A free boundary problem associated with the isoperimetric inequality

0.43 MB·

by  Ar. Abanov

A FREE BOUNDARY PROBLEM ASSOCIATED WITH THE
ISOPERIMETRIC INEQUALITY
ARTEMABANOV,CATHERINEBE´NE´TEAU,DMITRYKHAVINSON,
6
ANDRAZVANTEODORESCU
1
0
2 Abstract. Thispaperprovesa30yearoldconjecturethatdisksandannuli
aretheonlydomainswhereanalyticcontent-theuniformdistancefromz¯to
n
analytic functions - achieves its lower bound. This problem is closely related
a
J toseveralwell-knownfreeboundaryproblems,inparticular,Serrin’sproblem
aboutlaminaryflowofincompressibleviscousfluidformultiply-connecteddo-
5
mains, and Garabedian’s problem on the shape of electrified droplets. Some
1
further ramifications and open questions, including extensions to higher di-

mensions,arealsodiscussed.
]
V
C
.
h 1. Introduction
t
a
This paper solves a long-standing conjecture associated with the isoperimet-
m
ric inequality, rational approximation, analytic content, and related free boundary
[
problems. Let K be a compact set in the complex plane. Let C(K) be the algebra

1 of continuous complex-valued functions on K equipped with the standard uniform
v norm (cid:107)f(cid:107) =sup |f(z)|, and let R(K) be the subalgebra of the closure inside
∞ z∈K
5
C(K) of rational functions with poles off K. The analytic content of K ([10]) is
8
defined by
8
3
Definition 1.1.
0
. λ(K):= inf (cid:107)z¯−φ(cid:107)C(K).
1 φ∈R(K)
0
AnalyticcontentservesasoneofthepossibleindicatorsforK tocarryareason-
6
1 able complex-analytic structure. For example, it follows at once from the Stone-
: Weierstrass theorem that λ(K) = 0 if and only if R(K) = C(K), in other words,
v
the elements of this rather special algebra of rational functions on K are simply
i
X generic continuous functions on K. The concept of analytic content can be read-
r ily extended to deal with other spaces of “nice” functions, such as the space of
a
harmonicfunctions(seetheconceptofharmoniccontentin[12,18]),orspacescon-
sisting of solutions of more general elliptic equations ([19]). For more information
and for a comprehensive literature review, we refer the reader to the survey [2].
The analytic content of a set enjoys nice estimates in terms of simple geometric
characteristics of K. Indeed,
(cid:114)
2Area(K) Area(K)
(1.1) ≤λ(K)≤ .
P(K) π
1991 Mathematics Subject Classification. Primary: 30D05,Secondary: 30E10,30E25.
Key words and phrases. isoperimetric inequality, free boundary problem, droplets, quadratic
differentials,Schwarzfunction.
1
2 AR.ABANOV,C.BE´NE´TEAU,D.KHAVINSON,ANDR.TEODORESCU
𝝘
𝝘 3
2
𝝘
1
Ω
𝝘
4
Figure 1. The domain Ω and its boundary components, shown
withtheirorientationsrelativetoΩ(clockwisefortheinteriorcon-
tours, counterclockwise for the exterior one).
Here, P(K) is the perimeter of K, which is of course finite if for example K is
the closure of a finitely-connected domain with a rectifiable boundary. Note that
otherwise,ifP(K)=∞,thelefthandsideoftheinequalityistrivial. Wesometimes
denote the right hand side by R (K), which is the radius of the disk having the
vol
same area as K. The upper bound of (1.1) was obtained by H. Alexander in [1]
and independently by D. Khavinson in [15, 16]. The lower bound is from [15, 16].
Note that (1.1) implies the isoperimetric inequality P(K)2 ≥ 4πArea(K). The
most elementary proof of (1.1) is obtained in [10], where it is also noted that λ(K)
attains its upper bound if and only if K is a disk, modulo a “negligeable” set, that
is, a compact set E for which R(E) = C(E). Since in this paper, we will only
concernourselveswithdomainswithreasonableboundaries,wecansafelysayfrom
now on that λ(K) = R (K) if and only if K is a disk. The question that was
vol
raised in [17] is then natural:
Question. For which sets K is λ(K)=2Area(K)/P(K)?
