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Method of characteristics

  1. Sep 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of the following equation
    ut + x2ux = t, u(x,0) = f(x), -inf < x < inf, t > 0

    2. Relevant equations

    3. The attempt at a solution
    Using method of characteristics I get
    du/dt = ut + dX/dtux = ux(dX/dt - x^2) + t
    so along the curve dX/dt = x^2 with x(0) = x0 we get x = -1/(t+x0) and x0 = -1/x - t
    So du/dt = t = -1/x - x0 so
    u(x,t) = -t/x - x0t + f(x0)
    so u(x,t) = t2 + f(-1/x - t)
    but when I check this I don't get the original PDE.
    Someone help.
  2. jcsd
  3. Oct 5, 2011 #2


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    so Im only just starting to understand method of characteristics, but Im gona try to help.

    The first step I believe is to get your characteristic equations




    solving these equations gives

    x=[itex]x^{2}[/itex]s +[itex]x_{0}[/itex]



    from the initial condition [itex]t_{0}[/itex]=0

    so t=s





    so [itex]u_{0}[/itex]=f(x-[itex]x^{2}[/itex]t)
  4. Oct 14, 2011 #3
    The system of ODEs for the characteristic lines is


    The first ODE can be integrated by separation of variables:


    The second ODE is straightforward:


    The third ODE can be solved using the value of [itex]t[/itex] just computed:


    Note that the constants of integration are functions of [itex]r[/itex] since we are integrating w.r.t. [itex]s[/itex]. The general solution is:


    The initial conditions are:


    which we substitute into the general solution


    getting the value for the constants of integration


    that we can substitute into [itex]u[/itex] to get it as a function of [itex]s,r[/itex]:


    In order to get [itex]u[/itex] as a function of [itex]x,t[/itex], we need to invert [itex]x(r,s),t(r,s)[/itex] first


    and then substitute them into the above expression for [itex]u[/itex]:

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