@dharr I have to work on it more for full understanding the topic and thus finding is just begin I will try find all solution I hope to not stuck... 

again thank a lot....

@dharr  i will test for more pdes and for sure i am go to next topic after i got expert in finding thus but i think some problem will come up it seems becuase i need find all function, i will do the rest for sure if i have a question i will ask again , and is a gate for backlund transformation also there is another way for finding same step for shcrodinger equation which after this i will go apply for that but  undrestanding this is mean undrestanding all the gate to other topic.

i don't know how say thank you without you is imposible to done this you  made my night again 

@dharr in a lot of place  but very small place they talk about this please watch thie yellow place of step3 of this article i think if we do the change of variable like that we can solve it, in a some other place they talk about this replacement

e-kdv-burger-equation.mw

i try to solve but i got error, and we use the series wrongly i think, becuase phi(x,t)* series to all term must have that part

@dharr  i did by old version of code for some pdes the a give me wrong i will use this now for clerify, i can solve byy hand but code is something else,and i am try to figure out this test is good for what i hope  untill then you can figure out the step 3, and i am sure this test give us some pdes solution too 

there is any different between 2024-2025? if i know is make a prolem for your code i will change to 2025

@dharr  i try to fix it but not working, and t is very compact 

@dharr I am kepp searching for more example, but in all paper which I saw they didn't tell what they did just I saw in that paper I mention up, tomorrow I will search more , any idea come in my mind  I will update here ...

@dharr  i don't know how your code is work, you deserve to be called by god of maple...

some code give me error in first step but i replaced by your last file to fixed, it work , and it is emazing you have done the step two i will do test for more of them tommorow, about the step 3 your step is true but i think when you replacing w=W in eqs2 i think all of them not change maybe at there is mistake ,even if you watch the indets you will figure out, and maybe i am wrong but i determine that point , also i change that series you used,  have three function we just have two function which one of them we take for get coiefecient i determined in file, there is something in paper i am not confortable with it  which i am don't know what is mean of that (By substituting Eq. (14) into Eq. (4) and collecting the terms of the same powers of and equating them to zero,) same power what mean of this

GOM-Dr.D-test-1-2.mw

@dharr is so good not just corect, and in end of the work i will compact more...

@dharr if you have time please fix that part step two and three are the same i think i try to fix but i can't see the error ,but you do

@dharr  step one done completly- i will go combine your code i am busy with it a lot and i am in step two i can't figure out  how they j point why is not appear is not hard but  something is there about coding i can't hundle it, 

for your question about nonlinear term

you have to chose  the big power or order or multiply of them together for example if we have something like that in one pde term
pde1=u_xx+u^5+u*u_x+u_x*u_xx...

highorder=u_xx=a-2

nonlinear term=u^5=5*a is highest in pde1

if we done have u^5 term in pde1

pde2=u_xx+u*u_x+u_x*u_xx...

highorder=u_xx=a-2

nonlinear term=u_x*u_xx=a-1+a-2

i need your help for finishing this emazing job and this 3 steps  is just begin, is a gate for two other topic like backlund transformation and lax pair but in begin i have to do this , and if you have any confusing in pdes about nonlinear term mention me , and i did for two pdes in step one by the old code but stucked in the step2, if need anything please mention 

pdes-s-1-2.mw

step1-2.mw

i will go combine your code and keep work on it, thank you so much your job is valuable 

@dharr in this writing i will show you  if you can write the program it will be good and go to next step it will be so better i did find a for each pdes is like the odes it take to to undrestand but is work i will show you with hand example 

