- Research
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On the Wiener criterion in higher dimensions
- Tariq Ismaeel^{1, 2}Email author
- Received: 9 July 2017
- Accepted: 13 November 2017
- Published: 21 November 2017
Abstract
The Wiener criterion is a sufficient and necessary condition for the solvability of the Dirichlet problem. However, its geometric interpretation is not clear. In the case that the domain satisfies an exterior spine condition, the requirement for the spine is clear in dimension 3. In this note, we intend to obtain the condition that the exterior spine should satisfy in higher dimensions.
Keywords
- Wiener criterion
- Laplace equation
- Dirichlet problem
- boundary regularity
MSC
- 35A01
- 35B65
- 35J05
- 35J25
1 Introduction
In 1851, Riemann [1] proposed the famous Dirichlet principle, which states that there always exists a harmonic function continuous up to the boundary and coinciding with g on the boundary. However, Lebesgue [2] constructed a bounded domain on which the Dirichlet problem is not always solvable in 1912. By Perron’s method [3, 4], there always exists a harmonic function u in Ω with respect to g. If \(x_{0}\in\partial\Omega \) is a regular point (see [4, p.25] for the definition), then u is continuous up to \(x_{0}\). Hence, the solvability of (1.1) reduces to the problem whether the boundary points are regular. In 1924, Wiener [5] provided a sufficient and necessary condition for the regularity of \(x_{0}\). This is the famous Wiener criterion, which solves the Dirichlet problem completely.
What condition φ should satisfy in higher dimensions is not known. Besides, the following boundary regularity results are well known. Suppose that g is identically zero on a portion of the boundary near 0. If \(\varphi(r)\geq Cr^{1/2}\), which implies that Ω satisfies the exterior sphere condition, then the solution is Lipschitz continuous at 0. If \(\varphi(r)\geq Cr\), i.e., Ω satisfies the exterior cone condition, then the solution is Hölder continuous at 0. A natural question is whether the condition \(\varphi (r) \geq Cr^{a}\) for some constant C and \(a>1\) can guarantee the continuity (even Hölder continuity) of the solution at 0. The geometric meaning of this condition is that Ω satisfies the exterior Hölder spine condition. It should be pointed out that for the dimension \(n=3\), the continuity of the solution in this case is guaranteed (see (1.4)).
This note is devoted to deriving the sufficient and necessary condition for φ in a higher dimension and answer the above question.
Our main result is the following theorem.
Theorem 1.1
An immediate consequence is the following.
Corollary 1.2
Let \(n\geq4\) and \(0\in\partial\Omega\). Suppose that Ω satisfies the exterior spine condition with φ at 0 and (1.6) holds.
Then 0 is a regular point with respect to (1.1).
Remark 1.3
From (1.6), the special dimensions \(n=2,3\) should be noted. In addition, for \(n\geq4\) and \(a>1\), \(\varphi(r) \geq Cr^{a}\) is not enough to guarantee that 0 is a regular point, which is an essential difference to dimensions 2 and 3.
2 Proof of Theorem 1.1
Now, we give the proof of Theorem 1.1.
Proof
3 Conclusion
Let Ω be a bounded domain in \(R^{n}\) and \(x_{0}\in\partial\Omega \). The Wiener criterion provides a necessary and sufficient condition on Ω to solve the Dirichlet problem of the Laplace equation (i.e., (1.1)). However, its geometric meaning is not clear and it is not easy to verify whether a domain satisfies the Wiener criterion at some boundary point. One interesting case is that Ω is constructed by a spine near \(x_{0}\) (see (1.2)). It is natural to ask what condition the spine should satisfy to guarantee that \(x_{0}\) is a regular point. When the dimension \(n=2\) or 3, the results are well known. In this paper, we consider \(n>3\) and prove that \(x_{0}\) is a regular point with respect to the Dirichlet problem if and only if the spine satisfies (1.6).
Declarations
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Funding
This research is supported by NSFC 11671316.
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The author declares that they have no competing interests.
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