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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

    syj

    User Avatar

    hey,
    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

    [itex]\frac{dx}{ds}[/itex]=x[itex]^{2}[/itex]

    [itex]\frac{dt}{ds}[/itex]=1

    [itex]\frac{du}{ds}[/itex]=t

    solving these equations gives

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

    t=s+[itex]t_{0}[/itex]

    u=ts+[itex]u_{0}[/itex]

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

    so t=s

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

    u=[itex]t^{2}[/itex]+[itex]u_{0}[/itex]

    u([itex]x_{0}[/itex],0)=f([itex]x_{0}[/itex])

    [itex]x_{0}[/itex]=x-[itex]x^{2}[/itex]t

    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

    [itex]\frac{dx}{ds}(r,s)=x^2[/itex]
    [itex]\frac{dt}{ds}(r,s)=1[/itex]
    [itex]\frac{du}{ds}(r,s)=t[/itex].

    The first ODE can be integrated by separation of variables:

    [itex]\frac{dx}{ds}(r,s)=x^2[/itex]
    [itex]∫\frac{dx}{x^2}=∫ds[/itex]
    [itex]-\frac{1}{x(r,s)}=s+g_1(r)[/itex]
    [itex]x(r,s)=-\frac{1}{s+g_1(r)}[/itex].

    The second ODE is straightforward:

    [itex]\frac{dt}{ds}(r,s)=1[/itex]
    [itex]t=s+g_2(r)[/itex].

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

    [itex]\frac{du}{ds}(r,s)=t=s+g_2(r)[/itex]
    [itex]u(r,s)=(s^2)/2+s*g_2(r)+g_3(r)[/itex].

    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:

    [itex]x(r,s)=-\frac{1}{s+g_1(r)}[/itex].
    [itex]t=s+g_2(r)[/itex]
    [itex]u(r,s)=(s^2)/2+s*g_2(r)+g_3(r)[/itex].

    The initial conditions are:

    [itex]x(r,0)=r[/itex]
    [itex]t(r,0)=0[/itex]
    [itex]u(r,0)=f(r)[/itex]

    which we substitute into the general solution

    [itex]x(r,0)=-\frac{1}{g_1(r)}=r[/itex]
    [itex]t(r,0)=g_2(r)=0[/itex]
    [itex]u(r,0)=g_3(r)=f(r)[/itex]

    getting the value for the constants of integration

    [itex]g_1(r)=-\frac{1}{r}[/itex]
    [itex]g_2(r)=0[/itex]
    [itex]g_3(r)=f(r)[/itex].

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

    [itex]u(r,s)=s^2/2+f(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

    [itex]t=s[/itex]
    [itex]x=-\frac{1}{t-1/r}=\frac{r}{1-rt}[/itex]
    [itex]1-rt=\frac{r}{x}[/itex]
    [itex]1=r(t+\frac{1}{x})[/itex]
    [itex]r=\frac{x}{1+tx}[/itex]

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

    [itex]u(x,t)=t^2/2+f(\frac{x}{1+tx})[/itex].
     
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