For the rest of this paper, we will only consider a set K that is the closure of
a finitely connected domain Ω with a real analytic boundary Γ= ∪n Γ , and we
k=1 k
write λ:=λ(Ω):=λ(K). The following conjecture goes back to [16, 17]:
FREE BOUNDARY PROBLEM AND THE ISOPERIMETRIC INEQUALITY 3
Conjecture 1. We have λ(Ω) = 2Area(Ω)/P(Γ) if and only if Ω is a disk or an
annulus.
Forsimply-connectedΩ,theextremaldomainsareknowntobeonlydisks. This
was proved in [17]. Conjecture 1 can be expressed in several equivalent forms, as
follows.
Theorem 1.2 ([17, 12]). Let Ω and Γ be as above. The following are equivalent:
(i) λ= 2Area(Ω);
P(Γ)
(ii) There is ϕ analytic in Ω such that z¯(s)−iλdz¯ = ϕ(z(s)) on Γ, where s is
ds
the arc-length parameter;
(iii) The following quadrature identity
1 (cid:90) 1 (cid:90)
(1.2) fdA= fds
Area(Ω) P(Γ)
Ω Γ
holds for all bounded analytic functions f in Ω, where dA denotes area measure in
C.
(iv) There exist constants c ,c ,...,c ∈R such that the overdetermined bound-
1 2 n
ary value problem

∆u=1 in Ω

(1.3) ∂u = Area(Ω) on Γ
∂n P(Γ)
u| =c k =1,2,...,n
Γk k
has a smooth solution in Ω. (Here, ∂ denotes the exterior normal derivative on
∂n
Γ.)
In view of the fact that (1.1) yielded a new proof of the isoperimetric inequality
andbecauseofthequadratureidentity(1.2),Conjecture1anditsramificationshave
attracted the attention of a number of mathematicians (see [2, 20] and references
therein). Condition (iv) in the above theorem allows one to connect problems
involvinganalyticcontenttoaclassoffreeboundaryproblemsknownasJ.Serrin’s
problem (see, e.g., [25, 33, 2, 12, 20]).
Therestofthepaperisorganizedasfollows. InSection2,wediscussfourphysical
free boundary problems associated with Conjecture 1. We then turn to a proof of
Conjecture 1: in Section 3, we reduce the problem to domains of connectivity at
most 2, and in Section 4, we prove the conjecture for doubly-connected domains.
We close with some final remarks and open questions.
Acknowledgements. Theworkonthispaperbeganduringthespecialsemester
on“ComplexAnalysisandIntegrableSystems”attheMittag-LefflerInstitute. The
authors would like to thank the Institute and the organizers for their support.
Ar. A. is also very grateful for the warm hospitality of the INSPIRE group in
Johannes Gutenberg-Universit¨at, Mainz, Germany and its supporting staff. D. K.
was partially supported by NSF grant DMS-0855597.
2. Four physical problems
In this section, we consider four physical problems related to Conjecture 1, two
concerning Newtonian fluid flow, and the other two concerning classical and quan-
tum electrically-charged liquids in two dimensions.
4 AR.ABANOV,C.BE´NE´TEAU,D.KHAVINSON,ANDR.TEODORESCU
2.1. Serrin’s Problem. J.Serrin’sproblemconcernsalaminaryflowofaviscous,
Newtonian fluid in a pipe with cross-section Ω. (For an expository description of
thisproblem,see,forexample,[2].) Letthezaxisbedirectedalongthelengthofthe
pipe while (x,y) represent the coordinates in the pipe’s cross-section. Neglecting
gravity, in the limit of viscous flows with low Reynolds numbers, the Navier-Stokes
equations reduce to the Stokes equations, pressure can be taken to be a linear
function of z, and velocity has only one non-vanishing component, (cid:126)v = (0,0,u).
