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: Wave equation help

  1. Mar 3, 2010 #1
    how do i show that v(x,t)=u(x,t)-ue(x) satisfies the wave equation? =(

    i get that ue(x)=gx2/2c2 + ax + b where a and x are just constants but how does this satisfy the wave equation?
  2. jcsd
  3. Mar 15, 2010 #2


    User Avatar

    Can you clarify your question? I don't understand what ue(x) actually is in terms of the function u(x,t)? Is u(x,t) an arbitrary function, or one that itself satisfies the wave equation? Perhaps I'm missing something obvious. :)

    Generally to show that a function satisfies a DE, you'll need to show that its derivatives actually have the relationship in question. So I'd start by differentiating your definition of v twice with respect to each variable.
  4. Mar 16, 2010 #3
    thanks for taking a look. still stumped. Ue(x) is the sagged equilibrium position ( when Q(x,t)=-g and the boundary conditions are u(0)=0 and u(L)=0 or fixed boundary/ends of the string). i hope that clarifies it a bit?
  5. Mar 16, 2010 #4


    User Avatar

    Ah, I see. :) So you've got a string with a uniform loading (which you've called Q) along the x-direction, sagging in the shape of a parabola because of that (note that in reality, if this loading were due to the weight of the string, the equilibrium shape would be a catenary, not a parabola).

    And you're trying to prove that if u(x,t) is a solution to the wave equation on an equivalent (same boundary conditions) unloaded string, then:

    [tex]u(x,t) - u_e(x)[/tex]

    Will be a solution to the wave equation on the loaded string.

    Did I understand the question correctly?

    If so, then look up the superposition theorem for linear differential equations (such as the wave equation). This states that the sum of two solutions to a DE will also be a solution to the DE - So in this case, if both ue(x) and u(x,t) are solutions then it immediately follows that v(x,t) is a solution.

    If you also need to show that it's the solution that you're looking for, then you'll need to check it satisfies the appropriate boundary/initial conditions.
  6. Mar 16, 2010 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are we supposed to be psychic? Why don't you give us the equation you are trying to satisfy? How does Q enter in to it?
  7. Mar 16, 2010 #6
    wow you totally got the question. thanks alot. i see it clearly now :)

    LCKurtz: Q(x,t) is just force acting on the string, so gravity in most cases. i got the answer now, sorry for not being more precise.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook