Non-smoothness of the fundamental solutions for Schrödinger equations with super-quadratic and spherically symmetric potential

Kato, Keiichi; Nakahashi, Wataru; Tadano, Yukihide

doi:10.1063/5.0184443

We study non-smoothness of the fundamental solution for the Schrödinger equation with a spherically symmetric and super-quadratic potential in the sense that V(x) ≥ C|x|^2+ɛ at infinity with constants C > 0 and ɛ > 0. More precisely, we show the fundamental solution E(t, x, y) does not belong to C¹ as a function of (t, x, y), which partially solves Yajima’s conjecture.

I. INTRODUCTION

We consider the following initial value problem for the Schrödinger equation on

R^{d}

with a super-quadratic and spherically symmetric potential V(x)

\{\begin{cases} i \frac{\partial u}{\partial t} (t, x) = - Δ u (t, x) + V (x) u (t, x), (t, x) \in R \times R^{d}, \\ u (0, \cdot) = u_{0} \in L^{2} (R^{d}), \end{cases}

(1.1)

where d ≥ 3,

i = \sqrt{- 1}

⁠,

Δ = \sum_{j = 1}^{d} \partial^{2} / \partial {x_{j}}^{2}

⁠, and

u : R \times R^{d} \to C

⁠. The aim of this paper is to show non-smoothness of the fundamental solutions under the following assumption, which partially solves Yajima’s conjecture.¹²

Assumption 1.1.

The potential

V (x) \in C^{3} (R^{d})

is a real valued spherically symmetric function. Moreover, letting

V (x) = \tilde{V} (| x |)

, there are constants R > 0 and c > 1 such that for r ∈ [R, ∞)

{\tilde{V}}^{''} (r) > 0,

(1.2)

r {\tilde{V}}^{'} (r) \geq 2 c \tilde{V} (r) > 0,

(1.3)

{\tilde{V}}^{(j)} (r) = O (\frac{1}{r}) {\tilde{V}}^{(j - 1)} (r), j = 1, 2, 3 .

(1.4)

It follows from that there exists C > 0 such that V(x) ≥ C|x|^2c for |x| ≫ 1, which implies H = −Δ + V is essentially self-adjoint on

C_{0}^{\infty} (R^{d})

⁠. Thus the equation generates a unique unitary propagator U(t) = e^−itH on

L^{2} (R^{d})

and u(t, x) = (U(t)u₀) (x) is the unique solution of . We denote the integral kernel of U(t) by E(t, x, y), that is,

(U (t) u_{0}) (x) = \int_{R^{d}} E (t, x, y) u_{0} (y) d y,

and we call E(t, x, y) the fundamental solution of .

The main theorem of this paper is the following, which claims that E(t, x, y) is generically nowhere C¹.

Theorem 1.2.

Suppose that d ≥ 3 and that V(x) satisfies Assumption 1.1. Then, for any $t_{0} \in R$ and r₁, r₂ > 0, there exist $x_{0}, y_{0} \in R^{d}$ , satisfying |x₀| = r₁ and |y₀| = r₂, such that the fundamental solution of (1.1) does not belong to C¹ near (t₀, x₀, y₀).

Moreover, we can obtain the following proposition which is (almost) stronger than Theorem 1.2.

Proposition 1.3.

Suppose that d ≥ 3 and that V(x) satisfies Assumption 1.1. Let Y_n be a spherical harmonic of arbitrary degree n with

‖ Y_{n} ‖_{L^{2} (S^{d - 1})} = 1

and let

E^{Y_{n}} (t, x, y) = {(E (t, | x | \cdot, y), Y_{n})}_{L^{2} (S^{d - 1})} Y_{n} (\frac{x}{| x |}) .

Then, for any

(t_{0}, x_{0}, y_{0}) \in R \times R^{d} \times R^{d}

,

E^{Y_{n}}

is not in C¹ near (t₀, x₀, y₀).

Here ${(\cdot, \cdot)}_{H}$ denotes the inner product of a Hilbert space $H$ ⁠, and we suppose ${(\cdot, \cdot)}_{H}$ is linear with respect to the first entry and anti-linear with respect to the second entry.

We give some comments on what Theorem 1.2 and Proposition 1.3 mean and how our proof works. We denote the self-adjoint extension of H = −Δ + V by the same symbol. Since V(x) → ∞ as |x| → ∞ under Assumption 1.1, the spectrum σ(H) of H is discrete. Thus we have, at least formally,

E (t, x, y) = \sum_{l} e^{- i λ_{l} t} u_{l} (x) \bar{u_{l} (y)},

where

{u_{l}}_{l}

is an arbitrary complete orthonormal system of

L^{2} (R^{d})

with each u_l being an eigenfunction of H: Hu_l = λ_lu_l.

It is known that H is decomposed by the partial wave expansion; let {Y_nm∣n = 0, 1, 2, …, m = 1, 2, …, d_n} be a complete orthonormal system of

L^{2} (S^{d - 1})

such that each Y_nm is a spherical harmonic of degree n, where

d_{n} = (\binom{d + n - 1}{d - 1}) - (\binom{d + n - 3}{d - 1})

is the dimension of the space of the spherical harmonics of degree n. Then we have

E (t, x, y) = \sum_{n = 0}^{\infty} \sum_{m = 1}^{d_{n}} \sum_{l = 0}^{\infty} e^{- i λ_{n, l} t} | x |^{- \frac{d - 1}{2}} f_{n, l} (| x |) Y_{n m} (\frac{x}{| x |}) \bar{| y |^{- \frac{d - 1}{2}} f_{n, l} (| y |) Y_{n m} (\frac{y}{| y |})},

(1.5)

where f_n,l ∈ L²((0, ∞)) is the normalized λ_n,l-eigenfunction of the Schrödinger operator on the half line (0, ∞):

- f_{n, l}^{″} (r) + (\frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r)) f_{n, l} (r) = λ_{n, l} f_{n, l} (r), r \in (0, \infty),

(1.6)

with the boundary condition f_n,l(0) = 0. Hence, if Y_n is a spherical harmonic of degree n, we have

\begin{array}{l} E^{Y_{n}} (t, x, y) & = \sum_{l = 0}^{\infty} e^{- i λ_{n, l} t} | x |^{- \frac{d - 1}{2}} f_{n, l} (| x |) Y_{n} (\frac{x}{| x |}) \bar{| y |^{- \frac{d - 1}{2}} f_{n, l} (| y |) Y_{n} (\frac{y}{| y |})} \\ = | x |^{- \frac{d - 1}{2}} | y |^{- \frac{d - 1}{2}} Y_{n} (\frac{x}{| x |}) \bar{Y_{n} (\frac{y}{| y |})} \sum_{l = 0}^{\infty} e^{- i λ_{n, l} t} f_{n, l} (| x |) \bar{f_{n, l} (| y |)} . \end{array}

Since

\sum_{l = 0}^{\infty} e^{- i λ_{n, l} t} f_{n, l} (r) \bar{f_{n, l} (s)}

is the fundamental solution associated to the above one-dimensional Schrödinger operator, we can employ the argument of Yajima .¹²

One might expect that Proposition 1.3 and the representation “imply” that our E(t, x, y), the formal summation over spherical harmonics

E (t, x, y) = \sum_{n = 0}^{\infty} \sum_{m = 1}^{d_{n}} E^{Y_{n m}} (t, x, y),

(1.7)

would be nowhere in C¹, which is, however, still not clear. Instead, we can obtain Theorem 1.2.

Remark 1.4.

(i)

E^{Y_{n}} (t, x, y)

is actually defined as a distribution on

R^{2 d + 1}

by

\begin{array}{l} ⟨ E^{Y_{n}}, F ⟩ & = \int_{R} \int_{R^{d}} \int_{R^{d}} \int_{S^{d - 1}} E (t, | x | ω, y) \bar{Y_{n} (ω)} Y_{n} (\frac{x}{| x |}) F (t, x, y) d ω d y d x d t \\ = \int_{R} \int_{R^{d}} \int_{S^{d - 1}} \bar{Y_{n} (ω)} Y_{n} (\frac{x}{| x |}) [U (t) (F (t, x, \cdot))] (| x | ω) d ω d x d t . \end{array}

The representation (1.5) implies the identity

E^{Y_{n}} (t, x, y) = {(E (t, x, | y | \cdot), \bar{Y_{n}})}_{L^{2} (S^{d - 1})} \bar{Y_{n} (\frac{y}{| y |})} .

The smoothness of the fundamental solution is related to the growth rate of V. Fujiwara (Ref. 2, Theorem 1.1) has given the construction of the fundamental solution E(t, x, y) with the classical orbit if V is at most of quadratic growth, which shows that E(t, x, y) is smooth with respect to (x, y) for any t ≠ 0 small enough as a corollary.

We can give an alternative proof for the smoothness of the fundamental solution by Ref. 4, which shows characterization of wave front set of solutions to Schrödinger equations with sub-quadratic potentials via wave packet transform. Theorem 1.2 in Ref. 4 yields that (x₀, ξ₀) is not in the wave front set of E(t₀, ·) if and only if

| e^{- i t_{0} △} φ_{λ} (x (0; t_{0}, x_{0}, λ ξ_{0})) | \leq C λ^{- N} for λ \geq 1,

where φ is a Schwartz function, φ_λ(x) = λ^d/2φ(λ^1/2x) and x(t; t₀, x₀, λξ₀) is the solution of

\dot{x} (t) = ξ (t), \dot{ξ} (t) = - \nabla V (t, x), x (t_{0}) = x_{0}, ξ (t_{0}) = λ ξ_{0}

⁠, which implies the smoothness of the fundamental solutions for Schrödinger equations with sub-quadratic potentials.

On the other hand, Yajima (Ref. 12, Theorem 1.2) has studied that if V is super-quadratic and the spatial dimension is one, then E(t, x, y) is not smooth anywhere with respect to (t, x, y). The proof is given by the estimates of eigenvalues and eigenfunctions of H. Yajima has conjectured that the same result as in Yajima (Ref. 12, Theorem 1.2) is true even for higher dimensional cases.

