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ANALYTIC ASPECTS OF THE TODA SYSTEM: I.A MOSERTRUDINGER INEQUALITY
JURGEN JOST AND GUOFANG WANG
Abstract. In this paper, we analyze solutions of the open Toda
system and establish an optimal MoserTrudinger type inequality
for this system. Let be a closed surface with area 1 and K =
(aij)NN the Cartan matrix for SU(N + 1), i.e.,
2 1 0 0
1 2 1 0 0
0 1 2 1 0
0 1 2 1
0 0 1 2
.
We show that
M (u) =1
2
N
i,j=1
aij(uiuj+2Miuj)N
i=1
Mi log
exp(N
j=1
aijuj)
has a lower bound in (H1())N if and only if
Mj 4, for j = 1, 2, , N.
As a direct consequence, if Mj < 4 for j = 1, 2, , N , M has
a minimizer u which satisfies
ui = Mi(exp(
N
j=1 aijuj)
exp(
N
j=1 aijuj) 1), for 1 i N.
Date: Nov. 23, 2000.1
http://arxiv.org/abs/mathph/0011039v1
2 JURGEN JOST AND GUOFANG WANG
1. Introduction
Let be a closed surface with area 1. The MoserTrudinger inequal
ity is
u2 + 8
u 8 log
eu > C, for any u H1(),
(1.1)
for some constant C > 0. ((1.1) is a slightly weaker form of the original
MoserTrudinger inequality , see [41, 43, 25, 15, 38].) The inequality
(1.1) has been extensively used in many mathematical and physical
problems, for instance, in the problem of prescribing Gaussian curva
ture [7, 8, 11, 9], the mean field equation [5, 6, 31, 18, 42], the model of
chemotaxis [46, 28] and the relativistic Abelian ChernSimons model
[27, 29, 4, 44, 16, 17, 19, 38], etc. In all such problems, the correspond
ing equation is similar to the Liouville equation
u =M0(eu
eu 1), (1.2)
for some prescribed constant M0 > 0.
The systemanalog of (1.2) is the following system of equations
ui =Mi(exp(
nj=1 aijuj)
exp(n
j=1 aijuj) 1), 1 i n, (1.3)
for a coefficient matrix A = (aij)nn. Here Mi > 0 ( i = 1, 2, , n)
are prescribed constants. When the coefficient matrix admits only
nonnegative entries, there is a generalized MoserTrudinger inequality
obtained in [13, 45]
Theorem 1.1. [13, 45] Let the coefficient matrix A be a positive def
inite matrix with nonnegative entries. If for any subset J I :=
{1, 2, , n}
J := 8
jJ
Mj
i,jJ
aijMiMj > 0, (1.4)
then there is a constant C > 0 such that for any u = (u1, u2, , un)
(H1())n
1
2
n
i=1
aij(uiuj + 2Miuj)n
i=1
Mi log
exp(
n
j=1
aijuj) C.
(1.5)
ANALYTIC ASPECTS OF THE TODA SYSTEM I 3
In fact, the condition that A is positive definite can be removed by
using another formulation of the functional JM , see [13, 45]. Theorem
1.1 is sharp in the sense that if there is a subset J I with J(M) < 0,
then infu(H1))n JM(u) = . When n = 1, the condition (1.4) is
equivalent to a11M1 < 8. Therefore, it is natural to conjecture [45]
that Theorem 1.1 holds if and only if
J(M) 0, for any J I. (1.6)
In [45], a proof of this conjecture was sketched for a special, but inter
esting case.
Theorem 1.2. Let A be a symmetric, positive definite row stochastic
matrix, i.e.,
aij 0 andn
j=1
aij = 1 for any i I. (1.7)
Then there is a constant C > 0 such that
JM(u) C for any u (H1())n
if and only if (1.6) holds.
The only if part was first proved in [13] in a more general case. It is
clear that in general, such an inequality cannot be true if the coefficient
matrix A admits some negative entries. However, in many interesting
systems arising in Physics and Differential Geometry, there are negative
coefficients, for instance, in the Toda system and the relativistic and
nonrelativistic nonAbelian ChernSimons models [32, 20, 24, 26].
In this paper, we want to generalize Theorem 1.2 to the Toda system.
Let K denote the Cartan matrix for SU(N + 1), i.e.,
2 1 0 0
1 2 1 0 0
0 1 2 1 0
0 1 2 1
0 0 1 2
.
4 JURGEN JOST AND GUOFANG WANG
The open SU(N + 1) Toda system (or, 2dimensional Toda lattice) is
ui =N
j=1
aijeuj , for i = 1, 2, , N. (1.8)
The popular interpretation of the (onedimensional) Toda lattice is a
Hamiltonian system which describes the motion of N particles moving
in a straight line, with exponential interaction. The twodimensional
Toda system has a much closer relationship with differential geometry.
It can be seen as the Frenet frame of holomorphic curves into CPN .
For the Toda system and its geometric interpretation see, e.g.,[26] and
references therein. See also [20].
In this paper, we establish the following MoserTrudinger type in
equality for (1.8).
Theorem 1.3. Let be a closed surface with area 1 and A = K.
Define a functional N : (H1())N R by
M(u) =1
2
N
i,j=1
aij(uiuj + 2Miuj)
N
i=1
Mi log
exp(
N
j=1
aijuj).
Then, the functional has a lower bound if and only if
Mj 4, for j = 1, 2, , N. (1.9)
We remark that the condition (1.9) is equivalent to (1.6) in this
case. Since the coefficient matrix K admits negative entries, we might
encounter the problem that the maximum principle fails. This is the
reason why we cannot classify all entire solutions of (1.8) with finite
energy yet (see Sections 2 and 7). Fortunately, when proving Theorem
1.3 we can avoid this problem.
We outline our main idea of the proof of Theorem 1.3 for N = 2. We
first show that M has a lower bound ifMj < 4 for j = 1, 2 (Theorem
4.1). The idea is as follows. Let us define
= {M = (M1,M2) R+ R+ inf M > }.
