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\newcommand{\un}{\underline}
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\textbf{Question}

BRIEFLY explain the difference between the $Lagrangian$ and $Eulerian$
dscripitons of flow. A fluid occupies the region $D(t)$, and at a time
$t=)$ position vectores in $D(0)$ are described by
$\un{X}=(X,Y,Z)$. At a later time $t$ position vectors in the fluid
are given by $\un{x}=\un{r}(\un{X}, t)$ where
\begin{eqnarray*}
x &amp; = &amp; X + Yt^2\\
y &amp; = &amp; Y + Y t^3\\
z &amp; = &amp; Z + 2tZ
\end{eqnarray*}

Determine bothe the Eulerian description of the flow (in the form
$\un{q} = \un{q}(\un{x},t)$ where $\un{q} = \partial \un{x} / \partial
t$) and the ``inverse Lagrangian'' description of the flow in the form
$\un{X} = \un{r}^{-1}(\un{x},t)$. Is this an incompressible flow?

Prove that, if $d/dt$ denotes time derivative with $\un{X}$ fixed and
$\partial / \partial t$ denotes a time derivatice with $\un{x}$ fixed,
the the ``convective differentiation'' formula
$$\frac{d\un{g}}{dt} = \frac{\partial \un{g}}{\partial t} +
(\un{q}.\nabla )\un{g}$$
holds for any suitably differentiable vector function $\un{g}$.

Verify the convective derivative formula for the motion considered in
the first part of the question when
$$\un{g} = \left ( \begin{array}{c} xt \\ yt^2 \\ z \end{array} \right
).$$

\newpage
\textbf{Answer}

\begin{description}

\item{LAGRANGIAN} Move with the flow, hold $\underline{X}$ constant. Seek to determine $\underline{X}
=\underline{X} (\underline{x} , t)$.

\item{EULERIAN} Fix attention on one spot in space; seek to determine velocity
$\underline{q}=\underline{q}(\underline{x} ,t)$

Now $\begin{array}{l} \displaystyle x=X+Yt^2\\
y=Y(1+t^3)\\z=Z(1+2t)
\end{array}$ \ \ \ $\underline{q}=\displaystyle \frac{\partial \underline{x}}{\partial t} = \left. \frac{\partial{\underline{x}}}{\partial t}
\right |_{\underline{X}}=\left ( \begin{array}{c} 2Yt\\ 3t^2Y\\ 2Z
\end{array} \right ) $

But we need $\underline{q}(\underline{x},t)$ not
$\underline{q}(\underline{X},t)$.

Now also we have

$\displaystyle Y=\frac{y}{1+t^3}$,\ \ $\displaystyle
Z=\frac{z}{1+2t}$,\ \ $\displaystyle \Rightarrow
X=x-Yt^2=x-\frac{yt^2}{1+t^3}$

Thence $$\underline{q}=\left( \begin{array}{c} 2yt/(1+t^3)\\
3yt^2/(1+t^3)\\ 2z/(1+2t) \end{array} \right)$$ $$rm{and\ \ }
\left( \begin{array}{c} X\\ Y\\ Z\end{array}
\right)=\left(\begin{array}{c} x-yt^2/(1+t^3)\\ y/(1+t^3)\\
z/(1+2t) \end{array} \right)$$

Now $\displaystyle
div(\underline{q})=0+\frac{3t^2}{1+t^3}+\frac{2}{1+2t}\ne 0$

Now if $\displaystyle \frac{d}{dt}$ means \lq fix $\underline{X}$'
we have, for any suitably differentiable $\underline{g}$,
\begin{eqnarray*} \frac{\partial \underline{g}}{\partial t} &amp; = &amp; \frac{\partial
\underline{g}}{\partial t}\frac{\partial t}{\partial t} +
\frac{\partial \underline{g}}{\partial x}\frac{\partial
x}{\partial t} + \frac{\partial \underline{g}}{\partial
y}\frac{\partial y}{\partial t} + \frac{\partial
\underline{g}}{\partial z}\frac{\partial z}{\partial t}\\ &amp; = &amp;
\underline{g}_t+\underline{g}_xu+\underline{g}_yv+\underline{g}_zw
\end{eqnarray*}

(where, as usual, $\displaystyle
\underline{q}=(u,v,w)=\frac{\partial \underline{x}}{\partial t}$)

Thus $\displaystyle
\frac{d\underline{g}}{dt}=\underline{g}_t+(\underline{q}.\nabla)\underline{q}$

Now we have $\underline{g}_t=\left ( \begin{array}{c} x\\ 2yt\\ 0
\end{array} \right )$,\ \ \
$(\underline{q}.\nabla)\underline{g}=\left (
\begin{array}{c} 2yt^2/(1+t^3)\\ 3yt^4/(1+t^3)\\ 2z/(1+2t)
\end{array} \right )$ and so

\begin{eqnarray*}
\underline{g}_t+(\underline{q}.\nabla)\underline{g} &amp; = &amp; \left (
\begin{array}{c} x+2yt^2/(1+t^3)\\ 2yt+3yt^4/(1+t^3)\\ 2z/(1+2t)
\end{array} \right )\\
&amp; = &amp; \left ( \begin{array}{c} x+2yt^2/(1+t^3)\\
(2yt+5yt^4)/(1+t^3)\\ 2z/((1+2t) \end{array} \right )
\end{eqnarray*}

Now
\begin{eqnarray*}
\underline{g} &amp; = &amp; \left ( \begin{array}{c} Xt+Yt^3\\
Y(1+t^3)t^2\\ Z(1+2t) \end{array} \right )\\ \Rightarrow
\frac{d\underline{g}}{dt} &amp; = &amp; \left(
\begin{array}{c} X+3Yt^2\\ Y(2t+5t^4)\\ 2Z \end{array} \right )
\end{eqnarray*}

$=\left( \begin{array}{c} \displaystyle
x-\frac{yt^2}{1+t^3}+\frac{3yt^2}{1+t^3}\\ \displaystyle
\frac{2ty}{1+t^3}+\frac{5yt^4}{1+t^3}\\ \displaystyle
\frac{2z}{1+2t}
\end{array} \right ) = \left ( \begin{array}{c} \displaystyle x+\frac{2yt^2}{1+t^3}\\
 \displaystyle \frac{2ty+5yt^4}{1+t^3}\\ \displaystyle \frac{2z}{1+2t} \end{array} \right )
=\underline{q}_t +(\underline{q}.\nabla)\underline{g}$

\end{description}

\end{document}
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