 2 1 −1 (k) · b 2 G d − − F √ (t ) 0 p
(k) 0 L b t 2 1 1 1 −1 C t − 2 − (x−) x b 4(t−) F (k) 2 (t − )− 2 d − = d e √
(k) Lp 0 1 √ x −y z 2 dz 1 t − 21 1− p1 − ( 4(1−z)) b (k) x = Cεt F e dy − (y) 1 p z tz 0 (1 − z) 2 L
I5 =
Cεt
− 21 1− p1 −
.
Hence by induction (38) is true for any k 0 uniformly with respect to k. Taking b = 0 and a limit k → ∞ in (38) we get 1 1 w (t) − e1 t − 21 V √· Cεt − 2 1− p − t Lp
for all t > 0.
(41)
That is by virtue of (40) and (41) we have the asymptotics w (t) = t − 2 e1 V 1
· √
t
1 + O t − 2 −
(42)
and g (t) = t 1− 2 1 + O t −
(43)
for t → ∞ uniformly with respect to x ∈ R. Therefore via the formula u (t, x) = e−(t) w (t, x) and (42), (43) we obtain the asymptotics of the solution. Theorem 1.1 is proved.
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
193
Acknowledgments This work of P.I.N. and E.I.K. is partially supported by CONACYT. We are grateful to an unknown referee for many useful suggestions and comments.
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Damped wave equation in the subcritical case Nakao Hayashia,∗ , Elena I. Kaikinab , Pavel I. Naumkinc a Department of Mathematics, Graduate School of Science, Osaka University, Osaka,
Toyonaka 560-0043, Japan b Departamento de Ciencias Básicas, Instituto Tecnológico de Morelia, Morelia CP 58120,
Michoacán, Mexico c Instituto de Matemáticas, Unam Campus Morelia, AP 61-3 (Xangari), Morelia CP 58089,
Michoacán, Mexico Received 22 July 2003; revised 31 March 2004 Available online 12 August 2004
Abstract We study large time asymptotics of small solutions to the Cauchy problem for the one dimensional nonlinear damped wave equation vtt + vt − vxx + v 1+ = 0, x ∈ R, t > 0, (1) v (0, x) = εv0 (x) , vt (0, x) = εv1 (x) −1 v1 ∈ in the sub critical case ∈ 2 − ε 3 , 2 . We assume that the initial data v0 , 1 + *x L∞ ∩ L1,a , a ∈ (0, 1) where
L1,a = ∈ L1 ; L1,a = · a L1 < ∞ , x = 1 + x 2 . Also we suppose that the mean value of initial data (v0 (x) + v1 (x)) dx > 0. R
∗ Corresponding author. E-mail addresses: [email protected] (N. Hayashi), [email protected] (P.I. Naumkin).
0022-0396/$ - see front matter © 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jde.2004.06.018
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Then there exists a positive value ε such that the Cauchy problem (1) has a unique global ∞ 1,a , satisfying the following time decay estimate: solution v (t, x) ∈ C [0, ∞) ; L ∩ L v (t) L∞ Cε t −
1
for large t > 0, here 2 − ε 3 < < 2. © 2004 Elsevier Inc. All rights reserved. MSC: 35B33; 35B40 Keywords: Damped wave equation; Subcritical nonlinearity; Asymptotic expansion; Large time behavior
1. Introduction We study the one-dimensional nonlinear damped wave equation
vtt + vt − vxx + v 1+ = 0, x ∈ R, t > 0, v (0, x) = εv0 (x) , vt (0, x) = εv1 (x) , x ∈ R
(2)
in the sub critical case ∈ 2 − ε 3 , 2 , where ε > 0. Note that the case = 2 is critical for the asymptotic behavior of solutions to (2). Recently, much attention was drawn to nonlinear wave equations with dissipative terms. The blow-up results were proved in [28] for the case of nonlinearity − |v|1+ , with < 2, when the initial data are such that R v0 (x) dx > 0, R v1 (x) dx > 0. Blow-up results for the critical and sub critical cases 2 were obtained in [21,28,30]. In paper [28], it was proved global existence and large time decay estimates of solutions to the Cauchy problem for the damped wave equation (2) with nonlinearities ± |v|1+ or ± |v| v, for the super critical case > 2, if the initial data are sufficiently small and have a compact support. The large time asymptotic behavior of solutions to (2) for the super critical case > 2 was obtained in paper [15] in the framework of the usual L2 -theory. When the initial data are in the usual Sobolev space * u0 ∈ L1 ∩ L∞ , || 1, u1 ∈ L1 ∩ L∞ , problem (2) was considered in [24,27]. Via the energy-type estimates obtained in papers [22,19] it was proved in [17] that solutions of the nonlinear damped wave equation (2) in the super critical cases > 3 with arbitrary initial data u0 ∈ H1 ∩ L1 , u1 ∈ L2 ∩ L1 (i.e. without smallness assumption on the initial data) have the same large time asymptotics 2 as that for the linear heat equation *t − *x , that is u (t) − MG0 (t) Lp
1 1 − 1− p =o t 2 |x|2
as t → ∞, where 2 p ∞, and G0 (t, x) = (4t)− 2 e− 4t is the heat kernel, M is a constant. Recently the critical case = 2 was considered in paper [14], where it was 1
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
163
proved that solutions of (2) have an additional large time decay v (t) L∞ C t − 2 (1 + log t )− 2 , 1
1
√ where t = 1 + t 2 . Note that similar behavior first was discovered for the nonlinear heat equation vt − vxx − v 1+ = 0 in the critical case = 2, comparing with the linear heat equation, (see [6,11]). For blow-up results we refer [5,8,20]. Large time behavior of solutions to the nonlinear heat equations in the sub critical cases ∈ (1, 2) was obtained in papers [3,4,9,18,29]. −1 −1 −1 Taking v = u1 and 1 + *x vt = u2 , with 1 + *x = F→x 1 + i Fx→ = x e−x −∞ dx ex , we rewrite Eq. (2) in the form of a system of nonlinear evolutionary equations ut + N (u) + Lu = 0
(3)
for the vector u (t, x) =
u1 (t, x) u2 (t, x)
.
