SSMB

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MaplePrimes Activity


These are replies submitted by SSMB

@Carl Love i did a lot stuff but is not what i am looking i want systematically do all stuff but i can't in each step i stuck and i have to do it by hand  i want to do get this and i am near but like that i don't like it , 

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(P(x, t)); declare(q(x, t))

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

 

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

 

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

(2)

pde := I*(diff(u(x, t), t))+diff(u(x, t), `$`(x, 2))+abs(u(x, t))^2*u(x, t) = 0

I*(diff(u(x, t), t))+diff(diff(u(x, t), x), x)+abs(u(x, t))^2*u(x, t) = 0

(3)

S := u(x, t) = (sqrt(a)+P(x, t))*exp(I*a*t)

u(x, t) = (a^(1/2)+P(x, t))*exp(I*a*t)

(4)

S1 := conjugate(u(x, t)) = (sqrt(a)+conjugate(P(x, t)))*exp(-I*a*t)

conjugate(u(x, t)) = (a^(1/2)+conjugate(P(x, t)))*exp(-I*a*t)

(5)

Q := abs(u(x, t))^2 = u(x, t)*conjugate(u(x, t))

abs(u(x, t))^2 = u(x, t)*conjugate(u(x, t))

(6)

F1 := expand(simplify(subs({S, S1}, rhs(Q))))

abs(P(x, t))^2+P(x, t)*a^(1/2)+conjugate(P(x, t))*a^(1/2)+a

(7)

F2 := abs(u(x, t))^2 = remove(has, F1, abs(P(x, t))^2)

abs(u(x, t))^2 = P(x, t)*a^(1/2)+conjugate(P(x, t))*a^(1/2)+a

(8)

FF := collect(F2, sqrt(a))

abs(u(x, t))^2 = a+(conjugate(P(x, t))+P(x, t))*a^(1/2)

(9)

F3 := abs(u(x, t))^2*u(x, t) = (a+(P(x, t)+conjugate(P(x, t)))*sqrt(a))*rhs(S)

abs(u(x, t))^2*u(x, t) = (a+(conjugate(P(x, t))+P(x, t))*a^(1/2))*(a^(1/2)+P(x, t))*exp(I*a*t)

(10)

F4 := remove(has, F3, P(x, t)*conjugate(P(x, t)))

abs(u(x, t))^2*u(x, t) = (a+(conjugate(P(x, t))+P(x, t))*a^(1/2))*(a^(1/2)+P(x, t))*exp(I*a*t)

(11)

expand(%)

abs(u(x, t))^2*u(x, t) = 2*exp(I*a*t)*P(x, t)*a+exp(I*a*t)*P(x, t)^2*a^(1/2)+exp(I*a*t)*conjugate(P(x, t))*a+exp(I*a*t)*conjugate(P(x, t))*a^(1/2)*P(x, t)+exp(I*a*t)*a^(3/2)

(12)

pde_linear, pde_nonlinear := selectremove(proc (term) options operator, arrow; not has((eval(term, P(x, t) = T*P(x, t)))/T, T) end proc, expand(%))

() = (), abs(u(x, t))^2*u(x, t) = 2*exp(I*a*t)*P(x, t)*a+exp(I*a*t)*P(x, t)^2*a^(1/2)+exp(I*a*t)*conjugate(P(x, t))*a+exp(I*a*t)*conjugate(P(x, t))*a^(1/2)*P(x, t)+exp(I*a*t)*a^(3/2)

(13)

F6 := abs(u(x, t))^2*u(x, t) = exp(I*a*t)*a^(3/2)+2*exp(I*a*t)*a*P(x, t)+exp(I*a*t)*a*conjugate(P(x, t))

abs(u(x, t))^2*u(x, t) = exp(I*a*t)*a^(3/2)+2*exp(I*a*t)*P(x, t)*a+exp(I*a*t)*conjugate(P(x, t))*a

(14)

subs({F6, S}, pde)

I*(diff((a^(1/2)+P(x, t))*exp(I*a*t), t))+diff(diff((a^(1/2)+P(x, t))*exp(I*a*t), x), x)+exp(I*a*t)*a^(3/2)+2*exp(I*a*t)*P(x, t)*a+exp(I*a*t)*conjugate(P(x, t))*a = 0

(15)

eval(%)

I*((diff(P(x, t), t))*exp(I*a*t)+I*(a^(1/2)+P(x, t))*a*exp(I*a*t))+(diff(diff(P(x, t), x), x))*exp(I*a*t)+exp(I*a*t)*a^(3/2)+2*exp(I*a*t)*P(x, t)*a+exp(I*a*t)*conjugate(P(x, t))*a = 0

(16)

expand(%)

I*(diff(P(x, t), t))*exp(I*a*t)+exp(I*a*t)*P(x, t)*a+(diff(diff(P(x, t), x), x))*exp(I*a*t)+exp(I*a*t)*conjugate(P(x, t))*a = 0

(17)

expand(%/exp(I*a*t))

