2 years, 55 days

Thanks

## Thank you very much...

I appreciate your taking the time.

Just how can I adjust the colors to be brighter?

## Guide and help, thanks...

I want to write a code for the general state of the attached file with Maple, but unfortunately I could not, can anyone help me in this field?

GER.pdf

## @C_R  okey ...

okey

How is the MATLAB code? Of course, my MATLAB version is 2014

Okey

## guidance and help...

It is possible to change the program in such a way that the number of EQ equations is reduced, for example to 6, which is equal to the size of our variables.

ddd3_ac.mwddd3_ac.mw

## I appreciate your guidance and time....

Does this mean that the answers to these 16 equations are self-evident? The last two commands are not wrong?

## Thank you very much...

@acer

Thank you very much.

Yes, it was done. Can you tell me what adjustments, because I have several programs that I want to implement a series of changes in them, I would appreciate it.

Ok, thanks.

## @mmcdara  Sorry, It can be said t...

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...

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)))
 (1)
 > eq_28 := n -> exp(-beta*u[n](t)) = 1+v[n](t)/alpha
 (2)
 > aux_1 := isolate(eq_28(n), u[n](t));
 (3)
 > eq_29 := eval(convert(lhs(eq_27), Diff), aux_1)          =          simplify(eval(rhs(eq_27), {seq(eq_28(n+k), k=-1..1)}));
 (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);
 (5)
 > aux_2 := select(has, indets(rhs(aux_1), function), ln)[] = H(t);
 (6)
 > eq_29_a := simplify(isolate(convert(eval(eq_29, aux_2), diff), diff(H(t), t\$2)));
 (7)
 > eq_30_1 := lhs_27=rhs(eq_29_a)
 (8)
 > eq_30 := eval(convert(eq_30_1, Diff), u[n](xi[n])=lhs(aux_2))
 (9)
 > eq_30_a := eval(eq_30, {seq(v[n+k](t)=V[n+k](xi[n]), k=-1..1)})
 (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
 Is it not this that you should do to account for consistency?
 Or more simply
 (11)
 > fin0 := eval(eq_30_a, Replacements_Vth)
 (12)
 > # To force the derivation uncomment the next line fin0 := convert(fin0, diff)
 (13)
 > # look at this to understand why you get an error. p/q=P/Q; numer(%);
 > # or to this p=P/Q; numer(%);
 > # if you want to get the numerators of both sides do this p/q=P/Q; map(numer, %)
 (14)
 > # APPLICATION: fin1 := map(numer, fin0)
 (15)
 > # Use this correction and do a step worward
 >

 (16)
 >

 (17)

## .Thank you for everything you do...

Thank you for everything you do.

Unfortunately no.

## @mmcdara Thanks Sorry, I was n...

Thanks

Sorry, I was not familiar with vote.

## @mmcdara  Hello,Can you see wh...

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)))
 (1)
 > eq_28 := n -> exp(-beta*u[n](t)) = 1+v[n](t)/alpha
 (2)
 > aux_1 := isolate(eq_28(n), u[n](t));
 (3)
 > eq_29 := eval(convert(lhs(eq_27), Diff), aux_1)          =          simplify(eval(rhs(eq_27), {seq(eq_28(n+k), k=-1..1)}));
 (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);
 (5)
 > aux_2 := select(has, indets(rhs(aux_1), function), ln)[] = H(t);
 (6)
 > eq_29_a := simplify(isolate(convert(eval(eq_29, aux_2), diff), diff(H(t), t\$2)));
 (7)
 > eq_30_1 := lhs_27=rhs(eq_29_a)
 (8)
 > eq_30 := eval(convert(eq_30_1, Diff), u[n](xi[n])=lhs(aux_2))
 (9)
 > eq_30_a := eval(eq_30, {seq(v[n+k](t)=V[n+k](xi[n]), k=-1..1)})
 (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
 Is it not this that you should do to account for consistency?
 Or more simply
 (11)
 > fin0 := eval(eq_30_a, Replacements_Vth)
 (12)
 > # To force the derivation uncomment the next line # convert(fin0, diff)

 > fin := simplify(subs(tanh(xi[n]) = Psi, fin1));
 (13)
 > degree(fin,Psi)
 (14)
 > FF:=convert(series(fin,Psi,7),polynom):
 > degree(%,Psi)
 (15)
 > for i from 0 to degree(FF, Psi) do     EQ[i] := simplify(coeff(FF, Psi, i)); end do
 (16)

 >