Contradiction Argument for Exterior Stationary Navier–Stokes Flow

October 12, 2025

Assume $\Omega$ is an exterior domain in $\mathbb{R}^2$, which we simplify to $\Omega = \mathbb{R}^2 \setminus B$ where $B$ is the unit ball. Let $u$ be a solution to the steady Navier-Stokes equations:

$$ \begin{cases} -\nu \Delta u + u\cdot\nabla u + \nabla p = 0 & \text{in } \Omega \\ \text{div } u = 0& \text{in } \Omega \end{cases} \label{PDE1} $$

We assume the solution has finite Dirichlet energy :

$$ \int_\Omega |\nabla u |^2 < \infty $$

We define the total pressure $\Phi$ and vorticity $\omega$ as

$$ \begin{aligned} & \Phi = \frac{1}{2}|u|^2 + p \\ & \omega = \nabla^\perp \cdot u = -\partial_2u_1+\partial_1u_2 \end{aligned} $$

The total pressure $\Phi$ satisfies the following equations derived from \eqref{PDE1} and the fact that $\Delta u = \nabla^\perp \omega$:

$$ \begin{aligned} \Delta \Phi - \frac{1}{\nu} \text{div}(u\Phi) =\omega^2 \\ \nabla\Phi = \nu\nabla^\perp\omega +\omega u^\perp \end{aligned} $$

Setup for Contradiction. Suppose $\Phi=1$ on $\partial B$, and $\Phi(x)$ is increasing to $+\infty$ as $|x| \to \infty$. Assume the level sets $S_t = \{x:\Phi(x)=t\}$ are "good circles", i.e.

(1) $t$ is a regular value of $\Phi$ for every $t>1$.
(2) $S_t$ is a $C^1$ circle separating 0 and $\infty$ for every $t>1$.
(3) $S_t$ is unique for every $t>1$.

Our Main Theorem is

Theorem 1. The $\Phi$ with its level set $S_t$ satisfying the above setup will lead to a contradiction.

Proof. Let $\Omega(t_1,t_2)$ be the annulus region bounded by $S_{t_1}$ and $S_{t_2}$, with $t_1<t_2$. Integrating equation \eqref{PDE1} on $\Omega(t_1,t_2)$ and applying the divergence theorem yields:

$$ \int_{S_{t_2}} |\nabla\Phi| - \int_{S_{t_1}} |\nabla \Phi| = (t_2-t_1) \frac{\mathcal{F}}{\nu} + \int_{\Omega(t_1,t_2)} \omega^2 \label{esti1} $$

where $\mathcal{F} = \int_{S_t} u \cdot n$ is the constant flux through $S_t$. Hereafter we use $\hat{\mathcal{F}} = \mathcal{F}/\nu$ for simplicity. The proof proceeds by examining the three possible cases for the flux $\mathcal{F} = \int_{\partial B} u \cdot n$

Case 1: Negative Flux ($\mathcal{F}<0$)

Let $t_2 = t \to +\infty$ and $t_1=1$ in the equation $\eqref{esti1}$. Since $\int_{\Omega} \omega^2 < \infty$ is bounded, we have:

$$ \int_{S_{t}} |\nabla\Phi | = (t-1) \hat{\mathcal{F}} + \int_{S_1} |\nabla \Phi| + \int_{\Omega(1,t)} \omega^2 $$

As $t \to +\infty$, the right-hand side goes to $-\infty$ because $\hat{\mathcal{F}} < 0$. However, the left-hand side must be non-negative. This is an immediate contradiction.

Case 2: Positive Flux ($\mathcal{F}>0$)

The estimate $\eqref{esti1}$ shows that $\int_{S_t} |\nabla\Phi|$ grows linearly with $t$, as we fix $t_1=1$ and choose $t_2=t\to +\infty$

First Estimate:
$$ \left|\int_{S_t} |\nabla \Phi| - \hat{\mathcal{F}} t\right| \le M_1 \quad \text{for every } t \text{ and some }M_1 \in \mathbb{R_+}. \label{estim1} $$
Second Estimate (Claim 1):
$$ \int_{S_\tau} |\nabla \Phi| \lesssim o(1) \cdot \sqrt{\tau} \quad \text{for some } \tau \in [t,2t], \forall t \text{ large} $$

Using the identity $\nabla \Phi = \nu \nabla^\perp \omega + \omega u^\perp$ and the coarea formula gives a contradiction with the first estimate.

Case 3: Zero Flux ($\mathcal{F}=0$)

First Estimate:
$$ 0< m \coloneqq \int_{S_{1}} |\nabla \Phi|\le \int_{S_t} |\nabla \Phi| \le \int_{S_1}|\nabla\Phi|+\int_\Omega \omega^2 \coloneqq M < +\infty $$
Second Estimate (Claim 2):
$$ \int_{S_\tau}|\nabla\Phi| \lesssim o(1) \quad \text{for some }\tau \in [t,2t], \ \forall \ t\text{ large} $$

Repeating the coarea argument again leads to a contradiction since $\int_{S_\tau} |\nabla\Phi| \to 0$ while the first estimate requires $\int_{S_t} |\nabla\Phi| \ge m > 0$.

All cases lead to a contradiction. This finishes the proof.

</html>