Self-intersecting Interfaces for Stationary Solutions of the Two-Fluid Euler Equations

Abstract

We prove that there are stationary solutions to the 2D incompressible free boundary Euler equations with two fluids, possibly with a small gravity constant, that feature a splash singularity. More precisely, in the solutions we construct the interface is a $\mathcal {C}^{2,\alpha }$ smooth curve that intersects itself at one point, and the vorticity density on the interface is of class $\mathcal {C}^\alpha$. The proof consists in perturbing Crapper’s family of formal stationary solutions with one fluid, so the crux is to introduce a small but positive second-fluid density. To do so, we use a novel set of weighted estimates for self-intersecting interfaces that squeeze an incompressible fluid. These estimates will also be applied to interface evolution problems in a forthcoming paper.

A. Appendix

1.1 A.1 The variable change $\tau \rightarrow h(\tau )$

Let $\rho (\nu ) := \kappa _+(\nu ) - \kappa _-(\nu )$, where, by making $\delta$ smaller if necessary, we assume $\rho$ is strictly monotonically increasing on $[0,\delta ]$. We implicitly define

$$\begin{aligned} h^{-1}(u) := \int ^\delta _{u} \frac{d\nu }{\rho (\nu )}, \qquad u\in (0,\delta ) \end{aligned}$$

(cf. [22]). Then $h^{-1}:(0,\delta ) \rightarrow (0, \infty )$ is strictly monotonically decreasing and it is three times continuously differentiable ($\rho$ is at least two times continuously differentiable on $(0,\delta )$). In particular, the inverse h exists and satisfies

$$\begin{aligned} h'(\tau ) = -\rho (h(\tau )). \end{aligned}$$

(49)

Regularity assumptions (14) imply (after possibly making $\delta$ smaller) that

$$\begin{aligned} \frac{1}{\rho (\nu )} \sim \frac{1}{k\nu ^{\mu + 1}} \end{aligned}$$

and therefore

$$\begin{aligned} h^{-1}(u) \sim \frac{\mu }{k}\big (u^{-\mu } - \delta ^{-\mu } \big ). \end{aligned}$$

In particular,

$$\begin{aligned} h(\tau ) \sim (1 + \tau )^{-\frac{1}{\mu }} \end{aligned}$$

with constants depending only on $k,\delta$ and $\mu$. By taking derivatives of formula (49), we see the asymptotic formula can be differentiated three times, i.e.

$$\begin{aligned} |h^{(k)}(\tau )|\sim (1 + \tau )^{-k - \frac{1}{\mu }}, \qquad k=1,2,3. \end{aligned}$$

1.2 A.2 Hardy Operator and Compactness

We will need the following lemma on continuity and compactness of Hardy operator in weighted Lebesgue spaces (cf. [21]):

Lemma 15

Let $p>1$ and let $R < \infty$.

• 1. Let $\beta + p^{-1} < 1$, then

  $$\begin{aligned} f\longrightarrow \int _0^x f(t)dt \,:\, \mathcal {L}_{p,\beta }([0,R]) \longrightarrow \mathcal {L}_{p,\gamma }([0,R]) \end{aligned}$$

  is continuous for all $\gamma \ge \beta -1$ and compact for all $\gamma > \beta -1$.

• 2. Let $\lambda + p^{-1} >0$, then

  $$\begin{aligned} f\longrightarrow \int _x^R f(t)dt \,:\, \mathcal {L}_{p,\gamma }([0,R]) \longrightarrow \mathcal {L}_{p,\beta }([0,R]) \end{aligned}$$

  is continuous for all $\gamma \le \beta + 1$ and compact for all $\gamma < \beta + 1$.

In order to show compactness for certain singular integral operators, we will also need the following result:

Lemma 16

Let $a:[0,\delta ]\rightarrow \mathbb {R}$ be continuous with $a(0) = 0$ and let

$$\begin{aligned} A\omega (x):= & {} \int _0^\delta \omega (u)\frac{1}{(x-u) + i\rho (u)}\,du; \qquad \\ A_*\omega (x):= & {} \int _0^\delta \omega (u)\frac{1}{(x-u) + i\rho (x)}\,du. \end{aligned}$$

Then A and $A_*$ are continuous as operators $\mathcal {L}_{p,\beta }\longrightarrow \mathcal {L}_{p,\beta }$. Moreover $aA,\,Aa,\,aA_*,\,A_*a$ and their complex conjugates are compact.

