LIMITING DIRECTION AND BAKER WANDERING DOMAIN OF ENTIRE SOLUTIONS OF DIFFERENTIAL EQUATIONS∗

2016-11-24JunWANGSchoolofMathematicalSciencesFudanUniversityShanghai200433Chinamailmajwangfudaneducn

Acta Mathematica Scientia(English Series) 2016年5期

Jun WANGSchool of Mathematical Sciences,Fudan University,Shanghai 200433,China E-mail:majwang@fudan.edu.cn

Zongxuan CHEN†School of Mathematical Sciences,South China Normal University,Guangzhou 510631,China E-mail:chzx@vip.sina.com

LIMITING DIRECTION AND BAKER WANDERING DOMAIN OF ENTIRE SOLUTIONS OF DIFFERENTIAL EQUATIONS∗

In this paper,we mainly investigate entire solutions of complex differential equations with coefficients involving exponential functions,and obtain the dynamical properties of the solutions,their derivatives and primitives.With some conditions on coefficients,for the solutions,their derivatives and their primitives,we consider the common limiting directions of Julia set and the existence of Baker wandering domain.

limiting direction;Baker wandering domain;entire function;linear differential equation

2010 MR Subject Classification34M10;37F10;30D35

1　Introduction and Main Results

The Nevanlinna theory is an important tool in this paper,its usual notations and basic results come mainly from[9,11,15,22].We use λ(f)andµ(f)to denote the order and the lower order of f respectively,which are defined as[22,Definition 1.6]

For a transcendental entire function f(z),Baker[2]firstly observed that J(f)can not lie in finitely many rays emanating from the origin.Qiao[18]introduced the limiting direction of J(f),which means a limit θ of the set{argzn|zn∈J(f)is an unbound sequence}.For such limit θ,Zheng also say that the Julia set has the radial distribution with respect to the ray argz=θ[23].Define

Clearly,Δ(f)is closed,by mesΔ(f)we stand for its linear measure.

If the transcendental entire function f(z)satisfiesµ(f)<∞,then mesΔ(f)≥min{2πsee[18].Furthermore,Qiao[19,Theorem 1]investigated the common limiting directions of all f’s derivatives and primitives,and his result is stated as follows.

Theorem ALet f be a transcendental entire function of lower orderµ<∞.Then there exists a closed interval I⊂ℝ such that all θ∈I are the common limiting directions of J(f(n)),n=o,±1,±2,···,and mesI≥min{2π,π/µ},here f(n)denotes the n-th derivative or the n-th integral primitive of f for n≥o or n

The example in[2]shows that there exists an entire function of infinite lower order whose Julia set has only one limiting direction θ=o.Later some observations for a transcendental meromorphic function f(z)were made by[2o,23]:ifµ(f)<∞and δ(∞,f)>o,then

For a connected component U of F(f),we know that fn(U)must be contained in a component Unof F(f).If all Unare different,then U is called a wandering domain of f.By Sullivan’s famous theorem,rational functions have no wandering domains.For transcendental entire functions,it has been shown earlier by Baker[4]that such domains may exists.The wandering domain in Baker’s example is multiply connected,then such wandering domain was named after his name.For the convenience of the readers,we still state the definition.

Definition 1For the wandering domain U,if all Unare multiply connected component of F(f)which surrounds o,and the Euclidean distance dist(o,Un)→+∞as n→+∞,then U is called Baker wandering domain.

If f is a transcendental entire function,then each multiply connected component of F(f) must be a Baker wandering domain,see[3].There are some criterions of non-existence of the Baker wandering domains[5,7],which also determine whether there exists only simply connected Fatou component for given entire functions.For example,there is an interesting result in[5,Corollary].

Theorem BIf the entire function f has path to∞on which f is bounded,then all components of F(f)are simply-connected.

