Can someone help explain a proof from Feller paper: The Parabolic Diffusion Equation and associated Semi-groups for transformations? When I read the proof of lemma 8.2 and lemma 9.1 written in the paper. I do not feel comfortable about the proofs.
For Lemma 8.2, it states that if
(a) If there exists a solution $p_{1}(x)$ of the ODE $a(x)p^{''}+b(x)p^{'}+c(x)p=kp$ with $p_1(x_{0})=p_1(x_{1})=0$, then every solution has a zero somewhere in $[x_{0},x_{1}]$
(b) If no solution of the ODE $a(x)p^{''}+b(x)p^{'}+c(x)p=kp$ with $p_1(x_{0})=p_1(x_{1})=0$, then there exists at least one positive solution
The proofs are listed below:

For Lemma 9.1, it consider the ODE $\lambda u+\Omega u=0$, where $\Omega=\frac{d^2}{dx^2}+b(x)\frac{d}{dx}$ and $\lambda>0$, $b(x)$ is continuous in $(r_{1},r_{2})$ where $(-\infty\leq r_{1}<0<r_{2}\leq \infty)$

Indeed, I do not understand quite well about the proofs. i.e. how one can make the corresponding conclusion. Here, I would like to emphasize which parts make me feel uncomfortable.
lemma 8.2 (a) How one can assume that $p_{1}(x)>0$? There should be three cases in total: $p_{1}(x)>0$, $p_{1}(x)<0$ and $p_{1}(x)>0$ & $p_{1}(x)<0$ for some intervals. The proof only shows the first case, how about the third case? Also, how can he come up with the last statement?
lemma 8.2 (b) How one can assume that $p_{1}(\zeta)=0$? Why does the author say that if it is not the case, nothing is to be proved? Also, how to come up with $p_{1}(x)>0$ for $x>\zeta$ and $p_{1}(x)<0$ for $x<\zeta$? Also, why the last statement make sense?
Lemma 9.1 What does the term $u_{s}(x)$? This term is not clear which make the whole proof not clear to me.
Could someone explain the proof in details?
 A: 8.2a
First, if $p_1(x)$ vanishes at some $x_2$ between $x_0 < x_2 < x_1$, then we can focus on the interval $[x_0,x_2]$ instead of the original $[x_0,x_1]$. For if we can show every other solution vanishes somewhere between $[x_0, x_2]$, the conclusion of the original statement also holds. 
By iterating the above argument, we can restrict to an interval in which $p_1(x)$ does not vanish in the interior. Notice next that if $p_1(x)$ solves the solution, so does $-p_1(x)$, we can assume that $p_1(x)$ takes non-negative values (otherwise replace it by its negative). 
The final statement is basically the same as when you consider the (real) roots of the quadratic equation $x^2 + b$ as you change $b$. For $b > 0$ you have no roots, for $b < 0$ you have two of them, and in between you have one root, and the graph is tangent to the $x$ axis at that root. 
8.2b
If for every $\xi$ it holds that $p_1(\xi) \neq 0$, then $p_1$ is a signed function. As both $p_1$ and $-p_1$ solve the ODE, at least one of them is positive. That is the desired conclusion (there exists a positive solution), so nothing is left to be proven. 
By hypothesis there does not exist any solutions with two distinct zeros. So $p_1(x)$ must have exactly one zero. But $p_1'(\xi) \neq 0$ there, by the uniqueness theorem (the same thing used for concluding part (a)), so $p_1(x)$ must change sign. Therefore on one side $p_1$ is positive, on the other $p_1$ is negative. Again, by possibly replacing $p_1$ by its negative you can arrange for the case you want. 
The argument is classic shooting method. You look at solutions to the ODE with initial data $p_m(0) = 1$ and $p_m'(0) = m$. For $m$ very large and positive, you have that $p_m$ has a negative zero. For $m$ very negative, you have that $p_m$ has a positive zero. But for no $m$ can there be a zero at $x = 0$, since you have prescribed $p_m(0) \neq 0$. So to transition from there being a positive zero to there being a negative zero, there are only two possibilities: either for some intermediate values of $m$ there are two zeros (one positive and one negative), or that for some intermediate values of $m$ there are no zeros. (We use the fact that the set of $m$ for which there is a negative zero is an open set and the set for which there is a positive zero is also an open set, and so there must be some left-over set with not exactly one zero.) The first alternative is ruled out by the hypothesis. 
9.1
As stated in the proof, $u_s$ is the unique solution that satisfies
$$ u_s(s) = 0 \quad \text{and} \quad u_s(0) = 1 $$
In general solving a second order ODE with these types of conditions are "boundary value problems", and there are many standard textbooks covering such results. In the case at hand, however, the existence and uniqueness follows from the maximum principle. 
