analyticity of solutions of the generalized korteweg-de ... - CiteSeerX
ANALYTICITY OF SOLUTIONS OF THE GENERALIZED KORTEWEG-DE VRIES EQUATION WITH RESPECT TO THEIR INITIAL VALUES Bing-Yu Zhang
y
Institute of Mathematics and Its Applications University of Minnesota Minneapolis, Minnesota 55455
Abstract
It is shown that the initial value problem (IVP) of the generalized KdV equation @t u + @x (a(u)) + @x u = 0;
u(x; 0) = (x) is well posed in the classical Sobolev space H s (R) with s > 3=4, which thus establishes a nonlinear map K from H s (R) to C ([?T; T ]; H s(R)). Then it is proved that (i) if a = a(x) is a C 1 function on R to R, then K is in nitely many times Frechet di erentiable; (ii) if a = a(x) is a polynomial, then K is analytic, i.e. for any 2 H s (R), K has a Taylor series expansion 1 1 X K ( + h) = K n ( )[hn ] n ! n 3
(
)
=0
Supported partially by 1992's Taft Summer Research Grant On leave of absence from: Department of Mathematical Sciences, University of Cincinnati, Ohio 45221 y
1
Each term yn = K n [hn ] in the series solves a linearized KdV equation. Thus any \small" perturbation K ( + h) of K ( ) can be obtained by solving a series of linear problems. The proof of these results relies on various smoothing properties of the associated linear KdV equation. (
)
1 Introduction This paper is mainly concerned with the initial value problem (IVP) for the generalized Korteweg-de Vries (KdV) equation: 8 > < @tu + @x(a(u)) + @xu = 0; t; x 2 R (1:1) > : u(x; 0) = (x) 3
in which a(x) is assumed to be a C 1 function on R to R , though a weaker di erentiablity suces for most results below. The KdV equation (i.e., a(u) = u =2 in (1.1) ) and its generalized form (1.1) have been studied by many authors [1, 2, 4 - 6, 10 - 17, 19 - 23, 27 37, 42 - 44, 47]. For a complete list of references, see, for example, [ 13, 17, 22, 30 ]. In particular, it is well-known that the IVP (1.1) is locally well posed in the classical Sobolev space H s (R) with s > 3=2 (cf. [17]) and is globally well posed with some restrictions on a(u) or the size of the initial data when s 2 (see [5], [14], [17], [19] and [28] for the well-posedness of the IVP (1.1) in other function spaces). In the case that a0(u) = uk in (1.1) with k being a positive integer, Kenig, Ponce and Vega [23] recently proved that the IVP (1.1) is locally well posed in the space H s (R) with 8 > s > 3=4 if k = 1 > > > > > s 1=4 if k = 2 < (1:2) > > if k = 3 s > > > > : s k?k if k 4 2
1
12
4
2
2
and is global well-posedness when 1 k 3 and s 1. Their proof is based on careful analysis of various smooth properties of the associated linear problem together with the contraction mapping principle. In this paper, we rst continue to study the well-posedness of the IVP (1.1) in the space H s (R). While only assuming that a(0) = a0(0) = 0, we shall prove that the IVP (1.1) is locally well posed in the space H s(R) with s > 3=4 by using Kenig, Ponce and Vega's argument [22], [23] with a few modi cation. In order to state our result precisely, we introduce the following Banach spaces as Kenig, Ponce and Vega did in [23]. Let s > 0 and T > 0 be given. For
w : R [?T; T ] ! R; de ne
(T; w) = sup kw(:; t)ks; 1
?T;T T s [
]
!= jD @xw(x; t)j dt ; (T; w) = sup x ?T != ZT (T; w; l) = ?T kJ l@xw(:; t)k1dt Z
2
1 2
2
1 4
4
3
with l 2 [0; s ? 3=4] where J s = (1 ? @x)s= , != Z ? r (T; w; r) = (1 + T ) sup jJ w(x; t)j dx 2
2
1 2
R ?T;T
4
[
2
]
with r 2 [0; s ? 3=4) and > 3=4 being a xed constant, and sl;r (T ; w) = max f (T; w); (T; w); (T; w; l); (T; w; r)g : (1:3) 1
2
3
Denote by
4
n o Xl;rT;s = w 2 C ([?T; T ]; H s(R)) j sl;r (T ; w) < 1 (1:4) with (l; r) 2 [0; s ? ] [0; s ? ). This is a Banach space equipped with the norm kwkXl;rT;s := sl;r(T ; w): 3
3
4
4
3
Clearly, Xl;rT;s is a subspace of C ([?T; T ]; H s(R)) with stronger topology. We shall prove that if a(0) = a0(0) = 0, then for any 2 H s (R), there exists a T > 0 depending only on k ks such that he IVP (1.1) has a unique solution u 2 Xl;rT;s where s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ). The global result is also obtained while the same restrictions are enforced on a(u) in (1.1) or the size of the initial value as those in Kato [17] but with s 1 instead of s 2. Thus the IVP (1.1) establishes a nonlinear map K from H s (R) to Xl;rT;s (or C ([?T; T ]; H s(R))). In the second part of this paper, we study di erentiability of the map K . It has been known for many years that the map K is continuous from H s (R) to C ([?T; T ]; H s(R)) [2], [17] and is proved recently by Kenig, Ponce and Vega [23] being Lipschitz continuous in the case that a0(u) = uk . In our early paper [47], we proved the map K corresponding to the classical KdV equation, i.e. a0(u) = u in (1.1), is in nitely many times Frechet di erentiable from H s(R) to Xl;rT;s and it has Taylor series expansion at any given 2 H s (R). That is to say, the map K is analytic from H s(R) to Xl;rT;s. In this paper we shall show that the nonlinear map K established by the IVP (1.1) is also in nitely many times Frechet di erentiable from H s (R) to Xl;rT;s. For any n 1, its n-th derivative K n ( ) at 2 H s (R), a n-linear map from the n-fold product space (H s(R))n into Xl;rT;s, can be constructed by solving a system of inhomogeneous linearized KdV equations. More precisely, for any n 1 and hk 2 H s(R) (k = 1; 2; :::; n), denote by w n;:::;n := K n ( )[h ; :::; hn]; (1:5) then it solves 8 > < @tw + @x(a0(u)w ) + @xw = 0 (1:6) > : w (x; 0) = h(x) (
(
)
[1
(
3
3
4
4
)
)
1
]
(1)
(1)
[1]
[1]
(1)
3
[1]
(1)
[1]
for n = 1 and 8 n n n > < @tw ;:::;n + @x(a0(u)w ;:::;n ) + @xw ;:::;n = ?@x(Hn ) > : w n;:::;n (x; 0) = 0 (
)
(
[1
(
[1
]
)
3
[1
]
)
]
4
(
[1
)
]
(1:7)
for n 2 where u = K ( ) and Hn is a polynomial of w ij ;:::;ij with 1 i ; :::; ij n and 1 j n ? 1 (see section 3 for the structure of Hn ). If we choose h = h = ::: = h and denote by yn = K n ( )[hn]; which is a homogeneous polynomial of degree n from H s (R) to Xl;rT;s, then 8 > @ty + @x(a0(u)y ) + @xy = 0 < (1:8) > : y (x; 0) = h(x) ( )
[ 1
]
1
1
2
(
)
1
3
1
1
1
for n = 1 and
8 > < @tyn + @x(a0(u)yn) + @xyn = ?@x fMn g > : yn(x; 0) = 0 3
for n 2 where
(1:9)
n j X X n! y :::y Mn = a j(! u) k kj j k ::: kj n k !:::kj ! and the summation Pk ::: kj n is over all possible (k ; :::; kj ) with 1 k ; :::kj n and k + :::kj = n. Thus we can de ne the n-th Taylor polynomial Pn of K at 2 H s (R): n n X Pn( )[h] := K ( ) + K k!( ) [hk ] n X (1.10) = u + ykk! k for any h 2 H s(R). Let zn = K ( + h) ? Pn ( )[h]: It is called the n-th Taylor remainder of K at and we shall see that zn solves 8 > < @tz + @x(F (u; v)z ) + @xz = 0 (1:11) > : z (x; 0) = h(x) ( )
1
=2
1+
1
1+
+
+
1
=
1
=
1
(
)
1
=1
0
1
0
0
5
3
0
for n = 0 and
8 > < @tzn + @x(F (u; v)zn) + @xzn = ?@x(Gn) > : zn (x; 0) = 0 3
1
(1:12)
for n 1 where
u = K ( ); v = K ( + h); nX ?m nX X zk Gn = Fm(u; v) qk :::qkm? +1
+1
m
with and
Fm(u; v) =
k
=2
k1 ::: km?1 n?k
=0
+
+
1
1
(1:13)
=
ym ; qm = m !
Z 0
1
m? :::: 1
1
Z
1
m?
Z
1
(
)
1
1
0
1 0m m Y Y a m @ j v + (1 ? j )uA dm :::d j j (1:14)
0
=1
=1
for m = 1; :::; n + 1. Letting n ! 1 in (1.10), we obtain a formal Taylor series of K at 2 H s(R). Assuming that a(u) is a polynomial, we shall prove that for any 2 H s (R), there exists a > 0 such that if h 2 H s(R) with khks , then 1 K n ( ) X n ]; K ( + h) = [ h (1:15) n! (
k
)
=0
the series conversing uniformly about h with khks in the space Xl;rT;s. In another word, the map K is analytic from H s (R) to Xl;rT;s . The paper is organized as follows. { In section 2, we rst list the estimates concerning the IVP @tu + @xu = f (x; t); u(x; 0) = u (x) (1:16) which are needed to establish the nonlinear results. Then we consider the IVP for the following linear equation 8 > < @tu + @x(a(v)u) + @xu = f (x; t); x; t 2 R : (1:17) > : u(x; 0) = (x) 3
0
3
6
Let s > 3=4, (l; r) 2 [0; s ? ] [0; s ? ) and T > 0 be given. We shall s s show that if v 2 X T;s ; , then for any 2 H (R) and f 2 L ([?T; T ]; H (R)), T;s the IVP (1.17) has a unique solution u 2 Xl;r and ! ZT (1:18) kukXl;rT;s kvkX T;s; k ks + ?T kf (:; t)ksdt 3
3
4
4
1
0 0
00
This would be a key estimate to obtain di erentiability of the map K . { In section 3, we show that the map K de ned by the IVP (1.1) is in nitely many times Frechet di erentiable from H s (R) to Xl;rT;s. { In section 4, assuming that a(u) in (1.1) is a polynomial, we show that the nonlinear map K is analytic from H s(R) to Xl;rT;s.
Notations:
{ The norm in L (R) will be denoted by k:k and the norm in H s(R) will be denoted by k:ks. The notation k:k1 is used to denote the norm in L1 (R). { Ds = (?@x)s= and J s = (1 ? @x)s= denote the Riesz and the Bessell potential of order s respectively. { [A; B ] = AB ? BA, where A; B are operators. Thus [J s; f ]g = J s(fg) ? fJ s g in which f is regarded as a multiplication operator. { H 1(R) := \s> H s(R) { For 1 p; q 1 and f : R [?T; T ] ! R; pq ! q Z T Z 1 p jf (x; t)j dx dt kf kLqT Lpx = 2
2
2
2
2
0
1
?T
and
?1
0Z Z ! qp 1 p T 1 j f (x; t)jqdt dxA : =@ ?1 ?T 1
kf kLpxLqT
7
2 Linear estimates 1 to denote the unitary group which de nes the solution We use fW (t)g?1 of the IVP associated to 8 > < @tv + @xv = 0; for x; t 2 R (2:1) > : v(x; 0) = v (x) +
3
0
where
v(t) = W (t)v = St v with St(:) de ned by the oscillatory integral Z 1 eix eit d: St(x) = c 0
0
+
3
?1
Then the solution of the inhomogeneous equation 8 > < @tv + @x v = f (x; t); x; t 2 R > : v(x; 0) = 0 3
is expressed as
v(t) =
Zt
Lemma 2.1 For any s 0,
0
(2:2)
W (t ? )f (:; )d :
= Z1 s sup jD @xW (t)v j dt ckv ks ; x ?1
(2:3)
= W (t)v k1 dt ckv ks:
(2:4)
1 2
2
0
and
Z 1 ?1
kDs =
+1 4
1 4
4
0
In addition, if s > 3=4, then
Z
1
+
sup
?1 ?T;T [
]
jJ lW (t)v
0
j (x)dx 2
0
0
!=
1 2
c(1 + T )kv ks 0
where l 2 [0; s ? 3=4) and is a xed constant larger than 3=4.
8
(2:5)
Proof: see Kenig, Ponce and Vega ([22], Lemma 2.1, Theorem 2.4 and
Corollary 2.9). Remark 2.1 (2.3) is a stronger version of local smoothing e ect of Kato type and (2.4) is the global smoothing e ect of Strichartz type [38] present in solutions of (2.1). The estimate (2.5), which gives a bound for the associated maximal function sup ?T;T jW (t):j, is due to Vega [46]. [
]
Lemma 2.2 For any s 0 and T > 0, kW (t)v ks = kv ks 0
and
sup k
?T;T
[
]
Zt 0
(2:6)
0
W (t ? )f (:; )d ks
ZT ?T
kf (:; )ksd :
(2:7)
Proof: (2.6) and (2.7) follow easily from Kato ([17], Lemma 3.1). Lemma 2.3 For any s 0 and T > 0, Zt
kDxs @x W (t ? )f (:; )d kL1x LT ckf kL
?T;T H s R
(2:8)
kf (:; )ksd :
(2:9)
kf (:; )ksd
(2:10)
1 ([
2
0
and
kDs
+
1 4
Zt
If s > 3=4, then
W (t ? )f (:; )d kL1x LT c 4
0
Zt
kJ l W (t ? )f (:; )d kLxL1T c(1 + T ) 2
0
ZT ?T
ZT ?T
];
(
))
where l 2 [0; s ? 3=4) and is a xed constant larger than 3=4.
Proof: It follows from Lemma 2.1 by using Minkowski's integral inequality (see [47]).
Lemma 2.4 Let s > 1=2 and T > 0 be given. Then there is a constant c > 0 such that ZT
?T
ku@xvksdt cT = (1 + T )kukX T;s; kvkX T;s; 1 2
00
9
00
(2:11)
and
ZT ?T
k@x(uv)ksdt cT = (1 + T )kukX T;s; kvkX T;s; :
(2:12)
1 2
00
00
for any u; v 2 X T;s ; . 0 0
Proof: see Lemma 2.4 in [47]. Lemma 2.5 Let s > 0 be given. Then there is a constant c > 0 such that for any yk 2 H s (R), k = 1; 2; :::; m, k
m Y k
1 0m m Y X kyk k1A yk ks cm @ kyj ks j
=1
and if s > 1=2,
k
m Y k
(2:13)
k ; k6 j
=1
=1
yk ks cm
=1
m Y k
=
kyk ks
(2:14)
=1
for any m 2 where c in (2.14) may be di erent from c in (2.13).
Proof: According to Kato and Ponce ([18], Lemma X4 ), ky y ks c fky ks ky k1 + ky ksky k1g 1
2
1
2
2
1
which is (2.13) with m = 2. Suppose that (2.13) is true for m = N . Then for m = N + 1, NY N Y k yk ks = kyN yk ks k k ( ) N N Y Y c kyN ksk yk k1 + kyN k1k yk ks +1
+1
=1
=1
+1
ckyN ks +1
cN
NX
+1
+1
j
=1
N Y k
k
+1
=1
kyk k1 + cN
=1
kyj ks
NY
+1
k ; k6 j =1
+1
N X j
=1
k
=1
kyj ks
NY
+1
k ; k6 j =1
kyk k1
=
kyk k1
=
Thus (2.13) is proved by induction. As for (2.14), it follows directly from (2.13) since kyk k1 kyk ks if s > 1=2. The proof is completed. 2 10
Lemma 2.6 Let b 2 C 1(R; R) with b(0) = 0. Then kb(u)ks ~b(kuks); s > 1=2 where ~b(:) is a monotone increasing function depending only on b.
