Sharp bounds for Gauss Lemniscate functions and Lemniscatic means

1.   Introduction

Gauss's arc lemniscate sine and the hyperbolic arc lemniscate sine functions are defined by as follows:

respectively (cf. [1,p.259] or [2,(2.5) and (2.6)]).

Another pair of arc lemniscate functions, Gauss's arc lemniscate tangent and the hyperbolic arc lemniscate tangent functions are defined in terms of the arc lemniscate sine and the hyperbolic arc lemniscate sine functions, respectively (cf. [3,(3.5) and (3.6)]):

It is not difficult to verify that

and

where

is the complete elliptic integral of the first kind ^{[4,5,6,7,8,9,10]}, and

is the classical Euler gamma function ^{[11,12,13,14]}.

For $a, b > 0$ with $a\neq b$, the arithmetic mean $\mathcal{A}(a, b)$, geometric mean $\mathcal{G}(a, b)$, harmonic mean $\mathcal{H}(a, b)$, quadratic mean $\mathcal{Q}(a, b)$, contra-harmonic mean $\mathcal{C}(a, b)$ and the lemniscatic mean $\mathcal{LM}(a, b)$ ^[3,15] of $a$ and $b$ are respectively defined by

and

Recall that the lemniscatic mean $\mathcal{LM}(a, b)$ of two positive real $a$ and $b$ is an iterative mean, i.e.,

where

for $n = 0, 1, 2, \cdots$. And the lemniscatic mean is non-symmetric and homogeneous of degree one with respect to its variables $a$ and $b$. Let

Then one has four symmetric and homogeneous means given by explicitly as follows: (c.f. ^[3,15])

with $z = \sqrt{|a-b|/(a+b)}$. Moreover, [3,(6.10)] showed that the derived means satisfy the following inequality chain:

for all $a, b > 0$ with $a\neq b$.

In the recent years, the arc lemniscate functions and lemniscatic means have attracted the attention of many researchers because they are closely related to special functions. In particular, many remarkable properties and inequalities involving them can be found in the literature ^{[16,17,18,19,20,21,22]}. For example, Neuman ^[3] obtained

Proposition 1.1. ([3,Proposition 6.1]) Inequalities

hold true.

Let $a, b > 0$, $p\in[1, \infty)$, $q\in[1/2, +\infty)$, $t\in[0, 1/2]$ and

and

Then it is easy to check that

and the function $t\rightarrow \mathcal{GA}_{t, p}(a, b)$ is strictly increasing on the interval $[0, 1/2]$ for fixed $a, b > 0$ with $a\neq b$ and $p\in[1, +\infty)$, while $t\rightarrow \mathcal{CA}_{t, q}(a, b)$ is strictly decreasing on $[0, 1/2]$ for fixed $a, b > 0$ with $a\neq b$ and $q\in[1/2, +\infty)$. For more inequalities involving $\mathcal{GA}_{t, p}(a, b)$, $\mathcal{CA}_{t, q}(a, b)$ and thier applications in special functions, the readers can refer to the literature ^{[8,23,24,25,26,27,28,29,30,31,32,33,34,35]}.

Since

hold for all $a, b > 0$ with $a\neq b$, then it is nature to ask that: what are the best possible parameters $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, $\lambda_{4}$, $\mu_{1}$, $\mu_{2}$, $\mu_{3}$, $\mu_{4}\in[0, 1/2]$ such that the double inequalities

hold for all $p\in[1, \infty)$, $q\in[1/2, +\infty)$ and $a, b > 0$ with $a\neq b$.

2.   Lemmas

In order to prove our main results, we need the derivative formulas of the arc lemniscate functions and several lemmas, which we present in this section.

