ESI
The Erwin Schrodinger International
Institute for Mathematical Physics
Boltzmanngasse 9
A-1090 Wien, Austria
Existence and Semi{Classical Limit
of the Least Energy Solution to a Nonlinear
Schrodinger Equation with Electromagnetic Fields
Kazuhiro Kurata
Vienna, Preprint ESI 562 (1998)
Supported by Federal Ministry of Science and Transport, Austria
Available via http://www.esi.ac.at
June 16, 1998
Existence and semi-classical limit
of the least energy solution to a nonlinear
Schrodinger equation with electromagnetic
elds
Kazuhiro Kurata
July 2, 1998
Keywords: nonlinear Schrodinger equation, magnetic eld, least energy
solution, existence, semi-classical limit
1 Introduction and Main Result
We study a standing wave (x; t) = e?iEth? u(x) to a time-dependent nonlinear Schrodinger equation with electromagnetic potentials:
(1)
ih@t = ( hi r ? A(x))2 + V (x) ? f (jj2);
where h is positive constant, A(x) is a real-valued magnetic vector potential,
V (x) is a real-valued electric potential, and f (jj2) is a nonlinear term. For
simplicity, we treat the case n 3 and a special nonlinearity f (t) = t(p?1)=2
for some 1 < p < (n + 2)=(n ? 2), although we can also treat 2-dimensional
case with some modications( see, Remark 2 and Remark 5). Then u satises
(2)
( hi r ? A(x))2u + (V (x) ? E )u ? f (juj2)u = 0 in Rn:
1
Department of Mathematics, Tokyo Metropolitan University, Minami-Ohsawa 1-1,
Hachioji-shi, Tokyo, Japan, e-mail: [email protected],
1
Put W (x) = V (x) ? E and assume that W is strictly positive;
W (x) inf W W 0 > 0; x 2 Rn:
Let F (t) = 0t f (s) ds and
J (u) = 12 fjDh uj2 + W (x)juj2g dx ? F (juj2) dx;
R
R
where Dh u = (D1h u; ; Dnh u) and Djh = i?1h@j ? Aj (x).The purpose of this
paper is to show existence of the least energy solution (see Lemma 2) to (2)
and to study a semi-classical limit of uh as h ! 0. We denote by H the
Hilbert space dened by the closure of C01(Rn ) with respect to the norm
R
Z
Z
n
n
kuk2H =
Z
R
n
jDh uj2 + W (x)juj2 dx:
Because of the super-nonlinearity of f (juj2)u = jujp?1u, there exists u0 2 H
such that J (u0) 0. We look for the least energy solution by the mountain
pass theorem. Let ? = f 2 C ([0; 1]; H ); (0) = 0; J ( (1)) 0g and let
bh = inf
sup J ( (t)):
2?
t2[0;1]
Let B = (Bjk ) be a magnetic eld dened by Bjk = @j Ak ? @k Aj . We say
U 2 (RH )q ; 1 < q < +1, if U 2 Lqloc (Rn) and there exists a constant C such
that
1=q
1
1
q dy
j
U
(
y
)
j
C
jB (x; r)j B(x;r)
jB (x; r)j B(x;r) jU (y)j dy;
and say U 2 (RH )1 if U 2 L1loc(Rn) and
1
sup jU (y)j C jB (x;
r)j B(x;r) jU (y)j dy
B (x;r)
Z
Z
Z
for every ball B (x; r) in Rn , respectively. For jB j + W 2 (RH )n=2 , we have
jB j + W 2 (RH )n=2+ for some > 0 and hence we can dene m(x; jB j + W )
by
2
1
= inf fr > 0; r
m(x; jB j + W )
jB (x; r)j B(x;r)(jB (y)j + W (y)) dy 1g:
Z
2
We refer to [Sh1] for these facts and the properties of m(x; jB j+W ). Throughout this paper we assume the following conditions on A, B and W : A 2
2 (Rn ); W 2 C (Rn ) for some 2 (0; 1) and there exists a constant C
Cloc
loc
such that
jB j + W 2 (RH )n=2 ; jrB (x)j Cm(x; jB j + W )3:
(3)
Theorem 1 Under the assumption (3) and m(x; jB j+W ) ! 1 as jxj ! 1,
there exists a least energy solution uh 2 H such that bh = J (uh ).
