Electronic Journal of Differential Equations, Vol. 2004(2004), No. 71, pp. 1–24.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
GAIN OF REGULARITY FOR A KORTEWEG - DE VRIES KAWAHARA TYPE EQUATION
OCTAVIO PAULO VERA VILLAGRÁN
Abstract. We study the existence of local and global solutions, and the gain
of regularity for the initial value problem associated to the Korteweg - de Vries
- Kawahara (KdVK) equation perturbed by a dispersive term which appears in
several fluids dynamics problems. The study of gain of regularity is motivated
by the results obtained by Craig, Kappeler and Strauss [8].
1. Introduction
In 1976, Saut and Temam [25] remarked that a solution u of a Korteweg-de
Vries type equation cannot gain or lose regularity. They showed that if u(x, 0) =
ϕ(x) ∈ H s (R) for s ≥ 2, then u(·, t) ∈ H s (R) for all t > 0. The same result was
obtained independently by Bona and Scott [3] through a different method. For the
Korteweg-de Vries equation on the line, Kato [17] motivated by work of Cohen [7]
showed that if u(x, 0) = ϕ(x) ∈ L2b ≡ H 2 (R) ∩ L2 (ebx dx) (b > 0) then the solution
u(x, t) of the KdV equation becomes C ∞ for all t > 0. A main ingredient in the
3
proof was the fact that formally the semi-group S(t) = e−t∂x in L2b is equivalent
3
to Sb (t) = e−t(∂x −b) in L2 when t > 0. One would be inclined to believe that this
was a special property of the KdV equation. This is not however the case. The
effect is due to the dispersive nature of the linear part of the equation. Kruzkov and
Faminskii [21] proved that for u(x, 0) = ϕ(x) ∈ L2 such that xα ϕ(x) ∈ L2 ((0, +∞))
the weak solution of the KdV equation has l-continuous space derivatives for all
t > 0 if l < 2α. The proof of this result is based on the asymptotic behavior of the
Airy function and its derivatives, and on the smoothing effect of the KdV equation
which was found in [17, 21]. Similar work for some special nonlinear Schrödinger
equations was done by Hayashi et al. [13, 14] and Ponce [23]. While the proof of
Kato appears to depend on special a priori estimates, some of its mystery has been
resolved by the result of local gain of finite regularity for various others linear and
nonlinear dispersive equations due to Constantin and Saut [11], Sjolin [26], Ginibre
and Velo [12] and others. However, all of them require growth conditions on the
nonlinear term.
2000 Mathematics Subject Classification. 35Q53, 47J35.
Key words and phrases. Evolution equations, weighted Sobolev space.
c
2004
Texas State University - San Marcos.
Submitted December 15, 2003. Published May 17, 2004.
Supported by MECESUP 9903, Universidad Católica de la Santı́sima Concepción, Chile.
The author dedicates this work to Dr. Patricio Sierralta Standen in the Iquique Hospital.
1
2
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
All the physically significant dispersive equations and systems known to us have
linear parts displaying this local smoothing property. To mention only a few, the
KdV, Benjamin-Ono, intermediate long wave, various Boussinesq, and Schrödinger
equations are included. Continuing with the idea of Craig, Kappeler and Strauss
[10] we study a equation of Korteweg - de Vries - Kawahara type (KdVK) which
appears in fluids dynamics (see [24] and references therein).
ut + ηuxxxxx + uxxx + uux = 0
(1.1)
with −∞ < x < +∞, t > 0 and η ∈ R. It is shown that C ∞ solutions u(x, t) are
obtained for all t > 0 if the initial data u(x, 0) decays faster than polynomially on
R+ = {x ∈ R : x > 0} and has certain initial Sobolev regularity. In section three
we prove the main inequality. In section 4 we prove an important a priori estimate.
In section 5 we prove a basic local-in-time existence and uniqueness theorem. In
section 6 we prove a basic global existence theorem. In section 7 we develop a
series of estimates for solutions of equation (1.1) in weighted Sobolev norms. These
provide a starting point for the a priori gain of regularity. In section 8 we prove
the following theorem.
Theorem 1.1. Let T > 0 and u(x, t) be a solution of (1.1) in the region R × [0, T ]
such that
u ∈ L∞ ([0, T ]; H 5 (W0L0 ))
(1.2)
for some L ≥ 2 and all σ > 0. Then
\
u ∈ L∞ ([0, T ]; H 5+l (Wσ,L−l,l )) L2 ([0, T ]; H 6+l (Wσ,L−l,l ) ∩ H 7+l (Wσ,L−l−1,l ))
for all 0 ≤ l ≤ L − 1.
2. Preliminaries
We consider the initial-value problem
ut + ηuxxxxx + uxxx + uux = 0
(2.1)
with −∞ < x < +∞, t ∈ [0, T ], T an arbitrary positive time, and η ∈ R is a
constant.
As a notation, we use ∂ = ∂/∂x, ∂t = ∂/∂t and uj = ∂ j u, ∂j = ∂/∂uj .
Definition. A function ξ(x, t) belongs to the weight class Wσ i k if it is a positive
C ∞ function on R × [0, T ], ξx > 0 and there are constant cj , 0 ≤ j ≤ 5 such that
0 < c1 ≤ t−k e−σx ξ(x, t) ≤ c2
∀x < −1, 0 < t < T,
(2.2)
x ξ(x, t) ≤ c4 ∀x > 1, 0 < t < T,
t|ξt | + |∂ ξ| /ξ ≤ c5 ∀(x, t) ∈ R × [0, T ], ∀j ∈ Z+ .
(2.3)
0 < c3 ≤ t
−k −i
j
(2.4)
We remark that, we will always consider σ ≥ 0, i ≥ 1 and k ≥ 0. For example, let
(
1 + e−1/x for x > 0
ξ(x) =
1
for x ≤ 0 ;
then ξ ∈ W0i0 .
Definition. Let s be a positive integer. We define the space
s Z +∞
X
H s (Wσik ) = {v : R → R : kvk2 =
|∂ j v(x)|2 ξ(x, ·)dx < +∞}
j=0
−∞
EJDE-2004/71
GAIN OF REGULARITY
3
with ξ ∈ Wσik fixed. Note that H s (Wσik ) depends on t because ξ = ξ(x, t)).
Lemma 2.1 ([6]). For ξ ∈ Wσi0 and σ ≥ 0, i ≥ 0, there exists a constant c > 0
such that, for u ∈ H 1 (Wσi0 ),
Z +∞
sup |ξu2 | ≤ c
|u|2 + |∂u|2 ξdx
x∈R
−∞
Definition. For fixed ξ ∈ Wσik , we define the spaces
Z T
L2 ([0, T ]; H s (Wσik )) = {v(x, t) : |||v|||2 =
kv(·, t)k2 dt < +∞}
0
L∞ ([0, T ]; H s (Wσik )) = {v(x, t) : |||v|||∞ = ess sup kv(·, t)k < +∞},
t∈[0,T ]
where s is a positive integer. Note that The usual Sobolev space H s (R) is H s (W000 ),
i.e., without weight.
We shall derive the a priori estimates assuming that the solution is C ∞ , bounded
as x → −∞, and rapidly decreasing together with all of its derivatives as x → +∞.
We consider the following KdVK equation
ut + ηu5 + u3 + uu1 = 0
(2.5)
with η ∈ R constant. This equation will be studied for −∞ < x < +∞, t ∈ [0, T ]
with T an arbitrary positive time.
3. Main Inequality
Lemma 3.1. Let u be a solution to (2.5) with enough Sobolev regularity (for instance, u ∈ H N (R), N ≥ α + 5), then
Z
Z
Z
Z
Z
2
2
2
2
∂t ξuα dx + µ1 uα+1 dx + µ2 uα+2 dx + θuα dx + Rα dx ≤ 0
R
R
R
R
R
with
µ1 = −c5 (5η + 3)ξ
η < −3/5
for
(Natural Condition)
µ2 = −5η∂ξ
θ = −ξt − η∂ 5 ξ − ∂ 3 ξ − ∂(ξu)
Rα = O(uα , . . . )
Proof. Taking α-derivatives of (2.5) (for α ≥ 3) over x ∈ R
∂t uα + ηuα+5 + uα+3 + uuα+1 + Rα (uα , uα−1 , . . . ) = 0
Multiply this equation by 2ξuα and, integrate over x ∈ R to have
Z
Z
Z
2 ξuα ∂t uα dx + 2η ξuα uα+5 dx + 2 ξuα uα+3 dx
R
R
ZR
Z
+2 ξuuα uα+1 dx + 2 ξuα Rα dx = 0
R
R
Integrating by parts,
Z
ξu2α dx +
Z
(5η∂ 3 ξ + 3∂ξ)u2α+1 dx
Z
Z
Z
2
2
+ −5η∂ξuα+2 dx + θuα dx + Rα dx = 0
∂t
R
R
R
R
R
(3.1)
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OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
with θ = −ξt − η∂ 5 ξ − ∂ 3 ξ − ∂(ξu). Using (2.4), it follows for c5 > 0 that
Z
Z
Z
Z
Z
∂t ξu2α dx − c5 (5η + 3) ξu2α+1 dx − 5η ∂ξu2α+2 dx + θu2α dx + Rα dx ≤ 0 ,
R
R
R
R
R
from where we obtain the main inequality.
