notes

chap1.tex (25268B)

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 \section{Governing Equations of Fluid Mechanics}
 We first start off with a fluid with a density given by
 \begin{align}
     \rho(\mathbf{x}, t),
 \end{align}
 in three dimensional Cartesian coordinates $\mathbf{x} = (x, y, z)$ at time
 $t$. For water-wave applications, we should note that we take
 $\rho=\text{constant}$, but we will go into this fact later. The fluid moves
 in time and space with a velocity field
 \begin{align}
     \mathbf{u}(\mathbf{x}, t) = (u, v, w).
 \end{align}
 Additionally it is also described by its pressure
 \begin{align}
     P(\mathbf{x}, t),
 \end{align}
 generally depending on time and position. When thinking of e.g. water the
 pressure increases the deeper we go, that is with decreasing or increasing $z$
 direction (depending on how we set up our system $z$ pointing up or down
 respectively).
 
 The general assumption in fluid mechanics is the \textbf{Continuum
 Hypothesis}, which assumes continuity of $\textbf{u}, \rho$ and $P$ in
 $\mathbf{x}$ and $t$. In other words, we premise that the velocity field,
 density and pressure are ''nice enough`` functions of position and time, such
 that we can do all the differential operations we desire in the framework of
 fluid mechanics.
 \subsection{Mass Conservation}
 Our aim is to derive a model of the fluid and its mechanics, with respect to
 time and position, in the most general way. This is usually done thinking
 of the density of a given fluid, which is a unit mass per unit volume,
 intrinsically  an integral representation to derive these equations suggests
 by itself.
 
 Let us now think of an arbitrary fluid. Within this fluid we define a fixed
 volume $V$ relative to a chosen inertial frame and bound it by a surface $S$
 within the fluid, such that the fluid motion $\mathbf{u}(\mathbf{x}, t)$ may
 cross the surface $S$. The fluid density is given by $\rho(\mathbf{x}, t)$,
 thereby the mass of the fluid in the defined Volume $V$ is an integral
 expression
 \begin{align}
     m = \int_V \rho(\mathbf{x}, t) dV.
 \end{align}
 The figure bellow \ref{fig:volume}, expresses the above described picture.
 \begin{figure}[H]
     \centering
   \begin{tikzpicture}[>=latex,scale=1, xscale=1, opacity=.8]
 % second sphere
     \begin{scope}[rotate=10, xscale=3, yscale=2, shift={(2.3,-0.2)}]
       \coordinate (O) at (0,0);
       \shade[ball color=gray!10!] (0,0) coordinate(Hp) circle (1) ;
 
       \draw[thick] (O) circle (1);
       \draw[rotate=5] (O) ellipse (1cm and 0.66cm);
       \draw[rotate=90] (O) ellipse (1cm and 0.33cm);
 \node[circle, fill=black, inner sep=1pt] at (0.15, 0.25) {} ; \draw[-latex, thick] (0.15, 0.25) -- (1, 1) ;
       \node[right] at (1, 1) {$\mathbf{u}(\mathbf{x}, t)$};
 
       \node[] at (O) {$V$};
       \node[] at (0.55, -0.25) {$\rho(\mathbf{x}, t)$};
 
       \draw[-] (0.76, -0.66) -- (1.2, -0.7);
       \node[right] at (1.2, -0.7) {$S$};
 
       \draw[-latex, thick] (-0.25, -0.65) -- (-1, -1);
       \node[left] at (-1, -1) {$\mathbf{n}$};
 
     \end{scope}
 
 % axis
   \end{tikzpicture}
   \caption{Volume bounded by a surface in a fluid with density and momentum,
   with a surface normal vector $\mathbf{n}$ \label{fig:volume}}
 \end{figure}
 
