A FUNCTION THEORETIC VIEW OF THE MEAN FIELD EQUATIONS ON TORI
CHANG-SHOU LIN AND CHIN-LUNG WANG
ABSTRACT. The mean field equations on flat tori is analyzed purely in terms of the theory of elliptic functions. Some results which were known previously through PDE techniques are reproved here using more el-ementary methods. Also a general method to construct explicit solu-tions is developed when the topological degree is known to be non-zero. Hopefully our method will lead to all solutions. This article is a supple-mentary reading to our recent work [7].
1. INTRODUCTION
1.1. A brief historical review. The mean field equation on a flat torus T = C/Zω1+Zω2takes the form
(1.1) 4u+ρeu =ρδ0, ρ∈R+.
It is originated from the prescribed curvature problem in geometry. It also comes from statistical physics as the mean field limits of the Euler flow. Recently it was shown to be related to the self dual condensation of Chern-Simons-Higgs model. We refer to [2], [3], [4], [5], [8] and [9] for the recent development of this subject.
When ρ 6= 8mπ for m ∈ Z, it has been recently proved in [3], [4], [5]
that the Leray-Schauder degree is non-zero, so the equation always has solutions, regardless on the actual shape of T.
For the first dangerous value ρ = 8π where the degree theory fails, the following results were recently obtained by the authors.
Theorem 1.1([7]). For ρ =8π, the mean field equation on a flat torus has solu-tions if and only if the Green function G(z)has critical points other than the three half period points. Moreover, each extra pair±z0 of critical points corresponds to an one parameter scaling family of solutions.
It is known that for rectangular tori G(z)has precisely the three obvious critical points, namely the half periods12ωi, hence for ρ =8π equation (1.1) has no solutions. However for ω1 = 1 and τ = ω2 = eπi/3, it is shown in [7] that there are five critical points and solutions of (1.1) exist. (The two others are±1
3(ω1+ω2).)
Theorem 1.2([7]). The Green function has at most five critical points. Equiva-lently, for ρ=8π, the mean field equation on a flat torus has at most one solution up to scaling.
In fact no direct proof of the first statement is known. Instead, the unique-ness theorem was proved using PDE methods first, and then Theorem 1.1 implies that there are at most 5 critical points.
The critical point equation for z0 = tω1+sω2is a very simple equation in elliptic functions (c.f. Lemma 2.3):
(1.2) ζ(tω1+sω2) =tη1+sη2.
However, it remains a challenge to analyze it within function theory. The proof of Theorem 1.2 in [7] is based on the continuity method ap-plied to the one parameter family of mean field equations
4u+ρeu=ρδ0, ρ∈ [4π, 8π]
on T within even solutions, where for the starting point ρ = 4π the so-lution is uniquely constructed through integration of Weierstrass elliptic functions.
1.2. The content of the paper. As the title suggests, this paper concerns the function theoretic aspect of the mean field equations and Green functions on tori without appealing to the techniques of partial differential equations. Specifically, we give a proof of Theorem 1.1 following the same line in [7] with one technical difference: For the proof in [7] we used the fact:
Theorem 1.3. For ρ = 8π, the blow-up points of solutions to the mean field equation4u+ρeu =ρδ0on a torus T are precisely those non half-period critical points of the Green function G.
This now follows as a consequence (c.f. Corollary 5.1) of our proof. Secondly we explore the construction of solutions for ρ=4π and ρ=8π in a more systematic manner aiming at investigating the general situations for ρ = 4πl, l ∈ N. As in [7], we use the classical Liouville theory to find
local developing maps f ’s of a solution u:
(1.3) u=c1+log
|f0|2
(1+ |f|2)2,
and glue them to a global meromorphic function onC (c.f. Lemma 3.1). The developing maps are locally unique up to a PSU(1)action. By com-paring f(z+ωi)with f(z)for i = 1, 2, we achieve two types of solutions.
To describe them, it is convenient to consider the logarithmic derivative
(1.4) g= (log f)0 = f
0 f in our discussions. Then we have (see §3)
Type I: Topological solutions, where g is elliptic on T0 =C/Zω1+Z2ω2: f(z+ω1) = −f(z), f(z+ω2) = − ¯q2 f(z). (1.5)
Type II: Blow-up solutions, where g is elliptic on T: f(z+ω1) =e2iθ1f(z), f(z+ω2) =e2iθ2f(z). (1.6)
The type I solutions are called topological since a Leray-Schauder degree counting formula is known when ρ6=8mπ [3], [4], [5]. The answer is 1 for
ρ=4π and 2 for ρ=12π.
To warm up, we repeat in §5 the proof in [7] showing that there is a unique solution for ρ=4π and no solutions for ρ=8π of Type I.
In §6 we proceed to construct two explicit solutions for ρ = 12π by in-vestigating in details the properties satisfied by g and f . It is reasonable to believe that solutions constructed from the holomorphic data will con-tribute a positive degree in the degree counting. Thus the two solutions constructed should then be all the solutions.
The type II solutions are called blow-up solutions since if f is a develop-ing map of u then eλf , λ ∈ R will also be a developing map for another
solution
(1.7) uλ(z):=c1+log
e2λ|f0(z)|2 (1+e2λ|f(z)|2)2.
