Theorem(*): Let $p\in C^1(\bar I)$ with $I=(0,1)$ and $p\ge\alpha>0$ on $I$;let $q\in C(\bar I).$ Then there exits a sequence $(\lambda_n)$ of real numbers and a Hilbert basis $(e_n)$ of $L^2(I)$ such that $e_n\in C^2(\bar I)$ for any n and $$-(pe'_n)'+qe_n=\lambda_ne_n\\e_n(0)=e_n(0)=0$$ on I. Furthermore, we have $\lambda_n\to \infty$ as $n\to \infty$. (we say that $\lambda_n$ are the eigenvalues of the differential operator $Au=-(pu')'+qu$ with Dirichlet boundary condition and that $(e_n)$ are the associated eigenfunctions. ) This theorem is from therem 8.22 in the book called functional analysis,Sobolev Spaces and Partial Differential equations written by Haim Brezis
Question: Assume we have the regular Sturm-Liouville eigenvalue problem$$L\phi:=(p(x)\phi'(x))'+q(x)\phi(x)=\lambda\phi(x) on [a,b],\\\phi(a)=0,\phi(b)=0,$$where we have $p\in C^1[a,b],p>=\alpha>0,q\in C^0[a,b].$ Look at the theorem(*),Show that as for eigenvalue $\lambda_n$,its corresponding engienspace is one-dimensional.
This problem has a hint:if $\phi_1 $and $\phi_2$ are two eigenfunctions for the same eignevalue then ${d\over dx}(p(\phi_1' \phi_2-\phi_1 \phi_2'))=0$
My thought: actually, I already proved ${d\over dx}(p(\phi_1' \phi_2-\phi_1 \phi_2'))=0$(just based on definitions above). however, how can I prove $\lambda_n$'s eigenspace only has one dimension? This is where I am confused about.
Suppose you have two solutions $\phi_1,\phi_2$ of the eigenvalue problem. Then $\phi_1(a)=0=\phi_2(a)$ and $\phi_1(b)=0=\phi_2(b)$. The Wronksian result gives \begin{align} &p(x)\{\phi_1(x)\phi_2'(x)-\phi_1'(x)\phi_2(x)\}\\ = &p(a)\{\phi_1(a)\phi_2'(a)-\phi_1'(a)\phi_2(a)\}=0,\;\; x\in[a,b]. \end{align} So, $$ \frac{d}{dx}\left(\frac{\phi_2}{\phi_1}\right)=0. $$ You have to deal with the special cases where $\phi_1$ vanishes in the interior of $(a,b)$, but you can see that the left-most zero of $\phi_1$ inside $(a,b)$ must also be a zero of $\phi_2$ because $\phi_2/\phi_1$ is constant between $a$ and that first $0$, if any.