Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 48, 1-9; http://www.math.u-szeged.hu/ejqtde/

Fixed points of the derivative and k-th power of

solutions of complex linear differential equations in

the unit disc

Guowei Zhang and Ang Chen^{∗}

Abstract

In this paper we consider the question of the existence of fixed points of the derivatives of solutions of complex linear differential equations in the unit disc. This work improves some very recent results of T.-B. Cao. Keywords: fixed points, solutions, complex linear differential equations, unit disc. 2000 MR subject classification: 30D35.

1 Introduction and main results

In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna’s theory on the complex plane and in the unit disc D = {z ∈ ℂ : |z| < 1} (see [6, 5, 15]). Many authors have investigated the growth and oscillation of the solutions of complex linear differential equations in ℂ. In the unit disc, there already exist many results [11, 7, 8], but the study is more difficult than that in the complex plane, because the efficient tool, Wiman-Valiron theory, in the complex plane doesn’t hold in the unit disc.

Many important results have been obtained on the fixed points of general transcendental meromorphic functions for almost four decades, see [4]. However, there are few studies on the fixed points of solutions of differential equations, specially in the unit disc. In [3], Z.-X. Chen firstly studied the problem on the fixed points and hyper-order of solutions of second order linear differential equations with entire coefficients. After that, there were some results which improve those of Z.-X. Chen, see [10, 13, 14, 9]. Recently, T.-B. Cao [1] firstly investigated the fixed points of solutions of linear complex differential equations in the unit disc. In the present paper, we continue to study the problem in the unit disc. In order to be read more clearly, we give some definitions as following.

∗Department of Mathematics, Shandong University, Jinan 250100, P. R. China

E-mail address: zhirobo@yahoo.com.cn (G. W. Zhang); ang.chen.jr@gmail.com (A. Chen)

The second author is the corresponding author.

The authors were supported by the NNFC of China (No.10771121), the NSF of Shang-dong Province, China (No Z2008A01)

For n ∈ ℕ, the iterated n-order of a meromorphic function f in D is defined by

σ_{n}(f) = lim sup_{r→1^{−}} \frac{log_{n}^{+} T (r, f)}{− log(1 − r)},

where log_{1}^{+} = log^{+} = max{log x, 0}, log_{n+1}^{+} x = log^{+} log_{n}^{+} x. If f is analytic in D, then the iterated n-order is defined by

σ_{M,n}(f) = lim sup_{r→1^{−}} \frac{log_{n+1}^{+} M(r, f)}{− log(1 − r)}.

If f is analytic in D, it is well known that σ_{M,1}(f) and σ_{1}(f) satisfy the inequalities σ_{1}(f) ≤ σ_{M,1}(f) ≤ σ_{1}(f) + 1 which are the best possible in the sense, see [12]. However, it follows by Proposition 2.2.2 in [2] that σ_{M,n}(f) = σ_{n}(f) for n ≥ 2.

For n ∈ ℕ and a ∈ ℂ ⋃{∞}, the iterated n-convergence exponent of the sequence of a-points in D of a meromorphic function f in D is defined by

λ_{n}(f − a) = lim sup_{r→1^{−}} \frac{log_{n}^{+} N(r, \frac{1}{f−a})}{− log(1 − r)}.

Similarly, \overline{λ}_{n}(f −a), the iterated n-convergence exponent of the sequence of distinct a-points in D of a meromorphic function f in D is defined by

\overline{λ}_{n}(f − a) = lim sup_{r→1^{−}} \frac{log_{n}^{+} \overline{N}(r, \frac{1}{f−a})}{− log(1 − r)}.

For n ∈ ℕ, the iterated n-convergence exponent of the sequence of fixed points in D of a meromorphic function f in D is defined by

τ_{n}(f) = lim sup_{r→1^{−}} \frac{log_{n}^{+} N(r, \frac{1}{f−z})}{− log(1 − r)}.

