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Unsolvable PDE?

  1. Sep 30, 2011 #1
    I just had a 'quiz' in my PDE class today and there was a problem my friends and I are convinced has no explicit solution. I want to know if maybe we are doing something wrong?

    1. The problem statement, all variables and given/known data
    [tex] (x+y)u_{x} + yu_{y} = 0 [/itex]

    [tex] u(1,y) = \frac{1}{y} + ln(y) [/itex]


    2. Relevant equations
    ...


    3. The attempt at a solution
    Here was my approach via method of characteristics:

    [tex] \frac{dy}{dt} = y \to y(t) = Se^{t} [/tex]
    [tex] y(t=0) = S [/tex]

    [tex] \frac{dx}{dt} = x+y [/tex]
    [tex] x'-x = Se^{t} [/tex]
    Via integrating factor we arrive at:
    [tex] x(t) = Ste^{t} + g(s)e^{t} [/tex]
    [tex] x(t=0) = 1 \to 1 = g(s) [/tex]

    Thus we are left with:

    [tex] x(t) = Ste^{t} + e^{t} [/tex]
    and
    [tex] y(t) = Se^{t} [/tex]

    You can check for yourself that the following is true:

    [tex] \frac{dU(x(t),y(t))}{dt} = (x+y)u_{x} + yu_{y} = 0 [/tex]

    Thus our general solution is:

    [tex] U(t) = f(s) [/tex]

    A general function of s.

    However, we have found that it is impossible to explicitly solve for s. Am I missing something?
     
  2. jcsd
  3. Oct 16, 2011 #2
    I suspect the problem cannot be solved. Along the initial line [itex]y=1[/itex] you can differentiate the initial conditions and get

    [itex]u_y(1,y)=-1/y^2+1/y=(y-1)/y^2=0[/itex]

    that you can substitute back into the PDE in order to obtain [itex]u_x[/itex]

    [itex](x+1)u_x=0 → u_x=0 [/itex]

    which means [itex]u(x,y)=u(y)[/itex]. Hence, [itex]u(x,y)[/itex] should be equal to the initial conditions everywhere. But this is not true, as one can check substituting
    [itex]u_x=0[/itex] into the PDE:

    [itex]yu_y=y(1-y)/y^2=0[/itex]

    true only when [itex]y=0[/itex] or [itex]y=1[/itex].

    I am not an expert of the method of characteristics, but I believe the curve on which the initial conditions are given is "noncharacteristic".
     
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