@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 

@dharr i will update some information from other paper, really  that one i provided from top is easiest one for explanation but i have other  paper and keep lookong for some thing better becuase i have to figure out and find this without this my papers gets rejected, so is important and in 2004 someone write the package for this i will put pictures of him too and another paper steps from 1993, i hope this is make to undrestand becuase if you get notion of topic you can solve anything , i can send you thus pdf in case your are intrested and the picture confuse you 

Moderator: papers removed for copyright reasons

https://doi.org/10.1016/j.amc.2004.10.055 (PDEPtest paper)
https://doi.org/10.1016/0895-7177(93)90224-M (Vlieg-Hulstman paper)
https://doi.org/10.1016/j.cpc.2004.04.005 (wkptest paper)

@acer  thanks a lot, in future i will use this if something come up i will ask again thank you so much....

@acer  but if all part be together  it will be better just linear and non-linear not seperate the one derivative two derivate and non-derivative so on..  and seperate by comma is becuase of that i needed to use in huge equation if seperated automatically like that i didn't use any other extra code 

Download L-N-L.mw

@janhardo the v is also a function so two  term in non linear v*u is non linear

@sand15  i try to explain in here , i hope find a way ,  just need a tactic of coding to arrange and trail what i explained i think it will be answer to this equation

same-problem.mw

@sand15  it is becuase of this term

(p[i]^2 - 1)*(diff(diff(f(x, y, z, w, t), x), x)/f(x, y, z, w, t) - diff(f(x, y, z, w, t), x)^2/f(x, y, z, w, t)^2)

in u if you watch is p[i] this p[i], i must be one or two
i think beside theta[1] is one and beside theta[2] is two but how arrange this

@sand15  your way is so good and really still i am watching your approach is not easy to do that, i already solve more than 30 pdes in this way and this is just begin when we find a[12],  thus problem are not solved before i have another equation  have same problem but we have to find this a[12] is not zero be sure about that is imposible to be zero,  in both of thus equation in top equation and in this one if you watch the R when we found it is not number  in other when i solve  is number and i didn't have problem , there must be a way for finding this a[12] 

same-problem.mw

@dharr  the result is same like before, i think this approach is not suatable for this equation i have to change it, Thanks Dr.

@dharr i did that, but answer is not what i  i am looking i am looking for a[0],a[1] and w,  i did a for a lot but this is problem come up i don't know how fix that

Thanks Dear Dr.David(@dharr) you are very helpfull for me on this topic, still i am work on this topic, and i did a great achivment and keep improving the work, this mehod  need a super computer for a long range of variable of (t) i saw your last graph which is so far from exact but if you make the range of t smaller it will be so small error or maybe zero,  your code is great like always, but i don't undrestand about this part when you write (infolevel[LAD] := 2) what mean of this 2 in the first example and we can change to 3 or more? i have mine code and i have to improve it, and i am intresting in your which is so compact, i don't know this method can hundle the hard one?  and if  need help  i will ask for help