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I'm trying to plot an implicit function satisfying a complex transcendental equation. The code is the following:

ContourPlot[(e (-1 + e^2) l (-3 + 2 e^2 - l^2) Cosh[2 Sqrt[1 - e^2]] + 
  (-1 + e^2) (-1 + l^2) (e l Cos[2] - I (1 + l^2) Sin[2]) + Sqrt[1 - e^2]
  (I - I l^4 - 2 e (-2 + e^2) l Sqrt[-e^2 + l^2]) Sinh[2 Sqrt[1 - e^2]]) == 0, 
  {l, 1, 2}, {e, -2, 2}]

I need a real solution of e as a function of l and also e<=Min{1,l}. I don't know why it's empty in the plot. I don't think it's because the solution is out of the range since I tried l=1, and there are several solutions, one of which is e = 0. Thus I think there must be something wrong with the plot.

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1 Answer

Your expression is not real, but complex. Think about how you want to turn your expression real. The documentation states

At positions where f does not evaluate to a real number, holes are left so that the background to the contour plot shows through.

So what you probably want is to test Abs[f[..]]==0 which gives you the lines where the whole left side vanishes. Unfortunately, they are not in your plot region. Therefore, take a look at this and hover with the mouse over the region borders to see the iso-value.

f[l_, e_] := (e (-1 + e^2) l (-3 + 2 e^2 - l^2) Cosh[
     2 Sqrt[1 - e^2]] + (-1 + e^2) (-1 + l^2) (e l Cos[2] - 
      I (1 + l^2) Sin[2]) + 
   Sqrt[1 - e^2] (I - I l^4 - 2 e (-2 + e^2) l Sqrt[-e^2 + l^2]) Sinh[
     2 Sqrt[1 - e^2]]);

ContourPlot[Abs[f[l, e]], {l, 1, 2}, {e, -2, 2}]

Mathematica graphics

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Thanks for answering. I don't quite understand why only the real part of function is zero. The equation is complex does not mean the solution of e has to be complex, right? –  Knightq Oct 18 '13 at 4:18
    
I see what you mean, but this does not equivalent to the original equation anymore. I try Im[f]==0, the solution is different. Seems like my equation is problematic.. –  Knightq Oct 18 '13 at 4:24
1  
I just wanted to point out where the issue lies. You don't have to (and probably shouldn't) use Re. What you want is probably Abs[f[l,e]] but this is for you to decide. –  halirutan Oct 18 '13 at 4:26
    
Thanks a lot, I see where I got wrong. And it seems to me that ContourPlot ignores isolated solution, doesn't it? –  Knightq Oct 18 '13 at 6:52
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