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Sketching the solution to the IVP (Characteristic curves)

  1. Oct 2, 2012 #1

    Leb

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    1. The problem statement, all variables and given/known data
    Sketch the solution to the IVP
    [itex]u_{t}+uu_{x}=0 \\ u(x,0) = e^{-x^{2}}[/itex]


    2. Relevant equations
    Monge's equations [itex]\frac{dt}{d\tau} = 1 \\ \frac{dx}{d\tau}=u \\ \frac{du}{d\tau}=0[/itex]


    3. The attempt at a solution
    I think I do not really need the Monge's equation, but I have no idea, how to sketch the wave profile. Could someone please explain it to me step by step ?
    P.S.
    I also got stuck on solving the Monge's equations. Particularly in the part where [itex]dx=u(x,t)d\tau[/itex] I tried taking [itex]t_{0} = 0[/itex] which gave me [itex]t=\tau[/itex] and also taking [itex]u(x,t)=u_{0}=F({\sigma})=const.[/itex] which makes it independent of t. Pluging it back in [itex]dx=u_{0}dt --> x = u_{0}t + x_{0}[/itex] where [itex]x_{0} =x'*u_{0}=const. [/itex] Then I blindly assume [itex]x_(0) = \sigma[/itex] which then gives me that the Characteristic Curves should be of the form [itex]x=u_{0}t+\sigma[/itex]. Now, since we are given that at t=0 [itex]u(x,0)=e^{-x^{2}}[/itex] and [itex]u(x,0)=u_{0}(\sigma)=e^{-\sigma^{2}}[/itex]. So from this, I naively deduce, that each characteristic passes through [itex](\sigma,0)[/itex] with gradient [itex]e^{\sigma^{2}}[/itex]. Can someone point out which assumptions/steps are wrong ? Thanks.
     
  2. jcsd
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