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How solve this Special ODE ?...

salim-barzani 190
ode differential-equations laplace
November 04 2024
1 6

I upload picture of solution and i try to solve but i fail i don't know how maple can do that just take laplace of one sidehow posible ?


 

> 

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

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

 (2)
> 

eq1 := laplace(eq, x, s)

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

 (3)
> 

NULL
> 

NULL
> 

NULL
> 

NULL
> 

NULL
> 

NULL
> 

NULL
> 

NULL
> 

> 

NULL
> 

restart
> 

with(inttrans)

[addtable, fourier, fouriercos, fouriersin, hankel, hilbert, invfourier, invhilbert, invlaplace, invmellin, laplace, mellin, savetable, setup]

 (4)
> 

eq := diff(y(x, t), x)+y(x, t)^2 = 1

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

 (5)
> 

eq1 := laplace(eq, x, s)

s*laplace(y(x, t), x, s)-y(0, t)+laplace(y(x, t)^2, x, s) = 1/s

 (6)
> 

eq2 := subs(y(0, t) = 3, eq1)

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

 (7)
> 

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

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

 (8)
> 

eq3 := lap/s

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

 (9)
> 

expand(%)

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

 (10)
> 

Geq := y(x, t) = invlaplace(1/s^2+3/s, s, x)-invlaplace(laplace(y(x, t)^2, x, s)/s, x, s)

y(x, t) = x+3-invlaplace(laplace(y(x, t)^2, x, s), x, s)/s

 (11)
> 

NULL
> 

k := K

K

 (12)
> 

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

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

 (13)
> 

for j from 0 to 4 do A[j] := subs(lambda = 0, (diff(f(sum(lambda^i*y[i](x), i = 0 .. 20)), [`$`(lambda, j)]))/factorial(j)) end do

y[0](x)^2

  

2*y[0](x)*y[1](x)

  

y[1](x)^2+2*y[0](x)*y[2](x)

  

2*y[1](x)*y[2](x)+2*y[0](x)*y[3](x)

  

y[2](x)^2+2*y[1](x)*y[3](x)+2*y[0](x)*y[4](x)

 (14)
> 

" y[0](x):=x+3"

proc (x) options operator, arrow, function_assign; 3+x end proc

 (15)
> 

lapy[1] := -laplace(A[0], x, s)/s

-(9*s^2+6*s+2)/s^4

 (16)
> 

simplify(%)

(-9*s^2-6*s-2)/s^4

 (17)
> 

invlaplace(%, s, x)

-(1/3)*x*(x^2+9*x+27)

 (18)
> 

simplify(-(1/3)*x*(x^2+9*x+27))

-(1/3)*x*(x^2+9*x+27)

 (19)
> 

normal(-(1/3)*x*(x^2+9*x+27), ':-expanded')

-(1/3)*x^3-3*x^2-9*x

 (20)
> 

"y[1](x):=-1/3 x^3-3 x^2-9 x"

proc (x) options operator, arrow, function_assign; -(1/3)*x^3-3*x^2-9*x end proc

 (21)
> 

lapy[2] := -laplace(A[1], x, s)/s

-(-72/s^3-48/s^4-16/s^5-54/s^2)/s

 (22)
> 

simplify(-(-72/s^3-48/s^4-16/s^5-54/s^2)/s)

(54*s^3+72*s^2+48*s+16)/s^6

 (23)
> 

expand(%)

54/s^3+72/s^4+48/s^5+16/s^6

 (24)
> 

invlaplace(%, s, x)

(1/15)*x^2*(2*x^3+30*x^2+180*x+405)

 (25)
> 

expand(%)

(2/15)*x^5+2*x^4+12*x^3+27*x^2

 (26)
> 

NULL
> 

NULL


 

Download laplace_of_special_ode.mw

example 1 i could't solve but example to i did

Moderator..................................

salim-barzani 190 MaplePrimes
deleted
November 04 2024
0 8

i am not jobles to post in here and you delete them one by one, when i post i am waiting for response not you delete them, with this cuase my question are repeat or something like that 
this time delete my post i will upload 1000 of nonsense question and post  and go delete all of them ...

