On the Local Existence and Uniqueness for the 3D Euler Equation with a Free Interface

Abstract

We address the local existence and uniqueness of solutions for the 3D Euler equations with a free interface. We prove the local well-posedness in the rotational case when the initial datum $u_0$ satisfies $u_0\in H^{2.5+\delta }$ and , where $\delta >0$ is arbitrarily small, under the Taylor condition on the pressure.

Appendix

For convenience, we provide here two proofs of the Cauchy invariance identity

$$\begin{aligned} \epsilon _{ijk} \partial _{j} v^{m} \partial _{k}\eta ^{m} = \omega _0^i \mathrm{,\qquad {}} t\ge 0 \end{aligned}$$

(6.1)

The first proof, which establishes also the Weber formula, is from [10, 19, 51], however rewritten in the coordinate notation used in the present paper. The second proof, which we believe is new, is shorter and bypasses the Weber formula.

Proof 1 We start with the Weber formula [12, 44]

$$\begin{aligned} \partial _{t} ( v^{j}\partial _{k}\eta ^{j} ) = \partial _{k} \left( \frac{1}{2} |v|^2 - q \right) \mathrm{,\qquad {}} i=1,2,3 \end{aligned}$$

(6.2)

which is proven as follows. The left side equals

$$\begin{aligned}&v_t^{j} \partial _{k}\eta ^{j} + v^{j}\partial _{k}\eta _t^{j} = - a_{j}^{m}\partial _{m}q \partial _{k}\eta ^{j} + v^{j}\partial _{k}v^{j} \nonumber \\&\qquad {}\!\!\!\!= - \partial _{k} q +\frac{1}{2} \partial _{k}(|v|^2) = \partial _{k} \left( \frac{1}{2} |v|^2-q \right) \end{aligned}$$

(6.3)

where we used

$$\begin{aligned} a_{j}^{m} \partial _{k}\eta ^j = \delta _{jk} \mathrm{,\qquad {}} j,k=1,2,3 . \end{aligned}$$

(6.4)

Applying the curl operator to the identity (6.2), rewritten as

$$\begin{aligned} \partial _{t} \left( \begin{array}{c} v^{j}\partial _{1}\eta ^{j} \\ v^{j}\partial _{2}\eta ^{j} \\ v^{j}\partial _{3}\eta ^{j} \end{array} \right) = \nabla \left( \frac{1}{2} |v|^2-q \right) , \end{aligned}$$

(6.5)

we get

$$\begin{aligned} \partial _{t} \left( \epsilon _{ijk} \partial _{j}(v^{m}\partial _{k}\eta ^{m}) \right) =0 . \end{aligned}$$

(6.6)

Note that

$$\begin{aligned} \partial _{t} \left( \epsilon _{ijk} v^{m} \partial _{jk}\eta ^{m} \right) =0 \end{aligned}$$

(6.7)

since

$$\begin{aligned} \epsilon _{ijk} v^{m}\partial _{jk}\eta ^{m} = \epsilon _{ijk} v^{m}\partial _{kj}\eta ^{m} = - \epsilon _{ikj} v^{m}\partial _{kj}\eta ^{m} \end{aligned}$$

(6.8)

using $\partial _{jk}=\partial _{kj}$ in the first equality and $\epsilon _{ijk}=-\epsilon _{ikj}$ in the second. By (6.6) and (6.7), we obtain

$$\begin{aligned} \partial _{t} \left( \epsilon _{ijk} \partial _{j}v^{m}\partial _{k}\eta ^{m} \right) =0 . \end{aligned}$$

(6.9)

The expression in parentheses at $t=0$ equals

$$\begin{aligned}&\epsilon _{ijk} \partial _{j}v^{m}(0)\partial _{k}\eta ^{m}(0) =\omega _0^i \mathrm{,\qquad {}} t\ge 0 \end{aligned}$$

(6.10)

and thus (6.1) follows.

Proof 2 Taking the time derivative, we obtain

$$\begin{aligned} \partial _{t}(\epsilon _{ijk} \partial _{j} v^{m} \partial _{k} \eta ^{m})&= \epsilon _{ijk} \partial _{j} v^{m} \partial _{k} v^{m} + \epsilon _{ijk} \partial _{j} v_{t}^{m} \partial _{k} \eta ^{m} \nonumber \\&= 0- \epsilon _{ijk} \partial _{j} (a^{l}_{m}\partial _{l} q) \partial _{k} \eta ^{m} \end{aligned}$$

(6.11)

where we replaced $v^{m}_{t}$ by $-a^{l}_{m}\partial _{l} q$ using the Euler equation. Now, by $\partial _{j} a^{l}_{m} = a^{l}_{s} \partial _{jr}\eta ^{s} a^{r}_{m}$, which follows by differentiating $a:\nabla \eta =I$, we get

$$\begin{aligned} \partial _{t}(\epsilon _{ijk} \partial _{j} v^{m} \partial _{k} \eta ^{m})&= - \epsilon _{ijk} a^{l}_{m}\partial _{jl} q \partial _{k} \eta ^{m} - \epsilon _{ijk} a^{l}_{s} \partial _{jr}\eta ^{s} a^{r}_{m} \partial _{l} q \partial _{k} \eta ^{m} \nonumber \\&= - \epsilon _{ijk} \partial _{jl} q \delta _{kl} - \epsilon _{ijk} a^{l}_{s} \partial _{jr} \eta ^{s} \partial _{l} q \delta _{kr} \nonumber \\&=- \epsilon _{ijk} \partial _{jk} q - \epsilon _{ijk} a^{l}_{s} \partial _{jk} \eta ^{s} \partial _{l} q \nonumber \\&= 0+0 =0 \end{aligned}$$

(6.12)

where we used in the second equality $a:\nabla \eta =I$. Hence,

$$\begin{aligned} \epsilon _{ijk} \partial _{j} v^{m} \partial _{k} \eta ^{m} = \epsilon _{ijk} \partial _{j} v^{m}_{0} \partial _{k} \eta ^{m}(0) = \omega _{0}^i \end{aligned}$$

and the proof of (6.1) is concluded.