Equilibrium points of logarithmic potentials on convex domains

0.14 MB·English

by  J. K. Langley

EQUILIBRIUM POINTS OF LOGARITHMIC POTENTIALS ON
CONVEX DOMAINS
6
0
0
2 J.K.LANGLEY

n
a Abstract. Let D be a convex domain in C. Let ak > 0 be summable con-
J stants andlet zk ∈D. If the zk converge sufficiently rapidlyto η∈∂D from
0 within an appropriate Stolz angle then the function ∞k=1ak/(z −zk) has
3 infinitelymanyzeros inD. Anexampleshowsthat the hypotheses onthe zk
P
arenotredundant,andthattworecentlyadvanced conjectures arefalse.
] M.S.C.2000classification: 30D35,31A05,31B05.
V Keywords: criticalpoints,potentials,zerosofmeromorphicfunctions.
C
.
h
t
a 1. Introduction
m
A number of recent papers [4, 5, 9, 10] have concerned zeros of functions
[

∞ a
1 k
(1) f(z)= ,
v z z
k
9 kX=1 −
2 and in particular the following conjecture [4].
7
1 Conjecture 1.1 ([4]). Let f be given by (1), where ak >0 and
0
a
6 (2) z C, lim z = , k < .
k k
/0 ∈ k→∞ ∞ zXk6=0(cid:12)(cid:12)zk(cid:12)(cid:12) ∞
th Then f has infinitely many zeros in C. (cid:12)(cid:12) (cid:12)(cid:12)
a
m TheassumptionsofConjecture1.1implythatf ismeromorphicintheplaneand,
assuming that all z are non-zero, f(z) is the complex conjugate of the gradient
: k
v of the associated subharmonic potential u(z) = ∞ a log 1 z/z . Moreover,
i k=1 k | − k|
X Conjecture1.1hasaphysicalinterpretationintermsoftheexistenceofequilibrium
P
pointsoftheelectrostaticfieldarisingfromasystemofinfinitewires,eachcarryinga
r
a chargedensitya andperpendiculartothecomplexplaneatz [8,p.10]. Conjecture
k k
1.1 is known to be true when a =o(√r) as r [4, Theorem 2.10](see
|zk|≤r k →∞
also [6, p.327]), and when inf a >0 [5] (see also [9]).
{Pk}
An analogue of Conjecture 1.1 for a disc was advanced in [3, Conjecture 2].
Conjecture 1.2 ([3]). Let 0<ρ< and θ R. Let f be given by (1), where
∞ ∈
∞
(3) z C, z <ρ, lim z =ρeiθ, a >0, a < .
k k k k k
∈ | | k→∞ ∞
k=1
X
Then f has infinitely many zeros in z <ρ.
| |
If f satisfies the assumptions of Conjecture 1.2 then f = u in z < ρ, where
u(z) = ∞ a log z z . Obviously there is no loss of ge∇neralit|y|in assuming
k=1 k | − k|
1
P
2 J.K.LANGLEY
that ρ=1 and θ =0 in Conjecture 1.2. Writing
1 1
(4) w = , w = , f(z)=wF(w),
k
1 z 1 z
k
− −
where
∞
a w
k k
(5) F(w)= ,
w w
k
k=1 −
X
it is easy to verify that Conjecture 1.2 is equivalent to the following.
Conjecture 1.3. Let F be given by (5), where
∞
1
(6) w C, Rew > , lim w = , a >0, a < .
k k k k k
∈ 2 k→∞ ∞ ∞
k=1
X
Then F has infinitely many zeros in Rew >1/2.
With the assumptions(6), the function F(w) in (5)is evidently meromorphicin
the plane. In 2 an example satisfying (5) and (6) will be constructed, such that
F(w) has no z§eros in C. Thus Conjectures 1.2 and 1.3 are false, and there is no
direct analogue of Conjecture 1.1 for the unit disc.
On the other hand the following theorem shows in particular that if the z
k
convergetoρeiθ sufficientlyrapidly,andifallbutfinitelymanyz lieinasufficiently
k
smallStolz angle, then the conclusionof Conjecture 1.2 does hold. It is convenient
tostateandprovetheresultwhenthez lieinaconvexdomainDandtheboundary
k
point ρeiθ is 1. There then exists (see 4) an open half-plane H such that D H
§ ⊆
and 1 lies on the boundary ∂H, and there is no loss of generality in assuming that
H is the half-plane Rez <1.
Theorem 1.1. Let D z C:Rez <1 be a convex domain such that 1 ∂D.
⊆{ ∈ } ∈
Let f be given by (1), where
∞
(7) z D, a >0, a < .
k k k
∈ ∞
k=1
X
Assume that 1 is a limit point of the set z : k N , and that there exist real
k
{ ∈ }
numbers ε>0 and λ 0 such that
≥
(8) 1 z τ < for all τ >λ,
k
| − | ∞
|1−Xzk|≤ε
and
π
(9) sup arg(1 z ) :k N, 1 z ε <C(λ)= .
k k
{| − | ∈ | − |≤ } 2λ
Then there exists a sequence (η ) of zeros of f satisfying η D,lim η =1.
j j j→∞ j
∈
Note that (8) implies that z : k N has no limit points z in the punctured