 Since we want to figure out the fluid's mechanics, we can consider the rate
 of change in the completely arbitrary volume $V$, by
 \begin{align}
     \frac{d}{dt}\left( \int_V \rho(\mathbf{x}, t)\ dV \right) = \int_V
     \frac{\partial \rho(\mathbf{x}, t)}{\partial t} \ dV
 \end{align}
 On the other hand we have that density is mass over volume, meaning
 \begin{align}
     dm = \rho \cdot dV.
 \end{align}
 The infinitesimal volume has the base area $dS$ with hight $h$, which is the
 distance in the direction perpendicular to to the base area, leaving us with
 $dV = h\ dS$. By definition $\mathbf{n}$ is perpendicular to $dS$, we have
 that $h = l \mathbf{n}$. Where $l$ is the unit of length, or velocity times
 time $l = \mathbf{u}\ dt$, since mass is flowing out of the surface we
 change the sign of the flow leading us to
 \begin{align}
    dm  = -\rho dV = -\rho \mathbf{u} \cdot \mathbf{n}\ dt\ dS,
 \end{align}
 All together we have
 \begin{align}
     \frac{dm}{dt} = -\int_{\partial V} \rho(\mathbf{x},t)
     \mathbf{u}\cdot\mathbf{n}\ dS.
 \end{align}
 Putting both equations on one side leaves us with the equation
 \begin{align}\label{eq:mass balance}
      \int_V \frac{\partial \rho(\mathbf{x}, t)}{\partial t}\ dV
     +\int_{\partial V} \rho(\mathbf{x}, t) \mathbf{u}\cdot\mathbf{n}\ dS
     = 0.
 \end{align}
 The above equation in \ref{eq:mass balance} is an underlying equation, describing that the rate of
 change of mass in V is brought about, only by the rate of mass flowing into
 V across S, and thus the mass does not change.
 
 For the second integral in \ref{eq:mass balance} we utilize the Gaussian
 integration law to acquire an integral over the volume
 \begin{align}
     \int_{\partial V} \rho(\mathbf{x}, t) \mathbf{u} \cdot \mathbf{n} \ dS =
     \int_V \nabla (\rho \mathbf{u})\ dV.
 \end{align}
 Thereby we can put everything inside the volume integral
 \begin{align}
     \int_V \left(\partial_t \rho + \nabla(\rho \mathbf{u}) \right) \ dV = 0.
 \end{align}
 Everything under the integral sign needs to be zero, thus we obtain
 the \textbf{Equation of Mass Conservation} or in the general sense also
 called the \textbf{Continuity Equation}
 \begin{align}\label{eq:continuity}
     \partial_t \rho + \nabla(\rho \mathbf{u}) = 0
 \end{align}
 
 In light of the results of the equation of mass conservation
 in \ref{eq:continuity}, the product rule gives
 \begin{align}
     \partial_t \rho + (\nabla \rho)\mathbf{u} + \rho(\nabla \mathbf{u}),
 \end{align}
 for notational purposes, we define the \textbf{material/convective derivative}
 as follows
 \begin{align}
     \frac{D}{Dt} = \frac{\partial }{\partial t}  + \mathbf{u}\nabla.
 \end{align}
 With the material derivative the equation of mass conservation reads
 \begin{align}
     \frac{D\rho}{Dt} + \rho \nabla\mathbf{u} = 0
 \end{align}
 We may undertake the first case separation, initiating $\rho = \text{cosnt.}$
 called \textbf{incompressible flow} causes the material derivative of $\rho$ to
 be zero, and thereby
 \begin{align}
     \frac{D\rho}{Dt} = 0 \quad \Rightarrow \quad \nabla \mathbf{u} = 0,
 \end{align}
 following that the divergence of the velocity field is zero, in this case
 $\mathbf{u}$ is called \textbf{solenoidal}.
 \subsection{Euler's Equation of Motion}
 Additional consideration we undertake is the assumption of an
 \textbf{inviscid} fluid, that is we set viscosity to zero. Otherwise we would
 get a viscous contribution under the integral which results in the
 Navier-Stokes equation. In this regard we apply Newton's second law to our
 fluid in terms of infinitesimal pieces $\delta V$ of the fluid. The
 acceleration divides into two terms, a \textbf{body force} given by gravity
 of earth in the $z$ coordinate $\mathbf{F} = (0, 0, -g)$ and a
 \textbf{local/short-rage force} described by the stress tensor in the fluid.
 In the inviscid case, the local force retains the pressure $P$, producing a
 normal force, with respect to the surface, acting onto the infinitesimal
 element in the fluid. Summing over all infinitesimal pieces of the fluid,
 gives us an integral formulation of the force
 \begin{align}
     \int_V \rho \mathbf{F}\ dV - \int_S P\mathbf{n}\ dV.
 \end{align}
 Now applying the Gaussian rule of integration on the second integral over the
 surface, the resulting force in per unit volume is
 \begin{align}
     \int_V \left(\rho \mathbf{F} - \nabla P\right)\ dV.
 \end{align}
 The acceleration of the fluid particles is given by $\frac{D\mathbf{u}}{Dt}$,
 and thus the total force per unit volume on the other hand is
 \begin{align}
     \int_V \rho \frac{D\mathbf{u}}{Dt}\ dV =
     \int_V \left(\rho \mathbf{F} - \nabla P\right)\ dV.
 \end{align}
 Newton's Second Law for a fluid in a Volume is essentially saying that the
 rate of change of momentum of the fluid in the fixed volume $V$, which is the particle
 acceleration is the resulting force acting on V together with the rate of
 flow of momentum across the surface $S$ into the volume $V$. Hence we arrive
 at the \textbf{Euler's Equation(s) of Motion}
 \begin{align}
     \frac{D\mathbf{u}}{Dt} = \left(\frac{\partial \mathbf{u}}{\partial t}
     +(\mathbf{u}\nabla)\mathbf{u}\right)  =
     -\frac{1}{\rho}\nabla P + \mathbf{F}.
 \end{align}
 As a side note we have mentioned that there is another contribution if the
 fluid is viscid. Indeed there is a tangential force due to the velocity
 gradient, which into introduces the additional term
 \begin{align}
     \mu \nabla^2 \mathbf{u}, \qquad
     \mu = \text{viscosity of the Fluid}.
 \end{align}
 Thereby the equations become
 \begin{align}
     \rho\frac{D\mathbf{u}}{Dt}
     =  -\nabla P + \rho \mathbf{F} + \mu \nabla^2 \mathbf{u}.
 \end{align}
 
