Positive Self-adjoint Operator Extensions with Applications to Differential Operators

Abstract

In this paper we consider extensions of positive operators. We study the connections between the von Neumann theory of extensions and characterisations of positive extensions via decompositions of the domain of the associated form. We apply the results to elliptic second order differential operators and look in particular at examples of the Laplacian on a disc and the Aharonov–Bohm operator.

1 Introduction

Let A be a closed strictly positive symmetric operator with dense domain D(A) and range R(A) in a Hilbert space H. In [11, 12], Krein proved that there is a one to one correspondence between the set of positive self-adjoint extensions $A_B$ of A and a set of pairs $\{N_B, B\}$, where $N_B$ is a subspace of the kernel N of $A^{*}$ and B is a positive self-adjoint operator with domain and range in $N_B$. Krein’s result was subsequently developed further by Visik [15] and Birman [3]; this work of the three authors will be referred to as the KVB theory. An important extension of the KVB theory was made in [8] to a pair of closed densely defined operators $A, A'$, which form a dual pair in the sense that $A \subset (A')^{*}$ and are such that $A \subset A_{\beta } \subset (A')^{*}$ for an operator $A_{\beta }$ with a bounded inverse. The results in [8] include those of KVB when $A=A'$. Of particular interest to us in [8] is the application of the abstract theory to the case when A is generated by an elliptic differential expression acting in a bounded smooth domain $\Omega$ in $\mathbb {R}^n$. In this case the self-adjoint extensions of A are determined by boundary conditions on the boundary $\partial \Omega$ of $\Omega$.

In [5], results in Rellich [13], Kalf [9] and Rosenberger [14] were applied to the KVB theory to determine all the positive self-adjoint extensions of a positive Sturm–Liouville operator with minimal conditions on the coefficients. Our objective in this paper is to investigate what can be achieved by applying similar methods to two problems on bounded domains in $\mathbb {R}^n, n \ge 2$; in the first A is generated by a second order elliptic expression, and in the second it is the Aharonov–Bohm operator on a punctured disc. Our analysis depends on an abstract result which incorporates the von Neumann theory concerning all the self-adjoint extensions of any symmetric operator.

Denote by $A_F,~ a_F[\cdot ,\cdot ]$ the Friedrichs extension and associated sesquilinear form of A. Then for all $u \in D(A_F)$ and $v \in Q(A_F)$ we have

$$\begin{aligned} a_F[u,v] = (A_Fu,v), \end{aligned}$$

where $(\cdot ,\cdot )$ is the inner product of H, and $D(A_F)$ is dense as a subspace of $Q(A_F)$ with inner product $a_F[\cdot ,\cdot ]$ (see [7, Chapter IV] for more on the relation between sesquilinear forms, operators and their Friedrichs extension). By the KVB theory, $\hat{A}$ is a positive self-adjoint extension of A if and only if, $\hat{A} = A_B$, where B is a positive self-adjoint operator acting in a subspace $N_B$ of N and $A_B,~B$ have associated forms $a_B,~b$, respectively which satisfy

$$\begin{aligned} a_B = a_F + b,\quad Q(A_B) = Q(A_F) \dotplus Q(B). \end{aligned}$$

(1.1)

Thus any $u \in Q(A_B)$ can be uniquely written as $u=u_F + u_N$, where $u_F \in Q(A_F),~ u_N \in Q(B)$. There are two distinguished positive self-adjoint extensions of A, namely the Friedrichs (or strong) extension $A_F$ and the Krein–von Neumann (or weak) extension $A_K$. These are extremal in the sense that any positive self-adjoint extension $\hat{A}$ of A satisfies $A_K \le \hat{A} \le A_F$ in the form sense. In (1.1), the Krein–von Neumann extension $A_K$ corresponds to $B=0,~ N_B = N$, and the Friedrichs extension $A_F$ to $B= \infty ,~Q(B) = 0,$ that is, B acts trivially on a zero dimensional space.

2 Positive Extensions and the Von Neumann Theory

The von Neumann theory characterises the self-adjoint extensions of any closed densely defined symmetric operator T. Denoting the deficiency spaces $\text {ker}(T^{*}\mp iI)$ by $N_{\pm }$, we have

$$\begin{aligned} D(T^{*}) = D(T) \dotplus N_+ \dotplus N_-, \end{aligned}$$

(2.1)

and $T_S$ is a self-adjoint extension of T if and only if there is a unitary operator $U(T_S){:}\,N_+ \rightarrow N_-$ such that

$$\begin{aligned} D(T_S) = D(T) \dotplus (I + U(T_S)) N_+. \end{aligned}$$

(2.2)

Let $u, v \in D(T^{*})$. Then by the von Neumann theory, there exist unique $u_0,v_0 \in D(T)$ and $u_\pm ,v_\pm \in N_\pm$ such that $u= u_0+u_+ + u_-$ and $v= v_0+v_+ + v_-$.

It follows that

$$\begin{aligned} (T^{*}u,v)-(u,T^{*}v)= & {} (Tu_0+iu_+-iu_-, v_0+v_++v_-) \\&-\,(u_0+u_++u_-, Tv_0+i v_+-iv_-) \\= & {} (Tu_0, v_++v_-)+i(u_+-u_-,v_0+v_++v_-)\\&+\,i (u_0+u_++u_-, v_+-v_-) -(u_++u_-, T v_0)\\= & {} -i(u_0, v_+-v_-)+ i (u_+-u_-, v_0+v_++v_-)\\&+\,i ( u_0+u_++u_-, v_+-v_-)-i (u_+-u_-,v_0)\\= & {} 2i \big [(u_+,v_+)_{N_+}-(u_-,v_-)_{N_-} \big ]{.} \end{aligned}$$

Let $P_+$ and $P_-$ denote the projections from $D(T^{*})$ to $N_+$ and $N_-$ with respect to the decomposition (2.1) and let $U{:}\,N_+\rightarrow N_-$ be unitary. Set $\tilde{\Lambda }_0 =UP_++P_-$ and $\tilde{\Lambda }_1=-iUP_++iP_-$. Then, for any $u,v\in D(T^{*})$

$$\begin{aligned} \left( T^{*}u,v\right) -\left( u,T^{*}v\right) = \left( \tilde{\Lambda }_0 u, \tilde{\Lambda }_1 v\right) -\left( \tilde{\Lambda }_1 u, \tilde{\Lambda }_0 v\right) \end{aligned}$$

(2.3)

(see [10, Theorem 3]). The triple $(N_+, \tilde{\Lambda }_0,\tilde{\Lambda }_1)$ is a boundary triple (also known as a space of boundary values) for T.