From the ordinary MoserTrudinger inequality (1.1), we know 6= .
Theorem 4.1 now is equivalent to (0, 4) (0, 4) . If it were false
then there existsM0 = (M01 ,M02 ) such thatM
= (M01,M02) ,
We are able to classify all such solutions now.
ANALYTIC ASPECTS OF THE TODA SYSTEM I 5
but (M01 + ,M02 + ) 6 for any small > 0. It is easy to show
that M admits a minimizer u = (u1, u
2) which satisfies (1.3) for
M = (M01 ,M02 ). If u
blows up, there are three cases (see
Section 4 below). For each case, after rescaling we obtain a bubble
which is an entire solution of the Liouville equation (3.6) or the Toda
system (1.8) with finite energy. A classification result of [12] for the
Liouville equation and Corollary 2.6 below imply that in any case one
of M01 and M02 is larger than or equal to 4, a contradiction. However,
u may not blow up. Ding [14] introduced a trick to deal with such
a problem in his study of the ordinary MoserTrudinger inequality .
Following his trick, we perturb the functional M a little bit such that
the resulting functional FM also admits a minimizer u which does
blow up as 0. u satisfies a similar system as (1.3) and has the
same bubble. Then we are able to use the argument above to get a
contradiction.
To prove Theorem 1.3, we consider the sequence of minimizers u =
(u1, u2) of M with M
= M0 (, ) = (4 , 4 ) for small
> 0. (The existence of the u follows from Theorem 4.1.) u satisfies
a Toda type system. If u converges u0 in H2 := H1 H1, then
it is clear that infuH2 M0(u) = M0(u0) > . Hence, we may
assume u does not converge in H2. Using the analysis developed in
Sections 2 and 3, we know there are three possibilities: (a) S11 and
S2 = , (b) S1 = S2 = 1 and S1 = S2 and (c) S1 = S2 = 1
and S1 S2 = . Here Sj (j = 1, 2) is the blowup set defined in
Section 5 below and S is the number of points of the finite set S.
We use a local Pohozaev identity to exclude case (b). This is the
crucial point to avoid the aforementioned problem that the maximum
principle fails, since the remaining cases are essentially scalar problems.
In fact, we can reduce case (a) directly to the corresponding problem
of one functionthe ordinary MoserTrudinger inequality. For case (b),
we can apply the method developed in [15, 38] to give a lower bounded
of M0 .
We shall apply Theorem 1.3 in a forthcoming paper [30] to study the
relativistic SU(N + 1) nonAbelian ChernSimons model ([32, 34, 35,
6 JURGEN JOST AND GUOFANG WANG
20]), which can be seen as a nonintegrable perturbation of the inte
grable Toda system (1.8). For mathematical aspects of the relativistic
nonAbelian ChernSimons model, see [49, 47]. We hope Theorem 1.3
will become a powerful tool for studying problems arising from higher
rank models, as the ordinary MoserTrudinger inequality has become
in the Abelian case.
For simplicity we only give detailed proofs for the case N = 2. The
proofs in case N > 2 are completely analogous and just require a more
complicated notation. In Section 2, we analyze the solutions of (1.8)
in R2 and obtain a relation between
R2eu1 and
R2eu2 for any entire
solutions of (1.8). In Section 3, we analyze the convergence of solutions
as in [3]. In Section 4, we first show M has a lower bound if Mj < 4
for all j. Then we show that if M has a lower bound, Mj 4 for all
j. We prove Theorem 1.3 in Section 5.
2. Analysis of the Toda system
In this section, we consider the analysis of solutions of the following
system
u1 = 2eu1 eu2 ,
u2 = eu1 + 2eu2 ,
(2.1)
which is equivalent to system (1.8) with N = 2. Similar results were
obtained in [3, 12] for the Liouville equation and in [10, 13] for the
Liouville type systems, see also [36].
Lemma 2.1. Let u C2(B3) C1(B3) satisfy
u1 = 2eu1 eu2 in B3,
u2 = eu1 + 2eu2 in B3,
uj(xj) = bj , for some xj B1, j = 1, 2,
uj a0, in B3, for j = 1, 2.(2.2)
Then there is a constant C = C(a0, b1, b2) such that
minxB1
{u1, u2} C.
See also [40].
ANALYTIC ASPECTS OF THE TODA SYSTEM I 7
Proof. Let w = (w1, w2) defined by
w1(x) =1
2
B3log x y(2eu1(y) eu2(y))dy
w2(x) =1
2
B3log x y(2eu2(y) eu1(y))dy
Set u = (u1, u2) = u w. Obviously,
uj = 0 in B2, for j = 1, 2 (2.3)
and
wj(x) c1(a0), for y B2. (2.4)
(2.4) and (2.2) imply that uj c2(a0) and uj(xj) c3(bj). Now from
the Harnack inequality, we have
minxB1
uj(x) c(a0, b1, b2). (2.5)
The Lemma follows from (2.4) and (2.5).
Lemma 2.2. There exists 0 > 0 such that for any u C2(B+2 )
C1(B+2 ) satisfying
u1 = 2eu1 eu2 in B3,
u2 = eu1 + 2eu2 in B3,
B2eu1 < 0,
B2eu2 < 0,
(2.6)
we have
maxB1/4
max{u1, u2} C1,
for some positive constant C1.
Proof. Choosing 0 1. The Lemma follows
from the standard elliptic estimates.
Note that Lemma 2.2 is true for any 0 < 4, see Lemma 3.2 below.
8 JURGEN JOST AND GUOFANG WANG
Proposition 2.3. Let u = (u1, u2) H1loc(R
2)H1loc(R2) be a solution
of (2.1) on R2 with
R2eu1 4 and 2 := 22 1 > 4.
Proof.
The previous Lemma implies that
(21 2 + ) log x c 2w1(x) w2(x) (21 2 ) log x+ c(2.9)
Since u1 = 2w1w2+c1, from (2.9) and (2.8), we deduce that 1 > 4.
