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

[Jonsson]

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\\<br /> x\,y\,\mathrm{d}y = y\,\mathrm{d}x\\<br /> \mathrm{d}y = \frac{1}{x}\,\mathrm{d}x\\<br /> \int\,\mathrm{d}y = \int \frac{1}{x}\,\mathrm{d}x\\<br /> y = log(x) + C<br />

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?

Thank you for your time.

Kind regards,
Marius

 

[mfb]

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.

 

[Jonsson]

mfb, thank you so much :)

Kind regards,
Marius