We now analyze the eigenstructure of the operator \(\mathcal{K}_{l, \alpha, k}\) by proving Theorem 1. It is precisely the solution structure of the equation \(\det{\mathbf{Q}} = 0\) in λ, which is equivalent to that of (25) in λ. Remember that we only need to consider the case when \(0 < \lambda< 1/k\), which is equivalent to \(\kappa> 0\) by (8).
By Lemma 2, (25) has a solution only when \(0 < \varphi_{+}(\kappa) < 1\) or \(0 < \varphi_{-}(\kappa) < 1\). By (27), (28), and Lemma 3(a), the set of \(\kappa> 0\) satisfying \(0 < \varphi_{+}(\kappa) < 1\) is contained in the union of the intervals
$$A_{n}^{+} := \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr), h^{-1} ( 2\pi n ) \biggr), \quad n=1,2,3, \ldots $$
Similarly, the set of \(\kappa> 0\) satisfying \(0 < \varphi_{-}(\kappa) < 1\) is contained in the union of the intervals
$$A_{n}^{-} := \biggl( h^{-1} ( 2\pi n ), h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) \biggr), \quad n=0,1,2,\ldots $$
In fact, by the intermediate value theorem, there exists at least one κ in each \(A_{n}^{+}\), for \(n = 1,2,3,\ldots\) , satisfying \(p(\kappa) = \varphi_{+}(\kappa)\), since
$$ \begin{aligned} &p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr) > 0 = \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr), \\ &p \bigl( h^{-1} ( 2\pi n ) \bigr) < 1 < \varphi_{+} \bigl( h^{-1} ( 2\pi n ) \bigr) \end{aligned} $$
(35)
for \(n = 1,2,3,\ldots\) , by Lemma 2 and (27), (28). Similarly, there exists at least one κ in each \(A_{n}^{-}\), for \(n = 1,2,3,\ldots\) , satisfying \(p(\kappa) = \varphi_{-}(\kappa)\), since
$$ \begin{aligned} &p \bigl( h^{-1} ( 2\pi n ) \bigr) < 1 < \varphi_{-} \bigl( h^{-1} ( 2\pi n ) \bigr), \\ &p \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) \biggr) > 0 = \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) \biggr) \end{aligned} $$
(36)
for \(n = 1,2,3,\ldots\) Note that we cannot apply the intermediate value theorem to \(A_{0}^{-}\), since \(p(0) = 1 = \varphi_{-}(0)\). In fact, it will be shown in Lemma 5 that \(A_{0}^{-}\) contains no κ satisfying \(p(\kappa) = \varphi_{-}(\kappa)\).
Since the functions \(p(\kappa)\) and \(\varphi_{\pm}(\kappa)\) are real-analytic (and different), the set of κ satisfying (25) is discrete. Thus we can take the smallest \(\beta_{n}\) in \(A_{n}^{+}\) satisfying \(p(\kappa) = \varphi_{+}(\kappa)\), and the largest \(\gamma_{n}\) in \(A_{n}^{-}\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\) for \(n = 1,2,3,\ldots\) Then we have
$$ h^{-1} \biggl( 2 n \pi- \frac{\pi}{2} \biggr) < \beta_{n} < h^{-1} ( 2 n \pi ) < \gamma_{n} < h^{-1} \biggl( 2 n \pi+ \frac{\pi}{2} \biggr), \quad n = 1,2,3, \ldots $$
(37)
Lemma 5
The set of
κ
satisfying the characteristic equation (25) is
$$\{ \beta_{n} \mid n = 1,2,3,\ldots \} \cup \{ \gamma_{n} \mid n = 1,2,3,\ldots \}. $$
Proof
It is sufficient to show that there is no κ in \(A_{0}^{-}\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\), and there is at most one κ in \(A_{n}^{+}\) (respectively, \(A_{n}^{-}\)) satisfying \(p(\kappa) = \varphi_{+}(\kappa)\) (respectively, \(p(\kappa) = \varphi_{-}(\kappa)\)) for \(n = 1,2,3,\ldots\)
