@dharr  it is true but for phase portrait specially for chaotic this make a problem becuase it chose a complex number  and when i used complex number chaotic not appear anything 

i tested on that S[1] and S[2] in complex equation which totally satisfy thanks for code

true.mw

@mmcdara right now i am reach on this plotting, and i am notic this which for work better we have to do something is ok if we don't make 3D explor option we can use 2D explor option for outcome of 3D shape becuase they related, so i need thus two part in after explor option i need the system and parameter outcome,  for plotting, also in case if we want to convert for other language of programing How we can do that i want plot that shape of 2D when i construct by explor option?

i can't jump in your code they sensitive a little bit

pp.mw

@dharr  let say S[1]=x+y+z
i want find S[1]=1, what is x+y+z which x,y,z are not be zero , let say x=1,y=1, so z=-1 becuase parameter sometime are a lot i want something automatically and randomly find them if x,y,z are be rond it will be better 

i think this time is clear, in the paper he did like that is totally clear 

@dharr  i just change letter and i solve another problem is not like this but idea is same  you say mu[1] mu[2] any letter just find element iside them which make them 1 or -1 it is easier for showing and for phase portrait it is best idea  which be 1,

you can use the picture system in case the exact solution and result are clear.

about plot i don't know how plot imaginary it mean  must have another structure or what i don't have idea  i can't go to global bifurcation while i am stuck in local , 

@mmcdara i hope you be able to to explore option for phase portrait too , i have work on it and with your comunity maple i think i can do the best job ever possible  thank you 

@mmcdara  Wow, you just explode  it is so emazing  really i have to work on it to improve this work i have 1000 question in my mind, i saw the explore option for 3D too, i have the code for easy function in case needed mention, mr accr did for me a long time ago, but for this it is in my mind i say i will ask in future but you did it now and it is far amazing this is made my day, 

i agree with you i saw the rung kutta method they used for sensitivity but  i am not sure how and i am not involve that part yet but for sensity Mr.rubien did another way which yet i didn't have much time for work on that too i have to write a paper and part by part i am stuck  becuase this is my first work about dynamic system, 
additionally, i have a lot question  and i will ask after , please if you can do for 3D it will be something new  my future work and i don't think now one done this 

@dharr  about point 1: if you watch S[1] in fact contain p[1],p[2],p[3] and k[3] inside S[1] i want set all of p and k and other parameter inside S[1] also for S[2] which the result be 1, if you watch the picture i mentioned the different picture, if you watch case1: you will see in system we have mu[1] and mu[2] but inside of them we have other parameter which author set them up arrange them some how somekind of solve all parameter use  to find mu[1] and mu[1] in equalibriuom point i have to say is (1,0) instead of (sqrt(S[1]/S[2],0) aso i have to say what is parameter inside S[1] and S[2] for getting this (1,0) 

point2: if you watch the equalibriom point we have three one which two of them are complex i don't know why not shown this is my first time i write a paper so i am looking for information stil about that i know first integral plot are great  but just show the point (0,0) how about (i,0) and (-i,0) i try ask Ai he say we need global phase portrait which i don't have idea how  can show imaginary part of phase portrait  i didn't see that, in paper when they have this case they just say we have (0,00 but i want to know why other not include

please if don't undrestand mention i will explain more and if need example 

@sand15 i can't upload why i don't know but i take a screenshot

@mmcdara i don't beliave that which you done this, this is amazing but i have work on it and fully undrestand the coding also i have a lot new idea about getting new shape of thus chaotic, and the code is well runing untill here 

YOUR MATLAB EXAMPLE

is not give me outcome i waited for 5 minute at least but no outcoming 

i have to work on it and used for other example also new idea if you have a time i can give you a example with new idea 

i just got a new article from my friend which really outcome are amazing and i want to ask a new question about it but in here i think is better to present becuase your outcome are great and quite close let change the system and new using function 

@dharr  i don't know why the picture in here are so different from matlab, i try to use matlab but at there i didn't give xi and otehr parameter i just write the system it self with k[1] and k[2] with thus option

K1 = 1.0;      % Cubic nonlinearity coefficient
K2 = -1.0;     % Linear restoring force coefficient  
epsilon = 1.0;  % Forcing amplitude
theta = 1.2;   % Forcing frequency

% Time span
tspan = [0 200];  % Long integration time for chaotic behavior
dt = 0.01;
t = 0:dt:tspan(2);

% Initial conditions
u0 = 0.1;  % Initial position
v0 = 0.1;  % Initial velocity
initial_conditions = [u0; v0];
%% Define the system of ODEs
duffing_system = @(t, y) [
    y(2);                                    % du/dt = v
    -K1*y(1)^3 + K2*y(1) + epsilon*sin(theta*t)  % dv/dt
];

and i get the good result  

@mmcdara  thank you so much i will wait , i don't undrestand the first coment where i should add and i paste your code there but nothing happen, and when i change the name of color in cyan to cyan or gold to Gold is give me error so colors are still have same name 

