salim-barzani

simplify the multiply exponential term ...

Asked: salim-barzani 165 Product: Maple 2021

July 28 2024

1 2

i am looking for simplify this type of simplifying assume beta is Real and there is any stuf package for working with complex and conjugate automaticaly

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(-2*beta^2*exp((2*I)*x)*conjugate(beta*exp(I*x))+(I*conjugate(beta*exp(I*x))+1+I)*(conjugate(beta*exp(I*x))+(-1+I))*exp((3*I)*x)*beta^3)/s

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Download simplify.mw

calculate adomian polynomial ...

Asked: salim-barzani 165 Product: Maple 2021

July 25 2024

0 11

Hi
i write my code for calculate this type of function but the result is so different from mine i will post here i hope someone tell me where is problem

i have this

i want this

Download EX1.mw

How Making loop for collecting solution?...

Asked: salim-barzani 165 Product: Maple 2021

July 24 2024

1 3

Hi
i did calculation part by part of adomian laplace method but if we can make a loop for it is gonna be so great and take back a lot of time

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pde := diff(u(x, t), t)+u(x, t)*(diff(u(x, t), x)) = t^2*x+x

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s*laplace(u(x, t), t, s)-u(x, 0)+laplace(u(x, t)*(diff(u(x, t), x)), t, s) = x*(s^2+2)/s^3

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s*laplace(u(x, t), t, s)+laplace(u(x, t)*(diff(u(x, t), x)), t, s) = x*(s^2+2)/s^3

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lap := s^alpha*laplace(u(x, t), t, s) = x*(s^2+2)/s^3-laplace(u(x, t)*(diff(u(x, t), x)), t, s)

s^alpha*laplace(u(x, t), t, s) = x*(s^2+2)/s^3-laplace(u(x, t)*(diff(u(x, t), x)), t, s)

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laplace(u(x, t), t, s) = (x*(s^2+2)/s^3-laplace(u(x, t)*(diff(u(x, t), x)), t, s))/s^alpha

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lap3 := u(x, t) = t^alpha*x/GAMMA(alpha+1)+2*x*t^(alpha+2)/GAMMA(alpha+3)-invlaplace(laplace(u(x, t)*(diff(u(x, t), x)), t, s)/s^alpha, s, t)

u(x, t) = t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha)-invlaplace(laplace(u(x, t)*(diff(u(x, t), x)), t, s)/s^alpha, s, t)

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u[1](x, t) = -invlaplace(laplace(u[0](x, t)*(diff(u[0](x, t), x)), t, s)/s^alpha, s, t)

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"u[0](x,t):=(t^alpha x)/(GAMMA(1+alpha))+(2 x t^(alpha+2))/(GAMMA(3+alpha))"

proc (x, t) options operator, arrow, function_assign; t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha) end proc

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for j from 0 to 3 do A[j] := subs(lambda = 0, (diff(f(seq(sum(lambda^i*u[i](x, t), i = 0 .. 20), m = 1 .. 2)), [`$`(lambda, j)]))/factorial(j)) end do

(t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha))*(t^alpha/GAMMA(1+alpha)+2*t^(alpha+2)/GAMMA(3+alpha))

u[1](x, t)*(t^alpha/GAMMA(1+alpha)+2*t^(alpha+2)/GAMMA(3+alpha))+(t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha))*(diff(u[1](x, t), x))

u[2](x, t)*(t^alpha/GAMMA(1+alpha)+2*t^(alpha+2)/GAMMA(3+alpha))+u[1](x, t)*(diff(u[1](x, t), x))+(t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha))*(diff(u[2](x, t), x))

u[3](x, t)*(t^alpha/GAMMA(1+alpha)+2*t^(alpha+2)/GAMMA(3+alpha))+u[2](x, t)*(diff(u[1](x, t), x))+u[1](x, t)*(diff(u[2](x, t), x))+(t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha))*(diff(u[3](x, t), x))

