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How Making loop for collecting solution?...

salim-barzani 1555 Maple 2021
homework pde differential-equations
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

> 

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]

Download solving_example_1.mw

Same equation but different parameter? ...

salim-barzani 1555 Maple 2021
ode parameters homework
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!

> 

restart
> 

with(PDEtools)
> 

undeclare(prime)

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

 (1)
> 

with(DEtools)
> 

``
> 

with(LinearAlgebra)
> 

with(sumtools)
> 

eq0 := 2*beta*g[1]^3*r[0]^3+2*p^2*sigma*g[1]*r[0]^3 = 0
> 

eq1 := 6*beta*g[1]^3*r[0]^2*r[1]+3*p^2*sigma*g[1]*r[0]^2*r[1]+6*beta*f[0]*g[1]^2*r[0]^2 = 0
> 

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
> 

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
> 

eq4 := 6*beta*g[1]^3*r[0]*r[2]^2+6*beta*g[1]^3*r[1]^2*r[2]+2*p^2*sigma*g[1]*r[0]*r[2]^2+p^2*sigma*g[1]*r[1]^2*r[2]+12*beta*f[0]*g[1]^2*r[1]*r[2]+12*beta*f[1]*g[1]^2*r[0]*r[2]+6*beta*f[1]*g[1]^2*r[1]^2+2*p^2*sigma*f[1]*r[0]*r[2]+p^2*sigma*f[1]*r[1]^2+6*beta*f[0]^2*g[1]*r[2]+12*beta*f[0]*f[1]*g[1]*r[1]+6*beta*f[1]^2*g[1]*r[0]-k^2*sigma*g[1]*r[2]+6*beta*f[0]^2*f[1]-k^2*sigma*f[1]-2*w*g[1]*r[2]-2*w*f[1] = 0
> 

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
> 

eq6 := 2*beta*g[1]^3*r[2]^3+2*p^2*sigma*g[1]*r[2]^3+6*beta*f[1]*g[1]^2*r[2]^2+2*p^2*sigma*f[1]*r[2]^2+6*beta*f[1]^2*g[1]*r[2]+2*beta*f[1]^3 = 0
> 

NULL
> 

NULL
> 

COEFFS := solve({eq0, eq1, eq2, eq3, eq4, eq5, eq6}, {p, f[0], f[1], g[1]}, explicit)
> 

NULL
> 

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

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

 (2)
> 

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)

 (3)
> 

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))}

 (4)
> 

NULL

``

 (5)
> 

K := diff(R(xi), xi) = r[0]+r[1]*R(xi)+r[2]*R(xi)^2

diff(R(xi), xi) = r[0]+r[1]*R(xi)+r[2]*R(xi)^2

 (6)
> 

S1 := subs(K, P)

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

 (7)

NULL

> 

C1 := subs(case1, S1)

-(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))

 (8)
> 

f := U(xi) = C1

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))

 (9)

NULL

> 

SO := subs(case1, ode)

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

 (10)

NULL

> 

odetest(f, SO)


same_equation_different_parameter.mw

test solution of PDE...

salim-barzani 1555 Maple 2021
pde differential-equations pdetest
July 21 2024
1 9
> 

restart
> 

with(PDEtools)
> 

undeclare(prime)

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

 (1)
> 

with(DEtools)
> 

NULL
> 

with(DifferentialAlgebra)
> 

"with(Student[ODEs][Solve]): "
> 

with(IntegrationTools)
> 

with(inttrans)
> 

with(PDEtools)
> 

with(Physics)
> 

with(PolynomialTools)
> 

with(RootFinding)
> 

with(SolveTools)
> 

with(LinearAlgebra)
> 

with(sumtools)
> 

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/(18*alpha*beta[4]+18*gamma*beta[4]))*beta[3], w = -(9*alpha*k^2*beta[4]+2*beta[3]^2)/(9*beta[4]), A[0] = -beta[3]/(3*beta[4]), A[1] = beta[3]/(3*beta[4]), B[1] = 0}
> 

" psi(x,t):=U(xi)*exp(I*(-k*x+w*t+theta))"

proc (x, t) options operator, arrow, function_assign; U(xi)*exp(I*(-k*x+w*t+theta)) end proc

