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A MINIMIZATION PROBLEM ON RN WITH NONZERO DATA
1995 .1S
.e.at
A MINIMIZATION PROBLEM ON R N . WITH NONZERO DATAYan Shusen("'~A) Dept. of Appl. Math. , South China University of T'echnology sGuangzhou 510641 -China.
Zhang Zhengjie(*.iE..~) Wuhan Inst: of Math; Sci. , Academia Sinica ,Wuhan 430071 -China.
We first establish a general framework to study a minimization problem on R N with
Abstract
nonzero data. Then, we use this framework to solve the eigenvalue problem associated with this minimization problem. And finally, we check the crucial strict inequality S 1 in the general framework for a typical case.
Key words
1
Concentration compactness, Unbounded domain.
Introduction
~ 3 and 1 < q < p <
Let N
2*
=
N
2N
2' In this paper, we consider the following
minimization problem
I).
=
inf{I(u)
IIDu I E
V(R N )
,
U
where
E LQ+1(RN )
-f
,
f~ lu +
9'11'+1
=
0,1)
il},
I(u)=lf IDuI 2 + q+lR -1 luI + 1 , 2R N
(1.2)
Q
N
and q; E LI'+l(R N ) satisfies certain conditions which will be stated later. The minimization problem at infinity associated with (1. 1) is the following
1';:
= inf{I(u) IIDu l E V(R N )
It is not difficult to check that
Ir satisfies I;;'
< Ole;
,
u E LQ+1(RN )
,
L.v 1u l
1'+ 1
for 0 < 1.
Thus the concentration compactness principle of P. L. Lions [4J tells us that tained for each A > -
0.3)
= il},
o.
Received Mar. 17,1992. Supported by Youth Fundanon- NSFC.
Ie;
is at-
Supp.
Van
s, Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
17
Basing on the idea of Brezis and Nirenberg [lJ, we have studied the case q = 1 in paper [5J. The basic inequality used in [lJ is
LI Dvl 2~S(LlvI2·)2/2·,
v
E H~({2),
0.4)
while the counterpart of (1.4) for (1.1) in the case q = 1 is
f RN IDulz+u2~I(f RN luI where 1= inf{f
RN
+ uzju E
IDui z
H 1(RN
),f
RN
P+ 1 ) 21 (P+ l> ,
lul P+ 1
=
(1.5)
I}. Since we don't have such
simple relation for (1. 3), we have to develop other methods in order to solve problem (1.1). The aim of this paper is to show that the concentration compactness method fits
quite well to solve problem (1. 1) and we have the following result:
Theorem 1. 1 Suppose that 1 < q < min(p, N N 2) and rp(x) Co
N O,max(-, p N - 2)
>
~ 1
Co Ixl- l Then
I~
<
If there are C1 ~
2 - 1 ' such that q .
~ cp(x) ~ Clxl-
is achieved for each it >
~ o.
for Ixllarge .
l
(1.6)
o.
NN 2' we have --L > max(N, N - 2). q - 1 P This paper is arranged as follows. In Section 2, we develop a general framework Remark 1.1
Since q
<
for the minimization problem with nonzero data. In Section 3, we apply the result of Section 2 to study the related eigenvalue problem. And finally in Section 4, we first get the decay estimate for the solution of the corresponding Euler-Lagrange equation of (1.
1), and then use this decay estimate to check the crucial strict inequality 51 and thus complete the proof of Theorem 1.1.
2
General Framework Let M be the space consisting all the functions
Ft.u)
=
u with IDu I E L 2 (R N )
, U
E Lq(R N )
,
f:f(s)ds. In this Section, we develop a general framework to study the follow-
ing minimization problem
I.
=
inf{l(u) lu E M, L.vF(u
The minimization problem at infinity is
r; =
inf{I(u) lu
E M,
f
RN
+ ffJ)
(2.1)
= il}.
(2.2)
F(u) = it}.
Suppose that !(t) E C(R 1 ) satisfies ao It IP+l ~ !(t)t ~ alit IP+1
Let Ao = fR'iF(ffJ)
> o.
The case 0
for some at
~ ao ~
< ,l ~ Ao is quite simple.
o.
(2.3)
We have the following
18
Vol. 15
ACTA MATHEMATICA SCIENTIA
result. Lemma 2. 1 Proof
If 0 < A ~ Ao
then 1). is attained.
,
We first prove that the set A). = {u
-E
I
M,
RN
F(u
+ rp) =
(2.4)
A},
is not empty for 0 < A < Ao• In fact, since I C~ (R
N
) ,
RN
rp + rp) = 0, we have that for every e > 0, there exists
F( -
u
E
such that
L.vF(- u + rp) < e
(2.5)
I F(- tu + rp) satisfies G
Hence, the continuous function G (t) =
RN
(0) = Ao
> IF(- u +
rp) = G(l). Thus, for each A E [£,Ao ) , there is t E (0,1) such that
I
RN
So, - tu E A)..
F(- tu
+ rp)
=
ceo
=
A.
(2.6)
Let us point out that A o contains only one element - rp. Thus, if rp EM, then A o is nonempty,
Now we show that 1;. is attained for each A E (O,Jlo].
=
It is obvious that u
0 attains 1).0 =
o.
Let A E (0, Ao ) , and {u".} be a minimizing se-
quence of (2. 1). Then {u",} is bounded in ."'11, we assume that t.here is u EM, such that Diu",
We assert that
I
RN
u; F(u
Indeed, if LNF(U
+ rp)
-+
strongly in Lfoc(R
u
+ rp) < A, then u:f: 0 and RN
Consequently 2
= 2t
I
Remark 2. 1
I RN as
a
-+
I
RN
F(u
(2.7)
,
(2.8)
) .
q
IDu
+ rp)
F(tu
1
I
1 + q t ++ 1 2
there is t E [0,1), such that
=
lu Iq+1
+
(2. 9)
A.
< 1(u)::;;;' ..!!~J(u,,) =
I.,
+ rp) ~
F (u".
which is a contradiction. On the other hand, since I
A, we conclude that
N
= Aand thus u attains 1). •
I
I;.::;;;' 1(tu)
weakly in L 2(R N )
Diu
-+
R
N
F (u
+ rp)
rp) = A.
Since
F(au+rp)~af)I ltV lau+rpIP+l~aloaP+lI lu IP+
1
RN
co , it is easy to see A). is nonempty for A E
(Jlo,
- c I'
+ co) .
In the following, we concentrate our attention to the case A > Lemma 2.2
lim I
.~ RN
1). is continuous on A E [Jlo,
+ 00).
RN
IrplJ>+I-=
Jlo.
=
Supp.
19
Yan &. Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
We first prove that I;. is nondecreasing on [tlo, + 00). Let A~
Proof
each e > 0, there is u E A).=, such that
> Al ~ Ao•
< I;. z + E.
ICu)
For
C2.10)
Besides, there is t E [0,1) , such that tu E A;.1 . Hence
I;.
1
~
ICtu)
Next, we claim that For every
E
>0,
All
c; CR
N
) ,
V
a
< I;, + E. Z
E
C2.11)
[~,AJ.
be such that
< III + E.
ICui) Choose u~ E
I(u)
+ tz.:
I;. ~ III let u~ E
<
C2. 12)
such that
I
RN
FCu;
+
cp) = a
+ p(e),
C2. 13)
and
C2. 14) where p(E) --. 0 as e --. Fix
a > o.
o.
Similarly, we can choose
=
LNF(uD
I Cu;)
u; E C: CR N ) , such that
A- a
<
+ 8 + p(e),
I~~a+o
-+-
(2. 15) C2.16)
€.
Since t: is translation invariant, we assume supp U C { Ix C
ly large, such that supp
u; cc { Ia: I ~ R o } . Thus
~F(u~ + u; + cp)
I R' =
=
I
Izl~RQ
FCu~ + cp) + I
LvF(u; + cp) -lZI~RoF(cp)
=A
+ a + pee)
I ~ R o } with s, sufficient-
- I
1.II~Ro
Izl~RQ
FCu; + cp)
+ L.vF(u D + IIZI~Ro (F(u; + cp)
FCcp)
+I
-
F(U2))
FCu; + cp)-- FCu;).
C2.17)
!zl;;;:R o
Since
II II
Izl~Ro
1.rI~Ro
FCu; + cp) -
FCCP)I~CI
F(u~) I =I
Izl~RI)
Izl~Ro
IcpIP+l=e(R o )
IfCu~ +
C2.18)
,
8q»q>1
~CIIZI;;'Ro lui IP Icpl + ClZI~Ro IcplP+! ~C(I.
Ixl~Ro
lu; IP+l)l-p~l
cI
:.rI;;;:Ro
IcpIP+l)p~l
+ E(Ro)
~ccI RoYFCu;))p:h(I Ixl ?;R) IcpIP+l)p~l + ECR o) =£'(R o )
,
(2.19)
20 where
ACTA MATHEMATICA SCIENTIA E'
(R o ) - . 0 as R;
-.+
Vol. 15
00.
