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salim-barzani

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1 years, 74 days
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These are questions asked by salim-barzani

problem in replacing mu in equation ?...

salim-barzani 1640 Maple 2024
2dmath subscript indexed
November 24 2024
1 3

when i replace mu=0 in my equation mu[3] and mu[4] also become zero why ?

> 

restart
> 

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)
> 

NULL
> 

B[1] := 0; mu := 0

0

  

0

 (2)
> 

U1 := lambda*(1+sqrt(-lambda)*(cosh(xi*sqrt(-lambda))*B[2]+sinh(xi*sqrt(-lambda))*B[1])*`&+-`(sqrt(lambda/(lambda^2*B[1]^2-lambda^2*B[2]^2-mu^2))))*beta[0]*exp(I*(d*x+e*y+f*t^beta/beta))/(B[1]*cosh(xi*sqrt(-lambda))*lambda+B[2]*sinh(xi*sqrt(-lambda))*lambda+mu)

(1+(-lambda)^(1/2)*cosh(xi*(-lambda)^(1/2))*B[2]*`&+-`((-1/(lambda*B[2]^2))^(1/2)))*beta[0]*exp(I*(d*x+e*y+f*t^beta/beta))/(B[2]*sinh(xi*(-lambda)^(1/2)))

 (3)
> 

NULL
> 

V1 := -mu[4]*b*lambda^2*(1+sqrt(-lambda)*(cosh(xi*sqrt(-lambda))*B[2]+sinh(xi*sqrt(-lambda))*B[1])*`&+-`(sqrt(lambda/(lambda^2*B[1]^2-lambda^2*B[2]^2-mu^2))))^2*beta[0]^2/((B[1]*cosh(xi*sqrt(-lambda))*lambda+B[2]*sinh(xi*sqrt(-lambda))*lambda+mu)^2*a*mu[3])

Error, numeric exception: division by zero

  

Download tester.mw

Find ODE solution with condition paramet...

salim-barzani 1640 Maple 2024
ode assumptions homework
November 23 2024
0 11

How we can calculate solution of this ODE by give assumption to the equation , we have otehr case too, like lambda>0&mu<0

How collect coefficient of term of ODE ...

salim-barzani 1640 Maple 2024
differential-equations coefficients power
November 08 2024
1 9

I want collect 1/G(xi) & G'(xi)/G(xi) also 1/G(xi)*G(xi)/G(xi)  when they have power ,and give me what i looking for, i can do by hand but it take time any one can do this by maple code? like this picture below, and if possible find some arbitrary parameter

collect.mw

How make a table for numerical results b...

salim-barzani 1640 Maple
table numeric exact
November 05 2024
1 1

If we calculating it take to much time but if we make a procedure it will be more effectable for such example, i want the exact and approximat and error

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(inttrans)
> 

with(SolveTools)
> 

undeclare(prime)

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

 (1)
> 

NULL
> 

eq := diff(y(x, t), `$`(t, 2))+(1+x)*(diff(y(x, t), x))-y(x, t) = 2*y(x, t)^3

diff(diff(y(x, t), t), t)+(1+x)*(diff(y(x, t), x))-y(x, t) = 2*y(x, t)^3

 (2)
> 

eq1 := laplace(eq, t, s)

s^2*laplace(y(x, t), t, s)-s*y(x, 0)+laplace(diff(y(x, t), x), t, s)*x-laplace(y(x, t), t, s)-(D[2](y))(x, 0)+laplace(diff(y(x, t), x), t, s) = 2*laplace(y(x, t)^3, t, s)

 (3)
> 

eq2 := subs({y(x, 0) = 1, (D[2](y))(x, 0) = 1}, eq1)

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

 (4)
> 

eq3 := s^2*laplace(y(x, t), t, s) = s-laplace(diff(y(x, t), x), t, s)*x+1+laplace(diff(y(x, t), x), t, s)+2*laplace(y(x, t)^3, t, s)+laplace(y(x, t), t, s)

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

 (5)
> 

eq4 := expand(eq3/s^2)

laplace(y(x, t), t, s) = 1/s-laplace(diff(y(x, t), x), t, s)*x/s^2+1/s^2+laplace(diff(y(x, t), x), t, s)/s^2+2*laplace(y(x, t)^3, t, s)/s^2+laplace(y(x, t), t, s)/s^2

 (6)
> 

NULL
> 

"u[0](x):=invlaplace(1/s+1/(s^2),s,x)"

proc (x) options operator, arrow, function_assign; invlaplace(1/s+1/s^2, s, x) end proc

 (7)
> 

u[0](x)

1+x

 (8)
> 

n := N

N

 (9)
> 

k := K

K

 (10)
> 

f := proc (u) options operator, arrow; u^3 end proc

proc (u) options operator, arrow; u^3 end proc

 (11)
> 

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

(1+x)^3

  

