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Math Help - Simple PDE equation

  1. #1
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    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
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  2. #2
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    Re: Simple PDE equation

    Let me ask, how did you solve the two dimensional problem? The one with F1.
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    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...
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  4. #4
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    Re: Simple PDE equation

    What's \theta_1 , \theta_2 , \omega_1 and \omega_2 ?
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  5. #5
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    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=
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  6. #6
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    Re: Simple PDE equation

    Do you know how to solve the PDE

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

    in general? BTW - what is c_0(x,y)?
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  7. #7
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    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.
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  8. #8
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    Re: Simple PDE equation

    So what you have is

    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 c = 0 by letting

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

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

    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 d_0?
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  9. #9
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    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...
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  10. #10
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    Re: Simple PDE equation

    Let's go back to your original PDE in 2D. Are we to assume that a_0, b_0, p_1,q_1 and p_2 are all not zero. If so, do you actually have a solution to this problem?
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  11. #11
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    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?
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  12. #12
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    Re: Simple PDE equation

    Yes you are right, as you say not zero, a0, bo,...
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  13. #13
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    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

    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

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

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

    From the first, we obtain G'(0) = 0 and from the second p_1 = 0 but this leads to a contradiction since p_1 \ne 0.
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  14. #14
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    Re: Simple PDE equation

    How can I attached the .jpg file from my computer? Direct solving od 2D PDE?
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  15. #15
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    Re: Simple PDE equation

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