Global well-posedness of fluid-solid interaction system

Zhuo Xu

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Zhuo Xu. Global well-posedness of fluid-solid interaction system. 2026. hal-05484267

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# Global well-posedness of fluid-solid interaction system

Zhuo Xu

Institut de Mathématiques de Bordeaux UMR 5251, Université de
Bordeaux/Bordeaux INP/CNRS, 351 Cours de la Libération, 33 405 TALENCE,
France

## Abstract

This paper studies a two-dimensional fluid-solid interaction system consisting of a rigid unit disk moving inside a larger circular container filled with a viscous incompressible fluid. The fluid motion is governed by the Navier-Stokes equations with no-slip boundary conditions imposed both on the container and on the disk, while the disk motion obeys Newton's second law. Our main result establishes global-in-time existence of strong solutions to this system and, more importantly, proves the absence of collisions: for any admissible initial data and for any moving trajectory of the solid, the inner disk never touches the boundary of the outer container. The key idea is to introduce a time-dependent rigid rotation that continuously aligns the center of the inner disk with a fixed axis passing through the center of the outer boundary. In these new coordinates, we construct a divergence-free thin-gap test function tailored to estimate the minimal distance between the moving disk and the container boundary. By transferring this construction back to the original coordinates and applying a contradiction argument, we get the non-collision property.

**Key words:** Navier-Stokes equation; fluid-solid interaction system; strong solution; well-posedness.

**AMS subject classification:** 35Q35, 35R35, 74F10

## Contents

<table><tr><td>1</td><td>Introduction</td><td>2</td></tr><tr><td>2</td><td>Notation and Main Result</td><td>5</td></tr><tr><td>3</td><td>Change of Variables and Invariance Properties</td><td>7</td></tr><tr><td>4</td><td>Construction of the Test Function</td><td>8</td></tr><tr><td>5</td><td>Some estimates of the test function</td><td>12</td></tr></table>
<figure><img src="image_2.png" /><figcaption>Figure 1: Fluid-solid interaction system</figcaption></figure>

For all $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \in \mathbb{R}^2$, we denote by $x^{\perp}$ the vector

$$x^{\perp} = \begin{pmatrix} x_2 \\ -x_1 \end{pmatrix}.$$

We have also denoted by $\dot{w}$ and $\ddot{w}$ the first and second derivatives of a function $w$ depending only on time $t$.

If $x, y \in \mathbb{R}^2$, then $x \cdot y$ stands for the Euclidean inner product of $x$ and $y$, and $|x|$ stands for the corresponding Euclidean norm.

Moreover, we denote by $\partial S(t)$ the boundary of the inner rigid body at time $t$, and by $n(x, t)$ the unit normal to $\partial S(t)$ at the point $x$, directed toward the interior of the rigid body. The positive constant $\nu$ denotes the viscosity of the fluid. The Cauchy stress tensor is defined by

$$\sigma(x, t) = -p(x, t)I_d + 2\nu D(u), \qquad (1.9)$$

where $I_d$ is the identity matrix in $\mathbb{R}^2$ and

$$D(u) = \frac{1}{2} (\nabla u + (\nabla u)^{\top}) \qquad (1.10)$$

is the symmetric part of the velocity gradient. Equivalently, for all $i, j \in \{1, 2\}$,

$$D_{ij}(u) = \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right).$$

For fluid-solid interaction system where the fluid moves in a bounded domain, a key issue is global well-posedness of this system. In two dimensions, for general shapes of the container and with the solid being a disk, Takahashi [14] proved that the strong solution exists globally as long as the solid never touches the boundary; conversely, contact implies a finite-time breakdown of the strong solution to the system. Then Glass and Sueur [5] showed uniqueness of weak solutions (in both viscous and inviscid cases) as long as no collision occurs. For this reason, collision problems are worth studying.

For one-dimensional fluid-particle interaction systems, where particles move in a viscous fluid governed by a Burgers equation. In the whole-space case, Vázquez and Zuazua [16] consider finitely
many point particles in $\mathbb{R}$ and prove that inter-particle collisions cannot occur provided the initial
fluid velocity is sufficiently regular (e.g. $u_0 \in H^1(\mathbb{R})$), relying on a backward-uniqueness argument
for the associated ODEs. In the bounded-domain case, Tucsnak and Xu [15] study the same model
on a finite interval and establish well-posedness for initial data $u_0 \in L^2(-1, 1)$. They further show
that the particle trajectory remains uniformly away from the boundary by introducing a suitably
designed test function. Another approach to prevent boundary contact is to incorporate feedback
controls that keep the particle inside the domain; see, for instance, Cîndea, Micu, Rovenţa and
Tucsnak [1].

In higher space dimensions, fluid-solid interaction is typically modeled by the incompressible
Navier-Stokes or Euler equations in a time-dependent fluid region, coupled with Newton's laws for
the motion of a rigid body.

For weak solutions to these systems, early works such as Conca-San Martín-Tucsnak [2], Desjardins-Esteban [3], and Tucsnak [12] established global-in-time existence results for fluid-rigid body systems. Beyond existence, these papers also clarified the correct energy structure of the coupled dynamics: they identify a natural total energy (fluid kinetic energy plus rigid-body kinetic energy) and show that it satisfies an energy inequality compatible with the viscous dissipation. Concerning contact situation, Starovoitov [13] showed that if collisions are allowed (either between solids or with the container boundary), then weak solutions need not be unique after the impact time: different continuations may satisfy the same weak formulation and energy bounds, reflecting the loss of determinism at contact within the weak-solution framework.

For the well-posedness of strong solutions to the interaction, a classical approach is given in [14]: firstly, by introducing a smooth change of variables maps the moving fluid domain $\Omega(t)$ back to a fixed reference domain; osing some estimates for the linearized system and a fixed point theorem, we get that strong solutions exist (and are unique) locally in time. Moreover, the only obstruction to global continuation is the possible occurrence of contact with the boundary. The collision problem itself has been investigated from several complementary angles. On the one hand, in no-slip viscous settings one may observe “no-collision” mechanisms driven by smooth effects: for instance, Hillairet [7] proved that, in two dimensions, if a rigid disk moves in the the upper half-plane in $\mathbb{R}^2$, then the disk cannot reach the container boundary $y = 0$ in any finite time, and then Hillairet-Takahashi [8] obtained an analogous no-contact result for a rigid ball above a plane in three dimensions. Sabbagh [11] proved that, for a system of several rigid disks moving in the whole space $\mathbb{R}^2$, the disks never come into contact with one another in finite time. On the other hand, collisions may occur and can be accompanied by singular behavior of strong solutions: see Hillairet-Takahashi [9] for finite-time contact together with blow-up in a viscous cavity problem under symmetry assumptions. Moreover, the possibility of contact crucially depends on boundary conditions: Gérard-Varet-Hillairet-Wang [4] show that introducing slip may circumvent the classical no-collision paradox, and Hillairet-Takahashi [10] prove finite-time contact for the model with Tresca-type boundary conditions.

For the disk-in-disk geometry (1.1)-(1.8), Helsa [6] introduced a fruitful variational viewpoint
on the contact problem. The guiding principle is that, if one can represent the physical gap—more
precisely, an “effective” or “equivalent” distance—between the rigid body and the fluid boundary by
an explicit scalar quantity, then one can build a *special test function* by minimizing an appropriate
energy functional among admissible fields. Thanks to the specific structure of the minimizer, the
viscous dissipation term can be related—up to controlled error terms—to the time variation of this
distance. Using this minimizer as a test function in the weak formulation of the coupled system
yields a differential inequality that governs the evolution of the minimal distance between the inner
disk and the outer boundary.

Within this approach, Hesla proved in particular that no collision occurs for *purely vertical
motions* (with respect to the center of the outer disk), by choosing a test function which will help
us get the estimate from the inner disk to the solid. However, the case of *general trajectories* in
the disk-in-disk setting remains open.

The test function is given by a stationary Stokes system depends on the solid position $h(t)$ in
two coupled ways (see [6, Section 10]). Firstly, the fluid domain $\Omega(t)$ changes with $h(t)$. Secondly,
the boundary datum prescribed on the inner boundary $\partial S(t)$ is itself $h(t)$-dependent: it is given by the unit vector pointing from the center of the outer disk $(0,0)$ to the center of the inner disk, namely

$$ \mathbf{e}(t) := \frac{h(t)}{|h(t)|}. $$

In the vertical-motion regime, one has $\mathbf{e}(t) \equiv (0, 1)^{\top}$, hence the inner boundary condition is constant in time. This special structure makes it possible to derive an explicit representation of the corresponding stationary Stokes solution, relying on two-dimensional complex-analytic techniques.

This paper resolves this open problem. The main difficulty in constructing the test function for a general trajectory lies in the time-dependent boundary condition $\mathbf{e}(t) = h(t)/|h(t)|$ on the inner disk. Our key idea is to introduce a time-dependent rotation of the coordinate system so that, for each time $t$, the vector $\mathbf{e}(t)$ is mapped to the fixed vector $(0, 1)^{\top}$. This strategy is inspired by [11]. In this rotated frame, we can adapt the stationary thin-gap test functions from [7, 11]. The invariance properties of the Navier-Stokes equations under rigid rotations (Proposition 3.1) then allow us to pull back this test function to the original coordinates and use it in the weak formulation. The rest of the proof follows the approach of [7, 11], leading to a differential inequality for $d(t)$.

## 2 Notation and Main Result

In this section, we mainly use the same notation as in [14, Section 2], which will help us define the well-posedness of this interaction system (1.1)–(1.8). In the sequel, we denote by $\Omega = \Omega(0) = B(0, R) \setminus S(0)$ the initial fluid domain and by $S(0)$ the initial inner rigid body. For $t \ge 0$, we denote by $\Omega(t)$ and $S(t)$ the corresponding domains at time $t$. Throughout this paper we shall use Sobolev spaces defined on a smooth diffeomorphism

$$ \psi : \Omega \times [0, T] \to \bigcup_{t \in [0, T]} \Omega(t) \times \{t\}, \quad y \mapsto \psi(y, t), $$

such that $\psi(\cdot, t)$ is a $C^{\infty}$ diffeomorphism from $\Omega$ onto $\Omega(t)$ for each $t$, and

$$ \frac{\partial^{i+\alpha_1+\alpha_2} \psi}{\partial t^i \partial y_1^{\alpha_1} \partial y_2^{\alpha_2}} \quad (i \le 1, \alpha_1, \alpha_2 \ge 0) $$

exist and are continuous; more details we could find in [14, Section 3].

For all functions $u(\cdot, t) : \Omega(t) \to \mathbb{R}^2$, we denote by

$$ U(y, t) = u(\psi(y, t), t), \quad y \in \Omega, $$

their pullback to the reference domain. With this identification, the spaces above can be equivalently defined on $\Omega$ by

$$ L^2(0, T; H^2(\Omega(t))) = \{u : U \in L^2(0, T; H^2(\Omega))\}, $$

$$ H^1(0, T; L^2(\Omega(t))) = \{u : U \in H^1(0, T; L^2(\Omega))\}, $$

$$ C([0, T]; H^1(\Omega(t))) = \{u : U \in C([0, T]; H^1(\Omega))\}, $$

$$ L^2(0, T; H^1(\Omega(t))) = \{u : U \in L^2(0, T; H^1(\Omega))\}. $$

For any domain $\mathcal{O}$, we denote by $H^m(\mathcal{O})$ the Sobolev space of functions with weak derivatives up to order $m$ in $L^2(\mathcal{O})$. Finally, we introduce the space

$$ \mathcal{U}(0, T; \Omega(t)) = L^2(0, T; H^2(\Omega(t))) \cap H^1(0, T; L^2(\Omega(t))) \cap C([0, T]; H^1(\Omega(t))). \quad (2.1) $$
**Definition 2.1** (Strong solution). Suppose that $T > 0$. A quadruplet $(u, p, h, \omega)$ is called a strong solution of system (1.1)-(1.8) on $[0, T]$ if

$$ (u, p, h, \omega) \in \mathcal{U}(0, T; \Omega(t)) \times L^2(0, T; H^1(\Omega(t))) \times H^2(0, T; \mathbb{R}^2) \times H^1(0, T; \mathbb{R}), $$

and if the distance between $S(t)$ and $\partial B(0, R)$ remains strictly positive for all $t \in [0, T]$, while the equations (1.1)-(1.8) are satisfied almost everywhere in $(0, T) \times \Omega(t)$ (or in the trace sense on the boundaries).

We recall the local in time well-posedness of system (1.1)-(1.8).

**Proposition 2.2** (Theorem 2.2, [14]). Let the initial data $(u, p, h, \dot{h}, \omega)$ satisfies

$$ \left\{ \begin{array}{ll} u_0 \in H^1(\Omega_0), \quad \dot{h}_0 \in \mathbb{R}^2, \quad \omega_0 \in \mathbb{R}, \\ \operatorname{div} u_0 = 0 & \text{in } \Omega, \\ u_0(x) = 0 & \text{on } \partial B(0, R), \\ u_0(x) = h_1 + \omega_0(x - h_0)^\perp & \text{on } \partial S(0), \\ \operatorname{dist}(S(0), \partial B(0, R)) > 0. & \end{array} \right. $$

Then there exists a maximal time $T_0 > 0$ ($T_0$ could be infinity) such that system (1.1)-(1.8) admits a unique strong solution

$$ (u, p, h, \omega) \in \mathcal{U}(0, T; \Omega(t)) \times L^2(0, T; H^1(\Omega(t))) \times H^2(0, T; \mathbb{R}^2) \times H^1(0, T; \mathbb{R}), $$

for all $T \in (0, T_0)$ with the following estimate

$$ \begin{aligned} & \frac{1}{2} \int_{\Omega(t)} |u(x,t)|^2 dx + \frac{M}{2} |\dot{h}(t)|^2 + \frac{J}{2} |\omega(t)|^2 + 2\nu \int_0^t \int_{\Omega(s)} D(u) : D(u)(x,s) dx ds \\ & = \frac{1}{2} \int_{\Omega(0)} |u(x,0)|^2 dx + \frac{M}{2} |\dot{h}(0)|^2 + \frac{J}{2} |\omega(0)|^2. \end{aligned} \quad (2.2) $$

Moreover, one of the following alternatives holds:

1. $T_0 = +\infty$, i.e., the solution is global;

2. There exists a finite $T_0$, such that $\lim_{t \to T_0} \operatorname{dist}(S(t), \partial B(0, R)) = 0$.

Now, we state the main result in this paper:

**Theorem 2.3.** For any initial data as above, system (1.1)-(1.8) admits a unique global strong solution. In particular, for every $t > 0$, we have

$$ d(t) = \operatorname{dist}(S(t), \partial B(0, R)) > 0. \quad (2.1) $$

In view of Proposition 2.2, the question of global well-posedness reduces to excluding the collision scenario. To this end, we will exploit the weak formulation of the coupled system (1.1)-(1.8), testing it with a carefully chosen time-dependent test function.

