Abstract

Let $\Gamma$ be a closed smooth Jordan curve in the complex plane $\mathbb{C}$ ,G be a bounded domain in $\mathbb{C}$ with the boundary $\Gamma$ , and let $\overline{G}=G\cup\Gamma$ . We study functions that are continuous in $\mathbb{C}\setminus G$ and harmonic in $\mathbb{C}\setminus\overline{G}$ that grow more slowly than the function $|z|^2$ at $z\to\infty$ . It is shown that, if in the class of such functions there exists a solution to the overdetermined Neumann boundary value problem in which the function equals zero on $\Gamma$ and $\mu$ -almost everywhere on $\Gamma$ there exists and equals one the normal derivative of this function, then the domain G is a disk (Theorem 1). In this case, the solution is unique and it coincides up to a constant with the fundamental solution for the Laplace operator in $\mathbb{C}$ with a singularity in the center of the disk G. The proof of Theorem 1 is based on the application of the conformal mapping of the exterior of the unit disk onto the domain $\mathbb{C}\setminus \overline{G}$ . This mapping allows us to reduce the original problem for the domain $\mathbb{C}\setminus \overline{G}$ to overdetermined boundary value problem for the exterior of the disk in which the main difficulty is the heterogeneity of the boundary condition for the normal derivative. To study this condition some subtle results were required on the boundary properties of a function that performs the indicated conformal mapping as well as some properties of the Hardy classes Hp in the unit disk. Theorem 2 of the paper shows that in the general case the conditions in Theorem 1 cannot be relaxed. It states the existence of a bounded domain $G\subset\mathbb{C}$ different from a disk with a smooth Jordan boundary $\Gamma$ and functions $f_1,f_2,f_3\in C(\mathbb{C}\setminus G)$ harmonic in $\mathbb{C}\setminus\overline{G}$ for each of which exactly one of the conditions of Theorem 1 is not satisfied.

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