Proof

We only need to show compactness, continuity follows as in Proposition 2. For $k\in \mathbb {N}$ let $\chi _k:[0,\delta ]\rightarrow [0,1]$ be a smooth cut-off function such that $\chi _k(u) = 1$ if $u<\frac{\delta }{k}$ and $\chi _k(u) = 0$ if $u\ge \frac{2\delta }{k}$. We set

$$\begin{aligned} a_k:=(1-\chi _k)a;\quad B:=Aa. \end{aligned}$$

The operators $B_k\omega (x) :=A(a_k\sigma )(x)$ are compact since $\frac{a_k(u)}{(x-u) + i\rho (u)}$ are bounded kernels ($a_k$ are identically vanishing near 0). It is not difficult to see that

$$\begin{aligned} \sup _{\Vert \sigma \Vert = 1}\Vert (B_k-B)\sigma \Vert _{p,\beta }\longrightarrow 0 \end{aligned}$$

and therefore B is compact as the limit of a sequence of compact operators. Others follow similarly.

1.3 A.3 Fourier Multipliers on Weighted Sobolev Spaces

We now state few important lemmas related to the Fourier multiplier theorems on certain weighted Lebesgue spaces. The first, proof of which can be found in [22], gives continuity properties of an integral operator with prescribed decay at infinity. More precisely, let us define

$$\begin{aligned} \phi \in \mathcal {L}_{p, \gamma }(\mathbb {R}) \, \Leftrightarrow \, (1+\tau ^2)^{\gamma /2}\phi \in L^p(\mathbb {R})\,; \end{aligned}$$

then one has:

Lemma 17

Let T be an integral operator on $\mathbb {R}$ with kernel K(x, y), satisfying, for some $J\ge 0$ the estimate

$$\begin{aligned} |K(x,y)| \lesssim \frac{1}{|x-y|}\frac{1}{(1 + |x-y|^J)}. \end{aligned}$$

If $0<\gamma + p^{-1}<1 + J$ and $T:L^p(\mathbb {R})\longrightarrow L^p(\mathbb {R})$ is continuous, then

$$\begin{aligned} T: \mathcal {L}_{p, \gamma }(\mathbb {R}) \longrightarrow \mathcal {L}_{p, \gamma }(\mathbb {R}) \end{aligned}$$

is continuous.

In the construction of harmonic functions on $\Omega$ we work with weighted Lebesgue spaces with exponential weights. We make use of the following Fourier multiplier result (cf. [27]):

Theorem 18

Let $1<p<\infty$ and $u(x)=\exp (\pm d|x|)$ with $d>0$. Assume there exists a constant $c>0$ such that

$$\begin{aligned} \sup _{\xi \in \mathbb {R}} \langle \xi \rangle ^{|\beta + \gamma |} |D^{\beta + \gamma }_\xi a(\xi )| \le c \frac{\beta ^\beta }{(ed)^{|\beta |}}; \qquad \forall \xi \in \mathbb {R}^n,\, \beta \in \mathbb {N}^n, \, \gamma \le 2n +2 \end{aligned}$$

(50)

where $\langle \xi \rangle :=\sqrt{1 + \xi ^2}$. Then, linear operator $T_a$, defined via

$$\begin{aligned} \widehat{T_a \phi }(\xi ):= a(\xi )\hat{\phi }(\xi ) \end{aligned}$$

on $L^2\cap L^p$ is bounded as an operator $\mathcal {L}^p(u)\rightarrow \mathcal {L}^p(u)$.

We need to verify that (50) is satisfied for:

Lemma 19

Let $d<1$ and $n=1$. Then

• 1. $a_1(\xi ):= \tanh (\frac{\pi \xi }{2})$

• 2. $a_2(\xi ):= \coth (\frac{\pi \xi }{2}) - \frac{2}{\pi \xi }$

satisfy (50).

