Uniqueness of the potential flow past a cylinder

I have a question regarding the uniqueness of the potential flow past a cylinder.

Consider a two dimensional uniform potential flow in $$x_1$$-direction past the cylinder $$B_R = \{ x = (x_1, x_2) \in \mathbb{R}^2 : \vert x \vert < R \}$$ for some radius $$R > 0$$. The velocity potential $$\varphi(x)$$ is then subject to the boundary value problem

\begin{equation} \begin{aligned} \Delta_x \, \varphi(x) &= 0 \qquad &&\text{in} \quad \mathbb{R}^2 \setminus \overline{B_R} \; , \\ \frac{x}{\vert x \vert} \cdot \nabla_x \, \varphi(x) &= 0 \qquad &&\text{on} \quad \partial B_R \; , \\ \end{aligned} \end{equation}

together with the limiting behaviour $$\nabla_x \, \varphi(x) \to (v_\infty, 0)$$ as $$\vert x \vert \to \infty$$ for some $$v_\infty > 0$$. In most hydrodynamic lectures the general solution to this problem is given as the superposition of a uniform flow and doublet in $$x_1$$-direction,

$$\begin{equation} \varphi(x) = v_\infty \left( x_1 + \frac{R^2 x_1}{x_1^2 + x_2^2} \right) \; . \end{equation}$$

While I see that the given solution solves the stated problem I do not see how to proof uniqueness or wether this is even possible.

As far as I know the exterior Neumann problem of the Laplace equation for some bounded (and for simplicity connected) domain $$\Omega \subset \mathbb{R}^2$$ with outer normal $$\nu$$,

\begin{equation} \begin{aligned} \Delta u&= 0 \qquad &&\text{in} \quad \mathbb{R}^2 \setminus \overline{\Omega} \; , \\ \frac{\partial u}{\partial \nu} &= 0 \qquad &&\text{on} \quad \partial \Omega \; , \\ \end{aligned} \end{equation}

together with the limiting behaviour $$\vert u(x) \vert / \ln(\vert x \vert) \to 0$$ as $$\vert x \vert \to \infty$$ has the unique harmonic solution $$u = const.$$ on $$\mathbb{R}^2 \setminus \Omega$$ up to an additive constant.

Since for the above solution $$\vert \varphi(x) \vert / \ln(\vert x \vert)$$ diverges as $$\vert x \vert \to \infty$$ I wonder wether it is the unique harmonic solution and how to proof/contradict it.