As the flow is laminar, the velocity u of the fluid particle is the same along every
streamline, that is, u = u(x,y) does not depend on z. Since the rate of change
of pressure p along the pipe is constant dp = C, the Stokes equations reduced
dz
to this situation yield that −∆u = ν−1dp = C/ν, where ν is the coefficient of
dz
dynamic viscosity. The tangential stress of the viscous fluid on the pipe walls is
proportional to the normal derivative of the velocity. One can imagine that along
the pipe walls (the boundary components of Γ), the fluid is either at rest (u = 0
on Γ, or no-slip condition) or moving with perhaps different velocities (u = c on
k
Γ ). For the simply-connected case (n=1, c =0), Serrin ([25]) proved that if the
k 1
tangential stress on the pipe is constant, then Ω is a disk. This, as noted in [17],
proves Conjecture 1 under the additional assumption that Ω is simply-connected:
in that case, Ω must be a disk. An independent proof of Serrin’s theorem for R2
that is based solely on the use of (ii) in Theorem 1.2 is due to Gustafsson and
can be found in [17], or in [2]. Various partial cases of Conjecture 1 in the form
of (iv) in Theorem 1.2, with assumptions on the constants c , k = 1,...,n, were
k
treated by many authors (see the references in [2, 20]). The physical requirement
corresponding to the third equality in (1.3) is that the “drag” force on the pipe is
constant along the perimeter. Most extra assumptions are reduced to having c be
1
thelargestofalltheconstantssothattheSerrin-Alexandrovmovingplanemethod
can be applied, yielding spherical symmetry of Ω. This forces Ω to be a spherical
shell. (See [12, 2] for multi-dimensional analogues of conditions (i) through (iii) of
Theorem1.2andrelevantdiscussions.) Yet,withoutadditionalassumptionsonthe
boundary values c , k =1,...,n, of u on Γ (as in (iv) of Theorem 1.2), Conjecture
k
1 remained open.
2.2. The shape of an electrified droplet. If we consider a droplet of perfectly
conducting fluid in the plane, with given electrostatic potential Φ, there are three
forces acting on the free boundary of the droplet: the electrostatic force F(cid:126) and
el
the force due to pressure F(cid:126) , both trying to tear the droplet apart, and the force
pr
due to surface tension, F(cid:126) , trying to keep the droplet together. Let us sketch a
sf
derivation of the equation for the free boundary Γ of the droplet in equilibrium.
See [11, 2, 20] for more details and references.
The equilibrium electrostatic force acting on a piece of the boundary of Γ of
infinitesimal length ds is F(cid:126) ∼ |E(cid:126)|2(cid:126)nds, where s is arc-length, (cid:126)n is the outward
el
unit normal vector, and E(cid:126) = ∇U is the electrostatic field (since the linear charge
density in equilibrium is proportional to the normal component of the electrostatic
field, dρ ∼(cid:126)n·E(cid:126) =|E(cid:126)|). HeretheharmonicfunctionU isthe electrostaticpotential.
ds
Set Φ=U +iV to be the analytic potential corresponding to U. Then
∂ (cid:18)Φ+Φ¯(cid:19) ∂Φ
E(cid:126) =2 = .
∂z¯ 2 ∂z
FREE BOUNDARY PROBLEM AND THE ISOPERIMETRIC INEQUALITY 5
(cid:16) (cid:17)
Here, ∂ = 1 ∂ +i ∂ so that ∇U = 2∂U. Assuming Γ to be real-analytic, it
∂z¯ 2 ∂x ∂y ∂z¯
can be parametrized by its Schwarz function S, analytic in a neighborhood of Γ so
that Γ = {z : z¯ = S(z)} (see [4, 26]). Then, since 1 = dz¯dz = S(cid:48)(z)(cid:0)dz(cid:1)2, the
dsds ds
normal (cid:126)n=−i dz =−i/(cid:112)S(cid:48)(z). Thus,
|dz|
−i
F(cid:126) ∼ |∂Φ|2ds.
el (cid:112)
S(cid:48)(z)
Moreover, the electric field
(cid:18) (cid:19)
i
E(cid:126) =∂Φ=|∂Φ| −√ .