Remark 1 When A 0, previous results require some condition for W
at innity (e.g., liminf jxj!1 W (x) > inf W ) or the smallness of h(cf., e.g.,
[Oh1],[Ra], [DF1,2], [GU]). Note that Theorem 1 need to assume neither any
condition at innity for W itself nor the smallness of h. If jB j+W 2 (RH )1 ,
then jB (x)j + W (x) Cm(x; jB j + W )2 holds for some constant C . It
is known that U (x) = jP (x)j belongs to (RH )1 for any polynomial P (x)
and > 0 ([Sh1,2]). Hence, if A(x) is polynomial and jB (x)j ! 1 for
instance, we can apply Theorem 1. Moreover, we note that the condition
m(x; jB j) ! 1 as jxj ! 1 holds in many cases, even if jB j itself do not
have an homogeneous growth at innity.
Next, we study the prole of the solution uh in the semi-classical limit of
h ! 0. For the case A(x) 0, many authors studied the prole of the least
energy solution uh, when h tends to zero( see, e.g., [FW], [Oh1], [Wa], [DN],
[BW], [DF1], [Ra].) and showed that uh concentrates at the global minimum
of W . Several authors ( [Oh2], [Gu], [DF2]) also studied the existence of
multi-bump solutions which concentrate at local minimums of W . In such a
case, the study of uh is decoupled into the scalar case and is reduced to the
one of positive solutions of ?h2u + Wu = up in Rn : Now, we consider
the general case. First, we observe that the transformation y = h?1 (x ? x0)
for some x0 2 Rn and vh(y) = uh (hy + x0), vh satises
? y vh ? 2i A(hy + x0) ry vh(y) ? hi (divxA)(hy + x0)vh(y)
+ jA(hy + x0)j2vh(y) + W (hy + x0)vh(y) = jvh(y)jp?1vh(y):
Since A(hy + x0) ! A(x0); W (hy + x0) ! W (x0) as h ! 0 on fjyj M g for
each M > 0, vh might tend to some w in some sense and w satises
( 1i r ? A(x0))2w + W (x0)w = jwjp?1w:
3
Let (y) =
es
n A (x )y .
j =1 i 0 j
P
Then A(x0) = r and w~(y) = e?i(y)w(y) satis-
?w~ + W 0w~ = jw~jp?1w:
~
Hence, one might conjecture that vh(y) ! ei(y)w~(y) as h ! 0. Actually, we
prove an analogous result due to [Wa] in the case of A(x) 0. Namely, we
show that fvhg has a subsequence fvk g, in which fvk g converges to a certain
2 (Rn ) and fjv jg converges to the least energy solution U
function v~ in Cloc
k
for the scalar case in H 1(Rn ) and concentrates at a global minimum point
of W . The magnetic potential A(x) contributes the phase factor of v~ (see
also the part (c) of Theorem 2). Let U be the unique positive solution
to ?U + W 0U = U p in Rn with max U = U (0). We use the notation
A0 = A(x0) for the point x0 which will appear in the next theorem.
Theorem 2 In addition to the assumptions in Theorem 1, suppose W =
fx 2 Rn; W (x) = min W ( W 0)g is not empty and liminf jxj!1 W (x) >
min W .
(a) Let be a arbitrary small number. Then, there exist a sequence fxk g
and a subsequence fhk g; hk ! 0(k ! 1), and positive constants C and R0
which do not depend on k such that fxk g converges to a point x0 2 W and
?
pW 0 ?jx?x j
k
juh (x)j Ce
holds on fjx ? xk j hk R0 g: In the estimate above, as xk we can take a
maximum point of juh j and choose a large R0 to assure
max ju (x)j > (W 0)1=(p?1):
jx?x jh R0 h
k
hk
k
k
k
k
(b) There exist a constant ! 2 R such that for each R > 0
uh (x) ? U ( x ?h xk )e?i(!+h
k
k
?1 Pn A0 (x?xk )j ) k
j =1 j
!0
as hk ! 0 on fjx ? xk j Rhk g for each R > 0.
(c) Fix the R0 as in the part (a). Suppose maxx2 jA(x)jR0 < 2. Then
the maximum point xk of juh j is unique.
Actually, we do not use the assumptions (3) and m(x; jB j + W ) ! 1 at
innity in the proof of Theorem 2. We just assumed these assumptions to
W
k
4
assure the existence of the least solution, although we would obtain the existence of the least energy solution for small h, once we assume the condition
liminf jxj!1 W (x) > min W or its local version (see [Ra], [DF1,2], [Gu]).
Remark 2 Although we considered in Theorem 1 and 2 the simple model
case f (juj2 )u = jujp?1u for the sake of simplicity, these results would be true
for certain general nonlinear term f (juj2)u which was treated in [NT], [Gu],
[DF1,2].
