$$V=\bigg[ \begin{array}{ccc} P\left(x,y\right) \\ Q\left(x,y\right) \end{array} \bigg] $$
$P$ gives you the x component at all points in space.
$Q$ gives you the y component at any point in space.
vortexes at $(6,0)$, $(-6,0)$, $(0,6)$, $(0,-6)$
and a source and a sink at the origin.
symmetrical
Hints, links welcome
On
Working through Andrew D. Hwang's answer above:
$$F\left(x,y\right)=\left(\frac{2(-y)}{1+\sqrt{x^2+y^2}},\frac{2(x)}{1+\sqrt{x^2+y^2}}\right)$$
$$G\left(x,y\right)=\left(\frac{-2(y)}{1+\sqrt{(x-6)^2+(y)^2}}+\frac{-2(y)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(y-6)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(y+6)}{1+\sqrt{(x)^2+(y+6)^2}},\\ \\ \\ \frac{-2(x-6)}{1+\sqrt{(x-6)^2+(y)^2}}\\ \\ +\frac{-2(x+6)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y+6)^2}}\right)$$
It's difficult to be certain of the exact formula, but here's a general strategy for constructing this type of field.
Pick a vector field having a vortex at the origin, such as $$ F(x, y) = \phi(x, y)(-y, x) $$ for some real-valued function $\phi$. For example:
If $(x_{0}, y_{0})$ is an arbitrary point, the field $$ F_{(x_{0}, y_{0})}(x, y) = F(x - x_{0}, y - y_{0}) \tag{2} $$ has a vortex at $(x_{0}, y_{0})$.
Form the sum or difference of fields of the form (2), summing over each vortex location, and using a negative sign to change the direction of flow.
In the diagram, $$ F(x, y) = \frac{2(-y, x)}{1 + \sqrt{x^{2} + y^{2}}}, $$ so the field plotted is $$ G(x, y) = F(x - 6, y) + F(x + 6, 0) - F(x, y - 6) - F(x, y + 6). $$ (Writing this out as a pair of component functions is left as a mildly masochistic exercise.)