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Quasi-Linear PDE

  1. Sep 16, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex](Z+e^x)Z_x + (Z+e^y)Z_y = Z^2 - e^{x+y} [/tex]
    Where [tex]Z = Z(x,y)[/tex]

    2. Relevant equations

    Equations of the form

    [tex]PZ_x + QZ_y = R[/tex]
    Where [tex] P = P(x,y,z) [/tex] , [tex]Q=Q(x,y,z)[/tex] , [tex]R=R(x,y,z)[/tex]
    Are solved with the Lagrange method.
    It is possible to write this in the form:
    [tex] \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R} [/tex]
    Where from here we get two equations. We Solve these equations like ODE's, and the Final Solution would be [tex]F(c_1,c_2)=0 [/tex] And [tex]C_1, C_2[/tex] are the integrating constants of the ODE's so [tex]C_1=C_1(x,y,z,c_2)[/tex] and [tex]C_2=C_2(x,y,z,c_1)[/tex]
    It is possible to use the multiplier method, and try to find [tex]\alpha , \beta , \gamma[/tex] (which can be functions)
    To satisfy: [tex] \alpha P + \beta Q + \gamma R =0 [/tex]
    Which would imply
    [tex] (\alpha)dx + (\beta)dy + (\gamma)dz = 0[/tex]
    And to make this simple we look for [tex] \alpha = \alpha(x) , \beta=\beta(y) , \gamma=\gamma(z) [/tex]

    3. The attempt at a solution

    [tex](Z+e^x)Z_x + (Z+e^y)Z_y = Z^2 - e^{x+y} [/tex]

    [tex] \frac{dx}{z+e^x} = \frac{dy}{z+e^y} = \frac{dz}{z^2 - e^{x+y}} [/tex]
    But we can't get 2 solvable equations from here, because it would always involve the third variable, and we couldn't treat it like an ODE.
    So we go to The Multiplier Mehod:
    [tex] \alpha(z+e^x) + \beta(z+e^y) + \gamma(z^2 - e^{x+y}) =0 [/tex]
    Which would imply:
    [tex] \alpha + \beta + \gamma z = 0[/tex]
    [tex] \alpha e^x + \beta e^y - \gamma e^{x+y} = 0[/tex]

    But I haven't been able to find any [tex]\alpha , \beta , \gamma[/tex] that do that...
    I found [tex]\alpha = (z+e^y), \beta = (-z-e^x) , \gamma = 0[/tex]
    Fullfills, but then the equation
    [tex] (z+e^y)dx - (z+e^x)dy = 0 [/tex]
    Cannot be solved (we can't treat the z as a constant, because it is a function of x,y ...

    Thank you
    Last edited: Sep 16, 2009
  2. jcsd
  3. Sep 18, 2009 #2
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