The initial data are u (0, x) = u (x) ≡
εv0 (x) −1 ε 1 + *x v1 (x)
.
The linear part of system (3) is a pseudodifferential operator defined by the Fourier transformation as follows: Lu = F→x L Fx→ u, with a matrix-symbol L = Lj k j,k=1,2 =
− 1 + i
0 2
1
1+i
The nonlinearity is defined by
N (u) = 1 + *x
−1
0
u1+ 1
.
We define the direct Fourier transformation Fx→ by 1 uˆ ≡ Fx→ u = (2)− 2
R
e−i x u (x) dx,
.
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
then the inverse Fourier transformation F→x is − 21
uˇ (x) ≡ F→x u = (2)
eix u d . R
Denote by 1 1 1 − 4 2 , 1 = + 2 2
1 1 2 = − 1 − 4 2 2 2
the eigenvalues of the matrix L . Note that the matrix
Q =
1 + i 1 + i Q11 Q12 = −1 −2 Q21 Q22
and Q−1 =
1 1 + i 1 − 2
−2 − 1 + i 1 1 + i
diagonalize the matrix L , i.e. Q
−1
L Q =
0 1 . 0 2
Consider the system of ordinary differential equations with constant coefficients depending on the parameter ∈ R d u t, = 0. u t, + L dt
(4)
Multiplying system (4) by Q−1 from the left and changing u t, = Q w t, we diagonalize system (4) d dt
0 w1 t, 1 w1 t, =− , w2 t, 0 2 w2 t,
whence integrating with respect to time t 0 we find
−t w1 t, 0 w1 0, e 1( ) = w2 t, w2 0, 0 e−t 2 ()
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165
Returning to the solution u t, we get w1 t, u1 (t, x) u t, = =Q u2 (t, x) w2 t, e−t 1 () u0 0 −1 =Q Q u1 0 e−t 2 () u0 = e−tL() , u1
where the fundamental Cauchy matrix has the form e
−tL()
0 e−t 1 () Q−1 −t ( ) 2 0 e − 2 − 1 + i 1 = e−t 1 () 2 1 2 1+i 1 − 4 1 1 + i 1 e−t 2 () . + 2 − − 1+i 2 2 1 − 4
=Q
We rewrite the Cauchy problem (3) in the form of the integral equation u (t) = G (t) u−
t
G (t − ) N (u) () d ,
(5)
0
ˆ . So by the solution of the where the Green operator G (t) = F→x e−tL() Cauchy problem (3) we always understand the solution u (t, x) of the corresponding integral equation (5), belonging to C0 ([0, ∞); X) with an appropriate choice of a functional space X. In the present paper we prove the following result. Denote
L1,a = ∈ L1 : L1,a = · a L1 < ∞ . 2 Theorem 1.1. We assume that the initial data u ∈ L∞ ∩ L1,a , a ∈ (0, 1), and the mean value
=ε
u1 (x) + u2 (x)) dx > 0. ( R
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Then there exists a positive ε such that the Cauchy problem for Eq. (3) has a unique 2 satisfying the following time decay mild solution u (t, x) ∈ C [0, ∞) ; L∞ ∩ L1,a estimate: u (t) L∞ Cε t − 1
for large t > 0 and any ∈ 2 − ε 3 , 2 . Furthermore, the asymptotic formula
− 1
u (t, x) = e1 (t )
V
x √
+O t
t
− 1 −
,
is valid for t → ∞ uniformly with respect to x ∈ R, where = 1 e1 = , V ∈ L1,a ∩ L∞ is the solution of the integral equation 0 V =
1
2
e− 4 − 1
(4) 2
1 1
(4) 2
dz
1
1
z (1 − z) 2
0
e−
√ 2 z) 4(1−z)
(−y
1 2
min a, 1 − 2 ,
F (y) dy,
(6)
R
where
=
1−
2
V 1+ (y) dy R
and F (y) = V 1+ (y) − V (y)
V 1+ d . R
Remark 1.1. As a consequence of Theorem 1.1 we have the following asymptotics for the damped wave equation (2): v (t, x) = (t )− V 1
x √
t
1 + O t − −
−1 v1 ∈ for t → ∞ uniformly with respect to x ∈ R if the initial data v0 , 1 + *x 2 ∞ 1,a , a ∈ (0, 1), are such that the mean value L ∩L R
−1 v0 (x) + 1 + *x v1 (x) dx > 0.
(7)
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167
We believe that condition (7) is necessary for the global existence of solutions for the −1 v1 (x) dx 0 solutions damped wave equation. In the case R v0 (x) + 1 + *x could blow up in a finite time. Remark 1.2. The restriction ∈ 2 − ε 3 , 2 with ε sufficiently small is only technical. It comes from the application of the contraction mapping principle in the proof of Lemma 3.1 below. The asymptotic behavior of solutions obtained in Theorem 1.1 is general whenever ∈ (0, 2) with some decay restrictions on the initial data, as it happens for the nonlinear heat equation (see [4]). Indeed in [26], a sharp time decay of solutions in Lp norm is obtained recently under the condition such that the data decay exponentially at infinity but do not have any restriction on the size, where 2 2 p n−2 for the space dimension n 3, 2 p < ∞ for n = 2 and 2 p ∞ for n = 1. For higher space dimension with super critical nonlinearities see [16,23–25] in which asymptotic expansion of small solutions in uniform norm was studied. We organize the rest of the paper as follows. In Section 2 we obtain some preliminary estimates of the Green operator solving the linearized Cauchy problem corresponding to (3). Section 3 is devoted to the proof of Theorem 1.1. 2. Preliminary lemmas We first collect some preliminary estimates for the Green operator G (t) = F→x e−tL() ˆ , in the weighted Lebesgue norms Lp and L1,a , where L1,a = · a L1 , a ∈ (0, 1), 1 p ∞. Also we show that G (t) asymptotically behaves as a Green function G0 (t) for the heat equation 2 G0 (t) = F→x e−t ˆ = G0 (t, x − y) (y) dy R
with a kernel 2 1 1 x2 G0 (t, x) = (2)− 2 F→x e−t = (4t)− 2 e− 4t . Denote a matrix A0 =
1 1 0 0
and a vector
ϑ=
(x) dx. R
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
2 Lemma 2.1. Suppose that the vector-function ∈ L∞ ∩ L1,a , where a ∈ (0, 1). Then the estimates G (t) p C p , L L and 2 |·| G (t) − (4t)− 21 e− x4t A0 ϑ
Lp
Ct
− 21 1− p1 − a− 2
L1,a
are valid for all t > 0, 1 p ∞, 0 a. Proof. We represent the fundamental Cauchy matrix e−tL() in the form e−tL()
cos t = e− 2
t 2
−
√ sin 2t 42 −1 4 − 1 + √ 2 4 −1 √ 2 t 2
22
(1+i )
sin
2 √
4 −1
42 −1
√ sin 2t 42 −1 √ 2 2 1 + i 4 −1 √ sin 2t 42 −1 2 t cos 2 4 − 1 − √ 2
.
