# RETRACTED ARTICLE: Growth properties at infinity for solutions of modified Laplace equations

## Abstract

Let $$\mathscr{F}$$ be a family of solutions of Laplace equations in a domain D and for each $$f\in\mathscr{F}$$, f has only zeros of multiplicity at least k. Let n be a positive integer and such that $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$. Let a be a complex number such that $$a\neq0$$. If for each pair of functions f and g in $$\mathscr{F}$$, $$f^{n}f^{(k)}$$ and $$g^{n}g^{(k)}$$ share a value in D, then $$\mathscr{F}$$ is normal in D.

## Introduction

Let D be a domain in $$\mathbb{C}$$. Let $$\mathscr{F}$$ be a solution of certain Laplace equations defined in the domain D. $$\mathscr{F}$$ is said to be normal in D, in the sense of Montel, if for any sequence $$\{f_{n}\}\subset\mathscr{F}$$, there exists a subsequence $$\{f_{n_{j}}\}$$ such that $$f_{n_{j}}$$ converges spherically locally uniformly in D to a meromorphic function or ∞.

Let $$g(z)$$ be a solution of certain Laplace equations and a be a finite complex number. If $$f(z)$$ and $$g(z)$$ have the same zeros, then we say that they share a IM (ignoring multiplicity) (see ).

In 1998, Wang and Fang  proved the following result.

### Theorem A

Let f be a transcendental meromorphic function in the complex plane. Let n and k be two positive integers such that $$n\geq k+1$$, then $$(f^{n})^{(k)}$$ assumes every finite non-zero value infinitely often.

Corresponding to Theorem A, there are the following theorems about normal families in .

### Theorem B

Let $$\mathscr{F}$$ be a family of meromorphic functions in D, n, k be two positive integers such that $$n\geq k+3$$. If $$(f^{n})^{(k)}\neq1$$ for each function $$f\in\mathscr{F}$$, then $$\mathscr{F}$$ is normal in D.

Recently, corresponding to Theorem B, Yang  proved the following result.

### Theorem C

Let $$\mathscr{F}$$ be a family of meromorphic functions in D. Let n, k be two positive integers such that $$n\geq k+2$$. Let $$a\neq0$$ be a finite complex number. If $$(f^{n})^{(k)}$$ and $$(g^{n})^{(k)}$$ share a in D for each pair of functions f and g in $$\mathscr{F}$$, then $$\mathscr{F}$$ is normal in D.

Recently, Zhang and Li  proved the following theorem.

### Theorem D

Let f be a transcendental meromorphic function in the complex plane. Let k be a positive integer. Let $$L[f]=a_{k}f^{(k)}+a_{k-1}f^{(k-1)}+\cdots+a_{0}f$$, where $$a_{0}, a_{1},\ldots, a_{k}$$ are small functions and $$a_{j}$$ (0) ($$j=1,2,\ldots,k$$). For $$c\neq0, \infty$$, let $$F=f^{n}L[f]-c$$, where n is a positive integer. Then, for $$n\geq2$$, $$F=f^{n}L[f]-c$$ has infinitely many zeros.

From Theorem D, we immediately obtain the following result.

### Corollary D

Let f be a transcendental meromorphic function in the complex plane. Let c be a finite complex number such that $$c\neq0$$. Let n, k be two positive integers. Then, for $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$, $$f^{n}f^{(k)}-c$$ has infinitely many zeros.

It is natural to ask whether Corollary D can be improved by the idea of sharing values similarly with Theorem C. In this paper we investigate the problem and obtain the following result.

### Theorem 1

Let $$\mathscr{F}$$ be a family of meromorphic functions in D. Let n, k be two positive integers such that $$n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$$. Let a be a complex number such that $$a\neq0$$. For each $$f\in\mathscr{F}$$, f has only zeros of multiplicity at least k. If $$f^{n}f^{(k)}$$ and $$g^{n}g^{(k)}$$ share a in D for every pair of functions $$f, g\in\mathscr{F}$$, then $$\mathscr{F}$$ is normal in D.

### Remark 1

From Theorem 1, it is easy to see $$\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}\geq2$$ for any positive integer k.

