Mathematica Plot Real Portion (Contour Plot)

  • #1
member 428835
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:
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}]
 
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  • #2
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?
 
  • #3
joshmccraney said:
In Mathematica and given the code below, I am trying to the real part of ##\sigma##, ##Re(\sigma)##, in stead of simply ##\sigma##.
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.
 
  • #4
DrClaude said:
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.
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!
 
  • #5
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?
 
  • #6
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##?
 

Attachments

  • correctplot.pdf
    25 KB · Views: 367
  • incorrectplot.pdf
    15.9 KB · Views: 372

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