Singular eigenvalue problems for the equation \(y^{(n)}+\lambda p(x)y=0\)
This paper deals with the eigenvalue problem \[ y^{(n)}+\lambda p(x)y=0\quad\text{in}\quad (a,\infty), \] \[ y^{(i)}(a)=0,\quad i=0,\ldots, k-1, \] \[ \lim_{x\to\infty}y^{(i)}(x)=0,\quad i=k,\ldots,n-1, \] where \(p(x)\) is a one-sided, continuous function on \([a,\infty)\), \(1\leq k\leq n-1\), and \(\roman{sgn} [-\lambda p]=(-1)^{n-k}\). The main results are as follows. (1) (Discrete spectrum case) When \(\int^{\infty}s^{n-1}|p(s)|ds<\infty\) holds, then the eigenvalue problem has a sequence of eigenvalues \(\{\lambda_{i}\}\) which satisfies \[ 0<(-1)^{n-k-1}\lambda_{1}\roman{sgn}[p]<(-1)^{n-k-1}\lambda_{2}\roman{sgn}[p]< \cdots,\quad|\lambda_{i}|\to\infty\quad\text{as}\quad i\to\infty. \] To each eigenvalues \(\lambda_{i}\) there corresponds an essentially unique eigenfunction, and it has precisely \(i-1\) simple zeros in \((a,\infty)\). (2) (Continuous spectrum case) Suppose that \[ \lim_{x\to\infty} x^{n-\alpha-1}\int^{\infty}_{x}s^{\alpha}|p(s)|ds=0 \] holds for some \(\alpha<n-1\), but \(\int^{\infty}s^{n-1}|p(s)|ds<\infty\) fails to hold. Then, for every \(\lambda\neq 0\), such that \(\roman{sgn}[-\lambda p]=(-1)^{n-k}\), there exists an essentially unique eigenfunction \(y(x,\lambda)\) of the eigenvalue problem. Moreover, there exists a sequence of numbers \(\lambda_{i}\), \(\roman{sgn}[-\lambda_{i}p]=(-1)^{n-k}\), such that for \(|\lambda_{i-1}|<|\lambda|\leq |\lambda_{i}|\), \(y(x,\lambda)\) has exactly \(i-1\) simple zeros in \((a,\infty)\).
scientific article; zbMATH DE number 2123204