Let \(n = 1,2,3,\ldots\) Note that, by (35) and the definition of \(\beta_{n}\), we have \(p(\kappa) > \varphi_{+}(\kappa)\) for every \(\kappa\in ( h^{-1} ( 2\pi n - \pi/2 ), \beta_{n} )\). Suppose there exists another κ in \(A_{n}^{+}\) satisfying \(p(\kappa) = \varphi_{+}(\kappa)\), which we denote \(\tilde{\beta}_{n}\). By the definition of \(\beta_{n}\), we have \(\beta_{n} < \tilde{\beta}_{n}\). We can assume \(\tilde{\beta}_{n}\) is chosen such that there is no κ between \(\beta_{n}\) and \(\tilde{\beta}_{n}\) satisfying \(p(\kappa) = \varphi_{+}(\kappa)\), since the set of solutions of (25) is discrete. So we have either \(p(\kappa) > \varphi_{+}(\kappa)\) for every \(\kappa\in ( \beta_{n}, \tilde{\beta}_{n} )\), or \(p(\kappa) < \varphi_{+}(\kappa)\) for every \(\kappa\in ( \beta_{n}, \tilde{\beta}_{n} )\). Suppose the former. Then the graphs of \(p(\kappa)\) and \(\varphi_{+}(\kappa)\) should be tangent to each other at \(\kappa= \beta_{n}\), which implies \(p^{\prime}(\beta_{n}) = \varphi_{+} ^{\prime}(\beta_{n})\). Since \(p(\beta_{n}) = \varphi_{+}(\beta_{n})\), this contradicts Lemma 4(a), and it follows that \(p(\kappa) < \varphi_{+}(\kappa)\) for every \(\kappa\in ( \beta_{n}, \tilde{\beta}_{n} )\). Then by Lemma 4(a) again, we have \(p^{\prime}(\kappa) <\varphi_{+}^{\prime}(\kappa)\) for every \(\kappa\in ( \beta_{n}, \tilde{\beta}_{n} )\). Applying the mean value theorem to the function \(p(\kappa) - \varphi_{+}(\kappa)\) on \([ \beta_{n}, \tilde{\beta}_{n} ]\), we have
$$0 = \bigl\{ p ( \tilde{\beta}_{n} ) - \varphi_{+} ( \tilde{ \beta}_{n} ) \bigr\} - \bigl\{ p ( \beta_{n} ) - \varphi_{+} ( \beta_{n} ) \bigr\} = \bigl\{ p^{\prime}( \tilde{\kappa} ) - \left . \varphi_{+} \right .^{\prime}( \tilde{\kappa} ) \bigr\} \cdot ( \tilde{\beta}_{n} - \beta_{n} ) $$
for some \(\tilde{\kappa} \in ( \beta_{n}, \tilde{\beta}_{n} )\). Then we have \(p^{\prime}( \tilde{\kappa} ) = \varphi_{+} ^{\prime}( \tilde{\kappa} )\), which is a contradiction. Thus we conclude that there is no κ in \(A_{n}^{+}\) other than \(\beta_{n}\), which satisfies \(p(\kappa) = \varphi_{+}(\kappa)\).
Let \(n = 1,2,3,\ldots\) Note that, by (36) and the definition of \(\gamma_{n}\), we have \(p(\kappa) > \varphi_{-}(\kappa)\) for every \(\kappa\in ( \gamma_{n}, h^{-1} ( 2\pi n + \pi/2 ) )\). Suppose there exists another κ in \(A_{n}^{-}\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\), which we denote \(\tilde{\gamma}_{n}\). By the definition of \(\gamma_{n}\), we have \(\tilde{\gamma}_{n} < \gamma_{n}\). We can assume \(\tilde{\gamma}_{n}\) is chosen such that there is no κ between \(\tilde{\gamma}_{n}\) and \(\gamma_{n}\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\), since the set of solutions of (25) is discrete. So we have either \(p(\kappa) > \varphi_{-}(\kappa)\) for every \(\kappa\in ( \tilde{\gamma}_{n}, \gamma_{n} )\), or \(p(\kappa) < \varphi_{-}(\kappa)\) for every \(\kappa\in ( \tilde{\gamma}_{n}, \gamma_{n} )\). Suppose the former. Then the graphs of \(p(\kappa)\) and \(\varphi_{-}(\kappa)\) should be tangent to each other at \(\kappa= \gamma_{n}\), which implies \(p^{\prime}(\gamma_{n}) = \varphi_{-} ^{\prime}(\gamma_{n})\). Since \(p(\gamma_{n}) = \varphi_{-}(\gamma_{n})\), this contradicts Lemma 4(b), and it follows that \(p(\kappa) < \varphi_{-}(\kappa)\) for every \(\kappa\in ( \tilde{\gamma}_{n}, \gamma_{n} )\). Then by Lemma 4(b) again, we have \(p^{\prime}(\kappa) > \varphi_{-} ^{\prime}(\kappa)\) for every \(\kappa\in ( \tilde{\gamma}_{n}, \gamma_{n} )\). Applying the mean value theorem to the function \(p(\kappa) - \varphi_{-}(\kappa)\) on \([ \tilde{\gamma}_{n}, \gamma_{n} ]\), we have
$$0 = \bigl\{ p ( \gamma_{n} ) - \varphi_{-} ( \gamma_{n} ) \bigr\} - \bigl\{ p ( \tilde{\gamma}_{n} ) - \varphi_{-} ( \tilde{ \gamma}_{n} ) \bigr\} = \bigl\{ p^{\prime}( \tilde{\kappa} ) - \varphi_{-} ^{\prime}( \tilde{\kappa} ) \bigr\} \cdot ( \gamma_{n} - \tilde{\gamma}_{n} ) $$
for some \(\tilde{\kappa} \in ( \tilde{\gamma}_{n}, \gamma_{n} )\). Then we have \(p^{\prime}( \tilde{\kappa} ) = \varphi_{-} ^{\prime}( \tilde{\kappa} )\), which is a contradiction. Thus we conclude that there is no κ in \(A_{n}^{-}\) other than \(\gamma_{n}\), which satisfies \(p(\kappa) = \varphi_{-}(\kappa)\).
