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delvin

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@C_R   Ok, thanks....

delvin 55
June 22 2023
0 0

@C_R 

 Ok, thanks.

@mmcdara  Sorry, It can be said t...

delvin 55
June 21 2023
0 0

@mmcdara 

Sorry,

It can be said that there is a disadvantage of software 1 because I previously executed 2 and 2 of these commands without error, but for other equations.

error...

delvin 55
June 21 2023
0 0

@delvin 

My dear friend @mmcdara, they worked hard.

But I have a problem with the loop again, I don't know why it gives an error

> 

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)))):
der    := [select(has, indets(%), diff)[]]:
lhs_27 := eval(lhs_27, der =~ eval(der, 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)
> 

eq_30_a := eval(eq_30, {seq(v[n+k](t)=V[n+k](xi[n]), k=-1..1)})

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

 (10)

 

eq_30_a is consistent for all the V are functions of xi[n] and you take derivatives wrt xi[n].
eq_30_a is the true equation (30) in the paper you refer to.

But eq_30 is not consistent (funstions of t and derivative wrt xi[n]).

 

 

> 

printf("\n\nWhat you do\n\n"):

Replacements_v := {
  v[n](t) = a[0]+sum(-a[i]*(tanh(t[n]))^i, i = 1 .. 2)+sum(-b[i]*(tanh(t[n]))^(-i), i = 1 .. 2)
  , 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))^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
  , 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
}:
print~(Replacements_v):


# Is it not this that you should do?

printf("\n\nIs it not this that you should do to account for consistency?\n\n"):

Replacements_V := eval(Replacements_v, {t[n]=xi[n], seq(v[n+k](t)=V[n+k](xi[n]), k=-1..1)}):
print~(Replacements_V):


# Or more simply

printf("\n\nOr more simply\n\n"):

Replacements_Vth := {
  V[n](xi[n]) = a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2
  , V[n-1](xi[n]) = a[0]-a[1]*tanh(xi[n]-d)-a[2]*tanh(xi[n]-d)^2-b[1]/tanh(xi[n]-d)-b[2]/tanh(xi[n]-d)^2
  , V[n+1](xi[n]) = a[0]-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2
}:
print~(Replacements_Vth):



What you do
 

  

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

  

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))^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

  

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))^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

  



Is it not this that you should do to account for consistency?
 

  

V[n](xi[n]) = a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2

  

V[n-1](xi[n]) = a[0]-a[1]*(tanh(xi[n])-tanh(d))/(1-tanh(xi[n])*tanh(d))-a[2]*(tanh(xi[n])-tanh(d))^2/(1-tanh(xi[n])*tanh(d))^2-b[1]*(1-tanh(xi[n])*tanh(d))/(tanh(xi[n])-tanh(d))-b[2]*(1-tanh(xi[n])*tanh(d))^2/(tanh(xi[n])-tanh(d))^2

  

V[n+1](xi[n]) = a[0]-a[1]*(tanh(xi[n])+tanh(d))/(1+tanh(xi[n])*tanh(d))-a[2]*(tanh(xi[n])+tanh(d))^2/(1+tanh(xi[n])*tanh(d))^2-b[1]*(1+tanh(xi[n])*tanh(d))/(tanh(xi[n])+tanh(d))-b[2]*(1+tanh(xi[n])*tanh(d))^2/(tanh(xi[n])+tanh(d))^2

  



Or more simply
 

  

V[n](xi[n]) = a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2

  

V[n-1](xi[n]) = a[0]+a[1]*tanh(-xi[n]+d)-a[2]*tanh(-xi[n]+d)^2+b[1]/tanh(-xi[n]+d)-b[2]/tanh(-xi[n]+d)^2

  

V[n+1](xi[n]) = a[0]-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2

 (11)
> 

fin0 := eval(eq_30_a, Replacements_Vth)

(Diff(Diff(ln((a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2+alpha)/alpha), xi[n]), xi[n]))*c__1^2 = (a[1]*tanh(-xi[n]+d)-a[2]*tanh(-xi[n]+d)^2+b[1]/tanh(-xi[n]+d)-b[2]/tanh(-xi[n]+d)^2+2*a[1]*tanh(xi[n])+2*a[2]*tanh(xi[n])^2+2*b[1]/tanh(xi[n])+2*b[2]/tanh(xi[n])^2-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2)*beta

 (12)
> 

# To force the derivation uncomment the next line

fin0 := convert(fin0, diff)