@dharr when in pdes we look for balance number  always i find in odes i didn't find it in pdes but in odes we have a formulla but i do without formula just by watching and do a simple calculation i can find balancing term is like that
if we have a pde=diff(u,x$2)+u*diff(u,x$2)+u+diff(u,x)+u*diff(u,x)

if watch this pdes choose the high order derivative which is diff(u,x$2) and chose high term of nonlinear term  we have two of them u*diff(u,x$2) & u*diff(u,x) but we have to chose one of them which is the begist one is u*diff(u,x$2) so   

highorder=nonlinear term
m+2=m+(m+2) in here we have to find m which we called this balace number each derivative is + and multiply by two term is again +  in left hand side we have m+2 is mean order of derivative is two in right hand side we have one function is m and plut one derivative of order two which is m+2 when he substittue he just use the high order and high nonlinear term for finding alpha, but you did a better job from them which remove other term and did something different  by using the code of 

termexps[1] = termexps[-2], so my openion is this for finding alpha which you use remove i think without using remove we can find it by chose the high order and high nonlinear term  lets give you example if high order is like that is clear in pdes and after substitute function into pdes and do eval we get something like that (phi^{2alpha-4}) and nonlinear term is phi^{alpha-3} then  we have 2alpha-4=alpha-3   in paper that mean this i hope this time undrestand the step one

tommorow i will use this for other equation too and i will be sure is true or not , and really i saw another code is first time i saw this is another  emazing work again
for step two please us that series in eq11 in the papers and subs into pdes and collect the coieficcent of phi^{-5} to zero 

@dharr i did some new thing and remve some part of that but still i didn't found that  step2 and 3

new-p.mw

@dharr  THIS PAINLEVE test have three steps  i will explain one by one as in  first paper say it and he explain in best shape, but need a little bit undrestanding if you watch with more focus you will undrestand more and i am keep looking for other explanation  ... i am  explain that equation you find it the alpha[1]

1-step one: Leading order and coefficient: you find the alpha[1] untill there we dont have problem, then we have a function which is u=X(x,y,z,t)*phi(x,y,z,t)^alpha[1] but that is u=u(x,y,z,t)[0]*phi(x,y,z,t)^alpha[1], you find alpha replace in it and then we have to find  u(x,y,z,t)[0] how just in eq7 replave a=-1 i did this part and solve for (X)  we can find that u(x,y,z,t)[0]=X by equating the coefficients of leading terms(high power in negative sense)  he use (phi(x,y,z,t))^{-5}  this is end of step one  i hope make is full automatic for generate this  specially in seperating high order derivative and high non linear term

in fact is for me a question how you find it is hard even imagin you did that but it is totally  true we just take two term of pde for finding alpha and you did it and is totally correct 

2-Resonance point : this is more easier becuase as i seen alwasy they use this function as in eq(10)-eq(11) you can see it, so we use the same and already we found half of that series  is two part,  u(x, y, z, t) = u(x, y, z, t)[0]*phi(x, y, z, t)^a + u(x, y, z, t)[j]*phi(x, y, z, t)^(j + alpha[1]),  which we found first term in step one replace it, in here in eq(11) we use this and we just need to find the (j) points, how replace  the eq(11) after replacing a and first term then again use equating the coefficients of leading terms(high power in negative sense)  he use (phi(x,y,z,t))^{-5}  again becuase is high power as in step one then you will find that (j)s .

3-compatiblity condition: this is my openion and us i undrestand i will explain  in eq(5) which we have this series 

u(x, y, z, t) = sum(u(x, y, z, t)[k]*psi(x, y, z, t)^(a + k), k = 0 .. 6) , why 0 to 6 in k becuase the last point in step two for j is 6 i think always we use that high number  and in step one we find a=-1 replace in the series then  we found u(x,y,z,t)[0] in step 1 again  so how we can find u[1]...u[6]  like that replace that series which in paper is eq(14) into our pdes and then collecting the  term with phi(x,y,z,t)^i how many obtain then  by equating them to zero we can find u[1]...u[6] which 3 of them at my equation are free and he find other three

i just did one step for finding u[0], becuase of coding i am not involve the other part  if need anything else please let me know?
i hope is help full

i have to finish this coding then i apply to all other equation to see it true or not..

p1_DAH.mw

@janhardo I did some of it in first just I need to undrestand how they do it by maple in steps is clear what they did but how we do the same