Similarly, \overline{λ}_{n}(f − z), the iterated n-convergence exponent of the sequence of distinct fixed points in D of a meromorphic function f in D is defined by

\overline{τ}_{n}(f) = lim sup_{r→1^{−}} \frac{log_{n}^{+} \overline{N}(r, \frac{1}{f−z})}{− log(1 − r)}.

Finally, we give the definition about the degree of small growth order of functions in D as polynomials on the complex plane. Let f be a meromorphic function in D and

D(f) = lim sup_{r→1^{−}} \frac{T (r, f)}{− log(1 − r)} = b.

If b < ∞, we say that f is non-admissible; if b = ∞, we say that f is admissible. Moreover, for F ⊂ [0, 1), the upper and lower densities of F are defined by

\overline{dens}_{D}F = lim sup_{r→1^{−}} \frac{m(F ∩ [0, r))}{m([0, r))} , \underline{dens}_{D}F = lim inf_{r→1^{−}} \frac{m(F ∩ [0, r))}{m([0, r))},

respectively, where m(G) = ∫_{G} \frac{dt}{1−t} for G ⊂ [0, 1).

In [1], T.-B. Cao investigated the fast growth of the solutions of high order complex differential linear equation with analytic coefficients of n-iterated order in the unit disc. For using the results of T.-B. Cao conveniently, we write them in the following form. T.-B. Cao considered the equation

f^{(k)} + A(z)f = 0  (1.1)

where A(z) is analytic function in D, and proved the following theorems;

Theorem A. [1] Let H be a set of complex numbers satisfying \overline{dens}_{D}\{|z| : z ∈ H ⊆ D} > 0, and let A(z) be an analytic function in D such that σ_{M,n}(A) = σ < ∞ and for constant α we have, for all ε > 0 sufficiently small,

|A(z)| ≥ exp_{n}\{α(\frac{1}{1 − |z|})^{σ−ε}\}

as |z| → 1^{−} for z ∈ H. Then every nontrivial solution f of (1.1) satisfies σ_{n}(f) = ∞ and σ_{n+1}(f) = σ.

Theorem B. [1] Let H be a set of complex numbers satisfying \overline{dens}_{D}\{|z| : z ∈ H ⊆ D} > 0, and let A(z) be an analytic function in D such that σ_{n}(A) = σ < ∞ and for constant α we have, for all ε > 0 sufficiently small,

T (r, A(z)) ≥ exp_{n−1}\{α(\frac{1}{1 − |z|})^{σ−ε}\}

as |z| → 1^{−} for z ∈ H. Then every nontrivial solution f of (1.1) satisfies σ_{n}(f) = ∞ and σ_{M,n}(A) ≥ σ_{n+1}(f) ≥ σ.

In [1], T.-B. Cao also investigated the fixed points of the solutions of high order complex differential linear equation with analytic coefficients in the unit disc and proposed that: How about the fixed points and iterated order of differential polynomials generated by solutions of linear differential equations in the unit disc. In the present paper we consider the derivatives of the solutions of the equations and get some theorems as following:

Theorem 1.1. Let the assumptions of Theorem A hold and assume also that f is a nontrivial solution of Equation (1.1). Then

τ_{n}(f^{(i)}) = \overline{τ}_{n}(f^{(i)}) = λ_{n}(f^{(i)} − z) = \overline{λ}_{n}(f^{(i)} − z) = σ_{n}(f) = ∞,  (1.2)
τ_{n+1}(f^{(i)}) = \overline{τ}_{n+1}(f^{(i)}) = λ_{n+1}(f^{(i)} − z) = \overline{λ}_{n+1}(f^{(i)} − z)
= σ_{n+1}(f) = σ.  (1.3)

Theorem 1.2. Let the assumptions of Theorem B hold and assume also that f is a nontrivial solution of Equation (1.1). Then

τ_{n}(f^{(i)}) = \overline{τ}_{n}(f^{(i)}) = λ_{n}(f^{(i)} − z) = \overline{λ}_{n}(f^{(i)} − z) = σ_{n}(f) = ∞,  (1.4)

