What is Wrong with the Generalized Green Function Equation?

[Dustinsfl]

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
\[
\int_0^{\ell}u_m(x)G_g(x, x')dx = 0.
\]
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?

 

[gtoguy97]

It appears that you have not evaluated the left-hand side of the equation correctly. The expression $\mathcal{L} + k_m^2$ should be evaluated first, before applying it to $G_g(x, x')$. This can be done using the definition of $\mathcal{L}$ provided in the problem statement:\[(\mathcal{L} + k_m^2)G_g(x, x') = \sum_{n\neq m}\frac{-k_n^2u_n(x)u_n(x')}{k_m^2 - k_n^2} + k_m^2\sum_{n\neq m}\frac{u_n(x)u_n(x')}{k_m^2 - k_n^2}.\]The right-hand side of the equation should then be evaluated using the definition of $\delta(x-x')$ provided in the problem statement:\[\delta(x - x') - u_m(x)u_m(x') = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx)\sin(k_nx') - u_m(x)u_m(x').\]Once these expressions are evaluated, they should match, giving you a proof that $G_g$ satisfies the equation.Finally, the condition that $\int_0^{\ell}u_m(x)G_g(x, x')dx = 0$ can be satisfied by noting that $u_m(x)$ is an eigenfunction of $\mathcal{L}$ with eigenvalue $k_m^2$, and so $u_m(x)$ is orthogonal to all other eigenfunctions (which appear in the sum of $G_g(x, x')$). Therefore, the integral of $u_m(x)G_g(x, x')$ over the interval from 0 to $\ell$ is zero.