@mmcdara i can't see it seem something new there is no error just not showing 

>  restart; >  with(plots): >  (1) >  (2) >  de1 := diff(u(t),t) = v(t); de2 := diff(v(t),t) = - K[1]*u(t)^3 + K[2]*u(t) + epsilon*sin(theta*t); (3) >  ic := u(0)=0, v(0)=0; (4) >  rt := 0, 200: dsol := dsolve(           {de1,de2,ic}           , numeric           , parameters=[epsilon, theta]           , range=rt[1]..rt[2]           , maxfun=10^5         ); (5) A correct Maple 2015 procedure >  figures := proc(__theta, __epsilon, col::{name, string}, planes::list)   local p2, p3, data, ru, rv, i, j, k, t_plane, u_plane, v_plane, black_back:   dsol(parameters=[__theta, __epsilon]):   p2 := odeplot(           dsol           , [u(t),v(t)]           , t=rt[1]...rt[2]           , color=col           , background=black           , axes=frame           , refine=1         ):     # u and v ranges   data := plottools:-getdata(p2)[3]:   ru   := (min, max)(data[.., 1]);   rv   := (min, max)(data[.., 2]);   # The 3 selected background planes.      i, j, k := op(planes):   t_plane := [ [rt[i], ru[1], rv[1]], [rt[i], ru[2], rv[1]], [rt[i], ru[2], rv[2]], [rt[i], ru[1], rv[2]] ]:   u_plane := [ [rt[1], ru[j], rv[1]], [rt[2], ru[j], rv[1]], [rt[2], ru[j], rv[2]], [rt[1], ru[j], rv[2]] ]:   v_plane := [ [rt[1], ru[1], rv[k]], [rt[2], ru[1], rv[k]], [rt[2], ru[2], rv[k]], [rt[1], ru[2], rv[k]] ]:   black_back := display(     PLOT3D(       POLYGONS(t_plane, COLOR(RGB, 0$3)),       POLYGONS(u_plane, COLOR(RGB, 0$3)),       POLYGONS(v_plane, COLOR(RGB, 0$3))     )   ):   p3 := odeplot(           dsol           , [t, u(t), v(t)]           , t=rt[1]...rt[2]           , color=col           , axes=frame           , refine=1         ):   return p2, p3, black_back end proc: Can you run these commands to see what function behaves differently with you present Maple version? >  # Command 1    dsol(parameters=[1, 1]): dsol(Pi); (6) >  # Command 2    p2 := odeplot(           dsol           , [u(t),v(t)]           , t=rt[1]..rt[2]           , color=red           , axes=frame         ): p2;   Warning, extending a solution obtained using the range argument with 'maxfun' large or disabled is highly inefficient, and may consume a great deal of memory. If this functionality is desired, it is suggested to call dsolve without the range argument   >  # Command 3    odeplot(           dsol           , [u(t),v(t)]           , t=rt[1]..rt[2]           , color=red           , axes=frame           , refine=1         );   >  # Command 4    odeplot(           dsol           , [u(t),v(t)]           , t=rt[1]..rt[2]           , color=red           , background=black           , axes=frame           , refine=1         );   >  # Command 5    data := plottools:-getdata(p2)[3]: ru   := (min, max)(data[.., 1]); rv   := (min, max)(data[.., 2]); (7) >  # Command 6    plot_1 := figures(1, 1, cyan, [1$3]): plot_1[1]; EXAMPLE argument 1 : value of θ argument 2 : value of ε argument 3 : color argument 4 : a list of three integers in {1, 2}:                      integer 1 = 1 means "plane [u, v] for t = min(t range)                      integer 1 = 1 means "plane [u, v] for t = min(t range)                      integer 2 = 2 means "plane [t, v] for t = max(u range)                      integer 2 = 1 means "plane [t, v] for t = min(u range)                      integer 3 = 2 means "plane [t, u] for t = max(v range)                      integer 3 = 2 means "plane [t, u] for t = max(v range) >  plot_1 := figures(2.73, 2.7, cyan, [1$3]): plot_2 := figures(3.43, 9.17, gold, [1$3]): >  display(   <     plot_1[1]   |     display(plot_1[2..3])   > ) >  display(   <     plot_2[1]   |     display(plot_1[2..3])   > ) >  ru_1 := max(map2(op, 2, op([1, 1], plot_1[3]))); rv_1 := max(map2(op, 3, op([1, 1], plot_1[3]))); ru_2 := max(map2(op, 2, op([1, 1], plot_2[3]))); rv_2 := max(map2(op, 3, op([1, 1], plot_2[3]))); # What follows might not be robust bur seems sufficient backgrounds := plot_1[3]: if ru_2 > ru_1 then   backgrounds := plot_2[3] end if: (8) >  display(   plot_1[2]   , plot_2[2]   , backgrounds )

Download chaotic-2_mmcdara_2.mw

@dharr in this two equation  we have three different  coeficent which lead us to get three different condition  like that
#u^3=$u^3 & @u=%u & #u''=@u''  when we set up we have to get the condition which there is parameter in code i saw that you mark  duble term as one  which i think this is gonna be mistake i have to check it analytically by solution

Download C11.mw

@mmcdara i can't see any effect in thus plot, i don't know why 

@acer i think maybe no one answer i post easy question in begin then go to main problem, and you are right i should post question in begin next time i will , sorry, and your code is good but becuase problem are too much and if do  like i mentioned it will be great