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[SOLVED] Harmonic oscillator no solution

dwsmith

Well-known member
Feb 1, 2012
1,673
Given \((\mathcal{L} + k^2)y = \phi(x)\) with homogeneous boundary conditions \(y(0) = y(\ell) = 0\) where
\begin{align}
y(x) & = \frac{2}{\ell}\sum_{n = 1}^{\infty} \frac{\sin(k_nx)}{k^2 - k_n^2},\\
\phi(x) & = \frac{2}{\ell}\sum_{n = 1}^{\infty}\sin(k_nx),\\
u_n(x) &= \sqrt{\frac{2}{\ell}}\sum_{n = 1}^{\infty}\frac{\sin(k_nx)}{k^2 - k_n^2},
\end{align}
\(\mathcal{L} = \frac{d^2}{dx^2}\), and \(k_n = \frac{n\pi}{\ell}\).
If \(k = k_m\), there is no solution unless \(\phi(x)\) is orthogonal to \(u_m(x)\).

Why is this?
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
I end up getting
\[
\sum_n\frac{k_m^2 - k_n^2}{k_m^2 - k_n^2}\sin(k_nx)=\sum_n\sin(k_nx)
\]
If \(k_m = k_n\), I have an indeterminant form. If \(k_m\neq k_n\), equality holds. How does orthgonality play a role? For \(k_m\) different from \(k_n\), it seems to not matter unless I am missing someting.