Lemma 2.1. ([36,Theorem 1.25]). Let $-\infty < a < b < \infty$, $f, g:[a, b]\rightarrow {\mathbb{R}}$ be continuous on $[a, b]$, and differentiable on $(a, b)$. Let $g'(x)\neq 0$ on $(a, b)$. Then, if $f'(x)/g'(x)$ is increasing (decreasing) on $(a, b)$, so are

If $f'(x)/g'(x)$ is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.2. ([22,Lemma 3.3]). $(1)$ The function $x\rightarrow \frac{\sqrt{1-x^4}{ {\rm{arcsl}}(x)}}{x}$ is strictly decreasing from $(0, 1)$ onto $(0, 1)$;

$(2)$ The function $x\rightarrow \frac{\sqrt{1+x^4} {\rm{arcslh}}(x)}{x}$ is strictly increasing from $(0, 1)$ onto $(1, \omega)$;

$(3)$ The function $x\rightarrow \frac{\sqrt[4]{1+x^4} {\rm{arctl}}(x)}{x}$ is strictly increasing from $(0, 1)$ onto $(1, 2^{1/4}\tau)$;

$(4)$ The function $x\rightarrow \frac{\sqrt[4]{1-x^4}{ {\rm{arctlh}}}(x)}{x}$ is strictly decreasing from $(0, 1)$ onto $(0, 1)$.

Lemma 2.3. For $u\in[0, 1]$, $p\in\left[{1, \infty}\right)$. Let

Then the following statements are true:

$(1)$ $f\left({u, p;x}\right) > 0$ for $x\in\left({0, 1}\right)$ if and only if $u \leq 3/\left({5p} \right)$;

$(2)$ $f\left({u, p;x}\right) < 0$ for $x\in\left({0, 1}\right)$ if and only if $u \geq 1-{\left[{1/({4\omega^4})}\right]^{1/p}}$.

Proof. It follows from (2.1) that

where

Let $\zeta_{1}(x) = x/(1-x^4)^{3/4}- {\rm{arctlh}}(x)$ and $\zeta_{2}(x) = x^5/(1-x^4)^{3/4}+(p-1)x^4 {\rm{arctlh}}(x)$. Then simple computations lead to

Eq (2.6) and Lemma 2.2(4) show that ${\zeta_{1}'(x)}/{\zeta_{2}'(x)}$ is strictly increasing on $(0, 1)$, so is $f_{p}(x)$ by (2.5) and Lemma 2.1. Moreover,

Following we divide the proof into three cases.

Case 1 $0\leq u \leq 3/(5p)$. Then from (2.4) and (2.7) together with the monotonicity of $f_{p}(x)$ lead to the conclusion that the function $x \mapsto f\left({u, p;x} \right)$ is strictly increasing on $(0, 1)$. Therefore, $f\left({u, p;x} \right) > 0$ for all $x \in \left({0, 1} \right)$ follows from (2.2).

Case 2 $1-\left[1/(4\omega^4)\right]^{1/p}\leq u\leq 1$(Since the function $u\rightarrow f(u, p;1^-)$ is strictly decreasing on $(0, 1)$, and $f(3/(5p), p;1^-) > 0$, $f(1, p;1^-) = -\infty$, then the function $u\rightarrow f(u, p;1^-)$ has a unique zero-point $1-\left[1/(4\omega^4)\right]^{1/p}\in(3/(5p), 1)$ for any fixed $p\in[1, \infty)$). Then it follows from (2.3), (2.4) and (2.7) together with the monotonicity of $f_{p}(x)$ that

and there exists ${x_0} \in \left({0, 1} \right)$ such that the function $x \mapsto f\left({u, p;x} \right)$ is strictly decreasing on $\left({0, {x_0}}\right)$ and strictly increasing on $\left({{x_0}, 1}\right)$. Therefore, together with (2.2), $f\left({u, p;x}\right) < 0$ for all $x\in\left({0, 1}\right)$.

Case 3 $3/(5p) < u < 1-\left[1/(4\omega^4)\right]^{1/p}$. Then it is easy to check that the function $x \mapsto f\left(u, p;x\right)$ has the same piecewise monotone property in Case 2, and while

Combining with (2.2), we conclude that there exists $x_{0}^*\in \left(0, 1\right)$ such that $f\left(u, p;x\right) < 0$ for $x\in\left(0, x_{0}^*\right)$ and $f\left(u, p;x\right) > 0$ for $x\in\left(x_{0}^*, 1\right)$.

Lemma 2.4. For $v\in[0, 1]$, $p\in\left[{1, \infty}\right)$. Let

Then the following statements are true:

$(1)$ $g\left({v, p;x}\right) > 0$ for $x\in\left({0, 1}\right)$ if and only if $v \leq 2/\left({5p} \right)$;

$(2)$ $g\left({v, p;x}\right) < 0$ for $x\in\left({0, 1}\right)$ if and only if $v \geq 1-{\left({1/{\omega^4}}\right)^{1/p}}$.

