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D 'Alembert's solution

  1. Apr 16, 2008 #1
    What is d 'Alembert's solution? (in simple terms)

    What does it mean physically
  2. jcsd
  3. Apr 16, 2008 #2
    You have two waves moving at eachother and passing by.
  4. Apr 16, 2008 #3
    Just to clarify, do you mean the D'Alembert solution for the wave equation or the D'Alembert's method for Newtonian dynamics?

    For the wave equation, I think the intuition is basically like what Cyrus said. D'Alembert's solution says that you get two disturbances which propagate outwards in space and time, but in opposite directions. Imagine taking a jump rope and having two people hold it on either side. If another person comes and kicks the jump rope in the center, you will see that the rope moves outwards towards both people, in two different directions. That's kind of the intuition I think. However, I'd like to hear more people speak about the intuition as far as the general energy/momentum transfer goes.
  5. Apr 17, 2008 #4
    D'Alemberts solution isnt saying anything about momentum transfer. Its simply saying that two waves move past eachother. Its a solution to a PDE.

    It can be shown that energy is conserved starting from the wave equation, but that has nothing to do with D'Alemberts solution.
  6. Apr 17, 2008 #5


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    it's a clean and simple solution to a partial differential equation, called the wave equation, and in one spatial dimension (along with time) the wave equation:

    [tex] \frac{\partial^2 f(x,t) }{ \partial t^2 } = c^2 \frac{ \partial^2 f(x,t) }{ \partial x^2 } [/tex]

    is the physical description of what the displacement of a string, f(x,t), is at point x and at time t. this is a 2nd order differential equation so it will have two linearly independent solutions to it, f+(x,t) and f-(x,t) to it. the subscripts will be explained below.

    the d 'Alembert's solution or " d 'Alembert's formula" is

    [tex] f(x,t) = f_{+}(x,t) + f_{-}(x,t) [/tex]


    [tex] f_{+}(x,t) = f_1(x-ct) [/tex]


    [tex] f_{-}(x,t) = f_2(x+ct) [/tex]

    or more simply

    [tex] f(x,t) = f_1(x-ct) + f_2(x+ct) [/tex]

    and where f1(x) and f2(x) are any two continuous functions of x. f+(x,t) or f1(x-ct) represents a wave moving in the +x direction and f-(x,t) or f2(x+ct) is a wave on the string moving in the -x direction. so pick any two f1(x) and f2(x) and you have a solution to the differential equation above. so from that, there are still an infinite number of solutions available to you and you need more information to get to a particular solution.

    now, if there are boundary conditions on f(x,t), like the string is terminated (like a guitar string) so that f(0,t)=0 and f(L,t)=0 for all t, (L is the length of the string between the termination points (the bridge and nut or fret of a guitar), then some relationship between f1(x) and f2(x) is indicated:

    [tex] f(0,t) = f_{+}(0,t) + f_{+}(0,t) = 0 = f_1(-ct) + f_2(ct) [/tex]

    or [tex] f_1(-x) = -f_2(x) [/tex]


    [tex] f(L,t) = f_{+}(L,t) + f_{+}(L,t) = 0 = f_1(L-ct) + f_2(L+ct) [/tex]

    or [tex] f_1(L-x) = -f_2(L+x) [/tex]

    That gives you some symmetry properties. If it's a guitar string, and it's a slow, careful pluck (at t=0) where the string is deflected by the pick but is at rest at t=0 when the pick is released, the string's initial shape is assumed known

    [tex] f(x,0) = f_1(x) + f_2(x) [/tex]

    it turns out that if the string's velocity at time 0 is zero,

    [tex] \frac{\partial f(x,t) }{\partial t}\Big|_{t=0} = 0 [/tex]

    then f1(x) = f2(x) and

    [tex] f(x,t) = ( f(x-ct,0) + f(x+ct,0) )/2 [/tex]

    it's not the simple answer. but reasonably complete.

    it means that the wave equation can be solved into a form of adding two wave functions together each representing waves going in opposite directions on the string and having equal wave speed.
  7. Apr 23, 2008 #6
    Thanks for your clarification
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