Let $u(s,\lambda)$ solves $$\frac{d}{ds} \left( p(s) \frac{du}{ds}(s) \right) - q(s)u(s) + \lambda u(s) = 0, \ s \geq 0$$ with boundary condition $$u(0,\lambda) = -\sin(h), \qquad p(0)u_s(0,\lambda) = \cos(h)$$ Here, we assume that $p(s) > 0$ on $[0,+\infty)$, $p \in C^1([0,+\infty))$ and $q \in C^0([0,+\infty))$.
If $p \equiv 1$, one can show that for $\lambda \in \mathbb C \setminus \{0\}$ and $\alpha \in \mathbb C \setminus \{0\}$ a square-root of $\lambda$, then $$\phi(s,\lambda) = -\cos(\alpha s) \sin(h) + \sin(\alpha s) \dfrac{\cos(h)}{\alpha} + \dfrac{1}{\alpha} \int_0^s \sin(\alpha(s-t))q(t)\phi(t,\lambda)dt, \ s \geq 0$$
Here, no other solution and no Green function appear and we have a factor $1/\alpha$, which is quite important to me.
If $p \in C^2([0,+\infty))$, then one can transform the above Sturm-Liouville equation in a canonical form where $p \equiv 1$ and get a similar integral equation.
But do we have a similar integral equation when we only assume $p \in C^1([0,+\infty))$ ?
I've tried replicating the proof of the above formula, but we don't get as good cancellations.
Let $$ u(x) = v\left(\int_0^x\frac{1}{p(y)}dy\right). $$ This is reasonable because $\int_0^x\frac{1}{p(y)}dy$ is well-defined for $x \ge 0$, and is a strictly increasing function of $x$, which will allow $u$ to be eliminated in favor of $v$ with the interval of definition for $v$ being $[0,\int_0^{\infty}\frac{1}{p(y)}dy)$. Then $u'(x)=\frac{1}{p(x)}v'(\int_0^x 1/p(y)\,dy)$ and $$ \frac{d}{dx}\left(p(x)\frac{du}{dx}\right)=\frac{d}{dx}\left(v'(\int_0^x1/p(y)dy)\right)=\frac{1}{p(x)}v''(\int_0^x1/p(y)dy) $$ The new equation for $v$ becomes $$ v''(\int_0^x1/p(y)dy)-p(x)q(x)v(\int_0^x 1/p(y)dy)+\lambda p(x)v(\int_0^x1/p(y) dy) = 0. $$ Let $r(x)$ be the inverse function of $\int_0^x1/p(y)dy$. Then $$ v''(w)-p(r(w))q(r(w))v(w)+\lambda p(r(w))v(w)=0. $$