https://doi.org/10.1007/s40315-021-00373-w
On the Uniqueness Problem for Quadrature Domains
Yacin Ameur1·Martin Helmer2 ·Felix Tellander3
Received: 18 June 2020 / Revised: 15 September 2020 / Accepted: 29 September 2020 / Published online: 4 May 2021
© The Author(s) 2021
Abstract
We study questions of existence and uniqueness of quadrature domains using com- putational tools from real algebraic geometry. These problems are transformed into questions about the number of solutions to an associated real semi-algebraic system, which is analyzed using the method of real comprehensive triangular decomposition.
Keywords Quadrature domain·Conformal mapping·Real comprehensive triangular decomposition
Mathematics Subject Classification 30C20·31A25·14P10·68W30
1 Introduction
This note is the result of investigations into an open uniqueness question for quadra- ture domains in the complex planeC, which appears in papers such as [21,24,44].
After describing the problem and reviewing some known results, we will suggest and explore an approach based on methods from real algebraic geometry and symbolic computation.
To get started, it is convenient to fix some notation that will be used throughout.
Communicated by Darren Crowdy.
B
Felix Tellander felix@tellander.se Yacin AmeurYacin.Ameur@math.lu.se Martin Helmer
martin.helmer@anu.edu.au
1 Department of Mathematics, Lund University, 22100 Lund, Sweden
2 Mathematical Sciences Institute, The Australian National University, Canberra, Australia 3 Department of Astronomy and Theoretical Physics, Lund University, 22362 Lund, Sweden
General notation. By a “domain”we mean an open and connected subset ofC;
we writefor its closure,∂for its boundary, ande =C\for its exterior. A bounded domainis said to be “solid” ifeis connected and∂=∂(e). Note that what we call solid domains are also referred to as “Carathéodory domains” in some sources.
We write “d A” for the normalized Lebesgue measure in the planed A(z)= π1d x d y (so the unit disc has area 1).
We denote byL1()the usualL1-space of functions onthat are integrable with respect tod A, and we writeAL1(),H L1(), andS L1()for the subsets ofL1() consisting of all analytic, harmonic, and subharmonic functions, respectively.
Standard sets:D(a,r)= {z∈C; |z−a|<r},D=D(0,1),T=∂D.
Differential operators:∂= 12(∂x−i∂y),∂¯ = 12(∂x+i∂y),=4∂∂¯ =∂x2+∂y2. Given an open set⊂C, a subspace (or cone)F ⊂L1(), and a linear functional μ:F→C, we consider quadrature identities of the form
f d A=μ(f), f ∈F. (1)
The linear functionalμwill always be a fixed measure or distribution of appropriate type with compact support in(and defined on the appropriate test-classF). Since we will frequently takeμ to be a combination of point-evaluations, we stress that statements likeF ⊂ L1()should not be taken literally, but rather in terms of the natural injective mapsF →L1().
If the above conditions hold, we say that is a quadrature domain (or “q.d.”) with data(μ,F), and we write∈ Q(μ,F). We are mainly interested in the case F= AL1(), but alsoF =H L1(),F=S L1()will play a role. In the last case, (1) must be replaced by the inequality
μ(f)≤
f d A, f ∈ S L1().
It is easy to see that Q(μ,S L1) ⊂ Q(μ,H L1) ⊂ Q(μ,AL1)and that a solid domain belongs simultaneously to the classesQ(μ,H L1)andQ(μ,AL1).
The above classes are conveniently interpreted in terms of the logarithmic potentials Uμ:=∗μ, U:=U1d A, where (z):=1
2log 1
|z|.
For example, we have that∈ Q(μ,AL1)if and only if∂Uμ=∂Uoneand ∈ Q(μ,S L1)if and only ifUμ=UoneandUμ≥UonC.
Given these proviso, we can formulate our basic problem in a succinct way (cf.
[21]).
(Q). Determine whether or not there exists a functional μ such that the class Q(μ,AL1)contains two distinct, solid domains.
Theorem 1.1 The following uniqueness results are known.
(i) If1, 2 ∈ Q(μ,AL1)are star-shaped with respect to a common point, then 1=2.
(ii) If there exists a solid domain∈ Q(μ,S L1), then thisis the unique solid quadrature domain, even within the class Q(μ,AL1).
(iii) Ifμis a positive measure of total mass m and ifsuppμis contained in a disc of radius r where r2 < m, then each solid domainof class Q(μ,AL1)is obtainable fromμby partial balayage, and so it belongs to Q(μ,S L1).
Remark on the proof Part (i) is due to Novikov [33], cf. also [17,29]. Part (ii) was proved in Sakai’s book [36], using the technique of partial balayage. An alternative proof is found in the paper [21] by Gustafsson. The statement (iii) was likewise proved by Sakai using partial balayage, see the papers [27,35,38].