Proof: see Kato ([17], Lemma A.3). Lemma 2.7 Let s > 1=2 and T > 0 be given and assume a 2 C 1(R; R) with a(0) = 0. Then there is a constant c > 0 such that for any u 2 X T;s ; T;s and y 2 X ; , 0 0
0 0
ZT
k@x(a(u)y)ksdt c kukX T;s; kykX T;s; ?T
(2:15)
00
00
where (:) is a continuous monotone increasing function only depending on a.
Proof: First of all, k@x(a(u)y)ks ka0(u)y@xuks + ka(u)@xyks and it is easy to see by using Lemma 2.5 and Lemma 2.6 that
ka0(u)y@xuks k(a0(u) ? a0(0))y@xuks + ja0(0)jky@xuks c fka0(u) ? a0(0)ks ky@xuk1 + ka0(u) ? a0(0)k1ky@xuksg + +ja0(0)jky@xuks c fka0(u) ? a0(0)ks + ja0(0)jg ky@xuks c (kuks)ky@xuks where (:) : R ! R is a continuous monotone increasing function only 1
1
+
+
depending on a. Using Lemma 2.4 yields ZT ZT k a0(u)y@xuksdt c sup (kuks) ky@xuksdt ?T ?T ?T;T = cT (1 + T ) kukX T;s; kukX T;s; kykX T;s; : [
]
1 2
1
11
00
00
00
In addition, applying Lemma 2.10 in [22], we obtain
ka(u)@xyks = kJ s(a(u)@xy)k = ka(u)Ds@xy + a(u)(J s ? Ds )@xy + [J s; a(u)]@xyk ka(u)Ds@xyk + ka(u)k1kyks + c fk@xyk1ka(u)ks+ + ka0(u)@xuk1kyksg : Note that ZT ZT ka(u)k1kyksdt sup ka(u)ks kyksdt ?T
?T;T
[
]
?T (kuks)
cT sup sup kyks (by Lemma 2.6) ?T;T ?T;T cT = (1 + T ) (kukX T;s; )kykX T;s; 2
[
]
[
]
1 2
2
00
00
where (:) : R ! R is a continuous monotone increasing function depending only on a; ZT ZT ka(u)ksk@xyk1dt sup ka(u)ks ?T k@xyk1dt ?T ?T;T ZT sup (kuks) ?T k@xyk1dt ?T;T != ZT = k@xyk1dt (kukX T;s; )(2T ) 2
+
+
[
]
[
]
2
1 4
3 4
2
cT
00
=
1 2
(1 + T )
2
?T
4
(kukX T;s; )kykX T;s; ; 00
00
ZT
ZT
ka0(u)@xuk1kyksdt sup (ka0(u)k1kyks) ?T k@xuk1dt ?T ?T;T cT = (1 + T ) (kukX T;s; )kukX T;s; kykX T;s; ; [
]
1 2
3
where
(r) = sup ja0()j; 3
and
ZT
ka(u)Ds @xykdt T ?T
00
=
1 2
jjr
ZTZ ?T R
ja(u)Ds@xuj dxdt
12
2
00
00
Z ZT
s @ uj dxdt sup k a(u) k j u j j D x u 1 ?T
R
2
2
?T;T
!= != ZT s sup ju(x; t)j dx sup jD @xyj dt x ?T [
Z
]
1 2
1 2
T (kukX T;s; ) R ?T;T = cT (1 + T ) (kukX T;s; )kukX T;s; kykX T;s; =
1 2
4
00
2
[
2
]
1 2
4
where
00
00
00
(r) = sup a() : jjr 4
Therefore,
ZT ?T
k@x(a(u)y)ksdt cT = (1 + T ) (kukX T;s; )kykX T;s; 1 2
00
00
for some constant c > 0 where
(r) = max fr (r); (r); r (r); r (r)g : 1
2
3
4
The proof is completed. 2 >From the proof of the above lemma we may draw the following corollary. Corollary 2.1 Let s > 1=2 and T > 0 be given and assume a 2 C 1(R; R) with a0(0) = 0. Then for any u 2 X T;s ; , ZT (2:16) k@x(a(u))ksdt c kukX T;s; T = (1 + T )kukX T;s; 0 0
1 2
?T
00
00
where (:) is a continuous monotone increasing function only depending on a.
Lemma 2.8 Let s > 1=2, T > 0 and a 2 C 1(R; R) be given. Then there T;s exists a constant c > 0 such that for any u 2 X T;s ; and yk 2 X ; with k = 1; 2; :::; m and m 2, 0 0
ZT
k@x(a(u) ?T
m Y
k
=1
yk )ksdt cm (kukX T;s; ) 00
m Y
k
=1
0 0
kyk kX T;s; 00
(2:17)
where (:) is a continuous monotone increasing function only depending on a.
13
Proof: By applying (2.14) we have k@x(a(u)
m Y
yk )ks ka0(u)@xu
m Y
yk ks +
m X
k
m Y
a(u)yj @xyk ks j ; j6 k m Y cm (ka0(u) ? a0(0)ks + ja0(0)j) ky @xuks kyk ks + k 0m m X Y +cm (ka(u) ? a(0)ks + ja(0)j) @ ky @xyk ks kyj ks+ k j ; j6 k 1 m Y + ky @xy ks kyj ksA : j k
=1
k
=1
k
=1
=1
=
+1
1
=2
+1
1
=2
2
=2
=
1
=3
Thus, using Lemma 2.6 and Lemma 2,4, we have ZT m Y k @x(a(u) yk )ks dt cm sup (ka0(u) ? a0(0)ks + ja0(0)j) ?T ?T;T k Z m T Y sup kyk ks ?T ky @xuksdt + cm sup (ka(u) ? a(0)ks + ja(0)j) ?T;T k0 ?T;T 1 ZT m m m X Y Y @ sup kyj ks ky @xyk ks dt + sup kyj ks sup ky @xy ksA +1
1
[
=1
]
+1
1
=2 [
k
1
]
[
?T
?T;T j ; j 6 k
=2 [
]
=2
=
1
cm (kukX T;s; )T = (1 + T )kukX T;s; 1 2
+1
1
1
00
m Y
j
?T;T
=3 [
]
?T;T
[
2
1
]
kyk kX0T;s;0 + k m = (1 + T ) Y ky k T;s k X0;0 k 00
+mcm (kukX T;s; )T m Y m c (kukX T;s; ) kyk kX T;s;
=1
1 2
+1
1
]
1
00
00
k
=1
=1
00
for some c > 0. The proof is completd. 2 To end this section, we consider the following linear problem: 8 > < @tu + @x(a(v)u) + @xu = f (x; t); x; t 2 R > : u(x; 0) = (x) 3
where a(:) 2 C 1(R; R) with a(0) = 0. 14
(2:18)
Theorem 2.1 Let s > 3=4, (l; r) 2 [0; s? ][0; s? ), T > 0 and v 2 X T;s; be given. Then for any f 2 L ([?T; T ]; H s(R)) and 2 H s (R), there exists a unique solution u 2 Xl;rT;s to (2.18) such that 3
3
4
4
0 0
1
kukXl;rT;s (kvkX T;s; ) k ks +
ZT
00
?T
! kf (:; t)ksdt
(2:19)
where is a continuous monotone increasing function only depending on a. Proof: We use the contraction principle argument that Kenig, Ponce and Vega used in [23]. For any given 2 H s (R) and f 2 L ([?T; T ]; H s(R)), denote by u = (w), the solution of the following IVP 8 > < @tu + @xu = f ? @x(a(v)w) (2:20) > : u(x; 0) = (x) where n o s w 2 SbT = w 2 X T;s ; j ; (T ; w) b for some b > 0 to be determined. We shall show that there exists a b = b(k ks; kf kL ?T;T H s R ) > 0 and a T > 0 such that u = (w) 2 SbT if w 2 SbT and 1
3
0 0
0 0
1 ([
];
(
))
: SbT ! SbT
is a contraction map. Consider the integral equation form of the IVP (2.20), Zt u(t) = W (t) + W (t ? )@x(a(v)w)(:; )d : 0
(2:21)
Applying (2.3) - (2.10) and (2.15) to (2.21) leads to Zt Zt s l;r (t; u) c k ks + ?t kf (:; )ksd + c ?t k@x(a(v)w)ksd ! ZT c k ks + kf (:; )ksd +
ct
=
1 2
?T (1 + t) (kvkX0T;s;0 )s;
0 0
15
(t; w)
(2.22)
for (l; r) 2 [0; s ? ] [0; s ? ). In particular, ! ZT s ; (t; u) c k ks + kf ksd + c (kvkX T;s; )t = (1 + t)s; (t; w): 3
3
4
4
1 2
?T
0 0
Choosing
0 0
00
b = 2 k ks +
ZT
and 0 < T < T such that
?T
kf ks d
!
(2:23)
cT (1 + T ) (kvkX T;s; ) = 21 ;
(2:24)
00
we obtain
s; (T ; u) b: Thus, is a map from SbT to SbT . For any w ; w 2 SbT , let
(2:25)
0 0
1
2
z = (w ) ? (w ): 1
Then and
z(t) =
Zt
2
W (t ? )@x (a(v)(w ? w )) d 1
0
2
Z T s; (T ; z) c k@x (a(v)(w ? w )) ks d ?T p c (kvkX T;s; ) T (1 + T )s; (T ; w ? w ) 1
0 0
2
0 0
00
1
2
It follows from (2.24) that
s; (T ; z) 21 s; (T ; w ? w ): 0 0
0 0
1
2
(2:26)
Consequently, by the contraction principle, there exists a unique solution u 2 SbT such that (u) = u i.e. Zt Zt u(t) = W (t) ? W (t ? ) (@x(a(v)u))( )d + W (t ? )f (:; )d (2:27) 0
0
16
for ?T < t < T : In addition, it follows from (2.22), (2.23) and (2.25) that ! ZT s (2:28) l;r (T ; u) c k ks + kf ksd ?T
for (l; r) 2 [0; s ? ] [0; s ? ). Finally, note that T determined by (2.24) only depends on (kvkX T;s; ) and, in particular, it does not depend on f and . Thus, a standard argument shows that T can be extended to T = T and (2.28) hold for T = T with another c depending only on kvkX T;s; and a. The proof is completed. 3
3
4
4
00
2
00
Remark 2.2 Theorem 2.1 is still true if a(v) in (2.18) is replaced by Z
1
a(v + (1 ? )v )d 1
0
2
with v ; v 2 X T;s ; . 1
2
0 0
3 Well-posedness In this section we consider the IVP of the generalized KdV equation 8 > < @tu + @x(a(u)) + @xu = 0; x; t 2 R (3:1) > : u(x; 0) = (x) 3
where a = a(x) is assumed to be a C 1 function from R to R and
a(0) = a0(0) = 0: The following result is due to Kato [17]
Proposition 3.1 Let s > 3=2 be given. Then (i) For each 2 H s (R), there exists a T > 0 depending only on k ks, and a unique solution u to (3.1) such that
u 2 C ([?T; T ]; H s(R)): 17
The map ! u is continuous from H s (R) to C ([?T; T ]; H s(R)). (ii). If, in addition, 2 H s0 (R) for some s0 > s, then (i) is true with same T and with s replaced by s0. (iii). There is a number a > 0 depending only on a, which is called the ceiling for the given a, such that if 2 H s (R) with s 2 and
k k a ;
(3:2)
1
then there is a unique solution u to (3.1) such that
u 2 C ([0; 1); H s(R)): Based on Kato's results, we shall prove the following thoerems by using Kenig, Ponce and Vega's argument in [22] with a few modi cation. Theorem 3.1 Let s > 3=4 be given. Then (i). For any 2 H s(R), there exists a T > 0 depending only on k ks, and a unique solution u 2 C ([?T; T ]; H s(R)) to the IVP (1.1) satisfying != ZT k@xu(:; t)k1dt < 1 1 4
4
?T
and moreover,
kukXl;rT;s (k ks) where (l; r) 2 [0; s ? ] [0; s ? ) and (:) is a continuous monotone 3
3
increasing function with (0) = 0. (ii). For any T 0 < T , there exists a neighborhood U of in H s (R) such that the map K : ! u(:; t) from U to Xl;rT;s is Lipschitz continuous. 4
4
Theorem 3.2 Let s 1 be given. Then Theorem 1.1 is true with T arbitrarily large provided that
k k a 1
where a is the ceiling of a de ned in Theorem 3.1.
18
Remark 3.1 According to Kato [17]
a = 1 if
lim sup jj? a () = 0
where
(3:3)
6
jj!1
2
a () = 2
Z
2
0
Note that (3.3) is true if
( ? )a0()d:
lim sup jj? a0() = 0: 4
jj!1
or with p 3 as juj ! 1.
a0(u) jujp
Remark 3.2 Theorem 3.2 follows from Theorem 3.1 and the global a priori
estimates for solutions of the IVP (3.1) due to Kato [17] by a standard argument.
We start to prove Theorem 3.1 by establishing the following a priori estimate for solutions of (3.1).
Proposition 3.2 Let s > 3=4, T > 0 and (l; r) 2 [0; s ? ] [0; s ? ) be given. If u 2 X T;s ; is a solution of (3.1), then there exists a T > 0 depending only on k ks such that sl;r (T ; u) < ck ks (3:4) 3
3
4
4
0
0 0
0
where c > 0 is a constant independent of u.
Proof: Consider the integral equation form of (3.1), u(t) = W (t) ?
Zt 0
W (t ? )@x(a(u))( )d : 19
(3:5)
Applying (2.3) - (2.10) to (3.1) yields
Zt s l;r(t; u) ck ks + c k@x(a(u))( )ksd ?t
(3:6)
for any t T . In particular, using (2.16), we have Zt s ; (t; u) ck ks + c k@x(a(u))( )ksd ?t ck ks + c (s; (t; u))t = (1 + t)s; (t; u): 0 0
1 2
0 0
0 0
(3.7)
Note that (s; (t; u)) is a continuous increasing function of t. There exists a t = T such that 0 0
0
c (s; (T ; u))T = (1 + T ) = 1=2
(3:8)
1 2
0
0 0
0
0
and it follows from (3.7) that s; (T ; u) 2ck ks ; 0 0
(3:9)
0
Thus the following inequality must hold
c (2ck ks)T = (1 + T ) 1=2; 1 2
0
0
which implies that T > M > 0 with M > 0 depending only on k ks. (3.4) follows from (3.6), (2.16) and (3.9). The proof is completed. 2 0
1
1
Proposition 3.3 Let s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ) be given. For any 2 H s (R), suppose 2 H 1 , 2 (0; 1), with lim = ! 0
3
3
4
4
in H s(R).
0
Then there exist T > 0 such that for any 2 (0; 1), (3.1) has a unique solution u 2 C ([?T; T ]; H 1(R)) with u(x; 0) = (x) satisfying
sl;r(T ; u) < ck ks for 0 < < 1 where c is the same constant as that in (3.4).