Moreover any zero of f will be a blow-up point for the sequence uλ.
The following result is well-known. Here we give a purely elementary proof of it through elliptic functions (c.f. Theorem 4.4).
Theorem 1.4. For ρ = 4πl with l being odd, there are no type II, i.e. blow-up, solutions to the mean field equation
4u+ρeu =ρδ0 in T.
We hope that the strategy developed here from §4 to §6 will be useful for later studies on the mean field equations on tori with ρ = 4πl, l ≥ 3, especially on the explicit constructions of solutions.
2. GREEN FUNCTIONS VIA ELLIPTIC FUNCTIONS
2.1. Green functions via theta functions. Let T=C/Zω1+Zω2be a flat torus. As usual we let ω3 = ω1+ω2. The Green function G(z, w)is the unique function on T which satisfies
(2.1) −4zG(z, w) =δw(z) − 1
andR
TG(z, w)dA = 0. It has the property that G(z, w) = G(w, z)and it is smooth in(z, w)except along the diagonal z= w, where
(2.2) G(z, w) = − 1
2π log|z−w| +O(|z−w| 2) +C
for a constant C which is independent of z and w. Moreover, due to the translation invariance of T we have that G(z, w) = G(z−w, 0). Hence it is also customary to call G(z) := G(z, 0)the Green function. It is an even function with the only singularity at 0.
The Green function can be explicitly solved in terms of elliptic functions. It takes a simple form using theta functions.
Consider a torus T =C/Λ with Λ = Z+Zτ, a lattice with ω1 = 1 and
ω2=τ= a+bi, b>0. Let q=eπiτwith|q| =e−πb <1. The theta function ϑ1(z; τ)is the exponentially convergent series
ϑ1(z; τ) = −i ∞
∑
n=−∞ (−1)nq(n+21)2e(2n+1)πiz =2 ∞∑
n=0 (−1)nq(n+21)2sin(2n+1)πz. (2.3)For simplicity we also write it as ϑ1(z). It is entire with
ϑ1(z+1) = −ϑ1(z),
ϑ1(z+τ) = −q−1e−2πizϑ1(z), (2.4)
which has simple zeros at the lattice points (and no others). As usual we denote z=x+iy.
Lemma 2.1. Up to a constant C(τ), the Green function G(z, w)for the Laplace operator4on T is given by (2.5) G(z, w) = − 1 2πlog|ϑ1(z−w)| + 1 2b(Im(z−w)) 2+C(τ). Proof. Let R(z, w) be the right hand side. Clearly for z 6= w we have
4zR(z, w) =1/b which integrated over T gives 1. Near z= w, R(z, w)has the correct behavior. So it remains to show that R(z, w)is indeed a function on T. From the quasi-periodicity, R(z+1, w) =R(z, w)is obvious. Also
R(z+τ, w) −R(z, w) = − 1
2π log e
πb+2πy+ 1
2b((y+b)
2−y2) =0. These properties characterize the Green’s function up to a constant.
By the translation invariance of G, it is enough to consider w=0. Let G(z) =G(z, 0) = − 1
2π log|ϑ1(z)| + 1 2by
2+C(τ).
If we represent the torus T as centered at 0, then the symmetry z 7→ −z shows that G(z) = G(−z). By differentiation, we get∇G(z) = −∇G(−z). If−z0= z0in T, that is 2z0 =0 (mod Λ), then we get∇G(z0) =0. Hence
we obtain the half periods12,12τand12(1+τ)as three obvious critical points
of G(z)for any T.
By computing ∂G/∂z= 12(Gx−iGy)we find
Corollary 2.2. The equation of critical points z=x+iy of G(z)is given by
(2.6) ∂G ∂z ≡ −1 4π (log ϑ1)z+2πi y b =0.
2.2. Weierstrass elliptic functions and periods integrals. Now we trans-late these to Weierstrass’ elliptic functions. Recall that ζ = −R ℘ = z−1+ · · · and σ=expR ζ =z+ · · ·. From
(log σ(z))0 =ζ(z),
it is known that σ(z) is entire, odd with a simple zero on lattice points. Moreover,
(2.7) σ(z±ωi) = −e±ηi(z± 1
2ωi)σ(z).
This is similar to the theta function transformation law, indeed (2.8) σ(z) =eη1z2/2ϑ1(z) ϑ01(0). Hence (2.9) ζ(z) −η1z= logϑ1(z) ϑ01(0) z = (log ϑ1(z))z.
It is extremely important to understand the geometric meaning of Fi(z):=ωiζ(z) −ηiz, i=1, 2,
due to the following
Proposition 2.3. Write z=tω1+sω2, then
(2.10) Gz = −
1
4π(ζ(z) −η1t−η2s). In particular, z is a critical point of G if and only if (2.11) ζ(tω1+sω2) =tη1+sη2. Moreover, it is also equivalent to that
(2.12) Fi(z) ∈iR, i=1, 2.
Proof. Since z= x+yi =tω1+sω2= t+sa+sbi, we have
(log ϑ1)z+2πiy
b = ζ−η1(t+sω2) +2πis=ζ−tη1−sη2, where the Legendre relation η1ω2−η2ω1=2πi is used.
