Can anyone tell me the formula for this vector field? 3Blue1Brown's video on curl

$$V=\bigg[ \begin{array}{ccc}
P\left(x,y\right) \\
Q\left(x,y\right) \end{array} \bigg] $$


*

*$P$ gives you the x component at all points in space.

*$Q$ gives you the y component at any point in space.
What is the explicit formula for the above vector field?


*

*vortexes at $(6,0)$, $(-6,0)$, $(0,6)$, $(0,-6)$

*and a source and a sink at the origin.

*symmetrical
Hints, links welcome
 A: 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:


*

*Taking $\phi \equiv 1$ gives the velocity field of a plane rotating counterclockwise about the origin at constant angular speed.

*Taking $\phi(x, y) = \dfrac{1}{\sqrt{x^{2} + y^{2}}}$ gives a vortex whose flow has unit velocity at each point (except the origin).

*Taking $\phi(x, y) = \dfrac{1}{x^{2} + y^{2}}$ gives a vortex whose flow is incompressible.


*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.)

A: Working through Andrew D. Hwang's answer above:
$$F\left(x,y\right)=\left(\frac{2(-y)}{1+\sqrt{x^2+y^2}},\frac{2(x)}{1+\sqrt{x^2+y^2}}\right)$$

$$G\left(x,y\right)=\left(\frac{-2(y)}{1+\sqrt{(x-6)^2+(y)^2}}+\frac{-2(y)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(y-6)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(y+6)}{1+\sqrt{(x)^2+(y+6)^2}},\\
\\
\\
\frac{-2(x-6)}{1+\sqrt{(x-6)^2+(y)^2}}\\
\\
+\frac{-2(x+6)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y+6)^2}}\right)$$



*

*first four terms give the x component 

*the second four terms compute the y component



