CC BY
26DOI: 10.15393/j3.art.20014.2689 Issues of Analysis. Vol. 3(21), No. 2, 2014
K. F. Amqzqva, E. G. Ganenkqya
ABOUT PLANAR (a, ft)-ACCESSIBLE DOMAINS1
Abstract. The article is devoted to the class Aof all (a, ¡3)— accessible with respect to the origin domains D, a, 3 € [0,1), possessing the property p = min where p € (0, is a fi-
p£&D
xed number. We find the maximal set of points a such that all domains D € Aare (y, 5)-accessible with respect to a, Y € [0; a], 5 € [0; 3]. This set is proved to be the closed disc of center 0 and radius p sin where p = min {a — y, 3 — 5} .
Keywords: a-accessible domain, (a, 3)-accessible domain, cone condition.
2010 Mathematical Subject Classification: 52A30, 03E15.
In [1] the notion of a-accessible domain was introduced. Let a E [0,1) be a fixed number. A domain D C Rn, 0 E D, is called a-accessible if for every point p E dD there exists a number r = r(p) > 0 such that the cone
( ( p \ an 1
K+(p, a, r) = < x E Rn : IIxN < r, [ x — p,^—^ > ^x — p\\ cos — >
I V m\J 2 j
is included in Rn\D.
In the case n = 2 these domains have been studied earlier by J. Stankie-wicz [2—4], D. A. Brannan and W. E. Kirwan [5], W. Ma and D. Minda [6], T. Sugawa [7] and others as a generalization of starlike domains. It was noted (see, for example, [7—10]) that in the planar case it is possible to consider domains, possessing a more general property. In [9, 10] such domains were called (a, [3)-accessible.
1 This work was supported by the Russian Foundation for Basic Research (project No. 14-01-00510) and by Program of Strategic Development of Petrozavodsk State University.
© Amozova K. F., Ganenkova E. G., 2014
Definition 1. [9], [10] Let a, ft £ [0,1), D c C, a £ D. A domain D is called (a, ft)-accessible with respect to a if for every point p £ dD there exists a number r = r(p) > 0 such that the cone
K + (p, a, a, ft, r) =
f ftn an 1
= < z £ C : —— < Arg(z — p) — Arg(p — a) < , \z — P\ < r >
is contained in C\D.
The notion of (a, ft)-accessible domain is a generalization of the notion of a-accessible domain. They are equal if a = ft.
In the article we choose values of arguments so that their difference belongs to (—n; n].
It was shown in [1] and [9, 10] that a- and (a, ft)-accessible domains satisfy the so-called "cone condition", i. e. such domains are also conically accessible from the interior.
The problem of characterization all domains with the "cone condition" is very hard. a- and (a, ft)-accessible domains are only special, but important cases of such domains. For a-accessible domains the axis of the symmetry of the cone is radial, for (a, ft)-accessible domains D the angle between this axis and the vector p — a, p £ dD, is fixed.
The following criterion of (a, ft)-accessibility is needed for the sequel.
Theorem A. [9, 10] Let D c C, dD be smooth, n(p) be an outward normal to the domain D at a point p £ dD, a, ft £ (0; 1). Then the domain D is (a, ft)-accessible with respect to the origin if and only if
(1 — ft )n . , ... (1 — a)n , .
< Arg(p) — Arg(n(p)) < -—(1)
for every p £ dD.
Let us note that Theorem A can be applied locally. In particular, if r is an open smooth curve, r c dD, a point p £ r, then from the proof of Theorem A it follows that the unbounded cone
K+ (p, 0, a, ft) = jz £ C : — < Arg(z — p) — Arg p < ^
is included in C\D.
Consider the class A^, containing all (a, fl)-accessible with respect to the origin domains D, possessing the property min \p\ = p, where
pedD
p e (0, to) is a fixed number. Let 7 e [0; a], S e [0; fl], D e Aa'3. By
iiD)S denote the set of all points a e D such that the domain D is (7, S)-accessible with respect to a.
In the present paper we find the maximal set, containing in WD6 for all domains D from A.
