@acer  i want exactly sketch this plot like that which right now you did it, the problem is this i sketch the plot with explor but is not give me this plot, even  i want looking the perfect shape then plotted like yours , do you think 2D plot explor is better or 3D explor option for this plot ?

g1.mw

@dharr  i am stuck at case of non-schrodinger equation which jetnotation not working  i write equation and did some but at there i am stuck, for non schrodinger equation substitution is different is easier but in coding how remove that part which power is >1

non-schrodinger.mw

@acer  this plot is different i ploted explor for 3D in first example but when i ploted and used that parameter is unlike what i want specially 3D of plot but in that paper he did something like 2D plot i need 3D plot but  iam sure there is a physicall trick for this ploting 

@dharr i will keep try using this for new equation which not apply, if i got problem i will notify but the programing is so good i don't think problem come up , thanks a lot you are angel of maple

@dharr  in thus paper i can see we have conjugate but in our i can't see that which i am not sure which one is true perhaps they are true  this is easy pdes for solving and i think if we can hundle this we can do for other too what is your idea?
please also watch the extra note in top comment

p3.mw

@dharr i don't know why i got error and why result  is so different , extra note
when we have |U|^2 i thin we have to use property of complext which say |U|^2=U*U^* where U^* is conjugate

p1.mw

@dharr i will work on it right now untill i finish this, i will try for other papers too, if problem come up i will update here , you did a great job thanks a lot... 

@dharr you did a great job again, just solution 6 have a problem now but i have to write my review report for this paper   , you give me a new idea regarding to convert the function like you did to convert(S15, sec) and convert(S7, sinh) thus are so new for me which they change i=sqrt(-1) inside bracket is the same but different type of solution , thanks a lot , and i don't know  a lot about this two function `Weierstrass elliptic function and eliptic jacobi function which one of them have more than 52 solution for pdes my next work after penlvy test is thus, but still i didn't got time for doing work on penlve test i am so sorry for that i will work on it as soon as possible

@dharr  there is not any problem is exactly what i think is true, i have to work on it but in thus day i am busy with market and some revision i got in two future day i will work on it more and i will give my openion in here.

@dharr  i am intrested a lot and i don't know where to start is so pretty, i have to work on it harder but each time you do something better and better which make me more intrested and confused by thus amazing code and pretty outcome which, i am intrested i have more than 4 paper of that author which is one of the best author, beside this i am intrested about the u[0],v[0] i think they are not the same which i am not sure which is corect  i don't have issue with yours one of g[x] is extra if you watch you will see, the other point is about that condition how he got that condition in eq((16),  and when N=2 we have to find the u[1],u[2],..u[6] and v[1],v[2],...v[6]  i din't see that which part is thus which out comes,i want to test by myself but you are 100 time faster and intalegent than me the other pdes which  you mentioned too that have same result like this or it is different?

and about step one is not about appearing twice i think, is about non linear term which we chose which a-N is bigger and non linear term is about a+b-N which N is begger and have same a then we can find b like the non couple equation .

and in this equation is more clear u[0],v[0],.... and resonance point and third step too

@dharr 

I’ve been studying this carefully, and in your code (and in the paper) they obtained three specific functions, u2,u3,u5 while the others remained free. I’m trying to deepen my understanding of the Painlevé test and its steps. I’m still not entirely clear why it is useful, or why certain positions are called “resonances” (some of which are zero and some not).

The main reason I’m focusing on the Painlevé test is that I recently came across a new method (with only one published paper on it so far). I’m convinced that if I can fully understand this test, I’ll be able to publish several papers, but at the moment I’m stuck.

I’ll update you with the link soon; if you can’t download the paper, let me know and I can send it by email. I can also share a picture of the equations, I haven’t skipped to the end, but after doing the transformation, some very complicated equations appeared. I don’t fully understand how they derived them, so maybe your version is different.

I know you’re busy, but I just wanted to let you know in case you’re also interested.

specially becuase of this i study penlevy test 

i did transform but i didn't penlevy yet

paper

 F-ode-Fpaper.mw

@dharr is more than perfect and in some place really i have to work on it more specially of thus plot you did is novel idea in a=-1 and a=1 have two intersection but in other place not , i have to check for other pdes too i need some time for that, beside this i add one part in third step which you did for finding u[1] and u[2] in last file but in here you didn't mention it i add that part for finding u[1]-u[6] when we have u[1]=() it mean free, and how you know at j=1 when simplify is zero why in j=4 is not zero at that resonance point?

also did you work on couple system is so importan if we complete that part, i have problem at there a lot and today i didn't have much time to work on it sorry for that i will keep looking 

all-steps-Dr.D.mw

@dharr  it seems in solving schrodingers he use the coupled system like the paper you mentioned, i  didn't work on couple system but i wil search more  and i will try to solve 

@dharr new update for step one

schrodinger-test-1.mw

@dharr this is penleve test again but more complicated with same steps, How we can find r and p of substitution is like before when we try to find (a) in non shcrodinger equation but this time we have two function i saw in a paper how they find but we need find all posiblety  or just high order and nonlinear term for two function which one get u=a[1] and v=a[2] and i think after substitute u=u[0]F(x,t)^p in real part and calculate p and r at there also in imaginary but i think both will be equal, beside this how find u[0] and v[0] is the same of step but have two function i try to substitute but i don't know my code why not work specially for finding p and r u(0) v(0) also j point 

i got the idea of p and r from this paper 

schrodinger-test-1.mw