2 A Lagrangian functional for the coupled fluid–body dynamics

Consider a three-dimensional (3-D) rigid body which contains an inviscid and incompressible fluid, with free surface, undergoing three-dimensional rotational and translational motions. Using the calculus of variations, the Euler–Poincaré equations for the dynamic coupling between the rigid body motion and its interior fluid sloshing can be derived from a Lagrangian action functional. The derivation of this action functional is given below.

For the study of dynamic coupling between the rigid body motion and its interior rotating and translating fluid, three frames of reference are used. One spatial frame and two body frames. The spatial frame, which is fixed in space, has coordinates denoted by $\boldsymbol{X}=(X,Y,Z)$ . The first body frame, which is placed at the centre of rotation of the moving body and used for the analysis of the rigid body motion, has coordinates denoted by $\boldsymbol{x}_{b}=(x_{b},y_{b},z_{b})$ . The second body frame, which is attached to the moving body and used for the analysis of the fluid motion inside the tank, has coordinates denoted by $\boldsymbol{x}=(x,y,z)$ . The distance between the origin of the body frame $\boldsymbol{x}$ to the point of rotation, i.e. the origin of the body frame $\boldsymbol{x}_{b}$ , is denoted by $\boldsymbol{d}=(d_{1},d_{2},d_{3})$ which is a constant vector. So the position of a fluid particle relative to the body frame $\boldsymbol{x}_{b}$ is $\boldsymbol{x}_{b}=\boldsymbol{x}+\boldsymbol{d}$ . The fluid–tank system has a uniform translation $\boldsymbol{q}(t)=(q_{1}(t),q_{2}(t),q_{3}(t))$ relative to the spatial frame $\boldsymbol{X}$ , which is the vector from the origin of the spatial frame $\boldsymbol{X}$ to the origin of the body frame $\boldsymbol{x}_{b}$ . The configuration of the fluid in a rotating–translating vessel is schematically shown in figure 1.

The Lagrangian action for the coupled fluid–vessel dynamics takes the form

(2.1) $$\begin{eqnarray}\mathscr{L}=\int _{t_{1}}^{t_{2}}\left(\text{KE}^{fluid}-\text{PE}^{fluid}+\text{KE}^{vessel}-\text{PE}^{vessel}\right)\,\text{d}t,\end{eqnarray}$$

where $\text{KE}^{fluid}$ is the kinetic energy of the fluid, $\text{KE}^{vessel}$ is the kinetic energy of the vessel, $\text{PE}^{fluid}$ is the potential energy of the fluid and $\text{PE}^{vessel}$ is the potential energy of the vessel. The expressions are derived below. For the kinetic and potential energies, the velocity vector and the displacement vector should be relative to the spatial frame $\boldsymbol{X}$ , but the analysis of the fluid motion is carried out in the body frame $\boldsymbol{x}$ . The relationship between the two velocities is developed using the kinetic theory of rigid bodies (Murray, Lin & Sastry Reference Murray, Lin and Sastry1994; O’Reilly Reference O’Reilly2008).

Figure 1. Schematic showing a configuration of the fixed coordinate system $\boldsymbol{X}=(X,Y,Z)$ relative to the moving coordinate systems $\boldsymbol{x}_{b}=(x_{b},y_{b},z_{b})$ and $\boldsymbol{x}=(x,y,z)$ , attached to the vessel. The distance between the origin of the spatial frame $\boldsymbol{X}$ and the point of rotation is denoted by $\boldsymbol{q}$ . The distance from the origin of the body frame $\boldsymbol{x}$ to the point of rotation is denoted by $\boldsymbol{d}$ .

Let $\unicode[STIX]{x1D64C}(t)\in \text{SO}(3)$ be a proper rotation in $\mathbb{R}^{3}$ ,

(2.2) $$\begin{eqnarray}\unicode[STIX]{x1D64C}^{\text{T}}\unicode[STIX]{x1D64C}=\unicode[STIX]{x1D644}\quad \text{and}\quad \text{det}(\unicode[STIX]{x1D64C})=1.\end{eqnarray}$$

Then the relation between the spatial displacement and the body displacement for a fluid particle is

(2.3) $$\begin{eqnarray}\boldsymbol{X}=\unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d})+\boldsymbol{q}.\end{eqnarray}$$

To clarify, we write $\boldsymbol{x}=(x(a,b,c,t),y(a,b,c,t),z(a,b,c,t))$ as Cartesian coordinates of a fluid particle marked by Lagrangian labels $\boldsymbol{a}=(a,b,c)$ at time $t$ . This formulation is consistent with the theory of rigid body motion, where an arbitrary motion can be described by the pair $(\unicode[STIX]{x1D64C}(t),\boldsymbol{q}(t))$ .

The body angular velocity is a time-dependent vector

(2.4) $$\begin{eqnarray}\unicode[STIX]{x1D734}(t)=(\unicode[STIX]{x1D6FA}_{1}(t),\unicode[STIX]{x1D6FA}_{2}(t),\unicode[STIX]{x1D6FA}_{3}(t)),\end{eqnarray}$$

relative to the body coordinate system $\boldsymbol{x}_{b}$ with entries determined from $\unicode[STIX]{x1D64C}$ by

(2.5) $$\begin{eqnarray}\unicode[STIX]{x1D64C}^{\text{T}}\dot{\unicode[STIX]{x1D64C}}=\left[\begin{array}{@{}ccc@{}}0 & -\unicode[STIX]{x1D6FA}_{3} & \unicode[STIX]{x1D6FA}_{2}\\ \unicode[STIX]{x1D6FA}_{3} & 0 & -\unicode[STIX]{x1D6FA}_{1}\\ -\unicode[STIX]{x1D6FA}_{2} & \unicode[STIX]{x1D6FA}_{1} & 0\end{array}\right]:=\widehat{\unicode[STIX]{x1D734}},\end{eqnarray}$$

where $\widehat{\unicode[STIX]{x1D734}}\in \mathfrak{so}(3)$ is the Lie algebra of the group of rotations of $\mathbb{R}^{3}$ , i.e. $\unicode[STIX]{x1D64C}(t)$ . The convention for the entries of the skew–symmetric matrix $\widehat{\unicode[STIX]{x1D734}}$ is such that

(2.6) $$\begin{eqnarray}\text{Hat map:}\quad \widehat{\unicode[STIX]{x1D734}}\boldsymbol{r}=\unicode[STIX]{x1D734}\times \boldsymbol{r},\quad \text{for any }\boldsymbol{r}\in \mathbb{R}^{3},\quad \unicode[STIX]{x1D734}:=(\unicode[STIX]{x1D6FA}_{1},\unicode[STIX]{x1D6FA}_{2},\unicode[STIX]{x1D6FA}_{3}).\end{eqnarray}$$

The body angular velocity is to be contrasted with the spatial angular velocity, the angular velocity viewed from the spatial frame, which is

(2.7) $$\begin{eqnarray}\widehat{\unicode[STIX]{x1D734}}^{spatial}:=\dot{\unicode[STIX]{x1D64C}}\unicode[STIX]{x1D64C}^{\text{T}}.\end{eqnarray}$$

As vectors the spatial and body angular velocities are related by $\unicode[STIX]{x1D734}^{spatial}=\unicode[STIX]{x1D64C}\unicode[STIX]{x1D734}$ . Either representation for the angular velocity can be used. But the body representation is the sensible choice leading to great simplification of the differential equations.

The relation between the body velocity and space velocity is

(2.8) $$\begin{eqnarray}\dot{\boldsymbol{X}}=\unicode[STIX]{x1D64C}(\dot{\boldsymbol{x}}+\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}).\end{eqnarray}$$

If $\boldsymbol{u}=\boldsymbol{u}(\boldsymbol{x},t)$ denotes the Eulerian velocity of a fluid particle relative to the body frame with $\boldsymbol{x}=\boldsymbol{x}(\boldsymbol{a},t)$ the corresponding flow map, the fluid particle initially at position $\boldsymbol{a}$ is at position $\boldsymbol{x}=\boldsymbol{x}(\boldsymbol{a},t)$ at time $t$ , then the Lagrangian velocity of the fluid particle is $\dot{\boldsymbol{x}}(\boldsymbol{a},t)=\boldsymbol{u}(\boldsymbol{x}(\boldsymbol{a},t),t)$ , and the Lagrangian acceleration of the fluid particle is $\ddot{\boldsymbol{x}}(\boldsymbol{a},t)=\text{D}\boldsymbol{u}/\text{D}t=\boldsymbol{u}_{t}+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\boldsymbol{u}$ . For an incompressible fluid the Jacobian $J$ of the label-to-particle mapping $(a,b,c)\rightarrow (x,y,z)$ is the motion invariant, i.e.  $\unicode[STIX]{x2202}J/\unicode[STIX]{x2202}t=0$ which is the continuity equation $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0$ in the Lagrangian particle-path formulation. The Lagrangian labels $(a,b,c)$ can be chosen such that

(2.9) $$\begin{eqnarray}J=\frac{\unicode[STIX]{x2202}(x,y,z)}{\unicode[STIX]{x2202}(a,b,c)}=x_{a}(y_{b}z_{c}-y_{c}z_{b})+x_{b}(y_{c}z_{a}-y_{a}z_{c})+x_{c}(y_{a}z_{b}-y_{b}z_{a})=1.\end{eqnarray}$$

This means that at an initial time $t_{0}$ the Lagrangian labels $(a,b,c)$ are physically possible coordinates, i.e.  $(a,b,c)=(x_{0},y_{0},z_{0})$ . So in Eulerian coordinates (2.8) takes the form

(2.10) $$\begin{eqnarray}\boldsymbol{U}=\unicode[STIX]{x1D64C}(\boldsymbol{u}+\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}),\end{eqnarray}$$

where $\boldsymbol{U}(\boldsymbol{X},t)$ is the Eulerian velocity of a fluid particle in the spatial frame $\boldsymbol{X}$ . Using this expression the kinetic energy of the fluid is

(2.11) $$\begin{eqnarray}\displaystyle \text{KE}^{fluid} & = & \displaystyle \displaystyle \int _{{\mathcal{V}}}\frac{1}{2}\left\Vert \boldsymbol{U}\right\Vert ^{2}\unicode[STIX]{x1D70C}\,\text{d}{\mathcal{V}}\nonumber\\ \displaystyle & = & \displaystyle \displaystyle \int _{{\mathcal{V}}}\left(\frac{1}{2}\left\Vert \boldsymbol{u}\right\Vert ^{2}+\boldsymbol{u}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\right)+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\frac{1}{2}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}+\frac{1}{2}\left\Vert \unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})\right\Vert ^{2}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x},\end{eqnarray}$$

where $\unicode[STIX]{x1D70C}$ is the density of the fluid which is considered to be inviscid and incompressible, and the integral is over the volume ( ${\mathcal{V}}$ ) of the fluid inside the container. This expression can be simplified using the definition of the fluid mass moment of inertia, noting that

(2.12) $$\begin{eqnarray}\displaystyle \int _{{\mathcal{V}}}\frac{1}{2}\left\Vert \unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})\right\Vert ^{2}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x} & = & \displaystyle \int _{{\mathcal{V}}}\frac{1}{2}(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\nonumber\\ \displaystyle & = & \displaystyle \int _{{\mathcal{V}}}\frac{1}{2}((\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D734})((\boldsymbol{x}+\boldsymbol{d})\boldsymbol{\cdot }(\boldsymbol{x}+\boldsymbol{d}))-(\unicode[STIX]{x1D734}\boldsymbol{\cdot }(\boldsymbol{x}+\boldsymbol{d}))^{2})\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\nonumber\\ \displaystyle & = & \displaystyle \displaystyle \unicode[STIX]{x1D734}\boldsymbol{\cdot }\left(\left(\frac{1}{2}\int _{{\mathcal{V}}}\left(\left\Vert \boldsymbol{x}+\boldsymbol{d}\right\Vert ^{2}\unicode[STIX]{x1D644}-(\boldsymbol{x}+\boldsymbol{d})\otimes (\boldsymbol{x}+\boldsymbol{d})\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\right)\unicode[STIX]{x1D734}\right)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734},\end{eqnarray}$$

with

(2.13) $$\begin{eqnarray}\unicode[STIX]{x1D644}_{f}=\int _{{\mathcal{V}}}(\Vert \boldsymbol{x}+\boldsymbol{d}\Vert ^{2}\unicode[STIX]{x1D644}-(\boldsymbol{x}+\boldsymbol{d})\otimes (\boldsymbol{x}+\boldsymbol{d}))\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x},\end{eqnarray}$$

where $\otimes$ denotes the tensor product, $\unicode[STIX]{x1D644}$ is the $3\times 3$ identity matrix and $\unicode[STIX]{x1D644}_{f}$ is the mass moment of inertia of the fluid relative to the point of rotation, i.e. the origin of the body frame $\boldsymbol{x}_{b}$ . So the kinetic energy of the fluid can be written as

(2.14) $$\begin{eqnarray}\displaystyle \text{KE}^{fluid} & = & \displaystyle \displaystyle \int _{{\mathcal{V}}}\left(\frac{1}{2}\left\Vert \boldsymbol{u}\right\Vert ^{2}+\boldsymbol{u}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))+\frac{1}{2}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}.\end{eqnarray}$$

The potential energy of the fluid in (2.1) is

(2.15) $$\begin{eqnarray}\text{PE}^{fluid}=\displaystyle \int _{{\mathcal{V}}}\unicode[STIX]{x1D70C}g\left(\unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d})+\boldsymbol{q}\right)\boldsymbol{\cdot }\hat{\boldsymbol{z}}\,\text{d}\boldsymbol{x},\end{eqnarray}$$

where $\hat{\boldsymbol{z}}$ is the unit vector in the $Z$ direction, and $g$ is the acceleration due to gravity.

