Remarks on the Solution Map for Yudovich Solutions of the Euler Equations

Abstract

Consider Yudovich solutions to the incompressible Euler equations with bounded initial vorticity in bounded planar domains. We present a purely Lagrangian proof that the solution map is strongly continuous in $L^p$ for all $p\in [1, \infty )$ and is weakly-$*$ continuous in $L^\infty$.

1 Introduction

Let $\Omega$ be a bounded domain in $\mathbb {R}^2$ with $C^2$ boundary $\partial \Omega$. Let $\Delta _D$ denote the Dirichlet Laplacian associated to $\Omega$. The vorticity formulation of the 2D incompressible Euler equations is

$$\begin{aligned} \partial _t \omega +u\cdot \nabla \omega =0,\quad (x, t)\in \Omega \times (0, \infty ), \end{aligned}$$

(1.1)

where the velocity u is recovered from the vorticity $\omega$ through the Biot-Savart law

$$\begin{aligned} u=\nabla ^\perp \Delta _D^{-1}\omega =K*\omega . \end{aligned}$$

(1.2)

Note that u given by (1.2) is parallel to the boundary $\partial \Omega$. In the celebrated work [6], Yudovich proved the existence and uniqueness of global solutions to (1.1) with bounded initial vorticity. This theory includes the important class of vortex patches [3].

Theorem 1.1

[4, 6]. Let $\omega _0\in L^\infty (\Omega )$. There exists a unique triple $(\omega , u, X_t)$ solution to (1.1) such that $\omega \in L^\infty (\mathbb {R}; L^\infty (\Omega ))$, $u(t)=K*\omega (t)$, $X_t:\Omega \rightarrow \Omega$ measure-preserving, invertible and

$$\begin{aligned}&\frac{d}{dt}X_t(x)=u(X_t(x), t),\quad X_0(x)=x\quad \forall x\in \overline{\Omega }, \end{aligned}$$

(1.3)

$$\begin{aligned}&\omega (x, t)=\omega _0(X^{-1}_t(x)). \end{aligned}$$

(1.4)

Moreover, the flow $X_t:\overline{\Omega }\rightarrow \overline{\Omega }$ is Hölder continuous on $\overline{\Omega }$ with exponent $\exp (-C|t|\Vert \omega _0\Vert _{L^\infty (\Omega )})$ for some $C=C(\Omega )$.

The preceding version of Yudovich theory is taken from [4] and is elegant in that the notion of solution is naturally defined in terms of the Lagrangian flow and does not involve test functions. The purpose of this note is to present proofs of folklore about the continuity of the solution map for Yudovich solutions in this purely Lagrangian framework.

To define the inverse of the flow $X_t$, we let $X_{s, t}(x)$ be the solution of

$$\begin{aligned} \frac{d}{dt}X_{s, t}(x)=u(X_{s, t}(x), t),\quad X_{s, s}(x)=x. \end{aligned}$$

(1.5)

In view of (1.3), we denote $X_{0, t}\equiv X_t$. Then we have

$$\begin{aligned} X_t^{-1}=X_{t, 0}, \end{aligned}$$

and (1.4) becomes

$$\begin{aligned} \omega (x, t)=\omega _0(X_{t, 0}(x)). \end{aligned}$$

(1.6)

We first state the continuity in time of Yudovich solutions.

Lemma 1.2

For all initial data $\omega _0\in L^\infty (\Omega )$, the unique solution $\omega$ given by Theorem 1.1 belongs to $C(\mathbb {R}; L^p(\Omega ))\cap C_w(\mathbb {R}; L^\infty (\Omega ))$ for all $p\in [1, \infty )$. Here $C_w(\mathbb {R}; L^\infty (\Omega ))$ denotes the space of functions that are continuous in time with values in the weak-$*$ topology of $L^\infty (\Omega )$.

Proof

We first note that since the velocity field u is Log-Lipschitz (see (2.5)), $X_t(x)\in C(\overline{\Omega }\times \mathbb {R})$ (see [4]). Therefore, if $\omega _0\in C(\overline{\Omega })$ then it is clear that $\omega \in C(\mathbb {R}; L^p(\Omega ))$ for all $p\in [1, \infty ]$. For $p\in [1, \infty )$ and $\omega _0\in L^\infty (\Omega )\subset L^p(\Omega )$, using the fact that $C(\overline{\Omega })$ is dense in $L^p(\Omega )$ and $X_t$ is measure-preserving, we obtain $\omega \in C(\mathbb {R}; L^p(\Omega ))$.

For any $f\in C(\overline{\Omega })$, (1.4) yields

$$\begin{aligned} g(t):=\int _\Omega \omega (x, t)f(x)dx=\int _\Omega \omega _0(X^{-1}_t(x))f(x)dx=\int _\Omega \omega _0(x)f(X_t(x))dx. \end{aligned}$$

Thus $g\in C(\mathbb {R})$ since $X_t(x)\in C(\overline{\Omega }\times \mathbb {R})$. Since $C(\overline{\Omega })$ is dense in $L^1(\Omega )$ and $X_t$ is measure-preserving, it follows that $t\mapsto \int _\Omega \omega (x, t)f(x)dx$ is continuous for all $f\in L^1(\Omega )$. Therefore, $\omega \in C_w(\mathbb {R}; L^\infty (\Omega ))$. $\square$

By virtue of Lemma 1.2, for every $t>0$, the solution map

$$\begin{aligned} S_t: L^\infty (\Omega )\ni \omega _0\mapsto \omega (t) \in L^\infty (\Omega ) \end{aligned}$$

(1.7)

is well defined. We prove that $S_t$ is strongly continuous in $L^p(\Omega )$ for all $p\in [1, \infty )$.