The calculation indeed shows that
And similarly
ω2ζ−η2z = (ζ−tη1−sη2)ω2+2πit.
If z is a critical point then it is clear that both Fi(z)’s are purely imaginary. Conversely if both Fi(z)’s are imaginary, then the above equalities implies that(ζ−tη1−sη2)ωi’s are linearly dependent. This is possible only if ζ−
tω1−sω2 =0. The lemma is proved.
Finally, Fi(z)can be interpreted as certain period integrals of the mero-morphic one form
Ω(z, ξ)dξ := ℘
0(z)
℘(ξ) − ℘(z)dξ. Lemma 2.4. Up to2πiN, 2Fi(z)is the period integral
Z
Li Ω dξ
along the i-th fundamental cycle Li(the i-th complete period). Proof. Recall the addition law (c.f. [1])
(2.13) ℘(z) − ℘(y) = −σ(z+y)σ(z−y) σ2(z)σ2(y) . Then Ω= d dzlog(℘(z) − ℘(ξ)) =2ζ(z) −ζ(z+ξ) −ζ(z−ξ). Hence Z Li Ω dξ=2ωiζ(z) +log σ(ξ−z) σ(ξ+z) a+ωi a =2ωiξ(z) −2ηiz+2πim for some m∈ {−1, 0, 1}, where the last equality uses (2.7).
3. LIOUVILLE THEORY WITH SINGULAR DATA
3.1. Theorem of Liouville. We start with a quick review of the Liouville theory. A classical theorem of Liouville says that any solution u of
4u+ρeu=0
in a simply connected domain D ⊂C can be represented by
(3.1) u=c1+log
|f0|2
(1+ |f|2)2,
where f is a meromorphic function in D and c1 = log 4/ρ. Such an f is called a developing map of u and can be selected to be holomorphic.
Developing maps are not unique, different developing maps are related by M ¨obius transformations. Indeed it is easily checked that u and f satisfy
(3.2) uzz− 1 2u 2 z = f000 f0 − 3 2 f00 f0 2 .
The right hand side of (3.2) is the Schwartz derivative of f . Thus for any two developing maps f and ˜f of u, there exists S = p −¯q
q ¯p
∈ PSU(1)
(i.e. p, q∈C and|p|2+ |q|2=1) such that
(3.3) ˜f=S f := p f − ¯q
q f+ ¯p.
Here is a useful observation: By conjugating a matrix U ∈ PSU(1)to ˜f and f with USU−1=e
iθ 0 0 e−iθ
for some θ∈ R, we may achieve that
(3.4) U ˜f=e2iθU f .
3.2. Liouville theory on tori with isolated singular data. It is a challeng-ing problem to extend the Liouville theory to Riemann surfaces and with singular source. Here we consider the simplest genus one case with iso-lated singularities, namely the mean field equation
(3.5) 4u+ρeu=ρδ0, ρ∈R+
in a flat torus T, with δ0being the Dirac measure at 0 and the volume of T is normalized to be 1.
Lemma 3.1. For T = C/Zω1+Zω2 and ρ = 4πl, l ∈ N, by gluing the
f ’s among simply connected domains, (3.1) holds on the whole C with f being a meromorphic function.
Proof. This is stated and proved in [5] for rectangular tori with l = 2, but the proof works for the general case which is included below.
It is sufficient to prove the lemma for any chosen f . Let z = e2πiw be the universal covering map H → ∆× from the upper half plane to the punctured disk (of some small radius which is assumed to be 1 to save notations) and let F(w) = f(z) = f(e2πiw). Then
F(w+1) =SF(w)
for some S ∈ PSU(1). By the above observation, up to a conjugation, we may start with another f so that F(w+1) =e−2iθF(w)for some θ∈ [0, π).
Now letΨ(w) =e−2iθwF(w). Then
Ψ(w+1) =e−2iθ(w+1)F(w+1) =e−2iθwF(w) =Ψ(w). HenceΨ(w)comes from a meromorphic function ψ(z)on∆×.
By [6], §2, Lemma 4, if ψ has essential singularity at z=0 then f = zθ/πψ
takes any value inC infinitely many times except at most one value. In our case ∞> Z ∆×e udA=Z ∆× |f0|2 (1+ |f|2)2dA
with the RHS being the spherical area, under the inverse stereographic pro-jections, covered by f(∆×). This implies that ψ extends to a meromorphic function on the whole disk∆.
Let n=ordz=0ψ∈ Z and ψ=zng. Then f =zag with a=n+θ/π and
(3.6) u=c1+2 log
|z|a−1|ag+zg0| 1+ |zag|2 .
If a =0 then n= 0 and θ = 0 (since 0≤ θ < π). In this case f = g= ψ
is holomorphic at 0. So we may assume that a6=0.
For ρ=4πl with l∈N, the asymptotic of u at z=0 is given by u(z) ∼2l log|z|.
On the other hand, by (3.6) the asymptotic is given by (3.7) u(z) ∼2(|a| −1)log|z|.
In particular, |a| = l+1 ∈ N, which forces θ = 0 because 0 ≤ θ < π.
Moreover f =z±(l+1)g is meromorphic at z=0.