Combining (2.17)-(2.19) we get
LNF(U; + u; +
+ ~ + p.(e) + e (R o ) . (2.20) Choose e > 0 small and R, > 0 large, such that A + a + p(E) + e' (R o ) > A. -Then I;. ~ 1;'+~+p(C)+C(Ro) ~ l(u~ + u;) = l(u~) + leu;) < i. + tz:«, + 2E. = A
Letting e -. 0 , and then a-. 0 in the above inequality, by the continuity of see that (2. 11) holds. Now we prove I). is continuous. For any A1 E
[Ao, A1J,
we see I). ~ lim 1
cr-A
i..
+
from I;. 1 ~ L,
00),
+ I~ -a,a E 1
Since I;. is nondecreasing , we have I). ~ lim i: Thus I). =
1-O
lim L. Besides, from I). ~ I;.
cr-,41-0
[tlo,
I: , we
1
+ l~_;.
1
1
a-;'1- 0
1
, V A> A1 , we get lim 1). ~ I). , which gives lim I). ;'-).1 +0
1
_ ;'-).1 +0
= I;. • 1
The framework to solve (2. 1) is the following.
Theorem 2. 1 Suppose that
qJ E L,+1 (R
8 1 :1,4 < t,
N
+ ti..,
)
and A > Ao
V
a
,
if
E [Ao ,A),
then every minimizing sequence of (2.1) is relatively compact in M. In particular, I;. is attained. If S1 holds, we have
Remark 2. 2
I).
< t, + tz..
V a E [O,A).
In fact, since t, ~ 0 and I: is nondecreasing , we have I;.
< tz.; ~ t, + tz;
(2. 21)
V a E [0,
AoJ. Proof of Theorem 2. 1
Let {u",} be a minimizing sequence of (1. 1). Clearly,
{ IDu,..1 } and {u,..} are bounded in LZ(R N ) and Lq(R N ) respectively. Hence, we may assume
Diu,.. -. Diu U,.. - .
u
weakly in L Z (R N ) strongly in Lroc (R
N
(2.22)
,
(2.23)
) .
Set (2.24) Then
(2. 25) (i) Vanishing does not occur. Suppose sup
yE R N
J
BR(y)
e: (z ) -. 0 as m -+ +
00
for fixed R
> o.
Then by Lemma I. 1 of P.
L. Lions [4J, we have u; --. 0
and consequently,
strongly in L"(R N )
,
q
2* ,
Supp.
Yan &. Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
21
which is a contradiction. Hence, vanishing does not occur.
Suppose that dichotomy occurs. Then there is a E (O,k) such that for any there are R,
>
0, R". -++ oo,y", E R N
II II
,
£
m o > 0, such that for m ~ m o
al < e;
p",(x) -
Iz-y",I~Ro
(k -
p",(x) -
a)
Iz-y",I~2R",
>
0,
(2.26)
I<
e,
(2.27) (2. 28)
Let ~,C/J E C:(R N ) be such that ~
=
1,ep =
ofor
[z ] ~ 1;~
=
O,ep
=
1 for
Ixl
~ 2.
Let
(!J
0) _
u;
-
<;
l
x -
R
y",
II
:
o
(2)
U".,
U".
=
11,( x Y'
y",)" U"..
(2.29)
R".
(2. 30) where p.(£)
-+ 0, as £-+ o.
Similarly (2. 31) Consequently II(u".) -
~CI
-
I(u~l»
Ro~lz-y".I~2R".
I(u~2»)
(IDu m 1 2
I
+ u~ ID~(x -
s, Y"') 1 + u~ ID¢(X -R",Y"') 1 + u:.) 2
2
(2.32)
~p.(£).
Suppose that {y".} is bounded in R IL.v[F(u..
+ rp)
-
1
2,
Then
+ rp) - F(u~»)JI
rp) - F(u..)
IF(u".
Ro~ Iz- y".1 ~2R",
•
F(u~J)
~II%-Y.I;'ZR. (IF(u.. +
+I 6.J + J
N
+
cp) -
I+
F(rp))
F(u~l> +
cp) -
F(U~2») 1 (2.33)
22
ACTA MATHEMATICA SCIENTIA
IJII
= lZ-Y*';;'ZR* (If(u.. + 8/p)/p1
~! f'Z-Y*';;"ZR* I/pI
p+ I )
P~I
(
Vol. 15
+ F(/p»
L)
U ..
I
H I
+ I/pI
HI )
Ph + lZ-Y*';;'ZR,. F(/p) (2.34)
~JLw, ,
where
p.".
-++
00
as m
-++
00.
(2.35) Thus A=
LNF(u~1l + /p) + LNF(u~2» + p(E) + /4..
+
Suppose that, as m .....
(2. 36)
00
LNF(u~1l + /p) -
AI (E),
LNF(u~Z»
(2.37)
- A2( E) ,
and as e -+ 0, (2. 38) Then from (3. 36) ,
(2. 39) If Al
=
0, then
which implies a contradiction. Thus Al
>
0.
Suppose Az = 0. We first claim that
LN IDu~2) 1 + IU~2) Iq;;: ~ > o. 2
(2.40)
In fact, since
by (2. 27) (k -
a) - e~f Ir-y",I~2R", p".(x) ~f. (IDu~2)12 + IU~2)lq) + C(f IDu~~)12)2·/2, R
N
R
N
which implies (2.40). From (2. 32), we get I(u".) = I(u~l»
So
+ I(u~2») +
;.t(e) ~ I;.1 (.)
+ /-l". + 8 + 1
/-l(e).
Supp.
Yan &. Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
I;.
>
which is a contradiction. Thus A~
~I;.
23
+ s.,
0 , and consequently 0
<
At, A2
<
A. Again by (2.
32)
Thus which contradicts to (2.21). Similarly, if Ym -.
+
00,
we can also get a contradiction. So dichotomy does not oc-
cur. Thus we have {p".(.+ y".)} is tight. We claim that {y".} is bounded, and consequently {p". (z ) } is tight. Suppose Ym -.
+
00.
Since {Pm ( •
+ y".) } is tight, for any e> 0, there is R >
0 , such
that
f
p".(x)
IX- Ylll
l;;-:R
< e.
(2.41)
On the other hand A= f
R
N
rc«: + cp) =
f
IX-YlIIl~R
rc«: + cp) + f IX-Yllll~R r«; + '1'),
(2.42)
Ilz-YM1";R (F(u.. + q» - F(u... » I ~ f'Z-YM1";R If(u.. + Oq»q>1
~ (Lv(I u ..
1 p+1
+ Iq>1 HI)\ pfr ( f,z-yM,;,) q>1 HI) ) ~ ~ P.. ,
(2.43)
If'Z-YM,;'R (F(u.. + q» - F(q») I ~ f,z_yM,;,)u ..f(Ou .. + q» I
~ c( f'Z-YM1;'R lu.. IP+I») ~ ~ (f'Z-YM I;') u ..
o 1q )
q(
1-0
fIZ-Y I;'R I u .. 1
2 • ) -;;-
M
~ p(e).
(2.44)
Combining (2.41)-(2.44), we get ). = LNF(U..)
+ LNF(q» + P.. + p(e).
(2.45)
Then l(u",) ~ IJoo _-11(. ROY-
,..
)
=
Ir'-;.0
+ Pm + p(e),
which gives L, ~ 1~;'Q ' a contradiction. Hence {p", (z ) } is tight, and consequently, u; --. u
strongly in LP. So
A= Remark 2. 3
f
RN
F (u".
+
'1') -.
f
RN
F (u
+ '1').
We can study the following minimization problem without any change
24
ACTA MATHEMATICA SCIENTIA
in the presentation:
If
inf f IDu I'" lmR N U
2<
3
+ q+1R - l - f Iu Iq+ II Du I E
Lq-I(R N )
I
f,rVF(u +
,
L'"(R N )
1
N
< q + 1< p + 1<
m
NN::: m'
r:p)
Coltl H
= I
,
A},
:;;;;;' !(t)t:;;;;;'
Ctltl H
I
•
Eigenvalue Problem
o.
Let rp~ 0 but rp~ p
E
Vol. 15
E R+,
U
EM,
>
u
We are concerned with the following eigenvalue problem: find
0, such that - 6,u
+u
=
q
pf(u
+ rp).
(3.1)
We know that if I;. is attained by some u EM, u must satisfy (3. 1) for some p E R+. By Lemma 2. 1, I;.is attained if 0< A~ Ao• Unfortunately, the minimizer of 1;. in this case is nonpositive. Indeed, if A = a minimizer of I;. with u+ =1= A= f
>
f
RN
Ao , then u =
°, then
F(u
+ rp) =
F(rp)
_>0
f
°is the unique minimizer of 1;'0. If
A<
Ao,
u is
rc« + rp) + f -~o F(u + rp)
_>0
+ f rc« + rp) = f _~o
R
N
F ( - u-+ rp),
(3.2)
whereu-= max(- U,O). Thus, there ist E (0,1) such that
f
RN
F(- tu-+ rp)
=
A.
