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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?


  2. jcsd
  3. Apr 12, 2007 #2


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    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.

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

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


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    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)?
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