To derive the kinetic energy of the dry vessel note that the relation between the body velocity and the space velocity is

(2.16a,b ) $$\begin{eqnarray}\dot{\boldsymbol{X}}=\dot{\unicode[STIX]{x1D64C}}\boldsymbol{x}_{b}+\dot{\boldsymbol{q}}\quad \text{or}\quad \unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{X}}=(\unicode[STIX]{x1D734}\times \boldsymbol{x}_{b})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}.\end{eqnarray}$$

Then the kinetic energy of the vessel is

(2.17) $$\begin{eqnarray}\displaystyle \text{KE}^{vessel} & = & \displaystyle \displaystyle \int _{\mathbb{V}}\frac{1}{2}\left\Vert \dot{\boldsymbol{X}}\right\Vert ^{2}\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V}=\displaystyle \int _{\mathbb{V}}\frac{1}{2}\left\Vert \unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{X}}\right\Vert ^{2}\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V}\nonumber\\ \displaystyle & = & \displaystyle \displaystyle \int _{\mathbb{V}}\left(\frac{1}{2}\left\Vert \unicode[STIX]{x1D734}\times \boldsymbol{x}_{b}\right\Vert ^{2}+\left(\unicode[STIX]{x1D734}\times \boldsymbol{x}_{b}\right)\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}+\frac{1}{2}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}\right)\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V},\end{eqnarray}$$

where $\unicode[STIX]{x1D70C}_{v}$ is the density of the vessel and the integral is over the volume $\mathbb{V}$ of the vessel. Taking into account that

(2.18) $$\begin{eqnarray}\displaystyle & & \displaystyle \displaystyle \frac{1}{2}\int _{\mathbb{V}}\left\Vert \unicode[STIX]{x1D734}\times \boldsymbol{x}_{b}\right\Vert ^{2}\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V}=\frac{1}{2}\int _{\mathbb{V}}\left(\left(\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D734}\right)\left(\boldsymbol{x}_{b}\boldsymbol{\cdot }\boldsymbol{x}_{b}\right)-\left(\unicode[STIX]{x1D734}\boldsymbol{\cdot }\boldsymbol{x}_{b}\right)^{2}\right)\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V}\nonumber\\ \displaystyle & & \displaystyle \qquad =\unicode[STIX]{x1D734}\boldsymbol{\cdot }\left(\left(\frac{1}{2}\int _{\mathbb{V}}\left(\Vert \boldsymbol{x}_{b}\Vert ^{2}\unicode[STIX]{x1D644}-\boldsymbol{x}_{b}\otimes \boldsymbol{x}_{b}\right)\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V}\right)\unicode[STIX]{x1D734}\right)=\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734},\end{eqnarray}$$

with

(2.19) $$\begin{eqnarray}\unicode[STIX]{x1D644}_{v}=\int _{\mathbb{V}}(\Vert \boldsymbol{x}_{b}\Vert ^{2}\unicode[STIX]{x1D644}-\boldsymbol{x}_{b}\otimes \boldsymbol{x}_{b})\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V},\end{eqnarray}$$

and that

(2.20) $$\begin{eqnarray}\displaystyle \int _{\mathbb{V}}(\unicode[STIX]{x1D734}\times \boldsymbol{x}_{b})\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\unicode[STIX]{x1D70C}_{v}\,\text{d}\mathbb{V} & = & \displaystyle (\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v})\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}},\end{eqnarray}$$

then the kinetic energy of the vessel simplifies to

(2.21) $$\begin{eqnarray}\text{KE}^{vessel}={\textstyle \frac{1}{2}}m_{v}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}+\left(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v}\right)\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734},\end{eqnarray}$$

where $\unicode[STIX]{x1D644}_{v}$ is the mass moment of inertia of the dry vessel relative to the point of rotation, $m_{v}$ is the mass of the dry vessel and $\overline{\boldsymbol{x}}_{v}=(\overline{x}_{v},\overline{y}_{v},\overline{z}_{v})$ is the centre of mass of the dry vessel relative to the body frame $\boldsymbol{x}_{b}$ . The potential energy of the dry vessel in (2.1) is

(2.22) $$\begin{eqnarray}\text{PE}^{vessel}=\displaystyle m_{v}g(\unicode[STIX]{x1D64C}\overline{\boldsymbol{x}}_{v}+\boldsymbol{q})\boldsymbol{\cdot }\hat{\boldsymbol{z}}.\end{eqnarray}$$

Now substitution of (2.14), (2.15), (2.21) and (2.22) into the action integral (2.1) gives an explicit form of the Lagrangian functional for the dynamics of a rigid body coupled to its interior fluid motion

(2.23) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathscr{L}\left(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{u}\right)=\int _{t_{1}}^{t_{2}}\left(\int _{{\mathcal{V}}}\left(\frac{1}{2}\left\Vert \boldsymbol{u}\right\Vert ^{2}+\boldsymbol{u}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\left.\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))+\frac{1}{2}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}-g\left(\unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d})+\boldsymbol{q}\right)\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\frac{1}{2}m_{v}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}+\left(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v}\right)\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}-m_{v}g\left(\unicode[STIX]{x1D64C}\overline{\boldsymbol{x}}_{v}+\boldsymbol{q}\right)\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\,\text{d}t.\end{eqnarray}$$

Transformation of the Lagrangian functional (2.23) from the Eulerian setting to the Lagrangian particle-path setting gives

(2.24) $$\begin{eqnarray}\displaystyle & & \displaystyle \displaystyle \mathscr{L}\left(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}}\right)=\displaystyle \int _{t_{1}}^{t_{2}}\left(\int _{{\mathcal{V}}^{\prime }}\left(\frac{1}{2}\left\Vert \dot{\boldsymbol{x}}\right\Vert ^{2}+\dot{\boldsymbol{x}}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\left.\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))+\frac{1}{2}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}-g\left(\unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d})+\boldsymbol{q}\right)\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\frac{1}{2}m_{v}\left\Vert \dot{\boldsymbol{q}}\right\Vert ^{2}+\left(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v}\right)\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}-m_{v}g\left(\unicode[STIX]{x1D64C}\overline{\boldsymbol{x}}_{v}+\boldsymbol{q}\right)\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\,\text{d}t,\end{eqnarray}$$

where the integral is over the volume ${\mathcal{V}}^{\prime }$ of the reference space, and

(2.25) $$\begin{eqnarray}\unicode[STIX]{x1D644}_{f}=\int _{{\mathcal{V}}^{\prime }}(\Vert \boldsymbol{x}+\boldsymbol{d}\Vert ^{2}\unicode[STIX]{x1D644}-(\boldsymbol{x}+\boldsymbol{d})\otimes (\boldsymbol{x}+\boldsymbol{d}))\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}.\end{eqnarray}$$

Taking the first variations of the Lagrangian functional (2.24) with respect to $\unicode[STIX]{x1D734}$ , $\unicode[STIX]{x1D64C}$ , $\boldsymbol{q}$ and $\dot{\boldsymbol{q}}$ yields the Euler–Poincaré equations for the angular momentum and linear momentum of the rigid body containing fluid. Moreover, taking the first variation of the action integral (2.24) with respect to $\boldsymbol{x}$ and $\dot{\boldsymbol{x}}$ , after the addition of a constraint term to this functional to enforce incompressibility of the fluid (see § 4), gives the Euler–Poincaré equation for the motion of the interior fluid relative to the body frame $\boldsymbol{x}$ . In the action integrals (2.23) and (2.24), $\boldsymbol{q}(t)$ is relative to the spatial frame $\boldsymbol{X}$ .

Using a similar approach, the Lagrangian functional (2.23) can be derived for the case where $\boldsymbol{q}(t)$ is relative to the body frame $\boldsymbol{x}_{b}$ , i.e. $\boldsymbol{q}_{b}(t)=\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q}$ . The action integral for this case reads

(2.26) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathscr{L}\left(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q}_{b},\dot{\boldsymbol{q}}_{b},\boldsymbol{u}\right)=\int _{t_{1}}^{t_{2}}\left(\int _{{\mathcal{V}}}\left(\frac{1}{2}\Vert \boldsymbol{u}+\dot{\boldsymbol{q}}_{b}\Vert ^{2}+(\boldsymbol{u}+\dot{\boldsymbol{q}}_{b})\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}+\boldsymbol{q}_{b})\right)\right.\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\left.(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times \boldsymbol{q}_{b})+\frac{1}{2}\left\Vert \unicode[STIX]{x1D734}\times \boldsymbol{q}_{b}\right\Vert ^{2}-g\unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d}+\boldsymbol{q}_{b})\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}+\frac{1}{2}m_{v}\Vert \dot{\boldsymbol{q}}_{b}\Vert ^{2}+\dot{\boldsymbol{q}}_{b}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times m_{v}\left(\overline{\boldsymbol{x}}_{v}+\boldsymbol{q}_{b}\right)\right)+\frac{1}{2}m_{v}\Vert \unicode[STIX]{x1D734}\times \boldsymbol{q}_{b}\Vert ^{2}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left.\left(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v}\right)\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times \boldsymbol{q}_{b}\right)+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}-m_{v}g\unicode[STIX]{x1D64C}\left(\overline{\boldsymbol{x}}_{v}+\boldsymbol{q}_{b}\right)\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\,\text{d}t,\end{eqnarray}$$

where $\unicode[STIX]{x1D734}$ and $\boldsymbol{q}_{b}$ are both relative to the body frame $\boldsymbol{x}_{b}$ . Similarly, to obtain the equations of motion for the interior fluid, a constraint term should be added to the functional (2.26) to enforce incompressibility of the fluid (see § 4).