The “classical” setting corresponds to point-functionals, i.e., functionalsμof the form
μ(f)= m
i=1 ni−1
j=0
ci jf(j)(ai), f ∈ AL1()
whereaiare some points inandci j some complex numbers. A quadrature domain of this type is said to be of ordern1+ · · · +nm. Whenμcontains no derivatives, i.e., whenμ(f)=n
i=1ci f(ai)we speak of a pure point-functional. (Cf. [15].) Example 1.1 Theorem1.1(ii) completely settles the uniqueness problem for subhar- monic quadrature domains. The following example due to Gustafsson shows that the question (Q) for analytic test functions is of a different kind.
It is shown in [20, Sect. 4] that there exists a quadrature domain having the appearance in Fig.1, satisfying a three-point identity
f d A=c1f(a1)+c2f(a2)+
c3f(a3)wherec1,c2,c3>0.
It is known (see e.g. [27, Thm. 2.1]) that if∈ Q(μ,S L1)and if∂has cusps, then those cusps must be contained in the convex hull of the nodes ai. Hence the quadrature domain in Fig.1is not subharmonic.
Now fix a solid quadrature domain ∈ Q(μ,AL1)containing the origin 0. Let ϕ:D→be the conformal map normalized byϕ(0)=0 andϕ(0) >0. Recall that is uniquely determined byϕvia Riemann’s mapping theorem.
Fig. 1 Gustafsson’s example of a domain in
Q(μ,H L1)\Q(μ,S L1). The triangle depicts the convex hull of the nodesa1,a2,a3
-2 -1.5 -1 -0.5 0 0.5 1 1.5 -2
-1.5 -1 -0.5 0 0.5 1 1.5 2
The following theorem, which gives a non-trivial relation forϕ, will be the main tool in our subsequent investigations.
Theorem 1.2 Suppose thatμhas compact support inand letνbe the pullback of μtoD, i.e.,ν(g)=μ(g◦ϕ−1)for g∈ AL1(D). Thenϕsatisfies the relation
ϕ(z)=νλ∗
z ϕ(λ)(1−zλ)¯
, z∈D, (2)
whereνλ∗(g):=νλ(g)¯ acts on integrable anti-analytic functions g(λ).
Conversely, ifϕ is any univalent solution to(2), then the domain=ϕ(D)is of class Q(μ,AL1).
This result is not very easy to spot in the literature, but it has in fact been noticed earlier in somewhat different guises. The first proof might be due to Davis, see [15, Ch. 14], cf. also [14, Sect. 5]. Since the result will be central for what follows, we include an alternative proof (that we have found independently) in Sect.2.
In the special case of a pure point-functionalμ(f)=n
1ci f(ai), the relation (2) takes the form
ϕ(z)= n i=1
¯ ci
¯ wi
z
1− ¯λiz, wi =ϕ(λi), ϕ(λi)=ai, (3) which appears implicitly in e.g. the books [25,40,43].
The main idea behind our approach is to “solve” functional relations such as (3) by using techniques from algebraic geometry. To set up a suitable system of polynomial equations we differentiate (3) and substitutez=λj, giving
wj = n i=1
¯ ci
¯ wi
1
(1−λjλ¯i)2, aj = n i=1
¯ ci
¯ wi
1 1−λjλ¯i
, j=1, . . . ,n, (4)
where the unknown complex numbersλiandwiare subject to the constraints λ1=0, |λ2|<1, . . . , |λn|<1, w1>0. (5) The appearance of inequalities and complex-conjugates means that we are consid- ering the real semi-algebraic geometry of a particular system of rational functions.
Such semi-algebraic systems, i.e. those having the special structure of (4), (5) have, to the best of our knowledge, not been systematically studied before.
It is of course possible that no univalent solution to (4), (5) exists; for example the quadrature identity
f d A= f(a1)+f(a2)implies thatis the disjoint union D(a1,1)∪D(a2,1)if|a1−a2| ≥2. However, after having studied exact solutions for many examples of lower order quadrature domains, we find “empirically” the pattern that the system tends to have at most one solution which may or may not give rise to a univalent mappingϕ. The non-univalent solutions fail to be locally univalent, i.e.,
they (still, empirically) satisfyϕ =0 somewhere in the discD. For a different type of quadrature domains, not described by (4) and (5), a similar observation was made by Ullemar in [42].
Remark The non-univalent solutionsϕare believed to represent quadrature domains on Riemann surfaces with branch points. Such q.d.’s are studied in the references [23,37,41].