Proof. It follows from (2.8) that

where

Let $\xi_{1}(x) = x/\sqrt{1-x^4}- {\rm{arcsl}}(x)$ and $\xi_{2}(x) = x^5/\sqrt{1-x^4}+(p-1)x^4 {\rm{arcsl}}(x)$. Then simple computations lead to

Eq (2.13) and Lemma 2.2(1) imply that ${\xi_{1}'(x)}/{\xi_{2}'(x)}$ is strictly increasing on $(0, 1)$. Therefore, the conclusion that $g_{p}(x)$ is strictly increasing on $(0, 1)$ follows from Lemma 2.1 and (2.12). Moreover,

Following we divide the proof into three cases.

Case 1 $0 < v\leq 2/(5p)$. Then (2.11), (2.14) and the monotonicity of the function $g_{p}(x)$ lead to the conclusion that the function $x\mapsto g(v, p;x)$ is strictly increasing on $(0, 1)$. Therefore, $g(v, p;x) > 0$ for all $x \in (0, 1)$ follows from (2.9).

Case 2 $1-\left(1/\omega^4\right)^{1/p}\leq v < 1$ (With the similar argument in the proof of Lemma 2.3, we claim that $1-\left(1/\omega^4\right)^{1/p}\in(2/(5p), 1)$). Then it follows from (2.10), (2.11) and (2.14) together with the monotonicity of $g_{p}(x)$ that

and there exists $x_{1}\in\left(0, 1\right)$ such that the function $x \mapsto g(v, p;x)$ is strictly decreasing on $\left({0, {x_1}}\right)$ and strictly increasing on $\left(x_{1}, 1\right)$. Therefore, together with (2.9), $g(v, p;x) < 0$ for all $x\in\left({0, 1}\right)$.

Case 3 $2/(5p) < v < 1-\left(1/\omega^4\right)^{1/p}$. Then the function $x \mapsto g(v, p;x)$ also first decreases and then increases on $(0, 1)$, and from (2.10) we get

Combining with (2.9), we conclude that there exists $x_{1}^*\in(0, 1)$ such that $g(v, p;x) < 0$ for $x \in \left(0, x_{1}^*\right)$ and $g(v, p;x) > 0$ for $x\in\left(x_{1}^*, 1\right)$.

Lemma 2.5. For $u\in[0, 1]$, $q\in\left[1/2, \infty\right)$. Let

Then the following statements are true:

$(1)$ $F\left({u, q;x}\right) > 0$ for $x\in\left({0, 1}\right)$ if and only if $u \geq 1/\left({5q} \right)$;

$(2)$ $F\left({u, q;x}\right) < 0$ for $x\in\left({0, 1}\right)$ if and only if $u \leq \left(\sqrt{2}/\omega\right)^{2/q}-1$.

Proof. It follows from (2.15) that

where

Let $\zeta_{1}^*(x) = {\rm{arcslh}}(x)-x/\sqrt{1+x^4}$ and $\zeta_{2}^*(x) = x^5/\sqrt{1+x^4}+(2q-1)x^4 {\rm{arcslh}}(x)$. Then simple computations lead to

Eq (2.20) and Lemma 2.2(2) show that ${{\zeta_{1}^*}'(x)}/{{\zeta_{2}^*}'(x)}$ is strictly decreasing on $(0, 1)$, so is $F_{q}(x)$ by (2.19) and Lemma 2.1. Moreover,

Following we divide the proof into three cases.

Case 1 $1/(5q)\leq u\leq 1$. Then (2.18) and (2.21) together with the monotonicity of $F_{q}(x)$ imply that the function $x \mapsto F\left(u, q;x\right)$ is strictly increasing on $(0, 1)$. Therefore, $F\left(u, q;x\right) > 0$ for all $x \in(0, 1)$ follows from (2.16).

Case 2 $0\leq u\leq (\omega-1)/[1+(2q-1)\omega]$. Then the function $x \mapsto F\left(u, q;x\right)$ is strictly decreasing on $(0, 1)$, and consequently $F\left(u, q;x\right) < 0$ for all $x \in(0, 1)$.

