What is the correct way to calculate streamlines of a vector fieldby Jonsson Tags: correct, field, streamlines, vector 

#1
Feb2513, 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 mirrorimage 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 



#2
Feb2513, 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 xvalues as well. 



#3
Feb2613, 04:25 PM

P: 15

mfb, thank you so much :)
Kind regards, Marius 


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