Theorem 1.3

Let $p\in [1, \infty )$. Let $\omega _0$, $\omega _0^n\in L^\infty (\Omega )$ such that $(\omega _0^n)_n$ converges to $\omega _0$ in $L^p(\Omega )$. Then for all $T>0$ we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in [-T, T]}\Vert S_t(\omega _0^n)-S_t(\omega _0)\Vert _{L^p(\Omega )}=0. \end{aligned}$$

(1.8)

Moreover, $S_t$ is continuous in the weak-$*$ topology of $L^\infty (\Omega )$.

Theorem 1.4

If $\omega _0^n\overset{*}{\rightharpoonup } \omega _0$ in $L^\infty (\Omega )$, then $S_\cdot (\omega _0^n)\overset{*}{\rightharpoonup }S_\cdot (\omega _0)$ in $L^\infty (\Omega \times (-T, T))$ for all $T>0$ and $S_t(\omega _0^n)\overset{*}{\rightharpoonup }S_t(\omega _0)$ in $L^\infty (\Omega )$ for all $t\in \mathbb {R}$.

It was obtained in [2, Corollary 1] that for the torus $\mathbb {T}^2$, the solution map for Yudovich solutions is continuous in $L^p$ when restricted to bounded sets of $L^\infty$. Theorem 1.3 dispenses with the restriction to bounded sets of $L^\infty$ and holds for domains with boundary. The proof in [2] is Eulerian and relies on $L^2$ energy estimates for the velocity and vorticity differences. On the other hand, our proof of Theorem 1.3 is purely Lagrangian: $L^p$ estimates for the vorticity difference is deduced from an $L^1$ estimate for the difference of the flow maps. The latter is established by employing an idea in [4] for the uniqueness of Yudovich solutions. We remark that Theorem 1.4 is stated without proof in [5] and is used to deduce properties of the omega-limit set of the 2D Euler equations.

On the whole space $\Omega =\mathbb {R}^2$, the same statement in Theorem 1.1 holds with $L^\infty (\Omega )$ replaced by $L^\infty _c(\mathbb {R}^2)$, the space of $L^\infty (\mathbb {R}^2)$ functions with compact support. Note however that the flow map $X_t$ is then only locally Hölder continuous with exponent $\exp (-C|t|\Vert \omega _0\Vert _{L^1\cap L^\infty })$, where C is a universal constant and $L^1\cap L^\infty \equiv L^1(\mathbb {R}^2)\cap L^\infty (\mathbb {R}^2)$ is equipped with the norm

$$\begin{aligned} \Vert \cdot \Vert _{L^1\cap L^\infty }=\Vert \cdot \Vert _{L^1(\mathbb {R}^2)}+\Vert \cdot \Vert _{L^\infty (\mathbb {R}^2)}. \end{aligned}$$

We have the following version of Theorem 1.4.

Theorem 1.5

Assume that $\omega _0 \in L^\infty _c(\mathbb {R}^2)$, $(\omega _0^n)_n\subset L^\infty _c(\mathbb {R}^2)$ is bounded in $L^1(\mathbb {R}^2)$ and $\omega _0^n\overset{*}{\rightharpoonup }\omega _0$ in $L^\infty (\mathbb {R}^2)$. Then $S_\cdot (\omega _0^n)\overset{*}{\rightharpoonup }S_\cdot (\omega _0)$ in $L^\infty (\Omega \times (-T, T))$ for all $T>0$ and $S_t(\omega _0^n)\overset{*}{\rightharpoonup }S_t(\omega _0)$ in $L^\infty (\mathbb {R}^2)$ for all $t\in \mathbb {R}$.

Under the hypothesis of Theorem 1.5, for all $p\in [1, \infty ]$ we have that $\Vert S_t(\omega ^n_0)\Vert _{L^p}=\Vert \omega _0^n\Vert _{L^p}$ is uniformly bounded by interpolation. Therefore, the conclusion in Theorem 1.5 implies $S_t(\omega _0^n)\overset{*}{\rightharpoonup }S_t(\omega _0)$ in ${\mathcal {M}}(\mathbb {R}^2)$, the space of signed Radon measures on $\mathbb {R}^2$, and $S_t(\omega _0^n)\rightharpoonup S_t(\omega _0)$ in $L^p(\Omega )$ for all $p\in (1, \infty )$.