Given a self-adjoint extension $T_S$ of T, we now choose

$$\begin{aligned} \Lambda _0(T_S)= & {} U(T_S)P_+ +P_-,\end{aligned}$$

(2.4)

$$\begin{aligned} \Lambda _1(T_S)= & {} -iU(T_S)P_+ +iP_-. \end{aligned}$$

(2.5)

Then, from (2.2), $\ker \Lambda _1(T_S) = \mathcal D(T_S)$ and we obtain, for all $u,v \in D(T^{*})$

$$\begin{aligned} (T^{*}u,v) -(u, T^{*}v)= & {} (\Lambda _0(T_S) u,\Lambda _1(T_S)v) -(\Lambda _1(T_S)u,\Lambda _0(T_S) v). \end{aligned}$$

(2.6)

Let $T=A$ be positive and B a positive self-adjoint operator on a subspace $N_B$ of the kernel of $A^{*}$ with domain D(B). By [2, Theorem 3.1], the domain of the self-adjoint extension $A_B$ of A corresponding to B is

$$\begin{aligned} \mathcal {D}(A_B) = \left\{ u_0 + A_F^{-1}\left( Bv +f\right) +v{:}\,u_0 \in \mathcal {D}(A), v \in \mathcal {D}(B), f \in N \cap \mathcal {D}(B)^{\bot }\right\} . \end{aligned}$$

(2.7)

Remark 2.1

The special case $B=0, N_B =N$ gives the domain of the Krein–von Neumann extension $A_K$, namely

$$\begin{aligned} \mathcal {D}(A_K) = \mathcal {D}(A) \dotplus N, \end{aligned}$$

(2.8)

the sum being a direct sum since A is strictly positive. It follows that

$$\begin{aligned} \ker (A_K) = N. \end{aligned}$$

(2.9)

The Friedrichs extension is characterised by the choice of B as acting trivially on $N_B=\{0\}$. Following the approach of [2], we can set $b[u] = \infty$ for $u \in N{\setminus } Q(B)$. It follows from (1.1) that $Q(A_B) =Q(A_F )$ if and only if $Q(B) = \{0\}$. Since $A_F$ is the only self-adjoint extension of A with domain in $Q(A_F)$ it follows that its domain is determined by $b[u] = \infty$ for all $u \in N {\setminus }\{0\}$.

Theorem 2.2

Let $A_B$ be a positive self-adjoint extension of the positive operator A associated with the pair $\{B,N_B\}$. Let $u \in D(A_B)$, where $u=u_F + w$, $u_F = u_0 +A_F^{-1}(Bw+v), u_0 \in \mathcal {D}(A), w \in \mathcal {D}(B), v \in N \cap \mathcal {D}(B)^{\bot }$. Then

$$\begin{aligned} b[w,\zeta ] = \left( \Lambda _0(A_B) u,\Lambda _1(A_B)\zeta \right) , \ \ \forall \ \zeta \in Q(B), \end{aligned}$$

(2.10)

where $\Lambda _0(A_B) = U(A_B)P_+ +iP_-$ and $\Lambda _1(A_B) =-iU(A_B)P_+ +i P_-$.

Proof

Let $\varphi =\theta +\zeta \in Q(A_B)$ with $\theta \in Q(A_F)$ and $\zeta \in Q(B)$. Then on the one hand, we have

$$\begin{aligned} (A_B u,\varphi )= (A^{*} u,\varphi ) = (A^{*} u_F,\varphi ) \end{aligned}$$

(2.11)

since $w\in N$, and on the other hand,

$$\begin{aligned} ( A_B u,\varphi )= & {} a_B[u,\varphi ] = a_F[u_F,\theta ] + b[w,\zeta ] \nonumber \\= & {} (A_Fu_F, \theta ) + b[w,\zeta ] \nonumber \\= & {} (A^{*}u_F, \theta ) + b[w,\zeta ]. \end{aligned}$$

(2.12)

On combining (2.11) and (2.12) we get

$$\begin{aligned} b[w,\zeta ] = (A^{*} u_F,\varphi -\theta ) = (A^{*} u_F,\zeta ) = (A^{*} u, \zeta ), \end{aligned}$$

and as $A^{*}\zeta =0$, Eq. (2.6) yields

$$\begin{aligned} b[w,\zeta ]= & {} \left( \Lambda _0(A_B) u,\Lambda _1(A_B)\zeta \right) -\left( \Lambda _1(A_B) u,\Lambda _0(A_B)\zeta \right) . \end{aligned}$$

(2.13)

Since $\ker \Lambda _1(A_B) = D(A_B)$, (2.10) follows. $\square$

Let $\{\psi _k\}$ be an orthonormal-basis of Q(B), where B is a positive self-adjoint operator in $N_B\subset N$, and let $w =\sum _j w_j \psi _j,~ \zeta =\sum \zeta _k\psi _k$ and $b_{jk} =b[\psi _j,\psi _k]$. Then $b[w,\zeta ] = \sum _{j,k} b_{jk}w_j \overline{\zeta _k}$ and from (2.10) and the fact that $\ker \Lambda _1(A_B) = D(A_B)$, $u = u_F + w \in D(A_B)$ if and only if

$$\begin{aligned} \forall k.\ \left( \Lambda _0(A_B) u,\Lambda _1(A_B)\psi _k\right) =\sum _j b_{jk}w_j \overline{\zeta _k}. \end{aligned}$$

(2.14)

3 Elliptic Differential Operators of Second Order

In this section we shall apply the above abstract theory to the case when A is the closure of a symmetric second-order differential operator in $L^2(\Omega )$ defined by

$$\begin{aligned} A'u := (-\nabla \cdot p \nabla +q)u = \left( -\sum _{i,j =1}^n D_i p_{ij}D_j + q\right) u,\ \ u \in C_0^{\infty }(\Omega ), \end{aligned}$$

(3.1)

subject to conditions on the coefficients $p_{ij},~q$ and the domain $\Omega$. The assumptions are the ones made in [1] which weaken the smoothness requirements on the coefficients and the boundary of $\Omega$ made by Grubb [8]. In the following definition of a boundary regularity class, $B^{s}_{p,q}$ is the Besov space of order s (see [1, Section 2]), and we set $x=(x',x_n), x'\in \mathbb {R}^{n-1}, x_n \in \mathbb {R}$.

Definition 3.1

The boundary $\partial \Omega$ is said to be of class $B^{M-\frac{1}{2}}_{p,q}$ if for each $x\in \partial \Omega$ there exist an open neighbourhood U satisfying the following: for a suitable choice of coordinates on $\mathbb {R}^n$, there is a function $\gamma \in B^{M-\frac{1}{2}}_{p,q}(\mathbb {R}^{n-1})$ such that $U\cap \Omega = U\cap \mathbb {R}^n_\gamma$ and $U\cap \partial \Omega = U\cap \partial \mathbb {R}^n_{\gamma }$, where $\mathbb {R}^{n}_{\gamma } = \{x \in \mathbb {R}^n{:}\,x_n > \gamma (x')\}$.

In the list of assumptions to be made, we shall denote the boundary of $\Omega$ by $\Sigma$, and $H^s_t$ is a Bessel potential space (a Sobolev space for $s \in \mathbb {N}$), which we write as $H^s$ when $t=2$; see [1, Section 2] for definitions of $H^s_t(\Omega )$ and $H^s_t(\Sigma )$.