20
(3:10)
Proof: According to Theorem 3.1, u 2 C ([?T; T]; H 1(R)) where T only depends on k ks. Since k ks is uniformly bounded for 2 (0; 1), we may assume that
T > T for some T > 0 and therefore
1
for any 2 (0; 1)
1
u 2 X T: ;s: Thus it follows from Proposition 3.2 that sl;r (T ; u) ck ks where T > 0 is a constant independent of since k ks is bounded. The proof is completed. 2 Proof of Theorem 3.1 For 2 H s (R), choose 2 H 1(R) such that lim = in H s (R): ! 1
0 0
0
By Proposition 3.3, there exists a T > 0 and K > 0 such that for any 2 (0; 1) the IVP 8 > < @tu + @x(a(u)) + @xu = 0 > : u(x; 0) = (x) 3
has a unique solution u 2 C ([?T; T ]; H 1(R)) satisfying
sl;r(T ; u) < K: It suces to show that u is a Cauchy sequence in Xl;rT;s and then the limit u of u as ! 0 is the desired solution of the IVP (3.1) corresponding to the initial value . Let 0 < and w = u ? u0 : Then w solves 8 > < @tw + @x(A(u; u0 )w) + @xw = 0 > : w(x; 0) = ? 0 3
21
where
Z A(u; v) = a0 (u ? (1 ? )(v)) d: According to Theorem 2.1 and its remark, 1
0
sl;r(T ; w) ck ? 0 ks for some c > 0 independent of and therefore u is a Cauchy sequence in Xl;rT;s. Finally, if T 0 < T , (3.1) de nes a map k from a neighborhood U of in H s (R) to Xl;rT;s. For any 2 H s (R) with k ? ks for some > 0, let u and v be the solutions of the IVP (3.1) with u(x; 0) = (x) and v(x; 0) = (x), respectively. Then, similarly, we have 0
0
sl;r (T 0; u ? v) ck ? ks where c depends only on k ks . Therefore the map ! u is Lipschitz continuous. The proof is completed. 2
4 Di enrentiability Let X; Y be two Banach spaces. An n-linear map from X to Y is a map from the n-fold product space X n into Y such that
xk ! f (x ; :::; xn) 1
is linear for each xed (x ; x ; :::; xk? ; xk ; :::; xn), for k = 1; 2; :::; n. A homogeneous polynomial of degree n from X to Y is a map of the form x ! f (x; :::; x), for some n-linea map f . A map f : X ! Y is Frechet di erentiable at a point x 2 X if there exists a continuous linear map f 0(x ) : X ! Y so that 1
2
1
+1
0
0
kf (x) ? f (x ) ? f 0(x )(x ? x )kY = o(kx ? x kX ) as x ! x where f 0(x ) 2 L(X; Y ) is called the Frechet derivative of f at 0
0
0
0
0
0
x . f is said twice di erentiable at x if f is di erentiable at each point in 0
0
22
a neighborhood of x and x ! f 0(x) 2 L(X; Y ) is di erentiable at x ; and so on. A map f : U X ! Y , U open, is analytic in U if f is in nitely often di erentiable at each point of U and if, for each x 2 U , there exists = (x) > 0 so that whenever khkX , 1 X f (x + h) = k1! f k (x)[hk ]; k 0
0
( )
=0
the series converging in Y -norm uniformly in khkX . Consider the IVP for the generalized KdV equation 8 > < @tu + @x(a(u)) + @xu = 0; x; t 2 R > : u(x; 0) = (x) 3
(4:1)
where a 2 C 1(R; R) with a(0) = a0(0) = 0. According to Theorem 3.1, for any 2 H s (R), s > 3=4, there is a T > 0 and a neighborhood U of in H s (R) such that (4.1) de nes a nonlinear map K from U to Xl;rT;s u := K ( ) for any 2 U where u is the solution of (3.1). We shall show in this section that K is in nitely many times Frechet di erentiable from U to Xl;rT;s . If we suppose that the map K is n times Frechet di erentiable, then its n-th order derivative K n ( ) at 2 U is a symmetric n-linear map from H s (R) to Xl;rT;s and for any h ; :::; hn 2 H s (R), ( n ) n X @ n K ( )[h ; :::; hn] = @ :::@ K ( + k hk ) : (
)
1
(
)
1
n
1
k
;:::;
=1
0
0
As for the homogeneous polynomial K n ( )[hn] of degree n induced by K n ( ), it is given by ( n ) d n n K ( )[h ] = dn K ( + h) (
(
)
)
(
)
23
=0
for any h 2 H s(R). Denote by w n;:::;n = K n ( )[h ; :::; hn] and yn = K n ( )[hn]: (
)
(
[1
)
(
)
1
]
Then direct computation show that w n;:::;n solves 8 > < @tw + @x(a0(u)w ) + @xw = 0 > : w (x; 0) = h for n = 1 and 8 n n n > < @tw ;:::;n + @x(a0(u)w ;:::;n ) + @xw ;:::;n = ?@x(Hn ) > : w n (x; 0) = 0 (
)
[1
]
(1)
(1)
[1]
[1]
(1)
3
[1]
(4:2)
(1)
1
[1]
(
)
(
[1
(
]
[1
)
(
3
]
)
[1
]
(4:3)
)
;:::;n
[1
]
for n 2 with u = K ( ) and n j X X k X w i ;:::;ik w ik ;:::;ik :::w ikjj;:::;ijk (4:4) Hn = aj ! j j k ::: kj n 0 where P0 is the summation over all (i ; :::; ik ; :::ij ; :::; ijkj ) satisfying 1 im < im < ::: < imkm n for m = 1; 2; :::; j and [j k[m m fil g = f1; 2; :::; ng: ( )
( 1)
[
=2
1+
+
=
1 1
1
m
[
2 1
(
2 ] 2
[
)
1
]
1
1
1
( 2)
1 ] 1
1
1
2
l
=1 =1
As for yn, it solves
8 > < @ty + @x(a0(u)y ) + @xy = 0 > : y (x; 0) = h
(4:5)
8 > < @tyn + @x(a0(u)yn) + @xyn = ?@x(Mn ) > : yn(x; 0) = 0
(4:6)
1
3
1
1
1
for n = 1 and
3
24
for n 2 where
Mn =
n a j (u) X X n! y ::y : j ! k ::: kj n k !:::kj ! k kj j ( )
1
=2
1+
+
1
=
On the other hand, according to Theorem 2.1, given u 2 X T;s ; with T;s s s > 3=4, (4.2) de nes a linear map K (u) from H (R) to Xl;r with (l; r) 2 [0; s ? ] [0; s ? ), K (u)[hk ] = w k where w k is the solution of (4.2) with the initial value hk 2 H s (R), k = 1; 2:::. Inductively, (4.2)-(4.3) de nes an n-linear map K n (u) from the n-fold space (H s(R))n to Xl;rT;s for any n 2. 0 0
(1)
3
3
4
4
(1)
(1)
[ ]
(1)
[ ]
(
)
Proposition 4.1 Let s > 3=4, (l; r) 2 [0; s ? ] [0; s ? ), T > 0 and s u 2 X T;s ; be given. Then for any h ; :::; hn 2 H (R), (4.2)-(4.3) has a unique solution w n;::;n , which de nes an n-linear map K n (u) from the n3
3
4
4
1
0 0
(
)
[1
(
)
]
fold space (H s(R))n to Xl;rT;s . Moreover,
kK n (u)[h ; :::; hn]kXl;rT;s c(n; kukX T;s; ) (
)
1
00
n Y k
khk ks
(4:7)
=1
for any n 1 and h ; :::; hn 2 H s (R) where c(n; :) is a continuous monotone increasing function from R to R with c(n; 0) = 0. 1
+
+
Proof: It is easy to see that (4.7) is true with n = 1 by applying (2.19) to
(4.2). Similarly, for n = 2, applying (2.19) to (4.3) yields ZT kw ; kXl;rT;s (kukX T;s; ) k@xH ksd : (2)
[1 2]
00
?T
2
Then it follows from Lemma 2.8 that ! ZT ZT a ( u ) k@xH ks dt = ?T k@x 2 w w ksdt ?T c (kukX T;s; )kw kX T;s; kw kX T;s; c (kukX T;s; )c (1; kukX T;s; )kh kskh ks: (2)
2
(1)
(1)
[1]
[2]
(1)
2
2
[1]
00
2
(1)
[2]
00
00
2
2
00
25
00
1
2
Thus,
kw ; kXl;rT;s c (kukX T;s; ) (kukX T;s; )c (1; kukX T;s; )kh ks kh ks := c(2; kukX T;s; )kh kskh kX T;s; : (2)
2
2
[1 2]
00
00
2
00
1
2
1
00
2
00
Assume that (4.7) is true for 1 k N ? 1. Applying (2.19) to (4.3) with n = N yields ZT kw N;:::;N kXl;rT;s (kukX T;s; ) ?T k@xHN ksdt N cj X Yj kl X kw il ;:::;ilkl kX T;s; (kukX T;s; ) j ! j (kukX T;s; ) j k ::: kj N l N j X Yj X c(kl; kukX T;s; )kh il ;::;ilkl ks (kukX T;s; ) cj ! j (kukX T;s; ) j k ::: kj N l N Y := c(N; kukX T;s; ) khk ks: (
[1
)
]
00
(
00
00
=2
00
00
=2
00
k
[
1+
+
=
=1
1+
+
=
=1
)
1
]
00
00
[
1
]
=1
The proof is completed by induction. 2
Corollary 4.1 Let s > 3=4, (l; r) 2 [0; s ? ] [0; s ? ), T > 0 and u 2 X T;s ; be given. Then (4.5)-(4.6) de nes a homogeneous polynomial n K (u)[hn] of degree n from H s (R) to Xl;rT;s and kK n (u)[hn]kXl;rT;s c(n; kukX T;s; )khkns (4:8) 3
3
4
4
0 0
(
)
(
)
00
for any h 2 H s (R).
Now we may de ne formally the n-th Taylor polynomial Pn (h) of the map K at some 2 H s (R) as n X Pn (h) = K ( ) + k1! K k (u)[hk ] k n 1 X := u + k! yk k ( )
=1
=1
where u = K ( ). 26
Proposition 4.2 Let zn denote the n-th Taylor remainder of K at 2
H s (R), i.e.
zn = K (( + h) ? Pn(h):
Then it solves
8 > < @tz + @x(F (u; v)z ) + @xz = 0 > : z (x; 0) = h(x)
(4:9)
8 > < @tzn + @x(F (u; v)zn) + @xzn = ?@x(Dn ) > : zn (x; 0) = 0
(4:10)
0
1
3
0
0
0
for n = 0 and
3
1
for n 1 where and
Dn =
u = K ( );
nX
+1
m
Fm(u; v)
=2
v = K ( + h);
nX ?m +1
k
zk
=0
k1 ::: km?1 n?k +
with Fm(u:v) de ned by (1.14) and for m = 1; :::; n + 1.
X +
qk :::qkm? 1
1
(4:11)
=
ym qm = m !
Proof: Direct computation shows easily that (4.9) and (4.10) with n = 1 are true. Assume that (4.10) is true for n = N . Then, for n = N + 1, by de nition, zN = zN ? (N +1 1)! yN = zN ? qN +1
+1
+1
where qN solves 8 > < @tqN + @x(a0(u)qN ) + @xqN = ?@x(EN ) > : qN = 0 +1
+1
3
+1
+1
with 27
+1
+1
(4:12)
EN = +1
NX
+1
m
=2
X
a m (u) m! k (
)
::: km N
1+
=
+
qk :::qkm : 1
+1
Hence
@tzN + @xzN = ?@x F (u; v)zN ? a (u)qN ? @x(GN ) where N X?m NX X qk :::qkm? ? zk FM (u; v) GN = (1)
3
+1
1
+1
+1
+1
+1
m NX
k
=2
?
+1
m
=1
+1
+1
a m (u) m! k (
+
X
)
::: km N
1+
+
1
k1 :::: kM1 N ?k
=0
=
+
1
=
qk :::qkm : 1
+1
Note that
F (u; v)zN ? a0(u)qN 1
= F (u; v) (zN ? qN ) + (F (u; v) ? a0(u)) qN = F (u; v)zN + F (u; v)z qN ;
+1
1
+1
1
NX
+1
Fm(u; v)
N X?m +1
+1
1
2
0
+1
+1
X
zk
qk1 :::qkm?1 k1 ::: km?1 N ?k N X?m NX X (zk ? qk ) qk1 :::qkm?1 + = Fm(u; v) m k k1 ::: km?1 Nk N X?m NX X qk1 :::qkm?1 qk + Fm(u; v) m k1 ::: km?1 N ?k k N X?m NX X zk = Fm(u; v) qk1 :::qkm?1 + m k k1 ::: km?1 N ?k NX X qk1 :::qkm + Fm(u; v) m k1 ::: km N
m
k
=2
=0
+
+
=
+1
+1
+1
=2
=0
+
+
=
+1
+1
+1
=2
+
=0
+
=
+2
+1
=2
=1
+
+
=
+1
+1
=2
and
NX
+1
m
=2
Fm(u; v)
+
X k1 ::: km N +
+
=
+
=
qk :::qkm ? 1
+1
NX
+1
m
=2
+1
28
a m (u) m! k (
X
)
::: km N
1+
+
=
qk :::qkm 1
+1
m (u) ! a Fm(u; v) ? m! = m k NX X = Fm (u; v)z NX
+1
(
X
)
=2
::: km N
1+
+
=
+1
m NX
+1
0
=2
=
+2
m
k1 k2 ::: km N +
Fm(u; v)z
0
=3
+
+
X
=
k1 ::: km?1 N +
+
=
qk :::qkm 1
+1
qk :::qkm 1
+1
qk :::qkm? : 1
1
+1
Thus
F (u; v)zn ? a0(u)qN ? GN = F (u; v)zN + F (u; v)z qN + N X?m NX X qk :::qkm? + zk + Fm(u; v) 1
+1
+2
=2
k
m
+
NX
+2
m
= F (u; v)zN + +1
NX
+2
m
=2
1
+1
2
+ +
=
X
0
1
k1 :: km?1 N ?k
=1
+1
1
+1
qk1 :::qkm?1 k1 ::: km?1 N N X?m X Fm(u; v) zk qk1 :::qkm?1 k k1 ::: km?1 N ?k
Fm(u; v)z
=3
1
+1
+1
0
+
+
=
+1
+2
=0
+
+
=
+1
and we conclude that 8 > @tzN + @x(F (u; v)zN ) + @xzN = ?@x(DN ) < > : zN (x; 0) = 0 +1
1
3
+1
+1
+1
+1
which is (4.10) with n = N + 1. The proof is completed by induction. 2
Theorem 4.1 Let s > and (l; r) 2 [0; s? ][0; s? ) be given. Then, for any 2 H s (R), there exist a T > 0 and a neighborhood U of in H s (R) 3
3
3
4
4
4
such that the nonlinear map K de ned by the IVP (3.1) is in nitely many time Frechet di erentiable in U from H s (R) to Xl;rT;s . Its n-th derivative K n at 2 U is given by (
)
K n ( )[h ; :::; hn] = K n (u)[h ; :::; hn] (
)
(
1
)
1
for any h ; :::; hn 2 H s (R) where K n (u) is de ned by (4.2) and (4.3) and u = K ( ). 1
(
)
29
Proof: We only need to prove that for any 2 U , K ( + h) =
n 1 X K k ( )[hk ] + o(khkns ) k ! k ( )
=0
as h ! 0 in H s (R) uniformly for k ? ks khks by the Converse Taylor Theorem (see [8]). Let v = K ( + h); u = K ( ); y k = K k (u)[hk ] for 1 k n and z = v ? u; z = z ? y ; zn = zn? ? n1! yn: Then, by Proposition 4.2, 8 > < @tz + @x(F (u; v)z ) + @xz = 0 (4:13) > : z (x; 0) = h ( )
0
1
0
0
( )
0
1
1
3
0
0
0
and for n 1, 8 > < @tzn + @x(F (u; v)zn) + @xznF (u; v) = ?@xDn > : zn(x; 0) = 0 3
1
1
(4:14)
Choose > 0 such that 1
S ( ) = f 2 H s(R); k ? ks g U: 1
1
Obviously, K ( ) is bounded on S ( ) since K is a continuous map. By Corollary 4.1, 1
kqkkXl;rT;s c(k; kukX T;s; )khkks ;
k = 1; 2; :::; n
00
where u = K ( ) and c(k; kukX T;s; ) is uniformly bounded on S ( ). It suces to prove that kznkXl;rT;s n khkns (4:15) 1
00
+1
30
for n 0 where n is uniformly bounded for 2 S ( ). Indeed, it is easy to obtain (4.15) for n = 0 by using Theorem 2.1 to (4.13) and if we suppose (4.15) is true for n N , then applying Theorem 2.1 and Lemma 2.8 to (4.14) with n = N + 1, we have that 1
kzN kXl;rT;s +1
c
NX N X?m +2
m k k1 ::: km?1 N ?k NX N X?m X =2
c
+2
=0
+
+
=
m k k1 ::: km?1 N ?k NX N X?m X +1
m
=2
=0
+
+
=
k
k1 ::: km?1 N ?k
=0
:= N khkNs
+
+
=
1
cm ms;
0 0
+1
+2
k@x Fm(u; v)zkqk :::qkm? ks dt ?T
+1
+2
=2
c
ZT
X
+2
cm m m
+1
(T ; zk)
mY ?