S(cid:48)
Hence,
√ √
|∂Φ|=i S(cid:48)∂Φ=−i/ S(cid:48)∂Φ
as |S(cid:48)|=1 on Γ. Thus
(cid:18) (cid:19)
−i 1 i
(2.1) F(cid:126) ∼ √ − (∂Φ)2ds= (∂Φ)2ds.
el S(cid:48) S(cid:48) (S(cid:48))3/2
Now, the surface tension is proportional to the curvature, that is, F(cid:126) ∼ dτds,
sf ds
where τ = dz = dz = √1 is the unit tangent vector to Γ. Since d = √1 d , we
|dz| ds S(cid:48) ds S(cid:48)dz
arrive at
(cid:18) (cid:19)
ds d 1
(2.2) F(cid:126) ∼ √ √ .
sf S(cid:48)dz S(cid:48)
Now the force due to pressure, we simplify to be
i
(2.3) F(cid:126) ∼(cid:126)nds=−√ ds.
pr
S(cid:48)
If the droplet is in equilibrium, the sum of the forces (2.1),(2.2), (2.3) must be 0,
and we obtain
(cid:18)∂Φ(cid:19)2 i 1 d (cid:18) 1 (cid:19) −i
c ds+c √ √ ds+c √ ds=0,
1 dz (S(cid:48))3/2 2 S(cid:48)dz S(cid:48) 3 S(cid:48)
where c ,c , and c are real constants. Equivalently,
1 2 3
(cid:18)∂Φ(cid:19)2 d (cid:18) 1 (cid:19)
c −ic S(cid:48) √ −c S(cid:48) =0.
1 dz 2 dz S(cid:48) 3
(cid:16) (cid:17) √
Noticing that S(cid:48) d √1 =− √1 dS(cid:48) while d S(cid:48) = √1 dS(cid:48), we obtain
dz S(cid:48) 2 S(cid:48) dz dz 2 S(cid:48) dz
(cid:18)∂Φ(cid:19)2 d (cid:16)√ (cid:17)
(2.4) c +ic S(cid:48) −c S(cid:48) =0.
1 dz 2dz 3
Now define F(z)=c (cid:82) (cid:0)∂Φ(cid:1)2dz. Then (2.4) becomes, after integration,
1 dz
(cid:112)
(2.5) F(z)+ic S(cid:48)(z)−c S(z)=0.
2 3
Dividing by c , and renaming F(z)/c = ϕ(z) and c /c = λ, and using the fact
3 3 2 3
that S(z)=z¯on Γ and (cid:112)S(cid:48)(z)= dz¯, we arrive at
ds
dz¯
(2.6) z¯(s)−iλ =ϕ(z),
ds
which is precisely the equation (ii) given in Theorem 1.2.
6 AR.ABANOV,C.BE´NE´TEAU,D.KHAVINSON,ANDR.TEODORESCU
Several remarks are in order.
(i) If the potential U has a point charge at z , then ϕ(z) ∼ const near z , that
0 z−z0 0
is, ϕ has a pole.
(ii) Usually, for a physical droplet, the fluid is assumed to be incompressible.
Then either the area is assumed to be fixed, or the area, the pressure, and the
temperature are connected by the “equation of state”. In particular, for an in-
compressible fluid, the pressure has to be adjusted each time the area is fixed. If
we amend the problem with this requirement, the physical picture is the following.
Consider a plane with a system of charges on it. We throw a droplet of fluid onto
the plane and see where it will come to rest and what shape it will have. For ex-
ample, if there is only one charge, this charge will induce a dipole moment on the
droplet, and the dipole will move to “swallow” the charge. Then, there will be no
charge outside, and the charge inside will redistribute itself over the surface, while
at∞westillhaveϕ∼ c. Thus,asinexample(i)withz =0,(2.6)wouldbecome,
z 0
for some constant c∈R,
dz¯ c
z¯(s)−iλ = ,
ds z
or
(cid:112) c
(2.7) S(z)−iλ S(cid:48)(z)= .
z
(cid:112)
Denoting u(z)= S(cid:48)(z) and differentiating with respect to z, we reduce (2.7) to
c
u2−iλu(cid:48) =− ,
z2
the Ricatti equation, and the unique solution u = const/z is easily found. This
implies that S(cid:48) = const/z and Γ is a circle centered at the origin. Note that
a “physical” solution yields the same result without any calculation, merely by
noticingthatconst/z isradiallysymmetric(U =log|z|),andthereforetheproblem
musthavearadiallysymmetricsolution,hence,acircle. TheRicattiequationplays
a crucial role in the proof of Conjecture 1 in subsequent sections, see also [17].
(iii) Let us look again at (2.6), where ϕ(z)=const(cid:82) (cid:0)∂Φ(cid:1)2dz, and Φ=U +iV
dz
is the analytic potential. Then (2.6) enforces an extra condition on the problem,
√
(cid:82)
namelythat ϕ(cid:48) isasingle-valuedfunction. Ingeneral,ifU(z)= log|z−ζ|dµ(ζ)
C
is an arbitrary potential of a charge distribution µ, then
(cid:112) ∂Φ (cid:90) dµ(ζ)
ϕ(cid:48) =const = ,
∂z ζ−z
√
a single-valued function. We will call the solution to the problem (2.6) with ϕ(cid:48)
single-valued a physical droplet versus a mathematical droplet if not (see the dis-
cussion in [20]).
(iv) Note that the free boundary problem (2.6) is extremely restrictive. As
was already noted in [17], if the free boundary Γ contains a circular arc, then the
extremal domain must either be a disk (of radius λ) or an annulus. Indeed, if say
Γ contains a circular arc centered at the origin of radius R, then (2.6) implies that
either ϕ ≡ 0 if λ = R and Γ = {z : |z| = R}, or ϕ = const/z, so that every
connected component of Γ is a circle centered at the origin, and therefore Ω must
be an annulus.
FREE BOUNDARY PROBLEM AND THE ISOPERIMETRIC INEQUALITY 7
(v) Finally, we mention that a slightly more general free boundary problem
dz¯
(2.8) pz¯−it =F(z),
ds
where F is a given analytic or meromorphic function and t is a real parameter,
was discussed in [20] in detail. In particular, choosing p = 0 and F analytic in
C−Ω (the complement of the droplet) with a simple pole at infinity gives rise to
an interesting family of non-circular algebraic droplets depending on the value of
the parameter t (see [20] for details). The easier version of the latter problem with
F analytic in C−Ω (including infinity) was considered in [7] in connection with
the study of the first eigenvalue of the spectrum of the single layer potential.
2.3. Incompressible flows in 2D and generalized Rankine vortices. Incom-
pressible flow dynamics in two dimensions with non-vanishing vorticity have a dis-
tinguished history [28, 13, 21], with some important open problems relevant to the
field-theoreticextensionapplicabletoQuantumHallsystemsandother2Dstrongly-
interactingquantumelectronicsystems[34]. Webrieflyreviewheretheconnections
betweenthisclassofproblemsandtheisoperimetricinequality,referringthereader
to [30] for a more in-depth quantum-field theoretic discussion.
2.3.1. Classical2Dincompressiblevortexflows. In2Dclassicalincompressibleflows,
the problem of equilibrium distribution of vorticity is particularly relevant because
of its connection to the onset of turbulence (hence, to regularity of solutions for
the Navier-Stokes equations). Incompressible 2D velocity fields(cid:126)v can be expressed
in complex notation as (cid:126)v = v +iv = 2i∂¯ψ, where the stream function ψ(x,y) is
x y
real-valued,solvingthePoissonequation∇2ψ =ω(x,y),andω(x,y)=∂ v −∂ v
x y y x
is the 2D vorticity field of the flow. Flow incompressibility follows directly from
(2.9) i∇2ψ =2∂(v +iv )=∇(cid:126) ·(cid:126)v+iω ∈iR.
x y
This formulation is useful because it allows to express time-independent solutions
toincompressibleflowsin2Dentirelyviathestreamfunctionψ(x,y). Forexample,
irrotational,incompressibleflowsareequivalenttoboundary-valueproblemsforthe
Laplaceoperator, sinceψ isaharmonicfunctioninthedomainofirrotationalflow.