((2*a[1]*tanh(xi[n])*(1-tanh(xi[n])^2)-2*a[2]*(1-tanh(xi[n])^2)^2+4*a[2]*tanh(xi[n])^2*(1-tanh(xi[n])^2)-2*b[1]*(1-tanh(xi[n])^2)^2/tanh(xi[n])^3-2*b[1]*(1-tanh(xi[n])^2)/tanh(xi[n])-6*b[2]*(1-tanh(xi[n])^2)^2/tanh(xi[n])^4-4*b[2]*(1-tanh(xi[n])^2)/tanh(xi[n])^2)/(a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2+alpha)-(-a[1]*(1-tanh(xi[n])^2)-2*a[2]*tanh(xi[n])*(1-tanh(xi[n])^2)+b[1]*(1-tanh(xi[n])^2)/tanh(xi[n])^2+2*b[2]*(1-tanh(xi[n])^2)/tanh(xi[n])^3)^2/(a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2+alpha)^2)*c__1^2 = (a[1]*tanh(-xi[n]+d)-a[2]*tanh(-xi[n]+d)^2+b[1]/tanh(-xi[n]+d)-b[2]/tanh(-xi[n]+d)^2+2*a[1]*tanh(xi[n])+2*a[2]*tanh(xi[n])^2+2*b[1]/tanh(xi[n])+2*b[2]/tanh(xi[n])^2-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2)*beta

 (13)
> 

# look at this to understand why you get an error.

p/q=P/Q;
numer(%);

p/q = P/Q

  

Error, invalid input: numer expects its 1st argument, x, to be of type {algebraic, list, set}, but received p/q = P/Q

  
> 

# or to this

p=P/Q;
numer(%);

p = P/Q

  

Error, invalid input: numer expects its 1st argument, x, to be of type {algebraic, list, set}, but received p = P/Q

  
> 

# if you want to get the numerators of both sides do this

p/q=P/Q;
map(numer, %)

p/q = P/Q

  

p = P

 (14)
> 

# APPLICATION:

fin1 := map(numer, fin0)

(tanh(xi[n])^2-1)*tanh(xi[n])^4*(-10*tanh(xi[n])^3*a[1]*b[2]-2*tanh(xi[n])^3*b[1]*b[2]+6*tanh(xi[n])^2*alpha*b[2]+6*tanh(xi[n])^2*a[0]*b[2]-4*tanh(xi[n])*b[1]*b[2]-2*tanh(xi[n])^7*a[0]*a[1]+2*tanh(xi[n])^7*a[1]*a[2]+10*tanh(xi[n])^7*a[2]*b[1]+2*tanh(xi[n])^6*alpha*a[2]+2*tanh(xi[n])^6*a[0]*a[2]+4*tanh(xi[n])^6*a[1]*b[1]+16*tanh(xi[n])^6*a[2]*b[2]+8*tanh(xi[n])^5*a[1]*b[2]-8*tanh(xi[n])^5*a[2]*b[1]-2*tanh(xi[n])^4*alpha*b[2]-2*tanh(xi[n])^4*a[0]*b[2]-4*tanh(xi[n])^4*a[1]*b[1]-16*tanh(xi[n])^4*a[2]*b[2]+2*tanh(xi[n])^3*alpha*b[1]+2*tanh(xi[n])^3*a[0]*b[1]+4*tanh(xi[n])^9*a[1]*a[2]-6*tanh(xi[n])^8*alpha*a[2]-6*tanh(xi[n])^8*a[0]*a[2]-2*tanh(xi[n])^7*alpha*a[1]+2*tanh(xi[n])^10*a[2]^2+tanh(xi[n])^8*a[1]^2+2*tanh(xi[n])^8*a[2]^2+tanh(xi[n])^6*a[1]^2-tanh(xi[n])^4*b[1]^2-tanh(xi[n])^2*b[1]^2-2*tanh(xi[n])^2*b[2]^2-2*b[2]^2)*c__1^2 = -(a[2]*tanh(-xi[n]+d)^4*tanh(xi[n])^2*tanh(xi[n]+d)^2+a[2]*tanh(xi[n]+d)^4*tanh(-xi[n]+d)^2*tanh(xi[n])^2-2*a[2]*tanh(xi[n])^4*tanh(-xi[n]+d)^2*tanh(xi[n]+d)^2-a[1]*tanh(-xi[n]+d)^3*tanh(xi[n])^2*tanh(xi[n]+d)^2+a[1]*tanh(xi[n]+d)^3*tanh(-xi[n]+d)^2*tanh(xi[n])^2-2*a[1]*tanh(xi[n])^3*tanh(-xi[n]+d)^2*tanh(xi[n]+d)^2-2*b[1]*tanh(-xi[n]+d)^2*tanh(xi[n])*tanh(xi[n]+d)^2+b[1]*tanh(-xi[n]+d)^2*tanh(xi[n])^2*tanh(xi[n]+d)-b[1]*tanh(-xi[n]+d)*tanh(xi[n])^2*tanh(xi[n]+d)^2-2*b[2]*tanh(-xi[n]+d)^2*tanh(xi[n]+d)^2+b[2]*tanh(-xi[n]+d)^2*tanh(xi[n])^2+b[2]*tanh(xi[n])^2*tanh(xi[n]+d)^2)*beta