1
j
=1
mY ?
1
j
1
s; (T ; qkj ) 0 0
=1
c(kj ; kukX T;s; )khkNs
+2
00
+2
+1
The proof is completed by induction. 2 Corollary 4.2 (Taylor's Formula) For any 2 U and h 2 H s (R) satisfying + h 2 U; for any 2 (0; 1); Z nX ? n? K ( + h) = j1! K j ( )[hj ] + (1 ?n!) K n ( + h)[hn ]d j with any n 1. Proof: See [7, Theorem 8.14.3]. 1
=0
1
1
( )
(
)
0
5 Analyticity In the previous section we proved the map K is in nitely many times Frechet di erentiable from H s (R) to Xl;rT;s. Naturally the further question is whether the map K is an analytic map, i.e., whether it has Taylor series expansion at any 2 H s (R): 1 n X K ( + h) = K n!( ) [hn]; (5:1) n (
=0
31
)
where the series coverges in Xl;rT;s uniformly for khks with some > 0 only depending on . For that purpose we need a better estimate of the n-th derivative yn = K n ( )[hn] of the map K at and the n-th Taylor remainder zn. Proposition 5.1 Let u 2 X T;s; be given and yn is the solution of (4.5)(4.6). Then there exists a sequence (n) given by (
)
0 0
(1) = 1; and
(5:2)
n X X (n) = j !j (k ) (k )::: (kj ) j k ::: kj n
(5:3)
kynkXl;rT;s cn n! (n)khkns
(5:4)
1
for n 2 such that
=2
1+
+
2
=
for any n 1 where c > 0 is a constant independt of n and h,
j = j (kukX T;s; ) 00
is a continuous monotone function depending only on a j , and j 0 if j 0. Proof: Denote by qn = yn=n!. It solves 8 > < @tq + @x(a0(u)q ) + @xq = 0 (5:5) > : q (x; 0) = h(x) for n = 1 and 8 Pn a j u P 0 > @ q + @ ( a ( u ) q ) + @ q = ? @ q :::q n x kj < tn x k ::: kj n k x n j j ( )
( )
1
3
1
1
1
( )
3
=2
> : qn(x; 0) = 0
( )
!
1+
+
=
1
(5:6)
for n 2. Applying (2.19) to (5.5) yields
kq kXl;rT;s ckhks: 1
32
Assume that
kqmkXl;rT;s c m? (m)khkms; 2
for 1 m N
1
Then applying (2.19) to (5.6) with n = N + 1, we obtain NX X ZT j k @x a (u)qk :::qkj ks dt kqN kXl;rT;s c j1! ?T j k ::: kj N NX X Yj c j !j kqkl kX T;s; j k ::: kj N l NX X Yj kl? kl c khks (kl ) c j !j j k ::: kj N l c N ? (N + 1)khkNs : Thus we have proved by induction that kynkXl;rT;s c n? n! (n)khkns +1
( )
+1
1
=2
1+
+
=
+1
+1
=2
1+
+
=
00
+1 =1
+1
2
=2
2(
+1)
1+
+
=
1
+1 =1
1
+1
2
1
which is (5.4) with a di erent c > 0. The proof is completed by induction.
2
Proposition 5.2 Let s > 3=4, T > 0 , (l; r) 2 [0; s ? ] [0; s ? ) and u; v 2 X T;s ; be given. If zn is the solution of (4.9)-(4.10) with yn being the 3
3
4
4
0 0
solution of (4.5)-(4.6), then there exists a sequence (n) given by
(0) =
(5:7)
1
(n) =
nX
+1
1
m
m
=2
nX ?m +1
k
(k)
=0
mY ?
X
1
k1 ::: km?1 n?k i +
+
=
for n 1 with (n) given by (5.2)-(5.3) such that
kznkXl;rT;s (n)khkns ;
(5:8)
=1
n0
+1
where
ci (i)
(5:9)
j = j (kukX T;s; ; kvkX T;s; ); 00
00
depending only on a j , is a continuous function of kukX T;s; and kvkX T;s; with j 0 if j 0, and c > 0 is a constant independent of n and h. ( )
00
( )
33
00
Proof: It follows from (4.9) by applying (2.19) that kz kXl;rT;s khks : Assume that (5.9) is true for all 0 n N ? 1. Then applying (2.19) and 0
1
(2.17) to (4.10) with n = N , we obtain that ! ZT NX N X?m mY ? X kzN kXl;rT;s k@x Fmzk qki ksdt +1
+1
1
m i k k1 ::: km1 N ?k ?T NX N X?m mY ? X mkzk kX0T;s;0 kqki ks m i k k1 ::: km?1 N ?k NX N X?m mY ? X m (k)khkNs ci (i) m i k k1 ::: km?1 N ?k N
(N )khks : 1
=2
+1
+
+
=1
=
+1
1
1
=2
=0
+1
=0
+
+
=1
=
+1
1
+1
1
=2
=0
+
+
=1
=
+1
The proof is completed by induction. 2 Now we consider the map K de ned by (3.1) from an open set U in s H (R) to Xl;rT;s where s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ). Proposition 5.3 Let 2 U . If there is a > 0 and c > 0 such that 3
3
4
4
1
S ( ) = f 2 H s (R)j k ? ks < g U 1
1
and
(n) cn; n0 (5:10) uniformly for 2 S ( ) where (n) is de ned by (5.7)-(5.8) with u = K ( ) and v = K ( ), then there exists a > 0 such that the series (5.1) coverges in Xl;rT;s uniformly for khks . Proof: Consider the n-th Taylor remainder n 1 X zn = K ( + h) ? j ! K j ( )[hj ]: j According to Proposition 5.2 and hypothesis (5.10), kznkXl;rT;s (n)khkns cnkhkns 1
( )
=0
+1
34
+1
where c > 0 is independent of n and h 2 H s (R) with khks < . Thus if we choose > 0 such that < 21c ; then kznkXl;rT;s ( 21 )n ; n 1 for any h 2 H s(R) with khks . The proof is completed. 2 In the following we shall show that if a(u) in (3.1) is a polynomial of degree N , i.e. N X (5:11) a(u) = bj uj 1
1
j
=2
with bj , j = 1; 2:::; N being real constants, then hypothesis (5.10) is satis ed for any 2 U and therefore the map K is an analytic map from U H s(R) to Xl;rT;s. First we prove a technical lemma. Lemma 5.1 Let N 1 be a given integer and n be a sequence given by N (1) = 1; (5:12) n X X N (n) = bjj! N (k )::: N (kj ); for 2 n N ? 1; (5:13) j k ::: kj n and N b X X N (k ):::: N (kj ); for n N . (5:14) N (n) = jj! j k ::: kj n where bj , 1 j N are given constants. Then there exists a constant c > 0 such that N (n) cn (5:15) for any n 1. Remark 5.1 In the case N = 2 and b = 1, the lemma is Proposition 3.4 in [47] where it is shown that n? (n) = 2 (2nn! ? 3)!! for any n 2: 1
=2
1+
+
=
1+
+
1
=2
=
2
1
2
35
Proof of Lemma 5.1: Note that that N (n), for any n 1, is uniquelly determined by (5.12) and (5.13) inductively. In particular, we may obtain
N (1); N (2); :::; N (N ? 1) explicitly by computation. Let PN (x) be a polynomial of degree N as follows !j N b NX ? X j k PN (x; y) = y ? j ! N (k)x + y ? 1
j k N ? N X bj X k X x N (k j j ! k j k1 ::: kj k =2
?
=1
1
1
=2
=
+
+
)::: N (kj ):
(5.16)
=
It is easy to check that
@ P (0; 0) = 1: @y N Thus, according to the implicit function theorem, PN (0; 0) = 0;
PN (x; y) = 0
(5:17)
has a unique solution
y = f (x)
with f (0) = 0
in the neighborhood of x = 0 such that for any jxj
PN (x; f (x)) = 0;
where > 0 is a constant. Besides, direct computation shows that f j (0) = 0; for j = 1; 2; :::; N ? 1: Moreover, y = f (x) is a real analytic function in a neighborhood of x = 0, i.e. f (x) has a Taylor series expansion at x = 0, 1 X y = f (x) = dj xj (5:18) ( )
j N =
36
which is uniformly convergent for jxj < with some > 0. To see this, let NX ? g = N (k)xk + y 1
k
and
z=
NX ?
1
k
=1
N b NX ? X X j n N (k)x + j ! xk N (k j k j k1 ::: kj k 1
1
=1
=2
=
+
+
)::: N (kj ) := h(x):
=
Then, equation (5.17) may be written as
g?
N b X j j g =z j j! =2
which obviously has an analytic solution g(z) in a neiborhood of z = 0 such P N that g(0) = 0. Thus y = f (x) = g(h(x)) ? k ? N (k)xk is an analytic function in a neighborhood of x = 0 sicne h(x) is a polynomial of x. Plugging (5.18) into (5.17) and denoting by 1
=1
for j = 1; 2; :::; N ? 1;
dj = N (j ); we have
y =
1 X
bj xj
j N !j 1 ? N b NX X X j k k N (k)x + dk x ? j j! k k N N N ? X X X ? bjj! xk N (k )::: N (kj ) j k j k1 ::: kj k !j X N b X 1 N b NX ? X X j j k k ? d x dk1 :::dkj kx j j! k k j ! k j k1 ::: kj k 1 N b X X X j dk1 xk1 dk2 xk2 :::dkj xkj ? j ! j k j k1 ::: kj k =
=
1
=2
=
=1
1
1
=2
=
+
+
=
1
=2
=
=
=2
=1
=
=2
+
+
=
37
=
+
+
=
N b NX ? X X j dk :::dkj xk j ! j k j k ::: kj k 0 1 N b X 1 X X j@ = dk :::dkj A xk j ! j k N k ::: kj k 1 0 1 X N b X X k @ j d = K :::dkj A x k N j j ! k ::: kj k
?
1
1
=2
=
1+
+
1+
+
=
1+
+
=
=
1
=2
=
1
=
=2
for any jxj < . Thus we have dj = N (j ); and N b X dk = jj! j
=2
j = 1; 2; :::; N ? 1
X k1 :::: kj k +
+
dk :::dkj 1
=
for any k N . That is to say, dj for j 1 also satisfy the induction relation (5.12)-(5.13). By uniqueness we have N (k) = dk ; for any k 1. On the other hand, dk , k 1 are coecients of Taylor series (5.18) and therefore we must have dn cn ; for all n 1 for some c > 0 independent of n. The proof is completed. 2 Theorem 5.1 Let s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ) be given and suppose that a(u) is (3.1) is a polynomial of degree N . Then for any 2 H s (R), there is a T > 0 and a neighborhood U of in H s (R) such that the IVP (3.1) de ne an analytic map K from U to Xl;rT;s , i.e., for any 2 U , there is a > 0 such that the Taylor series 1 X K ( + h) = K ( ) + n1! K n ( )[hn] n 3
3
4
4
(
)
=1
uniformly converges for khks in the space Xl;rT;s . Moreover, if denote by
u = K ( );
yn = K n ( )[hn]; n 1 (
38
)
then
8 > < @ty + @x(a0(u)y ) + @xy = 0 > : y (x; 0) = h 1
3
1
1
1
for n = 1 and
8 > < @tyn + @x(a0(u)yn) + @xyn = ?@xGN (n) > : yn(x; 0) = 0 3
for n 2 where
8 Pn > > < j GN (n) = > > : PN
=2
j
=2
a(j) u j
P
a(j) u j
P
( )
n k1 ::: kj n k1 :::kj yk1 :::ykj
!
( )
!
+
+
=
!
!
n k1 :::kj n k1 :::kj yk1 :::ykj
!
!
+
=
!
!
for 2 n N ? 1 for n N .
Proof: Since U is an open subset in H s (R), there is a > 0 such that if h 2 H s (R) with khks , then + h 2 U: 1
1
Denote by
v = K ( + h);
u = K( )
and
z = v ? u; zn = zn? ? n1! yn; for n 1: Then, by proposition 5.1 and 5.2, 0
1
kynkXl;rT;s n!cn (n)khkns and
kznkXl;rT;s (n)khkns : Note that a j (u) 0 for j N + 1 since a(u) is a polynomial of degree N +1
( )
by the assumption. Thus (n) is given by
(1) = 1; 39
(n) =
n X X j (k )::: (kj ) j j ! k :::: kj n 1
=2
for 2 n N ? 1 and
(n) =
1+
+
=
N X X j (k )::: (kj ) j ? j ! k ::: kj n 1
2
1+
for n N . Moreover,
+
=
(0) = ; 1
(n) =
n X 1
j
j
=2
n X?j +1
k
X
(k)
+
j
k
=2
=0
jY ?
1
k1 :::kj?1 n?k i +
=
cki (ki )
=1
=
X
+1
1
1
k1 :::kj?1 n?k i
=0
for 1 n N ? 1 and N n X?j X
(k)
(n) = j
jY ?
cki (ki ):
=1
According to Proposition 5.3, we need to show that there is a c > 0 such that
(n) cn for any n N with c independent of n and khks . To this end, we rst see from Lemma 5.1 that 1
(n) cn;
nN
1
with some c > 0 independent of n and h. Thus, for 0 j N , jY ? X X cki (ki ) (cc )n?k 1
1
k1 :::kj?1 n?k i +
=
=1
1
k1 ::: kj?1 n?k (n ? k ? 1)N (cc )n?k +
cn?k 2
40
+
=
1
for some c > 0 independent of n and N n X?j X
(n) c
(k) 2
X
+1
3
c
j
=2
k
=0
+1
3
j
=2
k
3
nX ?
1
k
+
cki (ki )
=1
=
cn?k (k) 2
=0
c (N ? 2)
1
k1 ::: kj?1 n?k i +
N n X?j X
jY ?
nX ?
1
k
cn?k (k) 2
=0
cn?k (k) 4
=0
for some c > 0 independent of n where 4
c = max f ; ::::; N g 3
1
2
1
Assume that
(k) ck 2k? ; for 1 k n ? 1: 1
4
Then,
(n)
nX ?