Assumptions

• 1. There exists $c_0>0$ such that for all $x\in \Omega$ and $\xi \in \mathbb {R}^{n}$

  $$\begin{aligned} \sum _{i,j=1}^n p_{ij}(x)\xi _i\xi _j dx \ge c_0 \Vert \xi \Vert ^2. \end{aligned}$$

• 2. There exists $c >0$ such that

  $$\begin{aligned} \Vert u\Vert ^2_1 = \int _{\Omega } \left( p |\nabla u|^2 + q|u|^2 \right) dx \ge c \Vert u\Vert ^2,\ \ \ u \in C_0^{\infty }(\Omega ). \end{aligned}$$

  The completion of $C_0^{\infty }(\Omega )$ with respect to the norm $\Vert \cdot \Vert _1$ is the form domain $Q(A_F)$ of A.

• 3. The boundary $\Sigma$ is of class $B^{\frac{3}{2}}_{r,2}$ and the coefficients p and q of A lie in $H^1_t(\Omega )$ and $L_t(\Omega )$, respectively, under the constraints $n\ge 2$, $2< r<\infty$, $2< t \le \infty$, and

  $$\begin{aligned} 1-\tfrac{n}{t}\ge \tfrac{1}{2}-\tfrac{n-1}{r}> 0. \end{aligned}$$

  (3.2)

Remark 3.2

Our third assumption is Assumption 2.18 in [1]. Therefore, we have that for $v \in Q(A_F)$, $\gamma _0 v =0$, where $\gamma _0$ is the trace operator which maps v into its value on $\Sigma$ (see [1, Theorem 2.11]). Moreover, in the notation of [1, 6], denote the solution of

$$\begin{aligned} A w = 0 \ \text {in}\ \ \Omega ,\ \ w = u \ \ \text {on}\ \ \Sigma . \end{aligned}$$

(3.3)

by

$$\begin{aligned} w = K^0_{\gamma } \gamma _0 u. \end{aligned}$$

(3.4)

Then by [1, Theorem 5.4], for all $s \in [0,2]$,

$$\begin{aligned} K^0_{\gamma }{:}\,H^{s-1/2}(\Sigma ) \rightarrow H^s(\Omega ) \end{aligned}$$

(3.5)

is continuous,

$$\begin{aligned} K^0_{\gamma }{:}\,H^{s-1/2}(\Sigma ) \rightarrow Z_0^s(A) := \{u \in H^s(\Omega ){:}\,Au=0\} \end{aligned}$$

(3.6)

is a homeomorphism, and

$$\begin{aligned} \gamma _0: Z_0^s(A) \rightarrow H^{s-1/2}(\Sigma ) = (K^0_{\gamma })^{-1}. \end{aligned}$$

(3.7)

We remark that under the more restrictive assumptions that $\Omega$ is a bounded domain whose boundary is an $(n-1)$-dimensional $C^{\infty }$ manifold, and the coefficients $p_{jk},~q$ of $A'$ in (3.1) lie in $C^{\infty }(\overline{\Omega })$ these properties were already shown by Grubb in [8].

Theorem 3.3

Let the above assumptions hold and let $A_B$ be a positive self-adjoint extension of A. For $u \in D(A_B)$, we have $u=u_F+w$ for some $u_F \in D(A_F),~w \in Q(B)$, and for all $\zeta \in Q(B)$

$$\begin{aligned} b[w,\zeta ] = \int _{\Omega } (-\nabla p \nabla +q ) u_F\ \overline{\zeta }\ dx. \end{aligned}$$

(3.8)

If $\{\psi _k\}$ is an orthonormal basis of Q(B) then, with $b_{jk}$ as in (2.14),

$$\begin{aligned} \forall k. \sum _j b_{jk}w_j+ \int _\Omega (\nabla p \nabla -q) u_F\ \overline{\psi _k} dx=0. \end{aligned}$$

(3.9)

Proof

Let $a_B[\cdot ,\cdot ], a_F[\cdot ,\cdot ], b[\cdot ,\cdot ]$ denote the forms associated with $A_B, A_F, B$, respectively. For $u, \varphi \in Q(A_B)$ we have the decompositions

$$\begin{aligned} u= & {} u_F + w,\quad (u_F \in Q(A_F),~w = K^0_{\gamma } \gamma _0 u \in Q(B)), \nonumber \\ \varphi= & {} \varphi _F + \zeta , \quad (\varphi _F \in Q(A_F), ~\zeta { = K^0_{\gamma }\gamma _0 \varphi }\in Q(B)). \end{aligned}$$

(3.10)

If $u \in D(A_B)$, it has the decomposition $u=A_F^{-1}A^{*}u+(u-A_F^{-1}A^{*}u)$, i.e., $u_F=A_F^{-1}A^{*}u$ and $w=u-A_F^{-1}A^{*}u,$ since $u_F\in D(A^{*})\cap Q(A_F) = D(A_F)$ and $w\in Q(A_B)\cap N=Q(B)$. Now, let $\varphi =\varphi _F+\zeta \in Q(A_B)$. Then

$$\begin{aligned} (A_B u,\varphi ) = \int _{\Omega } (-\nabla p \nabla + q) u \ \overline{\varphi } dx \end{aligned}$$

(3.11)

and furthermore,

$$\begin{aligned} ( A_B u,\varphi )= & {} (u,\varphi )_1 = a_B[u,\varphi ] = a_F[u_F,\varphi _F] + b[w,\zeta ] \nonumber \\= & {} \int _{\Omega } (-\nabla p \nabla + q ) u_F \ \overline{\varphi _F} dx + b[w,\zeta ] \nonumber \\= & {} \int _{\Omega } (-\nabla p \nabla + q ) u \ \overline{\varphi _F} dx + b[w,\zeta ]. \end{aligned}$$

(3.12)

Combining (3.11) and (3.12), we get

$$\begin{aligned} b[w,\zeta ] = \int _{\Omega } (-\nabla p \nabla +q )u \ (\overline{\varphi }-\overline{\varphi _F})dx = \int _{\Omega } (-\nabla p \nabla +q ) u_F\ \overline{\zeta }dx. \end{aligned}$$

(3.13)

Now let $\{\psi _k\}$ be an orthonormal-basis of Q(B), $w=\sum w_j\psi _j$ and $\zeta =\sum \zeta _k\psi _k$. Then

$$\begin{aligned} b[w,\zeta ]=\sum _{j,k} b_{jk}w_j\overline{\zeta _k}, \end{aligned}$$

while the right-hand side of (3.13) is

$$\begin{aligned} \sum _k \int _{\Omega }(-\nabla p \nabla +q ) u\ \overline{\zeta _k\psi _k}dx. \end{aligned}$$

Consequently

$$\begin{aligned} \sum _k \overline{\zeta _k} \left[ \sum _j b_{jk}w_j+ \int _\Omega (\nabla p \nabla -q ) u\ \overline{\psi _k}dx\right] =0, \end{aligned}$$

and equivalently,

$$\begin{aligned} \sum _k \overline{\zeta _k} \left[ \sum _j b_{jk}w_j+\int _\Omega (\nabla p \nabla -q ) u_F\ \overline{\psi _k}dx\right] =0. \end{aligned}$$

As $\{\zeta _k\}$ is an arbitrary sequence in $\ell ^2$, the ‘boundary condition’ associated with $A_B$ is given by

$$\begin{aligned} \forall k. \sum _j b_{jk}w_j+ \int _\Omega (\nabla p \nabla -q) u_F\ \overline{\psi _k}=0. \end{aligned}$$

(3.14)

$\square$

Corollary 3.4

The boundary condition associated with $A_B$ is given by

$$\begin{aligned} \forall k\cdot \sum _j b_{jk}w_j= (\nu _1 u_F, \gamma _0 \psi _k), \end{aligned}$$

(3.15)

where $\nu _1 u = \sum _{j,k =1}^n n_j \gamma _0[p_{jk}D_k u]$ and $\mathbf{n} = (n_1,\ldots ,n_n)$ is the interior unit normal.