Suppose there exists κ in \(A_{0}^{-}\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\). Since the set of solutions of (25) is discrete, we can take \(\gamma_{0}\) to be the largest among such κ. Then we have \(p(\kappa) > \varphi_{-}(\kappa)\) for every \(\kappa\in ( \gamma_{0}, h^{-1} ( \pi/2 ) )\), since \(p ( h^{-1} ( \pi/2 ) ) > 0 = \varphi_{-} ( h^{-1} ( \pi/2 ) )\) by Lemma 2 and (28). Let \(\tilde{\gamma}_{0}\) be the largest in \([ 0, \gamma_{0} )\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\). Note that \(\tilde{\gamma}_{0}\) exists, since \(p(0) = \varphi_{-}(0) = 1\). Replacing \(\tilde{\gamma}_{n}\), \(\gamma_{n}\) by \(\tilde{\gamma}_{0}\), \(\gamma_{0}\), respectively, and applying the same argument in the above paragraph again, results in a contradiction. Thus we conclude that there is no κ in \(A_{0}^{-}\) satisfying \(p(\kappa) = \varphi_{-}(\kappa)\), and the proof is complete. □
Note that the inverse function \(h^{-1}\) of h is strictly increasing from \([0,\infty)\) onto \([0,\infty)\) by Lemma 1(a). Putting \(t = h(\kappa)\), (17) can be written as
$$ L \cdot h^{-1}(t) = t + \hat{h} \bigl( h^{-1}(t) \bigr) \quad \text{for } t \geq0. $$
(38)
Lemma 6
-
(a)
\(1/ ( L + 2 + \sqrt{2} ) \leq ( h^{-1} )^{\prime}(t) < 1/L\)
for
\(t \geq0\).
-
(b)
\(h^{-1}(t) \sim t\)
and
\(h^{-1}(t) - (t - 2\pi)/L \sim t^{-1}\).
Proof
(a) follows immediately from Lemma 1(b), since \(( h^{-1} )^{\prime}(t) = 1/ \{ h^{\prime}( h^{-1}(t) ) \} = 1/h^{\prime}(\kappa)\), where we put \(t = h(\kappa)\).
By (38), we have
$$\begin{aligned}& \lim_{t \to\infty} t \biggl( h^{-1}(t) - \frac{t - 2\pi}{L} \biggr) \\& \quad = \lim_{t \to\infty} t \biggl\{ \frac{t + \hat{h} ( h^{-1}(t) )}{L} - \frac{t - 2\pi}{L} \biggr\} \\& \quad = \frac{1}{L} \lim_{t \to\infty} t \bigl\{ \hat{h} \bigl( h^{-1}(t) \bigr) + 2\pi \bigr\} = \frac{1}{L} \lim _{\kappa\to\infty} h(\kappa) \bigl\{ \hat{h}(\kappa) + 2\pi \bigr\} \\& \quad = \frac{1}{L} \lim_{\kappa\to\infty} \frac{h(\kappa)}{\kappa} \cdot \lim _{\kappa\to\infty} \kappa \bigl\{ \tilde{h}(\kappa) + 2\pi \bigr\} = \frac{1}{L} \cdot L \cdot \lim_{\kappa\to\infty} \frac { \hat{h}(\kappa) + 2\pi }{ \frac{1}{\kappa}}, \end{aligned}$$
where the last equality comes from Lemma 1(b). Since \(\lim_{\kappa\to\infty}{\hat{h}(\kappa)} = -2\pi\), we can use l’Hôspital’s rule to get
$$ \lim_{t \to\infty} t \biggl( h^{-1}(t) - \frac{t - 2\pi}{L} \biggr) = \lim_{\kappa\to\infty} \frac { \hat{h}^{\prime}(\kappa) }{ -\frac{1}{\kappa^{2}}} = \lim _{\kappa\to\infty} \frac{2\sqrt{2} \kappa^{2} ( \kappa^{2} + 1 )}{ \kappa^{4} + 1} = 2\sqrt{2} $$
(39)
by (16). This shows \(\vert h^{-1}(t) - (t - 2\pi)/L \vert \sim t^{-1}\), which also implies \(h^{-1}(t) \sim t\). □
