Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mathematica Plot Real Portion (Contour Plot)

  1. Jun 21, 2017 #1

    joshmccraney

    User Avatar
    Gold Member

    Hi PF!

    In Mathematica and given the code below, I am trying to the real part of ##\sigma##, ##Re(\sigma)##, in stead of simply ##\sigma##. Any help would be awesome! Here's what I have:
    Code ( (Unknown Language)):

    n = 3;
    ContourPlot[(\[Sigma]^2 -
          2 (n - 1) (2 n + 1) \[Epsilon] \[Sigma] + (n - 1) n (n + 2)) ((
          Sqrt[\[Sigma]/\[Epsilon]]
            BesselJ[n + 1/2, Sqrt[\[Sigma]/\[Epsilon]]])/
          BesselJ[n + 3/2, Sqrt[\[Sigma]/\[Epsilon]]] - 2) +
       4 (n - 1)^2 (n + 1) \[Epsilon] \[Sigma] == 0, {\[Epsilon], 0,
      1.5}, {\[Sigma], 0, 7}]
     
    Last edited: Jun 21, 2017
  2. jcsd
  3. Jun 25, 2017 #2

    joshmccraney

    User Avatar
    Gold Member

    If no one knows this, does anyone know how I can get a numerical solution for sigma as a function of epsilon values for integers n=2,3,4,5,6?
     
  4. Jun 26, 2017 #3

    DrClaude

    User Avatar

    Staff: Mentor

    I don't understand what you mean. σ is one of the plotting variable, so I don't see why ot wouldn't be real.
     
  5. Jun 27, 2017 #4

    joshmccraney

    User Avatar
    Gold Member

    For all ##\epsilon\in(0,1.5)## there are associated ##\sigma## values, some of which have non-zero imaginary components. Something like ##y=1+\sqrt{x+1}:x\in(-2,2)##. I would like to plot the real component of ##y## for listed values of ##x##, but obviously for the more difficult problem in post 1 (as opposed to this toy problem).

    Thanks for taking an interest!
     
  6. Jun 28, 2017 #5

    DrClaude

    User Avatar

    Staff: Mentor

    Sorry, but I'm still having problems understanding. Would you mind explaining it in mathematical terms first? What is the relation between σ and ε and what is the function you want to plot?
     
  7. Jun 28, 2017 #6

    joshmccraney

    User Avatar
    Gold Member

    Sorry for being unclear. I'll reduce this problem to a much simpler one, which I think will make much more sense, and if you can answer it then you've answered my initial question. Consider the problem
    $$ y^2 - x y + 1 = 0$$

    I am trying to plot ##Re(y)## given a set of ##x## values in ##[0,1.5]##. One way to do this is use the quadratic formula and explicitly plot $$Re(y) = Re\left(\frac{x\pm\sqrt{x^2-4}}{2}\right)$$ This is shown in the attached pdf as "correctplot". Notice the linear region from the origin corresponds to solutions of ##y## whose imaginary components are non-zero.

    The method above gives the correct plot, but I had to solve for ##y## explicitly. However, since I can't solve for the independent variable explicitly in more complicated equations (such as that in post 1), I am unsure how to implicitly plot the real component of ##y## for given ##x## values. The attached pdf "incorrectplot" plots ##y^2 - x y + 1 = 0## via ContourPlot. Notice the linear portion is not shown.

    Any idea how to plot ##Re(y)## implicitly for any value of ##x##?
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Mathematica Plot Real Portion (Contour Plot)
  1. Mathematica plot (Replies: 1)

  2. Plot in mathematica (Replies: 2)

Loading...