Iterate problem for a complex function and plot it.

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Now I have a complex function, for example
1.8633 - 0.00576465 s1^3 + s1^2 (0.160853 - 0.050155 s2) - 1.20076 s2 + 0.161107 s2^2 - 0.00576465 s2^3 + s1 (-1.1984 + 0.631511 s2 - 0.050155 s2^2);
  1. It contains two variables s1 and s2, I can use contour plot the lines that equal to zero.
  2. Give a map T, such that s1 -> s1^2+s1*s2^2; s2 -> s1+s1^2*s2+s2;
  3. After the map, repeat the contour plot for the new function.
Matlab codes:
S1=0:0.01:10;
S2=0:0.01:10;
[s1,s2] = meshgrid(S1,S2);
Z=1.8633 - 0.00576465 .* s1.^3 + s1.^2 .* (0.160853 - 0.050155 .* s2) - 1.20076 .* s2 + 0.161107 .* s2.^2 - 0.00576465 .* s2.^3 + s1 .* (-1.1984 + 0.631511 .* s2 - 0.050155 .* s2.^2);
contour(s1,s2,Z,[0,0])
This figure is Contour when the equation equal to zero, then I want all the points on these lines will iterate by a rule (as above 2).
My question is how to plot it more convenient, my way is solve the funciton first (solve s1), then I can get s1 equal a formula, then do the iterate separately, and plot (s1, s2), but it will be difficult, because there are some solution for s1 at the same time, so I want to know is there any other method do it.
% Since it's a cubic function, there are three solutions,
s1 = f1(s2)
s1 = f2(s2)
s1 = f3(s2)
% Apply the mapping to the three formulas
(news1,news2) = T[s1, f1(s2)]
....(another two are same as this one)
% Then use plot(news1, news2) get new lines which apply the mapping
% But the solution is complicated, so is there any convenient way.
  7 Comments
darova
darova on 14 Jun 2020
Are you asking how to extract data from contour?
zongxian
zongxian on 14 Jun 2020
Edited: zongxian on 14 Jun 2020
if we can extract data that will be a good way. Then I can apply map T on these data. @darova

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Accepted Answer

darova
darova on 14 Jun 2020
To extract data from contour use get
[C,h] = contour(...);
xdata = get(h,'xdata');
  1 Comment
zongxian
zongxian on 14 Jun 2020
You gave me some inspiration, and then I iterated through the points I got. Thank you.

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