<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt

\textbf{Question}

\begin{description}

%Question 1i
\item{(i)}
Using tensorial notation, or otherwise, prove the vector
identities $$\nabla . (\phi \un{a} ) = \phi \nabla . \un{a} +
(\un{a} . \nabla) \phi$$ $$\nabla \times (\nabla\times\un{a}) = -
\nabla^2\un{a} + \nabla (\nabla . \un{a} )$$ $$\un{a} \times (
\nabla\times\un{a} ) = \nabla ( \un{a}^2/2) - ( \un{a} . \nabla )
\un{a}$$ for a general suitably differentiable scalar function
$\phi$ and vector $\un{a}$.

%Question 1ii
\item{(ii)}
Explain briefly the major differences between the 'Eulerian' and
'Lagrangian' descriptions of flow. What does it mean to say that a
fluid flow is 'steady'? For a fluid flow with velocity $\un{q}$ define
$\un{\omega}$, the $vorticity$, and give the condition that the flow
is $irrotational$.

%Question 1iii
\item{(iii)}
YOU MAY ASSUME that the Navier-Stokes equations for an unsteady
viscous flow are given by
\begin{eqnarray*}
\un{q}_t+(\un{q}.\nabla) \un{q} &amp; = &amp; - \frac{1}{\rho} \nabla \un{p} + \nu
\nabla^2 \un{q} + \un{F}\\
\nabla . \un{q} &amp; = &amp; 0
\end{eqnarray*}
where $\rho$ denotes the (constant) density, $\nu$ the (constant)
kinematic viscosity, $\un{p}$ the pressure and $\un{F}$ the body
force. Show that for an inviscid irrotational flow with a conservative
body force the quantity
$$\phi_t+\frac{1}{2} |\un{q}|^2 + \chi$$
is a function of time alone, identifying the quantities $\phi$ and
$\chi$. Show also that for a steady (but not necessarily irrotational)
inviscid flow the quantity
$$\frac{\un{p}}{\rho} + \frac{1}{2} | \un{q}|^2 + \chi$$
is constant along both streamlines and vortex lines in the flow.

\end{description}

\newpage
\textbf{Answer}
%Question 1

\begin{description}
%Question 1i
\item{(i)}
\begin{eqnarray*} \displaystyle
div(\phi\underline{a})=\frac{\partial}{\partial a_j}(\phi a_j) &amp; =
&amp; \displaystyle \frac{\partial \phi}{\partial x_j}a_j +\phi
\frac{\partial a_j}{\partial x_j}\\ &amp; = &amp;
(\underline{a}.\nabla)\phi+\phi div(\underline{a}) \end{eqnarray*}

$\displaystyle \nabla\times(\nabla\times\underline{a})
=\epsilon_{ijk}\frac{\partial}{\partial
x_j}(\nabla\times\underline{a})_k=\epsilon_{ijk}\frac{\partial}{\partial
x_j}\epsilon_{kpq}\frac{\partial}{\partial x_p}a_q$

$\displaystyle =\epsilon_{ijk}\epsilon_{kpq}\frac{\partial^2
a_q}{\partial x_j \partial
x_p}=\epsilon_{kij}\epsilon_{kpq}\frac{\partial^2 a_q}{\partial
x_j \partial x_p}= (\delta_{ip}\delta_{jq}-
\delta_{jp}\delta_{iq})\frac{\partial^2 a_q}{\partial x_j \partial
x_p}$

$\displaystyle =\frac{\partial^2}{\partial x_j \partial x_i}-
\frac{\partial^2a_i}{\partial x_j \partial x_j} =
\nabla(div(\underline{a}))-\nabla^2\underline{a}$

$\displaystyle \underline{a}\times (\nabla
\times\underline{a})=\epsilon{ijk}
a_j(\nabla\times\underline{a})_k=\epsilon_{ijk}a_j\epsilon_{kpq}\frac{\partial
a_q}{\partial x_p}$

$\displaystyle =\epsilon_{kij}\epsilon_{kpq}a_j\frac{\partial
a_q}{\partial x_p}=(\delta_{ip}\delta_{jq}-
\delta_{jp}\delta_{iq})a_j\frac{\partial a_q}{\partial x_p}$

$\displaystyle = a_j\frac{\partial a_j}{\partial x_i}-
a_j\frac{\partial a_i}{\partial x_j} = \nabla
(\frac{1}{2}\underline{a}^2) -(\underline{a}\nabla)\underline{a}$

%Question 1ii
\item{(ii)}
EULERIAN: We try to find the fluid velocity $\underline{q}$ as a
function of $\underline{x}$ and $t$. Attention is focussed on a
particular position in the flow rather than on a particular fluid
particle.

LAGRANGIAN: We try to find the fluid motion $\underline{x}$ as a
function of $\underline{X}$ and $t$, where $\underline{X}$ denotes
the positions of fluid particles at $t=0$. Attention is focussed
on a given fluid particle following the flow.

A flow is STEADY if $\underline{q}$ does not depend on $t$.

If a flow has velocity $\underline{q}$ then $\underline{\omega}
=curl(\underline{q})$

For IRROTATIONAL FLOW $\underline{\omega}=0$


%Question 1iii
\item{(iii)}
We have $$\displaystyle \underline{q}_t+
(\underline{q}.\nabla)\underline{q}= -\frac{1}{\rho} \nabla p +\nu
\nabla^2\underline{q} +\underline{F},\ \ \nabla .\underline{q}=0$$

For irrotational flow, define $\underline{q}=\nabla \phi$ and for
a conservative body force define $\underline{F}=-\nabla \psi$.
Then

$(\nabla \phi)_t+(\underline{q}.\nabla)\underline{q}=-\nabla \left
( \frac{p}{\rho} \right ) -\nabla \psi$ since the fluid is
inviscid and $\rho$ is constant.

$\Rightarrow (\nabla\phi)_t +\nabla \left (
\frac{\underline{q}^2}{2} \right ) = \underline{q}
\times(\nabla\times\underline{q})+ \nabla\left ( \frac{p}{\rho}
\right ) +\nabla (\psi) =0$

But $\nabla\times\underline{q}=0$ so

$\nabla (\phi_t + \frac{1}{2}\underline{q}^2 + \frac{p}{\rho}
+\psi )=0$. The quantity in brackets is independent of $x,y$ and
$z$ and thus $$\displaystyle \phi_t +\frac{1}{2}|\underline{q}|^2
+ \frac{p}{\rho} + \psi = f(t) \ \ \rm{only}$$

($\phi = $velocity potential, $\psi =$ force potential)

For a steady inviscid flow $(\underline{q}.\nabla)\underline{q}=
-\nabla (p/\rho ) -\nabla\psi$

$\Rightarrow \nabla(\underline{q}^2/2)-\underline{q}\times
(\nabla\times\underline{q})=-\nabla (p/\rho ) -\nabla (\psi)$

$\Rightarrow
\underline{q}\times\underline{\omega}=\nabla(\underline{q}^2/2 +
p/\rho + \psi)$

Thence $\underline{q}.(\underline{q}\times\underline{\omega})= 0
=(\underline{q}.\nabla)(\underline{q}^2/2 +p/\rho +\psi)$ and so
$p/\rho +\frac{1}{2}|\underline{q}|^2+ \psi$ is constant along
streamlines in the flow.

Also
$\underline{\omega}.(\underline{q}\times\underline{\omega})=0$ and
so

$\underline{q}^2/2+ p/\rho + \psi$ is constant along vortex lines
in the flow.
\end{description}

\end{document}
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