I have just start studying about theory of Sturm-Liouville problem and I have a doubt:
Given the Sturm-Liouville problem : $$(p(t)y'(t))'+q(t)y(t)+{\lambda}w(t)y(t)=0$$ such that $a_1y(a)+a_2y'(a)=0$ ; $b_1y(b)+b_2y'(b)=0$.
with $p\in{C^1[a,b]}, p(t)>0 $ $q,w\in{C[a,b]}$ and $w>0$.
Is true that each eigenvalue is simple?
My attempt:
I know that the number of functions linearly independent associated to an eigenvalue of this problem can´t be more than two because they are solutions of a second order differential equation. So, I have to prove that if I have two eigenfunctions of the same eigenvalue they are linearly dependant, but I don't know how to do it.
I'm a student myself and I only just started a course on this topic so my method may not be the most rigorous or fully thought-out but if we assume two solutions, U and V that satisify the sturm-liouville problem then we must have that:
$$a_1U(a)+a_2U′(a)=0 \implies U'(a)=-\frac{a_1}{a_2}U(a)$$
Similarly, we have: $$a_1V(a)+a_2V′(a)=0 \implies V'(a)=-\frac{a_1}{a_2}V(a)$$
Here we are assume $$a_2\neq 0 $$ solely for the purpose that we will plug these two solutions into the wronskian and evaluate at $$x=a$$, showing these solutions are linearly dependent and thus scalar multiples of each other.
If we have $$a_2=0$$, then trivially we get $$U(a)=0=V(a)$$ which we can plug into the wronskian and get linear dependence