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arXiv:math-ph/0011039v1 23 Nov 2000 (Revised version) ANALYTIC ASPECTS OF THE TODA SYSTEM: I. A MOSER-TRUDINGER INEQUALITY J ¨ URGEN JOST AND GUOFANG WANG Abstract. In this paper, we analyze solutions of the open Toda system and establish an optimal Moser-Trudinger type inequality for this system. Let Σ be a closed surface with area 1 and K = (a ij ) N×N 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 Φ M (u)= 1 2 N i,j=1 Σ a ij (u i u j +2M i u j )N i=1 M i log Σ exp( N j=1 a ij u j ) has a lower bound in (H 1 (Σ)) N if and only if M j 4π, for j =1, 2, ··· ,N. As a direct consequence, if M j < 4π for j =1, 2, ··· ,N M has a minimizer u which satisfies Δu i = M i ( exp( N j=1 a ij u j ) Σ exp( N j=1 a ij u j ) 1), for 1 i N. Date : Nov. 23, 2000. 1
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Transcript of arXiv:math-ph/0011039v1 23 Nov 2000 · Date: Nov. 23, 2000. 1. ... It can be seen as the Frenet...

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    (Revised version)

    ANALYTIC ASPECTS OF THE TODA SYSTEM: I.A MOSER-TRUDINGER INEQUALITY

    JURGEN JOST AND GUOFANG WANG

    Abstract. In this paper, we analyze solutions of the open Toda

    system and establish an optimal Moser-Trudinger 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/math-ph/0011039v1

  • 2 JURGEN JOST AND GUOFANG WANG

    1. Introduction

    Let be a closed surface with area 1. The Moser-Trudinger inequal-

    ity is

    |u|2 + 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

    Moser-Trudinger 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 Chern-Simons 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 system-analog 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 Moser-Trudinger 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 non-Abelian Chern-Simons 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, 2-dimensional Toda lattice) is

    ui =N

    j=1

    aijeuj , for i = 1, 2, , N. (1.8)

    The popular interpretation of the (one-dimensional) Toda lattice is a

    Hamiltonian system which describes the motion of N particles moving

    in a straight line, with exponential interaction. The two-dimensional

    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 Moser-Trudinger 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 Moser-Trudinger 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 Moser-Trudinger 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) |S1|1 and

    S2 = , (b) |S1| = |S2| = 1 and S1 = S2 and (c) |S1| = |S2| = 1

    and S1 S2 = . Here Sj (j = 1, 2) is the blow-up 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 function-the ordinary Moser-Trudinger 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) non-Abelian Chern-Simons model ([32, 34, 35,

  • 6 JURGEN JOST AND GUOFANG WANG

    20]), which can be seen as a non-integrable perturbation of the inte-

    grable Toda system (1.8). For mathematical aspects of the relativistic

    non-Abelian Chern-Simons 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 Moser-Trudinger 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

    2|u1|

    2) = 2

    BRReu1 4

    BReu1

    BReu2xu1,

    R

    BR(|u2r

    |2 1

    2|u2|

    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 12|uj|

    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 non-negative 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 blow-up 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 2-dimensional 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 g|U = 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 blow-up point there is a flat neighborhood.

    4. A Moser-Trudinger 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 Moser-Trudinger 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

    {2|v1|2 + 2|v2|

    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

    {|u1|2 + |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 Moser-Trudinger

  • 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

    Moser-Trudinger 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+)

    (|u1|2 + |u2|

    2 +u1u2)

    6(1+)

    (|u1|2 + |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 semi-continuous.

    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 non-increasing. Choose another se-

    quence ck withckak

    0 andbkck

    0.

  • ANALYTIC ASPECTS OF THE TODA SYSTEM I 17

    It is easy to find a non-increasing 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 + |uk2|2 +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 + |u2|2 +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

    (|u1|2 + |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+2|x|2)2in B0

    log 2

    (1+2|0|2)2in \B0

    u2 =

    12log

    2

    (1+2|x|2)2in B0 ,

    12log

    2

    (1+2|0|2)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 Moser-Trudinger 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

    {2|v1|2 + 2|v2|

    2 2v1v2 + 8v1+

    8v2} 4 log

    e2v1v2 4 log

    e2v2v1

    C, (5.1)

    for any v H2, or equivalently,

    M0(u) =13

    {|u1|2 + |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 blow-up 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 Moser-Trudinger

    inequality. Case (a) is easy to handle, while case (c) is more delicate.

    Case (a). Reduction to the ordinary Moser-Trudinger 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 Moser-Trudinger

    inequality . By Lemma 5.3, we have

    1

    3

    (|u1|

    2 + |u2|2 +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 Moser-Trudinger 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 12|uj|

    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

    n|2 1

    2|G|2) + 2

    Brr(|G

    n|2 1

    2|G|2)

    = 3

    Br r|Gr|2,

    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 blow-up 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 + |u2|q) 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 + |u2|2 +u1 u

    2) + (4 )

    1(u1 + u

    2)

    = 14

    1|u1|

    2 + 12(4 )

    1u1 +

    12(4 )(2u2 + u

    1)|1|+O(1)

    = 14

    1|u1|

    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 + |u2|2 +u2 u

    2) + (4 )

    1(u1 + u

    2)

    = 14

    2|u2|

    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

    1|u1|

    2 +1

    4

    2|u2|

    2 + (4 )(u1 + u2) > C.

    Unlike case (a), we cannot use the ordinary Moser-Trudinger in-

    equality directly. Fortunately, the ideas in the proof of the ordinary

    Moser-Trudinger 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

    1|u1|

    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 + |x|2) 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 Chen-Li [12] implies that

    0(x) = 2 log(1 + |x|2).

    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

    1|u1|

    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

    1|u1|

    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

    2|u2|

    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 x1|fdx+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 x1|f(1y + x

    1)dx

    = 1

    2

    B2(1)1(0)

    log|x y|

    |x|f(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 x1|f(x)dx+ u

    1 +O(1)

    = 1

    2

    B2(1)1(0)

    log |1x|f(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 |x|f(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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  • ANALYTIC ASPECTS OF THE TODA SYSTEM I 35

    Max-Planck institute for Mathematics in the Sciences, Leipzig, Ger-

    many

    E-mail address : [email protected]

    Institute of Mathematics, Academic Sinica, Beijing, China

    E-mail address : [email protected]

    Max-Planck institute for Mathematics in the Sciences, Leipzig, Ger-

    many

    E-mail address : [email protected]