Remark 1.6

As we have mentioned earlier, the notion of solution in Theorem 1.1 does not involve test functions. On the other hand, if $\omega$ is such a solution, then for any $\phi \in C^1(\overline{\Omega }\times [t_1, t_2])$ we have

$$\begin{aligned} \begin{aligned}&\int _{t_1}^{t_2}\int _\Omega \omega (x, t)\partial _t\phi (x, t)dxdt\\&\quad =\int _{t_1}^{t_2}\int _\Omega \omega _0(x)(\partial _t\phi )(X_t(x), t)dxdt\\&\quad =\int _{t_1}^{t_2}\int _\Omega \omega _0(x)\partial _t[\phi (X_t(x), t)]dxdt-\int _{t_1}^{t_2}\int _\Omega \omega _0(x)\nabla \phi (X_t(x), t)\cdot \partial _t X_t(x)dxdt\\&\quad =\int _\Omega \omega _0(x)[\phi (X_{t_2}(x), {t_2})-\phi (X_{t_1}(x), {t_1})]dx-\int _{t_1}^{t_2}\int _\Omega \omega _0(x)\nabla \phi (X_t(x), t)\cdot u(X_t(x), t)dxdt\\&\quad =\int _\Omega \omega (x, t_2)\phi (x, t_2)-\omega (x, t_1)\phi (x, {t_1})dx-\int _{t_1}^{t_2}\int _\Omega \omega (x, t) u(x, t)\cdot \nabla \phi (x, t)dxdt. \end{aligned} \end{aligned}$$

Thus $\omega$ obeys the weak form

$$\begin{aligned} \int _{t_1}^{t_2}\int _\Omega \omega (x, t)\left[ \partial _t\phi (x, t)+ u(x, t)\cdot \nabla \phi (x, t)\right] dxdt=\int _\Omega \omega (x, t_2)\phi (x, t_2)-\omega (x, t_1)\phi (x, {t_1})dx. \end{aligned}$$

(1.9)

2 Proof of Theorem 1.3

We first recall the following estimates for the Biot-Savart kernel K.

Lemma 2.1

[6]. There exists C depending only on $\Omega$ such that for all x, y, a, $b\in \Omega$, we have

$$\begin{aligned}&|K(x, y)|\le C|x-y|^{-1}, \end{aligned}$$

(2.1)

$$\begin{aligned}&\int _\Omega |K(x, a)-K(x, b)|dx\le C\varphi (|a-b|), \end{aligned}$$

(2.2)

where $\varphi$ is the Log-Lipschitz modulus of continuity

$$\begin{aligned} \varphi (r)=r(1-\ln r)\quad \text {if } 0<r\le 1,\quad \varphi (r)=1\quad \text {if } r> 1. \end{aligned}$$

(2.3)

As a direct consequence of (2.1) and (2.2), if $\omega \in L^\infty (\Omega )$ then $u=K*\omega$ is bounded and Log-Lipschitz:

$$\begin{aligned}&\Vert u\Vert _{L^\infty (\Omega )}\le C\Vert \omega \Vert _{L^\infty (\Omega )}, \end{aligned}$$

(2.4)

$$\begin{aligned}&|u(x)-u(y)|\le C\Vert \omega \Vert _{L^\infty }\varphi (|x-y|)\quad \forall x,~y\in \Omega . \end{aligned}$$

(2.5)

Let $(\omega ^j, u^j, X^j_t)$, $j=1, 2$ be two solutions of (1.1) with initial data $\omega _0^j\in L^\infty (\Omega )$. For notational simplicity we shall write $L^p\equiv L^p(\Omega )$. Fix $p\in [1, \infty )$. We have the elementary inequalities

$$\begin{aligned} (a+b)^p\le 2^{p-1}(a^p+b^p),\quad (a+b+c)^p\le 2^{p-1}a^p+2^{2p-2}(b^p+c^p)\quad \forall a, b, c\in \mathbb {R}_+. \end{aligned}$$

(2.6)

Since the flow maps $X^j_{s, t}$ are measure-preserving, we have

$$\begin{aligned} \begin{aligned} \Vert \omega ^1(t)-\omega ^2(t)\Vert _{L^p}^p&=\int _\Omega |\omega _0^1(X^1_{t, 0}(x))-\omega _0^2(X^2_{t, 0}(x))|^pdx\\&\le 2^{p-1}\int _\Omega |\omega _0^1(X^1_{t, 0}(x))-\omega _0^1(X^2_{t, 0}(x))|^pdx\\&\quad +2^{p-1}\int _\Omega |\omega _0^1(X^2_{t, 0}(x))-\omega _0^2(X^2_{t, 0}(x))|^pdx\\&\le 2^{p-1}\int _\Omega |\omega _0^1(X^1_{t, 0}(x))-\omega _0^1(X^2_{t, 0}(x))|^pdx+ 2^{p-1}\Vert \omega ^1_0-\omega ^2_0\Vert _{L^p}^p. \end{aligned} \end{aligned}$$

(2.7)

We extend $\omega _0^1$ to zero outside $\Omega$ and approximate $\omega _0^1$ by $\omega _0^1*\rho _\varepsilon$, where $\rho _\varepsilon$ is the standard mollifier. It follows from (2.7) that

$$\begin{aligned} \begin{aligned} \Vert \omega ^1(t)-\omega ^2(t)\Vert _{L^p}^p&\le 2^{p-1}\Vert \omega ^1_0-\omega ^2_0\Vert _{L^p}^p+2^{3p-2}\Vert \omega _0^1*\rho _\varepsilon -\omega _0^1\Vert _{L^p}^p\\&\quad +2^{2p-2}\int _\Omega |(\omega _0^1*\rho _\varepsilon )(X^1_{t, 0}(x))-(\omega _0^1*\rho _\varepsilon )(X^2_{t, 0}(x))|^pdx\\&\le 2^{p-1}\Vert \omega ^1_0-\omega ^2_0\Vert _{L^p}^p+2^{3p-2}\Vert \omega _0^1*\rho _\varepsilon -\omega _0^1\Vert _{L^p}^p\\&\quad +2^{2p-2}\Vert \omega _0^1*\rho _\varepsilon \Vert ^p_{\dot{C}^{1/p}}\int _\Omega |X^1_{t, 0}(x)-X^2_{t, 0}(x)|dx. \end{aligned} \end{aligned}$$

