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Engine on the net that will solve PDEs

  1. Nov 25, 2008 #1
    Hi

    Does anybody know if there is an engine on the net that will solve PDEs.

    I'm looking for the solution of a linear first order PDE

    Thanks
     
  2. jcsd
  3. Nov 25, 2008 #2

    jambaugh

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    Re: Pde

    Could you post the form of the PDE?
     
  4. Nov 25, 2008 #3
    Re: Pde

    yzx+xzy+z=y z=2 @ (3,1)
     
  5. Nov 25, 2008 #4
    Re: Pde

    Is it not that we first solve the auxiliary equation ?
    [tex] \frac{dx}{y} = \frac{dy}{x} = \frac{dz}{y - z} [/tex]

    Let the two solutions be u=c1 and v=c2. Then the general solution is u=f(v).

    But not sure how to include the given initial condition z=2 @ (3,1) .
     
  6. Nov 25, 2008 #5
    Re: Pde

    Thank you for your input. I will consider this
     
  7. Nov 25, 2008 #6
    Re: Pde

    I had sent the following to convode@riemann.physmath.fundp.ac.be

    NMAX:=3;
    SMAX:=4;
    LPARTI:={};
    ARGSTOP:=2;
    DEPEND Z,X;
    DEPEND Z,Y;
    L2:={Z};
    L3:={X,Y};
    L4:={2,{3,1}};
    L5:={};
    L1:={Y*DF(Z,X)+X*DF(Z,Y)+Z-Y=0};
    CONVODE(L1,L2,L3,L4,L5);

    and received the reply for the solution as follows (if I retype it correctly)
    [tex]z = (2x+2y)^{-1} (2\sqrt{- x^2 +y^2}f(- x^2 + y^2)+(y^2-x^2)\log(\frac{x+y}{\sqrt{-x^2 + y^2}})+x^2 +xy)[/tex]

    where f is an arbitrary function.
    Can we believe the solution given?
     
  8. Nov 26, 2008 #7
    Re: Pde

    Thanks

    Convode sends it back in a weird asss format in the email. Difficult to read no !?

    I think the question might be faulty or at least the initial conditions. I don't know is it solvable by just being given the value of z at a POINT rather than along a curve/line
     
    Last edited: Nov 26, 2008
  9. Nov 26, 2008 #8
    Re: Pde

    Never mind about that messy result given by convode. But it do gave something.

    Let say there is a mistake giving a point. A curve or a line given instead as you said. How do we determine that arbitrary function?
     
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