Green's function for the wave equation

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SUMMARY

The discussion centers on the application of Green's function in the context of the 1-dimensional wave equation as presented in "Wave Physics" by S. Nettel. The key takeaway is that for the spatial solution ##u(x)## to satisfy boundary conditions, the Green's function ##G(x|x_0)## must vanish at the specified endpoint. This ensures that the integral solution form, $$u(x) = \int G(x|x_0) \kappa(x_0) dx_0 + f(x)$$, adheres to the necessary boundary conditions, where ##\kappa## represents the inhomogeneity and ##f(x)## is the solution to the homogeneous equation.

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dRic2
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Hi, I'm reading "Wave Physics" by S. Nettel and in chapter 3 he introduces the Green's function for the 1-dimensional wave equation. Using the separation of variables method he restricts his attention to the spatial component only. Let ##u(x)## be the spatial solution to the wave equation and ##G(x|x_0)## the Green's function. The author now states the following:

Let us now suppose that it is ##u(x)## that is specified at an end point. Then we must choose ##G(x|x_o)## to vanish at that end point.

I don't really understand this connection. Can someone help me ?

Thanks
Ric
 
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You are looking to write down the solution on the form
$$
u(x) = \int G(x|x_0) \kappa(x_0) dx_0 + f(x)
$$
where ##\kappa## is the inhomogeneity in your wave equation and ##f(x)## solves the homogeneous wave equation and satisfies your boundary conditions. If ##G(x|x_0)## does not vanish at the boundary, then the integral will generally not be zero there and your function will not satisfy the boundary conditions.
 
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In the book I'm reading there is a slightly different version of the solution:

244190


(forget about the ##\frac 1 {\rho}##)

BTW I also do not understand your answer. If, for example, ##k## vanishes at the end point the integral could also be zero and the boundary condition would be verified.
 
dRic2 said:
If, for example, ##k## vanishes at the end point the integral could also be zero and the boundary condition would be verified.
No, this is incorrect. The point is to evaluate ##x## at the boundary, not ##x_0##, which is an integration variable.
 
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Sorry, dumb mistake. Now I get your argument, thanks.
 

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