4 −1
We can see that the components of the matrix e−tL() are C∞ -functions. Define the ∞ 2 and = 0 for ∈ C = 1 for such that
cut off function
(R) 1 1 1 3, = 1 − . Denote 2 1 Gj (t) = F→x j e−tL() ˆ = Gj (t, x − y) (y) dy.
(8)
R
1 with matrix-kernels Gj (t, x) = (2)− 2 F→x j e−tL() . Note that the kernel 1 G1 (t, x) = (2)− 2 F→x 1 e−tL() in representation (8) is a smooth matrixfunction G1 (t, x) ∈ (C∞ (R × R))4 . In view of the estimates e
− 2t
m k 2 * cosh t 1 − 42 C t m−k 2 e −Ct 1 2
and
t
e− 2
sinh 2t 1 − 42 m k 2 C t m−k * 2 e −Ct 1 1 − 4 2
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
169
for all ∈ R, t > 0, where m, k 0, we get 1 m m k x *x G1 (t, x) = (2)− 2 F→x * 1 k e−tL() m C * 1 k e−tL() 1
C t
m−k 2
−Ct 2 e
L
L1
(|| 3)
C t
m−1−k 2
for all x ∈ R, t > 0. Therefore G1 (t, x) have the following estimate: 1 sup x t − 2
x∈R
3 k *x G1 (t, x)
C t −
1+k 2
(9)
for all t > 0, where k 0. By virtue of (9) with k = 0 we find − 21
G1 (t) L1 C t
x t − 2
1
−3
dx C,
R
whence applying the Young inequality we obtain G1 (t) p C G1 (t) L1 p C p L L L for all t > 0, where 1 p ∞. Then applying the Lagrange finite differences Theorem, in view of (9) we obtain |x| |G1 (t, x − y) − G1 (t, x)| a |x| |y|a *x G1 (t, x ) (|G1 (t, x − y)| + |G1 (t, x)|)1−a 1 a− 1 −2 C t − 2 − 2 |y|a x t − 2 for all x, y ∈ R, if |y| |x| 2 or if
|x| 2
√ √ |y| t . When |y| max |x| t we get , 2
|x| |G1 (t, x − y) − G1 (t, x)| C |y| (|G1 (t, x − y)| + |G1 (t, x)|) −3 −3 − 21 − a− a − 21 − 21 2 |y| . C t + x t (x − y) t
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Thus we obtain |x| (|G1 (t, x − y) − G1 (t, x)|) 1 − 21 − a− a 2 |y| C t (x − y) t − 2
−2
+ x t
− 21
−2
for all x, y ∈ R, whence |x| (y) dy x − y) − G x)) (G (t, (t, 1 1 p R L −2 1 a− −1 · a 1 C t − 2 − 2 t 2 (·) L
Lp
− 21 1− p1 − a− 2
C t
a ·
L1
for all t > 0, 1 p ∞, ∈ [0, a]. Furthermore in view of the estimates m 2 2 * e− 2t cosh t 1 − 42 − 1 e−t C t m−1 2 e −Ct 1 2 2 and sinh 2t 1 − 42 m −t 2 2 m−1 1 −t * e 2 − e C t 2 e−Ct 1 2 1 − 4 2 for all ∈ R, t > 0, m 0, we estimate the difference 2 m − 21 − x4t x G t) e A x) − (t, (4 1 0 2 1 m = (2)− 2 F→x * 1 e−tL() − e−t A0 2 m C * 1 e−tL() − e−t A0 1
C t
m−1 2
−Ct 2 e
L1
L
m
Ct 2 −1
for all x ∈ R, t > 0, 0 m 2. Thus we obtain
sup xt
x∈R
− 21
3−
2 − 21 − x4t |x| G1 (t, x) − (4t) e A0 Ct 2 −1
(10)
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171
for all t > 0. By virtue of (10) we find 2 |x| G1 (t, x) − (4t)− 21 e− x4t A0 −2 − 21 −1 2 Ct (·) t
Lp
Ct
Lp − 21 1− p1 − 1− 2
.
Therefore the second estimate of the lemma follows 2 |x| G1 (t) − (4t)− 21 e− x4t A0 ϑ
Lp
2 − 21 − x4t |x| G t) e A (y) dy x − y) − = (t, (4 1 0 p R L |x| (G1 (t, x − y) − G1 (t, x)) (y) dy Lp
R
2 − 21 − x4t + |x| G1 (t, x) − (4t) e A0 ϑ
Ct
− 21
1− p1
− a− 2
a ·
L1
Lp
.