To make this strategy precise, we first recall the weak formulation of the fluid-solid interaction system, which serves as the starting point for all subsequent estimates. Since the problem involves a moving boundary (the surface of the rigid disk), the class of admissible test functions must respect the coupling conditions between the fluid and the solid (see, for instance, [6, Chapter 5, Page 52]). The formulation is expressed in terms of test triples $(\varphi, \ell, r)$, where $\varphi$ is a divergence-free vector field defined in the fluid domain and $(\ell, r)$ are functions of time that represent, respectively, a translational and a rotational velocity imposed on the disk.
**Definition 2.4** (Admissible test functions.) Let $T > 0$. We say that $(\varphi, \ell, r)$ is an admissible test triple of system (1.1)-(1.8) on $(0, T)$ if

• $\ell \in H^1((0, T); \mathbb{R}^2)$ and $r \in H^1((0, T); \mathbb{R})$,

• $\varphi \in H^1((0, T); L^2(B(0, R))) \cap L^2((0, T); H^1(B(0, R)))$,

• for a.e. $t \in (0, T)$,

$$ \operatorname{div} \varphi(t, \cdot) = 0 \text{ in } \Omega(t), \quad \varphi(t, \cdot) = 0 \text{ on } \partial B(0, R), $$

and $\varphi$ matches a rigid velocity on the disk:

$$ \varphi(t, x) = \ell(t) + r(t) (x - h(t))^{\perp} \quad \text{for a.e. } x \in S(t). $$

The result below can be obtained by simple integrations by parts, so we state it without proof.

**Proposition 2.5** (Weak formulation). The strong solution of system (1.1)-(1.8) satisfies the following weak identity for every admissible test triple $(\varphi, \ell, r)$:

$$ \begin{aligned} & \int_{\Omega(T)} u(T, x) \cdot \varphi(T, x) \,dx - \int_{\Omega(0)} u_0(x) \cdot \varphi(0, x) \,dx \\ & + M(\dot{h}(T) \cdot \ell(T) - \dot{h}_0 \cdot \ell(0)) + J(\omega(T) r(T) - \omega_0 r(0)) \\ & + \int_0^T \int_{\Omega(t)} \left( -u \cdot \partial_t \varphi - (u \otimes u) : \nabla \varphi + 2\nu D(u) : D(\varphi) \right) \,dx \,dt \\ & - \int_0^T \left( M \dot{h}(t) \cdot \dot{\ell}(t) + J \omega(t) \dot{r}(t) \right) \,dt = 0. \end{aligned} \quad (2.2) $$

The remaining parts of this work are organized as follows. In Section 3 we present the rotational change of variables, while the construction of the test function is carried out in Section 4. The necessary estimates of the test function are established in Sections 5 and 6, and the proof of the main theorem is completed in the last Section.

## 3 Change of Variables and Invariance Properties

In this section, we recall some basic properties from [11] of the relevant quantities under rotations of the coordinate frame.

For any given point $\tau = (\tau_1, \tau_2) \in B(0, R)$ and $\tau \neq 0$, we introduce an orthonormal basis $(\vec{e}_1, \vec{e}_2)$ of $\mathbb{R}^2$ defined by

$$ e_2 = -\frac{\tau}{|\tau|}, \quad e_1 = -\vec{e}_2^{\perp}. \quad (3.1) $$

We then define the rotation map $Y_{\tau}$ by

$$ Y_{\tau}(x) = (y_1(x), y_2(x)) = (x \cdot \vec{e}_1, x \cdot \vec{e}_2) = \left( x \cdot \frac{\tau^{\perp}}{|\tau|}, -x \cdot \frac{\tau}{|\tau|} \right). \quad (3.2) $$

Moreover, the inverse mapping of $Y_{\tau}$ is given by

$$ Y_{\tau}^{-1}(y) = y_1 \vec{e}_1 + y_2 \vec{e}_2 = y_1 \frac{\tau^{\perp}}{|\tau|} - y_2 \frac{\tau}{|\tau|}. \quad (3.3) $$

In particular, $Y_{\tau}$ is an isometry on $B(0, R)$ (since it is induced by an orthogonal rotation matrix $E \in SO(2)$, so $Y_{\tau}^{-1} = (Y_{\tau})^T$).
For every function $u \in H^1(B(0, R))$, we define its pullback under the transformation $Y_\tau$ by

$$ \bar{u}(y) := J_y Y_{\tau}^{-1}(y) u(Y_{\tau}^{-1}(y)), \quad y \in B(0, R). \qquad (3.4) $$

Similarly, for $\bar{v} \in H^1(B(0, R))$ we set

$$ v(x) = J_x Y_{\tau}(x) \bar{v}(Y_{\tau}(x)), \quad x \in B(0, R). \qquad (3.5) $$

Here $J_x Y_\tau(x)$ and $J_y Y_\tau^{-1}(y)$ denote the Jacobian matrices of $Y_\tau$ and $Y_\tau^{-1}$, respectively. In explicit coordinates, writing $\tau = (\tau_1, \tau_2)$, we have

$$ J_y Y_{\tau}^{-1}(y) = \frac{1}{|\tau|} \begin{pmatrix} -\tau_2 & -\tau_1 \\ \tau_1 & -\tau_2 \end{pmatrix}, \quad J_x Y_{\tau}(x) = \frac{1}{|\tau|} \begin{pmatrix} -\tau_2 & \tau_1 \\ -\tau_1 & -\tau_2 \end{pmatrix}. \qquad (3.6) $$

In this way, $\bar{u}$ represents the same vector field as $u$, but expressed in the rotated coordinate system associated with the basis $(\bar{e}_1, \bar{e}_2)$.

For the transformation $Y_\tau$ defined in (3.2), we recall the following invariance properties proved in [11, Lemma 3.2].

**Proposition 3.1.** Let $u, v \in H^1(B(0, R))$. For every $p \in [1, \infty)$,

$$ \|u\|_{L^p(B(0, R))} = \|\bar{u}\|_{L^p(B(0, R))}, \qquad (3.7) $$

and

$$ D(u) : D(v) = D(\bar{u}) : D(\bar{v}). \qquad (3.8) $$

Moreover, if $\mathrm{div}_x u = 0$ in $B(0, R)$, then also $\mathrm{div}_y \bar{u} = 0$ in $B(0, R)$.

Here $D(u) := \frac{1}{2}(\nabla u + (\nabla u)^T)$ is defined in (1.10), and $\bar{u}, \bar{v}$ are defined in (3.4). For two matrices $A, B \in \mathbb{R}^{2 \times 2}$, their Frobenius inner product is

$$ A : B := \sum_{i,j=1}^{2} A_{ij} B_{ij}. \qquad (3.9) $$

**Remark 3.2** (Rotation invariance of global norms). Geometrically, one may view $u$ as a height function over $\Omega$ (i.e. consider its graph in $\mathbb{R}^3$); rotating the coordinates corresponds to a rigid rotation of this graph, which does not alter global norms. See Figure 2 for an illustration.

<figure><img src="image_3.png" /><figcaption>Figure 2: Effect of a rigid rotation</figcaption></figure>

# 4 Construction of the Test Function

In this section we construct an admissible test function $v(t, \cdot)$ for system (1.1)–(1.8). It is required to satisfy
1. $\operatorname{div} v(t, \cdot) = 0$ in $\Omega(t)$;

2. $v(t, x) = e_{h(t)}$ for $x \in \partial S(t)$, where $e_{h(t)} := h(t)/|h(t)|$ is the unit vector pointing from the origin towards the disk centre;

3. $v(t, x) = 0$ for $x \in \partial B(0, R)$.

**Main idea** The main difficulty comes from the narrow fluid gap when the inner disk approaches the outer boundary. To handle this geometry, we follow the classical strategy of [7, 11]:

* we first rotate the coordinates so that the disk centres lie on the vertical axis;

* in the rotated frame, the gap becomes a simple strip-like region described by two graphs;

* we then build a scalar profile which varies smoothly from the outer boundary to the disk boundary;

* to enforce incompressibility, we introduce an auxiliary scalar potential $\bar{g}_d$ and set $\bar{v}_d = \nabla^\perp \bar{g}_d$;

* finally we rotate the vector field back to the original variables.

We define the (time-dependent) distance between the inner disk and the outer boundary by

$$d(t) := dist(S(t), \partial B(0, R)) = R - |h(t)| - 1, \quad (4.1)$$

where $S(t) = B(h(t), 1)$ denotes the position of the inner disk at time $t$, $B(0, R)$ is the fluid domain, and $h(t)$ is the disk centre. Throughout this section we fix a time $t$ such that $d(t) > 0$, and for simplicity we write

$$d := d(t).$$

**Step 1. Rotating the coordinate system.** For each time $t$, we apply the rotation map $Y_{h(t)}$ introduced in (3.2), which sends the direction of $h(t)$ to the negative $y_2$-axis:

$$Y_{h(t)}(x) = (y_1(t, x), y_2(t, x)) = \left( x \cdot \frac{h(t)^\perp}{|h(t)|}, -x \cdot \frac{h(t)}{|h(t)|} \right). \quad (4.2)$$

In these rotated coordinates, the centers of the two disks lie on the $y_2$-axis, and the gap becomes symmetric with respect to the vertical direction (see Figure 3).

<figure><img src="image_4.png" /><figcaption>Figure 3: Change of coordinates. Left: original configuration. Right: rotated frame where the disk centres are aligned vertically.</figcaption></figure>

**Step 2. The gap region in rotated coordinates.**

In the rotated frame, the minimal distance between the disk and the outer boundary is attained near $y_1 = 0$, and it equals $d$. Moreover, the gap opens up quadratically as $|y_1|$ increases. Therefore, the most delicate part of the construction is concentrated in a small strip around $y_1 = 0$.
We fix a small constant $\delta > 0$ depending only on $R$:

$$ \delta = \min \left\{ \frac{1}{8}, \frac{1}{2} \sqrt{\frac{1}{2} R^3 \left(1 - \frac{1}{R}\right)} \right\}, \qquad (4.3) $$

and we focus on the strip $|y_1| \le 2\delta$. Inside this strip, both boundaries can be represented as graphs $y_2 = \psi(y_1)$.

**Boundary graphs.** For $|y_1| \le 2\delta$, the outer boundary (lower graph) is

$$ \psi_{\text{low}}(y_1) = -\sqrt{R^2 - y_1^2} \qquad (\text{outer circle}), \qquad (4.4) $$

and the part of the disk boundary facing the outer boundary (upper graph) is

$$ \psi_{\text{up}}(y_1) = -\sqrt{1 - y_1^2} - R + d + 1 \qquad (\text{inner disk}). \qquad (4.5) $$

Note that $\psi_{\text{up}}(0) - \psi_{\text{low}}(0) = d$, so $d$ is indeed the minimal gap width.

**Gap subregions.** We define the strip-like gap region

$$ A_{\delta,d} = \{(y_1, y_2) : |y_1| \le 2\delta, \psi_{\text{low}}(y_1) \le y_2 \le \psi_{\text{up}}(y_1)\}, \qquad (4.6) $$

$$ A_{0,d} = \{(y_1, y_2) : |y_1| \le \delta, \psi_{\text{low}}(y_1) \le y_2 \le \psi_{\text{up}}(y_1)\}. \qquad (4.7) $$

The core region $A_{0,d}$ is where the gap is the narrowest, and it is where we will build the main profile explicitly.

**Local gap width.** For each $y_1$, we denote the vertical thickness of the gap by

$$ \mathcal{G}(y_1) := \psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1). \qquad (4.8) $$

**Lemma 4.1 (Gap estimate).** Let $R > 1$ and $d > 0$. Then for every $|y_1| \le 2\delta$, the width $\mathcal{G}(y_1)$ satisfies

$$ d + \left(\frac{1}{4} - \frac{1}{4R}\right)y_1^2 \le \mathcal{G}(y_1) \le d + \left(1 - \frac{1}{R}\right)y_1^2. \qquad (4.9) $$

A complete proof of the above is provided in Appendix 8.

### Step 3. Constructing a scalar profile across the gap.

We now build a scalar function $\bar{\varphi}_d$ satisfying:

$$ \bar{\varphi}_d = 0 \text{ on the outer boundary, } \quad \bar{\varphi}_d = 1 \text{ on the disk boundary,} $$

with a smooth transition inside the fluid region. This scalar profile will later be converted into a divergence-free vector field.

(a) Core region $A_{0,d}$. We introduce the normalised vertical coordinate

$$ F(y) = \frac{y_2 - \psi_{\text{low}}(y_1)}{\mathcal{G}(y_1)}, \qquad (4.10) $$

so that $F = 0$ on $\psi_{\text{low}}$ and $F = 1$ on $\psi_{\text{up}}$. We set

$$ \bar{\varphi}_d(y) = 3F(y)^2 - 2F(y)^3 \quad (y \in A_{0,d}). \qquad (4.11) $$

The cubic polynomial is chosen because it yields a $C^1$ matching at the endpoints: $\bar{\varphi}_d(0) = 0$, $\bar{\varphi}_d(1) = 1$, and $(\bar{\varphi}_d)'(0) = (\bar{\varphi}_d)'(1) = 0$.
(b) Region away from the gap. Outside the strip $A_{\delta,d}$, we use a smooth cut-off function $\phi \in C^\infty(B(0, R))$ such that

$$ \phi(y) = 1 \text{ in a neighbourhood of } S(t), \quad \phi(y) = 0 \text{ in a neighbourhood of } \partial B(0, R). \quad (4.12) $$

For $y \in B(0, R) \setminus A_{\delta,d}$, we define $\bar{\varphi}_d(y) := \phi(y)$.