In the Lagrangian action (2.24) the rotation tensor $\unicode[STIX]{x1D64C}(t)$ can be parameterised using the 3–2–1 Euler angles in three dimensions (O’Reilly Reference O’Reilly2008). This gives an expression for the body angular velocity $\widehat{\unicode[STIX]{x1D734}}=\unicode[STIX]{x1D64C}^{\text{T}}\dot{\unicode[STIX]{x1D64C}}$ . After substituting for $\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D64C}$ , in terms of the Euler angles and their derivatives, into the Lagrangian action, the Euler–Lagrange equations for the rotational and translational motion of the rigid body, containing fluid, can be provided by Hamilton’s principle $\unicode[STIX]{x1D6FF}\mathscr{L}\left(\unicode[STIX]{x1D64C},\dot{\unicode[STIX]{x1D64C}},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}}\right)=0$ . But the problem in using the Euler angles is that the resulting Euler–Lagrange equations are unwieldy to work with directly to determine the motion of the rigid body as the differential equations become very lengthy. However, we can follow an alternative path, taking variations of the action functional (2.24) with respect to $\unicode[STIX]{x1D734}$ , $\unicode[STIX]{x1D64C}$ , $\boldsymbol{q}$ , $\dot{\boldsymbol{q}}$ , applying similar calculus of variations used by Holm et al. (Reference Holm, Schmah and Stoica2009) for the Euler–Poincaré reduction of the non-free rigid body motion. See chapter $13$ of Marsden & Ratiu (Reference Marsden and Ratiu1999), and chapters $7$ and $11$ of Holm et al. (Reference Holm, Schmah and Stoica2009) for the background history and mathematics on the Euler–Poincaré reduction framework in Lagrangian mechanics. So our strategy is to adapt the Euler–Poincaré framework, first developed for rigid body dynamics, in order to derive the exact differential equations for the dynamic coupling between a rigid body and its interior fluid sloshing in three-dimensional rotating and translating coordinates.

In the presented derivations in this paper, the calculus of variations for the Lagrangian functional of the coupled (interior fluid $+$ body) system (2.24) is done in the Lagrangian particle-path setting using the Euler–Poincaré framework, and then the equations of motion are transformed to Eulerian coordinates. Alternatively, we could take the variations of the Lagrangian functional (2.23) in Eulerian coordinates, after the addition of a constraint term to this functional to enforce incompressibility of the fluid, by using the so-called Lin constraint (Bretherton Reference Bretherton1970; Oliver Reference Oliver2006) for the variations $\unicode[STIX]{x1D6FF}\boldsymbol{u}$ of the form

(2.27) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\boldsymbol{u}=\dot{\boldsymbol{w}}+\unicode[STIX]{x1D735}\boldsymbol{w}\boldsymbol{u}-\unicode[STIX]{x1D735}\boldsymbol{u}\boldsymbol{w},\end{eqnarray}$$

where $\boldsymbol{w}$ is a vector-valued free variation.

3 The Euler–Poincaré equations for the rigid body motion

The Euler–Poincaré reduction theorem for the motion of a free rigid body and a heavy top is given by Holm et al. (Reference Holm, Marsden and Ratiu1998a ). Here, we apply this framework to dynamic coupling between rigid body motion and its interior inviscid and incompressible fluid motion in the Lagrangian particle-path formulation. In this section the detailed derivations of the Euler–Poincaré equations for the body angular velocity $\unicode[STIX]{x1D734}(t)$ and translational motion $\boldsymbol{q}(t)$ of the rigid body are given. The aim is to recover equations (1.1) and (1.2), which are in Eulerian coordinates, from a variational principle with precise definitions of dependent variables with respect to the spatial and body coordinate systems.

3.1 The Euler–Poincaré equation for $\unicode[STIX]{x1D734}(t)$

The equation of motion for the body angular velocity $\unicode[STIX]{x1D734}(t)$ is provided by Hamilton’s variational principle:

(3.1) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\mathscr{L}(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}})=0,\end{eqnarray}$$

where the Lagrangian action $\mathscr{L}\left(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}}\right)$ is defined in (2.24) and the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ are taken among paths $\unicode[STIX]{x1D64C}(t)\in \text{SO}(3)$ , $t\in [t_{1},t_{2}]$ , with fixed endpoints, so that $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}(t_{1})=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}(t_{2})=\mathbf{0}$ . The variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ are induced by the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ via (Holm et al. Reference Holm, Schmah and Stoica2009)

(3.2) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\widehat{\unicode[STIX]{x1D734}}=\frac{\text{d}\widehat{\unicode[STIX]{x1D71E}}}{\text{d}t}+[\widehat{\unicode[STIX]{x1D734}},\widehat{\unicode[STIX]{x1D71E}}]=\frac{\text{d}\widehat{\unicode[STIX]{x1D71E}}}{\text{d}t}+\widehat{\unicode[STIX]{x1D734}}\widehat{\unicode[STIX]{x1D71E}}-\widehat{\unicode[STIX]{x1D71E}}\widehat{\unicode[STIX]{x1D734}},\quad\end{eqnarray}$$

where $\widehat{\unicode[STIX]{x1D71E}}\in \mathfrak{so}(3)$ is defined by

(3.3) $$\begin{eqnarray}\widehat{\unicode[STIX]{x1D71E}}=\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}.\end{eqnarray}$$

Since $[\widehat{\unicode[STIX]{x1D734}},\widehat{\unicode[STIX]{x1D71E}}]=\widehat{\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E}}$ , the equivalent vector representation of (3.2) is

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}=\dot{\unicode[STIX]{x1D71E}}+\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E}.\end{eqnarray}$$

Also it can be proved that (Marsden & Ratiu Reference Marsden and Ratiu1999; Holm et al. Reference Holm, Schmah and Stoica2009)

(3.5a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}^{-1}=-\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}\unicode[STIX]{x1D64C}^{-1}\quad \text{and}\quad \frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D64C}^{-1}\right)=-\unicode[STIX]{x1D64C}^{-1}\dot{\unicode[STIX]{x1D64C}}\unicode[STIX]{x1D64C}^{-1}.\end{eqnarray}$$

Now the Euler–Poincaré equation for $\unicode[STIX]{x1D734}(t)$ can be obtained by taking the first variation of the action integral $\mathscr{L}(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}})$ in (2.24) with respect to $\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D64C}$ . Due to lengthy derivations, here we calculate the first variation of each term in (2.24) with respect to $\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D64C}$ separately. For the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ of the second term in the Lagrangian action (2.24), assuming that $\dot{\boldsymbol{q}}$ , $\boldsymbol{x}$ and $\dot{\boldsymbol{x}}$ are constants, we have

(3.6) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\langle \dot{\boldsymbol{x}},\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\langle \dot{\boldsymbol{x}},\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \dot{\boldsymbol{x}},\underbrace{\left(\dot{\unicode[STIX]{x1D71E}}+\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E}\right)}_{=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d})\underbrace{-\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}\unicode[STIX]{x1D64C}^{-1}}_{=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}^{-1}}\dot{\boldsymbol{q}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \dot{\boldsymbol{x}},-(\boldsymbol{x}+\boldsymbol{d})\times \dot{\unicode[STIX]{x1D71E}}-(\boldsymbol{x}+\boldsymbol{d})\times \left(\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E}\right)\underbrace{-\widehat{\unicode[STIX]{x1D71E}}\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}}_{\text{using (3.3)}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \dot{\boldsymbol{x}},-(\boldsymbol{x}+\boldsymbol{d})\times \dot{\unicode[STIX]{x1D71E}}-(\boldsymbol{x}+\boldsymbol{d})\times (\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E})\underbrace{+\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D71E}}_{\substack{ \text{using the} \\ \text{hat map}}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left(\langle \dot{\unicode[STIX]{x1D71E}},-\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\rangle +\langle \unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E},-\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\rangle \right.\nonumber\\ \displaystyle & & \displaystyle \qquad \quad \left.+\,\langle \unicode[STIX]{x1D71E},\dot{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\rangle \right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\rightarrow \left\{\begin{array}{@{}l@{}}\text{using the vector identity}\quad \\ \boldsymbol{a}\boldsymbol{\cdot }(\boldsymbol{b}\times \boldsymbol{ c})=\boldsymbol{ c}\boldsymbol{\cdot }\left(\boldsymbol{a}\times \boldsymbol{ b}\right)\quad \end{array}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{{\mathcal{V}}^{\prime }}\left(\left.\langle \unicode[STIX]{x1D71E},-\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\rangle \right|_{t_{2}}-\left.\langle \unicode[STIX]{x1D71E},-\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\rangle \right|_{t_{1}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left(\left\langle \unicode[STIX]{x1D71E},\frac{\text{d}}{\text{d}t}(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d}))\right\rangle +\langle \unicode[STIX]{x1D71E},((\boldsymbol{x}+\boldsymbol{d})\times \dot{\boldsymbol{x}})\times \unicode[STIX]{x1D734}\rangle \right.\nonumber\\ \displaystyle & & \displaystyle \qquad \quad \left.+\,\langle \unicode[STIX]{x1D71E},\dot{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\rangle \right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\quad \rightarrow \text{(integrating by parts)}\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D71E},\frac{\text{d}}{\text{d}t}\left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\unicode[STIX]{x1D734}\times \left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\dot{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t,\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where the terms $\langle \unicode[STIX]{x1D71E},-\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\rangle |_{t_{1,2}}$ vanish if we apply the inverse of the hat map to $\widehat{\unicode[STIX]{x1D71E}}=\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ , noting that $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ vanishes at the endpoints $t_{1}$ and $t_{2}$ .

Now taking the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ of the third term in (2.24) using (3.3), (3.4) and (3.5a ), assuming that $\dot{\boldsymbol{q}}$ and $\boldsymbol{x}$ are constants, gives

(3.7) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\langle \unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}},\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D71E},\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+(\boldsymbol{x}+\boldsymbol{d})\times \left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D734}\right)\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t.\qquad\end{eqnarray}$$

See appendix A for the proof of (3.7). To take the variation of the potential energy of the fluid in (2.24) with respect to $\unicode[STIX]{x1D64C}$ , assuming that $\boldsymbol{q}$ and $\boldsymbol{x}$ are constants, set

(3.8) $$\begin{eqnarray}\unicode[STIX]{x1D72E}=\unicode[STIX]{x1D64C}^{-1}\hat{\boldsymbol{z}},\end{eqnarray}$$

which gives

(3.9) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D72E}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}^{-1}\hat{\boldsymbol{z}}=-\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}\unicode[STIX]{x1D64C}^{-1}\hat{\boldsymbol{z}}=-\widehat{\unicode[STIX]{x1D71E}}\unicode[STIX]{x1D72E}=\unicode[STIX]{x1D72E}\times \unicode[STIX]{x1D71E},\end{eqnarray}$$

and hence

(3.10) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}-g\underbrace{\langle \unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d})+\boldsymbol{q},\hat{\boldsymbol{z}}\rangle }_{\times \unicode[STIX]{x1D64C}^{-1}}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}-g\left\langle (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q},\underbrace{\unicode[STIX]{x1D64C}^{-1}\hat{\boldsymbol{z}}}_{=\unicode[STIX]{x1D72E}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}-g\left(\langle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q},\unicode[STIX]{x1D72E}\rangle +\langle (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q},\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D72E}\rangle \right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}-g\left\langle \underbrace{-\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}\unicode[STIX]{x1D64C}^{-1}}_{=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}^{-1}}\boldsymbol{q},\unicode[STIX]{x1D72E}\right\rangle -g\left\langle (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q},\underbrace{\unicode[STIX]{x1D72E}\times \unicode[STIX]{x1D71E}}_{=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D72E}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}-g\left\langle \underbrace{-\widehat{\unicode[STIX]{x1D71E}}\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q}}_{\text{using (3.3)}},\unicode[STIX]{x1D72E}\right\rangle -g\langle (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q},\unicode[STIX]{x1D72E}\times \unicode[STIX]{x1D71E}\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}-g\left\langle \underbrace{\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q}\times \unicode[STIX]{x1D71E}}_{\substack{ \text{using the} \\ \text{hat map}}},\unicode[STIX]{x1D72E}\right\rangle -g\langle (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q},\unicode[STIX]{x1D72E}\times \unicode[STIX]{x1D71E}\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\underbrace{-g(\langle \unicode[STIX]{x1D71E},\unicode[STIX]{x1D72E}\times \unicode[STIX]{x1D64C}^{-1}\boldsymbol{q}\rangle +\langle \unicode[STIX]{x1D71E},((\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q})\times \unicode[STIX]{x1D72E}\rangle )}_{{\hookrightarrow}\left\{\substack{ \text{using the vector identity} \\ \boldsymbol{a}\boldsymbol{\cdot }(\boldsymbol{b}\times \boldsymbol{c})=\boldsymbol{c}\boldsymbol{\cdot }(\boldsymbol{a}\times \boldsymbol{b})}\right.}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\langle \unicode[STIX]{x1D71E},-g(\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D72E}\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t.\end{eqnarray}$$

Taking the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ of the mass moment of inertia of the fluid in (2.24), assuming that $\boldsymbol{x}$ is constant, gives

(3.11) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\left\langle \frac{1}{2}\unicode[STIX]{x1D734},\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right\rangle \,\text{d}t=\int _{t_{1}}^{t_{2}}\langle \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734},\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\rangle \,\text{d}t\rightarrow (\text{noting that }\unicode[STIX]{x1D644}_{f}\text{ is symmetric})\nonumber\\ \displaystyle & & \displaystyle \qquad =\int _{t_{1}}^{t_{2}}\langle \dot{\unicode[STIX]{x1D71E}}+(\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D71E}),\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\rangle \,\text{d}t=\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right)+\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}\right\rangle \,\text{d}t,\qquad\end{eqnarray}$$

where, when integrating by parts, we used the condition that the variations vanish at the endpoints in time.