Note that, for \(0 < t < \pi/2\), we have
$$\begin{aligned}& \frac{d}{dt} \biggl( \frac{1 - \cos{t}}{\sin{t}} \biggr) = \frac {\sin{t} \cdot\sin{t} - ( 1 - \cos{t} ) \cdot\cos{t}}{ \sin^{2}{t}} = \frac{1 - \cos{t}}{\sin^{2}{t}} > 0, \\& \frac{d^{2}}{dt^{2}} \biggl( \frac{1 - \cos{t}}{\sin{t}} \biggr) = \frac { \sin{t} \cdot\sin^{2}{t} - ( 1 - \cos{t} ) \cdot2 \sin{t} \cos{t}}{ \sin^{4}{t}} \\& \hphantom{\frac{d^{2}}{dt^{2}} \biggl( \frac{1 - \cos{t}}{\sin{t}} \biggr)}= \frac{1 + \cos^{2}{t} - 2\cos{t}}{\sin^{3}{t}} = \frac{ ( 1 - \cos{t} )^{2}}{\sin^{3}{t}} > 0. \end{aligned}$$
This implies that the function \(( 1 - \cos{t} )/\sin{t}\) is increasing and convex on \(( 0, \pi/2 )\), and hence \(t/2 < ( 1 - \cos{t} )/\sin{t} < 2t/\pi \) for \(0 < t < \pi/2\), since \(\lim_{t \to0} \{ ( 1 - \cos{t} )/\sin{t} \} = 0\), \(( 1 - \cos ( \pi/2 ) ) /\sin ( \pi/2 ) = 1\), and \(\lim_{t \to0} \{ ( 1 - \cos{t} )/\sin{t} \}^{\prime}= \lim_{t \to0} \{ ( 1 - \cos{t} )/\sin^{2}{t} \} = 1/2\). It follows that
$$ \frac{t}{2} < \frac {1 + \sin ( 2\pi n - \frac{\pi}{2} + t )}{ \cos ( 2\pi n - \frac{\pi}{2} + t )} = \frac {1 - \sin ( 2\pi n + \frac{\pi}{2} - t )}{ \cos ( 2\pi n + \frac{\pi}{2} - t )} < \frac{2t}{\pi} \quad \text{for } 0 < t < \frac{\pi}{2}, $$
(40)
since
$$\frac {1 + \sin ( 2\pi n - \frac{\pi}{2} + t )}{ \cos ( 2\pi n - \frac{\pi}{2} + t )} = \frac {1 - \sin ( \frac{\pi}{2} - t )}{ \cos ( \frac{\pi}{2} - t )} = \frac{1 - \cos{t}}{\sin{t}}. $$
Note that \(0 < p(\kappa) < 1\) for \(\kappa> 0\) by Lemma 2. For each \(n = 1,2,3,\ldots\) , we can take \(0 < \epsilon_{n}^{+} < \delta_{n}^{+} < \pi/2\) such that
$$\begin{aligned}& \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) \biggr) = p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr), \end{aligned}$$
(41)
$$\begin{aligned}& \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) \biggr) = 1, \end{aligned}$$
(42)
since \(\varphi_{+}\) is strictly increasing on \(A_{n}^{+}\) from \(\varphi_{+} ( h^{-1} ( 2\pi n - \pi/2 ) ) = 0\) to \(\varphi_{+} ( h^{-1} ( 2\pi n ) ) > 1\) by (27), (28), Lemma 3(a). Similarly, we can take \(0 < \epsilon_{n}^{-} < \delta_{n}^{-} < \pi/2\) for each \(n = 1,2,3,\ldots\) , such that
$$\begin{aligned}& \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) \biggr) = 1, \end{aligned}$$
(43)
$$\begin{aligned}& \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \biggr) = p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr), \end{aligned}$$
(44)
since \(\varphi_{-}\) is strictly decreasing on \(A_{n}^{-}\) from \(\varphi_{+} ( h^{-1} ( 2\pi n ) ) > 1\) to \(\varphi_{+} ( h^{-1} ( 2\pi n + \pi/2 ) ) = 0\) by (27), (28), Lemma 3(a).
