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Homework Help: First Order PDE Cauchy problem Using Method of Characteristics

  1. Jul 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Ok, so I can get through most of this but I can't seem to get the last part... Here is the problem

    [tex]xU_x + (y^2+1)U_y = U-1; U(x,x) = e^x[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Characteristic equations are:

    [tex]\frac{dx}{x} = \frac{dy}{y^2+1} = \frac{dU}{U-1}[/tex]

    Solving the first and third gives:

    [tex]\frac{U-1}{x} = c_1[/tex]

    The first and second equation yield:

    [tex]tan^{-1}(y) - lnx = c_2[/tex]

    Put the two together in the form

    [tex]c_1 = f(c_2)[/tex]

    [tex]\frac{U-1}{x} = f(tan^{-1}(y) - lnx)[/tex]

    Sub in the Cauchy data and you get

    [tex]\frac{e^x-1}{x} = f(tan^{-1}(x) - lnx)[/tex]

    Now how do I find what my arbitrary function f is? I have spent hours on this. Is there something that relates inverse tan to natural log? Arrggghhhh!

    Thanks for any help.
     
  2. jcsd
  3. Jul 16, 2008 #2
    Where did the function f come from? The way I see it, you end up with three equations:

    [tex]\begin{cases}
    x=C_1 e^{\tan^{-1}y}\\
    U=C_2 x+1\\
    U=C_3 e^{\tan^{-1}y}+1
    \end{cases}[/tex]

    Reconciling those, you get
    [tex]U=Cxe^{\tan^{-1}y}+1[/tex]



    Oh, and there is a relation between inverse tan and natural log: http://functions.wolfram.com/ElementaryFunctions/ArcTan/02/0001/

    [tex]\tan^{-1}z={i \over 2}\ln{1-iz \over 1+iz}[/tex]
     
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