1
k
cn?k (k) 4
=0
nX ?
cn + cn?k ck 2k? k n n ? c2 1
1
4
4
=1
1
4
Thus we proved by induction that
(n) 2n? cn cn 1
4
for some c > 0 independent of n and h. The proof is completed. 2 Corollary 5.1 Assume that a(u) in (3.1) is a polynomial. Then for any T > 0 and s 1, the map K de ned by the IVP (3.1) is analytic from H s (R) to Xl;rT;s with (l; r) 2 [0; s ? ] [0; s ? ). 3
3
4
4
41
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[36] A. C. Scott, F. Y. Chu & D. W. McLaughin, The soliton: A new concept in applied sciences, Proc. IEEE, 61(1973), 1443 { 1483 [37] A. Sjoberg, On the Korteweg-de Vries equations: existence and uniqueness, J. Math. Anal. Appl., 29(1970), 569 { 579 [38] P. Sjolin, Regularity of solutions to the Schrodinger equations, Duke Math. J. , 55(1987), 699 { 715 [39] W. A. Strauss, Nonlinear scattering at lower energy, J. Funct. Anal., 41(1981), 110 { 133 [40] R. S. Strichartz, Multipliers in fractional Sobolev spaces, J. Math. Mech., 16(1967), 1031 { 1060 [41] |||-, Restriction of Fourier transform to quadratic surface and decay of solutions of wave equations, Duke Math. J., 44(1977), 705 { 714 [42] S. Tanaka, Korteweg-de Vries equation: construction of solution in terms of scattering data, Osaka J. Math., 11(1974), 49 { 59 [43] R. Temam, Sur un probleme non lineaire, J. Math. Pures Appl., 48(1969), 159 { 172 [44] Y. Tsutsumi, The Cauchy problem for the Korteweg-de Vries equation with measure as initial data, SIAM J. Math. Anal., 20(1989), 582 { 588 [45] L. Vega , The Schrodinger equation: pointwise convergence to the initial data, Proc. Amer. Math. Soc., 102(1988), 874 { 878 [46] L. Vega, Doctoral Thesis, University Autonoma de Madrid, Spain (1987) [47] B.-Y. Zhang, Taylor series expansion for solutions of the KdV equation with respect to their initial values, the IMA preprint, Series # 1015, Aug. 1992 45
y
Institute of Mathematics and Its Applications University of Minnesota Minneapolis, Minnesota 55455
Abstract
It is shown that the initial value problem (IVP) of the generalized KdV equation @t u + @x (a(u)) + @x u = 0;
u(x; 0) = (x) is well posed in the classical Sobolev space H s (R) with s > 3=4, which thus establishes a nonlinear map K from H s (R) to C ([?T; T ]; H s(R)). Then it is proved that (i) if a = a(x) is a C 1 function on R to R, then K is in nitely many times Frechet di erentiable; (ii) if a = a(x) is a polynomial, then K is analytic, i.e. for any 2 H s (R), K has a Taylor series expansion 1 1 X K ( + h) = K n ( )[hn ] n ! n 3
(
)
=0
Supported partially by 1992's Taft Summer Research Grant On leave of absence from: Department of Mathematical Sciences, University of Cincinnati, Ohio 45221 y
1
Each term yn = K n [hn ] in the series solves a linearized KdV equation. Thus any \small" perturbation K ( + h) of K ( ) can be obtained by solving a series of linear problems. The proof of these results relies on various smoothing properties of the associated linear KdV equation. (
)
1 Introduction This paper is mainly concerned with the initial value problem (IVP) for the generalized Korteweg-de Vries (KdV) equation: 8 > < @tu + @x(a(u)) + @xu = 0; t; x 2 R (1:1) > : u(x; 0) = (x) 3
in which a(x) is assumed to be a C 1 function on R to R , though a weaker di erentiablity suces for most results below. The KdV equation (i.e., a(u) = u =2 in (1.1) ) and its generalized form (1.1) have been studied by many authors [1, 2, 4 - 6, 10 - 17, 19 - 23, 27 37, 42 - 44, 47]. For a complete list of references, see, for example, [ 13, 17, 22, 30 ]. In particular, it is well-known that the IVP (1.1) is locally well posed in the classical Sobolev space H s (R) with s > 3=2 (cf. [17]) and is globally well posed with some restrictions on a(u) or the size of the initial data when s 2 (see [5], [14], [17], [19] and [28] for the well-posedness of the IVP (1.1) in other function spaces). In the case that a0(u) = uk in (1.1) with k being a positive integer, Kenig, Ponce and Vega [23] recently proved that the IVP (1.1) is locally well posed in the space H s (R) with 8 > s > 3=4 if k = 1 > > > > > s 1=4 if k = 2 < (1:2) > > if k = 3 s > > > > : s k?k if k 4 2
1
12
4
2
2
and is global well-posedness when 1 k 3 and s 1. Their proof is based on careful analysis of various smooth properties of the associated linear problem together with the contraction mapping principle. In this paper, we rst continue to study the well-posedness of the IVP (1.1) in the space H s (R). While only assuming that a(0) = a0(0) = 0, we shall prove that the IVP (1.1) is locally well posed in the space H s(R) with s > 3=4 by using Kenig, Ponce and Vega's argument [22], [23] with a few modi cation. In order to state our result precisely, we introduce the following Banach spaces as Kenig, Ponce and Vega did in [23]. Let s > 0 and T > 0 be given. For
w : R [?T; T ] ! R; de ne
(T; w) = sup kw(:; t)ks; 1
?T;T T s [
]
!= jD @xw(x; t)j dt ; (T; w) = sup x ?T != ZT (T; w; l) = ?T kJ l@xw(:; t)k1dt Z
2
1 2
2
1 4
4
3
with l 2 [0; s ? 3=4] where J s = (1 ? @x)s= , != Z ? r (T; w; r) = (1 + T ) sup jJ w(x; t)j dx 2
2
1 2
R ?T;T
4
[
2
]
with r 2 [0; s ? 3=4) and > 3=4 being a xed constant, and sl;r (T ; w) = max f (T; w); (T; w); (T; w; l); (T; w; r)g : (1:3) 1
2
3
Denote by
4
n o Xl;rT;s = w 2 C ([?T; T ]; H s(R)) j sl;r (T ; w) < 1 (1:4) with (l; r) 2 [0; s ? ] [0; s ? ). This is a Banach space equipped with the norm kwkXl;rT;s := sl;r(T ; w): 3
3
4
4
3
Clearly, Xl;rT;s is a subspace of C ([?T; T ]; H s(R)) with stronger topology. We shall prove that if a(0) = a0(0) = 0, then for any 2 H s (R), there exists a T > 0 depending only on k ks such that he IVP (1.1) has a unique solution u 2 Xl;rT;s where s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ). The global result is also obtained while the same restrictions are enforced on a(u) in (1.1) or the size of the initial value as those in Kato [17] but with s 1 instead of s 2. Thus the IVP (1.1) establishes a nonlinear map K from H s (R) to Xl;rT;s (or C ([?T; T ]; H s(R))). In the second part of this paper, we study di erentiability of the map K . It has been known for many years that the map K is continuous from H s (R) to C ([?T; T ]; H s(R)) [2], [17] and is proved recently by Kenig, Ponce and Vega [23] being Lipschitz continuous in the case that a0(u) = uk . In our early paper [47], we proved the map K corresponding to the classical KdV equation, i.e. a0(u) = u in (1.1), is in nitely many times Frechet di erentiable from H s(R) to Xl;rT;s and it has Taylor series expansion at any given 2 H s (R). That is to say, the map K is analytic from H s(R) to Xl;rT;s. In this paper we shall show that the nonlinear map K established by the IVP (1.1) is also in nitely many times Frechet di erentiable from H s (R) to Xl;rT;s. For any n 1, its n-th derivative K n ( ) at 2 H s (R), a n-linear map from the n-fold product space (H s(R))n into Xl;rT;s, can be constructed by solving a system of inhomogeneous linearized KdV equations. More precisely, for any n 1 and hk 2 H s(R) (k = 1; 2; :::; n), denote by w n;:::;n := K n ( )[h ; :::; hn]; (1:5) then it solves 8 > < @tw + @x(a0(u)w ) + @xw = 0 (1:6) > : w (x; 0) = h(x) (
(
)
[1
(
3
3
4
4
)
)
1
]
(1)
(1)
[1]
[1]
(1)
3
[1]
(1)
[1]
for n = 1 and 8 n n n > < @tw ;:::;n + @x(a0(u)w ;:::;n ) + @xw ;:::;n = ?@x(Hn ) > : w n;:::;n (x; 0) = 0 (
)
(
[1
(
[1
]
)
3
[1
]
)
]
4
(
[1
)
]
(1:7)
for n 2 where u = K ( ) and Hn is a polynomial of w ij ;:::;ij with 1 i ; :::; ij n and 1 j n ? 1 (see section 3 for the structure of Hn ). If we choose h = h = ::: = h and denote by yn = K n ( )[hn]; which is a homogeneous polynomial of degree n from H s (R) to Xl;rT;s, then 8 > @ty + @x(a0(u)y ) + @xy = 0 < (1:8) > : y (x; 0) = h(x) ( )
[ 1
]
1
1
2
(
)
1
3
1
1
1
for n = 1 and
8 > < @tyn + @x(a0(u)yn) + @xyn = ?@x fMn g > : yn(x; 0) = 0 3
for n 2 where
(1:9)
n j X X n! y :::y Mn = a j(! u) k kj j k ::: kj n k !:::kj ! and the summation Pk ::: kj n is over all possible (k ; :::; kj ) with 1 k ; :::kj n and k + :::kj = n. Thus we can de ne the n-th Taylor polynomial Pn of K at 2 H s (R): n n X Pn( )[h] := K ( ) + K k!( ) [hk ] n X (1.10) = u + ykk! k for any h 2 H s(R). Let zn = K ( + h) ? Pn ( )[h]: It is called the n-th Taylor remainder of K at and we shall see that zn solves 8 > < @tz + @x(F (u; v)z ) + @xz = 0 (1:11) > : z (x; 0) = h(x) ( )
1
=2
1+
1
1+
+
+
1
=
1
=
1
(
)
1
=1
0
1
0
0
5
3
0
for n = 0 and
8 > < @tzn + @x(F (u; v)zn) + @xzn = ?@x(Gn) > : zn (x; 0) = 0 3
1
(1:12)
for n 1 where
u = K ( ); v = K ( + h); nX ?m nX X zk Gn = Fm(u; v) qk :::qkm? +1
+1
m
with and
Fm(u; v) =
k
=2
k1 ::: km?1 n?k
=0
+
+
1
1
(1:13)
=
ym ; qm = m !
Z 0
1
m? :::: 1
1
Z
1
m?
Z
1
(
)
1
1
0
1 0m m Y Y a m @ j v + (1 ? j )uA dm :::d j j (1:14)
0
=1
=1
for m = 1; :::; n + 1. Letting n ! 1 in (1.10), we obtain a formal Taylor series of K at 2 H s(R). Assuming that a(u) is a polynomial, we shall prove that for any 2 H s (R), there exists a > 0 such that if h 2 H s(R) with khks , then 1 K n ( ) X n ]; K ( + h) = [ h (1:15) n! (
k
)
=0
the series conversing uniformly about h with khks in the space Xl;rT;s. In another word, the map K is analytic from H s (R) to Xl;rT;s . The paper is organized as follows. { In section 2, we rst list the estimates concerning the IVP @tu + @xu = f (x; t); u(x; 0) = u (x) (1:16) which are needed to establish the nonlinear results. Then we consider the IVP for the following linear equation 8 > < @tu + @x(a(v)u) + @xu = f (x; t); x; t 2 R : (1:17) > : u(x; 0) = (x) 3
0
3
6
Let s > 3=4, (l; r) 2 [0; s ? ] [0; s ? ) and T > 0 be given. We shall s s show that if v 2 X T;s ; , then for any 2 H (R) and f 2 L ([?T; T ]; H (R)), T;s the IVP (1.17) has a unique solution u 2 Xl;r and ! ZT (1:18) kukXl;rT;s kvkX T;s; k ks + ?T kf (:; t)ksdt 3
3
4
4
1
0 0
00
This would be a key estimate to obtain di erentiability of the map K . { In section 3, we show that the map K de ned by the IVP (1.1) is in nitely many times Frechet di erentiable from H s (R) to Xl;rT;s. { In section 4, assuming that a(u) in (1.1) is a polynomial, we show that the nonlinear map K is analytic from H s(R) to Xl;rT;s.
Notations:
{ The norm in L (R) will be denoted by k:k and the norm in H s(R) will be denoted by k:ks. The notation k:k1 is used to denote the norm in L1 (R). { Ds = (?@x)s= and J s = (1 ? @x)s= denote the Riesz and the Bessell potential of order s respectively. { [A; B ] = AB ? BA, where A; B are operators. Thus [J s; f ]g = J s(fg) ? fJ s g in which f is regarded as a multiplication operator. { H 1(R) := \s> H s(R) { For 1 p; q 1 and f : R [?T; T ] ! R; pq ! q Z T Z 1 p jf (x; t)j dx dt kf kLqT Lpx = 2
2
2
2
2
0
1
?T
and
?1
0Z Z ! qp 1 p T 1 j f (x; t)jqdt dxA : =@ ?1 ?T 1
kf kLpxLqT
7
2 Linear estimates 1 to denote the unitary group which de nes the solution We use fW (t)g?1 of the IVP associated to 8 > < @tv + @xv = 0; for x; t 2 R (2:1) > : v(x; 0) = v (x) +
3
0
where
v(t) = W (t)v = St v with St(:) de ned by the oscillatory integral Z 1 eix eit d: St(x) = c 0
0
+
3
?1
Then the solution of the inhomogeneous equation 8 > < @tv + @x v = f (x; t); x; t 2 R > : v(x; 0) = 0 3
is expressed as
v(t) =
Zt
Lemma 2.1 For any s 0,
0
(2:2)
W (t ? )f (:; )d :
= Z1 s sup jD @xW (t)v j dt ckv ks ; x ?1
(2:3)
= W (t)v k1 dt ckv ks:
(2:4)
1 2
2
0
and
Z 1 ?1
kDs =
+1 4
1 4
4
0
In addition, if s > 3=4, then
Z
1
+
sup
?1 ?T;T [
]
jJ lW (t)v
0
j (x)dx 2
0
0
!=
1 2
c(1 + T )kv ks 0
where l 2 [0; s ? 3=4) and is a xed constant larger than 3=4.
8
(2:5)
Proof: see Kenig, Ponce and Vega ([22], Lemma 2.1, Theorem 2.4 and
Corollary 2.9). Remark 2.1 (2.3) is a stronger version of local smoothing e ect of Kato type and (2.4) is the global smoothing e ect of Strichartz type [38] present in solutions of (2.1). The estimate (2.5), which gives a bound for the associated maximal function sup ?T;T jW (t):j, is due to Vega [46]. [
]
Lemma 2.2 For any s 0 and T > 0, kW (t)v ks = kv ks 0
and
sup k
?T;T
[
]
Zt 0
(2:6)
0
W (t ? )f (:; )d ks
ZT ?T
kf (:; )ksd :
(2:7)
Proof: (2.6) and (2.7) follow easily from Kato ([17], Lemma 3.1). Lemma 2.3 For any s 0 and T > 0, Zt
kDxs @x W (t ? )f (:; )d kL1x LT ckf kL
?T;T H s R
(2:8)
kf (:; )ksd :
(2:9)
kf (:; )ksd
(2:10)
1 ([
2
0
and
kDs
+
1 4
Zt
If s > 3=4, then
W (t ? )f (:; )d kL1x LT c 4
0
Zt
kJ l W (t ? )f (:; )d kLxL1T c(1 + T ) 2
0
ZT ?T
ZT ?T
];
(
))
where l 2 [0; s ? 3=4) and is a xed constant larger than 3=4.
Proof: It follows from Lemma 2.1 by using Minkowski's integral inequality (see [47]).
Lemma 2.4 Let s > 1=2 and T > 0 be given. Then there is a constant c > 0 such that ZT
?T
ku@xvksdt cT = (1 + T )kukX T;s; kvkX T;s; 1 2
00
9
00
(2:11)
and
ZT ?T
k@x(uv)ksdt cT = (1 + T )kukX T;s; kvkX T;s; :
(2:12)
1 2
00
00
for any u; v 2 X T;s ; . 0 0
Proof: see Lemma 2.4 in [47]. Lemma 2.5 Let s > 0 be given. Then there is a constant c > 0 such that for any yk 2 H s (R), k = 1; 2; :::; m, k
m Y k
1 0m m Y X kyk k1A yk ks cm @ kyj ks j
=1
and if s > 1=2,
k
m Y k
(2:13)
k ; k6 j
=1
=1
yk ks cm
=1
m Y k
=
kyk ks
(2:14)
=1
for any m 2 where c in (2.14) may be di erent from c in (2.13).
Proof: According to Kato and Ponce ([18], Lemma X4 ), ky y ks c fky ks ky k1 + ky ksky k1g 1
2
1
2
2
1
which is (2.13) with m = 2. Suppose that (2.13) is true for m = N . Then for m = N + 1, NY N Y k yk ks = kyN yk ks k k ( ) N N Y Y c kyN ksk yk k1 + kyN k1k yk ks +1
+1
=1
=1
+1
ckyN ks +1
cN
NX
+1
+1
j
=1
N Y k
k
+1
=1
kyk k1 + cN
=1
kyj ks
NY
+1
k ; k6 j =1
+1
N X j
=1
k
=1
kyj ks
NY
+1
k ; k6 j =1
kyk k1
=
kyk k1
=
Thus (2.13) is proved by induction. As for (2.14), it follows directly from (2.13) since kyk k1 kyk ks if s > 1=2. The proof is completed. 2 10
Lemma 2.6 Let b 2 C 1(R; R) with b(0) = 0. Then kb(u)ks ~b(kuks); s > 1=2 where ~b(:) is a monotone increasing function depending only on b.