Remark 3 In the part (b), we show actually that vk (y) = uh (xk + hk y)
A y ) in C 2 (Rn ) as h ! 0. We do not know in
converges to U (y)e?i(!+
k
loc
part (c) whether we can remove the smallness condition on maxx2 jA(x)j
Pn
j =1
k
0
j j
W
or not. We have no further information on ! and xk .
The construction of multi-bump solutions which concentrate near at the local
minimum of W is not treated in this paper. This is a natural question and
will be studied in near future.
2 The proof of Theorem 1
In this section, we give the proof of Theorem 1. Throughout this section, we
use the notation D = Dh for the sake of simplicity, since the dependence on
h is irrelevant here. The key step in the proof of Theorem 1 is the following
Proposition.
Proposition 1 Under the same assumption as in Theorem 1, the functional
J (u) satises the Palais-Smale(PS) condition.
Once we obtain this Proposition 1, we can prove Theorem 1 by the standard
mountain pass theorem(see, e.g., [AR]). To show Proposition 1, we use the
following inequality due to Shen[Sh2].
Lemma 1 Suppose n 3 and jB j+W 2 (RH )n=2 and jrB (x)j Cm(x; jB j+
W )3. There exists a constant C (which depends on h) such that
Z
Rn
m(x; jB j + W juj dx C
)2
2
for every u 2 C01(Rn).
5
Z
Rn
jDuj
2+
W juj dx
2
Remark 4 Even if W 0, we have the inequlity
2juj2 dx C
2 dx
m
(
x;
j
B
j
)
j
Du
j
R
R
for every u 2 C01(Rn)([Sh3]). Hence, we can show still the exsitence of the
least energy solution to (i?1 hr ? A(x))2u ? f (juj2)u = 0 under the condition
that m(x; jB j) > 0 for some positive constant and m(x; jB j) ! +1
as jxj ! 1 in the same way as in the proof of Theorem 1. Here, we take
the Hilbert space H as the completion of C01 (Rn ) with respect to the norm
kukH = kDukL . However, we do not know the semiclassical limit in this
Z
Z
n
case.
n
2
Remark 5 When n = 2 and B (x) 0, we know the following inequality due
to Avron, Herbst and Simon [AHS, Theorem 2.9]:
Z
Rn
B (x)ju(x)j2 dx Z
Rn
j(ir ? A(x))uj2 dx
for every u 2 C01 (Rn ). Therefore, in the 2-dimensional case we can obtain
existence of the least energy solution under the conditions B (x) B0 > 0
and jB (x)j ! +1 as jxj ! 1.
Before giving a proof of Proposition 1, we mention the meaning of the
least energy solution. We say u is the least energy solution, if u is a solution
and satises J (u) J (v) for every solutions v. Let M = fu 2 H ; u 6=
0; jDuj2 + W juj2 dx = jujp+1 dxg. Note that any non-trivial solution to
(2) belongs to M. The next lemma is rather well-known (see, e.g., [NT], [Wa],
[DF1]) and shows that the montain pass solution obtained by Theorem 1 is
the least energy solution.
Lemma 2 We have bh = inf v2M J (v) > 0.
Proof: Let bh = inf u2H (supt>0 J (tu)): It is easy to see that for each u 6=
0 2 H there exists a unique positive number t(u) such that supt>0 J (tu) =
J (t(u)u). Actually, t(u) is determined by
R
R
Z
Z
t(u)2 jDuj2 + W juj2 dx = t(u)p+1 jujp+1 dx:
Now, we can see M = fv = t(u)u; u 6= 0 2 H g and bh = inf v2M J (v). We
claim bh = bh which conclude Lemma 2. For each v 2 M, since J (tv) ! ?1
6
as t ! 1, there exists t0 > 0 such that J (t0v) 0. Hence, (t) = tt0v
belongs to ? and it follows that bh bh. On the other hand, we claim that
for each 2 ?, there exists 2 [0; 1] such that ( ) 2 M. This claim yields
bh bh and complete the proof. We show the claim by contradiction. If
(t) 62 M for every t 2 [0; 1]. Then
Z
Z
jD (t)j2 + W j (t)j2 dx > j (t)jp+1 dx
(4)
holds for every t 2 [0; 1]. By the Sobolev's inequality, we see
Z
Z
Z
jDuj2 + W juj2 dx = jujp+1 dx C ( jDuj2 + W juj2 dx)(p+1)=2
for some positice constant C and for v 2 M. Hence, there exists a constant
such that if kvkH and v =
6 0, then v 62 M and jDvj2 + W jvj2 dx >
p
+1
jvj dx hold. We may assume (t) =6 0 for t =6 0 and hence (t) 62 M for
small t and (4) holds. By the continuity of (t), (4) holds for all t 2 [0; 1]. It
follows that J ( (t)) > 0 for all t 2 [0; 1]. This contradicts to J ( (1)) 0. 2
Proof of Proposition 1: Take a sequence fun g H such that
J (un) ! c and J 0(un) ! 0 as n ! 1 for some constant c. Then, we have
R
R
hJ 0(un); i = Re
Z
Rn
Dun D + Wun ? junjp?1un
dx = o(kkH ) (5)
for any 2 H and
1 jDu j2 + W ju j2 dx ? 1
p+1
n
n
2 R
p + 1 R junj dx ! c:
Putting = un in (5), we obtain
Z
Z
n
n
kunkH ?