(2.8)

Set $F(x, t, r):=|X^1_{t, r}(x)-X^2_{t, r}(x)|$. Integrating (1.5) with respect to t we deduce

$$\begin{aligned} \begin{aligned} F(x, t, r)&\le \left| \int _t^r |u^1(X^1_{t, s}(x), s)-u^1(X^2_{t, s}(x), s)|ds\right| +\left| \int _t^r|u^1(X^2_{t, s}(x), s)-u^2(X^2_{t, s}(x), s)|ds\right| \\&:=I_1(x, t, r)+I_2(x, t, r). \end{aligned} \end{aligned}$$

(2.9)

The Log-Lipschitz bound (2.5) yields

$$\begin{aligned} |I_1(x, t, r)|\le C\Vert \omega _0^1\Vert _{L^\infty }\left| \int _t^r\varphi (F(x, t, s))ds\right| . \end{aligned}$$

(2.10)

As for $I_2$ we use the definition $u^j=K*\omega ^j$, (1.6) together the fact that the maps $X^j_{s, t}$ are measure-preserving, giving

$$\begin{aligned} |I_2(x, t, r)|&\le \left| \int _t^r\left| \int _\Omega K(X^2_{t, s}(x), y)\omega _0^1(X^1_{s, 0}(y))-K(X^2_{t, s}(x), y)\omega _0^2(X^2_{s, 0}(y))dy\right| ds\right| \\&=\left| \int _t^r\left| \int _\Omega K(X^2_{t, s}(x), X^1_{s}(y))\omega _0^1(y)-K(X^2_{t, s}(x), X^2_{s}(y))\omega _0^2(y)dy\right| ds\right| \\&\le \left| \int _t^r\int _\Omega |K(X^2_{t, s}(x), X^1_{s}(y))-K(X^2_{t, s}(x), X^2_{s}(y))||\omega _0^1(y)|dyds\right| \\&\quad +\left| \int _t^r\int _\Omega |K(X^2_{t, s}(x), X^2_{s}(y))|\omega _0^1(y)-\omega _0^2(y)|dyds\right| :=I_2^a+I_2^b. \end{aligned}$$

Integrating $I_2$ in x and using the fact that $X^2_{t, s}$ is measure-preserving, we deduce

$$\begin{aligned} \begin{aligned} \int _\Omega |I_2^a(x, t, r)|dx&=\left| \int _t^r\int _\Omega \int _\Omega |K(x, X^1_{s}(y))-K(x, X^2_{s}(y))|dx|\omega _0^1(y)|dyds\right| \\&\le C\Vert \omega _0^1\Vert _{L^\infty }\left| \int _t^r\int _\Omega \varphi (F(y, 0, s))dyds\right| , \end{aligned} \end{aligned}$$

where we have used (2.2) in the second estimate. Since $\varphi$ is concave, Jensen’s inequality implies

$$\begin{aligned} \frac{1}{|\Omega |}\int _\Omega |I_2^a(x, t, r)|dx\le C\Vert \omega _0^1\Vert _{L^\infty }\left| \int _t^r\varphi \left( \frac{1}{|\Omega |}\int _\Omega F(y, 0, s)dy\right) ds\right| . \end{aligned}$$

(2.11)

On the other hand, (2.1) gives

$$\begin{aligned} \begin{aligned} \int _\Omega |I_2^b(x, t, r)|dx&\le C\left| \int _t^r\int _\Omega \int _\Omega \frac{1}{|X_{t, s}^2(x)-X^2_{s}(y)|}dx |\omega _0^1(y)-\omega _0^2(y)|dyds\right| \\&=C\left| \int _t^r\int _\Omega \int _\Omega \frac{1}{|x-X^2_{ s}(y)|}dx |\omega _0^1(y)-\omega _0^2(y)|dyds\right| \\&\le C|t-r| \Vert \omega _0^1-\omega _0^2\Vert _{L^1}. \end{aligned} \end{aligned}$$

(2.12)

A combination of (2.9), (2.10), (2.11) and (2.12) implies that $\eta (t, r):=|\Omega |^{-1}\int _\Omega F(x, t, r)dx$ satisfies

$$\begin{aligned} \eta (t, r)\le C|t-r|\Vert \omega _0^1-\omega _0^2\Vert _{L^1} +C\Vert \omega _0^1\Vert _{L^\infty }\left\{ \left| \int _t^r \varphi (\eta (t, s))ds\right| +\left| \int _t^r \varphi (\eta (0, s))ds\right| \right\} \end{aligned}$$

(2.13)

for all $t, r\in \mathbb {R}$, where $C=C(\Omega )$.