(3.3)
Hence I
A :;;;;;'
I( -
tu-) = t 2
~ L)Du- 1 + /;11f,rV lu2
Iq+1 < I(u) =
IH
which is a contradiction. Hence, to find a positive solution of (3.1) we have to show that there is at least one A >
Ao , such that 1;. is attained. In Section 2, we get a general
framework to solve the minimization problem (2. 1). But generally speaking 51 is quite difficult to check. So Theorem 2. 1 does not give us any positive result unless we can check 51. As to the eigenvalue problem, one need only to show that there exists a A > Ao , such that I;. is attained. Hence, if we can show that there exists a A> Ao holds with respect to this
u
>
such that 51
then (3.1) is solved.
Suppose that rp~ 0 but rp~ O,f(t) satisfies (2.3) and F(u
Theorem 3.1
F(u)
A> Ao under some simple conditions,
,
+ Ft;u) for u > 0 and v > o.
+ v) >
Then there exists at least one pair (p,u) with p> 0,
0, satisfying (3.1).
Remark 3. 1
+ F (v) for u >
If 'F' (t) 0, v
>
=
f (t) is strictly inceasing in t
0.
Proof of Theorem 3. 1
First we claim that 1).
>
0 , Then F (u
< Ir-;.o for each A >
+ v) > F (zz')
Ao. Let v be the
Supp.
25
Yan &,Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
positive minimizer of Ir'-Af). Since
LNF(V
+ rp) > LNF(V) + LvF(rp) = A,
there is t E (0,1), such that LNF(tV
+ rp) =
I;. ~ I(tv)
Fix X> Ao• Set
=
Al
sup{AII~
>
Then by (3. 4), we have Al
<
t,
< I(v) = tz;
11 < t,
I;.
=
(3.4)
o
+ If-a,
Ao and I
A. Then.
I A1
+ Ii-;.
+ tc..
V a E [Ao,A]}
1
,
(3. 5)
V a E [Ao,A1 ) .
(3.6)
We now show that for this Al ,51 holds. Indeed, since I: ~ I:- fJ (3.5) and (3.6), we get
I;. 1 - 11
= If-;. < I, + Ir.-a - If-;. < I, 1
1
Thus, by Theorem 2. 1, I;'I is attained by some have u
~
o.
+ l~_« 1
+ I p , V f3 E
V a E- [Ao,A1 ) .
E lvI. Obviously, since Al
u
[O,a], by
>
Ao
,
we
If u- ~ 0, then Al
=
L.vF(u
+ rp) < Lv F( IuI + rp).
(3.7)
Hence, there is t E (0,1), such that
L.vF(t Iu But 1;.1
~ l(t
luI) <
I(u)
=
I + rp)
(3. 8)
= AI.
I;.1 , so we get a contradiction. Thus, u ~
On the other hand, since
u ~
°and
- 6u
+ u" =
p./(u
o.
+ cp),
(3.9)
we get (3. 10)
and by the strong maximum principle, we get u
> o.
4 Proof of Theorem 1.1 First, we prove a lemma which is useful in the later arguement. Lemma 4. 1
of a E R
N
,
J
+ h > N + L, then there is C> 0, independent
such that
RN
Proof
If k > L, h > Land k
(1+
Ixl)-"(l+ Ix-al)-'~Clal-L(l+O\)
for some 01 > 0.
26
ACTA MATHEMATICA SCIENTIA
+f
=(f rl _Ial I
7T
+
Ixl)-"(l
+
Ixl)-"+o+cJ
laj)(l
Irl~T
::(Clal-o+ctl)Lf
Irl~I~1
(1
+ Clal-o+ctl)Lf Irl~I;1 (1 +
+
Ix - al)-!
1)L(1
Ixl)-"(l
Vol. 15
+
Ix - al)-!
+ Ix -
al)-4+':1+ct
1)L
:::;;;C/al-OHt)LCLN(I + Ixj)-HOHt}L(I + Ix - al)-A
+ f RN (1 + We claim that if h, of a , such that
~
Ix 1)-"(1 + Ix - a I )-!+O+ctt)L].
+ k > N.
0 and k 1 ~ 0, hI
l
(4.1)
Then there is C
JI=f RN (1+ Ixl)-"1(1+ Ix-al)-4
>
0 independent
(4.2)
1::(C.
Clearly, (4.2) together with (4.1), implies that the Lemma holds. Proof of (4.2). Set D.
=
{xllxl
Ix Indeed, if Ix
Ix
- a
aI~
I ~ 21 a I, then Ix
I 2~ 1!l 2 2~ Isl 4 •
Hence
JI =(f
+f
lal
IZ'~T
::(f Iz'~ ';1 (1 +
+
Iz-a'~
~ I~I, Ix - al ~ I~I}.
l!1 2
Ixl V x E a; 4'
- aI
+f
First, we claim
DtJ
Ix 1)-"1- 41 +
~ Ix I
)(1
+
Ia I ~ I~ I ;
if I I :::;;; 21 x
Ixl)-"1(1 + Ix -
f Iz-a'~ I;' (1 +
(4.3)
a I, then
al)-4 t
Ix - a I )-"1- 41
f (I + Ixl )-ht(I + Ixl )-At 4
DoJ
:::;;;CLN(I + IXj)-(htHt) <+ 00. Remark 4.1
Using the same mothod , we can prove easily that if k > max(N ,L) ,
then
LN O+ Ixl)-L(I+ Ix-aj)-A:::;;;Clal- L.
(4.4)
Lemma 4. 2 Let u E M be the nonnegative solution of -
Suppose that there is C I
>
+u
q
= p(u
N- 2
+ rp)P.
0, 1 > - 2 - ' such that
Then, for any 8> 0, there is C' u(x)
L,u
>
I9'< x ) I ::( C I (1 + Ix I)
0, such that
rC' (I + Ix j)-Z/(q-DH if P... z ~ 2/(q ~~ q CO + IXj)-J>ljqH Z ~ 2/(q - D,
l
(4.5)
if:
1),
-I.
Supp.
Yan
Proof
s, Zhang:
27
A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
Since -
Lu ~ /-l(u
+ cp)P ~ CuP + Ci9'IP,
by Harnack inequality, we have that there is C
>
sup(y) u(x) ~ C( lu ILz - (B Z ~'/» -
zE 3
0, independent of y, such that
+
19'IL z • (B (v»).
(4.6)
Z -
1
Thus, we have (4.7)
and because luILZ·(Bz(y»
+
-
6,u
s, > 0, such that for Ix I ;:: s.,
+ (1 -
Set v = CO I~ 1-"', then there is a -
6,v
+ (1 -
e)v Cl = Co(m(N -
q
>
e)u ~ Cip " ~ Clxl-
0, such that
2- m)
Ixl-
(for Ixllarge, if Case 1. If
(4.8)
00,
y--.+oo.
as
19'ILZ·(Bz(Y»~O
Hence, for any e > 0, there is
Ix I --.+
as
u~0
m
m
-
Z
+ (1 -
pl
(4.9)
•
e)
Ixl-·
q
)
;::
a
!.rl-
mq
,
(4.10)
kq z;:: q ~1' then for any B"> 0, choosing m = --L - ~, by (4.10), we q1
get -
6,v
+ (1 -
e)v
Consequently -
u)
t6,(u -
q = alxl-~+8q
+ (1 -
e)(u
q
> Clal-
(for [z ] ;:: R 1 )
o") ~ 0
-
(for Ixllarge).
Pl
.
(4.11)
(4. 12)
Choose Co large enough such that
u~Colxl-(q':1-8) =v, for Ixl =R 1 • Since u - v
~
0 as Ix
I ~+
00,
by the maximum principle, (4. 13) implies u~v
for Ixl ~Rl.
< q--L ' then choosing m = E-Z - ~, - 1 q t6,v + (1 - e)v ~ a Ix l- l+q8 > C Ix
Case 2. If.P...Z q -
(4.13)
q
p
(4.14)
by (4. 10), we get
I-Pl
for Ix I large.
(4.15)
Thus, similar to the Case 1, we get u ~ v for Ia: I large. Remark 4. 2
If q
>
N N 2' then it is not difficult to prove that the solution
u
of
(4. 5) satisfies u(x)
~Colxl-(N-2+8),
8> 0,
no matter how fast the function 9' decays at infinity. Let V E 1\1 be a nonnegative solution of
- LV +
v = q
pV P
x ERN.
As a direct application of Lemma 4. 2, we know that Vex) ~ C(l
+
Ix I )-q':1+8.
(4.16)
Before we prove Theorem 1. 1, let us recall a Lemma of Brezis and Nirenbergjl ].
ACTA MATHEMATICA SCIENTIA
28
Lemma 4.3
a,p r C Ia 11,81 q- 1 if Ia I ~ IPI, 1,81"-Z) I ~ ~ LClal 1 lPI if lal ~ IPI.