By B [0, R] denote the disc {z e C : \z\ < R}.
Theorem 1. If a, fl e [0; 1), 7 e [0; a], S e [0;fl] then
n WD6 = B
deAp'3
n • Pn 0, p sin —
. ,H 2 J
where p = min {a — y, fl — S} .
Proof. Let us show that the set P| WD is a disc. If D e Aap'3, then
DeA°p'3
a domain U(D), obtaining by a rotation of D with respect to the origin, also belongs to A^'3. Therefore the set U| WUd) , where the intersection
extends over all rotation transformations of D, is a disc with center at the origin. Hence,
n WD6 = n OU'6
U (D)
DeAp'3 U
is a disc too.
Let a,fl e [0; 1), D e Ap3. Fix p e dD. Since the domain D is (a,fl)-accessible with respect to the origin, then K+ (p, 0,a,fl,r) C C\D for some r > 0. Let us show that for all points a from the intersection of the domain D and the cone
K_(p, 0, a — y, fl — S) =
= {z e C : - < Arg(z - p) — Arg(—p) < }
the following inclusion holds
K+ (p, a,Y, S,r) C K+(p, 0,a,fl,r).
p
Take a G K-(p, 0,a — y, fi — S) H D, then (fi — S)n
— Arg(a — p) — Arg(—p) —
(a — y )n
Let ^ G K+ (p, a, y,S,t), z = p, this means that \z — p\ < r and
Sn Yn
—2 < Arg(z — p) — Arg(p — a) < -j-■
fifl.^
(2)
(3)
Figure 1: The domain D, case | Arg(z — p) — Argp\ > | Arg(p — a) — Argp\
If \ Arg(z — p) — Argp\ > \ Arg(p — a) — Argp\ (see fig. 1), then, by (2)
and (3),
fin Sn (fi — S)n
2
2
< Arg(z — p) — Arg p =
= (Arg(z — p) — Arg(p — a)) + (Arg(p — a) — Argp) < Yn (a — y )n an
- ~2 + 2 = T"'
and therefore z G K+(p, 0, a, fi, r). In the case
\ Arg(z — p) — Argp\ < \ Arg(p — a) — Argp\ (see fig. 2), we have z G K+ (p, 0,a — y, fi — S,r) and therefore z G K+(p, 0, a, fi, r).
Figure 2: The domain D, case \ Arg(z — p) — Argp\ < \ Arg(p — a) — Argp\
Consequently for both variants
K+ (p,a,Y,S,r) C K+(p, 0,a,fl,r) C C\D. Inscribe a disc with center at the origin into the set
K-(p, 0, a — - S) n D
and find its radius R(p). Denote by y a point from dK- (p, 0,a — y, fl — S) such that
\y\ = dist(0, dK- (p, 0, a — j,fl — S)).
Then, from the right triangle 0,y,p, we obtain R(p) = \y\ = \p\ sin P—,
2
where p = min{a — y, fl — S}. Put
R = min R(p) = p sin P—.
pedD 2
Then the disc B [0, R] is contained in K- (p, 0,a — Y,fl — S) n D for all p e dD. Thus B [0, R] c &Df for all D e Aa?.
Let us prove that it is impossible to enlarge the constant R in the last inclusion. For this aim we find a domain D0 e Ap? such that B [0, R] c
C &DS, but for every e > 0
B [0, R + e] £ Wd6 .
a) Let us begin with the case a, / £ (0; 1), y £ [0; a], S £ [0; /3]. Consider the simply connected domain D0, 0 £ D0, bounded by the logarithmic spirals
, n • j- (1 — a)n n
la (p) = pelLp e* tg-^-, 0 < p < 2,
, , • i (1 — n
lp(p) = pel* e— , -2 < p < 0,
and the circle
l(p) = pe2 tg ^ e^, -n<p < n, if a > / or l(p) = pe2 tg (1 2))T eiif, -n < p < n, if a < / (see fig. 3).