Proof: see Kato ([17], Lemma A.3). Lemma 2.7 Let s > 1=2 and T > 0 be given and assume a 2 C 1(R; R) with a(0) = 0. Then there is a constant c > 0 such that for any u 2 X T;s ; T;s and y 2 X ; , 0 0
0 0
ZT
k@x(a(u)y)ksdt c kukX T;s; kykX T;s; ?T
(2:15)
00
00
where (:) is a continuous monotone increasing function only depending on a.
Proof: First of all, k@x(a(u)y)ks ka0(u)y@xuks + ka(u)@xyks and it is easy to see by using Lemma 2.5 and Lemma 2.6 that
ka0(u)y@xuks k(a0(u) ? a0(0))y@xuks + ja0(0)jky@xuks c fka0(u) ? a0(0)ks ky@xuk1 + ka0(u) ? a0(0)k1ky@xuksg + +ja0(0)jky@xuks c fka0(u) ? a0(0)ks + ja0(0)jg ky@xuks c (kuks)ky@xuks where (:) : R ! R is a continuous monotone increasing function only 1
1
+
+
depending on a. Using Lemma 2.4 yields ZT ZT k a0(u)y@xuksdt c sup (kuks) ky@xuksdt ?T ?T ?T;T = cT (1 + T ) kukX T;s; kukX T;s; kykX T;s; : [
]
1 2
1
11
00
00
00
In addition, applying Lemma 2.10 in [22], we obtain
ka(u)@xyks = kJ s(a(u)@xy)k = ka(u)Ds@xy + a(u)(J s ? Ds )@xy + [J s; a(u)]@xyk ka(u)Ds@xyk + ka(u)k1kyks + c fk@xyk1ka(u)ks+ + ka0(u)@xuk1kyksg : Note that ZT ZT ka(u)k1kyksdt sup ka(u)ks kyksdt ?T
?T;T
[
]
?T (kuks)
cT sup sup kyks (by Lemma 2.6) ?T;T ?T;T cT = (1 + T ) (kukX T;s; )kykX T;s; 2
[
]
[
]
1 2
2
00
00
where (:) : R ! R is a continuous monotone increasing function depending only on a; ZT ZT ka(u)ksk@xyk1dt sup ka(u)ks ?T k@xyk1dt ?T ?T;T ZT sup (kuks) ?T k@xyk1dt ?T;T != ZT = k@xyk1dt (kukX T;s; )(2T ) 2
+
+
[
]
[
]
2
1 4
3 4
2
cT
00
=
1 2
(1 + T )
2
?T
4
(kukX T;s; )kykX T;s; ; 00
00
ZT
ZT
ka0(u)@xuk1kyksdt sup (ka0(u)k1kyks) ?T k@xuk1dt ?T ?T;T cT = (1 + T ) (kukX T;s; )kukX T;s; kykX T;s; ; [
]
1 2
3
where
(r) = sup ja0()j; 3
and
ZT
ka(u)Ds @xykdt T ?T
00
=
1 2
jjr
ZTZ ?T R
ja(u)Ds@xuj dxdt
12
2
00
00
Z ZT
s @ uj dxdt sup k a(u) k j u j j D x u 1 ?T
R
2
2
?T;T
!= != ZT s sup ju(x; t)j dx sup jD @xyj dt x ?T [
Z
]
1 2
1 2
T (kukX T;s; ) R ?T;T = cT (1 + T ) (kukX T;s; )kukX T;s; kykX T;s; =
1 2
4
00
2
[
2
]
1 2
4
where
00
00
00
(r) = sup a() : jjr 4
Therefore,
ZT ?T
k@x(a(u)y)ksdt cT = (1 + T ) (kukX T;s; )kykX T;s; 1 2
00
00
for some constant c > 0 where
(r) = max fr (r); (r); r (r); r (r)g : 1
2
3
4
The proof is completed. 2 >From the proof of the above lemma we may draw the following corollary. Corollary 2.1 Let s > 1=2 and T > 0 be given and assume a 2 C 1(R; R) with a0(0) = 0. Then for any u 2 X T;s ; , ZT (2:16) k@x(a(u))ksdt c kukX T;s; T = (1 + T )kukX T;s; 0 0
1 2
?T
00
00
where (:) is a continuous monotone increasing function only depending on a.
Lemma 2.8 Let s > 1=2, T > 0 and a 2 C 1(R; R) be given. Then there T;s exists a constant c > 0 such that for any u 2 X T;s ; and yk 2 X ; with k = 1; 2; :::; m and m 2, 0 0
ZT
k@x(a(u) ?T
m Y
k
=1
yk )ksdt cm (kukX T;s; ) 00
m Y
k
=1
0 0
kyk kX T;s; 00
(2:17)
where (:) is a continuous monotone increasing function only depending on a.
13
Proof: By applying (2.14) we have k@x(a(u)
m Y
yk )ks ka0(u)@xu
m Y
yk ks +
m X
k
m Y
a(u)yj @xyk ks j ; j6 k m Y cm (ka0(u) ? a0(0)ks + ja0(0)j) ky @xuks kyk ks + k 0m m X Y +cm (ka(u) ? a(0)ks + ja(0)j) @ ky @xyk ks kyj ks+ k j ; j6 k 1 m Y + ky @xy ks kyj ksA : j k
=1
k
=1
k
=1
=1
=
+1
1
=2
+1
1
=2
2
=2
=
1
=3
Thus, using Lemma 2.6 and Lemma 2,4, we have ZT m Y k @x(a(u) yk )ks dt cm sup (ka0(u) ? a0(0)ks + ja0(0)j) ?T ?T;T k Z m T Y sup kyk ks ?T ky @xuksdt + cm sup (ka(u) ? a(0)ks + ja(0)j) ?T;T k0 ?T;T 1 ZT m m m X Y Y @ sup kyj ks ky @xyk ks dt + sup kyj ks sup ky @xy ksA +1
1
[
=1
]
+1
1
=2 [
k
1
]
[
?T
?T;T j ; j 6 k
=2 [
]
=2
=
1
cm (kukX T;s; )T = (1 + T )kukX T;s; 1 2
+1
1
1
00
m Y
j
?T;T
=3 [
]
?T;T
[
2
1
]
kyk kX0T;s;0 + k m = (1 + T ) Y ky k T;s k X0;0 k 00
+mcm (kukX T;s; )T m Y m c (kukX T;s; ) kyk kX T;s;
=1
1 2
+1
1
]
1
00
00
k
=1
=1
00
for some c > 0. The proof is completd. 2 To end this section, we consider the following linear problem: 8 > < @tu + @x(a(v)u) + @xu = f (x; t); x; t 2 R > : u(x; 0) = (x) 3
where a(:) 2 C 1(R; R) with a(0) = 0. 14
(2:18)
Theorem 2.1 Let s > 3=4, (l; r) 2 [0; s? ][0; s? ), T > 0 and v 2 X T;s; be given. Then for any f 2 L ([?T; T ]; H s(R)) and 2 H s (R), there exists a unique solution u 2 Xl;rT;s to (2.18) such that 3
3
4
4
0 0
1
kukXl;rT;s (kvkX T;s; ) k ks +
ZT
00
?T
! kf (:; t)ksdt
(2:19)
where is a continuous monotone increasing function only depending on a. Proof: We use the contraction principle argument that Kenig, Ponce and Vega used in [23]. For any given 2 H s (R) and f 2 L ([?T; T ]; H s(R)), denote by u = (w), the solution of the following IVP 8 > < @tu + @xu = f ? @x(a(v)w) (2:20) > : u(x; 0) = (x) where n o s w 2 SbT = w 2 X T;s ; j ; (T ; w) b for some b > 0 to be determined. We shall show that there exists a b = b(k ks; kf kL ?T;T H s R ) > 0 and a T > 0 such that u = (w) 2 SbT if w 2 SbT and 1
3
0 0
0 0
1 ([
];
(
))
: SbT ! SbT
is a contraction map. Consider the integral equation form of the IVP (2.20), Zt u(t) = W (t) + W (t ? )@x(a(v)w)(:; )d : 0
(2:21)
Applying (2.3) - (2.10) and (2.15) to (2.21) leads to Zt Zt s l;r (t; u) c k ks + ?t kf (:; )ksd + c ?t k@x(a(v)w)ksd ! ZT c k ks + kf (:; )ksd +
ct
=
1 2
?T (1 + t) (kvkX0T;s;0 )s;
0 0
15
(t; w)
(2.22)
for (l; r) 2 [0; s ? ] [0; s ? ). In particular, ! ZT s ; (t; u) c k ks + kf ksd + c (kvkX T;s; )t = (1 + t)s; (t; w): 3
3
4
4
1 2
?T
0 0
Choosing
0 0
00
b = 2 k ks +
ZT
and 0 < T < T such that
?T
kf ks d
!
(2:23)
cT (1 + T ) (kvkX T;s; ) = 21 ;
(2:24)
00
we obtain
s; (T ; u) b: Thus, is a map from SbT to SbT . For any w ; w 2 SbT , let
(2:25)
0 0
1
2
z = (w ) ? (w ): 1
Then and
z(t) =
Zt
2
W (t ? )@x (a(v)(w ? w )) d 1
0
2
Z T s; (T ; z) c k@x (a(v)(w ? w )) ks d ?T p c (kvkX T;s; ) T (1 + T )s; (T ; w ? w ) 1
0 0
2
0 0
00
1
2
It follows from (2.24) that
s; (T ; z) 21 s; (T ; w ? w ): 0 0
0 0
1
2
(2:26)
Consequently, by the contraction principle, there exists a unique solution u 2 SbT such that (u) = u i.e. Zt Zt u(t) = W (t) ? W (t ? ) (@x(a(v)u))( )d + W (t ? )f (:; )d (2:27) 0
0
16
for ?T < t < T : In addition, it follows from (2.22), (2.23) and (2.25) that ! ZT s (2:28) l;r (T ; u) c k ks + kf ksd ?T
for (l; r) 2 [0; s ? ] [0; s ? ). Finally, note that T determined by (2.24) only depends on (kvkX T;s; ) and, in particular, it does not depend on f and . Thus, a standard argument shows that T can be extended to T = T and (2.28) hold for T = T with another c depending only on kvkX T;s; and a. The proof is completed. 3
3
4
4
00
2
00
Remark 2.2 Theorem 2.1 is still true if a(v) in (2.18) is replaced by Z
1
a(v + (1 ? )v )d 1
0
2
with v ; v 2 X T;s ; . 1
2
0 0
3 Well-posedness In this section we consider the IVP of the generalized KdV equation 8 > < @tu + @x(a(u)) + @xu = 0; x; t 2 R (3:1) > : u(x; 0) = (x) 3
where a = a(x) is assumed to be a C 1 function from R to R and
a(0) = a0(0) = 0: The following result is due to Kato [17]
Proposition 3.1 Let s > 3=2 be given. Then (i) For each 2 H s (R), there exists a T > 0 depending only on k ks, and a unique solution u to (3.1) such that
u 2 C ([?T; T ]; H s(R)): 17
The map ! u is continuous from H s (R) to C ([?T; T ]; H s(R)). (ii). If, in addition, 2 H s0 (R) for some s0 > s, then (i) is true with same T and with s replaced by s0. (iii). There is a number a > 0 depending only on a, which is called the ceiling for the given a, such that if 2 H s (R) with s 2 and
k k a ;
(3:2)
1
then there is a unique solution u to (3.1) such that
u 2 C ([0; 1); H s(R)): Based on Kato's results, we shall prove the following thoerems by using Kenig, Ponce and Vega's argument in [22] with a few modi cation. Theorem 3.1 Let s > 3=4 be given. Then (i). For any 2 H s(R), there exists a T > 0 depending only on k ks, and a unique solution u 2 C ([?T; T ]; H s(R)) to the IVP (1.1) satisfying != ZT k@xu(:; t)k1dt < 1 1 4
4
?T
and moreover,
kukXl;rT;s (k ks) where (l; r) 2 [0; s ? ] [0; s ? ) and (:) is a continuous monotone 3
3
increasing function with (0) = 0. (ii). For any T 0 < T , there exists a neighborhood U of in H s (R) such that the map K : ! u(:; t) from U to Xl;rT;s is Lipschitz continuous. 4
4
Theorem 3.2 Let s 1 be given. Then Theorem 1.1 is true with T arbitrarily large provided that
k k a 1
where a is the ceiling of a de ned in Theorem 3.1.
18
Remark 3.1 According to Kato [17]
a = 1 if
lim sup jj? a () = 0
where
(3:3)
6
jj!1
2
a () = 2
Z
2
0
Note that (3.3) is true if
( ? )a0()d:
lim sup jj? a0() = 0: 4
jj!1
or with p 3 as juj ! 1.
a0(u) jujp
Remark 3.2 Theorem 3.2 follows from Theorem 3.1 and the global a priori
estimates for solutions of the IVP (3.1) due to Kato [17] by a standard argument.
We start to prove Theorem 3.1 by establishing the following a priori estimate for solutions of (3.1).
Proposition 3.2 Let s > 3=4, T > 0 and (l; r) 2 [0; s ? ] [0; s ? ) be given. If u 2 X T;s ; is a solution of (3.1), then there exists a T > 0 depending only on k ks such that sl;r (T ; u) < ck ks (3:4) 3
3
4
4
0
0 0
0
where c > 0 is a constant independent of u.
Proof: Consider the integral equation form of (3.1), u(t) = W (t) ?
Zt 0
W (t ? )@x(a(u))( )d : 19
(3:5)
Applying (2.3) - (2.10) to (3.1) yields
Zt s l;r(t; u) ck ks + c k@x(a(u))( )ksd ?t
(3:6)
for any t T . In particular, using (2.16), we have Zt s ; (t; u) ck ks + c k@x(a(u))( )ksd ?t ck ks + c (s; (t; u))t = (1 + t)s; (t; u): 0 0
1 2
0 0
0 0
(3.7)
Note that (s; (t; u)) is a continuous increasing function of t. There exists a t = T such that 0 0
0
c (s; (T ; u))T = (1 + T ) = 1=2
(3:8)
1 2
0
0 0
0
0
and it follows from (3.7) that s; (T ; u) 2ck ks ; 0 0
(3:9)
0
Thus the following inequality must hold
c (2ck ks)T = (1 + T ) 1=2; 1 2
0
0
which implies that T > M > 0 with M > 0 depending only on k ks. (3.4) follows from (3.6), (2.16) and (3.9). The proof is completed. 2 0
1
1
Proposition 3.3 Let s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ) be given. For any 2 H s (R), suppose 2 H 1 , 2 (0; 1), with lim = ! 0
3
3
4
4
in H s(R).
0
Then there exist T > 0 such that for any 2 (0; 1), (3.1) has a unique solution u 2 C ([?T; T ]; H 1(R)) with u(x; 0) = (x) satisfying
sl;r(T ; u) < ck ks for 0 < < 1 where c is the same constant as that in (3.4).
