**2. Main Results**

To prove our desired results, we need the following lemmas.

**Lemma 1.** *If p*(*z*) ∈ P*, then exists some x*, *z with* |*x*| ≤ 1(*see* [*28*]), |*z*| ≤ 1*, such that*

$$2c\_2 = c\_1^2 + \mathbf{x}(4 - c\_1^2),$$

$$4c\_3 = c\_1^3 + 2c\_1\mathbf{x}(4 - c\_1^2) - (4 - c\_1^2)c\_1\mathbf{x}^2 + 2(4 - c\_1^2)(1 - |\mathbf{x}|^2)z.$$

**Lemma 2.** *Let p*(*z*) ∈ P *(see [29]), then*

$$|c\_n| \le 2, \ n = 1, 2, \dots, \infty$$

We now state and prove the main results of our present investigation.

**Theorem 1.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form Equation (1), then*

$$|a\_2| \le 1, \ |a\_3| \le \frac{1}{2}, \ |a\_4| \le \frac{5}{9}, \ |a\_5| \le \frac{47}{72}.\tag{3}$$

**Proof.** Since *f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup>* , according to subordination relationship, so there exists a Schwarz function *ω*(*z*) with *ω*(0) = 0 and |*ω*(*z*)| < 1, such that

$$\frac{zf'(z)}{f(z)} = 1 + \sin(\omega(z)).$$

Now,

$$\frac{zf'(z)}{f(z)} = \frac{z + \sum\_{n=2}^{\infty} n a\_n z^n}{z + \sum\_{n=2}^{\infty} a\_n z^n}$$

$$= (1 + \sum\_{n=2}^{\infty} n a\_n z^{n-1}) [1 - a\_2 z + (a\_2^2 - a\_3) z^2 - (a\_2^3 - 2a\_2 a\_3 + a\_4) z^3]$$

$$+ (a\_2^4 - 3a\_2^2 a\_3 + 2a\_2 a\_4 + a\_3^2 - a\_5) z^4 + \cdots \ ]$$

$$= 1 + a\_2 z + (2a\_3 - a\_2^2) z^2 + (a\_2^3 - 3a\_2 a\_3 + 3a\_4) z^3$$

$$+ (4a\_5 - a\_2^4 + 4a\_2^2 a\_3 - 4a\_2 a\_4 - 2a\_3^2) z^4 + \cdots \ . \tag{4}$$

Define a function

$$\wp(z) = \frac{1 + \omega(z)}{1 - \omega(z)} = 1 + c\_1 z + c\_2 z^2 + \dotsb \dotsb$$

Clearly, we have *p*(*z*) ∈ P and

$$
\omega(z) = \frac{p(z) - 1}{1 + p(z)} = \frac{c\_1 z + c\_2 z^2 + c\_3 z^3 + \dotsb}{2 + c\_1 z + c\_2 z^2 + c\_3 z^3 + \dotsb}. \tag{5}
$$

On the other hand,

$$1 + \sin(\omega(z)) = 1 + \frac{1}{2}c\_1 z + (\frac{c\_2}{2} - \frac{c\_1^2}{4})z^2 + (\frac{5c\_1^3}{48} + \frac{c\_3 - c\_1c\_2}{2})z^3$$

$$+ (\frac{c\_4}{2} + \frac{5c\_1^2c\_2}{16} - \frac{c\_2^2}{4} - \frac{c\_1c\_3}{2} - \frac{c\_1^4}{32})z^4 + \dotsb \, . \tag{6}$$

Comparing the coefficients of *z*, *z*2, *z*3, *z*<sup>4</sup> between Equations (4) and (6), we obtain

$$a\_2 = \frac{c\_1}{2},\ a\_3 = \frac{c\_2}{4},\ a\_4 = \frac{c\_3}{6} - \frac{c\_1 c\_2}{24} - \frac{c\_1^3}{144},\ a\_5 = \frac{c\_4}{8} - \frac{c\_1 c\_3}{24} + \frac{5c\_1^4}{1152} - \frac{c\_1^2 c\_2}{192} - \frac{c\_2^2}{32}.\tag{7}$$

By using Lemma 2, we thus know that

$$|a\_2| \le 1, \quad |a\_3| \le \frac{1}{2}, \quad |a\_4| \le \frac{5}{9}, \quad |a\_5| \le \frac{47}{72}.$$

The proof of Theorem 1 is completed.

