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Answer need checking!

  1. Apr 28, 2006 #1
    I think I've solved this problem, but just need someone to check my answers.

    [tex]u_{x}u+u_{t}=2[/tex]

    With initial data

    [tex]u(x,0)=f(x)[/tex]

    Use the method of charactersitics [tex]u(x,t)=u(\xi,\tau)[/tex], we get

    [tex]\tau_{t}=t[/tex], [tex]x_{u}=u[/tex] and [tex]u_{\tau}=2[/tex].

    So, by choice of our initial condition,

    [tex]\tau=t[/tex] and [tex]u=2t + f(\xi)[/tex].

    Since [tex]\xi=\xi(x,t)[/tex]. We have

    [tex]u(x,t)=u(\xi(x,t),t)[/tex]

    At [tex]t=0[/tex] we have [tex]u(x,0)=u(\xi(x,0),0)[/tex], therefore

    [tex]x=\xi(x,0)[/tex]

    Since we do not know what [tex]f(x)[/tex] is, our solution for [tex]u(x,t)[/tex] is just

    [tex]u(x,t)=2t+f(\xi(x,t))[/tex]

    with initial condition

    [tex]\xi(x,0)=x[/tex]

    Is this right??
     
  2. jcsd
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