Now we look for the constraints. The first type of constraints are im-posed by the double periodicity of the equation. By applying (3.3) to f(z+ ω1)and f(z+ω2), we find S1and S2in PSU(1)with
f(z+ω1) =S1f , f(z+ω2) =S2f . (3.8)
These relations also force that S1S2 = ±S2S1since A≡ −A in PSU(1). After conjugating a matrix in PSU(1), we may and will assume that S1 = eiθ 0
0 e−iθ
for some θ∈R. Let S2 = pq −¯p¯q
then (3.8) becomes f(z+ω1) =e2iθf(z),
f(z+ω2) =S2f(z). (3.9)
Since S1S2 = ±S2S1, a direct computation shows that there are three possibilities:
(1) p=0 and eiθ = ±i; (2) q=0; and
(3) eiθ = ±1.
In case (3) we get that S1 is the identity. So by another conjugation we may assume that S2is diagonal and it is reduced to case (2).
Explicitly the two cases read as Type I: f(z+ω1) = −f(z), f(z+ω2) = − ¯q2 f(z). (3.10)
Type II:
f(z+ω1) =e2iθ1f(z), f(z+ω2) =e2iθ2f(z). (3.11)
The second type of constraints are imposed by the Dirac singularity of (3.5) at 0. By the argument from (3.6) to (3.7), we have
Lemma 3.2. (1) If f(z)has a pole at z0 ≡ 0 (mod ω1, ω2), then the order k=l+1.
(2) If f(z) =a0+akzk+ · · · is regular at z0 ≡0 (mod ω1, ω2)with ak 6= 0 then k=l+1.
(3) If f(z)has a pole at z0 6≡0 (mod ω1, ω2), then the order is 1.
(4) If f(z) = a0+ak(z−z0)k+ · · · is regular at z0 6≡ 0 (mod ω1, ω2) with ak 6=0 then k=1.
In particular, modulo ω1, ω2, f has only simple zeros and poles outside 0.
4. THE GENERAL STRUCTURE OF SOLUTIONS
In the study of mean field equations with ρ = 4πl, l ∈ N, the
over-all important object to be considered is the logarithmic derivative g of the developing map f :
g := (log f)0 = f
0 f .
It is clear that any zero or pole of f contributes a simple pole of g and the order k∈Z is simply the residue.
Thus the zeros of g must occur at a regular point z0of f with f(z0) 6=0. By Lemma 3.2, if z06≡0 then f0(z0) 6=0 and z0is not a zero of g. Hence we must have z0 ≡0 and it contributes a zero of g of order l.
4.1. Type I: Topological solutions. In Type I, we have g(z+ω1) =g(z),
g(z+ω2) = −g(z). (4.1)
Hence g is an elliptic function on T0 =C/Zω1+Z2ω2of the form
(4.2) g(z) = A σ
l(z)σl(z−
ω2) ∏l
i=1σ(z−pi)∏li=1σ(z−qi)
for some constant A. The location of poles pi’s and qi’s are 2l distinct points in T0 by Lemma 3.2. They are those (simple) zeros and poles of f which are constrained by
(4.3)
∑
pi+∑
qi =lω2.Since g(z+ω2) = −g(z), after some reordering we may assume that (4.4) qi ≡ pi+ω2 (mod ω1, 2ω2).
(This gives a correspondences between zeros and poles of f .) Combining with (4.3) we find
(4.5) p1+ · · · +pl ≡ m
2ω1+nω2 (mod ω1, 2ω2) for m, n∈ {0, 1}. This gives rise to the first equation of pi’s.
Lemma 4.1. We may assume that p1+ · · · +pl = 12ω1.
Proof. By modifying pi’s and qi’s by elements in Zω1+Z2ω2, there are only four cases of(m, n)in (4.5) needs to be considered. Namely
(m, n) = (i) (0, 0), (ii) (1, 0), (iii) (0, 1) and (iv) (1, 1). During the proof we may assume that qi = pi+ω2for i=1, . . . , l−1. For (i), we may further set ql = pl+ω2. Then
(4.6) g(z+ω2) = A
σl(z+ω2)σl(z) ∏l
i=1σ(z+ω2−pi)∏li=1σ(z+ω2−qi)
By transforming g(z+ω2)into g(z)using the periodicity property (2.7) on T0, we get an automorphic factor
(−1)lexp lη2z
(−1)lexp η
2∑li=1(z−pi)
=1.
That is g(z+ω2) =g(z), which is the not the case we want. For (ii), we may set
ql = pl+ω2−ω1
to match the sum constraint (4.3). Again by transforming g(z+ω2)as in (4.6) into g(z)in (4.2) we get the automorphic factor
(−1)lexp lη 2z
(−1)l+1exp η
2∑li=−11(z−pi) + (η2−η1)(z−pl− 12ω1) +η1(z−pl− 12ω1) which is−1 since∑ pi = 12ω1. Hence g(z+ω2) = −g(z)as desired.
For (iii), we may set
ql = pl−ω2.
By changing the role of pl and qlthis is reduced to case (i). Similarly for (iv) we may set
ql = pl−ω2−ω1.
Remark 4.2. For convenience sometimes we would like to assume that pi’s are simple zeros of f while qi’s are simple poles of f . In doing so we may also need to consider the possibility that
p1+ · · · +pl =
ω1 2 +ω2.