Remark Note that our method relies on knowledge of all solutions to (4), (5). To find one or several approximate solutions, one can of course try to apply numerical methods, such as Newton’s iterative method. By appropriately choosing different initial data, we may indeed obtain solutions by such methods in a relatively short time fornup to 10.
However, since the number of solutions to the system is unknown, it is impossible to know when one has found all solutions, so this kind of information is of no use when studying the uniqueness question for quadrature domains. Another problem with a numerical approach is that systems such as (4), (5) tend to be quite sensitive to small perturbations of the quadrature data{ci,ai}n1.
Remark In connection with uniqueness problems in the gravi-equivalent sense, it is pertinent to point (besides the sources already mentioned) to the early works of Zidarov and Zhelev in the context of geophysics, see e.g. [45,46]. (We thank one of the referees for this remark.)
The literature on quadrature domains is vast, and we have at this stage omitted to mention several important aspects. A somewhat fuller picture is given in Sect.7, where we briefly compare a few of the more well-known techniques that have been developed over the years, such as Laplacian growth and Schottky-Klein functions.
To illustrate the challenges involved in studying the uniqueness of quadrature domains, we now give an example demonstrating the subtlety of the problem even for a q.d of order 2.
Example 1.2 Let1be the solid quadrature domain obtained from a monopole with charge 1/2 and a dipole with strength√
3/18 placed at the origin, i.e.
1
f d A=1 2 f(0)+
√3
18 f(0), f ∈ AL1(1).
For quadrature domains of this type, Aharonov and Shapiro have proved uniqueness in [2]; in fact1is determined as the image ofDunder the conformal map p(z)=
√3(2z+z2)/6. The boundary∂1is a cardioid with a cusp atp(−1)= −1/√ 12, see Fig.2.
Let us now construct a similar q.d.˜2but only using point charges,
˜2
f d A= −1
2f(0)+ f(a2), f ∈ AL1(˜2).
Clearly both1and˜2have area 1/2. We shall choose the parametera2real, such that the boundary of˜2has a cusp.
-0.5 -0.25 0 0.25 0.5 0.75 -0.75
-0.5 -0.25 0 0.25
0.5 0.75
1 2
0.84 0.85 0.86 0.87 0.88
-0.1 -0.05 0 0.05
0.1
Fig. 2 The domains1and2
Since the parametersci, wi, λi in (3) must be real in this case, the conformal map
˜
ϕ:D→ ˜2takes the form
˜
ϕ:z→c1z w1 + c2
w2
z
1−λ2z (6)
wherec1= −1/2 andc2=1,λ2the pre-image ofa2, andw1= ˜ϕ(0), w2= ˜ϕ(λ1).
Writingν= −9+18i√
2, we find that there is a unique choice ofa2producing a cusp, namely
a2= 1 2ν1/3
−ν1/3 3i√
3ν2/3−27i√
3+3ν2/3+4ν1/3+27
≈0.1316.
The remaining parameters in the mapping function may be obtained by solving the system corresponding to (6) using the method of “RCTD” described in Sect.4. The result is
λ2= 3a2
3a22+3
≈0.2261,
w2= −c2λ23
−λ22c1+λ22c2+a22 a2
−c1λ26+c2λ26+λ24a22+2c1λ24−2λ22a22−λ22c1−λ22c2+a22≈0.6674, w1= c1λ2
λ22−1 w2
w2λ22a2−w2a2+λ2c2 ≈0.5016.
This gives a cusp at
˜
ϕ(−1)= −c1
w1 − c2
w2
1
1+λ2 ≈ −0.2253.
A translate of˜2byα:= p(−1)− ˜ϕ(−1)≈ −0.06333 leads to a domain2= ˜2+αhaving a cusp at the point p(−1)and satisfying the quadrature identity
2
f d A= −1
2 f(α)+ f(a2+α), f ∈ AL1(2).
The resemblance between1and2(Fig.2) is striking, even though they admit completely different quadrature identities. The similarity between1and2indicates that their potentials should be similar, and in terms of numerical values they are.
But there is one essential difference between the two: the potential of1is exactly determined by two terms in its multipole expansion while the potential for2needs the entire infinite series. In detail we have
U1(z)=
1 2
2π log 1
|z| +
√3 18
2πRe 1
z
U2(z)≈
1 2
2π log 1
|z| +0.09998
2π Re
1 z
+ · · · and for comparison note√
3/18≈0.09623. From this example we see that two very similar domains may have fundamentally different potentials.
2 The Master Formula
As previously stated, Theorem1.2appears (in equivalent form) in [15, Eq. (14.12)].
We shall here give a different derivation.
Consider the univalent mapϕ : D→normalized byϕ(0)=0 andϕ(0) >0 where ∈ Q(μ,AL1)and where the distributionμ is assumed to be of compact support in. Our point of departure is Poisson’s equation
Uμ= −μ.