3*(1+x)^2*u[1](x)

  

3*(1+x)*u[1](x)^2+3*(1+x)^2*u[2](x)

  

u[1](x)^3+6*(1+x)*u[1](x)*u[2](x)+3*(1+x)^2*u[3](x)

 (12)
> 

A[0]

(1+x)^3

 (13)
> 

y[0] := 1+x

1+x

 (14)
> 

y[1] := invlaplace(2*laplace(A[0], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[0], x), x, s)/s^2, s, x)+invlaplace(laplace(y[0], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[0], x), x, s)/s^2, s, x)

(1/10)*x^2*(x^3+5*x^2+10*x+10)-(1/2)*x^3+(1/6)*x^2*(x+3)-(1/2)*x^2

 (15)
> 

y[1] := expand((1/10)*x^2*(x^3+5*x^2+10*x+10)-(1/2)*x^3+(1/6)*x^2*(x+3)-(1/2)*x^2)

(1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2

 (16)
> 

"u[1](x) :=y[1] "

proc (x) options operator, arrow, function_assign; y[1] end proc

 (17)
> 

NULL
> 

A[1]

3*(1+x)^2*((1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2)

 (18)
> 

y[2] := invlaplace(2*laplace(A[1], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[1], x), x, s)/s^2, s, x)+invlaplace(laplace(y[1], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[1], x), x, s)/s^2, s, x)

(1/840)*x^4*(7*x^5+63*x^4+212*x^3+476*x^2+672*x+420)-(1/60)*x^4*(x^3+6*x^2+10*x+20)+(1/420)*x^4*(x^3+7*x^2+14*x+35)-(1/60)*x^3*(x^3+6*x^2+10*x+20)

 (19)
> 

y[2] := expand(%)

(1/120)*x^9+(3/40)*x^8+(5/21)*x^7+(7/15)*x^6+(17/30)*x^5+(1/12)*x^4-(1/3)*x^3

 (20)
> 

" u[2](x):=y[2]"

proc (x) options operator, arrow, function_assign; y[2] end proc

 (21)
> 

A[2]

3*(1+x)*((1/10)*x^5+(1/2)*x^4+(2/3)*x^3+x^2)^2+3*(1+x)^2*((1/120)*x^9+(3/40)*x^8+(5/21)*x^7+(7/15)*x^6+(17/30)*x^5+(1/12)*x^4-(1/3)*x^3)

 (22)
> 

y[3] := invlaplace(2*laplace(A[2], x, s)/s^2, s, x)-invlaplace(x*laplace(diff(y[2], x), x, s)/s^2, s, x)+invlaplace(laplace(y[2], x, s)/s^2, s, x)-invlaplace(laplace(diff(y[2], x), x, s)/s^2, s, x)

(1/10810800)*x^5*(7623*x^8+99099*x^7+518778*x^6+1634490*x^5+3647930*x^4+5167305*x^3+4221360*x^2+900900*x-1081080)-(1/25200)*x^5*(21*x^6+210*x^5+750*x^4+1680*x^3+2380*x^2+420*x-2100)+(1/831600)*x^5*(63*x^6+693*x^5+2750*x^4+6930*x^3+11220*x^2+2310*x-13860)-(1/25200)*x^4*(21*x^6+210*x^5+750*x^4+1680*x^3+2380*x^2+420*x-2100)

 (23)
> 

y[3] := expand(y[3])

(286/945)*x^9+(131/336)*x^8+(1/7)*x^10+(17/70)*x^7-(1/40)*x^6-(1/20)*x^5+(1/12)*x^4+(1091/23100)*x^11+(11/15600)*x^13+(11/1200)*x^12

 (24)
> 

NULL
> 

addingterm := y[0]+y[1]+y[2]+y[3]

1+x+(37/60)*x^5+(2/3)*x^4+(1/3)*x^3+x^2+(2351/7560)*x^9+(781/1680)*x^8+(101/210)*x^7+(53/120)*x^6+(1/7)*x^10+(1091/23100)*x^11+(11/15600)*x^13+(11/1200)*x^12

 (25)


 

Download aproximate_and_exact_solution.mw

a table like that

 

How calculate Adomian polynomial by diff...

salim-barzani 1640 Maple 2024
ode pde differential-equations
November 04 2024
0 21

there is four formula for calculate them which i know them by name of author the first one is adomian second one is (BiazarShafiofAdomian) which one member of mableprimes write code for me,but i don't know how use for all kind function maybe in future i upload this program for fix this issue, the third one is by zhao which is i think i easy for calculate just  i need someone one to wite the program and do some test for some example i  upload some picture in case for getting algorithm to writting and have some example for testing  so  lets see who can do this algorithm is very usfule when we solve ODE or PDE by LDM, also last method is by taking integral have a good method, in this question this algorithm is zhao which is usfull one

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