Similar calculations show that taking the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ of the remaining terms in (2.24) due to the kinetic and potential energies of the vessel, assuming that $\boldsymbol{q}$ and $\dot{\boldsymbol{q}}$ are constants, gives

(3.12) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v},\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\right\rangle \,\text{d}t=\displaystyle \int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-\frac{\text{d}}{\text{d}t}\left(m_{v}\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\right)\right\rangle \,\text{d}t\\ \displaystyle \hspace{142.26378pt}+\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},m_{v}\overline{\boldsymbol{x}}_{v}\times \left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D734}\right)\right\rangle \,\text{d}t,\\ \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\left\langle \frac{1}{2}\unicode[STIX]{x1D734},\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}\right\rangle \,\text{d}t=\displaystyle \int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-\unicode[STIX]{x1D644}_{v}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}\right\rangle \,\text{d}t,\\ \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\left\langle -m_{v}g\left(\unicode[STIX]{x1D64C}\overline{\boldsymbol{x}}_{v}+\boldsymbol{q}\right),\hat{\boldsymbol{z}}\right\rangle \,\text{d}t=\displaystyle \int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-m_{v}g\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D72E}\right\rangle \,\text{d}t.\end{array}\right\}\end{eqnarray}$$

Now from (3.6), (3.7), (3.10), (3.11) and (3.12) it can be concluded that Hamilton’s variational principle (3.1) for the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ and $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ reads

(3.13) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D71E},\frac{\text{d}}{\text{d}t}\left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\unicode[STIX]{x1D734}\times \left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\dot{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D71E},\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+(\boldsymbol{x}+\boldsymbol{d})\times \left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D734}\right)\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D71E},-g(\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D72E}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t+\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right)+\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}\right\rangle \,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-\frac{\text{d}}{\text{d}t}\left(m_{v}\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\right)+m_{v}\overline{\boldsymbol{x}}_{v}\times \left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D734}\right)\right\rangle \,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-\unicode[STIX]{x1D644}_{v}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}\right\rangle \,\text{d}t+\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-m_{v}g\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D72E}\right\rangle \,\text{d}t=\mathbf{0}.\end{eqnarray}$$

Therefore, since the variational principle (3.13) holds for any curve $\unicode[STIX]{x1D71E}(t)$ in $\mathfrak{so}(3)$ such that $\unicode[STIX]{x1D71E}(t_{1})=\unicode[STIX]{x1D71E}(t_{2})=\mathbf{0}$ , we find that the body angular velocity of the rigid body containing fluid is governed by the equation:

(3.14) $$\begin{eqnarray}\displaystyle & & \displaystyle \displaystyle \int _{{\mathcal{V}}^{\prime }}\left(\frac{\text{d}}{\text{d}t}\left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\unicode[STIX]{x1D734}\times \left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\dot{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{{\mathcal{V}}^{\prime }}\left(\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+(\boldsymbol{x}+\boldsymbol{d})\times \left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D734}\right)\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\int _{{\mathcal{V}}^{\prime }}g(\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D72E}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}-\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right)+\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}-\frac{\text{d}}{\text{d}t}\left(m_{v}\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,m_{v}\overline{\boldsymbol{x}}_{v}\times \left(\unicode[STIX]{x1D64C}^{-1}\dot{\boldsymbol{q}}\times \unicode[STIX]{x1D734}\right)-\unicode[STIX]{x1D644}_{v}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}-m_{v}g\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D72E}=\mathbf{0},\end{eqnarray}$$

which is the Euler–Poincaré equation for $\unicode[STIX]{x1D734}(t)$ . Equation (3.14) after differentiating with respect to time and simplifying using (3.5b ) and the hat map (2.6) reduces to

(3.15) $$\begin{eqnarray}\displaystyle & & \displaystyle \displaystyle \int _{{\mathcal{V}}^{\prime }}\left(\ddot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D734}\times \left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)+\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}\times (\boldsymbol{x}+\boldsymbol{d})\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.-\,g(\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D72E}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}-\dot{\unicode[STIX]{x1D644}}_{f}\unicode[STIX]{x1D734}-\left(\unicode[STIX]{x1D644}_{f}+\unicode[STIX]{x1D644}_{v}\right)\dot{\unicode[STIX]{x1D734}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(\unicode[STIX]{x1D644}_{f}+\unicode[STIX]{x1D644}_{v})\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}-m_{v}\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-m_{v}g\overline{\boldsymbol{x}}_{v}\times \unicode[STIX]{x1D72E}=\mathbf{0}.\end{eqnarray}$$

Now set the mass moment of inertia of the coupled system (body $+$ fluid) as

(3.16) $$\begin{eqnarray}\unicode[STIX]{x1D644}_{t}=\unicode[STIX]{x1D644}_{f}+\unicode[STIX]{x1D644}_{v},\end{eqnarray}$$

and note that

(3.17) $$\begin{eqnarray}m_{f}\overline{\boldsymbol{x}}_{f}=\displaystyle \int _{{\mathcal{V}}}(\boldsymbol{x}+\boldsymbol{d})\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}=\int _{{\mathcal{V}}^{\prime }}(\boldsymbol{x}+\boldsymbol{d})\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a},\end{eqnarray}$$

where $\overline{\boldsymbol{x}}_{f}(t)$ is the centre of mass of the fluid relative to the body frame, and

(3.18) $$\begin{eqnarray}m_{f}=\int _{{\mathcal{V}}}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}=\int _{{\mathcal{V}}^{\prime }}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a},\end{eqnarray}$$

is the mass of the fluid which is time independent. Also by setting

(3.19) $$\begin{eqnarray}m=m_{f}+m_{v},\end{eqnarray}$$

which is the total mass of the coupled system, we have

(3.20) $$\begin{eqnarray}m\overline{\boldsymbol{x}}(t)=m_{f}\overline{\boldsymbol{x}}_{f}+m_{v}\overline{\boldsymbol{x}}_{v},\end{eqnarray}$$

where $\overline{\boldsymbol{x}}$ is the centre of mass of the coupled system which is time dependent. Now the $\unicode[STIX]{x1D734}$ -equation (3.15) simplifies to

(3.21) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}^{\prime }}\left(\ddot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D734}\times \left(\dot{\boldsymbol{x}}\times (\boldsymbol{x}+\boldsymbol{d})\right)\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\dot{\unicode[STIX]{x1D644}}_{f}\unicode[STIX]{x1D734}-\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}=\mathbf{0}.\end{eqnarray}$$

Now, transforming this equation from the Lagrangian particle-path setting to Eulerian coordinates, replacing the Lagrangian variables $\dot{\boldsymbol{x}}$ and $\ddot{\boldsymbol{x}}$ by their respective Eulerian quantities $\boldsymbol{u}$ and $\text{D}\boldsymbol{u}/\text{D}t$ respectively, the Euler–Poincaré equation (3.21) takes the form

(3.22) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D734}\times \left(\boldsymbol{u}\times (\boldsymbol{x}+\boldsymbol{d})\right)\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\dot{\unicode[STIX]{x1D644}}_{f}\unicode[STIX]{x1D734}-\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}=\mathbf{0}.\end{eqnarray}$$

Note that $\ddot{\boldsymbol{q}}(t)$ in this equation is relative to the spatial frame $\boldsymbol{X}$ .

The Euler–Poincaré equation (3.22) for $\unicode[STIX]{x1D734}(t)$ can be written in a form where $\boldsymbol{q}(t)$ is relative to the body frame $\boldsymbol{x}_{b}$ , i.e.  $\boldsymbol{q}_{b}(t)=\unicode[STIX]{x1D64C}^{-1}\boldsymbol{q}$ . It can be proved that

(3.23) $$\begin{eqnarray}\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}=\ddot{\boldsymbol{q}}_{b}+\dot{\unicode[STIX]{x1D734}}\times \boldsymbol{q}_{b}+\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{q}_{b})+2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{q}}_{b}.\end{eqnarray}$$

Hence substitution of (3.23) into (3.22) gives

(3.24) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D734}\times \left(\boldsymbol{u}\times (\boldsymbol{x}+\boldsymbol{d})\right)\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,m\overline{\boldsymbol{x}}\times \left(\ddot{\boldsymbol{q}}_{b}+\dot{\unicode[STIX]{x1D734}}\times \boldsymbol{q}_{b}+\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times \boldsymbol{q}_{b}\right)+2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{q}}_{b}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\dot{\unicode[STIX]{x1D644}}_{f}\unicode[STIX]{x1D734}-\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D734}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}=\mathbf{0},\end{eqnarray}$$

which is the Euler–Poincaré equation for $\unicode[STIX]{x1D734}(t)$ with both $\unicode[STIX]{x1D734}$ and $\boldsymbol{q}_{b}$ relative to the body frame $\boldsymbol{x}_{b}$ .

Gerrits & Veldman (Reference Gerrits and Veldman2003) and Veldman et al. (Reference Veldman, Gerrits, Luppes, Helder and Vreeburg2007) studied dynamics of a liquid-filled spacecraft and presented the equation governing the rotational motion of a rigid body containing fluid based on the balance of angular momentum. Their $\unicode[STIX]{x1D734}$ -equation is equation (1.2). To compare the derived Euler–Poincaré equation for $\unicode[STIX]{x1D734}(t)$ with the equation for angular momentum of the rigid body (1.2) given by Gerrits & Veldman (Reference Gerrits and Veldman2003) and Veldman et al. (Reference Veldman, Gerrits, Luppes, Helder and Vreeburg2007), we rewrite equation (3.22) in the following form

(3.25) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}-(\boldsymbol{x}+\boldsymbol{d})\times \left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}=\displaystyle m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}+\dot{\unicode[STIX]{x1D644}}_{f}\unicode[STIX]{x1D734}+\int _{{\mathcal{V}}}\boldsymbol{u}\times ((\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D734})\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x},\end{eqnarray}$$

using the vector identity

(3.26) $$\begin{eqnarray}\unicode[STIX]{x1D734}\times (\boldsymbol{u}\times (\boldsymbol{x}+\boldsymbol{d}))+(\boldsymbol{x}+\boldsymbol{d})\times (\unicode[STIX]{x1D734}\times \boldsymbol{u})+\boldsymbol{u}\times ((\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D734})=\mathbf{0}.\end{eqnarray}$$

Differentiating the mass moment of inertia of the fluid (2.13) with respect to time gives

(3.27) $$\begin{eqnarray}\displaystyle \dot{\unicode[STIX]{x1D644}}_{f} & = & \displaystyle \int _{{\mathcal{V}}^{\prime }}(2\dot{\boldsymbol{x}}\boldsymbol{\cdot }(\boldsymbol{x}+\boldsymbol{d})\unicode[STIX]{x1D644}-(\dot{\boldsymbol{x}}\otimes (\boldsymbol{x}+\boldsymbol{d})+(\boldsymbol{x}+\boldsymbol{d})\otimes \dot{\boldsymbol{x}}))\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\nonumber\\ \displaystyle & = & \displaystyle \displaystyle \int _{{\mathcal{V}}}(2\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{x}+\boldsymbol{d})\unicode[STIX]{x1D644}-(\boldsymbol{u}\otimes (\boldsymbol{x}+\boldsymbol{d})+(\boldsymbol{x}+\boldsymbol{d})\otimes \boldsymbol{u}))\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x},\end{eqnarray}$$

and so

(3.28) $$\begin{eqnarray}\dot{\unicode[STIX]{x1D644}}_{f}\unicode[STIX]{x1D734}=\int _{{\mathcal{V}}}(2\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{x}+\boldsymbol{d})\unicode[STIX]{x1D644}-(\boldsymbol{u}\otimes (\boldsymbol{x}+\boldsymbol{d})+(\boldsymbol{x}+\boldsymbol{d})\otimes \boldsymbol{u}))\unicode[STIX]{x1D734}\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}.\end{eqnarray}$$