As a first step for generalization of Yajima’s result, we treat the case that V is spherically symmetric with the dimension d ≥ 3. We use the unitary equivalence of H = −Δ + V to ⨁_n,mH_nm, where the operator H_nm is defined on (0, ∞) (see Lemma 2.3), and the estimates of eigenvalues and eigenfunctions which are shown by the same way as in Ref. 12. Our main theorem partially solves Yajima’s conjecture. In the case that d = 2, the projection of H onto the subspace of spherically symmetric functions is unitarily equivalent to $H_{0} = - \frac{d^{2}}{d r^{2}} - \frac{1}{4 r^{2}} + \tilde{V} (r)$ on L²((0, ∞)) (see Sec. 2). The potential $- \frac{1}{4 r^{2}} + \tilde{V} (r)$ is not bounded from below near 0, which requires additional argument of self-adjoint extensions. We shall discuss the two-dimensional case in the forthcoming paper.

For a Schrödinger operator −Δ_g on a complete Riemannian manifold (M, g), the smoothness of the fundamental solution depends on whether M is compact or not. Kapitanski and Rodnianski (Ref. 3, Theorem I–III) has studied that E(t, x, y) is not smooth if M is the circle. Taylor (Ref. 8, Sec. 1) has mentioned that E(t, x, y) is not smooth if M is the sphere. Yajima (Ref. 12, Remark 4) has pointed out that E(t, x, y) is not smooth if M is the bounded interval [0, π] with the Dirichlet condition. Taira (Ref. 7, Remark 3.3) has studied that E(t, x, y) is not smooth if M is compact. When M is non-compact, Doi (Ref. 1, Theorem 1.5) has studied the smoothness of the fundamental solution in terms of the wave front set. Taira (Ref. 7, Theorems 1.1) has given a sufficient condition under which E(t, x, y) is smooth.

We introduce some notation. For sets U and V, we write U ⋐ V if U is relatively compact with respect to V. We write $C_{0}^{\infty} (U) = {f \in C^{\infty} (U) ∣ s u p p f ⋐ U}$ ⁠, where suppf denotes the support of f. For an open interval I, we denote the Sobolev space on I of order $k \in N$ by H^k(I) = {f ∈ L²(I)∣f^(m) ∈ L²(I) for m = 1, …, k}. We denote by $H_{0}^{1} (I)$ the closure of $C_{0}^{\infty} (I)$ in H¹(I). We denote the Schwartz space by $S (R^{d})$ ⁠. For any $f \in S (R)$ ⁠, the Fourier transform $\hat{f}$ of f is defined by $\hat{f} (λ) = \int_{R} e^{- i t λ} f (t) d t$ ⁠.

This paper is organized as follows. In Sec. II, we introduce the polar coordinates to decompose the operator H = −Δ + V into the direct sum of operators on the half line. In Sec. III, we observe estimates of eigenvalues and eigenfunctions. In Sec. IV, we prove Theorem 1.2 and Proposition 1.3.

II. SCHRÖDINGER OPERATOR IN THE POLAR COORDINATES

In the following, space dimension d is larger than or equal to 3. We call a function Y on $S^{d - 1}$ spherical harmonic of degree n if Y is the restriction to $S^{d - 1}$ of a homogeneous harmonic polynomial of degree n. We denote by $H_{n}$ the space of spherical harmonics of degree n. The following lemma is well-known [for more details and proofs, see Stein and Weiss (Ref. 6, Sec. 2 in Chap. IV) or Yajima (Ref. 13, Sec. 13 in Chap. 4)].

Lemma 2.1.

Let $Δ_{S^{d - 1}}$ be the Laplace-Beltrami operator on $S^{d - 1}$ . Then:

$- Δ_{S^{d - 1}} Y = n (n + d - 2) Y$ for $Y \in H_{n}$ ⁠.
$L^{2} (S^{d - 1}) = ⨁_{n = 0}^{\infty} H_{n}$ ⁠.
$d_{n} = \dim H_{n} = (\binom{d + n - 1}{d - 1}) - (\binom{d + n - 3}{d - 1})$ ⁠.

Let {Y_nm∣m = 1, …, d_n} be an orthogonal basis of

H_{n}

⁠, then {Y_nm∣n = 0, 1, … and m = 1, …, d_n} is a complete orthonormal system of

L^{2} (S^{d - 1})

⁠. If we define

J_{n m} : L^{2} (R^{d}) \to L^{2} ((0, \infty))

by

J_{n m} u (r) = r^{\frac{d - 1}{2}} {(u (r \cdot), Y_{n m})}_{L^{2} (S^{d - 1})},

we easily see that the operator

J : L^{2} (R^{d}) ∋ u ⟼ {\{J_{n m} u (r)\}}_{n m} \in ⨁_{n, m} L^{2} ((0, \infty))

is a unitary operator, and that the adjoint is given by

J^{*} ({g_{n m}}) (x) = \sum_{n, m} J_{n m}^{*} (g_{n m}) (x) = \sum_{n, m} | x |^{- \frac{d - 1}{2}} g_{n m} (| x |) Y_{n m} (\frac{x}{| x |}) .

Since

Δ = \partial_{r}^{2} + \frac{d - 1}{r} \partial_{r} + \frac{1}{r^{2}} Δ_{S^{d - 1}}

holds in the polar coordinates

(r, ω) \in (0, \infty) \times S^{d - 1}

⁠, we have

J_{n m} H J_{n m}^{*} g (r) = - g^{″} (r) + \{\frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r)\} g (r) .

Since the above operators are independent of m, we let

H_{n} = J_{n m} H J_{n m}^{*} = - \frac{d^{2}}{d r^{2}} + \frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r) .

Lemma 2.2.

If (d, n) ≠ (3, 0), H_n is essentially self-adjoint on $C_{0}^{\infty} ((0, \infty))$ . In particular, the domain of its closure is the maximal domain {g ∈ L²((0, ∞))∣H_ng ∈ L²((0, ∞))}. If (d, n) = (3, 0), H₀ is essentially self-adjoint on $J_{00} C_{0}^{\infty} (R^{d})$ , and the domain of its closure is ${g \in H_{0}^{1} ((0, \infty)) ∣ H_{n} g \in L^{2} ((0, \infty))}$ ⁠.

Proof.

Weyl’s limit point-limit circle criterion (Ref. 5, Theorem X.7) implies that H_n is essentially self-adjoint on $C_{0}^{\infty} ((0, \infty))$ if and only if H_n is in the limit point case at both 0 and ∞. It follows from Ref. 5 (Theorem X.8) that H_n is in the limit point case at ∞ for any (d, n), since the potential of H_n is bounded from below. Reference 5 (Theorem X.10) implies that H_n is in the limit point case at 0 if and only if the coefficient $\frac{(d - 1) (d - 3)}{4} + n (n + d - 2)$ of r⁻² is not smaller than $\frac{3}{4}$ ⁠, i.e., (d, n) ≠ (3, 0). Thus we have the assertion in the case (d, n) ≠ (3, 0).

On the other hand, if (d, n) = (3, 0), H_n is in the limit point case at ∞ and is in the limit circle case at 0. In this case the self-adjoint extensions of

H_{0} = - \frac{d^{2}}{d r^{2}} + \tilde{V} (r)

on

C_{0}^{\infty} ((0, \infty))

are those with the domain

D_{θ} = {g \in H^{1} ((0, \infty)) ∣ - g^{″} + \tilde{V} g \in L^{2} ((0, \infty)), g (0) \cos θ + g^{'} (0) \sin θ = 0},

where θ ∈ [0, 2π). We can see that

C_{0}^{\infty} ((0, \infty)) \subset J_{00} C_{0}^{\infty} (R^{d})

and that

J_{00} C_{0}^{\infty} (R^{d}) \subset D_{θ}

only if θ = 0, since J₀₀u(0) = 0 and (J₀₀u)′(0) = 1 if

u \in C_{0}^{\infty} (R^{d})

equals to

| S^{2} |^{- 1 / 2} = {(4 π)}^{- 1 / 2}

near 0. Therefore, the self-adjoint extension of H₀ on

J_{00} C_{0}^{\infty} (R^{d})

is unique and the domain of its closure is

D_{0}

⁠.□

We denote the above-mentioned self-adjoint extensions by the same symbols.

Lemma 2.3.

One has

J H J^{*} = ⨁_{n, m} H_{n m}, where H_{n m} = H_{n} .

More precisely,

J^{*} ({g_{n m}}_{n m}) \in D (H)

if and only if g_nm ∈ D(H_n) for any n, m and

\sum_{n, m} ‖ H_{n} g_{n m} ‖^{2} < \infty

, and

J H J^{*} ({g_{n m}}_{n m}) = {H_{n} g_{n m}}_{n m}, J^{*} ({g_{n m}}_{n m}) \in D (H) .

III. ESTIMATES OF EIGENVALUES AND EIGENFUNCTIONS

In this section, we fix n, and consider the eigenvalue problem

- \frac{d^{2}}{d r^{2}} f (r) + \{\frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r)\} f (r) = λ f (r)

(3.1)

with the boundary condition associated to the domain of H_n. The following argument is based on Ref. 12 (Sec. 3).

A. Asymptotic behavior in bounded regions

We first study the asymptotic behavior of the solutions of in the subset

Ω_{λ} = \{r \in [λ^{- 1 / 4}, \infty) |\frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r) \leq \frac{λ}{2}\}

as λ → ∞. Let λ₀ be sufficiently large so that Ω_λ is an interval for λ ≥ λ₀. We write

U_{d, n} (r) = \frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r)

for simplicity. We set

S (r) = \int_{1}^{r} \sqrt{λ - U_{d, n} (s)} d s, a (r) = {(λ - U_{d, n} (r))}^{- 1 / 4},

and make change of variables y = S(r) and f(r) = a(r)w(y). Then implies

\frac{d^{2} w}{d y^{2}} + w + U (y, λ) w = 0,

(3.2)

where

U (y, λ) = \frac{U_{d, n}^{''} (r (y))}{4 {(λ - U_{d, n} (r (y)))}^{2}} + \frac{5 {(U_{d, n}^{'} (r (y)))}^{2}}{16 {(λ - U_{d, n} (r (y)))}^{3}} .

It follows that w satisfies the integral equation

w (y) = w (0) \cos y + w^{'} (0) \sin y - \int_{0}^{y} \sin (y - z) U (z, λ) w (z) d z .