How collect series solution of PDE?...

salim-barzani 190 Maple 2024
pde differential-equations laplace
November 02 2024
1 30

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

Download explananing_of_get_solution.mw

Confusing of methods for odetest?...

salim-barzani 190 Maple 2024
ode series differential-equations
October 28 2024
1 9

Almost i did 10 method for this ode equation all of them are succes but this one is giving me some confusing and i am looking for  get my answer, the mothod say if we have the auxilary equation if substitute the solution of this auxilary equation in our series solution then substitute in ode equation must be satisfy but it is not satisfy so when he did assumption for the auxilary equation he say it satisfy if we sabstitute this assumption in our series solution!

My question is this how we get thus assumption ? and why finding exact  solution of auxilary equation not satisfy?

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

with(Physics)
> 

with(SolveTools)
> 

undeclare(prime)

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

 (1)
> 

NULL
> 

Fode := (-delta*eta^2+alpha*eta)*(diff(diff(U(xi), xi), xi))-U(xi)*(2*eta*gamma*theta*(delta*eta-alpha)*U(xi)^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(-delta*eta^2+alpha*eta)*(diff(diff(U(xi), xi), xi))-U(xi)*(2*gamma*eta*theta*(delta*eta-alpha)*U(xi)^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

 (2)
> 

NULL
> 

F := sum(a[i]*G(xi)^i, i = 0 .. 1)

a[0]+a[1]*G(xi)

 (3)

``

 (4)
> 

D1 := diff(F, xi)

a[1]*(diff(G(xi), xi))

 (5)
> 

NULL
> 

S := (diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

 (6)
> 

S1 := diff(G(xi), xi) = sqrt(G(xi)^4+A[2]*G(xi)^2+A[1])

diff(G(xi), xi) = (G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (7)
> 

E1 := subs(S1, D1)

a[1]*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (8)
> 

D2 := diff(E1, xi)

(1/2)*a[1]*(4*G(xi)^3*(diff(G(xi), xi))+2*A[2]*G(xi)*(diff(G(xi), xi)))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (9)
> 

E2 := subs(S1, D2)

(1/2)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (10)
> 

K := U(xi) = F

U(xi) = a[0]+a[1]*G(xi)

 (11)
> 

K1 := diff(U(xi), xi) = E1

diff(U(xi), xi) = a[1]*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (12)
> 

K2 := diff(U(xi), xi, xi) = E2

diff(diff(U(xi), xi), xi) = (1/2)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (13)
> 

L := eval(Fode, {K, K1, K2})

(1/2)*(-delta*eta^2+alpha*eta)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)-(a[0]+a[1]*G(xi))*(2*gamma*eta*theta*(delta*eta-alpha)*(a[0]+a[1]*G(xi))^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

 (14)
> 

L1 := normal((1/2)*(-delta*eta^2+alpha*eta)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)-(a[0]+a[1]*G(xi))*(2*gamma*eta*theta*(delta*eta-alpha)*(a[0]+a[1]*G(xi))^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0)
> 

 

> 

collect(L1, {G(xi)})

(-2*delta*eta^2*gamma*theta*a[1]^3+2*alpha*eta*gamma*theta*a[1]^3-2*delta*eta^2*a[1]+2*alpha*eta*a[1])*G(xi)^3+(-6*delta*eta^2*gamma*theta*a[0]*a[1]^2+6*alpha*eta*gamma*theta*a[0]*a[1]^2)*G(xi)^2+(-6*delta*eta^2*gamma*theta*a[0]^2*a[1]+6*alpha*eta*gamma*theta*a[0]^2*a[1]-delta*eta^2*k^2*a[1]+alpha*eta*k^2*a[1]-delta*eta^2*A[2]*a[1]+alpha*eta*A[2]*a[1]+2*delta*eta*k*a[1]-2*alpha*k*a[1]-delta*a[1])*G(xi)-2*gamma*delta*eta^2*theta*a[0]^3+2*gamma*alpha*eta*theta*a[0]^3-delta*eta^2*k^2*a[0]+alpha*eta*k^2*a[0]+2*delta*eta*k*a[0]-2*alpha*k*a[0]-delta*a[0] = 0