Case 3 $(\omega-1)/[1+(2q-1)\omega] < u < 1/(5q)$. Then it follows from (2.18) and (2.21) together with the monotonicity of $F_{q}(x)$ that and there exists ${x_2} \in \left(0, 1\right)$ such that the function $x \mapsto F\left(u, q;x\right)$ is strictly decreasing on $\left({0, {x_2}}\right)$ and strictly increasing on $\left({{x_2}, 1}\right)$.

Since the function $u\rightarrow F(u, q;1^-)$ is strictly increasing on $[0, 1]$, and $F(1/(5q), q;1^-) > 0$, $F((\omega-1)/[1+(2q-1)\omega], q;1^-) < 0$, then the function $u\rightarrow F(u, q;1^-)$ has a unique zero-point $\left(\sqrt{2}/\omega\right)^{2/q}-1\in((\omega-1)/[1+(2q-1)\omega], 1/(5q))$ for any fixed $q\in[1/2, +\infty)$. Finally we investigate two subcases $\left(\sqrt{2}/\omega\right)^{2/q}-1 < u < 1/(5q)$ and $(\omega-1)/[1+(2q-1)\omega] < u\leq \left(\sqrt{2}/\omega\right)^{2/q}-1$.

Subcase 3.1 $\left(\sqrt{2}/\omega\right)^{2/q}-1 < u < 1/(5q)$. Then

and thereby there exists $x_{2}^*\in(0, 1)$ such that $F(u, q;x) < 0$ for $x\in (0, x_{2}^*)$ and $F(u, q;x) > 0$ for $x\in\left(x_{2}^*, 1\right)$.

Subcase 3.2 $(\omega-1)/[1+(2q-1)\omega] < u\leq \left(\sqrt{2}/\omega\right)^{2/q}-1$. Then

and therefore $F(u, q;x) < 0$ for $x\in (0, 1)$.

Lemma 2.6. For $v\in[0, 1]$, $q\in\left[1/2, \infty\right)$. Let

Then the following statements are true:

$(1)$ $G\left({v, q;x}\right) > 0$ for $x\in\left({0, 1}\right)$ if and only if $v \geq 3/(10q)$;

$(2)$$G\left({v, q;x}\right) < 0$ for $x\in\left({0, 1}\right)$ if and only if $u \leq (1/\tau)^{2/q}-1$.

Proof. It follows from (2.22) that

where

Let $\xi_{1}^*(x) = {\rm{arctl}}(x)-x/(1+x^4)^{3/4}$ and $\xi_{2}^*(x) = x^5/(1+x^4)^{3/4}+(2q-1)x^4 {\rm{arctl}}(x)$. Then simple computations lead to

Eq (2.27) and Lemma 2.2 (3) show that ${{\xi_{1}^*}'(x)}/{{\xi_{2}^*}'(x)}$ is strictly decreasing on $(0, 1)$, so is $G_{q}(x)$ by (2.26) and Lemma 2.1. Moreover,

Following we divide the proof into three cases.

Case 1 $3/(10q)\leq v\leq 1$. Then (2.25) and (2.28) together with the monotonicity of $G_{q}(x)$ imply that the function $x \mapsto G\left(v, q;x\right)$ is strictly increasing on $(0, 1)$. Therefore, $G\left(v, q;x\right) > 0$ for all $x \in(0, 1)$ follows from (2.23).

Case 2 $0\leq v\leq (2^{3/4}\tau-1)/[1+2^{3/4}(2q-1)\tau]$. Then the function $x \mapsto G\left(v, q;x\right)$ is strictly decreasing on $(0, 1)$, and consequently $G\left(v, q;x\right) < 0$ for all $x \in(0, 1)$.

Case 3 $(2^{3/4}\tau-1)/[1+2^{3/4}(2q-1)\tau] < v < 3/(10q)$. Then it follows from (2.25) and (2.28) together with the monotonicity of $G_{q}(x)$ that and there exists ${x_3} \in \left(0, 1\right)$ such that the function $x \mapsto G\left(v, q;x\right)$ is strictly decreasing on $\left({0, {x_3}}\right)$ and strictly increasing on $\left({{x_3}, 1}\right)$.

With the similar argument in the proof of Lemma 2.5, we also obtain that $(1/\tau)^{2/p}-1\in((2^{3/4}\tau-1)/[1+2^{3/4}(2q-1)\tau], 3/(10q))$. We divide the proof into two Subcases.