Ingeneral,theproblemrequiresfindingtheequilibriumdistributionofthevorticity
field,decomposableasthesumofanabsolutely-continuouspartandasingularpart.
Inthisformulation,theproblemdiscussedinthispaperrequiresfindingabounded
domain Ω of connectivity n≥1, and a stream function ψ(x,y)
(cid:90)
(2.10) ψ(x,y)=|z|2−2(cid:60) ϕ(z)dz,
such that ω(x,y) = ∇2ψ = constant in Ω, and the velocity field (cid:126)v = 2i∂¯ψ =
2i(z−ϕ(z)) = 2λτ on ∂Ω, that is the boundary of Ω consists of streamlines with
constant (tangent) velocity |(cid:126)v| = λ, where τ represents the unit tangent vector, λ
is the analytic content, and ϕ is the best approximation to z¯, as in Theorem 1.2.
Thesingularitysetwhereϕisnotanalytic(insidethecomplementofΩ)willcorre-
spond to the singular distribution of vorticity, while inside Ω vorticity is constant,
ω(x,y) = 4. Applying Green’s theorem to the vector field (cid:126)v on Ω leads to the
expected identity
(cid:90) (cid:73)
(2.11) 4Area(Ω)= ωdxdy = (cid:126)v·τds=2λP.
Ω ∂Ω
8 AR.ABANOV,C.BE´NE´TEAU,D.KHAVINSON,ANDR.TEODORESCU
It is instructive to notice that the simply-connected case (n = 1) was shown
long ago to correspond to a disk domain, and the associated vorticity distribution
is known as the Rankine vortex [24].
2.3.2. Chiral fields in conformal theories with several boundary components. Not
surprisingly, the 2D equilibrium distribution vorticity problem described in §2.3.1
has a magneto-static counterpart, in which we require finding a domain Ω such
that the total magnetic field is oriented along the direction perpendicular to the
(x,y) plane, and whose intensity B(x,y) is constant in Ω (more precisely, we can
take B =4 in Ω to make explicit the analogy with the vorticity field from §2.3.1),
c
as well as having singularities in the complement Ω , corresponding to infinitely-
narrow magnetic flux tubes, and given by the singularity set of ϕ(z).
Thetime-independentvectorpotentialA(cid:126)(x,y)=(A ,A ,0)isfixedbythegauge
x y
condition ∇(cid:126) ·A(cid:126) =0, so that
(2.12) 2∂(A +iA )=∇(cid:126) ·A(cid:126)+iB(x,y).
x y
Just as in §2.3.1, the choice A +iA = 2i(z −ϕ¯(z)) solves all the constraints,
x y
withtheadditionalrequirementthatA(cid:126) =λτ on∂Ω,whereagainλ,τ representthe
analytic content, and the tangent unit vector, respectively. This means that the
boundary components of Ω can be identified with closed loops of electrical current,
and the vector potential has constant magnitude on ∂Ω,|A(cid:126)|=λ. Green’s theorem
for the field A(cid:126) provides again the expected identity (2.11), in fact the condition (i)
of Theorem 1.2.