Z
Z
Z
Z
Z
∂t ξu2α dx + µ1 u2α+1 dx + µ2 u2α+2 dx + θu2α dx + Rα dx ≤ 0
R
R
R
R
(3.2)
R
with
µ1 = −c5 (5η + 3)ξ
for η < −3/5
(Natural Condition)
µ2 = −5η∂ξ
θ = −ξt − η∂ 5 ξ − ∂ 3 ξ − ∂(ξu)
Rα = O(uα , . . . )
Lemma 3.2. For µ2 ∈ Wσik an arbitrary weight function and η < −3/5, there
exists ξ ∈ Wσ,i+1,k that satisfies
µ2 = −5η∂ξ
Indeed, we have
1
ξ=−
5η
Z
(3.3)
x
µ2 (y, t)dy
(3.4)
−∞
Lemma 3.3. The expression Rα in the inequality of Lemma 3.1 is a sum of terms
of the form
ξuν1 uν2 uα
(3.5)
where 1 ≤ ν1 ≤ ν2 ≤ α.
ν1 + ν2 = α + 1
(3.6)
Proof. Differentiating (2.5) once with respect to x and multiplying by 2ξu1 we have
2ξu1 ∂t u1 + 2ηξu1 u6 + 2ξu1 u4 + 2ξu1 uu2 + ξu1 u1 u1 = 0 ,
where R1 ξu1 = ξu1 u1 u1 . Taking 2-xderivatives of the equation (2.5), and multiplying by 2ξu2 we have
2ξu2 ∂t u2 + 2ηξu2 u7 + 2ξu2 u5 + 2ξu2 uu3 + 6ξu2 u1 u2 = 0
where R2 ξu2 = 6ξu1 u2 u2 . Taking 3-xderivatives of the equation (2.5), and multiplying by 2ξu3 we have
2ξu3 ∂t u3 + 2ηξu3 u8 + 2ξu3 u6 + 2ξu3 uu4 + 8ξu3 u1 u3 + 6ξu3 u2 u2 = 0
where R3 ξu3 = 8ξu1 u3 u3 + 6ξu2 u2 u3 . Taking 4-xderivatives of (2.5), and multiplying by 2ξu4 we have
2ξu4 ∂t u4 + 2ηξu4 u9 + 2ξu4 u7 + 2ξu4 uu5 + 10ξu4 u1 u4 + 20ξu4 u2 u3 = 0
where R4 ξu4 = 10ξu1 u4 u4 + 20ξu2 u3 u4 . Taking 5-xderivatives of (2.5), and multiplying by 2ξu5 we have
2ξu5 ∂t u5 +2ηξu5 u10 +2ξu5 u8 +2ξu5 uu6 +12ξu5 u1 u5 +30ξu5 u2 u4 +20ξu5 u3 u3 = 0
where R5 ξu5 = 12ξu1 u5 u5 + 30ξu2 u4 u5 + 20ξu3 u3 u5 . Throw away the first terms
in each derivative and the result follows.
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GAIN OF REGULARITY
5
4. An a priori estimate
We show a fundamental a priori estimate used for a basic local-in-time existence
theorem. We construct a mapping Z : L∞ ([0, T ]; H s (R)) → L∞ ([0, T ]; H s (R))
with the following property: Given u(n) = Z(u(n−1) ) and ku(n−1) ks ≤ c0 then
ku(n) ks ≤ c0 , where s and c0 > 0 are constants. This property tells us, in fact,
that Z : Bc0 (0) → Bc0 (0) where Bc0 (0) = {v(x, t); kv(·, t)ks ≤ c0 } is a ball in
L∞ ([0, T ]; H s (R)). To guarantee this property, we will appeal to an a priori estimate
which is the main object of this section. Differentiating (2.5) four times leads to
∂t u4 + ηu9 + u7 + uu5 + 5u1 u4 + 10u2 u3 = 0
(4.1)
Let u = ∧v where ∧ = (I − ∂ 4 )−1 . Then ∂t u4 = −vt + ut by replacing in (4.1) we
have
−vt +η∧v 9 +∧v 7 +∧v∧v 5 +5∧v 1 ∧v 4 +10∧v 2 ∧v 3 −[η∧v 5 +∧v 3 −∧v∧v 1 ] = 0 (4.2)
The (4.2) is linearized by substituting a new variable w in each coefficient;
−vt +η∧v 9 +∧v 7 +∧w∧v 5 +5∧w1 ∧v 4 +10∧w2 ∧v 3 −[η∧v 5 +∧v 3 −∧w∧v 1 ] = 0 (4.3)
Equation (4.3) is a linear equation at each iteration which can be solved in any
interval of time in which the coefficients are defined. This equation has the form
(n)
(n)
(n)
(n)
∂t v = η∧v 9 + ∧v 7 + b(1) ∧v 5 + b(2) ∧v 4 + b(3)
(4.4)
We consider the following lemma that will help us setting up the iteration scheme.
T
Lemma 4.1. Let η < −3/5. Given initial data ϕ ∈ H ∞ (R) = N ≥0 H N (R)
there exists a unique solution of (4.4) where b(1) = b(1) (∧w), b(2) = b(2) (∧w1 ) and
b(3) = b(3) (∧w3 , . . . , ∧w) are smooth bounded coefficients with w ∈ H ∞ (R). The
solution is defined in any time interval in which the coefficients are defined.
Proof. Let T > 0 be arbitrary and M > 0 a constant. Let
L = 2ξ(∂t − η∧∂ 9 − ∧∂ 7 − b(1) ∧∂ 5 − b(2) ∧∂ 4 )
where 0 < c6 ≤ ξ ≤ c7 . We consider the bilinear form B : D × D 7→ R,
Z TZ
B(u, v) = hu, vi =
e−M t uv dx dt
0
R
where D = {u ∈ C0∞ (R × [0, T ]) : u(x, 0) = 0}. We have
Z
Z
Z
Z
Lu · udx = 2 ξuut dx − 2η ξu∧u9 dx − 2 ξu∧u7 dx
R
R
R
R
Z
Z
− 2 ξb(1) u∧u5 dx − 2 ξb(2) u∧u4 dx
R
R
Each term is treated separately. The first term yields
Z
Z
Z
2
2 ξuut dx = ∂t ξu dx − ξt u2 dx
R
R
R
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OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
In the second term, by integrating by parts we obtain
Z
− 2η ξu∧u9 dx
R
Z
Z
Z
4
= −2η ξ ∧ (I − ∂ )u∧u9 dx − 2η ξ∧u∧u9 dx + 2η ξ∧u4 ∧u9 dx
R
R
Z R
Z
Z
9
2
7
2
5
= η ∂ ξ(∧u) dx − 9η ∂ ξ(∧u1 ) dx + 27η ∂ ξ(∧u2 )2 dx
R
R
R
Z
Z
Z
3
2
5
2
− 30η ∂ ξ(∧u3 ) dx − η (∂ ξ − 9∂ξ)(∧u4 ) dx + 5η ∂ 3 ξ(∧u5 )2 dx
R
R
Z R
2
− 5η ∂ξ(∧u6 ) dx .
R
All the others terms are calculated of the same way. We have
Z
Lu · udx
R
Z
Z
Z
Z
= ∂t ξu2 dx − ξt u2 dx + η ∂ 9 ξ(∧u)2 dx − 9η ∂ 7 ξ(∧u1 )2 dx
R
R
Z
ZR
ZR
+ 27η ∂ 5 ξ(∧u2 )2 dx − 30η ∂ 3 ξ(∧u3 )2 dx − η (∂ 5 ξ − 9∂ξ)(∧u4 )2 dx
R
Z R
Z R
Z
Z
3
2
2
7
+ 5η ∂ ξ(∧u5 ) dx − 5η ∂ξ(∧u6 ) dx + ∂ ξ(∧u)2 dx − 7 ∂ 5 ξ(∧u1 )2 dx
ZR
Z R
Z R
ZR
3
2
2
3
2
+ 14 ∂ ξ(∧u2 ) dx − 7 ∂ξ(∧u3 ) dx − ∂ ξ(∧u4 ) dx + 3 ∂ξ(∧u5 )2 dx
R
R
R
Z R
Z
Z
+ ∂ 5 (ξb(1) )(∧u)2 dx − 5 ∂ 3 (ξb(1) )(∧u1 )2 dx + 5 ∂(ξb(1) )(∧u2 )2 dx
ZR
Z R
Z R
(1)
2
4
(2)
2
− ∂(ξb )(∧u4 ) dx − ∂ (ξb )(∧u) dx + 4 ∂ 2 (ξb(2) )(∧u1 )2 dx
R
ZR
Z R
(2)
2
(2)
2
− ξb (∧u4 ) dx + 2 ξb (∧u4 ) dx .