(c) Transition region $A_{\delta,d} \setminus A_{0,d}$. We glue the two definitions using a cut-off depending only on $y_1$. Let $\mathcal{X} \in C^\infty(\mathbb{R})$ satisfy

$$ \mathcal{X}(y_1) = \begin{cases} 1, & |y_1| \le \delta, \\ 0, & |y_1| \ge 2\delta. \end{cases} \quad (4.13) $$

Then for $y \in A_{\delta,d} \setminus A_{0,d}$ we define

$$ \bar{\varphi}_d(y) = \mathcal{X}(y_1)(3F(y)^2 - 2F(y)^3) + (1 - \mathcal{X}(y_1))\phi(y). $$

**Remark 4.2.** For $\delta \le |y_1| \le 2\delta$, Lemma 4.1 guarantees $\mathcal{G}(y_1) \ge c_\delta > 0$, so the transition region has a thickness bounded uniformly with respect to the gap size $d$.

### Step 4. From the scalar profile to a divergence-free vector field.

To enforce incompressibility, we introduce an auxiliary scalar potential $\bar{g}_d$ and define the vector field through the perpendicular gradient (which automatically yields zero divergence). More precisely, we set

$$ \bar{g}_d(y) = y_1 \bar{\varphi}_d(y), \quad (4.14) $$

and define

$$ \bar{v}_d(y) = \nabla^\perp \bar{g}_d(y) = \begin{pmatrix} -\partial_{y_2} \bar{g}_d \\ \partial_{y_1} \bar{g}_d \end{pmatrix}. \quad (4.15) $$

**Proposition 4.3** (Properties of $\bar{v}_d$). For each $d > 0$, the vector field $\bar{v}_d$ satisfies:

1. $\text{div } \bar{v}_d = 0$;

2. $\bar{v}_d(y) = \bar{e}_2$ for every $y \in \partial S(t)$, where $\bar{e}_2 = (0, 1)^\top$;

3. $\bar{v}_d(y) = 0$ for every $y \in \partial B(0, R)$.

A detailed proof of these properties can be found in [7].

### Step 5. Rotating back to the original coordinates.

Finally, we return to the physical variables by applying the inverse rotation. Let $J_x Y_{h(t)}$ be the Jacobian matrix of $Y_{h(t)}$ (defined in (3.6)). We define the test function in the original $(x_1, x_2)$-coordinates by

$$ v(t, x) = J_x Y_{h(t)} \bar{v}_d(Y_{h(t)}(x)). \quad (4.16) $$

**Proposition 4.4** (Properties of $v$). For every $t \in (0, T)$ with $d(t) > 0$, the vector field $v(t, \cdot)$ defined by (4.16) satisfies:

1. $\text{div } v(t, \cdot) = 0$ in $\Omega(t)$;

2. $v(t, x) = e_{h(t)}$ for $x \in \partial S(t)$;

3. $v(t, x) = 0$ for $x \in \partial B(0, R)$,

where $T > 0$ is the maximal existence time provided by Proposition 2.2.

*Proof.* These properties follow directly from Proposition 4.3 and the fact that $Y_{h(t)}$ is an isometry (see Proposition 3.1). □
<figure><img src="image_5.png" /><figcaption>Figure 4: Test function.</figcaption></figure>

## 5 Some estimates of the test function

In this section, we derive several preliminary estimates that will be used throughout the remainder of the paper. These estimates were proved in [11]. However, since our geometric setting is different, we include the argument for completeness, indicating the minor modifications needed.

We begin by recalling the definition of $\bar{v}_d(y)$ from (4.15) and we establish $L^p$ estimation.

**Lemma 5.1.** Let $d \in (0, R-1]$. Then, for every $p \in [2, 3)$, there exists a constant $C = C(R) > 0$, depending only on $R$, such that

$$ \|\bar{v}_d\|_{L^p(B(0,R))} \le C(R). \qquad (5.1) $$

*Proof.* We work first on the core strip $A_{0,d}$ from (4.7). By the definition of $\bar{v}_d$ in (4.11), it follows that:

$$ \bar{v}_d = \nabla^{\perp}(y_1 \bar{\varphi}_{\text{core}}) = (\nabla^{\perp} y_1) \bar{\varphi}_{\text{core}} + y_1 \nabla^{\perp} \bar{\varphi}_{\text{core}} = \bar{\varphi}_{\text{core}} e_2 + y_1 \nabla^{\perp} \bar{\varphi}_{\text{core}} \quad \text{in } A_{0,d}. $$

Here $\bar{\varphi}_{\text{core}}(y) = F(y)^2(3-F(y))$ and $F(y)$ is defined in (4.10), hence $0 \le \bar{\varphi}_{\text{core}} \le 2$ and

$$ \partial_{y_i} \bar{\varphi}_{\text{core}} = 6F(y)(1-F(y)) \partial_{y_i} F(y), \quad i=1,2. $$

Since

$$ F(y) = \frac{y_2 - \psi_{\text{low}}(y_1)}{\mathcal{G}(y_1)}, \quad \mathcal{G}(y_1) = \psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1), $$

we have $\partial_{y_2} F(y) = \frac{1}{\mathcal{G}(y_1)}$. Using the facts $|y_2 - \psi_{\text{low}}(y_1)| \le \mathcal{G}(y_1)$ together with

$$ |\psi'_{\text{up}}(y_1)| + |\psi'_{\text{low}}(y_1)| \le C(R) |y_1| \quad (|y_1| \le 2\delta), \qquad (5.2) $$

$$ |\partial_{y_1} F(y)| = \frac{|-\psi'_{\text{low}}\mathcal{G}(y_1) - (y_2 - \psi_{\text{low}})(y_1)(\psi'_{\text{up}} - \psi'_{\text{low}})(y_1)|}{\mathcal{G}^2(y_1)} \le \frac{C(R)|y_1|}{\mathcal{G}(y_1)}. $$

Consequently, for $y \in A_{0,d}$,

$$ |\bar{v}_d(y)| \le C(R) + C(R) \frac{|y_1|}{\mathcal{G}(y_1)} + C(R) \frac{|y_1|^2}{\mathcal{G}(y_1)}, $$
and hence, for any $p \in [1, \infty)$,

$$|\bar{v}_d(y)|^p \le C(R) + C(R) \frac{|y_1|^p}{\mathcal{G}(y_1)^p} + C(R) \frac{|y_1|^{2p}}{\mathcal{G}(y_1)^p}.$$

Integrating over $A_{0,d}$ and using Toneli's theorem, we have

$$\int_{A_{0,d}} |\bar{v}_d|^p dy \le C(R) + C(R) \int_{\psi_{low}(y_1)}^{\psi_{up}(y_1)} \int_{-\delta}^{\delta} \frac{|y_1|^p}{\mathcal{G}(y_1)^p} + \frac{|y_1|^{2p}}{\mathcal{G}(y_1)^p} dy_1 dy_2 \\ \le C(R) + C(R) \int_{-\delta}^{\delta} \left( \frac{|y_1|^p}{\mathcal{G}(y_1)^{p-1}} + \frac{|y_1|^{2p}}{\mathcal{G}(y_1)^{p-1}} \right) dy_1.$$

By the lower bound in Lemma 4.1, we have

$$\frac{1}{\mathcal{G}(y_1)^{p-1}} \le \frac{C(R)}{|y_1|^{2(p-1)}} \quad (|y_1| \le 2\delta),$$

and therefore

$$\int_{A_{0,d}} |\bar{v}_d|^p dy \le C(R) + C(R) \int_{-\delta}^{\delta} (|y_1|^{p-2(p-1)} + |y_1|^{2p-2(p-1)}) dy_1 \\ = C(R) + C(R) \int_{-\delta}^{\delta} (|y_1|^{2-p} + |y_1|^2) dy_1.$$

The second integral is always finite, and the first one is finite provided $p < 3$. Hence,

$$\int_{A_{0,d}} |\bar{v}_d|^p dy \le C(R) \quad \text{for any } p \in [1, 3). \qquad (5.3)$$

On the exterior region $B(0, R) \setminus A_{\delta,d}$ we have, by definition, $\bar{\varphi}_d = \phi$, where $\phi$ defined in (4.12) is a fixed radial cut-off independent of $d$. Hence

$$\bar{v}_d = \nabla^{\perp}(y_1 \phi)$$

and all derivatives of $\phi$ are only depending on $R$; thus, for any $p \in [1, \infty)$,

$$\int_{B(0,R) \setminus A_{\delta,d}} |\bar{v}_d|^p dy \le C(R). \qquad (5.4)$$

On the transition strip $A_{\delta,d} \setminus A_{0,d}$, the definition

$$\bar{\varphi}_d = (1 - \mathcal{X}(y_1)) \phi + \mathcal{X}(y_1) F^2(y)(3 - F(y))$$

matches the core and exterior expressions. Using that $\mathcal{X}$ defined in (4.13) depends only on $R$, that $\mathcal{G}(y_1) \ge c_R \delta^2$ on this strip, and the same computations as above, we obtain a bound uniform in $d$:

$$\int_{A_{\delta,d} \setminus A_{0,d}} |\bar{v}_d|^p dy \le C(R) \quad \text{for all } p \in [1, \infty). \qquad (5.5)$$

Combining (5.4), (5.5), and the estimate on the core strip $A_{0,d}$, we conclude that

$$\int_{B(0,R)} |\bar{v}_d(y)|^p dy \le C(R) \quad \text{for all } p \in [1, 3),$$

with a constant $C(R)$ independent of $d$.

□
**Lemma 5.2.** Let $d \in (0, R-1]$. Then there exists a constant $C = C(R) > 0$, depending only on $R$, such that

$$ \left\| \frac{\partial \bar{g}_d}{\partial d} \right\|_{L^2(A_{\delta,d})} \le C(R), \qquad (5.6) $$

where $\bar{g}_d$ has been defined in (4.14).

*Proof.* Recall that

$$ \bar{g}_d(y) = y_1 \left( (1 - \mathcal{X}(y_1)) \phi(y) + \mathcal{X}(y_1) F^2(y) (3 - F(y)) \right) \quad (y \in A_{\delta,d}), \qquad (5.7) $$

where $F(y)$ and $\mathcal{X}(y_1)$ are defined by (4.10), (4.13), respectively.

Since only $\psi_{\text{up}}$ in (4.5) depends on $d$ (while $\psi_{\text{low}}$, $\phi$, and $\mathcal{X}$ do not), differentiating $F(y)$ with respect to $d$ yields:

$$ \frac{\partial F(y, d)}{\partial d} = - \frac{y_2 - \psi_{\text{low}}(y_1)}{\mathcal{G}(y_1)^2}, $$

where $\mathcal{G}$ is the gap defined in (4.8).

With a simple calculation, we get

$$ \frac{\partial \bar{g}_d}{\partial d}(y) = y_1 \mathcal{X}(y_1) \left( - \frac{6(y_2 - \psi_{\text{low}}(y_1))^2}{\mathcal{G}(y_1)^3} + \frac{3(y_2 - \psi_{\text{low}}(y_1))^3}{\mathcal{G}(y_1)^4} \right), \quad y \in A_{\delta,d}. $$

Using $0 \le y_2 - \psi_{\text{low}}(y_1) \le \mathcal{G}(y_1)$, we get the pointwise bound

$$ \left| \frac{\partial \bar{g}_d}{\partial d}(y) \right| \le C(R) \frac{|y_1|}{\mathcal{G}(y_1)} \quad (y \in A_{\delta,d}), \qquad (5.8) $$

with a constant $C(R)$ independent of $d$.

Integrating (5.8) over $A_{\delta,d}$ and using Tonelli's theorem, we have

$$ \int_{A_{\delta,d}} \left| \frac{\partial \bar{g}_d}{\partial d} \right|^2 dy \le C \int_{-2\delta}^{2\delta} \frac{y_1^2}{\mathcal{G}(y_1)^2} \left( \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{up}}(y_1)} dy_2 \right) dy_1 = C(R) \int_{-2\delta}^{2\delta} \frac{y_1^2}{\mathcal{G}(y_1)} dy_1. $$

By Lemma 4.1, there is a $c_R > 0$, such that

$$ \mathcal{G}(y_1) \ge d + c_R y_1^2, $$

with the fact $d \ge 0$, it follows that

$$ \int_{-2\delta}^{2\delta} \frac{y_1^2}{\mathcal{G}(y_1)} dy_1 \le \int_{-2\delta}^{2\delta} \frac{y_1^2}{d + c_R y_1^2} dy_1 \le C(R). $$

Therefore

$$ \left\| \frac{\partial \bar{g}_d}{\partial d} \right\|_{L^2(A_{\delta,d})} \le C(R). $$

we completes the proof. $\square$

Now we give some derivative bound of $\bar{v}$.

**Lemma 5.3.** Let $d \in (0, R-1]$. Then there exists a constant $C = C_i(R) > 0$, depending only on $R$, such that

$$ \| \partial_1 \bar{v}_{d,2} \|_{L^2(A_{\delta,d})} = \| \partial_{11} \bar{g}_d \|_{L^2(A_{\delta,d})} \le C(R) \qquad (5.9) $$

$$ \frac{C_1(R)}{d^{3/4}} \le \| \partial_2 \bar{v}_{d,1} \|_{L^2(A_{\delta,d})} \le \frac{C_2(R)}{d^{3/4}}, \qquad (5.10) $$

where $\bar{v}_d$ is defined in (4.15).