Suppose n is sufficiently large, so that \(h^{-1} ( 2\pi n - \pi/2 ) > 1\). This is possible, since \(h^{-1}\) is one-to-one and onto from \([0,\infty)\) to \([0,\infty)\) by Lemma 1(a). Then, since p is strictly increasing on \((1,\infty)\) by Lemma 2, we have
$$ p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr) < p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) \biggr) < p \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \biggr), $$
and hence by (41), (42), (43), (44),
$$\begin{aligned}& \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) \biggr) < p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) \biggr), \\& \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) \biggr) > p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) \biggr), \\& \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) \biggr) > p \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) \biggr), \\& \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \biggr) < p \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \biggr). \end{aligned}$$
It follows from the intermediate value theorem that, for sufficiently large n,
$$\begin{aligned}& h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) < h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) < \beta_{n} < h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr), \end{aligned}$$
(45)
$$\begin{aligned}& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) < \gamma_{n} < h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) < h^{-1} \biggl( 2 \pi n + \frac{\pi}{2} \biggr), \end{aligned}$$
(46)
since \(\beta_{n}\) (respectively, \(\gamma_{n}\)) is the only κ in \(A_{n}^{+}\) (respectively, \(A_{n}^{-}\)) satisfying \(p(\kappa) = \varphi_{+}(\kappa)\) (respectively, \(p(\kappa) = \varphi_{-}(\kappa)\)).
Lemma 7
\(\beta_{n} \sim\gamma_{n} \sim n\), and
\(\beta_{n} - h^{-1} ( 2 \pi n - \pi/2 ) \sim h^{-1} ( 2 \pi n + \pi/2 ) - \gamma_{n} \sim e^{-2\pi n}\), \(\beta_{n} - ( 2 \pi(n-1) - \pi/2 )/L \sim \gamma_{n} - ( 2 \pi(n-1) + \pi/2 )/L \sim n^{-1}\).
Proof
Suppose n is sufficiently large so that (45), (46) hold. The fact \(\beta_{n} \sim\gamma_{n} \sim n\) immediately follows from (45), (46), since \(h^{-1}(t) \sim t\) by Lemma 6(b). By (45), (46), we have
$$\begin{aligned}& \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) > h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr), \end{aligned}$$
(47)
$$\begin{aligned}& \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) < h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr), \end{aligned}$$
(48)
$$\begin{aligned}& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} > h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr), \end{aligned}$$
(49)
$$\begin{aligned}& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} < h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr). \end{aligned}$$
(50)
By applying the mean value theorem to \(h^{-1}\), we have
$$\begin{aligned}& h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) = \bigl( h^{-1} \bigr)^{\prime}\biggl( 2\pi n - \frac{\pi}{2} + \tilde{\epsilon}_{n}^{+} \biggr) \cdot \epsilon_{n}^{+}, \\& h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) = \bigl( h^{-1} \bigr)^{\prime}\biggl( 2\pi n - \frac{\pi}{2} + \tilde{\delta}_{n}^{+} \biggr) \cdot \delta_{n}^{+}, \\& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) = \bigl( h^{-1} \bigr)^{\prime}\biggl( 2\pi n + \frac{\pi}{2} - \tilde{\epsilon}_{n}^{-} \biggr) \cdot \epsilon_{n}^{-}, \\& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) = \bigl( h^{-1} \bigr)^{\prime}\biggl( 2\pi n + \frac{\pi}{2} - \tilde{\delta}_{n}^{-} \biggr) \cdot \delta_{n}^{-} \end{aligned}$$
for some \(0 \leq\tilde{\epsilon}_{n}^{+} \leq\epsilon_{n}^{+}\), \(0 \leq\tilde{\delta}_{n}^{+} \leq\delta_{n}^{+}\), \(0 \leq\tilde{\epsilon}_{n}^{-} \leq\epsilon_{n}^{-}\), \(0 \leq\tilde{\delta}_{n}^{-} \leq\delta_{n}^{-}\). So by Lemma 6(a), we have
$$\begin{aligned}& h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \geq \frac {\epsilon_{n}^{+}}{ L + 2 + \sqrt{2}}, \\& h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) < \frac {\delta_{n}^{+}}{ L}, \\& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \geq \frac {\epsilon_{n}^{-}}{ L + 2 + \sqrt{2}}, \\& h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) < \frac {\delta_{n}^{-}}{ L}, \end{aligned}$$
and hence by (47), (48), (49), (50),
