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


by Jonsson
Tags: correct, field, streamlines, vector
Jonsson
Jonsson is offline
#1
Feb25-13, 10:42 AM
P: 15
Hello there,

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

[itex]\vec{v} = x\,y\,\vec{i} + y\,\vec{j}[/itex]

You can see a plot here: http://kevinmehall.net/p/equationexp...0,10,-10,10%5D

It appears as if the stream lines could be [itex]y = log(x) + C[/itex].

I proceed to find out:

[itex]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
[/itex]

This looks about right. However there is a problem, when I look back at my vector field (http://kevinmehall.net/p/equationexp...0,10,-10,10%5D), 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
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mfb
mfb is offline
#2
Feb25-13, 05:16 PM
Mentor
P: 10,836
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
Jonsson is offline
#3
Feb26-13, 04:25 PM
P: 15
mfb, thank you so much :)

Kind regards,
Marius


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