Potential flow near physical boundary

In a 2D or 3D channel flow, the vorticity is created at the physical boundary and diffuses into the body of the flow. Far away from the boundary, the flow can be assumed to be irrotational, i.e. having zero vorticity [1], [2, Chapter 14]. Let $u=u(x,t)$ be the velocity field. The vorticity is $\omega=\nabla\times u.$ Here, $\nabla$ is the differential operator with respect to spatial variables only. The assumption that $\omega=0$ is equivalent to the assumption that the flow is potential, i.e. $u=\nabla h$ for some scalar function $h=h(x,t).$ Again, this assumption/approximation is only good in the body of the flow far away from the boundary. For $u$ to satisfy the Navier-Stokes equations

$u_t-\Delta u+u\nabla u+\nabla p=0,\ \ \nabla\cdot u=0,$

$h$ must be a harmonic function and $p=-h_t-\frac{|\nabla h|^2}{2}.$ J. Serrin [3] pointed out that a weak solution of this kind can have no regularity in time. Solutions of the form $u(x,t)=a(t)\nabla h(x)$ is usually referred to as Serrin’s example.

Near the physical boundary, potential flow is no longer a good approximation for the true flow due to large vorticity. Interestingly, this fact is also confirmed at the mathematical level: one cannot impose a non-slip boundary condition $u=0$ on the solution $u=\nabla h$ without forcing it to be identically zero (static flow).

Proposition: Let $\Omega$ be an open connected subset of $\mathbb{R}_+^n$ such that $\Gamma= \partial \Omega\cap \{x_n=0\}$ is the closure of an open subset of $\{x_n=0\}$ . Let $h:\bar{\Omega}\to\mathbb{R}$ be a function that is continuous on $\bar{\Omega}$ and harmonic on $\Omega.$ If $\nabla h=0$ on $\Gamma$ then $h$ is a constant function.

Proof: Because $\nabla h=0$ and $\Gamma$ is path-connected, $h$ must be constant on $\Gamma.$ Subtracting a constant from $h$ if necessary, we can assume that $h=0$ on $\Gamma.$ By translation, one can assume that $0\in\mathbb{R}^n$ is an interior point of $\Gamma.$ One can extend $h$ by Schwarz reflection principle to the domain $D=\Omega\cup (-\Omega),$ where $-\Omega=\{(x',-x_n):\ x=(x',x_n)\in\Omega\}$ as follows:

$\tilde{h}(x',{{x}_{n}})=\pm h(x',\pm x_n).$

With a little abuse of notation, let us denote $\tilde{h}$ as $h.$ Note that points in the interior of $\Gamma$ are also in the interior of $D.$ Because $\Gamma\subset\{x_n=0\},$ one can prove by induction on $\alpha_1 + \ldots + \alpha_{n-1}\ge 0$ that the $\alpha$ -partial derivative

$\frac{{{\partial }^{|\alpha| }}h}{\partial x_{1}^{{{\alpha }_{1}}}\partial x_{2}^{{{\alpha }_{2}}}\ldots\partial x_{n}^{{{\alpha }_{n}}}}(0)=0\quad\quad (1)$

for any multi-index $\alpha=(\alpha_1,\ldots,\alpha_{n-1},\alpha_n)\in\mathbb{Z}_{\ge 0}^{n-1}\times\{0\}.$ From the fact that $\frac{\partial h}{\partial x_n}=0$ on $\Gamma,$ one can show that (1) is true for $\alpha_n=1.$ Then from the fact that

$\frac{\partial^2h}{\partial x_n^2}=-\frac{\partial^2h}{\partial x_1^2}-\ldots-\frac{\partial^2h}{\partial x_{n-1}^2}=0\ \ \text{on}\ \ \Gamma$

one can show that (1) is true for any $\alpha_n\ge 2.$ Therefore, all partial derivatives of $h$ vanish at 0. Together with the fact that $h$ is analytic on $D,$ which is an open connected subset of $\mathbb{R}^n,$ one concludes that $h$ is identically zero.

Remark: for $n=2$ , there is a simple proof using Complex Analysis. Namely, $h$ is the real part of a holomorphic function on $D.$ Let $k$ be the imaginary part. The Riemann-Cauchy equations read $\partial h/\partial x=\partial k/\partial y$ and $\partial h/\partial y=-\partial k/\partial x.$ The condition $\nabla h=0$ and $\Gamma$ implies that both $h$ and $k$ must be constant on $\Gamma.$ Subtracting a complex constant from $h+ik$ if necessary, we can assume that $h=k=0$ on $\Gamma.$ Because the holomorphic function $h+ik$ vanishes on $\Gamma,$ it must be identically zero.