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a example , but M is not exact calculated , so the solution is not fitting


 

Download berekening_coeiff_M_is_onjuist_en_niet_berekend_3-5-2025.mw

@mmcdara 

Thanks for the nice proof , it is an ingenious proof though ..could follow it with some help

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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) >    >

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