20
(3:10)
Proof: According to Theorem 3.1, u 2 C ([?T; T]; H 1(R)) where T only depends on k ks. Since k ks is uniformly bounded for 2 (0; 1), we may assume that
T > T for some T > 0 and therefore
1
for any 2 (0; 1)
1
u 2 X T: ;s: Thus it follows from Proposition 3.2 that sl;r (T ; u) ck ks where T > 0 is a constant independent of since k ks is bounded. The proof is completed. 2 Proof of Theorem 3.1 For 2 H s (R), choose 2 H 1(R) such that lim = in H s (R): ! 1
0 0
0
By Proposition 3.3, there exists a T > 0 and K > 0 such that for any 2 (0; 1) the IVP 8 > < @tu + @x(a(u)) + @xu = 0 > : u(x; 0) = (x) 3
has a unique solution u 2 C ([?T; T ]; H 1(R)) satisfying
sl;r(T ; u) < K: It suces to show that u is a Cauchy sequence in Xl;rT;s and then the limit u of u as ! 0 is the desired solution of the IVP (3.1) corresponding to the initial value . Let 0 < and w = u ? u0 : Then w solves 8 > < @tw + @x(A(u; u0 )w) + @xw = 0 > : w(x; 0) = ? 0 3
21
where
Z A(u; v) = a0 (u ? (1 ? )(v)) d: According to Theorem 2.1 and its remark, 1
0
sl;r(T ; w) ck ? 0 ks for some c > 0 independent of and therefore u is a Cauchy sequence in Xl;rT;s. Finally, if T 0 < T , (3.1) de nes a map k from a neighborhood U of in H s (R) to Xl;rT;s. For any 2 H s (R) with k ? ks for some > 0, let u and v be the solutions of the IVP (3.1) with u(x; 0) = (x) and v(x; 0) = (x), respectively. Then, similarly, we have 0
0
sl;r (T 0; u ? v) ck ? ks where c depends only on k ks . Therefore the map ! u is Lipschitz continuous. The proof is completed. 2
4 Di enrentiability Let X; Y be two Banach spaces. An n-linear map from X to Y is a map from the n-fold product space X n into Y such that
xk ! f (x ; :::; xn) 1
is linear for each xed (x ; x ; :::; xk? ; xk ; :::; xn), for k = 1; 2; :::; n. A homogeneous polynomial of degree n from X to Y is a map of the form x ! f (x; :::; x), for some n-linea map f . A map f : X ! Y is Frechet di erentiable at a point x 2 X if there exists a continuous linear map f 0(x ) : X ! Y so that 1
2
1
+1
0
0
kf (x) ? f (x ) ? f 0(x )(x ? x )kY = o(kx ? x kX ) as x ! x where f 0(x ) 2 L(X; Y ) is called the Frechet derivative of f at 0
0
0
0
0
0
x . f is said twice di erentiable at x if f is di erentiable at each point in 0
0
22
a neighborhood of x and x ! f 0(x) 2 L(X; Y ) is di erentiable at x ; and so on. A map f : U X ! Y , U open, is analytic in U if f is in nitely often di erentiable at each point of U and if, for each x 2 U , there exists = (x) > 0 so that whenever khkX , 1 X f (x + h) = k1! f k (x)[hk ]; k 0
0
( )
=0
the series converging in Y -norm uniformly in khkX . Consider the IVP for the generalized KdV equation 8 > < @tu + @x(a(u)) + @xu = 0; x; t 2 R > : u(x; 0) = (x) 3
(4:1)
where a 2 C 1(R; R) with a(0) = a0(0) = 0. According to Theorem 3.1, for any 2 H s (R), s > 3=4, there is a T > 0 and a neighborhood U of in H s (R) such that (4.1) de nes a nonlinear map K from U to Xl;rT;s u := K ( ) for any 2 U where u is the solution of (3.1). We shall show in this section that K is in nitely many times Frechet di erentiable from U to Xl;rT;s . If we suppose that the map K is n times Frechet di erentiable, then its n-th order derivative K n ( ) at 2 U is a symmetric n-linear map from H s (R) to Xl;rT;s and for any h ; :::; hn 2 H s (R), ( n ) n X @ n K ( )[h ; :::; hn] = @ :::@ K ( + k hk ) : (
)
1
(
)
1
n
1
k
;:::;
=1
0
0
As for the homogeneous polynomial K n ( )[hn] of degree n induced by K n ( ), it is given by ( n ) d n n K ( )[h ] = dn K ( + h) (
(
)
)
(
)
23
=0
for any h 2 H s(R). Denote by w n;:::;n = K n ( )[h ; :::; hn] and yn = K n ( )[hn]: (
)
(
[1
)
(
)
1
]
Then direct computation show that w n;:::;n solves 8 > < @tw + @x(a0(u)w ) + @xw = 0 > : w (x; 0) = h for n = 1 and 8 n n n > < @tw ;:::;n + @x(a0(u)w ;:::;n ) + @xw ;:::;n = ?@x(Hn ) > : w n (x; 0) = 0 (
)
[1
]
(1)
(1)
[1]
[1]
(1)
3
[1]
(4:2)
(1)
1
[1]
(
)
(
[1
(
]
[1
)
(
3
]
)
[1
]
(4:3)
)
;:::;n
[1
]
for n 2 with u = K ( ) and n j X X k X w i ;:::;ik w ik ;:::;ik :::w ikjj;:::;ijk (4:4) Hn = aj ! j j k ::: kj n 0 where P0 is the summation over all (i ; :::; ik ; :::ij ; :::; ijkj ) satisfying 1 im < im < ::: < imkm n for m = 1; 2; :::; j and [j k[m m fil g = f1; 2; :::; ng: ( )
( 1)
[
=2
1+
+
=
1 1
1
m
[
2 1
(
2 ] 2
[
)
1
]
1
1
1
( 2)
1 ] 1
1
1
2
l
=1 =1
As for yn, it solves
8 > < @ty + @x(a0(u)y ) + @xy = 0 > : y (x; 0) = h
(4:5)
8 > < @tyn + @x(a0(u)yn) + @xyn = ?@x(Mn ) > : yn(x; 0) = 0
(4:6)
1
3
1
1
1
for n = 1 and
3
24
for n 2 where
Mn =
n a j (u) X X n! y ::y : j ! k ::: kj n k !:::kj ! k kj j ( )
1
=2
1+
+
1
=
On the other hand, according to Theorem 2.1, given u 2 X T;s ; with T;s s s > 3=4, (4.2) de nes a linear map K (u) from H (R) to Xl;r with (l; r) 2 [0; s ? ] [0; s ? ), K (u)[hk ] = w k where w k is the solution of (4.2) with the initial value hk 2 H s (R), k = 1; 2:::. Inductively, (4.2)-(4.3) de nes an n-linear map K n (u) from the n-fold space (H s(R))n to Xl;rT;s for any n 2. 0 0
(1)
3
3
4
4
(1)
(1)
[ ]
(1)
[ ]
(
)
Proposition 4.1 Let s > 3=4, (l; r) 2 [0; s ? ] [0; s ? ), T > 0 and s u 2 X T;s ; be given. Then for any h ; :::; hn 2 H (R), (4.2)-(4.3) has a unique solution w n;::;n , which de nes an n-linear map K n (u) from the n3
3
4
4
1
0 0
(
)
[1
(
)
]
fold space (H s(R))n to Xl;rT;s . Moreover,
kK n (u)[h ; :::; hn]kXl;rT;s c(n; kukX T;s; ) (
)
1
00
n Y k
khk ks
(4:7)
=1
for any n 1 and h ; :::; hn 2 H s (R) where c(n; :) is a continuous monotone increasing function from R to R with c(n; 0) = 0. 1
+
+
Proof: It is easy to see that (4.7) is true with n = 1 by applying (2.19) to
(4.2). Similarly, for n = 2, applying (2.19) to (4.3) yields ZT kw ; kXl;rT;s (kukX T;s; ) k@xH ksd : (2)
[1 2]
00
?T
2
Then it follows from Lemma 2.8 that ! ZT ZT a ( u ) k@xH ks dt = ?T k@x 2 w w ksdt ?T c (kukX T;s; )kw kX T;s; kw kX T;s; c (kukX T;s; )c (1; kukX T;s; )kh kskh ks: (2)
2
(1)
(1)
[1]
[2]
(1)
2
2
[1]
00
2
(1)
[2]
00
00
2
2
00
25
00
1
2
Thus,
kw ; kXl;rT;s c (kukX T;s; ) (kukX T;s; )c (1; kukX T;s; )kh ks kh ks := c(2; kukX T;s; )kh kskh kX T;s; : (2)
2
2
[1 2]
00
00
2
00
1
2
1
00
2
00
Assume that (4.7) is true for 1 k N ? 1. Applying (2.19) to (4.3) with n = N yields ZT kw N;:::;N kXl;rT;s (kukX T;s; ) ?T k@xHN ksdt N cj X Yj kl X kw il ;:::;ilkl kX T;s; (kukX T;s; ) j ! j (kukX T;s; ) j k ::: kj N l N j X Yj X c(kl; kukX T;s; )kh il ;::;ilkl ks (kukX T;s; ) cj ! j (kukX T;s; ) j k ::: kj N l N Y := c(N; kukX T;s; ) khk ks: (
[1
)
]
00
(
00
00
=2
00
00
=2
00
k
[
1+
+
=
=1
1+
+
=
=1
)
1
]
00
00
[
1
]
=1
The proof is completed by induction. 2
Corollary 4.1 Let s > 3=4, (l; r) 2 [0; s ? ] [0; s ? ), T > 0 and u 2 X T;s ; be given. Then (4.5)-(4.6) de nes a homogeneous polynomial n K (u)[hn] of degree n from H s (R) to Xl;rT;s and kK n (u)[hn]kXl;rT;s c(n; kukX T;s; )khkns (4:8) 3
3
4
4
0 0
(
)
(
)
00
for any h 2 H s (R).
Now we may de ne formally the n-th Taylor polynomial Pn (h) of the map K at some 2 H s (R) as n X Pn (h) = K ( ) + k1! K k (u)[hk ] k n 1 X := u + k! yk k ( )
=1
=1
where u = K ( ). 26
Proposition 4.2 Let zn denote the n-th Taylor remainder of K at 2
H s (R), i.e.
zn = K (( + h) ? Pn(h):
Then it solves
8 > < @tz + @x(F (u; v)z ) + @xz = 0 > : z (x; 0) = h(x)
(4:9)
8 > < @tzn + @x(F (u; v)zn) + @xzn = ?@x(Dn ) > : zn (x; 0) = 0
(4:10)
0
1
3
0
0
0
for n = 0 and
3
1
for n 1 where and
Dn =
u = K ( );
nX
+1
m
Fm(u; v)
=2
v = K ( + h);
nX ?m +1
k
zk
=0
k1 ::: km?1 n?k +
with Fm(u:v) de ned by (1.14) and for m = 1; :::; n + 1.
X +
qk :::qkm? 1
1
(4:11)
=
ym qm = m !
Proof: Direct computation shows easily that (4.9) and (4.10) with n = 1 are true. Assume that (4.10) is true for n = N . Then, for n = N + 1, by de nition, zN = zN ? (N +1 1)! yN = zN ? qN +1
+1
+1
where qN solves 8 > < @tqN + @x(a0(u)qN ) + @xqN = ?@x(EN ) > : qN = 0 +1
+1
3
+1
+1
with 27
+1
+1
(4:12)
EN = +1
NX
+1
m
=2
X
a m (u) m! k (
)
::: km N
1+
=
+
qk :::qkm : 1
+1
Hence
@tzN + @xzN = ?@x F (u; v)zN ? a (u)qN ? @x(GN ) where N X?m NX X qk :::qkm? ? zk FM (u; v) GN = (1)
3
+1
1
+1
+1
+1
+1
m NX
k
=2
?
+1
m
=1
+1
+1
a m (u) m! k (
+
X
)
::: km N
1+
+
1
k1 :::: kM1 N ?k
=0
=
+
1
=
qk :::qkm : 1
+1
Note that
F (u; v)zN ? a0(u)qN 1
= F (u; v) (zN ? qN ) + (F (u; v) ? a0(u)) qN = F (u; v)zN + F (u; v)z qN ;
+1
1
+1
1
NX
+1
Fm(u; v)
N X?m +1
+1
1
2
0
+1
+1
X
zk
qk1 :::qkm?1 k1 ::: km?1 N ?k N X?m NX X (zk ? qk ) qk1 :::qkm?1 + = Fm(u; v) m k k1 ::: km?1 Nk N X?m NX X qk1 :::qkm?1 qk + Fm(u; v) m k1 ::: km?1 N ?k k N X?m NX X zk = Fm(u; v) qk1 :::qkm?1 + m k k1 ::: km?1 N ?k NX X qk1 :::qkm + Fm(u; v) m k1 ::: km N
m
k
=2
=0
+
+
=
+1
+1
+1
=2
=0
+
+
=
+1
+1
+1
=2
+
=0
+
=
+2
+1
=2
=1
+
+
=
+1
+1
=2
and
NX
+1
m
=2
Fm(u; v)
+
X k1 ::: km N +
+
=
+
=
qk :::qkm ? 1
+1
NX
+1
m
=2
+1
28
a m (u) m! k (
X
)
::: km N
1+
+
=
qk :::qkm 1
+1
m (u) ! a Fm(u; v) ? m! = m k NX X = Fm (u; v)z NX
+1
(
X
)
=2
::: km N
1+
+
=
+1
m NX
+1
0
=2
=
+2
m
k1 k2 ::: km N +
Fm(u; v)z
0
=3
+
+
X
=
k1 ::: km?1 N +
+
=
qk :::qkm 1
+1
qk :::qkm 1
+1
qk :::qkm? : 1
1
+1
Thus
F (u; v)zn ? a0(u)qN ? GN = F (u; v)zN + F (u; v)z qN + N X?m NX X qk :::qkm? + zk + Fm(u; v) 1
+1
+2
=2
k
m
+
NX
+2
m
= F (u; v)zN + +1
NX
+2
m
=2
1
+1
2
+ +
=
X
0
1
k1 :: km?1 N ?k
=1
+1
1
+1
qk1 :::qkm?1 k1 ::: km?1 N N X?m X Fm(u; v) zk qk1 :::qkm?1 k k1 ::: km?1 N ?k
Fm(u; v)z
=3
1
+1
+1
0
+
+
=
+1
+2
=0
+
+
=
+1
and we conclude that 8 > @tzN + @x(F (u; v)zN ) + @xzN = ?@x(DN ) < > : zN (x; 0) = 0 +1
1
3
+1
+1
+1
+1
which is (4.10) with n = N + 1. The proof is completed by induction. 2
Theorem 4.1 Let s > and (l; r) 2 [0; s? ][0; s? ) be given. Then, for any 2 H s (R), there exist a T > 0 and a neighborhood U of in H s (R) 3
3
3
4
4
4
such that the nonlinear map K de ned by the IVP (3.1) is in nitely many time Frechet di erentiable in U from H s (R) to Xl;rT;s . Its n-th derivative K n at 2 U is given by (
)
K n ( )[h ; :::; hn] = K n (u)[h ; :::; hn] (
)
(
1
)
1
for any h ; :::; hn 2 H s (R) where K n (u) is de ned by (4.2) and (4.3) and u = K ( ). 1
(
)
29
Proof: We only need to prove that for any 2 U , K ( + h) =
n 1 X K k ( )[hk ] + o(khkns ) k ! k ( )
=0
as h ! 0 in H s (R) uniformly for k ? ks khks by the Converse Taylor Theorem (see [8]). Let v = K ( + h); u = K ( ); y k = K k (u)[hk ] for 1 k n and z = v ? u; z = z ? y ; zn = zn? ? n1! yn: Then, by Proposition 4.2, 8 > < @tz + @x(F (u; v)z ) + @xz = 0 (4:13) > : z (x; 0) = h ( )
0
1
0
0
( )
0
1
1
3
0
0
0
and for n 1, 8 > < @tzn + @x(F (u; v)zn) + @xznF (u; v) = ?@xDn > : zn(x; 0) = 0 3
1
1
(4:14)
Choose > 0 such that 1
S ( ) = f 2 H s(R); k ? ks g U: 1
1
Obviously, K ( ) is bounded on S ( ) since K is a continuous map. By Corollary 4.1, 1
kqkkXl;rT;s c(k; kukX T;s; )khkks ;
k = 1; 2; :::; n
00
where u = K ( ) and c(k; kukX T;s; ) is uniformly bounded on S ( ). It suces to prove that kznkXl;rT;s n khkns (4:15) 1
00
+1
30
for n 0 where n is uniformly bounded for 2 S ( ). Indeed, it is easy to obtain (4.15) for n = 0 by using Theorem 2.1 to (4.13) and if we suppose (4.15) is true for n N , then applying Theorem 2.1 and Lemma 2.8 to (4.14) with n = N + 1, we have that 1
kzN kXl;rT;s +1
c
NX N X?m +2
m k k1 ::: km?1 N ?k NX N X?m X =2
c
+2
=0
+
+
=
m k k1 ::: km?1 N ?k NX N X?m X +1
m
=2
=0
+
+
=
k
k1 ::: km?1 N ?k
=0
:= N khkNs
+
+
=
1
cm ms;
0 0
+1
+2
k@x Fm(u; v)zkqk :::qkm? ks dt ?T
+1
+2
=2
c
ZT
X
+2
cm m m
+1
(T ; zk)
mY ?
