1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Solve PDE Method of Characteristics help

  1. Sep 20, 2011 #1
    The problem: Solve for u(x,y,z) such that

    [tex] xu_x+2yu_y+u_z=3u\; \;\;\;\;u(x,y,0)=g(x,y) [/tex]

    So I write

    [itex]\frac{du}{ds}=3u \implies \frac{dx}{ds}=x,\; \frac{dy}{ds}=2y\;\frac{dz}{ds}=1 .[/itex]

    Thus [tex]u=u_0e^{3s},\;\;x=x_0e^{s}\;\;y=y_0e^{2s}\;\;z=s+z_0 [/tex]
    but from here I can't figure out what to do, there are several ways I can write s....I have only done the method of characteristics before with two variables, and those are pretty much the only examples I can find. Can someone help please?
  2. jcsd
  3. Sep 20, 2011 #2


    User Avatar
    Science Advisor

    If you have done the method of characteristiced with two variables it pretty much generalizes to three variables directly- write out the characteristics in terms of x, y, and z and use those to define new variables.

    If [itex]x= x_0e^s[/itex] and [itex]y= y_0e^{2s}[/itex], you have [itex]s= ln(x/x_0)= (1/2)ln(y/y_0)[/itex] so that ln(x/x_0)- ln(y^{1/2}/y_0^{1/2})= ln((y_0^{1/2}/x_0)(x/y^{1/2}= 0 which then gives [itex]x/y^{1/2}= constant[/itex]. Each curve with different constant is a characteristic. With [itex]z= s+ z_0[/itex] that gives also [itex]s= z-z_0= ln(x/x_0)[/itex], have [itex]x/x_0= e^{z-z_0}= e^{-z_0}e^z[/itex] so [itex]x= (constant)e^z[/itex] or [itex]xe^{-z}= constant[/itex]. Putting [itex]s= z-z_0= ln(y^{1/2}/y_0^{1/2})[/itex], we get [itex]y^{1/2}e^{-z}= constant[/itex].

    We want to use those "characteristics" or "characteristic curves" as axes: let [itex]u= x/y^{1/2}[/itex], [itex]v= xe^{-z}[/itex], and [itex]w= y^{1/2}e^{z}[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook