Electronic Journal of Differential Equations, Vol. 2008(2008), No. 71, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
POWER SERIES SOLUTION FOR THE MODIFIED KDV
EQUATION
TU NGUYEN
Abstract. We use the method developed by Christ [3] to prove local wellposedness of a modified Korteweg de Vries equation in F Ls,p spaces.
1. Introduction
The modified Korteweg de Vries (mKdV) equation on a torus T has the form
∂t u + ∂x3 u + u2 ∂x u = 0
u(·, 0) = u0
(1.1)
where (x, t) ∈ T × R, u is a real-valued function. If u is a smooth solution of (1.1),
1
then ku(·, t)kL2 (T) = ku0 kL2 (T) for all t; therefore, u
e(x, t) = u(x + 2π
ku0 k2L2 (T) t, t)
is a solution of
Z
1
u2 (x, t)dx ∂x u = 0
∂t u + ∂x3 u + u2 −
2π T
(1.2)
u(·, 0) = u0
Thus, (1.2) and (1.1) are essentially equivalent. Using Fourier restriction norm
method, Bourgain [1] proved that (1.2) is locally well-posed for initial data u0 ∈
H s (T) when s ≥ 1/2, and the solution map is uniformly continuous. In [2], he
also showed that the solution map is not C 3 in H s (T) when s < 1/2. Takaoka and
Tsutsumi [10] proved local-wellposedness of (1.2) when 1/2 > s > 3/8, and they
showed that solution map is not uniformly continuous for this range of s. For (1.1),
Kappeler and Topalov [8] used inverse scattering method to show wellposedness
when s ≥ 0 and Christ, Colliander and Tao [4] showed that uniformly continuous
dependence on the initial data does not hold when s < 1/2. Thus, there is a gap
between known local well-posedness results and the space H −1/2 (T) suggested by
the standard scaling argument.
Recently, Grünrock and Vega [7] showed local well-posedness of the mKdV equation on R with initial data in
csr (R) := {f ∈ D0 (R) : kf k cr := kh·is fˆ(·)k r0 < ∞},
H
L
H
s
2000 Mathematics Subject Classification. 35Q53.
Key words and phrases. Local well-posedness; KdV equation.
c
2008
Texas State University - San Marcos.
Submitted April 10, 2008. Published May 13, 2008.
1
2
T. NGUYEN
EJDE-2008/71
1
when 2 ≥ r > 1 and s ≥ 12 − 2r
. (for r > 43 , this was obtained by Grünrock
[5]). This is an extension of the result of Kenig, Ponce and Vega [9] that localcsr scales like H σ
wellposedness holds in H s (R) when s ≥ 1/4. Furthermore, as H
1
1
with σ = s + 2 − r , this result covers spaces that have scaling exponent − 21 +.
There is also a related recent work of Grünrock and Herr [6] on the derivative
nonlinear Schrödinger equation on T. Both [7] and [6] used a version of Bourgain’s
method.
In this paper, we apply the new method of solution developed by Christ [3] to
investigate the local well-posedness of (1.2) with initial data in
FLs,p (T) := {f ∈ D0 (T) : kf kF Ls,p := kh·is fˆ(·)klp < ∞}.
Let B(0, R) be the ball of radius R centered at 0 in FLs,p (T). Our main result is
the following.
Theorem 1.1. Suppose s ≥ 1/2, 1 ≤ p < ∞ and p0 (s + 1/4) > 1. Let W be the
solution map for smooth initial data of (1.2). Then for any R > 0 there is T > 0
such that the solution map W extends to a uniformly continuous map from B(0, R)
to C([0, T ], FLs,p (T)).
We note that the FLs,p (T) spaces that are covered by Theorem 1.1 have scaling
index 41 +. The restriction s ≥ 1/2 is due to the presence of the derivative in the
nonlinear term, and is only used to bound the operator S2 in section 3. The same
restriction on s is also required in the work on the derivative nonlinear Schrödinger
equation on T by Grünrock and Herr [6]. We believe that the range of p in Theorem
1.1 is not sharp.
Concerning (1.1), we have the following result.
f be the
Corollary 1.2. Suppose s ≥ 1/2, 1 ≤ p < ∞ and p0 (s + 1/4) > 1. Let W
solution map for smooth initial data of (1.1). Then for any R > 0 there is T > 0
f extends to a uniformly continuous
such that for any c > 0, the solution map W
s,p
map from B(0, R) ∩ {ϕ : kϕkL2 = c} ⊂ FL (T) to C([0, T ], FLs,p (T)).
