1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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 partial wave eqaution, one boundary and one initial condit

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data


    Solve

    [tex] \frac{\partial{w}}{\partial{t}} + c \frac{\partial{w}}{\partial{x}} =0 \hspace{3 mm} (c>0) [/tex]



    for x>0 and t>0 if

    [tex] w(x,0) = f(x) [/tex]
    [tex] w(0,t) = h(t) [/tex]


    2. Relevant equations

    3. The attempt at a solution
    I know how to solve for the conditions separately and that would give
    [tex] w(x,t) = f(x-ct) [/tex] and
    [tex] w(x,t) = h(t-\frac{1}{c}x) [/tex]

    but how do you solve it for both? And when does x>0 or t>0 matter?
     
    Last edited: Oct 13, 2013
  2. jcsd
  3. Oct 13, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Let's not call your solution ##w(x,t) = f(x-ct)## because you are using ##f## in the initial condition. So begin by stating your general solution is ##w(x,t) = g(x-ct)## for arbitrary, but as yet unknown, ##g##. Now what happens when you apply your first initial condition to that. Does it tell you what ##g## must be? Then continue...
     
  4. Oct 13, 2013 #3
    Now I am confused, if I start out with [tex] g(x-ct) [/tex] then applying the first initial condition [tex] w(x,0) = f(x) =g(x-0)=g(x) [/tex]. So [tex]g(x-ct)[/tex] must be [tex]f(x-ct)[/tex], right?
     
    Last edited: Oct 13, 2013
  5. Oct 13, 2013 #4
    I might have overlooked the fact that I have to give a solution for : [itex] x>0[/itex] and [itex] t>0[/itex]
    So [itex] g(x-ct)=f(x-ct) [/itex] is only valid for [itex]x-ct>0[/itex] and [itex]g(t-\frac{1}{c}x)=h(t-\frac{1}{c}x )[/itex] is only valid for [itex]t-\frac{1}{c}x>0[/itex]

    I think I have the right visual picture in my head now. So, in the [itex]x,t[/itex] plane my solution only considers the first quadrant of the plane.

    And the triangle made by the lines [itex] x=ct [/itex], the [itex]x[/itex]-axis and the line [itex]x=\infty[/itex] is defined by the initial condition at the positive [itex]x[/itex]-axis.
    And the triangle made by the lines [itex] x=ct [/itex], the [itex]t[/itex]-axis and the line [itex]t=\infty[/itex] is given by the boundary condition at the positive [itex]t[/itex]-axis.

    So the solution would be:
    [tex]w(x,t) = f(x-ct) \hspace{3 mm} (x>ct)[/tex]
    [tex]w(x,t) = h(t-\frac{1}{c}x) \hspace{3 mm} (x<ct)[/tex]
    This would be ok right?
     
    Last edited: Oct 13, 2013
  6. Oct 13, 2013 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, I think that does it.

    Disclaimer: I'm not a PDE expert, so if any such experts disagree, speak up now....:uhh:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted