Abstract

We establish sharp estimates for Green’s functions of cone-type planar domains. Our work generalizes all estimates given by Zhao in 1988 and Selmi in 2000. Our principal idea is to use conformal mappings.

1. Introduction

We work in the Euclidean space . By , we denote the Green function for the Laplacian in a domain . We write for the distance from to the Euclidean boundary of . For two positive functions and on a set , we say that is comparable to on and we denote , if there exists , such that , . The constant will be called the constant of comparison. For , we denote and . denote the open disk of center and radius . For , we denote and . In [1], Zhao established estimates for the Green function on bounded domains in . More precisely, he proved thatThis result was generalized later by Selmi in [2]. He established inequalities for the Green function of a class of Dini-smooth Jordan domains in . He proved that if is a bounded multiply connected Dini-smooth Jordan domain in , then inequality (1) holds. In [3], under restricted conditions on , Sweers established an implicit formula for the estimate of the Green function; the estimate is expressed as a function of the first eigenfunction. The aim of the present paper is to give estimates for the Green function in some particular bounded and unbounded nonsmooth domains in . Our work generalizes all estimates given by Zhao in [1], Selmi in [2], and Sweers in [3]. Our principal idea is to use conformal mappings. Note that, by the Riemann Theorem (see [4], pages 133–136 and pages 229–233), if is a nonempty open simply connected domain in , distinct of , then there exists a conformal mapping from onto the unit disk . Consequently, for all Thus, to give estimates for the Green function, it is sufficient to give estimates for and on . The problem reduces to understanding distortion introduced by . The paper is organized as follows. In Section 2, we establish some preliminary results that will be necessary throughout this paper. In Section 3, we give estimates for the Green function in , . We put and For , we denote (resp., ), the distance from to (resp., from to ). If , we define the bracket . We prove the following results.

Theorem 1. Let , . Then

Corollary 2. Let , . Then

In Section 4, we give estimates for the Green function in , We put , , and . For , we denote (resp., ; resp., ), the distance from to (resp., from to ; resp., from to ). For , we define the brackets , , , and . We prove the following.

Theorem 3. Let , . Then

Corollary 4. Let , . Then, for all ,

In Section 5, we give estimates for the Green function in . To study the intersection of two open disk, let us remark that using translations, rotations, and homotheties we reduce the problem to study the situation of and , where , , and . The two boundary circles and intersect nontangentially at with and . The interior angle of the intersection is given by and the boundary of is formed by the arcs of circlesWe denote by the distance of to the boundary . For , we write We prove the following.

Theorem 5. For all ,where .

Corollary 6. For all ,

In Section 6, we give estimates for the Green function in . We reduce the problem as in Section 5. We prove the following.

Theorem 7. For all , where .

In Section 7, we give estimates for the Green function in . We reduce the problem as in Section 5. We prove the following.

Theorem 8. For all , where .

In Section 8, we give estimates for the Green function in , . We prove the following.

Theorem 9. For all ,where , , and .

2. Preliminaries Results

In this section, the following notations will be adopted. For and , we denote

Lemma 10. Let , . Then,

Proof. It suffices to see that is continuous and does not vanish on .

Lemma 11. Let . Then

Proof. Let . The functionis holomorphic on and continuous and does not vanish on . Then, by maximum and minimum principles, it is sufficient to control on . We have to discuss three cases.
Case 1. If , then The result follows from Lemma 10.
Case 2. If , then Case 3. If , thenSo, we get Since the constants are independent of , we obtain, for all ,

Lemma 12. Let . Then

Proof. The proof is similar to that of Lemma 11. Let . The function is holomorphic on and continuous and does not vanish on . By maximum and minimum principles, it is sufficient to control on . We have to discuss four cases.
Case 1. If , thenwhere . The result follows from Lemma 10.
Case 2. If , thenCase 3. If , then Case 4. If , then As the constants are independent of , we obtain for all Now let ; then . By using the previous relation, we get, for all ,

By using the conjugate expression in the last lemma, we obtain the following.

Lemma 13. Let . Then

Lemma 14. Let , . Then,

Proof. We have to discuss two cases.
Case 1. If , assume that and put . Then, and . Sincethen the result follows by Lemma 11.
Case 2. If , assume that and put . Then, and , or . Since then the result follows by Lemmas 12 and 13.

Lemma 15. Let , . Then

Proof. The function is continuous and even and does not vanish on . To study on a neighborhood of , we put . Hence As , the function can be extended by continuity on and the compacity argument finishes the proof.