As in [3], the solution map W obtained in Theorem 1.1 gives a weak solution
of (1.2) in the following
sense.
Let TN be defined by TN u = (χ[−N,N ] u
b)∨ . Let
R
1
N u := u2 − 2π
u2 (x, t)dx ∂x u be the limit in C([0, T ], D0 (T)) of N (TN u) as
T
N → ∞, provided it exists.
Proposition 1.3. Let s and p be as in Theorem 1.1. Let ϕ ∈ FLs,p and u := W ϕ ∈
C([0, T ], FLs,p ). Then N u exists and u satisfies (1.2) in the sense of distribution
in (0, T ) × T.
To prove these results, we formally expand the solution map into a sum of
multilinear operators. These multilinear operators are described in the section 2.
Then we will show that if u(·, 0) ∈ FLs,p then the sum of these operators converges
in FLs,p for small time t, when s and p satisfy the conditions of Theorem 1.1.
Furthermore, this gives a weak solution of (1.2), justifying our formal derivation.
EJDE-2008/71
POWER SERIES SOLUTION
3
2. Multilinear operators
We rewrite (1.2) as a system of ordinary differential equations of the spatial
Fourier series of u (see [10, formula (1.9)], and [1, Lemma 8.16]).
dû(n, t)
− in3 û(n, t)
dt
X
X
= −i
û(n1 , t)û(n2 , t)n3 û(n3 , t) + i
û(n1 , t)û(−n1 , t)nû(n, t)
=
n1 +n2 +n3 =n
∗
X
−in
3 n
n1
(2.1)
û(n1 , t)û(n2 , t)û(n3 , t) + inû(n, t)û(−n, t)û(n, t),
1 +n2 +n3 =n
where the star means the sum is taken over the triples satisfying nj 6= n, j = 1, 2, 3.
We note that these are precisely the triples with σ(n1 , n2 , n3 ) 6= 0.
3
Let a(n, t) = e−in t û(n, t), then an (t) satisfy
in
da(n, t)
=−
dt
3 n
∗
X
eiσ(n1 ,n2 ,n3 )t a(n1 , t)a(n2 , t)a(n3 , t)
1 +n2 +n3 =n
+ ina(n, t)a(−n, t)a(n, t),
where
σ(n1 , n2 , n3 ) = n31 + n32 + n33 − (n1 + n2 + n3 )3 = −3(n1 + n2 )(n2 + n3 )(n3 + n1 ).
Or, in integral form,
Z
∗
X
in t
a(n, t) = a(n, 0) −
eiσ(n1 ,n2 ,n3 )s a(n1 , s)a(n2 , s)a(n3 , s)ds
3 0 n +n +n =n
1
2
3
(2.2)
Z t
2
+ in
|a(n, s)| a(n, s)ds.
0
If, a is sufficiently nice, say a ∈ C([0, T ], l1 ) (which is the case if u ∈ C([0, T ], H s (T))
for large s) then we can exchange the order of the integration and summation to
obtain
Z t
∗
X
in
eiσ(n1 ,n2 ,n3 )s a(n1 , s)a(n2 , s)a(n3 , s)ds
a(n, t) = a(n, 0) −
3 n +n +n =n 0
1
2
3
(2.3)
Z t
+ in
|a(n, s)|2 a(n, s)ds.
0
Replacing the a(nj , s) in the right hand side by their equations obtained using (2.3),
we get
Z t
∗
X
in
a(n1 , 0)a(n2 , 0)a(n3 , 0)
eiσ(n1 ,n2 ,n3 )s ds
a(n, t) = a(n, 0) −
3 n +n +n =n
0
1
2
3
Z t
+ in|a(n, 0)|2 a(n, 0)
ds + additional terms
(2.4)
0
∗
X
n
a(n1 , 0)a(n2 , 0)a(n3 , 0) iσ(n1 ,n2 ,n3 )t
= a(n, 0) −
(e
− 1)
3 n +n +n =n
σ(n1 , n2 , n3 )
1
2
3
+ int|a(n, 0)|2 a(n, 0) + additional terms
4
T. NGUYEN
EJDE-2008/71
The additional terms are those which depend not only on a(·, 0). An example of
the additional terms is
Z s
Z t
∗
∗
X
X
0
nn3
−
eiσ(m1 ,m2 ,m3 )s
a(n1 , 0)a(n2 , 0)
eiσ(n1 ,n2 ,n3 )s
9 n +n +n =n
0
0
m +m +m =n
1
2
3
1
2
3
3
× a(m1 , s0 )a(m2 , s0 )a(m3 , s0 )ds0 ds
Then we can again use (2.3) for each appearance of a(m, ·) in the additional terms,
and obtain more and more complicated additional terms. We refer to section 2 of
[3] for more detailed description of these additional terms. Continuing this process
indefinitely, we get a formal expansion of a(n, t) as a sum of multilinear operators
of a(·, 0).
