salim-barzani

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These are replies submitted by salim-barzani

@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.

@mmcdara  thank you and @janhardo  this is emazing i can't beliave that which i found this , i will try to get solve the pde now , just try to change your version of maple i hope the B[i,j] not make any problem in coding but today you made my day i will check them later for solution thank you so much....

@janhardo  you did a good job i check for f[2] and f[4] and half of f[6] by hand determined each term one by one all true but the fraction number are extra How remove thus fractional number beside each term? i know in series is determine this number but i don't know why they remove it !

@dharr you did the limit so pretty just one step remain, and equation (11) is true i find exactly same!, i need to get the series without series we can't go further in all paper they mention this which from N-soliton reach the M-Lump by applying this , what they see which we can't see 

in your paper which linked mention this note too in below of (22)

for w(i)

m-lump.mw

if needed other paper please mention to upload more and more 

@mmcdara i just mention the second restart for know how the series work is strange series and the second series is making like that  but contain a different structure i want to make that series 

@mmcdara  we have two series the second series which contain product multiplier  we need that series watch lump you will see, but instead of N we must put 2M which m is number of lump ,for F[N]=F[2M] so in  series we have two define b[12] which we have it from N soliton is a[12] after that  in a[12] we have two take limit from a[12] when k[i]->0 then we get B[12] and then we can build all B[12] in 1 and 2 lump and 3 lump  i just need that series which contain product  and if possible apply k[i]->0 in a[12]=B[12] 

@mmcdara  why in function you just replace b[12] where i should add the b[13] and b[23] where  add code for substitute other b[ij] substitute 

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