 (2)
> 

" 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

 (3)
> 

convert(U(xi), trig)

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

 (4)
> 

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)

 (5)
> 

S := psi(x, t)

-(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)))

 (6)
> 

solution := subs(case1, S)

-(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)))

 (7)
> 

pdetest(psi(x, t) = -beta[3]*(cosh(sqrt(-1/(18*alpha*beta[4]+18*gamma*beta[4]))*beta[3]*(2*alpha*k+x))-sinh(sqrt(-1/(18*alpha*beta[4]+18*gamma*beta[4]))*beta[3]*(2*alpha*k+x)))*exp(I*(-k*x-(9*alpha*k^2*beta[4]+2*beta[3]^2)*t/(9*beta[4])+theta))/(3*beta[4]*cosh(sqrt(-1/(18*alpha*beta[4]+18*gamma*beta[4]))*beta[3]*(2*alpha*k+x)))*exp(I*(-k*x-(9*alpha*k^2*beta[4]+2*beta[3]^2)*t/(9*beta[4])+theta)), pde)

Error, (in pdetest) unable to determine the indeterminate function

  
> 

NULL

 

 

 

 

Download pde-solve.mw

set up equations and find parameter ...

salim-barzani 1555 Maple
loop coefficients
July 20 2024
1 1 
restart;
with(PolynomialTools);
with(RootFinding);
with(SolveTools);
with(LinearAlgebra);
NULL;
NULL;
E1 := (-alpha*k^2*A[1] - alpha*k^2*B[1] + 3*A[0]^2*A[1]*beta[4] + 3*A[0]^2*B[1]*beta[4] + A[1]^3*beta[4] + 3*A[1]^2*B[1]*beta[4] + 3*A[1]*B[1]^2*beta[4] + B[1]^3*beta[4] + 2*A[0]*A[1]*beta[3] + 2*A[0]*B[1]*beta[3] - w*A[1] - w*B[1])*cosh(xi)^6 + (-alpha*k^2*A[0] + A[0]^3*beta[4] + 3*A[0]*A[1]^2*beta[4] + 6*A[0]*A[1]*B[1]*beta[4] + 3*A[0]*B[1]^2*beta[4] + A[0]^2*beta[3] + A[1]^2*beta[3] + 2*A[1]*B[1]*beta[3] + B[1]^2*beta[3] - w*A[0])*sinh(xi)*cosh(xi)^5 + (2*alpha*k^2*A[1] + alpha*k^2*B[1] - 2*alpha*lambda^2*A[1] + 2*alpha*lambda^2*B[1] - 2*gamma*lambda^2*A[1] + 2*gamma*lambda^2*B[1] - 6*A[0]^2*A[1]*beta[4] - 3*A[0]^2*B[1]*beta[4] - 3*A[1]^3*beta[4] - 6*A[1]^2*B[1]*beta[4] - 3*A[1]*B[1]^2*beta[4] - 4*A[0]*A[1]*beta[3] - 2*A[0]*B[1]*beta[3] + 2*w*A[1] + w*B[1])*cosh(xi)^4 + (alpha*k^2*A[0] - A[0]^3*beta[4] - 6*A[0]*A[1]^2*beta[4] - 6*A[0]*A[1]*B[1]*beta[4] - A[0]^2*beta[3] - 2*A[1]^2*beta[3] - 2*A[1]*B[1]*beta[3] + w*A[0])*sinh(xi)*cosh(xi)^3 + (-alpha*k^2*A[1] + 4*alpha*lambda^2*A[1] + 4*gamma*lambda^2*A[1] + 3*A[0]^2*A[1]*beta[4] + 3*A[1]^3*beta[4] + 3*A[1]^2*B[1]*beta[4] + 2*A[0]*A[1]*beta[3] - w*A[1])*cosh(xi)^2 + (3*A[0]*A[1]^2*beta[4] + A[1]^2*beta[3])*sinh(xi)*cosh(xi) - 2*alpha*lambda^2*A[1] - 2*gamma*lambda^2*A[1] - A[1]^3*beta[4] = 0;
N := 6;
for i from 0 to N do
    equ[1][i] := coeff(E1, {cosh(xi)^i, sinh(xi)^i}, i) = 0;
end do;
             //        2               2     
equ[1][0] := \\-alpha k  A[1] - alpha k  B[1]