We will now describe these operators and then show that their sum converges.
Again, we refer to section 3 of [3] for more detailed explanations. Each of our
multilinear operators will be associated to a tree, which has the property that each
of its node has either zero or three children. We will only consider trees with this
property. If a node v of T has three children, they will be denoted by v1 , v2 , v3 . We
denote by T 0 the set of non-terminal nodes of T , and T ∞ the set of terminal nodes
of T . Clearly, if |T | = 3k + 1 then |T 0 | = k and |T ∞ | = 2k + 1.
Definition 2.1. Let T be a tree. Then J (T ) is the set of j ∈ ZT such that if
v ∈ T 0 then
jv = jv1 + jv2 + jv3 ,
and either jvi 6= jv for all i, or jv1 = −jv2 = jv3 = jv . We will denote by v(T ) be
the root of T and j(T ) = j(v(T )). For j ∈ J (T ) and v ∈ T 0 ,
σ(j, v) := σ(j(v1 ), j(v2 ), j(v3 )).
Also define
0
R(T, t) = {s ∈ RT+ : if v < w then 0 ≤ sv ≤ sw ≤ t}.
Using the above definitions, we can rewrite (2.4) as
X
X
a(n, t) = a(n, 0) +
ωT
na(j(v(T )1 ), 0)a(j(v(T )2 ), 0)
|T |=4
j∈J (T ),j(T )=n
Z
× a(j(v(T )3 ), 0)
c(j, v(T ), s)ds + additional terms,
R(T,t)
here c(j, v, s) = eiσ(j,v)s , and ωT is a constant with |ωT | ≤ 1.
Continuing this replacement process, it leads to
Z
X
X
Y
Y
a(n, t) = a(n, 0) +
ωT
ju
a(jv , 0)
|T |<3k+1
j∈J (T ),j(T )=n u∈T 0
c(j, s)ds
R(T,t)
v∈T ∞
+ additional terms
where
c(j, s) =
Y
c(j, v, sv )
v∈T 0
We will show that the series
X
X
a(n, 0) +
ωT
T
Y
j∈J (T ),j(T )=n u∈T 0
converges in C([0, T ], lp ) when a(·, 0) ∈ lp .
ju
Y
v∈T ∞
Z
a(jv , 0)
c(j, s)ds
R(T,t)
EJDE-2008/71
POWER SERIES SOLUTION
5
3. lp convergence
Let T be a tree and j ∈ J (T ). We define
Z
IT (t, j) =
c(j, s)ds,
R(T,t)
and
X
ST (t)(av )v∈T ∞ (n) = ωT
Y
Y
ju
av (jv )IT (t, j).
v∈T ∞
j∈J (T ),j(T )=n u∈T 0
We first give an estimate for IT (t, j) which allows us to bound ST .
Lemma 3.1. For 0 ≤ t ≤ 1,
|IT (j, t)| ≤ (Ct)|T
0
Y
|/2
hσ(j, v)i−1/2 .
v∈T 0
Proof. For v ∈ T 0 , define the level of v, denoted l(v), to be the length of the unique
path connecting v(T ) and v. Let O be the set of v ∈ T 0 for which l(v) is odd, and
E those v for which l(v) is even.
First we fix the variables sv with v ∈ E, and take the integration in the variables
sv with v ∈ O. For each v ∈ O, the result of the integration is
1 iσ(j,v)sṽ
e
− eiσ(j,v) max{sv(1) ,sv(2) ,sv(3) }
σ(j, v)
if σ(j, v) 6= 0, and
sṽ − max{sv(1) , sv(2) , sv(3) }.
if σ(j, v) = 0. Here ve is the parent of v. Thus, we obtain the factor
Y
hσ(j, vi−1
v∈O
and an integral in sv , v ∈ E where the integrand is bounded by 2|O| . As the domain
of integration in sv with v ∈ E has measure less than t|E| , we see that
Y
0
|IT (j, t)| ≤ 2|T | t|E|
hσ(j, v)i−1 .
v∈O
By switching the role of O and E, we get
0
|IT (j, t)| ≤ 2|T | t|O|
Y
hσ(j, v)i−1 .
v∈E
Combining these two estimates, we obtain the lemma.
