If $u_m, u_n$ are orthogonal solutions to the Sturm-Liouville problem, then $u'_m, u'_n$ are ortogonal

Question

$u_n, u_m$ are functions such that: $$\frac{d}{dx}(p(x)\frac{du_n}{dx})+\lambda_nw(x)u_n(x)=0$$ $$\frac{d}{dx}(p(x)\frac{du_m}{dx})+\lambda_mw(x)u_m(x)=0$$ for $x \in [a,b]$, $u_m,u_n$ are orthogonal, $\lambda_m \ne \lambda_n$. Show that for adequate initial conditions, $u'_m(x)$ and $u'_n(x)$ are orthogonal with the weight function $p(x)$.

Answer 1

Suppose we have boundary conditions $u(a) = u(b) = 0$. Then, multiplying the first equation by $u_m$ and through integration by parts, we obtain

$$\int_{a}^{b} p(x) \frac{du_n}{dx} \frac{du_m}{dx}~\mathrm d x = \lambda_n \int_{a}^{b} w(x)u_n(x)u_m(x)~ \mathrm d x .$$ Analogously, by multiplying the second equation by $u_n$, we obtain $$\int_{a}^{b} p(x) \frac{du_n}{dx} \frac{du_m}{dx}~\mathrm d x = \lambda_m \int_{a}^{b} w(x)u_n(x)u_m(x)~ \mathrm d x .$$ As $\lambda_n \neq \lambda_m$, this is only possible if $\int_{a}^{b} w(x)u_n(x)u_m(x)~ \mathrm d x = 0$.