$$\begin{aligned}& \frac {\epsilon_{n}^{+}}{ L + 2 + \sqrt{2}} < \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) < \frac {\delta_{n}^{+}}{ L}, \end{aligned}$$
(51)
$$\begin{aligned}& \frac {\epsilon_{n}^{-}}{ L + 2 + \sqrt{2}} < h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} < \frac {\delta_{n}^{-}}{ L}. \end{aligned}$$
(52)
Using (40), (41), (42), (43), (44), and the definition (24) of \(\varphi_{\pm}\), we have
$$\begin{aligned}& p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr) \\& \quad = \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) \biggr) \\& \quad = \exp \biggl\{ L \cdot h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} \biggr) \biggr\} \cdot \frac { 1 + \sin ( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} ) }{ \cos ( 2\pi n - \frac{\pi}{2} + \epsilon_{n}^{+} ) } \\& \quad < \exp \bigl\{ L \cdot h^{-1} ( 2\pi n ) \bigr\} \cdot \frac{2}{\pi} \epsilon_{n}^{+}, \\& p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr) \\& \quad = \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \biggr) \\& \quad = \exp \biggl\{ L \cdot h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} \biggr) \biggr\} \cdot \frac { 1 - \sin ( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} ) }{ \cos ( 2\pi n + \frac{\pi}{2} - \epsilon_{n}^{-} ) } \\& \quad < \exp \biggl\{ L \cdot h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) \biggr\} \cdot \frac{2}{\pi} \epsilon_{n}^{-} \end{aligned}$$
and
$$\begin{aligned}& 1= \varphi_{+} \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) \biggr) \\& \hphantom{1}= \exp \biggl\{ L \cdot h^{-1} \biggl( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} \biggr) \biggr\} \cdot \frac { 1 + \sin ( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} ) }{ \cos ( 2\pi n - \frac{\pi}{2} + \delta_{n}^{+} ) } \\& \hphantom{1}> \exp \biggl\{ L \cdot h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} \cdot \frac{1}{2} \delta_{n}^{+}, \\& 1= \varphi_{-} \biggl( h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) \biggr) \\& \hphantom{1}= \exp \biggl\{ L \cdot h^{-1} \biggl( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} \biggr) \biggr\} \cdot \frac { 1 - \sin ( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} ) }{ \cos ( 2\pi n + \frac{\pi}{2} - \delta_{n}^{-} ) } \\& \hphantom{1}> \exp \bigl\{ L \cdot h^{-1} ( 2\pi n ) \bigr\} \cdot \frac{1}{2} \delta_{n}^{-}, \end{aligned}$$
and hence
$$\begin{aligned}& \epsilon_{n}^{+} > \frac{\pi}{2} \cdot p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr) \exp \bigl\{ -L \cdot h^{-1} ( 2\pi n ) \bigr\} , \end{aligned}$$
(53)
$$\begin{aligned}& \epsilon_{n}^{-} > \frac{\pi}{2} \cdot p \biggl( h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr) \exp \biggl\{ -L \cdot h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) \biggr\} , \end{aligned}$$
(54)
$$\begin{aligned}& \delta_{n}^{+} < 2 \exp \biggl\{ -L \cdot h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} , \end{aligned}$$
(55)
$$\begin{aligned}& \delta_{n}^{-} < 2 \exp \bigl\{ -L \cdot h^{-1} ( 2\pi n ) \bigr\} . \end{aligned}$$
(56)
Note that, for any constant c, we have \(\lim_{n \to\infty} p ( h^{-1} ( 2\pi n + c ) ) = 1 \) by Lemma 2 and
$$\begin{aligned}& \lim_{n \to\infty} \bigl[ e^{2\pi n} \cdot \exp \bigl\{ -L \cdot h^{-1} ( 2\pi n + c ) \bigr\} \bigr] \\& \quad= \lim_{n \to\infty} \exp \bigl\{ 2\pi n -L \cdot h^{-1} ( 2\pi n + c ) \bigr\} \\& \quad= \lim_{t \to\infty} \exp \bigl\{ t - c - L \cdot h^{-1} ( t ) \bigr\} = \lim_{t \to\infty} \exp \bigl\{ t - 2 \pi + 2\pi- c - L \cdot h^{-1} ( t ) \bigr\} \\& \quad= \lim_{t \to\infty} \exp \biggl[ L \cdot \biggl\{ \frac{t - 2\pi}{L} - h^{-1} ( t ) \biggr\} + ( 2\pi- c ) \biggr] = e^{2\pi- c} \end{aligned}$$
by Lemma 6(b). So by combining (51), (52), and (53), (54), (55), (56), we have
$$\begin{aligned}& \frac{\pi e^{2\pi}}{2 ( L + 2 + \sqrt{2} )} \leq \lim_{n \to\infty} \biggl[ e^{2\pi n} \cdot \biggl\{ \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} \biggr] \leq \frac{2 e^{2\pi+ \frac{\pi}{2}}}{L}, \end{aligned}$$
(57)
$$\begin{aligned}& \frac{\pi e^{2\pi- \frac{\pi}{2}}}{2 ( L + 2 + \sqrt{2} )} \leq \lim_{n \to\infty} \biggl[ e^{2\pi n} \cdot \biggl\{ h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} \biggr\} \biggr] \leq \frac{2 e^{2\pi}}{L}, \end{aligned}$$
(58)
which shows \(\beta_{n} - h^{-1} ( 2 \pi n - \pi/2 ) \sim h^{-1} ( 2 \pi n + \pi/2 ) - \gamma_{n} \sim e^{-2\pi n}\).