Proof

For $a_1$, note that hyperbolic tangent can be written as a fractional series

$$\begin{aligned} \tanh \Big (\frac{\pi \xi }{2}\Big )&= \frac{4\xi }{\pi } \sum _{k=1}^\infty \frac{1}{(2k-1)^2 + \xi ^2} \\&=\frac{2}{\pi } \sum _{k=1}^\infty \bigg [\frac{1}{\xi + i\xi _k} + \frac{1}{\xi - i\xi _k}\bigg ], \qquad \xi _k:=2k-1 \\&=\frac{4}{\pi } \sum _{k=1}^\infty \frac{\cos \theta _k(\xi )}{\rho _k(\xi )}; \qquad \rho _k(\xi ):=\sqrt{\xi _k^2 + \xi ^2},\,\,\tan \theta _k(\xi )=\frac{\xi _k}{\xi } \end{aligned}$$

In particular, for $n\in \mathbb {N}_0$, we have

$$\begin{aligned} D^n_\xi \tanh \Big (\frac{\pi \xi }{2}\Big )&= (-1)^n n! \frac{2}{\pi } \sum _{k=1}^\infty \Bigg [\frac{1}{(\xi + i\xi _k)^{n+1}} + \frac{1}{(\xi - i\xi _k)^{n+1}}\Bigg ] \\&=(-1)^n n! \frac{2}{\pi } \sum _{k=1}^\infty \frac{\cos \big ((n+1)\theta _k(\xi )\big )}{\rho ^{n+1}_k(\xi )} \end{aligned}$$

and

$$\begin{aligned} \frac{\pi }{2}\langle \xi \rangle ^n\Big |D^n_\xi&\tanh \big ((\pi \xi )/2\big )\Big | \le n! \sum _{k=1}^\infty \frac{\rho _1(\xi )}{\rho _k(\xi )^2} \\&\le n! \rho _1(\xi )\,\int _1^\infty \frac{dy}{ (2y-1)^2 + \xi ^2} \end{aligned}$$

where the right-hand side

$$\begin{aligned} f(\xi ) := \rho _1(\xi )\,\int _1^\infty \frac{dy}{ (2y-1)^2 + \xi ^2} = \frac{\rho _1(\xi )}{2\xi }\bigg (\frac{\pi }{2}- \arctan \frac{1}{\xi }\bigg ) \end{aligned}$$

is monotonically increasing for $\xi \ge 0$ with $\lim _{\xi \rightarrow \infty }|f(\xi )|=\frac{\pi }{4}$, so in particular

$$\begin{aligned} \sup _{\xi \in \mathbb {R}}\langle \xi \rangle ^n\big |D^n_\xi \tanh \big ((\pi \xi )/2\big )\big |\le \frac{n!}{2}. \end{aligned}$$

For $\gamma \le 4$, we have

$$\begin{aligned} \sup _{\xi \in \mathbb {R}}\langle \xi \rangle ^{n+\gamma }\big |D^{n+\gamma }_\xi \tanh \big ((\pi \xi )/2\big )\big | \le \frac{n!}{2d^n}P(n)d^n \end{aligned}$$

where $P(n) = (n+4)!/n!$ is a polynomial of order 4 in n. The Stirling formula implies

$$\begin{aligned} n! \sim \sqrt{2\pi n}\Big (\frac{n}{e}\Big )^n, \end{aligned}$$

so we are finished if we can show $\sqrt{n} P(n)d^n$ is bounded, but this is true provided $d<1$.

Similarly, for $a_2$ we can write

$$\begin{aligned} a_2(\xi )=\coth (\frac{\pi \xi }{2}) - \frac{2}{\pi \xi } = \frac{4\xi }{\pi } \sum _{k=1}^\infty \bigg [\frac{1}{\xi + i\xi _k} + \frac{1}{\xi - i\xi _k}\bigg ]; \qquad \xi _k = 2k, \end{aligned}$$

and the proof follows analogously.