When taking the curl of the Navier Stokes Equation, the Equation takes the form $${\rm rot} \left( \vec{v} \times {\rm rot} \vec{v} \right) + \nu \Delta {\rm rot} \vec{v} = 0 \, ,$$ where $\nu$ is the kinematic viscosity, which is supplemented by the incompressibility condition $${\rm div}\vec{v}=0 \, .$$
When looking for solutions which are constant at infinity, one typically writes $\vec{v}=\vec{v}_0 + \vec{v}_1$ where $\vec{v}_0$ is the constant flow-field at infinity. In a perturbational manner, it is then assumed that terms containing squares of $\vec{v}_1$ are small and the Equation linearizes $${\rm rot} \left( \vec{v}_0 \times {\rm rot} \vec{v}_1 \right) + \nu \Delta {\rm rot} \vec{v}_1 = 0 \, .$$ Using the vector identity $${\rm rot} \left( \vec{v}_0 \times {\rm rot} \vec{v}_1 \right) = - \vec{v}_0 \cdot {\rm grad} \, {\rm rot} \vec{v}_1$$ it goes over in $$- \vec{v}_0 \cdot {\rm grad} \, {\rm rot} \vec{v}_1 + \nu \Delta {\rm rot} \vec{v}_1 = {\rm rot} \left( \nu \Delta \vec{v}_1 - \vec{v}_0 \cdot {\rm grad} \vec{v}_1 \right) = 0 \, . \tag{1}$$
Now I have two questions:
Since $\vec{v}_1$ vanishes at infinity, also its derivatives vanish at infinity sufficiently strong. Hence, I can write $$\nu \Delta \vec{v}_1 - \vec{v}_0 \cdot {\rm grad} \vec{v}_1 = {\rm grad} \phi $$ with some undetermined scalar function $\phi$. Does this help me somehow? I can solve for the homogenous solution, and possibly write the solution in terms of Greensfunctions? Is it possibly true that $\phi=0$?
Suppose I have a two dimensional flow in the $x,y$-plane and $\vec{v}_0=v_0 \vec{e}_x$. Since ${\rm div}\vec{v}_1=0$ it follows $\vec{v}_1={\rm rot} \left( \psi \vec{e}_z \right)$ with some scalar function $\psi$. Equation (1) then becomes $$\left(\nu \Delta^2 - v_0 \partial_x \Delta \right) \psi=\Delta\left(\nu \Delta - v_0 \partial_x \right) \psi=0 \, . \tag{2}$$ It is not a biharmonic, and not a Laplace Equation. Without the $\partial_x$ it would be a Helmholtz-Laplace Equation for which solutions are known, but with the $\partial_x$? PS: One idea I had was to write as independent variables $$z=x+iy \qquad \bar{z}=x-iy \\ 2\partial_z = \partial_x - i\partial_y \qquad 2\partial_\bar{z} = \partial_x + i\partial_y \\ \partial_x = \partial_z + \partial_\bar{z}$$ and so the above equation reads $$\left\{16\nu \, \partial_z^2 \partial_\bar{z}^2 - 4v_0 \left( \partial_z^2 \partial_\bar{z} + \partial_z \partial_\bar{z}^2 \right) \right\} \psi = 0$$ which is fully symmetric in $z,\bar{z}$, and has the solution $$\psi = \psi_1(z) + \psi_2(\bar{z}) + c_0 {\rm e}^{c_1 z + \frac{c_1 v_0 \bar{z}}{4\nu c_1 - v_0}} \, .$$ $\psi_1,\psi_2$ are arbitrary functions of its argument, and $c_0,c_1$ are constants.
edit/just a quick remark: I think the problem with the solution is that it is not sufficiently general. $\psi_1,\psi_2$ are solutions to $\Delta \psi=0$ and the third term is only a solution to $\left(\nu \Delta - v_0 \partial_x \right) \psi=0$, but the entire solution space is much larger for equation (2).