Assume 1 ~ q ~ 3. Then there is C such that for all
Iia + Plq - lal q - IPlq - qap( lal q- Z + When q
Vol. 15
q
-
3, we have
~
Iia + ,Blq - lal q - 1,8l q - qap( lal q- Z+ IPlq-Z) I ~ C( lal q- zl,8l:= Proof of Theorem 1. 1
J.1~IP+l.
+ laI
As we have pointed out, I;. is achieved for
21,8l
°<
q -
Z
) .
A ~ Ao
=
We only need to prove that Ilis achieved forA>A o• Now we check that I;. sat-
RN
isfies S 1. Thus, by Theorem 2. 1, I;. is achieved. Set a = SUp{Al II;. < IfJ
+ Ir'-fJ'V P E
Ao <
}.
By Theorem 3.1, we have
(4.17)
a~;\,
< t, + tz.; I;. = I. + tz.:
V P E [Ao,a),
I;.
I.
[Ao,A1 )
(4.18) (4.19)
is attained by some
uE
M,
u;;::
0.
Now we show a = A. We argue by contradiction. Suppose a tive function which attains Ir':-•• Set Va(x) = Vex - a) ,a E R each a E R
N
N
•
(4.20)
<;\.
Let V be the posi-
It is easy to see that for
there is C; E (0,1) , such that
,
J~ lu + 9'+ C.V.IJ>+1
(4.2D
= A.
On the other hand, since ;\1/(1'+1>
~
lu +
+ Ca IVa IL + 1(R N) =
9'ILP+1(Kv)
P
a1/(p+l>
+ Ca(" -
a)I/(P+1>,
we get
c. ~ Co > 0, Set w = A=
u
+
9'.
L)W I
Now we prove as
<
+ C:+
1L,vV:+l
+ (p + DL,v(w/>C.V. + w(C.V.)/»
laI -.+ 00 IRal =o(lalCJ
+ a)l > l , 1 < q - 2, -l
rna 4.1 we get
+ S. (4.23)
l
ISal =o(lal-
) ,
wV: W~V4
~CJ
2a - q--1 ~ N
-:.... R.
l
(4.24)
) .
P ~ 2. IRa 1 ~
. Since (1
(4.22)
Applying Lemma 4. 3, we get
J>+ 1
Case 1, 1
V a ERN.
- q) + 2(p q- 1 -
RN
(1
~ CJ NWl+·V~:-· R
+ I·_!'I)-lo+«)(l + Ix -
2 2q (p - a) ( - - - a) = - q-l q-l
2c:.
al)-(P-·)(q:t- c)') .
- q) + 2(p q-l
a(p - a)
a(p - a) - q--1 > N 1£ a,a small enough, applying Lem-
Supp.
Yan
s, Zhang:
A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
29
Similarly,
Case 2, P > 2
IR.I ~ CL.v W1'-1V; ~ CL.v (l clearly, I(p - 1)
> 1.
~ 4,
then
(i) If N
--.i- >
+
u+
Ix 1)-1<1'-0
Ix - a i)-Z
2(N - 2) ~ N and consequently
q- 1 is small enough. By Lemma 4. 1, we get
IRa 1 = (ii) If N = 3, then 1 ~ N -
. 4
q _ 1
+ (p -
1)[
=
2
- 28> N, if 8
I-I).
1, and
4
=q
0 ( Ia
--.i-
q- 1
_ 1
+ (p -
>---.L+ q- 1
(q -
2)[ 2)
+l
+ 1 ~ 3 + 1 = N + 1,
since
_4_
q-1
+ (q
_ 2) - 3 = _1_[q2 q-1
6q
+ 9J ~ o.
Applying Lemma 4. 1, we get
IRal
=
o(lal- 1 )
+
Ixi)-Zl(l
+ (q -
1) q~1
15.1 ~Cf~WZV~-1 ~CLN(l Since 21
+ (p -
1) q~1
>N
- 2
,
+
Ix - ai)-<1'-IH q": I -
8>.
+ 1 = N + 1, applying Lemma 4.
1, we get
ISal = o(lal-
1
) .
Hence we have proved (4. 24). By (4.23)
+ C~+I(A -
a)
+
~a + C~+I(A -
a)
+ (p + 1)LN(W(C.V.}/,) + o(lal-
A =a
(p
+ 1)LN(WC.V. + w(C.v.)1') + o(lal1
1
)
) .
Hence
C~+l ~ 1 - ~ + ;L.vw(C.V.)1') + Now we deal with L.v lu
f RN lu + c.v, Iq+l.
o( la I-I).
(4.25)
We have
+ C.V.lq+l = f~ lu Iq+l + ~+lfR'vV~+l
+ (q + 1) f~(u(C.V.)q + uqC.V.) + R~ + S~. By Lemma 4.3, we have
(4. 26)
30
ACTA MATHEMATICA SCIENTIA
f UhV~,
IR~ I, IS~ I ~ By Lemma 4. 2,
f
RN
with h
>
l,k
>
Vol. 15
1 and h
+ k = q + 1.
jf NO + Ixl)-(q':1-8lhO+ Ix-al)-(q':l-m
UhV!a ""'.-::::: ~ v
l'f
R'
R
+
(1
~
Ixl)-(~I-et)It(1
~ Z < --L , we
Since max (N - 2, N)
P
q- 1
+ Ix -
al)-
(4.27)
if :l~q 2 if.£.Z q
<
1,
_2_. q-l
have if.£.Z ~ --L , then --L1h q q- 1 q-
+ q--L1k =
2(q+l) =~+_2_>N+Z;ifP..Z<_2_,thenJ!...Zh+_2_k>J!...Z( + 1) q-l q-l q-l q q-l q q-l q q
= pI +
+ Z.
.£.Z> N q
f
RN
Hence
By Lemma 4.1, we get
J lu + cavalq+l J RN
+q~ ~I;. +
1L
N
tz;
< RN
+ q~ 1(C:+
1
1J 2
IDu I2+CaJRN DuDVa +
1
~v
+k=q+
1.
J~u(C.V.)P + o(lal-
(DuDV a
+ uV:) +
1 2
(4.28)
q+1 + C:+lf vr: RN
1, we have
G:+ V :+1 +
+ CaJ
lul
RN
1 and h
+ (q + 1) L.v u (C.V . ) 9 + o( la I-I).
1J 2
=
=
.
So by (4. 29) and C,
I).~I(u+CaVa)
>
oClal- /), h "> l,k
=
UhV:
(C; -
l
~v
(4. 29)
C;IDVaI 2
+ q +IIJ
~v
u q+ 1
)
I)J
~v
IDVa1 2
-1)L.vV :+ +o(la\-I). 1
(4.30)
On the other hand, since (4.31) with (4.32) we have CaJ
RN
(DVaDu+V:u)=~(J A-a ~,l-J
RN
A - a RN
CIDVaI2+V:+l))J V V:u R'
(IDValz+V:+l)J N V:u. R
(4.33)
Besides, by (4.25)
C; - 1 ~=-
~J a R A -
_2_J
N
A-aJli
w(C"V a ) 1'
+ o(lal- /)
wV~ + _2_(1 - C~)J N wV~ A-a R
+ o(lal-').
(4.34)
Supp.
Van &,Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
31
But,
(4.35) Hence
(4.36) Similarly
C:+
1
1
=- q
-
+ If
A- a
RN
wV~
+ o( lal-
l
(4.37)
) .
Combining (4. 30), (4.33), (4. 36) and (4. 37), we obtain
I;.
+.,J-f + )f V~u --l-f wv~f IDValz--I-f wv~f
~ItI + tz:
A-aRN
=1;.
+ ti: -
A-a
RN
RN
(IDV a 1 2
V:+ 1
A-aRN
-,---Lf .,(IDVI2 + vQ+l)f A
-
RN
a e"
RN
RN
soV:
V:+ 1
+ o( lal- l)
+ o(lal-
l
) .
(4. 38)
Since
f ~¢V: ~f R"
Iz-al~l
soV: ~Colal-lf
Izl~l
v- = Colal-
l
,
by (4. 38), we get
I;. ~ I;. + tz: -
Co Ia I-l + o( Ia I-l) < 1;. + tz: for Ia I sufficiently large,
which con-
tradicts to (4. 19). Reference 1
Brezis H, Nirenberg L. A minimization problem with critical exponent and nonzero data. To appear.
2
Gilbarg D, Trudinger N S. Elliptic partial differential equations of second order. Second edition, Springer-Verlag, 1983
3 Li Gongbao , Yan Shusen. Eigenvalue problem for quasilinear elliptic equations oti R": Comm , P. D. E. ,1989,14:1291--1314 4
Lions P L. The concentration compactness principle in the calculus of variations. The locally compact case, Ann. Inst. H. Poincare Analyse nonlineaire , Part 1, 1984,1 (2) :109-145 ;Part 2, 1984, 1(4) :223--283
5
Yan Shusen , Zhang Zhengjie. A minimization problem on exterior domain with nonzero data. To appear in Acta Math Sci.