 For now we have separated two simplifications, that define an
 \textbf{idealized/perfect fluid}
 \begin{enumerate}
     \item \textbf{inviscid} $\qquad \mu=0$
     \item  \textbf{incompressible} $\quad \rho = \text{const.},\ \nabla \mathbf{u}=
         0$
 \end{enumerate}
 \subsection{Vorticity and irrotational Flow}
 The curl of the velocity field $\mathbf{\omega} = \nabla \times \mathbf{u}$
 of a fluid (i.e. the vorticity), describes a ``spinning'' motion of the fluid
 near a position $\mathbf{x}$ at time $t$. The vorticity is an important
 property of a fluid. Flows or regions of flows where $\mathbf{\omega}=0$ are
 \textbf{irrotational}, and thus can be modeled and analyzed following well
 known routine methods. Even though real flows are rarely irrotational
 anywhere (!), in water wave theory wave problems, from the classical aspect
 of vorticity has only minor contributions. Hence we can assume irrotational flow
 modeling water waves. To arrive at the vorticity in the equations of motions
 derived in the last section we resort to a differential identity derived in appendix
 \ref{appendix:diff identity}, rewriting the motion derivative in terms of
 \begin{align}
     \frac{D\mathbf{u}}{Dt} = \frac{\partial \mathbf{u}}{\partial t}
     +\nabla(\frac{1}{2}\mathbf{u}\mathbf{u)}
     - \left( \mathbf{u}\times (\nabla \times  \mathbf{u} \right).
 \end{align}
 Thus the equations of motion become
 \begin{align}
     \frac{\partial \mathbf{u}}{\partial t} + \nabla\left(
     \frac{1}{2}\mathbf{u}\mathbf{u} + \frac{P}{\rho} + \Omega \right)
     = \mathbf{u} \times  \mathbf{\omega},
 \end{align}
 where $\Omega$ is the force potential per
 unite mass given by $\mathbf{F} = -\nabla \Omega$.
 
 At this point we may differentiate between \textbf{steady and unsteady flow}.
 For \textbf{Steady Flow} we assume that $\mathbf{u}, P$ and $\Omega$ are time
 independent, thus we get
 \begin{align}
       \nabla\left( \frac{1}{2}\mathbf{u}\mathbf{u} + \frac{P}{\rho} + \Omega
       \right)  = \mathbf{u} \times  \mathbf{\omega}.
 \end{align}
 It is general knowledge that the gradient of a function $\nabla f$ is
 perpendicular the level sets of $f(\mathbf{x})$, where $f(\mathbf{x}) =
 \text{const.}$. Thus $\mathbf{u} \times  \mathbf{\omega}$ is orthogonal to
 the surfaces  where
 \begin{align} \label{eq:bernoulli}
     \frac{1}{2}\mathbf{u}\mathbf{u} + \frac{P}{\rho} + \Omega =
     \text{const.},
 \end{align}
 The above equation is called \textbf{Bernoulli's Equation}.
 