Taking the distributional∂-derivative of−4Uμwe obtain theCauchy transform Cμ(z):=μ∗k(z)=μ(kz),
wherek(λ):=λ−1denotes the Cauchy kernel andkz(λ):=k(z−λ). Since−4∂Uμ= Cμ, (2) says that∂Cμ¯ =μ. In particularCμis holomorphic onC\suppμ. Taking
f =kz(z∈/)in the quadrature identity (1) we see that
∂U=∂Uμ on C\. (7)
In fact, an application of Bers’ approximation theorem from [4] shows that (7) is equivalent to (1). Now consider the “Schwarz potential”udefined by
u=C(1−μ)= −4∂(U−Uμ),
which is zero forz ∈ e. Using the continuity of∂Uwe get u = 0 also on∂. Moreover, Poisson’s equation gives∂¯u=1onC\suppμ. Hence the function
S(z):= ¯z−u(z)
is holomorphic on\suppμand continuous up to the boundary∂, while satisfying S(z) = ¯zforz ∈ ∂. This determines Sas the Schwarz functionfor the boundary curve∂, cf. [15,40].
The following lemma is well-known, see e.g. [15,40].
Lemma 2.1 The conformal mappingϕ : D→ extends holomorphically acrossT to an analytic function on the disk D(0,R)for some R>1.
Proof As we saw above, the functionS◦ϕis defined and holomorphic in some annulus 1− <|z|<1, continuous up to the boundary and satisfiesS(ϕ(z))=ϕ(z)when z ∈T. Likewise, the functionϕ∗(z)=ϕ(z)is holomorphic inDand continuous up to the boundary, and we have the relation
S(ϕ(z))=ϕ∗(1/z), z∈T.
Now,ϕ∗(1/z)is holomorphic in the exterior ofD, so the above formula shows that the functionsS◦ϕ(z)andϕ∗(1/z)are analytic continuations of each other across the circleT. In particular,ϕ∗(1/z)is analytically continuable inwards acrossT, which means thatϕ∗as well asϕ are analytically continuable outwards acrossT, to some discD(0,R)withR>1.
Lemma 2.2 We have thatC ϕ·1D
(z)=ϕ(z),z∈T.
Remark For an absolutely continuous measure ν = f d A, we prefer to denote its Cauchy transform byCf rather thanCν.
Proof of Lemma2.2 Fix a pointz=eiθ ∈Tand a positive numberr<1 and put Ir =C
ϕ·1D
(r z)=
D
ϕ(λ)
r eiθ−λ d A(λ)=r e−iθ
rD
ϕ(rζeiθ) 1−ζ d A(ζ ).
Set ϕ(λ) = ∞
j=0cjλj, where ∞
j=0|cj| < ∞ by Lemma 2.1. Inserting the expansion 1/(1−ζ )=
ζj we find that
Ir = ∞
j=1
¯ cj r e−iθ
j
r2j.
Since
|cj|<∞we may pass to the limit asr 1, leading to
rlim→1Ir =lim
r→1
∞ j=1
¯ cj r e−iθ
j
r2j = ∞
j=1
¯ cj e−iθ
j
=ϕ(z).
The proof of the lemma is complete.
Proof of Theorem1.2 The quadrature identity (1) pulls back to
f d A=
D
(f ◦ϕ)· |ϕ|2d A=ν(f ◦ϕ) (8)
whereν(g)=μ(g◦ϕ−1).
Given an arbitrary f ∈ AL1()we define a functiong∈ AL1(D)by g =(f ◦ϕ)·ϕ.
The identity (8) can be written as
D
gϕd A=ν g
ϕ
. (9)
Now fix a pointz∈Tand choosegto be the Cauchy-kernelg=kz. With this choice, (9) takes the form
C ϕ·1D
(z)=C 1
ϕ ·ν
(z), z∈T.
By Lemma2.2this is equivalent to ϕ(z)=C
1 ϕ ·ν
(z)=νλ
1 ϕ(λ)(1/¯z−λ)
, z∈T.
Taking complex-conjugates and considering the analytic continuation toDwe obtain ϕ(z)=νλ∗
z ϕ(λ)(1−zλ)¯
, z∈D, (10)
as desired.
Conversely, if ϕ is univalent (and normalized) in D and satisfies (10), we may read backwards and deduce (8), so=ϕ(D)belongs toQ(μ,AL1)whereμis the push-forward ofν.
We conclude this section with three examples of applications of Theorem1.2, which are known from the literature on quadrature domains.