σ_{M,n}(A) ≥ τ_{n+1}(f^{(i)}) = \overline{τ}_{n+1}(f^{(i)}) = λ_{n+1}(f^{(i)} − z)
= \overline{λ}_{n+1}(f^{(i)} − z) = σ_{n+1}(f) ≥ σ.  (1.5)

In addition, we study the fixed points of f^{k}, here f is a nontrivial solution of equation

f′′ + A(z)f = 0,  (1.6)

where A(z) is an analytic function in D. We get our theorems as following:

Theorem 1.3. Let the assumptions of Theorem A hold and assume also that f is a nontrivial solution of Equation (1.6). Then

τ_{n}(f^{k}) = \overline{τ}_{n}(f^{k}) = λ_{n}(f^{k} − z) = \overline{λ}_{n}(f^{k} − z) = σ_{n}(f) = ∞,  (1.7)
τ_{n+1}(f^{k}) = \overline{τ}_{n+1}(f^{k}) = λ_{n+1}(f^{k} − z) = \overline{λ}_{n+1}(f^{k} − z) = σ_{n+1}(f) = σ.  (1.8)

Theorem 1.4. Let the assumptions of Theorem B hold and assume also that f is a nontrivial solution of Equation (1.6). Then

τ_{n}(f^{k}) = \overline{τ}_{n}(f^{k}) = λ_{n}(f^{k} − z) = \overline{λ}_{n}(f^{k} − z) = σ_{n}(f) = ∞,  (1.9)
σ_{M,n}(A) ≥ τ_{n+1}(f^{k}) = \overline{τ}_{n+1}(f^{k}) = λ_{n+1}(f^{k} − z) = \overline{λ}_{n+1}(f^{k} − z)
= σ_{n+1}(f) ≥ σ.  (1.10)

2 Preliminary lemmas

Lemma 2.1. [6] Let f be a meromorphic function in the unit disc, and let k ∈ ℕ. Then

m(r, \frac{f^{(k)}}{f}) = S(r, f),

where S(r, f) = O(log^{+} T (r, f))+O(log(\frac{1}{1−r})), possibly outside a set E ⊂ [0, 1) with ∫_{E} \frac{dr}{1−r} < ∞. If f is of finite order (namely, finite iterated 1-order) of growth, then

m(r, \frac{f^{(k)}}{f}) = O(log(\frac{1}{1 − r})).

Lemma 2.2. [10] Suppose that f(z) is a nonzero solution of equation (1.1) and k ≥ 2. Let ω_{i} = f^{(i)} − z, i = 0, 1, ⋯, k − 2. Then w_{i} satisfy the following equations

∑_{j=0}^{i} H_{ij}(A)ω_{i}^{(k−j)} − A^{i+1}ω_{i} = zA^{i+1},  (2.1)

where H_{ij}(A) (j = 0, 1, ⋯, i) are differential polynomials of A.

Lemma 2.3. [10] Suppose that f(z) is a nonzero solution of equation (1.1) and k ≥ 2. Let ω_{k−1} = f^{(k−1)} − z. Then w_{k−1} satisfies the following equation

∑_{j=0}^{k−1} H_{(k−1)j}(A)ω_{k−1}^{(k−j)} − A^{k}ω_{k−1} = zA^{k} − H_{(k−1)(k−1)}(A),  (2.2)

where H_{(k−1)j}(A) (j = 0, 1, ⋯, k − 1) are differential polynomials of A.

Lemma 2.4. [3] Suppose that f(z) is a nonzero solution of equation (1.1) and k ≥ 2. Let ω_{k} = f^{(k)} − z. Then w_{k} satisfies the following equation

∑_{j=0}^{k−1} H_{kj}(A)ω_{k}^{(k−j)}(H_{kk}(A) − A^{k+1})ω_{k} = zA^{k+1}
−H_{k(k−1)}(A) − zH_{kk}(A),  (2.3)

where H_{kj}(A) (j = 0, 1, ⋯, k − 1) and H_{kk}(A) are differential polynomials of A.