2
Z
ju jp+1 dx = o(kun kH ):
Rn n
(6) and (7) imply
p?1
p+1
p + 1 R junj dx = 2c + o(1) + o(kunkH ):
Substituting (8) to (7), we obtain the boundedness of fung in H and
Z
n
kun kH =
2
Z
Rn
junjp+1 dx + o(1):
7
(6)
(7)
(8)
(9)
Hence, we can take a subsequence fun g fung such that un converges
weakly to a u 2 H . We write uk = un for the sake of simplicity. Taking
= uk ? u in (5), we get
k
k
k
Z
Re Duk D(uk ? u) + Wuk (uk ? u) ? juk jp?1uk (uk ? u) dx = o(1):
This yields
Z
kuk ? uk2H + Re uD(uk ? u) + Wu(uk ? u) dx
? Re jukjp?1uk (uk ? u) dx = o(1):
Z
The second term converges to zero, because of the weak convergence of uk
to u in H . We claim the following which concludes the strong convergence
kuk ? ukH ! 0.
Claim: Jk juk jp juk ? uj dx ! 0 as k ! 1.
Holder's inequality and (8) yield
R
Jk Z
C
p=(p+1)Z
Rn
Z
ju ? ujp+1 dx
Rn k
Rn
juk ? ujp+1 dx
=(p+1)
1
=(p+1)
1
juk jp+1 dx
:
Now, by Lemma 1 we have
Z
Rn
m(x; jB j + W )juk (x)j2 dx M
for some constant M . Since m(x; jB j + W ) ! 1, for any > 0 there exists
R > 0 such that m(x; jB j + W ) M= on jxj R. This imples
Z
fjxjRg
for every k and hence
Z
also holds. Since 2 < p + 1
that
kuk ? ukL
juk (x)j2 dx ju(x)j2 dx 2n=(n ? 2), there exists 2 (0; 1) such
fjxjRg
< 2 =
p+1
kuk ? ukL kuk ? uk1L? :
2
8
2
By Sobolev's inequality, we have
kuk ? ukL C krjuk ? ujkL kD(uk ? u)kL M:
Here we also used the inequality jrjuj(x)j jDu(x)j a.e. x (see, e.g., [RS]).
Note that, since jruj jDuj + jAuj, fuk g is bounded also on H 1(K ) for any
bounded domain K , and hence uk converges strongly to u in L2(K ). Thus,
we obtain
lim sup juk ? uj2 dx lim sup
juk ? uj2 dx 2:
2
2
Z
k!1
2
Z
Rn
k!1
fjxjRg
Therefore, we have kuk ? ukL ! 0 and Jk ! 0. 2
2
3 Proof of Theorem 2
First, we give an estimate of the energy of uh (see, e.g., [NT], [Wa], [DF1]).
Let W 0 = min W and let U be the least energy solution associated to the
energy on M0;r;
(10)
J0(v) = 21 jrvj2 + W 0jvj2 dx ? p +1 1 jvjp+1 dx;
where M0;r = fv 6= 0 2 H 1(Rn; R); jrvj2 + W 0jvj2 dx = jvjp+1 dxg. It
is well-known that U is a radially symmetric and unique (up to translation)
positive solution to the equation:
?U + W 0U = jU jp?1U in Rn:
(11)
We put b0;r = J0(U ). In a similar way, we dene the class M0;c = fv 6= 0 2
H 1(Rn; C); jrvj2 + W 0jvj2 dx = jvjp+1 dxg and denote by b0;c the least
energy of J0(v) on M0;c.
Lemma 3 We have hnb0;r bh = J (uh) hnb0;r(1 + o(1)) as h ! 0.