Similarly, we have 2 > 4.
Corollary 2.6. j > 4 for j = 1, 2.
ANALYTIC ASPECTS OF THE TODA SYSTEM I 9
Lemma 2.7. Let u be a solution of (2.1) and (2.7). We have,
uj j2
log(x+ 1) c,
limr rujr
= j2,
limruj
= 0,
(2.10)
where x = (r, ).
Proof. From above, we have
u1 =1
2
R2[log x y log(y+ 1)](2eu1(y) eu2(y))dy + c1,
u2 =1
2
R2[log x y log(y+ 1)](2eu2(y) eu1(y))dy + c2.
The Lemma follows from potential analysis, see for instance [12].
Now we are in the position to give a relation between
eu1 and
eu2 .
Proposition 2.8. Let j =
R2euj for j = 1, 2. We have
21 + 22 12 = 4(1 + 2). (2.11)
Proof. From equation (2.1), we have the Pohozaev identities as follows:
R
BR(u1r
2 1
2u1
2) = 2
BRReu1 4
BReu1
BReu2xu1,
R
BR(u2r
2 1
2u2
2) = 2
BRReu2 4
BReu2
BReu1xu2
and
BRRu1
nu2n
+
BRu1u2 +
BRx2
j=1(ju2)ju1=
BRReu2 + 2
BReu2 + 2
BReu1xu2,
BRRu1
nu2n
+
BRu1u2 +
BRx2
j=1(ju1)ju2=
BRReu1 + 2
BReu1 + 2
BReu2xu1.
(2.12)
From above, we get the Pohozaev identity for the Toda system (2.1)
3
BRR(eu1 + eu2) 6
BR(eu1 + eu2) =
22
j=1
BRR(nuj
2 12uj
2) 2
BRR(u1
nu2n
12u1u2).
(2.13)
Applying Lemmas 2.5 and 2.7 in (2.13) and letting R , we get
21 + 22 + 12 = 12(1 + 2),
10 JURGEN JOST AND GUOFANG WANG
which is equivalent to (2.11). This proves the Lemma.
Similar results for systems with nonnegative entries were obtained
in [10] and [13]. We conjecture that 1 = 2 = 8. It is also very
interesting to classify all solutions of (2.1) with finite energy
R2euj 0 and p > 1. Let
Sj = {x there is a sequence y x such that uj(y
) +}.
(3.3)
Then, one of the following possibilities happens: (after taking subse
quences)
(1) uk is bounded in Lloc() Lloc().
ANALYTIC ASPECTS OF THE TODA SYSTEM I 11
(2) For some j {1, 2}, uki in Lloc(), but u
kj uniformly on
any compact subset of for j 6= i.
(3) For some i {1, 2}, Si 6= , but Sj = , for j 6= i. In this
case, uki on any compact subset of \Si, and either, ukj is
bounded in Lloc(), or ukj on any compact subset of .
(4) S1 6= and S2 6= . Moreover, ukj is either bounded or on
any compact subset of \(S1 2) for j = 1, 2.
Proof. Here, for simplicity we only give a proof of the Theorem when
k1 = k2 = 0.
In view of (3.2), we may assume that there exist two nonnegative
bounded measures 1 and 2 such that
eukj
dj as k ,
for every smooth function with support in and j = 1, 2. A point
x is called a regular point with respect to j if there is a function
Cc() , 0 1, with = 1 in a neighborhood of x such that
dj < .
We define
j() = {x  x is not a regular point with respect to j}.
By definition and (3.2), it is clear that 1() and 2() are finite. And
j() is independent of for small > 0, see below. We divide the
proof into 3 steps.
Step 1. For j = 1, 2, Sj = j() provided 0. Note
12 JURGEN JOST AND GUOFANG WANG
that uk1 = 2euk1 eu
k2 2eu
k1 . Define w : B(x0) R by
w = 2euk1 , in B(x0),
w = uk1, on B(x0).
The maximum principle implies that uk1 w. Since S1 is finite, we
may assume that uk1+is uniformly bounded in L(B(x0)). In view
of
B(x0)eu
k1 < 4
3, a result of Brezis and Merle [3] implies that
w+ L(B 2(x0)), which in turn implies that u
k1+
L(B 2(x0)),
a contradiction. Hence, S1 1(). 1() S1 follows from the
argument in [3]. Similarly, we have S2 = 2().
Step 2. S1 S2 = implies (1) and (2). S1 6= and S2 6= imply (4).
S1 S2 = means that u1+ and u2
+ are bounded in Lloc(). Thus,
eu1 and eu
2 are bounded in Lploc() for any p > 1, which implies that
1, 2 L1()Lploc(). Applying the Harnack inequality as in [3], we
have (1) or (2). The second statement follows similarly.
Step 3. S1 6= and S2 = imply (3).
We need the following lemma.
Lemma 3.2. Lemma 2.2 is true for any 0 < 4.
Proof. We use a blowup argument to prove this Lemma. Assume by
contradiction that Lemma 2.2 were false for some 0 < 4, i.e., there
exists a sequence of solutions uk = (uk1, uk2) of (2.7) with
euj 0
(j = 1, 2) for some 0 < 4 such that
maxB1/4
max{uk1, uk2} , as k .
Without loss of generality, we may assume that S1 = {x0} and S2
(B1/4\{x0}) = . We may also assume that there exists a sequence of
points {xk} B1/4 such that
uk1(xk) = maxB1/4
max{uk1, uk2}.
ANALYTIC ASPECTS OF THE TODA SYSTEM I 13
Let mk = uk1(xk). Define u
kj (x) = u
kj (kx+ xk)mk for j = 1, 2 with
k = e 1
2mk . Clearly, k 0 as k and u
k = (uk1, uk2) satisfies
uk1 = 2euk1 eu
k2 in k
uk2 = 2euk2 eu
k1 in k
keu
kj 0, j = 1, 2,
ukj (x) 0 = uk1(0),
(3.5)
where k = 1k (B1/4 xk). Applying Lemma 2.1, Lemma 2.2 and
(1)(2) in Theorem 3.1, we have two possibilities:
(i). uk converges to u0 = (u01, u02) in H
1loc(R
2) H1loc(R2) which satis
fies the Toda system (2.1) in R2. (In this case uk2 is bounded in
Lloc(R2).)
