# Eerste expressie definiëren
expr1 := (3*q*B[1]^2*A[1]*sin(W) + 3*q*B[1]*A[1]^2*cos(W) + 6*q*B[1]*A[1]*A[0])*sin(W)*cos(W)
         + q*A[1]^3*cos(W)^3 + 3*q*A[1]^2*cos(W)^2*A[0] 
         + (3*q*A[0]^2*A[1] + p*A[1] - s*A[1])*cos(W) 
         + q*B[1]^3*sin(W)^3 + 3*q*B[1]^2*sin(W)^2*A[0] 
         + (3*q*A[0]^2*B[1] + p*B[1] - s*B[1])*sin(W) 
         + q*A[0]^3 + p*A[0] - s*A[0]:

# Tweede expressie definiëren
expr2 := (A[1]*cos(W) + B[1]*sin(W) + A[0])*(q*A[1]^2*cos(W)^2 
         + 2*q*A[1]*(B[1]*sin(W) + A[0])*cos(W) 
         + q*B[1]^2*sin(W)^2 + 2*q*A[0]*B[1]*sin(W) 
         + q*A[0]^2 + p - s):

# Vereenvoudig het verschil om te controleren op gelijkheid
check_equiv := simplify(expr1 - expr2);

# Controleer of het verschil gelijk is aan nul
is_zero := evalb(check_equiv = 0);
is_zero;

@dharr , thanks for the solution, yes an error I have seen several times and thought I was going to solve it with with (package) but apparently that is not the right way for a procedure.
Why not, that's another specialist question....  

Error, (in Odegenerator) `DEtools` is not a module or member
Seems that uses not accept DEtools ?

@Carl Love , Thanks for correction!
It creeps in anyway ...

@WD0HHU 
Yes , the procedure can now calculate iteratively or manually.
In both calculations the odeadvisor has to determine the type of ode .
For iterative it is correct, but for manual I think not yet ,but not a big issue to solve

restart;
params := {alpha = 1, gg = 0.1, k = 1, mu = 10, sigma = 5, w = 2};
M := 2*sqrt(-(gg*(gg*k*mu - 2*k*sigma)*k/(gg*sigma - 1) + k^2)/alpha)*exp((-(4*gg*k*w + 4*k^2 - sigma^2)*t/(gg*sigma - 1) + sigma*x)*I)/sinh(-2*k*x + (-gg*k*(4*gg*k*w + 4*k^2 - sigma^2)/(gg*sigma - 1) - 2*k*sigma)*t/(gg*sigma - 1));
Explore(plot3d([Re, Im](M), t = 0 .. 5, x = 0 .. 5, view = -100 .. 100, grid = [150, 150], color = [red, blue], style = surface, adaptmesh = false, size = [500, 500]), alpha = 0.1 .. 2.0, gg = 0.1 .. 0.3, w = 1 .. 5, mu = 1.0 .. 10.0, sigma = 1.0 .. 4.0, k = 1.0 .. 2.0, placement = right);

Explore example plot of complex valued function
Can be used for trying to get a explore plot with Wiener function 

@salim-barzani      A.D. Polyanin

@salim-barzani , there is some direction now , but did you succeed in your code to get this reduced pde form (ode)  ?
I did, but  now how to go further ..well it seems there is enough probably to find the right information for me  : handbooks and so on
i do have another solution to your problem try contact : AndreiD.Polyanin or
AlexeiI.Zhurov from Russia they are experts pdes and odes 

 pde

u(x,t) is pde value of U(z) and is plot from a ode , it all revolves about a reduceded pde , to solve this 
How to solve this non lineair complex valued third orde ode , right?

Another second way of defining pde with a table U 

with(PDEtools): 
declare(u(x,t)); 
U := diff_table(u(x,t));
interface(showassumed = 0); 
assume(k > 0);
PDE1 := I*U[t] + U[x,x] + k*abs(U[])^2*U[] = 0;
Sol1 := u(x,t) = C1*exp(I*(C2*x + (k*C1^2 - C2^2)*t + C3));
Test1 := pdetest(Sol1, PDE1); 
Test11 := simplify(evalc(Test1));
Sol2 := u(x,t) = A*sqrt(2/k)*(exp(I*B*x + I*(A^2 - B^2)*t + I*C1))/(cosh(A*x - 2*A*B*t + C2));
Test2 := pdetest(Sol2, PDE1); 
Test21 := simplify(evalc(Test2));# jd complexe getallen

a third way is : direct 

with(PDEtools):(u(x,y));
PDE1:=x*diff(u(x,y),y)-diff(u(x,y),x)=g(x)*u(x,y)^2;
Sol1:=pdsolve(PDE1);

pdetest(Sol1,PDE1);

a fourth way example is with declare 

with(PDEtools): declare(u(x,y)); # hier word declare gebruikt
PDE1:=u(x,y)*diff(u(x,y),y)=diff(u(x,y),x);
sysCh:=charstrip(PDE1,u(x,y)); funcs:=indets(sysCh,Function);
Sol1:=dsolve(sysCh,funcs,explicit);

As notation seems te be handier in this example an note how this is done 
with(PDEtools): with(plots): tr1:=x-c*t=z; tr2:={a=1,b=1,c=1}; # JD traveling wave 
Eq1:=u->diff(u,t)+a*u*diff(u,x)+b*diff(u,x$3)=0;
Eq2:=expand(Eq1(U(lhs(tr1)))); Eq3:=algsubs(tr1,Eq2);
Eq4:=map(convert,Eq3,diff); Eq5:=map(int,lhs(Eq4),z)-C1=0;
Eq6:=expand(Eq5*2*diff(U(z),z)); Eq7:=map(int,Eq6,z);
Eq8:=lhs(Eq7)=C2; Eq9:=subs({C1=0,C2=0},Eq8);
Sol1:=[dsolve(Eq9,U(z))]; Sol11:=subs(_C1=0,simplify(Sol1[2]));
Sol12:=convert(Sol11,sech); Sol2:=eval(subs(z=x-c*t,Sol11),tr2);
convert(Sol2,sech);
animate(plot,[rhs(Sol2),x=-20..20,color=blue],t=0..20,numpoints=100,
frames=50,thickness=2);
pdetest(u(x,t)=rhs(Sol2),subs(tr2,Eq1(u(x,t))));


your pde

Eq1 := u -> (diff(u, t, t) - c^2*diff(u, x, x))*I + diff(U(-t*tau + x)^2*u, t) - lambda*c*diff(U(-t*tau + x)^2*u, x) + 1/2*diff(u, t, x, x) - 1/2*epsilon*c*diff(u, x, x, x) = 0

PDE
ans := pdsolve(PDE)
pdetest(ans, PDE)

Ai says separation of variables is possible in a resulting ODE ?

Have made a few more attempts to make the function T3 work via explore plot, unfortunately not successful yet.
The function T3 is a complex function and should the explore plot be able to show the real part of the function T3 ?   

By induction can this proved ?
no.. thats only for one variabele

this example is a intersection in implicit 2D functions
Add any two implicit function you can think of in the intersections ( )  proc 
let's look how intersections( ) can handle this :-)  

Its about polynome expressions, so all functions as polynomes expressed can be used


 

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