.e.at
A MINIMIZATION PROBLEM ON R N . WITH NONZERO DATAYan Shusen("'~A) Dept. of Appl. Math. , South China University of T'echnology sGuangzhou 510641 -China.
Zhang Zhengjie(*.iE..~) Wuhan Inst: of Math; Sci. , Academia Sinica ,Wuhan 430071 -China.
We first establish a general framework to study a minimization problem on R N with
Abstract
nonzero data. Then, we use this framework to solve the eigenvalue problem associated with this minimization problem. And finally, we check the crucial strict inequality S 1 in the general framework for a typical case.
Key words
1
Concentration compactness, Unbounded domain.
Introduction
~ 3 and 1 < q < p <
Let N
2*
=
N
2N
2' In this paper, we consider the following
minimization problem
I).
=
inf{I(u)
IIDu I E
V(R N )
,
U
where
E LQ+1(RN )
-f
,
f~ lu +
9'11'+1
=
0,1)
il},
I(u)=lf IDuI 2 + q+lR -1 luI + 1 , 2R N
(1.2)
Q
N
and q; E LI'+l(R N ) satisfies certain conditions which will be stated later. The minimization problem at infinity associated with (1. 1) is the following
1';:
= inf{I(u) IIDu l E V(R N )
It is not difficult to check that
Ir satisfies I;;'
< Ole;
,
u E LQ+1(RN )
,
L.v 1u l
1'+ 1
for 0 < 1.
Thus the concentration compactness principle of P. L. Lions [4J tells us that tained for each A > -
0.3)
= il},
o.
Received Mar. 17,1992. Supported by Youth Fundanon- NSFC.
Ie;
is at-
Supp.
Van
s, Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
17
Basing on the idea of Brezis and Nirenberg [lJ, we have studied the case q = 1 in paper [5J. The basic inequality used in [lJ is
LI Dvl 2~S(LlvI2·)2/2·,
v
E H~({2),
0.4)
while the counterpart of (1.4) for (1.1) in the case q = 1 is
f RN IDulz+u2~I(f RN luI where 1= inf{f
RN
+ uzju E
IDui z
H 1(RN
),f
RN
P+ 1 ) 21 (P+ l> ,
lul P+ 1
=
(1.5)
I}. Since we don't have such
simple relation for (1. 3), we have to develop other methods in order to solve problem (1.1). The aim of this paper is to show that the concentration compactness method fits
quite well to solve problem (1. 1) and we have the following result:
Theorem 1. 1 Suppose that 1 < q < min(p, N N 2) and rp(x) Co
N O,max(-, p N - 2)
>
~ 1
Co Ixl- l Then
I~
<
If there are C1 ~
2 - 1 ' such that q .
~ cp(x) ~ Clxl-
is achieved for each it >
~ o.
for Ixllarge .
l
(1.6)
o.
NN 2' we have --L > max(N, N - 2). q - 1 P This paper is arranged as follows. In Section 2, we develop a general framework Remark 1.1
Since q
<
for the minimization problem with nonzero data. In Section 3, we apply the result of Section 2 to study the related eigenvalue problem. And finally in Section 4, we first get the decay estimate for the solution of the corresponding Euler-Lagrange equation of (1.
1), and then use this decay estimate to check the crucial strict inequality 51 and thus complete the proof of Theorem 1.1.
2
General Framework Let M be the space consisting all the functions
Ft.u)
=
u with IDu I E L 2 (R N )
, U
E Lq(R N )
,
f:f(s)ds. In this Section, we develop a general framework to study the follow-
ing minimization problem
I.
=
inf{l(u) lu E M, L.vF(u
The minimization problem at infinity is
r; =
inf{I(u) lu
E M,
f
RN
+ ffJ)
(2.1)
= il}.
(2.2)
F(u) = it}.
Suppose that !(t) E C(R 1 ) satisfies ao It IP+l ~ !(t)t ~ alit IP+1
Let Ao = fR'iF(ffJ)
> o.
The case 0
for some at
~ ao ~
< ,l ~ Ao is quite simple.
o.
(2.3)
We have the following
18
Vol. 15
ACTA MATHEMATICA SCIENTIA
result. Lemma 2. 1 Proof
If 0 < A ~ Ao
then 1). is attained.
,
We first prove that the set A). = {u
-E
I
M,
RN
F(u
+ rp) =
(2.4)
A},
is not empty for 0 < A < Ao• In fact, since I C~ (R
N
) ,
RN
rp + rp) = 0, we have that for every e > 0, there exists
F( -
u
E
such that
L.vF(- u + rp) < e
(2.5)
I F(- tu + rp) satisfies G
Hence, the continuous function G (t) =
RN
(0) = Ao
> IF(- u +
rp) = G(l). Thus, for each A E [£,Ao ) , there is t E (0,1) such that
I
RN
So, - tu E A)..
F(- tu
+ rp)
=
ceo
=
A.
(2.6)
Let us point out that A o contains only one element - rp. Thus, if rp EM, then A o is nonempty,
Now we show that 1;. is attained for each A E (O,Jlo].
=
It is obvious that u
0 attains 1).0 =
o.
Let A E (0, Ao ) , and {u".} be a minimizing se-
quence of (2. 1). Then {u",} is bounded in ."'11, we assume that t.here is u EM, such that Diu",
We assert that
I
RN
u; F(u
Indeed, if LNF(U
+ rp)
-+
strongly in Lfoc(R
u
+ rp) < A, then u:f: 0 and RN
Consequently 2
= 2t
I
Remark 2. 1
I RN as
a
-+
I
RN
F(u
(2.7)
,
(2.8)
) .
q
IDu
+ rp)
F(tu
1
I
1 + q t ++ 1 2
there is t E [0,1), such that
=
lu Iq+1
+
(2. 9)
A.
< 1(u)::;;;' ..!!~J(u,,) =
I.,
+ rp) ~
F (u".
which is a contradiction. On the other hand, since I
A, we conclude that
N
= Aand thus u attains 1). •
I
I;.::;;;' 1(tu)
weakly in L 2(R N )
Diu
-+
R
N
F (u
+ rp)
rp) = A.
Since
F(au+rp)~af)I ltV lau+rpIP+l~aloaP+lI lu IP+
1
RN
co , it is easy to see A). is nonempty for A E
(Jlo,
- c I'
+ co) .
In the following, we concentrate our attention to the case A > Lemma 2.2
lim I
.~ RN
1). is continuous on A E [Jlo,
+ 00).
RN
IrplJ>+I-=
Jlo.
=
Supp.
19
Yan &. Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
We first prove that I;. is nondecreasing on [tlo, + 00). Let A~
Proof
each e > 0, there is u E A).=, such that
> Al ~ Ao•
< I;. z + E.
ICu)
For
C2.10)
Besides, there is t E [0,1) , such that tu E A;.1 . Hence
I;.
1
~
ICtu)
Next, we claim that For every
E
>0,
All
c; CR
N
) ,
V
a
< I;, + E. Z
E
C2.11)
[~,AJ.
be such that
< III + E.
ICui) Choose u~ E
I(u)
+ tz.:
I;. ~ III let u~ E
<
C2. 12)
such that
I
RN
FCu;
+
cp) = a
+ p(e),
C2. 13)
and
C2. 14) where p(E) --. 0 as e --. Fix
a > o.
o.
Similarly, we can choose
=
LNF(uD
I Cu;)
u; E C: CR N ) , such that
A- a
<
+ 8 + p(e),
I~~a+o
-+-
(2. 15) C2.16)
€.
Since t: is translation invariant, we assume supp U C { Ix C
ly large, such that supp
u; cc { Ia: I ~ R o } . Thus
~F(u~ + u; + cp)
I R' =
=
I
Izl~RQ
FCu~ + cp) + I
LvF(u; + cp) -lZI~RoF(cp)
=A
+ a + pee)
I ~ R o } with s, sufficient-
- I
1.II~Ro
Izl~RQ
FCu; + cp)
+ L.vF(u D + IIZI~Ro (F(u; + cp)
FCcp)
+I
-
F(U2))
FCu; + cp)-- FCu;).
C2.17)
!zl;;;:R o
Since
II II
Izl~Ro
1.rI~Ro
FCu; + cp) -
FCCP)I~CI
F(u~) I =I
Izl~RI)
Izl~Ro
IcpIP+l=e(R o )
IfCu~ +
C2.18)
,
8q»q>1
~CIIZI;;'Ro lui IP Icpl + ClZI~Ro IcplP+! ~C(I.
Ixl~Ro
lu; IP+l)l-p~l
cI
:.rI;;;:Ro
IcpIP+l)p~l
+ E(Ro)
~ccI RoYFCu;))p:h(I Ixl ?;R) IcpIP+l)p~l + ECR o) =£'(R o )
,
(2.19)
20 where
ACTA MATHEMATICA SCIENTIA E'
(R o ) - . 0 as R;
-.+
Vol. 15
00.
