salim-barzani

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These are replies submitted by salim-barzani

@dharr in here i looking for equations, for finding a[12] the founded one is  the other equation i am looking for the a[12] after removing each indets and using algsub then when i get equation i want find a[12] , but in indets in equation p2a the unknown exp is too much and undetectable really the  more than 15 of them and each one of them have a small different how make them the number of them be decrease or something like that  you can  do some magic 

@dharr  i have a small question dctor in here there is too much number of indets what i should do label all of them?

Dr.D-pde.mw

@dharr i have to solve a lot of them and reach a lot of problem for solving equation which no one solved before becuase of that i  have try all kind of them, by your method i can hundle all but  i have to figure out use that series for findding all is just for more modified work, there is any way for finding that series and replacing to get our trail function just this step remain ,ofcurse i will use yours is more easier and more simple ,is just for my knowledge if can find series of trail 

@mmcdara your numerical work is so valuable, i don't know the mean of this dot in figure but i thing if all of them are in one line is mean is totally true but the dot place is solution and theyseperate it mean we have singularity and is not continuos , i am agree in a lot place is singular solution the lump in some rare place is not singulare i must to find thus place which my function don't give me singulare plot i have this issue in plotting such function is big problem, also give me persmisin to tell you you are wrong about saying is not a solution my source of of paper a lot of them did the same i can't check it becuase the equation is to long and not run in my pc , but for the first one when i did transfor is satisfy and for second one when i did the pc not give me answer but keep runing for finding solution , check this you will figure out it must be true this is for lump 1 and 2  the graph i plot it will be change and i am have to looking for non singular plot by using paper i give the same parameter as he did but is lump but is singular so i am looking for nonsigular , check my file in below please

check this please 

1-2-p.mw

@mmcdara  i need check eqt3 and eq33 in pde they have to  be check  in pdetest

@dharr  i don't have idea why they do that but the figure is emazing when they did that but i don't know the pattern of variable yet

@acer i try to do like you but i ma fail i think each time i am stuck like that i should post question here i think i will delete this and repost a good one with more option can i do that don't be mad at me please but i don't want dublicate question i  want getting some shape like kink wave and dark bright mixed dark bright can i told explore give me that parameter for searching such graph for the function? i want do that by iteration anythink like that is possible?

@mmcdara  i will repost this after  i mix all part of equation and also the that paper i take it from i have to work on it thank you 

@janhardo  my odetest is zero know my orgin pde must be zero i use this trick i use transfromation instead of pde i can used pde too but is a little bit hard i did two or three time but i lost the code and don't remember it, this is the pde which is  by my method when changed to real and imaginary part is satisfy but when i back to my orgind pde is not i don't know why, when you upload code please upload the shit i can't use the up code where i should write it 

i have the u(x,t) can you check it i will upload here 

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

declare(u(x, t)); declare(U(xi)); declare(V(xi)); declare(F(xi)); declare(V(xi))*declare(H(xi))

u(x, t)*`will now be displayed as`*u

 

U(xi)*`will now be displayed as`*U

 

V(xi)*`will now be displayed as`*V

 

F(xi)*`will now be displayed as`*F

 

V(xi)*`will now be displayed as`*V

 

H(xi)*`will now be displayed as`*H

 

()^2

(2)

case1 := [beta = 2*RootOf(3*_Z^2-3*_Z-1)*(n+2)/(B[1]*n^2), delta = 2*B[1]*(RootOf(3*_Z^2-3*_Z-1)+1)*(3*n+2)/(3*n^2), eta = (k^2*n^2*B[1]^2-n^2*w*B[1]^2-1)/(n^2*B[1]^2), gamma = -6*RootOf(3*_Z^2-3*_Z-1)*(n+1)/n^2, lambda = B[1]^2*(3*RootOf(3*_Z^2-3*_Z-1)-7)*(2*n+1)/(9*n^2), A[0] = RootOf(3*_Z^2-3*_Z-1)*B[1], A[1] = 0, B[1] = B[1]]

p := 2*k

2*k

(3)

NULL

K := q(x, t) = (B[1]*(RootOf(3*_Z^2-3*_Z-1)+coth(-p*t+x)))^(-1/n)*exp(I*(k*x-t*w))

q(x, t) = (B[1]*(RootOf(3*_Z^2-3*_Z-1)-coth(2*k*t-x)))^(-1/n)*exp(I*(k*x-t*w))

(4)
 

NULL

Download s1.mw

@dharr thanks Dear Dr

@mmcdara just if possible remove this [] from B[ij] from the code which is so ugly john did but your code is shorter if possible upload here thanks

@janhardo  thank you john you did a perfect job

@dharr @janhardo @mmcdara  thanks for all of you, all of you made my day  and have a big part of my PDE solution i am very greatfull to have all of you in this website. proud of all of you 

@dharr How i can tell it? is that like this ::n>0,  How reach this point i am shock when i saw this and when i saw your answer i got duble shock ...thank you so much dear Dr. David

@mmcdara  in john code we have this  for f[4]

theta[1]*theta[2]*theta[3]*theta[4] + 1/2*B[1, 2]*theta[3]*theta[4] + 1/2*B[1, 3]*theta[2]*theta[4] + 1/2*B[1, 4]*theta[2]*theta[3] + 1/2*B[2, 3]*theta[1]*theta[4] + 1/2*B[2, 4]*theta[1]*theta[3] + 1/2*B[3, 4]*theta[1]*theta[2] + 1/8*B[1, 2]*B[3, 4] + 1/8*B[1, 3]*B[2, 4] + 1/8*B[1, 4]*B[2, 3]

How they remove is multiply by something or what? is so confusing for me i didn't work with such series with combinations but i have too , also i used your code and for sure i will try to upgrad it i can't do more with it but i will make merg with other code for getting the best result for my solution.

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