I*(diff(P(x, t), t))+P(x, t)*a+diff(diff(P(x, t), x), x)+conjugate(P(x, t))*a = 0

(18)

PP := collect(%, a)

(conjugate(P(x, t))+P(x, t))*a+I*(diff(P(x, t), t))+diff(diff(P(x, t), x), x) = 0

(19)

U1 := P(x, t) = r[1]*exp(I*(l*x-m*t))+r[2]*exp(-I*(l*x-m*t))

P(x, t) = r[1]*exp(I*(l*x-m*t))+r[2]*exp(-I*(l*x-m*t))

(20)

U2 := conjugate(P(x, t)) = r[1]*exp(I*(l*x-m*t))+r[2]*exp(I*(l*x-m*t))

conjugate(P(x, t)) = r[1]*exp(I*(l*x-m*t))+r[2]*exp(I*(l*x-m*t))

(21)

eval(subs({U1, U2}, PP))

(2*r[1]*exp(I*(l*x-m*t))+r[2]*exp(I*(l*x-m*t))+r[2]*exp(-I*(l*x-m*t)))*a+I*(-I*r[1]*m*exp(I*(l*x-m*t))+I*r[2]*m*exp(-I*(l*x-m*t)))-r[1]*l^2*exp(I*(l*x-m*t))-r[2]*l^2*exp(-I*(l*x-m*t)) = 0

(22)

simplify((2*r[1]*exp(I*(l*x-m*t))+r[2]*exp(I*(l*x-m*t))+r[2]*exp(-I*(l*x-m*t)))*a+I*(-I*r[1]*m*exp(I*(l*x-m*t))+I*r[2]*m*exp(-I*(l*x-m*t)))-r[1]*l^2*exp(I*(l*x-m*t))-r[2]*l^2*exp(-I*(l*x-m*t)) = 0)

r[2]*(-l^2+a-m)*exp(-I*(l*x-m*t))+2*exp(I*(l*x-m*t))*((1/2)*r[2]*a+r[1]*(-(1/2)*l^2+a+(1/2)*m)) = 0

(23)

J := eval(%)

r[2]*(-l^2+a-m)*exp(-I*(l*x-m*t))+2*exp(I*(l*x-m*t))*((1/2)*r[2]*a+r[1]*(-(1/2)*l^2+a+(1/2)*m)) = 0

(24)

expand(%)

-r[2]*exp(I*m*t)*l^2/exp(I*l*x)+r[2]*exp(I*m*t)*a/exp(I*l*x)-r[2]*exp(I*m*t)*m/exp(I*l*x)+exp(I*l*x)*r[2]*a/exp(I*m*t)-exp(I*l*x)*l^2*r[1]/exp(I*m*t)+2*exp(I*l*x)*r[1]*a/exp(I*m*t)+exp(I*l*x)*m*r[1]/exp(I*m*t) = 0

(25)

indets(%)

{a, l, m, t, x, r[1], r[2], exp(I*l*x), exp(I*m*t)}

(26)

subs({exp(-I*(l*x-m*t)) = Y, exp(I*(l*x-m*t)) = X}, J)

r[2]*(-l^2+a-m)*Y+2*X*((1/2)*r[2]*a+r[1]*(-(1/2)*l^2+a+(1/2)*m)) = 0

(27)

collect(%, {X, Y})

(r[2]*a+2*r[1]*(-(1/2)*l^2+a+(1/2)*m))*X+r[2]*(-l^2+a-m)*Y = 0

(28)

eq := r[2]*a+2*r[1]*(-(1/2)*l^2+a+(1/2)*m) = 0

r[2]*a+2*r[1]*(-(1/2)*l^2+a+(1/2)*m) = 0

(29)

eq1 := r[2]*(-l^2+a-m) = 0

r[2]*(-l^2+a-m) = 0

(30)

solve({eq, eq1}, m)

Download conjugate.mw

@janhardo this is wrong even not related to mine

@Aixleft math this is the easiest one i have here, the other is more complecated 

@Kitonum  i have f which by this f i found 3 soliton solution , i can get this by a series which already we define it i will upload the series,

we have to change this f[3] to equation (14) as mention in paper , by do something like long wave limit to k[1] &k[2]->0 as mention in paper and eta[1] and eta[2]=I*pi

i can use the (14) for getting my solution directly but i want to find out how reach this point ,this is more importan for me 
the key point is just how we act with k[1] and k[2] also eta[1] and eta[2], to reach eq(14) also we have for f[4] for f[5] we have for all of them but this is easiest one i upload all information in paper and mw fild and this is a series of reaching N-soliton

N-sol.mw

@Scot Gould is so nice, i am looking for  generate the best shape for 3D of this kind of graph by changing all parameter not just time like explore for all parameter but automatically search and use iteration for all parameter, maybe it take a lot time for looking but it will be emazing for geometrically present, is this posible ?

@mmcdara thanks a lot!

@Scot Gould  thank you so much

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