# [SOLVED]Harmonic oscillator no solution

#### dwsmith

##### Well-known member
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
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.