Proof: Let x0 2 W be a point and 2 C01 (B (x0; 2R)) satisfy (x) 1
on B (x0; R), where R > 0 is a some xed number. Put u(x) = (x)U ((x ?
x0)=h)ei((x?x )=h). By Lemma 2, we have
bh = J (uh) sup J (tu) = J (thu):
(12)
Z
Z
R
R
R
0
t>0
9
R
Here, the positive constant th is uniquely determined by
Z
Z
t2h jDuj2 + W juj2 dx = tph+1 jujp+1 dx:
It follows that
(13)
J (thu) = 21 ? p +1 1 tph+1 jujp+1 dx:
By using the transformation y = (x ? x0)=h and the exponential decay of U ,
it is easy to see
Z
and
jujp+1 dx = hn
Z
Z
U p+1 dy(1 + o(1))
Z
Z
W juj2 dx = hn W 0 U 2 dy(1 + o(1))
as h ! 0. Let M = max(kkL1 ; krkL1 ). We see
jDuj2 dx = h(r)(x)U ( x ?h x0 )ei((x?x )=h) + (x)(rU )( x ?h x0 )ei((x?x )=h)
2
+ (x)(A(x0) ? A(x0 + hx))U ( x ?h x0 )ei((x?x )=h) dx
Z Z
0
0
0
Z
= hn (x0 + hy)jrU (y)j2 dy + J;
where
J 2hn+1 M 2
+ 2hn M 2
+
+
Z
Z
Z
f2R=hjyjR=hg
f2R=hjyjg
jU (y)jjrU (y)j dy + hn+2M 2
jU (y)j2 dy
U (y)jA(x0) ? A(x0 + hy)jjrU (y)j dy
j
U (y)j2jA(x0) ? A(x0 + hy)j dy
f2R=hjyjg
2hn+2 M 2
Z
Z
j
U (y)j2jA(x0) ? A(x0 + hy)j dy:
f2R=hjyjg
Note that jA(x0) ? A(x0 + hy)j maxz2B(x ;2R) jrA(z)jhjyj Chjyj on
jyj 2R=h. Hence, by using the exponential decay of U we have
0
Z
jU (y)j2jA(x0) ? A(x0 + hy)j dy Ch
f2R=hjyjg
10
Z
Rn
jU (y)j2jyj dy O(h):
By controlling other terms in a similar way by using also the exponential
decay of rU , we obtain
Z
Z
jDuj2 dx = hn jrU j2 dy(1 + o(1)):
Hence, by substituting these into (13) we have
Z
Z
t2h jrU j2 + W 0U 2 dy(1 + o(1)) = tph+1 U p+1 dy(1 + o(1)):
Since jrU j2 + W 0U 2 dy = U p+1 dy, we obtain th ! 1 as h ! 0. Hence, it
follows
J (thu) = ( 21 ? p +1 1 )hn U p+1 dy(1 + o(1)) = b0;rhn (1 + o(1)) (14)
and (12) and (14) yield the desired upper bound. Since it is easy to see
hn J0(jvj) J (u) for each u 2 H and v(y) = u(hy), we have hnb0;r bh by
the characterization bh = bh = inf u2H (supt>0 J (tu)) (see Lemma 2). 2
Lemma 4 There exist a sequence fxk g converges to some point x0 2 W
a subsequence fvh g; hk ! 0(k ! 1) and U~ such that vh (y) = uh (xk +
2 (Rn ) as h ! 0. Here (y ) = n A (x )y and
hk y) ! ei(y)U~ (y) in Cloc
k
j =1 j 0 j
~U 2 H 1(Rn ) is a solution to ?U~ + W (x0)U~ = jU~ jp?1U~ in Rn. Moreover,
v~ = jU~ j is a positive solution to ?~v + W 0v~ = v~p in Rn .
Proof: We follow the argument as in [Wa].
(Step 1) By Lemma 3, we have
( 21 ? p +1 1 )kuhk2H hn b0;r(1 + o(1)):
Let wh (y) = uh(hy). Then wh satises
(r ? iA(hy))2wh (y) ? W (hy)wh(y) + jwh(y)jp?1wh(y); y 2 Rn:
Since kuhk2H = hn( j(r?iA(hy))wh(y)j2 dy + W (hy)jwh(y)j2 dy), we obtain
( 21 ? p +1 1 ) jrjwh(y)jj2 dy + W (hy)jwh(y)j2 dy
( 12 ? p +1 1 ) j(r ? iA(hy))wh(y)j2 dy + W (hy)jwh(y)j2 dy
b0;r(1 + o(1)):
R
R
Z
k
Pk
R
R
Z
Z
Z
Z
11
k
In particular, we have the boundedness of H 1(Rn)-norm of jwhj. Note that
since W 0 = min W we have
b0;r 21 jrjwhj(y)j2 + W 0jwh(y)j2 dy ? p +1 1 jwhj(y)p+1 dy
21 j(r ? iA(hy))wh(y)j2 + W (hy)jwh(y)j2 dy ? p +1 1 jwh j(y)p+1 dy
= ( 12 ? p +1 1 ) jwhj(y)p+1 dy = b0;r(1 + o(1)):
(15)
Then there exists a sequence fyhg and positive constants L and such that
Z
Z
Z
Z
Z
lim hinf
!0
Z
B (yh ;L)
jwh(x)j2 dx > 0:
(16)
Otherwise, by using Lemma 2.18 in [CR] (see also [Li]) we have kwhkL (R ) !