Proof

Since $u,\psi _k\in D(A^{*})$, we can use [1, Theorem 6.1] to write

$$\begin{aligned}&\int _\Omega (\nabla p \nabla -q ) u_F\ \overline{\psi _k}dx -\int _\Omega u_F\ (\nabla p \nabla -q )\overline{\psi _k} dx\nonumber \\&\quad =( \gamma _0 u_F,\Gamma _1 \psi _k)_{\Sigma } -(\Gamma _1 u_F, \gamma _0\psi _k)_{\Sigma }, \end{aligned}$$

(3.16)

where $\Gamma _1$ is the “regularised” Neumann operator given by $\Gamma _1 u = \nu _1 u - P_{\gamma _0, \nu _1} \gamma _0 u$ and $P_{\gamma _0, \nu _1} \gamma _0$ is the Dirichlet-to-Neumann map $P_{\gamma _0, \nu _1} \gamma _0 = \nu _1 K_{\gamma }^0$. Under the smoothness conditions assumed, the trace maps $\gamma _0,~ \Gamma _1$, map $D(A^{*})$ continuously into $H^{-1/2}(\Sigma ),~ H^{1/2}(\Sigma )$, respectively. The terms on the right-hand side of (3.16) therefore represent in fact, $H^{-1/2},H^{1/2}$-duality products over the boundary $\Sigma$, which are extensions of the $L^2(\Sigma )$ inner products (see [1, Theorem 6.1]).

Since $(\nabla p \nabla -q )\psi _k=0$ and $u_F\in Q(A_F)$, two of the four terms in (3.16) vanish and, as $\Gamma _1 u_F=\nu _1 u_F$, we get that the boundary condition in (3.14) becomes

$$\begin{aligned} \forall k\cdot \sum _j b_{jk}w_j= (\nu _1 u_F, \gamma _0\psi _k)_{\Sigma }. \end{aligned}$$

(3.17)

$\square$

Remark 3.5

The Friedrichs extension is determined by the boundary condition $\gamma _0 u=0$. Under the additional smoothness assumptions on $\Omega$ and the coefficients of $A'$ in (3.1) in [8], the Friedrichs extension has domain $H_0^1(\Omega ) \cap H^2(\Omega )$.

Remark 3.6

The Krein–von Neumann extension corresponds to $B=0, Q_B = N_B = N =\text {ker}(A^{*})$ and so

$$\begin{aligned} Q(A_B) = Q(A_F) \dotplus N,\ \ a_B [u] = a_F[u_F], \end{aligned}$$

when $u=u_F+w, u_F \in Q(A_F), w \in N$. Thus in (3.15), $b_{jk} =0$ for all j, k and $\nu _1 u_F = \Gamma _1 u_F$. Since $\nu _1$ maps $D(A^{*})$ continuously into $H^{-1/2}(\Sigma )$ and $\gamma _0$ is a homeomorphism of N onto $H^{1/2}(\Sigma )$, it follows from (3.15) that the boundary condition satisfied by the Krein–von Neumann extension is

$$\begin{aligned} \nu _1 u_F = 0. \end{aligned}$$

Since $w = K_{\gamma }^0 \gamma _0 u$ we have

$$\begin{aligned} \nu _1 u_F = (\nu _1 - \nu _1 K_{\gamma }^0 \gamma _0) u = \Gamma _1 u. \end{aligned}$$

Remark 3.7

On combining (2.10) and (3.17) we have

$$\begin{aligned} \left( \Lambda _0(A_B) u,\Lambda _1(A_B)\psi _k\right) =(\nu _1 u_F,\gamma _0\psi _k)_{\Sigma }. \end{aligned}$$

(3.18)

For the Krein–von Neumann extension $\Lambda _K\psi _k=0$, so we again get

$$\begin{aligned} \Gamma _1 u = \nu _1 u_F= 0 \end{aligned}$$

as the Krein–von Neumann boundary condition.

Example 3.8

We consider extensions of the positive operator $A=-\Delta +1$ when $\Omega$ is the unit disc in $\mathbb {R}^2$. According to (3.15), $v=v_F+w$ lies in the domain of an extension $A_B$ if and only if

$$\begin{aligned} \left( \nu _1 v_F,\gamma _0\psi _k\right) _{\Sigma }=\sum _j b_{jk}\left( w,\psi _j\right) \end{aligned}$$

(3.19)

for all k, where $\{\psi _k\}$ is an orthonormal-basis of the subspace Q(B) in $N=\ker A^{*}$.

Let $-\,\Delta \psi + \psi =0$ and put $\psi (r, \theta )=R(r) \Theta (\theta )$, where $x =(r,\theta )$ are polar co-ordinates. Then since

$$\begin{aligned} \Delta \psi = R''\Theta + \dfrac{1}{r} R' \Theta +\dfrac{1}{r^2} R \Theta '' =R\Theta =\psi , \end{aligned}$$

we get

$$\begin{aligned} r^2 \dfrac{ R''}{R} + r \dfrac{R'}{R} - r^2 = n^2 \end{aligned}$$

and $\dfrac{\Theta ''}{\Theta }=-n^2$ with constant n and $\Theta (0)=\Theta (2\pi )$; thus $\Theta _n(\theta ) = e^{in\theta }, n\in \mathbb {Z}$ and we seek the $L^2(0,1;rdr)$ solutions of

$$\begin{aligned} r^2 R''+ r R'- r^2 R =n^2 R, \quad r \in [0,1]. \end{aligned}$$

These solutions are given by the modified Bessel functions $I_n(r)$ and $K_n(r)$.

For $n \ge 1$, $K_n(r)$ does not lie in $L^2(0,1;rdr)$. The function $K_0(r)$ has a logarithmic singularity at 0, which means that $\Delta K_0$ is not zero in the sense of distributions, excluding $K_0$ from N. Therefore

$$\begin{aligned} \psi _k(r,\theta )= I_k(r) e^{i k \theta }, \quad k \in \mathbb {Z}\end{aligned}$$

is a basis for N; note that $I_{-k}=I_k$.