3. Estimates for the Green Function on the Sector

It is clear that

Lemma 16. Let , , Then

Proof. First, we remark that On the other hand, the bracket satisfiesLetObviously, is continuous and . Assume that ; we see that and . In consequence, . This implies the contradiction ( and ). Also, implies the contradiction ( and ). By compacity argument, with . This leads us to

Lemma 17. Let , . Then

Proof. It is clear that for all , . MoreoverThe functionis continuous and even and does not vanish. In fact allows us to conclude that with , where the contradiction is clear enough. Then .

For , , , we denote by the distance of to the boundary of .

Proof of Theorem 1. If , then : As, for , the function is a conformal mapping from onto , then we have By Lemma 15, we remark that and . Then, On the other hand, by Lemmas 11, 14, and 16, we getSo, we obtain

Now we present two proofs of Corollary 2.

First Proof of Corollary 2. By using Lemma 17, we get, for all ,By substituting the previous expression in relation (3), we get

Second Proof of Corollary 2. As, for , the conformal mapping , from onto , is the only singular point of order and in a neighborhood of 0. Moreover for We observe thatand for all Consequently,This implies that for all

4. Estimates of the Green Function on

It is clear thatand . We see that on and on In the same way in Lemmas 11, 12, and 13, we prove the following lemma.

Lemma 18. Consider

By using the last Lemma, we get the following.

Lemma 19. Let , . Then

Lemma 20. Let , . Then

Proof. By symmetry, it is sufficient to prove the result for . We see that, for all and , Moreover, if ,If , then for Since is positive and for , then . This implies for that Finally, for , we consider the function is continuous and does not vanish in . We will prove that is continuous at and does not vanish. Let . Every can be written as with . Besides, if and only if . This implies that , and so and . This proves that for and in a neighborhood of . This enables us to extend continuously on by 1. The compacity argument ensures the proof.

Lemma 21. Let , . Then

Proof. The proof follows from Lemmas 19 and 20.

Proposition 22. Let . Then

Proof. We equip with the norm , . We have Let . The functionis continuous and differentiable in two points , andIt follows thatHence, the function defined on is continuous, with Moreover, the numerator and the denominator of do not vanish on , . In fact, implies and then and . In consequence,For the denominator the argument is clear. Finally, is continuous on the compact subset and does not vanish. In consequence, , on .

Remark 23. Let , . Then

Proof. We have to discuss three cases.
Case 1. If , then , which implies thatUsing the same argument, we obtain the result if .
Case 2. If , then and , which implies In the same way, we prove the result for .
Case 3. If , then , which implies

Remark 24. If , then for all Moreover, since , thenThis implies that

Proof of Theorem 3. The Green function of is given, for all , by Hence, the Green function of isSince, the function is a conformal mapping from onto , then for all On the other hand, if , then . It follows that By Lemma 15, for all , we have . By Lemma 14, . By Lemma 21, . By Lemma 16, . Hence, we obtainRelation (6) follows by using (87), Lemmas 14 and 15, and Proposition 22.

Proof of Corollary 4. The results follows by using (78) and (6) and the fact that .

Remark 25. By using relations (83) and (86) of Proposition 22, we get

5. Estimates of the Green Function on

We can see that

Lemma 26. Let such that and . Then,

Proof. () It is clear that . Inversely, we must prove that is lower bounded.(a)If , (or , , then , . This implies (b)If , (or , ), then , . This implies , , and so For (2), we use the fact that .

Proof of Theorem 5. The conformal mapping is a one to one application from onto the sector , where . More precisely,The Green function of is given for by From (4), the Green function estimates on , we have But, Besides, by Lemma 26,Moreover, we can verify that, for , . Thus the results follow.

Proof of Corollary 6. From the property of Lemma 17, we deduce for all For , we have Now, we see that, for ,Thus the results follow.

6. Estimates of the Green Function on

The conformal mapping is a one to one application from onto , where . The Green function is given for by The same analysis leads to analogous estimates. The boundary of is formed by the two arcs of circles We denote by the distance of to the boundary , . For Thus, we obtain Theorem 7.

7. Estimates of the Green Function on

The conformal mapping is a one to one application from onto the sector , where . The Green function is given for by The boundary of is formed by two arcs of circles We denote by the distance of to the boundary , . ConsiderWe find using the same analysis Theorem 8.

8. Estimates of the Green Function on

The conformal mapping is a one to one application from this domain onto the sector , where . The Green function is given for by The boundary of this domain is formed by the arc of the unit circle and the segment . For , let us putWe find using the same analysis Theorem 9.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.