# What is the correct way to calculate streamlines of a vector field

1. ### Jonsson

34
Hello there,

What is wrong with my way of finding stream lines of a vector field? Say I have this vector field:

$\vec{v} = x\,y\,\vec{i} + y\,\vec{j}$

You can see a plot here: http://kevinmehall.net/p/equationexplorer/vectorfield.html#xyi+yj|[-10,10,-10,10]

It appears as if the stream lines could be $y = log(x) + C$.

I proceed to find out:

$v_y \, \mathrm{d}x = v_x \, \mathrm{d}y\\ x\,y\,\mathrm{d}y = y\,\mathrm{d}x\\ \mathrm{d}y = \frac{1}{x}\,\mathrm{d}x\\ \int\,\mathrm{d}y = \int \frac{1}{x}\,\mathrm{d}x\\ y = log(x) + C$

This looks about right. However there is a problem, when I look back at my vector field (http://kevinmehall.net/p/equationexplorer/vectorfield.html#xyi+yj|[-10,10,-10,10]), for values of x less than zero, it appears as if the streamlines should be a mirror-image of y = log(x) + C.

So my question, does the above streamline calculation have more solutions which I have missed? Or is there something else which is wrong, which is causing me only to find the streamlines for x values greater than 0?

Kind regards,
Marius

Last edited: Feb 25, 2013

### Staff: Mentor

Your derivation simply does not care about the orientation of the field. You could see y=0 as region where field lines from both sides end, as the field vanishes.

The correct integral of ##\frac{1}{x}## is ##ln(|x|)+C## (where you can use a different C for positive and negative x), this allows to use negative x-values as well.

3. ### Jonsson

34
mfb, thank you so much :)

Kind regards,
Marius