Combining (2.17)-(2.19) we get
LNF(U; + u; +
+ ~ + p.(e) + e (R o ) . (2.20) Choose e > 0 small and R, > 0 large, such that A + a + p(E) + e' (R o ) > A. -Then I;. ~ 1;'+~+p(C)+C(Ro) ~ l(u~ + u;) = l(u~) + leu;) < i. + tz:«, + 2E. = A
Letting e -. 0 , and then a-. 0 in the above inequality, by the continuity of see that (2. 11) holds. Now we prove I). is continuous. For any A1 E
[Ao, A1J,
we see I). ~ lim 1
cr-A
i..
+
from I;. 1 ~ L,
00),
+ I~ -a,a E 1
Since I;. is nondecreasing , we have I). ~ lim i: Thus I). =
1-O
lim L. Besides, from I). ~ I;.
cr-,41-0
[tlo,
I: , we
1
+ l~_;.
1
1
a-;'1- 0
1
, V A> A1 , we get lim 1). ~ I). , which gives lim I). ;'-).1 +0
1
_ ;'-).1 +0
= I;. • 1
The framework to solve (2. 1) is the following.
Theorem 2. 1 Suppose that
qJ E L,+1 (R
8 1 :1,4 < t,
N
+ ti..,
)
and A > Ao
V
a
,
if
E [Ao ,A),
then every minimizing sequence of (2.1) is relatively compact in M. In particular, I;. is attained. If S1 holds, we have
Remark 2. 2
I).
< t, + tz..
V a E [O,A).
In fact, since t, ~ 0 and I: is nondecreasing , we have I;.
< tz.; ~ t, + tz;
(2. 21)
V a E [0,
AoJ. Proof of Theorem 2. 1
Let {u",} be a minimizing sequence of (1. 1). Clearly,
{ IDu,..1 } and {u,..} are bounded in LZ(R N ) and Lq(R N ) respectively. Hence, we may assume
Diu,.. -. Diu U,.. - .
u
weakly in L Z (R N ) strongly in Lroc (R
N
(2.22)
,
(2.23)
) .
Set (2.24) Then
(2. 25) (i) Vanishing does not occur. Suppose sup
yE R N
J
BR(y)
e: (z ) -. 0 as m -+ +
00
for fixed R
> o.
Then by Lemma I. 1 of P.
L. Lions [4J, we have u; --. 0
and consequently,
strongly in L"(R N )
,
q
2* ,
Supp.
Yan &. Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
21
which is a contradiction. Hence, vanishing does not occur.
Suppose that dichotomy occurs. Then there is a E (O,k) such that for any there are R,
>
0, R". -++ oo,y", E R N
II II
,
£
m o > 0, such that for m ~ m o
al < e;
p",(x) -
Iz-y",I~Ro
(k -
p",(x) -
a)
Iz-y",I~2R",
>
0,
(2.26)
I<
e,
(2.27) (2. 28)
Let ~,C/J E C:(R N ) be such that ~
=
1,ep =
ofor
[z ] ~ 1;~
=
O,ep
=
1 for
Ixl
~ 2.
Let
(!J
0) _
u;
-
<;
l
x -
R
y",
II
:
o
(2)
U".,
U".
=
11,( x Y'
y",)" U"..
(2.29)
R".
(2. 30) where p.(£)
-+ 0, as £-+ o.
Similarly (2. 31) Consequently II(u".) -
~CI
-
I(u~l»
Ro~lz-y".I~2R".
I(u~2»)
(IDu m 1 2
I
+ u~ ID~(x -
s, Y"') 1 + u~ ID¢(X -R",Y"') 1 + u:.) 2
2
(2.32)
~p.(£).
Suppose that {y".} is bounded in R IL.v[F(u..
+ rp)
-
1
2,
Then
+ rp) - F(u~»)JI
rp) - F(u..)
IF(u".
Ro~ Iz- y".1 ~2R",
•
F(u~J)
~II%-Y.I;'ZR. (IF(u.. +
+I 6.J + J
N
+
cp) -
I+
F(rp))
F(u~l> +
cp) -
F(U~2») 1 (2.33)
22
ACTA MATHEMATICA SCIENTIA
IJII
= lZ-Y*';;'ZR* (If(u.. + 8/p)/p1
~! f'Z-Y*';;"ZR* I/pI
p+ I )
P~I
(
Vol. 15
+ F(/p»
L)
U ..
I
H I
+ I/pI
HI )
Ph + lZ-Y*';;'ZR,. F(/p) (2.34)
~JLw, ,
where
p.".
-++
00
as m
-++
00.
(2.35) Thus A=
LNF(u~1l + /p) + LNF(u~2» + p(E) + /4..
+
Suppose that, as m .....
(2. 36)
00
LNF(u~1l + /p) -
AI (E),
LNF(u~Z»
(2.37)
- A2( E) ,
and as e -+ 0, (2. 38) Then from (3. 36) ,
(2. 39) If Al
=
0, then
which implies a contradiction. Thus Al
>
0.
Suppose Az = 0. We first claim that
LN IDu~2) 1 + IU~2) Iq;;: ~ > o. 2
(2.40)
In fact, since
by (2. 27) (k -
a) - e~f Ir-y",I~2R", p".(x) ~f. (IDu~2)12 + IU~2)lq) + C(f IDu~~)12)2·/2, R
N
R
N
which implies (2.40). From (2. 32), we get I(u".) = I(u~l»
So
+ I(u~2») +
;.t(e) ~ I;.1 (.)
+ /-l". + 8 + 1
/-l(e).
Supp.
Yan &. Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
I;.
>
which is a contradiction. Thus A~
~I;.
23
+ s.,
0 , and consequently 0
<
At, A2
<
A. Again by (2.
32)
Thus which contradicts to (2.21). Similarly, if Ym -.
+
00,
we can also get a contradiction. So dichotomy does not oc-
cur. Thus we have {p".(.+ y".)} is tight. We claim that {y".} is bounded, and consequently {p". (z ) } is tight. Suppose Ym -.
+
00.
Since {Pm ( •
+ y".) } is tight, for any e> 0, there is R >
0 , such
that
f
p".(x)
IX- Ylll
l;;-:R
< e.
(2.41)
On the other hand A= f
R
N
rc«: + cp) =
f
IX-YlIIl~R
rc«: + cp) + f IX-Yllll~R r«; + '1'),
(2.42)
Ilz-YM1";R (F(u.. + q» - F(u... » I ~ f'Z-YM1";R If(u.. + Oq»q>1
~ (Lv(I u ..
1 p+1
+ Iq>1 HI)\ pfr ( f,z-yM,;,) q>1 HI) ) ~ ~ P.. ,
(2.43)
If'Z-YM,;'R (F(u.. + q» - F(q») I ~ f,z_yM,;,)u ..f(Ou .. + q» I
~ c( f'Z-YM1;'R lu.. IP+I») ~ ~ (f'Z-YM I;') u ..
o 1q )
q(
1-0
fIZ-Y I;'R I u .. 1
2 • ) -;;-
M
~ p(e).
(2.44)
Combining (2.41)-(2.44), we get ). = LNF(U..)
+ LNF(q» + P.. + p(e).
(2.45)
Then l(u",) ~ IJoo _-11(. ROY-
,..
)
=
Ir'-;.0
+ Pm + p(e),
which gives L, ~ 1~;'Q ' a contradiction. Hence {p", (z ) } is tight, and consequently, u; --. u
strongly in LP. So
A= Remark 2. 3
f
RN
F (u".
+
'1') -.
f
RN
F (u
+ '1').
We can study the following minimization problem without any change
24
ACTA MATHEMATICA SCIENTIA
in the presentation:
If
inf f IDu I'" lmR N U
2<
3
+ q+1R - l - f Iu Iq+ II Du I E
Lq-I(R N )
I
f,rVF(u +
,
L'"(R N )
1
N
< q + 1< p + 1<
m
NN::: m'
r:p)
Coltl H
= I
,
A},
:;;;;;' !(t)t:;;;;;'
Ctltl H
I
•
Eigenvalue Problem
o.
Let rp~ 0 but rp~ p
E
Vol. 15
E R+,
U
EM,
>
u
We are concerned with the following eigenvalue problem: find
0, such that - 6,u
+u
=
q
pf(u
+ rp).
(3.1)
We know that if I;. is attained by some u EM, u must satisfy (3. 1) for some p E R+. By Lemma 2. 1, I;.is attained if 0< A~ Ao• Unfortunately, the minimizer of 1;. in this case is nonpositive. Indeed, if A = a minimizer of I;. with u+ =1= A= f
>
f
RN
Ao , then u =
°, then
F(u
+ rp) =
F(rp)
_>0
f
°is the unique minimizer of 1;'0. If
A<
Ao,
u is
rc« + rp) + f -~o F(u + rp)
_>0
+ f rc« + rp) = f _~o
R
N
F ( - u-+ rp),
(3.2)
whereu-= max(- U,O). Thus, there ist E (0,1) such that
f
RN
F(- tu-+ rp)
=
A.