(ii). uk1 converges to 0 in H1loc(R
2) and uk2 tends to uniformly in
any compact subset in R2. Moreover, 0 satisfies
0 = 2e0 (3.6)
with
R2e0 0. Now we can follow the argument in [3] to exclude
(i).
Remark 3.3. We believe that in case (4) ukj on any compact
subset. From the argument of Step 3, one can show this if S1 6= S2.
In our application (the proof of Theorems 5.1 and 1.3), we can exclude
case (4).
Before we start to prove our main results, we remark that
(1) 0 always means some sequence n > 0 such that n 0 as
n .
14 JURGEN JOST AND GUOFANG WANG
(2) C denotes a constant independent of , which may vary from line
to line.
(3) Any 2dimensional surface (, g) is locally conformally flat, i.e.,
for any x there is a neighborhood x U and a conformal
factor : U R such that gU = e(dx2 + dy2) in local coordinates.
Instead of considering the equation g = f in U , we can consider
0 = ef in a domain of the Euclidean plane. Hence, wlog, we can
assume =0, i.e., U is a flat domain. In the following sections, we will
assume that near a blowup point there is a flat neighborhood.
4. A MoserTrudinger inequality for the SU(3) Toda
system
Let K be the Cartan matrix for SU(3), i.e.,
K =
2 1
1 2
.
We have the following MoserTrudinger inequality.
Theorem 4.1. Let be a closed surface with area 1 and A = K. For
any M = (M1,M2) (0, 4) (0, 4), there is a constant C > 0 such
that
M (v) =12
{2v12 + 2v2
2 2v1v2 + 2(2M1 M2)v1+
2(2M2 M1)v2} M1 log
e2v1v2 M2 log
e2v2v1
C, (4.1)
for any v H2, or equivalently,
M (u) =13
{u12 + u2
2 +u1u2 + 3M1u1 + 3M2u2}
M1 log
eu1 M2 log
eu2 C,(4.2)
for any u H2.
The equivalence between (v) and (u) can be seen easily from the
following equation
u1 = 2v1 v2u2 = 2v2 v1
(4.3)
Here we use the method of Ding [14] to prove Theorem 4.1. This
method was introduced in his study of the ordinary MoserTrudinger
ANALYTIC ASPECTS OF THE TODA SYSTEM I 15
inequality and was applied in [45] to obtain a similar inequality for a
system of functions.
Proof. Set
= {(M1,M2) R+ R+M is bounded from below on H2}.
Since A is positive definite, it is easy to see that 6= from the ordinary
MoserTrudinger inequality [41, 43]. In fact, one can show easily that
(83, 8
3) and (16
3+ , 16
3+ ) 6 ( > 0) from the ordinary Moser
Trudinger inequality and the Holder inequality. Clearly, preserves a
partial order of R+ R+, namely, if (M1,M2) , then (M1,M
2)
provided that M 1 M1 and M2 M2. The Theorem is equivalent to
(0, 4) (0, 4) . (4.4)
Assume by contradiction that (4.4) is false. We may assume that there
is a point
M0 = (M01 ,M02 ) (0, 4) (0, 4) (4.5)
such that
1. For any > 0, M0 = (M01 ,M02 ) ,
2. For any > 0, M0 + = (M01 + ,M02 + ) 6 .
We first need several lemmas.
Lemma 4.2. For any M with M1 < M01 and M2 < M
02 , there exists a
constant c > 0 such that
M (u) c1(u1
2 + u22) c, for any u H2.
(4.6)
Moreover, M admits a minimizer u = (u1, u2).
Proof. Choose a small number > 0 such that M1(1 + ) < M01 and
M2(1 + ) < M02 . By the definition of M
0, we know that (1 + )M =
((1 + )M1, (1 + )M2) , i.e., there is a constant C > 0 such that
(1+)M (u) C, for any u H2.
It follows that
M (u) =1
1+(1+)M +
3(1+)
(u12 + u2
2 +u1u2)
6(1+)
(u12 + u2
2) C.
16 JURGEN JOST AND GUOFANG WANG
This inequality means that M satisfies the coercivity condition. Now
it is standard to show that M admits a minimizer, for M is weakly
lower semicontinuous.
Lemma 4.3. There exists a sequence uk H2 such that limk uk2
and
limk
M0(uk)
uk2 0, (4.7)
where u2 = u12 + u2
2.
Proof. Assume by contradiction that for any sequence uk H2 with
limk uk2 ,
limk
M0(uk)
uk2> > 0.
Then we can show that M0 satisfies the coercivity condition,
M0(u)
2u2 + c, u H2,
for some constant c > 0, which implies that there is a small > 0 such
that M0 + , a contradiction.
Lemma 4.4. ([14]) For any two sequences ak and bk satisfying
limk
ak + and d0 = limk
bkak
0, (4.8)
there exists a smooth function F : [1,) [0, 12] satisfying
F (t) < 1/2 and F (t) 0 as t (4.9)
and
F (ank) bnk + as k , (4.10)
for some subsequence {nk}.
Proof. We give the proof for completeness, though it is rather elemen
tary. If there is a subsequence bnk of bk with property that bkn 0, we
can choose F (t) = log t. So we may assume that bk 0 and d0 = 0.
Wlog, we assume more that bkak
is nonincreasing. Choose another se
quence ck withckak
0 andbkck
0.
ANALYTIC ASPECTS OF THE TODA SYSTEM I 17
It is easy to find a nonincreasing smooth function F : [0,) R with
the property that
F (ak) = ck and F(t) 0 as t .
