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delvin

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correct...

delvin 50
June 19 2023
0 0

@mmcdara 

 I appreciate your taking the time.

 Yes, I think it is correct, as if I wrote the variables wrong, I want to get the answer to equation 30 with these placements. .I very much appreciate your help

solve equation 30 Or the same equation 9...

delvin 50
June 18 2023
0 0

The first part was answered by dear friend mmcdara 4975.

But I used a simpler method to solve it, but it didn't work. Can anyone help me?

> 

restart
> 

eq_27 := diff(u[n](t), t$2) = alpha*(- exp(-beta*u[n-1](t)) + 2*exp(-beta*u[n](t)) - exp(-beta*u[n+1](t)))

diff(diff(u[n](t), t), t) = alpha*(-exp(-beta*u[n-1](t))+2*exp(-beta*u[n](t))-exp(-beta*u[n+1](t)))

 (1)
> 

eq_28 := n -> exp(-beta*u[n](t)) = 1+v[n](t)/alpha

proc (n) options operator, arrow; exp(-beta*u[n](t)) = 1+v[n](t)/alpha end proc

 (2)
> 

aux_1 := isolate(eq_28(n), u[n](t));

u[n](t) = -ln((v[n](t)+alpha)/alpha)/beta

 (3)
> 

eq_29 := eval(convert(lhs(eq_27), Diff), aux_1)
         =
         simplify(eval(rhs(eq_27), {seq(eq_28(n+k), k=-1..1)}));

Diff(Diff(-ln((v[n](t)+alpha)/alpha)/beta, t), t) = -v[n-1](t)+2*v[n](t)-v[n+1](t)

 (4)
> 

lhs_27 := eval(lhs(eq_27), u[n]=(t -> u[n](xi[n](t)))):
d      := [select(has, indets(%), diff)[]]:
lhs_27 := eval(lhs_27, d =~ eval(d, xi[n](t) = d__1*n + c__1*t + zeta__1)):
lhs_27 := convert(eval(lhs_27, xi[n](t)=xi[n]), diff);

(diff(diff(u[n](xi[n]), xi[n]), xi[n]))*c__1^2

 (5)
> 

aux_2 := select(has, indets(rhs(aux_1), function), ln)[] = H(t);

ln((v[n](t)+alpha)/alpha) = H(t)

 (6)
> 

eq_29_a := simplify(isolate(convert(eval(eq_29, aux_2), diff), diff(H(t), t$2)));

diff(diff(H(t), t), t) = (v[n-1](t)-2*v[n](t)+v[n+1](t))*beta

 (7)
> 

eq_30_1 := lhs_27=rhs(eq_29_a)

(diff(diff(u[n](xi[n]), xi[n]), xi[n]))*c__1^2 = (v[n-1](t)-2*v[n](t)+v[n+1](t))*beta

 (8)
> 

eq_30 := eval(convert(eq_30_1, Diff), u[n](xi[n])=lhs(aux_2))

(Diff(Diff(ln((v[n](t)+alpha)/alpha), xi[n]), xi[n]))*c__1^2 = (v[n-1](t)-2*v[n](t)+v[n+1](t))*beta

 (9)

"v[n](t):=a[0]+sum(-a[i]*(tanh)^(i)(t[n]),i=1..2)+sum(-b[i]*(tanh)^(-i)(t[n]),i=1..2);"

proc (t) options operator, arrow, function_assign; a[0]+sum(-a[i]*(tanh^i)(t[n]), i = 1 .. 2)+sum(-b[i]*(tanh^(-i))(t[n]), i = 1 .. 2) end proc

 (10)

"v[n+1](t):=a[0]-a[1]*(tanh(t[n])+tanh(d))/(1+tanh(t[n])*tanh(d))-a[2]*((tanh(t[n])+tanh(d))/(1+tanh(t[n])*tanh(d)))^(2)-b[1]*(1+tanh(t[n])*tanh(d))/(tanh(t[n])+tanh(d))-b[2]*((1+tanh(t[n])*tanh(d))/(tanh(t[n])+tanh(d)))^(2);"

proc (t) options operator, arrow, function_assign; a[0]-a[1]*(tanh(t[n])+tanh(d))/(1+tanh(t[n])*tanh(d))-a[2]*(tanh(t[n])+tanh(d))^2/(1+tanh(t[n])*tanh(d))^2-b[1]*(1+tanh(t[n])*tanh(d))/(tanh(t[n])+tanh(d))-b[2]*(1+tanh(t[n])*tanh(d))^2/(tanh(t[n])+tanh(d))^2 end proc

 (11)

NULL

"v[n-1](t):=a[0]-a[1]*(tanh(t[n])-tanh(d))/(1-tanh(t[n])*tanh(d))-a[2]*((tanh(t[n])-tanh(d))/(1-tanh(t[n])*tanh(d)))^(2)-b[1]*(1-tanh(t[n])*tanh(d))/(tanh(t[n])-tanh(d))-b[2]*((1-tanh(t[n])*tanh(d))/(tanh(t[n])-tanh(d)))^(2);"

proc (t) options operator, arrow, function_assign; a[0]-a[1]*(tanh(t[n])-tanh(d))/(1-tanh(t[n])*tanh(d))-a[2]*(tanh(t[n])-tanh(d))^2/(1-tanh(t[n])*tanh(d))^2-b[1]*(1-tanh(t[n])*tanh(d))/(tanh(t[n])-tanh(d))-b[2]*(1-tanh(t[n])*tanh(d))^2/(tanh(t[n])-tanh(d))^2 end proc