By Lemma 3.1,
|ST (t)(av )v∈T ∞ (n)| ≤ (Ct)|T
0
X
|/2
Y
hσ(j, u)i−1/2 |ju |
Y
|av (jv )|.
v∈T ∞
j∈J (T ):j(T )=n u∈T 0
Let
X
SeT (av )v∈T ∞ (n) =
Y
hσ(j, u)i−1/2 |ju |
Y
|av (jv )|,
v∈T ∞
j∈J (T ):j(T )=n u∈T 0
and
e 1 , a2 , a3 )(n) =
S(a
∗
X
n1 +n2 +n3 =n
−1/2
|n|hσ(n1 , n2 , n3 )i
3
Y
i=1
|ai (ni )| + |n|
3
Y
i=1
|ai (n)|.
6
T. NGUYEN
EJDE-2008/71
It is clear that
e SeT (av )v∈T ∞ , SeT (av )v∈T ∞ , SeT (av )v∈T ∞ ).
SeT (av )v∈T ∞ = S(
1
2
3
1
2
3
where Ti is the subtree of T that contains all nodes u such that u ≤ v(T )i (recall
e For this
that v(T ) is the root of T ). Hence, to bound ST , it suffices to bound S.
purpose, we will use the following simple lemma.
Lemma 3.2. Let S be the multilinear operator defined by
S(a1 , a2 , a3 )(n) =
X
m(n1 , n2 , n3 )
n1 +n2 +n3 =n
3
Y
aj (nj ),
j=1
Let 1 ≤ p ≤ ∞. Then for any pair of indices i 6= j ∈ {1, 2, 3},
kS(a1 , a2 , a3 )klp ≤ sup km(n1 , n2 , n3 )klp0
i,j
n
3
Y
kak klp .
k=1
Proof. By Hölder’s inequality, for any n,
|S(a1 , a2 , a3 )(n)| ≤ km(n1 , n2 , n3 )klp0 k
i,j
3
Y
≤ sup km(n1 , n2 , n3 )klp0 k
n
p
ak kli,j
k=1
i,j
3
Y
p
ak kli,j
k=1
Taking lp -norm in n we obtain the lemma.
s
Showing that Se is a bounded multilinear map on ls,p := {a : h·i a ∈ lp } is
equivalent to showing that S is bounded on lp where S is the operator with kernel
hnis |n|
m(n1 , n2 , n3 ) =
Q3
hσ(n1 , n2 , n3 )i1/2 k=1 hnk is
where n1 + n2 + n3 = n. We split S into sum of two operators S1 and S2 where S1
has kernel
hnis |n|
if n = n1 + n2 + n3 , ni 6= n
m1 (n1 , n2 , n3 ) = Q3
s
1/2
k=1 hnk i hn − nk i
and S2 has kernel
m2 (n1 , n2 , n3 ) = n/hni2s
if n1 = −n2 = n3 = n.
p
Clearly, S2 is bounded on l if and only if s ≥ 1/2.
It remains to bound S1 , for which we have the following result.
Proposition 3.3. S1 is bounded in lp ×lp ×lp to lp when s ≥ 1/4 and p0 (s+ 41 ) > 1.
Proof. In the proof, all the sums are taken over the triples (n1 , n2 , n3 ) that satisfy
the additional property that ni 6= n, for all 1 ≤ i ≤ 3. Clearly, we can assume
n > 0. Note that if say |n1 | ≥ 5n then as |n2 + n3 | = |n − n1 | ≥ 4n, at least one of
n2 and n3 has absolute value bigger than 2n. Also, we cannot have |ni | ≤ n/4 for
all i. Thus, up to permutation, there are four cases.
