[DRAFT] On a rigid body inside an incompressible viscous flow
I want to understand the behaviour of weak solutions to the Navier-Stokes system coupled with a rigid body, and the limits of solutions as the size of the rigid body vanishes.
- Introduction
- Boundary conditions on the rigid body.
- The two-dimensional Navier-Stokes with periodic boundary conditions
- The three-dimensional Navier-Stokes with periodic boundary conditions
- The three-dimensional axisymmetrical Navier-Stokes on a cylinder
Introduction
The spatial domain of the fluid-body system is denoted by $\mathcal{D}$. In the two-dimensional case we have $\mathcal{D} \subseteq \mathbb{R}^2$, and in the three-dimensional case we have $\mathcal{D} \subseteq \mathbb{R}^3$.
I will try to consider the following cases:
- The two-dimensional case $\mathcal{D} = \mathbb{R}^2$ with periodic boundary conditions.
- The three-dimensional case $\mathcal{D} = \mathbb{R}^3$ with periodic boundary conditions.
In sake of simplicity, we assume the body is a rigid disc or a rigid sphere with radius $\varepsilon > 0$, and constant density $\rho_\varepsilon$. For every $t \geq 0$, the center of mass of the body is denoted by $h(t)$, and we assume it starts centered at the origin, i.e., $h(0) = 0$.
The space occupied by the body at time $t \geq 0$ is denoted by $\mathcal{S}_ \varepsilon(t) \subset \mathcal{D}$, and the space filled by the fluid at time $t \geq 0$ is denoted by $\mathcal{F}_ \varepsilon(t)$. Hence, \begin{aligned} \mathcal{S}_ \varepsilon (t) &= B_\varepsilon(h(t)) \\[10pt] \mathcal{F}_ \varepsilon (t) &= \mathcal{D} \setminus B_\varepsilon(h(t)) \,.\end{aligned} where $$B_r( c ) = \left\{x : |x - c| \leq r \right\}\,.$$
Because the body is rigid, its position at any time $t \geq 0$ can be fully described by its center $h(t)$ and a rotation $R(t) \in SO(n)$, where $SO(n)$ is the group of rotations on $\mathbb{R}^n$, also called the special orthogonal group.
Let $x_0$ be a point in the rigid body at time $t=0$. Its position at $t > 0$ is described by $$x(t) = h(t) + R(t) x_0\,.$$ Its velocity is \begin{aligned}\dot x(t) &= \dot h(t) + \dot R(t) x_0 \\ &= \dot h(t) + \dot R(t) R(t)^{-1} (x(t) - h(t)) \\ &= \dot h(t) + \dot R(t) R(t)^{T} (x(t) - h(t)) \end{aligned}
Because $R(t)$ is an element of $SO(n)$ it satisfies $R(t)R(t)^T = I$, then
$$\begin{aligned}
\frac{d}{dt} R(t)R(t)^T &= \dot R(t) R(t)^T + R(t) \dot R(t)^T = 0 \\
&= \dot R(t) R(t)^T + \left(\dot R(t) R(t)^T\right)^{T} = 0\,.
\end{aligned}$$ Therefore, the matrix $\dot R(t)R(t)^T$ is a skew-symmetric matrix.
In the three-dimensional case the skew-symmetric matrix $\dot R(t) R(t)^T$ has three distinct values, then we can define $\omega(t)$ to be the vector satisfying $$\dot R(t) R(t)^T u = \omega(t) \times u \quad \forall u \in \mathbb{R}^3$$
In the two-dimensional case the matrix $\dot R(t) R(t)^T$ has only one distinct value and it acts on vectors $u = (x, y) \in \mathbb{R}^2$ as follows: $$\dot R(t) R(t)^T u = \omega(t) (y, -x) = \omega(t) u^\perp$$
The Cauchy stress tensor
In the body-fluid interface the body is subject to forces from the fluid, and, thanks to Newton’s third law, the fluid is subject to the same forces from the body in the opposite direction.
First, the usual description of the Navier-Stokes equations includes a pressure term. For the incompressible case ($\nabla \cdot u = 0$) the pressure term is not really a thermodynamical pressure, as it is in the compressible case. In any case, pressure is a quantity of force per unit of area as is the total stress tensor.
Consider a point $x$ in the interface at time $t \geq 0$, that is, $x \in \partial \mathcal{S}(t)$, the forces per unit area acting upon the body are described by the Cauchy stress tensor $\Sigma(t, x)$ acting on the unitary vector $\mathbf{n}$ normal to the interface, pointing outward the body. In our case we have $\mathbf{n} = \varepsilon^{-1} (x - h(t))$, hence the forces per unit area acting upon the body at the point $x$ is $\varepsilon^{-1}\Sigma(t, x) \cdot (x - h(t))\,.$ The stress tensor depends on the pressure and the viscous stress tensor, $$\Sigma = - p I + 2 \nu D(u) = - p I + \nu \left(\nabla u + \nabla u^T\right)$$
The total force extered on the body by the fluid is $$\varepsilon^{-1} \int_{\partial \mathcal{S}_ \varepsilon(t)} \Sigma(t, x) \cdot (x - h(t)) dx$$
The torque exerted on the body can be computed as well from the stress tensor $$\varepsilon^{-1} \int_{\partial \mathcal{S}_ \varepsilon(t)} (x - h(t)) \times \left( \Sigma(t, x) \cdot (x - h(t)) \right) dx$$
The velocity field
Since we are only interested in the incompressible case, we will take the density of the fluid to be $1$, and the density of the rigid body to be a constant depending on its radius $\rho_\varepsilon$. In a slight abuse of notation we will also set $\rho$ to be the function $$\rho(t, x) = \begin{cases}1 &\quad |x - h(t)| > \varepsilon \\[5pt] \rho_\varepsilon &\quad | x - h(t)| \leq \varepsilon \end{cases}$$
The function $u_f: \mathcal{F} \to \mathbb{R}^n$ denotes the velocity of the flow on its moving domain. In the other hand, the function $u_s$ denotes the velocity of the points on the rigid body.
