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\textbf{Question}

\begin{description}
%Question 1a
\item{(a)}
Give a brief explanation of the difference between $Eulerian$ and
$Lagrangian$ descriptions of a fluid flow. Denoting the Eulerian
coordinates as usual by $\un{x}$, prove that for any suitably smooth
function $\phi (\un{x},t)$
$$\frac{d\phi}{dt} = \phi_t + (\un{q}.\nabla) \phi$$
where $\un{q}$ is the flow velocity and $d/dt$ denotes the Lagrangian
derivative.

YOU MAY ASSUME that the equation of motion of a fluid with constant
density $\rho$, stress tensor $T$ and acceleration $\un{a}=d\un{q}/dt$
is
$$\textrm{div}T + \rho \un{b} = \rho \un{a} \longrightarrow (1)$$
where $\un{b}$ is the body force per unit mass.

\begin{description}
\item{(i)}
State from which of Newton's laws (1) is derived and identify the
physical quantity that is being conserved.

\item{(ii)}
Name the physical principle that may be used to show that $T=T^T$, and
therefore that the stress temsor is symmetric.

\item{(iii)}
Assuming that the stress tensor for an incompressible linear viscous
fluid is given by
$$T_{ij} = -p\delta_{ij}+ 2\mu e_{ij},$$
derive the Navier-Stokes equations for the flow of a viscous fluid.
\end{description}

%Question 1b
\item{(b)}
Using tensorial notation or otherwise, prove, for suitably smooth
vectors $\un{u}$ and $\un{v}$, the vector identities
\begin{description}

\item{(i)}
$$\nabla\times(\un{v}\times\un{v}) = (\un{v}.\nabla)\un{u}
-(\un{u}.\nabla)\un{v} +\un{u}\nabla.\un{v} - \un{v}\nabla.\un{u}$$

\item{(ii)}
$$\nabla\times(\nabla\times\un{u}) =
\nabla(\nabla.\un{u})=\nabla^2\un{u}$$
\end{description}

\end{description}

\newpage
\textbf{Answer}

\begin{description}

\item{EULERIAN} $\frac{\partial}{\partial x}$, fix attention on
point $\underline{x}$ fixed in space; seek to determine velocity
$\underline{q} (\underline{x} ,t)$

\item{LAGRANGIAN} $\frac{\partial}{\partial X}$, fix attention on
and follow a given fluid particle with position $\underline{X}$ at
$t=0$; seek to determine the motion via $\underline{x}
=\underline{x} (\underline{X} , t)$
\end{description}

Now $\displaystyle \left. \frac{d\phi}{dt}
\right|_{\underline{X}}$ means fix $\underline{X}$. Relative to
$\underline{x}$ instead, use chain rule. $$\frac{\partial
\phi}{\partial t} =\frac{\partial \phi}{\partial t}\frac{dt}{dt} +
\frac{\partial \phi}{\partial x}\frac{dx}{dt} + \frac{\partial
\phi}{\partial y}\frac{dy}{dt} + \frac{\partial \phi}{\partial
z}\frac{dz}{dt}$$ But by definition $\frac{\partial x}{\partial t}
=u$, $\frac{\partial y}{\partial t} = v$, $\frac{\partial
z}{\partial t} = w$.

$\Rightarrow \displaystyle \left. \frac{d\phi}{dt}
\right|_{\underline{X}} = \frac{\partial \phi}{\partial t} +
u\phi_x + v\phi_y + w\phi_t = \phi_t + (\underline{q}
.\nabla)\phi$

Now we have $div(T) + \rho\underline{b}=\rho \underline{\alpha}$

\begin{description}
\item{(i)} This is derived from Newtons 2nd law; linear momentum
is being conserved.
\item{(ii)} Conservation of angular (moment of) momentum leads to
the symmetry of the stress tensor.
\item{(iii)} With $$T_{ij}=-p\delta_{ij}+2\mu e_{ij}$$

\begin{eqnarray*} div(T)=\frac{\partial T_{ij}}{\partial x_j} & =
& -\delta_{ij}\frac{\partial p}{\partial
x_j}+\mu\frac{\partial}{\partial x_j}(\frac{\partial q_i}{\partial
x_j}+\frac{\partial q_j}{\partial x_i})\\ & = & -\frac{\partial
p}{\partial x_i} + \mu\nabla^2\underline{q}+\mu\nabla(\nabla
.\underline{q})
\end{eqnarray*}

So $div{T}=-\nabla p + \mu\nabla^2\underline{q} +
\mu\nabla(\nabla.\underline{q})$

\end{description}

Now $\nabla.\underline{q} = 0$ since incompressible.

$\Rightarrow -\nabla p + \mu\nabla^2q+ \rho\underline{b}=
\rho\underline{a}=\rho(\underline{q}_t+(\underline{q}.\nabla)\underline{q})$

$\Rightarrow \underline{q}_t + (\underline{q}.\nabla)\underline{q}
= -\frac{1}{\rho}\nabla p + \nu\nabla^2\underline{q}$
(Navier-Stokes)


\begin{description}

%Question 1b
\item{(b)}

\begin{eqnarray*} \nabla\times(\underline{u}\times\underline{v})
& = & \epsilon_{ijk}\frac{\partial}{\partial
x_j}(\underline{u}\times\underline{v})_k=\epsilon_{ijk}\frac{\partial}{\partial
x_j}(\epsilon_{klm}u_lv_m)\\ & = &
\epsilon_{kij}\epsilon_{klm}(\frac{\partial u_l}{\partial x_j}v_m
+ u_l\frac{\partial v_m}{\partial x_j})\\ & = &
(\delta_{il}\delta_{jm}-\delta_{jl}\delta_{mi})(v_m\frac{\partial
u_l}{\partial x_j}+u_l\frac{\partial v_m}{\partial x_j})\\ & = &
v_j\frac{\partial u_i}{\partial x_j}+ u_i\frac{\partial
v_j}{\partial x_j}-v_i\frac{\partial u_j}{\partial x_j}-
u_j\frac{\partial v_m}{\partial x_j}\\ & = &
(\underline{v}.\nabla)\underline{u}+
\underline{u}div(\underline{v}) -
\underline{v}div(\underline{u})-(\underline{u}.\nabla)\underline{v}
\end{eqnarray*}

\end{description}

\begin{eqnarray*} \nabla\times(\nabla\times\underline{u}) & = & \epsilon_{ijk}\frac{\partial}{\partial
x_j}(\nabla\times\underline{u})_k=\epsilon_{ijk}\frac{\partial}{\partial
x_j}(\epsilon_{klm}\frac{\partial u_m}{\partial x_l})\\ & = &
\epsilon_{kij} \epsilon_{klm}\frac{\partial^2 u_m}{\partial x_j
\partial x_l} = (\delta_{il}\delta_{jm}-\delta_{jl}\delta_{im})\frac{\partial^2 u_m}{\partial x_j \partial x_l}\\ & = & \frac{\partial^2
u_j}{{\partial x_j} \partial x_i} - \frac{\partial^2 u_i}{\partial
x_j \partial x_j} = \nabla(\nabla
.\underline{u})-\nabla^2\underline{u}
\end{eqnarray*}


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