Alternatively we may switch the role of pland ql and still use p1+ · · · +pl =
ω1 2 .
Other equations come from the residue consideration. Let
(4.7) rj =
σl(pj)σl(pj−ω2) ∏l
i=1,6=jσ(pj−pi)∏li=1σ(pj−qi) .
Then the residue of g at z= pjis Arjand we must have, under the arrange-ment that pi’s are (simple) zeros of f except perhaps pl, that
(4.8) r1 =r2 = · · · =rl−1= ±rl.
There are (at most) two sets of l−1 independent equations from it. To-gether with (4.3) we obtain two systems of non-linear system of l equations in l variables.
Since this system is holomorphic in the variables pi ∈T0. The solution set actually defines an analytic subvariety V ∈ (T0)n(which is in fact algebraic by Chow’s theorem). It is expected that V consists of a finite number of points.
Naively the solution structure in this case is topological because the in-tersection number N of the l hypersurfaces defined by the l equations is independent of the shape of the torus T.
However more care needs to be taken since some of the solutions may lead to pi = pj for some i 6= j and need to be excluded. Futhermore we need to integrate
f =exp
Z
g
to check whether (3.10) is satisfied. In fact f(z+ω1) = −f(z)corresponds to the statement that
(4.9)
Z
L1
g(z)dz≡πi (mod 2πi).
Later we will check these technical issues through explicit determina-tions for small values of l up to l = 3. In all the solutions we found, there must be some pi = 12ω1 and the function g is odd. This will lead to (4.9) (c.f. §5.1) and so the solution is topological (independent of the shape of T). For general odd l we expect this to be true but we do not know how to prove it in the full generality.
4.2. Type II: Scaling families and blow-up solutions. In Type II, it follows from (3.11) that g is an elliptic function on T. And by the discussion at the beginning of this section g must take the form
(4.10) g(z) = f 0(z) f(z) = A σl(z) ∏l i=1σ(z−pi) with∑ pi =0.
As in Type I, there are other l−1 equations arising from the residues. Let
(4.11) rj =
σl(pj) ∏l
i=1,6=jσ(pj−pi)
so that the residue of g at z = pj is given by Arj. Since f has only simple zeros and poles outside 0, we have Arj = ±1 and this leads to equations (4.12) r1= ±rj, j=2, . . . , l.
At this point the solutions seem to be topological. But in fact they are not. Recall that
f(z) = f(0)exp
Z z
0 g
(w)dw.
Lemma 4.3. In order for f to verify (3.11), it is equivalent to require that the
periods integrals are purely imaginary: (4.13)
Z
Li
g(z)dz∈iR, i=1, 2.
Later we will see how this lemma for ρ=8π links to the critical points of Green functions via Lemma 2.4. It is known [7] that the structure (or even the number) of critical points depends heavily on the geometry of T.
Another characteristic feature for Type II is that any solution must exist in an one parameter scaling family of solutions.
To see this, notice that if f is a developing map of solution u then eλf also
satisfies (3.11) for any λ∈ R. In fact eλf is a developing map of u
λdefined
by
(4.14) uλ(z) =c1+log
e2λ|f0(z)|2
(1+e2λ|f(z)|2)2 and it is clear that uλis a scaling family of solutions of (3.5).
To understand the quantitative behavior of uλ as λ varies, we notice
also that f must have zero in T. For otherwise (3.11) implies that 1/ f is a bounded entire function onC which by Liouville’s elementary theorem f reduces to a constant, which is absurd.
Let z0be a zero of f . We know that z0 6≡0 and f0(z0) 6=0. Thus
while if f(z) 6=0 then
(4.16) uλ(z) ∼ −2λ→ −∞ as λ→ −∞.
It is customary to call points like z0blow-up points.
We close this section by noticing an elementary consequence.
Theorem 4.4(=Theorem 1.4). For ρ=4πl with l being odd, there are no type II, i.e. blow-up, solutions to the mean field equation
4u+ρeu =ρδ0 in T.
Proof. If there is a solution u with developing map f , then g = f0/ f is elliptic on T with residues at pi, i = 1, . . . , l, to be±1. Since l is odd, the sum of residues is non-zero which is a contradiction.
5. RESULTS IN[7]REVISITED: ρ=4π OR8π
To illustrate how the general principle outlined in the last section works in practice, we will work out in this section solutions of the mean field equations for ρ = 4π and ρ = 8π. These results are contained in [7], but here we will treat the 8π case slightly differently without using the knowl-edge that blow-up points are critical points of Green functions. In fact this will follow from our current treatment.
5.1. The case ρ = 4π. In Type II, g is elliptic on T. Since l = 1, g has a simple zero at z = 0, hence also a simple pole at z= p1. However there is no such elliptic functions (or using p1 =0 to get a contradiction).
In Type I, g is elliptic on T0 and
(5.1) g(z) = A σ(z)σ(z−ω2)
σ(z−p1)σ(z−q1)
with p1+q1 = ω2. By Lemma 4.1, there is a unique solution of(p1, q1)up to order modulo ω1, 2ω2given by
(5.2) p1 =
ω1
2 ; q1= −
ω1 2 +ω2.
It is easily checked, using (2.7), that g(z+ω1) = g(z), g(z+ω2) = −g(z) and
(5.3) g(−z) = −g(z).