Example 2.1 Let us compare Theorem1.2with the computations in Shapiro’s book [40, Prop. 3.2]. For this purpose we fix a pure point functionalμ=n
i=1ciδai, which pulls back to
ν(g)=μ g◦ϕ−1
= n i=1
cig ϕ−1(ai)
= n i=1
cig(λi), λi =ϕ−1(ai).
By Theorem1.2we know that an arbitrary solid domain ∈ Q(μ,AL1)is of the form=ϕ(D)whereϕis univalent and normalized and satisfies
ϕ(z)= n i=1
ci
ϕ(λi) z 1−zλi
(11)
Given such aϕ, we putS(ϕ(z)):=ϕ∗(1/z)and note thatSis the Schwarz function for∂. In view of (11),Sis a meromorphic function with simple poles atz=aj. As z→λjthe dominant term inS◦ϕsatisfies
S(ϕ(z))∼ cj
ϕ(λj) 1 z−λj. From this we get that (asz→aj)
S(z)∼ cj
ϕ(λj) 1
ϕ−1(z)−λj = cj
ϕ(λj)
z−aj
ϕ−1(z)−ϕ−1(aj) 1
z−aj ∼ cj
z−aj. The residues ofSare thus just Res(S;aj)=cj for all j. SinceS(z)= ¯zon∂, an application of Green’s theorem and the Residue theorem now gives
f d A= 1 2πi
f(z)z d z= 1 2πi
f(z)S(z)d z=
cj f(aj)
where is the positively oriented boundary of. We have shown again that ∈ Q(μ,AL1).
We remark that a similar proof applied to a more general point functionalμ(f)= ci jf(j)(ai)gives the well known result (see [26, Thm. 3.3.1]) that a solid domain is a quadrature domain of finite order if and only if each conformal mapϕ :D→is a rational function, if and only if the Schwarz function of∂extends to a meromorphic function in.
Example 2.2 Letμbe a linear combination of a monopole and a dipole at the origin, i.e.μ(f)=M0f(0)+M1f(0). The action of the pullback is then given by
ν(g)=μ(g◦ϕ−1)=M0g(0)+ M1
ϕ(0)g(0), g∈ AL1(D).
Applying Theorem1.2, we find that a normalized conformal mapϕ:D→, where ∈ Q(μ,AL1), must satisfy
ϕ(z)= 1
ϕ(0)M0− ϕ(0) ϕ(0)3M1
¯ z+ M1
ϕ(0)2z¯2.
Henceϕ(z)is a polynomial of degree two. To determine this polynomial, we need to determine the derivativesϕ(0),ϕ(0). The computation is postponed to Subsect.5.1, after we have discussed some algebraic prerequisites.
Fig. 3 The q.d.generated by the linear density
dμ(x)=1[−1,1](x)d x
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
-1.5 -1 -0.5
0 0.5
1 1.5
Example 2.3 Following Davis [15, pp. 162-166] we now takeμbe a line charge with linear densityhon the segment[−a,a]of thex-axis, i.e.,
μ(f)= a
−a
f(x)h(x)d x.
Suppose that=ϕ(D)∈ Q(μ,AL1); the pullbackνbyϕis then given by ν(g)=μ(g◦ϕ−1)=
a
−a
g(ϕ−1(x))h(x)d x= λ
−λg(w)h(ϕ(w))ϕ(w)dw, λ=ϕ−1(a).
Applying Theorem1.2we see thatϕmust satisfy the functional equation (cf. [15, Eq.
(14.25)])
ϕ(z)= ¯z λ
−λ
h(ϕ(w)) 1− ¯zw dw.
In particular, if we specialize to a uniform line chargeh ≡1, we obtain ϕ(z)=log
1+zλ 1−zλ
.
This relation was found by Davis (see Eq. (14.13)), where it is also shown that ifa =1 thenλ=√
tanh(1/2)=.6798· · ·<1. It is then easy to see that the above mapϕis well-defined by choosing the standard branch of the logarithm. We have shown that there exists a unique solid domainin the classQ(1[−1,1](x)d x,AL1); a picture is given in Fig.3.
3 Schur–Cohn’s Test
As stated earlier, the mapping problem (3) may have non-univalent solutionsϕ. The Schur–Cohn test provides a convenient way of discarding suchϕthat fail to be locally
univalent. One might hope that this procedure would leave us with at most one univalent ϕ, thus settling the uniqueness problem. As we will see later, this is indeed the case for a large class of examples.
The Schur–Cohn test is well known and quite elementary, see [28]. For reasons of completeness, we have found it convenient to briefly review the main ideas behind it here. In what follows, we letPndenote the space of all polynomials pof degree at mostn,
p(z)=a0+a1z+ · · · +anzn, a0, . . . ,an∈C.
Lemma 3.1 If n ≥1then
(i) if|a0|>|a1| + · · · + |an|then p(z)=0for all z with|z| ≤1, (ii) if p(z)=0 for allzwith|z| ≤1then|a0|>|an|.