Lemma 2.5. [13] Suppose that f(z) is a nonzero solution of equation (1.6) and k ≥ 2. Let g = f^{k} − z. Then g satisfies the following equation

g′′g + zg′′ − \frac{k − 1}{k}(g′)^{2} − 2\frac{k − 1}{k}(g′) + kAg^{2} + 2kAzg = \frac{k − 1}{k} − kAz^{2}.

3 Proof of Theorems

Proof of Theorem 1.1 and 1.2

Proof. We will prove Theorem 1.1 and 1.2 together. Suppose that f(z) is a nontrivial solution of equation (1.1). Set ω_{i} = f^{(i)} − z, i = 0, 1, ⋯, k, then for every i, a point z_{0} is a fixed points of f^{(i)} if and only if z_{0} is a zero of ω_{i}, and

σ_{n}(ω_{i}) = σ_{n}(f^{(i)}) = σ_{n}(f), \overline{τ}_{n}(f^{(i)}) = \overline{λ}_{n}(ω_{i}),
σ_{n_{1}}(ω_{i}) = σ_{n+1}(f^{(i)}) = σ_{n+1}(f), \overline{τ}_{n+1}(f^{(i)}) = \overline{λ}_{n+1}(ω_{i}),

By Lemma 2.2, Lemma 2.3 and Lemma 2.4, we know that ω_{i}, i = 0, 1, ⋯, k satisfy the following equations :

∑_{j=0}^{i} H_{ij}(A)ω_{i}^{(k−j)} − A^{i+1}ω_{i} = zA^{i+1}, (i = 0, 1, ⋯, k − 2).  (3.1)

∑_{j=0}^{k−1} H_{(k−1)j}(A)ω_{k−1}^{(k−j)} − A^{k}ω_{k−1} = zA^{k} − H_{(k−1)(k−1)}(A).  (3.2)

∑_{j=0}^{k−1} H_{kj}(A)ω_{k}^{(k−j)}(H_{kk}(A) − A^{k+1})ω_{k} = zA^{k+1}
−H_{k(k−1)}(A) − zH_{kk}(A).  (3.3)

where H_{ij}(A) are differential polynomials of A. Since A is admissible analytic functions in D, we have zA^{i+1} ≢ 0, i = 0, 1, ⋯, k − 2. We claim that

zA^{k} − H_{(k−1)(k−1)}(A) ≢ 0,

and

zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A) ≢ 0.

In fact, if zA^{k} − H_{(k−1)(k−1)}(A) ≡ 0, rewrite it as

A = \frac{H_{(k−1)(k−1)}(A)}{zA^{k−1}}.

Since A is analytic in D, we have T (r, A) = m(r, A) = S(r, A) in D, which is impossible. If zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A) ≢ 0, rewrite it as

A = \frac{H_{k(k−1)}(A)}{zA^{k}} + \frac{H_{kk}(A)}{A^{k}}.

Similarly, we obtain that T (r, A) = m(r, A) = S(r, A) in D. Hence our claim holds. Rewrite the above equations (3.1)-(3.3) as

\frac{1}{ω_{i}} = \frac{1}{zA^{i+1}} (∑_{j=0}^{i} H_{ij}(A)\frac{ω_{i}^{(k−j)}}{ω_{i}} − A^{i+1}), (i = 0, 1, ⋯, k − 2).  (3.4)

\frac{1}{ω_{k−1}} = \frac{1}{zA^{k} − H_{(k−1)(k−1)}(A)}(∑_{j=0}^{k−1} H_{(k−1)j}(A)\frac{ω_{k−1}^{(k−j)}}{ω_{k−1}} − A^{k}),  (3.5)

\frac{1}{ω_{k}} = \frac{1}{zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A)} ·

(∑_{j=0}^{k−1} H_{kj}(A)\frac{ω_{k}^{(k−j)}}{ω_{k}} + H_{kk}(A) − A^{k+1}).  (3.6)