The residues of g at p1 and q1 are equal to Ar and −Ar respectively, where r = σ( 1 2ω1)σ(12ω1−ω2) σ(ω1−ω2) .
Let A=1/r, then f(z)is uniquely defined up to a factor f(0). There is an unique choice of f(0)up to a norm one factor such that c := f(ω2)f(0)has
By integrating g(z+ω1) =g(z)we get f(z+ω1) =c0f(z)where c0 = f(ω1) f(0) =exp Z ω1 0 g(z)dz.
To evaluate the period integral, notice that the following anti-symmetric property at ω1/2 holds: gω1 2 +u = g−ω1 2 −u = −gω1 2 −u .
By using the Cauchy principal value integral and the fact that the residue of g at 12ω1is±1, we get (5.4) Z ω1 0 g (z)dz=0± 1 2×2πi= ±πi and so c0 = −1.
Thus f gives rise to a topological solution of (3.5) for ρ = 4π. The de-veloping map for the other choice A = −1/r is 1/ f(z)which leads to the same solution. Since equation (3.5) is invariant under z 7→ −z, the unique solution is necessarily even. This can also be seen directly from our con-struction.
5.2. The case ρ =8π. In Type I, let℘(z), ζ(z)and σ(z)be the Weierstrass elliptic functions on T0. Recall that for ˜ω1 = ω1, ˜ω2 = 2ω2 and ˜ω3 =
ω1+2ω2,
(5.5) σ(z±ω˜i) = −e±ηi(z± 1
2ω˜i)σ(z). Now l=2 and (4.10) reads as
(5.6) g(z) = A σ
2(z)σ2(z−ω 2)
σ(z−a)σ(z−b)σ(z−c)σ(z−d)
with four distinct simple poles such that a+b+c+d =2ω2. We will show that such a function g(z)does not exist.
By Lemma 4.1, we have (a, b, c, d) = (a,−a+ 12ω1, a+ω2,−a− 12ω1+
ω2).
The residues of g at a, b, c and d are equal to −Ar, Ar0, Ar and −Ar0 respectively, where r= σ 2(a+ ω2)σ2(a) σ(ω2)σ(2a− 1 2ω1+ω2)σ(2a+ 12ω1) and r0 = σ 2(a−1 2ω1)σ2(a− 1 2ω1+ω2) σ(2a−1 2ω1)σ(2a− 12ω1+ω2)σ(ω1−ω2) . We must have r= ±r0. Using (5.5), this is equivalent to
(5.7) σ
2(a+
ω2)σ2(a)
σ2(a+12ω1)σ2(a−12ω1+ω2)
To solve a from this equation, by substituting y = 12ω˜i into the addition formula (2.13) and using (5.5), we get
(5.8) ℘(z) −ei =
σ2(z+12ω˜i)
σ2(z)σ2(12ω˜i) e−ηiz. With (5.8), the “+” case in equation (5.7) simplifies to
℘a− ω1
2 +ω2
−e1 = ℘(a) −e1. That is, 2a≡ 1
2ω1−ω2. But this implies that b≡c, a contradiction.
Similarly, the “−” case in (5.7) simplifies to (using the Legendre relation)
℘a+ ω1
2
−e3 = ℘(a) −e3.
That is, 2a+12ω1 ≡0. This leads to c≡ d, which is again a contradiction. In Type II, g is elliptic on T and we have
(5.9) g(z) = A σ
2(z)
σ(z−z0)σ(z+z0)
where f has a zero at z0and a pole at−z0. In particular z06= −z0in T and we conclude that z0 6=ωk/2 for any k∈ {1, 2, 3}.
Now the residue of g at z0is 1, hence
(5.10) Aσ 2(z 0) σ(2z0) =1=⇒ A= σ(2z0) σ2(z0) . Then we may transform g into Weierstrass functions as
g(z) = σ(2z0) σ2(z0) σ2(z) σ(z−z0)σ(z+z0) = σ(2z0) σ4(z0) σ2(z)σ2(z0) σ(z−z0)σ(z+z0) = ℘ 0(z 0) ℘(z) − ℘(z0) (5.11)
where the addition law (2.13) and the duplication formula
℘0(z) = −σ(2z)/σ4(z) are used. Then f(z) = f(0) Z z 0 g (ξ)dξ= f(0) Z z 0 ℘0(z0) ℘(ξ) − ℘(z0) dξ is a developing map for a solution precisely when
Z
Li
℘0(z0)
℘(ξ) − ℘(z0)
dξ ∈iR, i =1, 2.
By Proposition 2.3 and Lemma 2.4, this is equivalent to that z0 is a critical point of the Green function G(z)on T.
Together with §4.2, we have recovered the following known result which was first proved in [2] using PDE methods.
Corollary 5.1. For ρ = 8π, the blow-up points of solutions to the mean field equation4u+ρeu =ρδ0on a torus T are precisely those non-half period critical points of the Green function G.
