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\begin{center}
\textbf{Vector Fields}

\textit{\textbf{Conservative Fields}}
\end{center}

\textbf{Question}

A source of strength 2 is placed at $(0,0,0)$ and a sink of strength 1
is placed at $(0,0,1)$. Find the velocity field for this system. Show
that the velocity is vertical at all points of a certain
sphere. Sketch the streamlines of the flow.


\textbf{Answer}

The scalar potential of the system is
$$\phi (x,y,z) = \phi(\un{R}) = -\frac{2}{|\un{r}|}+\frac{1}{|\un{r} -
\un{k}|}.$$
This gives the velocity field
\begin{eqnarray*}
\un{v} & = & \nabla \phi = \frac{2\un{r}}{|\un{r}|^3} -
\frac{\un{r}-\un{k}}{ |\un{r} - \un{k}|^3}\\
& = & \frac{2(x\un{i} +y\un{j} + z\un{k})}{(x^2+y^2+z^2)^{3/2}} -
\frac{x\un{i}+y\un{j} +(z-1) \un{k}}{(x^2+y^2 +(z-1)^2)^{3/2}}.
\end{eqnarray*}
In order to have vertical velocity
$$\frac{2x}{(x^2+y^2+z^2)^{3/2}} = \frac{x}{(x^2+y^2+(z-1^2)^{3/2}}$$
with a similar equation for $y$. Both of these equation will be
satisfied at all points of the $z$-axis, and where
\begin{eqnarray*}
2(x^2+y^2+(z-1)^2)^{3/2} & = & (x^2+y^2+z^2)^{3/2}\\
2^{2/3}(x^2+y^2+(z-1)^2) & = & x^2+y^2+z^2\\
x^2+y^2+(z-K)^2 & = & K^2 - K,
\end{eqnarray*}
with $K=2^{2/3}/(2^{2/3}-1)$. 

The latter equation represents a sphere $S$, as $K^2-K>0$. The
velocity is vertical on all point of $S$, as well as at all points of
the $z$-axis.

As the source (at $(0,0,0)$) is twice as strong as the sink (at
$(0,0,1)$), half of the fluid that the sources emits will go to the
sink. By a process of symmetry, this will be the upper half plane
$z>0$. All of the fluid emitted below $z=0$ will flow out to infinity.

There exists one point with $\un{v}=\un{0}$. This point
($(0,0,2+\sqrt{2})$) lies inside $S$. Any streamlines that emerge from
the origin parallel to the $xy$-plane lead to this point. Any
streamlines that emerge into $z>0$ cross $S$ and approach the
sink. Any streamlines that emerge into $z<0$ flow to infinity.

Some cross $S$ twice, some are tangent to $S$ and some do not
intersect $S$ at all.

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