### Example 1

Let $$D=\{z:|z|<1\}$$, $$n, k\in N$$ with $$n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$$ and n be a positive integer; for $$k=2$$, let

$$\mathscr{F}=\bigl\{ f_{m}(z)=mz^{k-1}, z\in D, m=1,2,\ldots \bigr\} .$$

Obviously, for any functions $$f_{m}$$ and $$g_{m}$$ in $$\mathscr{F}$$, we have $$f_{m}^{n}f^{(k)}_{m}=0$$, obviously $$f_{m}^{n}f^{(k)}_{m}$$ and $$g_{m}^{n}g^{(k)}_{m}$$ share any $$a\neq0$$ in D. But $$\mathscr{F}$$ is not normal in D.

### Example 2

Let $$D=\{z:|z|<1\}$$, $$n, k\in N$$ with $$n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$$ and n is a positive integer, and let

$$\mathscr{F}=\bigl\{ f_{m}(z)=e^{mz}, z\in D, m=1,2,\ldots \bigr\} .$$

Obviously, for any $$f_{m}$$ and $$g_{m}$$ in $$\mathscr{F}$$, we have $$f_{m}^{n}f^{(k)}_{m}=m^{k}e^{(mn+m)z}$$, obviously $$f_{m}^{n}f^{(k)}_{m}$$ and $$g_{m}^{n}g^{(k)}_{m}$$ share 0 in D. But $$\mathscr{F}$$ is not normal in D.

### Example 3

Let $$D=\{z:|z|<1\}$$, $$n, k\in N$$ with $$n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$$, and n be a positive integer, let

$$\mathscr{F}=\biggl\{ f_{m}(z)=\sqrt{m}\biggl(z+\frac{1}{m} \biggr), z\in D, m=1,2,\ldots\biggr\} .$$

For functions $$f_{m}$$ and $$g_{m}$$ in $$\mathscr{F}$$, we have $$f_{m}f'_{m}=mz+1$$. Obviously $$f_{m}f'_{m}$$ and $$g_{m}g'_{m}$$ share 1 in D. But $$\mathscr{F}$$ is not normal in D.

### Remark 2

Example 1 shows that the condition that f has only zeros of multiplicity at least k is necessary in Theorem 1. Example 2 shows that the condition $$a\neq0$$ in Theorem 1 is inevitable. Example 3 shows that Theorem 1 is not true for $$n=1$$.

## Lemmas

In order to prove our theorem, we need the following lemmas.

### Lemma 2.1

(Zalcman’s lemma, see )

Let $$\mathcal{F}$$ be a family of meromorphic functions in the unit disc with the property that, for each $$f\in\mathcal{F}$$, all zeros of multiplicity are at least k. Suppose that there exists a number $$A\geq1$$ such that $$|f^{(k)}(z)|\leq A$$ whenever $$f\in\mathcal{F}$$ and $$f=0$$. If $$\mathcal{F}$$ is not normal in Δ, then for $$0\leq\alpha\leq k$$, there exist:

1. 1.

a number $$r\in(0,1)$$;

2. 2.

a sequence of complex numbers $$z_{n}$$, $$|z_{n}|< r$$;

3. 3.

a sequence of functions $$f_{n}\in\mathcal{F}$$;

4. 4.

a sequence of positive numbers $$\rho_{n}\rightarrow0^{+}$$

such that $$g_{n}(\xi)=\rho_{n}^{-\alpha}f_{n}(z_{n}+\rho_{n}\xi)$$ locally uniformly converges (with respect to the spherical metric) to a non-constant meromorphic function $$g(\xi)$$ on $$\mathbb{C}$$, and, moreover, the zeros of $$g(\xi)$$ are of multiplicity at least k, $$g^{\sharp}(\xi)\leq g^{\sharp}(0)=kA+1$$. In particular, g has order at most 2.

### Lemma 2.2

Let n, k be two positive integers such that $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$, and let $$a\neq0$$ be a finite complex number. If f is a rational but not a polynomial meromorphic function and f has only zeros of multiplicity at least k, then $$f^{n}f^{(k)}-a$$ has at least two distinct zeros.

### Proof

If $$f^{n}f^{(k)}-a$$ has zeros and has exactly one zero.