1
j
=1
mY ?
1
j
1
s; (T ; qkj ) 0 0
=1
c(kj ; kukX T;s; )khkNs
+2
00
+2
+1
The proof is completed by induction. 2 Corollary 4.2 (Taylor's Formula) For any 2 U and h 2 H s (R) satisfying + h 2 U; for any 2 (0; 1); Z nX ? n? K ( + h) = j1! K j ( )[hj ] + (1 ?n!) K n ( + h)[hn ]d j with any n 1. Proof: See [7, Theorem 8.14.3]. 1
=0
1
1
( )
(
)
0
5 Analyticity In the previous section we proved the map K is in nitely many times Frechet di erentiable from H s (R) to Xl;rT;s. Naturally the further question is whether the map K is an analytic map, i.e., whether it has Taylor series expansion at any 2 H s (R): 1 n X K ( + h) = K n!( ) [hn]; (5:1) n (
=0
31
)
where the series coverges in Xl;rT;s uniformly for khks with some > 0 only depending on . For that purpose we need a better estimate of the n-th derivative yn = K n ( )[hn] of the map K at and the n-th Taylor remainder zn. Proposition 5.1 Let u 2 X T;s; be given and yn is the solution of (4.5)(4.6). Then there exists a sequence (n) given by (
)
0 0
(1) = 1; and
(5:2)
n X X (n) = j !j (k ) (k )::: (kj ) j k ::: kj n
(5:3)
kynkXl;rT;s cn n! (n)khkns
(5:4)
1
for n 2 such that
=2
1+
+
2
=
for any n 1 where c > 0 is a constant independt of n and h,
j = j (kukX T;s; ) 00
is a continuous monotone function depending only on a j , and j 0 if j 0. Proof: Denote by qn = yn=n!. It solves 8 > < @tq + @x(a0(u)q ) + @xq = 0 (5:5) > : q (x; 0) = h(x) for n = 1 and 8 Pn a j u P 0 > @ q + @ ( a ( u ) q ) + @ q = ? @ q :::q n x kj < tn x k ::: kj n k x n j j ( )
( )
1
3
1
1
1
( )
3
=2
> : qn(x; 0) = 0
( )
!
1+
+
=
1
(5:6)
for n 2. Applying (2.19) to (5.5) yields
kq kXl;rT;s ckhks: 1
32
Assume that
kqmkXl;rT;s c m? (m)khkms; 2
for 1 m N
1
Then applying (2.19) to (5.6) with n = N + 1, we obtain NX X ZT j k @x a (u)qk :::qkj ks dt kqN kXl;rT;s c j1! ?T j k ::: kj N NX X Yj c j !j kqkl kX T;s; j k ::: kj N l NX X Yj kl? kl c khks (kl ) c j !j j k ::: kj N l c N ? (N + 1)khkNs : Thus we have proved by induction that kynkXl;rT;s c n? n! (n)khkns +1
( )
+1
1
=2
1+
+
=
+1
+1
=2
1+
+
=
00
+1 =1
+1
2
=2
2(
+1)
1+
+
=
1
+1 =1
1
+1
2
1
which is (5.4) with a di erent c > 0. The proof is completed by induction.
2
Proposition 5.2 Let s > 3=4, T > 0 , (l; r) 2 [0; s ? ] [0; s ? ) and u; v 2 X T;s ; be given. If zn is the solution of (4.9)-(4.10) with yn being the 3
3
4
4
0 0
solution of (4.5)-(4.6), then there exists a sequence (n) given by
(0) =
(5:7)
1
(n) =
nX
+1
1
m
m
=2
nX ?m +1
k
(k)
=0
mY ?
X
1
k1 ::: km?1 n?k i +
+
=
for n 1 with (n) given by (5.2)-(5.3) such that
kznkXl;rT;s (n)khkns ;
(5:8)
=1
n0
+1
where
ci (i)
(5:9)
j = j (kukX T;s; ; kvkX T;s; ); 00
00
depending only on a j , is a continuous function of kukX T;s; and kvkX T;s; with j 0 if j 0, and c > 0 is a constant independent of n and h. ( )
00
( )
33
00
Proof: It follows from (4.9) by applying (2.19) that kz kXl;rT;s khks : Assume that (5.9) is true for all 0 n N ? 1. Then applying (2.19) and 0
1
(2.17) to (4.10) with n = N , we obtain that ! ZT NX N X?m mY ? X kzN kXl;rT;s k@x Fmzk qki ksdt +1
+1
1
m i k k1 ::: km1 N ?k ?T NX N X?m mY ? X mkzk kX0T;s;0 kqki ks m i k k1 ::: km?1 N ?k NX N X?m mY ? X m (k)khkNs ci (i) m i k k1 ::: km?1 N ?k N
(N )khks : 1
=2
+1
+
+
=1
=
+1
1
1
=2
=0
+1
=0
+
+
=1
=
+1
1
+1
1
=2
=0
+
+
=1
=
+1
The proof is completed by induction. 2 Now we consider the map K de ned by (3.1) from an open set U in s H (R) to Xl;rT;s where s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ). Proposition 5.3 Let 2 U . If there is a > 0 and c > 0 such that 3
3
4
4
1
S ( ) = f 2 H s (R)j k ? ks < g U 1
1
and
(n) cn; n0 (5:10) uniformly for 2 S ( ) where (n) is de ned by (5.7)-(5.8) with u = K ( ) and v = K ( ), then there exists a > 0 such that the series (5.1) coverges in Xl;rT;s uniformly for khks . Proof: Consider the n-th Taylor remainder n 1 X zn = K ( + h) ? j ! K j ( )[hj ]: j According to Proposition 5.2 and hypothesis (5.10), kznkXl;rT;s (n)khkns cnkhkns 1
( )
=0
+1
34
+1
where c > 0 is independent of n and h 2 H s (R) with khks < . Thus if we choose > 0 such that < 21c ; then kznkXl;rT;s ( 21 )n ; n 1 for any h 2 H s(R) with khks . The proof is completed. 2 In the following we shall show that if a(u) in (3.1) is a polynomial of degree N , i.e. N X (5:11) a(u) = bj uj 1
1
j
=2
with bj , j = 1; 2:::; N being real constants, then hypothesis (5.10) is satis ed for any 2 U and therefore the map K is an analytic map from U H s(R) to Xl;rT;s. First we prove a technical lemma. Lemma 5.1 Let N 1 be a given integer and n be a sequence given by N (1) = 1; (5:12) n X X N (n) = bjj! N (k )::: N (kj ); for 2 n N ? 1; (5:13) j k ::: kj n and N b X X N (k ):::: N (kj ); for n N . (5:14) N (n) = jj! j k ::: kj n where bj , 1 j N are given constants. Then there exists a constant c > 0 such that N (n) cn (5:15) for any n 1. Remark 5.1 In the case N = 2 and b = 1, the lemma is Proposition 3.4 in [47] where it is shown that n? (n) = 2 (2nn! ? 3)!! for any n 2: 1
=2
1+
+
=
1+
+
1
=2
=
2
1
2
35
Proof of Lemma 5.1: Note that that N (n), for any n 1, is uniquelly determined by (5.12) and (5.13) inductively. In particular, we may obtain
N (1); N (2); :::; N (N ? 1) explicitly by computation. Let PN (x) be a polynomial of degree N as follows !j N b NX ? X j k PN (x; y) = y ? j ! N (k)x + y ? 1
j k N ? N X bj X k X x N (k j j ! k j k1 ::: kj k =2
?
=1
1
1
=2
=
+
+
)::: N (kj ):
(5.16)
=
It is easy to check that
@ P (0; 0) = 1: @y N Thus, according to the implicit function theorem, PN (0; 0) = 0;
PN (x; y) = 0
(5:17)
has a unique solution
y = f (x)
with f (0) = 0
in the neighborhood of x = 0 such that for any jxj
PN (x; f (x)) = 0;
where > 0 is a constant. Besides, direct computation shows that f j (0) = 0; for j = 1; 2; :::; N ? 1: Moreover, y = f (x) is a real analytic function in a neighborhood of x = 0, i.e. f (x) has a Taylor series expansion at x = 0, 1 X y = f (x) = dj xj (5:18) ( )
j N =
36
which is uniformly convergent for jxj < with some > 0. To see this, let NX ? g = N (k)xk + y 1
k
and
z=
NX ?
1
k
=1
N b NX ? X X j n N (k)x + j ! xk N (k j k j k1 ::: kj k 1
1
=1
=2
=
+
+
)::: N (kj ) := h(x):
=
Then, equation (5.17) may be written as
g?
N b X j j g =z j j! =2
which obviously has an analytic solution g(z) in a neiborhood of z = 0 such P N that g(0) = 0. Thus y = f (x) = g(h(x)) ? k ? N (k)xk is an analytic function in a neighborhood of x = 0 sicne h(x) is a polynomial of x. Plugging (5.18) into (5.17) and denoting by 1
=1
for j = 1; 2; :::; N ? 1;
dj = N (j ); we have
y =
1 X
bj xj
j N !j 1 ? N b NX X X j k k N (k)x + dk x ? j j! k k N N N ? X X X ? bjj! xk N (k )::: N (kj ) j k j k1 ::: kj k !j X N b X 1 N b NX ? X X j j k k ? d x dk1 :::dkj kx j j! k k j ! k j k1 ::: kj k 1 N b X X X j dk1 xk1 dk2 xk2 :::dkj xkj ? j ! j k j k1 ::: kj k =
=
1
=2
=
=1
1
1
=2
=
+
+
=
1
=2
=
=
=2
=1
=
=2
+
+
=
37
=
+
+
=
N b NX ? X X j dk :::dkj xk j ! j k j k ::: kj k 0 1 N b X 1 X X j@ = dk :::dkj A xk j ! j k N k ::: kj k 1 0 1 X N b X X k @ j d = K :::dkj A x k N j j ! k ::: kj k
?
1
1
=2
=
1+
+
1+
+
=
1+
+
=
=
1
=2
=
1
=
=2
for any jxj < . Thus we have dj = N (j ); and N b X dk = jj! j
=2
j = 1; 2; :::; N ? 1
X k1 :::: kj k +
+
dk :::dkj 1
=
for any k N . That is to say, dj for j 1 also satisfy the induction relation (5.12)-(5.13). By uniqueness we have N (k) = dk ; for any k 1. On the other hand, dk , k 1 are coecients of Taylor series (5.18) and therefore we must have dn cn ; for all n 1 for some c > 0 independent of n. The proof is completed. 2 Theorem 5.1 Let s > 3=4 and (l; r) 2 [0; s ? ] [0; s ? ) be given and suppose that a(u) is (3.1) is a polynomial of degree N . Then for any 2 H s (R), there is a T > 0 and a neighborhood U of in H s (R) such that the IVP (3.1) de ne an analytic map K from U to Xl;rT;s , i.e., for any 2 U , there is a > 0 such that the Taylor series 1 X K ( + h) = K ( ) + n1! K n ( )[hn] n 3
3
4
4
(
)
=1
uniformly converges for khks in the space Xl;rT;s . Moreover, if denote by
u = K ( );
yn = K n ( )[hn]; n 1 (
38
)
then
8 > < @ty + @x(a0(u)y ) + @xy = 0 > : y (x; 0) = h 1
3
1
1
1
for n = 1 and
8 > < @tyn + @x(a0(u)yn) + @xyn = ?@xGN (n) > : yn(x; 0) = 0 3
for n 2 where
8 Pn > > < j GN (n) = > > : PN
=2
j
=2
a(j) u j
P
a(j) u j
P
( )
n k1 ::: kj n k1 :::kj yk1 :::ykj
!
( )
!
+
+
=
!
!
n k1 :::kj n k1 :::kj yk1 :::ykj
!
!
+
=
!
!
for 2 n N ? 1 for n N .
Proof: Since U is an open subset in H s (R), there is a > 0 such that if h 2 H s (R) with khks , then + h 2 U: 1
1
Denote by
v = K ( + h);
u = K( )
and
z = v ? u; zn = zn? ? n1! yn; for n 1: Then, by proposition 5.1 and 5.2, 0
1
kynkXl;rT;s n!cn (n)khkns and
kznkXl;rT;s (n)khkns : Note that a j (u) 0 for j N + 1 since a(u) is a polynomial of degree N +1
( )
by the assumption. Thus (n) is given by
(1) = 1; 39
(n) =
n X X j (k )::: (kj ) j j ! k :::: kj n 1
=2
for 2 n N ? 1 and
(n) =
1+
+
=
N X X j (k )::: (kj ) j ? j ! k ::: kj n 1
2
1+
for n N . Moreover,
+
=
(0) = ; 1
(n) =
n X 1
j
j
=2
n X?j +1
k
X
(k)
+
j
k
=2
=0
jY ?
1
k1 :::kj?1 n?k i +
=
cki (ki )
=1
=
X
+1
1
1
k1 :::kj?1 n?k i
=0
for 1 n N ? 1 and N n X?j X
(k)
(n) = j
jY ?
cki (ki ):
=1
According to Proposition 5.3, we need to show that there is a c > 0 such that
(n) cn for any n N with c independent of n and khks . To this end, we rst see from Lemma 5.1 that 1
(n) cn;
nN
1
with some c > 0 independent of n and h. Thus, for 0 j N , jY ? X X cki (ki ) (cc )n?k 1
1
k1 :::kj?1 n?k i +
=
=1
1
k1 ::: kj?1 n?k (n ? k ? 1)N (cc )n?k +
cn?k 2
40
+
=
1
for some c > 0 independent of n and N n X?j X
(n) c
(k) 2
X
+1
3
c
j
=2
k
=0
+1
3
j
=2
k
3
nX ?
1
k
+
cki (ki )
=1
=
cn?k (k) 2
=0
c (N ? 2)
1
k1 ::: kj?1 n?k i +
N n X?j X
jY ?
nX ?
1
k
cn?k (k) 2
=0
cn?k (k) 4
=0
for some c > 0 independent of n where 4
c = max f ; ::::; N g 3
1
2
1
Assume that
(k) ck 2k? ; for 1 k n ? 1: 1
4
Then,
(n)
nX ?
1
k
cn?k (k) 4
=0
nX ?
cn + cn?k ck 2k? k n n ? c2 1
1
4
4
=1
1
4
Thus we proved by induction that
(n) 2n? cn cn 1
4
for some c > 0 independent of n and h. The proof is completed. 2 Corollary 5.1 Assume that a(u) in (3.1) is a polynomial. Then for any T > 0 and s 1, the map K de ned by the IVP (3.1) is analytic from H s (R) to Xl;rT;s with (l; r) 2 [0; s ? ] [0; s ? ). 3
3
4
4
41
References [1] J. L. Bona & R. Scott, Solutions of the Korteweg-de Vries equation in fractional order Sobolev spaces, Duke Math. J., 43(1976), 87 { 99 [2] J. L. Bona & R. Smith, The initial value problem for the Korteweg-de Vries equation, Roy. Soc. London, Ser. A, 278(1978), 555 { 601 [3] L. Carleson, Some analytical problems related to statistical mechanics, Euclidean Harmonic Analysis, Lecture Notes in Math., SpringerVerlag, 779(1979), 9 { 45 [4] A. Cohen, Solutions of the Korteweg-de Vries equation from irregular data, Duke Math. J., 45(1978), 149 { 181 [5] A. Cohen & T. Kappeler, Solution to the Korteweg-de Vries equation with initial pro le in L (R)\LN (R ), SIAM J. Math. Anal. , 18(1987), 991 { 1025 1
1
+
1
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