Generalized Green function of harmonic oscillator

Dustinsfl
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Homework Statement


The generalized Green function is
$$
G_g(x, x') = \sum_{n\neq m}\frac{u_n(x)u_n(x')}{k_m^2 - k_n^2}.
$$
Show G_g satisfies the equation
$$
(\mathcal{L} + k_m^2)G_g(x, x') = \delta(x - x') - u_m(x)u_m(x')
$$
where \delta(x - x') = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx')
and the condition that
$$
\int_0^{\ell}u_m(x)G_g(x, x')dx = 0.
$$


Homework Equations





The Attempt at a Solution


From a previous problem, I found
$$
u_n(x) = \sqrt{\frac{2}{\ell}}\sin(k_nx).
$$
I then end up with
\begin{gather}
(\mathcal{L} + k_m^2)G_g(x, x') = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx') - u_m(x)u_m(x') \\
\sum_{n\neq m}^{\infty}\sin(k_nx)\sin(k_nx') = \left(\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx')\right) - \sin(k_mx)\sin(k_mx')
\end{gather}
What is going wrong?

 
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Dustinsfl said:

Homework Statement



The generalized Green function is
$$G_g(x, x') = \sum_{n\neq m}\frac{u_n(x)u_n(x')}{k_m^2 - k_n^2}.$$ Show G_g satisfies the equation
$$(\mathcal{L} + k_m^2)G_g(x, x') = \delta(x - x') - u_m(x)u_m(x')$$ where \delta(x - x') = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx') and the condition that
$$\int_0^{\ell}u_m(x)G_g(x, x')dx = 0.$$


Homework Equations





The Attempt at a Solution



From a previous problem, I found
$$u_n(x) = \sqrt{\frac{2}{\ell}}\sin(k_nx).$$ I then end up with
\begin{gather}
(\mathcal{L} + k_m^2)G_g(x, x') = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx') - u_m(x)u_m(x') \\
\sum_{n\neq m}^{\infty}\sin(k_nx)\sin(k_nx') = \left(\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx')\right) - \sin(k_mx)\sin(k_mx')
\end{gather}
What is going wrong?
Why do you think anything is going wrong? Please explain what you did. I can't tell if you just plugged in the expression for ##\delta(x-x')## to get the first line or if you evaluated the lefthand side and got the righthand side or did something else.
 
The Fourier method to solve differential equations is always to make an ansatz of the unknown function in terms of a Fourier series. In your case
G(x,x')=\sum_{n=1}^{\infty} A_n(x') u_n(x).
Then you apply the operator, \mathcal{L}+k_m^2 to this ansatz, expand the inhomogeneity of the equation also in terms of a Fourier series and then compare the coefficients on both sides, which let's you solve for A_n(x').

Note that in this case, where \lambda=-k_m^2 is an eigenvalue of \mathcal{L}, the inhomoeneity must be perpendicular to the eigenfunction u_m for consistency. That's why you have to subtract the part parallel to this eigenmode. Note also that any solution of the inhomogeneous equation is only determined up to a function proportional to this eigenmode!
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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