By Lemma 2.1, there exists E ⊂ [0, 1) with ∫_{E} \frac{dr}{1−r} < ∞, such that for r ∉ E, we have,

m(r, \frac{1}{ω_{i}}) ≤ O(m(r, \frac{1}{A})) + C(log^{+} T (r, ω_{i}) + log \frac{1}{1 − r}),  (3.7)

here i = 0, 1, ⋯, k. By the assumption, we know that A and H_{ij}(A) are analytic in D. Now we split three cases to discuss the zeros of ω_{i}.

Case 1: for i = 0, 1, ⋯, k − 2, if ω_{i} has a zero at z_{0} ∈ D of order m(> k), then from (3.4) we know z_{0} is a zero of zA^{i+1} of order at least m − k. Hence we have

N(r, \frac{1}{ω_{i}}) ≤ k\overline{N}(r, \frac{1}{ω_{i}}) + N(r, \frac{1}{zA^{i+1}} ).  (3.8)

Case 2: for i = k − 1, if ω_{i} has a zero at z_{0} ∈ D of order m(> k), then from (3.5) we see z_{0} is a zero of zA^{k} − H_{(k−1)(k−1)}(A) of order at least m − k. Hence we have

N(r, \frac{1}{ω_{i}}) ≤ k\overline{N}(r, \frac{1}{ω_{i}}) + N(r, \frac{1}{zA^{k} − H_{(k−1)(k−1)}(A)}).  (3.9)

Case 3: for i = k, if ω_{i} has a zero at z_{0} ∈ D of order m(> k), then we get from (3.6) that zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A) has zeros at z_{0} of order at least m − k. Hence we have

N(r, \frac{1}{ω_{i}}) ≤ k\overline{N}(r, \frac{1}{ω_{i}}) + N(r, \frac{1}{zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A)}).  (3.10)

Note that for r → 1^{−}, C(log^{+} T (r, ω_{i}) + log \frac{1}{1−r}) ≤ \frac{1}{2}T (r, ω_{i}), where C is a constant. Thus we have

T (r, ω_{i}) ≤ 2k\overline{N}(r, \frac{1}{ω_{i}}) + O(T (r, A)), (i = 0, 1, ⋯, k.)  (3.11)

Under the hypotheses of Theorem 1.1, we know that σ_{n}(f) = ∞ by Theorem A. Then, for any given sufficiently large positive number N > σ + 1, there exists {r′_{n}}(r′_{n} → 1^{−}) such that

σ_{n}(f) = lim sup_{r′_{n}→1^{−}} \frac{log_{n}^{+} T (r′_{n}, f)}{− log(1 − r′_{n})} ≥ N.

Set ∫_{E} \frac{dr}{1−r} = log δ < ∞. Since ∫_{r′_{n}}^{1− \frac{1−r′_{n}}{δ+1}} \frac{dr}{1−r} = log(δ + 1), then there exists r_{n} ∈ [r′_{n}, 1 − \frac{1−r′_{n}}{δ+1} ] \ E ⊂ [0, 1), such that

\frac{log_{n}^{+} T (r_{n}, f)}{− log(1 − r_{n})} ≥ \frac{log_{n}^{+} T (r′_{n}, f)}{log \frac{δ+1}{1−r′_{n}}} = \frac{log_{n}^{+} T (r′_{n}, f)}{log \frac{1}{1−r′_{n}} + log(δ + 1)}.

Hence, we have

lim inf_{r′_{n}→1^{−}} \frac{log_{n}^{+} T (r_{n}, f)}{− log(1 − r_{n})} ≥ lim sup_{r′_{n}→1^{−}} \frac{log_{n}^{+} T (r′_{n}, f)}{log \frac{1}{1−r′_{n}} + log(δ + 1)} ≥ N.

It yields

lim_{r′_{n}→1^{−}} \frac{log_{n}^{+} T (r_{n}, f)}{− log(1 − r_{n})} ≥ N.

