The Helmholtz–Weyl decomposition of $$L^r$$ vector fields for two dimensional exterior domains

Abstract

Let $$\Omega $$ be a two-dimensional exterior domain with smooth boundary $$\partial \Omega $$ and $$1< r < \infty $$ . Then $$L^r(\Omega )^2$$ allows a Helmholtz–Weyl decomposition, i.e., for every $$\mathbf{u}\in L^r(\Omega )^2$$ there exist $$\mathbf{h} \in X^r_{\tiny {\text{ har }}}(\Omega )$$ , $$w \in {\dot{H}}^{1,r}(\Omega )$$ and $$p \in {\dot{H}}^{1,r}(\Omega )$$ such that $$\begin{aligned} \mathbf{u} = \mathbf{h} + \mathrm{rot}\, w + \nabla p. \end{aligned}$$ The function $$\mathbf{h}$$ can be chosen alternatively also from $$V^r_{\tiny {\text{ har }}}(\Omega )$$ , another space of harmonic vector fields subject to different boundary conditions. These spaces $$X^r_{\tiny {\text{ har }}}(\Omega )$$ and $$V^r_{\tiny {\text{ har }}}(\Omega )$$ of harmonic vector fields are known to be finite dimensional. The above decomposition is unique if and only if $$1< r \leqq 2$$ , while in the case $$2< r < \infty $$ , uniqueness holds only modulo a one dimensional subspace of $$L^r(\Omega )^2$$ . The corresponding result for the three dimensional setting was proved in our previous paper, where in contrast to the two dimensional case, there are two threshold exponents, namely $$r=3/2$$ and $$r =3$$ . In our two dimensional situation, $$r=2$$ is the only critical exponent, which determines the validity of a unique Helmholtz–Weyl decomposition.