R
R
It follows that
Z
Lu · udx
R
Z
Z
Z
= ∂t ξu2 dx − 5η ∂ξ(∧u6 )2 dx + (5η∂ 3 ξ + 3∂ξ)(∧u5 )2 dx
R
R
ZR
+ (−η∂ 5 ξ − ∂ 3 ξ + 9η∂ξ − ∂(ξb(1) ) + 2ξb(2) )(∧u4 )2 dx
ZR
Z
+ (−30η∂ 3 ξ − 7∂ξ)(∧u3 )2 dx − ξt u2 dx
R
ZR
5
3
3
(1)
+ (27η∂ ξ + 14∂ ξ + 5∂ (ξb ) − 2ξb(2) )(∧u2 )2 dx
ZR
+ (−9η∂ 7 ξ − 7∂ 5 ξ − 5∂ 3 (ξb(1) ) + 4∂ 2 (ξb(2) ))(∧u)2 dx
R
EJDE-2004/71
GAIN OF REGULARITY
7
Using (2.4), ∧un = (I − (I − ∂ 4 ))∧un−4 = ∧un−4 − un−4 for n positive integer and
standard estimates it follows that
Z
Z
Z
Lu · udx ≥ ∂t ξu2 dx − c ξu2 dx
(4.5)
R
R
R
−M t
Multiply this equation by e
, and integrate with respect to t for t ∈ [0, T ] and
u ∈ D.
Z TZ
e−M t Lu · udxdt
0
R
T
Z
−M t
≥
e
0
Z
∂t
2
ξu dx dt − c
Z
R
= e−M t
Z
ξu2 (x, t)dx|T0 + M
Z
T
Z
ξe−M t u2 dxdt
R
R
Z
= e−M T
ξ(x, T )u2 (x, T )dx + M
Z
−c
0
Z
T
Z
0
R
T
Z
ξe−M t u2 dxdt
R
ξe−M t u2 dxdt
−c
0
Z
0
T Z
0
R
Z
T
ξe−M t u2 dxdt
R
ξe−M t u2 dx dt.
R
Thus
Z
T
Z
0
e−M t Lu · udxdt
R
≥ e−M T
Z
ξ(x, T )u2 (x, T )dx + (M − c)
Z
T
Z
0
T
0
R
≥
Z
Z
ξe−M t u2 dxdt
R
ξe−M t u2 dx dt
R
provided M is chosen large enough. Then hLu, ui ≥ hu, ui, for all u ∈ D. Let
L∗ = 2ξ(−∂t + η∧∂ 9 + ∧∂ 7 + b(1) ∧∂ 5 − b(2) ∧∂ 4 ) be the formal adjoint of L. Let
D∗ = {w ∈ C0∞ (R × [0, T ]) : w(x, L) = 0}. In the same way we prove that
hL∗ w, wi ≥ hw, wi
∀w ∈ D∗
(4.6)
From this equation, we have that L∗ is one-one. Therefore hL∗ w, L∗ vi is an inner
product on D∗ . We denote by X the completion of D∗ with respect to this inner
product. By the Riesz Representation Theorem, there exists a unique solution V ∈
X, such that for any w ∈ D∗ , hξb(3) , wi = hL∗ V, L∗ wi where we use that ξb(3) ∈ X.
Then if v = L∗ V we have hv, L∗ wi = hξb(3) , wi or hL∗ w, vi = hw, ξb(3) i. Hence
v = L∗ V is a weak solution of Lv = ξb(3) with v ∈ L2 (R×[0, T ]) ' L2 ([0, T ]; L2 (R)).
Remark To obtain higher regularity of the solution, we repeat the proof with
higher derivatives. It is a standard approximation procedure to obtain a result for
general initial data.
The next step is to estimate the corresponding solutions v = v(x, t) of the equation (4.3) via the coefficients of that equation.
Lemma 4.2. Let v, w ∈ C k ([0, +∞); H N (R)) for all k, N which satisfy (4.3). Let
0 < c8 ≤ ξ ≤ c9 and η < −3/5. For each integer α there exist positive nondecreasing
8
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
functions G and F such that for all t ≥ 0,
Z
ξvα2 dx ≤ G(kwkλ )kvk2α + F (kwkα )
∂t
(4.7)
R
where k · kα is the norm in H α (R) and λ = max {1, α}.
Proof. Differentiating α-times the equation (4.3), for some α ≥ 0, we obtain
−∂t vα + η∧v α+9 + ∧v α+7 +
α+5
X
h(j) ∧v j + c10 ∧v 3 ∧wα+2 + p(∧wα+1 , . . . ) = 0 (4.8)
j=6
where h(j) is a smooth function depending on ∧wi , . . . with i = 5 + α − j. We
multiply equation (4.8) by 2ξvα , and integrate over R,
Z
−2
Z
ξvα ∧v α+9 dx
ξvα ∂t vα dx + 2η
R
R
α+5
XZ
Z
ξvα ∧v α+7 dx + 2
+2
R
j=6
Z
R
Z
ξvα ∧v 3 ∧wα+2 dx + 2
+2c10
ξh(j) vα ∧v j dx
ξvα p(∧wα+1 , . . . )dx = 0
R
R
Each of these terms is treated separately. The first term yields
Z
−2
Z
ξvα ∂t vα dx = −∂t
R
ξvα2 dx
Z
+
R
ξt vα2 dx
R
In the second term we have, by integrating by parts
Z
ξvα ∧v α+9 dx
Z
= 2η ξ∧(I − ∂ 4 )vα ∧v α+9 dx
ZR
Z
= 2η ξ∧v α ∧v α+9 dx − 2η ξ∧v α+4 ∧v α+9 dx
ZR
ZR
Z
9
2
= −η ∂ ξ(∧v α ) dx + 9η ∂ 7 ξ(∧v α+1 )2 dx − 27η ∂ 5 ξ(∧v α+2 )2 dx
R
R
R
Z
Z
3
2
5
+ 30η ∂ ξ(∧v α+3 ) dx + η (∂ ξ − 9∂ξ)(∧v α+4 )2 dx
Z R
ZR
− 5η ∂ 3 ξ(∧v α+5 )2 dx + 5η ∂ξ(∧v α+6 )2 dx
2η
R
R
R
(4.9)
EJDE-2004/71
GAIN OF REGULARITY
9
The others terms are treated similarly. Replacing the equations obtained, on (4.9),
we have
Z
ξvα2 dx
Z
Z
+
R
+2
−η
R
R
α+4
XZ
2
∂ ξ(∧v α ) dx + 9η
R
R
R
ξh(j) vα ∧v j dx + 2c10
R
j=6
9
Z
∂ 7 ξ(∧v α+1 )2 dx
Z
Z
Z
− 27η ∂ 5 ξ(∧v α+2 )2 dx + 30η ∂ 3 ξ(∧v α+2 )2 dx − 3δ ∂ 3 ξ(∧v α+3 )2 dx
R
R
Z R
Z
Z
5
2
3
2
+ η (∂ ξ − 9∂ξ)(∧v α+4 ) dx − 5η ∂ ξ(∧v α+5 ) dx + 5η ∂ξ(∧v α+6 )2 dx
R
R
Z R
Z
Z
− ∂ 7 ξ(∧v α )2 dx + 7 ∂ξ(∧v α+1 )2 dx − 14 ∂ 3 ξ(∧v α+2 )2 dx
R
R
R
Z
Z
Z
2
3
2
+ 7 ∂ξ(∧v α+3 ) dx + ∂ ξ(∧v α+4 ) dx − 3 ∂ξ(∧v α+5 )2 dx
R
R
Z R
Z
5
(α+5)
2
3
(α+5)
− ∂ (ξh
)(∧v α ) dx − ∂ (ξh
)(∧v α+1 )2 dx
R
R
Z
Z
(α+5)
2
− 5 ∂(ξh
)(∧v α+2 ) dx + ∂(ξh(α+5) )(∧v α+4 )2 dx
− ∂t
ξt vα2 dx
Z
ξvα ∧v 3 ∧wα+2 dx
R
Z
+2
ξvα p(∧wα+1 , . . . )dx = 0
R
and
Z
ξvα2 dx
Z
Z
Z
= 5η ∂ξ(∧v α+6 )2 dx − (5η∂ 3 ξ + 3∂ξ)(∧v α+5 )2 dx + ξt (∧v α )2 dx
R
R
ZR
5
3
(α+5)
2
+ (η∂ ξ + ∂ ξ − 9η∂ξ + ∂(ξh
))(∧v α+4 ) dx
ZR
+ (30η∂ 3 ξ + 7∂ξ)(∧v α+3 )2 dx
ZR
+ (−27η∂ 5 ξ − 14∂ 3 ξ − 5∂(ξh(α+5) ))(∧v α+2 )2 dx
ZR
+ (9η∂ 7 ξ + 7∂ξ + ∂ 3 (ξh(α+5) ))(∧v α+1 )2 dx
∂t
R
R
Z
+
(−η∂ 9 ξ − ∂ 7 ξ − ∂ 5 (ξh(α+5) ))(∧v α )2 dx + 2
R
α+4
X
ξh(j) vα ∧v j dx
j=6
Z
Z
ξvα ∧v 3 ∧wα+2 dx + 2
+ 2c10
R
ξvα p(∧wα+1 , . . . )dx
R
10
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
Using (2.4) we have that the first and the second term in the right hand side of the
above expression are nonpositive. Hence
Z
Z
Z
∂t ξvα2 dx ≤ (η∂ 5 ξ + ∂ 3 ξ − 9η∂ξ + ∂(ξh(α+5) ))(∧v α+4 )2 dx + ξt (∧v α )2 dx
R
R
R
Z
+ (30η∂ 3 ξ + 7∂ξ)(∧v α+3 )2 dx
ZR
+ (−27η∂ 5 ξ − 14∂ 3 ξ − 5∂(ξh(α+5) ))(∧v α+2 )2 dx
ZR
+ (9η∂ 7 ξ + 7∂ξ + ∂ 3 (ξh(α+5) ))(∧v α+1 )2 dx
R
Z
+
(−η∂ 9 ξ − ∂ 7 ξ − ∂ 5 (ξh(α+5) ))(∧v α )2 dx + 2
R
α+4
X
ξh(j) vα ∧v j dx
j=6
Z
Z
ξvα ∧v 3 ∧wα+2 dx + 2
+ 2c10
R
ξvα p(∧wα+1 , . . . )dx
R
Using that ∧v n = ∧v n−4 − vn−4 and standard estimates, the Lemma follows.