The proof involves routine but lengthy computations; we put it in Appendix 8.
# 6 The main estimate for the test function

In this section we establish the key estimates needed for the proof of Theorem 2.3. The argument follows [11]; we adapt it to our geometric setting and provide full details for the reader's convenience.

**Standing notation.** Fix $t \in (0, T)$ such that $d(t) > 0$, and set

$$d := d(t) \in (0, R - 1).$$

All objects in this section are written in the rotated variables $y = (y_1, y_2)$. Recall that $\bar{g}_d$ and $\bar{v}_d = \nabla^\perp \bar{g}_d = (\bar{v}_{d,1}, \bar{v}_{d,2})^\top$ are defined in (4.14)-(4.15). We denote by $A_{\delta,d}$ the gap region defined in (4.6), bounded by the graphs $y_2 = \psi_{\text{low}}(y_1)$ and $y_2 = \psi_{\text{up}}(y_1)$ introduced in (4.4)-(4.5). We also define the inner disk in the rotated frame by

$$S_d := B(y_C, 1), \quad y_C = (0, -R + d + 1)^\top,$$

and we write $n = (n_1, n_2)^\top$ for the outward unit normal to $\partial A_{\delta,d}$. Finally, $\bar{e}_2 = (0, 1)^\top$, and $C = C(R) > 0$ denotes a generic constant depending only on $R$. For brevity, we write

$$\partial_i := \frac{\partial}{\partial y_i} \quad (i = 1, 2).$$

*Roadmap of this section.* The goal of this section is to prove Propositions 6.7 and 6.8. More precisely, we will first establish a comparison estimate between the viscous term $\int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{u})$ and a scalar boundary quantity involving $\tilde{m}(d)$. We then show that $\tilde{m}(d)$ blows up as $d \to 0$, which is the key mechanism leading to the non-collision argument. Since the proofs require additional notation and preparatory results, we begin by introducing an auxiliary pressure and a boundary coefficient.

## 6.1 Auxiliary pressure and boundary coefficient

We introduce an auxiliary pressure $q_d$ and a scalar coefficient $\tilde{m}(d)$, which will capture the boundary contribution on $\partial S_d$.

**Definition 6.1.** We define $q_d : A_{\delta,d} \to \mathbb{R}$ by

$$q_d(y) = \partial_{12} \bar{g}_d(y) + \int_{-2\delta}^{y_1} \frac{12 s \chi(s)}{(\psi_{\text{up}}(s) - \psi_{\text{low}}(s))^3} \mathrm{d}s, \quad (6.1)$$

where $\delta$ is defined in (4.3), $\chi$ is defined in (4.13), and $\psi_{\text{up}}$, $\psi_{\text{low}}$ are given in (4.5)-(4.4) (with parameter $d$). This choice of $q_d$ is tailored so that the defect $-\Delta \bar{v}_d + \nabla q_d$ admits a convenient explicit form and enjoys good symmetry properties.

We also define

$$\tilde{m}(d) := \int_{\partial A_{\delta,d} \cap \partial S_d} (2D(\bar{v}_d) - q_d I) n \cdot \bar{e}_2 \mathrm{d}\Gamma. \quad (6.2)$$

*Comment.* The pressure $q_d$ is only used as an auxiliary quantity in the thin region $A_{\delta,d}$. It is introduced so that the defect $-\Delta \bar{v}_d + \nabla q_d$ can be written explicitly and, crucially, so that it satisfies convenient parity properties with respect to the reflection $y_1 \mapsto -y_1$. Moreover, $\tilde{m}(d)$ depends only on the gap parameter $d$ and will later provide the effective “dissipation coefficient” in the contradiction argument.

Throughout the rest of this section, we keep the above notation and do not repeat it in each statement.
## 6.2 A decomposition formula

The next lemma is the basic integration-by-parts identity used in the proof of Proposition 6.7.

**Lemma 6.2 (Integration by parts identity).** Let $\bar{u} \in H_0^1(B(0, R))$ be divergence-free. Then

$$ \int_{A_{\delta,d}} (\Delta \bar{v}_d - \nabla q_d) \cdot \bar{u} \, dy = - \int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{u}) \, dy + \int_{\partial A_{\delta,d}} (2D(\bar{v}_d) - q_d I)n \cdot \bar{u} \, d\Gamma. \quad (6.3) $$

*Proof.* This follows from integration by parts, the definition of $D(\cdot)$ in (1.10), and the incompressibility $\text{div} \, \bar{u} = 0$. □

## 6.3 Estimate of the left term in (6.3)

We first compute an explicit expression for the defect $-\Delta \bar{v}_d + \nabla q_d$ in $A_{\delta,d}$, and then bound its action on divergence-free test fields.

**Lemma 6.3 (Expression of the defect).** From the definitions of $\bar{v}_d$ and $\bar{g}_d$ in (4.14)-(4.15), we have

$$ \Delta \bar{v}_d(y) - \nabla q_d(y) = \left( \frac{-2 \partial_{112} \bar{g}_d(y) - y_1(1 - \mathcal{X}(y_1)) \partial_{222} \varphi(y)}{\partial_{111} \bar{g}_d(y)} \right), \quad y \in A_{\delta,d}. \quad (6.4) $$

and,

$$ \int_{A_{\delta,d}} (-\Delta \bar{v}_d + \nabla q_d) \cdot \bar{u} \, dy = \int_{A_{\delta,d}} (2\partial_{112}\bar{g}_d \bar{u}_1 - \partial_{111}\bar{g}_d \bar{u}_2) \, dy \\ + \int_{A_{\delta,d}} y_1(1 - \mathcal{X}(y_1)) \partial_{222} \varphi(y) \bar{u}_1 \, dy. \quad (6.5) $$

Here $\mathcal{X}$ is defined in (4.13) and $\varphi$ is defined in (4.12).

*Proof.* The result follows by a direct computation from the definitions of $v_d$ in (4.15) and (6.1). □

**Lemma 6.4.** There exists $C(R) > 0$ such that for any divergence-free $\bar{u} \in H_0^1(B(0, R))$, we have

$$ \left| \int_{A_{\delta,d}} (-\Delta \bar{v}_d + \nabla q_d) \cdot \bar{u} \, dy \right| \le C(R) \left( \|\bar{u}\|_{L^2(B(0,R))} + \|\nabla \bar{u}\|_{L^2(B(0,R))} + \|\bar{u}\|_{L^\infty(\partial S_d)} \right). $$

*Proof.* Starting from (6.5), we first estimate

$$ \int_{A_{\delta,d}} (2\partial_{112}\bar{g}_d \bar{u}_1 - \partial_{111}\bar{g}_d \bar{u}_2) \, dy. $$

Integrating by parts gives

$$ \int_{A_{\delta,d}} (2\partial_{112}\bar{g}_d \bar{u}_1 - \partial_{111}\bar{g}_d \bar{u}_2) \, dy = - \int_{A_{\delta,d}} \partial_{11}\bar{g}_d (2\partial_2\bar{u}_1 - \partial_1\bar{u}_2) \, dy \\ + \int_{\partial A_{\delta,d}} \partial_{11}\bar{g}_d (2\bar{u}_1 n_2 - \bar{u}_2 n_1) \, d\Gamma. \quad (6.6) $$

Using (5.9) and Hölder's inequality,

$$ \left| \int_{A_{\delta,d}} \partial_{11} \bar{g}_d (2 \partial_2 \bar{u}_1 - \partial_1 \bar{u}_2) \, dy \right| \le C(R) \left( \| \partial_2 \bar{u}_1 \|_{L^2(A_{\delta,d})} + \| \partial_1 \bar{u}_2 \|_{L^2(A_{\delta,d})} \right). \quad (6.7) $$
**Boundary decomposition.** We now estimate the boundary term in (6.6). We split $\partial A_{\delta,d}$ into the inner boundary, the two vertical segments $|y_1| = 2\delta$, and the outer circle. This allows us to use (i) $\bar{u} = 0$ on $\partial B(0, R)$, (ii) uniform smoothness on the vertical parts, and (iii) symmetry properties on $\partial S_d$. We decompose

$$ \partial A_{\delta,d} = (\partial A_{\delta,d} \cap \partial S_d) \cup (\partial A_{\delta,d} \cap \{|y_1| = 2\delta\}) \cup (\partial A_{\delta,d} \cap \partial B(0, R)). $$

Since $\bar{u} = 0$ on $\partial B(0, R)$, the outer boundary does not contribute. Using the symmetry in $y_1$ and boundary parametrizations, we obtain

$$ \left| \int_{\partial A_{\delta,d}} \partial_{11} \bar{g}_d \bar{u}_2 n_1 d\Gamma \right| \le C(R) \left( \|\bar{u}\|_{L^{\infty}(\partial S_d)} \int_{-2\delta}^{2\delta} \frac{|y_1| |\partial_1 \psi_{\text{up}}(y_1)|^3}{(\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1))^2} dy_1 \right. \\ \left. + \left| \int_{\partial A_{\delta,d} \cap \{|y_1| = 2\delta\}} \partial_{11} \bar{g}_d \bar{u}_2 n_1 d\Gamma \right| \right). \quad (6.8) $$

For the first term on the right-hand side, using (4.9) and (5.2) gives

$$ \|\bar{u}\|_{L^{\infty}(\partial S_d)} \int_{-2\delta}^{2\delta} \frac{|y_1| |\partial_1 \psi_{\text{up}}(y_1)|^3}{(\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1))^2} dy_1 \le C(R) \|\bar{u}\|_{L^{\infty}(\partial S_d)}. $$

For the vertical parts, we use the parametrization and the fact that $\partial_{11} \bar{g}_d$ is odd in $y_1$ to write

$$ \int_{\partial A_{\delta,d} \cap \{|y_1|=2\delta\}} \partial_{11} \bar{g}_d \bar{u}_2 n_1 d\Gamma = \int_{\psi_{\text{low}}(2\delta)}^{\psi_{\text{up}}(2\delta)} \partial_{11} \bar{g}_d(2\delta, y_2) (\bar{u}_2(2\delta, y_2) - \bar{u}_2(-2\delta, y_2)) dy_2 \\ = \int_{\psi_{\text{low}}(2\delta)}^{\psi_{\text{up}}(2\delta)} \partial_{11} \bar{g}_d(2\delta, y_2) \int_{-2\delta}^{2\delta} \partial_1 \bar{u}_2(s, y_2) ds dy_2. \quad (6.9) $$

Applying (5.9) and Hölder's inequality yields

$$ \left| \int_{\partial A_{\delta,d} \cap \{|y_1|=2\delta\}} \partial_{11} \bar{g}_d \bar{u}_2 n_1 d\Gamma \right| \le C(R) \|\partial_1 \bar{u}_2\|_{L^2([-2\delta, 2\delta] \times [\psi_{\text{low}}(2\delta), \psi_{\text{up}}(2\delta)])} \quad (6.10) $$

Combining the above bounds gives

$$ \left| \int_{\partial A_{\delta,d}} \partial_{11} \bar{g}_d \bar{u}_2 n_1 d\Gamma \right| \le C(R) \left( \|\bar{u}\|_{L^{\infty}(\partial S_d)} + \|\partial_1 \bar{u}_2\|_{L^2([-2\delta, 2\delta] \times [\psi_{\text{low}}(2\delta), \psi_{\text{up}}(2\delta)])} \right). \quad (6.11) $$

**Cancellation on the top boundary.** On the upper boundary $y = (y_1, \psi_{\text{up}}(y_1))$, the integration is taken over a symmetric interval $[-2\delta, 2\delta]$. Since $\partial_{11} \bar{g}_d$ is odd in $y_1$ while $\psi_{\text{up}}$ is even, the corresponding boundary integrand becomes odd and the contribution cancels.

On the upper boundary $y = (y_1, \psi_{\text{up}}(y_1))$, we have $n_2 d\Gamma = dy_1$, hence

$$ \int_{\partial A_{\delta,d}} \partial_{11} \bar{g}_d \bar{u}_2 n_2 d\Gamma = \int_{-2\delta}^{2\delta} \partial_{11} \bar{g}_d(y_1, \psi_{\text{up}}(y_1)) \bar{u}_2(y_1, \psi_{\text{up}}(y_1)) dy_1. \quad (6.12) $$

Since $\partial_{11} \bar{g}_d$ is odd in $y_1$ while $\psi_{\text{up}}$ is even, the integrand is odd, and therefore

$$ \int_{\partial A_{\delta,d}} \partial_{11} \bar{g}_d \bar{u}_2 n_2 d\Gamma = 0. \quad (6.13) $$

Finally, since $y_1(1 - \mathcal{X}(y_1))$ and $\partial_{222}\varphi$ are uniformly bounded on $A_{\delta,d}$, we have

$$ \left| \int_{A_{\delta,d}} y_1(1 - \mathcal{X}(y_1)) \partial_{222}\varphi(y) \bar{u}_1 dy \right| \le C(R) \|\bar{u}\|_{L^2(A_{\delta,d})}. $$

Collecting the above bounds together with (6.7), (6.11), and (6.13) completes the proof. $\square$
## 6.4 Boundary reduction by symmetry

We now control the boundary contributions appearing in Lemma 6.2. The vertical parts $|y_1| = 2\delta$ are estimated by standard trace inequalities, and the part on $\partial S_d$ is reduced to $\tilde{m}(d)$ by symmetry.

**Lemma 6.5 (Vertical boundary contribution).** There exists $C(R) > 0$ such that

$$ \left| \int_{\partial A_{\delta,d} \cap \{|y_1|=2\delta\}} (2D(\bar{v}_d) - q_d I) n \cdot \bar{u} \, d\Gamma \right| \le C(R) \|\bar{u}\|_{H^1(A_{\delta,d})}. \quad (6.14) $$

*Proof.* Since $\nabla \bar{v}_d$ and $q_d$ are smooth up to the boundary, they are uniformly bounded on $\partial A_{\delta,d} \cap \{|y_1|=2\delta\}$. Moreover, $\delta > 0$ depends only on $R$, and

$$ \psi_{\text{up}}(2\delta) - \psi_{\text{low}}(2\delta) \le C(R). $$

Hence

$$ |(2D(\bar{v}_d) - q_d I)n| \le C(R) \quad \text{for } y \in \partial A_{\delta,d} \cap \{|y_1|=2\delta\}. \quad (6.15) $$

The claim follows from the trace theorem and Hölder's inequality. $\square$

*Key step.* The next lemma shows that the boundary integral on $\partial S_d$ depends only on the translational component $V(\bar{u})$ of the rigid motion inside $S_d$, and it reduces to the scalar coefficient $\tilde{m}(d)$. This reduction relies on parity cancellations with respect to $y_1 \mapsto -y_1$.

