On estimates of a fractional counterpart of the logarithmic derivative of a meromorphic function

УДК 517.53

I. E. Chyzhykov, N. S. Semochko

ON ESTIMATES OF A FRACTIONAL COUNTERPART OF THE

LOGARITHMIC DERIVATIVE OF A MEROMORPHIC FUNCTION

I. E. Chyzhykov, N. S. Semochko.On estimates of a fractional counterpart of the logarithmic derivative of a meromorphic function, Mat. Stud.39_{(2013), 107–112.}

We consider the problem of finding lower bounds for growth of solutions of a fractional differential equation in the complex plane. We estimate a fractional integral of the logarithmic derivative of a meromorphic function.

И. Э. Чижиков, Н. С. Семочко.К оценкам дробного аналога логарифмической производной

// Мат. Студiї. – 2013. – Т.39, №1. – C.107–112.

Рассматривается проблема оценки снизу роста решений дифференциального уравне-ния дробного порядка в комплексной плоскости. Оценивается дробный интеграл логариф-мической производной мероморфной функции.

Letf be a meromorphic function inC_{,f(0)6= 0,∞. We use the standard notation of the} Nevanlinna theory ([7]).

There are two standard ways to estimate the growth of solutions of ordinary differential equations in a complex domain. The first one is based on the Wiman-Valiron theory (see monographs [20], [19], [14]), which is very effective for the complex plane when solutions are of finite order of growth. However, there are some principal restrictions for application of the Wiman-Valiron method when the complex domain is not the whole plane, e.g. the unit disk, an angle, as well as when solutions are of infinite order. Concerning recent development of the Wiman-Valiron method we address the reader to the papers [15], [17], [18], [1], [6].

On the other hand, estimates of the logarithmic derivative f_f(z)′(z) have been successfully applied in all mentioned cases for obtaining lower bounds for growth of solutions of ordinary differential equations (see e.g. [9, 2, 3, 4]). One of the typical, but not the best possible estimate is (cf. [8])

kr

Z

r

dt

2π

Z

0

f′(reiϕ)

f(reiϕ₎

dϕ=O(T(k

2_{r, f)), k >1, r}

→+_∞,

whereT(r, f)is the Nevanlinna characteristic of a meromorphic functionf. Note that logari-thmic derivative estimates have various applications in the theory of entire and meromorphic functions ([7]).

2010Mathematics Subject Classification:26A33, 34M05, 34M10.

Keywords:Riemann-Liouville operator, fractional derivative, fractional differential equation, logarithmic deri-vative, meromorphic function, Nevanlinna’s characteristic, growth.

c

In contrast to ordinary differential equations, the analytic theory of fractional differential equations with variable coefficients was initiated only recently ([11], [13]). Such equations are widely used for modeling of diffusion phenomena and anomalous relaxation ([5, 12]).

Let h_∈L(0, a), a >0. The Riemann-Liouville fractional integral of order α >0 for h is defined as

D−αh(x) = 1 Γ(α)

x

Z

0

(x₋t)α−1h(t)dt, x_∈(0, a),

D0h(x)_≡h(x), Dαh(x) = d p

dxp{D

−(p−α)_h(x)

}, α_∈(p₋1, p], p_∈N_,

where Γ(α)is the Gamma function.

In the papers [11] and [13] the following equation is studied

Dαu(t) =a(t)u(t), t >0, (1)

where a(t) =A(tα_{), A(z) is analytic in |z|< r}

0, Dαu(t) =Dαu(t)−_tα_Γ(1u(0)_−α) is the

Caputo-Dzhrbashyan fractional derivative. In particular, in [13] the following theorem is proved.

Theorem A ([13]). Suppose that A is a polynomial of degree m _≥0. The solution u(t) of equation (1) satisfying the initial condition u(0) = u0 has the form u(t) =v(tα), where v is an entire function whose order does not exceed 1+m_α .

Essentially, lack of a fractional analogue of the logarithmic derivative estimate did not allow to obtain a lower bound for the growth of the solution of (1) in Theorem A.

Problem 1. Find a reasonable estimate for D_fαf(z) in the class of meromorphic (analytic) functions in C _{of finite order of growth.}

Remark 1. Writing Dαf(z) we mean that the operator is taken of variable r =_|z_|. Since

Dα_f

f (z) equals infinity at zeros (and poles) of f, it makes sense to consider the integral

means with respect to the circles centered at the origin. Deriving uniform estimates for a meromorphic functionf we have to omit even rays emanating from the origin and containing poles, because the Riemann-Liouville operatorDα_{f is represented by a divergent integral in}

this case.