(3.3)
Hence I
A :;;;;;'
I( -
tu-) = t 2
~ L)Du- 1 + /;11f,rV lu2
Iq+1 < I(u) =
IH
which is a contradiction. Hence, to find a positive solution of (3.1) we have to show that there is at least one A >
Ao , such that 1;. is attained. In Section 2, we get a general
framework to solve the minimization problem (2. 1). But generally speaking 51 is quite difficult to check. So Theorem 2. 1 does not give us any positive result unless we can check 51. As to the eigenvalue problem, one need only to show that there exists a A > Ao , such that I;. is attained. Hence, if we can show that there exists a A> Ao holds with respect to this
u
>
such that 51
then (3.1) is solved.
Suppose that rp~ 0 but rp~ O,f(t) satisfies (2.3) and F(u
Theorem 3.1
F(u)
A> Ao under some simple conditions,
,
+ Ft;u) for u > 0 and v > o.
+ v) >
Then there exists at least one pair (p,u) with p> 0,
0, satisfying (3.1).
Remark 3. 1
+ F (v) for u >
If 'F' (t) 0, v
>
=
f (t) is strictly inceasing in t
0.
Proof of Theorem 3. 1
First we claim that 1).
>
0 , Then F (u
< Ir-;.o for each A >
+ v) > F (zz')
Ao. Let v be the
Supp.
25
Yan &,Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
positive minimizer of Ir'-Af). Since
LNF(V
+ rp) > LNF(V) + LvF(rp) = A,
there is t E (0,1), such that LNF(tV
+ rp) =
I;. ~ I(tv)
Fix X> Ao• Set
=
Al
sup{AII~
>
Then by (3. 4), we have Al
<
t,
< I(v) = tz;
11 < t,
I;.
=
(3.4)
o
+ If-a,
Ao and I
A. Then.
I A1
+ Ii-;.
+ tc..
V a E [Ao,A]}
1
,
(3. 5)
V a E [Ao,A1 ) .
(3.6)
We now show that for this Al ,51 holds. Indeed, since I: ~ I:- fJ (3.5) and (3.6), we get
I;. 1 - 11
= If-;. < I, + Ir.-a - If-;. < I, 1
1
Thus, by Theorem 2. 1, I;'I is attained by some have u
~
o.
+ l~_« 1
+ I p , V f3 E
V a E- [Ao,A1 ) .
E lvI. Obviously, since Al
u
[O,a], by
>
Ao
,
we
If u- ~ 0, then Al
=
L.vF(u
+ rp) < Lv F( IuI + rp).
(3.7)
Hence, there is t E (0,1), such that
L.vF(t Iu But 1;.1
~ l(t
luI) <
I(u)
=
I + rp)
(3. 8)
= AI.
I;.1 , so we get a contradiction. Thus, u ~
On the other hand, since
u ~
°and
- 6u
+ u" =
p./(u
o.
+ cp),
(3.9)
we get (3. 10)
and by the strong maximum principle, we get u
> o.
4 Proof of Theorem 1.1 First, we prove a lemma which is useful in the later arguement. Lemma 4. 1
of a E R
N
,
J
+ h > N + L, then there is C> 0, independent
such that
RN
Proof
If k > L, h > Land k
(1+
Ixl)-"(l+ Ix-al)-'~Clal-L(l+O\)
for some 01 > 0.
26
ACTA MATHEMATICA SCIENTIA
+f
=(f rl _Ial I
7T
+
Ixl)-"(l
+
Ixl)-"+o+cJ
laj)(l
Irl~T
::(Clal-o+ctl)Lf
Irl~I~1
(1
+ Clal-o+ctl)Lf Irl~I;1 (1 +
+
Ix - al)-!
1)L(1
Ixl)-"(l
Vol. 15
+
Ix - al)-!
+ Ix -
al)-4+':1+ct
1)L
:::;;;C/al-OHt)LCLN(I + Ixj)-HOHt}L(I + Ix - al)-A
+ f RN (1 + We claim that if h, of a , such that
~
Ix 1)-"(1 + Ix - a I )-!+O+ctt)L].
+ k > N.
0 and k 1 ~ 0, hI
l
(4.1)
Then there is C
JI=f RN (1+ Ixl)-"1(1+ Ix-al)-4
>
0 independent
(4.2)
1::(C.
Clearly, (4.2) together with (4.1), implies that the Lemma holds. Proof of (4.2). Set D.
=
{xllxl
Ix Indeed, if Ix
Ix
- a
aI~
I ~ 21 a I, then Ix
I 2~ 1!l 2 2~ Isl 4 •
Hence
JI =(f
+f
lal
IZ'~T
::(f Iz'~ ';1 (1 +
+
Iz-a'~
~ I~I, Ix - al ~ I~I}.
l!1 2
Ixl V x E a; 4'
- aI
+f
First, we claim
DtJ
Ix 1)-"1- 41 +
~ Ix I
)(1
+
Ia I ~ I~ I ;
if I I :::;;; 21 x
Ixl)-"1(1 + Ix -
f Iz-a'~ I;' (1 +
(4.3)
a I, then
al)-4 t
Ix - a I )-"1- 41
f (I + Ixl )-ht(I + Ixl )-At 4
DoJ
:::;;;CLN(I + IXj)-(htHt) <+ 00. Remark 4.1
Using the same mothod , we can prove easily that if k > max(N ,L) ,
then
LN O+ Ixl)-L(I+ Ix-aj)-A:::;;;Clal- L.
(4.4)
Lemma 4. 2 Let u E M be the nonnegative solution of -
Suppose that there is C I
>
+u
q
= p(u
N- 2
+ rp)P.
0, 1 > - 2 - ' such that
Then, for any 8> 0, there is C' u(x)
L,u
>
I9'< x ) I ::( C I (1 + Ix I)
0, such that
rC' (I + Ix j)-Z/(q-DH if P... z ~ 2/(q ~~ q CO + IXj)-J>ljqH Z ~ 2/(q - D,
l
(4.5)
if:
1),
-I.
Supp.
Yan
Proof
s, Zhang:
27
A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
Since -
Lu ~ /-l(u
+ cp)P ~ CuP + Ci9'IP,
by Harnack inequality, we have that there is C
>
sup(y) u(x) ~ C( lu ILz - (B Z ~'/» -
zE 3
0, independent of y, such that
+
19'IL z • (B (v»).
(4.6)
Z -
1
Thus, we have (4.7)
and because luILZ·(Bz(y»
+
-
6,u
s, > 0, such that for Ix I ;:: s.,
+ (1 -
Set v = CO I~ 1-"', then there is a -
6,v
+ (1 -
e)v Cl = Co(m(N -
q
>
e)u ~ Cip " ~ Clxl-
0, such that
2- m)
Ixl-
(for Ixllarge, if Case 1. If
(4.8)
00,
y--.+oo.
as
19'ILZ·(Bz(Y»~O
Hence, for any e > 0, there is
Ix I --.+
as
u~0
m
m
-
Z
+ (1 -
pl
(4.9)
•
e)
Ixl-·
q
)
;::
a
!.rl-
mq
,
(4.10)
kq z;:: q ~1' then for any B"> 0, choosing m = --L - ~, by (4.10), we q1
get -
6,v
+ (1 -
e)v
Consequently -
u)
t6,(u -
q = alxl-~+8q
+ (1 -
e)(u
q
> Clal-
(for [z ] ;:: R 1 )
o") ~ 0
-
(for Ixllarge).
Pl
.
(4.11)
(4. 12)
Choose Co large enough such that
u~Colxl-(q':1-8) =v, for Ixl =R 1 • Since u - v
~
0 as Ix
I ~+
00,
by the maximum principle, (4. 13) implies u~v
for Ixl ~Rl.
< q--L ' then choosing m = E-Z - ~, - 1 q t6,v + (1 - e)v ~ a Ix l- l+q8 > C Ix
Case 2. If.P...Z q -
(4.13)
q
p
(4.14)
by (4. 10), we get
I-Pl
for Ix I large.
(4.15)
Thus, similar to the Case 1, we get u ~ v for Ia: I large. Remark 4. 2
If q
>
N N 2' then it is not difficult to prove that the solution
u
of
(4. 5) satisfies u(x)
~Colxl-(N-2+8),
8> 0,
no matter how fast the function 9' decays at infinity. Let V E 1\1 be a nonnegative solution of
- LV +
v = q
pV P
x ERN.
As a direct application of Lemma 4. 2, we know that Vex) ~ C(l
+
Ix I )-q':1+8.
(4.16)
Before we prove Theorem 1. 1, let us recall a Lemma of Brezis and Nirenbergjl ].
ACTA MATHEMATICA SCIENTIA
28
Lemma 4.3
a,p r C Ia 11,81 q- 1 if Ia I ~ IPI, 1,81"-Z) I ~ ~ LClal 1 lPI if lal ~ IPI.
