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Homework Help: Method of Characteristics

  1. Jan 13, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm trying to solve the following via the method of characteristics.

    [tex] \frac{\partial P}{\partial t} + k\omega \frac{\partial P}{\partial\omega} = -\frac12 D\omega^2 P [/tex]


    2. Relevant equations
    By the method of characteristics, we assume that [itex] P(\omega, t) [/itex] can be parameterized as [itex] P(\omega(s), t(s) ) [/itex]. Then using the chain rule, we get that

    [tex] \frac{dP}{ds} = \frac{\partial P}{\partial t} \frac{ dt}{ds} + \frac{\partial P}{\partial \omega} \frac{ d\omega}{ds} [/tex]

    So if we have a PDE of the form

    [tex] a(\omega, t)\frac{\partial P}{\partial t} + b(\omega, t) \frac{\partial P}{\partial\omega} = c(\omega, t) [/tex]

    we can rewrite this as a system of first order ODEs

    [tex] \frac{dt}{ds} = a(\omega, t), \quad \frac{d\omega}{ds} = b(\omega, t), \quad \frac{ dP}{ds} = c(\omega, t) [/tex]

    3. The attempt at a solution
    So I allow [itex] \frac{dt}{ds} = 1, t(0) = 0 [/itex] so that [itex] s=t [/itex]. Next I take [itex] \frac{d\omega}{ds} = k \omega [/itex] so that [itex] \omega(s) = \omega(t) = C_1 e^{kt} [/itex]. Finally, I have [itex] \frac{dP}{ds} = -\frac12 D\omega^2 P [/itex] which implies that

    [tex] P(s) = P(t) = C_2 e^{-\frac12 D \omega^2 t} [/tex]

    Now I could substitute my value for [itex] \omega(t) [/itex] into here, but it doesn't give me a correct answer. If I leave [itex] \omega [/itex] as a variable, it still doesn't work. So I figured that [itex] \omega [/itex] is actually a function of s and that should be accounted for, so I got [itex] \frac{dP}{ds} = -\frac12 D\omega^2 P = -\frac12 D C_1^2 e^{2ks} P[/itex] which gives me a solution of

    [tex] P(s) = -C_2 e^{-\frac{DC_1^2 e^{2ks} }{4k} } [/itex]

    But then not only is this not correct, it seems to me that there should also be an [itex] \omega [/itex] dependency somewhere, and I've removed that.
     
  2. jcsd
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