(3.3)

Note that w(0) and w′(0) are given by f(1), f′(1), U_d,n(1),

U_{d, n}^{'} (1)

and λ; in fact,

\begin{array}{l} w (0) & = {(λ - U_{d, n} (1))}^{1 / 4} f (1), \\ w^{'} (0) & = {\frac{1}{S^{'} (r)} \frac{d}{d r} ({(λ - U_{d, n} (r))}^{1 / 4} f (r))|}_{r = 1} . \end{array}

We set C_λ = (w(0) − iw′(0)) so that w(0) cos y + w′(0) sin y = Re(C_λe^iy).

Lemma 3.1.

Suppose that a real valued function f satisfies (3.1) on Ω_λ. Then, there exists a constant C > 0 independent of λ ≥ λ₀ and r ∈ Ω_λ such that

|a {(r)}^{- 1} f (r) - Re (C_{λ} e^{i S (r)})| \leq C |C_{λ}| λ^{- 1 / 2} .

Proof.

The assertion is proved similarly to Ref. 12 (Lemma 3.1). Let R be the same as in Assumption 1.1. Since

U_{d, n}^{'} (r) \leq C U_{d, n} (r) \leq C (λ - U_{d, n} (r))

on Ω_λ ∩ [R, ∞), we have, for any r₀ ∈ Ω_λ ∩ (R, ∞),

\int_{S (R)}^{S (r_{0})} | U (z, λ) | d z \leq \frac{C}{λ^{3 / 2}} \int_{R}^{r_{0}} U_{d, n}^{''} (r) d r + \int_{R}^{r_{0}} \frac{C U_{d, n}^{'} (r)}{{(λ - U_{d, n} (r))}^{3 / 2}} d r \leq \frac{C}{λ^{1 / 2}} .

On the other hand, for any r₁ ∈ Ω_λ ∩ (0, 1), we have

\int_{S (r_{1})}^{S (1)} | U (z, λ) | d z \leq \frac{C}{λ^{3 / 2}} \int_{λ^{- 1 / 4}}^{1} | U_{d, n}^{''} (r) | d r + \frac{1}{λ^{5 / 2}} \int_{λ^{- 1 / 4}}^{1} C | U_{d, n}^{'} (r) |^{2} d r \leq \frac{C}{λ^{1 / 2}} .

Applying to , we have the assertion.□

We note that

S (r) = \sqrt{λ} (r - 1) + O (λ^{- 1 / 2})

and a(r) = λ^−1/4(1 + O(λ⁻¹)) uniformly on any compact set of (0, ∞). Hence, for any compact set I ⊂ (0, ∞), we have I ⊂ Ω_λ for sufficiently large λ and

f (r) = λ^{- 1 / 4} (R e {C_{λ} e^{i \sqrt{λ} (r - 1)}} + O (C_{λ} λ^{- 1 / 2})),

(3.4)

uniformly in r ∈ I.

The next lemma gives a lower bound of |C_λ| in the previous lemma. The assertion is proved by the same way as in Ref. 12 (Lemma 3.2) and we omit the proof.

Lemma 3.2.

Suppose that a real valued function f ∈ L²((0, ∞)) satisfies (3.1) and

‖ f ‖_{L^{2} ((0, \infty))} = 1

, and let C_λ be as in Lemma 3.1. Then there is a constant C > 0 such that

| C_{λ} | \geq C λ^{(c - 1) / 4 c}, λ \geq λ_{0},

where c > 1 is the constant in Assumption 1.1.

B. Gaps of the eigenvalues

Let λ_n,0 < λ_n,1 < ⋯ be the eigenvalues of H_n. We show that the gap λ_n,l+1 − λ_n,l of eigenvalues increases polynomially with respect to λ_n,l.

Lemma 3.3.

There exists a constant C > 0 which satisfies

|λ_{n, l \pm 1} - λ_{n, l}| \geq C λ_{n, l}^{(c - 1) / 2 c}

(3.5)

for any n and large l.

The proof of the above lemma is found in Ref. 12 (Lemma 3.3), for which we use.

Proposition 3.4.

Let d ≥ 3 and n be fixed. Then as l → ∞

\frac{1}{π} \int_{0}^{X_{n, l}} {\{λ_{n, l} - \tilde{V} (r)\}}^{1 / 2} d r = l + \frac{n}{2} + \frac{d}{4} + o (1),

(3.6)

where

\tilde{V} (X_{n, l}) = λ_{n, l}

⁠.

We show the Proof of Proposition 3.4 in the rest of this section. Note that it has been proved for the case d = 3, see Titchmarsh (Ref. 9, p. 151).

Proposition 3.4 is proved by the same method as in Ref. 9 (p. 158, Sec. 7.13). In Subsection III C, we observe the asymptotic behavior of solutions of (3.1) which are L² near ∞ and 0. We then prove Proposition 3.4 in the last subsection.

C. Asymptotic behavior of eigenfunctions

Lemma 3.5.

For sufficiently large λ, there is a real-valued function ψ_λ ∈ H²((1/2, ∞)) such that

{f \in L^{2} ((1 / 2, \infty)) ∣ H_{n} f = λ f on (1 / 2, \infty)} = {α ψ_{λ} ∣ α \in C}

(3.7)

and

ψ_{λ} (1) = 2 {\{λ - U_{d, n} (1)\}}^{- \frac{1}{4}} \{\cos (Z - \frac{1}{4} π) + O (Z^{- 1})\},

(3.8)

ψ_{λ}^{'} (1) = 2 {\{λ - U_{d, n} (1)\}}^{\frac{1}{4}} \{\sin (Z - \frac{1}{4} π) + O (Z^{- 1})\},

(3.9)

where

Z = \int_{1}^{T} \sqrt{λ - U_{d, n} (s)} d s,

and T = T_λ > 1 is the value satisfying U_d,n(T) = λ.

Proof.

We first remark that the dimension of the left hand side of is one, since H_n is the limit point case at ∞, see the Proof of Lemma 2.2. We give an instant proof for the readers’ convenience. Suppose that there are linearly independent L²-eigenfunctions f₁ and f₂ of H_n with eigenvalue λ. Then the ellipticity of H_n implies

f_{1}, f_{2} \in H_{loc}^{2} ((0, \infty))

⁠, and the Wronskian

W (r) = f_{1} (r) f_{2}^{'} (r) - f_{1}^{'} (r) f_{2} (r) \neq 0

is independent of r. Thus we learn

\int_{1}^{\infty} | W (r) | d r = \infty,

which contradicts

‖ W ‖_{L^{1} ((1, \infty))} \leq ‖ f_{1} ‖_{L^{2} ((1, \infty))} ‖ f_{2}^{'} ‖_{L^{2} ((1, \infty))} + ‖ f_{1}^{'} ‖_{L^{2} ((1, \infty))} ‖ f_{2} ‖_{L^{2} ((1, \infty))} < \infty

⁠.

Let λ be sufficiently large, and let T = T_λ > 1 be as above. We set

ζ (r) = \int_{T}^{r} \sqrt{λ - U_{d, n} (s)} d s,

(3.10)

where the branch is chosen such that arg ζ(r) = π/2 for r > T, and arg ζ(r) = −π for r < T.

We employ Langer’s method in Chap. 22.27 of Ref. 10, where the term

\frac{5}{36 ζ^{2}}

plays the essential role in computations of iterations near ζ = 0, i.e., r = T. Changing variables r → ζ and η = a(r)⁻¹ψ(r), where

a (r) = {(λ - U_{d, n} (r))}^{- 1 / 4}

⁠, imply, at least formally,

\frac{d^{2} η}{d ζ^{2}} + (1 + \frac{5}{36 ζ^{2}}) η = g (ζ) η,

(3.11)

where

g (ζ) = \frac{5}{36 ζ^{2}} - U (r (ζ), λ)

and

U (r, λ) = \frac{U_{d, n}^{''} (r)}{4 {(λ - U_{d, n} (r))}^{2}} + \frac{5 {(U_{d, n}^{'} (r))}^{2}}{16 {(λ - U_{d, n} (r))}^{3}} .

It is well-known that the functions

{(\frac{1}{2} π ζ)}^{\frac{1}{2}} J_{\frac{1}{3}} (ζ)

and

{(\frac{1}{2} π ζ)}^{\frac{1}{2}} H_{\frac{1}{3}}^{(1)} (ζ)

⁠, where J_ν is the Bessel function of the first kind and

H_{ν}^{(j)}

is the Hankel function, are the solutions of

\frac{d^{2} η}{d ζ^{2}} + (1 + \frac{5}{36 ζ^{2}}) η = 0 .

(3.12)

We write

η_{1} (ζ) = e^{\frac{5}{6} π i} {(\frac{1}{2} π ζ)}^{\frac{1}{2}} J_{\frac{1}{3}} (ζ), η_{2} (ζ) = e^{\frac{2}{3} π i} {(\frac{1}{2} π ζ)}^{\frac{1}{2}} H_{\frac{1}{3}}^{(1)} (ζ),

for simplicity. We remark that

η_{1} (ζ), η_{2} (ζ) \in \{\begin{cases} e^{\frac{π}{4} i} R, & ζ \in e^{\frac{π}{2} i} (0, \infty), \\ R, & ζ \in e^{- π i} (0, \infty), \end{cases}

(3.13)

since

H_{\frac{1}{3}}^{(1)} (ζ) = \frac{2}{\sqrt{3}} (e^{\frac{π}{6} i} J_{\frac{1}{3}} (ζ) + e^{- \frac{π}{2} i} J_{- \frac{1}{3}} (ζ))

and

J_{ν} (ζ) = {(\frac{ζ}{2})}^{ν} \sum_{m = 0}^{\infty} \frac{{(- 1)}^{m}}{m! Γ (m + ν + 1)} {(\frac{ζ}{2})}^{2 m} \in ζ^{ν} R, \arg ζ \in \frac{π}{2} Z

for

ν \in R

⁠. In particular, ψ_j(r) = a(r)η_j(ζ(r)), j = 1, 2, are real valued. We also remark that

ψ_{j} \in H^{2} ((1 / 2, \infty)), j = 1, 2,

(3.14)

see Appendixes A and B for the proof.