Next we estimate G2 (t) = F→x 2 e−tL() ˆ 1+ 2*t R F2 (t,x − y)1 (y) dy t +2 R F2 (t, x − y) 1 + *y 2 (y) dy, = e− 2 2 F2 (t, x − y) *y − 1 1 (y) dy − F3 (t, x − y) 1 (y) dy R R + 2*t − 1 R F2 (t, x − y) 2 (y) dy, with kernels
1 F2 (t, x) = (2)− 2 F→x 2
sinh
t 2
1 − 4
1 − 42
2
and sinh t 1 − 42 2 2 2 . F→x 1 + i 2 1 − 4
F3 (t, x) = (2)− 2
1
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
We use the asymptotic formula e
it 2
√
42 −1
=e
it ||
t eit || + O t 2 −2 , + 2i 1 + 2
(11)
whence
k *
sin
t 2
4 − 1
t cos t − sin t ! "2 − 2 2 8 1 + 2
2
4 2 − 1
−3 = O t 2+k
(12)
for → ∞, k = 0, 1, 2, therefore
1 F2 (t, x) = √ 2
eix
sin
t 2
R
4 − 1 2
4 2 − 1
2 d
= H1 (t, x) + H2 (t, x) + R1 (t, x) , where 1 H1 (t, x) = √ 2 2 H2 (t, x) =
t √
e R
8 2
ix
d sin t ! "2 ,
d eix cos t ! "2 R
and d e sin t 1 ! "2 R sin t 2 − 1 1 ix +√ e 2 R 2 − 1
1 R1 (t, x) = √ 2 2
ix
t cos t 2 d . − sin t ! "2 − 2 2 8 1+
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
173
Computing the Fourier transform we get (see [1,7]) ∞ d 1 cos x sin t H1 (t, x) = √ 2 0 1 + 2 1 e−t cosh x, |x| < t, 2 2 = − 1 e−|x| sinh t, |x| > t, 2 2 H2 (t, x) =
t √
∞
cos x cos t
d
4 2 0 1 + 2 t e−t cosh x, |x| < t, 8 2 = t e−|x| cosh t, |x| > t. 8 2
In view of asymptotics (12) we have the estimate l k *t *x R1 (t, x) C t 2 x −2 for all t > 0, x ∈ R, where k + l = 0, 1. Therefore by the Young inequality we get F2 (t, x − y) 1 (y) dy
L∞
R
C t 2 1 L∞
and F2 (t, x − y) 1 (y) dy
L1,a
R
C t 4 1 L1,a .
In the same manner we estimate the kernel F3 (t, x) |F3 (t, x)| C t 2 x −2 for all t > 0, x ∈ R, hence F x − y) dy (t, (y) 3 1
L∞
R
C t 2 1 L∞
and F3 (t, x − y) 1 (y) dy R
L1,a
C t 4 1 L1,a .
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
We now estimate terms with derivatives *t and *y . Since *t F2 (t, x − y) 1 (y) dy R
=
#
4 1− H2 (t, x − y) 1 (y) dy 1 (x − t) − 1 (x + t) + 2 t R 1 − H1 (t, x − y) 1 (y) dy + *t R1 (t, x − y) 1 (y) dy, 2 R R 1 2
whence we get estimates *t x − y) dy F (t, (y) 2 1
L∞
R
C t 2 1 L∞
and * t F2 (t, x − y) 1 (y) dy
L1,a
R
C t 4 1 L1,a .
Similarly
1 F2 (t, x − y) *y 1 (y) dy = 2 R
#
1 (x − t) − 1 (x + t) 2
+
R
R
+ + R
*x H1 (t, x − y) 1 (y) dy *x H2 (t, x − y) 1 (y) dy *x R1 (t, x − y) 1 (y) dy,
whence the estimates x − y) dy F * (t, (y) 2 y 1
L∞
R
C t 2 1 L∞
and F2 (t, x − y) *y 1 (y) dy
L1,a
R
follow. Lemma 2.1 is proved.
C t 4 1 L1,a
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
175
Next lemma will be used for estimates of the nonlinearity via contraction mapping principle. Lemma 2.2. Let the vector-function (t, x) be such that A0 norm
R
(t, x) dx = 0 and the
1 a sup t + 2 (t) L∞ + sup t − 2 · a (t) L1 ≡ F t>0
t>0
be finite, where a ∈ (0, 1), ∈ (0, 1). We also suppose that the function g (t) is such that g (t) t for all t > 0, where > 0 is such that + − a2 < 1. Then the following inequalities are valid: t g −1 () G (t − ) () d
L∞
0
1 C t 2 −− F
and a t −1 · g () G (t − ) () d
L1
0
a C t 1+ 2 −− F
for all t > 0.
Proof. Since A0
R
(t, x) dx = 0, we have by the estimates of Lemma 2.1 G (t − ) () ∞ C () ∞ , L L
(13)
G (t − ) () ∞ C (t − )− 1+a 2 · a () L L1
(14)
a · G (t − ) () 1 C · a () 1 L L
(15)
and
for all 0 < < t. Therefore by virtue of (13) and (14) we get t g −1 () G (t − ) () d
L∞
0
C
t 2
0
a 1 (t − )− 2 − 2 · a () L1 − d + C
t t 2
()
L∞
− d
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C F
t 2
− a2 − 21
(t − )
a 2 −−
d +
t t 2
0
−−− 21
d
1 C t 2 −− F for all t > 0. In the same manner via (15) we find a t −1 · g G − d ( ) (t ) ( )
L1
0
C
t
− d C F L1
a · ()
0
1+ a2 −−
C t
for all t > 0. Lemma 2.2 is proved. Denote = that 1.