Let $T>0$ be arbitrary. We first consider $t=0$ in (2.13). We choose $\omega _0^j$ such that $CT\Vert \omega _0^1-\omega _0^2\Vert _{L^1}<1$. Since $\eta (0, 0)=0$ and $\eta (0, \cdot )$ is continuous, there exists a maximal time $T_1\in (0, T]$ such that $\eta (0, s)<1$ for all $s\in [0, T_1)$. Consequently, in (2.13) we have $\varphi (\eta (0, s))=\eta (0, s)[1-\ln (\eta (0, s))]$ provided that $s\in [0, T_1)$. An application of Osgood’s lemma [1, Lemma 3.4] yields

$$\begin{aligned} \eta (0, r)\le e^{1-\exp (-C|r|\Vert \omega _0^1\Vert _{L^\infty })}\big (CT\Vert \omega _0^1-\omega _0^2\Vert _{L^1}\big )^{\exp (-C|r|\Vert \omega _0^1\Vert _{L^\infty })}\quad \forall r\in [0, T_1]. \end{aligned}$$

(2.14)

Using (2.14) with $r=T_1$ we find that if

$$\begin{aligned} CT\Vert \omega _0^1-\omega _0^2\Vert _{L^1}<e^{1-\exp (CT\Vert \omega _0^1\Vert _{L^\infty })}, \end{aligned}$$

(2.15)

then $\eta (0, T_1)<1$, and hence $T_1=T$ and (2.14) holds for all $r\in [0, T]$. By the same argument, we obtain (2.14) for all $r\in [-T, T]$. Since $\varphi$ is increasing, inserting (2.14) into the right-hand side of (2.13), we deduce

$$\begin{aligned} \eta (t, r)\le \Phi _{T, \Vert \omega _0^1\Vert _{L^\infty }}(\Vert \omega _0^1-\omega _0^2\Vert _{L^1})+C\Vert \omega _0^1\Vert _{L^\infty }\left| \int _t^r \varphi (\eta (t, s))ds\right| \end{aligned}$$

(2.16)

for all $t, r\in [-T, T]$, where

$$\begin{aligned} \Phi _{T, \Vert \omega _0^1\Vert _{L^\infty }}(z)=CTz +CT\Vert \omega _0^1\Vert _{L^\infty }\varphi \left( e\big (CTz\big )^{\exp (-CT\Vert \omega _0^1\Vert _{L^\infty })}\right) \end{aligned}$$

with C depending only on $\Omega$. Clearly $\Phi (z)\rightarrow 0$ as $z\rightarrow 0$. Similarly to (2.14), we can apply Osgood’s lemma to (2.16) and obtain

$$\begin{aligned} \eta (t, r)\le e^{1-\exp (-C|t-r|\Vert \omega _0^1\Vert _{L^\infty })}\Phi _{T, \Vert \omega _0^1\Vert _{L^\infty }}(\Vert \omega _0^1-\omega _0^2\Vert _{L^1})^{\exp (-C|t-r|\Vert \omega _0^1\Vert _{L^\infty })} \end{aligned}$$

(2.17)

for all $t, r\in [-T, T]$ provided that

$$\begin{aligned} \Phi _{T, \Vert \omega _0^1\Vert _{L^\infty }}(\Vert \omega _0^1-\omega _0^2\Vert _{L^1})<e^{1-\exp (2CT\Vert \omega _0^1\Vert _{L^\infty })}. \end{aligned}$$

(2.18)

By virtue of (2.17), (2.8) yields

$$\begin{aligned} \begin{aligned}&\Vert \omega ^1(t)-\omega ^2(t)\Vert _{L^p}^p\le 2^{p-1}\Vert \omega ^1_0-\omega ^2_0\Vert _{L^p}^p+2^{3p-2}\Vert \omega _0^1*\rho _\varepsilon -\omega _0^1\Vert _{L^p}^p\\&\quad +2^{2p-2}\Vert \omega _0^1*\rho _\varepsilon \Vert ^p_{\dot{C}^{1/p}}|\Omega |e^{1-\exp (-CT\Vert \omega _0^1\Vert _{L^\infty })}\Phi _{T, \Vert \omega _0^1\Vert _{L^\infty }}(\Vert \omega _0^1-\omega _0^2\Vert _{L^1})^{\exp (-CT\Vert \omega _0^1\Vert _{L^\infty })} \end{aligned} \end{aligned}$$

(2.19)

for all $t\in [-T, T]$.

To obtain Theorem 1.3, let $\omega _0$, $\omega _0^n\in L^\infty (\Omega )$ such that $(\omega _0^n)_n$ converges to $\omega _0$ in $L^p(\Omega )$. For $n\ge N$ sufficiently large, the smallness conditions (2.15) and (2.18) hold for $\omega _0-\omega _0^n$, so that (2.19) holds for $S_t(\omega _0)-S_t(\omega _0^n)$. In (2.19), taking the supremum over $t\in [0, T]$, then letting $n\rightarrow \infty$ followed by $\varepsilon \rightarrow 0$ , we conclude that

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in [-T, T]}\Vert S_t(\omega _0^n)-S_t(\omega _0)\Vert _{L^p(\Omega )}=0. \end{aligned}$$

(2.20)

The proof of Theorem 1.3 is complete.