We set

$$f=\frac{A(z-\alpha_{1})^{m_{1}}(z-\alpha_{2})^{m_{2}}\cdots (z-\alpha_{s})^{m_{s}}}{(z-\beta_{1})^{n_{1}}(z-\beta_{2})^{n_{2}}\cdots (z-\beta_{t})^{n_{t}}},$$
(2.1)

where A is a non-zero constant. Because the zeros of f are at least k, we obtain $$m_{i}\geq k$$ ($$i=1,2,\ldots,s$$), $$n_{j}\geq 1$$ ($$j=1,2,\ldots,t$$).

For simplicity, we denote

\begin{aligned}& m_{1}+m_{2}+\cdots+m_{s}=m\geq ks, \end{aligned}
(2.2)
\begin{aligned}& n_{1}+n_{2}+\cdots+n_{t}=n\geq t. \end{aligned}
(2.3)

From (2.1), we obtain

$$f^{(k)}=\frac{(z-\alpha_{1})^{m_{1}-k}(z-\alpha_{2})^{m_{2}-k}\cdots (z-\alpha_{s})^{m_{s}-k}g(z)}{(z-\beta_{1})^{n_{1}+k}(z-\beta_{2})^{n_{2}+k}\cdots (z-\beta_{t})^{n_{t}+k}},$$
(2.4)

where g is a polynomial of degree at most $$k(s+t-1)$$.

From (2.1) and (2.4), we obtain

$$f^{n}f^{(k)}=\frac{A^{n}(z-\alpha_{1})^{M_{1}}(z-\alpha_{2})^{M_{2}}\cdots (z-\alpha_{s})^{M_{s}}g(z)}{(z-\beta_{1})^{N_{1}}(z-\beta_{2})^{N_{2}}\cdots (z-\beta_{t})^{N_{t}}}=\frac{p}{q}.$$
(2.5)

Here p and q are polynomials of degree M and N, respectively. Also p and q have no common factor, where $$M_{i}=(n+1)m_{i}-k$$ and $$N_{j}=(n+1)n_{j}+k$$. By (2.2) and (2.3), we deduce $$M_{i}=(n+1)m_{i}-k\geq k(n+1)-k=nk$$, $$N_{j}=(n+1)n_{j}+k\geq n+k+1$$. For simplicity, we denote

\begin{aligned}& \deg P=M=\sum_{i=1}^{s} M_{i}+ \deg(g)\geq nks+ k(s+t-1) \\ & \hphantom{\deg P}=(nks+ks)+k(t-1)\geq(nk+k)s, \end{aligned}
(2.6)
\begin{aligned}& \deg q=N=\sum_{j=1}^{t} N_{j} \geq(k+1+n)t. \end{aligned}
(2.7)

Since $$f^{n}f^{(k)}-a=0$$ has just a unique zero $$z_{0}$$, from (2.5) we obtain

$$f^{n}f^{(k)}=a+\frac{B(z-z_{0})^{l}}{(z-\beta_{1})^{N_{1}}(z-\beta_{2})^{N_{2}}\cdots (z-\beta_{t})^{N_{t}}}=\frac{p}{q}.$$
(2.8)

By $$a\neq0$$, we obtain $$z_{0}\neq\alpha_{i}$$ ($$i=1,\ldots,s$$), where B is a non-zero constant.

From (2.5), we obtain

$$\bigl[f^{n}f^{(k)}\bigr]'=\frac{(z-\alpha_{1})^{M_{1}-1} (z-\alpha_{2})^{M_{2}-1}\cdots(z-\alpha_{s})^{M_{s}-1}g_{1}(\xi)}{(z-\beta _{1})^{N_{1}+1}\cdots(z-\beta_{t})^{N_{t}+1}},$$
(2.9)

where $$g_{1}(\xi)$$ is a polynomial of degree at most $$(k+1)(s+t-1)$$.

From (2.8), we obtain

$$\bigl[f^{n}f^{(k)}\bigr]'=\frac{(z-z_{0})^{l-1}g_{2}(z)}{(z-\beta_{1})^{N_{1}+1}\cdots +(z-\beta_{t})^{N_{t}+1}},$$
(2.10)

where $$g_{2}(\xi)=B(l-N)z^{t}+B_{1}z^{t-1}+\cdots+B_{t}$$ is a polynomial ($$B_{1},\ldots, B_{t}$$ are constants).

