1. ## Simple PDE equation

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

F1(0,0)=F1(0,2Pi)=T1(2Pi,0)=T1(2Pi,2Pi)=0;
dF1(0,0)/dx=dF1(2Pi,0)/dx;
dF1(0,0)/dy=dF1(0,2Pi)/dy

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?

Thank you

2. ## Re: Simple PDE equation

Let me ask, how did you solve the two dimensional problem? The one with F1.

3. ## Re: Simple PDE equation

I supposed the solution in form C0 = A + B Sin[2 \[Theta]2 - (2 \[Theta]1 \[Omega]2)/\[Omega]1] +
C Sin[4 \[Theta]2 - (4 \[Theta]1 \[Omega]2)/\[Omega]1] because Mathematica give me general form...

4. ## Re: Simple PDE equation

What's $\displaystyle \theta_1$, $\displaystyle \theta_2$, $\displaystyle \omega_1$ and $\displaystyle \omega_2$?

5. ## Re: Simple PDE equation

Sorry.. Mistake.

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

F1(0,0)=F1(0,2Pi)=F1(2Pi,0)=T1(2Pi,2Pi)=0;
dF1(0,0)/dx=dF1(2Pi,0)/dx;
dF1(0,0)/dy=dF1(0,2Pi)/dy.

Mathematica gives general solution + C[[1]][(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.
F=

6. ## Re: Simple PDE equation

Do you know how to solve the PDE

$\displaystyle a_0 \dfrac{\partial F}{\partial x} + b_0 \dfrac{\partial F}{\partial y} + c_0(x,y) = 0$

in general? BTW - what is $\displaystyle c_0(x,y)$?

7. ## Re: Simple PDE equation

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.

8. ## Re: Simple PDE equation

So what you have is

$\displaystyle a_0 \dfrac{\partial F}{\partial x} + b_0 \dfrac{\partial F}{\partial y} + q_1 \sin 2x + p_1 \cos y + p_2 \sin 4x = 0$
You know, you can transform this problem to one where $\displaystyle c = 0$ by letting

$\displaystyle F = G + k_1 \sin 2x + k_2 sin 4x + k_3 \cos y$

and picking $\displaystyle k_1, k_2$ and $\displaystyle k_3$ so that your new PDE is

$\displaystyle a_0 \dfrac{\partial G}{\partial x} + b_0 \dfrac{\partial G}{\partial y} = 0.$

This new PDE is easily solved.

As for your original question - what isthe form of $\displaystyle d_0$?

9. ## Re: Simple PDE equation

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...

10. ## Re: Simple PDE equation

Let's go back to your original PDE in 2D. Are we to assume that $\displaystyle a_0, b_0, p_1,q_1$ and $\displaystyle p_2$ are all not zero. If so, do you actually have a solution to this problem?

11. ## Re: Simple PDE equation

How to simple solve PDE where figure new G function? I put it in Mathematica and it got me solution in for C[y-x*(b0/a0)]? What is the analytic for of solution?

12. ## Re: Simple PDE equation

Yes you are right, as you say not zero, a0, bo,...

13. ## Re: Simple PDE equation

If all these constant are all not zero, I believe there is no solution to this problem.

First off, the general solution is

$\displaystyle F = G\left(b_0 x - a_0y\right) + \dfrac{q_1}{2 a_0} \cos 2x + \dfrac{p_2}{4 a_0} \cos 4x - \dfrac{p_1}{b_0} \sin y$

where G is an arbitrary function of its argument. Imposing your BC

$\displaystyle F_x(0,0) = 0 \; \text{and}\; F_y(0,0) = 0$ gives

$\displaystyle b_0G'(0) = 0$ and $\displaystyle -a_0G'(0) - \dfrac{p_1}{b_0} = 0$.

From the first, we obtain $\displaystyle G'(0) = 0$ and from the second $\displaystyle p_1 = 0$ but this leads to a contradiction since $\displaystyle p_1 \ne 0$.

14. ## Re: Simple PDE equation

How can I attached the .jpg file from my computer? Direct solving od 2D PDE?

15. ## Re: Simple PDE equation

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