(4) 2
1+
1
(1− 2 )(1+) 2
t
a
2 −− d
0
F
. Since 2 − ε 3 < < 2 and Cε, we may suppose
2 with a ∈ (0, 1) and ≡ Lemma ∈ L1,a ∩ L∞ 2.3. We assume that w (x) dx 1 Cε > 0. Let the function (w (t, x))1 = w1 (t, x) satisfy the ε A0 R w estimates
w1 (t) L1+ Cε t − 2(1+) and
w1 (t) − (G (t) ε w )1 L1+ Cε 1+ t − 2(1+) for all t > 0. Then there exists ε > 0 such that the following inequality is valid: 1+
t 0
for all t > 0.
d R
w11+ (, x) dx
1 1− t 2 2
(16)
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177
Proof. By the Hölder inequality with r ∈ (1, 1 + a), a > 0, we obtain choosing 1 − 1 = · a a 1 1a > 0 L
1 L
L
a − a 2r 2r (x) dx = 2 + x 2 2 + x 2 R
1 1− 1 a − a r r r 2 2(r−1) 2 + x 2 dx 2 + x 2 (x) dx R
C
R
1− 1r
1 1 1− 1r 1r ∞ 1 + C 1− 1r − ar 1−∞r · a r 1 L L L L
r−1 1 − r−1 r−1 C Lr∞ · a Lar1 Lr 1 ar , whence
L1
1 a a 1+a · 1 . C L1+a ∞ L
Again applying the Hölder inequality we have 1− 1 1 p C ∞p(1+a) · a p(1+a) L L L1
(17)
for all 1 p ∞. Then via Lemma 2.1 we get (G (t) ε w )1 − G0 (t, x)
L1+
a
Cε t − 2(1+) − 2 ,
and via the assumption
w1 (t) − (G (t) ε w )1 L1+ Cε 1+ t − 2(1+) we have applying the Hölder inequality 1+ 1+ 1+ G − ε w (t) ( (t) ) w1 − 1+ G1+ w 0 1 1 L1 L1 1+ 1+ w )1+ x) + (G (t) ε − G (t, 1 1 0 L
w )1 L1+ w1 (t) − (G (t) ε w)1 L1+ C w1 L1+ + (G (t) ε +C (G (t) ε w )1 L1+ + G0 L1+ (G (t) ε w)1 − G0 (t, x) L1+
a
Cε2+ t − 2 + Cε1+ t − 2 − 2
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for all t > 0. By a direct computation we obtain for the heat kernel G0 (t, x) = x2
(4t)− 2 e− 4t 1
(G0 (t, x))1+ dx = (1 + )− 2 (4t)− 2 . 1
R
Therefore we get 1+ − 2 t w11+ (t, x) dx − 1 R (4) 2 (1 + ) 2 = w11+ (t, x) dx − 1+ (G0 (t, x))1+ dx R R a 1+ 2+ t − 2 + Cε 1+ t − 2 − 2 C w1 − 1+ G1+ 0 1 Cε L
for all t > 0, whence t 1+ t 1− 2 1+ w1 (t, x) dx − d 1 0 2 2 R (4) 1 − 2 (1 + ) t d v (, x) dx − t 1− 2 = 0
Cε2+
R
t
− 2 d + Cε 1+
0
Cεt
1− 2
t
a
− 2 − 2 d
0
+ Cε 1+ t
1− 2 − a2
1 t 1− 2 2
for all t > 0. Thus we get
t 0
d R
w11+ (t, x) dx t 1− 2 .
Estimate (18) implies (16), since 1. Lemma 2.3 is proved.
(18)
Lemma 2.4. Let the function f (x) have the zero mean value fˆ(0) = 0 and the norm f L∞ + · a f L1
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
179
be finite, where a ∈ (0, 1), 1 p ∞. Then the following inequalities are valid: √ ! " 1 −y z 2 dz a − ( 4(1−z)) dye f (y) C · a f L1 + C f L∞ 1 p R 0 z (1 − z) 2 L and √ ! " 1 −y z 2 dz 1 t a − ( 4(1−z)) dye f (y) − 1 p tz z R 0 (1 − z) 2 L − a2 a · f L1 + f L∞ C t for all t > 0, where 1 p ∞.
Proof. By the Young inequality for convolutions we get √ −y z 2 ! "a − ( 4(1−z)) e f (y) dy p R L √ 2 √ C −y z 2 ! √ "a − C (−y z) − ( 1−z ) a 1−z |f (y)| dy + e y |f (y)| dy −y z e p Lp L 2 2 ! "a − C − C a 1−z · f p (1 − z) 21 · a f p 1−z e f Lp + e L L L1
for all z ∈
$1
L1
%
2, 1
, since
· − y √z f (y) dy
y = z (· − y) f √ dy p z R L 1 · Cz− 2p L1 f √z p C L1 f Lp . L 1 − 2p
Lp
Therefore ! " a
1
1 2
z (1 − z) 2
C
1
e
1
1 1 2
√
−y z 2 − ( 4(1−z))
R
(1 − z)− 2
1
f (y) dy dz
Lp
√ −y z 2 ! "a − ( 4(1−z)) e f dy (y) R
Lp
dz C · a f Lp
(19)
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
and ! " a
√ −y z)2 ( 1 t − 4(1−z) dye f − (y) 1 1 p z tz R 2 (1 − z) 2 L 1 √ 2 (−y z) 1 ! "a e− 4(1−z) f (y) dy C t −1 (1 − z)− 2 dz dz
1
1 2
C t −1 · a f
Lp
Lp
,
(20)
where 1 p ∞. Via the condition R f (y) dy = 0, we write √ −y z 2 ! "a − ( 4(1−z)) e f dy (y)
Lp
R
(−y √z)2 ! "a 2 − 4(1−z) − 4(1−z) e f (y) dy −e =
Lp
R
√ ! "a ! √ "a − (−y z)2 4(1−z) f (y) dy − −y z e
Lp
R
√ ! √ "a − (−y z)2 ! "a − 2 4(1−z) − e 4(1−z) f dy + − y z e (y) p R L √ 2 a e−C (−y z) y a |f (y)| dy Cz 2 p R L √ 2 a −C (−y z) −C 2 a |f y dy +Cz 2 + e e (y)| R
2 a Cz 2 · a f L1 e−C
Lp
a Cz 2 · a f L1
Lp
for all z ∈ 0, 21 , since √ · − y z f (y) dy p R L 1 y − 2p =z (· − y) f √ dy p z R L 1 1 1 · Cz 2 1− p p f L1 f Cz− 2p Lp √ L z L1 C Lp f L1 .
