Question: How Making loop for collecting solution?

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

restart

with(inttrans)

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

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

(1)

eq := laplace(pde, t, s)

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

(2)

eq2 := subs({u(x, 0) = 0}, eq)

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

(3)

NULL

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)

(4)

lap1 := 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

(5)

NULL

lap2 := invlaplace(lap1, s, t)

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

(6)

NULL

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)

(7)

NULL

NULL

NULL

NULL

``

(8)

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

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

(9)

"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

(10)

n := N

N

(11)

k := K

K

(12)

f := proc (u) options operator, arrow; u*(diff(u, x)) end proc

proc (u) options operator, arrow; u*(diff(u, x)) end proc

(13)

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))

(14)

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)

(15)

NULL

NULL

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))

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))

(16)

NULL

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

NULL

NULL


for get definition use this pdf for fractional derivation

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

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