The classical vortex flow problem has a quantum correspondent [30], related to
open problems in conformal field theory (CFT). It is a boundary CFT problem
requiring finding a domain Ω (as indicated above), with (holomorphic) energy-
momentum tensor density T = ϕ(cid:48)(z). Since ϕ(z) is analytic in Ω, T dz2 =
zz zz
ϕ(cid:48)(z)dz2mustbeaquadraticdifferentialinΩ(aswillbeindeeddiscussedinthenext
section). The chiral fields v (z) = exp[−iλ−1(cid:82)zu (ζ)dζ],z,ζ ∈ Γ , analytically
k k k
continued into Ω, satisfy the projective connection [8] null condition
(cid:20)d2 ϕ(cid:48)(z)(cid:21)
(2.13) + v =0, z ∈Ω,
dz2 λ2 k
whilethegaugefieldsu (z)(whichreducetothevectorpotentialfieldsA(cid:126) on∂Ω)are
k
consistentlyrelatedtotheenergy-momentumtensorT viathecovariantderivative
zz
(or momentum) constraint
(cid:18) (cid:19)
d i i
(2.14) ∇ (u)= − u u= ϕ(cid:48)(z), z ∈Ω,
u dz λ λ
which is equivalent to differentiating (ii) in Theorem 1.2 with respect to z.
3. Reduction to the doubly-connected case
Let us now turn to a proof of Conjecture 1. Assume Ω is a finitely-connected
extremal domain, that is, a domain such that λ(Ω) = 2Area(Ω), with boundary
P(∂Ω)
components Γ ,k = 1,2,...,n with n ≥ 2. In this section, we will show that Ω
k
must be doubly-connected (n=2).
FREE BOUNDARY PROBLEM AND THE ISOPERIMETRIC INEQUALITY 9
Denote by {Ω }n the domains defined by Ω ∩Ω=∅, ∂Ω =Γ , and choose
k k=1 k k k
Ω for the one which is unbounded. Recall that τ = dz is the unit tangent vector
1 ds
at z ∈Γ, and define κ to be the signed curvature at z ∈Γ, that is
dτ¯ d2z¯
κ=−iτ · =−ids2 .
ds dz¯
ds
Notice that κ is real. We then have the following.
Theorem 3.1. Let Ω be an extremal domain, let ϕ be the best approximation of
z¯, and let λ be the analytic content of Ω. Then ϕ(cid:48)(z)dz2 is a quadratic differential
that is real-valued on ∂Ω, and
(3.1) ϕ(cid:48)(z)dz2 =(1+λκ)ds2
(cid:72)
along each component Γ of ∂Ω. Moreover, on every component Γ of ∂Ω, (1+
k k Γk
λκ)ds>0.
Proof. By Theorem 1.2, ϕ satisfies
dz¯
(3.2) z¯(s)−iλ =ϕ(z(s))
ds
onΓ,wheresisthearc-lengthparameter. Differentiatingwithrespecttoarc-length
gives
dz¯ d2z¯ dz
(3.3) −iλ =ϕ(cid:48)(z) .
ds ds2 ds
Dividingby dz¯, usingthefactthatsisarc-length, andbydefinitionofκ, wearrive
ds
at
(cid:18)dz(cid:19)2
(3.4) 1+λκ=ϕ(cid:48)(z) ,
ds
or, equivalently,
(3.5) ϕ(cid:48)(z)dz2 =(1+λκ)ds2, z ∈Γ .
k
Since ϕ(cid:48) is analytic and since the right hand side of (3.5) is real, ϕ(cid:48)(z)dz2 is a
quadratic differential that is real-valued on ∂Ω.
Now notice that for any contour Γ ,
k
(cid:90) (cid:90) d2z¯
κds = −i ds2 ds
dz¯
Γk Γk ds
(cid:18) (cid:18) (cid:19)(cid:19)
dz¯
= −i∆ log
Γk ds
(cid:18) (cid:19)
dz¯
= ∆ arg .
