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Intuitively d'Alembert's solution to 1D wave equation

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ralqs
#1
Oct10-11, 07:29 PM
P: 99
D'Alembert's solution to the wave equation is
[tex]u(x,t) = \frac{1}{2}(\phi(x+ct) + \phi(x-ct)) + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(\xi)d\xi[/tex] where [itex]\phi(x) = u(x,0)[/itex] and [itex]\psi(x) = u_t (x,0)[/itex]. I'm trying to understand this intuitively. The first term I get: a function like f = 0 (x/=0), = a (x=0) will "break into two functions" and become f = a/2 (x = +/- ct), = 0 (x /= +/- ct). But I can't see how the integral term comes about. Does anyone here have a good physical intuition about this? Thanks.
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lurflurf
#2
Oct11-11, 08:48 AM
HW Helper
P: 2,264
The integral comes from the initial conditions. In particular the fact that we are given a derivative of our solution so our solution must be an integral of what we are given. Think of it in two steps.
We have the operator

(Dt)2-(c Dx)2=(Dt+c Dx)(Dt-c Dx)

In the factored form it is easy to see our solution is the sum of two parts. In one x+ct is held constant and in the other c-ct is constant.

f(x+ct)+g(x-ct)

Then apply the Cauchy boundary condition and preform some elementary algebra to express the solution in terms of the Cauchy data.


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