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Method of characteristics for linear PDE's (variable coefficients)

  1. Jul 30, 2008 #1
    I was going through an inroductory book on PDE's and at one point they proceed with little show of work. I have problem with equation [itex] -yu_x + xu_y = u [/itex].
    Characteristics for this equation are [itex] x_t = -y, y_t = x, u_t = u [/itex].

    So far it is clear, but now books states that solution of first characteristic is [itex] x(t,s) = f_1(s)sin(t) + f_2(s)cos(t) [/itex], which is perplexing to me, I would just integrate righthand side treating x or y as constants (we are integrating with respect to t). Any suggestion?
     
  2. jcsd
  3. Jul 31, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Defconist! Welcome to PF! :smile:

    Nooo … x and y depend on t, so if you vary t, then you must vary x and y … they're not constants!

    Hint: xt = -y, yt = x,

    means that xtt = -x. :smile:
     
  4. Jul 31, 2008 #3
    Oh, I get it. It is a system od ODE's because the in y the second equation is the same as y in the first one... It's easy to see why I missed that. It is a possibility I feared from the very beginning. Anyway, thanks for getting me from this predicament. :)
     
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