Clearly, this function F satisfies all conditions of the Lemma.
Applying Lemma 4.4 to sequences
ak =1
3
[uk1
2 + uk22 +uk1u
k2] and bk = M0(u
k),
where uk is obtained in Lemma 4.3, we can find a function F satisfying
(4.9) and (4.10). For any small 0, define a perturbed functional by
I(u) = M0(u) F (1
3
[u1
2 + u22 +u1u2]),
where M0 = (M01 ,M02 ).
Lemma 4.5. Let = infuH2 I(u). We have
1. for > 0, the infimum > , and it is achieved by u H2
satisfying
(1 F (g))u1 = 2(M
01 )e
u1 2M01 ,
+(M02 )eu2 +M02
(1 F (g))u2 = 2(M
02 )e
u2 2M02
+(M01 )eu1 +M01 ,
(4.11)
with
R2eu
1 =
R2eu
2 = 1. (4.12)
where
g =1
3
(u12 + u2
2 +u1 u32).
2. for = 0, I0 has no lower bound, i.e.,
0 = . (4.13)
Proof. 1. As in Lemma 4.2, we can show that I ( > 0) satisfies the
coercivity condition from the construction of F . It is also easy to check
that I is weakly lower semicontinuous. Therefore, I has a minimizer
which satisfies (4.11).
18 JURGEN JOST AND GUOFANG WANG
2. From the construction of F , we have
I0(uk) = M0(u
k) F (uk2) = (F (ak) bk) ,
as k , where the sequence uk was obtained in Lemma 4.3.
We now continue to prove Theorem 4.1 by considering the sequence
u obtained in the previous Lemma.
We claim that u blows up, i.e.,
maxx
max{u1(x), u2(x)} + as 0. (4.14)
Otherwise, we can show that u converges to some u0 in H2 that is a
minimizer of I0. On the other hand, by Lemma 4.5, I0 has no minimizer,
a contradiction.
Let m = maxx max{u1(x), u
2(x)} = max{u
1(x
), u2(x)}. We
have m as 0. Define a new sequence v by
u1 = u1(e
12mx+ x)m and u
2 = u
2(e
12mx+ x)m
near a small, but fixed neighborhood U of x0 , where x0 = lim x.
Here, we have abused a little bit the notation. (For simplicity, we
consider U as a domain in R2.) Clearly, u = (u1, u2) satisfies on
U = m12 U( x) that
(1 F (u2))u1 = 2(M01 )e
u1 + 2e12m(M01 )
+(M02 )eu2 e
12m(M02 ),
(1 F (u2))u2 = 2(M02 )e
u2 + e12m(M02 )
+(M01 )eu1 e
12m(M01 ),
(4.15)
Recall that u2 as 0 and limt F(t) = 0. By using the
same argument as in the proof of Lemma 3.2, we have that one of M1and M2 is larger than or equal to 4, which is a contradiction. This
completes the proof of Theorem 4.1.
Corollary 4.6. Let be a closed surface with area 1. For any M =
(M1,M2) with M1 < 4 and M2 < 4, M admits a minimizer u =
(u1, u2) which satisfies
u1 = 2M1(eu1 1)M2(e
u2 1),
u2 = 2M2(eu2 1)M1(e
u1 1).
ANALYTIC ASPECTS OF THE TODA SYSTEM I 19
Proof. Theorem 4.1 implies that M not only has a lower bound but
also satisfies the coercivity condition. Since M is weakly lower semi
continuous, there is a minimizer that satisfies (4.6) which is the Euler
Lagrange equation of M .
To end this section, we observe that 4 is the best constant.
Proposition 4.7. If one of Mi is greater that 4, then M has no
lower bound in H2.
Proof. Wlog, assume that M1 > 4 and contains a flat disk B0 for
a small constant 0 > 0. Let
u1 =
log 2
(1+2x2)2in B0
log 2
(1+202)2in \B0
u2 =
12log
2
(1+2x2)2in B0 ,
12log
2
(1+202)2in \B0 .
We estimate
u1 2 = 32 log +O(1),
u2 2 = 8 log +O(1),
u1 u2 = 16 log +O(1),
u1 = 2 log +O(1),
u2 = log +O(1),
log
eu1 = O(1),
log
eu2 = log +O(1),
(4.16)
which implies that
M(u) = 2(4 M1) log +O(1).
The Proposition follows from the previous formula by letting .
5. The optimal MoserTrudinger inequality
In this section, we prove Theorem 1.3 for N = 2. LetM0 = (4, 4).
20 JURGEN JOST AND GUOFANG WANG
Theorem 5.1. Let be a closed surface with area 1. There is a con
stant C > 0 such that
M0(v) =12
{2v12 + 2v2
2 2v1v2 + 8v1+
8v2} 4 log
e2v1v2 4 log
e2v2v1
C, (5.1)
for any v H2, or equivalently,
M0(u) =13
{u12 + u2
2 +u1u2 + 12u1 + 12u2}
4 log
eu1 4 log
eu2 C,(5.2)
for any u H2.
To prove Theorem 5.1, we only have to prove that there is a constant
C > 0 independent of such that
M(u) C, for any u H2, (5.3)
where M = (4 , 4 ) for small constant > 0.
To show (5.3), first applying Corollary 4.6 we obtain a minimizer
u = (u1, u2) of M with M
= (4 , 4 ). Recall that u satisfies
u1 = (8 2)eu1 (4 )eu
2 (4 )
u2 = (8 2)eu2 (4 )eu
1 (4 ),
(5.4)
with
eu
1 =
eu
2 = 1. (5.5)
Then, we will show that
M(u) C, (5.6)
for some constant C > 0. (5.6) is equivalent to (5.3). If u is bounded
from above uniformly, by using the analysis developed in Section 3
we can show that u converges (by taking subsequences) to u0 in H2
strongly. This implies that infuH2 M0(u) = M0(u0) > , and
hence (5.6). Hence, we assume that
max{m1, m2} + as 0. (5.7)
ANALYTIC ASPECTS OF THE TODA SYSTEM I 21
where mj = maxuj . Assume that m
1 m
2. As in Section 3, we may
assume that there exist two nonnegative bounded measures 1 and 2such that
euj
dj as 0,
for every smooth function : R and j = 1, 2. We define, for
j = 1, 2 and small > 0,
Sj = {x there is a sequence y x such that uj(y
) +}
and
j() = {x  x is not a regular point with respect to j}.
