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These are replies submitted by delvin





@Carl Love 

Thank you for taking the time to do this.

@Carl Love 

I change 𝛯(𝜑), but it gives an error, what is the reason?


@Carl Love 



The program does not run and the system hangs.


I wrote Jacobi correctly in the program.

what does it mean?

you can help me?

@Carl Love 

Will it be fixed later?


No, I just wanted to enter in the attached program instead of definition.

@Carl Love 

If you intended the tanh question to be independent of the error, I can attempt to answer it.


Your pde2 is not explicitly an equation because it doesn't consist of two expressions separated by =

It is the same image above the program. I wrote wrong?


I changed jacoby to Jacoby but it didn't work?

@Carl Love 

it is the Partial differential equation that becomes an ordinary differential equation in the second step.







eq1 := 12*beta^3*rho[3]^2*(diff(w(psi), `$`(psi, 2)))+(-3*beta*rho[2]^2+4*omega*rho[3]^2)*w(psi)+beta*rho[3]^2*(rho[1]+2*rho[3])*w(psi)^3

12*beta^3*rho[3]^2*(diff(diff(w(psi), psi), psi))+(-3*beta*rho[2]^2+4*omega*rho[3]^2)*w(psi)+beta*rho[3]^2*(rho[1]+2*rho[3])*w(psi)^3


proc (psi) options operator, arrow, function_assign; kappa[0]+sum(kappa[i]*(diff(E(psi), psi))^i/E(psi)^i, i = 1 .. 1)+sum(h[i]*((diff(E(psi), psi))/E(psi))^(-i), i = 1 .. 1) end proc

"E(psi):=((epsilon[1]*jacobiCN(Zeta[1]*psi))+(epsilon[2]*jacobiSN(Zeta[2]*psi)))/((epsilon[3]*jacobiCN(Zeta[3]*psi))+(epsilon[4]*jacobiSN(Zeta[4]*psi))) ;"

proc (psi) options operator, arrow, function_assign; (varepsilon[1]*jacobiCN(Zeta[1]*psi)+varepsilon[2]*jacobiSN(Zeta[2]*psi))/(varepsilon[3]*jacobiCN(Zeta[3]*psi)+varepsilon[4]*jacobiSN(Zeta[4]*psi)) end proc


fin1 := simplify(eq1)

#Extremely long output that neither displays well on MaplePrimes nor contributes to
#understanding the Question deleted by Moderator. -- Carl Love

Sol := solve(fin1, {beta, omega, Zeta[1], Zeta[2], Zeta[3], Zeta[4], epsilon[1], epsilon[2], epsilon[3], epsilon[4], h[1], kappa[0], kappa[1]})

for i to 2 do Case[i] := allvalues(Sol[i]) end do





Thanks for your hard work on this.

I waited a little to upload quickly, but it was long yesterday.

Now the answers are different!!!


It didn't work, I had to stop.

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