 (15)
> 

eq0 := -2*delta*eta^2*gamma*theta*a[0]^3+2*alpha*eta*gamma*theta*a[0]^3-delta*eta^2*k^2*a[0]+alpha*eta*k^2*a[0]+2*delta*eta*k*a[0]-2*alpha*k*a[0]-delta*a[0] = 0
> 

eq1 := -6*delta*eta^2*gamma*theta*a[0]^2*a[1]+6*alpha*eta*gamma*theta*a[0]^2*a[1]-delta*eta^2*k^2*a[1]+alpha*eta*k^2*a[1]-delta*eta^2*A[2]*a[1]+alpha*eta*A[2]*a[1]+2*delta*eta*k*a[1]-2*alpha*k*a[1]-delta*a[1] = 0
> 

eq2 := -6*delta*eta^2*gamma*theta*a[0]*a[1]^2+6*alpha*eta*gamma*theta*a[0]*a[1]^2 = 0
> 

eq3 := -2*delta*eta^2*gamma*theta*a[1]^3+2*alpha*eta*gamma*theta*a[1]^3-2*delta*eta^2*a[1]+2*alpha*eta*a[1] = 0
> 

COEFFS := solve({eq0, eq1, eq2, eq3}, {alpha, eta, a[0], a[1]}, explicit)
> 

case1 := COEFFS[4]

{alpha = delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k), eta = eta, a[0] = 0, a[1] = 1/(-gamma*theta)^(1/2)}

 (16)
> 

NULL
> 

S

(diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

 (17)
> 

S1

diff(G(xi), xi) = (G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

 (18)
> 

S2 := dsolve(S, G(xi))

G(xi) = -(1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2), G(xi) = (1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2), G(xi) = -(1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2), G(xi) = (1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2), G(xi) = JacobiSN((1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2)*xi+c__1, (-2*(A[2]*(A[2]^2-4*A[1])^(1/2)-A[2]^2+2*A[1])*A[1])^(1/2)/(A[2]*(A[2]^2-4*A[1])^(1/2)-A[2]^2+2*A[1]))*A[1]*2^(1/2)/(A[1]*(-A[2]+(A[2]^2-4*A[1])^(1/2)))^(1/2)

 (19)
> 

K

U(xi) = a[0]+a[1]*G(xi)

 (20)
> 

K4 := subs(case1, K)

U(xi) = G(xi)/(-gamma*theta)^(1/2)

 (21)
> 

NULL
> 

K5 := subs(S2, K4)

U(xi) = -(1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2)/(-gamma*theta)^(1/2)

 (22)
> 

NULL
> 

F2 := subs(case1, Fode)

(-eta^2*delta+delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)*eta/(eta*k^2+eta*A[2]-2*k))*(diff(diff(U(xi), xi), xi))-U(xi)*(2*gamma*eta*theta*(delta*eta-delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k))*U(xi)^2+eta^2*delta*k^2+(-k^2*delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k)-2*k*delta)*eta+2*k*delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k)+delta) = 0

 (23)

``

 (24)
> 

NULL
> 

odetest(K5, F2)

-(1/2)*delta*eta*(A[2]^2-4*A[1])^(1/2)*(-2*(A[2]+(A[2]^2-4*A[1])^(1/2))/gamma)^(1/2)/((eta*k^2+eta*A[2]-2*k)*(-theta)^(1/2))

 (25)


and i hope mapleprimes don't delete this question becuase of this pictures also it help for undrestanding

 

there is other picture for different auxilary equation just  add one multiply term for G(xi)^4 in case anyone needed i will upload

Download odetest.mw

Why not apear all term when i export fil...

salim-barzani 190 Maple 2024
matrix vector pdf
October 27 2024
1 2

how fixed that?

> 

NULL
> 

restart
> 

NULL
> 

A := Vector[row]([4, 0, 5])

Vector[row](%id = 36893490313998292860)

 (1)
> 

B := Vector[row]([3, 1, 0])

Vector[row](%id = 36893490313998279372)

 (2)
> 

C := Vector[row]([2, 1, 1])

Vector[row](%id = 36893490313998273820)

 (3)
> 

C.`&x`(A, B)

9

 (4)
> 

NULL

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