Essential Self-Adjointness of Even-Order, Strongly Singular, Homogeneous Half-Line Differential Operators

Abstract

AbstractWe consider essential self-adjointness on the space $$C_0^{\infty }((0,\infty ))$$ C 0 ∞ ( ( 0 , ∞ ) ) of even-order, strongly singular, homogeneous differential operators associated with differential expressions of the type $$\begin{aligned} \tau _{2n}(c) = (-1)^n \frac{d^{2n}}{d x^{2n}} + \frac{c}{x^{2n}}, \quad x > 0, \; n \in {{\mathbb {N}}}, \; c \in {{\mathbb {R}}}, \end{aligned}$$ τ 2 n ( c ) = ( - 1 ) n d 2 n d x 2 n + c x 2 n , x > 0 , n ∈ N , c ∈ R , in $$L^2((0,\infty );dx)$$ L 2 ( ( 0 , ∞ ) ; d x ) . While the special case $$n=1$$ n = 1 is classical and it is well known that $$\tau _2(c)\big |_{C_0^{\infty }((0,\infty ))}$$ τ 2 ( c ) | C 0 ∞ ( ( 0 , ∞ ) ) is essentially self-adjoint if and only if $$c \ge 3/4$$ c ≥ 3 / 4 , the case $$n \in {{\mathbb {N}}}$$ n ∈ N , $$n \ge 2$$ n ≥ 2 , is far from obvious. In particular, it is not at all clear from the outset that $$\begin{aligned} \begin{aligned}&\textit{there exists }c_n \in {{\mathbb {R}}}, n \in {{\mathbb {N}}}\textit{, such that} \\&\quad \tau _{2n}(c)\big |_{C_0^{\infty }((0,\infty ))} \, \textit{ is essentially self-adjoint}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (*)\\ {}&\quad \textit{ if and only if } c \ge c_n. \end{aligned} \end{aligned}$$ there exists c n ∈ R , n ∈ N , such that τ 2 n ( c ) | C 0 ∞ ( ( 0 , ∞ ) ) is essentially self - adjoint ( ∗ ) if and only if c ≥ c n . As one of the principal results of this paper we indeed establish the existence of $$c_n$$ c n , satisfying $$c_n \ge (4n-1)!!\big /2^{2n}$$ c n ≥ ( 4 n - 1 ) ! ! / 2 2 n , such that property (*) holds. In sharp contrast to the analogous lower semiboundedness question, $$\begin{aligned} \textit{for which values of }c\textit{ is }\tau _{2n}(c)\big |_{C_0^{\infty }((0,\infty ))}{} \textit{ bounded from below?}, \end{aligned}$$ for which values of c is τ 2 n ( c ) | C 0 ∞ ( ( 0 , ∞ ) ) bounded from below ? , which permits the sharp (and explicit) answer $$c \ge [(2n -1)!!]^{2}\big /2^{2n}$$ c ≥ [ ( 2 n - 1 ) ! ! ] 2 / 2 2 n , $$n \in {{\mathbb {N}}}$$ n ∈ N , the answer for (*) is surprisingly complex and involves various aspects of the geometry and analytical theory of polynomials. For completeness we record explicitly, $$\begin{aligned} c_{1}&= 3/4, \quad c_{2 }= 45, \quad c_{3 } = 2240 \big (214+7 \sqrt{1009}\,\big )\big /27, \end{aligned}$$ c 1 = 3 / 4 , c 2 = 45 , c 3 = 2240 ( 214 + 7 1009 ) / 27 , and remark that $$c_n$$ c n is the root of a polynomial of degree $$n-1$$ n - 1 . We demonstrate that for $$n=6,7$$ n = 6 , 7 , $$c_n$$ c n are algebraic numbers not expressible as radicals over $${{\mathbb {Q}}}$$ Q (and conjecture this is in fact true for general $$n \ge 6$$ n ≥ 6 ).