(1) |n1 |, |n2 |, |n3 | ∈ [n/4, 5n]
(2) |n1 |, |n2 | ∈ [n/4, 5n], |n3 | ≤ n/4
(3) |n1 | ∈ [n/4, 5n], |n2 |, |n3 | ≤ n/4
(4) |n1 |, |n2 | ≥ 2n
EJDE-2008/71
POWER SERIES SOLUTION
7
By Lemma 3.2, it suffices to show that in each of these four regions, for some i 6= j
p0
the li,j
-norm of m is bounded.
P
Case 1. As 3n = (n − ni ) for some index i, say i = 3, we must have |n − n3 | ∼ n.
Since we also have |n1 |, |n2 | & n,
|m(n1 , n2 , n3 )| .
hni1/2−s
.
hn3 is |(n − n1 )(n − n2 )|1/2
We will use the inequality
|
1
1 1
11
1 1
|≤
.
|=|
−
+
n3 (n − n2 )
n1 n3
n − n2
|n1 | |n3 | |n − n2 |
0
0
(1) If 1/4 ≤ s ≤ 1/2: then hn3 ip (1/2−s) . hnip (1/2−s) , so
0
0
X
kmkpp0 .
l1,2
|n1 |≤5n
hnip (1/2−s)
|n − n1 |p0 /2
0
p0 /2
|n2 |≤5n
0
X
.
|n1 |≤5n
(hn3 i|n − n2 |)
X hnip0 (1/2−s) 1
1
+
|n1 |p0 /2
|n − n2 |p0 /2
|n − n1 − n2 |p0 /2
|n2 |≤5n
p0 (1−2s)
X
.
hnip (1/2−s)
|n − n1 |p0 /2
hn3 ip (1/2−s)
X
|n1 |≤5n
hni
An
|(n − n1 )n1 |p0 /2
0
. hnip (1−2s) An
0
X 1
1
1 p /2
(
+
)
n |n − n1 | |n1 |
|n1 |≤5n
p0 (1/2−2s)
. hni
P
where
A2n .
0
0<j<5n
j −p /2 = An . As

1−p0 /2

n
An . loghni


1
if p0 < 2
if p0 = 2
if p0 > 2
0
we easily check that hni(1/2−2s)p A2n is bounded by a constant, under our hypothesis
on s and p0 .
0
0
(2) If s > 1/2: then hn − n2 ip (s−1/2) . hnip (s−1/2) , so
0
0
kmkpp0 .
l1,2
X
|n1 |≤5n
hnip (1/2−s)
|n − n1 |p0 /2
0
X
p0 s
|n2 |≤5n
0
.
X
|n1 |≤5n
.
X
|n1 |≤5n
hnip (1/2−s)
|n − n1 |p0 /2
0
X
|n2 |≤5n
0
|n − n1 |p (s−1/2)
|n1 |≤5n
−p0 /2
. hni
(hn3 i|n − n2 |)
1
hnip (s−1/2) 1
+
|n1 |p0 s
|n − n2 |p0 s
|n − n1 − n2 |p0 s
Bn
|n − n1 |p0 /2 |n1 |p0 s
X
. Bn
hn − n2 ip (s−1/2)
Bn2 .
0
1
1 p s
+
)
n |n − n1 | |n1 |
1
(
8
T. NGUYEN
where Bn =
EJDE-2008/71
0
P
0<j<5n
j −p s . As

1−p0 s

n
Bn . loghni


1
if p0 s < 1
if p0 s = 1
if p0 s > 1
0
we easily check that hni−p /2 Bn2 is bounded by a constant, under our hypothesis on
s and p0 .
Case 2 This case can be treated in exactly the same way as the first case, except
when n3 = 0. In the region n3 = 0,
1
X
X hnip0 (1/2−s)
0
1
−p0 s
kmkpp0 .
≤
hni
+
l1,3
|n1 (n − n1 )|p0 /2
|n1 |p0 /2
|n − n1 |p0 /2
n
n
1
1
−p0 s
. hni
An . 1
Case 3 As |n1 |, |n − n2 |, |n − n3 | ∼ n,
|m(n1 , n2 , n3 )| .
1
.
hn2 is hn3 is |n2 + n3 |1/2
Without loss of generality, we assume that |n3 | ≥ |n2 |.
(1) If |n2 | < |n3 |/2:
X
X
0
1
1
kmkpp0 .
0
l2,3
hn2 ip s
hn3 ip0 (s+1/2)
n/4≥|n3 |>2n2
0≤|n2 |≤n/4
.