**Lemma 6.6 (Reduction on $\partial S_d$ by symmetry).** Assume in addition that $\bar{u}$ coincides with a rigid motion in $S_d$, namely

$$ \bar{u}(y) = V(\bar{u}) + \omega (y - y_C)^{\perp}, \quad y \in S_d, \quad (6.16) $$

where $V(\bar{u}) \in \mathbb{R}^2$, $\omega \in \mathbb{R}$, and $y_C = (0, -R + d + 1)^{\top}$. Then

$$ \int_{\partial A_{\delta,d} \cap \partial S_d} (2D(\bar{v}_d) - q_d I) n \cdot \bar{u} \, d\Gamma = \tilde{m}(d) (V(\bar{u}) \cdot \bar{e}_2). $$

*Proof.* Using the explicit formulas for $(\bar{v}_d, q_d)$ and the reflection $y_1 \mapsto -y_1$, one checks that

$$ q_d, \quad \partial_{y_1} \bar{v}_{d,1}, \quad \partial_{y_2} \bar{v}_{d,2} \text{ are even in } y_1, \quad \partial_{y_2} \bar{v}_{d,1}, \quad \partial_{y_1} \bar{v}_{d,2} \text{ are odd in } y_1. \quad (6.17) $$

Therefore, when integrating over the symmetric boundary piece $\partial A_{\delta,d} \cap \partial S_d$, all odd contributions cancel.

*Reduction to translation.* Since the boundary piece $\partial A_{\delta,d} \cap \partial S_d$ is symmetric with respect to $y_1 \mapsto -y_1$, the rotational part $\omega (y - y_C)^{\perp}$ yields only odd contributions and thus vanishes after integration. Consequently, only the vertical component of $V(\bar{u})$ survives.

Using the rigid-motion trace (6.16), the rotational part does not contribute by symmetry, and only the vertical component of $V(\bar{u})$ remains. This gives the desired identity. $\square$

## 6.5 Main estimate

We now combine the estimate (6.4) and the boundary reductions to prove the main inequality.

**Proposition 6.7 (Main inequality for the test function).** Let $d \in (0, R-1)$ and let $\bar{u} \in H_0^1(B(0, R))$ be divergence-free. Assume moreover that $\bar{u}$ satisfies the rigid-motion trace condition (6.16) in $S_d$. Then there exists a constant $C(R) > 0$ such that

$$ \left| \int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{u}) \, dy - \tilde{m}(d) (V(\bar{u}) \cdot \bar{e}_2) \right| \le C(R) \left( \|\bar{u}\|_{L^2(B(0,R))} + \|\bar{u}\|_{L^{\infty}(\partial S_d)} + \|\nabla \bar{u}\|_{L^2(B(0,R))} \right). \quad (6.18) $$
Proof. We start from the integration-by-parts identity in Lemma 6.2. The right term of (6.2) is controlled by Lemma 6.4, the vertical boundary terms by Lemma 6.5, and the inner boundary term on $\partial S_d$ is reduced to $\tilde{m}(d)$ by Lemma 6.6.

Starting from Lemma 6.2, we rewrite (6.3) as

$$ \int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{u}) dy = - \int_{A_{\delta,d}} (\Delta \bar{v}_d - \nabla q_d) \cdot \bar{u} dy + \int_{\partial A_{\delta,d}} (2D(\bar{v}_d) - q_d I)n \cdot \bar{u} d\Gamma. $$

The first term in the right equality is estimated by Lemma 6.4. For the boundary term, we decompose

$$ \partial A_{\delta,d} = (\partial A_{\delta,d} \cap \{|y_1| = 2\delta\}) \cup (\partial A_{\delta,d} \cap \partial B(0, R)) \cup (\partial A_{\delta,d} \cap \partial S_d). $$

Since $\bar{u} = 0$ on $\partial B(0, R)$, the outer boundary contribution vanishes. The vertical contribution is controlled by Lemma 6.5. Finally, the inner boundary term is reduced by Lemma 6.6, yielding

$$ \int_{\partial A_{\delta,d} \cap \partial S_d} (2D(\bar{v}_d) - q_d I)n \cdot \bar{u} d\Gamma = \tilde{m}(d) (V(\bar{u}) \cdot \bar{e}_2). $$

Combining the above estimates proves (6.18). □

## 6.6 Lower bound on $\tilde{m}(d)$

We finally prove that $\tilde{m}(d)$ blows up as $d \to 0$, which is crucial for the dissipation mechanism.

**Proposition 6.8 (Lower bound on $\tilde{m}(d)$).** There exists $\eta = \eta(R) > 0$, independent of $d$, such that for all $0 < d \le \eta$,

$$ \tilde{m}(d) \ge \frac{C(R)}{d^{3/2}}. \qquad (6.19) $$

Proof. Main mechanism. The blow-up of $\tilde{m}(d)$ as $d \to 0$ comes from the dominant shear inside the thin gap, captured by $\partial_2 \bar{v}_{d,1}$, whose $L^2$-norm scales like $d^{-3/4}$. This yields the scaling $\tilde{m}(d) \gtrsim \|\partial_2 \bar{v}_{d,1}\|_{L^2}^2 \sim d^{-3/2}$.

Taking $\bar{u} = \bar{v}_d$ in Proposition 6.7, we obtain

$$ \left| \int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{v}_d) dy - \tilde{m}(d) (V(\bar{v}_d) \cdot \bar{e}_2) \right| \le C(R) \left( \|\bar{v}_d\|_{L^2(B(0,R))} + \|\bar{v}_d\|_{L^\infty(\partial S_d)} + \|\nabla \bar{v}_d\|_{L^2(B(0,R))} \right). \qquad (6.20) $$

Since $\bar{v}_d = \bar{e}_2$ in $S_d$ (by Proposition 4.3), we have $V(\bar{v}_d) = \bar{e}_2$ and hence $V(\bar{v}_d) \cdot \bar{e}_2 = 1$. Therefore, (6.20) implies

$$ \tilde{m}(d) \ge \int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{v}_d) dy - C(R) \left( \|\bar{v}_d\|_{L^2(A_{\delta,d})} + \|\bar{v}_d\|_{L^\infty(S_d)} + \|\partial_{y_2} \bar{v}_{d,1}\|_{L^2(A_{\delta,d})} + \|\partial_{y_1} \bar{v}_{d,2}\|_{L^2(A_{\delta,d})} \right). \qquad (6.21) $$

From (5.9) we have

$$ \|\partial_1 \bar{v}_{d,2}\|_{L^2(A_{\delta,d})} = \|\partial_{11} \bar{g}_d\|_{L^2(A_{\delta,d})} \le C(R), \qquad \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})} \ge \frac{C(R)}{d^{3/4}}. \qquad (6.22) $$

Using the elementary inequality

$$ (a+b)^2 \ge \frac{1}{2}a^2 - b^2 \quad (a,b \in \mathbb{R}), $$
together with the explicit structure of $D(\bar{v}_d)$, we obtain (for $d$ small enough)

$$ \int_{A_{\delta,d}} 2D(\bar{v}_d) : D(\bar{v}_d) \, dy \geq \frac{1}{2} \|\partial_2 \bar{v}_{d,1} + \partial_1 \bar{v}_{d,2}\|_{L^2(A_{\delta,d})}^2 \\ \geq \frac{1}{4} \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})}^2 - \frac{1}{2} \|\partial_1 \bar{v}_{d,2}\|_{L^2(A_{\delta,d})}^2 \geq \frac{C(R)}{d^{3/2}}. \quad (6.23) $$

This proves (6.19) for all $0 < d \leq \eta(R)$. $\square$

## 7 Proof of Theorem 2.3

In this section, we prove Theorem 2.3 by a contradiction argument. The strategy is the same as in [11, Section 3.2]; for completeness, we provide the details.

To begin with, we first restrict to the case where the disk stays away from the origin. This is only for technical convenience: the rotation map $Y_\tau$ introduced in (3.2) involves the normalization $\tau/|\tau|$ and becomes singular as $\tau \to 0$. More precisely, we assume that

$$ |h(t)| > \frac{1}{8}(R-1) \quad \text{for } t \in (0, T^*). \quad (7.1) $$

Here, $T^*$ denotes the maximal existence time of system (1.1)-(1.8) in Proposition 2.2; it may be finite or equal to $+\infty$. (We will explain at the end of the proof why this assumption does not restrict the contradiction argument.)

We justify that the test function constructed in Section 4 is admissible in the weak formulation in Definition 2.4.

**Lemma 7.1 (Regularity of the test function).** Assume (7.1). Let $v$ be the test function defined by (4.16). Then the associated rigid-body coefficients can be chosen as

$$ \ell(t) := e_{h(t)} = \frac{h(t)}{|h(t)|}, \quad r(t) := 0, $$

and they satisfy

$$ \ell \in H^1((0, T); \mathbb{R}^2), \quad r \in H^1((0, T); \mathbb{R}). $$

Moreover,

$$ v \in H^1((0, T); L^2(B(0, R))) \cap L^2((0, T); H^1(B(0, R))). $$

*Proof sketch. Step 1: Regularity of $\ell$ and $r$. By (7.1) there exists $c_0 > 0$ such that $|h(t)| \geq c_0$ for all $t \in (0, T)$. Since the map $z \mapsto z/|z|$ is smooth on $\{z \in \mathbb{R}^2 : |z| \geq c_0\}$, the chain rule yields*

$$ \dot{\ell}(t) = \left( \frac{1}{|h(t)|} I - \frac{h(t) \otimes h(t)}{|h(t)|^3} \right) \dot{h}(t), $$

hence $|\dot{\ell}(t)| \leq C(c_0)|\dot{h}(t)|$. Therefore $\ell \in H^1((0, T); \mathbb{R}^2)$. Clearly $r(t) \equiv 0 \in H^1((0, T); \mathbb{R})$.

**Step 2: Regularity of the gap function.** Recall that $d(t) = R - 1 - |h(t)|$. Differentiating gives

$$ \dot{d}(t) = -\frac{h(t)}{|h(t)|} \cdot \dot{h}(t) = \dot{h}(t) \cdot e_2(t), \quad e_2(t) := -\frac{h(t)}{|h(t)|}, $$

and therefore $|\dot{d}(t)| \leq |\dot{h}(t)|$, so $d \in H^1(0, T)$.

**Step 3: Regularity of $v$ before contact.** Fix $T_0 \in (0, T^*)$ such that the disk remains at a positive distance from the outer boundary on $[0, T_0]$, namely

$$ d_0 := \min_{t \in [0, T_0]} d(t) > 0. \quad (7.2) $$
We claim that on $(0, T_0)$ the test function $v$ defined by (4.16) is admissible, i.e.

$$v \in L^2((0, T_0); H^1(B(0, R))) \cap H^1((0, T_0); L^2(B(0, R))).$$

Recall that

$$v(t, x) = J_x Y_{h(t)} \bar{v}_{d(t)}(Y_{h(t)}(x)),$$

where $Y_{h(t)}$ is a rotation, hence an isometry. In particular, the change of variables $y = Y_{h(t)}(x)$ preserves Lebesgue measure and $L^2$-norms. Therefore,

$$\|\nabla v(t)\|_{L^2(B(0,R))} = \|\nabla \bar{v}_{d(t)}\|_{L^2(B(0,R))}.$$

*Spatial regularity.* We split the domain into the thin region $A_{\delta,d(t)}$ and its complement. Outside $A_{\delta,d(t)}$, the field $\bar{v}_{d(t)}$ is built using smooth cut-off functions independent of $d(t)$, hence

$$\|\nabla \bar{v}_{d(t)}\|_{L^2(B(0,R)\setminus A_{\delta,d(t)})} \le C(R).$$

Inside $A_{\delta,d(t)}$, Lemma 5.3 yields

$$\|\partial_1 \bar{v}_{d(t),2}\|_{L^2(A_{\delta,d(t)})} \le C(R), \quad \|\partial_2 \bar{v}_{d(t),1}\|_{L^2(A_{\delta,d(t)})} \le \frac{C(R)}{d(t)^{3/4}} \le \frac{C(R)}{d_0^{3/4}}.$$

Combining these bounds, we obtain for a.e. $t \in (0, T_0)$,

$$\|\nabla v(t)\|_{L^2(B(0,R))}^2 = \|\nabla \bar{v}_{d(t)}\|_{L^2(B(0,R))}^2 \le C(R) + \frac{C(R)}{d_0^{3/2}}.$$

Hence $v \in L^2((0, T_0); H^1(B(0, R)))$.

*Time regularity.* Differentiating $v(t, x)$ with respect to $t$ and using the chain rule, we obtain

$$\partial_t v(t, x) = (\partial_t J_x Y_{h(t)}) \bar{v}_{d(t)}(Y_{h(t)}(x)) \\ + J_x Y_{h(t)} (\dot{d}(t) \partial_d \bar{v}_{d(t)}(Y_{h(t)}(x)) + (\nabla \bar{v}_{d(t)})(Y_{h(t)}(x)) \partial_t Y_{h(t)}(x)).$$

Using the isometry of $Y_{h(t)}$, the Cauchy–Schwarz inequality, and the estimate $\|\partial_t Y_{h(t)}\|_{L^\infty(B(0,R))} \le C(R) |\dot{h}(t)|$, we infer that for a.e. $t \in (0, T_0)$,

$$\|\partial_t v(t)\|_{L^2(B(0,R))} \le C(R) (|\dot{h}(t)| + |\dot{d}(t)|) \left( \|\bar{v}_{d(t)}\|_{H^1(B(0,R))} + \|\partial_d \bar{v}_{d(t)}\|_{L^2(B(0,R))} \right). \quad (7.3)$$

Moreover, since $d(t) = R - 1 - |h(t)|$, we have $|\dot{d}(t)| \le |\dot{h}(t)|$. Using again the uniform lower bound $d(t) \ge d_0$ on $[0, T_0]$ and the estimates collected in Lemmas 5.1, 5.2, and 5.3, we obtain

$$\|\partial_t v(t)\|_{L^2(B(0,R))} \le \frac{C(R)}{d_0^{3/4}} |\dot{h}(t)| \quad \text{for a.e. } t \in (0, T_0). \quad (7.4)$$

Finally, the energy inequality (2.2) implies $\dot{h} \in L^2(0, T_0)$. Therefore (7.4) yields $\partial_t v \in L^2((0, T_0); L^2(B(0, R)))$, and hence $v \in H^1((0, T_0); L^2(B(0, R)))$.