k
{ ∈ }
disc A given by 0< 1 z <ε, and that f is meromorphic on A. Moreover,(9) is
| − |
obviously satisfied if λ<1.
EQUILIBRIUM POINTS OF LOGARITHMIC POTENTIALS ON CONVEX DOMAINS 3
2. A counterexample to Conjecture 1.3
Let
1
(10) g(w)= .
w(w 2)(ew−1+1)
−
Then g has no zeros, but has simple poles at 0, 2 and
(11) u =1+(2k+1)πi, k Z.
k
∈
Straightforwardcomputations give
1 1
(12) Res(g,0)= − = a, Res(g,2)= =b,
2(e−1+1) − 2(e+1)
and, using (11),
1 1 1
(13) Res(g,u )= − = − = =c .
k u (u 2) (u 1)2 1 (2k+1)2π2+1 k
k k k
− − −
Then b and the c evidently satisfy
k
(14) b>0, c >0, c < .
k k
∞
k∈Z
X
Next, let
a b c
k
(15) h(w)= + + , L(w)=h(w) g(w).
−w w 2 w u −
− k∈Z − k
X
By (10), (11), (12), (13) and (14) the function h(w) is meromorphic in the plane,
and L(w) is an entire function.
Let m be a large positive integer, let R = 4mπ, and use c to denote positive
constants independent of m. Then simple estimates give
c
(16) g(w) for w 1 =R
| |≤ R2 | − |
and, as m ,
→∞
c c
k
(17) h(w) +c c +c =o(1) for w 1 =R.
k
| |≤ R R | − |
k∈ZX,|k|≥m k∈ZX,|k|<m
Combining (16) and (17) shows that L(w) 0 in (15), so that h=g has no zeros,
≡
and applying the residue theorem in conjunction with (16) now gives
(18) a=b+ c .
k
k∈Z
X
Hence h(w) may be expressed using (18) in the form
1 1 1 1
h(w) = b + c
k
w 2 − w w u − w
(cid:18) − (cid:19) k∈Z (cid:18) − k (cid:19)
X
1 2b c u
k k
(19) = + .
w w 2 w u
− k∈Z − k!
X
By (11), (14) and (19) the function F(w)=wh(w) may be written in the form
∞ ∞
d v
k k
(20) F(w)= , Rev 1, v , d >0, d < .
k k k k
w v ≥ →∞ ∞
k
k=1 − k=1
X X
4 J.K.LANGLEY
HereF evidently satisfiesthe requirementsof(5)and(6), butF hasno zerosinC,
since h has no zeros and h(0)= .
∞
Remark. It is conjectured further in [3, Conjecture 6] that if f satisfies (1) and
(2) with a z > 0 for each k then f has infinitely many zeros in C. The example
k k
(20), with a = d v and a v = d v 2 > 0, shows that this conjecture is also
k k k k k k k
| |
false.
3. An auxiliary result needed for Theorem 1.1
The proof of Theorem 1.1 rests upon the following proposition, which concerns
functions in the plane of the form (5), and uses standard notation from [7, p.42].
Proposition 3.1. Let 0<σ 1. Let F be given by (5), where
≤
∞
(21) w C, Rew >0, a >0, a < .
k k k k
∈ ∞
k=1
X
Assume that the set w : k N is unbounded and that there exist real numbers
k
{ ∈ }
R>0 and λ 0 such that
≥
(22) w −τ < for all τ >λ,
k
| | ∞
|wXk|≥R
and
2 σ
(23) s=sup argw :k N, w R <C(λ,σ)= arcsin .
k k
{| | ∈ | |≥ } λ 2
r
Then there exists a transcendental meromorphic function G with
(24) F(w)=G(w)(1+o(1)) as w ,
→∞
and the Nevanlinna deficiency δ(0,G) of the zeros of G satisfies δ(0,G) < σ. In
particular, F(w) has a sequence of zeros tending to infinity.
The zero-free example of (20) has λ = 1 and δ(0,F) = σ = 1, and all its poles
lie in Rew 1, so that Proposition 3.1 is essentially sharp.
≥
To prove Proposition 3.1, assume that F is as in the statement of Proposition
3.1. It follows from (22) that the set w : k N has no limit points w with
k
{ ∈ }
R < w < . In particular, F is meromorphic in the region 2R w < with
| | ∞ ≤ | | ∞
an essential singularity at infinity. The existence of a transcendental meromorphic
function G satisfying (24) then follows from a result of Valiron [12, p.15] (see also
[2, p.89]). In particular,Gis constructed[12]sothatF andGhavethe same poles
and zeros in w 2R. If w 4R then (21) gives
| |≥ | |≥
a w
k k
F(w) F (w) +O(1), F (w)= ,
1 1
| |≤| | w w
k
|wkX|≥2R −
so that
m(r,G) m(r,F )+O(1)=O(1)
1
≤
as r , by [6, p.327]. Since the poles w of G have exponent of convergence at
k
→∞
most λ by (22), it follows that G has lower order µ λ.
≤
Choose s ,s ,s with
0 1 2
(25) s<s <s <s <min π,C(λ,σ) ,
0 1 2
{ }
EQUILIBRIUM POINTS OF LOGARITHMIC POTENTIALS ON CONVEX DOMAINS 5