 Secondly \textbf{Unsteady Flow} but irrotational (+ incompressible), first of
 all gives us the condition for the existence of a velocity potential $\phi$
 in the sense
 \begin{align}
     \mathbf{\omega} = \nabla \times  \mathbf{u} = 0  \quad \Rightarrow \quad
     \mathbf{u} = \nabla \phi,
 \end{align}
 where $\phi$ needs to satisfy the Laplace equation
 \begin{align}
     \Delta \phi = 0.
 \end{align}
 According to the Theorem of Schwartz we may exchange $\frac{\partial
 }{\partial t}$ and $\nabla$, giving us the expression
 \begin{align}
     \nabla\left( \frac{\partial \phi}{\partial t} +\frac{1}{2}
     \mathbf{u}\mathbf{u} + \frac{P}{\rho}  + \Omega \right) = 0
 \end{align}
 Thus the expression differentiated by the $\nabla$ operator is an arbitrary
 function $f(\mathbf{x}, t)$, writing
 \begin{align}
      \frac{\partial \phi}{\partial t} +\frac{1}{2}
     \mathbf{u}\mathbf{u} + \frac{P}{\rho}  + \Omega = f(\mathbf{x}, t).
 \end{align}
 The function $f(\mathbf{x}, t)$ can be removed by gauge transformation of
 $\phi \rightarrow \phi + \int f(\mathbf{x}, t)\ dt$.
 \subsection{Boundary Conditions for water waves}
 The boundary conditions for water-wave problems vary, generally on the
 simplification we undertake. At the surface, called the free surface as in
 free from the velocity conditions, we have the atmospheric stress on the
 fluid. The stress component would again have a viscid component, this however
 is only relevant when modeling surface wind, in this review we model the
 fluid within reason as inviscid. The atmosphere employs
 only a pressure on the surface, this pressure is taken to be the atmospheric
 pressure, dependent on time and point in space. Thereby  any surface tension
 effects can also include a scenario at a curved surface (e.g. wave), giving
 rise to the pressure difference across the surface. A more precise
 description would use Thermomechanics to derive boundary conditions coupling
 water surface and the air above it, yet the density component of air
 compared to that of water makes our ansatz viable. The described conditions
 are called the \textbf{dynamic conditions}
 
 An additional condition revolves around the fluid particles on the moving
 surface, called the \textbf{kinematic condition}. This condition bounds
 the vertical velocity component on the surface.
 
 The logical step now is to define boundary conditions on the bed of the
 fluid. In the viscid case bottom is impermeable, we a no
 slip condition to all fluid particles $\mathbf{u}_\text{bottom}= 0$. If we
 assume that the fluid is inviscid then the bottom becomes a surface of the
 fluid in the sense that the fluid particles in contact with the bed move in
 the surface, we more or less mirror the kinematic condition of the surface.
 For many problems the condition is going to vary, in most cases the bottom
 will be rigid and fixed not necessarily horizontal. This condition is simply
 called the \textbf{bottom condition}.
 \subsubsection{Kinematic Condition}
 Obtaining the free surface is the primary objective in the theory of modeling
 water waves, represented by
 \begin{align}
     z = h(\mathbf{x}_\perp, t),
 \end{align}
 where $\mathbf{x}_\perp = (x, y)$ in Cartesian, or $\mathbf{x}_\perp = (r,
 \theta)$ in cylindrical coordinates. A surfaces that moves with the fluid,
 always contains the same fluid particles, described as
 \begin{align}
     \frac{D}{Dt}\left(z - h(\mathbf{x}_\perp, t ) \right) = 0.
 \end{align}
 Upon expanding the derivative we get
 \begin{align}
     \frac{Dz}{Dt} - \frac{Dh}{Dt}
     &= \frac{\partial z}{\partial t}+
     (\mathbf{u}\nabla)z - \frac{\partial h}{\partial t} -(\mathbf{u}\nabla)\\
     &= w - \left(h_t - (\mathbf{u}_\perp \nabla_\perp) h\right) = 0,
 \end{align}
 where the subscript $\perp$ describes the components with regard to
 $\mathbf{x}_\perp$. The \textbf{kinematic condition} reads
 \begin{align}
     w = h_t - (\mathbf{u}_\perp \nabla_\perp) h \qquad \text{on}\;\;
     z=h(\mathbf{u}_\perp, t).
 \end{align}
 