5. Uniqueness and Existence of a Local Solution
In this section, we study the uniqueness and the existence of local strong solutions
in the Sobolev space H N (R) for N ≥ 5 for the problem (2.5). To establish the
existence of strong solutions for (2.5) we use the a priori estimate together with an
approximation procedure.
Theorem 5.1 (Uniqueness). Let η < −3/5, ϕ ∈ H N (R) with N ≥ 5 and 0 < T <
+∞. Then there is at most one strong solution u ∈ L∞ ([0, T ]; H N (R)) of (2.5)
with initial data u(x, 0) = ϕ(x).
Proof. Assume that u, v ∈ L∞ ([0, T ]; H N (R)) are two solutions of (2.5) with ut , vt ∈
L∞ ([0, T ]; H N −5 (R)) and with the same initial data. Then
(u − v)t + η(u − v)5 + (u − v)3 + uu1 − vv1 = 0
(5.1)
with (u − v)(x, 0) = 0. By (5.1),
(u − v)t + η(u − v)5 + (u − v)3 + (u − v)u1 + (u − v)1 v = 0 .
Multiplying (5.2) by 2ξ(u − v) and integrating with respect to x over R,
Z
Z
2 ξ(u − v)(u − v)t dx + 2η ξ(u − v)(u − v)5 dx
R
R
Z
Z
+ 2 ξ(u − v)(u − v)3 dx + 2 ξuu1 (u − v)2 dx
R
ZR
+ 2 ξv(u − v)(u − v)1 dx = 0
R
Each term is treated separately. In the first term we obtain
Z
Z
Z
2
2 ξ(u − v)(u − v)t dx = ∂t ξ(u − v) dx − ξt (u − v)2 dx
R
R
R
(5.2)
(5.3)
EJDE-2004/71
GAIN OF REGULARITY
11
In the others terms, we also integrate by parts,
Z
Z
Z
2η ξ(u − v)(u − v)5 dx = −η ∂ 5 ξ(u − v)2 dx + 5η ∂ 3 ξ(u − v)21 dx
R
R
R
Z
2
= −5η ∂ξ(u − v)2 dx
Z
Z R
Z
3
2
2 ξ(u − v)(u − v)3 dx = − ∂ ξ(u − v) dx + 3 ∂ξ(u − v)21 dx
R
R
R
Z
Z
2 ξv(u − v)(u − v)1 dx = − ∂(ξv)(u − v)2 dx
R
R
Replacing these expression in (5.3), we have
Z
Z
Z
Z
2
2
5
2
∂t ξ(u − v) dx − ξt (u − v) dx − η ∂ ξ(u − v) dx + 5η ∂ 3 ξ(u − v)21 dx
R
R
R
R
Z
Z
Z
2
3
2
− 5η ∂ξ(u − v)2 dx − ∂ ξ(u − v)dx + 3 ∂ξ(u − v)1 dx
R
Z R
ZR
+ 2 ξu1 (u − v)2 dx − ∂(ξv)(u − v)2 dx = 0
R
R
then
Z
ξ(u − v)2 dx +
∂t
Z
R
Z
+
(5η∂ 3 ξ + 3∂ξ)(u − v)21 dx − 5η
R
Z
∂ξ(u − v)22 dx
R
(−ξt − η∂ 5 ξ − ∂ 3 ξ + 2ξu1 − ∂(ξv))(u − v)2 dx = 0
R
By using (2.4), we obtain for c5 > 0 and η < −3/5 that
Z
Z
Z
∂t ξ(u − v)2 dx − c5 (5η + 3)ξ(u − v)21 dx − 5η ∂ξ(u − v)22 dx
R
R
ZR
≤ (ξt + η∂ 5 ξ + ∂ 3 ξ − 2ξu1 + ∂(ξv))(u − v)2 dx
R
and using Gagliardo-Nirenberg’s inequality and standard estimates, we have
Z
Z
2
∂t ξ(u − v) dx ≤ c ξ(u − v)2 dx
R
R
By Gronwall’s inequality and the fact that (u − v) vanishes at t = 0, it follows that
u = v. This proves the uniqueness of the solution.
We construct the mapping Z : L∞ ([0, T ]; H s (R)) → L∞ ([0, T ]; H s (R)) by
u(0) = ϕ(x)
u(n) = Z(u(n−1) ) n ≥ 1,
where u(n−1) is in place of w in equation (4.3) and u(n) is in place of v which
is the solution of equation (4.3). By Lemma 4.1, u(n) exists and is unique in
C((0, +∞); H N (R)). A choice of c0 and the use of the a priori estimate in §4 show
that Z : Bc0 (0) → Bc0 (0) where Bc0 (0) is a bounded ball in L∞ ([0, T ]; H s (R)) Theorem 5.2 (Local solution). Let η < −3/5 and N an integer ≥ 5. If ϕ ∈
H N (R), then there is T > 0 and u such that u is a strong solution of (2.5),
u ∈ L∞ ([0, T ]; H N (R)), and u(x, 0) = ϕ(x)
12
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
T
Proof. We prove that for ϕ ∈ H ∞ (R) = k≥0 H k (R) there exists a solution u ∈
L∞ ([0, T ]; H N (R)) with initial data u(x, 0) = ϕ(x) which time of existence T > 0
only depends on the norm of ϕ. We define a sequence of approximations to equation
(4.3) as
(n)
(n)
(n)
(n)
(n)
− vt + η∧v 9 + ∧v 7 + ∧v (n−1) ∧v 5 − η∧v 5
(5.4)
(n−1)
(n)
(n−1)
(n−1)
+ 5∧v 1
∧v 4 + O(∧v 3
, ∧v 1
,...) = 0
where the initial condition v (n) (x, 0) = ϕ(x) − ∂ 4 ϕ(x). The first approximation is
given by v (0) (x, 0) = ϕ(x) − ∂ 4 ϕ(x). Equation (5.4) is a linear equation at each
iteration which can be solved in any interval of time in which the coefficients are
defined. This is shown in Lemma 4.1. By Lemma 4.2, it follows that
Z
∂t ξ[vα(n) ]2 dx ≤ G(kv (n−1) kλ )kv (n) k2α + F (kv (n−1) kα )
(5.5)
R
Choose α = 1 and let c ≥ kϕ − ∂ 4 ϕk1 ≥ kϕk5 . For each iterate n, kv (n) (·, t)k is
(n)
c9 2
continuous in t ∈ [0, T ] and kv (n) (·, 0)k ≤ c. Define c0 = 2c
c + 1. Let T0 be the
8
(n)
maximum time such that kv (k) (·, t)k1 ≤ c3 for 0 ≤ t ≤ T0 , 0 ≤ k ≤ n. Integrating
(n)
(5.5) over [0, t] we have for 0 ≤ t ≤ T0 and j = 0, 1.