For the definition of regular point, see Section 3 above. From Section
3, we know that
Sj = j() and Sj 0.
Lemma 5.2. S1 = 1 and S2 1.
Proof. By (5.7), we have S1 1. Let y S1 and choose > 0 so small
that (B(y)\{y}) (S1 S2) = . One can use the blowup argument
as in the previous section to show that
lim0
B(y)eu
1 1,
which implies that S1 = {y} by (5.8), because of (5.4).
Assume by contradiction that S2 2. Then there exists z S2but z 6= y. Similarly, we can show that there is a small constant > 0
satisfying (B(z)\{z}) (S1 S2) = and
lim0
B(z)eu
2 1.
It follows, together with (5.5) that
lim0
\B(z)eu
2 = 0.
By (5.8), we have S2 = {z}, a contradiction.
Proof of Theorem 5.1. In view of Lemma 5.2, there are three possibil
ities:
22 JURGEN JOST AND GUOFANG WANG
(a). S2 = 0.
(b). S2 = 1 and S1 = S2,
(c). S2 = 1 and S1 S2 = ,
We discuss the sequence u case by case. We shall show that case
(b) cannot occur, which is crucial to establish our MoserTrudinger
inequality. Case (a) is easy to handle, while case (c) is more delicate.
Case (a). Reduction to the ordinary MoserTrudinger inequality .
Let S1 = {p}. In this case, by Theorem 3.1 and similar arguments
given in the proof of Lemma 5.6 in case (c) below, we can show that
(a1). u2 converges to G2 in H1,q() (q (1, 2)) and in C2loc(\{p}),
where G2 satisfies
G2 = 8eG2 4p 4
with
eG2 = 1.
(a2). u1 u1 converges to G1 in H
1,q() (q (1, 2)) and in C2loc(\{p}),
where G1 satisfies
G1 = 8p 4eG2 4
with
G1 = 0.
Furthermore, we have the following Lemma.
Lemma 5.3. For any small > 0 there exists a function w satisfying
w = 2(4 )ehew 2(4 ) in (5.9)
such that
(i). u1 w and h are bounded in C1(),
(ii). u2 +12w
12w is bounded in C
1().
Proof. For small > 0, let w be a function defined by
w = (4 )eu2 (4 )
w = 0
Since u2 is uniformly bounded from above, w is bounded in C1() by
elliptic estimates. Let w = u1 + w. Clearly, w satisfies (5.9) with
h = w, which is bounded in C1(). By definition, u1 w = w.
ANALYTIC ASPECTS OF THE TODA SYSTEM I 23
Thus u1 w is also bounded in C1(). In view of (5.9), the function
u2 +12w 2w is harmonic, thus
u2 +1
2w 2w = c,
for some constant c. By (a1), u2 is uniformly bounded. Hence, c
12w
is bounded. This proves the Lemma.
We now can reduce our problem to the ordinary MoserTrudinger
inequality . By Lemma 5.3, we have
1
3
(u1
2 + u22 +u1u
2) =
1
4
w
2 +O(1)
and
(u1 + u
2) =
Sw +O(1).
Thus,
M(u)
1
4
(w
2 + 4(4 )w) +O(1),
which has a lower bound due to the ordinary MoserTrudinger inequal
ity , see e.g. [25] or [15]. This completes the proof of Theorem 5.1 in
case (a).
Case (b). We show that case (b) does not happen.
Let S1 = S2 = {p}. Wlog, we assume that B(p) is a flat disk for
small > 0. By Lemma 3.2, in this case, we have that lim0
B(p)eu
1 =
1 for any small > 0. By the argument in Step 3 in the proof of The
orem 3.1, we have that u1 (as 0) on any compact subset
of \{p}. We also have either
(i) u2 (as 0) on any compact subset of \{p}, or
(ii) u2 is uniformly bounded on any compact subset of \{p}.
We first consider case (i). In this case, the same argument given in
the proof of Lemma 5.6 implies
Lemma 5.4. For any q (1, 2) and j = 1, 2
uj uj converges to Gp in H
1,q(),
24 JURGEN JOST AND GUOFANG WANG
where Gp satisfies
Gp = 4p 4, in ,
Gp = 0(5.10)
Moreover, uj uj converges to Gp in C
2loc(\{p}).
By this Lemma, we know that for any small, but fixed number r > 0,
we have
uj uj Gp 0, in C
1(Br(p)), (5.11)
as 0. For any small > 0, there exist 0 > 0 and r0 > 0 such that
for any < 0 and r < r0
Br(p)eu
j > 1 . (5.12)
As in Proposition 2.8, we have the following Pohozaev identity for (5.1)
6(4 )
Br(eu1 + eu2) = 2
2j=1
Br r(ujn
2 12uj
2)
+2
Br r(u1n
u2n
12u1u
2)
+3(4 )
Br r(eu1 + eu2). (5.13)
It is clear that, letting 0, the left hand side of (5.13) tends to a
number 48(1 ), while the right hand side of (5.13) tends to
4
Brr(G
n2 1
2G2) + 2
Brr(G
n2 1
2G2)
= 3
Br rGr2,
which tends to 24 as r 0, a contradiction. Hence, case (i) does not
happen.
Now we consider case (ii), i.e., u2 is bounded on any compact subset
of \{p}. Let
2 = lim0
lim0
B(p)eu
2.