The summation of the last two terms on the right-hand side of (3.25) gives

(3.29) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(\left(2\boldsymbol{u}\boldsymbol{\cdot }(\boldsymbol{x}+\boldsymbol{d})\unicode[STIX]{x1D644}-\left(\boldsymbol{u}\otimes (\boldsymbol{x}+\boldsymbol{d})+(\boldsymbol{x}+\boldsymbol{d})\otimes \boldsymbol{u}\right)\right)\unicode[STIX]{x1D734}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.+\,\boldsymbol{u}\times \left((\boldsymbol{x}+\boldsymbol{d})\times \unicode[STIX]{x1D734}\right)\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}=\int _{{\mathcal{V}}}(\boldsymbol{x}+\boldsymbol{d})\times \left(\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}.\end{eqnarray}$$

Substitution of (3.29) into (3.25) gives

(3.30) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}-(\boldsymbol{x}+\boldsymbol{d})\times \left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}\nonumber\\ \displaystyle & & \displaystyle \qquad =m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}+\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734},\end{eqnarray}$$

which recovers equation (1.2), neglecting the torque due to viscosity of the fluid in (1.2). However, it should be noted that in the works by Gerrits & Veldman (Reference Gerrits and Veldman2003) and Veldman et al. (Reference Veldman, Gerrits, Luppes, Helder and Vreeburg2007) it is stated that $\dot{\unicode[STIX]{x1D666}}$ in (1.2) is the acceleration of the moving origin with respect to an inertial reference frame, but our calculations show precisely that $\dot{\unicode[STIX]{x1D666}}$ is actually $\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}$ which is relative to the body frame (see (3.23)). Also it should be noted that the terms involving $\boldsymbol{F}$ in (1.2) are gathered in (3.30) as $-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}$ with a precise definition for $\unicode[STIX]{x1D72E}$ in the body coordinate system. Using (3.23) equation (3.30) can be written in terms of $\boldsymbol{q}_{b}$ as

(3.31) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}-(\boldsymbol{x}+\boldsymbol{d})\times \left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}\nonumber\\ \displaystyle & & \displaystyle \qquad =m\overline{\boldsymbol{x}}\times (\ddot{\boldsymbol{q}}_{b}+\dot{\unicode[STIX]{x1D734}}\times \boldsymbol{q}_{b}+\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times \boldsymbol{q}_{b})+2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{q}}_{b})+\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}+\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}.\qquad\end{eqnarray}$$

3.2 The Euler–Poincaré equation for $\boldsymbol{q}(t)$

The governing equation for the translational motion of the rigid body $\boldsymbol{q}(t)$ is provided by Hamilton’s variational principle (3.1) by taking the variations $\unicode[STIX]{x1D6FF}\boldsymbol{q}$ and $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{q}}$ of the Lagrangian action (2.24) with fixed endpoints $\unicode[STIX]{x1D6FF}\boldsymbol{q}(t_{1})=\unicode[STIX]{x1D6FF}\boldsymbol{q}(t_{2})=\mathbf{0}$ , and assuming that $\unicode[STIX]{x1D734}$ , $\unicode[STIX]{x1D64C}$ , $\boldsymbol{x}$ and $\dot{\boldsymbol{x}}$ are constants. Applying similar calculations to the calculus of variations presented in § 3.1, it can be proved that Hamilton’s principle leads to

(3.32) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}^{\prime }}\left(-\ddot{\boldsymbol{x}}-\unicode[STIX]{x1D734}\times \dot{\boldsymbol{x}}-\frac{\text{d}}{\text{d}t}(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-g\unicode[STIX]{x1D72E}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,m_{v}\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\frac{\text{d}}{\text{d}t}\left(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v}\right)-\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v}\right)-m_{v}g\unicode[STIX]{x1D72E}=\mathbf{0},\end{eqnarray}$$

which is the Euler–Poincaré equation for the translational motion $\boldsymbol{q}(t)$ of the rigid body in the spatial frame $\boldsymbol{X}$ . This equation, after differentiating with respect to time and applying (3.17), (3.18) and (3.20), simplifies to

(3.33) $$\begin{eqnarray}\int _{{\mathcal{V}}^{\prime }}\left(-\ddot{\boldsymbol{x}}-2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{x}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}-m\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\dot{\unicode[STIX]{x1D734}}\times m\overline{\boldsymbol{x}}-\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times m\overline{\boldsymbol{x}}\right)-mg\unicode[STIX]{x1D72E}=\mathbf{0}.\end{eqnarray}$$

Transforming this equation from the Lagrangian particle-path setting to Eulerian coordinates, replacing the Lagrangian variables $\dot{\boldsymbol{x}}$ and $\ddot{\boldsymbol{x}}$ by their respective Eulerian quantities $\boldsymbol{u}$ and $\text{D}\boldsymbol{u}/\text{D}t$ respectively, the Euler–Poincaré equation (3.33) reduces to

(3.34) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-m\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\dot{\unicode[STIX]{x1D734}}\times m\overline{\boldsymbol{x}}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times m\overline{\boldsymbol{x}}\right)-mg\unicode[STIX]{x1D72E}=\mathbf{0}.\end{eqnarray}$$

The $\boldsymbol{q}$ -equation (3.34) can be written in terms of the translational acceleration in the body frame $\ddot{\boldsymbol{q}}_{b}$ using (3.23) as

(3.35) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-m\left(\ddot{\boldsymbol{q}}_{b}+\dot{\unicode[STIX]{x1D734}}\times \boldsymbol{q}_{b}+\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times \boldsymbol{q}_{b}\right)+2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{q}}_{b}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\dot{\unicode[STIX]{x1D734}}\times m\overline{\boldsymbol{x}}-\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times m\overline{\boldsymbol{x}}\right)-mg\unicode[STIX]{x1D72E}=\mathbf{0},\end{eqnarray}$$

or in the form

(3.36) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-m\left(\ddot{\boldsymbol{q}}_{b}+2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{q}}_{b}\right)-m\dot{\unicode[STIX]{x1D734}}\times \left(\overline{\boldsymbol{x}}+\boldsymbol{q}_{b}\right)\nonumber\\ \displaystyle & & \displaystyle \qquad -\,m\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\overline{\boldsymbol{x}}+\boldsymbol{q}_{b}))-mg\unicode[STIX]{x1D72E}=\mathbf{0}.\end{eqnarray}$$

The final form of the Euler–Poincaré equation for the linear momentum of the rigid body is (3.34) in the spatial frame $\boldsymbol{X}$ or (3.36) in the body frame $\boldsymbol{x}_{b}$ .

The Euler–Poincaré equation (3.34) for $\boldsymbol{q}(t)$ recovers equation (1.1) for linear momentum of the rigid body, neglecting the force due to the viscosity of the fluid in (1.1), given by Gerrits & Veldman (Reference Gerrits and Veldman2003) and Veldman et al. (Reference Veldman, Gerrits, Luppes, Helder and Vreeburg2007). However, it should be noted that in the works by Gerrits & Veldman (Reference Gerrits and Veldman2003) and Veldman et al. (Reference Veldman, Gerrits, Luppes, Helder and Vreeburg2007) it is stated that $\dot{\unicode[STIX]{x1D666}}$ in (1.1) is the acceleration of the moving origin with respect to an inertial reference frame, but from our calculations it is obvious that $\dot{\unicode[STIX]{x1D666}}$ is actually $\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}$ which is relative to the body frame. The terms involving $\boldsymbol{F}$ in (1.1) are gathered in (3.34) as $-mg\unicode[STIX]{x1D72E}$ with $\unicode[STIX]{x1D72E}$ defined in (3.8).

3.3 Reconstruction of $\unicode[STIX]{x1D64C}(t)\in \text{SO}(3)$

The Euler–Poincaré equations (3.30) and (3.34) determine the body angular velocity $\unicode[STIX]{x1D734}(t)$ and translational motion $\boldsymbol{q}(t)$ of the rigid body, respectively. The tangent vectors $\dot{\unicode[STIX]{x1D64C}}(t)\in \text{TSO}(3)$ along the integral curve in the rotation group $\unicode[STIX]{x1D64C}(t)\in \text{SO}(3)$ may be retrieved via the reconstruction formula (Holm et al. Reference Holm, Schmah and Stoica2009)

(3.37) $$\begin{eqnarray}\dot{\unicode[STIX]{x1D64C}}=\unicode[STIX]{x1D64C}\widehat{\unicode[STIX]{x1D734}}.\end{eqnarray}$$

The solution of (3.37) yields the integral curve $\unicode[STIX]{x1D64C}(t)\in \text{SO}(3)$ for the orientation of the rigid body.

Finally, differentiating the constraint equation $\unicode[STIX]{x1D72E}(t)=\unicode[STIX]{x1D64C}^{-1}(t)\hat{\boldsymbol{z}}$ gives

(3.38) $$\begin{eqnarray}\dot{\unicode[STIX]{x1D72E}}(t)=\unicode[STIX]{x1D72E}(t)\times \unicode[STIX]{x1D734}(t)\quad \text{with }\unicode[STIX]{x1D72E}\left(0\right)=\unicode[STIX]{x1D64C}^{-1}(0)\hat{\boldsymbol{z}}.\end{eqnarray}$$

So the evolutionary system for the rigid body motion (3.30) and (3.34) is completed by (3.37) and (3.38).

4 The Euler–Poincaré equation for the interior fluid motion

In Eulerian coordinates, the interior fluid occupies the region ${\mathcal{V}}$

(4.1) $$\begin{eqnarray}0\leqslant x\leqslant L_{1},\quad 0\leqslant y\leqslant L_{2},\quad 0\leqslant z\leqslant h(x,y,t),\end{eqnarray}$$

where $L_{1}$ and $L_{2}$ are given positive constants, and $z=h(x,y,t)$ is the position of the free surface. The boundary conditions are

(4.2) $$\begin{eqnarray}\left.\begin{array}{@{}rl@{}}u=0 & \text{at }x=0\quad \text{and}\quad x=L_{1},\\ v=0 & \text{at }y=0\quad \text{and}\quad y=L_{2},\\ w=0 & \text{at }z=0,\end{array}\right\}\end{eqnarray}$$

which are the no-flow boundary condition on the rigid walls, and at the free surface, the kinematic and dynamic boundary conditions are respectively

(4.3a,b ) $$\begin{eqnarray}w=h_{t}+uh_{x}+vh_{y}\quad \text{and}\quad p=0\quad \text{at }z=h(x,y,t),\end{eqnarray}$$

where the surface tension is neglected in the boundary condition for the pressure $p$ .