Proof (i) If|a0|>|a1| + · · · + |an|then|a1z+ · · · +anzn| ≤ |a1| + · · · + |an|<|a0| for|z| ≤1, so|p(z)| ≥ |a0| − |a1z+ · · · +anzn|>0.
(ii) Assume p(z) = 0 for all z with|z| ≤ 1. Then p(0) = a0 = 0 which leads to two cases: (1) ifan = 0 then obviously|a0| > |an|; (2) ifan = 0 we have p(z)=an(z−z1)· · ·(z−zn)whereznare the zeros ofp. But thenp(0)=a0= an(−z1)· · ·(−zn)and thus|a0|>|an|since|zk|>1 fork=1, . . . ,n.
For eachp∈Pnthe reciprocal polynomialp#∈Pnis defined by p#(z)=zn·p(1/z)=an+an−1z+ · · · +a1zn−1+a0zn. Let us now define theSchur transform,Sn: Pn→Pn−1by
(Snp)(z)=a0p(z)−anp#(z)=
n−1
k=0
(a0ak−anan−k)zk, p∈Pn.
We note a few simple facts pertaining to these objects.
First,|p#| = |p|onT. Moreover, every zero of ponTis also a zero of p#and is thus a zero ofSnp. Finally,
(Snp)(0)= |a0|2− |an|2∈R.
We now construct a chain of polynomialsp0,p1, . . . ,pnstarting withp0=pand then taking successive Schur transforms,
p1=Snp0, p2=Sn−1p1, . . . , pn=S1pn−1. The last polynomialpnis an element inP0and thus is a constant.
Lemma 3.2 If pkhas no zeros on the unit circleTand pk+1(0) >0, then pkand pk+1
have equally many zeros inD.
Proof Letpk(z)=b0+b1z+ · · · +bn−kzn−k. Then pk+1(z)=b0pk−bn−kp#k and pk+1(0) = |b0|2− |bn−k|2 > 0 so|b0| > |bn−k|. Since pk(z) = 0 onTwe have
|b0pk| > |bn−kpk| = |bn−kpk#|onTand thus Rouché’s theorem implies that pk+1
andpkhave equally many zeros inD.
We are now ready to formulate Schur–Cohn’s test (e.g. [28]).
Theorem 3.1 A polynomial p ∈ Pn has no zeros inDif and only if pk(0) > 0 for k=1, . . . ,n.
Proof Assume thatp=p0has no zeros inD. By Lemma3.1,p1(0)= |a0|2−|an|2>
0. Sincep0has no zeros onT, Lemma3.2implies thatp1andp0have the same number of zeros inD, i.e., none. Ifn≥2 we can repeat the reasoning with p1∈Pn−1instead of p0and deduce p2(0) >0 and so on, and after a finite number of steps we finally get pn(0) >0.
To prove the reverse implication, assume thatpk(0) >0 fork=1, . . . ,n. Then pn
is a non-zero constant, and especially has no zeros onT. Since pn =S1pn−1, pn−1
has no zeros onTand by Lemma3.2,pn−1and pnhas the same number of zeros in
|z| ≤1, i.e., none. We may now repeat this withpn−1∈P1instead of pnand deduce that pn−2has no zeros inDand so on, and after a finite number of steps we finally get that p= p0has no zeros inD.
4 Real Comprehensive Triangular Decomposition
The method of real triangular decompositions, introduced recently in [7, §4], [5, §10]
and [6], provides a suitable framework to deal with the the mapping problem for quadrature domains, in the form of systems such as (4), (5). These methods in turn make use of the idea of a Cylindrical Algebraic Decomposition [3, §5], for an overview of this and other applicable methods from real algebraic geometry see, for example, the book [3].
The objects we consider here aresemi-algebraic sets, given a list of polynomial equations and inequalities inR[x1, . . . ,xm]a basic semi-algebraic setis the set of all points inRmwhich simultaneously satisfy all these equations and inequalities, [3,
§3]. In this section we will specifically consider semi-algebraic systems defined by polynomials in the polynomial ring
Q[c,x] =Q[c1, . . . ,cd][x1, . . . ,xn], where we think of thecias parameters and thexjas variables.
More precisely, given polynomials f1, . . . , fr andp1, . . . ,psinQ[c,x], we define thesemi-algebraic system[f=0,p>0]to be the following set of equations and inequal- ities
f1(c,x)=0, . . . , fr(c,x)=0, p1(c,x) >0, . . . , ps(c,x) >0. (12) The set of real solutions(c,x)∈Rd×Rnto (12) is called the(parameterized) semi- algebraic setgenerated by the system, denoted byS([f=0,p>0]). Moreover, for fixed
c ∈ Rd we define thespecialized semi-algebraic set S(c)([f=0,p>0])as the set of pointsx∈Rnwhich satisfy the system (12) for the particular parameter-valuec.