A single function $u: \mathbb{R} \times \mathbb{R}^3 \to \mathbb{R}^3$ denotes a joint velocity of fluid and rigid body $$u(t, x) = \begin{cases}
u_f(t, x) & \text{if } |h(t) - x | > \varepsilon \\[5pt]
u_s(t, x) & \text{otherwise} \end{cases}$$
The advantage of this simplification is that $$ \frac12 \int_{\mathcal{D}} \rho(t, x) |u(t,x)|^2 dx = \frac12 \int_{\mathcal{F}(t)} |u_f(t, x)|^2 dx + \frac12 m_\varepsilon |\dot h(t)|^2 + \frac12 I_\varepsilon |\omega(t)|^2$$ where the last two terms are the linear kinetic energy and the angular kinetic energy of the rigid body. Here, we denote the mass of ther rigid body by $m_\varepsilon$ and its moment of rotational inertia by $I_\varepsilon$ $$ m_\varepsilon = \int_{B_\varepsilon} \rho(t, x) dx = \frac43 \pi \varepsilon^2 \rho_\varepsilon $$ and $$I_\varepsilon = \int_{B_\varepsilon} \rho(t, x) |x|^2 dx = \frac25 m_\varepsilon \varepsilon^2$$
The velocity of the flow is denoted by , this is a function defined for pairs $(t, x)$ as long as the point $x$ is not an element of the rigid body at time $t$, i.e. $$u_f(t, \cdot): \mathbb{R}^3 \setminus S(t) \to \mathbb{R}^3 \quad \forall t \geq 0$$
A vector field $u_s : \mathbb{R}^3 \to \mathbb{R}^3$
Let $u_f
If $x$ is a point of the rigid body at time $t$,
The velocity of the points on the rigid body is denoted by $u_s$, defined only over the points $S(t)$
For any point of the rigid body,
The velocity of the fluid $u_f$ is defined on the domain $\Omega(t) = \mathbb{R}^n \setminus S(t)$ where $S(t)$ is the location of the rigid body at time $t$.
Because the rigid body is disc in $\mathbb{R}^2$ and a sphere in $\mathbb{R}^3$ with uniform density its center of mass is the geometric center.
The intial conditions
The usual setup for weak solutions of the Navier-Stokes starts with initial velocity as a function $u : \mathcal{D} \to \mathbb{R}^3$ such that the fluid is incompressible $\nabla \cdot u = 0$ and such that the total kinetic energy is finite $$\int_\mathcal{D} satisfying an energy constraint and the incompressibility cons
Given $u0 \in L^2{loc}(\mathcal{D})$ and $\varepsilon > 0$, define the function $u^{(\varepsilon)}_0$
Boundary conditions on the rigid body.
For the viscous case we consider the no-penetration/no-slip conditions, otherwise we use no-penetration/slip conditions.
$$u_f(t, x) = u_s(t, x) \quad \forall x \in \partial \Omega_t$$
The two-dimensional Navier-Stokes with periodic boundary conditions
Functional spaces
Let $J^\infty(\mathbb{T}^2)$ be the space of all divergence free smooth vector fields with periodic boundary conditions, and let $J(\mathbb{T}^2)$ be the closure of $J^\infty(\mathbb{T}^2)$ in $L^2(\mathbb{T}^2)$ with the norm $$\|f\| = \left( \int_{\mathbb{T}^2} |f(x)|^2 dx \right)^\frac12$$
In this case we consider the system of equations \begin{aligned}\partial_t u + (u \cdot \nabla) u + \nabla p &= \Delta u \quad &\text{on } \left\{(t, x) : x \in Q_t\right\} \\[5pt] \nabla \cdot u &= 0 \quad &\text{on } \left\{(t, x) : x \in Q_t\right\} \end{aligned} together with the boundary condition on the interface of the disc $$ u(t, x) = \dot h(t) + \omega(t) (x - h(t))^\perp \quad \text{on } \left\{(t, x) : x \in \partial B_\varepsilon(h(t)) \right\}$$ the linear forces acting on the disc $$m_\varepsilon \ddot h(t) = - \int_{\partial B_\varepsilon(h(t))} \Sigma \mathbf{n} dS$$ where $\mathbf{n}$ is the vector normal to the surface of the disc pointing inward the disc.