The boundary value problem
\begin{align} &\frac{\mathrm{d} }{\mathrm{d}x } \left( p(x) \frac{\mathrm{d} y(x)}{\mathrm{d}x } \right) + q(x) y(x) = f(x), \quad a \leq x \leq b \nonumber\\ &y(a) = 0,\quad y(b) = 0 \end{align}
can be solved by Green’s function
\begin{align} y(x) = \int_a^b G(x,\xi) f(\xi) \mathrm{d}\xi. \end{align}
The Green’s function can be constructed from two independent solutions of the homogeneous problem $y_1(x)$ and $y_2(x)$ satisfying $y_1(a)=0$, $y_2(b)=0$ and $y_1(b)\neq 0$, $y_2(a) \neq 0$
\begin{align} \label{eq:green} G(x,\xi) = \begin{cases} \frac{y_1(\xi) y_2(x)}{p(\xi)W(\xi)}, & a \leq \xi \leq x, \\ \frac{y_1(x) y_2(\xi)}{p(\xi)W(\xi)}, & x \leq \xi \leq b. \end{cases} \end{align}
$W(x) = y_1(x)y'_2(x) - y_2(x) y'_1(x)$ is the Wronskian. On the other hand, the Green’s function can be expressed in the form of an eigenfunction expansion. Denote
\begin{align} M = \frac{\mathrm{d} }{\mathrm{d}x } \left( p(x) \frac{\mathrm{d} }{\mathrm{d}x } \right) + q(x). \end{align}
The Green’s function is
\begin{align} G(x,\xi) = \sum_{n=1}^{\infty} \frac{\phi_n(x) \phi_n(\xi)}{\lambda_{n}}, \end{align}
where $M \phi_n(x) = \lambda_n \phi_n(x)$. Assume all the eigenvalues are positive $\lambda_n > 0$. Consider the square root of the Green’s function in the sense of
\begin{align} H(x,\xi) = \sum_{n=1}^{\infty} \frac{\phi_n(x) \phi_n(\xi)}{\sqrt{\lambda_{n}}}. \end{align}
The question is that can $H(x,\xi)$ be expressed by $y_1(x)$ and $y_2(x)$ like the case of $G(x,\xi)$?
