Finding flow lines of a velocity vector field Find and draw the flow lines of the velocity vector fields $\vec{F}(x, y) = (-2y, \frac{1}{2}x)$
Solution:
$x' = -2y$
$y' = \frac{1}{2}x$
$x'' = -2y$
$x'' = -2 \frac{1}{2}x = -x$
No idea how to proceed. Is there a general formula to get the flow line of velocity vector fields? thank you
 A: You're interested in solving for the contours of some function $\psi$ such that $\psi = \text{const.}$ $\psi$ is called a streamline (flow line) of the velocity field $\vec{V} = u(x,y)\hat{i} + v(x,y)\hat{j}.$ Note that $\vec{V}$ must be divergence free, which it is. The equations for $\psi$ are then
\begin{cases}
\displaystyle \frac{\partial \psi}{\partial y} = u(x,y) \\
\displaystyle \frac{\partial \psi}{\partial x} = -v(x,y).
\end{cases}
For this case, we have that $u = -2y$ and $v = x/2$. It doesn't matter which equation you start with, so lets say we start with the first one.
$$
\int \psi_y \,\mathrm{d}y = -2\int y \, \mathrm{d}y \implies \psi(x,y) = -y^2 + f(x).
$$
We have $\psi$, but in terms of an arbitrary function $f$; however, we have another equation. Using it then, we have
$$
\psi_x = f'(x) = -\frac{1}{2}x.
$$
$\psi$ will be known up to a constant if $f$ is found. What results is the stream function $\psi$, and the streamlines to $\vec{V}$ are given by $\psi = \text{const.}$
