Let me ask, how did you solve the two dimensional problem? The one with F1.
a0*dF(x,y,z)/dx+b0*dF(x,y,z)/dy+c0*dF(x,y,z)/dz+d0=0, F(x,y,z) must be positive and periodic. a0,b0,c0,d0-known constants.
For case of two variables a0*dF1(x,y)/dx+b0*dF1(x,y)/dy+c0=0
We know conditions
I must find analog conditions for F(x,y,z) and how to determine the "constant" which is function for my F(x,y,z) case?
a0*dF1(x,y)/dx+b0*dF1(x,y)/dy+c0(x,y)=0 This special case of partial equation can be solved if b0/c0 is integer. b0 and c0 are constants and I know function c0(x,y). I already know conditions
Mathematica gives general solution + C[][(x,y). I suppose this particular solution in the form. C=A0+B0*Sin[2y-2(b0/a0)x]+C0*Sin[4y-4(b0/a0)x]. But I can not find all constants. Many equations from the conditions are the same. b0/a0 if integer, can help me to vanish some of participants Sin[(b0/a0)*Pi] and some of them are 1 (Cos[(b0/a0)*Pi]). I find conditions for equation of third order (x,y,z) but I have problem to find constants because some of equations from conditions are annulment or same.
Yes, I explained in my last answer. c is the known function of x, y (trigonometric). c(x,y)=q1*sin[2x]+p1*cos[y]+p2*sin[4x]... long form. qi, pi known constants. But from conditions, I can not determine A1, B1, C1 ... Function F must be positive and periodic.
d0 is also trigonometric function of many participants in form d0(x,y,z)= Sin[2x],Sin[2y],Sin[2z],Cos[2x],Cos[2y],Cos[2z],Sin[4x],Sin[4y],Sin[4z],Cos[4x],Cos[4y],Cos[4z] and also have some free constants and some constants multiple this trigonometric functions...
If all these constant are all not zero, I believe there is no solution to this problem.
First off, the general solution is
where G is an arbitrary function of its argument. Imposing your BC
From the first, we obtain and from the second but this leads to a contradiction since .