0 for every 2 < q < 2n=(n ? 2). This contradicite to (15).
(Step 2) Let vh(y) = wh(yh + y). Then fjvhjg is bounded in H 1(Rn) and
satises
jvhj W (hyh + hy)jvhj ? jvhjp
(17)
in the distribution ( and weak ) sense by Kato's inequality. By taking a
subsequence fvh g, we may assume that jvk j = jvh j converges to a nonnegative v~ 2 H 1(Rn) weakly in H 1(Rn) and converges strongly in Lqloc (Rn)
for 2 q < 2n=(n ? 2). By (16) v~ 6= 0. We now claim that fhk yh g is a
bounded sequence. Otherwise, there exists a sequence hk ! 0 and such that
xk hk yh ! 1 as k ! 1. Multiplying v~ to (17), we have
q
k
n
k
k
k
Z
? rjvkj rv~ dy Z
Z
fjyjRg
W (xk + hk y)jvkjv~ dy ? jvkjpv~ dy
for any xed R > 0. By the assumption W~ 0 liminf jxj!1 W (x) > min W ,
there exsits a constant > 0 (e.g., = (W~ 0 ? W 0)=2) such that
Z
jrjv~jj dy + (W
2
0+
)
Z
fjyjRg
Z
jv~j dy jv~jp+1 dy:
2
By letting R ! 1, we obtain
Z
jrjv~jj dy + (W
2
0+
Z
Z
) jvj dy jv~jp+1 dy:
12
~2
This means there exists 2 (0; 1) such that v~ 2 M0;r. Now we have
b0;r J0(v~) = ( 21 ? p +1 1 ) (v~)p+1 dy
< ( 21 ? p +1 1 ) (~v)p+1 dy
Z
Z
Z
lim kinf
jvk jp+1 dy = b0;r(1 + o(1));
!1
and this is a contradiction.
(Step 3) Taking a subsequence again, we may assume that fvkg converges
to v~(6= 0) 2 H 1(Rn ) and xk = hk yk ! x0 for some x0. We claim x0 2 W .
By the argument above, we have
Z
Z
Z
jrv~j2 + W (x0)~v2 dy v~p+1 dy:
Since W 0 = min W , it follows that
Z
Z
Z
jrv~j2 + W 0v~2 dy v~p+1 dy:
Hence, there exists 2 (0; 1] such that v~ 2 M0;r. Now we have
?n b = lim ( 1 ? 1 ) jv jp+1 dy
b0;r = klim
h
k
!1 k h
k!1 2 p + 1
( 21 ? p +1 1 ) jv~jp+1 dy
( 21 ? p +1 1 ) jv~jp+1 dy = J0(v~) b0;r:
This implies = 1 (i.e., v~ 2 M0;r) and W (x0) = W 0, i.e., x0 2 W , and
J0(~v) = b0;r. Thus v~ is a non-negative and non-trivial solution to ?~v +
W 0v~ = v~p in Rn, and hence v~ is positive by the strong maximum principle.
(Step 4) For each compact set K Rn, there exists a constant CK such
that supy2K jA(xk + hk y)j2 CK for every 0 < h < 1 and large k. Therefore,
we obtain the boundedness of H 1(K )-norm of fvk g for each K . By the
subsolution estimate, we can also obtain the boundedness L1 (K )-norm, and
2; n
hence Schauder's estimate leads to the uniform boundedness in Cloc
(R ).
Z
k
Z
Z
13
Hence, taking a subsequence again if necessary, fvh g converges to certain v
2 and also converges weakly in Lp+1 (Rn ), and v satises
in Cloc
?v ? 2i A0 rv + jA0j2v + W 0v = jvjp?1v in Rn ;
where A0 = A(x0) and W 0 = W (x0). Let U~ (y) = e?i(y)v(y). Then U~
satises jU~ j = v~ 6= 0 and
?U~ + W 0U~ = jU~ jp?1U~ in Rn :
k
2
Lemma 5 Let be arbitrary small and fvh g be the subsequence as above.
k
Then there exist positive constants C; R0 which do not depend on k such that
jvh (y)j Ce?
k
p 0
W ?jyj
on jyj R0 .