Now suppose that T = {T1, . . . ,T} is a collection of semi-algebraic systems inQ[c,x]. We extend the definitions of semi-algebraic set and specialization to a parameter valuecby
S(T)=
j=1
S(Tj)⊂Rd×Rn, S(c)(T)=
j=1
S(c)(Tj)⊂Rn.
Moreover, given a semi-algebraic system T = [f=0,p>0] we define the con- structible set ofT to be the set of complex solutions(c,x)∈Cd×Cnto the system of equations and inequalities
f1(c,x)=. . .= fr(c,x)=0, p1(c,x)=0, . . . , ps(c,x)=0. (13) We denote byCS(T)the set of solutions(c,x)∈Cd×Cnto (13); givenc∈Cdwe define the associated specializationCS(c)(T)to be the set of pointsx∈Cnsuch that (c,x)∈CS(T).
IfT= {T1, . . . ,T}is a finite collection of semi-algebraic systems, it should now be obvious how to extend our definitions of constructible set (CS(T)=
jCS(Tj)) and specializations (CS(c)(T)=
jCS(c)(Tj)⊂Cn).
A semi-algebraic systemT = [f=0,p>0]is calledsquare-freeif all polynomials fjandpioccurring inT are square-free. (A polynomialg ∈Q[c,x]is square-free if it has no factor of the formw2wherew∈Q[c,x]is non-constant.)
Definition 4.1 LetW= [f=0,p>0]be a semi-algebraic system defined by polynomi- als f1, . . . , fr andp1, . . . ,ps inQ[c,x]and letS(W)⊂Rd×Rnbe the associated semi-algebraic set.
Areal comprehensive triangular decomposition (RCTD)ofWis a pair(C,{TC ; C∈ C})whereCis a finite partition ofRd into non-empty semi-algebraic setsC(called
“cells”) and for eachC ∈C,TC is a finite set of square-free semi-algebraic systems such that exactly one of the following holds:
(1) TCis empty soS(TC)=Rd×RnandS(c)(TC)=Rnfor allc∈Rd, (2) The specialized constructible setCS(c)(TC)is infinite for allc∈C,
(3) TC= {T1, . . . ,T}is a finite set of semi-algebraic systems satisfying the following conditions:
• CS(c)(TC)is finite and has fixed cardinality for allc∈C,
• the specialized semi-algebraic setsS(c)(Tj)are finite and non-empty for allj and further for a fixedTjthe specialized semi-algebraic setS(c)(Tj)has fixed cardinality for allc∈C,
• S(c)(W)=
j=1S(c)(Tj)for allc∈C.
The following proposition summarizes the results about RCTD’s that we will apply in the sequel.
Fig. 4 The cellsCfrom the RCTD in Example4.1
Proposition 4.1 (§10 of [5]) Let f1, . . . , fr and p1, . . . ,ps be polynomials in the polynomial ringQ[c,x]defining a semi-algebraic systemW = [f=0,p>0]. Then a real comprehensive triangular decomposition ofW exists and may be computed by an explicit algorithm which is guaranteed to terminate in finite time. This algorithm is implemented in theRegularChainsMaple package [31].
Example 4.1 Define f =a·y2+b·x+3 andg=b·x y−y+2 wherex,ydenote real variables anda,bdenote real parameters. Consider the semi-algebraic system
W = [f =0,g=0,y≤1].
Using theRegularChainspackage, the parameter spaceR2is partitioned into five cellsC= {C0,C1,C2,C3,C∞}where
C∞=
−3 4,0
, C0=
(a,b)∈R2; a= −3
4, b=0
,
C1=
(a,b)∈R2; a <−64
27, b=0 (a,b)∈R2|a ≥0, b=0 ,
C2=
(a,b)∈R2; a = −64
27, b=0 (a,b)∈R2| −2<a<0, b=0 ,
C3=
(a,b)∈R2; −64
27 <a≤ −2, b=0
.
These cells are illustrated in Fig.4.
It is not hard to show that the specialized semi-algebraic setS(a,b)(W)consists ofj points for all(a,b)∈Cjwhere j ∈ {0,1,2,3}; for the cellC∞the RCTD guarantees only that the specialized constructible set associated to the parameter choiceC∞has infinitely many points (a corresponding specialized semi-algebraic set may be infinite, empty, or finite). HenceC, along with the associated semi-algebraic systems for each parameter cell, gives a RCTD ofW.