**Theorem 2.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|a\_3 - a\_2^2| \le \frac{1}{2}.\tag{8}$$

**Proof.** According to Equation (7), we have

$$|a\_3 - a\_2^2| = \left| \frac{c\_2}{4} - \frac{c\_1^2}{4} \right|.$$

By applying Lemma 1, we get

$$|a\_3 - a\_2^2| = \left| \frac{x(4 - c\_1^2)}{8} - \frac{c\_1^2}{8} \right|.$$

Let |*x*| = *t*, *t* ∈ [0, 1], *c*<sup>1</sup> = *c*, *c* ∈ [0, 2]. Then, using the triangle inequality, we obtain

$$|a\_3 - a\_2^2| \le \frac{t(4 - c^2)}{8} + \frac{c^2}{8}.$$

Suppose that

$$F(c, t) = \frac{t(4 - c^2)}{8} + \frac{c^2}{8}t$$

then ∀*t* ∈ (0, 1), ∀*c* ∈ (0, 2),

$$\frac{\partial F}{\partial t} = \frac{4 - c^2}{8} > 0,$$

which shows that *F*(*c*, *t*) is an increasing function on the closed interval [0,1] about *t*. Therefore, the function *F*(*c*, *t*) can get the maximum value at *t* = 1, that is, that

$$\max F(c, t) = F(c, 1) = \frac{(4 - c^2)}{8} + \frac{c^2}{8} = \frac{1}{2}.$$

Thus, obviously,

$$|a\_3 - a\_2^2| \le \frac{1}{2}.$$

The proof of Theorem 2 is thus completed.

**Theorem 3.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|a\_2 a\_3 - a\_4| \le \frac{1}{3}.\tag{9}$$

**Proof.** From Equation (7), we have

$$\begin{split} |a\_2 a\_3 - a\_4| &= |\frac{c\_1 c\_2}{8} + \frac{c\_1^3}{144} - \frac{c\_3}{6} + \frac{c\_1 c\_2}{24}| \\ &= |\frac{c\_1 c\_2}{6} - \frac{c\_3}{6} + \frac{c\_1^3}{144}|. \end{split}$$

Now, in view of Lemma 1, we get

$$|a\_2 a\_3 - a\_4| = \left| \frac{7c\_1^3}{144} + \frac{(4 - c\_1^2)c\_1 x^2}{24} - \frac{(4 - c\_1^2)(1 - |x|^2)z}{12} \right|.$$

Let |*x*| = *t*, *t* ∈ [0, 1], *c*<sup>1</sup> = *c*, *c* ∈ [0, 2]. Then, using the triangle inequality, we deduce that

$$|a\_2 a\_3 - a\_4| \le \frac{7c^3}{144} + \frac{(4 - c^2)ct^2}{24} + \frac{(4 - c^2)(1 - t^2)}{12}.$$

Assume that

$$F(c,t) = \frac{7c^3}{144} + \frac{(4-c^2)ct^2}{24} + \frac{(4-c^2)(1-t^2)}{12}.$$

Therefore, we have, ∀*t* ∈ (0, 1), ∀*c* ∈ (0, 2)

$$\frac{\partial F}{\partial t} = \frac{(4 - c^2)t(c - 2)}{12} < 0,$$

namely, *F*(*c*, *t*) is an decreasing function on the closed interval [0,1] about *t*. This implies that the maximum value of *F*(*c*, *t*) occurs at *t* = 0, which is

$$\max F(c, t) = F(c, 0) = \frac{(4 - c^2)}{12} + \frac{7c^3}{144}.$$

Define

$$G(c) = \frac{(4 - c^2)}{12} + \frac{7c^3}{144}i$$

clearly, the function *G*(*c*) has a maximum value attained at *c* = 0, also which is

$$|a\_2 a\_3 - a\_4| \le G(0) = \frac{1}{3}.$$

The proof of Theorem 3 is completed.