The Euler–Poincaré equation for the position $\boldsymbol{x}(t)$ of fluid particles in the body frame $\boldsymbol{x}$ can be provided by Hamilton’s variational principle (3.1) by taking the variations $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ and $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{x}}$ of the Lagrangian action (2.24), in Lagrangian coordinates, with fixed endpoints $\unicode[STIX]{x1D6FF}\boldsymbol{x}(t_{1})=\unicode[STIX]{x1D6FF}\boldsymbol{x}(t_{2})=\mathbf{0}$ , assuming that $\unicode[STIX]{x1D734}$ , $\unicode[STIX]{x1D64C}$ , $\boldsymbol{q}$ and $\dot{\boldsymbol{q}}$ are constants. However, a constraint term should be added to the Lagrangian functional (2.24) to enforce incompressibility of the fluid $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0$ . The action functional (2.24) is modified to

(4.4) $$\begin{eqnarray}\mathscr{L}_{t}=\displaystyle \mathscr{L}\left(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}}\right)+\mathscr{L}_{c}\quad \text{with}\quad \mathscr{L}_{c}=\displaystyle \int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(\boldsymbol{a},t)(J-1)\,\text{d}\boldsymbol{a}\,\text{d}t,\end{eqnarray}$$

where $p(\boldsymbol{a},t)$ is the pressure field of the interior fluid and $\mathscr{L}_{c}$ enforces incompressibility of the fluid in the Lagrangian particle-path formulation. Hence

(4.5) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\mathscr{L}_{t}=\unicode[STIX]{x1D6FF}\mathscr{L}(\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}})+\unicode[STIX]{x1D6FF}\mathscr{L}_{c}=\unicode[STIX]{x1D6FF}\mathscr{L}+\unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(\boldsymbol{a},t)(J-1)\,\text{d}\boldsymbol{a}\,\text{d}t=0,\end{eqnarray}$$

where the integral is over the volume ${\mathcal{V}}^{\prime }$ of the reference space

(4.6) $$\begin{eqnarray}0\leqslant a\leqslant \unicode[STIX]{x1D613}_{1},\quad 0\leqslant b\leqslant \unicode[STIX]{x1D613}_{2},\quad 0\leqslant c\leqslant \unicode[STIX]{x1D613}_{3},\end{eqnarray}$$

where $\unicode[STIX]{x1D613}_{1}$ , $\unicode[STIX]{x1D613}_{2}$ and $\unicode[STIX]{x1D613}_{3}$ are given positive constants. The free surface $z=h(x,y,t)$ is the map of the boundary surface $\{(a,b,c):c=\unicode[STIX]{x1D613}_{3}\}$ .

After some calculations, presented in appendix B, it is proved that taking the variations $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ and $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{x}}$ of the first component of the Lagrangian functional $\mathscr{L}_{t}$ , in the Lagrangian particle-path setting, leads to

(4.7) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\mathscr{L} & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D6FF}\boldsymbol{x},-\ddot{\boldsymbol{x}}-\dot{\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d})-2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{x}}-\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.-\,\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-g\unicode[STIX]{x1D72E}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}\,\text{d}t.\end{eqnarray}$$

Taking the variations $\unicode[STIX]{x1D6FF}p$ and $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ of the constraint component of $\mathscr{L}_{t}$ in (4.4b ) gives

(4.8) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}\mathscr{L}_{c} & = & \displaystyle \unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(J-1)\,\text{d}\boldsymbol{a}\,\text{d}t=\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\unicode[STIX]{x1D6FF}p(J-1)\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(\unicode[STIX]{x1D6FF}x_{a}(y_{b}z_{c}-y_{c}z_{b})+\unicode[STIX]{x1D6FF}x_{b}(y_{c}z_{a}-y_{a}z_{c})+\unicode[STIX]{x1D6FF}x_{c}(y_{a}z_{b}-y_{b}z_{a}))\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(\unicode[STIX]{x1D6FF}y_{a}(x_{c}z_{b}-x_{b}z_{c})+\unicode[STIX]{x1D6FF}y_{b}(x_{a}z_{c}-x_{c}z_{a})+\unicode[STIX]{x1D6FF}y_{c}(x_{b}z_{a}-x_{a}z_{b}))\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(\unicode[STIX]{x1D6FF}z_{a}(x_{b}y_{c}-x_{c}y_{b})+\unicode[STIX]{x1D6FF}z_{b}(x_{c}y_{a}-x_{a}y_{c})+\unicode[STIX]{x1D6FF}z_{c}(x_{a}y_{b}-x_{b}y_{a}))\,\text{d}\boldsymbol{a}\,\text{d}t.\qquad\end{eqnarray}$$

After integrating by parts and imposing the no-flow boundary conditions in the Lagrangian particle-path setting

(4.9) $$\begin{eqnarray}\left.\begin{array}{@{}rl@{}}x_{b}=x_{c}=0 & \text{at }a=0\quad \text{and}\quad a=\unicode[STIX]{x1D613}_{1},\\ y_{a}=y_{c}=0 & \text{at }b=0\quad \text{and}\quad b=\unicode[STIX]{x1D613}_{2},\\ z_{a}=z_{b}=0 & \text{at }c=0,\end{array}\right\}\end{eqnarray}$$

and the free-surface pressure boundary condition $p=0$ at $c=\unicode[STIX]{x1D613}_{3}$ , $\unicode[STIX]{x1D6FF}\mathscr{L}_{c}$ in (4.8) reduces to

(4.10) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}\mathscr{L}_{c}=\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\unicode[STIX]{x1D6FF}p(J-1)\,\text{d}\boldsymbol{a}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}\left\langle \unicode[STIX]{x1D6FF}\boldsymbol{x},\overbrace{\left[\begin{array}{@{}c@{}}\displaystyle (y_{c}z_{b}-y_{b}z_{c})p_{a}+(y_{a}z_{c}-y_{c}z_{a})p_{b}+(y_{b}z_{a}-y_{a}z_{b})p_{c}\\ \displaystyle (x_{b}z_{c}-x_{c}z_{b})p_{a}+(x_{c}z_{a}-x_{a}z_{c})p_{b}+(x_{a}z_{b}-x_{b}z_{a})p_{c}\\ \displaystyle (x_{c}y_{b}-x_{b}y_{c})p_{a}+(x_{a}y_{c}-x_{c}y_{a})p_{b}+(x_{b}y_{a}-x_{a}y_{b})p_{c}\end{array}\right]}^{=-\unicode[STIX]{x1D735}p\text{ in Eulerian coordinates }{\looparrowleft}}\right\rangle \,\text{d}\boldsymbol{a}\,\text{d}t.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Therefore, since $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ and $\unicode[STIX]{x1D6FF}p$ are arbitrary, from the variations (4.7) and (4.10) and Hamilton’s principle (4.5) we conclude that

(4.11) $$\begin{eqnarray}\displaystyle & & \displaystyle \ddot{\boldsymbol{x}}+\frac{1}{\unicode[STIX]{x1D70C}}\left[\begin{array}{@{}c@{}}(y_{b}z_{c}-y_{c}z_{b})p_{a}+(y_{c}z_{a}-y_{a}z_{c})p_{b}+(y_{a}z_{b}-y_{b}z_{a})p_{c}\\ (x_{c}z_{b}-x_{b}z_{c})p_{a}+(x_{a}z_{c}-x_{c}z_{a})p_{b}+(x_{b}z_{a}-x_{a}z_{b})p_{c}\\ (x_{b}y_{c}-x_{c}y_{b})p_{a}+(x_{c}y_{a}-x_{a}y_{c})p_{b}+(x_{a}y_{b}-x_{b}y_{a})p_{c}\end{array}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad =-\dot{\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d})-2\unicode[STIX]{x1D734}\times \dot{\boldsymbol{x}}-\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-g\unicode[STIX]{x1D72E},\quad J-1=0.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Now, transforming the equations of motion (4.11) from Lagrangian coordinates to Eulerian coordinates we obtain

(4.12) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \frac{\text{D}\boldsymbol{u}}{\text{D}t}+\frac{1}{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D735}p=-\dot{\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d})-2\unicode[STIX]{x1D734}\times \boldsymbol{u}-\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-g\unicode[STIX]{x1D72E},\\ \displaystyle \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0,\end{array}\right\}\end{eqnarray}$$

which are the Euler–Poincaré equations for the interior fluid motion relative to the body frame $\boldsymbol{x}$ . The Euler equations in rotating coordinates are presented by Alemi Ardakani & Bridges (Reference Alemi Ardakani and Bridges2011). Here we have recovered these equations using a variational principle in the Lagrangian particle-path formulation.

The momentum equation in (4.12) can be written in terms of the translational acceleration in the body frame $\ddot{\boldsymbol{q}}_{b}$ as

(4.13) $$\begin{eqnarray}\displaystyle \frac{\text{D}\boldsymbol{u}}{\text{D}t}+\frac{1}{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D735}p & = & \displaystyle -\dot{\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d}+\boldsymbol{q}_{b})-2\unicode[STIX]{x1D734}\times (\boldsymbol{u}+\dot{\boldsymbol{q}}_{b})\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times \left(\boldsymbol{x}+\boldsymbol{d}+\boldsymbol{q}_{b}\right)\right)-\ddot{\boldsymbol{q}}_{b}-g\unicode[STIX]{x1D72E}.\end{eqnarray}$$

Having developed a variational principle for rigid body dynamics with interior fluid motion in three-dimensional rotating and translating coordinates, this variational principle can be extended to the problem of interactions between water waves and a floating rigid body with interior inviscid fluid motion.

5 A variational principle for interactions between water waves and a floating rigid body containing fluid

Van Daalen et al. (Reference Van Daalen, Van Groesen and Zandbergen1993) extended Luke’s variational principle for the classical water-wave problem to the hydrodynamic interaction with a freely floating empty rigid body, i.e. without interior fluid, in three dimensions. However, they did not present the exact differential equations for the rigid body motion, due to the approximation used for the angular velocity in the kinetic energy of the rigid body. Alemi Ardakani (Reference Alemi Ardakani2017) derived a coupled variational principle for the two-dimensional interactions between ocean waves and a freely floating rigid body with interior fluid sloshing with uniform vorticity. The complete set of equations of motion for the exterior water waves, the exact hydrodynamic equations for the planar rigid body motion and the full set of equations of motion for the interior potential flow of the body are derived.

The classical water-wave problem in three dimensions is described by the equations

(5.1) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x0394}\unicode[STIX]{x1D6F7}:=\unicode[STIX]{x1D6F7}_{XX}+\unicode[STIX]{x1D6F7}_{YY}+\unicode[STIX]{x1D6F7}_{ZZ}=0\quad \text{for }-H\left(X,Y\right)<Z<\unicode[STIX]{x1D702}(X,Y,t),\\ \displaystyle \unicode[STIX]{x1D6F7}_{t}+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}+gZ=0\quad \text{on }Z=\unicode[STIX]{x1D702}(X,Y,t),\\ \displaystyle \unicode[STIX]{x1D6F7}_{Z}=\unicode[STIX]{x1D702}_{t}+\unicode[STIX]{x1D6F7}_{X}\unicode[STIX]{x1D702}_{X}+\unicode[STIX]{x1D6F7}_{Y}\unicode[STIX]{x1D702}_{Y}\quad \text{on }Z=\unicode[STIX]{x1D702}(X,Y,t),\\ \displaystyle \unicode[STIX]{x1D6F7}_{Z}+\unicode[STIX]{x1D6F7}_{X}H_{X}+\unicode[STIX]{x1D6F7}_{Y}H_{Y}=0\quad \text{on }Z=-H\left(X,Y\right),\end{array}\right\}\end{eqnarray}$$

where $(X,Y,Z)$ is the spatial coordinate system, $\unicode[STIX]{x1D6F7}(X,Y,Z,t)$ is the velocity potential of an irrotational fluid lying between $Z=-H\left(X,Y\right)$ and $Z=\unicode[STIX]{x1D702}(X,Y,t)$ with the gravity acceleration $g$ acting in the negative $Z$ direction. In the horizontal directions $X$ and $Y$ , the fluid domain is cut off by a cylindrical vertical surface ${\mathcal{S}}$ of infinite radius which extends from the bottom to the free surface. Then an extension of Luke’s variational principle, for 3-D water waves, as reported by Van Daalen et al. (Reference Van Daalen, Van Groesen and Zandbergen1993) reads

(5.2) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\mathscr{L}_{w}(\unicode[STIX]{x1D6F7},\unicode[STIX]{x1D702})=\unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{V(t)}-\unicode[STIX]{x1D70C}\left(\unicode[STIX]{x1D6F7}_{t}+\frac{1}{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}+gZ\right)\,\text{d}V\,\text{d}t=0,\end{eqnarray}$$

where the Bernoulli pressure, playing the role of the Lagrangian density, is integrated over the transient fluid domain $V(t)$ , with variations in $\unicode[STIX]{x1D6F7}(X,Y,Z,t)$ and $\unicode[STIX]{x1D702}(X,Y,t)$ subject to the restrictions $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}=0$ at the endpoints of the time interval, $t_{1}$ and $t_{2}$ . In (5.2) the gradient vector field is denoted by $\unicode[STIX]{x1D735}$ , and $\unicode[STIX]{x1D70C}$ is the water density. The variational principle (5.2) recovers the complete set of equations of motion for the water-wave problem described by (5.1).