Z t
Z
Z t
Z t
(n)
∂s ξ[vj ]2 dx ds ≤
G(kv (n−1) k1 )kv (n) k2j ds +
F (kv (n−1) kj )ds
0
0
R
0
It follows that
Z
(n)
ξ(x, t)[vj (x, t)]2 dx
R
Z
Z t
Z t
(n)
≤
ξ(x, 0)[vj (x, 0)]2 dx +
G(kv (n−1) k1 )kv (n) k2j ds +
F (kv (n−1) kj )ds
0
R
0
hence
Z
c8
(n)
[vj ]2 dx ≤
R
Z
(n)
ξ[vj ]2 dx
R
Z
≤
(n)
ξ(x, 0)[vj (x, 0)]2 dx +
R
Z t
+
F (kv (n−1) kj )ds
Z
t
G(kv (n−1) k1 )kv (n) k2j ds
0
0
and
Z
Z
c9
G(c3 ) 2
F (c3 )
(n)
[vj (x, 0)]2 dx +
c3 t +
t
c8 R
c8
c8
R
and we obtain for j = 0, 1 that
(n)
[vj ]2 dx ≤
kv (n) k1 ≤
(n)
c9 2 G(c0 ) 2
F (c0 )
c +
c t+
t
c8
c8 0
c8
Claim: T0 does not approach 0
(n)
On the contrary, assume that T0 → 0. Since kv (n) (·, t)k is continuous for t ≥ 0,
(n)
there exists τ ∈ [0, T ] such that kv (k) (·, τ )k1 = c0 for 0 ≤ τ ≤ T0 , 0 ≤ k ≤ n.
Then
c9
G(c0 ) 2 (n) F (c0 ) (n)
c20 ≤ c2 +
c T
+
T0 .
c8
c8 0 0
c8
EJDE-2004/71
GAIN OF REGULARITY
13
as n → +∞, we have
2
c9 2
c9
c + 1 ≤ c2
2c8
c8
c29 4
c +1≤0
4c28
=⇒
(n)
which is a contradiction. Consequently T0 6→ 0. Choosing T = T (c) sufficiently
small, and T not depending on n, one concludes that
kv (n) k1 ≤ C
(5.6)
(n)
that T0
(n)
for 0 ≤ t ≤ T . This shows
exists a subsequence v (nj ) := v
∗
v (n) * v
≥ T . Hence from (5.6) we imply that there
such that
weakly on L∞ ([0, T ]; H 1 (R))
(5.7)
Claim: u = ∧v is a solution.
In the linearized equation (5.4) we have
(n)
∧v 9
(n)
= ∧(I − (I − ∂ 4 ))v5
(n)
(n)
= ∧v 5 − v5
(n)
(n)
= ∂ 4 ( ∧v 1 ) − ∂ 4 (v1 )
| {z }
| {z }
∈L2 (R)
∈H −4 (R)
(n)
∧v 9
1
Since ∧ = (I − ∂ 4 )−1 is bounded in H 1 (R) so
belongs to H −4 (R). v (n) is still
∞
1
2
bounded in L ([0, T ]; H (R)) ,→ L ([0, T ]; H (R)) and since ∧ : L2 (R) → H 4 (R)
is a bounded operator,
(n)
(n)
(n)
k∧v 1 kH 4 (R) ≤ c11 kv1 kL2 (R) ≤ c12 kv1 kH 1 (R) .
(n)
Consequently ∧v 1 is bounded in L2 ([0, T ]; H 4 (R)) ,→ L2 ([0, T ]; L2 (R)). It follows
(n)
that ∂ 4 (∧v 1 ) is bounded in L2 ([0, T ]; H −4 (R)), and
(n)
∧v 9
is bounded in L2 ([0, T ]; H −4 (R))
(5.8)
(n)
vt
Similarly, the other terms are bounded. By (5.4),
is a sum of terms each
of which is the product of a coefficient, uniformly bounded on n and a func(n)
tion in L2 ([0, T ]; H −4 (R)) uniformly bounded on n such that vt is bounded in
c
1/2
1
L2 ([0, T ]; H −4 (R)). On the other hand, Hloc
(R) ,→ Hloc (R) ,→ H −4 (R). By LionsAubin’s compactness Theorem [22] there is a subsequence v (nj ) := v (n) such that
1/2
v (n) → v strongly on L2 ([0, T ]; Hloc (R)). Hence, for a subsequence v (nj ) := v (n) ,
1/2
(n)
we have v (n) → v a. e. in L2 ([0, T ]; Hloc (R)). Moreover, from (5.8), ∧v 9 * ∧v 9
2
−4
weakly in L ([0, T ]; H (R)).
(n)
Similarly, ∧v 5 * ∧v 5 weakly in L2 ([0, T ]; H −4 (R)). Since k∧v (n) kH 5 (R) ≤
1/2
c13 kv (n) kH 1 (R) ≤ c14 kv (n) kH 1/2 (R) and v (n) → v strongly on L2 ([0, T ]; Hloc (R))
5
4
(R)). Thus the
then ∧v (n) → ∧v strong in L2 ([0, T ]; Hloc
(R)) ,→ L2 ([0, T ]; Hloc
(n)
(n−1)
fourth term on the right hand side of (5.4), ∧v
∧v 5 * ∧v∧v 5 weakly in
(n)
L2 ([0, T ]; L1loc (R)) as ∧v 5 * ∧v 5 weakly in L2 ([0, T ]; H −4 (R)) and ∧v (n−1) → ∧v
4
strongly on L2 ([0, T ]; Hloc
(R)). Similarly, the other terms in (5.4) converge to their
(n)
limits, implying vt * vt weakly in L2 ([0, T ]; L1loc (R)). Passing to the limit
vt = ∂ 4 (η∧v 5 + ∧v 3 + ∧v∧v 1 ) − (η∧v 5 + ∧v 3 + ∧v∧v 1 )
= −(I − ∂ 4 )(η∧v 5 + ∧v 3 + ∧v∧v 1 )
thus vt +(I −∂ 4 )(η∧v 5 +∧v 3 +∧v∧v 1 ) = 0. This way, we have that (2.5) for u = ∧v.
Now, we prove that there exists a solution to (2.5) with u ∈ L∞ ([0, T ]; H N (R)) and
14
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
N ≥ 6, where T depends only on the norm of ϕ in H 5 (R). We already know
that there is a solution u ∈ L∞ ([0, T ]; H 5 (R)). It is suffices to show that the
approximating sequence v (n) is bounded in L∞ ([0, T ]; H N −4 (R)). Taking α = N −2
(n)
c9
and considering (5.5) for α ≥ 2, we define cN −5 = 2c
kϕ(·)kN + 1. Let TN −5 be
8
(n)
the largest time such that kv (k) (·, t)kα ≤ cN −5 for 0 ≤ t ≤ TN −5 , 0 ≤ k ≤ n.
(n)
Integrating (5.5) over [0, t], for 0 ≤ t ≤ TN −5 , we have
Z t Z
Z t
Z t
∂s ξ[vα(n) ]2 dx ds ≤
G(kv (n−1) kα )kv (n) k2α ds +
F (kv (n−1) kα )ds.
0
0
R
0
It follows that
Z
ξ(x, t)[vα(n) (x, t)]2 dx
R
Z
Z t
Z t
≤
ξ(x, 0)[vα(n) (x, 0)]2 dx +
G(kv (n−1) kα )kv (n) k2α ds +
F (kv (n−1) kα )ds
0
R
0
hence
Z
[vα(n) ]2 dx ≤
c8
R
Z
ξ[vα(n) ]2 dx
R
Z
≤
ξ(x, 0)[vα(n) (x, 0)]2 dx +
R
Z t
+
F (kv (n−1) kα )ds.
Z
t
G(kv (n−1) kα )kv (n) k2α ds
0
0
Then
Z
Z
c9
G(cN −5 ) 2
F (cN −5 )
[v (n) (x, 0)]2 dx +
cN −5 t +
t
c8 R α
c8
c8
c9
G(cN −5 ) 2
F (cN −5 )
≤ kv (n) (·, 0)k2α +
cN −5 t +
t
c8
c8
c8
G(cN −5 ) 2
F (cN −5 )
c9
≤ kϕ(·, 0)k2N +
cN −5 t +
t
c8
c8
c8
[vα(n) ]2 dx ≤
R
and we obtain
kv (n) (·, t)k2α ≤
c9
G(c3 ) 2
F (c3 )
kϕ(·, 0)k2N +
c t+
t.
c8
c8 3
c8
(n)
Claim: TN −5 does not approach 0.
(n)
On the contrary, assume that TN −5 → 0. Since kv (n) (·, t)k is continuous for t ≥ 0,
there exists τ ∈ [0, TN −5 ] such that kv (k) (·, τ )kα = cN −5 for 0 ≤ τ ≤ T (n) , 0 ≤ k ≤
n. Then
c2N −5 ≤
c9
G(cN −5 ) 2
F (cN −5 ) (n)
(n)
kϕ(·, 0)k2N +
cN −5 TN −5 +
TN −5 .
c8
c8
c8
as n → +∞ we have
2
c9
c9
c2
kϕ(·, 0)k2N + 1 ≤ kϕ(·, 0)k2N =⇒ 92 kϕ(·, 0)k4N + 1 ≤ 0
2c8
c8
4c8
EJDE-2004/71
GAIN OF REGULARITY
15
(n)
which is a contradiction. Then TN −5 6→ 0. By choosing TN −5 = TN −5 (kϕ(·, 0)k2N )
sufficiently small, and TN −5 not depending on n, we conclude that
kv (n) (·, t)k2α ≤ c2N −5
for all 0 ≤ t ≤ TN −5 .