**Theorem 4.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|a\_2 a\_4 - a\_3^2| \le \frac{1}{4}.\tag{10}$$

**Proof.** Suppose that *f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup>* , then from Equation (7), we have

$$|a\_2 a\_4 - a\_3^2| = \left| \frac{c\_1 c\_3}{12} - \frac{c\_1^2 c\_2}{48} + \frac{c\_1^4}{48} - \frac{c\_2^2}{16} \right|.$$

Now, in terms of Lemma 1, we obtain

$$\begin{aligned} \left| a\_2 a\_4 - a\_3^2 \right| &= \left| \frac{\frac{c\_1 c\_3}{12} - \frac{c\_1^2 c\_2}{48} - \frac{c\_1^4}{288} - \frac{c\_2^2}{16}}{48} \right| \\\\ &= \left| -\frac{5c\_1^4}{576} - \frac{x^2 c\_1^2 (4 - c\_1^2)}{48} - \frac{x^2 (4 - c\_1^2)^2}{64} + \frac{c\_1 (4 - c\_1^2) (1 - |x|^2) z}{24} \right|. \end{aligned}$$

Let |*x*| = *t*, *t* ∈ [0, 1], *c*<sup>1</sup> = *c*, *c* ∈ [0, 2]. Then, using the triangle inequality, we get

$$|a\_2 a\_4 - a\_3^2| \le \frac{t^2 c^2 (4 - c^2)}{48} + \frac{(1 - t^2) c (4 - c^2)}{24} + \frac{t^2 (4 - c^2)^2}{64} + \frac{5c^4}{576}.$$

Putting

$$F(c,t) = \frac{t^2c^2(4-c^2)}{48} + \frac{(1-t^2)c(4-c^2)}{24} + \frac{t^2(4-c^2)^2}{64} + \frac{5c^4}{576}i$$

then, ∀*t* ∈ (0, 1), ∀*c* ∈ (0, 2), we have

$$\frac{\partial F}{\partial t} = \frac{t(c^2 - 8c + 12)(4 - c^2)}{96} > 0,$$

which implies that *F*(*c*, *t*) increases on the closed interval [0,1] about *t*. That is, that *F*(*c*, *t*) have a maximum value at *t* = 1, which is

$$\max F(c, t) = F(c, 1) = \frac{c^2(4 - c^2)}{48} + \frac{(4 - c^2)^2}{64} + \frac{5c^4}{576}.$$

Setting

$$G(c) = \frac{c^2(4-c^2)}{48} + \frac{(4-c^2)^2}{64} + \frac{5c^4}{576}c$$

then we have

$$G'(c) = \frac{c(4-c^2)}{24} - \frac{c^3}{24} - \frac{c(4-c^2)}{16} + \frac{5c^3}{144}.$$

If *G* (*c*) = 0, then the root is *<sup>c</sup>* <sup>=</sup> 0. In addition, since *<sup>G</sup>*(0) = <sup>−</sup> <sup>1</sup> <sup>12</sup> < 0, so the function *G*(*c*) can take the maximum value at *c* = 0, which is

$$|a\_2a\_4 - a\_3^2| \le G(0) = \frac{1}{4}.$$

The proof of Theorem 4 is completed.

**Theorem 5.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|a\_2^2 - a\_3^2| \le \frac{5}{4}.\tag{11}$$

**Proof.** Suppose that *f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup>* , then, by using Equation (7), we have

$$|a\_2^2 - a\_3^2| = |\frac{c\_1^2}{4} - \frac{c\_2^2}{16}|.$$

Next, according to Lemma 1, we obtain

$$|a\_2^2 - a\_3^2| = \left| \frac{c\_1^2}{4} - \frac{c\_2^2}{16} \right|$$

$$= \left| \frac{c\_1^2}{4} - \frac{c\_1^4}{64} - \frac{xc\_1^2(4 - c\_1^2)}{32} - \frac{x^2(4 - c\_1^2)^2}{64} \right| \dots$$

Let |*x*| = *t*, *t* ∈ [0, 1], *c*<sup>1</sup> = *c*, *c* ∈ [0, 2]. Then, by applying the triangle inequality, we get

$$|a\_2^2 - a\_3^2| \le \frac{c^2}{4} + \frac{c^4}{64} + \frac{tc^2(4 - c^2)}{32} + \frac{t^2(4 - c^2)^2}{64}.$$