 (15)
> 

# Use this correction and do a step worward
> 

 

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

2*(Psi-1)*Psi^4*(Psi+1)*c__1^2*(Psi^10*a[2]^2+2*Psi^9*a[1]*a[2]+(a[2]^2+(-3*alpha-3*a[0])*a[2]+(1/2)*a[1]^2)*Psi^8+((a[1]+5*b[1])*a[2]-a[1]*(alpha+a[0]))*Psi^7+((alpha+a[0]+8*b[2])*a[2]+(1/2)*a[1]*(a[1]+4*b[1]))*Psi^6+(4*a[1]*b[2]-4*a[2]*b[1])*Psi^5+(-8*a[2]*b[2]+(-alpha-a[0])*b[2]-2*((1/4)*b[1]+a[1])*b[1])*Psi^4+((-5*a[1]-b[1])*b[2]+b[1]*(alpha+a[0]))*Psi^3+(-b[2]^2+(3*alpha+3*a[0])*b[2]-(1/2)*b[1]^2)*Psi^2-2*Psi*b[1]*b[2]-b[2]^2) = -(a[2]*tanh(-xi[n]+d)^4*Psi^2*tanh(xi[n]+d)^2-a[1]*tanh(-xi[n]+d)^3*Psi^2*tanh(xi[n]+d)^2+(a[2]*tanh(xi[n]+d)^4*Psi^2+a[1]*tanh(xi[n]+d)^3*Psi^2+(-2*Psi^4*a[2]-2*Psi^3*a[1]-2*Psi*b[1]-2*b[2])*tanh(xi[n]+d)^2+b[1]*Psi^2*tanh(xi[n]+d)+b[2]*Psi^2)*tanh(-xi[n]+d)^2-b[1]*tanh(-xi[n]+d)*Psi^2*tanh(xi[n]+d)^2+b[2]*Psi^2*tanh(xi[n]+d)^2)*beta

 (16)
> 

 

NULL

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

Error, final value in for loop must be numeric or character

  

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

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

 (17)

Download ddd3.mw

.Thank you for everything you do...

delvin 55
June 20 2023
0 0

@mmcdara 

 Thank you for everything you do.

Unfortunately no.

@mmcdara Thanks Sorry, I was n...

delvin 55
June 20 2023
0 0

@mmcdara 

Thanks 

Sorry, I was not familiar with vote.

Only one of the answers had a cup.

@mmcdara  Hello,Can you see wh...

delvin 55
June 19 2023
0 0

@mmcdara 

 Hello,

Can you see why the rest is wrong?

> 

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)))):
der    := [select(has, indets(%), diff)[]]:
lhs_27 := eval(lhs_27, der =~ eval(der, 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)
> 

eq_30_a := eval(eq_30, {seq(v[n+k](t)=V[n+k](xi[n]), k=-1..1)})

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

 (10)

 

eq_30_a is consistent for all the V are functions of xi[n] and you take derivatives wrt xi[n].
eq_30_a is the true equation (30) in the paper you refer to.

But eq_30 is not consistent (funstions of t and derivative wrt xi[n]).

