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

  1. Feb 25, 2013 #1
    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/equationexplorer/vectorfield.html#xyi+yj|[-10,10,-10,10]

    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/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
     
    Last edited: Feb 25, 2013
  2. jcsd
  3. Feb 25, 2013 #2

    mfb

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    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.
     
  4. Feb 26, 2013 #3
    mfb, thank you so much :)

    Kind regards,
    Marius
     
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