Taking

$$F(c,t) = \frac{c^2}{4} + \frac{c^4}{64} + \frac{tc^2(4-c^2)}{32} + \frac{t^2(4-c^2)^2}{64}.$$

Then, ∀*t* ∈ (0, 1), ∀*c* ∈ (0, 2), we have

$$\frac{\partial F}{\partial t} = \frac{c^2(4 - c^2)}{32} + \frac{t(4 - c^2)^2}{32} > 0,$$

which implies that *F*(*c*, *t*) increases on the closed interval [0,1] about *t*. Namely, the maximum value of *F*(*c*, *t*) attains at *t* = 1, which is

$$\max F(c, t) = F(c, 1) = \frac{c^2}{4} + \frac{c^4}{64} + \frac{c^2(4 - c^2)}{32} + \frac{(4 - c^2)^2}{64}.$$

Let

$$G(c) = \frac{c^2}{4} + \frac{c^4}{64} + \frac{c^2(4-c^2)}{32} + \frac{(4-c^2)^2}{64}i$$

then

*G* (*c*) = *<sup>c</sup>* <sup>2</sup> > 0, ∀*c* ∈ (0, 2).

Therefore, the function *G*(*c*) is an increasing function on the closed interval [0,2] about *c*, and thus *G*(*c*) has a maximum value attained at *c* = 2, which is

$$|a\_2^2 - a\_3^2| \le G(2) = \frac{5}{4}.$$

The proof of Theorem 5 is completed.

**Theorem 6.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|a\_2 a\_3 - a\_3 a\_4| \le \frac{13}{12}.\tag{12}$$

**Proof.** Assume that *f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup>* , then from Equation (7), we obtain

$$|a\_2a\_3 - a\_3a\_4| = |\frac{c\_1c\_2}{8} + \frac{c\_1^3c\_2}{576} - \frac{c\_2c\_3}{24} + \frac{c\_1c\_2^2}{96}|.|$$

Now, by using Lemma 1, we see that

$$\begin{aligned} \left| a\_2 a\_3 - a\_3 a\_4 \right| &= \left| \frac{\epsilon\_1 \epsilon\_2}{8} + \frac{\epsilon\_1^3 \epsilon\_2}{5^3 6} - \frac{\epsilon\_2 \epsilon\_3}{24} + \frac{\epsilon\_1 \epsilon\_2^2}{96} \right| \\\\ &= \left| \frac{\epsilon\_1^3}{16} - \frac{\epsilon\_1^5}{5^5 6} - \frac{11 \kappa\_1^3 (4 - \epsilon\_1^2)}{1152} + \frac{\kappa \epsilon\_1 (4 - \epsilon\_1^2)}{16} + \frac{\kappa^2 \epsilon\_1 (4 - \epsilon\_1^2) [\epsilon\_1^2 + \pi (4 - \epsilon\_1^2)]}{192} + \frac{\epsilon\_1 \kappa^2 (4 - \epsilon\_1^2)^2}{128} + \frac{(1 - |x|^2) \pi (4 - \epsilon\_1^2) [\pi (4 - \epsilon\_1^2) + \pi^2 (4 - \epsilon\_1^2)]}{96} + \frac{\pi^2 \epsilon\_1 (4 - \epsilon\_1^2)}{128} \right| \end{aligned}$$

If we let |*x*| = *t*, *t* ∈ [0, 1], *c*<sup>1</sup> = *c*, *c* ∈ [0, 2], then, using the triangle inequality, we have

<sup>|</sup>*a*2*a*<sup>3</sup> <sup>−</sup> *<sup>a</sup>*3*a*4| ≤ *<sup>c</sup>*<sup>3</sup> <sup>16</sup> <sup>+</sup> *<sup>c</sup>*<sup>5</sup> <sup>576</sup> <sup>+</sup> <sup>11</sup>*tc*3(4−*c*2) <sup>1152</sup> <sup>+</sup> *<sup>t</sup>*(4−*c*2) <sup>8</sup> <sup>+</sup> *<sup>t</sup>* <sup>2</sup>[*c*2+*t*(4−*c*2)](4−*c*2) <sup>96</sup> <sup>+</sup> *<sup>t</sup>* <sup>2</sup>(4−*c*2)<sup>2</sup> <sup>64</sup> <sup>+</sup> (4−*c*2)[*t*(4−*c*2)+*c*2] <sup>96</sup> .