In §§ 3 and 4, a variational principle is developed for dynamic coupling between rigid body motion and its interior inviscid and incompressible fluid sloshing in three-dimensional rotating and translating coordinates. This variational principle can be extended to the problem of 3-D water waves in hydrodynamic interaction with a freely floating rigid body containing fluid by the addition of the extended version of Luke’s variational principle (5.2) to Hamilton’s variational principle (4.5) as

(5.3) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}{\mathcal{L}}\left(\unicode[STIX]{x1D6F7},\unicode[STIX]{x1D702},\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}},p\right)=\unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{V(t)}-\unicode[STIX]{x1D70C}\left(\unicode[STIX]{x1D6F7}_{t}+\frac{1}{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}+gZ\right)\,\text{d}V\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\left(\int _{{\mathcal{V}}^{\prime }}\left(\frac{1}{2}\left\Vert \dot{\boldsymbol{x}}\right\Vert ^{2}+\dot{\boldsymbol{x}}\boldsymbol{\cdot }\left(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d})+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\right)+\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}\boldsymbol{\cdot }(\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))\right.\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.\left.+\,\frac{1}{2}\Vert \dot{\boldsymbol{q}}\Vert ^{2}-g(\unicode[STIX]{x1D64C}(\boldsymbol{x}+\boldsymbol{d})+\boldsymbol{q})\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{a}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{f}\unicode[STIX]{x1D734}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.+\,\frac{1}{2}m_{v}\Vert \dot{\boldsymbol{q}}\Vert ^{2}+(\unicode[STIX]{x1D734}\times m_{v}\overline{\boldsymbol{x}}_{v})\boldsymbol{\cdot }\unicode[STIX]{x1D64C}^{\text{T}}\dot{\boldsymbol{q}}+\frac{1}{2}\unicode[STIX]{x1D734}\boldsymbol{\cdot }\unicode[STIX]{x1D644}_{v}\unicode[STIX]{x1D734}-m_{v}g(\unicode[STIX]{x1D64C}\overline{\boldsymbol{x}}_{v}+\boldsymbol{q})\boldsymbol{\cdot }\hat{\boldsymbol{z}}\right)\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D6FF}\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}^{\prime }}p(\boldsymbol{a},t)(J-1)\,\text{d}\boldsymbol{a}\,\text{d}t=0,\end{eqnarray}$$

where $V(t)$ consists of a fluid bounded by the impermeable bottom $S_{b}$ defined by the equation $Z=-H(X,Y)$ , the free surface $S_{\unicode[STIX]{x1D702}}$ defined by the equation $Z=\unicode[STIX]{x1D702}(X,Y,t)$ , the vertical surface ${\mathcal{S}}$ and the wetted surface $S_{w}$ of the rigid body interacting with exterior water waves. The configuration of the fluid in a rotating–translating floating structure interacting with exterior water waves is schematically shown in figure 2.

Figure 2. Schematic showing a floating structure containing fluid in hydrodynamic interaction with exterior ocean waves.

In order to take the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}$ and $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}$ in (5.3), the variational Reynold’s transport theorem should be used, since the domain of integration $V$ is time dependent. The background mathematics on the variational analogue of Reynold’s transport theorem can be found in Flanders (Reference Flanders1973), Daniliuk (Reference Daniliuk1976) and Gagarina, van der Vegt & Bokhove (Reference Gagarina, van der Vegt and Bokhove2013). Then, according to the usual procedure in the calculus of variations, the variational principle (5.3) for the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}$ , $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}$ , $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ , $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ , $\unicode[STIX]{x1D6FF}\boldsymbol{q}$ , $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{q}}$ , $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ , $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{x}}$ and $\unicode[STIX]{x1D6FF}p$ becomes

(5.4) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}{\mathcal{L}}\left(\unicode[STIX]{x1D6F7},\unicode[STIX]{x1D702},\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\dot{\boldsymbol{q}},\boldsymbol{x},\dot{\boldsymbol{x}},p\right)=\int _{t_{1}}^{t_{2}}\int _{S_{\unicode[STIX]{x1D702}}}-\left.\left(\unicode[STIX]{x1D6F7}_{t}+\frac{1}{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}+gZ\right)\right|^{Z=\unicode[STIX]{x1D702}}\unicode[STIX]{x1D70C}\,\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\ell ^{-1}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{S_{w}}P(X,Y,Z,t)(\unicode[STIX]{x1D6FF}\boldsymbol{X}_{w}\boldsymbol{\cdot }\boldsymbol{n})\,\text{d}S\,\text{d}t-\int _{t_{1}}^{t_{2}}\int _{V(t)}(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}_{t}+\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7})\unicode[STIX]{x1D70C}\,\text{d}V\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\left\langle \unicode[STIX]{x1D71E},-(\boldsymbol{x}+\boldsymbol{d})\times \left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}-\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}\right\rangle \,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\left\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},-m\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\dot{\unicode[STIX]{x1D734}}\times m\overline{\boldsymbol{x}}-\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times m\overline{\boldsymbol{x}}\right)-mg\unicode[STIX]{x1D72E}\right\rangle \,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\left\langle \unicode[STIX]{x1D6FF}\boldsymbol{x},-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-\unicode[STIX]{x1D70C}^{-1}\unicode[STIX]{x1D735}p-\dot{\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d})-2\unicode[STIX]{x1D734}\times \boldsymbol{u}-\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.\qquad -\,\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-g\unicode[STIX]{x1D72E}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\,\text{d}t+\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\unicode[STIX]{x1D6FF}p\,\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}\,\text{d}\boldsymbol{x}\,\text{d}t=0,\end{eqnarray}$$

where the results of §§ 3 and 4 are applied in taking the variations of the second and third components of the variational principle (5.3). Note that in (5.4) after taking the variations $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D734}$ , $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}$ , $\unicode[STIX]{x1D6FF}\boldsymbol{q}$ , $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{q}}$ , $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ , $\unicode[STIX]{x1D6FF}\dot{\boldsymbol{x}}$ and $\unicode[STIX]{x1D6FF}p$ of the coupled (interior fluid + body) system in the Lagrangian particle-path setting, the results are transformed to Eulerian coordinates. In (5.4), $\boldsymbol{X}_{w}$ denotes the position of a point on the wetted body surface $S_{w}$ relative to the spatial frame $\boldsymbol{X}$ , $\boldsymbol{n}$ is the unit normal vector along $\unicode[STIX]{x2202}V\supset S_{w}$ in the spatial frame, $\ell =(1+\unicode[STIX]{x1D702}_{X}^{2}+\unicode[STIX]{x1D702}_{Y}^{2})^{1/2}$ giving $\text{d}S=\ell \,\text{d}X\text{d}Y$ , and $P$ is the pressure field of the exterior water waves defined by

(5.5) $$\begin{eqnarray}P(X,Y,Z,t)=-\unicode[STIX]{x1D70C}\left(\unicode[STIX]{x1D6F7}_{t}+{\textstyle \frac{1}{2}}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}+gZ\right)\quad \text{on }S_{w}.\end{eqnarray}$$

These variations are subject to the restrictions that they vanish at the endpoints of the time interval. Moreover, the variations in $\unicode[STIX]{x1D702}$ and $\unicode[STIX]{x1D6F7}$ vanish on the vertical boundary at infinity, i.e. on ${\mathcal{S}}$ .

The change in $\boldsymbol{X}_{w}$ due to the variations in $\unicode[STIX]{x1D64C}$ and $\boldsymbol{q}$ is given by

(5.6) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}\boldsymbol{X}_{w}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}\boldsymbol{x}_{w}+\unicode[STIX]{x1D6FF}\boldsymbol{q},\end{eqnarray}$$

where $\boldsymbol{x}_{w}$ is the position of a point on the wetted rigid body surface $S_{w}$ relative to the body frame $\boldsymbol{x}_{b}$ . Using the variations (5.6), the second integral on the right-hand side of (5.4) simplifies to

(5.7) $$\begin{eqnarray}\displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \unicode[STIX]{x1D6FF}\boldsymbol{X}_{w},\boldsymbol{n}\rangle \,\text{d}S\,\text{d}t & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{X}_{w},\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n}\rangle \,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D64C}\boldsymbol{x}_{w}+\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n}\rangle \,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \widehat{\unicode[STIX]{x1D71E}}\boldsymbol{x}_{w}+\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n}\rangle \,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \unicode[STIX]{x1D71E}\times \boldsymbol{x}_{w}+\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n}\rangle \,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \unicode[STIX]{x1D71E},-\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n}\times \boldsymbol{x}_{w}\rangle +P\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n}\rangle \,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & = & \displaystyle \int _{t_{1}}^{t_{2}}\int _{S_{w}}P\langle \unicode[STIX]{x1D71E},\boldsymbol{x}_{w}\times \boldsymbol{n}_{b}\rangle +P\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},\boldsymbol{n}_{b}\rangle \,\text{d}S\,\text{d}t,\end{eqnarray}$$

where

(5.8) $$\begin{eqnarray}\boldsymbol{n}_{b}=\unicode[STIX]{x1D64C}^{-1}\boldsymbol{n},\end{eqnarray}$$

is the unit normal vector along $S_{w}$ in the body frame $\boldsymbol{x}_{b}$ . Taking into account the motion of $V(t)$ , we may write

(5.9) $$\begin{eqnarray}\displaystyle & & \displaystyle -\frac{\text{d}}{\text{d}t}\int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}V\,\text{d}t=\left.-\int _{t_{1}}^{t_{2}}\int _{S_{\unicode[STIX]{x1D702}}}\unicode[STIX]{x1D702}_{t}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\right|^{Z=\unicode[STIX]{x1D702}}\unicode[STIX]{x1D70C}\ell ^{-1}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\int _{t_{1}}^{t_{2}}\int _{S_{w}}(\dot{\boldsymbol{X}}_{w}\boldsymbol{\cdot }\boldsymbol{n})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}S\,\text{d}t-\int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}_{t}\unicode[STIX]{x1D70C}\,\text{d}V\,\text{d}t.\end{eqnarray}$$

This is the same as the variational Reynold’s transport theorem but with variational derivatives replaced by time derivatives. Noting that the left-hand side vanishes due to the restriction $\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}=0$ at times $t=t_{1}$ and $t=t_{2}$ , this expression simplifies to

(5.10) $$\begin{eqnarray}-\int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}_{t}\unicode[STIX]{x1D70C}\,\text{d}V\,\text{d}t=\left.\int _{t_{1}}^{t_{2}}\int _{S_{\unicode[STIX]{x1D702}}}\unicode[STIX]{x1D702}_{t}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\right|^{Z=\unicode[STIX]{x1D702}}\unicode[STIX]{x1D70C}\ell ^{-1}\,\text{d}S\,\text{d}t+\int _{t_{1}}^{t_{2}}\int _{S_{w}}(\dot{\boldsymbol{X}}_{w}\boldsymbol{\cdot }\boldsymbol{n})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}S\,\text{d}t.\end{eqnarray}$$

With Green’s first identity, we may write

(5.11) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}V\text{d}t=-\int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x0394}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}V\text{d}t+\int _{t_{1}}^{t_{2}}\int _{\unicode[STIX]{x2202}V}\left(\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\boldsymbol{n}\right)\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad =-\int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x0394}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}V\text{d}t+\int _{t_{1}}^{t_{2}}\int _{S_{\unicode[STIX]{x1D702}}}\left(-\unicode[STIX]{x1D702}_{X}\unicode[STIX]{x1D6F7}_{X}-\unicode[STIX]{x1D702}_{Y}\unicode[STIX]{x1D6F7}_{Y}+\unicode[STIX]{x1D6F7}_{Z}\right)\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\bigg|^{Z=\unicode[STIX]{x1D702}}\unicode[STIX]{x1D70C}\ell ^{-1}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \quad \qquad \left.+\,\int _{t_{1}}^{t_{2}}\int _{S_{b}}(\unicode[STIX]{x1D6F7}_{X}H_{X}+\unicode[STIX]{x1D6F7}_{Y}H_{Y}+\unicode[STIX]{x1D6F7}_{Z})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\right|_{Z=-H}\unicode[STIX]{x1D70C}\,\text{d}S\text{d}t+\int _{t_{1}}^{t_{2}}\int _{S_{w}}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}\boldsymbol{n}}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}S\,\text{d}t.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Now, using the expressions (5.7), (5.10) and (5.11), the variational principle (5.4) simplifies to