There exists the radial growth property of the exponential functions A(z)eP(z),where P(z) is a non-constant polynomial,and A(z)is an entire function of order less than degP,see Lemma 2.6.Clearly by Theorem B,A(z)eP(z)only has simply connected Fatou component. Zheng continued Baker and Bergweiler’s work,and investigated transcendental meromorphic functions with at most finitely many poles[25].

For differential equations,the solutions are always controlled by the behavior of coefficients. When there is a dominate coefficient Aoin the sense T(r,Aj)=o(T(r,Ao))(j=1,2,···,n−1)as r→∞,the dynamical property of linear differential equations

where Aj(z)(j=o,1,2,···,n−1)are entire functions of finite lower order,are investigated in[12–14].For the non-trivial solution f of(1.1),Huang and Wang discussed mesΔ(f)and the existence of Baker wandering domain.We are interested in the dynamical property of(1.1) without the dominant condition,for example,the coefficients are of the same growth rate.

Consider the exponential function’s good growth property along the radial line,we want to investigate the differential equations with exponential coefficients.In this paper,without loss of generality,we mainly investigate the following second order linear differential equations

where Aj(z)(/≡o),Bj(z)(j=1,2)are entire functions,and

are two polynomials of degree kj≥o.We obtain the following two theorems on all derivatives and primitives of entire solutions.When B1=B2=o,the coefficient are periodic Xiao and Chen[21]already investigated the complex oscillation of the periodic linear differential equations.

Theorem 1.1Suppose that k1+k2/=o and max{λ(Aj),λ(Bj)}o.If kj=o,then Ajand Bjare constants.For every solution f(/≡o)of(1.2),we define

then we have

(1)if k1

(2)if k1=k2and a1/a2=b<1,thenµ(f)=∞and mesE(f)≥π;

(3)if k1=k2and a1/a2=b/∈ℝ,thenµ(f)=∞and

Remark 1.1Clearly,the case k1

By[14,Theorem 1.1],it is easy to see mesΔ(f)≥π/k2.In Theorem 1.1(1),we obtain mesE(f)≥π,which is better when k2≥2.

Remark 1.2There are many results on the order of solutions of complex differential equations.But the result on the lower order of solutions is few.Example 4 in[9,p.238]shows that there exists an entire function f withµ(f)=o and λ(f)=∞.This means that the lower order is quite different from the order,so we make sureµ(f)=∞in Theorem 1.1.Note that f′′+ezf′−ezf=o admits the particular solution fo=ez+1,it implies that the condition a1/a2<1 in Theorem 1.1(2)can not be omitted,that is,our results are sharp.

Theorem 1.2Suppose that Bj(j=1,2)are constants,and that Pj,Aj(i=1,2)are defined as in Theorem 1.1.Suppose that any one of the following two conditions holds:

(1)k1

(2)k1=k2and a1/a2=b/∈ℝ or b∈(o,1).

Then for every solution f(/≡o)of(1.2),all f(n)(n=o,±1,±2,···)have no Baker wandering domain,that is,they only have simply connected Fatou component.

Remark 1.3The existence of Baker wandering domain for E(z)=f1f2was discussed in [13],where f1,f2are two linearly independent solutions of f′+A(z)f=o,A(z)satisfies some radial growth condition.Our theorem treats all primitives and derivatives of entire solutions of (1.2)instead of E(z).

In the following examples,there are some special functions satisfying(1.2)in our case, which shows that our theorems may be useful.We assume that C1,C2are arbitrary constants.

Example 1.1The equation f′′−(2ez+1)f′+e2zf=o admits the general solution

Example 1.2The general solution of the equation f′′−f′+e2zf=o is of the form

Example 1.3[17,Section 2.1.3]We use Jn(z)and Yn(z)to denote Bessel functions of the first kind and the second kind,respectively,and a,b are real constants withn∈ℕ∪{o}.The equation f′′+(1+a)f′+(ez+b)f=o has the general solution

Finally,by the method in the proof of Theorem 1.1,we could also investigate the Julia set of Mathieu functions,which are discussed in the book by Abramowitz and Stegun[1]in more detail.The Mathieu functions satisfy the Mathieu differential equation

with given numbers a and q,which occurs in many applications in physics and engineering.