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S1 := u[1](x, t) = -invlaplace((t^alpha*x/GAMMA(1+alpha)+2*x*t^(alpha+2)/GAMMA(3+alpha))*(t^alpha/GAMMA(1+alpha)+2*t^(alpha+2)/GAMMA(3+alpha))/s^alpha, s, t)

u[1](x, t) = -x*(t^alpha)^2*invlaplace(s^(-alpha), s, t)*(1/GAMMA(1+alpha)^2+4*t^2/(GAMMA(3+alpha)*GAMMA(1+alpha))+4*t^4/GAMMA(3+alpha)^2)

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u[1](x, t) = x*GAMMA(2*alpha+1)*t^(3*alpha)/(GAMMA(1+alpha)^2*GAMMA(3*alpha+1))-4*x*GAMMA(2*alpha+3)*t^(3*alpha+2)/(GAMMA(1+alpha)*GAMMA(3+alpha)*GAMMA(3*alpha+3))-4*`xΓ`(2*alpha+5)/(GAMMA(3+alpha)^2*GAMMA(3*alpha+3))

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u[2](x, t) = -invlaplace(laplace(u[1](x, t)*(diff(u(x, t), x)), t, s)/s^alpha, s, t)

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for get definition use this pdf for fractional derivation

[Copyrighted material removed by moderator - see https://doi.org/10.4236/am.2018.94032]

Download solving_example_1.mw

Same equation but different parameter? ...

Asked: salim-barzani 165 Product: Maple 2021

July 24 2024

1 3

Hi

i use other code for equation too when i use allvalues(Root(...)) it is more near but question is this why not satisfy the ode equation this is my equation this parameter are find for this ODe why not satisfy otherwise my equestions must be wrong!

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eq2 := 6*beta*g[1]^3*r[0]^2*r[2]+6*beta*g[1]^3*r[0]*r[1]^2+2*p^2*sigma*g[1]*r[0]^2*r[2]+p^2*sigma*g[1]*r[0]*r[1]^2+12*beta*f[0]*g[1]^2*r[0]*r[1]+6*beta*f[1]*g[1]^2*r[0]^2+6*beta*f[0]^2*g[1]*r[0]-k^2*sigma*g[1]*r[0]-2*w*g[1]*r[0] = 0

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eq3 := 12*beta*g[1]^3*r[0]*r[1]*r[2]+2*beta*g[1]^3*r[1]^3+2*p^2*sigma*g[1]*r[0]*r[1]*r[2]+12*beta*f[0]*g[1]^2*r[0]*r[2]+6*beta*f[0]*g[1]^2*r[1]^2+12*beta*f[1]*g[1]^2*r[0]*r[1]+p^2*sigma*f[1]*r[0]*r[1]+6*beta*f[0]^2*g[1]*r[1]+12*beta*f[0]*f[1]*g[1]*r[0]-k^2*sigma*g[1]*r[1]+2*beta*f[0]^3-k^2*sigma*f[0]-2*w*g[1]*r[1]-2*w*f[0] = 0

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eq5 := 6*beta*g[1]^3*r[1]*r[2]^2+3*p^2*sigma*g[1]*r[1]*r[2]^2+6*beta*f[0]*g[1]^2*r[2]^2+12*beta*f[1]*g[1]^2*r[1]*r[2]+3*p^2*sigma*f[1]*r[1]*r[2]+12*beta*f[0]*f[1]*g[1]*r[2]+6*beta*f[1]^2*g[1]*r[1]+6*beta*f[0]*f[1]^2 = 0

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2*beta*U(xi)^3+(-k^2*sigma-2*w)*U(xi)+(diff(diff(U(xi), xi), xi))*p^2*sigma = 0

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P := f[0]+sum(f[i]*R(xi)^i, i = 1 .. 1)+sum(g[i]*((diff(R(xi), xi))/R(xi))^i, i = 1 .. 1)

f[0]+f[1]*R(xi)+g[1]*(diff(R(xi), xi))/R(xi)

