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How to plot the graph of the equation F[x,y]=0 ?

Artem Strashko
Posted 11 years ago
POSTED BY: Artem Strashko
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Use ContourPlot3D, it should give you your result; a 3d surface.
POSTED BY: Sander Huisman
Artem Strashko
Artem Strashko
Posted 11 years ago

Thank you. But can I further extract two 2D graphs from this 3D graph? I mean I can plot F[Re[x],Im[x],y]=0, but I would like to get two projections of this graph: Re[x],y and Im[x],y.
POSTED BY: Artem Strashko
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Sander, I try to use ContourPlot3D for a very simple problem, but it doesn't give a right answer.enter image description here

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POSTED BY: Artem Strashko
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Daniel, thank you so much for your help! Hovewer, I am not sure that this is exactly what I need, because I don't quietly understand how it works.

I can explain what I need in other words. I need the graph of the equation F[x,y]=0, x is a complex number, y is a real number, {y,0,8}. The graph of this equation is 3D, because we have 3 variables Re[x], Im[x], y. I need two 2D graphs y as a function of Re[x] and Im[x], I mean that I need two projection of this 3D graph.I mean smth like that
POSTED BY: Artem Strashko

You might do better to define a root-finding function that, given a value for y, finds Re[x] and Im[x]. Using the redefined form I showed earlier, with s and t as variables for the real and imaginary parts of x respectively, this might be done as below. I use random starting points so as to have a chance at sampling from different branches of the solution curve.

root[yval_?NumericQ] := {s, t} /. 
  FindRoot[{re == 0, im == 0} /. y -> yval, {s, 
    RandomReal[{-5, 5}]}, {t, RandomReal[{-5, 5}]}]

Here is how it looks.

pts = Quiet[Table[root[j], {j, .01, 8, .01}]];
ListPlot[pts]

enter image description here
POSTED BY: Daniel Lichtblau
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Daniel, could you clarify how do you do this? I try to use your code, but I don't get a result.g = 1/10; expr = x - Sqrt[y^2 + I*y]; e2 = expr /. x -> s + I*t; {re, im} = ComplexExpand[{Re[e2], Im[e2]}]; root[yval_?NumericQ] := {s, t} /. FindRoot[{re == 0, im == 0} /. y -> yval, {s, RandomReal[{-5, 5}]}, {t, RandomReal[{-5, 5}]}] pts = Quiet[Table[root[j], {j, .01, 8, .01}]]; ListPlot[pts]

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POSTED BY: Artem Strashko

Your nb gave a nice plot when I executed the top cell. Maybe you had other definitions active at the time that interfered in some way?
POSTED BY: Daniel Lichtblau
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Daniel, I am sorry for interruption. I don't know what was the problem, but now everything is fine. Could you answer a couple of other questions? Firstly, what shoul I do to consider this problem only in the region {Im[x],0,8}, {Re[x],0,8}, {y,0,8}. And how can I distinguish between the dependense of y on Re[x] and y on Im[x] depicted on the figure? Also, could you clarify what do I change when alter RandomReal[{0, 5}] and {j, .01, 5, .01}?

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POSTED BY: Artem Strashko
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Daniel. I have decided to check your method I have considered a very simple problem (x=f[y]) which can be easily solved by a very straightforward method (just using Plot). I don't know why but the results are essentially different.

I attach my file, but the general idea is very simple. If we have x = f[y] and x=x'+ix'', we can just plot x'=Re[f[y]] and x''=Im[f[y]] and compare these graphs with the graphs which we get using your method.

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POSTED BY: Artem Strashko

The plots look more similar if you recognize that the equivalent form for the second one should use ParametricPlot instead of Plot (such distinctions are very important). The ListPlot will need a range setting as well. Try the two below.

ListPlot[pts, PlotRange -> All]

ParametricPlot[{Re[y*Sqrt[(y^2 + 1/10*I*y - 4)/(2 y^2 + 1/10*2*I*y - 4)]], 
  Im[y*Sqrt[(y^2 + 1/10*I*y - 4)/(2 y^2 + 1/10*2*I*y - 4)]]}, {y,  0.0001, 4}, PlotRange -> All]
POSTED BY: Daniel Lichtblau
Artem Strashko
Artem Strashko
Posted 11 years ago

No, no, no. Dear Daniel, I am sorry that my explanations were probably unclear. I need exactly what I have shown you in my version (because I know the answer for this particular example). Now I attach a picture where I explain everything in symbols (and in a couple of words). I think that this picture makes everything much more clear without words (I am sorry that I have presented in just on paper, but I think it is clear enough).

However, it would good to try to explain again what I need by words. I have a function f[x,y], where y is real and x is complex. I need to solve this equation for y from 0 to 10, i.e. to find x for every y from 0 to 10. After getting the solutions (Re(x), Im(x), y)I need to plot two graphs: y as a function of Re(x) and y as a function of Im(x) (i.e. use the numbers from 1st and 3d column for the first graph and the numbers from 2nd and 3d column for the second graph)

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POSTED BY: Artem Strashko
POSTED BY: Daniel Lichtblau
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Daniel, thank you for you help. Could you tell me what version of Mathematica do you use (because I can't use this code in my version 7)?
POSTED BY: Artem Strashko

I'm using version 10.1. You can do similarly with NDSolve if you solve for both {y[rex],imx[rex]}.
POSTED BY: Daniel Lichtblau
Artem Strashko
Artem Strashko
Posted 11 years ago

Ok, thanks!
POSTED BY: Artem Strashko
Artem Strashko
Artem Strashko
Posted 11 years ago

Dear Daniel, I have installed Mathematica 10, but again I can't use your code.

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POSTED BY: Artem Strashko

You should do some debugging. Look hard at the messages, the NDSolveValue result, that sort of thing. Maybe even start in a fresh kernel.
POSTED BY: Daniel Lichtblau

This might be a start. Define the expression, then use ComplexExpand to form explicit real and imaginary parts.

g = 1/10;
expr = 
  Tanh[Sqrt[-x^2 - y^2]*3/10] (y^2 + I*g*y - 4) Sqrt[-x^2 - 
      y^2] + (y^2 + I*g*y) Sqrt[-x^2 - 
      y^2 ((y^2 + I*g*y - 4)/(y^2 + I*g*y))];

e2 = expr /. x -> s + I*t;
{re, im} = ComplexExpand[{Re[e2], Im[e2]}];

 ContourPlot[(re /. y -> 0.7) == 0, {s, -10, 10}, {t, -10, 10}]

enter image description here

ContourPlot[(im /. y -> 0.7) == 0, {s, -10, 10}, {t, -10, 10}]

enter image description here
POSTED BY: Daniel Lichtblau
Artem Strashko
Artem Strashko
Posted 11 years ago
POSTED BY: Artem Strashko

You will get better answers if you give a complete, very specific example of what you want.
POSTED BY: Daniel Lichtblau
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