Here we give a contribution, which will probably help to solve Problem 1. According to formula (17.11) in [16, p. 241] we have

Dαf

f =D

α−1

f′ f +R1

,

where the error term R1 has the form

R1(x) =₋ 1 Γ(1₋α)

x

Z

a

(x₋t)α−2Dαf(t) x

Z

t

1

f

′

(s)ds. (2)

In view of these relationships, we search for an estimate of the value

Iα[f](r) = 2π

Z

0

Dα−1|f

′

(reiϕ)_|

Theorem 1. Let f be a meromorphic function in C_{, f(0) = 06 ,∞, {cq} be the sequence of}

its zeros and poles, α_∈(0,1), β _∈(1 +_∞). Then for somer0 >0and a constant C(α, β)

Iα[f](r) = 2π

Z

0

Dα−1|f

′

(reiϕ)_|

|f(reiϕ_)|dϕ≤C(α, β)

T(βr, f)

rα +

1

rα−1 r/2

Z

0

n(t,0,_∞, f)

t2 dt

, r_≥r0,

where n(t,0,_∞, f) is the counting function of _{cq}.

Moreover, C(β, α) =O(_(β−1)(11 _−α)) asβ _→1+, α_→1₋.

Corollary 1. For every ε >0we have Iα[f](r) =O(r(ρ−α+ε)+

) as r_→+_∞.

The following problem is left open.

Problem 2. Find an estimate for Dα−1_{R(x, f), α∈(0,1), where R(x, f) is given by (2).} Proof of Theorem 1. Let1< k < 3₂. We have

Dα−1|f

′

(reiϕ_)|

|f(reiϕ_)| = 1 Γ(1₋α)

r

Z

0

|f′(teiϕ_)|

|f(teiϕ_)|(r−t)

−α_dt.

We use the following well-known estimate of the logarithmic derivative ([7])

|f′(z)_|

|f(z)_| ≤

4sT(s, f) (s_{− |}z_|)2 + 2

X

|cq|<s

1

|z₋cq|

, _|z_|< s.

Therefore

Iα[f](r) = 2π

Z

0 r

Z

0

1 Γ(1₋α)

|f′(teiϕ)_|

|f(teiϕ_)|(r−t)

−α_dtdϕ

≤

≤ _Γ(12 −α)

4π

r

Z

0

sT(s, f)(r₋t)−α (s₋t)2 dt+

X

|cq|<s 2π

Z

0 r

Z

0

(r₋t)−α

|teiϕ_−c q|

dtdϕ

=:

=: 2 Γ(1₋α)

I+ X

|cq|<s

Jq

. (3)

We puts =kr. Then

I = 4πsT(s, f)

r−2(s−r)

Z

0 +

r

Z

r−2(s−r)

!

(r₋t)−α (s₋t)2 dt≤

≤4πsT(s, f) 1 2α_(s−r)α

r−2(s−r)

Z

0

dt

(s₋t)2 + 1 (s₋r)2

r

Z

r−2(s−r)

(r₋t)−αdt

!

=

= 4πsT(s, f) 1 2α_(s−r)α

1 (s₋t)

r−2(s−r)

0 −

1

(1₋α)(s₋r)2(r−t) 1−α

r r−2(s−r)

!

≤

≤ 4₂πsα

1

3 + 2 1₋α

T(s, f)

(s₋r)α ≤C(k, α)

T(kr, f)

We have to estimate Jq =

Rr

0

R2π

0

(r−t)−α

|teiϕ_−cq|dϕdt. Without loss of generality we may assume

that cq>1. By Theorem 1.7 ([10, p. 7–8]) we have for some C >0 2π

Z

0

dϕ

|1₋ze−iϕ_| ≤Cln 2

1_{− |}z_|, |z|<1.

Hence

2π

Z

0

dϕ |teiϕ_−c

q| ≤

   C t ln 2 1−t

cq

, t > cq; C

cq ln

2 1−t

cq

, t < cq.

(5)

Note that the choice of k yields the inequality 2₃cq < 2₃kr < r. We split the integralJq in the

following way

Jq =

2cq/3

Z

0 +

r

Z

2cq/3

!

(r₋t)−αdt

2π

Z

0

dϕ |teiϕ_−c

q|

=:Jq1+Jq2. (6)

Using (5) we get

Jq1 ≤C 2cq/3

Z

0

(r₋t)−α

cq

ln 2 1_{− c}t

q

dt _≤ Cln 6 cq

2cq/3

Z

0

(r₋t)−αdt _≤ Cln 6

1₋α r1−α

cq

. (7)

In order to estimateJq2 we consider three cases: i)cq ≤ r₂;ii)₂r < cq ≤r, and iii)r < cq ≤kr.

