Does Helmholtz's decomposition give an over-determined rotational flow?

From Helmholtz's decomposition,
$v=v_{\scriptscriptstyle IR} +v_{\scriptscriptstyle R}$
where $\nabla\times v_{IR} =0$ and $\nabla\cdot v_R=0$

when apply this to the linearized Navier-Stokes equation,

i, imaginary unit
k, reduced frequency
$\gamma$, square root of the Prandtl number
s, shear wave number
$\xi$, viscosity ratio

it splits into two equations, namely,

$iv_{\scriptscriptstyle IR}-{1\over s^2}({4\over 3}+\xi)\nabla^2v_{\scriptscriptstyle IR}=-{1\over k\gamma}\nabla p$

$iv_{\scriptscriptstyle R}-{1\over s^2} \nabla ^2v_{\scriptscriptstyle R}=0$

now, just consider the rotational velocity. Does the following system is over-determined? (3 components of $v_{\scriptscriptstyle R}$, 4 equations)

$\left\{\begin{array}{cols} iv_{\scriptscriptstyle R}-{1\over s^2} \nabla ^2v_{\scriptscriptstyle R}=0 \\ \nabla\cdot v_R=0 \end{array} \right.$

• If $i^2=-1$ and $v$ is real... – Jean Duchon Mar 26 '18 at 7:56
• your term $iv$ should read $\partial v/\partial t$ – Carlo Beenakker Mar 26 '18 at 13:15

If you take the curl of your equation for $v$ (after the correction $iv\mapsto\partial v/\partial t$), you find
$$\nabla\times(\partial v/\partial t-s^{-2}\nabla^2 v)=0.$$
This partial differential equation is underdetermined , it does not have a unique solution because to any solution $v(r,t)$ you can add the gradient of a scalar field $\chi(r,t)$, and $v+\nabla\chi$ will still be a solution. The condition
$$\nabla\cdot v=0$$
• Yes, I totally agree with you. So it means on the right hand side of the rotational flow eqn, I'm free to add a scalar field, let's say,$\chi$. Then the eqn gives:$\partial v/\partial t - s^{-2} \nabla ^2 v = \nabla \chi$ Consequently, the right hand side of the irrotational flow should gives, $-{1\over {k\gamma}} \nabla p - \nabla \chi$. In other words, does that mean when one apply Helmholtz's decomposition to the linearised N-S eqn, the separation is not unique? – Huang Wei Mar 27 '18 at 6:30