Figure 3: The domain Do, case a, ft £ (0; 1)
We will consider the case a > / only. The proof for a < / is analogous. Let us check that D0 £ Aa>P. Note that min \p\ = \la(0)| = \lp(0)| = p.
p pedD
Show that domain D0 is (a, 3)-accessible with respect to the origin. Take p £ dD0. Prove that the unbounded cone K+(p, 0,a,3) is contained in C\D0. Denote by n(p) the outward normal to the domain D0 at the point p if such a normal exists. Divide the proof into six cases.
a 1) Let p = la(p), p G (0; J) ■ Then
Argp — Arg n(p) = Argp — (Arg l'a(p) — Jj
(1-a)n i (1 — a)n\\ n
= p — arg (pe^e*tg ^ ^ + tg (1 f^) J + J =
1 n ( an) n (1 — a)n
+ - = — arctg tg— + - = -—.
1 n (
= p — p — arctg —(1-a)n + j = — arctg (tg
tg (i-2")n 1 2 - —&VU6 2/2 2
By Theorem A, for such p we have K+ (p, 0, a, fi) C C\D0■
a 2) If p = lp (p), p G (p0; 0), where p0 is the solution of the equation l(po) = lp (po )■ Then
Argp — Arg n(p) = Argp — (Arg l'p(p) — = *>-««! "^ " tg )) + 2 =
1 n (1 — fi )n
= p — p — n + arctg ¡gn-p + 2 =--— ■ (4)
Consequently, by Theorem A, K+(p, 0,a,fi) C C\D0■
(n 1
a 3) Let p = l(p), p G (—n; p0) U ( —; n ■ In this case
2
K+(p, 0, a, fi) C K+ (p, 0,1,1) C C\D0■
a 4) Consider
(n) (n) . n (i-q)n
p = l\ 2/ = H 2/ = pie 2
Since arg l'a = n + 02", arg l^"2) = n (here and below in a 5), a 6),
b3), b 5) we consider one-sided derivatives), and arga l (p), arg l (p) are monotone we get
K+(p, 0, a, fi) C K+ (p, 0,1,1) C C\D0■
an
a 5) Let p = la (0) = lp (0) = p■ Since arg l'a (0) = —— and arg lp (0) =
2
Bn
= n — -—, then K+ (p, 0, a, fi) C C\D0 and for all £i ,£2 > 0, £i2 + £22 = 0, 2
K+(p, 0, a + £1 ,fi + £2) H D0 =
Moreover, both rays of dK+ (p, 0,a + ei, fl + e2) intersect the domain Do.
n
a 6) The last case is p = l(po) = l?(po). Here arg l'(po) = po + — and
2
arg l? (po) = po + n — Thus,
K+(p, 0,a,ß) c K+ (p, 0,1,1) c C\Do.
Consequently K+ (p, 0,a,fl) C C\Do for all p e dDo. Therefore Do e
rz Aa,ß ^ p
Now we will show that B
in .
Do
0, p sin —
. ,H 2 J
is the maximal disc, contained
Let t E (0,1 - p). Fix
a* £ dB
0, p sin
(p + t)n
n K-(p, 0, p + t,p + t),
such that Im a* > 0 if p = ß — S and Im a* < 0 if p = a — j (see fig. 4).
Figure 4: Intersection Do and K+(p,a* ,j,S), case a, ß £ (0; 1)
2
Further we consider the case p = ft — S only, because the proof for the
a,
case p = a — 7 is analogous. Let us show that a* £ . By definition of
n
Arg p — Arg(p — a*) = (p + t)2. (5)
Denote by l the ray, consisting of points w £ dK+ (p, a* ,Y,S), w = p, such that
Sn
Arg(w — p) — Arg(p — a* ) = ——. (6) By (5) and (6), for every w £ l
Arg(w — p) — Arg(p) = (Arg(w — p) — Arg(p — a* )) + (Arg(p — a*) — Arg p) =
Sn (p + t)n (ft + t)n
2 2 2
Consequently, l is one of the ray of dK+ (p, 0,a, ft + t). As it was proved above (see a5)) for every t > 0
l n Do =
Therefore a* £ fi^o. Since t is arbitrary positive, we obtain that
B [0, p sin — 2
is the maximal disc, contained in .
