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Homework Help: Hyperbolic PDE, Cauchy-type problem

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the equation
    [itex]4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy} - \frac{2y}{1+y^2} (2u_x - u_y) = 0[/itex]

    Find the solution [itex]u(x,y)[/itex] which satisfies [itex]u(x,0) = g(x)[/itex], and [itex]u_y(x,0) = f(x)[/itex], where [itex]f, g \in \mathcal{C}^2(\mathbb{R})[/itex] are arbitrary functions.

    2. Relevant equations

    I used the discriminant method to show that this PDE can be categorized as hyperbolic. I found that the characteristics are [itex]s= - \frac{2}{3}y^3 +x[/itex] and [itex]t = 2y + x[/itex]. I know the general solution is the form [itex]u(x,y) = F( - \frac{2}{3}y^3 +x) + G(2y + x)[/itex]. (Depending on algebra of finding the characteristics, this may be stated in a different way.)

    3. The attempt at a solution

    There are other problems I have done without such abstract conditions, so I followed the general solution style I have been using. NAMELY:

    Find [itex]u_y[/itex]: [itex]-2y^2 F'( - \frac{2}{3}y^3 +x) + 2G'(2y+x)[/itex]. Then, look at the initial values.

    [itex]u(x,0) = F(x) + G(x) = g(x)[/itex]
    [itex]u_y (x,0) = 0 + 2 G'(x) = f(x)[/itex]

    Since [itex]u_y[/itex] has already isolated one of the general functions, I work with it. I integrate with respect to [itex]x[/itex]. Ultimately, I get

    [itex]G(x) = \frac{1}{2} \int f(x) \; dx [/itex]

    Solving for [itex]F(x)[/itex] with this information, I get

    [itex]F(x) = g(x) - \frac{1}{2} \int f(x) \; dx [/itex]

    At this point I've reached a "SO WHAT?" stage. In the previous problems, the function was simply a multiplication factor that was easy to manipulate. I found the multiplication factor and then I multiplied the original information by it to get a [itex]u(x,y)[/itex].

    I feel that I should calculate:
    [itex]F(-\frac{2}{3} y^3 + x) = g(-\frac{2}{3} y^3 + x) + \frac{1}{2} \int -\frac{2}{3} y^3 + x \; d(-\frac{2}{3} y^3 + x)[/itex]
    [itex]G(2y+x) = \frac{1}{2} \int 2y + x \; d(2y + x)[/itex]

    Yet, I feel like this is going to get me nowhere.
  2. jcsd
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