6. THE CASEρ =4πlWITHl=3 By Theorem 4.4, there are no Type II solutions.
6.1. Constructing g. For Type I solutions, g(z) = (log f(z))0 = f0(z)/ f(z)
is elliptic on T0 =C/Zω1+Z2ω2and we may write
g(z) =A σ
3(z)σ3(z−
ω2)
σ(z−a)σ(z−b)σ(z−c)σ(z−d)σ(z−e)σ(z− f)
where a, b, c are simple zeros and d, e, f are simple poles of f(z). From
g(z+ω2) = −g(z),
Lemma 4.1 gives rise to two possibilities: Let a, b be free variables. Then d= a+ω2, e=b+ω2 and either (6.1) c= −a−b+ ω1 2 , f = −a−b− ω1 2 +ω2 or (6.2) c= −a−b+ ω1 2 +ω2, f = −a−b− ω1 2 .
The second case can be included into the first one if we change the role of c and f , namely to require that c is a pole and f is a zero of f(z).
We fist consider the former case (6.1). Denote the residues of g(z)at a, b, c by Ar1, Ar2and Ar3respectively. Then Ar1 = Ar2= Ar3=1. Here
r1= − σ3(a)σ3(a−ω2) σ(a−b)σ(2a+b− ω1 2 )σ(ω2)σ(a−b−ω2)σ(2a+b+ ω1 2 −ω2) , r2= − σ3(b)σ3(b−ω2) σ(b−a)σ(2b+a−ω1 2 )σ(ω2)σ(b−a−ω2)σ(2b+a+ ω1 2 −ω2) and r3 = σ3(a+b− ω21)σ3(a+b−ω21 +ω2)e−η1( ω1 2 −ω2) σ(2a+b− ω1 2 )σ(2b+a− ω1 2 )σ(2a+b− ω1 2 +ω2)σ(2a+b− ω1 2 +ω2)σ(ω2) . We consider the equation r1=r2, which simplifies to
σ3(a)σ3(a−ω2) σ(2a+b−ω1 2 )σ(2a+b+ ω1 2 −ω2) = −eη2(a−b) σ 3(b)σ3(b− ω2) σ(2b+a−ω1 2 )σ(2b+a+ ω1 2 −ω2) . (6.3)
The obvious candidate of solutions would be those coming from equat-ing the triple powers of sigma functions, namely to force that b = ±a or
b= ±(a−ω2). By our assumption b6=a and also b6=a−ω2(otherwise r1 is not defined). So the candidates are b= −a and b= −a+ω2.
Indeed at most one choice will do since g(z+ω2) = −g(z). It is easily checked, using (5.5), that for b= −a the LHS= RHS in (6.3) while (hence) for b= −a+ω2the LHS= −RHS.
Thus b= −a and c= 12ω1. Then g(z) =
A σ 3(z)σ3(z− ω2) σ(z−a)σ(z+a)σ(z− ω1 2 )σ(z−a−ω2)σ(z+a−ω2)σ(z+ ω1 2 −ω2) . It is very important to notice, and easy to verify, that
(6.4) g(−z) = −g(z).
Indeed the transformation factor from g(−z)to g(z)is
(−1)3exp 3η2z
exp(η1z+η2(z+a) +η2(z−a) + (−η1+η2)z)
= −1.
We proceed to solve r1=r3under b= −a, c = 12ω1. (Another possibility that r1 = −r3will be treated later.) The equation r1 =r3reads as
(6.5) σ 3(a)σ3(a−ω 2) σ(2a)σ(2a−ω2) = −e−η1(ω12 −ω2)+(η1−η2)a σ 3(ω1 2 )σ3( ω1 2 −ω2) σ(a+ ω1 2 )σ(a+ ω1 2 −ω2) . Recall the duplication formula
(6.6) σ(2a)σ(w1)σ(w2)σ(w3) 2σ(a)σ(a+w1)σ(a+w2)σ(a+w3)
=1 for any three half periods wiwith w1+w2+w3 =0 (c.f. [10]).
With this in mind, we may rewrite (6.5) as
σ2(a)σ2(a−ω2) σ(2a−ω2)σ(ω2) = −e−η1(ω12 −ω2)+(η1−η2)a× σ(2a)σ(ω1 2 )σ( ω1 2 −ω2)σ(ω2) σ(a)σ(a−ω2)σ(a+ ω21)σ(a+ω21 −ω2) .σ 2(ω1 2 )σ2( ω1 2 −ω2) σ2(ω2) . (6.7)
Using the addition law (2.13) and its consequence (5.8)
℘(z) −ei =
σ2(z+ ω˜2i) σ2(z)σ2(ω˜2i)
e−ηiz, together with (6.6), we simplify (6.7) to
−1 ℘(a) − ℘(a−ω2) = −2 1 ℘(ω2−ω1 2 ) − ℘( ω1 2 ) . That is (6.8) ℘(a) − ℘(a−ω2) = 1 2(e3−e1).
The LHS has 2 poles at a = 0 and a = ω2 with total order 4, hence the equation have 4 zeros a = ±p and a = ±p0. Notice that a is not a half
period. To see that p6=p0, i.e. they are simple zeros, we use the half period formula (c.f. [10]): (6.9) ℘a+ω˜2 2 =e2+ (e1−e2)(e3−e2) ℘(a) −e2 . By substituting this into (6.8), a simple calculation gives (6.10) ℘(a) =e2+ e3 −e1 4 ± 1 4 q (e3−e1)2+16(e1−e2)(e3−e2). It is then clear that there are 4 in-equivalent solutions. Since b = −a, these solutions give rise to 2 independent solutions of g.