Since σ_{n}(f) ≤ σ_{M,n}(f) = σ, for any given ε, we have T (r_{n}, A) ≤ exp_{n}(\frac{1}{1−r_{n}})^{σ+ε}. So, for r_{n} ∈ [r′_{n}, 1 − \frac{1−r′_{n}}{δ+1} ] \ E ⊂ [0, 1), we get

\frac{T (r_{n}, A)}{T (r_{n}, f)} ≤ \frac{exp_{n}(\frac{1}{1−r_{n}})^{σ+ε}}{exp_{n}(\frac{1}{1−r_{n}})^{N}} → 0.

So we have T (r, A) = o(T (r, f)). Since σ_{n}(ω_{i}) = σ_{n}(f), we have T (r, A) = o(T (r, ω_{i})). From (3.11), we have

\overline{λ}_{n}(ω_{i}) = λ_{n}(ω_{i}) = σ_{n}(ω_{i}),  (3.12)

and

\overline{λ}_{n+1}(ω_{i}) = λ_{n+1}ω_{i}) = σ_{n+1}(ω_{i}).  (3.13)

Combining (3.12), (3.13) with Theorem A, we can get (1.2) and (1.3). Thus we complete the proof of Theorem 1.1. Combining (3.12), (3.13) with Theorem B, we can get (1.4) and (1.5), which are the results of Theorem 1.2.

Proof of Theorem 1.3 and 1.4

Proof. Suppose that f(z) is a nonzero solution of equation (1.6) and k ≥ 2. Let g = f^{k} − z. By Lemma 2.5, g satisfies the following equation

g′′g + zg′′ − \frac{k − 1}{k}(g′)^{2} − 2\frac{k − 1}{k}(g′) + kAg^{2} + 2kAzg = \frac{k − 1}{k} −
kAz^{2}.  (3.14)

Obviously, we know that \frac{k−1}{k} − kAz^{2} ≢ 0. For z satisfying |g(z)| < 1, we have \frac{1}{|g(z)|} > 1. From (3.14), we have

\frac{\frac{k−1}{k} − kAz^{2}}{|g|} ≤ (1 + |z|)\frac{|g′′|}{|g|} + \frac{k − 1}{k}(\frac{|g′|}{|g|})^{2} + 2\frac{k − 1}{k} \frac{|g′|}{|g|}
+ k|A| + 2k|Az|.  (3.15)

From (3.15) we have

m(r, \frac{1}{g}) ≤ O(∑_{i=1}^{2} m(r, \frac{g^{(i)}}{g})) + O(m(r, A)).  (3.16)

From (3.14), we know that if z_{0} is a zero of g with multiplicity m, then z_{0} is a zero of \frac{k−1}{k} − kAz^{2} with multiplicity at least m − 2. So we have

N(r, \frac{1}{g}) ≤ 2\overline{N}(r, \frac{1}{g}) + N(r, \frac{1}{\frac{k−1}{k} − kAz^{2}} ).  (3.17)

From (3.16) and (3.17), we have

T (r, g) ≤ 2\overline{N}(r, \frac{1}{g}) + O(T (r, A)) + S(r, g).  (3.18)

From (3.18), we have

σ_{n}(g) = λ_{n}(g) = \overline{λ}_{n}(g) = τ_{n}(f^{k}) = \overline{τ}_{n}(f^{k}).  (3.19)

and

σ_{n+1}(g) = λ_{n}(g) = \overline{λ}_{n+1}(g) = τ_{n}(f^{k}) = \overline{τ}_{n+1}(f^{k}).  (3.20)

Since

σ_{n}(g) = σ_{n}(f^{k}) = σ_{n}(f)  (3.21)

From (3.19)- (3.21), and Theorem A (Theorem B), we can get Theorem 1.3 (Theorem 1.4).

Acknowledgement The authors would like to express their sincere thanks to the referee for helpful comments and suggestion.

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(Received February 26, 2009)