For $k\in \mathbb {Z}$

$$\begin{aligned} \gamma _0 \psi _k(\theta )=\psi _k(1,\theta )=I_k(1) e^{ik\theta }. \end{aligned}$$

and since $v_F\in D(A_F)$, we have $\nu _1 v_F=\frac{\partial v_F}{\partial \nu }$. On expanding $v_F$ in $\theta$ in terms of its Fourier series,

$$\begin{aligned} v_F(r,\theta )=\sum _{n\in \mathbb {Z}} v_{F,n}(r) e^{in\theta }, \end{aligned}$$

we derive

$$\begin{aligned} \frac{\partial v_F}{\partial \nu }\Big \vert _{\partial \Omega } =\sum _{n\in \mathbb {Z}} \frac{\partial v_{F,n}}{\partial r}(1) e^{in\theta }. \end{aligned}$$

Consequently $v=v_F+w\in D(A_B)$ if and only if for all $k\in \mathbb {Z}$

$$\begin{aligned} \sum _j b_{jk}\left( w,\psi _j\right)= & {} (\nu _1 v_F, \gamma _0\psi _k)_{\Sigma }\\= & {} \int _0^{2\pi } \sum _{n\in \mathbb {Z}} \frac{\partial v_{F,n}}{\partial r}(1) e^{in\theta } I_k(1) e^{-ik\theta } d\theta \\= & {} 2\pi \frac{\partial v_{F,k}}{\partial r}(1) I_k(1). \end{aligned}$$

Remark 3.9

• 1. For the Krein–von Neumann extension, $v=v_F+w\in D(A_K)$ if and only if for all $k\in \mathbb {Z}$ we have

  $$\begin{aligned} 0 = 2\pi \frac{\partial v_{F,k}}{\partial r}(1) I_k(1). \end{aligned}$$

  As $I_k(1)\ne 0$ for all $k\in \mathbb {Z}$, this implies that

  $$\begin{aligned} v_F(1,\theta ) = \frac{\partial v_{F}}{\partial r}(1,\theta ) = 0 \end{aligned}$$

  and hence $v_F\in D(A)$. As there are no restrictions on w, we get $D(A_K)= D(A) + N$, as expected. Also, the boundary condition satisfied by any $u \in D(A_K)$ is $\Gamma _1 u = 0$, where $\Gamma _1 = \nu _1 - P_{\gamma _0, \nu _1} \gamma _0$ is the regularised Neumann operator.

• 2. For the Friedrichs extension, we formally have $b_{jk}=\infty$ for all j, k in (3.19). This implies that w must be orthogonal to all the $\psi _k$. As $w\in N$, this gives $w=0$.

4 Aharonov–Bohm Operator

Let $\Omega = \{x{:}\,|x| <1\} {\setminus } \{0\} \subset \mathbb {R}^2$, and let A be the closure in $L^2(\Omega )$ of $A'\upharpoonleft _{C_0^{\infty }(\Omega )}$, where

$$\begin{aligned} A' := -(\nabla +iM)^2. \end{aligned}$$

Here, the Aharonov–Bohm magnetic potential

$$\begin{aligned} M:= \alpha \frac{1}{(x_1^2 + x_2^2)}(-x_2,x_1) = \alpha \frac{e_{\theta }}{r}, \quad \alpha \in (0,1), \end{aligned}$$

(4.1)

where $x=(r \cos ~\theta , r\sin ~\theta )$ in polar co-ordinates and $e_{\theta } = (-\sin ~\theta , \cos ~\theta )$ is the unit vector orthogonal to $e_r = x/r$. Then

$$\begin{aligned} \text {curl}~ M =0 \ \ \text {in}\ \Omega ,\ \text {and}\ M\cdot e_r =0. \end{aligned}$$

(4.2)

For $u \in C_0^{\infty }(\Omega )$ we have

$$\begin{aligned} (A'u,u)= & {} \int _{\Omega } |(\nabla +iM) u|^2 dx \nonumber \\= & {} \int _0^1 \int _0^{2\pi }\left( \left| \frac{\partial u}{\partial ~r}\right| ^2 + r^{-2} \left| i \frac{\partial u}{\partial ~\theta } +\alpha u \right| ^2 \right) r dr d\theta . \end{aligned}$$

(4.3)

The sequence $\{\varphi _k(\theta ): k \in \mathbb {Z}\}$, where $\varphi _k(\theta ) = \frac{e^{-ik\theta }}{\sqrt{2\pi }}$, is an orthonormal basis for $L^2(0,2\pi )$ and hence any $u \in L^2(\Omega )$ has the representation

$$\begin{aligned} u(r, \theta ) = \Sigma _k u_k(r) \varphi _k(\theta ), \end{aligned}$$

(4.4)

where

$$\begin{aligned} u_k(r) = \int _0^{2\pi } u(r,\theta ) \overline{\varphi _k(\theta )} d \theta . \end{aligned}$$

On substituting in (4.3), we have, with $\lambda _k=k+\alpha$

$$\begin{aligned} (A'u,u) = \sum _k \int _0^1 \left( |u_k'(r)|^2 +\frac{\lambda _k^2}{r^2} |u_k(r)|^2\right) r dr. \end{aligned}$$

Since $\min \{|\lambda _k|/r{:}\,k \in \mathbb {Z},\ 0<r<1 \} \ge \min \{\alpha ,1-\alpha \} >0$, it follows that A is strictly positive and its form domain $Q(A_F)$ is the completion of $C_0^{\infty }(\Omega )$ with respect to the norm given by the square root of

$$\begin{aligned} a_F[u]:= \sum _k \int _0^1 \left( |u_k'(r)|^2 + \frac{\lambda _k^2}{r^2} |u_k(r)|^2\right) r dr. \end{aligned}$$

(4.5)

Let B be a positive self-adjoint operator acting in a closed subspace $N_B$ of $N= \text {ker}~A^{*}$ which is associated with the self-adjoint extension $A_B$ of A in the KVB theory, and let $a_B[\cdot ,\cdot ], a_F[\cdot ,\cdot ], b[\cdot ,\cdot ]$ be the forms of $A_B, A_F, B$, respectively.

For $u, \varphi \in Q(A_B)$, we have

$$\begin{aligned} \begin{aligned} u =&v + w,\ v \in Q(A_F),\ w \in Q(B) \\ \varphi =&\vartheta + \zeta , \vartheta \in Q(A_F), \ \zeta \in Q(B). \end{aligned} \end{aligned}$$

(4.6)

and since $v(R,\theta ) = \vartheta (R,\theta ) = 0$ (which follows from the definition of $Q(A_F)$),

$$\begin{aligned} A^{*} w= & {} 0 \ \text {in} \ \Omega ,\ \ w(R) = u(R), \nonumber \\ A^{*} \zeta= & {} 0\ \text {in}\ \Omega ,\ \ \zeta (R) = \varphi (R). \end{aligned}$$

Remark 4.1

Since $v(1, \theta ) =0$ for any $v \in Q(A_F)$, $Q(A_F)$ coincides with Brasche and Melgaard’s form domain of $A_F$ in [4], and so $A_F$ is determined in their Theorem 4.5.