Since p S2 = 2() for small > 0, we have that 0 < 2 < 1. As in
the proof of (i), we first have
Lemma 5.5. There exists a function w C2() with
ew = 1 2
such that
1. u1u1 converges to G1 in H
1,q()C2loc(\{p}), where G1 satisfies
G1 = 4(2 2)p 4ew 4 and
G1 = 0.
ANALYTIC ASPECTS OF THE TODA SYSTEM I 25
2. u2 u2 converges to w + G2 in H
1,q() C2loc(\{p}), where G2satisfies
G2 = 8ew + 4(2 1)p 4 and
(G2 + w) = 0.
Then, we apply (5.13) again to get a contradiction. In fact, we can
show that in this case its left hand tends to 24(1 + 2) while its right
hand tends 24(22 2 + 1), which is impossible if 0 < 2 < 1. This
implies that case (ii), hence case (b), does not happen either.
Such an argument, using alocal Pohozaev identity, was used in [46]
and [50] for studying the blow up of Liouville type equations.
Case (c). This case is more delicate.
Set S1 = {p1} and S2 = {p2}. Note that p1 6= p2. In view of Theorem
3.1 and the blowup argument given above, uj (j = 1, 2) tends to
uniformly on any compact subset of \{p1, p2}. We first show the
following lemma.
Lemma 5.6. Let uj be the average of uj (j = 1, 2). For any q (1, 2),
we have
uj uj converges to Gj in H
1,q(),
where G1 and G2 satisfy
G1 = 8p1 4p2 4,
G2 = 8p2 4p1 4,
Gj = 0, for j = 1, 2
(5.14)
where y is the Dirac distribution. Moreover,
uj uj converges to Gj in C
2loc(\{p1, p2}). (5.15)
Proof. First we show that for any q (1, 2),
(u1
q + u2q) is bounded. (5.16)
Let q > 2 be determined by 1q+ 1
q= 1. By definition, we know
u1Lq sup{
u1   H
1,q(),
= 0 & H1,q = 1}.
(5.17)
26 JURGEN JOST AND GUOFANG WANG
The Sobolev embedding theorem implies that for any with Lq =
1,
L < c,
for some constant c > 0. Hence,

u1  = 
u1
c
{(8 )eu1 + (4 )eu
2 + (4 )}
c1. (5.18)
It follows that u1Lq c1. Similarly, we have u2Lq c2. This
proves (5.16).
By Theorem 3.1 and Remark 3.3, we can show that
eu1 p1 and e
u2 p2 (5.19)
in the sense of measures as 0. Like (5.18), we have

(u1 u
1 G1) c
(8 2)eu1 8p1
+c
(4 )eu1 4p2 +O()
O(). (5.20)
Therefore, we have
(u1 u1 G1)Lq 0 as 0.
It follows that u1 u1 G1Lq 0 as 0. Hence, we have
u1 u1 G1 in H
1,q(). Similarly, u2 u2 G2 in H
1,q(). Now it
is easy to show (5.15). This proves the Lemma.
Let be a smooth closed curve on with the properties that /
consists of two disjoint component 1 and 2 and p1 1 and p2 2.
Now we consider our system in 1 first. As above, we set
v1 =1
3(2u1 + u
2) and v
2 =
1
3(2u2 + u
1).
Clearly, (v1, v2) satisfies
v1 = (4 )eu1 (4 ),
v2 = (4 )eu2 (4 ).
(5.21)
Lemma 5.7. v2 12(2u2 + u
1) is bounded in C
1(1).
ANALYTIC ASPECTS OF THE TODA SYSTEM I 27
Proof. Define a function v2 satisfying
v2 = (4 )eu2 (4 ), in 1
v2 = 0, on .
Since u2 is bounded from above in 1, v2 is bounded in C
1(1). Now
it is easy to see that v2 v2
13(2u2 + u
1) is also bounded in C
1(1).
Hence the Lemma follows.
From Lemma 5.7, we have
13
1(u1
2 + u22 +u1 u
2) + (4 )
1(u1 + u
2)
= 14
1u1
2 + 12(4 )
1u1 +
12(4 )(2u2 + u
1)1+O(1)
= 14
1u1
2 + (4 )(u2 + u1)1+O(1), (5.22)
where we have used the fact that
1u1 u
1 is bounded, which was
implied by Lemma 5.3. Here 1 is the area of 1. Similarly, we can
get
13
2(u1
2 + u22 +u2 u
2) + (4 )
1(u1 + u
2)
= 14
2u2
2 + (4 )(u2 + u1)2+O(1). (5.23)
Hence, to prove Theorem 5.1 in this case, we only need to show the
following
Lemma 5.8. There exists a constant C > 0 independent of such that
1
4
1u1
2 +1
4
2u2
2 + (4 )(u1 + u2) > C.
Unlike case (a), we cannot use the ordinary MoserTrudinger in
equality directly. Fortunately, the ideas in the proof of the ordinary
MoserTrudinger inequality in [38, 15] can be applied to show Lemma
5.8. We claim that there is a constant C > 0 independent of such
that
1
4
1u1
2 + (4 )u1 > C. (5.24)
Let xj be one of the maximum points of uj, i.e., m
j = u
1(x
j) (j =
1, 2). Set
u1(x) = u1(
1x+ x
1)m
1 and u
2(x) = u
2(
1x+ x
1)m
1,
28 JURGEN JOST AND GUOFANG WANG
where (1)2 = em
1 . Since m1 ,
1 0 as 0. From the
discussion above, we know that u1(0) = max u1 = 0 and u
2
uniformly on any compact domain of 1.
Lemma 5.9. We have that u1 converges to 0 in H1loc(R
2), where 0 =
2 log(1 + x2) is a solution of the Liouville equation
= 8e.
Proof. Note that u = (u1, u2) satisfies
u1 = (8 2)eu1 (4 )eu
2 1(4 ), in (
1)
1B(x1)
u2 = (8 2)eu2 (4 )eu
1 1(4 ). in (
1)
1B(x1)(5.25)
Clearly, (1)1B(x
1) R
2 as 0. For any large, but fixed constant
R > 0, we consider u on BR(0). Define R by
R = (4 )eu2 (1)
2(4 ) := f in BR(0),
R = 0, on BR(0).
