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Download clarifying_solving_new_ode_33-5-2025mprimes.mw

Download ode_naar_riccati_type_ode3-5-2025_mprimes.mw

@vv 
I only raised A and B each to the second power, has nothing to do with : A^2 = B^2  does NOT imply A = B.

@mmcdara 
Is A=B ? 

@salim-barzani 
Glimpses of Soliton Theory (Second Edition) by Alex Kasman

Download ricatti_ode_3_oplossingen_mprimes_30-4-2025.mw

solve(x^2 = a, x); sqrt(a), -sqrt(a)


restart:
with(DEtools):

printf("📘 Step 1: Define the nonlinear ODE in Q(zeta):\n");
ode := (diff(Q(zeta), zeta))^2 - r^2*Q(zeta)^2*(a - b*Q(zeta) - l*Q(zeta)^2) = 0;

printf("\n📘 Step 2: Substitute Q = 1/u(zeta) to simplify the ODE:\n");
ode_u := simplify(eval(ode, Q(zeta) = 1/u(zeta)) * u(zeta)^4);

printf("\n📘 Step 3: Result after substitution and simplification:\n");
print(ode_u);

printf("\n📘 Step 4: Take the square root on both sides. Here, the ± symbol appears because (du/dzeta)^2 = ...:\n");
printf("When solving a square equation, you must take ± sqrt(...)\n");
printf("→ du/dzeta = ± r * sqrt(a*u^2 - b*u - l)\n");

printf("\n📘 Step 5: Separate variables and integrate, leading to a tanh-type solution:\n");
printf("∫ du / sqrt(a*u^2 - b*u - l) = ± r ∫ dzeta\n");

printf("\n📘 Step 6: After integration, you obtain a tanh-expression for u(zeta):\n");

# Specific solutions for + and -
u_specific_plus := b/(2*a) + sqrt(4*a*l + b^2)/(2*a)*tanh(r*sqrt(a)*zeta/2);
u_specific_minus := b/(2*a) - sqrt(4*a*l + b^2)/(2*a)*tanh(r*sqrt(a)*zeta/2);

printf("u_specific_plus (with + square root):\n");
print(u_specific_plus);
printf("u_specific_minus (with - square root):\n");
print(u_specific_minus);

printf("\n📘 Step 7: Solve for Q(zeta) = 1/u(zeta) for both cases:\n");

Q_sol_plus := simplify(1/u_specific_plus);
Q_sol_minus := simplify(1/u_specific_minus);

printf("Q_sol_plus (plus solution):\n");
print(Q_sol_plus);
printf("Q_sol_minus (minus solution):\n");
print(Q_sol_minus);

printf("\n📘 Step 8: Transform to exponential form using the tanh identity:\n");
printf("tanh(x) = (exp(x) - exp(-x)) / (exp(x) + exp(-x))\n");

Q_exp_plus := 4*a/((4*a*l + b^2)*exp(r*sqrt(a)*zeta) - exp(-r*sqrt(a)*zeta) + 2*b);
Q_exp_minus := 4*a/((4*a*l + b^2)*exp(r*sqrt(a)*zeta) + exp(-r*sqrt(a)*zeta) + 2*b);

printf("Q_exp_plus (minus between exponentials):\n");
print(Q_exp_plus);
printf("Q_exp_minus (plus between exponentials):\n");
print(Q_exp_minus);

printf("\n📘 Step 9: Summary: the ± sign appears when taking the square root of (du/dzeta)^2.\n");
printf("Taking a square root always yields two solutions: +sqrt(...) and -sqrt(...).\n");
printf("Thus, there are two solution families for Q(zeta).\n");

printf("\n✅ Full derivation completed.\n");

Needs further be enhanced..

Download contourdiagram_cirkel_-2cirkels_mprimes_28-4-2025.mw

@dharr 
Of course to fill in as function type the procedure input

I also thought of it , only it didn't occur to me to do the function call with lowercase ...

Well done!, and it the procedure outcome exactly matches what I cumbersomely did in another procedure : ToLowerFunc := proc(expr), but maybe this one is still of use ?

@salim-barzani 

 

Use ND procedure than output as expression ,then procedure ToLowerFunc , then >  restart; >  ToLowerFunc := proc(expr)     local replacements, vars, v, f_lower;          replacements := {};     vars := indets(expr, 'function');          for v in vars do         if type(op(0,v), name) then             f_lower := parse(StringTools:-LowerCase(convert(op(0,v), string)));             replacements := replacements union { op(0,v) = f_lower };         end if;     end do;          return subs(replacements, expr); end proc: >    >  expr2:= diff(diff(F(x, y, z, t), x), y)*G(x, y, z, t) - diff(F(x, y, z, t), x)*diff(G(x, y, z, t), y) - diff(F(x, y, z, t), y)*diff(G(x, y, z, t), x) + F(x, y, z, t)*diff(diff(G(x, y, z, t), x), y); (1) >  ToLowerFunc(expr2); (2) >  restart; with(PDEtools): with(LinearAlgebra): with(SolveTools): undeclare(prime): alias(F=F(x,y,z,t), G=G(x,y,z,t)): (3) >  diff(f(x, y, z, t), x, y)*g(x, y, z, t) - diff(f(x, y, z, t), x)*diff(g(x, y, z, t), y) - diff(f(x, y, z, t), y)*diff(g(x, y, z, t), x) + f(x, y, z, t)*diff(g(x, y, z, t), x, y); (4) >    >

Download kleine_letters_procedurND_uitvoer_omzettingmprines27-4-2025.mw

@salim-barzani 
It seems simple, but Maple processes these lowercase letters differently from the uppercase letters again as it seems?

A question for the maple specialists here

@salim-barzani 
To find  , a expression for  : L2, L3, L4, L5?..

@salim-barzani 
i substituted (32) into pde and it seems be correct. 
Substitute in L2  :after a11, a4, a9 are expressed in the other variables a
Substitute the numeric values , then we get ...

{a11 = -3*(a1^2 + a6^2)*(a1^3*a2 + a1^2*a6*a7 + a1*a2*a6^2 + a6^3*a7)/(a1^2*a7^2 - 2*a1*a2*a6*a7 + a2^2*a6^2), a4 = -(a1^3*alpha + a1^2*a2*beta + a1^2*a3*gamma + a1*a6^2*alpha + a2*a6^2*beta + a3*a6^2*gamma - 5*a1*a2^2 + 5*a1*a7^2 - 10*a2*a6*a7)/(9*(a1^2 + a6^2)), a9 = -(a1^2*a6*alpha + a1^2*a7*beta + a1^2*a8*gamma + a6^3*alpha + a6^2*a7*beta + a6^2*a8*gamma - 10*a1*a2*a7 + 5*a2^2*a6 - 5*a6*a7^2)/(9*(a1^2 + a6^2))}