 \subsubsection{Dynamic Condition}
 As described in the prescript of this section, the case of an inviscid fluid,
 requires that only the pressure $P$ needs to be described on the free surface
 $z = h(\mathbf{x}_\perp, t)$. Assuming incompressible, irrotational,
 unsteady flow and setting $P=P_a$ for atmospheric pressure and $\Omega =
 g\cdot z$ for the force per unit mass potential the equations of motion are
 \begin{align}
     \frac{\partial \phi}{\partial t} +\frac{1}{2}\mathbf{u}\mathbf{u}
     + \frac{P_a}{\rho}+gh = f(t) \qquad \text{on}\;\; z=h.
 \end{align}
 Somewhere $\|\mathbf{x}_\perp\| \rightarrow \infty$ the fluid reaches
 equilibrium and is thereby stationary, meaning it has no motion and the
 pressure is $P=P_a$, the surface is a constant $h = h_0$ and therefore
 \begin{align}
     f(t) = \frac{P_a}{\rho}+gh_0.
 \end{align}
 The \textbf{dynamic condition} may be written as
 \begin{align}
     \frac{\partial \phi}{\partial t}
     +\frac{1}{2}\mathbf{u}\mathbf{u}+g(h-h_0) = 0 \qquad \text{on}\;\; z=h.
 \end{align}
 
 Regarding the pressure difference on a curved surface, we may expand the
 dynamic condition by introducing the pressure difference known as the
 \textbf{Young-Laplace Equation}
 \begin{align}
     \Delta P = \frac{\Gamma}{R},
 \end{align}
 where $\Gamma>0$ is the coefficient of surface tension and $\frac{1}{R}$ is
 the curvature representing an implicit function, in our case the implicit
 function is $z - h(\mathbf{x}_\perp, t)=0$ for fixed time. The curvature in
 Cartesian coordinates takes the form
 \begin{align}
     \frac{1}{R} = \frac{(1+h_y^2)h_{x x}+(1+h_y^2)h_{yy} -
     2h_xh_yh_{xy}}{\left( h_x^2+h_y^2+1 \right)^{\frac{3}{2}} },
 \end{align}
 the derivation is precisely described in \ref{appendix:curvature}
 
 
 
 \subsubsection{The Bottom Condition}
 The representation for the bottom is
 \begin{align}
    z = b(\mathbf{x}_\perp, t),
 \end{align}
 where the fluid surface needs to satisfy
 \begin{align}
     \frac{D}{Dt} \left(z - b(\mathbf{x}_\perp) \right)  = 0.
 \end{align}
 Hence we arrive at the bottom boundary conditions
 \begin{align}
     w = b_t + (\mathbf{u}_\perp \nabla_\perp)b \qquad \text{on}\;\; z=b ,
 \end{align}
 where $b(\mathbf{x}_\perp, t)$ is already known for most water wave
 problems. If we consider a stationary bottom then the time derivative
 vanishes, leaving us with the following condition
 \begin{align}
     w = (\mathbf{u}_\perp \nabla_\perp)b \qquad \text{on}\;\; z=b
 \end{align}
 