The elliptic estimate implies that R 0 in L(BR(0)). Set w
1 =
u1 R. It is clear that
w1 = (8 2)eRew
1 . (5.26)
Since u1 0, u1(0) = 0 and
R is bounded, w
1 is bounded from above
and w1(0) is bounded. It is easy to show that w1, hence u
1, converges
in H1,2(BR). (See, for example, [3].) Now by a diagonal argument, we
have that u1 converges to 0 in H1,2loc (R
2), where 0 satisfies
0 = 8e0
with 0(0) = 0 = max0. A classification of ChenLi [12] implies that
0(x) = 2 log(1 + x2).
Now we need the following
Lemma 5.10. 1. For any small > 0, there exist constants R > 0,
> 0 and C such that
u1(x) (1 3)m1 (4
)(1 ) log x x1+ C,
ANALYTIC ASPECTS OF THE TODA SYSTEM I 29
for any x B(x1)\Br(x
1) with r = (
1)
1R and < .
2. m1 + u1 O(1).
Now from Lemma 5.10, we can finish the proof of our main theorem.
From (5.4), we have
1u1
2 =
1u1u1n
+ 2(4 )
1eu
1u1
(4 )
1eu
2u1 (4 )
1u1
=
1u1u1n
+ 2(4 )
1eu
1m1 (4 )
1eu
2u1
+2(4 )
1eu
1(u1 m
1) (4 )
1u1,
where n is the outer normal of 1. By Lemma 5.6 and equation (5.4),
we have
1u1u1n
=
1
u1n
u1 +
1
u1n
(u1 u1)
=
1
u1n
u1 +O(1)
= u1{2(4 )
1eu
1 (4 )
1eu
2 (4 )1}+O(1).
By Lemmas 5.9 and 5.10, we have
1eu
1(u1 m
1) =
1\B(x1)eu
1(u1 m
1) +
B(x1)eu
1(u1 m
1)
=
B(x1)eu
1(u1 m
1) +O(1)
=
B(1)1(0)
eu1 u1 +O(1)
B(1)1(0)
C1e(11)u1 +O(1) = O(1),
for a small fixed number 1 > 0. Here, we have used (5.27) below.
Hence, using Lemma 5.10 we get
1u1
2 = 2(4 )(m1 u1)
1eu
1 (4 )(
1u1 u
1) +O(1)
4(4 )u1
1eu
1 +O(1)
4(4 )u1 +O(1).
30 JURGEN JOST AND GUOFANG WANG
This is (5.24). Here we have used the fact that
m1
\B(x1)eu
1 o(1), (5.27)
which can be deduced as follows. From (5.4), we have
u1 (8 2)eu1 in \B(x
1).
Let h satisfy
h = (8 2)eu1 , in \B(x
1),
h = 0, on B(x1).
Since u1 is uniformly bounded from above in \B(x1), h is uniformly
bounded. Now applying the maximum principle to ue1 h, together
with Lemma 5.10, we can obtain that u1(x) (1 3)m1 + C for
any x 6 B(x1), which implies (5.27).
Similarly, we can show that
1
4
2u2
2 + 4( )u2 > C,
for some constant C > 0, which, together with (5.24), proves Lemma
5.8, hence Theorem 5.1.
Now it remains to prove Lemma 5.10.
Proof of Lemma 5.10. Wlog, we assume that B2(p1) is a flat disk,
for some small constant > 0, see Remark at the end of Section 3.
Consider in B2(x1)
u1 = 2(4 )eu1 (4 )eu
2 (4 ) =: f.
(5.28)
Recall that uj = uj(
1x+x
1)m
1 (for j = 1, 2). For any small > 0,
by Lemma 5.9 we choose R > 0 and > 0 such that
BR(0)eu
1 > 1
2and
1eu
2 0 is fixed, it is easy to check that
B2(x1)(x y)
u1n
= O(1)
and
B2(x1)u1(x)
(x y)
n= u1 +O(1).
Hence, we have
u1(y)m1 = u
1(y) u
1(x
1)
= 1
2
B2(x1)log
x y
x x1fdx+O(1).
Now it is convenient to write the previous equation as follows.
u1(y) = u1(
1y + x
1)m
1
= 1
2
B2(x1)log
x (1y + x1)
x x1f(1y + x
1)dx
= 1
2
B2(1)1(0)
logx y
xf(x)dx,
where
f 1(x) = 2(4 )eu1 (4 )eu
2 (1)
2(4 ) = (1)2f(1x+ x
1).
Applying the potential analysis, it is easy to show that there is a con
stant C > 0 such that
u1(y) (1 )(4
) log y+ C,
for y R. See, for instance, Lemma 2.4 or [38]. This implies state
ment 1 of the Lemma.
From (5.29) we have
m1 = u1(x
1) =
1
2
B2(x1)log x x1f(x)dx+ u
1 +O(1)
= 1
2
B2(1)1(0)
log 1xf(x)dx+ u1 +O(1)
= 1
2log 1
B2(x1)f(x)dx+ u
1 +O(1)
=m14
B2(x1)f(x)dx+ u
1 +O(1)
2m1 + u1 +O(1).
32 JURGEN JOST AND GUOFANG WANG
which implies Statement 2. Here we have used
B2(1)1(0)
log xf(x)dx C,
for some constant C, which is deduced from Statement 1. Hence, we
finish the proof of the Lemma, hence our main theorem.
Acknowledgement. We would like to thank the referee for his/or her care
ful and critical reading and for pointing out some inaccuracies in the first
version.
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MaxPlanck institute for Mathematics in the Sciences, Leipzig, Ger
many
Email address : [email protected]
Institute of Mathematics, Academic Sinica, Beijing, China
Email address : [email protected]
MaxPlanck institute for Mathematics in the Sciences, Leipzig, Ger
many
Email address : [email protected]