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Question: How collect series solution of PDE?

Question:How collect series solution of PDE?

salim-barzani 170 Maple 2024
pde differential-equations laplace adomian-decomposition duplicate-question
November 02 2024
1

this example is easiest one for getting solution but i can't collect each part and do like elite i can do each part seperatly but it take to much time i want collect solution and do by easier way if possible this is a laplace adomian decomposition methd which contain adomian polynomial too i want upgrade the code, can any one help the  process for get better vision of this topic 
i do upload some picture and my mw. for more undrestanding

and please can any one explan why when i take laplace why is write D[2](u)(x, 0) must be D[1]?

> 

restart
> 

with(inttrans)
> 

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)
> 

with(PDEtools)
> 

declare()

`Declared :`

  

u(x, t)*`to be displayed as`*u

  

`The prime differentiation variable has not been declared yet`

  

`Displayed derivatives and declared functions will be copied and pasted "as they were entered"`

 (2)
> 

declare(u(x, t))

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

 (3)
> 

eq := diff(u(x, t), `$`(t, 2))+u(x, t)^2-(diff(u(x, t), x))^2 = 0

diff(diff(u(x, t), t), t)+u(x, t)^2-(diff(u(x, t), x))^2 = 0

 (4)
> 

NULL
> 

eqs := laplace(eq, t, s)

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

 (5)
> 

solve({eqs}, {laplace(u(x, t), t, s)})

{laplace(u(x, t), t, s) = (s*u(x, 0)+(D[2](u))(x, 0)-laplace(u(x, t)^2, t, s)+laplace((diff(u(x, t), x))^2, t, s))/s^2}

 (6)
> 

subs({u(x, 0) = 0, (D[2](u))(x, 0) = exp(x)}, %)

{laplace(u(x, t), t, s) = (exp(x)-laplace(u(x, t)^2, t, s)+laplace((diff(u(x, t), x))^2, t, s))/s^2}

 (7)
> 

eq3 := invlaplace(%, s, t)

{u(x, t) = exp(x)*t-(int(u(x, _U1)^2*(t-_U1), _U1 = 0 .. t))+int((diff(u(x, _U1), x))^2*(t-_U1), _U1 = 0 .. t)}

 (8)
> 

NULL
> 

NULL
> 

NULL
> 

"u[0](x,t):=exp(x)*t"

proc (x, t) options operator, arrow, function_assign; exp(x)*t end proc

 (9)
> 

n := N

N

 (10)
> 

k := K

K

 (11)
> 

f := proc (u) options operator, arrow; u^2 end proc

proc (u) options operator, arrow; u^2 end proc

 (12)
> 

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

(exp(x))^2*t^2

  

2*exp(x)*t*u[1](x, t)

  

u[1](x, t)^2+2*exp(x)*t*u[2](x, t)

  

2*u[1](x, t)*u[2](x, t)+2*exp(x)*t*u[3](x, t)

 (13)
> 

NULL
> 

NULL
> 

n := N

N

 (14)
> 

k := K

K

 (15)
> 

f := proc (u) options operator, arrow; (diff(u, x))^2 end proc

proc (u) options operator, arrow; (diff(u, x))^2 end proc

 (16)
> 

for j from 0 to 3 do B[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

(exp(x))^2*t^2

  

2*exp(x)*t*(diff(u[1](x, t), x))

  

(diff(u[1](x, t), x))^2+2*exp(x)*t*(diff(u[2](x, t), x))

  

2*(diff(u[1](x, t), x))*(diff(u[2](x, t), x))+2*exp(x)*t*(diff(u[3](x, t), x))

 (17)
> 

"#` know we need find all term of u[0]=exp(x)*t` #` u`[1]=-invlaplace(1/(s^(2))(`A__0`+`B__0`)) u[2]=-invlaplace(1/(s^(2))(`A__1`+`B__1`)) ans so on ... at end i want collect all of them and find final result even if is aproximate and want do test of pde too "
> 

NULL

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