Now we distinguish two cases.

Case 1. If $$l\neq N$$, by (2.8), then we obtain $$\deg p\geq\deg q$$. So $$M\geq N$$. By (2.9) and (2.10), we obtain $$\sum_{i=1}^{s}(M_{i}-1)\leq\deg g_{2}=t$$. So $$M-s-\deg(g)\leq t$$, and $$M\leq s+t+\deg(g)\leq(k+1)(s+t)-k<(k+1)(s+t)$$. By (2.6) and (2.7), we obtain

$$M < (k+1) (s+t)\leq(k+1)\biggl[\frac{M}{nk+k}+\frac{N}{n+k+1}\biggr] \leq (k+1)\biggl[\frac{1}{nk+k}+\frac{1}{n+k+1}\biggr]M.$$

By $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$, we deduce $$M< M$$, which is impossible.

Case 2. If $$l= N$$, then we distinguish two subcases.

Subcase 2.1. If $$M\geq N$$, by (2.9) and (2.10), we obtain $$\sum_{i=1}^{s}(M_{i}-1)\leq\deg g_{2}=t$$. So $$M-s-\deg(g)\leq t$$, and $$M\leq s+t+\deg(g)\leq(k+1)(s+t)-k<(k+1)(s+t)$$, then we can proceed similarly to Case 1. This is impossible.

Subcase 2.2. If $$M< N$$, by (2.9) and (2.10), we obtain $$l-1\leq\deg g_{1}\leq(s+t-1)(k+1)$$, and then

\begin{aligned} N=l \leq&\deg g_{1}+1\leq(k+1) (s+t)-k< (k+1) (s+t) \\ \leq&(k+1)\biggl[\frac{1}{nk+k}+\frac{1}{n+k+1}\biggr]N\leq N. \end{aligned}

By $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$, we deduce $$N< N$$. This is impossible.

If $$f^{n}f^{(k)}-a\neq0$$ and we know f is rational but not a polynomial, then $$f^{n}f^{(k)}$$ also is rational but not a polynomial. At this moment, $$l=0$$ for (2.8), and proceeding as in Case 1, we have a contradiction.

Lemma 2.2 is proved. □

## Proof of Theorem 1

We may assume that $$D=\{|z|<1\}$$. Suppose that $$\mathscr{F}$$ is not normal in D. Without loss of generality, we assume that $$\mathscr{F}$$ is not normal at $$z_{0}=0$$. Then, by Lemma 2.1, there exist:

1. 1.

a number $$r\in(0,1)$$;

2. 2.

a sequence of complex numbers $$z_{j}$$, $$z_{j}\rightarrow0$$ ($$j\rightarrow\infty$$);

3. 3.

a sequence of functions $$f_{j}\in\mathcal{F}$$;

4. 4.

a sequence of positive numbers $$\rho_{j}\rightarrow0^{+}$$

such that $$g_{j}(\xi)=\rho_{j}^{-\frac{k}{n+1}}f_{j}(z_{j}+\rho_{j}\xi)$$ converges uniformly with respect to the spherical metric to a non-constant meromorphic function $$g(\xi)$$ in C. Moreover, $$g(\xi)$$ is of order at most 2.

By Hurwitz’s theorem, the zeros of $$g(\xi)$$ are at least k multiple.

On every compact subset of $$\mathbb{C}$$ which contains no poles of g, we see that

$$f^{n}_{j}(z_{j}+\rho_{j} \xi)f^{(k)}_{j}(z_{j}+\rho_{j} \xi)-a=g_{j}^{n}(\xi) \bigl(g^{(k)}_{j}( \xi )\bigr)-a$$
(3.1)

converges uniformly with respect to the spherical metric to $$g^{n}(\xi)(g^{(k)}(\xi))-a$$.