We look for a solution of in

(ζ (\frac{1}{2}), 0] \cup i [0, \infty)

with η(ix) ∼ η₂(ix) as x → ∞, since ψ₂(r) = a(r)η₂(ζ(r)) is L² near ∞. Thanks to the standard ODE calculus, we only have to observe the solution of the integral equation

\begin{align} η (ζ) & = η_{2} (ζ) + C_{0}^{- 1} \int_{ζ}^{i \infty} (η_{1} (ξ) η_{2} (ζ) - η_{2} (ξ) η_{1} (ζ)) g (ξ) η (ξ) d ξ \\ = η_{2} (ζ) + C_{0}^{- 1} \int_{r (ζ)}^{\infty} (η_{1} (ζ (s)) η_{2} (ζ) - η_{2} (ζ (s)) η_{1} (ζ)) g (ζ (s)) η (ζ (s)) \sqrt{λ - U_{d, n} (s)} d s, \end{align}

(3.15)

where

C_{0} = W (η_{1}, η_{2}) = η_{1} (ζ) η_{2}^{'} (ζ) - η_{1}^{'} (ζ) η_{2} (ζ)

is the Wronskian which is independent of ζ, and the contour of the above first integral is taken over the path connecting ζ and i∞ in (−∞, 0] ∪ i[0, ∞). We note that implies the integral in maps the functions η satisfying into themselves, which guarantees that the solution of satisfies and hence ψ_λ(r) = a(r)η(ζ(r)) is real valued.

We write the second term of by F_λη(ζ), and we define a function space X and its norm by

‖ η ‖_{X} = \sup_{ζ \in (ζ (\frac{1}{2}), 0] \cup i [0, \infty)} e^{I m ζ} | η (ζ) |, X = \{η \in C ((ζ (1 / 2), 0] \cup i [0, \infty); C) : ‖ η ‖_{X} < \infty\} .

We note that η₁(ζ) = O(e^Imζ), η₂(ζ) = O(e^−Imζ) [see Ref. 11 (Sec. 7.21) for the asymptotic behavior of

J_{\frac{1}{3}} (ζ)

and

H_{\frac{1}{3}}^{(1)} (ζ)

as |ζ| → ∞], and that

\int_{1 / 2}^{\infty} | g (ζ (r)) | | λ - U_{d, n} (r) |^{\frac{1}{2}} d r = O (λ^{- 1 / 2} T^{- 1}) (λ \to \infty)

(3.16)

(see Appendixes A and B for the proof). Then we have ‖η₂‖_X < ∞ and ‖F_λ‖_X→X = O(λ^−1/2T⁻¹), which implies

η (ζ) = {(I - F_{λ})}^{- 1} η_{2} (ζ) = η_{2} (ζ) + O (λ^{- \frac{1}{2}} T^{- 1} e^{- Im ζ})

(3.17)

for sufficiently large λ and

ζ \in (ζ (\frac{1}{2}), 0] \cup i [0, \infty)

⁠, and thus

ψ_{λ} (r) = a (r) η (ζ) = {\{λ - U_{d, n} (r)\}}^{- \frac{1}{4}} \{{(\frac{1}{2} π ζ)}^{\frac{1}{2}} H_{\frac{1}{3}}^{(1)} (ζ) + O (λ^{- \frac{1}{2}} T^{- 1} e^{- Im ζ})\} .

If r < T, we have

H_{\frac{1}{3}}^{(1)} (ζ) = \frac{2}{\sqrt{3}} e^{- \frac{1}{6} π i} {J_{\frac{1}{3}} (z) + J_{- \frac{1}{3}} (z)}

⁠, where ζ = e^−iπz [see Ref. 9 (Sec. 7.8)]. By using

J_{ν} (z) = {(\frac{2}{π z})}^{\frac{1}{2}} \{\cos (z - \frac{ν}{2} π - \frac{1}{4} π) + O (z^{- 1})\} as z \to \infty,

(3.18)

[see Ref. 11 (Sec. 7.21)], we have

J_{\frac{1}{3}} (z) + J_{- \frac{1}{3}} (z) = {(\frac{6}{π z})}^{\frac{1}{2}} \{\cos (z - \frac{1}{4} π) + O (z^{- 1})\} .

Since

Z = ζ (1) = \int_{1}^{T} \sqrt{λ - U_{d, n} (s)} d s \leq λ^{1 / 2} T

for sufficiently large λ, we obtain .

For the proof of , differentiating in ζ on

(ζ (\frac{1}{2}), 0) \cup i (0, \infty)

implies

η^{'} (ζ) = η_{2}^{'} (ζ) + C_{0}^{- 1} \int_{r (ζ)}^{\infty} (η_{1} (ζ (s)) η_{2}^{'} (ζ) - η_{2} (ζ (s)) η_{1}^{'} (ζ)) g (ζ (s)) η (ζ (s)) \sqrt{λ - U_{d, n} (s)} d s .

(3.19)

Thus, by , we have

η^{'} (ζ) = (1 - O (λ^{- \frac{1}{2}} T^{- 1})) η_{2}^{'} (ζ) + O (λ^{- \frac{1}{2}} T^{- 1}) η_{1}^{'} (ζ),

(3.20)

and therefore the identity

J_{ν}^{'} (z) = - J_{ν + 1} (z) + \frac{ν}{z} J_{ν} (z)

⁠, and

\frac{d ζ}{d r} = \sqrt{λ - U_{d, n} (r)}

imply .

For ψ_λ ∈ H²((1/2, ∞)), it suffices to confirm that

ψ_{λ}^{'}

is continuous at r = T. We compute

\begin{array}{l} φ_{λ}^{'} (r) & = a^{'} (r) η (ζ) + a (r) \frac{d ζ}{d r} η^{'} (ζ) \\ = \frac{1}{4} U_{d, n}^{'} (r) {(λ - U_{d, n} (r))}^{- \frac{5}{4}} η (ζ) + {(λ - U_{d, n} (r))}^{\frac{1}{4}} η^{'} (ζ) \\ = {(λ - U_{d, n} (r))}^{- \frac{5}{4}} (\frac{1}{4} U_{d, n}^{'} (r) η (ζ) + {(λ - U_{d, n} (r))}^{\frac{3}{2}} η^{'} (ζ)), \end{array}

It follows from and that

\begin{array}{l} φ_{λ}^{'} (r) & = - G_{2} (ζ) {(λ - U_{d, n} (r))}^{- \frac{5}{4}} (\frac{1}{4} U_{d, n}^{'} (r) η_{1} (ζ) + {(λ - U_{d, n} (r))}^{\frac{3}{2}} η_{1}^{'} (ζ)) \\ + (1 + G_{1} (ζ)) {(λ - U_{d, n} (r))}^{- \frac{5}{4}} (\frac{1}{4} U_{d, n}^{'} (r) η_{2} (ζ) + {(λ - U_{d, n} (r))}^{\frac{3}{2}} η_{2}^{'} (ζ)), \end{array}

where

G_{j} (ζ) = C_{0}^{- 1} \int_{r (ζ)}^{\infty} η_{j} (ζ (s)) g (ζ (s)) η (ζ (s)) \sqrt{λ - U_{d, n} (s)} d s

⁠, j = 1, 2, are continuous at r = T. The same argument for the proof of continuity of

φ_{\pm}^{'}

in Appendix A, where

φ_{\pm} (r) = a (r) ζ^{\frac{1}{2}} J_{\pm \frac{1}{3}} (ζ)

⁠, shows

{(λ - U_{d, n} (r))}^{- \frac{5}{4}} (\frac{1}{4} U_{d, n}^{'} (r) η_{j} (ζ) + {(λ - U_{d, n} (r))}^{\frac{3}{2}} η_{j}^{'} (ζ)), j = 1, 2,

are continuous at r = T, which implies ψ_λ is continuous at r = T.□

Lemma 3.6.

For sufficiently large λ, there is ϕ_λ ∈ H²((0, 2)) such that

\begin{array}{l} {f \in L^{2} ((0, 2)) ∣ H_{n} f = λ f on (0, 2)} = {α ϕ_{λ} ∣ α \in C}, (d, n) \neq (3, 0), \\ {f \in H^{1} ((0, 2)) ∣ H_{n} f = λ f on (0, 2), f (0) = 0} = {α ϕ_{λ} ∣ α \in C}, (d, n) = (3, 0), \end{array}

and

ϕ_{λ} (1) = \sqrt{\frac{2}{π}} λ^{- \frac{1}{4}} \{\cos (\sqrt{λ} - \frac{2 n + d - 1}{4} π) + O (λ^{- \frac{1}{2}})\},

(3.21)

ϕ_{λ}^{'} (1) = \sqrt{\frac{2}{π}} λ^{\frac{1}{4}} \{- \sin (\sqrt{λ} - \frac{2 n + d - 1}{4} π) + O (λ^{- \frac{1}{2}})\} .

(3.22)

Proof.

We suppose (d, n) ≠ (3, 0) for simplicity. It is easy to see that

ϕ_{+} (r) = r^{\frac{1}{2}} J_{(n + \frac{d - 2}{2})} (r \sqrt{λ})

and

ϕ_{-} (r) = r^{\frac{1}{2}} Y_{(n + \frac{d - 2}{2})} (r \sqrt{λ})

⁠, where Y_ν is the Bessel function of the second kind, are solutions of

\frac{d^{2}}{d r^{2}} f (r) + \{λ - \frac{(d - 1) (d - 3)}{4 r^{2}} - \frac{n (n + d - 2)}{r^{2}}\} f (r) = 0,

and

ϕ_{+} (r) = r^{n + \frac{d - 1}{2}} (C_{λ}^{'} + o (1)), ϕ_{-} (r) = r^{- n - \frac{d - 3}{2}} (- \frac{1}{π C_{λ}^{'}} + o (1)) as r \to 0,

where

C_{λ}^{'} = \frac{1}{Γ (n + \frac{d}{2})} {(\frac{\sqrt{λ}}{2})}^{n + \frac{d - 2}{2}}

[see Ref. 11 (Sec. 3)]. Thus it suffices to construct the solution ϕ of with ϕ ∼ ϕ₊, regarding the term

\tilde{V} (r) ϕ (r)

as a perturbation employing the norm

‖ f ‖ = \sup_{r \in (0, 2)} \min {(1, {(\sqrt{λ} r)}^{n + \frac{d - 1}{2}})}^{- 1} | f (r) |

⁠. Finally we have and by the asymptotic behavior .□

D. Proof of Proposition 3.4

Suppose that λ_n,l is the l + 1th eigenvalue of H_n and that f_n,l is the real valued eigenfunction for λ_n,l.

We denote by

Z_{n, l} = \int_{1}^{T} \sqrt{λ_{n, l} - U_{d, n} (s)} d s, U_{d, n} (T) = λ_{n, l},

the constant Z in Lemma 3.5 with λ = λ_n,l. Let

m_{n, l} \in Z

and

δ \in [- \frac{π}{2}, \frac{π}{2})

be the constants satisfying

Z_{n, l} + \sqrt{λ_{n, l}} = (\frac{2 n + d}{4} + m_{n, l}) π + δ .

(3.23)

Lemma 3.7.

Let X_n,l > 1 be the constant in Proposition 3.4 such that

\tilde{V} (X_{n, l}) = λ_{n, l}

. Then

\int_{0}^{X_{n, l}} {λ_{n, l} - \tilde{V} (r)}^{\frac{1}{2}} d r = (\frac{2 n + d}{4} + m_{n, l}) π + O (λ_{n, l}^{- \frac{1}{2}}) .

Proof.

Since Lemmas 3.5 and 3.6 imply

f_{n, l} = A ψ_{λ_{n, l}} = B ϕ_{λ_{n, l}}

on (1/2, 2) with some constants A and B, we have

ϕ_{λ_{n, l}} (1) ψ_{λ_{n, l}}^{'} (1) - ϕ_{λ_{n, l}}^{'} (1) ψ_{λ_{n, l}} (1) = 0 .

By , , , and , we have

\sin (Z_{n, l} + \sqrt{λ_{n, l}} - \frac{2 n + d}{4} π) = O (λ_{n, l}^{- \frac{1}{2}}) + O (Z_{n, l}^{- 1}) .

The notation implies

\sin δ = O (λ_{n, l}^{- \frac{1}{2}}) + O (Z_{n, l}^{- 1})

and so

δ = O (λ_{n, l}^{- \frac{1}{2}}) + O (Z_{n, l}^{- 1}) .

Then we have

Z_{n, l} + \sqrt{λ_{n, l}} = (\frac{2 n + d}{4} + m_{n, l}) π + O (λ_{n, l}^{- \frac{1}{2}}) + O (Z_{n, l}^{- 1}) .

(3.24)

On the other hand, it is proved in Ref. 9 (pp. 161–163) that

Z_{n, l} + \sqrt{λ_{n, l}} = \int_{0}^{X_{n, l}} {λ_{n, l} - \tilde{V} (r)}^{\frac{1}{2}} d r + O (λ_{n, l}^{- \frac{1}{2}}) .

(3.25)

Thus, using and , we obtain the assertion, noting that

Z_{n, l} = \int_{1}^{T} \sqrt{λ_{n, l} - U_{d, n} (s)} d s \geq C λ_{n, l}^{\frac{1}{2}} .

□

By Lemma 3.7, it suffices to show that m_n,l = l for sufficiently large l, which follows from counting the zeros of f_n,l in (0, ∞) by two different methods.

We easily see [e.g. Ref. 9 (Sec. 5.4)] that f_n,l has l zeros in (0, ∞), since f_n,l is the l + 1th eigenfunction of H_n. On the other hand, we can find that m_n,l is the number of the zeros of f_n,l, which concludes the proof. In the rest of the proof, we show that.

Lemma 3.8.

Let $p = [π^{- 1} \sqrt{λ_{n, l}} - \frac{2 n + d - 1}{4}] \in Z$ and $r_{0} = (p + \frac{2 n + d - 1}{4}) π / \sqrt{λ_{n, l}}$ (note that $1 - \frac{1}{π \sqrt{λ_{n, l}}} < r_{0} \leq 1$ ). Then

f_n,l has no zeros in [T, ∞).
f_n,l has m_n,l − p zeros in [r₀, T).
f_n,l has p zeros in (0, r₀).

Proof.

(i) is proved by contradiction. If there exists a zero of f_n,l in [T, ∞), f_n,l and $f_{n, l}^{'}$ are positive (or negative) on the right side of the zero. Since $f_{n, l}^{″} (r) = (U_{d, n} (r) - λ_{n, l}) f_{n, l} (r)$ ⁠, f_n,l is convex (or concave). Hence f_n,l tends to ∞ (or −∞). This contradicts that f_n,l ∈ L²((0, ∞)).

The proof is divided into four steps.
1. We remark that the Proof of Lemma 3.5 implies $f_{n, l} (r) = c ψ_{λ_{n, l}} (r) = c a (r) η (ζ (r))$ with some $c \in R \ {0}$ ⁠, where $a (r) = {(λ_{n, l} - U_{d, n} (r))}^{- \frac{1}{4}}$ ⁠, $ζ (r) = \int_{T}^{r} \sqrt{λ_{n, l} - U_{d, n} (s)} d s$ and η(ζ) is the solution of (3.15). Since ζ([r₀, T)) = [ζ(r₀), 0), the number of zeros of f_n,l(r) in [r₀, T) is same as that of zeros of η(ζ) in [ζ(r₀), 0).
2. We show.
  1. $- ζ (r_{0}) = (m_{n, l} - p + \frac{1}{4}) π + o (1)$ ⁠, and m_n,l − p → ∞ as l → ∞.
  2. $η_{2} (ζ) = 2 \cos (- ζ - \frac{1}{4} π) + O (| ζ |^{- 1})$ ⁠.
  3. Let ɛ = inf{maximal values of |η₂(ζ)| in (−∞, 0)}. Then ɛ > 0 holds.

Since
$- ζ (r_{0}) - Z_{n, l} = \int_{r_{0}}^{1} {λ_{n, l} - U_{d, n} (r)}^{\frac{1}{2}} d r = (1 - r_{0}) \sqrt{λ_{n, l}} + O (λ_{n, l}^{- \frac{1}{2}}),$
the definitions of m_n,l, p, r₀ show that
$- ζ (r_{0}) = (\frac{2 n + d}{4} + m_{n, l}) π - r_{0} \sqrt{λ_{n, l}} + δ = (m_{n, l} - p + \frac{1}{4}) π + o (1) .$
Since $- ζ (r_{0}) \geq \int_{1}^{T} \sqrt{λ_{n, l} - U_{d, n} (s)} d s > c \sqrt{λ_{n, l}} \to \infty$ as l → ∞, m_n,l − p → ∞ also holds.
is proved by $H_{\frac{1}{3}}^{(1)} (ζ) = \frac{2}{\sqrt{3}} e^{- \frac{1}{6} π i} {J_{\frac{1}{3}} (e^{i π} ζ) + J_{- \frac{1}{3}} (e^{i π} ζ)}$ in (−∞, 0) and the asymptotic behavior (3.18) of $J_{\pm \frac{1}{3}}$ ⁠.
η₂ is concave (resp. convex) if η₂(ζ) > 0 (resp. η₂(ζ) < 0) since η₂ solves $\frac{d^{2} η}{d ζ^{2}} + (1 + \frac{5}{36 ζ^{2}}) η = 0$ ⁠. Hence this and (2) imply ɛ > 0.
- Step 3. Let S = {ζ ∈ [ζ(r₀), 0)∣|η(ζ)| < ɛ/3}. Then η(ζ) is monotone on each segment of S. In fact, $S \subset η_{2}^{- 1} ((- 2 ε / 3, 2 ε / 3))$ for large l by (3.17), and thus (3.20) implies η′(ζ) ≠ 0 for ζ ∈ S. Moreover, we note that.
  - If ζ is between ζ(r₀) and $- (m_{n, l} - p + \frac{1}{4}) π$ ⁠, Step 2 implies
    $\begin{array}{l} | η (ζ) | \geq | η_{2} (ζ) | - O (λ_{n, l}^{- \frac{1}{2}} T^{- 1}) & \geq |η_{2} (- (m_{n, l} - p + \frac{1}{4}) π)| - o (1) \\ \geq 2 - o (1) > ε / 2 \end{array}$
    for sufficiently large l.
  - Since f_n,l is bounded and a(r)⁻¹ → 0 as r → T,
    $η (ζ) = c^{- 1} a {(r)}^{- 1} f_{n, l} (r) \to 0 as ζ \to - 0 .$
    Therefore each segment of S, except the one of the form (−a, 0), has exactly one zero, and the number of segments of S \ (−a, 0) is same as that of zeros of η₂ in $[- (m_{n, l} - p + \frac{1}{4}) π, 0)$ thanks to (3.17).
- Step 4. The number of the zeros of $J_{\frac{1}{3}} (z) + J_{- \frac{1}{3}} (z)$ in $(0, (m_{n, l} - p + \frac{1}{4}) π]$ is m_n,l − p for large l [Ref. 9 (Lemma 7.9b)]. Thus, since $H_{\frac{1}{3}}^{(1)} (ζ) = \frac{2}{\sqrt{3}} e^{- \frac{1}{6} π i} {J_{\frac{1}{3}} (e^{i π} ζ) + J_{- \frac{1}{3}} (e^{i π} ζ)}$ in (−∞, 0), $η_{2} (ζ) = {(\frac{1}{2} π ζ)}^{\frac{1}{2}} H_{\frac{1}{3}}^{(1)} (ζ)$ has exactly m_n,l − p zeros in $[- (m_{n, l} - p + \frac{1}{4}) π, 0)$ ⁠, which concludes the proof of (ii).
is proved by the same argument as above, noting that $ϕ_{+} (r) = r^{\frac{1}{2}} J_{(n + \frac{d - 2}{2})} (r \sqrt{λ})$ has p zeros in (0, r₀) [the number of the zeros of $J_{(n + \frac{d - 2}{2})} (r)$ is p in $(0, (p + \frac{2 n + d - 1}{4}) π)$ by Ref. 9 (Lemma 7.9a)].
□

IV. PROOF OF THEOREM 1.2 AND PROPOSITION 1.3

We first note the formula for

κ \in C_{0}^{\infty} (R)

and

Φ, Ψ \in C_{0}^{\infty} (R^{d})

\int_{R} \int_{R^{d}} \int_{R^{d}} E (t, x, y) κ (t) Φ (x) Ψ (y) d y d x d t = {(\hat{κ} (H) Ψ, \bar{Φ})}_{L^{2} (R^{d})} = {(\hat{κ} (H) Φ, \bar{Ψ})}_{L^{2} (R^{d})}

(4.1)

with the help of identity U(t) = e^−itH, the spectral decomposition theorem and Fubini’s theorem, where

\hat{κ} (λ) = \int_{R} e^{- i t λ} κ (t) d t

⁠. It follows from Lemma 2.3 that

\int_{R} \int_{R^{d}} \int_{R^{d}} E^{Y_{n m}} (t, x, y) κ (t) Φ (x) Ψ (y) d y d x d t = {(\hat{κ} (H_{n}) J_{n m} Φ, J_{n m} \bar{Ψ})}_{L^{2} ((0, \infty))} .