By (57), (58), we have
$$\begin{aligned}& 0 \leq \lim_{n \to\infty} n \biggl\{ \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} \\& \hphantom{0} = \lim_{n \to\infty} {n e^{-2\pi n}} \cdot \lim _{n \to\infty} \biggl[ e^{2\pi n} \cdot \biggl\{ \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} \biggr] \\& \hphantom{0} \leq { \frac{2 e^{2\pi+ \frac{\pi}{2}}}{L} } \cdot \lim_{n \to\infty} {n e^{-2\pi n}} = 0, \\& 0 \leq \lim_{n \to\infty} n \biggl\{ h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} \biggr\} \\& \hphantom{0} = \lim_{n \to\infty} {n e^{-2\pi n}} \cdot \lim _{n \to\infty} \biggl[ e^{2\pi n} \cdot \biggl\{ h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} \biggr\} \biggr] \\& \hphantom{0} \leq { \frac{2 e^{2\pi}}{L} } \cdot \lim_{n \to\infty} {n e^{-2\pi n}} = 0, \end{aligned}$$
and hence
$$\lim_{n \to\infty} n \biggl\{ \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} = \lim_{n \to\infty} n \biggl\{ h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \gamma_{n} \biggr\} = 0. $$
So by (39), we have
$$\begin{aligned}& \lim_{n \to\infty} n \biggl\{ \beta_{n} - \frac{1}{L} \biggl( 2\pi(n-1) - \frac{\pi}{2} \biggr) \biggr\} \\& \quad= \lim_{n \to\infty} n \biggl\{ \beta_{n} - h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) \biggr\} \\& \qquad {}+ \lim _{n \to\infty} n \biggl\{ h^{-1} \biggl( 2\pi n - \frac{\pi}{2} \biggr) - \frac{1}{L} \biggl( 2\pi n - \frac{\pi}{2} \biggr) + \frac{2\pi}{L} \biggr\} \\& \quad = \lim_{t \to\infty} \frac{t + \frac{\pi}{2}}{2\pi} \biggl( h^{-1}(t) - \frac{t - 2\pi}{L} \biggr) \\& \quad= \lim_{t \to\infty} \frac{t + \frac{\pi}{2}}{2\pi t} \cdot \lim _{t \to\infty} t \biggl( h^{-1}(t) - \frac{t - 2\pi}{L} \biggr) = \frac{1}{2\pi} \cdot2\sqrt{2} = \frac{\sqrt{2}}{\pi}, \\& \lim_{n \to\infty} n \biggl\{ \gamma_{n} - \frac{1}{L} \biggl( 2\pi(n-1) + \frac{\pi}{2} \biggr) \biggr\} \\& \quad= \lim_{n \to\infty} n \biggl\{ \gamma_{n} - h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) \biggr\} \\& \qquad {}+ \lim _{n \to\infty} n \biggl\{ h^{-1} \biggl( 2\pi n + \frac{\pi}{2} \biggr) - \frac{1}{L} \biggl( 2\pi n + \frac{\pi}{2} \biggr) + \frac{2\pi}{L} \biggr\} \\& \quad = \lim_{t \to\infty} \frac{t - \frac{\pi}{2}}{2\pi} \biggl( h^{-1}(t) - \frac{t - 2\pi}{L} \biggr) \\& \quad = \lim_{t \to\infty} \frac{t - \frac{\pi}{2}}{2\pi t} \cdot \lim _{t \to\infty} t \biggl( h^{-1}(t) - \frac{t - 2\pi}{L} \biggr) = \frac{1}{2\pi} \cdot2\sqrt{2} = \frac{\sqrt{2}}{\pi}, \end{aligned}$$
which shows \(\beta_{n} - ( 2 \pi(n-1) - \pi/2 )/L \sim \gamma_{n} - ( 2 \pi(n-1) + \pi/2 )/L \sim n^{-1}\), and the proof is complete. □
Lemma 8
Suppose positive sequences
\(\{ a_{n} \}_{n=1}^{\infty}\), \(\{ b_{n} \}_{n=1}^{\infty}\), \(\{ c_{n} \}_{n=1}^{\infty}\)
satisfy
\(a_{n} \sim b_{n} \sim n\)
and
\(a_{n} - b_{n} \sim c_{n}\). Then
\(1/ ( 1 + b_{n}^{4} ) - 1/ ( 1 + a_{n}^{4} ) \sim n^{-5} c_{n}\).