**Remark.** Estimate (7.4) shows explicitly that the admissibility of $v$ relies on the strict positivity of the gap. If $d(t) \to 0$, then the factor $d_0^{-3/4}$ may blow up, so the bounds on $\partial_t v$ (and similarly on $\nabla v$) degenerate. This is precisely why the test function is only used on time intervals where contact has not yet occurred. $\square$
*Proof of Theorem 2.3. Step 1. Extension of the velocity.* For later use, we extend the velocity to the whole ball $B(0, R)$ by

$$ \tilde{u}(t, x) = \begin{cases} u(t, x), & x \in \Omega(t), \\ \dot{h}(t) + \omega(t)(x - h(t))^{\perp}, & x \in S(t), \end{cases} \qquad (7.5) $$

where $u$ is the fluid velocity, $\dot{h}$ is the translational velocity of the disk center, and $\omega$ is the angular velocity. By the no-slip condition (1.4) and Proposition 2.2,

$$ \tilde{u} \in H^1(0, T; L^2(B(0, R))) \cap C([0, T]; H^1(B(0, R))). \qquad (7.6) $$

**Step 2. Weak formulation with the test function.** We choose the test function $v$ defined in (4.16) and integrate from 0 to $t$. Then, for a.e. $t \in (0, T)$,

$$ \int_{B(0,R)} \tilde{u}(t,x) \cdot v(t,x) \, dx - \int_{B(0,R)} \tilde{u}(0,x) \cdot v(0,x) \, dx \\ = \int_{0}^{t} \int_{B(0,R)} (\tilde{u} \cdot \partial_t v(s,x) + (\tilde{u} \otimes \tilde{u}) : D[v](s,x) - 2 D[\tilde{u}] : D[v](s,x)) \, dx \, ds. \qquad (7.7) $$

*Insertion of the coefficient term.* Since $\tilde{m}(d(t))V(\tilde{u}) \cdot \bar{e}_2$ is a scalar quantity depending only on time, we may add its time antiderivative to both sides of (7.7) without altering the equality. Thus we obtain

$$ P(t) - P(0) - \int_{0}^{t} A_{1}(s) \, ds + \int_{0}^{t} A_{2}(s) \, ds + \int_{0}^{t} A_{3}(s) \, ds = \int_{0}^{t} \tilde{m}(d(s)) V(\tilde{u}) \cdot \bar{e}_{2} \, ds, \qquad (7.8) $$

where

$$ P(t) = \int_{B(0,R)} \tilde{u}(t,x) \cdot v(t,x) \, dx, \qquad (7.9) $$

$$ A_1(t) = \int_{B(0,R)} \tilde{u}(t,x) \cdot \partial_t v(t,x) \, dx, \qquad (7.10) $$

$$ A_2(t) = \int_{B(0,R)} (\tilde{u}(t,x) \otimes \tilde{u}(t,x)) : D[v](t,x) \, dx, \qquad (7.11) $$

$$ A_3(t) = \int_{B(0,R)} 2 D[\tilde{u}](t,x) : D[v](t,x) \, dx - \tilde{m}(d(t)) V(\tilde{u}) \cdot \bar{e}_2. \qquad (7.12) $$

**Outline of the estimates.** We next show that $P(t)$ is uniformly bounded, that $\int_0^t A_1$, $\int_0^t A_2$ and $\int_0^t A_3 ds$ are controlled by the initial energy. These bounds yield a global upper bound on the right-hand side of (7.8).

## Preparatory estimates

To carry out the estimates below, we introduce the energy associated with the system (1.1)–(1.8):

$$ E(t) = \frac{1}{2} \int_{\Omega(t)} |u(x,t)|^2 dx + \frac{M}{2} |\dot{h}(t)|^2 + \frac{J}{2} |\omega(t)|^2. \qquad (7.13) $$

**Lemma 7.2 (Bound for the momentum term).** There exists $C(R) > 0$ such that for all $t \in (0, T)$,

$$ |P(t)| \le C(R) \sqrt{E(0)}. $$
*Proof.* By Proposition 3.1 and Lemma 5.1,

$$ \|v(t)\|_{L^2(B(0,R))} = \|\bar{v}_{d(t)}\|_{L^2(B(0,R))} \le C(R). $$

Hence,

$$ |P(t)| \le \|\tilde{u}(t)\|_{L^2(B(0,R))} \|v(t)\|_{L^2(B(0,R))} \le C(R) \|\tilde{u}(t)\|_{L^2(B(0,R))}. $$

Using the energy estimate (2.2) yields $\|\tilde{u}(t)\|_{L^2(B(0,R))}^2 \le C E(0)$, which proves the claim. $\square$

**Lemma 7.3** (Estimate of the time-derivative term). There exists $C(R) > 0$ such that for a.e. $t \in (0, T)$,

$$ |A_1(t)| \le C(R) \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))} |\dot{h}(t)| \le C(R) \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))}^2 + C(R) |\dot{h}(t)|^2. $$

Consequently, for all $t \in (0, T)$,

$$ \left| \int_{0}^{t} A_{1}(s) \, ds \right| \le C(R) E(0) + C(R) E(0) t. $$

*Proof.* Recall that

$$ v(t, x) = \nabla_{x}^{\perp}(\bar{g}_{d(t)}(Y_{h(t)}(x))). $$

Differentiating with respect to $t$ gives

$$ \partial_t v(t, x) = \nabla_x^\perp \left( \dot{d}(t) \partial_d \bar{g}_{d(t)}(Y_{h(t)}(x)) + \nabla_y \bar{g}_{d(t)}(Y_{h(t)}(x)) \cdot \partial_t Y_{h(t)}(x) \right). \quad (7.14) $$

As a consequence, integration by parts in the $x$-variables yields

$$ \begin{aligned} & \int_{B(0,R)} \tilde{u}(x,t) \cdot \partial_t v(x,t) \, dx \\ &= \int_{B(0,R)} (\partial_{x_1} \tilde{u}_2 - \partial_{x_2} \tilde{u}_1) \partial_t (\bar{g}_{d(t)}(Y(x,t))) \, dx \\ &= \int_{B(0,R)} (\partial_{x_1} \tilde{u}_2 - \partial_{x_2} \tilde{u}_1) (\dot{d}(t) \partial_d \bar{g}_{d(t)}(Y(x,t)) + \nabla_y \bar{g}_{d(t)}(Y(x,t)) \cdot \partial_t Y(x,t)) \, dx. \end{aligned} \quad (7.15) $$

Hence, by the Cauchy-Schwarz inequality,

$$ \left| \int_{B(0,R)} \tilde{u} \cdot \partial_t v \, dx \right| \le \|\nabla \tilde{u}\|_{L^2(B(0,R))} \left( |\dot{d}(t)| \|\partial_d \bar{g}_{d(t)}(Y)\|_{L^2(B(0,R))} + \|\nabla_y \bar{g}_{d(t)}(Y)\|_{L^2(B(0,R))} \|\partial_t Y\|_{L^\infty(B(0,R))} \right). \quad (7.16) $$

Moreover, since $d(t) = R - 1 - |h(t)|$, we have

$$ \dot{d}(t) = \dot{h}(t) \cdot e_2(t), \quad e_2(t) = - \frac{h(t)}{|h(t)|}, $$

and thus $|\dot{d}(t)| \le |\dot{h}(t)|$. Under (7.1), one also has $\|\partial_t Y_{h(t)}\|_{L^\infty(B(0,R))} \le C(R) |\dot{h}(t)|$. Finally, Lemma 5.2 yields

$$ \|\partial_d \bar{g}_{d(t)}\|_{L^2(B(0,R))} + \|\nabla_y \bar{g}_{d(t)}\|_{L^2(B(0,R))} \le C(R). $$

Therefore,

$$ \|\partial_t (\bar{g}_{d(t)}(Y_{h(t)}(\cdot)))\|_{L^2(B(0,R))} \le C(R) |\dot{h}(t)|. $$

This gives

$$ |A_1(t)| \le C(R) \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))} |\dot{h}(t)|. $$
The quadratic bound follows from Young's inequality.

Integrating in time and using (2.2), we obtain

$$ \int_0^t \|\nabla \tilde{u}(s)\|_{L^2(B(0,R))}^2 \mathrm{d}s \le C(R) E(0), \quad \int_0^t |\dot{h}(s)|^2 \mathrm{d}s \le C E(0) t, $$

which yields the integrated estimate.

□

**Lemma 7.4** (Estimate of the nonlinear term). There exists $C(R) > 0$ such that for a.e. $t \in (0, T)$,

$$ |A_2(t)| \le C(R) \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))}^2. $$

Consequently,

$$ \left| \int_0^t A_2(s) \mathrm{d}s \right| \le C(R) E(0) \quad \text{for all } t \in (0, T). $$

*Proof.* We split the domain into the thin gap region $\bar{A}_{\delta,d}$ and its complement:

$$ A_2(t) = \int_{B(0,R) \setminus \bar{A}_{\delta,d}} (\tilde{u} \otimes \tilde{u}) : D[v] \mathrm{d}x + \int_{\bar{A}_{\delta,d}} (\tilde{u} \otimes \tilde{u}) : D[v] \mathrm{d}x, $$

where $\bar{A}_{\delta,d} := \{x \in B(0, R) : Y_{h(t)}(x) \in A_{\delta,d}\}$.

(i) Outside the gap. On $B(0, R) \setminus \bar{A}_{\delta,d}$, the construction of $v$ involves only smooth cut-offs independent of $d$, hence $\|D[v]\|_{L^{\infty}(B(0,R)\setminus\bar{A}_{\delta,d})} \le C(R)$. Therefore,

$$ \left| \int_{B(0,R) \setminus \bar{A}_{\delta,d}} (\tilde{u} \otimes \tilde{u}) : D[v] \mathrm{d}x \right| \le C(R) \|\tilde{u}(t)\|_{L^2(B(0,R))}^2 \le C(R) E(0). $$

(ii) Inside the gap. Using incompressibility and integrating by parts, we write

$$ \int_{\bar{A}_{\delta,d}} (\tilde{u} \otimes \tilde{u}) : D[v] \mathrm{d}x = - \int_{\bar{A}_{\delta,d}} (u \cdot \nabla u) \cdot v \mathrm{d}x + \int_{\partial \bar{A}_{\delta,d}} (u \cdot v)(u \cdot n) \mathrm{d}\Gamma. $$

For the first term on the right-hand side of this equality, Hölder's inequality and the 2D Gagliardo–Nirenberg inequality yield

$$ \left| \int_{\bar{A}_{\delta,d}} (u \cdot \nabla u) \cdot v \mathrm{d}x \right| \le \|\nabla u\|_{L^2} \|u\|_{L^{10}} \|v\|_{L^{5/2}} \le C(R) \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))}^2, $$

where we used the uniform bound $\|v\|_{L^{5/2}(B(0,R))} \le C(R)$ (Lemma 5.1). For the boundary term, using $\|v\|_{L^{\infty}(\partial \bar{A}_{\delta,d})} \le C(R)$ and the trace theorem,

$$ \left| \int_{\partial \bar{A}_{\delta,d}} (u \cdot v)(u \cdot n) \mathrm{d}\Gamma \right| \le C(R) \|u\|_{L^2(\partial \bar{A}_{\delta,d})}^2 \le C(R) \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))}^2. $$

Combining the above bounds yields the desired pointwise estimate for $A_2(t)$. Integrating in time and using (2.2) gives the integrated bound.

□

**Lemma 7.5** (Estimate of the viscous term). There exists $C(R) > 0$ such that for all $t \in (0, T)$,

$$ \int_0^t A_3(s) \mathrm{d}s \le C(R) E(0) + C(R) E(0) t. $$
*Proof.* By Proposition 6.7 applied with $\bar{u} = \tilde{\bar{u}}$, we have for a.e. $t \in (0, T)$,

$$ \left| \int_{B(0,R)} 2 D[\tilde{u}](t,x) : D[v](t,x) \,dx - \tilde{m}(d(t)) V(\tilde{u}) \cdot \bar{e}_2 \right| \le C(R) \left( \|\tilde{u}\|_{L^2} + \|\tilde{u}\|_{L^\infty(\partial S(t))} + \|\nabla \tilde{u}\|_{L^2} \right). $$

Recalling the definition of $A_3(t)$, we obtain

$$ |A_3(t)| \le C(R) \left( \|\tilde{u}(t)\|_{L^2(B(0,R))} + \|\tilde{u}(t)\|_{L^\infty(\partial S(t))} + \|\nabla \tilde{u}(t)\|_{L^2(B(0,R))} \right). $$

Using Young's inequality and the fact that $\tilde{u}$ is a rigid motion on $S(t)$, $\|\tilde{u}(t)\|_{L^\infty(\partial S(t))} \le |\dot{h}(t)| + |\omega(t)|$, together with the energy bound (2.2), we obtain

$$ \begin{aligned} \int_0^t |A_3(s)| \, ds \le & C(R) \int_0^t \|\nabla \tilde{u}(s)\|_{L^2}^2 \, ds \\ & + C(R) \int_0^t \|\tilde{u}(s)\|_{L^2}^2 \, ds + C(R) E(0) t \le C(R) E(0) + C(R) E(0) t. \end{aligned} \quad (7.17) $$

□

## Completion of the proof

**Step 4. Derivation of the key inequality.** Combining (7.8) with Lemmas 7.2, 7.3, 7.4, 7.5, we obtain that for every $t \in (0, T)$,

$$ \int_0^t \tilde{m}(d(s)) V(\tilde{u}) \cdot \bar{e}_2 \, ds \le C(R)E(0) + C(R)\sqrt{E(0)} + C(R)E(0)t. \quad (7.18) $$

**Step 5. Contradiction and exclusion of contact.** Assume by contradiction that there exists $0 < T^* < \infty$ such that

$$ \lim_{t \to T^*} d(t) = 0. $$

Then there exists $T_1 \in (0, T^*)$ such that for all $t \in (T_1, T^*)$,

$$ 0 < d(t) < \eta(R), \quad (7.19) $$

where $\eta(R)$ is the constant in Proposition 6.8.