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$case1 := {p = -sqrt(2)*sqrt(sigma*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))/(sigma*(4*r[0]*r[2]-r[1]^2)), f[0] = -(k^2*sigma+2*w)*r[1]/sqrt(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w)), f[1] = -(2*(k^2*sigma+2*w))*r[2]/sqrt(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w)), g[1] = -sqrt(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))/(beta*(4*r[0]*r[2]-r[1]^2))}$

${p = -2^(1/2)*(sigma*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)/(sigma*(4*r[0]*r[2]-r[1]^2)), f[0] = -(k^2*sigma+2*w)*r[1]/(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2), f[1] = -2*(k^2*sigma+2*w)*r[2]/(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2), g[1] = -(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)/(beta*(4*r[0]*r[2]-r[1]^2))}$

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f[0]+f[1]*R(xi)+g[1]*(r[0]+r[1]*R(xi)+r[2]*R(xi)^2)/R(xi)

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-(k^2*sigma+2*w)*r[1]/(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)-2*(k^2*sigma+2*w)*r[2]*R(xi)/(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)-(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)*(r[0]+r[1]*R(xi)+r[2]*R(xi)^2)/(beta*(4*r[0]*r[2]-r[1]^2)*R(xi))

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U(xi) = -(k^2*sigma+2*w)*r[1]/(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)-2*(k^2*sigma+2*w)*r[2]*R(xi)/(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)-(-2*beta*(4*r[0]*r[2]-r[1]^2)*(k^2*sigma+2*w))^(1/2)*(r[0]+r[1]*R(xi)+r[2]*R(xi)^2)/(beta*(4*r[0]*r[2]-r[1]^2)*R(xi))

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2*beta*U(xi)^3+(-k^2*sigma-2*w)*U(xi)+2*(diff(diff(U(xi), xi), xi))*(k^2*sigma+2*w)/(4*r[0]*r[2]-r[1]^2) = 0

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same_equation_different_parameter.mw

test solution of PDE...

Asked: salim-barzani 165 Product: Maple 2021

July 21 2024

1 9

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pde := I*(diff(psi(x, t), t))+alpha*(diff(psi(x, t), `$`(x, 2)))+(beta[3]*abs(psi(x, t))+beta[4]*abs(psi(x, t))^2)*psi(x, t)+gamma*(diff(abs(psi(x, t))^2, `$`(x, 2)))*psi(x, t)/abs(psi(x, t)) = 0

>	$case1 := {k = k, lambda = sqrt(-1/(18alphabeta[4]+18gammabeta[4]))beta[3], w = -(9alphak^2beta[4]+2beta[3]^2)/(9beta[4]), A[0] = -beta[3]/(3beta[4]), A[1] = beta[3]/(3beta[4]), B[1] = 0}$

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" U(xi):=-(beta[3] (cosh(xi)-sinh(xi)))/(3 beta[4] cosh(xi))"

proc (xi) options operator, arrow, function_assign; -(1/3)*beta[3]*(cosh(xi)-sinh(xi))/(beta[4]*cosh(xi)) end proc

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-(1/3)*beta[3]*(cosh(xi)-sinh(xi))/(beta[4]*cosh(xi))

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xi := sqrt(-1/(18*alpha*beta[4]+18*gamma*beta[4]))*beta[3]*(2*alpha*kt+x)

(-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x)

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-(1/3)*beta[3]*(cosh((-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x))-sinh((-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x)))*exp(I*(-k*x+t*w+theta))/(beta[4]*cosh((-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x)))

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-(1/3)*beta[3]*(cosh((-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x))-sinh((-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x)))*exp(I*(-k*x-(1/9)*(9*alpha*k^2*beta[4]+2*beta[3]^2)*t/beta[4]+theta))/(beta[4]*cosh((-1/(18*alpha*beta[4]+18*gamma*beta[4]))^(1/2)*beta[3]*(2*alpha*kt+x)))