To complete our discussion we still needs to consider the case r1 = −r3. In exactly the same calculation with sign changed, we arrive at the equation (6.11) ℘(a) − ℘(a−ω2) = −1
2(e3−e1).
The important observation is that a7→ a+ω2exchanges equations (6.8) and (6.11). Hence the solutions of (6.11) is given by
±p+ω2 and ±p0+ω2.
By our general discussion of the structure of solutions in §4.1, this shows that in fact r1= −r3gives rise to the same set of solutions as r1=r3.
To be concrete, if in r1 =r3there is a solution of(a, b, c, d, e, f)to be
(6.12) (p,−p,ω1
2 , p+ω2,−p+ω2, ω2),
then a corresponding solution(a0, b0, c0, d0, e0, f0)in r1= −r3is (6.13) (p+ω2,−(p−ω2),
ω1
2 , p+2ω2,−p, ω2).
These two sets are equivalent modulo ω1, 2ω2hence they determine the same function g up to the choice of constants A and A0 in both cases. By definition of the constant the choice is only up to a sign.
Indeed, in (6.12) a, b, c are viewed as simple zeros of f . They correspond to d0, e0, c0 which according to Remark 4.2 are viewed as simple poles of, perhaps another, ˜f. A closer look shows that A0 = −A and hence that
˜f= 1
f. Clearly ˜f gives rise to the same u as f .
6.2. Constructing f and hence the solution u. This step is similar to the case ρ=4π in §5.1. Let A=1/ri, then
f(z):= f(0)exp
Z z
0 g
(w)dw
is uniquely defined up to a factor f(0). There is an unique choice of f(0)
up to a norm one factor such that c := f(ω2)f(0)has |c| = 1. Thus by integrating g(z+ω2) = −g(z)we get f(z+ω2) =c/ f(z).
By integrating g(z+ω1) =g(z)we get f(z+ω1) =c0f(z)where c0 =exp
Z
L1
g(z)dz.
We would like to show that c0 = −1. Notice that since g is odd (c.f. 6.4), the following anti-symmetric property at 12ω1still holds:
gω1 2 +u = g−ω1 2 −u = −gω1 2 −u .
By using the Cauchy principal value integral and the fact that the residue of g at 12ω1is±1, we get (6.14) Z L1 g(z)dz=0±1 2 ×2πi= ±πi
and so c0 = −1. In fact it follows that the integral of g along any line a+L1 is congruent to πi modulo 2πi since the residue at each pole is±1.
Thus f gives rise to a topological solution of the mean field equation (3.5) for ρ = 12π. The solutions as constructed, though not unique, are again even functions. Indeed g(−z) = −g(z) implies that f(−z) = f(z)
and then u(−z) =u(z).
It becomes clear that the key property is the oddness of g. Whether or not this is generally true for ρ = 4πl with l =2k+1 is a guiding problem for future studies.
We end this paper by raising the following
Conjecture 6.1. For ρ=4πl with odd l, any type I solution constructed by solv-ing the (holomorphic) residue equations contribute positively in the degree count-ing. Moreover all solutions are obtained in the way.
Since the Leray-Schauder degree is 2 for ρ = 12π. The validity of this conjecture would imply that the two solutions constructed in this section are all the solutions.
REFERENCES
[1] L. Alhfors; Complex Analysis, 2nd edition, McGraw-Hill Book Co., New York, 1966. [2] C.C. Chen and C.S. Lin; Sharp estimates for solutions of multi-bubbles in compact Riemann
surfaces, Comm. Pure Appl. Math., Vol. LV (2002), 0728-0771.
[3] ——; Topological degree for a mean field equation on a Riemann surfaces, Comm. Pure Appl. Math., Vol. LVI (2003), 1667-1727.
[4] ——; Topological degree for a mean field equation with singular data, in preparation. [5] C.C. Chen, C.S. Lin and G. Wang; Concentration phenomenon of two-vortex solutions in a
Chern-Simons model, Ann. Scuola Norm. Sup. Pisa CI. Sci. (5) Vol. III (2004), 367-379. [6] K.S. Chou and Tom Y.M. Wan; Asymptotic radial symmetry for solutions of∆u+eu=0 in
a punctured disk, Pacific J. Math. 163 (1994), 269-276.
[7] C.-S. Lin and C.-L. Wang; Elliptic functions, Green functions and the mean field equations on tori, to appear in Annals of Mathematics.
[8] M. Nolasco and G. Tarantello; Double vortex condensates in the Chern-Simons-Higgs the-ory, Calc. Var. and PDE 9 (1999), no. 1, 31-94.
[9] ——; Vortex condensates for the SU(3) Chern-Simons theory, Comm. Math. Phys. 213 (2000), no.3, 599-639.
[10] E.T. Whittaker and G.N. Watson; A Course of Modern Analysis, 4th edition, Cambridge University Press, 1927.
DEPARTMENT OFMATHEMATICS ANDTAIDAINSTITUTE OFMATHEMATICALSCIENCES
(TIMS), NATIONALTAIWANUNIVERSITY, TAIPEI