Proof: (Step 1) We write vk = vh . By Kato's inequality( see, e.g., [RS]),
we have
jvkj W (xk + hk y)jvkj ? jvkjp ?jvk jp?1jvkj
in the distribution sense. Boundedness of k(r ? iA (x0 + h))vhkL (R ) and
Sobolev's inequality imply the boundedness of L2 (Rn)-norm of jvhj. Let
ch(x) = jvh(x)jp?1, then there exists a constant C such that kch kL ? (R ) C . Since 2=(p ? 1) > n=2, by the subsolution estimate (see, e.g., [GT]) we
obtain
Z
1=2
2
jvk (y)j C0
jvkj dz
(18)
k
2
2 =(p 1)
B (y;1)
n
n
for jyj 1 and a constant C0 which does not depend on k. Now we claim
jvk j converges to v~ strongly in H 1(Rn), since
Z
Z
jrv~j2 + W 0v~2 dy lim kinf
jrjvkjj2 + W 0jvkj2 dy
!1
lim kinf
j(r ? iA(xk + hk y))vkj2 + W (xk + hk y)jvk j2 dy
!1
1 ? 1 )?1b = ( 1 ? 1 )?1b
= klim
(
h
!1 2 p + 1
2 p + 1 0;r
Z
k
=
Z
jrv~j2 + W 0v~2 dy:
14
Thus there exists a constant C1 which does not depend on k and R such that
Z
lim (
R!1
j
vkj2 dy)1=2 C1kjvk j ? v~k1L=2(R ):
fjyjRg
2
n
Now, take an arbitrary small positive constant a so that c0 = W 0 ? ap?1 > 0.
Choose k0 so that kjvkj ? v~k1L=2(R ) a(2C0C1)?1 holds for every k k0.
Then we can take a suciently large R1 so that
(
j
vkj2 dy)1=2 2Ca
fjyjR g
0
for k k0. Now we can take a sucientlly large R0 R1 such that
( fjyjR0 g jvkj2 dy)1=2 2Ca
0
for k 1. Combining with the subsolution estimate (18), we get jvk (y)j a
on R0 + 1 jyj for every k.
(Step 2) Let ?0(y) = ?0(y; 0) be a fundamental solution to ? + c0. We
choose ?0 so that jvk(y)j c0?0(y) holds on jyj = R0 . Let w = jvk j ? c0?0.
Then we have by Kato's inequality
w = jvk j ? c0?0
(sign vk )D2 (vk ) ? (c0)2?0
(W 0 ? jvkjp?1)jvk j ? (c0)2?0
c0(jvk j ? c0?0) = c0w:
By the maximum principle, we can conclude w(y) 0 on jyj R0 . pSince it
is well-known that there exists a constant C such that ?0(y) Ce? c jyj on
jyj 1 and we can take c0 to be arbitrary close to W 0, we obtain the desired
estimate. 2
Lemma 6 There exist a constant ! 2 R and a point y0 2 Rn such that
U~ (y) = ei! v~(y) = ei! U (y ? y0). Moreover, U~ is a non-trival least energy
solution for b0;c .
Proof: Since we already know jU~ j = v~ > 0. Let R00 be a xed number.
Then, we have
bh = J (uh )
n
2
Z
1
Z
0
k
k
15
1 j(r ? iA(x + h y))v (y)j2 + W (x + h y)jv (y)j2 dy ? 1
p+1
=
k
k
k
k
k
k
2
p + 1 jvk j dy
hnk 12 fjyjR00g j(r ? iA(xk + hk y))vk (y)j2 + W 0jvk (y)j2 dy ? p +1 1 R jvk jp+1 dy
hnk 12 fjyjR00g jrU~ (y)j2 + W 0jU~ (y)j2 dy
? p +1 1 fjyjR00 g jU~ (y)jp+1 dy ? p +1 1 fjyjR00 g jvk (y)jp+1 dy + o(1)
hnk 12 fjyjR00g jrU~ (y)j2 + W 0jU~ (y)j2 dy
? p +1 1 fjyjR00 g jU~ (y)jp+1 dy ? p +1 1 fjyjR00 g jvk (y)jp+1 dy + o(1) ;
where W 0 = W (x0) for some x0 2 W . For any > 0, by Lemma 5 we can
take large R00 enough to assure
1
p+1
p + 1 fjyjR00g jvk (y)j dy < for every k. By letting k ! 1, we have
1
~ (y)j2 + W 0jU~ (y)j2 dy ? 1
~ p+1
jr
U
2 fjyjR00g
p + 1 fjyjR00g jU (y)j dy ? b0;r:
Taking the limit of R00 ! 1, we obtain J0(U~ ) b0;r. On the other hand,
we know J0(jU~ j) = J0(~v) = b0;r. This implies J0(U~ ) = b0;r and jrjU~ j(x)j jrU~ (x)j a.e. x. Hence, we obtain the desired estimate. The last statement
in Lemma 6 follows from Lemma 7. 2
Although we do not need the following lemma in our proof, it is interesting. We do not know whether Lemma 7 is also true for general nonlinearity
f (t) or not.