(5.9)
(n)
This shows that TN −5 ≥ TN −5 . Thus,
v ∈ L∞ ([0, TN −5 ]; H α (R)) ≡ v ∈ L∞ ([0, TN −5 ]; H N −4 (R)).
Now, denote by 0 ≤ TN∗ −5 ≤ +∞ the maximal number such that for all 0 < t ≤
TN∗ −5 , u = ∧v ∈ L∞ ([0, t]; H N (R)). In particular TN −5 ≤ TN∗ −5 for all N ≥ 6.
Thus, T can be chosen depending only on the norm of ϕ in H 5 (R). Approximating
ϕ by {ϕj } ∈ C0∞ (R) such that kϕ−ϕj kH N (R) → 0 as j → +∞. Let uj be a solution
of (2.5) with uj (x, 0) = ϕj (x). According to the above argument, there exists T
which is independent on n but depending only on supj kϕj k such that uj exists on
[0, T ] and a subsequence uj
j→+∞
→
u in L∞ ([0, T ]; H N (R)).
As a consequence of Theorems 5.1 and 5.2 and its proof, one obtains the following
result.
Corollary 5.3. Let ϕ ∈ H N (R) with N ≥ 5 such that ϕ(γ) → ϕ in H N (R). Let
u and u(γ) be the corresponding unique solutions given by Theorems 5.1 and 5.2 in
L∞ ([0, T ]; H N (R)) with T depending only on supγ kϕ(γ) kH 5 (R) then
∗
u(γ) * u
weakly on L∞ ([0, T ]; H N (R)),
u(γ) → u
strongly on L2 ([0, T ]; H N +1 (R)),
u(γ) → u
strongly on L2 ([0, T ]; H N +2 (R))
6. Existence of Global Solutions
Here, we will try to extend the local solution u ∈ L∞ ([0, T ]; H N (W0i0 )) of (2.5)
obtained in Theorem 5.2 to t ≥ 0. A standard way to obtain these extensions
consists into deducing global estimations for the H N (W0i0 )-norm of u in terms
of the H N (W0i0 )-norm of u(x, 0) = ϕ(x). These estimations are frecuently based
on conservation laws which contain the L2 -norm of the solution and their spatial
derivatives. It is not possible to do the same to give a solution of the problem
of global existence because the difficulty here is that the weight depends on the
variables x and t variables. To solve our problem we follow a different method
using the Leibniz rule like in the proof of Theorem 3.1 of Bona and Saut, cf. [5].
Theorem
T 6.1. For η < −3/5 there exists a global solution to (2.5) in the space
H s (R) H N (W0i0 ) with N integer ≥ 5 and s ≥ 2.
Proof. The first part was proved in [1], with N ≥ 5 and a nonegative integer i.
Taking ∂ α derivatives of the equation (2.5)
∂t uα + ηuα+5 + uα+3 + (uu1 )α = 0.
(6.1)
We multiply (6.1) by 2ξuα and integrate over R.
Z
Z
Z
Z
2 ξuα ∂t uα dx+2η ξuα uα+5 dx+2 ξuα uα+3 dx+2 ξuα (uu1 )α dx = 0. (6.2)
R
R
R
R
16
OCTAVIO PAULO VERA VILLAGRÁN
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Each term is treated separately. The first term yields
Z
2
Z
ξuα ∂t uα dx = ∂t
R
ξu2α dx −
R
Z
ξt u2α dx.
R
In the second and third term, integrating by parts, we obtain
Z
2η
Z
Z
Z
ξuα uα+5 dx = −η ∂ 5 ξu2α dx + 5η ∂ 3 ξu2α+1 dx − 5η ∂ξu2α+2 dx ,
R
R
R
R
Z
Z
Z
3
2
2
2 ξuα uα+3 dx = − ∂ ξuα dx + 3 ∂ξuα+1 dx.
R
R
R
In the last term, using the Leibniz rule, we obtain
Z
2
ξuα (uu1 )α dx
Z
Z
Z
α(α − 1)
ξu2 uα−1 uα dx
= 2 ξuuα uα+1 dx + 2α ξu1 u2α dx + 2
2
R
R
Z
Z R
α!
α!
+2
ξu3 uα−2 uα dx + 2
ξu4 uα−3 uα dx
3!(α − 3)! R
4!(α − 4)! R
Z
+ · · · + 2 ξu1 u2α dx.
R
R
Integrating by parts it follows that
Z
2
ξuα (uu1 )α dx
Z
Z
Z
α(α − 1)
= − ∂(ξu)u2α dx + 2α ξu1 u2α dx −
∂(ξu2 )u2α−1 dx
2
R
R
Z
ZR
α!
α!
ξu4 uα−3 uα dx
+2
ξu3 uα−2 uα dx + 2
3!(α − 3)! R
4!(α − 4)! R
Z
+ · · · + 2 ξu1 u2α dx.
R
R
Substituting in (6.2), we have
Z
Z
Z
Z
ξu2α dx − ξt u2α dx − η ∂ 5 ξu2α dx + 5η ∂ 3 ξu2α+1 dx
R
R
R
R
Z
Z
Z
Z
2
3
2
2
− 5η ∂ξuα+2 dx − ∂ ξuα dx + 3 ∂ξuα+1 dx − ∂(ξu)u2α dx
R
R
R
R
Z
Z
Z
α!
α(α
−
1)
∂(ξu2 )u2α−1 dx + 2
ξu3 uα−2 uα dx
+ 2α ξu1 u2α dx −
2
3!(α − 3)! R
R
R
Z
Z
α!
+2
ξu4 uα−3 uα dx + · · · + 2 ξu1 u2α dx = 0
4!(α − 4)! R
R
∂t
EJDE-2004/71
GAIN OF REGULARITY
17
hence
Z
∂t
Z
ξu2α dx
3
+
R
(5η∂ ξ +
3∂ξ)u2α+1 dx
Z
− 5η
R
∂ξu2α+2 dx
R
Z
(−ξt − η∂ 5 ξ − ∂ 3 ξ − ∂(ξu) + 2αξu1 )u2α dx
R
Z
Z
α!
α(α − 1)
∂(ξu2 )u2α−1 dx + 2
ξu3 uα−2 uα dx
−
2
3!(α − 3)! R
R
Z
Z
α!
ξu4 uα−3 uα dx + · · · + 2 ξu1 u2α dx = 0
+2
4!(α − 4)! R
R
+
then using (2.4), Gagliardo - Nirenberg’s inequality and standard estimates we get
Z
∂t
ξu2α dx +
Z
R
(5η + 3)ξu2α+1 dx − 5η
Z
R
∂ξu2α+2 dx ≤ c
R
Z
ξu2α dx .
(6.3)
R
Integrating (6.3) in t ∈ [0, Tmax = T ] we obtain
Z
Z tZ
Z tZ
2
2
ξuα dx +
(5η + 3)ξuα+1 dxds − 5η
∂ξu2α+2 dxds
R
0
R
0
R
Z t Z
2
≤ kϕkα +
c ξu2α dx ds ,
0
R
where
Z
ξu2α dx ≤ kϕk2α +
Z
Using Gronwall’s inequality
Z
Z
ξu2α dx ds.
c
0
R
t
R
ξu2α dx ≤ kϕk2α ect ≤ kϕk2α ecT
R
it follows that
Z
ξu2α dx ≤ C = C(T, kϕk).
R
Then for any T = Tmax > 0, there exists C = C(T, kϕk) such that
Z tZ
Z tZ
kuk2α +
(5η + 3)ξu2α+1 dxds − 5η
∂ξu2α+2 dxds ≤ C.
0
R
0
R
This concludes the proof.
7. Persistence Theorem
As a starting point for the a priori gain of regularity results that will be discussed
in the next section, we need to develop some estimates for solutions of the equation
(2.5) in weighted Sobolev norms. The existence of these weighted estimates is
often called the
T persistence of a property of the initial data ϕ. We show that
if ϕ ∈ H 5 (R) H L (W0i0 ) for L ≥ 0, i ≥ 1 then the solution u(·, t) evolves in
H L (W0i0 ) for t ∈ [0, T ]. The time interval of such persistence is at least as long as
the interval guaranteed by the existence Theorem 5.2.
18
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
Theorem 7.1 (Persistence). Let i ≥ 1 and L ≥ 0 be non-negative integers, 0 <
T < +∞. Assume that u is the solution to (2.5) in L∞ ([0, T ]; H 5 (R)) with initial
data ϕ(x) = u(x, 0) ∈ H 5 (R). If ϕ(x) ∈ H L (W0i0 ) then
\
u ∈ L∞ ([0, T ]; H 5 (R) H L (W0i0 ))
(7.1)
Z TZ
|∂ L+1 u(x, t)|2 µ1 dxdt < +∞
(7.2)
Z
0
T
0
R
Z
|∂ L+2 u(x, t)|2 µ2 dxdt < +∞ ,
(7.3)
R
where σ is arbitrary, µ1 ∈ Wσ,i,0 and µ2 ∈ Wσ,i−1,0 for i ≥ 1.