(4.2)

Proof of Proposition 1.3.

It suffices to show that

E^{Y_{n}} (t, x, y)

is not in C¹ on

(0, \infty) \times {x \in R^{d} ∣ x \neq 0} \times {y \in R^{d} ∣ y \neq 0}

⁠, which we show by contradiction. Suppose that

E^{Y_{n}}

is in C¹ near (t₀, x₀, y₀) with t₀ > 0, x₀ ≠ 0, and y₀ ≠ 0. The definition of

E^{Y_{n}}

and Remark 1.4 imply that

E^{Y_{n}} \in C^{1} (I \times B_{r_{1}, R_{1}} \times B_{r_{2}, R_{2}})

(4.3)

with some interval I ⋐ (0, ∞) and 0 < r_j < R_j, j = 1, 2, where

B_{r, R} = {x \in R^{d} ∣ r < | x | < R}

⁠. We may assume Y_n = Y_n1 without loss of generality.

We fix

κ \in C_{0}^{\infty} (I)

⁠,

Φ \in C_{0}^{\infty} (B_{r_{1}, R_{1}})

and

Ψ \in C_{0}^{\infty} (B_{r_{2}, R_{2}})

so that

\hat{κ} (0) \neq 0

and that ϕ(r) = J_n1Φ(r) ≢ 0 and ψ(r) = J_n1Ψ(r) ≢ 0 are non-negative. We set

\begin{array}{l} G (τ, j, k) & = \int_{R} \int_{R^{d}} \int_{R^{d}} E^{Y_{n}} (t, x, y) κ (t) Φ (x) Ψ (y) e^{i t τ + i | x | j - i | y | k} d y d x d t \\ = {(\hat{κ} (H_{n} - τ) ϕ_{j}, \bar{ψ_{k}})}_{L^{2} ((0, \infty))} \end{array}

for

(τ, j, k) \in R \times R \times R

⁠, where we used and the notation

ϕ_{j} (r) = e^{irj} ϕ (r), ψ_{k} (r) = e^{- i r k} ψ (r) .

It suffices to show that there exist a constant C > 0 and a sequence (τ_l, j_l, k_l) such that

|τ_{l}| + |j_{l}| + |k_{l}| \to \infty

⁠, as l → ∞ such that

| G (τ_{l}, j_{l}, k_{l}) | \geq C {(1 + | τ_{l} | + | j_{l} | + | k_{l} |)}^{- 1 / 2 c},

(4.4)

which contradicts . In the following, we show by taking τ = λ_n,l and

j_{l} = k_{l} = \sqrt{λ_{n, l}}

⁠.

The spectral decomposition theorem implies

\hat{κ} (H_{n} - τ) u = \sum_{l = 0}^{\infty} \hat{κ} (λ_{n, l} - τ) {(u, f_{n, l})}_{L^{2} ((0, \infty))} f_{n, l}, u \in L^{2} ((0, \infty))

where f_n,l, l = 0, 1, …, is the normalized eigenfunction of H_n with eigenvalue λ_n,l. Then it follows from Lemma 3.3 and

\hat{κ} \in S (R)

that for any

N \in N

\begin{array}{l} ‖ \hat{κ} (H_{n} - τ_{l}) - \hat{κ} (0) {(\cdot, f_{n, l})}_{L^{2} ((0, \infty))} f_{n, l} ‖_{L^{2} ((0, \infty)) \to L^{2} ((0, \infty))} \\ = \sup_{l^{'} \in N_{+} \ {l}} | \hat{κ} (λ_{n, l^{'}} - λ_{n, l}) | \\ \leq C_{N} λ_{n, l}^{- N}, \end{array}

which implies

G (τ_{l}, j_{l}, k_{l}) = \hat{κ} (0) {(ϕ_{j_{l}}, f_{n, l})}_{L^{2} ((0, \infty))} {(ψ_{k_{l}}, \bar{f_{n, l}})}_{L^{2} ((0, \infty))} + O (λ_{n, l}^{- \infty}) .

We note that supp ϕ ⋐ Ω_λ for sufficiently large λ, where Ω_λ is as in Lemma 3.1. Thus, by using , we can compute

\begin{array}{l} {(ϕ_{j_{l}}, f_{n, l})}_{L^{2} ((0, \infty))} & = \int_{0}^{\infty} e^{i r j_{l}} ϕ (r) \cdot λ_{n, l}^{- 1 / 4} (R e {C_{λ_{n, l}} e^{i \sqrt{λ_{n, l}} (r - 1)}} + O (C_{λ_{n, l}} λ_{n, l}^{- 1 / 2})) d r \\ = \frac{1}{2 λ_{n, l}^{1 / 4}} \{\int_{0}^{\infty} (C_{λ_{n, l}} e^{- i \sqrt{λ_{n, l}}} e^{2 i \sqrt{λ_{n, l}} r} + \bar{C_{λ_{n, l}} e^{- i \sqrt{λ_{n, l}}}}) ϕ (r) d r + O (C_{λ_{n, l}} λ_{n, l}^{- 1 / 2})\} \\ = \frac{\bar{C_{λ_{n, l}} e^{- i \sqrt{λ_{n, l}}}}}{2 λ_{n, l}^{1 / 4}} (\int_{0}^{\infty} ϕ (r) d r + O (λ_{n, l}^{- 1 / 2})), \end{array}

where we used integration by parts once to show

\int_{0}^{\infty} e^{2 i \sqrt{λ_{n, l}} r} ϕ (r) d r = O (λ_{n, l}^{- 1 / 2})

⁠. The same computations imply

{(ψ_{k_{l}}, f_{n, l})}_{L^{2} ((0, \infty))} = \frac{C_{λ_{n, l}} e^{- i \sqrt{λ_{n, l}}}}{2 λ_{n, l}^{1 / 4}} (\int_{0}^{\infty} ψ (r) d r + O (λ_{n, l}^{- 1 / 2}))

⁠. Thus we obtain, taking Lemma 3.2 into account,

\begin{array}{l} G (τ_{l}, j_{l}, k_{l}) & = \hat{κ} (0) {(ϕ_{j_{l}}, f_{n, l})}_{L^{2} ((0, \infty))} {(ψ_{k_{l}}, f_{n, l})}_{L^{2} ((0, \infty))} + O (λ_{n, l}^{- \infty}) \\ = \frac{\hat{κ} (0) | C_{λ_{n, l}} |^{2}}{4 λ_{n, l}^{1 / 2}} (\int_{0}^{\infty} ϕ (r) d r \int_{0}^{\infty} ψ (r) d r + O (λ_{n, l}^{- 1 / 2})) + O (λ_{n, l}^{- \infty}) \\ \geq c_{0} λ_{n, l}^{- 1 / 2 c} \end{array}

for sufficiently large l with some c₀ > 0, which concludes the assertion.□

Proof of Theorem 1.2.

Let

G_{0} (τ, j, k) = \int_{R} \int_{R^{d}} \int_{R^{d}} E (t, x, y) κ (t) J_{00}^{*} ϕ (x) J_{00}^{*} ψ (y) e^{i t τ + i | x | j - i | y | k} d y d x d t,

where

κ \in C_{0}^{\infty} (R)

and

ϕ, ψ \in C_{0}^{\infty} ((0, \infty))

are arbitrary non-negative functions and

J_{00} : L^{2} (R^{d}) \to L^{2} ((0, \infty))

is as in Sec. II. By and Lemma 2.3 we have

G_{0} (τ, j, k) = {(\hat{κ} (H - τ) J_{00}^{*} ϕ_{j}, \bar{J_{00}^{*} ψ_{k}})}_{L^{2} (R^{d})} = {(\hat{κ} (H_{0} - τ) ϕ_{j}, \bar{ψ_{k}})}_{L^{2} ((0, \infty))},

where ϕ_j(r) = e^irjϕ(r) and ψ_k(r) = e^−irkψ(r). Thus the same argument as in the Proof of Theorem 1.3 implies

G (τ_{l}, j_{l}, k_{l}) \geq c_{0} λ_{n, l}^{- 1 / 2 c}

⁠. Therefore we obtain

E (t, x, y) \notin C^{1} (I \times B_{r_{1}, R_{1}} \times B_{r_{2}, R_{2}})

for any interval I ⋐ (0, ∞) and 0 < r_j < R_j, j = 1, 2, which completes the proof.□

ACKNOWLEDGMENTS

The authors would like to thank the referee for helpful comments. The authors would also like to thank Fumihito Abe, Ryo Muramatsu and Kouichi Taira for helpful discussions. K.K. was partially supported by JSPS KAKENHI Grant No. 22K03394. Y.T. was partially supported by JSPS KAKENHI Grant No. 23K12991. The authors appreciate useful comments made by the referee.

AUTHOR DECLARATIONS

Conflict of Interest

The authors have no conflicts to disclose.

Author Contributions

Keiichi Kato: Writing – original draft (equal); Writing – review & editing (equal). Wataru Nakahashi: Writing – original draft (equal); Writing – review & editing (equal). Yukihide Tadano: Writing – original draft (equal); Writing – review & editing (equal).

DATA AVAILABILITY

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

APPENDIX A: PROOF OF (3.14)

Since $H_{\frac{1}{3}}^{(1)} (ζ) = \frac{2}{\sqrt{3}} (e^{\frac{π}{6} i} J_{\frac{1}{3}} (ζ) + e^{- \frac{π}{2} i} J_{- \frac{1}{3}} (ζ))$ ⁠, we show φ_±(r) = a(r)η_±(ζ(r)) ∈ H²((1/2, ∞)) with $η_{\pm} (ζ) = ζ^{\frac{1}{2}} J_{\pm \frac{1}{3}} (ζ)$ ⁠. We easily see that φ_± are H² away from r = T, since η_± solve (3.12) on {ζ ≠ 0}. Thus it suffices to show that φ_± and $φ_{\pm}^{'}$ are continuous at r = T and that $φ_{\pm}^{″}$ is L² near r = T, which imply that φ_± are H² near r = T.

