@Alfred_F 
In complexe analyse the limit is ?

restart;

# Definieer j als positief
assume(j > 0):

# Definieer de som over k
S := sum(1 / (j^2 + k^2), k = 1 .. n):

# Neem de limiet als n → ∞
limit(S, n = infinity);

               1                                
               - I (Psi(1 - I j) - Psi(1 + I j))
               2                                
               ---------------------------------
                               j                

# restart;

# Definieer de uitdrukking na binnenlimiet
a := j -> (Pi/2 - arctan(1/j)):

# Buitensom van j = 1 .. n
S := sum(a(j), j = 1 .. n):

# Buitenlimiet (deel door n)
limit(S/n, n = infinity);

                              1   
                              - Pi
                              2   

how_number_of_definition_mprimes_9-5-2025.mw

@JaneCherrytree 
Symbolic integral seems to problematic to get 
ntegral_of_long_expression_mprimesDEFA_7-5-2025_(2).mw

simplifyingnonlinterms_involving_two_function_in_ode_mprimes6-5-5.mw  

Download ModulationInstabilityAnalysis_CompleteAllSteps_mprimes-4-5-2025.mw

Needs more correspondending results 

Modulation_Instability_and_Dispersion_Relation_Analysis_4-5-2025mprimes.mw

Download berekening_coeiff_M_is_onjuist_en_niet_berekend_3-5-2025.mw

Comparing A2 and A2  , shows that they are equal and simplify and is  is not neccessary 

Download gelijkheid_expressie_mpries_3-5-2025_A.mw

see under

restart;
with(DEtools):

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

# Step 2: Apply substitution Q = 1/u to transform the ODE
ode_u := simplify(eval(ode, Q(zeta) = 1/u(zeta)) * u(zeta)^4);
printf("Step 2: Substitute Q = 1/u to transform the ODE into a simpler rational form.\n");

# Step 3: Solve the transformed ODE for u(zeta)
u_sol := dsolve(ode_u, u(zeta));
printf("Step 3: Solve the transformed ODE in terms of u(zeta).\n");

# Step 4: Choose the specific solution involving tanh and revert back to Q(zeta)
u_specific := b/(2*a) + sqrt(4*a*l + b^2)/(2*a)*tanh(r*sqrt(a)*zeta/2);
Q_sol := 1/u_specific;
printf("Step 4: Choose the specific solution using tanh and convert back to Q(zeta).\n");

# Step 5: Define the exponential form of Q(zeta) as shown in the image
Q_exp := 4*a/((4*a*l + b^2)*exp(r*sqrt(a)*zeta) - exp(-r*sqrt(a)*zeta) + 2*b);
printf("Step 5: Define the exponential expression for Q(zeta).\n");

# Verify equivalence of hyperbolic and exponential forms
verify_exp := simplify(Q_sol - Q_exp);
printf("Step 6: Verify if the expression using tanh equals the exponential form. Difference = %a\n", verify_exp);

# Step 6: Define the hyperbolic version (sech² and tanh form) of the solution
Q_hyp := a*b*sech(r*sqrt(a)*zeta/2)^2/(b^2 + a*l*(1 - tanh(r*sqrt(a)*zeta/2))^2);
printf("Step 7: Define the hyperbolic solution (sech²/tanh form).\n");

# Check equivalence with exponential form
verify_hyp := simplify(Q_exp - Q_hyp);
printf("Step 8: Verify if exponential form equals the hyperbolic form. Difference = %a\n", verify_hyp);

# Step 9: Validate that both expressions satisfy the original ODE
ode_check_exp := simplify(eval(ode, Q = unapply(Q_exp, zeta)));
ode_check_hyp := simplify(eval(ode, Q = unapply(Q_hyp, zeta)));
printf("Step 9: Substitute Q_exp into ODE → Result = %a\n", ode_check_exp);
printf("Step 10: Substitute Q_hyp into ODE → Result = %a\n", ode_check_hyp);

?

 

# --------------------------------------
# 🚧 Fix to remove the extra U'(xi) term
# --------------------------------------

# This is the condition needed to remove the extra real part involving dU/dxi:
FixCond := 3*delta[3]*(b[1]*d[1]^2 + b[2]*d[2]^2 + b[3]*d[3]^2)
           - 2*delta[1]*(b[1]*d[1] + b[2]*d[2] + b[3]*d[3])
           - b[4];

# Set this condition to zero to eliminate the extra U'(xi) term
FixCondSol := solve(FixCond = 0, b[4]);

# Substitute this condition directly back into your main reduced ODE
FixedODE := subs(b[4] = FixCondSol, Q);

# ✅ Now continue analysis with FixedODE instead of Q

@Paras31 
look promising,, 

CORRECT ?

Download aanpassoing_ND_pro_mprimes27-4-2025.mw

@salim-barzani 

Download solve_a4-a9-a11_pde_6_orde_mprimes_27-4-2025.mw

DrawCleanDomain := proc(outerPolygon::list,
                        path1Pts::list, path2Pts::list,
                        z0::list, z::list,
                        path1Color::string := "red",
                        path2Color::string := "green",
                        outerColor::string := "LightBlue",
                        outerTransparency::numeric := 0.4)

    uses plots, plottools;

    local outer, curve1, curve2, pt1, pt2, label1, label2, finalPlot;
    local z0Label, z0Coords, zLabel, zCoords;

    # Labels en coördinaten splitsen
    z0Label := z0[1]; z0Coords := z0[2];
    zLabel := z[1]; zCoords := z[2];

    # Buitenste domein
    outer := polygon(outerPolygon, color = outerColor, transparency = outerTransparency);

    # Paden
    curve1 := pointplot(path1Pts, color = path1Color, linestyle = dot, thickness = 2, connect = true);
    curve2 := pointplot(path2Pts, color = path2Color, linestyle = dot, thickness = 2, connect = true);

    # Begin- en eindpunt
    pt1 := pointplot([z0Coords], symbol = solidcircle, symbolsize = 15, color = black);
    pt2 := pointplot([zCoords], symbol = solidcircle, symbolsize = 15, color = black);

    # Labels bij punten
    label1 := textplot([z0Coords[1], z0Coords[2] - 0.1, z0Label], font = [Helvetica, Bold, 14]);
    label2 := textplot([zCoords[1], zCoords[2] - 0.1, zLabel], font = [Helvetica, Bold, 14]);

    # Alles tonen in één plot
    finalPlot := display([outer, curve1, curve2, pt1, pt2, label1, label2],
                         scaling = constrained, axes = none,
                         title = "Domain with Two Paths and Custom Points");

    return finalPlot;
end proc:

outerPolygon := [[-1, 0], [-0.7, 1.2], [0, 1.5], [0.7, 1.2], [1.2, 0], [0.7, -1.2], [0, -1.4], [-0.7, -1.2], [-1, 0]];
path1 := [[-0.8, 0.6], [-0.3, 1.0], [0.4, 0.5], [0.9, -0.4]];
path2 := [[-0.8, 0.6], [-0.7, 0.0], [0.2, -0.6], [0.9, -0.4]];
a0 := ["a_0", [-0.8, 0.6]];
z := ["z", [0.9, -0.4]];

DrawCleanDomain(outerPolygon, path1, path2, a0, z);

example ; fig1 := [[0, 0], [4, 0], [4, 3], [0, 3], [5, 5],[8,1],[8,8],[-5,8]]
feature : a last point in the list will not be connected to another point to draw a line 

Download punten_tekeken_-laatste_punt_niet_in_lijst_mprimes_23-4-2025.mw