Proof. We use induction on α. Let
u ∈ L∞ ([0, T ]; H 5 (R)
\
H α (W0i0 ))
for 0 ≤ α ≤ L.
We derive formally some a priori estimate for the solution where the bound, involves
only the norms of u in L∞ ([0, T ]; H 5 (R)) and the norms of ϕ in H 5 (W0i0 ). We do
this by approximating u(x, t) through smooth solutions, and the weight functions
by smooth bounded functions. By Theorem 5.2, we have
u(x, t) ∈ L∞ ([0, T ]; H N (R)) with N = max{L, 5}.
In particular uj (x, t) ∈ L∞ ([0, T ] × R) for 0 ≤ j ≤ N − 1. To obtain (7.1)-(7.2)
and (7.3) there are two ways of approximation perform. We approximate general
solutions by smooth solutions, and we approximate general weight functions by
bounded weight functions. The first of these procedures has already been discussed,
so we will concentrate on the second.
Given a smooth weight function µ2 (x) ∈ Wσ,i−1,0 with σ > 0, we take a sequence
µβ2 (x) of smooth bounded weight functions approximating µ2 (x) from below, uniformly on any half line (−∞, c). Define the weight functions for the α-th induction
step as
Z x
1
ξβ (x, t) = −
1+
µβ2 (y, t)dy
5η
−∞
then the ξβ are bounded weight functions which approximate a desired weight
function ξ ∈ W0i0 from below, uniformly on a compact set. For α = 0, multiplying
(2.5) by 2ξβ u, and integrating over x ∈ R.
Z
Z
Z
Z
2 ξβ uut dt + 2η ξβ uu5 dx + 2 ξβ uu3 dx + 2 ξβ u2 u1 dx = 0.
(7.4)
R
R
R
R
Each term is treated separately. In the first term we have
Z
Z
Z
2
2 ξβ uut dx = ∂t ξβ u dx − ∂t ξβ u2 dx.
R
R
R
For the others terms, using integration by parts, we have
Z
Z
Z
Z
2η ξβ uu5 dx = −η ∂ 5 ξβ u2 dx + 5η ∂ 3 ξβ u21 dx − 5η ∂ξβ u2 dx.
R
R
R
R
Z
Z
Z
3
2
2
2 ξβ uu3 dx = − ∂ ξβ u dx + 3 ∂ξβ u1 dx ,
R
R
R
Z
Z
2
2 ξβ u2 u1 dx = −
∂ξβ u3 dx.
3 R
R
EJDE-2004/71
GAIN OF REGULARITY
19
Replacing in (7.4), we obtain
Z
Z
Z
Z
2
2
5
2
∂t ξβ u dx − ∂t ξβ u dx − η ∂ ξβ u dx + 5η ∂ 3 ξβ u21 dx
R
R
R
R
Z
Z
Z
Z
2
2
3
2
2
− 5η ∂ξβ u dx − ∂ ξβ u dx + 3 ∂ξβ u1 dx −
∂ξβ u3 dx = 0
3
R
R
R
R
then
Z
∂t
ξβ u2 dx +
Z
ZR
(5η∂ 3 ξβ + 3∂ξβ )u21 dx − 5η
R
Z
∂ξβ u22 dx
R
2
+ (−∂t ξβ − η∂ ξβ − ∂ ξβ − ξβ u)u2 dx = 0 .
3
R
5
3
Using (2.4), for c5 > 0 (η < −3/5),
Z
Z
Z
2
2
∂t ξβ u dx − c5 (5η + 3) ξβ u1 dx − 5η ∂ξβ u22 dx
R
R
R
Z
2
5
3
+ (−∂t ξβ − η∂ ξβ − ∂ ξβ − ξβ u)u2 dx ≤ 0 .
3
R
Using again (2.4) and Gagliardo-Nirenberg’s inequality, we obtain
Z
Z
Z
Z
2
2
2
∂t ξβ u dx − c5 (5η + 3) ξβ u1 dx − 5η ∂ξβ u2 dx ≤ c ξβ u2 dx
R
R
R
R
thus
Z
Z
2
ξβ u dx ≤ c
∂t
R
ξβ u2 dx.
R
We apply Gronwall’s lemma to conclude
Z
ξβ u2 dx ≤ C = C(T, kϕk)
(7.5)
R
for 0 ≤ t ≤ T and c not depending on β > 0, the weighted estimate remains true
for β → 0.
Now, we assume that the result is true for (α − 1) and we prove that it is true for
α. To prove this, we start from the main inequality (3.2) with µ1 , µ2 and ξ given
by µβ1 , µβ2 and ξβ respectively.
Z
Z
Z
Z
Z
∂t ξβ u2α dx + µβ1 u2α+1 dx + µβ2 u2α+2 dx + θβ u2α dx + Rα dx ≤ 0
R
R
R
R
R
with
µβ1 = −c5 (5η + 3)ξβ
for η < −3/5 (Natural Condition)
µβ2 = −5η∂ξβ
θβ = −∂t ξβ − η∂ 5 ξβ − ∂ 3 ξβ − ∂(ξβ u)
Rα = O(uα , . . . )
20
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
then
Z
ξβ u2α dx +
∂t
Z
R
µβ1 u2α+1 dx +
Z
R
µβ2 u2α+2 dx ≤ −
Z
R
θβ u2α dx −
Z
Rα dx
R
R
Z
Z
≤ − θβ u2α dx − Rα dx
R
Z
Z R
≤
|θβ |u2α dx + |Rα |dx .
R
R
Using (2.4) and Gagliardo-Nirenberg in the first term of the right side we obtain
Z
Z
|θβ |dx ≤ c ξβ u2α dx
R
R
Thus
Z
∂t
ξβ u2α dx +
Z
R
µβ1 u2α+1 dx +
R
R
µβ2 u2α+2 dx ≤ c
R
R
According to (3.5),
Z
Z
ξβ u2α dx +
R
Z
|Rα |dx.
R
Rα dx contains a term of the form
Z
ξβ uν1 uν2 uα dx.
(7.6)
R
Let ν2 ≤ α − 2. Integrating (7.6) by parts and using Hölder’s inequality we obtain
h Z
1/2 Z
1/2 i Z
1/2
c
ξβ u2ν2 +1 dx
+
ξβ u2ν2 dx
ξβ u2α−1 dx
(7.7)
R
R
R
where (7.7) is bounded by hypothesis. Now suppose that α − 1 = ν1 = ν2 , then in
(7.6) we obtain
Z
Z
1/2 Z
1/2
2
2
ξβ uα−1 uα−1 uα dx ≤ kuα−1 k ∞
ξ
u
dx
ξ
u
dx
β
β
α−1
α
L (R)
R
R
R
where kuα−1 kL∞ (R) is bounded by hypothesis, and the estimate is complete. Finally,
for ν1 = α − 2; ν2 = α − 1 we have
Z
Z
p
p
ξβ uα−2 uα−1 uα dx = ξβ uα−2 uα−1 ξβ uα dx
R
R
Z
p
p
≤
ξβ uα−2 L∞ (R)
uα−1 ξβ uα dx
R
Z
1/2
p
≤ ξβ uα−2 L∞ (R) uα−1 L2 (R)
ξβ u2α dx
R
Z
1/2
2
≤ c kuα−1 kL2 (R)
ξβ uα dx
.
R
Using these estimates in (7.5), and applying the Gronwall’s argument, we obtain
for 0 ≤ t ≤ T ,
Z
Z
Z
Z
∂t ξβ u2α dx + µβ1 u2α+1 dx + µβ2 u2α+2 dx ≤ c0 ec1 t
ξβ ϕ2α (x)dx + 1
R
R
R
R
where c0 and c1 are independent β such that letting the parameter β → 0 the
desired estimates (7.2) and (7.3) are obtained.
EJDE-2004/71
GAIN OF REGULARITY
21
8. Main Theorem
In this section we state and prove our main Theorem, which states that if the
initial data u(x, 0) decays faster than polinomially on R+ = {x ∈ R; x > 0} and
possesses certain initial Sobolev regularity, then the solution u(x, t) ∈ C ∞ for all
t > 0. For the main Theorem, we take 6 ≤ α ≤ L + 4. For α ≤ L + 4, we take
µ1 ∈ Wσ,L−α+5,α−5
=⇒
ξ ∈ Wσ,L−α+5,α−5
(8.1)
µ2 ∈ Wσ,L−α+4,α−5
=⇒
ξ ∈ Wσ,L−α+5,α−5
(8.2)
Lemma 8.1 (Estimate of error terms). Let 6 ≤ α ≤ L + 4 and the weight functions
be chosen as in (8.1)-(8.2), then
Z Z
T
θu2α + Rα dxdt ≤ c ,
(8.3)
0
R
where c depends only on the norms of u in
\
L∞ ([0, T ]; H β (Wσ,L−β+5,β−5 )) L2 ([0, T ]; H β+1 (Wσ,L−β+5,β−5 )
\
H β+2 (Wσ,L−β+4,β−5 ))
for 5 ≤ β ≤ α − 1, and the norms of u in L∞ ([0, T ]; H 5 (W0L0 )).