In more detail; consider the cellsC0andC∞. The disjoint union of these cells is the lineb=0 (i.e. the a-axis in Fig.4). Along this line the semi-algebraic systemW simplifies to
a·y2+3=2−y=0, y≤1. (14) Solving forywe obtainy=√
−3/aandy=2, so f =g =0 has a real solution if and only ifa= −3/4, so{(a,b)} =C∞. The corresponding specialized constructible set is
(x,y)∈C2; (y−2)(y+2)=(y+2)=0, y=1
= {(x,2); x∈C},
which is infinite. However sincey≤1 is never satisfied the semi-algebraic setS(IC∞) is empty. On the other hand in the cellC0it is clear that the semi-algebraic system (14) has no real solutions, sincea= −43 when(a,b)∈C0.
Now consider the subsetPof the cellC2consisting of all(a,b)wherea = −64/27 andb=0. For(a,b)∈P the systemWsimplifies to
x= 7
3b, y= −3
2 , y≤1
or
x= −5
3b, y= 3
4, y≤1
. (15)
Clearly the system (15) has precisely two solutions for anyb =0, namely(x,y)= (7/(3b),−3/2)and(x,y)=(−5/(3b),3/4). Similarly all other choices of(a,b)∈ C2yield associated specialized semi-algebraic sets with exactly two points.
5 Some Computational Results
In this section we apply the method of real comprehensive triangular decompositions to obtain (new) proofs of uniqueness for certain families of quadrature domains.
5.1 Aharonov–Shapiro 1976
In this subsection, we shall give an alternative proof of a theorem of Aharonov and Shapiro from [2], which states that a solid quadrature domain obeying a quadrature identity of the form
f d A=M0f(0)+M1f(0), f ∈ AL1() (16) is unique. Here M0 is the area of , so necessarily M0 > 0. The constant M1 is allowed to be an arbitrary complex number.
Recall that the computations in Example 2.2 show that a normalized mapping functionϕ : D→ is necessarily a polynomial of degree at most 2, which solves
the system
w13w¯1=M0w12−M1w2
w12w¯2=2M1
, w1:=ϕ(0) >0, w2:=ϕ(0). (17)
It remains to show that this system gives rise to a unique univalent solution. For this, we write
M1:=m1+i n1, w2:=u2+iv2.
We now obtain the following semi-algebraic system, which is equivalent to (17),
(W)
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
w14=M0w21−(m1u2−n1v2) 0=m1v2+n1u2
w12u2=2m1
−w21v2=2n1
M0>0 w1>0
(w21M0−(m1u2−n1v2))2−4w12(m21+n21)≥0 .
We must verify that the system W gives rise to at most one univalent solution for all relevant choices of quadrature data. For this, we first recognize that the last condition inWis just the Schur–Cohn constraint (see Theorem3.1), which ensures thatϕ(z)=0 inD. In general this is only necessary for univalence but for polynomials of degree two it is also sufficient. We shall treat c = (m1,n1,M0)as parameters.
Computing a real comprehensive triangular decomposition of the systemW, using theRegularChainslibrary, we obtain a partition of the parameter spaceR3into two cellsC0,C1having the following properties.
All points in the cellC0are such thatM0≤0; hence no point ofC0can correspond to a quadrature domain. Withω :=3
(27m21+27n21)/2, the cellC1is expressed as the disjoint union of the following six subsets ofR3,
(i)
⎧⎪
⎨
⎪⎩ M0=ω m1=m1
n1<0 ,
⎧⎪
⎨
⎪⎩ M0=ω m1=m1
n1>0 (ii)
⎧⎪
⎨
⎪⎩ M0=ω m1<0 n1=0
(iii)
⎧⎪
⎨
⎪⎩ M0=ω m1>0 n1=0 (iv)
⎧⎪
⎨
⎪⎩ M0> ω m1=m1
n1<0 ,
⎧⎪
⎨
⎪⎩ M0> ω m1=m1
n1>0 (v)
⎧⎪
⎨
⎪⎩ M0> ω m1<0 n1=0
,
⎧⎪
⎨
⎪⎩ M0> ω m1>0 n1=0 (vi)
⎧⎪
⎨
⎪⎩ M0>0 m1=0 n1=0
.
For(M0,m1,n1)in each of these sets, we now prove that the system (17) has a unique solution(w1, w2) = (w1,u2+iv2)withw1 > 0. Indeed, straightforward calculations show that in each of the parameter-domains, (i)–(vi) the semi-algebraic systemWsimplifies to, respectively,
(i)
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
v2w21+2n1=0 n1u2+v2m1=0
(9m21+9n21)v2+2M02n1=0 w1>0
(ii)
⎧⎪
⎨
⎪⎩
2u2w1−u22+2M0=0 9m1u2−2M02=0 v2=0