Proof
Let \(f(x) = 1/ ( 1 + x^{4} )\). By the mean value theorem, we have
$$\begin{aligned} \frac{1}{1 + b_{n}^{4}} - \frac{1}{1 + a_{n}^{4}} &= f ( b_{n} ) - f ( a_{n} ) = f^{\prime}( \xi_{n} ) \cdot ( b_{n} - a_{n} ) \\ &= \frac{4 \xi_{n}^{3}}{ ( 1 + \xi_{n}^{4} )^{2}} \cdot ( a_{n} - b_{n} ) \end{aligned}$$
for some \(b_{n} \leq\xi_{n} \leq a_{n}\) for \(n = 1,2,3,\ldots\) Note that \(\xi_{n} \sim a_{n} \sim b_{n} \sim n\). So we have
$$n^{5} c_{n}^{-1} \cdot \biggl( \frac{1}{1 + b_{n}^{4}} - \frac{1}{1 + a_{n}^{4}} \biggr) = \frac { 4 ( \frac{\xi_{n}}{n} )^{3} }{ \{ \frac{1}{n^{4}} + ( \frac{\xi_{n}}{n} )^{4} \}^{2} } \cdot \frac{a_{n} - b_{n}}{c_{n}}, $$
which is bounded below and above by some positive constants for every sufficiently large n, since \(\xi_{n} \sim n\) and \(a_{n} - b_{n} \sim c_{n}\). This implies \(1/ ( 1 + b_{n}^{4} ) - 1/ ( 1 + a_{n}^{4} ) \sim n^{-5} c_{n}\). □
Proof of Theorem 1
By Proposition 3, \(\mathcal{K}_{l, \alpha, k}\) has no eigenvalues outside the interval \(( 0, 1/k )\). By (8) and Lemma 5, the eigenvalues in \(( 0, 1/k )\) are \(\mu_{n}/k\), \(\nu_{n}/k\), \(n = 1,2,3,\ldots\) , where we put
$$ \mu_{n} := \frac{1}{1 + \beta_{n}^{4}}, \qquad \nu_{n} := \frac{1}{1 + \gamma_{n}^{4}} $$
(59)
for \(n = 1,2,3,\ldots\) Note that L is the only parameter involved with the characteristic equation (25). So its solutions \(\beta_{n}\), \(\gamma_{n}\), and hence \(\mu_{n}\), \(\nu_{n}\), depend only on L for \(n = 1,2,3,\ldots\) The bounds on \(\mu_{n}\), \(\nu_{n}\) in (a) follow from (37) and (59), and thus we showed (a).
Since \(\beta_{n} \sim\gamma_{n} \sim n\) by Lemma 7, it follows easily from (59) that \(\mu_{n} \sim\nu_{n} \sim n^{-4}\). Note that \(h^{-1} ( 2 \pi n - \pi/2 ) \sim h^{-1} ( 2 \pi n + \pi/2 ) \sim n\) by Lemma 6(b). So by Lemma 8 and (59), we have
$$\begin{aligned}& \frac {1}{ 1 + \{ h^{-1} ( 2 \pi n - \frac{\pi}{2} ) \}^{4}} - \mu_{n} = \frac {1}{ 1 + \{ h^{-1} ( 2 \pi n - \frac{\pi}{2} ) \}^{4}} - \frac{1}{1 + \beta_{n}^{4}} \sim n^{-5} e^{-2\pi n}, \\& \nu_{n} - \frac {1}{ 1 + \{ h^{-1} ( 2 \pi n + \frac{\pi}{2} ) \}^{4}} = \frac{1}{1 + \gamma_{n}^{4}} - \frac {1}{ 1 + \{ h^{-1} ( 2 \pi n + \frac{\pi}{2} ) \}^{4}} \sim n^{-5} e^{-2\pi n}, \\& \frac {1}{ 1 + \frac{1}{L^{4}} ( 2 \pi(n - 1) - \frac{\pi}{2} )^{4} } - \mu_{n} = \frac {1}{ 1 + \frac{1}{L^{4}} ( 2 \pi(n - 1) - \frac{\pi}{2} )^{4} } - \frac{1}{1 + \beta_{n}^{4}} \sim n^{-6}, \\& \frac {1}{ 1 + \frac{1}{L^{4}} ( 2 \pi(n - 1) + \frac{\pi}{2} )^{4} } - \nu_{n} = \frac {1}{ 1 + \frac{1}{L^{4}} ( 2 \pi(n - 1) + \frac{\pi}{2} )^{4} } - \frac{1}{1 + \gamma_{n}^{4}} \sim n^{-6}, \end{aligned}$$
since \(\beta_{n} - h^{-1} ( 2 \pi n - \pi/2 ) \sim h^{-1} ( 2 \pi n + \pi/2 ) - \gamma_{n} \sim e^{-2\pi n}\) and \(\beta_{n} - ( 2 \pi(n-1) - \pi/2 )/L \sim \gamma_{n} - ( 2 \pi(n-1) + \pi/2 )/L \sim n^{-1}\) by Lemma 7. This shows (b), and the proof is complete. □