If $$g^{n}(\xi)(g^{(k)}(\xi))\equiv a$$ ($$a\neq0$$) and g has only zeros of multiplicity at least k, then g has no zeros. From the $$g^{n}g^{(k)}$$ having no zeros and the $$g^{n}(\xi)(g^{(k)}(\xi))\equiv a$$, we know g has no poles. Because the $$g(\xi)$$ is a non-constant meromorphic function in $$\mathbb{C}$$ and g has order at most 2. We obtain $$g(\xi)=e^{d\xi^{2}+h\xi+c}$$, where d, h, c are constants and $$dh\neq0$$. So $$g^{n}(\xi)(g^{(k)}(\xi))\not\equiv a$$, which is a contradiction.

When $$g^{n}(\xi)(g^{(k)}(\xi))-a\neq0$$, ($$a\neq0$$), we distinguish three cases.

Case 1. If g is a transcendental meromorphic function, by Corollary D, this is a contradiction.

Case 2. If g is a polynomial and the zeros of $$g(\xi)$$ are at least k multiple, and $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$, then $$g^{n}(\xi)(g^{(k)}(\xi))-a=0$$ must have zeros, which is a contradiction.

Case 3. If g is a non-polynomial ration function, by Lemma 2.2, this is a contradiction.

Next we will prove that $$g^{n}g^{(k)}-a$$ has just a unique zero. To the contrary, let $$\xi_{0}$$ and $$\xi_{0}^{\ast}$$ be two distinct solutions of $$g^{n}g^{(k)}-a$$, and choose δ (>0) small enough such that $$D(\xi_{0}, \delta)\cap D(\xi_{0}^{\ast}, \delta)=\emptyset$$ where $$D(\xi_{0}, \delta)=\{\xi:|\xi-\xi_{0}|<\delta\}$$ and $$D(\xi_{0}^{\ast}, \delta)=\{\xi:|\xi-\xi_{0}^{*}|<\delta\}$$. From (3.1), by Hurwitz’s theorem, there exist points $$\xi_{j}\in D(\xi_{0}, \delta)$$, $$\xi_{j}^{*}\in D(\xi_{0}^{*}, \delta)$$ such that for sufficiently large j,

\begin{aligned}& f^{n}_{j}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{j}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0, \\& f^{n}_{j}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{j}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0. \end{aligned}

By the hypothesis that for each pair of functions f and g in $$\mathscr{F}$$, $$f^{n}f^{(k)}$$ and $$g^{n}g^{(k)}$$ share a in D, we know that for any positive integer m

\begin{aligned}& f^{n}_{m}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{m}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0, \\& f^{n}_{m}(z_{j}+\rho_{j} \xi_{j}) \bigl(f^{(k)}_{m}(z_{j}+ \rho_{j}\xi_{j})\bigr)-a=0. \end{aligned}

Fix m, take $$j\rightarrow\infty$$, and note $$z_{j}+\rho_{j}\xi_{j}\rightarrow0$$, $$z_{j}+\rho_{j}\xi_{j}^{*}\rightarrow0$$, then

$$f_{m}^{n}(0) \bigl(f^{(k)}_{m}(0) \bigr)-a=0.$$

Since the zeros of $$f_{m}^{n}(0)(f^{(k)}_{m}(0))-a$$ have no accumulation point, so $$z_{j}+\rho_{j}\xi_{j}= 0$$, $$z_{j}+ \rho_{j}\xi_{j}^{*}= 0$$.

Hence

$$\xi_{j}=-\frac{z_{j}}{\rho_{j}},\quad\quad \xi_{j}^{*}=- \frac{z_{j}}{\rho_{j}}.$$

This contradicts with $$\xi_{j}\in D(\xi_{0},\delta)$$, $$\xi_{j}^{*}\in D(\xi_{0}^{*},\delta)$$, and $$D(\xi_{0},\delta)\cap D(\xi_{0}^{*},\delta)=\emptyset$$. So $$g^{n}g^{(k)}-a$$ has just a unique zero, which can be denoted by $$\xi_{0}$$.

From the above, we know $$g^{n}g^{(k)}-a$$ has just a unique zero. If g is a transcendental meromorphic function, by Corollary D, then $$g^{n}g^{(k)}-a= 0$$ has infinitely many solutions, which is a contradiction.