' Do
b) Let a = 0, ft £ (0; 1), S £ [0;ft]. Consider the domain Do C C, 0 £ Do, bounded by the logarithmic spiral
, (1-SW n
lp(p) = pe^ , — 2 < p < 0,
the circle
l(p) = pe2 tg e1^. p £ (—n. — 2J
n 4. (1-^)n / ni
l(p)= pe 2 elif, p £ (—n, — U [0,n],
2
and the segment p; pe2 tg ( ^ (see fig. 5). Show that D0 £ A0'p.
p
Let us prove that domain D0 is (0, ft)-accessible with respect to the origin. Fix p £ dDo. Show that K+ (p, 0, 0, ft) C C\Do.
Figure 5: The domain Do, case a = 0, ft £ (0; 1)
b 1) If p = lp(p), p £ , then (4) is true (see a1)) and
K+(p, 0, 0,3) c K+ (p, 0, a, 3) C C\D0. b2) If p = l(p), p £ (—n; — 2) U (0; n) , then
K+ (p, 0, 0,3) C K+(p, 0,1,1) C C\D0.
b 3) Let p = lp (0) = p. In this case arg l^ (0) = n--. In addition
arg lp (p) is monotone. Hence, K+(p, 0, 0,3) C C\D0.
\ ( n t (1 — \ N
b 4) Consider the case p £ ( p; pe2 tg 2 j . By b3), we have
K+(p, 0, 0,3) C K+(p, 0, 0,3) C C\D0. b5) Let p = l (— 2) = lp (— 2) = —pie2tg (1 2>)n. Since
arg . (-|)=0andargl/ (-^^ - 1
and arg l/ (v) is monotone, we get
K+ (p, 0, 0,fl) C K+(p, 0,1,1) C C\Do.
Summarizing the proved above we obtain that K+ (p, 0, 0, fl) C C\D0 for all p G dD0. Since, in addition min \p\ = \l/(0)| = p, we conclude
pedDo
that Do G A03.
Let us check that B [0, 0] = {0} is the maximal disc, contained in • Suppose that for some r > 0
B [0, r] C q0dO •
Let z0 G B [0, r] and Im z0 < 0. Then, by construction of D0,
K+ (p,z0, 0, 0) n D0 = see fig. 3. Since K+(p,z0, 0, 0) C K+(p,z0, 0,5), then
K+(p,Z0, 0,5) n D0 =
Therefore, z0 G &D) • This contradiction shows that
B [0, r] g ftDO
for every r > 0.
c) In the case [ = 0, a G (0; 1), 7 G [0; a], we consider the domain D0 c C, 0 G D0, bounded by the logarithmic spiral
la(v) = peeV, 0 < v < ^,
the circle
n j- (1 — a)n
l(v)= pe2eiV, v G (-n, 0] U
7T
L2
and the segment p; pen tg ( 2 ) (see fig. 6).
Arguing as in case b), taking z0, Im z0 > 0, we prove that D0 G A and B [0, 0] is the maximal disc, containing in fi^j0.
a,0
p
Figure 6: The domain Do, case a € (0; 1), ( = 0
d) Let a = fl = 0. In this case, the class of (0, 0)-accessible domains coincides with the class of 0-accessible domains and the class of starlike with respect to the origin domains (see [11,12]).
Consider domain D0 = C\lp e A°p'°, where lp = {pt,t > 1}. Then the set W°D°0 consists of all points a e D0 such that K+ (p, a, 0, 0) C lp for every p = pr, t > 1. Consequently, W°D°0 = {pk, k < 1}. Therefore, for all £ > 0
B [0,£] ^ WD0 and n WD0 = {0}.
deAP'0
□
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Received September 3, 2014.
Petrozavodsk State University,
Lenin Avenue, 33, 185910 Petrozavodsk, Russia.
E-mail: [email protected], [email protected]