           2                      2                    3        
   + 3 A[0]  A[1] beta[4] + 3 A[0]  B[1] beta[4] + A[1]  beta[4]

           2                           2               3        
   + 3 A[1]  B[1] beta[4] + 3 A[1] B[1]  beta[4] + B[1]  beta[4]

                                                                \ 
   + 2 A[0] A[1] beta[3] + 2 A[0] B[1] beta[3] - w A[1] - w B[1]/ 

          6   /        2            3        
  cosh(xi)  + \-alpha k  A[0] + A[0]  beta[4]

                2                                   
   + 3 A[0] A[1]  beta[4] + 6 A[0] A[1] B[1] beta[4]

                2               2               2        
   + 3 A[0] B[1]  beta[4] + A[0]  beta[3] + A[1]  beta[3]

                               2                 \          
   + 2 A[1] B[1] beta[3] + B[1]  beta[3] - w A[0]/ sinh(xi) 

          5   /         2               2     
  cosh(xi)  + \2 alpha k  A[1] + alpha k  B[1]

                   2                      2     
   - 2 alpha lambda  A[1] + 2 alpha lambda  B[1]

                   2                      2     
   - 2 gamma lambda  A[1] + 2 gamma lambda  B[1]

           2                      2             
   - 6 A[0]  A[1] beta[4] - 3 A[0]  B[1] beta[4]

           3                 2             
   - 3 A[1]  beta[4] - 6 A[1]  B[1] beta[4]

                2                              
   - 3 A[1] B[1]  beta[4] - 4 A[0] A[1] beta[3]

                                            \         4   /      
   - 2 A[0] B[1] beta[3] + 2 w A[1] + w B[1]/ cosh(xi)  + \alpha 

   2            3                      2        
  k  A[0] - A[0]  beta[4] - 6 A[0] A[1]  beta[4]

                                    2                 2        
   - 6 A[0] A[1] B[1] beta[4] - A[0]  beta[3] - 2 A[1]  beta[3]

                                 \                  3   /
   - 2 A[1] B[1] beta[3] + w A[0]/ sinh(xi) cosh(xi)  + \
        2                      2                      2     
-alpha k  A[1] + 4 alpha lambda  A[1] + 4 gamma lambda  A[1]

           2                      3        
   + 3 A[0]  A[1] beta[4] + 3 A[1]  beta[4]

           2                                            \ 
   + 3 A[1]  B[1] beta[4] + 2 A[0] A[1] beta[3] - w A[1]/ 

          2
  cosh(xi) 

     /           2               2        \                  
   + \3 A[0] A[1]  beta[4] + A[1]  beta[3]/ sinh(xi) cosh(xi)

                   2                      2            3           
   - 2 alpha lambda  A[1] - 2 gamma lambda  A[1] - A[1]  beta[4] = 

   \    
  0/ = 0


                       equ[1][1] := 0 = 0

                       equ[1][2] := 0 = 0

                       equ[1][3] := 0 = 0

                       equ[1][4] := 0 = 0

                       equ[1][5] := 0 = 0

                       equ[1][6] := 0 = 0

NULL;
NULL;

Download loop_for_coeficent.mw

finding loop for a function when paramet...

salim-barzani 1555 Maple
syntax loop differential-equations
July 16 2024
1 6

i am looking for all outcome of this function how i can do it



phi := (`\`p_1,p__2`, q__1, q__2, xi) -> p__1*exp(q__1*xi) - p__2*exp(q__2*xi);
for p[1 ] in [0,1,-1,I,-I] do ;
for p[2 ] in [0,1,-1,I,-I] do ;

 for q[1 ] in [0,1,-1,I,-I] do ;
 for q[2 ] in [0,1,-1,I,-I] do ;
result1 := evalf(phi(`\`p_1,p__2`, q__1, q__2, xi));
print('result1);

my code is not correct and not run in fact but it is my try

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