Moreover, By using (7.18) and the fact that $V(\tilde{u}) \cdot \bar{e}_2 = \dot{d}(t)$, we obtain for all $t \in (T_1, T^*)$,

$$ \int_{T_1}^t \tilde{m}(d(s)) \dot{d}(s) \, ds \le C(R, T^*, E(0)). $$

Writing $H(s) = d(s)$, this yields

$$ \int_{d(T_1)}^{d(t)} \tilde{m}(H) \, dH \le C(R, T^*, E(0)). $$

Using Proposition 6.8, namely $\tilde{m}(H) \ge C(R)H^{-3/2}$ for $H \in (0, \eta]$, we deduce

$$ \int_{d(T_1)}^{d(t)} \frac{C(R)}{H^{3/2}} \, dH \le C(R, T^*, E(0)), $$

which implies

$$ \frac{1}{d^{1/2}(t)} \le \frac{1}{d^{1/2}(T_1)} + C(R, T^*, E(0)). $$

In particular $d(t)$ is bounded below by a strictly positive constant on $(T_1, T^*)$, contradicting $\lim_{t \to T^*} d(t) = 0$. Therefore, no finite-time contact can occur.
**Remark 7.6 (On the assumption (7.1)).** The lower bound (7.1) is imposed only to avoid the singularity of the rotation map $Y_\tau$, which involves the normalization $\tau/|\tau|$. It is not an essential restriction for the non-collision argument. Indeed, in the contradiction proof we assume $d(t) \to 0$ as $t \to T^*$. Since $d(t) = R - 1 - |h(t)|$, this implies $|h(t)| \to R - 1$ as $t \to T^*$. Hence there exists $T_0 < T^*$ such that (7.1) holds for all $t \in (T_0, T^*)$. Therefore, all estimates involving the rotated coordinates can be carried out on $(T_0, T^*)$, and the proof does not require any global assumption on $|h(t)|$.

Finally, by Proposition 2.2, the strong solution is global in time.

# 8 Appendix

In this section, we prove Lemma 4.1 and Lemma 5.3.

**Reminder of notation.** For convenience, we briefly recall the main objects used in this appendix and indicate where they are defined:

* The parameter $R > 1$ is the radius of the outer disk, and $d \in (0, R - 1]$ denotes the (instantaneous) gap parameter.

* The boundary profiles $\psi_{up}$ and $\psi_{low}$ of the gap region are introduced in (4.5)–(4.4).

* The (small) constant $\delta > 0$ is defined in (4.3), and the gap region $A_{\delta,d}$ is defined in (4.6). The core region $A_{0,d}$ is defined in (4.7).

* The cut-off functions $\chi$ and $\varphi$ are defined in (4.13) and (4.12).

* The stream function $\bar{g}_d$ and the associated divergence-free field $\bar{v}_d = \nabla^\perp \bar{g}_d$ are defined in (4.14)–(4.15).

Throughout the appendix, all computations are performed in the rotated variables $y = (y_1, y_2)$, and we use the shorthand $\partial_i = \frac{\partial}{\partial y_i}$.

**A note on the piecewise structure of $\bar{g}_d$ and $\bar{v}_d$.** Recall from (4.14) that the construction involves the cut-off $\chi(y_1)$, hence $\bar{g}_d$ (and therefore $\bar{v}_d = \nabla^\perp \bar{g}_d$) should be viewed as a piecewise-defined function:

* On the core region $A_{0,d}$ one has $\chi(y_1) \equiv 1$, so the expression of $\bar{g}_d$ reduces to the thin-gap polynomial profile depending on $F(y)$.

* On the transition strip $A_{\delta,d} \setminus A_{0,d}$ one has $0 \le \chi(y_1) \le 1$, and the definition of $\bar{g}_d$ includes the additional smooth term involving $\varphi$.

This is why, in the estimates below, we systematically split the domain into these two parts.

*Proof of Lemma 4.1.* The purpose of this lemma is to provide a two-sided estimate of the gap thickness $\psi_{up}(y_1) - \psi_{low}(y_1)$ for $|y_1| \le 2\delta$.

**Step 1: Exact expression of the gap.** From the definitions of $\psi_{up}$ and $\psi_{low}$ we obtain

$$ \psi_{up}(y_1) - \psi_{low}(y_1) = d + \left(1 - \sqrt{1 - y_1^2}\right) - \left(R - \sqrt{R^2 - y_1^2}\right), \quad (8.1) $$

valid for

$$ -2\delta \le y_1 \le 2\delta. \quad (8.2) $$
**Step 2: Elementary bounds for the square roots.** Using $|y_1| \le 1$, we have the standard estimates

$$1 - \frac{y_1^2}{2} - \frac{y_1^4}{2} \le \sqrt{1 - y_1^2} \le 1 - \frac{y_1^2}{2}. \quad (8.3)$$

Similarly, for $\sqrt{R^2 - y_1^2}$, we obtain

$$R - \frac{y_1^2}{2R} - \frac{y_1^4}{2R^3} \le \sqrt{R^2 - y_1^2} \le R - \frac{y_1^2}{2R}. \quad (8.4)$$

**Step 3: Two-sided estimate of the gap.** Combining the above estimates with (8.1), it follows that

$$d + \left(\frac{1}{2} - \frac{1}{2R}\right) y_1^2 - \frac{1}{2R^3} y_1^4 \le \psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1) \le d + \left(\frac{1}{2} - \frac{1}{2R}\right) y_1^2 + \frac{1}{2} y_1^4. \quad (8.5)$$

**Step 4: Simplifying the lower bound.** We use the fact

$$-\frac{1}{2R^3} y_1^4 \ge -\left(\frac{1}{4} - \frac{1}{R}\right) y_1^2 \qquad y_1^2 \le \frac{R^3}{2} \left(\frac{1}{4} - \frac{1}{4R}\right),$$

so that the negative quartic term can be absorbed into the quadratic term. Hence we get the left-hand side of (4.9):

$$\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1) \ge d + \left(\frac{1}{4} - \frac{1}{R}\right) y_1^2. \quad (8.6)$$

**Step 5: Upper bound.** From the right-hand side of (8.5) and using $|y_1| \le 1$, it is easy to check that

$$\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1) \le d + \left(1 - \frac{1}{R}\right) y_1^2. \quad (8.7)$$

**Conclusion.** Combining (8.6)-(8.7), for all $|y_1| \le 2\delta$ we finally obtain

$$d + \left(\frac{1}{4} - \frac{1}{4R}\right) y_1^2 \le \psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1) \le d + \left(1 - \frac{1}{R}\right) y_1^2. \quad (8.8)$$

This completes the proof. $\square$

*Proof of Lemma 5.3.* The goal is to estimate $\|\partial_1 \bar{v}_{d,2}\|_{L^2(A_{\delta,d})}$ and $\|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})}$ uniformly in $d$ (up to the expected singular scaling in $d$).

**Step 1: Estimate of $\partial_1 \bar{v}_{d,2}$.** From (4.14)-(4.15), we have

$$\partial_1 \bar{v}_{d,2} = \partial_{11} \bar{g}_d.$$

As explained above, we split the gap region into the core part $A_{0,d}$ and the transition strip $A_{\delta,d} \setminus A_{0,d}$, since $\bar{g}_d$ (hence $\bar{v}_d$) has different expressions on these two sets. We first estimate $\partial_{11} \bar{g}_d$ on $A_{0,d}$, where the cut-off satisfies $\mathcal{X}(y_1) = 1$.

On $A_{0,d}$ we have $\mathcal{X}(y_1) = 1$ by (4.13), and a direct calculation gives

$$\partial_{11} \bar{g}_d(y_1) = 3 \partial_1 F(y) - 6 y_1 (\partial_1 F(y))^2, \quad (8.9)$$

where $F(y)$ is given by (4.10).

Moreover,

$$\partial_1 F(y) = \frac{-\psi'_{\text{low}}(y_1) \mathcal{G}(y_1) - (y_2 - \psi_{\text{low}}(y_1)) \mathcal{G}'(y_1)}{\mathcal{G}(y_1)^2}, \quad \mathcal{G}'(y_1) = \psi'_{\text{up}}(y_1) - \psi'_{\text{low}}(y_1). \quad (8.10)$$

Using (5.2), we obtain, for $y \in A_{0,d}$,

$$|\partial_1 F(y)| \le C(R) \left( \frac{|y_1|}{\mathcal{G}(y_1)} + \frac{|y_1|^3}{\mathcal{G}(y_1)^3} \right). \quad (8.11)$$
Therefore, arguing as in the calculation leading to (5.3), we conclude that

$$ \|\partial_1 F\|_{L^2(A_{0,d})} \le C(R). \qquad (8.12) $$

Combining the above inequality with (8.9), we obtain

$$ \|\partial_1 \bar{v}_{d,2}\|_{L^2(A_{0,d})} \le C(R). $$

For the remaining part on $A_{\delta,d} \setminus A_{0,d}$, we proceed exactly as in the proof of Lemma 5.1: on this transition strip, $0 \le \mathcal{X} \le 1$ and the definition of $\bar{g}_d$ involves only smooth cut-off functions (namely $\mathcal{X}$ and $\varphi$), whose derivatives are uniformly bounded. Moreover, the gap thickness is uniformly bounded from below away from zero on this strip. Hence the corresponding contributions are uniformly bounded by $C(R)$. This completes the proof of the estimate (5.9).

### Step 2: Estimate of $\partial_2 \bar{v}_{d,1}$ (upper bound).

Recall that $\bar{v}_d = \nabla^\perp \bar{g}_d$, hence

$$ \bar{v}_{d,1} = -\partial_{y_2} \bar{g}_d. $$

Differentiating once more with respect to $y_2$ gives

$$ \partial_{y_2} \bar{v}_{d,1} = -\partial_{y_2 y_2} \bar{g}_d. $$

From the explicit formula of $\bar{g}_d$ in (4.14), we obtain the following expression:

$$ \partial_2 \bar{v}_{d,1}(y) = -y_1(1 - \mathcal{X}(y_1)) \partial_{22} \varphi(y) - 6y_1 \mathcal{X}(y_1) \left( \frac{1}{(\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1))^2} - \frac{2(y_2 - \psi_{\text{low}}(y_1))}{(\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1))^3} \right), \quad y \in A_{\delta,d}. \quad (8.13) $$

**Step 2.1: Pointwise bound.** We estimate each term in (8.13) separately. Note that on the core region $A_{0,d}$ we have $\mathcal{X} \equiv 1$, so the first term in (8.13) vanishes there; it only contributes on the transition strip.

(i) The cut-off part. Since $\mathcal{X} \in C^\infty(\mathbb{R})$ satisfies $0 \le \mathcal{X} \le 1$ and $\varphi \in C^\infty(\overline{B(0, R)})$ is fixed (independent of $d$), we have

$$ |\partial_{22} \varphi(y)| \le C(R) \quad \text{for all } y \in B(0, R). $$

Moreover, $|y_1| \le 2\delta$, hence

$$ |-y_1(1 - \mathcal{X}(y_1))\partial_{22}\varphi(y)| \le |y_1| \cdot 1 \cdot C(R) \le C(R). $$

(ii) The thin-gap part. Set for simplicity

$$ \mathcal{G}(y_1) := \psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1). $$

Then the second line in (8.13) becomes

$$ -6y_1 \mathcal{X}(y_1) \left( \frac{1}{\mathcal{G}(y_1)^2} - \frac{2(y_2 - \psi_{\text{low}}(y_1))}{\mathcal{G}(y_1)^3} \right). $$

Taking absolute values and using $0 \le \mathcal{X} \le 1$, we get

$$ \left| -6y_1 \mathcal{X}(y_1) \left( \frac{1}{\mathcal{G}(y_1)^2} - \frac{2(y_2 - \psi_{\text{low}}(y_1))}{\mathcal{G}(y_1)^3} \right) \right| \le 6|y_1| \left( \frac{1}{\mathcal{G}(y_1)^2} + \frac{2|y_2 - \psi_{\text{low}}(y_1)|}{\mathcal{G}(y_1)^3} \right). $$
Since $y \in A_{\delta,d}$ implies

$$ \psi_{\text{low}}(y_1) \le y_2 \le \psi_{\text{up}}(y_1) \implies 0 \le y_2 - \psi_{\text{low}}(y_1) \le \mathcal{G}(y_1), $$

we obtain

$$ \frac{|y_2 - \psi_{\text{low}}(y_1)|}{\mathcal{G}(y_1)^3} \le \frac{\mathcal{G}(y_1)}{\mathcal{G}(y_1)^3} = \frac{1}{\mathcal{G}(y_1)^2}. $$

Therefore,

$$ \left| -6y_1 \mathcal{X}(y_1) \left( \frac{1}{\mathcal{G}(y_1)^2} - \frac{2(y_2 - \psi_{\text{low}}(y_1))}{\mathcal{G}(y_1)^3} \right) \right| \le C(R) \frac{|y_1|}{\mathcal{G}(y_1)^2}. $$