Since

η_{\pm} (ζ) = \sum_{m = 0}^{\infty} c_{m}^{\pm} ζ^{2 m + \frac{1}{2} \pm \frac{1}{3}}, c_{m}^{\pm} = \frac{{(- 1)}^{m}}{m! Γ (m + ν + 1) 2^{2 m \pm \frac{1}{3}}},

we have

\begin{array}{l} φ_{+} (r) & = {(λ - U_{d, n} (r))}^{- \frac{1}{4}} ζ^{\frac{5}{6}} (c_{0}^{+} + O (| ζ |^{2})), \\ φ_{-} (r) & = {(λ - U_{d, n} (r))}^{- \frac{1}{4}} ζ^{\frac{1}{6}} (c_{0}^{-} + O (| ζ |^{2})) . \end{array}

L’Hôpital’s rule implies

\lim_{r \to T} \frac{ζ (r)}{{(λ - U_{d, n} (r))}^{\frac{3}{2}}} = \lim_{r \to T} \frac{\frac{d ζ}{d r} (r)}{- \frac{3}{2} U_{d, n}^{'} (r) {(λ - U_{d, n} (r))}^{\frac{1}{2}}} = - \frac{2 U_{d, n}^{'} (T)}{3},

(A1)

which yields that φ_± are continuous near r = T.

Since

\begin{array}{l} φ_{\pm}^{'} (r) & = a^{'} (r) η_{\pm} (ζ) + a (r) \frac{d ζ}{d r} η_{\pm}^{'} (ζ) \\ = \frac{1}{4} U_{d, n}^{'} (r) {(λ - U_{d, n} (r))}^{- \frac{5}{4}} η_{\pm} (ζ) + {(λ - U_{d, n} (r))}^{\frac{1}{4}} η_{\pm}^{'} (ζ) \\ = {(λ - U_{d, n} (r))}^{- \frac{5}{4}} (\frac{1}{4} U_{d, n}^{'} (r) η_{\pm} (ζ) + {(λ - U_{d, n} (r))}^{\frac{3}{2}} η_{\pm}^{'} (ζ)), \end{array}

we have

\begin{array}{l} φ_{+}^{'} (r) & = {(λ - U_{d, n} (r))}^{- \frac{5}{4}} ζ^{\frac{5}{6}} (c_{0}^{+} (\frac{U_{d, n}^{'} (r)}{4} + \frac{5 {(λ - U_{d, n} (r))}^{\frac{3}{2}}}{6 ζ}) + O (| ζ |^{2})), \\ φ_{-}^{'} (r) & = {(λ - U_{d, n} (r))}^{- \frac{5}{4}} ζ^{\frac{1}{6}} (c_{0}^{-} (\frac{U_{d, n}^{'} (r)}{4} + \frac{{(λ - U_{d, n} (r))}^{\frac{3}{2}}}{6 ζ}) + O (| ζ |^{2})) . \end{array}

Hence the continuity of

φ_{+}^{'}

at r = T follows from , while the continuity of

φ_{-}^{'}

at r = T is proved by, in addition to ,

\begin{array}{l} \lim_{r \to T} \frac{{(λ - U_{d, n} (r))}^{\frac{3}{2}} + \frac{3}{2} U_{d, n}^{'} (r) ζ (r)}{ζ {(r)}^{\frac{5}{3}}} \\ = \lim_{r \to T} \frac{- \frac{3}{2} U_{d, n}^{'} (r) {(λ - U_{d, n} (r))}^{\frac{1}{2}} + \frac{3}{2} U_{d, n}^{″} (r) ζ (r) + \frac{3}{2} U_{d, n}^{'} (r) {(λ - U_{d, n} (r))}^{\frac{1}{2}}}{\frac{5}{3} ζ {(r)}^{\frac{2}{3}} \frac{d ζ}{d r} (r)} \\ = \lim_{r \to T} \frac{9 U_{d, n}^{″} (r) ζ {(r)}^{\frac{1}{3}}}{10 {(λ - U_{d, n} (r))}^{\frac{1}{2}}} \\ = \frac{9 U_{d, n}^{″} (T)}{10} {(- \frac{2 U_{d, n}^{'} (T)}{3})}^{\frac{1}{3}} . \end{array}

Since η_± solve on {ζ ≠ 0}, we have

\begin{array}{l} φ_{\pm}^{″} (r) = (\frac{1}{4} U_{d, n}^{″} (r) {(λ - U_{d, n} (r))}^{- \frac{5}{4}} + \frac{5}{16} {(U_{d, n}^{'} (r))}^{2} {(λ - U_{d, n} (r))}^{- \frac{9}{4}}) η_{\pm} (ζ) + {(λ - U_{d, n} (r))}^{\frac{3}{4}} η_{\pm}^{″} (ζ) \\ = {(λ - U_{d, n} (r))}^{\frac{3}{4}} (\frac{1}{4} U_{d, n}^{″} (r) {(λ - U_{d, n} (r))}^{- 2} + \frac{5}{16} {(U_{d, n}^{'} (r))}^{2} {(λ - U_{d, n} (r))}^{- 3} - 1 - \frac{1}{36 ζ {(r)}^{2}}) η_{\pm} (ζ), \end{array}

which and the estimate

\frac{1}{ζ {(r)}^{2}} = \frac{9 U_{d, n}^{'} {(r)}^{2}}{4 {(λ - U_{d, n} (r))}^{3}} + \frac{9 U_{d, n}^{″} (r)}{5 {(λ - U_{d, n} (r))}^{2}} + O (T^{- 2} | λ - U_{d, n} (r) |^{- 1}) as r \to T

in the next subsection imply

φ_{\pm}^{″} (r) = {(λ - U_{d, n} (r))}^{\frac{3}{4}} (- 1 + O (T^{- 2} | λ - U_{d, n} (r) |^{- 1})) η_{\pm} (ζ) .

Hence

φ_{\pm}^{″} = O (| λ - U_{d, n} (r) |^{- \frac{1}{4}}) = O (| r - T |^{- \frac{1}{4}})

is L² near r = T.

APPENDIX B: PROOF OF (3.16)

We recall that the definitions of U_d,n, g, and

U

are the following,

\begin{array}{l} U_{d, n} (r) = \frac{(d - 1) (d - 3)}{4 r^{2}} + \frac{n (n + d - 2)}{r^{2}} + \tilde{V} (r), \\ g (ζ) = \frac{5}{36 ζ^{2}} & - U (r (ζ), λ), U (r, λ) = \frac{U_{d, n}^{''} (r)}{4 {(λ - U_{d, n} (r))}^{2}} + \frac{5 {(U_{d, n}^{'} (r))}^{2}}{16 {(λ - U_{d, n} (r))}^{3}} . \end{array}

We write U(r) instead of U_d,n(r) until the end of the proof. We divide the integral as follows,

\int_{1 / 2}^{\infty} | g (ζ (r)) | | λ - U (r) |^{\frac{1}{2}} d r = \int_{1 / 2}^{(1 - ε) T} + \int_{(1 - ε) T}^{(1 + ε) T} + \int_{(1 + ε) T}^{\infty} = I_{1} + I_{2} + I_{3},

where ɛ > 0 is a small constant which will be fixed later.

Estimate of I₂.

Integrating by part twice on the definition of ζ, we have

\begin{align} - ζ (r) & = - \int_{T}^{r} \sqrt{λ - U (s)} d s \\ = \frac{2}{3} \int_{T}^{r} {({(λ - U (s))}^{3 / 2})}^{'} \frac{1}{U^{'} (s)} d s \\ = \frac{2}{3} \frac{{(λ - U (r))}^{3 / 2}}{U^{'} (r)} + \frac{2}{3} \int_{T}^{r} {(λ - U (s))}^{3 / 2} \frac{U^{″} (s)}{U^{'} {(s)}^{2}} d s \\ = \frac{2}{3} \frac{{(λ - U (r))}^{3 / 2}}{U^{'} (r)} - \frac{4}{15} \int_{T}^{r} {({(λ - U (s))}^{5 / 2})}^{'} \frac{U^{″} (s)}{U^{'} {(s)}^{3}} d s \\ = \frac{2}{3} \frac{{(λ - U (r))}^{3 / 2}}{U^{'} (r)} \{1 - \frac{2}{5} \frac{U^{″} (r) (λ - U (r))}{U^{'} {(r)}^{2}} + S (r)\}, \end{align}

(B1)

where

S (r) = \frac{2 U^{'} (r)}{5 {(λ - U (r))}^{3 / 2}} \int_{T}^{r} {(λ - U (s))}^{5 / 2} \frac{d}{d s} (\frac{U^{″} (s)}{U^{'} {(s)}^{3}}) d s .

We claim that for T/2 ≤ r ≤ 2T,

\frac{U^{″} (r) (λ - U (r))}{U^{'} {(r)}^{2}} = O (\frac{λ - U (r)}{U (r)}), S (r) = O ({(\frac{λ - U (r)}{U (r)})}^{2}) .

(B2)

The first assertion is easy to show by and . For the second one, implies

\frac{d}{d s} (\frac{U^{″} (s)}{U^{'} {(s)}^{3}}) = \frac{U^{‴} (s)}{U^{'} {(s)}^{3}} - \frac{3 U^{″} {(s)}^{2}}{U^{'} {(s)}^{4}} = \frac{1}{U^{'} {(s)}^{2}} O (s^{- 2}) .

We note that and yield that U′(r) is monotonically increasing. By using this property and , we have, for any T ≤ r ≤ 2T,

\begin{array}{l} | S (r) | & = \frac{2 U^{'} (r)}{5 {(U (r) - λ)}^{3 / 2}} \int_{T}^{r} {(U (s) - λ)}^{5 / 2} |\frac{d}{d s} (\frac{U^{″} (s)}{U^{'} {(s)}^{3}})| d s \\ \leq C \frac{U^{'} (r)}{{(U (r) - λ)}^{3 / 2}} \int_{T}^{r} \frac{{(U (s) - λ)}^{5 / 2}}{s^{2} U^{'} {(s)}^{2}} d s \\ = \frac{2 C}{7} \cdot \frac{U^{'} (r)}{{(U (r) - λ)}^{3 / 2}} \int_{T}^{r} \frac{{({(U (s) - λ)}^{7 / 2})}^{'}}{s^{2} U^{'} {(s)}^{3}} d s \\ \leq C^{'} \frac{U^{'} (r)}{{(U (r) - λ)}^{3 / 2}} \cdot \frac{1}{T^{2} U^{'} {(T)}^{3}} \int_{T}^{r} {({(U (s) - λ)}^{7 / 2})}^{'} d s \\ \leq C^{″} \frac{U^{'} (r) {(λ - U (r))}^{2}}{U {(T)}^{2} U^{'} (T)} \\ = C^{″} {(\frac{λ - U (r)}{U (r)})}^{2} {(\frac{U (r)}{U (T)})}^{2} \frac{U^{'} (r)}{U^{'} (T)} . \end{array}

By the same argument, we have, for any T/2 ≤ r ≤ T,

| S (r) | \leq C {(\frac{λ - U (r)}{U (r)})}^{2} .

