- #1
Dustinsfl
- 2,217
- 5
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?
\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: