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A two Variable PDE

  1. Oct 25, 2011 #1

    lavinia

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    In an open domain in the plane

    (xU[itex]_{x}[/itex]-yU[itex]_{y}[/itex]-U)/U =

    (xV[itex]_{x}[/itex] - y V[itex]_{y}[/itex]+V)/V
     
  2. jcsd
  3. Oct 25, 2011 #2
    I'd like to know the answer myself. Does [itex]\frac{xU_x}{U}-\frac{xV_x}{V}=\frac{yU_y}{U}-\frac{yV_y}{V}[/itex] make it any easier? I've only dabbled in PDEs myself, but that seems like a separation of variables problem, when rephrased like that.

    EDIT: Mathematica gives an error saying the answer is indeterminate, because there are more dependent variables than equations.
     
    Last edited: Oct 25, 2011
  4. Nov 1, 2011 #3

    lavinia

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    U = x V = K/x sort of works but doesn't allow the origin or the y axis.
    Also the answer is a little trivial.
     
  5. Nov 1, 2011 #4

    MathematicalPhysicist

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    So you have:
    x d/dx ln (U) -y d/dy ln(U) -1 = xd/dx ln (V) -y d/dy ln(V) +1
    x d/dx (ln(U/V))=y d/dy (ln(U/V))+2

    Now make a guess of a function of the type U/V = exp(F(x,y))

    to get:
    x d/dx F = y d/dy F +2

    so you have x d/dx F -y d/dy F =2

    d/dx (xF) - d/dy (yF) = 2

    Don't see how to solve this, though.

    p.s the derivatives above are partial btw.
     
  6. Nov 1, 2011 #5

    Ben Niehoff

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    I would rewrite this as

    [tex]x \partial_x (\log U) - y \partial_y (\log U) - 1 = x \partial_x (\log V) - y \partial_y (\log V) + 1[/tex]

    or

    [tex]x \partial_x (\log U - \log V) - y \partial_y (\log U - \log V) = 2[/tex]

    Define some function [itex]e^f = U/V[/itex] and you have

    [tex]x \partial_x f - y \partial_y f = 2[/tex]

    which should be easy to solve by characteristics. Mathematica gives

    [tex]f = 2 \log x + C x y[/tex]

    for arbitrary C. Then U and V can be any functions satisfying

    [tex]\frac{U}{V} = x^2 e^{C x y}[/tex]
     
  7. Nov 1, 2011 #6

    lavinia

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    Thanks Ben

    What approach would you recommend to learning more about differential equations? Right now i am learning differential geometry.
     
  8. Nov 1, 2011 #7

    Ben Niehoff

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    I don't know, I taught myself the method of characteristics when I started running across problems in my research that needed it.
     
  9. Nov 1, 2011 #8

    MathematicalPhysicist

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    Sometime I wonder why I even bother answering. :-/
     
  10. Nov 12, 2011 #9

    lavinia

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    I thank you for this solution and feel that I should explain where this equation came from.

    To learn some differential geometry I posed the question of what vector fields on a surface can be tangent to geodesics for some Riemannian metric. Computation led to the condition that a unit length vector field V is tangent to geodesics if its Lie bracket with iV, the unit vector field orthogonal to it, is a multiple of iV (and that this multiple satisfies a differential equation along the geodesic that relates it to the Gauss curvature of the metric).

    For the vector field xd/dx - yd/dy in a neighborhood of the origin in the plane, one gets the differential equation in this post. I think your solution shows that the field of hyperbolas y = k/x can be geodesics and that iV can be chosen to be x[itex]^{2}[/itex]d/dx + e[itex]^{-xy}[/itex]d/dy.

    So it seems that a vector field of index -1 around a singularity can be a geodesic flow.

    I think the Gauss curvature is -1 for this example but I haven't yet done the ugly calculation.
     
    Last edited: Nov 12, 2011
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