Method of characteristics

  1. 1. The problem statement, all variables and given/known data

    [itex] x \frac{ \partial u}{ \partial x} + y \frac{ \partial u}{ \partial y}= -x^2u^2[/itex]

    2. Relevant equations

    3. The attempt at a solution

    characteristics are given by
    [itex] \frac{ dy}{ dx} = \frac{ y}{ x} [/itex] (a)


    [itex] \frac{ du}{ dx} = -\frac{x^2u^2 }{ x} [/itex] (b)

    So i integrate both equations

    but for (a) do i bring the y across which ends up giving ln(y) = ln(x) + k
    or leave it where it is and i get y = -yln(x) + k

  2. jcsd
  3. hunt_mat

    hunt_mat 1,621
    Homework Helper

    Hello, still doing characteristics? Why not write the characteristic equations as:
    You won't be able to write down a complete solution as you have no initial condition to work from. For your equation for characteristics, your first answer was correct, the characteristics are given by [itex]y=kx[/itex] for k constant.
  4. To expand on hunt_mat's answer:
    Don't we need to separate the variables before integrating?

    1/y dy = 1/x dx

    Hmmm, then again you're using partial deriv. notation, which I've only studied Calc. for, not yet Differential Equations.
  5. thanks for the replies lads. I guess i never total got to grips with methods of chars.!!!
    I also have initial conditions of u(x,1) = x for -infinity<x<infinity.

    y=kx becomes k=y/x

    then from (b) we have [itex] \frac{ du}{ dx} = -\frac{x^2u^2 }{ x}[/itex] which when differentiated gives [itex] u = \frac{ 2}{ x^2} + F(k) [/itex]

    and transfer in our earlier value of k gives [itex] u = \frac{ 2}{ x^2} + F(\frac{ y}{ x}) [/itex]
    im getting confused now i think in different methods?
  6. hunt_mat

    hunt_mat 1,621
    Homework Helper

    There are essentially two ways for the method of characteristics (I recognise your name from a number of MOC posts, did I post my notes on the subject?)
    From one of your calculations you have:
    Integrating this equation shows that:
    \frac{1}{u}=\frac{x^{2}}{2}+F(\xi )
    We now paramatrise the initial data, so take [itex](\xi ,1)[/itex] as the point which the characteristic passes through, this will give the initial values as [itex]u(\xi ,1)=\xi[/itex], evaluating the characteristic at this point yields [itex]1=k\xi[/itex], giving a value for k which can then be inserted back into the equation for the characteristic. We now evaluate [itex]u[/itex] at the point [itex](\xi ,1)[/itex] to obtain
    \frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi )
    From this you can compute [itex]F(\xi )[/itex] and then from there you can substitute for [itex]\xi[/itex] by using the equation of the characteristic.
    Now to find u you have the solution
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