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
181
Thus √ ! " 21 −y z 2 1 a − ( 4(1−z)) e f dy dz (y) 1 p 0 z (1 − z) 2 R L 1 √ 2 −y z ! "a 2 dz − ( 4(1−z)) e f (y) dy 1 p 0 z (1 − z) 2 R L 1 a 2 dz · f 1 C · a f 1 C 1 a L L 0 z1− 2 (1 − z) 2
(21)
and √ ! " 21 −y z 2 dz 1 t a − ( 4(1−z)) e f (y) dy − 1 p tz z R 0 (1 − z) 2 L 1 √ 2 −y z ! "a 2 1 t dz − ( 4(1−z)) f dy e − (y) 1 p z tz R 0 (1 − z) 2 L a 1 2 a 1 t z2 − C dz · a f L1 C t − 2 · a f L1 , 1 tz z 0 (1 − z) 2
(22)
where 1 p ∞. Collecting estimates (19)–(22), we get the results of the lemma. Lemma 2.4 is proved.
3. Proof of Theorem 1.1 Following the idea of papers [12,13], we make a change of the dependent variable u (t, x) = e−(t) w (t, x), then we get from (3) wt + Lw + e− N (w) − w = 0
(23)
or in the integral form − w (t) = G (t) w
t
G (t − ) e− N (w) − w d .
(24)
0
According to the large time asymptotic behavior described by Lemma 2.1 we need to demand that the real-valued function (t) satisfies the zero total
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mass condition
e− N (w) − w dx = 0,
A0 R
which is reduced to one equation e−(t)
1 + *x
R
= e−(t)
R
−1
w11+ (t, x) dx − (t)
w11+ (t, x) dx − (t)
R
R
(A0 w (t, x))1 dx
(A0 w (t, x))1 dx = 0.
We also can assume that (0) = 0. Since A0 L (0) = 0 via Eq. (23) we get d dt
R
(A0 w (t, x))1 dx = 0
that is
R
(A0 w (t, x))1 dx = ε =ε
R
u (x))1 dx (A0
R
−1 v0 (x) + 1 + *x v1 (x) dx
(v0 (x) + v1 (x)) dx ≡
=ε R
for all t > 0. Therefore we obtain the equation
(t) = e−(t)
1
R
w11+ (t, x) dx.
(25)
Multiplying Eq. (25) by the factor e(t) , then integrating with respect to time t > 0 and making a change of the dependent variable e(t) = g (t), we get
g (t) = 1 +
t 0
R
w11+ (, x) dx d .
(26)
Thus we need to solve the following integral equation: w = M (w) ,
(27)
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
183
where the operator
t
u− M(w) (t) = G (t) ε
g −1 () G (t − ) f () d
0
and 1
f (t) = N (w (t)) − w (t)
R
w11+ (t, x) dx.
We define w(0) = G (t) u and successive approximations w(k+1) = M w (k) for k = 0, 1, 2,... We prove that M is a contraction mapping in the set
w ∈ C [0, ∞); L1,a ∩ L∞ :
X =
& 1 a sup t 2 w (t) L∞ + t − 2 · a w (t) L1 2ε . t>0
We prove that the mapping M transforms the set X into itself, also we prove the estimates w(k) ∈ X; w (k) (t) − G (t) ε u
L1+
g (k) (t) A0 f
(k)
Cε1+ t − 2(1+) ,
1 1− t 2 , 2
(29)
A0 w (k) (t, x)
(t, x) dx = 0,
R
(28)
R
1
dx =
(30)
for all t > 0 and k 0, where f
(k)
(t) = N w
(k)
1+ 1 (k) (k) dx w1 (t, x) (t) − w (t) R
and g (k) (t) = 1 +
t 0
R
(k)
1+
w1 (, x)
dx d .
For k = 0 estimates (28) follow from Lemma 2.1. Then applying Lemma 2.3 we get estimate (29) with k = 0. Equalities (30) are true due to the definition of w (0) . Then by induction we suppose that (28)–(30) are valid for some k 0.
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Applying the identity
1 + *x
−1
(x) = e
−x
x −∞
ey (y) dy
we see that −1 1 + *x
L∞
−1 1 + *x
C L∞ ,
L1,a
C L1,a .
Therefore as a consequence of (28) via interpolation inequality (17) we get estimates for the function f (k) (t) (k) f (t)
L∞
1 (k) (k) 1+ C w (t) ∞ 1 + w (t) 1 L L
Cε1+ t −
1+ 2
and a (k) · f (t)
L1
C w(k) (t) ∞ · a w(k) (t)
1+
L1
L
Cε1+ t
1 (k) w (t) 1 L
a− 2
which imply 1+ sup t 2 f (k) (t)
L∞
t>0
+ t
−a 2
a (k) · f (t) 1 Cε 1+ . L
(31)
By (29)–(31) we can apply Lemma 2.2 with = 1 − 2 , = 2 , to obtain (k+1) u (t) − G (t) ε w
L∞
C t − 2 sup t 1
t>0
1+ 2
C
t
1
0
g (k) ()
(k) f (t)
L∞
+ t
G (t − ) f (k) () d
L∞
−a 2
1 a (k) · f (t) 1 Cε 1+ t − 2 L
and a u · w (k+1) (t) − G (t) ε
L1
a · C
1+ a C t 2 sup t 2 f (k) (t) t>0
L∞
+ t
t 0
−a 2
1
G (t − ) f (k) () d
g (k) () L1 a a (k) · f (t) 1 Cε 1+ t 2 L
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
185
for all t > 0. In particular, we see that estimates w (k) (t) L1+ Cε t − 2(1+) and (k) w (t) − G (t) ε u L1+ Cε 1+ t − 2(1+) are true. Then application of Lemma 2.3 yields g (k+1) (t) = 1 +
t 0
(k+1)
R
w1
1+
(, x)
dx d
1 1− t 2 2
for all t > 0. Therefore estimates (28) and (29) are valid with k replaced by k + 1. Integrating (k+1)
wt
+ Lw (k+1) + e− N w (k) − w (k) = 0
with respect to x ∈ R we get d dt
w (k+1) dx +
e− N w (k) − w (k) dx = 0
R
R
In view of (30) A0
w
(k+1)
u (x) dx = e1 ,
(t, x) dx = A0
R
(32)
R
where 1 e1 = . 0 Therefore by (32)
f (k+1) (t, x) dx
A0 R
N w
= A0
R
= A0 R
(k+1)
(t, x) − w
(k+1)
1 (t)
R
1+ (k+1) w1 (t, x)
dx dx
1+ (k+1) w1 N w(k+1) (t, x) dx − dx = 0. (t, x) R
By induction we see that properties (28)–(30) are true for all k 0. Thus the mapping M transforms the set X into itself. In the same manner we can estimate the differences M(w(k) ) − M(w(k−1) ) to see that M is the contraction mapping in X. Therefore there exists a unique solution w of integral equation (27) in the set X.