Subcase 3.1 $(1/\tau)^{2/p}-1 < v < 3/(10q)$. Then

and thereby there exists $x_{3}^*\in(0, 1)$ such that $G(v, q;x) < 0$ for $x\in (0, x_{3}^*)$ and $G(v, q;x) > 0$ for $x\in\left(x_{3}^*, 1\right)$.

Subcase 3.2 $(2^{3/4}\tau-1)/[1+2^{3/4}(2q-1)\tau] < v\leq (1/\tau)^{2/p}-1$. Then

and therefore $G(v, q;x) < 0$ for $x\in (0, 1)$.

3.   Main results

Theorem 3.1. Let $p\in[1, \infty)$ and $\lambda_{1}, \mu_{1}\in[0, 1/2]$. Then the double inequalities

hold for all $a, b > 0$ with $a\neq b$ if and only if $\lambda_{1}\leq 1/2-\sqrt{1-\left[1/\left(4\omega^4\right)\right]^{1/p}}/2$ and $\mu_{1}\geq 1/2-\sqrt{15p}/(10p)$.

Proof. Let $t\in[0, 1/2]$, since $\mathcal{GA}_{t, p}(a, b)$ and $\mathcal{LM}_{\mathcal{GA}}(a, b)$ are symmetric and homogeneous of degree one, without loss of generality, we assume that $a > b > 0$. Let $x = \sqrt{(a-b)/(a+b)}\in(0, 1)$. Then from (1.1) and (1.4) we get

where $f(\cdot, p;x)$ is defined in Lemma 2.3.

Therefore, Theorem 3.1 follows easily from Lemma 2.3 and (3.1).

Theorem 3.2. Let $p\in[1, \infty)$ and $\lambda_{2}, \mu_{2}\in[0, 1/2]$. Then the double inequalities

hold for all $a, b > 0$ with $a\neq b$ if and only if $\lambda_{2}\leq 1/2-\sqrt{1-\left(1/\omega^4\right)^{1/p}}/2$ and $\mu_{2}\geq 1/2-\sqrt{10p}/(10p)$.

Proof. Let $t\in[0, 1/2]$. Without loss of generality, we suppose that $x = \sqrt{(a-b)/(a+b)}\in(0, 1)$. Then from (1.1) and (1.4) we get

where $g(\cdot, p;x)$ is defined in Lemma 2.4.

Therefore, Theorem 3.2 follows easily from Lemma 2.4 and (3.2).

Theorem 3.3. Let $q\in[1/2, \infty)$ and $\lambda_{3}, \mu_{3}\in[0, 1/2]$. Then the double inequalities

hold for all $a, b > 0$ with $a\neq b$ if and only if $\lambda_{3}\geq 1/2-\sqrt{\left(\sqrt{2}/\omega\right)^{2/q}-1}/2$ and $\mu_{3}\leq 1/2-\sqrt{5q}/(10q)$.

Proof. Let $t\in[0, 1/2]$, since $\mathcal{CA}_{t, q}(a, b)$ and $\mathcal{LM}_{\mathcal{GA}}(a, b)$ are symmetric and homogeneous of degree one, without loss of generality, we assume that $a > b > 0$. Let $x = \sqrt{(a-b)/(a+b)}\in(0, 1)$. Then from (1.2) and (1.5) we get

where $F(\cdot, q;x)$ is defined in Lemma 2.5.

Therefore, Theorem 3.3 follows easily from Lemma 2.5 and (3.3).

Theorem 3.4. Let $q\in[1/2, \infty)$ and $\lambda_{4}, \mu_{4}\in[0, 1/2]$. Then the double inequalities

hold for all $a, b > 0$ with $a\neq b$ if and only if $\lambda_{4}\geq 1/2-\sqrt{\left(1/\tau\right)^{2/q}-1}/2$ and $\mu_{4}\leq 1/2-\sqrt{30q}/(20q)$.

Proof. Let $t\in[0, 1/2]$. Without loss of generality, we suppose that $x = \sqrt{(a-b)/(a+b)}\in(0, 1)$. Then from (1.2) and (1.5) we get

where $G(\cdot, q;x)$ is defined in Lemma 2.6.

