salim-barzani

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1 years, 15 days

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These are questions asked by salim-barzani

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

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

Download we.mw

In many papers, I've noticed that the solution to an ODE (ordinary differential equation) often emerges directly when there's only a single function involved. My question is: is there a way to generate solutions to an ODE by producing specific parameters?

 

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

S := diff(F(xi), xi) = sqrt(P*F(xi)^4+Q*F(xi)^2+R)

diff(F(xi), xi) = (P*F(xi)^4+Q*F(xi)^2+R)^(1/2)

(2)

S1 := dsolve(S, F(xi))

xi-Intat(1/(P*_a^4+Q*_a^2+R)^(1/2), _a = F(xi))+c__1 = 0

(3)

S2 := (diff(F(xi), xi))^2 = P*F(xi)^4+Q*F(xi)^2+R

(diff(F(xi), xi))^2 = P*F(xi)^4+Q*F(xi)^2+R

(4)

S3 := dsolve(S2, F(xi))

F(xi) = -(1/2)*(-2*P*(Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = (1/2)*(-2*P*(Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = -(1/2)*2^(1/2)*(P*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = (1/2)*2^(1/2)*(P*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = JacobiSN((1/2)*(2*(-4*P*R+Q^2)^(1/2)-2*Q)^(1/2)*xi+c__1, (-2*(Q*(-4*P*R+Q^2)^(1/2)+2*R*P-Q^2)*R*P)^(1/2)/(Q*(-4*P*R+Q^2)^(1/2)+2*R*P-Q^2))*R*2^(1/2)/(R*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)

(5)
 

NULL

Download get_all_solution_of_ode_by_generation.mw

i can do one by one for all case but i am intrested for this idea how we can do that for each equation automatically calculate all case without use one by one case ?
there is any other shorter way for get solution of this mw

all_case.mw

restart

with(PDEtools)

undeclare(prime)

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

(1)

with(DEtools)

S := diff(G(xi), xi) = G(xi)^2+A*G(xi)+B

diff(G(xi), xi) = G(xi)^2+A*G(xi)+B

(2)

S1 := dsolve(S, G(xi))

G(xi) = -(1/2)*A-(1/2)*tanh((1/2)*(A^2-4*B)^(1/2)*(c__1+xi))*(A^2-4*B)^(1/2)

(3)

A := 0

0

(4)

S

diff(G(xi), xi) = G(xi)^2+B

(5)

dsolve(S, G(xi))

G(xi) = tan(B^(1/2)*(c__1+xi))*B^(1/2)

(6)

restart

NULL

with(PDEtools)

undeclare(prime)

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

(7)

with(DEtools)

S := diff(G(xi), xi) = G(xi)^2+A*G(xi)+B

diff(G(xi), xi) = G(xi)^2+A*G(xi)+B

(8)

B := 0

0

(9)

S1 := dsolve(S, G(xi))

G(xi) = A/(-1+exp(-A*xi)*c__1*A)

(10)

NULL

restart

NULL

with(PDEtools)

undeclare(prime)

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

(11)

with(DEtools)

S := diff(G(xi), xi) = G(xi)^2+A*G(xi)+B

diff(G(xi), xi) = G(xi)^2+A*G(xi)+B

(12)

B := 0; A := 0

0

 

0

(13)

S1 := dsolve(S, G(xi))

G(xi) = 1/(-xi+c__1)

(14)
 

NULL

Download find_all_case_solution_of_ode_.mw

i did all the time like that and don't have any issue but i don't know why not take derivative by x and t have a problem when i remove t is ok but when i take duble derivative by x and t not run and did't give me error too what is problem with this?

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)

tr := {t = tau, x = xi+delta*(tau+1), u(x, t) = U(xi)*exp(I*(-kx+w*(t+1)))}

{t = tau, x = xi+delta*(tau+1), u(x, t) = U(xi)*exp(I*(-kx+w*(t+1)))}

(2)

pde := I*(diff(u(x, t), t))+alpha*(diff(u(x, t), `$`(x, 2)))+beta*(diff(u(x, t), x, t))+gamma*u(x, t)*V(xi) = 0

 

NULL

Download non_sense.mw

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