3 Proof of Theorem 1.4

Assume that $\omega _0^n \overset{*}{\rightharpoonup } \omega _0$ in $L^\infty (\Omega )$. Let $(\omega ^n, u^n, X^n_t)$ (resp. $(\omega , u, X_t)$) be the Yudovich solution of (1.1) with initial data $\omega _0^n$ (resp. $\omega _0$). Fix $T>0$ arbitrary. With $M=\sup _{n}\Vert \omega _0^n\Vert _{L^\infty }<\infty$ we have $\sup _n \Vert \omega ^n\Vert _{L^\infty (\Omega \times (-T, T))}\le M$. Thus there exists a subsequence $\omega ^{n_k}\overset{*}{\rightharpoonup }\omega ^\infty$ in $L^\infty (\Omega \times (-T, T))$. Define

$$\begin{aligned}&u^\infty (t)=K*\omega ^\infty (t),\\&\frac{d}{dt}X^\infty _t(x)=u^\infty (X^\infty _t(x), t),\quad X^\infty _0(x)=x\quad \forall x\in \overline{\Omega }. \end{aligned}$$

Note that $u^\infty$ is divergence-free and Log-Lipschitz, whence $X^\infty _t$ is measure-preserving. We claim that

$$\begin{aligned} \omega ^\infty (x)=\omega _0(X^\infty _{t, 0}(x)). \end{aligned}$$

(3.1)

1. To prove (3.1), we first use the $L^\infty$ bound (2.4) to have

$$\begin{aligned} \sup _n\left| \frac{d}{dt}X^n_t(x)\right| \le CM\quad \forall x\in \Omega ,\quad C=C(\Omega ). \end{aligned}$$

Recall in addition from Theorem 1.1 that each $X^n_t$ is Hölder continuous with exponent

$$\begin{aligned} \exp (-C|t|\Vert \omega _0\Vert _{L^\infty (\Omega )})\ge \exp (-CMT),\quad C=C(\Omega ). \end{aligned}$$

It follows that the sequence $X^n_t(x)$ is uniformly bounded in $C^{\alpha }(\overline{\Omega }\times [-T, T])$ for some $\alpha =\alpha (M, T, \Omega )$. By the Arzelà-Ascoli theorem, the subsequence $X^{n_k}$ has a subsequence $X^{n_{k_\ell }}\rightarrow Y$ in $C(\overline{\Omega }\times [-T, T])$. Using this strong convergence, we now prove that

$$\begin{aligned} \omega ^{n_k}_0(X^{n_{k_\ell }}_{t, 0}(x))\overset{*}{\rightharpoonup }\omega _0(Y_{t, 0}(x))\quad \text {in } L^\infty (\Omega \times (-T, T)),\quad Y_t(x)\equiv Y(x, t). \end{aligned}$$

(3.2)

Indeed, for any $f\in C(\overline{\Omega }\times [-T, T])$, we have

$$\begin{aligned} \int _{-T}^T\int _\Omega \omega _0^{n_{k_\ell }}(X^{n_{k_\ell }}_{t, 0}(x))f(x, t)dxdt&= \int _{-T}^T\int _\Omega \omega _0^{n_{k_\ell }}(x)f(X^{n_{k_\ell }}_t(x), t)dxdt\\&=\int _{-T}^T\int _\Omega \omega _0^{n_{k_\ell }}(x)f(Y_t(x), t)dxdt\\&\quad +\int _{-T}^T\int _\Omega \omega _0^{n_{k_\ell }}(x)[f(X^{n_{k_\ell }}_t(x), t)-f(Y_t(x), t)]dxdt\\&:=I_1+I_2. \end{aligned}$$

Since $\omega _0^n\overset{*}{\rightharpoonup }\omega _0$ in $L^\infty (\Omega )$ and $f(Y_t(\cdot ), t)\in L^1(\Omega )$,

$$\begin{aligned} \lim _{\ell \rightarrow \infty }\int _\Omega \omega _0^{n_{k_\ell }}(x)f(Y_t(x), t)dx=\int _\Omega \omega _0(x)f(Y_t(x), t)dx. \end{aligned}$$

In addition, we have

$$\begin{aligned} \left| \int _\Omega \omega _0^{n_{k_\ell }}(x)f(Y_t(x), t)dx\right| \le M\Vert f(\cdot , t)\Vert _{L^1}\in L^1((-T, T)), \end{aligned}$$

so that the dominated convergence theorem gives

$$\begin{aligned} \lim _{\ell \rightarrow \infty }I_1=\int _{-T}^T\int _\Omega \omega _0(x)f(Y_t(x), t)dxdt. \end{aligned}$$

Since $X^{n_{k_\ell }}\rightarrow Y$ in $C(\overline{\Omega }\times [-T, T])$ and $|\omega _0^{n_{k_\ell }}(x)|\le M$, $I_2$ converges to 0 by uniform convergence on the compact set $\overline{\Omega } \times [-T, T]$. Consequently,

$$\begin{aligned} \begin{aligned} \lim _{\ell \rightarrow \infty }\int _{-T}^T\int _\Omega \omega _0^{n_{k_\ell }}(X^{n_{k_\ell }}_{t, 0}(x))f(x, t)dxdt&=\int _{-T}^T\int _\Omega \omega _0(x)f(Y_t(x), t)dx\\&=\int _{-T}^T\int _\Omega \omega _0(Y_{t, 0}(x))f(x, t)dx. \end{aligned} \end{aligned}$$