Lemma 7 The equality b0;c = b0;r holds, and the least energy solution Uc for
b0;c has the form ei! U (y ? y0) for some point y0 and a constant ! 2 R.
Proof of Theorem 2: It is easy to show max jvk j > (W 0 )1=(p?1) by the
maximum principle. We already know that the part (a) of Theorem 2 holds
for some xk converges to xo 2 W . Because of this uniform decay estimate,
we may take R0 large enough to assure
max ju (x)j = fjmax
jv (y)j > (W 0)1=(p?1):
fjx?x jh R0 g h
yjR0 g k
hnk
Z
Z
Z
Z
Z
n
Z
Z
Z
Z
Z
Z
Z
Z
k
k
k
16
This and Lemma 5 give the part (a) except the last statement. The part (b)
is a consequence of Lemma 4 and Lemma 6. We prove the part (c) and the
last statement of (a) in Theorem 2. We may take x0k 2 B (xk; hk R0 ) as the
maximum point for juh j. Since xk ! x0, x0k ! x0 as k ! 1. Hence we
obtain
p ?j ? 0j
?
juh (x)j Ce
on fjx ? x0k j hk R0g by replaceing the constant R0 and C if necessary. Since
A y in C 2 (Rn ) and jv (y )j = v~(y ) > 0,
vk (y) converges v(y) = v~(y)ei!ei
loc
if maxx2 jA(x)j is small enough to assure supjyjR0 jA(x0) yj < 2 we
may assume that vk (y) can be written vk (y) = jvkj(y)e?i! (y) on jyj R0
and that satises j!k (y) ? !0j < 2 for some !0. Hence jvkj ! v~ and
!k (y) ! ! + nj=1 A0j yj in C 2(B (O; R0)). Then the argument by using wellknown properties of U as in [Wa] yields the uniqueness of the maximum point
x0k of juh j and y0 = O. However, y0 must be the origin without the smallness
assumption on maxx2 jA(x)j, because we can emply the same argument as
above in a region jyj R for suciently small R. 2
Proof of Lemma 7: We dene
jruj2 + W 0juj2 dx ;
c = u2H inf
(R ;C) ( jujp+1 dx)2=(p+1)
jruj2 + W 0juj2 dx :
r = u2H inf
(R ;R) ( jujp+1 dx)2=(p+1)
Since jrjuj(x)j jru(x)j a.e. x, we have c r , and the opposite inequality is easy. Hence, c = r holds. Similaly, b0;c b0;r is easy since
M0;c M0;r . Let Uc and ur be the least energy solution associated to b0;c
and b0;r, respectively. Suppose b0;c < b0;r. Then we have
k
W0 x x
k
hk
k
Pn
0
j j
j =1
W
k
P
k
W
R
1
R
n
R
1
Z
c =
jUc
R
Z
jUc jp+1 dx < jur jp+1 dx:
Thus
Z
n
p?1)=(p+1)
(
jp+1 dx
<
Z
jur
jp+1 dx
p?1)=(p+1)
(
= r
by the uniqueness to the real-valued case. This is a contradiction and hence
b0;c = b0;r holds. This means that for the minimizer Uc we have jrjUcj(x)j =
17
jrUc(x)j a.e. x and this implies that there exists a constant ! 2 R such
that Uc (x) = ei! ur (x) (see, e.g., [LT, p.103]). Since ur (x) can be writen by
ur (x) = U (x ? y0) for some y0, we complete the proof. 2
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ACKNOWLEDGEMENT
This paper was writen while the author was visiting the Erwin Scgrodinger
International Institute for Mathematical Physics in Vienna. The author
wishes to thank Professor T. Homann-Ostenhof for his invitation and the
members of the Schrodinger institute for their hospitality. This work is
partially supported by the Erwin Schrodinger institute for Mathematical
Physics.
19