Since implies

\frac{U (2 r)}{U (r)} = O (1)

and

\frac{U^{'} (2 r)}{U^{'} (r)} = O (1)

⁠, we have the desired estimate.

By using , we can fix ɛ > 0 so that

|\frac{2}{5} \frac{U^{″} (r) (λ - U (r))}{U^{'} {(r)}^{2}} - S (r)| < \frac{1}{2}

for (1 − ɛ)T ≤ r ≤ (1 + ɛ)T. In fact,

\log \frac{U (a r)}{U (r)} = \int_{r}^{a r} \frac{U^{'} (s)}{U (s)} d s

and , imply for a ≥ 1

0 \leq \log \frac{U (a T)}{U (T)} \leq C \int_{r}^{a r} \frac{d s}{s} = C \log a .

Thus we can make

|\frac{λ - U (r)}{U (r)}| = |1 - \frac{U (T)}{U (r)}|

sufficiently small as ɛ → 0. By using and , and Taylor’s expansion of (1 − x)⁻² with

x = \frac{2}{5} \frac{U^{″} (r) (λ - U (r))}{U^{'} {(r)}^{2}} - S (r) \in (- \frac{1}{2}, \frac{1}{2})

⁠, we have

\begin{align} \frac{1}{ζ {(r)}^{2}} & = \frac{9 U^{'} {(r)}^{2}}{4 {(λ - U (r))}^{3}} {(1 - x)}^{- 2} \\ = \frac{9 U^{'} {(r)}^{2}}{4 {(λ - U (r))}^{3}} \{1 + \frac{4}{5} \frac{U^{″} (r) (λ - U (r))}{U^{'} {(r)}^{2}} + O (\frac{{(λ - U (r))}^{2}}{U {(r)}^{2}})\} \\ = \frac{9 U^{'} {(r)}^{2}}{4 {(λ - U (r))}^{3}} + \frac{9 U^{″} (r)}{5 {(λ - U (r))}^{2}} + O (T^{- 2} | λ - U (r) |^{- 1}) . \end{align}

(B3)

Hence we obtain

g (ζ (r)) = \frac{5}{36 ζ {(r)}^{2}} - U (r, λ) = O (T^{- 2} | λ - U (r) |^{- 1}),

which implies that I₂ = O(λ^−1/2T⁻¹).

Estimate of I₁.

Let R be the constant as in Assumption 1.1. By , there exists a constant C > 0 such that λ = U(T) ≥ CT^2c, i.e.,

λ^{- 1} \leq λ^{- \frac{1}{2 c}} \leq C T^{- 1}

⁠. Then we have

\begin{array}{l} \int_{1 / 2}^{R} | U (t, λ) ‖ λ - U (t) |^{1 / 2} d t & \leq \int_{1 / 2}^{R} \frac{| U^{″} (t) |}{4 {(λ - U (t))}^{3 / 2}} + \frac{5 U^{'} {(t)}^{2}}{16 {(λ - U (t))}^{5 / 2}} d t \\ \leq C (λ^{- 3 / 2} + λ^{- 5 / 2}) \\ \leq C λ^{- 1 / 2} T^{- 1} . \end{array}

Integrating by part yields that

\begin{array}{l} \int_{R}^{(1 - ε) T} \frac{U^{″} (t)}{{(λ - U (t))}^{3 / 2}} d t & = {[\frac{U^{'} (t)}{{(λ - U (t))}^{3 / 2}}]}_{R}^{(1 - ε) T} - \frac{3}{2} \int_{R}^{(1 - ε) T} \frac{U^{'} {(t)}^{2}}{{(λ - U (t))}^{5 / 2}} d t \\ = \frac{U^{'} ((1 - ε) T)}{{(λ - U ((1 - ε) T))}^{3 / 2}} - \frac{U^{'} (R)}{{(λ - U (R))}^{3 / 2}} - \frac{3}{2} \int_{R}^{(1 - ε) T} \frac{U^{'} {(t)}^{2}}{{(λ - U (t))}^{5 / 2}} d t, \end{array}

which implies that, since and

\log \frac{U (a r)}{U (r)} = \int_{r}^{a r} \frac{U^{'} (s)}{U (s)} d s

imply U((1 − ɛ)T) ≤ c_ɛU(T) = c_ɛλ with some c_ɛ < 1,

\begin{array}{l} \int_{R}^{(1 - ε) T} U (t, λ) & {λ - U (t)}^{1 / 2} d t \\ = \frac{U^{'} ((1 - ε) T)}{4 {(λ - U ((1 - ε) T))}^{3 / 2}} - \frac{U^{'} (R)}{4 {(λ - U (R))}^{3 / 2}} - \frac{1}{16} \int_{R}^{(1 - ε) T} \frac{U^{'} {(t)}^{2}}{{(λ - U (t))}^{5 / 2}} d t \\ \leq \frac{U^{'} ((1 - ε) T)}{4 {(λ - U ((1 - ε) T))}^{3 / 2}} \leq C_{ε} λ^{- 1 / 2} T^{- 1} . \end{array}

By the definition of ζ, we have

\int_{1 / 2}^{(1 - ε) T} ζ {(t)}^{- 2} {λ - U (t)}^{1 / 2} d t = ζ {((1 - ε) T)}^{- 1} - ζ {(1 / 2)}^{- 1} \leq - ζ {(1 / 2)}^{- 1} \leq C λ^{- 1 / 2} T^{- 1} .

Hence we have I₁ = O(λ^−1/2T⁻¹).

Estimate of I₃.

Since yields that r⁻¹U(r) is monotonically increasing, we have

\begin{array}{l} | ζ ((1 + ε) T) | & = \int_{T}^{(1 + ε) T} \sqrt{U (s) - λ} d s \\ \geq \int_{T}^{(1 + ε) T} {\{(s - T) (\min_{r \in [T, (1 + ε) T]} U^{'} (r))\}}^{1 / 2} d s \\ \geq {\{\frac{U (T)}{T}\}}^{1 / 2} \int_{T}^{(1 + ε) T} {(s - T)}^{1 / 2} d s = \frac{2}{3} ε^{3 / 2} λ^{1 / 2} T, \end{array}

which shows that, by using ζ′(s) = (λ − U(s)),

\int_{(1 + ε) T}^{\infty} \frac{{(U (s) - λ)}^{1 / 2}}{ζ {(s)}^{2}} d s = \frac{1}{ζ ((1 + ε) T)} = O (λ^{- 1 / 2} T^{- 1}) .

Noting that and

\log \frac{U (a r)}{U (r)} = \int_{r}^{a r} \frac{U^{'} (s)}{U (s)} d s

imply U((1 + ɛ)T) ≥ c_ɛU(T) = c_ɛλ with some c_ɛ > 1, we have

\frac{U (s)}{U (s) - λ} = O (1)

and (U(s) − λ)^−1/2 = O(λ^−1/2) for s ≥ (1 + ɛ)T. By using these estimates and , we have

\begin{array}{l} \int_{(1 + ε) T}^{\infty} \frac{{(U^{'} (s))}^{2}}{{(U (s) - λ)}^{5 / 2}} d s & = \int_{(1 + ε) T}^{\infty} {(\frac{U^{'} (s)}{U (s)})}^{2} \times {(\frac{U (s)}{U (s) - λ})}^{2} \times \frac{1}{{(U (s) - λ)}^{1 / 2}} d s \\ \leq C_{ε} λ^{- 1 / 2} \int_{(1 + ε) T}^{\infty} s^{- 2} d s \leq C_{ε} λ^{- 1 / 2} T^{- 1} . \end{array}

We also have the following estimate

\int_{(1 + ε) T}^{\infty} \frac{U^{″} (s)}{{(U (s) - λ)}^{3 / 2}} d s = \int_{(1 + ε) T}^{\infty} \frac{U^{″} (s)}{U^{'} (s)} \frac{U^{'} (s)}{{(U (s) - λ)}^{3 / 2}} d s \leq C_{ε} λ^{- 1 / 2} T^{- 1} .

Hence we obtain I₃ = O(λ^−1/2T⁻¹).□

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Non-smoothness of the fundamental solutions for Schrödinger equations with super-quadratic and spherically symmetric potential

I. INTRODUCTION

II. SCHRÖDINGER OPERATOR IN THE POLAR COORDINATES

III. ESTIMATES OF EIGENVALUES AND EIGENFUNCTIONS

A. Asymptotic behavior in bounded regions

B. Gaps of the eigenvalues

C. Asymptotic behavior of eigenfunctions

D. Proof of Proposition 3.4

IV. PROOF OF THEOREM 1.2 AND PROPOSITION 1.3

ACKNOWLEDGMENTS

AUTHOR DECLARATIONS

Conflict of Interest

Author Contributions

DATA AVAILABILITY

APPENDIX A: PROOF OF (3.14)

APPENDIX B: PROOF OF (3.16)

REFERENCES

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Non-smoothness of the fundamental solutions for Schrödinger equations with super-quadratic and spherically symmetric potential

I. INTRODUCTION

II. SCHRÖDINGER OPERATOR IN THE POLAR COORDINATES

III. ESTIMATES OF EIGENVALUES AND EIGENFUNCTIONS

A. Asymptotic behavior in bounded regions

B. Gaps of the eigenvalues

C. Asymptotic behavior of eigenfunctions

D. Proof of Proposition 3.4

IV. PROOF OF THEOREM 1.2 AND PROPOSITION 1.3

ACKNOWLEDGMENTS

AUTHOR DECLARATIONS

Conflict of Interest

Author Contributions

DATA AVAILABILITY

APPENDIX A: PROOF OF (3.14)

APPENDIX B: PROOF OF (3.16)

REFERENCES

Citing articles via

Submit your article

Sign up for alerts

Resources

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pubs.aip.org

Connect with AIP Publishing

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