In the case i) applying (5) and using a property of the fractional integral ([16, (2.44)]) we obtain

r

Z

3cq/2

2π

Z

0

(r₋t)−α

|teiϕ_−c q|

dϕdt_≤C

r

Z

3cq/2

(r₋t)−α

t ln

2 1₋ cq

t

dt_≤Cln 6Γ(1₋α)Dα−11 r

≤ C_r(αα). (8)

Similarly

3cq

2

Z

2cq/3

2π

Z

0

(r₋t)−α

|teiϕ_−c q|

dϕdt_≤C

cq

Z

2cq/3

(r₋t)−α

cq

ln 2 1₋ t

cq

dt+C

3cq/2

Z

cq

(r₋t)−α

t ln

2 1₋cq

t

dt_≤

≤ C

(r₋cq)α 1

Z

2/3

ln 2

1₋τdτ + C

(r₋ 3cq

2 )α 3/2 Z 1 1 τ ln 2τ

τ₋1dτ ≤

C(α)

rα . (9)

It now follows from (6), (8), and (9) that

Jq ≤

C(α)r1−α

cq

, cq<

r

2. (10)

We then consider the case iii), i.e.r < cq≤kr. Making use of (5) and H¨older’s inequality,

where αp <1< p, 1_p + _p1′ = 1, we deduce

Jq2 ≤C

Z r

2 3cq

(r₋t)−α

cq

ln 2 1_{− c}t

q

dt= C

cα q

Z _cqr

2 3

r cq −

τ−α

ln 2

≤ _cC_α

q

Z _cqr

2 3

r cq −

τ−αpdτ

1_pZ _cqr

2 3

ln 2 1₋τ

p′

dτ

_p1′

≤

≤ _cC_α

q

(_cr

q −

2 3)1−αp 1₋αp

1_pZ 1

2 3

ln 2 1₋τ

p′

dτ

1

p′

≤ C(_rα, p_α ).

It remains to consider the case r

2 < cq ≤r. Similarly to the previous cases we obtain

Jq2 ≤C cq

Z

2cq/3

(r₋t)−α

cq

ln 2 1₋ t

cq

dt+C

r

Z

cq

(r₋t)−α

t ln

2 1₋cq

t dt= = C cα q 1 Z 2/3 r cq −

τ−α

ln 2

1₋τdτ+ C cα q r/cq Z 1 r cq −

τ−α

ln 2τ

τ ₋1

dτ

τ ≤

≤ _cCα q

1

Z

2/3

(1₋τ)−αln 2

1₋τdτ + C cα q

Zr/cq

1

r cq −

τ−αp

dτ 1_p r cq Z 1 ln 2τ

τ−1

τ

p′

dτ

_p′1

≤

≤ C_c(_αα)

q

+ C

cα q

(_cr

q −1)

1−αp 1₋αp

1_p

2

Z

1

ln 2τ

τ−1

τ

p′

dτ

_p1′

≤ C(_rα, p_α ), αp <1< p.

We see that

Jq2 =O(r−α), r →+∞,

r

2 ≤cq ≤kr. (11)

We now able to give an upper estimate for _Iα[f](r). It follows from (4), (7), (10), and

(11) that

Iα[f](r) = 2π

Z

0

Dα−1|f

′

(reiϕ_)|

|f(reiϕ_)|dϕ≤C

T(kr, f)

rα +

X

cq≤r

2 r1−α

cq

+ X

r

2<cq<kr

1 rα ≤ ≤C

T(kr, f) +n(kr,0,_∞, f)

rα ) +

1

rα−1 r/2

Z

0

dn(t,0,_∞, f)

t

≤

≤C

T(kr, f) +n(kr,0,_∞, f)

rα ) +

1

rα−1 r/2

Z

0

n(t,0,_∞, f)

t2 dt

≤C

_{T(kr, f) +} N(k2r,0,∞,f)

lnk

rα +

+ 1

rα−1 r/2

Z

0

n(t,0,_∞, f)dt t2

≤C(α, k)

T(k2r, f)

rα +

1

rα−1 r/2

Z

0

n(t,0,_∞, f)dt t2

.

Here we used inequality ([7])n(kr,0,_∞, f)_≤ N(k2r,0,_lnk∞,f). Choosingk =√β, we arrive to the assertion of the theorem whenβ _∈(1,9₄). The general case follows from the monotonicity of the Nevanlinna characteristic.

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Lviv Ivan Franko National University chyzhykov@yahoo.com

semochkons@mail.ru