(5.12) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6FF}{\mathcal{L}}\left(\unicode[STIX]{x1D6F7},\unicode[STIX]{x1D702},\unicode[STIX]{x1D734},\unicode[STIX]{x1D64C},\boldsymbol{q},\boldsymbol{x},p\right)=\int _{t_{1}}^{t_{2}}\int _{S_{\unicode[STIX]{x1D702}}}-\left(\unicode[STIX]{x1D6F7}_{t}+\frac{1}{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}+gZ\right)\bigg|^{Z=\unicode[STIX]{x1D702}}\unicode[STIX]{x1D70C}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D702}\ell ^{-1}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{S_{w}}P(X,Y,Z,t)\left\langle \unicode[STIX]{x1D71E},\boldsymbol{x}_{w}\times \boldsymbol{n}_{b}\right\rangle +P(X,Y,Z,t)\left\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},\boldsymbol{n}_{b}\right\rangle \,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad \left.+\,\int _{t_{1}}^{t_{2}}\int _{S_{\unicode[STIX]{x1D702}}}(\unicode[STIX]{x1D702}_{t}+\unicode[STIX]{x1D702}_{X}\unicode[STIX]{x1D6F7}_{X}+\unicode[STIX]{x1D702}_{Y}\unicode[STIX]{x1D6F7}_{Y}-\unicode[STIX]{x1D6F7}_{Z})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\right|^{Z=\unicode[STIX]{x1D702}}\unicode[STIX]{x1D70C}\ell ^{-1}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{S_{w}}\left(\dot{\boldsymbol{X}}_{w}\boldsymbol{\cdot }\boldsymbol{n}-\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}\boldsymbol{n}}\right)\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}S\,\text{d}t+\int _{t_{1}}^{t_{2}}\int _{V(t)}\unicode[STIX]{x0394}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\unicode[STIX]{x1D70C}\,\text{d}V\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad \left.-\,\int _{t_{1}}^{t_{2}}\int _{S_{b}}(\unicode[STIX]{x1D6F7}_{X}H_{X}+\unicode[STIX]{x1D6F7}_{Y}H_{Y}+\unicode[STIX]{x1D6F7}_{Z})\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6F7}\right|_{Z=-H}\unicode[STIX]{x1D70C}\,\text{d}S\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\left\langle \unicode[STIX]{x1D71E},-(\boldsymbol{x}+\boldsymbol{d})\times \left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D71E},-m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}-\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}\right\rangle \,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\left\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\left\langle \unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q},-m\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}-\dot{\unicode[STIX]{x1D734}}\times m\overline{\boldsymbol{x}}-\unicode[STIX]{x1D734}\times \left(\unicode[STIX]{x1D734}\times m\overline{\boldsymbol{x}}\right)-mg\unicode[STIX]{x1D72E}\right\rangle \,\text{d}t\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\left\langle \unicode[STIX]{x1D6FF}\boldsymbol{x},-\frac{\text{D}\boldsymbol{u}}{\text{D}t}-\unicode[STIX]{x1D70C}^{-1}\unicode[STIX]{x1D735}p-\dot{\unicode[STIX]{x1D734}}\times (\boldsymbol{x}+\boldsymbol{d})-2\unicode[STIX]{x1D734}\times \boldsymbol{u}-\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.-\,\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times (\boldsymbol{x}+\boldsymbol{d}))-g\unicode[STIX]{x1D72E}\right\rangle \unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}\,\text{d}t+\int _{t_{1}}^{t_{2}}\int _{{\mathcal{V}}}\unicode[STIX]{x1D6FF}p\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}\,\text{d}\boldsymbol{x}\,\text{d}t=0.\end{eqnarray}$$

From (5.12), we conclude that invariance of ${\mathcal{L}}$ with respect to a variation in the free-surface elevation $\unicode[STIX]{x1D702}$ yields the dynamic free-surface boundary condition in (5.1), invariance of ${\mathcal{L}}$ with respect to a variation in the velocity potential $\unicode[STIX]{x1D6F7}$ yields the field equation in (5.1) in the domain $V(t)$ , invariance of ${\mathcal{L}}$ with respect to a variation in the velocity potential $\unicode[STIX]{x1D6F7}$ at $Z=-H(X,Y)$ gives the bottom boundary condition in (5.1), invariance of ${\mathcal{L}}$ with respect to a variation in the velocity potential $\unicode[STIX]{x1D6F7}$ at $Z=\unicode[STIX]{x1D702}(X,Y,t)$ gives the kinematic free-surface boundary condition in (5.1) and invariance of ${\mathcal{L}}$ with respect to a variation in the velocity potential $\unicode[STIX]{x1D6F7}$ on $S_{w}$ gives the contact condition on the wetted surface of the rigid body,

(5.13) $$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}\boldsymbol{n}}=\dot{\boldsymbol{X}}_{w}\boldsymbol{\cdot }\boldsymbol{n}\quad \text{on }S_{w}.\end{eqnarray}$$

Invariance of ${\mathcal{L}}$ with respect to $\unicode[STIX]{x1D71E}$ gives the hydrodynamic equation of motion for the rotational motion $\unicode[STIX]{x1D734}(t)$ of the floating rigid body interacting with exterior water waves and dynamically coupled to its interior fluid motion

(5.14) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}-(\boldsymbol{x}+\boldsymbol{d})\times \left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}-mg\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D72E}-m\overline{\boldsymbol{x}}\times \unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\unicode[STIX]{x1D644}_{t}\dot{\unicode[STIX]{x1D734}}-\unicode[STIX]{x1D734}\times \unicode[STIX]{x1D644}_{t}\unicode[STIX]{x1D734}+\int _{S_{w}}P(X,Y,Z,t)(\boldsymbol{x}_{w}\times \boldsymbol{n}_{b})\,\text{d}S=\mathbf{0}.\end{eqnarray}$$

Invariance of ${\mathcal{L}}$ with respect to $\unicode[STIX]{x1D64C}^{-1}\unicode[STIX]{x1D6FF}\boldsymbol{q}$ gives the hydrodynamic equation of motion for the translational motion $\boldsymbol{q}(t)$ of the floating rigid body containing fluid and interacting with exterior water waves

(5.15) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{{\mathcal{V}}}\left(\frac{\text{D}\boldsymbol{u}}{\text{D}t}+2\unicode[STIX]{x1D734}\times \boldsymbol{u}\right)\unicode[STIX]{x1D70C}\,\text{d}\boldsymbol{x}+m\unicode[STIX]{x1D64C}^{-1}\ddot{\boldsymbol{q}}+\dot{\unicode[STIX]{x1D734}}\times m\overline{\boldsymbol{x}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\unicode[STIX]{x1D734}\times (\unicode[STIX]{x1D734}\times m\overline{\boldsymbol{x}})+mg\unicode[STIX]{x1D72E}-\int _{S_{w}}P(X,Y,Z,t)\boldsymbol{n}_{b}\,\text{d}S=\mathbf{0}.\end{eqnarray}$$

Finally, the invariance of ${\mathcal{L}}$ with respect to $\unicode[STIX]{x1D6FF}\boldsymbol{x}$ and $\unicode[STIX]{x1D6FF}p$ gives the Euler equations and continuity equation for the interior inviscid and incompressible fluid of the rigid body in (4.12), respectively. The terms including the pressure field $P(X,Y,Z,t)$ in the hydrodynamic equations of motion (5.14) and (5.15) are the moments and forces respectively acting on the rigid body due to interactions with exterior water waves.

In summary, the equations of motion for the exterior water waves in $V(t)$ are (5.1) with the contact boundary condition (5.13). The equations of motion for the interior fluid of the rigid body are (4.12) which are dynamically coupled to the hydrodynamic equations of motion for the floating rigid body (5.14) and (5.15). The evolutionary system for the rigid body motion is completed by (3.37) and (3.38).

6 Concluding remarks

The paper is devoted to the derivation of a variational principle for dynamic coupling between a rigid body undergoing three-dimensional rotational and translational motions and its interior inviscid and incompressible fluid motion. The Euler–Poincaré reduction framework of rigid body dynamics is used to derive the exact differential equations for the linear momentum and angular momentum of the rigid body containing fluid, and the Euler equations for the motion of the interior fluid of the body relative to the rotating–translating coordinate system attached to the moving rigid body. The variational principle is extended to the problem of three-dimensional interactions between gravity-driven water waves and a freely floating rigid body dynamically coupled to its interior fluid motion. The exact nonlinear hydrodynamic equations of motion for the angular momentum and linear momentum of the floating body are derived.

The presented variational principle (4.5) for the coupled fluid–body dynamics and the variational principle (5.3) for wave–structure–slosh interactions can be a starting point for further analytical and numerical analysis of the dynamics of a liquid-filled spacecraft, fluid sloshing dynamics in moving tanks, a freely floating ship with fluid-filled tanks in hydrodynamic interaction with exterior water waves and the dynamics of floating structures such as ducted wave energy converters (Leybourne et al. Reference Leybourne, Batten, Bahaj, Minns and O’Nians2014) and offshore wind turbines (Calderer et al. Reference Calderer, Guo, Shen and Sotiropoulos2018) interacting with ocean waves. The proposed variational principle (4.5) and the corresponding Euler–Poincaré equations can be reduced for mathematical modelling of a suspended container with interior fluid sloshing undergoing pendular oscillations and constrained to rotate on the surface of sphere.

Variational principles are useful in constructing structure-preserving numerical schemes. Instead of discretising the Euler–Lagrange or Euler–Poincaré equations, a discretisation of the Lagrangian formulation allows for the derivation of integrators that are symplectic, preserve energy over long-time integration and respect a discrete version of Kelvin–Noether theorem which holds for solutions of the Euler–Poincaré equations. Gagarina et al. (Reference Gagarina, Ambati, van der Vegt and Bokhove2014) developed a variational finite element method for nonlinear free-surface gravity water waves using the potential-flow approximation for an inviscid and incompressible fluid with an irrotational velocity field. Their formulation stems from Luke’s and Miles’ variational principle (Luke Reference Luke1967; Miles Reference Miles1977) together with a space-plus-time approach for finite element discretisations that are continuous in space and discontinuous in time. Pavlov et al. (Reference Pavlov, Mullen, Tong, Kanso, Marsden and Desbrun2011) developed variational Lie group integrators for incompressible perfect fluids based on the Hamilton–d’Alembert’s principle, which are structure-preserving, exhibit long-time energy behaviour and give rise to a discrete form of Kelvin’s circulation theorem. Desbrun et al. (Reference Desbrun, Gawlik, Gay-Balmaz and Zeitlin2014) proposed structure-preserving Euler–Poincaré variational discretisations for rotating Euler equations and 2-D stratified flow in the Boussinesq approximation, based on a finite-dimensional approximation of the group of volume-preserving diffeomorphisms. A direction of great interest is to extend the variational symplectic methods of Gagarina et al. (Reference Gagarina, Ambati, van der Vegt and Bokhove2014, Reference Gagarina, Ambati, Nurijanyan, van der Vegt and Bokhove2016) and Kalogirou & Bokhove (Reference Kalogirou and Bokhove2016), for the exterior potential water waves, and the geometric structure-preserving discretisations of continuum theories by Gawlik et al. (Reference Gawlik, Mullen, Pavlov, Marsden and Desbrun2011), Pavlov et al. (Reference Pavlov, Mullen, Tong, Kanso, Marsden and Desbrun2011) and Desbrun et al. (Reference Desbrun, Gawlik, Gay-Balmaz and Zeitlin2014), for the interior rotating Euler equations, to develop hybrid numerical discretisations for the proposed variational principle (5.3) for 3-D interactions between exterior surface waves and a floating structure with interior fluid sloshing.