Simplicity of eigenvalues

Question

Consider the Sturm-Liouville operator$$Au = -(pu')' + qu \text{ on }I = (0, 1),$$where $p \in C([0, 1])$, $p \ge \alpha > 0$ on $I$, and $q \in C([0, 1])$. No further assumptions are made; in particular, the associated bilinear form$$a(u, v) = \int_0^1 (pu'v' + quv)$$need not be coercive. Set$$N = \{u \in H^1(0, 1): \text{ }a(u, v) = 0 \text{ }v \in H_0^1(0, 1)\}.$$Do we have that all the eigenvalues of $A$ with zero Dirichlet condition are simple?

Answer 1

Yes they are simple. This is a peculiarity of the 1-dimensional eigenvalue problem. By shiftin $q$, we may always consider the eigenvalue $\lambda=0$.

Geometric simplicity : as mentionned by Remling, two eigenfunctions $u_1$ and $u_2$ must be linearly dependent. For we may find a constant $a$ such that $u_2'(0)=au_1'(0)$, and then $u:=u_2-au_1$ is a solution of a linear second order ODE, with homogeneous initial data $u(0)=u'(0)=0$. By Cauchy-Lipschitz, we have $u\equiv0$.

Algebraic simplicity : the operator $A$ is symmetric in $L^2(0,1)$ (this is already used to prove that the eigenvalues are real), hence the algebraic multiplicity equals the geometric multiplicity (in other words, $A$ is semi-simple). Proof: if $Av=0$ and $Au=v$, then $$0=\int_0^1uAv\,dx=\int_0^1vAu\,dx=\int_0^1v^2dx$$ hence $v\equiv0$.

Remark that the result still hold true if $A=-pD^2_x+bD_x+q$ is not symmetric, because it can always be symmetrized. This is false in higher dimension.