Theorem 1.3Let f(/≡o)be a solution of(1.3).If q/=o,thenµ(f)=∞and mesE(f)= 2π,where E(f)is defined as in Theorem 1.1.

2　Preliminary Lemmas

Before introducing the preliminary lemmas,we recall the Nevanlinna’s Characteristic in an angle,see[9,26].We set Ω(α,β)={z∈ℂ:argz∈(α,β)},and Ω(r,α,β)=Ω(α,β)∩{z∈ℂ:|z|>r}.Let g(z)be analytic on the closure of Ω(α,β),where β−α∈(o,2π],we define

where ω=π/(β−α).The Nevanlinna’s angular characteristic of g is defined by

and we use σα,β(g)to denote the order of Sα,β(r,g),that is

Lemma 2.1(see[23,Lemma 2.2])Let f(z)be analytic in Ω(ro;θ1,θ2),U a hyperbolic domain and f:Ω(ro,θ1,θ2)→U.If there exists a point a∈∂U{∞},such that CU(a)>o, then there exists a constant d>o such that for sufficiently small ε>o,we have

Remark 2.1(see[23,p.4])The open set W is hyperbolic if ℂW has at least three points.For any a∈ℂW,we define

where λW(z)is the hyperbolic density on W.Note that|z−a|≥δW(z)where δW(z)is the Euclidean distance of z∈W to∂W.It is well known that if every component of W is simply connected,then CW(a)≥1/2.

Lemma 2.2(see[26,Theorem 2.5.1])Let f(z)be a meromorphic function on Ω(α−ε,β+ε)for ε>o and o<α<β<2π.Then

for r>1 possibly except a set with finite linear measure.

Lemma 2.3(see[1o,Theorem 2])Let f(z)be a transcendental meromorphic function, and let α>1 be a given constant.Then there exist a set E⊂[o,2π)of linear measure zero,and positive constant B depending only on α and i,j(o≤i1 such that for all z satisfying argz=ψoand |z|=r≥Ro,we have

In the angular domain,there also exists some estimate for the logarithmic derivative outside an R-set.Defineis called R-set if

Lemma 2.4(see[13,Lemma 2.2])Let z=rexp(iψ),ro+1

there exist K>o and M>o only depending on g,ε1,···,εn−1and Ω(αn−1,βn−1),and not depending on z such that

for all z∈Ω(αn−1,βn−1)outside a R-set D,where k=π/(β−α)and kj=π/(βj−αj) (j=1,2,···,n−1).

Lemma 2.5Let f(z)be an entire function satisfying mesE(f)<2π,and n≥1 is a positive integer.If(α,β)/⊆E(f),then we have(αo,βo)⊆(α,β)such that

for all z∈Ω(αo,βo)with|z|=r outside a set of finite linear measure,where M and K are two constants not depending on z.

ProofSince all Δ(f(n))(n∈ℤ)are closed,then E(f)must be a closed set,so the components of E(f)are closed intervals or one-point sets on[o,2π).Define S:=(o,2π)E(f). Clearly,S is open,and consists of at most countably many open intervals.We take(α,β)∩S, so there exists an open interval(α∗,β∗)⊆(α,β)∩S.Thus,every θ∈(α∗,β∗)is not limiting direction of some f(nθ),where nθis an integer depending on θ.It follows from the definition of the limiting direction that there exist constants ξθand r(θ)dependent of θ such that (θ−ξθ,θ+ξθ)⊂(α∗,β∗)and(θ−ξθ,θ+ξθ)is an open covering of the closed interval[α∗+ε,β∗−ε]with o<ε<(β∗−α∗)/4.Therefore,by Heine-Borel theorem,we have

From(2.4),there exist the corresponding rjand unbounded Fatou component Ujof F(f(nθj)) such that Ω(rj,θj−ξθj,θj+ξθj)⊂Uj,see[3].We take a unbounded and connected closed section Γjon boundary∂Ujsuch that ℂΓjis simply connected.Clearly,ℂΓjis hyperbolic and open. By Remark 2.1,we have CℂΓj(a)≥1/2.Since the mapping f(nθj):Ω(rj,θj−ξθj,θj+ξθj)→ℂΓjis analytic for all j,by Lemma 2.1,there exists a positive constant d such that

for z∈Ω(rj,αj,βj),where αj:=θj−ξθj+ε,βj:=θj+ξθj−ε.

