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Solving S.O. PDE by transforming to normal form

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Transform to normal form and solve:
    1) [tex] u_{xx}+u_{xy}-2u_{yy} = 0[/tex]


    2. Relevant equations

    Normal form: [tex]Au_{xx}+2Bu_{xy}+Cu_{yy}[/tex], hence, [tex] A = 1, B = \frac{1}{2}, C = -2[/tex].
    Since [tex]AC-B^2 = -2.25 < 0[/tex] this is a hyperbolic equation.
    Want to transform it by setting

    [tex]v = \Phi(x,y), z = \Psi(x,y)[/tex] where [tex]\Phi = const[/tex] and [tex]\Psi = const[/tex] are solutions [tex]y=y(x)[/tex] of

    2) [tex]A(y')^2-2By'+C = 0[/tex]


    3. The attempt at a solution

    From 2) I find

    [tex](y'-2)(y'+1) = 0[/tex], giving me [tex]y_1 = 2x+C, y_2 = -x+C[/tex]. From this I choose choose [tex]v = 2x-y, z = x + y[/tex] as the transformations for the variables.

    Finding the new expressions:

    [tex]u_{xx} = (u_x)_x = ..... = 4u_{vv}+4u_{zv}+u_{zz}[/tex]
    [tex]u_{xy} = (u_x)_y = .... = -2u_{vv}+u_{zv}+u_{zz}[/tex]
    [tex]u_{yy} = (u_y)_y = .... = u_{vv}-2u_{zv}+u_{zz}[/tex]

    Inserting these into 1) I get that

    [tex]9u_{zv} = 0 \leftrightarrow u_{zv} = 0[/tex], yielding [tex]u(v,z) = f_1(v)+f_2(z) \rightarrow u(x,y) = f_1(2x-y)+f_2(x+y)[/tex].

    I'm not sure if this is correct, hence, I hope someone could let me know if I'm onto something or way off :)
     
    Last edited: Jun 24, 2010
  2. jcsd
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