1
X
0≤|n2 |≤n/4
hn2 ip0 (2s+1/2)−1
.1
if (s + 1/4)p0 > 1.
(2) If |n2 | ≥ |n3 |/2:
0
kmkpp0 .
l2,3
X
|n3 |≤n/4
.
X
|n3 |≤n/4
.
X
|n3 |≤n/4
0
1
hn3 i2p0 s
X
|n3 |≥n2 ≥|n3 |/2
1
hn3 + n2 ip0 /2
1
−p0 /2+1
}
0 s max{loghn3 i, hn3 i
2p
hn3 i
loghn3 i
+
hn3 i2p0 s
X
|n3 |≤n/4
1
.1
hn3 ip0 (2s+1/2)−1
0
as 2p s ≥ p (s + 1/4) > 1.
Case 4 |n1 |, |n2 | > 2n: Note that in this case, |n1 | ∼ |n − n1 | and |n2 | ∼ |n − n3 |.
(1) If |n3 |, |n − n3 | ≥ n/2 : we have
|m(n1 , n2 , n3 )| .
hni1/2
,
hn1 is+1/2 hn2 is+1/2
hence
0
0
l1,2
1
X
kmkpp0 . hnip /2
|n1 |,|n2 |>2n
0
hn1
hnip /2
.
. 1.
h2nip0 (2s+1)−2
ip0 (s+1/2) hn
p0 (s+1/2)
2i
EJDE-2008/71
POWER SERIES SOLUTION
9
(2) If |n3 | < n/2: then |n1 | ∼ |n2 | and |n − n3 | ≥ n/2, so
|m(n1 , n2 , n3 )| .
hence
ns+1/2
,
hn1 i2s+1 hn3 is
0
p0
X
kmk p0 . Bn
l1,3
|n1 |>2n
Bn
np (s+1/2)
. p0 (s+1/2)−1 . 1
hn1 ip0 (2s+1)
n
(3) If |n − n3 | < n/2: then |n1 | ∼ |n2 | and |n3 | ∼ n. Hence,
n
.
|m(n1 , n2 , n3 )| .
2s+1
hn1 i
hn − n3 i1/2
Therefore,
0
0
kmkpp0 .
l1,3
X
np
0 (2s+1)
p
hn1 i
hn − n3 ip0 /2
X
|n1 |≥2n n/2<n3 <3n/2
0
.
X
|n1 |≥2n
An
An np
. 2p0 s−1 . 1
n
hn1 ip0 (2s+1)
This concludes the proof of the proposition.
Proof of Theorem 1.1. Let u0 ∈ FLs,p and a(n) = u
c0 (n). By Proposition 3.3,
Y
|T 0 | |T 0 |/2
kST ((av )v∈T ∞ )kls,p ≤ C
t
kav kls,p .
v∈T ∞
Hence,
ka(n) +
X
Y
j∈J (T ),j(T )=n u∈T 0
T
≤ kakls,p +
X
ωT
X
ju
Y
Z
a(jv )
c(j, s)dskls,p
R(T,t)
v∈T ∞
kST (a, . . . , a)kls,p
(3.1)
T
≤
∞
X
(Ct)k/2 kak2k+1
ls,p =
k=0
ku k s,p
√ 0 FL
1 − Ctku0 k2F Ls,p
for all t . min{1, ku0 k−4
F Ls,p }.
Let T ∼ min{1, ku0 k−4
F Ls,p }, then for t ∈ [0, T ] we can define
Z
X
X
Y
Y
a(n, t) = a(n) +
ωT
ju
a(jv )
T
j∈J (T ),j(T )=n u∈T 0
v∈T ∞
c(j, s)ds
R(T,t)
and the solution map u = W u0 by
3
u
b(n, t) = e−in t a(n, t).
It follows from (3.1) that W is uniformly continuous. The same argument as that
of [3] shows that u is limit of classical solutions.
The proof of Proposition 1.2 is basically the same as that of [3, Proposition 1.4],
hence we omit it.
Acknowledgement. I would like to thank my advisor Prof. Carlos Kenig for suggesting this topic and for his helpful conversations. I would also like to thank Profs.
Axel Grünrock and Sebastian Herr for their valuable comments and suggestions.
10
T. NGUYEN
EJDE-2008/71
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Tu Nguyen
Department of Mathematics, University of Chicago, 5734 S. University Ave., Chicago,
IL 60637, USA
E-mail address: tu@math.uchicago.edu