(3.3)

Since $C(\overline{\Omega }\times [-T, T])$ is dense in $L^1(\Omega \times (-T, T))$ and $\omega _0^{n}(X^{n}_{t, 0}(x))$ is uniformly bounded in $L^\infty (\Omega \times (-T, T))$, (3.3) implies (3.2). On the other hand, $\omega ^{n_k}_0(X^{n_{k_\ell }}_{t, 0}(x))=\omega ^{n_{k_\ell }}(x, t) \overset{*}{\rightharpoonup }\omega ^\infty (x, t)$, so that (3.2) implies

$$\begin{aligned} \omega ^\infty (x, t)=\omega _0(Y_{t, 0}(x)). \end{aligned}$$

(3.4)

Thus (3.1) would follow from (3.4) provided that

$$\begin{aligned} Y_t(x)=X^\infty _t(x). \end{aligned}$$

(3.5)

To prove (3.5) we start with

$$\begin{aligned} \frac{d}{dt}X^n_t(x)=u^n(X^n_t(x), t),\quad u^n(t)=K*\omega ^n(t). \end{aligned}$$

(3.6)

Using that $\omega ^{n_k}\overset{*}{\rightharpoonup }\omega ^\infty$ in $L^\infty (\Omega \times (-T, T))$ and $\int _\Omega |K(x, y)|dy\le C(\Omega )$, we deduce $u^{n_k}\overset{*}{\rightharpoonup }u^\infty$ in $L^\infty (\Omega \times (-T, T))$. Arguing as in the proof of (3.2) we obtain

$$\begin{aligned} u^{n_k}(X^{n_k}_t(x), t)\overset{*}{\rightharpoonup }u^\infty (Y_t(x), t)\quad \text {in } L^\infty (\Omega \times (-T, T)), \end{aligned}$$

whence (3.6) gives $\frac{d}{dt}Y_t(x)= u^\infty (Y_t(x), t)$. Therefore, $Y_t(x)=X^\infty _t(x)$ by the uniqueness of trajectories generated by Log-Lipschitz velocity fields. This finishes the proof of (3.5) and hence of (3.1).

2. With (3.1) established, the triple $(\omega ^\infty , u^\infty , X^\infty _t)$ is a Yudovich solution of (1.1) with initial data $\omega ^\infty \vert _{t=0}=\omega _0$. By the uniqueness part in Theorem 1.1, $(\omega ^\infty , u^\infty , X^\infty _t)\equiv (\omega , u, X_t)$. In fact, the above argument shows that every subsequence of $\omega ^n$ has a subsequence converging weakly-$*$ to $\omega$ in $L^\infty (\Omega \times (-T, T))$. It follows that the entire sequence $\omega ^n\overset{*}{\rightharpoonup }\omega$ in $L^\infty (\Omega \times (-T, T))$.

Now for each $t\in \mathbb {R}$, $\omega ^n(t)$ is well-defined in $L^\infty (\Omega )$ by virtue of Lemma 1.2. Moreover, $\Vert \omega ^n(t)\Vert _{L^\infty }\le M$, whence $\omega ^{n_k}\overset{*}{\rightharpoonup }h(t)$ in $L^\infty (\Omega )$ for some subsequence $n_k$ which a priori depends on t. For any $f\in C(\overline{\Omega })$ we have

$$\begin{aligned} \int _\Omega \omega ^n(x, t)f(x)dx&= \int _\Omega \omega ^n_0(x)f(X^n_t(x))dx\rightarrow \int _\Omega \omega _0(x)f(X_t(x))dx \end{aligned}$$

in views of the facts that $\omega _0^n\overset{*}{\rightharpoonup }\omega _0$ in $L^\infty (\Omega )$ and $X^n_t\rightarrow X^\infty _t\equiv X_t$ in $C(\overline{\Omega })$. It follows that

$$\begin{aligned} \int _\Omega h(x, t)f(x)dx= \int _\Omega \omega _0(x)f(X_t(x))dx=\int _\Omega \omega _0(X_{t, 0}(x))f(x)dx, \end{aligned}$$

and thus $h(x, t)= \omega _0(X_{t, 0}(x))=\omega (x, t)$ a.e. $x\in \Omega$. In fact, we have proved that every subsequence of $\omega ^n(\cdot , t)$ has a subsequence converging weakly-$*$ to $\omega (\cdot , t)$ in $L^\infty (\Omega )$. Therefore, the entire sequence $\omega ^n(\cdot , t)\overset{*}{\rightharpoonup }\omega (\cdot , t)$ in $L^\infty (\Omega )$. This concludes the proof of Theorem 1.4.