Γk ds
For k =1, this value is equal to −2π, while for k ≥2, we get 2π. Therefore for any
interior contour Γ , k ≥2, we obtain
k
(cid:90)
(3.6) (1+λκ)ds=L +2πλ>0.
k
Γk
On Γ , we have
1
(cid:90) 4πA
(3.7) (1+λκ)ds=L −2πλ=L − ,
1 1 P
Γ1
10 AR.ABANOV,C.BE´NE´TEAU,D.KHAVINSON,ANDR.TEODORESCU
(cid:80)
with A = Area(Ω) and P = L + L its perimeter. Using P ≥ L , we see
1 k≥2 k 1
that
4πA 4πArea(Ω) 4π
(3.8) L − ≥L − ≥ [Area(Ωc)−Area(Ω)]>0,
1 P 1 L L 1
1 1
wherewehaveusedtheisoperimetricinequalityforthecomplementofΩ , Ωc, and
1 1
the fact that Ω⊆Ωc. (cid:3)
1
Now recall that as discussed in the introduction, if S (z) are the Schwarz func-
k
tions for Γ , that is, S is analytic in a neighborhood of Γ and satisfies S (z)=z¯
k k k k
(cid:112)
on Γ , then for u (z)= S(cid:48)(z), the functions u satisfy the Ricatti equation
k k k k
(3.9) u2 +iαλu(cid:48) =ϕ(cid:48)(z), z ∈Ω,
k k
where α = 1 for k ≥ 2 and α = −1 for k = 1. By a standard reduction, the
functions
(cid:20) iα(cid:90) z (cid:21)
(3.10) v (z):=exp − u (ζ)dζ , k =1,2,...,n,
k λ k
solve the linear second-order differential equation associated with (3.9)
ϕ(cid:48)
(3.11) v(cid:48)(cid:48) =− v.
λ2
Definition 3.2. Let Σ± be the union of Stokes and anti-Stokes graphs of (3.11)
in Ω [31, Lemma 9.2-1], i.e. the union of arcs {γ±} satisfying
j
(cid:90) z (cid:90) z
(cid:112) (cid:112)
(cid:61) ϕ(cid:48)(ζ)dζ =0, ζ ∈γ+ ⊂Σ+, (cid:60) ϕ(cid:48)(ζ)dζ =0, ζ ∈γ− ⊂Σ−,
j j
z0 z0
where z is any zero of ϕ(cid:48)(z) in Ω.
0
It is known [31, 5] that if ϕ(cid:48)(z) is analytic in Ω, then Σ+,Σ− have the same
numberofarcsγ±,theyintersectonlyatzerosofϕ(cid:48)(z),andeacharcγ± isanalytic,
j j
with one endpoint being a zero of ϕ(cid:48), and the other being either another zero, or a
point on ∂Ω (or possibly, both). Moreover, at a zero z ∈ Ω of ϕ(cid:48) of order m ≥ 1,
0
there are exactly m+2 arcs from Σ+ with local angle between adjacent arcs equal
to 2π/(m+2), and another m+2 arcs from Σ−, each of them bisecting the angle
between two consecutive arcs of Σ+.
Let z ∈Ω be a zero of order m of ϕ(cid:48). By elementary calculations, it is easy to
0
describethelocalpowerseriesexpansionofv aboutz ,butthelocalsolutionisnot
0
convenient to use when exploring global properties of solutions such as |v (z)| =
k Γk
constant, satisfied by (3.10). Instead, we will examine the asymptotic series repre-
sentations, validoutsideasmallneighborhoodofz . Definingthelocalcoordinates
0
ζ =(cid:15)(z−z ), with (cid:15) a scale parameter, arbitrarily small but strictly positive, then
0
c.f. [22, Ch. 6], [31, Ch. 3], [9], the general solution for Eq. (3.11) admits the
asymptotic series representation known as Liouville-Green (LG) in applied mathe-
matics and Jeffreys-Wentzell-Kramers-Brillouin (JWKB) in theoretical physics
√
(3.12) v(ζ,(cid:15))= (ϕ(cid:48))λ1/4 (cid:104)C1eλi(cid:15)(cid:82)0ζ√ϕ(cid:48)dξ+C2e−λi(cid:15)(cid:82)0ζ√ϕ(cid:48)dξ(cid:105)[1+o((cid:15))],
whereC areconstants,andζ belongstoadomainDhaving0asboundarypoint.
1,2
In particular, for z ∈Σ+, the domain of validity includes a wedge domain of angle
2π/(m+2), with Σ+ bisecting the angle. The solution is approximated by the