**(iii) Conclusion of the pointwise bound.** Combining (i) and (ii), we obtain that for every $y \in A_{\delta,d}$,

$$ |\partial_2 \bar{v}_{d,1}(y)| \le C(R) \left( 1 + \frac{|y_1|}{\mathcal{G}(y_1)^2} \right). \qquad (8.14) $$

*Step 2.2: From the pointwise bound to an $L^2$ bound.* Squaring (8.14) and using $(a+b)^2 \le 2a^2+2b^2$, we get

$$ |\partial_2 \bar{v}_{d,1}(y)|^2 \le C(R) \left( 1 + \frac{y_1^2}{\mathcal{G}(y_1)^4} \right), \quad y \in A_{\delta,d}. $$

Integrating over $A_{\delta,d}$ gives

$$ \begin{aligned} \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})}^2 &= \int_{A_{\delta,d}} |\partial_2 \bar{v}_{d,1}(y)|^2 dy \\ &\le C(R) \int_{A_{\delta,d}} 1 dy + C(R) \int_{A_{\delta,d}} \frac{y_1^2}{\mathcal{G}(y_1)^4} dy. \end{aligned} \qquad (8.15) $$

The first integral is uniformly bounded because $A_{\delta,d} \subset B(0, R)$, hence

$$ \int_{A_{\delta,d}} 1 dy \le |B(0, R)| \le C(R). $$

We now estimate the second integral in (8.15). Using the definition of $A_{\delta,d}$, we apply Tonelli's theorem:

$$ \int_{A_{\delta,d}} \frac{y_1^2}{\mathcal{G}(y_1)^4} dy = \int_{-2\delta}^{2\delta} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{up}}(y_1)} \frac{y_1^2}{\mathcal{G}(y_1)^4} dy_2 dy_1. $$

Since the integrand does not depend on $y_2$, the inner integral equals the thickness $\psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1) = \mathcal{G}(y_1)$:

$$ \begin{aligned} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{up}}(y_1)} \frac{y_1^2}{\mathcal{G}(y_1)^4} dy_2 &= \frac{y_1^2}{\mathcal{G}(y_1)^4} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{up}}(y_1)} 1 dy_2 \\ &= \frac{y_1^2}{\mathcal{G}(y_1)^4} \mathcal{G}(y_1) = \frac{y_1^2}{\mathcal{G}(y_1)^3}. \end{aligned} $$

Hence

$$ \int_{A_{\delta,d}} \frac{y_1^2}{\mathcal{G}(y_1)^4} dy = \int_{-2\delta}^{2\delta} \frac{y_1^2}{\mathcal{G}(y_1)^3} dy_1. \qquad (8.16) $$

*Step 2.3: Using the gap lower bound.* By Lemma 4.1, there exists $c_R > 0$ such that for all $|y_1| \le 2\delta$,

$$ \mathcal{G}(y_1) \ge d + c_R y_1^2. $$
Therefore,

$$ \frac{1}{\mathcal{G}(y_1)^3} \le \frac{1}{(d + c_R y_1^2)^3}, $$

and hence (8.16) implies

$$ \int_{A_{\delta,d}} \frac{y_1^2}{\mathcal{G}(y_1)^4} dy \le \int_{-2\delta}^{2\delta} \frac{y_1^2}{(d + c_R y_1^2)^3} dy_1. \quad (8.17) $$

We now estimate the right-hand side by a scaling argument. Using symmetry, it is enough to consider $y_1 \ge 0$ and set $y_1 = \sqrt{d} z$:

$$ \begin{aligned} \int_0^{2\delta} \frac{y_1^2}{(d + c_R y_1^2)^3} dy_1 &= \int_0^{2\delta/\sqrt{d}} \frac{(dz^2)}{(d + c_R dz^2)^3} \sqrt{d} dz \\ &= \int_0^{2\delta/\sqrt{d}} \frac{d^{3/2} z^2}{d^3 (1 + c_R z^2)^3} dz = d^{-3/2} \int_0^{2\delta/\sqrt{d}} \frac{z^2}{(1 + c_R z^2)^3} dz. \end{aligned} $$

Since $\frac{z^2}{(1+c_R z^2)^3} \sim z^{-4}$ as $z \to \infty$, the integral $\int_0^{+\infty} \frac{z^2}{(1+c_R z^2)^3} dz$ is finite, and therefore

$$ \int_0^{2\delta/\sqrt{d}} \frac{z^2}{(1 + c_R z^2)^3} dz \le C(R). $$

We conclude that

$$ \int_{-2\delta}^{2\delta} \frac{y_1^2}{(d + c_R y_1^2)^3} dy_1 \le \frac{C(R)}{d^{3/2}}. \quad (8.18) $$

**Step 2.4: Conclusion.** Plugging (8.18) into (8.15), we obtain

$$ \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})}^2 \le C(R) + \frac{C(R)}{d^{3/2}}. $$

Taking the square root yields

$$ \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})} \le C(R) + \frac{C_2(R)}{d^{3/4}}, $$

which is the desired upper bound.

**Step 3: Estimate of $\partial_2 \bar{v}_{d,1}$ (lower bound).** Fix constants $0 < \eta_1 < \eta_2$ independent of $d$. In this step we work in the core region $A_{0,d}$, where the cut-off satisfies $\mathcal{X}(y_1) \equiv 1$ and the field $\bar{v}_d$ has the explicit thin-gap form. We assume $d > 0$ is small enough such that $\eta_2 \sqrt{d} \le \delta$. Then $y_1 \in [\eta_1 \sqrt{d}, \eta_2 \sqrt{d}] \subset [-\delta, \delta]$ and therefore $\mathcal{X}(y_1) \equiv 1$.

Recall that on $A_{0,d}$ we have

$$ \bar{g}_d(y) = y_1(3F(y)^2 - 2F(y)^3), \quad F(y) = \frac{y_2 - \psi_{\text{low}}(y_1)}{\mathcal{G}(y_1)}, \quad \mathcal{G}(y_1) = \psi_{\text{up}}(y_1) - \psi_{\text{low}}(y_1). $$

Moreover,

$$ \bar{v}_{d,1} = -\partial_2 \bar{g}_d. $$

Since $\partial_2 F = 1/\mathcal{G}(y_1)$ and $\partial_{22} F = 0$, we compute

$$ \partial_2 (3F^2 - 2F^3) = (6F - 6F^2) \partial_2 F = \frac{6F(1-F)}{\mathcal{G}(y_1)}, $$

and hence

$$ \bar{v}_{d,1}(y) = -y_1 \frac{6F(1-F)}{\mathcal{G}(y_1)}. $$
Differentiating once more with respect to $y_2$, we obtain

$$ \partial_2 \bar{v}_{d,1}(y) = -y_1 \partial_2 \left( \frac{6F(1-F)}{\mathcal{G}(y_1)} \right) = -y_1 \frac{6(1-2F)}{\mathcal{G}(y_1)^2}. $$

Therefore

$$ |\partial_2 \bar{v}_{d,1}(y)| = \frac{6|y_1| |1-2F(y)|}{\mathcal{G}(y_1)^2}. \qquad (8.19) $$

Choice of a subregion where $|1-2F|$ is uniformly positive. Define the subregion

$$ A := \{ (y_1, y_2) \in A_{\delta,d} : \eta_1 \sqrt{d} \le y_1 \le \eta_2 \sqrt{d}, \psi_{\text{low}}(y_1) \le y_2 \le \psi_{\text{low}}(y_1) + \frac{1}{3} \mathcal{G}(y_1) \}. \quad (8.20) $$

For every $y \in A$, we have

$$ 0 \le F(y) = \frac{y_2 - \psi_{\text{low}}(y_1)}{\mathcal{G}(y_1)} \le \frac{1}{3}, $$

hence $1 - 2F(y) \ge 1/3$. Plugging this into (8.19) yields

$$ |\partial_{y_2} \bar{v}_{d,1}(y)| \ge \frac{2y_1}{\mathcal{G}(y_1)^2}, \quad y \in A. \qquad (8.21) $$

L² lower bound. Using (8.21) and integrating first in $y_2$ over an interval of length $\mathcal{G}(y_1)/3$, we start from the definition of the L² norm on the set A:

$$ \| \partial_{y_2} \bar{v}_{d,1} \|_{L^2(A)}^2 = \int_A |\partial_{y_2} \bar{v}_{d,1}(y_1, y_2)|^2 dy_1 dy_2. $$

By the definition of A in (8.20), this integral can be written as an iterated integral:

$$ \| \partial_2 \bar{v}_{d,1} \|_{L^2(A)}^2 = \int_{\eta_1 \sqrt{d}}^{\eta_2 \sqrt{d}} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{low}}(y_1) + \mathcal{G}(y_1)/3} |\partial_2 \bar{v}_{d,1}(y_1, y_2)|^2 dy_2 dy_1. $$

Moreover, for every $(y_1, y_2) \in A$, we proved in (8.21) that

$$ |\partial_2 \bar{v}_{d,1}(y_1, y_2)| \ge \frac{2y_1}{\mathcal{G}(y_1)^2}. $$

Squaring both sides yields

$$ |\partial_2 \bar{v}_{d,1}(y_1, y_2)|^2 \ge \left( \frac{2y_1}{\mathcal{G}(y_1)^2} \right)^2 = \frac{4y_1^2}{\mathcal{G}(y_1)^4}, \quad (y_1, y_2) \in A. $$

Therefore, inserting this pointwise lower bound into the integral, we obtain

$$ \begin{aligned} \| \partial_2 \bar{v}_{d,1} \|_{L^2(A)}^2 &= \int_{\eta_1 \sqrt{d}}^{\eta_2 \sqrt{d}} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{low}}(y_1) + \mathcal{G}(y_1)/3} |\partial_2 \bar{v}_{d,1}(y_1, y_2)|^2 dy_2 dy_1 \\ &\ge \int_{\eta_1 \sqrt{d}}^{\eta_2 \sqrt{d}} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{low}}(y_1) + \mathcal{G}(y_1)/3} \frac{4y_1^2}{\mathcal{G}(y_1)^4} dy_2 dy_1. \end{aligned} $$

Now we compute the inner integral with respect to $y_2$. Since the integrand $\frac{4y_1^2}{\mathcal{G}(y_1)^4}$ does not depend on $y_2$, it factors out:

$$ \begin{aligned} & \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{low}}(y_1) + \mathcal{G}(y_1)/3} \frac{4y_1^2}{\mathcal{G}(y_1)^4} dy_2 = \frac{4y_1^2}{\mathcal{G}(y_1)^4} \int_{\psi_{\text{low}}(y_1)}^{\psi_{\text{low}}(y_1) + \mathcal{G}(y_1)/3} 1 dy_2 \\ &= \frac{4y_1^2}{\mathcal{G}(y_1)^4} \left( \psi_{\text{low}}(y_1) + \frac{\mathcal{G}(y_1)}{3} - \psi_{\text{low}}(y_1) \right) \\ &= \frac{4y_1^2}{\mathcal{G}(y_1)^4} \cdot \frac{\mathcal{G}(y_1)}{3} = \frac{4}{3} \frac{y_1^2}{\mathcal{G}(y_1)^3}. \end{aligned} $$
Plugging this back into the outer integral, we conclude

$$ \|\partial_{y_2} \bar{v}_{d,1}\|_{L^2(A)}^2 \ge \frac{4}{3} \int_{\eta_1 \sqrt{d}}^{\eta_2 \sqrt{d}} \frac{y_1^2}{\mathcal{G}(y_1)^3} dy_1. \quad (8.22) $$

Now we use the upper bound of Lemma 4.1, namely

$$ \mathcal{G}(y_1) \le d + \left(1 - \frac{1}{R}\right) y_1^2. $$

For $y_1 \in [\eta_1 \sqrt{d}, \eta_2 \sqrt{d}]$ we have $y_1^2 \le \eta_2^2 d$, hence

$$ \mathcal{G}(y_1) \le \left(1 + \left(1 - \frac{1}{R}\right) \eta_2^2\right) d =: C_0 d. $$

Moreover $y_1^2 \ge \eta_1^2 d$ on the same interval. Therefore (8.22) implies

$$ \|\partial_{y_2} \bar{v}_{d,1}\|_{L^2(A)}^2 \ge \frac{4}{3} \int_{\eta_1 \sqrt{d}}^{\eta_2 \sqrt{d}} \frac{\eta_1^2 d}{(C_0 d)^3} dy_1 = \frac{4}{3} \cdot \frac{\eta_1^2}{C_0^3} \cdot (\eta_2 - \eta_1) d^{-3/2}. $$

Taking the square root gives

$$ \|\partial_{y_2} \bar{v}_{d,1}\|_{L^2(A)} \ge C_1(R) d^{-3/4}, $$

for some constant $C_1(R) > 0$ (depending only on $R$ and the fixed $\eta_1, \eta_2$). Finally, since $A \subset A_{\delta,d}$, we obtain

$$ \|\partial_{y_2} \bar{v}_{d,1}\|_{L^2(A_{\delta,d})} \ge \|\partial_{y_2} \bar{v}_{d,1}\|_{L^2(A)} \ge C_1(R) d^{-3/4}. $$

This concludes the proof of the lower bound.

**Step 4: Conclusion.** Combining the upper and lower bounds, we obtain that there exist constants $C_1(R), C_2(R) > 0$ such that

$$ \frac{C_1(R)}{d^{3/4}} \le \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})} \le C(R) + \frac{C_2(R)}{d^{3/4}}. \quad (8.23) $$

Since $d \in (0, R - 1]$, the additive constant $C(R)$ can be absorbed into the singular term, and we may write

$$ \frac{C_1(R)}{d^{3/4}} \le \|\partial_2 \bar{v}_{d,1}\|_{L^2(A_{\delta,d})} \le \frac{C_2(R)}{d^{3/4}}. \quad (8.24) $$

This finishes the proof of inequality (5.10). □

**Acknowledgments.** This work was funded by the European Union (Horizon Europe MSCA project ModConFlex, grant number 101073558).

**Conflict of interest.** On behalf of all authors, the corresponding author states that there is no conflict of interest.

**Data availability.** Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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