@dharr i have two ode equation which i used one of them as auxiliary ode and one of them is mine ode which i want to solve it, i use the solution of auxiliary ode  by using another series which contain   solution of auxliary ode,  i can use that table solution which they are solution of auxiliary ode, but i used them to put inside the series to get my solution of ode, i think this time must be clear, and by this solution i use transformation which it will be the solution of pdes in fact.

i know table give me the solution of auxiliary ode but my ode which i want to solve is so different and more complex, which i know if i use the condition it will be zero just i want to tested and see it will be zero or not 

if you watch this file F-P-J-E.mw you will see a series (F) that series is our solution which contain function G(xi) this G(xi) is our solution from table put inside the ode after finding the contant of series in F must be zero.

@dharr  you did somehow about appear jacobicn jacobidn jacobisn and so on i want my solution appear like that can we convert it from trigonometric to that shape? and there is any way to do automatically test all that solution in tht table for ode? becuase if there is not automatically i have to check it one by one by hand for the ode, but important part is that to change it to jacobi form and when m=1 and m=0 like  the paper did two different solution will come out

@dharr i didn't know even we can write jacobi inside equation like paper did, i am working on a method based on jacobi eliptic and i know each step except yours step which is intresting, i have a table which contain 52 solution , 46 are jacobi eliptic and other are weirstrass-elliptic which i can't use use the function at all, i apply for easy odes but i need help there are a lot of fucntion, but i want when i write in paper jacobicn or jacobidn appear in my solution like the paper which i show you in below, this paper  did a good job about written in three shape when we have all type with cn,dn,sn and so on also with m=1, and m=0.
i just didn't do this method about analytic methods,  i don't know how the outcome will be clear more with three type becuase i have to export to latex after the outcome is appear  my coding is a little complex sorry for that  i will put my file in below and link with some picture  also file of table solution 

F-P-J-E.mw

46-solution.mw

paper

@acer  I want 2D after draw 3D which all parameter are fixed, the main parameters are (a,L) L is fixed but (a) change in three different number like (a=0.3,0.5,1) can we do that

@acer 

this is good but i need to complet the work i add some part in below which beside the shape of 3D and 2D which must the result be smaller and smaller in three different of four level of changing parameter the 2D and 3D both importan like the first maple file   and yours which is so clear  there is anyway for for do something like that  

i need the best shape of 2D  while  parameter (a) is changing in different level also for 3D how we do that?

g1_ac-explor.mw

@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