where s is as in (23) and satisfies s π/2 by (21). The proof of Proposition 3.1
≤
requires the following two lemmas.
Lemma 3.1. The function F satisfies liminfr∈R,r→+∞rF( r) >0.
| − |
Proof. Let r >0 and write w =u +iv with u and v real. Let
k k k k k
w u (r+u )+v2
p (r)=Re k = k k k.
k r+w (r+u )2+v2
(cid:18) k(cid:19) k k
Then (21) gives p (r)>0 and there exists d>0 such that p (r)>d/r as r .
k 1
→∞
Hence, again as r ,
→∞
∞
rF( r) rReF( r)=r a p (r) ra p (r)>a d.
k k 1 1 1
| − |≥− − ≥
k=1
X
(cid:3)
Lemma 3.2. There exists M > 0 such that F(w) M for all large w lying
1 1
| | ≤
outside the region argw <s .
0
| |
Proof. This follows from (21), (23) and (25), since there exists a positive constant
M such that w w M w for all such w and all k N. (cid:3)
2 k 2 k
| − |≥ | | ∈
The proof of Proposition3.1 may now be completed using Lemmas 3.1 and 3.2.
Assume that δ(0,G) σ. Then Baernstein’s spread theorem [1] gives a sequence
≥
r and, for each m, a subset I of the circle w = r , of angular measure
m m m
→ ∞ | |
at least
4 σ
min 2π, arcsin o(1) min 2π,2C(λ,σ) o(1) 2s ,
2
µ 2 − ≥ { }− ≥
(cid:26) r (cid:27)
using (23) and (25), and such that
max log G(w) : w I
m
(26) lim { | | ∈ } = .
m→∞ logrm −∞
Letm be large,andconsiderthe function v(w)=log F(w), whichis subharmonic
| |
on the domain
Ω= w C:r /4< w <r ,s <argw <2π s .
m m 0 0
{ ∈ | | − }
Thenv is bounded aboveonΩ, by Lemma 3.2. But the intersectionJ ofI with
m m
the arc w C : w = r ,s < argz < 2π s has angular measure at least
m 1 1
{ ∈ | | − }
2(s s ), so that standard estimates for the harmonic measure of J at r /2
2 1 m m
− −
now give
(27) ω( r /2,J ,Ω) M >0,
m m 3
− ≥
where M is independent of m. Since (24) implies that (26) holds with G replaced
3
by F, combining Lemma 3.2 with (27) and the two-constants theorem [11, p.42]
leads to
(28) r F( r /2) 0 as m .
m m
− → →∞
But (28) contradicts Lemma 3.1, and this completes the proof of Proposition 3.1.
6 J.K.LANGLEY
4. Proof of Theorem 1.1
Assume that f and D satisfy the hypotheses of Theorem 1.1. Define F using
thetransformations(4)and(5). ThenF satisfiesthehypothesesofProposition3.1
with R = 1/ε and σ = 1. Thus F has a sequence of zeros tending to infinity, and
so f has a sequence (η ) of zeros with lim η =1.
j j→∞ j
Itremainsonlytoshowthatsuchasequence(η )existswith,inaddition,η D,
j j
∈
and this is done by a standard argument of Gauss-Lucas type. Let η = η with j
j
large, and assume that η D. Since D is convex the supremum and infimum of
6∈
arg(z η) on D differ by at most π. Hence there exist an open half-plane H, with
−
D H andη ∂H, anda linear transformationu=T(z)=(z η)/a mapping H
⊆ ∈ −
onto Reu>0. Writing u =T(z ) then gives
k k
∞
a
k
0=Re(af(η))= Re <0.
− uk!
k=1
X
This contradiction completes the proof of Theorem 1.1.
References
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418-434.
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1965.
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chargedistributionsI:generaliseddeBruijn-Springerrelations,Trans.Amer.Math.Soc.,to
appear.
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∞k=1 z−akzk,J.d’AnalyseMath.62(1994), 271-286.
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Moscow1970.
[7] W.K.Hayman,Meromorphicfunctions,OxfordattheClarendonPress,1964.
[8] O.D.Kellogg,Foundations ofpotential theory,Springer,Berlin1967.
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RevMat.Iberoamericana20(2004), 285-314.
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[10] J.K.LangleyandJohnRossi,Criticalpointsofcertaindiscretepotentials,ComplexVariables
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[12] G. Valiron, Lectures on the general theory of integral functions, Edouard Privat, Toulouse,
1923.
Schoolof MathematicalSciences,University of Nottingham,NG72RD, UK
E-mail address: [email protected]