From the above, we know $$g^{n}g^{(k)}-a$$ has just a unique zero. If g is a polynomial, then we set $$g^{n}g^{(k)}-a=K(z-z_{0})^{l}$$, where K is a non-zero constant, l is a positive integer. Because the zeros of $$g(\xi)$$ are at least k multiple, and $$n\geq\frac{1+\sqrt{1+2k(k+1)^{2}}}{2k}$$, we obtain $$l\geq3$$. Then $$[g^{n}g^{(k)}]'=Kl(z-z_{0})^{l-1}$$ ($$l-1\geq2$$). But $$[g^{n}g^{(k)}]'$$ has exactly one zero, so $$g^{n}g^{(k)}$$ has the same zero $$z_{0}$$ too. Hence $$g^{n}g^{(k)}(z_{0})=0$$, which contradicts $$g^{n}g^{(k)}(z_{0})=a\neq0$$.

If g is a rational function but not a polynomial, by Lemma 2.2, then $$g^{n}g^{(k)}-a=0$$ at least has two distinct zeros, which is a contradiction.

Theorem 1 is proved.

## Discussion

In 2013, Yang and Nevo  has proved the following.

### Theorem E

Let $$\mathscr{F}$$ be a family of meromorphic functions in D, n be a positive integer and a, b be two constants such that $$a\neq0,\infty$$ and $$b\neq\infty$$. If $$n\geq3$$ and for each function $$f\in\mathscr{F}$$, $$f'-af^{n}\neq b$$, then $$\mathscr{F}$$ is normal in D.

Recently, Zhang improved Theorem E by the idea of shared values. Meanwhile, Zhang  has proved the following.

### Theorem F

Let $$\mathscr{F}$$ be a family of meromorphic functions in D, n be a positive integer and a, b be two constants such that $$a\neq0,\infty$$ and $$b\neq\infty$$. If $$n\geq4$$ and for each pair of functions f and g in $$\mathscr{F}$$, $$f'-af^{n}$$ and $$g'-ag^{n}$$ share the value b, then $$\mathscr{F}$$ is normal in D.

By Theorem 1, we immediately obtain the following result.

### Corollary 1

Let $$\mathscr{F}$$ be a family of meromorphic functions in a domain D and each f has only zeros of multiplicity at least $$k+1$$. Let n, k be positive integers and $$n\geq\frac{1+\sqrt{1+4k(k+1)^{2}}}{2k}$$ and let $$a\neq0, \infty$$ be a complex number. If $$f^{(k)}-af^{-n}$$ and $$g^{(k)}-ag^{-n}$$ share 0 for each pair function of f and g in $$\mathscr{F}$$, then $$\mathscr{F}$$ is normal in D.

### Remark 4.1

Obviously, for $$k=1$$ and $$b=0$$, Corollary 1 occasionally investigates the situation when the power of f is negative in Theorem F.

Recently, Zhang  proved the following.

### Theorem G

Let $$\mathscr{F}$$ be a family of meromorphic functions in the plane domain D. Let n, be a positive integer such that $$n\geq2$$. Let a be a finite complex number such that $$a\neq0$$. If $$f^{n}f'$$ and $$g^{n}g'$$ share a in D for every pair of functions $$f, g\in\mathscr{F}$$, then $$\mathscr{F}$$ is normal in D.

### Question 1

It is natural to ask if the conclusion of Theorems G and 1 still holds for $$n\geq1$$.

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## Acknowledgements

This work was written while the corresponding author was at the Department of Mathematics of the University of Delaware as a visiting professor, and he is grateful to the department for their support.

This work was supported by the National Science Foundation of China under Grant Nos. 61202455, 61472096; the Doctoral Program of Higher Education Foundation of China under Grant No. 20112304120025; the National Science Foundation of Heilongjiang Province under Grant No. F201306; the Specialized Foundation for the Basic Research Operating expenses Program of Central College No. HEUCF100612, HEUCFT1202.

## Author information

Authors

### Corresponding author

Correspondence to Binghang He.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

BH and CP made the main contribution in conceiving the presented research. JS, BH, and CP worked jointly on each section, while ML and JS drafted the manuscript. All authors read and approved the final manuscript. 