 

 

> 

printf("\n\nWhat you do\n\n"):

Replacements_v := {
  v[n](t) = a[0]+sum(-a[i]*(tanh(t[n]))^i, i = 1 .. 2)+sum(-b[i]*(tanh(t[n]))^(-i), i = 1 .. 2)
  , 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))^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
  , 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
}:
print~(Replacements_v):


# Is it not this that you should do?

printf("\n\nIs it not this that you should do to account for consistency?\n\n"):

Replacements_V := eval(Replacements_v, {t[n]=xi[n], seq(v[n+k](t)=V[n+k](xi[n]), k=-1..1)}):
print~(Replacements_V):


# Or more simply

printf("\n\nOr more simply\n\n"):

Replacements_Vth := {
  V[n](xi[n]) = a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2
  , V[n-1](xi[n]) = a[0]-a[1]*tanh(xi[n]-d)-a[2]*tanh(xi[n]-d)^2-b[1]/tanh(xi[n]-d)-b[2]/tanh(xi[n]-d)^2
  , V[n+1](xi[n]) = a[0]-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2
}:
print~(Replacements_Vth):



What you do
 

  

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

  

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))^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

  

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))^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

  



Is it not this that you should do to account for consistency?
 

  

V[n](xi[n]) = a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2

  

V[n-1](xi[n]) = a[0]-a[1]*(tanh(xi[n])-tanh(d))/(1-tanh(xi[n])*tanh(d))-a[2]*(tanh(xi[n])-tanh(d))^2/(1-tanh(xi[n])*tanh(d))^2-b[1]*(1-tanh(xi[n])*tanh(d))/(tanh(xi[n])-tanh(d))-b[2]*(1-tanh(xi[n])*tanh(d))^2/(tanh(xi[n])-tanh(d))^2

  

V[n+1](xi[n]) = a[0]-a[1]*(tanh(xi[n])+tanh(d))/(1+tanh(xi[n])*tanh(d))-a[2]*(tanh(xi[n])+tanh(d))^2/(1+tanh(xi[n])*tanh(d))^2-b[1]*(1+tanh(xi[n])*tanh(d))/(tanh(xi[n])+tanh(d))-b[2]*(1+tanh(xi[n])*tanh(d))^2/(tanh(xi[n])+tanh(d))^2

  



Or more simply
 

  

V[n](xi[n]) = a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2

  

V[n-1](xi[n]) = a[0]+a[1]*tanh(-xi[n]+d)-a[2]*tanh(-xi[n]+d)^2+b[1]/tanh(-xi[n]+d)-b[2]/tanh(-xi[n]+d)^2

  

V[n+1](xi[n]) = a[0]-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2

 (11)
> 

fin0 := eval(eq_30_a, Replacements_Vth)

(Diff(Diff(ln((a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2+alpha)/alpha), xi[n]), xi[n]))*c__1^2 = (a[1]*tanh(-xi[n]+d)-a[2]*tanh(-xi[n]+d)^2+b[1]/tanh(-xi[n]+d)-b[2]/tanh(-xi[n]+d)^2+2*a[1]*tanh(xi[n])+2*a[2]*tanh(xi[n])^2+2*b[1]/tanh(xi[n])+2*b[2]/tanh(xi[n])^2-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2)*beta

 (12)
> 

# To force the derivation uncomment the next line

# convert(fin0, diff)

fin1 := simplify(numer(fin0))

Error, invalid input: numer expects its 1st argument, x, to be of type {algebraic, list, set}, but received Diff(Diff(ln((a[0]-a[1]*tanh(xi[n])-a[2]*tanh(xi[n])^2-b[1]/tanh(xi[n])-b[2]/tanh(xi[n])^2+alpha)/alpha),xi[n]),xi[n])*c__1^2 = (a[1]*tanh(-xi[n]+d)-a[2]*tanh(-xi[n]+d)^2+b[1]/tanh(-xi[n]+d)-b[2]/tanh(-xi[n]+d)^2+2*a[1]*tanh(xi[n])+2*a[2]*tanh(xi[n])^2+2*b[1]/tanh(xi[n])+2*b[2]/tanh(xi[n])^2-a[1]*tanh(xi[n]+d)-a[2]*tanh(xi[n]+d)^2-b[1]/tanh(xi[n]+d)-b[2]/tanh(xi[n]+d)^2)*beta

  

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_1_mmcdara.mw

correct...

delvin 55
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 55
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 55
June 18 2023
0 0

@mmcdara 

 Thank you for helping me.

@Christian Wolinski   ok &nb...

delvin 55
June 18 2023
0 0

@Christian Wolinski 

 ok

 Sorry, I didn't know you

Problem running the program!!!...

delvin 55
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 55
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 55
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 55
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 55
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

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