Assume 1 ~ q ~ 3. Then there is C such that for all
Iia + Plq - lal q - IPlq - qap( lal q- Z + When q
Vol. 15
q
-
3, we have
~
Iia + ,Blq - lal q - 1,8l q - qap( lal q- Z+ IPlq-Z) I ~ C( lal q- zl,8l:= Proof of Theorem 1. 1
J.1~IP+l.
+ laI
As we have pointed out, I;. is achieved for
21,8l
°<
q -
Z
) .
A ~ Ao
=
We only need to prove that Ilis achieved forA>A o• Now we check that I;. sat-
RN
isfies S 1. Thus, by Theorem 2. 1, I;. is achieved. Set a = SUp{Al II;. < IfJ
+ Ir'-fJ'V P E
Ao <
}.
By Theorem 3.1, we have
(4.17)
a~;\,
< t, + tz.; I;. = I. + tz.:
V P E [Ao,a),
I;.
I.
[Ao,A1 )
(4.18) (4.19)
is attained by some
uE
M,
u;;::
0.
Now we show a = A. We argue by contradiction. Suppose a tive function which attains Ir':-•• Set Va(x) = Vex - a) ,a E R each a E R
N
N
•
(4.20)
<;\.
Let V be the posi-
It is easy to see that for
there is C; E (0,1) , such that
,
J~ lu + 9'+ C.V.IJ>+1
(4.2D
= A.
On the other hand, since ;\1/(1'+1>
~
lu +
+ Ca IVa IL + 1(R N) =
9'ILP+1(Kv)
P
a1/(p+l>
+ Ca(" -
a)I/(P+1>,
we get
c. ~ Co > 0, Set w = A=
u
+
9'.
L)W I
Now we prove as
<
+ C:+
1L,vV:+l
+ (p + DL,v(w/>C.V. + w(C.V.)/»
laI -.+ 00 IRal =o(lalCJ
+ a)l > l , 1 < q - 2, -l
rna 4.1 we get
+ S. (4.23)
l
ISal =o(lal-
) ,
wV: W~V4
~CJ
2a - q--1 ~ N
-:.... R.
l
(4.24)
) .
P ~ 2. IRa 1 ~
. Since (1
(4.22)
Applying Lemma 4. 3, we get
J>+ 1
Case 1, 1
V a ERN.
- q) + 2(p q- 1 -
RN
(1
~ CJ NWl+·V~:-· R
+ I·_!'I)-lo+«)(l + Ix -
2 2q (p - a) ( - - - a) = - q-l q-l
2c:.
al)-(P-·)(q:t- c)') .
- q) + 2(p q-l
a(p - a)
a(p - a) - q--1 > N 1£ a,a small enough, applying Lem-
Supp.
Yan
s, Zhang:
A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
29
Similarly,
Case 2, P > 2
IR.I ~ CL.v W1'-1V; ~ CL.v (l clearly, I(p - 1)
> 1.
~ 4,
then
(i) If N
--.i- >
+
u+
Ix 1)-1<1'-0
Ix - a i)-Z
2(N - 2) ~ N and consequently
q- 1 is small enough. By Lemma 4. 1, we get
IRa 1 = (ii) If N = 3, then 1 ~ N -
. 4
q _ 1
+ (p -
1)[
=
2
- 28> N, if 8
I-I).
1, and
4
=q
0 ( Ia
--.i-
q- 1
_ 1
+ (p -
>---.L+ q- 1
(q -
2)[ 2)
+l
+ 1 ~ 3 + 1 = N + 1,
since
_4_
q-1
+ (q
_ 2) - 3 = _1_[q2 q-1
6q
+ 9J ~ o.
Applying Lemma 4. 1, we get
IRal
=
o(lal- 1 )
+
Ixi)-Zl(l
+ (q -
1) q~1
15.1 ~Cf~WZV~-1 ~CLN(l Since 21
+ (p -
1) q~1
>N
- 2
,
+
Ix - ai)-<1'-IH q": I -
8>.
+ 1 = N + 1, applying Lemma 4.
1, we get
ISal = o(lal-
1
) .
Hence we have proved (4. 24). By (4.23)
+ C~+I(A -
a)
+
~a + C~+I(A -
a)
+ (p + 1)LN(W(C.V.}/,) + o(lal-
A =a
(p
+ 1)LN(WC.V. + w(C.v.)1') + o(lal1
1
)
) .
Hence
C~+l ~ 1 - ~ + ;L.vw(C.V.)1') + Now we deal with L.v lu
f RN lu + c.v, Iq+l.
o( la I-I).
(4.25)
We have
+ C.V.lq+l = f~ lu Iq+l + ~+lfR'vV~+l
+ (q + 1) f~(u(C.V.)q + uqC.V.) + R~ + S~. By Lemma 4.3, we have
(4. 26)
30
ACTA MATHEMATICA SCIENTIA
f UhV~,
IR~ I, IS~ I ~ By Lemma 4. 2,
f
RN
with h
>
l,k
>
Vol. 15
1 and h
+ k = q + 1.
jf NO + Ixl)-(q':1-8lhO+ Ix-al)-(q':l-m
UhV!a ""'.-::::: ~ v
l'f
R'
R
+
(1
~
Ixl)-(~I-et)It(1
~ Z < --L , we
Since max (N - 2, N)
P
q- 1
+ Ix -
al)-
(4.27)
if :l~q 2 if.£.Z q
<
1,
_2_. q-l
have if.£.Z ~ --L , then --L1h q q- 1 q-
+ q--L1k =
2(q+l) =~+_2_>N+Z;ifP..Z<_2_,thenJ!...Zh+_2_k>J!...Z( + 1) q-l q-l q-l q q-l q q-l q q
= pI +
+ Z.
.£.Z> N q
f
RN
Hence
By Lemma 4.1, we get
J lu + cavalq+l J RN
+q~ ~I;. +
1L
N
tz;
< RN
+ q~ 1(C:+
1
1J 2
IDu I2+CaJRN DuDVa +
1
~v
+k=q+
1.
J~u(C.V.)P + o(lal-
(DuDV a
+ uV:) +
1 2
(4.28)
q+1 + C:+lf vr: RN
1, we have
G:+ V :+1 +
+ CaJ
lul
RN
1 and h
+ (q + 1) L.v u (C.V . ) 9 + o( la I-I).
1J 2
=
=
.
So by (4. 29) and C,
I).~I(u+CaVa)
>
oClal- /), h "> l,k
=
UhV:
(C; -
l
~v
(4. 29)
C;IDVaI 2
+ q +IIJ
~v
u q+ 1
)
I)J
~v
IDVa1 2
-1)L.vV :+ +o(la\-I). 1
(4.30)
On the other hand, since (4.31) with (4.32) we have CaJ
RN
(DVaDu+V:u)=~(J A-a ~,l-J
RN
A - a RN
CIDVaI2+V:+l))J V V:u R'
(IDValz+V:+l)J N V:u. R
(4.33)
Besides, by (4.25)
C; - 1 ~=-
~J a R A -
_2_J
N
A-aJli
w(C"V a ) 1'
+ o(lal- /)
wV~ + _2_(1 - C~)J N wV~ A-a R
+ o(lal-').
(4.34)
Supp.
Van &,Zhang: A MINIMIZATION PROBLEM ON R N WITH NONZERO DATA
31
But,
(4.35) Hence
(4.36) Similarly
C:+
1
1
=- q
-
+ If
A- a
RN
wV~
+ o( lal-
l
(4.37)
) .
Combining (4. 30), (4.33), (4. 36) and (4. 37), we obtain
I;.
+.,J-f + )f V~u --l-f wv~f IDValz--I-f wv~f
~ItI + tz:
A-aRN
=1;.
+ ti: -
A-a
RN
RN
(IDV a 1 2
V:+ 1
A-aRN
-,---Lf .,(IDVI2 + vQ+l)f A
-
RN
a e"
RN
RN
soV:
V:+ 1
+ o( lal- l)
+ o(lal-
l
) .
(4. 38)
Since
f ~¢V: ~f R"
Iz-al~l
soV: ~Colal-lf
Izl~l
v- = Colal-
l
,
by (4. 38), we get
I;. ~ I;. + tz: -
Co Ia I-l + o( Ia I-l) < 1;. + tz: for Ia I sufficiently large,
which con-
tradicts to (4. 19). Reference 1
Brezis H, Nirenberg L. A minimization problem with critical exponent and nonzero data. To appear.
2
Gilbarg D, Trudinger N S. Elliptic partial differential equations of second order. Second edition, Springer-Verlag, 1983
3 Li Gongbao , Yan Shusen. Eigenvalue problem for quasilinear elliptic equations oti R": Comm , P. D. E. ,1989,14:1291--1314 4
Lions P L. The concentration compactness principle in the calculus of variations. The locally compact case, Ann. Inst. H. Poincare Analyse nonlineaire , Part 1, 1984,1 (2) :109-145 ;Part 2, 1984, 1(4) :223--283
5
Yan Shusen , Zhang Zhengjie. A minimization problem on exterior domain with nonzero data. To appear in Acta Math Sci.