We now proceed as in the proof of Theorem 2.2. For $u=u_F+w \in D(A_B)$ and $\varphi = \vartheta + \zeta \in Q(A_B)$

$$\begin{aligned} (A_Bu,\varphi ) = \int _{\Omega } (Au_F) \overline{\varphi } dx \end{aligned}$$

(4.7)

and

$$\begin{aligned} (A_Bu,\varphi ) = \int _{\Omega } (Au_F) \overline{\vartheta } dx + b[w,\zeta ]. \end{aligned}$$

(4.8)

Consequently

$$\begin{aligned} b[w,\zeta ] = \int _{\Omega } (Au_F) \overline{\zeta } dx. \end{aligned}$$

(4.9)

If $\{\psi _k\}$ is an orthonormal basis of Q(B), then we have with the same notation as in Sect. 3, that $u = u_F + w \in D(A_B)$ if and only if

$$\begin{aligned} \forall k : \ \ \sum _j b_{jk} w_j = \int _{\Omega } (A u_F) \overline{\psi _k} dx. \end{aligned}$$

(4.10)

The transformation

$$\begin{aligned} W f(r) = r^{1/2} f(r),\ \ f \in L^2(0,1;rdr) \end{aligned}$$

is a unitary operator from $L^2(0,1;rdr)$ onto $L^2(0,1)$, and as $\{e^{im\theta }/\sqrt{2 \pi }\}_{m \in \mathbb {Z}}$ is an orthonormal basis of $L^2(\mathbb {S}^1)$ we have

$$\begin{aligned} L^2(\Omega ) = \bigoplus _{m \in \mathbb {Z}} W^{-1} L^2(0,1) \otimes \mathrm {Span}\left\{ e^{im\theta }/\sqrt{2 \pi }\right\} . \end{aligned}$$

In terms of this decomposition it follows that

$$\begin{aligned} A = \bigoplus _{m \in \mathbb {Z}} W^{-1} T^{(m)}W \otimes 1, \end{aligned}$$

(4.11)

where $T^{(m)}$ is the closure in $L^2(0,1)$ of the operator defined on $C_0^{\infty }(0,1)$ by the Sturm–Liouville expression

$$\begin{aligned} \tau ^m y :=-y''+\left( (m+ \alpha )^2 -1/4\right) r^{-2}y, \ m \in \mathbb {Z},\ 0<\alpha < 1, \end{aligned}$$

(4.12)

i.e., $T^{(m)}$ is the minimal operator in $L^2(0,1)$ generated by $\tau ^m$. With $\nu = m+\alpha$, the set $\{r^{1/2+\nu }, r^{1/2-\nu }\}$ is a fundamental system for $\tau ^m u=0$. The expression $\tau ^m$ is non-oscillatory. For $m= -1,~0$, it is in the limit-circle case at 0; for all other values of m, it is in the limit-point case at 0. It is regular at 1 for all values of m. Thus $T^{(m)}$ has deficiency indices (2, 2) for $m=-1,~0$ and (1, 1) otherwise. We shall now apply results from [5] to determine the positive self-adjoint extensions of $T^{(m)}$ in $L^2(0,1)$ for all $m \in \mathbb {Z}$. Note that the singular point here is at the left endpoint of the interval [0, 1], i.e., it is the point 0, unlike the analysis of [5], where it is at the right endpoint. If $S^{(m)}$ is one such extension, then

$$\begin{aligned} \bigoplus _{m \in \mathbb {Z}} W^{-1} S^{(m)}W \otimes 1 \end{aligned}$$

(4.13)

is a positive self-adjoint extension of A.

Remark 4.2

We note that it is unlikely that all positive self-extensions of A are obtained in this way. This assertion is based on the situation for $A_0 = -\Delta +1$ from Example 3.8. As in (4.11),

$$\begin{aligned} A_0= \bigoplus _{m \in \mathbb {Z}} W^{-1} T_{(m)}W \otimes 1 \end{aligned}$$

(4.14)

where now, $T_{(m)}$ is the minimal operator generated by

$$\begin{aligned} \tau _m y =-y''+\left( [m^2 -1/4] r^{-2} +1\right) y,\ m \in \mathbb {Z}. \end{aligned}$$

At 0, $\tau _m$ is non-oscillatory and in the limit-circle case for $m=0$ and is otherwise limit-point. As above for A, if $S_{(m)}$ is a positive self-adjoint extension of $T_{(m)}$ then

$$\begin{aligned} \bigoplus _{m \in \mathbb {Z}} W^{-1} S_{(m)}W \otimes 1 \end{aligned}$$

(4.15)

is a positive self-adjoint extension of $A_0$. All such extensions have boundary conditions which depend on behaviour at 0, in view of the presence of the extension $S_{(0)}$ of $T_{(0)}$ which has deficiency indices (1, 1). However in Remark 3.9 we saw that this is not so for the Krein–von Neumann extension of $A_0!$

We shall proceed to determine the extensions $T^{(m)}$ in (4.11).

4.1 The Case when $\tau ^m$ is Limit Point at 0 ($m \ne -1,0$)

Theorem 2.1 in [5] establishes a one-one correspondence between the positive self-adjoint extensions of $T^{(m)}$ in this case and the one-parameter family $\{T^{(m)}_l\}$, $0 \le l \le \infty$ of restrictions of $(T^{(m)})^{*}$ to the domains

$$\begin{aligned} \mathcal {D}(T^{(m)}_{l})= \left\{ v{:}\,v \in \mathcal {D}((T^{(m)})^{*}), v'(1) =[\psi '(1)-l\Vert \psi \Vert ^2]v(1)\right\} . \end{aligned}$$

(4.16)

Here $\psi$ is a real function in $L^2(0,1)$ which satisfies $\tau ^m \psi =0$ and $\psi (1) =1.$ We therefore have

$$\begin{aligned} \psi (r) = r^{1/2 +|\nu |}, \ \psi '(1)= 1/2 + |\nu |,\ \Vert \psi \Vert ^2 = \left[ 2(1+|\nu |)\right] ^{-1}. \end{aligned}$$

4.2 The Case when $\tau ^m$ is Limit-Circle at 0 ($m=-1,0$) and $\mathbf dim ~N_B =1$

From Theorem 2.2 in [5] and writing $T^{*}$ instead of $\left( T^{(m)}\right) ^{*}$ for simplicity, it follows that the positive self-adjoint extensions of the operator $T^{(m)}$ which correspond to the pair $\{B,N_B\}$ in the KVB theory with $\text {dim}N_B =1$ form a one-parameter family $T_{\beta }$ of restrictions of $T^{*}$ with domains

$$\begin{aligned} \mathcal {D}(T_{\beta }):= \left\{ v \in \mathcal {D}(T^{*}){:}\,\left[ g^2\left\{ \left( \frac{v}{g}\right) \left( \frac{\psi }{g}\right) '- \left( \frac{v}{g}\right) ' \left( \frac{\psi }{g}\right) \right\} \right] _0^1 = \beta v(1) \Vert \psi \Vert ^2 \right\} , \end{aligned}$$

(4.17)

where $\psi$ is a real function in $N_B$ with $\psi (1) =1$, g is the non-principal solution of $\tau ^m u=0$ and $\beta \ge 0$. The non-principal solution is $r^{1/2 - |\nu |},~ \nu = m+\alpha$. The Wronskian W is given by

$$\begin{aligned} W(v,\psi )= g^2\left\{ \left( \frac{v}{g}\right) \left( \frac{\psi }{g}\right) '- \left( \frac{v}{g}\right) ' \left( \frac{\psi }{g}\right) \right\} . \end{aligned}$$

(4.18)

