Rigidity of Three-Dimensional Internal Waves with Constant Vorticity

Abstract

This paper studies the structural implications of constant vorticity for steady three-dimensional internal water waves in a channel. It is known that in many physical regimes, water waves beneath vacuum that have constant vorticity are necessarily two dimensional. The situation is more subtle for internal waves traveling along the interface between two immiscible fluids. When the layers have the same density, there is a large class of explicit steady waves with constant vorticity that are three-dimensional in that the velocity field is pointing in one horizontal direction while the interface is an arbitrary function of the other horizontal variable. We prove the following rigidity result: every three-dimensional traveling internal wave with bounded velocity for which the vorticities in the upper and lower layers are nonzero, constant, and parallel must belong to this family. If the densities in each layer are distinct, then in fact the flow is fully two dimensional. The proof is accomplished using an entirely novel but largely elementary argument that draws connection to the problem of uniquely reconstructing a two-dimensional velocity field from the pressure.

Appendix A. Dimension Reduction for the Vorticity

For completeness, we give here the proof of the dimension reduction result for the vorticity, which generalizes Constantin’s argument for the single-fluid case in [4].

Proof of Proposition 2.1

Seeking a contradiction, suppose that one of $\gamma _i$ is not zero, say, $\gamma _1 \ne 0$; the argument for the other case $\gamma _2 \ne 0$ can be treated the same way. Then from the third component of the vorticity equation (1.5) we see that $w_1$ is constant in the direction of $\varvec{\omega }_1$, which is transverse to the lower boundary at $z = - h_1$. From the kinematic condition (1.1e), it follows that $w_1$ vanishes identically on the open neighborhood ${\mathcal {N}}:= \{ (x,y,z): -h_1< z < \inf \eta \}$ of the bed. As it is real analytic, this forces

$$\begin{aligned} w_1 \equiv 0 \quad \text {in }\ \Omega _1. \end{aligned}$$

Reconciling this with (1.4), (1.1b) and (1.1a), we then have

$$\begin{aligned}&\partial _z u_{1} = \beta _1, \quad \partial _z v_{1} = -\alpha _1, \end{aligned}$$

(A.1)

$$\begin{aligned}&\partial _x u_{1} + \partial _y v_{1} = 0, \end{aligned}$$

(A.2)

$$\begin{aligned}&\partial _z P_{1} = -\rho _1 g \end{aligned}$$

(A.3)

in $\Omega _1$. By integrating (A.1), we infer that

$$\begin{aligned} u_1 = {\bar{u}}_1(x, y) + \beta _1 z, \qquad v_1 = {\bar{v}}_1(x, y) - \alpha _1 z, \end{aligned}$$

(A.4)

in ${\mathcal {N}}$ for some functions ${\bar{u}}_1$ and ${\bar{v}}_1$. The reduced incompressibility condition (A.2) then implies that

$$\begin{aligned} \partial _x {\bar{u}}_{1} + \partial _y {\bar{v}}_{1} = 0, \end{aligned}$$

which ensures the existence of a reduced stream function $\bar{\psi }_1 = \bar{\psi }_1(x, y)$ defined on ${\mathcal {N}}$ such that $\nabla ^\perp \bar{\psi }_1 = (-\partial _y \bar{\psi }_1,\partial _x \bar{\psi }_1) = ({\bar{u}}_1, {\bar{v}}_1)$. Rewriting the two horizontal momentum equations (1.1a) in terms of $\bar{\psi }_1$, differentiating the result with respect to z and then using (A.3), we see that in ${\mathcal {N}}$, $\bar{\psi }_1$ satisfies

$$\begin{aligned} \left\{ \begin{aligned} \beta _1 \partial _x \partial _y \bar{\psi }_{1} - \alpha _1 \partial _y^2 \bar{\psi }_{1}&= 0 \\ -\beta _1 \partial _x^2 \bar{\psi }_{1} + \alpha _1 \partial _x \partial _y\bar{\psi }_{1}&= 0 \\ \Delta \bar{\psi }_1 - \gamma _1&= 0 \end{aligned} \right. \qquad \text {in } {\mathcal {N}}, \end{aligned}$$

(A.5)

where the last equation comes from (1.2). We consider two cases.

Case 1: $\alpha _1^2 + \beta _1^2 = 0$. From A.1 and (1.4) it follows that

$$\begin{aligned} \partial _z u_{1} = \partial _z v_{1} = 0, \qquad \partial _x v_{1} - \partial _y u_{1} = \gamma _1. \end{aligned}$$

(A.6)

We also find from (A.4) and (2.1) that in the neighborhood ${\mathcal {N}}$, $u_1 = {\bar{u}}_1$ and $v_1 = {\bar{v}}_1$ are harmonic functions with domain $\mathbb {R}^2$. The boundedness of ${\varvec{u}}_1$, and thus the boundedness of $({\bar{u}}_1, \bar{v}_1)$, allows one to appeal to the Liouville theorem for harmonic functions to conclude that $u_1$ and $v_1$ are constants. However this contradicts that fact that $\gamma _1 \ne 0$.

Case 2: $\alpha _1^2 + \beta _1^2 \ne 0$. In this case, direct computation from (A.5) yields that the second-order derivatives of $\bar{\psi }_1$ are all constant:

$$\begin{aligned} \partial _x^2 \bar{\psi }_{1} = -{\alpha _1^2 \gamma _1 \over \alpha _1^2 + \beta _1^2} =: A_1, \quad \partial _x \partial _y \bar{\psi }_{1} = -{\alpha _1\beta _1\gamma _1 \over \alpha _1^2 + \beta _1^2} =: B_1, \quad \partial _y^2 \bar{\psi }_{1} = -{\beta _1^2\gamma _1 \over \alpha _1^2 + \beta _1^2} =: C_1, \end{aligned}$$

(A.7)

from which one can solve for ${\bar{u}}_1$ and ${\bar{v}}_1$

$$\begin{aligned} {\bar{u}}_1 = -B_1 x - C_1 y + a_1, \qquad {\bar{v}}_1 = A_1 x + B_1 y + b_1 \end{aligned}$$

for some constants $a_1$ and $b_1$. Thus

$$\begin{aligned} u_1 = -B_1 x - C_1 y + \beta _1 z + a_1, \qquad v_1 = A_1 x + B_1 y - \alpha _1 z + b_1. \end{aligned}$$

(A.8)

Again boundedness of ${\varvec{u}}_1$ forces $A_1 = B_1 = C_1 = 0$, leading to $\alpha _1 = \beta _1 = 0$, a contradiction. $\square$