 (12)

NULL

NULL

NULL

NULL

fin1 := simplify(numer(eq_30))

Error, (in v[`+`(n, `-`(1))]) invalid input: tanh expects its 1st argument, x, to be of type algebraic, but received [diff(diff(xi[n](t),t),t), diff(xi[n](t),t)]

  

NULL

> 

fin := simplify(subs(tanh(xi[n]) = Psi, fin1));

fin1

 (13)
> 

degree(fin,Psi)

0

 (14)
> 

FF:=convert(series(fin,Psi,7),polynom):
> 

degree(%,Psi)

0

 (15)
> 

for i from 0 to degree(FF, Psi) do
    EQ[i] := simplify(coeff(FF, Psi, i));
end do

fin1

 (16)

NULL

Eqs := {seq(EQ[i], i = 0 .. 6)}

Sol := dsolve(Eqs, {a[0], a[1], a[2], b[1], b[2], c[1]})

Error, (in dsolve) not a system with respect to the unknowns {a[0], a[1], a[2], b[1], b[2], c[1]}

  

Download ddd.mw

@mmcdara   Thank you for help...

delvin 50
June 18 2023
0 0

@mmcdara 

 Thank you for helping me.

@Christian Wolinski   ok &nb...

delvin 50
June 18 2023
0 0

@Christian Wolinski 

 ok

 Sorry, I didn't know you

Problem running the program!!!...

delvin 50
June 18 2023
0 0

 I run the attached Maple code, but no how long I wait, I can't run it. Is the code wrong?!
Maple file is word command file.

shro.mw

 

shrodin01.docx

@acer   If I ask the question...

delvin 50
June 18 2023
0 0

@acer 

 If I ask the question again here are those who see?

Because I can't run with a series of changes I make in

That’s very kind....

delvin 50
June 17 2023
0 0

@mmcdara 

 I really appreciate it.

 First of all, thank you for your time. If it is possible to add the file that I attached to the program, you would be very kind to me.

 I desperately need this app just to get some work done. I do not intend to disturb.

sh01.docx 

I'm so thankful...

delvin 50
June 16 2023
0 0

@mmcdara 

Hi,

  I'm so grateful, This was very kind of you.

You can look again at the two codes below to see what is wrong with the fsh code? (Of course if you can)

  fsh.mw

Many thanks...

delvin 50
June 16 2023
0 0

@mmcdara 

 Thanks for your hard work on this,

  • I want to convert equation 9 to equation 12 with a series of changes and then solve it, i.e. equation 12.
  • By substituting equation 4 and using equation 14 (that is, assigning the values of p_i, q_i to 1 or -1 or i and i -) and then inserting the unknown values (i.e. a_0, a_1, b_1) into equations 12 obtained, then by placing them, we get the value of the unknown function.

sh.docx

@mmcdara  Thank you for your t...

delvin 50
June 15 2023
0 0

@mmcdara 

Thank you for your time to will solve this code.

thank you...

delvin 50
June 15 2023
0 0

@mmcdara 

  • This equation (Eq.9) is a differential-difference equations, (u_n are recursive relations).
  • i,It is for: It is a complex equation.
  • DNSE The name of the equation is: the discrete nonlinear Schrödinger equation (DNSE).
  • Equation 9 is converted into an equation with a real part and an complex part, written separately, using deMoivre's formula.
  • We consider that equation 4 is a hypothetical solution for equation 9.
  • And m is a balance number, which here has a value of 1.

Is the explanation better now?

Thankful...

delvin 50
June 10 2023
0 0

@mmcdara 

 

Thankful
I want to solve the equation eq with several substitutions (substitution of the functions u , u _n +1 and u _n -1).
And after placing and simplifying, I will reach a polynomial to get the unknown values and then by placing them in the u equation, I will get the solution of the equation.

I wrote the last two commands to get the coefficients.

Can you guide me in this field.

@mmcdara   Sorry, If the...

delvin 50
May 28 2023
0 0

@mmcdara 

 Sorry, If the number of equations is more, we call such devices overdeterminate and it is possible that they have a unique solution or that they do not have any solution.

@mmcdara   Well, only these 4...

delvin 50
May 28 2023
0 0

@mmcdara 

 Well, only these 4 are unknown? Of course, alpha and beta can also be written?

In other words, did I make a mistake and is it practically impossible?

Thank you very much...

delvin 50
May 25 2023
0 0

@mmcdara 

 Thank you very much
It is now running on version 2022.
One question, how can I get the values of a0, a1, the rest of the unknowns?
I only know the following command
Sol := solve(Eqs, {c, a[0], a[1], b[1]})
  I don't know how to write here!!!

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