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janhardo

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Checked the procedure with the classical...

janhardo 690
June 04 2025
0 0

Checked the procedure with the classical  non -lineair schrodinger eq 
Do you have a idea what are the lineair and non linear parts of your pde

splitsing_pde_conjugate_vorm_4-6-2025mprimes_A.mw

"Ai is trash in 100 time just one time work , when i am stuck i will post here and my work is rare i cant find it any where i have to do step by step,  many thanks for you"

You can't use ai , so this is not a good  attitude , rather simplistic 

@salim-barzani   ....

janhardo 690
June 03 2025
0 0

@salim-barzani 

> 

assume(A < 0, 0 < C); printf("Original ODE before rewriting (with B = -2AC):\n"); ode_orig := (diff(G(xi), xi))^2 = A^2+(2*A*(-2*A*C))*G(xi)+2*A*C*G(xi)^2+(2*(-2*A*C))*C*G(xi)^3+C^2*G(xi)^4; ode_orig; printf("\nRewritten ODE (after completing the square):\n"); ode := diff(G(xi), xi) = A+C*G(xi)^2-2*A*C*G(xi); ode; sol_general := dsolve(ode, G(xi)); printf("\nExplicit solution (after simplification):\n"); explicit_sol := `assuming`([simplify(sol_general)], [A < 0, 0 < C]); explicit_sol; printf("\nVerification using odetest:\n"); odetest_result := odetest(explicit_sol, ode); `assuming`([simplify(odetest_result)], [A < 0, 0 < C]); G1 := proc (xi) options operator, arrow; -((1/2)*sqrt(-2*A*C)+(1/2)*sqrt(-6*A*C)*tanh((1/2)*sqrt(-6*A*C)*xi))/C end proc; G2 := rhs(explicit_sol); G2fun := unapply(G2, xi); printf("\nDifference between both solutions (should be 0):\n"); diffG := `assuming`([simplify(G1(xi)-G2fun(xi))], [A < 0, 0 < C]); diffG

Original ODE before rewriting (with B = -2AC):

  

(diff(G(xi), xi))^2 = A^2-4*A^2*C*G(xi)+2*A*C*G(xi)^2-4*A*C^2*G(xi)^3+C^2*G(xi)^4

  


Rewritten ODE (after completing the square):

  

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

  


Explicit solution (after simplification):

  

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

  


Verification using odetest:

  

0

  


Difference between both solutions (should be 0):

  

(1/2)*(2*tanh(C^(1/2)*(-A*C+1)^(1/2)*(-A)^(1/2)*(c__1+xi))*C^(1/2)*(-A*C+1)^(1/2)*(-A)^(1/2)-2*A*C-2^(1/2)*C^(1/2)*(-A)^(1/2)-2^(1/2)*3^(1/2)*C^(1/2)*(-A)^(1/2)*tanh((1/2)*6^(1/2)*C^(1/2)*(-A)^(1/2)*xi))/C

 (1)

no simplification ..  doing by maple xperts ..
 

another proofmethod that both solutions functions are the same at G1,2(0)

> 

G1 := proc (xi) options operator, arrow; -(1/2)*(sqrt(-2*A*C)+sqrt(-6*A*C)*tanh((1/2)*sqrt(-6*A*C)*xi))/C end proc; G2 := proc (xi) options operator, arrow; (A*C-sqrt(A^2*C^2-A*C)*tanh(sqrt(A^2*C^2-A*C)*xi+_C1))/C end proc; eq := `assuming`([simplify(G1(0) = G2(0))], [A < 0, C > 0]); C1sol := solve(eq, _C1); C1sol

proc (xi) options operator, arrow; -(1/2)*(sqrt(-2*A*C)+sqrt(-6*A*C)*tanh((1/2)*sqrt(-6*A*C)*xi))/C end proc

  

proc (xi) options operator, arrow; (A*C-sqrt(A^2*C^2-A*C)*tanh(sqrt(A^2*C^2-A*C)*xi+_C1))/C end proc

  

-(1/2)*2^(1/2)*(-A)^(1/2)/C^(1/2) = (A*C-C^(1/2)*(-A*C+1)^(1/2)*(-A)^(1/2)*tanh(_C1))/C

  

arctanh((1/2)*(2*A*C^(3/2)+2^(1/2)*(-A)^(1/2)*C)/(C*(-A*C+1)^(1/2)*(-A)^(1/2)))

 (2)
> 

``

Download G25_via_maple_nu_-wel_in_mathematica_mprimes3-6-2025.mw

@salim-barzani Wolfram cloud confir...

janhardo 690
June 03 2025
0 0

@salim-barzani 

try again...


Wolfram cloud confirms that with this constraints this is the solution ? ..now in Maple 

@salim-barzani Mathematica and Mapl...

janhardo 690
June 02 2025
0 0

@salim-barzani Mathematica and Maple can't give a confirmation of G25  for this ode2 that it is a solution ?

@janhardo  "after equatio...

janhardo 690
June 02 2025
0 0

@janhardo 
 "after equation 24 the reds one is not satisfy the equation i have to use equation 4-5 condition but i don't know how they use and satisfy the ode "

what  ode for this ?
its not the ode from ode-test-36.mw 

@salim-barzani   Yo...

janhardo 690
June 01 2025
0 0

@salim-barzani 

Download G25_odetest_1-6-2025_mprimes.mw

@salim-barzani   ....

janhardo 690
June 01 2025
0 0

@salim-barzani 

Download G5_oplosssing_in_Riccati_ode_mprimes1-6-2025.mw

@salim-barzani  Makes no differenc...

janhardo 690
May 31 2025
0 0

@salim-barzani 

Makes no difference this solution for a odetest in the Ricatti ode, as it seems ?





 

@salim-barzani tanh solution e...

janhardo 690
May 30 2025
0 0

@salim-barzani 
tanh solution 

eq5_naar_class1_via_bernouilli_tanh_oplossing_mprimes_30-5-2025.mw

eq5_naar_class1_via_bernouilli_mprimes_3...

janhardo 690
May 30 2025
0 0

eq5_naar_class1_via_bernouilli_mprimes_30-5-2025.mw

Solving the Bernoulli equation:

@salim-barzani focus on equation(6)...

janhardo 690
May 30 2025
0 0

@salim-barzani 


start with equation(6) 
 

  > ....

janhardo 690
May 30 2025
0 0

G25 solution has a different ode then in paper ?

Download G25_heeft_andere_ode_dan_gegeven_mprimes_30-5-2025.mw

"Could there be a mistake in my ass...

janhardo 690
May 29 2025
0 0

"Could there be a mistake in my assumptions, or are some of these solutions only valid under specific conditions?

Also depends on some mathematical function type what the solution is , according to me with the scaling I suspect?

@salim-barzani Download geschaalde_...

janhardo 690
May 29 2025
0 0

@salim-barzani 
All those solutions for Z , have certain conditions to get them to 0 in an odetest.
What are they ?
Download geschaalde_riccatie_ode_mprimes_29-5-2025_A.mw

@nm  ...

janhardo 690
May 21 2025
0 0

@nm 

q0 and q1 can be 0 , but not q2 =0 ( no riccatie ode anymore) 


   is a Riccati form 

The combination of the 3 functions in the Riccation ode makes many ode forms possible

General form : 

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