A question about Green's function

  • Thread starter karlzr
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  • #1
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There is a second order nonhomogeneous equation of motion with nonzero initial condition given at ##t=-\infty##:
##D^2 y(x)=f(x)## with ##y(-\infty)=e^{-i x}##
where I have used the shorthand notation ##D^2## for the full differential operator. Also I have the two solutions ##y_1(x)## and ##y_2(x)## to the homogeneous equation with ##y_1(x \to -\infty) \approx e^{-ix}## and ##y_2(x \to -\infty) \approx e^{ix}##. So how do I construct the particular solution using ##y_1(x)## and ##y_2(x)##?

I know Green's function can be constructed using the two homogeneous solutions. So the naive solution I got is
##y(x)=y_1(x)+\int^x_{-\infty} dx' G(x,x')f(x')## with
##G(x,x')=\frac{y_1(x)y_2(x')-y_2(x)y_1(x')}{W(y_2(x'),y_1(x'))}##.

Is it wrong? since I get some ridiculous result (divergence) when I use this solution to do calculation, because the lower bound is ##-\infty##.
 

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  • #2
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I know Green's function can be constructed using the two homogeneous solutions. So the naive solution I got is
##y(x)=y_1(x)+\int^x_{-\infty} dx' G(x,x')f(x')## with
##G(x,x')=\frac{y_1(x)y_2(x')-y_2(x)y_1(x')}{W(y_2(x'),y_1(x'))}##.

Is it wrong? since I get some ridiculous result (divergence) when I use this solution to do calculation, because the lower bound is ##-\infty##.
Where are you getting this out of curiosity? There are other ways to obtain the Green's function, at least.
 
  • #3
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Where are you getting this out of curiosity? There are other ways to obtain the Green's function, at least.
Actually I obtained this expression from the approach of variation of parameters. It plays the role of Green's function in this case.
 

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