Proof. We must estimate both Rα and θ. We begin with a term of Rα of the form
ξuν1 uν2 uα
(8.4)
assuming that ν1 ≤ α − 2. By the induction hypothesis, u is bounded in
L∞ ([0, T ]; H β (Wσ,L−(β−5)+ ,(β−5)+ )) for 0 ≤ β ≤ α − 1. By Lemma 2.1,
sup sup ζu2β < +∞
(8.5)
t>0 x∈R
for 0 ≤ β ≤ α − 2 and ζ ∈ Wσ,L−(β−4)+ ,(β−4)+ . We estimate uν1 using (8.5). We
estimate uν2 and uα using the weighted L2 bounds
Z TZ
(8.6)
ζu2ν2 dxdt < +∞ for ζ ∈ Wσ,L−(ν2 −5)+ ,(ν2 −6)+
0
R
and the same with ν2 replaced by α. It suffices to check the powers of t, the powers
of x as x → +∞ and the exponential of x as x → −∞.
For x > 1. In the term (8.4), the factor ξ constributed according to (8.1)-(8.2)
ξ(x, t) = t(α−5) x(L−α+5) t−(α−5) x−(L−α+5) ξ(x, t) ≤ c2 t(α−5) x(L−α+5)
by (2.3). Then ξuν1 uν2 uα ≤ c2 t(α−5) x(L−α+5) uν1 uν2 uα . Moreover
uν1 uν2 uα = t
(ν1 −4)+
2
× uν1 t
×t
x
L−(ν1 −4)+
2
(ν2 −6)+
2
(α−6)+
2
x
x
t
−(ν1 −4)+
2
L−(ν2 −5)+
2
L−(α−5)+
2
t
t
x
−(L−(ν1 −4)+ )
2
−(ν2 −6)+
2
−(α−6)+
2
x
x
−(L−(ν2 −5)+ )
2
−(L−(α−5)+ )
2
uν2
uα .
It follows that
ξuν1 uν2 uα
≤ c2 tM xT t
(ν1 −4)+
2
x
L−(ν1 −4)+
2
uν1 t
(ν2 −6)+
2
x
L−(ν2 −5)+
2
uν2 t
(α−6)+
2
x
L−(α−5)
2
(8.7)
uα
22
OCTAVIO PAULO VERA VILLAGRÁN
EJDE-2004/71
where M = α − 5 − 12 (ν1 − 4)+ − 12 (ν2 − 6)+ − 12 (α − 6)+ and
1
1
1
T = (T − α + 5) − (T − (α − 5)+ ) − (T − (ν2 − 5)+ ) − (T − (ν1 − 4)+ ).
2
2
2
Claim M ≥ 0 is large enough, that the extra power of t can be omitted.
2M = 2α − 10 − (ν1 − 4)+ − (ν2 − 6)+ − (α − 6)+
= α − 4 − (ν1 − 4)+ − (ν2 − 6)+
= α − 4 − ν1 + 4 − ν2 + 6
= α + 6 − (ν1 + ν2 )
= α + 6 − (α + 1) = 5 ≥ 0.
Claim T ≤ 0 is such that the extra power xT can be bounded as x → +∞.
1
1
1
T = L − α + 5 − (L − (α − 5)+ ) − (L − (ν2 − 5)+ ) − (L − (ν1 − 4)+ ).
2
2
2
Thus
2T = 2L − 2α + 10 − (L − (α − 5)+ ) − L + (ν2 − 5)+ − L + (α − 4)+
= −L − α + ν1 + ν2 − 4
= −L − α + α + 1 − 4
= −(L + 3) ≤ 0.
Now, we study the behavior as x → −∞. Since each factor uνj (j = 1, 2) must grow
,
slower than an exponential eσ |x| and ξ decays as an exponential e−σ|x| , we simply
,
need to choose the appropriate relationship between σ and σ at each induction step.
The analysis of all the terms of Rα will be completed with the case of ν1 ≥ α − 1.
Then in (3.6) if 2(α + 1) ≤ α + 1, α ≤ 3, but α ≥ 5 so this possibility is impossible.
For x < 1 the estimate is similar, except for an exponential weight. This completes
the estimate of Rα .
Now we estimate the term θu2α where θ is given in (3.2). We have that θ involves
derivatives of u only up to order one and hence θu2α is a sum of terms of the same
type which we have already encountered in Rα . So, its integral can be bounded in
the same manner. Indeed (3.2) shows that θ depends on ξt , ∂ 5 ξ and derivatives of
lower order. By using (3.3) we have the claim.
Theorem 8.2 (Main Theorem). Let T > 0 and u(x, t) be a solution of (2.5) in
the region R × [0, T ] such that
u ∈ L∞ ([0, T ]; H 5 (W0L0 ))
∞
(8.8)
5+l
for some L ≥ 2 and all σ > 0. Then u is in L ([0, T ]; H (Wσ,L−l,l ))
∩ L2 ([0, T ]; H 6+l (Wσ,L−l,l ) ∩ H 7+l (Wσ,L−l−1,l )) for all 0 ≤ l ≤ L − 1.
Remark 8.3. If the assumption (8.8) holds for all L ≥ 2, the solution is infinitely
differentiable in the x-variable. From (2.5) we have that the solution is C ∞ in both
variables.
Proof. We use induction on α. For α = 5, let u be a solution of (2.5) satisfying
5
(8.8). Therefore, ut ∈ L∞ ([0, T ]; L2 (W0L0 )) where u ∈ L∞
T([0, T ]; H (W50L0 )) and
∞
2
2
ut ∈ L ([0, T ]; L (W0L0 )). Then u ∈ C([0, T ]; L (W0L0 )) Cw ([0, T ]; H (W0L0 )).
Hence u : [0, T ] 7→ H 5 (W0L0 ) is a weakly continuous function. In particular, u(·, t) ∈
H 5 (W0L0 ) for all t. Let t0 ∈ (0, T ) and u(·, t0 ) ∈ H 5 (W0L0 ), then there are {ϕ(n) } ⊂
EJDE-2004/71
GAIN OF REGULARITY
23
C0∞ (R) such that ϕ(n) (·) → u(·, t0 ) in H 5 (W0L0 ). Let u(n) (x, t) be a unique solution
of (2.5) with u(n) (x, t0 ) = ϕ(n) (x). Then by Theorems 5.1 and 5.2, there exists in
a time interval [t0 , t0 + δ] where δ > 0 does not depend on n and u is a unique
solution of (2.5) u(n) ∈ L∞ ([t0 , t0 + δ]; H 5 (W0L0 )) with u(n) (x, t0 ) ≡ ϕ(n) (x) →
u(x, t0 ) ≡ ϕ(x) in H 5 (W0L0 ). Now, by Theorem 7.1, we have
\
u(n) ∈ L∞ ([t0 , t0 + δ]; H 5 (W0L0 )) L2 ([t0 , t0 + δ]; H 6 (WσL0 ) ∩ H 7 (Wσ,L−1,0 ))
with a bound that depends only on the norm of ϕ(n) in H 5 (W0L0 ). Furthermore,
Theorem 7.1 guarantees the non-uniform bounds
sup sup(1 + |x+ |)k |∂ α u(n) (x, t)| < +∞
[t0 ,t0 +δ] x
for each n, k and α. The main inequality (3.2) and the estimate (8.3) are therefore
valid for each u(n) in the interval [t0 , t0 + δ]. µ2 may be chosen arbitrarily in its
weight class (8.1) and then ξ is defined by (3.4) and the constant c1 , c2 , c3 , c4 are
independent of n. From (3.2) and (8.1)-(8.2) we have
Z
Z t0 +δ Z
Z t0 +δ Z
(n) 2
(n)
2
]
dxdt+
µ2 [uα+2 ]2 dxdt ≤ c (8.9)
sup
ξ[u(n)
]
dx+
µ
[u
1 α+1
α
[t0 ,t0 +δ]
R
t0
t0
R
R
where by (8.3), c is independ of n. Estimate (8.9) is proved by induction for
α = 5, 6, . . . Thus u(n) is also bounded in
\
L∞ ([t0 , t0 + δ]; H α (Wσ,L−α+5,α−5 )) L2 ([t0 , t0 + δ]; H α+1 (Wσ,L−α+5,α−5 ))
\
L2 ([t0 , t0 + δ]; H α+2 (Wσ,L−α+4,α−5 ))
(8.10)
for α ≥ 5. Since u(n) −→ u in L∞ ([t0 , t0 + δ]; H 5 (W0L0 )). By Corollary 5.3 it
follows that u belongs to the space (8.10). Since δ is fixed, this result is valid over
the whole interval [0, T ].
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Facultad de Ingenierı́a, Universidad Católica de la Santı́sima Concepción, Paicavı́
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