4 Proof of Theorem 1.5

We follow closely the proof of Theorem 1.4 and use the same notation whenever possible. We note that $\omega ^n(t)$ has compact support for all $t\in \mathbb {R}$ but $X^n_t$ does not in general. Since $X^n$ is uniformly Hölder continuous on any compact set of $\mathbb {R}^2_x\times \mathbb {R}_t$, any subsequence $n_k$ has a subsequence $n_{k_\ell }$ such that

$$\begin{aligned} \forall R>0,\quad X^{n_{k_\ell }}\rightarrow Y\quad \text {in } C(\overline{B_R}\times [-T, T]) \end{aligned}$$

(4.1)

by the Arzelà-Ascoli theorem and a diagonal procedure. Here $B_R$ denotes the ball of radius R centered at $0\in \mathbb {R}^2$. To prove (3.2) we take $f\in C_c(\mathbb {R}^2\times (-T, T))$, a dense subspace of $L^1(\Omega \times (-T, T))$, and only consider $I_2$ since the above argument for $I_1$ does not make use of the boundedness of $\Omega$. If ${{\,\mathrm{supp}\,}}f\subset B_R\times (-T, T)$ then

$$\begin{aligned} |I_2|\le M\int _{-T}^T\int _{X^{n_{k_\ell }}_{t, 0}(B_R)\cup Y_{t, 0}(B_R)} |f(X^{n_{k_\ell }}_t(x), t)-f(Y_t(x), t)|dxdt, \end{aligned}$$

where $M=\sup _n\Vert \omega _0^n\Vert _{L^\infty }$. We have

$$\begin{aligned} \Vert u^n\Vert _{L^\infty }\le C\Vert \omega ^n\Vert _{L^1\cap L^\infty }= C\Vert \omega _0^n\Vert _{L^1\cap L^\infty }\le CN, \end{aligned}$$

where $N:=\sup _n\Vert \omega _0^n\Vert _{L^1\cap L^\infty }+\Vert \omega _0\Vert _{L^1\cap L^\infty }$. This implies

$$\begin{aligned} X^{n_{k_\ell }}_{t, 0}(B_R)\cup Y_{t, 0}(B_R) \subset B_{R+TCN}\quad \forall |t|\le T, \end{aligned}$$

whence

$$\begin{aligned} |I_2|\le M\int _{-T}^T\int _{B_{R+TCN}} |f(X^{n_{k_\ell }}_t(x), t)-f(Y_t(x), t)|dxdt. \end{aligned}$$

Therefore, $\lim _{\ell \rightarrow \infty } I_2=0$ by the uniform convergence (4.1) on the compact set $\overline{B_{R+TCN}}\times [-T, T]$. This yields (3.2).

Regarding (3.6), we prove that $\omega ^{n_k}\overset{*}{\rightharpoonup }\omega ^\infty$ in $L^\infty (\mathbb {R}^2\times (-T, T))$ implies $u^{n_k}\overset{*}{\rightharpoonup }u^\infty :=K*\omega ^\infty$ in $L^\infty (\mathbb {R}^2\times (-T, T))$. Note that the Biot-Savart kernel for $\mathbb {R}^2$ is

$$\begin{aligned} K(x, y)\equiv K(x-y),\quad K(x)=\frac{(-x_2, x_1)}{2\pi |x|^2}, \end{aligned}$$

and K does not belong to $L^1(\mathbb {R}^2)$. For any $f\in C_c(\mathbb {R}^2\times (-T, T))$ we have

$$\begin{aligned} \begin{aligned} \int _{-T}^T\int _{\mathbb {R}^2}u^{n_k}(x, t)f(x, t)dxdt&=\int _{-T}^T \int _{\mathbb {R}^2}\omega ^{n_k}(y, t)\int _{\mathbb {R}^2}K(x-y)f(x, t)dxdydt\\&:=\int _{-T}^T \omega ^{n_k}(y, t)g(y, t)dydt. \end{aligned} \end{aligned}$$

(4.2)

Since $K\notin L^1(\mathbb {R}^2)$, we do not have $g(t)\in L^1(\mathbb {R}^2)$ to use the weak-$*$ convergence of $\omega ^{n_k}$. On the other hand, upon splitting the x-integration in g into $|x-y|\le 1$ and $|x-y|>1$ and applying suitable Young inequalities, we deduce $\Vert g(t)\Vert _{L^3}\le C\Vert f(t)\Vert _{L^1\cap L^3}$ and thus $g\in L^\infty ((-T, T); L^3)$. Now, $\omega ^n$ is uniformly bounded in $L^\infty (\mathbb {R}; L^\frac{3}{2})$ by interpolation, whence $\omega ^{n_k}\rightharpoonup \omega ^\infty$ in $L^\frac{3}{2}(\mathbb {R}^2\times (-T, T))$. It then follows from (4.2) that

$$\begin{aligned} \lim _{k\rightarrow \infty }\int _{-T}^T\int _{\mathbb {R}^2}u^{n_k}(x, t)f(x, t)dxdt&=\int _{-T}^T\int _{\mathbb {R}^2} \omega ^\infty (y, t)g(y, t)dydt\\&=\int _{-T}^T\int _{\mathbb {R}^2}u^\infty (x, t)f(x, t)dxdt \end{aligned}$$

for all $f\in C_c(\mathbb {R}^2\times (-T, T))$. Using this, the fact that $u_n$ is uniformly bounded in $L^\infty (\mathbb {R}^2)$ and a density argument, we conclude $u^{n_k}\overset{*}{\rightharpoonup }u^\infty$ in $L^\infty (\mathbb {R}^2\times (-T, T))$. The remainder of the proof follows along the same lines of the proof of Theorem 1.4.