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Now let us compute the asymptotics of the solution. We first show the existence of solutions to the integral equation 2 1 V = (4)− 2 e− 4 −
1 1
(4) 2
dz
1
e
1
z (1 − z) 2
0
√
−y z 2 − ( 4(1−z))
F (y) dy,
(33)
R
where
= 1−
V 1+ d
2
R
and F (y) = V 1+ (y) − V (y)
V 1+ d . R
Lemma 3.1. Let ∈ 2 − ε 3 , 2 , where ε > 0 is sufficiently small. Then there exists an unique solution V ∈ L1,a ∩ L∞ to the integral equation (33). 2 1 Proof. Denote V (0) = (4)− 2 e− 4 and define the successive approximations 2 1 V (k+1) = (4)− 2 e− 4 −
1 1
(k) (4) 2
1 0
dz z (1 − z)
1 2
e−
√ 2 z) 4(1−z)
(−y
F (k) (y) dy
R
for k = 0, 1, 2, . . ., where (k)
= 1−
2
1+ d V (k)
R
and 1+ 1+ F (k) (y) = V (k) (y) − V (k) (y) d . V (k) R
First by the induction argument we prove the identities V R
(k)
F (k) (y) dy = 0
(y) dy = 1,
(34)
R
for all k 0. Since R V (0) (y) dy = 1 by the definition of F (k) (y), we see that (34) is true for k = 0. Now by induction we assume that equalities (34) hold for some k.
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
187
Then we have
V (k+1) d = 1 −
1
dz
1
1
1 2
0 z (1 − z) 2
(k) (4) √ (−y z)2 × e− 4(1−z) d F (k) (y) dy = 1,
R
R
R
hence it follows that
F (k+1) (y) dy = 0. R
Thus (34) are true for all k 0. Now let us prove the estimates
(k)
1 Ca ε 3
(35)
and sup
1p∞
a · V (k+1) − V (0)
Lp
Ca ε 2
(36)
for all k 0, where the constant Ca > 0 does not depend on k 0. For k = 0 we have
(0)
= 1+ 1 − 2 (4) 2 =
1−
2
e−
d
R
2
(1+)2 4
(4) (1 + )
1 2
1 , Ca ε 3
since ∈ 2 − ε 3 , 2 . For (36) with k = 0, by virtue of Lemma 2.4 we get sup · a V (1) − V (0) p
1p∞
L
√ ! " 1 −y z 2 dz C a − ( 4(1−z)) (0) = (0) sup dye F (y) 1 p
1p∞ R 0 z (1 − z) 2 L C (0) sup · a F (0) p L
1p∞ a (0) 1+ 1+ C (0) (0) · V V = (0) sup − V d (·) (·) p
1p∞ R L
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
CCa ε3
sup
1p∞
a (0) sup · V p
1p∞
L
(0) V
Lp
+
sup
1p∞
(0) V
+1
Lp
Ca ε 2 , since ε > 0 is sufficiently small. Now by induction we assume that inequalities (35) and (36) hold for some k. Then via (36) we have
(k+1) =
1− +
=
1+ d = V (k+1)
2
1 − 2
(0)
1−
1+ d V (0) 1− 2 R R 1+ (0) 1+ (k+1) V d − V
R
+ 1− 2
1+ (0) 1+ (k+1) V d − V R
2
1
1
− 4 V (0) ∞ + V (k+1)
L (4) 2 (1 + ) 2 −V (0) ∞ V (k+1) − V (0) 1 L L 1 1 2 − 4Ca ε , 1 1 − 2 (4) 2 (1 + ) 2 Ca ε 3
therefore (35) with k replaced by k + 1 is also true. Note that by (36) a (k+1) · V
Lp
· a V (k+1) − V (0)
Lp
+ · a V (0)
Lp
C.