Setting

$$F(c, t) = \frac{c^3}{16} + \frac{c^5}{576} + \frac{11t\epsilon^3(4 - \epsilon^2)}{1152} + \frac{t(4 - \epsilon^2)}{8} + \frac{t^2[\epsilon^2 + t(4 - \epsilon^2)](4 - \epsilon^2)}{96} + \frac{t^2(4 - \epsilon^2)^2}{64} + \frac{(4 - \epsilon^2)[t(4 - \epsilon^2) + \epsilon^2]}{96} + \frac{t^2[\epsilon^2 + t(4 - \epsilon^2)](4 - \epsilon^2)}{1152} + \frac{t^2[\epsilon^2 + t(4 - \epsilon^2)](4 - \epsilon^2)}{96} + \frac{t^2[\epsilon^2 + t(4 - \epsilon^2)](4 - \epsilon^2)}{1152}$$

Then, we easily see that, ∀*t* ∈ (0, 1), ∀*c* ∈ (0, 2),

$$\frac{\partial F}{\partial t} = \frac{11c^3(4 - c^2)}{1152} + \frac{(4 - c^2)}{8} + \frac{t[c^2 + t(4 - c^2)](4 - c^2)}{48} + \frac{t^2(4 - c^2)^2}{96} + \frac{t(4 - c^2)^2}{32} + \frac{(4 - c^2)^2}{96} > 0,$$

which implies that *F*(*c*, *t*) is an increasing function on the closed interval [0,1] about *t*. That is, that the maximum value of *F*(*c*, *t*) occurs at *t* = 1, which is

$$\max F(c, t) = F(c, 1) = \frac{c^3}{16} + \frac{c^5}{576} + \frac{11c^3(4 - c^2)}{1152} + \frac{(4 - c^2)}{8} + \frac{(4 - c^2)}{24} + \frac{(4 - c^2)^2}{64} + \frac{(4 - c^2)}{24}.$$

Taking

$$G(c) = \frac{c^3}{16} + \frac{c^5}{576} + \frac{11c^3(4 - c^2)}{1152} + \frac{(4 - c^2)}{8} + \frac{(4 - c^2)}{24} + \frac{(4 - c^2)^2}{64} + \frac{(4 - c^2)}{24},$$

then

$$G'(c) = \frac{3c^2}{16} + \frac{5c^4}{576} + \frac{11c^2(4 - c^2)}{384} - \frac{11c^4}{576} - \frac{c(4 - c^2)}{16} - \frac{c}{12},$$

$$G''(c) = \frac{3c}{8} + \frac{5c^3}{144} + \frac{11c(4 - 2c^2)}{192} - \frac{11c^3}{144} - \frac{(4 - c^2)}{16} + \frac{c^2}{8} - \frac{1}{12}.$$

We easily find that *c* = 0 is the root of the function *G* (*c*) = 0, since*G*(0) < 0, which implies that the function *G*(*c*) can reach the maximum value at *c* = 0, also which is

$$|a\_2a\_3 - a\_3a\_4| \le G(0) = \frac{13}{12}.$$

The proof of Theorem 6 is completed.

**Theorem 7.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|H\_3(1)| \le \frac{275}{432} \approx 0.637.\tag{13}$$

**Proof.** Since

$$H\_3(1) = a\_3(a\_2a\_4 - a\_3^2) - a\_4(a\_4 - a\_2a\_3) + a\_5(a\_3 - a\_2^2),$$

by applying the triangle inequality, we get

$$|H\_3(1)| \le |a\_3||a\_2a\_4 - a\_3^2| + |a\_4||a\_4 - a\_2a\_3| + |a\_5||a\_3 - a\_2^2|.\tag{14}$$

Now, substituting Equations (3), (8), (9) and (10) into Equation (14), we easily obtain the desired assertion (Equation (13)).