The limits at 0 of the first and the second terms in (4.17) exist separately. To see this, let

$$\begin{aligned} \frac{g''(r)}{g(r)} = (\nu ^2 - 1/4)r^{-2} =: q(r). \end{aligned}$$

Hence by the Jacobi identity [5, Equation (1.10)], for $v\in \mathcal {D}(T^{*})$ we get

$$\begin{aligned} -\frac{1}{g} \left[ g^2 \left( \frac{v}{g}\right) '\right] ' = -v'' +qv = T^{*} v \in L^2(0,1). \end{aligned}$$

(4.19)

Thus, since $g \in L^2(0,1)$,

$$\begin{aligned} - \left[ g^2 \left( \frac{v}{g}\right) '\right] ' = g \left( T^{*} v\right) \in L^1(0,1), \end{aligned}$$

which implies that

$$\begin{aligned} \lim _{r\rightarrow 0+} \left[ g^2 \left( \frac{v}{g}\right) '\right] (r) \quad \hbox { and } \quad \lim _{r \rightarrow 0+} g^2(r) \left( \frac{\psi }{g}\right) '(r) \end{aligned}$$

both exist. From [9, Remark 3] (see also [5, (2.9)]), $\lim _{r \rightarrow 0+}(v/g)(r)$ exists, which confirms our assertion that the separate limits exist.

We shall now determine the boundary conditions satisfied by the self-adjoint extensions of $T^{(m)}$ in the two cases corresponding to $\nu = m + \alpha ,\ m = -1, 0,\ \alpha \in (0,1)$.

4.2.1 The Case $m = -1,\ \nu =-1 +\alpha \in (-1,0)$

In this case, the non-principal solution is $g(r) = r^{1/2 +\nu }$ and $\psi (r) = \gamma \left( C_1 r^{1/2-\nu } + C_2 r^{1/2+\nu }\right)$, where $\gamma = (C_1+C_2)^{-1}$ for $C_1,C_2$ are constants and $C_1\ne 0$. Thus,

$$\begin{aligned} \left( \frac{\psi }{g}\right) (r) = \gamma \left( C_1r^{-2\nu } +C_2\right) ,\ \ \ g^2\left( \frac{\psi }{g}\right) '(r) = -2 \gamma \nu C_1, \end{aligned}$$

(4.20)

and so using (4.18)

$$\begin{aligned} W(r) =\left( \frac{v(r)}{g(r)}\right) (-2 \gamma \nu C_1) -\left[ g^2(r) \left( \frac{v}{g}\right) '(r)\right] \left[ \gamma (C_1r^{-2\nu } +C_2)\right] . \end{aligned}$$

(4.21)

The value at $r=1$ is

$$\begin{aligned} W(1) = -v'(1) - v(1)\left[ 2 \gamma \nu C_1 -1/2 - \nu \right] . \end{aligned}$$

(4.22)

By (4.20) and since $\nu < 0$, the limits at 0 of both terms in (4.21) exist and

$$\begin{aligned} \lim _{r \rightarrow 0+} W(r)= & {} -2\gamma \nu C_1 \lim _{r \rightarrow 0+} \frac{v(r)}{r^{1/2+\nu }} - \gamma C_2 \lim _{r \rightarrow 0+} g^2(r) \left( \frac{v}{g}\right) '(r) \nonumber \\= & {} -2\gamma \nu C_1 \lim _{r \rightarrow 0+}\frac{v(r)}{r^{1/2+\nu }} \nonumber \\&-\,\gamma C_2\lim _{r \rightarrow 0+}\left[ r^{1/2+\nu } v'(r) -(1/2+\nu ) r^{-1/2 +\nu } v(r)\right] .\qquad \end{aligned}$$

(4.23)

Thus the boundary condition for $A_B$ in this case is

$$\begin{aligned} 2 \gamma \nu C_1\left\{ \lim _{r \rightarrow 0+} \frac{v(r)}{r^{1/2+\nu }} - v(1)\right\} +\left\{ \gamma C_2\lim _{r \rightarrow 0+}f_1(r) - f_1(1)\right\} =\beta v(1) \Vert \psi \Vert ^2, \end{aligned}$$

(4.24)

where $f_1(r) := \left[ r^{1/2+\nu } v'(r) -(1/2+\nu ) r^{-1/2 +\nu } v(r)\right]$.

4.2.2 The Case $m =0,\ \nu = \alpha \in (0,1)$

This time, the non-principal solution is $g(r) = r^{1/2-\nu }$ and, with $\psi$ as above, we have

$$\begin{aligned} \left( \frac{\psi }{g}\right) (r) = \gamma \left( C_1 +C_2r^{2 \nu }\right) ,\ \ \ g^2\left( \frac{\psi }{g}\right) '(r) = 2 \gamma \nu C_2, \end{aligned}$$

(4.25)

giving

$$\begin{aligned} W(r) =\left( \frac{v(r)}{g(r)}\right) (2 \gamma \nu C_2) -\left[ g^2(r) \left( \frac{v}{g}\right) '(r)\right] \left[ \gamma (C_1 +C_2r^{2\nu })\right] . \end{aligned}$$

(4.26)

Therefore

$$\begin{aligned} W(1) = 2 \gamma \nu C_2 v(1) - \left[ v'(1) -(1/2-\nu ) v(1)\right] . \end{aligned}$$

(4.27)

By (4.25) and since $\nu >0$, both limits at 0 in (4.26) exist and

$$\begin{aligned} \lim _{r \rightarrow 0+} W(r)= & {} 2 \gamma \nu C_2\lim _{r \rightarrow 0+} \frac{v(r)}{r^{1/2-\nu }} -\gamma C_1 \lim _{r \rightarrow 0+} g^2(r) \left( \frac{v}{g}\right) '(r) \nonumber \\= & {} 2 \gamma \nu C_2\lim _{r \rightarrow 0+}\frac{v(r)}{r^{1/2-\nu }} \nonumber \\&-\,\gamma C_1 \lim _{r \rightarrow 0+}\left[ r^{1/2-\nu } v'(r) -(1/2-\nu ) r^{-1/2 -\nu } v(r)\right] .\qquad \end{aligned}$$

(4.28)

Thus the boundary condition in this case is

$$\begin{aligned} 2 \gamma \nu C_2\left\{ \lim _{r \rightarrow 0+} \frac{v(r)}{r^{1/2-\nu }} -v(1)\right\} - \left\{ \gamma C_1 \lim _{r \rightarrow 0+}f_2(r) -f_2(1)\right\} = \beta v(1)\Vert \psi \Vert ^2, \end{aligned}$$

(4.29)

where $f_2(r) := \left[ r^{1/2-\nu } v'(r) -(1/2-\nu ) r^{-1/2-\nu }v(r)\right]$.

