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Method of characteristics and second order PDE.

  1. Jan 19, 2014 #1
    This isn't a homework question per se. Am merely seeking an explanation how the method of characteristics may be applied to a second order PDE. For instance, how is it used to solve utt=uxx-2ut?
     
  2. jcsd
  3. Jan 22, 2014 #2

    HallsofIvy

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    That is a "hyperbolic" equation and, ignoring the lower order, first derivative is [itex]u_{tt}= u_{xx}[/itex] so has "characteristic equation"[itex]t^2= x^2[/itex] or [itex]t= \pm x[/itex] so the "characteristics" are [itex]t- x= constant[/itex] or [itex]t+ x= constant[/itex].

    That tells us that we can simplify the equation by taking p= t- x and q= t+ x as variables instead of x and t. By the chain rule, [itex]u_t= u_pp_t+ u_qq_t= u_p+ u_q[/itex] and then [itex]u_{tt}= (u_p+ u_q)_t= (u_p+ u_q)_p+ (u_p+ u_q)_q= u_{pp}+ 2u_{pq}+ u_{qq}[/itex].

    Similarly, [itex]u_x= u_pp_x+ u_qq_x= -u_p+ u_q[/itex] and then [itex]u_{xx}= (-u_p+ u_q)_x= -(-u_p+ u_q)_p+ (-u_p+ u_q)_q= u_pp- 2u_{pq}+ u_{qq}[/itex].

    So [itex]u_{tt}= u_{xx}+ u_t[/itex] becomes [itex] u_{pp}+ 2u_{pq}+ u_{qq}= u_pp- 2u_{pq}+ u_{qq}+ u_p+ u_q[/itex]. The "[itex]u_{pp}[/itex]" and "[itex]u_{qq}[/itex]" cancel, leaving [itex]4u_{pq}= u_p+ u_q[/itex].
     
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