Asymptotic behavior of Sturm-Liouville eigenvalues

Question

I have two questions.

Consider the operator $Av = -v'' + a(x)v$ on $I = (0, L)$, with zero Dirichlet condition and $a \in C([0, L])$.

1. Let $(\lambda_n)$ denote the sequence of eigenvalues of $A$. Do we have that$$\left|\lambda_n - {{\pi^2 n^2}\over{L^2}}\right| \le \|a\|_{L^\infty(0, L)} \text{ for all }n?$$Thoughts. We probably want to use the Courant-Fischer min-max principle somehow?

Consider the general Sturm-Liouville operator$$Bu = -(pu')' + qu \text{ on }(0, L)$$ with zero Dirichlet condition. Assume that $p \in C^2([0, L])$, $p \ge \alpha > 0$ on $(0, L)$, and $q \in C([0, L])$.

2. Do the eigenvalues $(\mu_n)$ of the operator $B$ satisfy$$\left|\mu_n - {{\pi^2n^2}\over{L^2}}\right|\le C?$$

Answer 1

The answer to the first question is Yes. It is a particular case of the inequality $$\lambda_n(B)+\lambda_{\min}(C)\le\lambda_n(B+C)\le\lambda_n(B)+\lambda_{\max}(C)$$ for self-adjoint operators, which follows from Courant-Fisher formula. Here the Hilbert space is $L^2(0,1)$ and $$B=-D_x^2,\quad C=a,\qquad A=B+C.$$ We have $\lambda_n(B)=\frac{p^2n^2}{L^2}$ and $$\lambda_{\min}(C)=\min_xa(x),\qquad\lambda_{\max}(C)=\max_xa(x).$$

The answer to the second question is No. Just take $q\equiv0$ and $p\ne1$ a positive constant. Then $$\lambda_n=p\frac{\pi^2n^2}{L^2}\,.$$

Answer 2

The answer to this question can be found on Wikipedia :

https://en.wikipedia.org/wiki/Spectral_theory_of_ordinary_differential_equations

In particular, you will see that after a change of variables the general case gets reduced to the first one (on a different interval).