4.3 The Case when $\tau ^m$ is Limit-Circle at 0 ($m=-1,0$) and $\hbox {dim}N_B=2$

From [5, Theorem 2.2], we have

$$\begin{aligned} \mathcal {D}(A_B):= & {} \left\{ v \in \mathcal {D}(A^{*}){:}\,\left[ g^2\left\{ \left( \frac{v}{g}\right) \left( \frac{\psi _k}{g}\right) '- \left( \frac{v}{g}\right) ' \left( \frac{\psi _k}{g}\right) \right\} \right] _0^1 \right. \nonumber \\&\qquad = \left. \sum _{j=1}^2 b_{jk} c_j,~j=1,2 \right\} {,} \end{aligned}$$

(4.30)

where $B:= (b_{jk})_{j,k=1,2}$ is a matrix of parameters which is non-negative, $\{\psi _1, \psi _2\}$ is a real orthonormal basis of $N_B$ and $c_1,~c_2$ are determined by the values of v / g at 0 and 1:

$$\begin{aligned} \frac{v}{g}(0) = \sum _{j=1}^2 c_j \frac{\psi _j}{g}(0), \ \ \frac{v}{g}(1) = \sum _{j=1}^2 c_j \frac{\psi _j}{g}(1). \end{aligned}$$

(4.31)

The main difference from the analysis of the previous section is that we now replace $\psi$ by an orthonormal basis $(\psi _1,\psi _2)$ obtained from the linearly independent basis elements

$$\begin{aligned} r^{1/2-|\nu |} \quad \hbox { and }\quad r^{1/2 +|\nu |},\ \nu = m+\alpha . \end{aligned}$$

On using the Gram–Schmidt procedure, we obtain the orthogonal vectors

$$\begin{aligned} r^{\frac{1}{2}-|\nu |} \; \mathrm{and}\; r^{|\nu |+\frac{1}{2}} -(1- |\nu | ) r^{\frac{1}{2}-|\nu |}, \end{aligned}$$

and the orthonormal system

$$\begin{aligned} \psi _1= \sqrt{ 2 (1-|\nu |) }r^{\frac{1}{2}-|\nu |}, \quad \psi _2 =\frac{ \sqrt{ 2 (1+|\nu |)}}{|\nu |} \left( r^{|\nu |+\frac{1}{2}} -(1- |\nu | ) r^{\frac{1}{2}-|\nu |}\right) . \end{aligned}$$

The non-principal solution is $g(r) = r^{1/2-| \nu |}$ and we have

$$\begin{aligned} \psi _1/g=\sqrt{2(1-|\nu |)},\quad (\psi _1/g)'=0 \end{aligned}$$

and

$$\begin{aligned} \psi _2/g= \frac{ \sqrt{ 2 (1+|\nu |)}}{|\nu |} ( r ^{2|\nu |}+|\nu |-1), \quad g^2 (\psi _2/g)'=2\sqrt{2 (|\nu |+1)}. \end{aligned}$$

Let

$$\begin{aligned} W_k = g^2\left\{ \left( \frac{v}{g}\right) \left( \frac{\psi _k}{g}\right) ' - \left( \frac{v}{g}\right) ' \left( \frac{\psi _k}{g}\right) \right\} ,\ \ k=1,2. \end{aligned}$$

Then

$$\begin{aligned} W_1(r) =-\sqrt{2(1-|\nu |)}\, g^2(r) \left( \dfrac{v}{g} \right) '(r) , \end{aligned}$$

and we set

$$\begin{aligned} \Theta _1 := W_1(1)-W_1(0)= - \sqrt{2(1-|\nu |)} \Bigg ( \left( \dfrac{v}{g} \right) '(1) - \lim _{r\rightarrow 0} g^2(r) \left( \dfrac{v}{g} \right) '(r) \Bigg ). \end{aligned}$$

(4.32)

Also

$$\begin{aligned} W_2(r) =2\sqrt{2(1+|\nu |)} \left( \dfrac{v}{g} \right) (r) -\frac{ \sqrt{ 2 (1+|\nu |)}}{|\nu |} (|\nu | -1 + r^{2|\nu |}) g^2 \left( \dfrac{v}{g} \right) '(r) \end{aligned}$$

giving

$$\begin{aligned} \Theta _2:= & {} W_2(1)-W_2(0)\nonumber \\= & {} \sqrt{2 (1+|\nu |)} \left[ 2 v(1) -\left( \dfrac{v}{g}\right) '(1) \right] \nonumber \\&-\,\sqrt{2(1+|\nu |)}\lim _{r \rightarrow 0}\left\{ 2\left( \dfrac{v}{g} \right) (r) -\frac{1}{|\nu |} (|\nu | -1 + r^{2|\nu |})g^2\left( \dfrac{v}{g} \right) '(r)\right\} .\nonumber \\ \end{aligned}$$

(4.33)

We also have

$$\begin{aligned} V:=\left( \begin{array}{c} v(1)/g(1)\\ v(0)/g(0) \end{array}\right) =\left( \begin{array}{ll} \sqrt{ 2 (1-|\nu |)} &{}\quad \sqrt{2(1+|\nu |)}\\ \sqrt{2(1-|\nu |) }&{}\quad \sqrt{ 2 (1+|\nu |)} (1-1/|\nu |) \end{array}\right) \left( \begin{array}{c} c_1\\ c_2\end{array}\right) . \end{aligned}$$

Setting $V=\Psi c$, where

$$\begin{aligned} \Psi := \left( \begin{array}{ll} \sqrt{ 2 (1-|\nu |)} &{}\quad \sqrt{2(1+|\nu |)} \\ \sqrt{2(1-|\nu |) }&{}\quad \sqrt{ 2 (1+|\nu |)}(1-1/|\nu |) \end{array}\right) \end{aligned}$$

is invertible and has inverse

$$\begin{aligned} \Psi ^{-1} = \frac{|\nu |}{2\sqrt{(1-|\nu |^2)}}\left( \begin{array}{ll} \sqrt{2(1+|\nu |)}(1/|\nu |-1) &{}\quad \sqrt{2(1+|\nu |)} \\ \sqrt{2(1-|\nu |)}&{}\quad -\sqrt{2(1-|\nu |)} \end{array}\right) . \end{aligned}$$

(4.34)

The boundary condition in (4.30) is therefore

$$\begin{aligned} \left( \begin{array}{l} \Theta _1\\ \Theta _2 \end{array}\right) =B\Psi ^{-1}V. \end{aligned}$$

(4.35)

For the Krein–von Neumann extension $\left( T^{(m)}\right) _K$ of the one-dimensional operator, the boundary condition is determined by $\Theta _1 = \Theta _2 = 0$: Hence

$$\begin{aligned} \lim _{r\rightarrow 0} g^2(r) \left( \dfrac{v}{g} \right) '(r) = \left( \dfrac{v}{g} \right) '(1) \end{aligned}$$

(4.36)

and

$$\begin{aligned} \lim _{r \rightarrow 0} 2\left( \dfrac{v}{g} \right) (r)=2 v(1) - \frac{1}{|\nu |}\left( \dfrac{v}{g}\right) '(1). \end{aligned}$$

(4.37)

Following Remark 2.1, the Friedrichs extension $\left( T^{(m)}\right) _F$ is obtained by $c_1=c_2=0$, so that the right hand side of (4.30) is finite. Thus, from (4.31), the boundary conditions are given by $(v/g)(0)=(v/g)(1)=0$, i.e.,

$$\begin{aligned} \lim _{r \rightarrow 0} \frac{v(r)}{r^{1/2-|\nu |}} = v(1)=0. \end{aligned}$$

(4.38)