Therefore, Theorem 3.4 follows easily from Lemma 2.6 and (3.4).

Corollary 3.5. Let $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$, $\alpha_{4}$, $\beta_{1}$, $\beta_{2}$, $\beta_{3}$, $\beta_{4}\in[0, 1/2]$. Then the double inequalities

hold for all $a, b > 0$ with $a\neq b$ with the best possible parameters

Corollary 3.6. Let $\alpha_{1}^*$, $\alpha_{2}^*$, $\alpha_{3}^*$, $\alpha_{4}^*$, $\beta_{1}^*$, $\beta_{2}^*$, $\beta_{3}^*$, $\beta_{4}^*\in[0, 1/2]$. Then the double inequalities

hold for all $a, b > 0$ with $a\neq b$ with the best possible parameters

4.   Inequalities for Lemniscate functions

Theorem 4.1. The double inequalities

hold for all $p\in[1, \infty)$ and $x\in(0, 1)$.

Proof. Substituting $\lambda_{1} = 1/2-\sqrt{1-\left[1/\left(4\omega^4\right)\right]^{1/p}}/2$, $\mu_{1} = 1/2-\sqrt{15p}/(10p)$, and $\lambda_{2} = 1/2-\sqrt{1-\left(1/\omega^4\right)^{1/p}}/2$, $\mu_{2} = 1/2-\sqrt{10p}/(10p)$ into Theorems 3.1 and 3.2, we obtain (4.1) and (4.2) immediately.

Theorem 4.2. The double inequalities

hold for all $q\in[1/2, \infty)$ and $x\in(0, 1)$.

Proof. Substituting $\lambda_{3} = 1/2-\sqrt{\left(\sqrt{2}/\omega\right)^{2/q}-1}/2$, $\mu_{3} = 1/2-\sqrt{5q}/(10q)$ and $\lambda_{4} = 1/2-$ $\sqrt{\left(1/\tau\right)^{2/q}-1}/2$, $\mu_{4} = 1/2-\sqrt{30q}/(20q)$ into Theorems 3.3 and 3.4, we obtain (4.3) and (4.4) immediately.

Lemma 4.3. $(1)$ The function $t\rightarrow \frac{\log\left(1-rx^4t\right)}{t}$ is strictly decreasing on $(0, 1)$ for any fixed $x\in(0, 1)$ and $r\in(-1, 0)\cup (0, 1)$;

$(2)$ The function $t\rightarrow \frac{\log\left[1-(1-r^t)x^4\right]}{t}$ is strictly increasing on $(0, 1)$ for any fixed $x\in(0, 1)$ and $r\in(0, +\infty)$.

Proof. For part (1), let $\phi_{1}(t) = \log(1-rx^4t)$, $\phi_{2}(t) = t$ and $\phi(t) = \phi_{1}(t)/\phi_{2}(t)$. Then $\phi_{1}(0) = \phi_{2}(0) = 0$,

It is apparent from (4.5) to see that ${\phi_{1}'(t)}/{\phi_{2}'(t)}$ is strictly decreasing on $(0, 1)$. By application of Lemma 2.1, part (1) is clear.

For part (2), let $\varphi_{1}(t) = \log\left[1-(1-r^t)x^4\right]$, $\varphi_{2}(t) = t$ and $\varphi(t) = \varphi_{1}(t)/\varphi_{2}(t)$. Then by simple computation we get $\varphi_{1}(0) = \varphi_{2}(0) = 0$, and

for all $t\in(0, 1)$. So that ${\varphi_{1}'(t)}/{\varphi_{2}'(t)}$ is strictly increasing on $(0, 1)$, applying Lemma 2.1, the assertion of part (2) follows.

Employing Lemma 4.3, it is easy to see that the best estimates in (4.1) and (4.2) are arrived at for $p = 1$, and while these in (4.3) and (4.4) for $q = 1/2$.

Theorem 4.4. The double sharp inequalities

hold for all $x\in(0, 1)$.

Acknowledgments

This research was supported by the Natural Science Foundation of China (11701176, 61673169, 11301127) and the Natural Science Foundation of Zhejiang Province (Grant YL19A010012), and the Key Project of the Scientific Research of Zhejiang Open University in 2019 (Grant no XKT-19Z02).

Conflict of interest

The authors declare that they have no competing interests.