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!

D'Alembert's Solution

  1. Apr 12, 2007 #1
    I have to solve the 1-dimensional wave equation using D'Alembert's solution.

    I know the solution has the forum 1/2[f(x-ct) + f(x+ct)] + 1/2C*integral(g(s)ds)

    and i have two intial conditions f(x) and g(x).... do i just plug in (x-ct) where i see x etc...?

    also what is g(s) in the integral?

    Thanks

    Dan
     
  2. jcsd
  3. Apr 12, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You've written this very confusingly! You have f used in at least two different ways and g used in two different ways.

    What you are giving as the "D'Alembert solution" is too specialized.

    The general D'Alembert solution to the wave equation is [itex]\phi (x,t)= F(x+ ct)+ G(x- ct)[/itex] where F and G can be any two twice differentiable functions of a single variable. Then [itex]\phi_t(x,t)= cF'(x+ct)- cG'(x- ct)[/itex]. If one initial condition is [itex]\phi_t(x,0)= 0[/itex], then it follows that [itex]F'(x)= G'(x)[/itex] so F and G differ by a constant- in that case, [itex]\phi (x,t)= (1/2)(F(x+ct)+ F(x-ct))+ constant[/itex].

    But that is not the case here. (It is also confusing to say just "initial values f(x) and g(x)" since functions are not "conditions". I assume you mean [itex]\phi (x,0)= f(x)[/itex] and [itex]\phi_t(x,0)= g(x)[/itex].)

    Then you have [itex]\phi (x,0)= F(x)+ G(x)= f(x)[/itex] and [itex]\phi_t(x,0)= cF'(x)- cG'(x)= g(x)[/itex]. Solve those two equations for F and G.
     
  4. Apr 12, 2007 #3
    Yes, sorry to be confusing but your assumptions were right...

    The initial conditions are #(x,0) = e^(-x^2) and #'(x,0) = 2cxe^(-x^2)...

    I wasn't sure how to use these to get to the solution but it seems like i need to solve those two equations.

    Thanks!
     
  5. Apr 12, 2007 #4
    I have got the answer #(x,t) = e^-((x-ct)^2)

    Am i right?
     
  6. Apr 12, 2007 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, it's easy to check, isn't it? Calculate #xx and #tt and see if it satisfies the differential equation. Is #(x,0)= e^(-x^2)?
    Is #_t(x,0)= -2cxe^(-x^2)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?