 
 \subsubsection{Integrated Mass Condition}
 In this section we want to combine the kinematics of both the free and the
 bottom surface with the mass conservation equation on the perpendicular
 components
 \begin{align}
     \nabla \mathbf{u} = \nabla_\perp \mathbf{u}_\perp + w_z = 0 .
 \end{align}
 Integrating the above expression from bottom to surface, i.e. from
 $z=b(\mathbf{x}_\perp,t)$ to $z = h (\mathbf{x},t)$ gives
 \begin{align}
     \int_b^h \nabla_\perp \mathbf{u}_\perp\ dz + w\bigg|_{z=b}^{z=h} = 0,
 \end{align}
 where we insert the conditions on the free surface and on the bottom surface
 \begin{align}
     w &= h_t + (\mathbf{u}_{\perp \text{s}} \nabla_\perp) h \quad
     \text{on}\;\; z = h\\
     w &= b_t + (\mathbf{u}_{\perp \text{b}} \nabla_\perp) h \quad
     \text{on}\;\; z =b,
 \end{align}
 with the subscript $s$ and $b$ indicating the evaluation of a quantity
 on the free surface and the bottom surface respectively. Inserting the
 boundary conditions we get
 \begin{align}
     \int_b^h \nabla_\perp \mathbf{u}_\perp
     + h_t + (\mathbf{u}_{\perp \text{s}} \nabla_\perp) h,
     - b_t - (\mathbf{u}_{\perp \text{b}} \nabla_\perp) b= 0.
 \end{align}
 To simplify the equation we resort to the Leibniz Rule of Integration
 \ref{appendix:leibniz},
 \begin{align}
      \int_b^h \nabla_\perp\mathbf{u}_\perp =
     \nabla_\perp \int_b^h \mathbf{u}_\perp\ dz - (\mathbf{u}_{\perp \text{s}}
     \nabla_\perp)h + (\mathbf{u}_{\perp \text{b}})b.
 \end{align}
 As a consequence the \textbf{Integrated Mass Condition} is given by
 \begin{align}
     \nabla_\perp \int_b^h \mathbf{u}_\perp\ dz  + \underbrace{h_t -
     b_t}_{=d_t} = 0.
 \end{align}
 %\subsection{Energy Equation}
 %To derive the energy equation we start off with Euler's Equation of Motion
 %\begin{align}
 %  \mathbf{u} _t + \nabla
 %  (\frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}+\Omega) = \mathbf{u}\times
 %  \mathbf{w},
 %\end{align}
 %multiplying the equation with $\mathbf{u}$ we get
 %\begin{align}
 %  &\mathbf{u}\mathbf{u} _t \label{eq:energy1} \\
 %  &+(\mathbf{u}\nabla)(\frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}+\Omega)\label{eq:energy2}\\
 %  &= \mathbf{u}(\mathbf{u}\times
 %  \mathbf{w})\label{eq:energy3}.
 %\end{align}
 %The first equation given in \ref{eq:energy1} can we rewritten using inverse
 %product rule of differentiation
 %\begin{align}
 %    \mathbf{u}\frac{\partial \mathbf{u}}{\partial t}
 %    &= \frac{\partial
 %    }{\partial t} (\mathbf{u}\mathbf{u}) - \frac{\partial \mathbf{u}}{\partial t}
 %    \mathbf{u} \\
 %    &= \frac{\partial
 %    }{\partial t} (\mathbf{u}\mathbf{u}) - \mathbf{u}\frac{\partial
 %    \mathbf{u}}{\partial t}\\
 %    \Rightarrow\quad & \mathbf{u} \frac{\partial \mathbf{u}}{\partial t}  =
 %    \frac{1}{2}\frac{\partial }{\partial t} (\mathbf{u}\mathbf{u}).
 %\end{align}
 %Then we may add
 %\begin{align}
 %    \left(\frac{1}{2} \mathbf{u}\mathbf{u}+\frac{P}{\rho} +\Omega  \right)
 %    \underbrace{(\nabla u)}_{=0} = 0,
 %\end{align}
 %to above not changing anything. Thereby getting
 %\begin{align}
 %    \frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}\right)
 %    +(\mathbf{u}\nabla \mathbf{u})\left(
 %    \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho} \right)
 %    +\left( \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho} + \Omega \right)
 %    (\nabla \mathbf{u}) = 0.
 %\end{align}
 %Applying the product rule we can simplify
 %\begin{align}
 %    \frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}\right)
 %    +\nabla \left(\mathbf{u}\left(\mathbf{u}(
 %    \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}\right) \right) = 0,
 %\end{align}
 %additionally adding $\frac{\partial \Omega}{\partial t}  =0$ leads us to
 %\begin{align}
 %    \underbrace{\frac{\partial }{\partial t} \left(\frac{1}{2}\mathbf{u}\mathbf{u}
 %    +\Omega\right)}_{\text{change of total energy density}}
 %    +\underbrace{\nabla \left(\mathbf{u}\left(\mathbf{u}(
 %    \frac{1}{2}\mathbf{u}\mathbf{u}+\frac{P}{\rho}\right)
 %\right)}_{\text{energy flow of the velocity field}} = 0.\label{eq:energy}
 %\end{align}
 %This is called the \textbf{energy equation} and is a general result for a
 %inviscid and incompressible fluids, which we can apply to study water waves.
 %We start off with replacing $\nabla = \nabla_\perp + \frac{\partial }{\partial
 %z} $ and $\Omega = g z$ and multiplying by $\rho$, then our energy equation
 %in \ref{eq:energy} becomes
 %\begin{align}
 %    \frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
 %    g z\right)  + \nabla_\perp\left( \mathbf{u}_\perp\left(
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right)  \right)
 %    \frac{\partial}{\partial z} \left( w\left(
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho g z \right)  \right)  = 0.
 %\end{align}
 %Integrating from bottom to top, i.e. from bed to free surface gets us to
 %\begin{align}
 %    &\int_b^h\frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
 %    g z\right)\ dz  \label{eq:e-int1}\\
 %    &+ \int_b^h \nabla_\perp\left( \mathbf{u}_\perp\left(
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right)  \right)\
 %    dz\label{eq:e-int2}\\
 %    &+ \left(\frac{\partial}{\partial z} \left( w\left(
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho g z \right)
 %\right)\right)\Bigg|_b^h \label{eq:e-int3}
 %    = 0.
 %\end{align}
 %For equation \ref{eq:e-int1} we use Leibniz Rule of Integration, leaving us
 %with
 %\begin{align}
 %    \int_b^h\frac{\partial }{\partial t}\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho
 %    g z\right)\ dz
 %    &= \frac{\partial }{\partial t} \int_b^h
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho gz \ dz\\
 %    &+ \left( \frac{1}{2}\rho \mathbf{u}_s \mathbf{u}_s  + \rho g h \right)
 %    h_t\\
 %    &- \left( \frac{1}{2}\rho \mathbf{u}_b \mathbf{u}_b  + \rho g b \right)
 %    b_t
 %\end{align}
 %For equation \ref{eq:e-int2} we again take note of the Leibniz Rule of
 %Integration, getting
 %\begin{align}
 %    \int_b^h \nabla_\perp\left( \mathbf{u}_\perp\left(
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u}+P+\rho gz \right)  \right)\
 %    dz
 %    &= \nabla_\perp \int_b^h \mathbf{u}_\perp\left(
 %    \frac{1}{2}\rho\mathbf{u}\mathbf{u} + P + \rho g z \right) \ dz\\
 %    &- \left( \frac{1}{2}\rho \mathbf{u}_s\mathbf{u}_s + P + \rho g h \right)
 %    \left( \mathbf{u}_{\perp s} \nabla_\perp \right) h\\
 %    &+\left( \frac{1}{2}\rho \mathbf{u}_b\mathbf{u}_b + P + \rho g b \right)
 %    \left( \mathbf{u}_{\perp b} \nabla_\perp \right) b
 %\end{align}
 %Thereby transforming our equation into
 %\begin{align}
 %    \frac{\partial }{\partial t} \underbrace{\int_b^h \frac{1}{2}\rho
 %    \mathbf{u}\mathbf{u}+\rho g z\ dz}_{=:\mathcal{E}}
 %    + \nabla_\perp&\underbrace{\int_b^h
 %    \mathbf{u}_\perp\left( \frac{1}{2}\rho\mathbf{u}\mathbf{u} + \rho g z
 %\right)\ dz}_{:=\mathcal{F}}
 %+ \underbrace{P_s h_t - P_b b_t}_{:=\mathcal{P}} = 0\\
 %\nonumber\\
 %    &\frac{\partial \mathcal{E}}{\partial t}
 %    + \nabla_\perp \mathcal{F} + \mathcal{P} = 0,
 %\end{align}
 %where $\mathcal{E}$ represents the energy in the flow per unit horizontal
 %area, since we are integrating from bed to free surface. Where $\mathcal{F}$
 %is the horizontal energy flux vector and lastly $\mathcal{P} = P_s h_t -
 %P_b b_t$ is the net energy input due to the pressure forces doing work on the
 %upper and lower boundaries, i.e. bottom and free surface of the fluid.
 %Assuming stationary rigid bottom condition and constant surface pressure, we
 %can set $P_s=0$, such that $\mathcal{P} =0$ leaving us with the equation
 %\begin{align}
 %    \frac{\partial \mathcal{E}}{\partial t}
 %    + \nabla_\perp \mathcal{F} = 0.
 %\end{align}
 %We note that the assumption $P_s=0$ is only possible if the coefficient of
 %surface tension is set to 0, which usually is not the case.