We now use Lemma 2.4 to obtain sup
1p∞
a · V (k+2) − V (0)
Lp
√ ! " 1 −y z 2 dz a − ( 4(1−z)) (k+1) = (k+1) sup dye F (y) 1
1p∞ p R 0 z (1 − z) 2 L C (k+1) sup · a F (k+1) p L
1p∞ a (k+1) 1+ 1+ C (k+1) (k+1) · V V (k+1) sup −V d (·) (·) p
R 1p∞ L C
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
CCa ε
3
sup
1p∞
+
sup
1p∞
a (k+1) · V
sup
Lp
(k+1) V
1p∞
+1
(k+1) V
189
Lp
Lp
Ca ε 2 , since ε > 0 is sufficiently small. Therefore (35) and (36) are true for any k. In the same manner we obtain the estimates sup
1p∞
(k+1) − V (k) V
Lp
1 sup V (k) − V (k−1) p L 2 1p∞
(37)
for any k 1. Thus the successive approximations V (k) converge to an unique solution V ∈ L1,a ∩ L∞ of integral equation (33). Lemma 3.1 is proved. We are now in the position to prove asymptotics of solutions. By induction let us prove that for all k 0 t
− b2
b · w (k) (t) − e1 t − 21 V (k) √· t
Lp
Cε t − 2 + 2p − 1
1
(38)
for all t > 0, 1 p ∞, b ∈ [0, a], where = 21 min a, 1 − 2 . Estimate (38) is true for k = 0 since by Lemma 2.1 we have · − b2 b − 21 (0) (0) t · w (t) − t e1 V √ p t L b −1+ 1 −a u − e1 G0 (t) p Cε t 2 2p 2 . = t − 2 · b G (t) (39) L
We assume that (38) is valid for some k. Then from (38) it follows that 1+ t (k) (k) dx d w1 (, x) g (t) − (k) t 1− 2 = 1 + 0 R 1+ (k) −t 1− 2 d V 1 − 2 R 1+ 1+ t (k) 1+ − 2 (k) = 1 + dx d − d d w1 (, x) V 0 R R 1+ x 1+ C t (k) 1+ − 2 − 21 (k) V 1 + − √ dx d w1 (, x) 0
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
1 C t (k) 1 + w ∞ + − 2 V (k) ∞ L L 0 (k) · − 1 (k) × d √ w (, ·) − e1 2 V L1 Cε 1+ t − − 2 d 1 + C (k) t 1− 2 − 1+ 0
(40)
for t > 0, where 1 p ∞. Changing variables such that = zt and √ = y we have 1
(k) =
=
t
2 −1
− 2 − 21
G0 (t − )
0
1 1
(k) (4) 2
1
(k) (4t) 2
= t− 2
1
d
−1− 21
· √
− 21
V (0)
d
(t − )
d e
2
− (x−) 4(t−)
F
(k)
R
0
1
t
F
(k)
dz
1
1 2
√
e−
(xt −1/2 −y z)2 4(1−z)
z (1 − z) x x (k+1) , √ −V √ t t 0
√
F (k) (y) dy
where the Green function G0 (t) for the heat equation
G0 (t) =
G0 (t, x − y) (y) dy R x2
have a kernel G0 (t, x) = (4t)− 2 e− 4t . Therefore we get 1
b · b − 21 (k+1) (k+1) t − 2 · e t V − w √ (t) 1 p t L t b 1 · b − 21 (k+1) (k) · u + − f d = t − 2 e t V G G − √ (t) (t ) ( ) 1 p (k) () t 0 g L b · b − 1 (0) C t − 2 √ − w (0) (t) · e1 t 2 V p t L t − 1 − b2 b −1 (k) 2 − +C t · G − f d (t ) ( ) p g (k) ()
(k) 0 L b t 2 d C t − 2 + (k) · b (G (t − ) − G0 (t − ) A0 ) f (k) () p
0 L
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194 b
C t − 2 + (k)
t 1+ (k) · 2 d b (k) G0 (t − ) A0 f () − e1 +1 F √ · 0 2
Lp
b t · 1 1 −1 (k) · G0 (t − ) F 2 d − √ p 0 L
b
+
191
C t − 2
(k)
≡ I1 + I2 + I3 + I4 + I5 . Via (39) we have I1 Cεt
− 21 1− p1
a
t − 2 .
By (31), (40) and Lemma 2.2 with = 2 , = 1 − 2 + , where =
1 2
min a, 1 − 2
t − (k) 1 b −1 1− (k) 2 2 − (k) g () (k) I2 t · G (t − ) f () d p g ()
0 L t b 1 − 1 1− p1 − b − (k) G (t − ) f (k) () d Cεt 2 . C t − 2 · g () 0 Lp − b2
We have by the estimates of Lemma 2.1 (G (t − ) − G0 (t − ) A0 ) () ∞ C () ∞ , L L (G (t − ) − G0 (t − ) A0 ) () ∞ C (t − )− 1+a 2 · a () , L L1 a a · (G (t − ) − G0 (t − ) A0 ) () 1 C · () 1 L L for all 0 < < t, hence as in the proof of Lemma 2.2 with = 2 , = 1 − 2 we have b
C t − 2 I3 = (k)
t 2 d b (k) (G (t − ) − G0 (t − ) A0 ) f () · 0
a
Cεt − 2 + 2p − 2 . 1
1
Lp
In the same manner via Lemma 2.2 with = 2 , = 1 − 2 + we get b
C t − 2 I4 = (k)
b t · 1+ (k) − +1 (k) · 2 G0 (t − ) A0 f () − e1 F √ 0 1− p1 −
1 b C − 21 1+− p1 + t − 2 sup sup t 2 t (k)
t>0 1 p ∞ b · 1+ (k) − 1+ (k) 2 F × · A0 f (t) − e1 t √ p t L
d 1− 2 Lp
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N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
C −1 (k) t 2
× sup
Cεt
1− p1 −
1
sup
t>0 1 p ∞ − 21 1− p1 −
t2
sup t w (k) (t)
t>0
1 2
∞
1− p1 +
b
t − 2
(k) + V ∞
b · w (k) (t) − t − 21 e1 V (k) √· t
Lp
.
Finally changing independent variables = zt and √ = y and applying Lemma 2.4 we obtain b t · 1 C t − 2 1 −1 (k) · b 2 G d − − F √ (t ) 0 p
(k) 0 L b t 2 1 1 1 −1 C t − 2 − (x−) x b 4(t−) F (k) 2 (t − )− 2 d − = d e √
(k) Lp 0 1 √ x −y z 2 dz 1 t − 21 1− p1 − ( 4(1−z)) b (k) x = Cεt F e dy − (y) 1 p z tz 0 (1 − z) 2 L
I5 =
Cεt
− 21 1− p1 −
.
Hence by induction (38) is true for any k 0 uniformly with respect to k. Taking b = 0 and a limit k → ∞ in (38) we get 1 1 w (t) − e1 t − 21 V √· Cεt − 2 1− p − t Lp
for all t > 0.
(41)
That is by virtue of (40) and (41) we have the asymptotics w (t) = t − 2 e1 V 1
· √
t
1 + O t − 2 −
(42)
and g (t) = t 1− 2 1 + O t −
(43)
for t → ∞ uniformly with respect to x ∈ R. Therefore via the formula u (t, x) = e−(t) w (t, x) and (42), (43) we obtain the asymptotics of the solution. Theorem 1.1 is proved.
N. Hayashi et al. / J. Differential Equations 207 (2004) 161 – 194
193
Acknowledgments This work of P.I.N. and E.I.K. is partially supported by CONACYT. We are grateful to an unknown referee for many useful suggestions and comments.
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