Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

First order linear PDE-the idea of characteristic curves

  1. Sep 19, 2009 #1
    "Consider a first order linear PDE. (e.g. y ux + x uy = 0)
    If u(x,y) is constant along the curves y2 - x2 = c, then this implies that the general solution to the PDE is u(x,y) = f(y2 - x2) where f is an arbitrary differentiable funciton of one variable. We call the curves along which u(x,y) is constant the characteristic curves."
    ===============================

    I don't understand the implication above highlighted in red. The idea of characteristic curves seems to be very important in solving first order linear PDEs, but I am never able to completely understand the idea of it. Why would finding the characteristic curves help us find the general solution to the PDE?

    Thanks for explaining!
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 20, 2009 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  4. Sep 20, 2009 #3
    For that link, they seem to be using a different approach. In particular, the characteristic equation is different and not matching that in my textbook...so I'm getting more confused(?) :(
     
  5. Sep 21, 2009 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I admit it's a little tricky to see. You understand that for a solution u(x,y), the level curves satisfy dy/dx = x/y in your example and these curves are of the form y2-x2=c.

    These happen to be hyperbolas in your example.

    Now any function that has those level curves will be a solution to the PDE because the surface is made up of its level curves. Think of a mountain being the aggregate of its level curves (contour lines). Of course u(x,y) = y2-x2 is such a solution. But that is just one.

    What about u(x,y) = exp(y2-x2)? It has the same type of level curves. Or sin(y2-x2).

    In fact you can take any differentiable function of (y2-x2):

    u(x,y) = f(y2-x2)

    To verify this take

    ux = f'(y2-x2)(-2x)
    uy = f'(y2-x2))(2y)

    dy/dx = - ux/uy = x/y

    so u(x,y) = f(y2-x2) is a solution for any f.

    Hope this helps. Sack time here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: First order linear PDE-the idea of characteristic curves
  1. First order linear PDE (Replies: 8)

Loading...