**Theorem 8.** *If the function f*(*z*) ∈ S<sup>∗</sup> *<sup>s</sup> and of the form in Equation (1), then we have*

$$|T\_3(2)| \le \frac{139}{72} \approx 1.931.\tag{15}$$

**Proof.** Because

$$T\_3(2) = a\_2(a\_2^2 - a\_3^2) - a\_3(a\_2a\_3 - a\_3a\_4) + a\_4(a\_3^2 - a\_2a\_4),$$

by using the triangle inequality, we obtain

$$|T\_3(2)| \le |a\_2||a\_2^2 - a\_3^2| + |a\_3||a\_2a\_3 - a\_3a\_4| + |a\_4||a\_3^2 - a\_2a\_4|.\tag{16}$$

Next, from Equations (3), (10), (11) and (12), we immediately get the desired assertion (Equation (15)).

Finally, we give two examples to illustrate our results obtained.

**Example 1.** *If we take the function f*(*z*) = *<sup>e</sup><sup>z</sup>* <sup>−</sup> <sup>1</sup> <sup>=</sup> *<sup>z</sup>* <sup>+</sup> <sup>∑</sup><sup>∞</sup> *<sup>n</sup>*=<sup>2</sup> *<sup>z</sup><sup>n</sup> <sup>n</sup>*! ∈ S<sup>∗</sup> *<sup>s</sup> , then we obtain*

$$|H\_3(1)| \le |a\_3||a\_2a\_4 - a\_3^2| + |a\_4||a\_4 - a\_2a\_3| + |a\_5||a\_3 - a\_2^2|$$

$$= \frac{1}{3!} \times |\frac{1}{2!} \times \frac{1}{4!} - \frac{1}{3!} \times \frac{1}{3!}| + \frac{1}{4!} \times |\frac{1}{4!} - \frac{1}{2!} \times \frac{1}{3!}| + \frac{1}{5!} \times |\frac{1}{3!} - \frac{1}{2!} \times \frac{1}{2!}|$$

$$\approx 0.004 < 0.637.$$

**Example 2.** *If we set the function f*(*z*) = <sup>−</sup> log(<sup>1</sup> <sup>−</sup> *<sup>z</sup>*) = *<sup>z</sup>* <sup>+</sup> <sup>∑</sup><sup>∞</sup> *<sup>n</sup>*=<sup>2</sup> *<sup>z</sup><sup>n</sup> <sup>n</sup>* ∈ S<sup>∗</sup> *<sup>s</sup> , then we get*

$$|T\_3(2)| \le |a\_2||a\_2^2 - a\_3^2| + |a\_3||a\_2a\_3 - a\_3a\_4| + |a\_4||a\_3^2 - a\_2a\_4|$$

$$= \frac{1}{2} \times |\frac{1}{2} \times \frac{1}{2} - \frac{1}{3} \times \frac{1}{3}| + \frac{1}{3} \times |\frac{1}{2} \times \frac{1}{3} - \frac{1}{3} \times \frac{1}{4}| + \frac{1}{4} \times |\frac{1}{3} \times \frac{1}{3} - \frac{1}{2} \times \frac{1}{4}|$$

$$\approx 0.107 < 1.931.$$


**Author Contributions:** conceptualization, H.T. and H.-Y.Z.; methodology, H.T. and H.-Y.Z.; software, H.-Y.Z.; validation, H.-Y.Z, R.S. and H.T.; formal analysis, R.S.; investigation, H.T.; resources, H.T.; data curation, H.T.; writing-original draft preparation, H.-Y.Z.; writing—review and editing, H.T. and R.S.; visualization, R.S.; supervision, H.T. and R.S.; project administration, H.T.; funding acquisition, H.T.

**Funding:** This research was funded by the Natural Science Foundation of the People's Republic of China under Grants 11561001 and 11271045, the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region under Grant NJYT-18-A14, the Natural Science Foundation of Inner Mongolia of the People's Republic of China under Grant 2018MS01026, the Higher School Foundation of Inner Mongolia of the People's Republic of China under Grants NJZY17300 and NJZY18217 and the Natural Science Foundation of Chifeng of Inner Mongolia.

**Conflicts of Interest:** The authors declare no conflict of interest.