For the case nθj>o,since f is entire,we have

where the integral path is the segment of a straight line from o to z.Combining this fact and(2.5),it is easy to deduce|f(nθj−1)(z)|=O(|z|d+1)for z∈Ω(rj,αj,βj).Repeating the discussion nθjtimes,we can obtain

It follows from the definition of Nevanlinna’s angular characteristic that

For the case nθj

From(2.5),we have Sαj,βj(r,f(nθj))=O(1).Then applying Lemma 2.2 yields,for|nθj|ε′=ε,

for r>1 possibly outside a set of finite linear maesure.Using the discussion|nθj|times yields

for r/∈E,mesE<∞.

From(2.6)and(2.7),whatever nθjis positive or not,we always have

Lemma 2.6(see[8,16])Suppose that P(z)=(α+βi)zn+···is a non-constant polynomial with degree n≥1,α,β are real constants,and that A(z)(/≡o)is an entire function with λ(A)o,there exists a set H1⊂[o,2π)of linear measure zero such that for any θ∈[o,2π)(H1∪H2),there is R>o such that for|z|=r>R,we have

(i)if δ(P,θ)>o,then

(ii)if δ(P,θ)

where H2={θ∈[o,2π);δ(P,θ)=o}.

Remark 2.2For the polynomial P(z),we define

for j=o,1,···,2n−1.From the basic property of polynomials[16],if θ∈Sj,then δ(P,θ)>o for even j,and δ(P,θ)

Lemma 2.7(see[25,Corollary 1])Let f(z)be a transcendental meromorphic function with at most finitely many poles.If J(f)has only bounded component,then for any complex number a,there exists a constant o

where M(r,a,f)=max{|f(z)|:|z−a|=r},L(r,a,f)=min{|f(z)|:|z−a|=r},andwhich has an infinite logarithmic measure.

Lemma 2.8(see[6])Let pj(x)(j=1,2,···,n)and f(x)be a continuous complexvalued functions on the interval[a,b],and let Pj(x)(j=1,2,···,n)and F(x)be non-negative continuous functions with|pj(x)|≤Pj(x)and|f(x)|≤F(x).Suppose that v(x)and V(x)are the solutions of differential equations

respectively.Then if V(k)(a)≥|v(k)(a)|(k=o,1,···,n−1),we have

Lemma 2.9(see[22,Theorem 4.2])If f(z)is meromophic in ℂ,then f(z)and its derivative f′(z)have the same order and lower order.

3　Proof of Theorem 1.1

By(1.2)and Lemma 2.3,for all z satisfying argz=θ∈[o,2π)H1and|z|=r≥Ro(θ), we obtain

where H1is a set of linear measure zero,

is a positive constant.

(1)Assume that k1

for all z with|z|=r≥ro.It follows from(3.2),Lemma 2.6 and Remark 2.2 that for all z=reiθwith θ∈Sj(P2,θ)H2and j=o,2,···,2k2−2,there is Ro(θ)such that

holds for r≥Ro(θ),where H2is a set of linear measure zero.Taking(3.2)and(3.3)into(3.1) yields out that for every θ∈Sj(P2,θ)(H1∪H2)with even j,

as r is sufficiently large.This immediately leadsµ(f)=∞.

According to Remark 2.2,Sj(P2,θ)with even j are k2open intervals of linear measure π/k2. Next,we would prove that the union of such Sj(P2)is contained in E(f),so mes(E(f))≥π. If the union is not contained in E(f),there must exist one Sjo(P2,θ)/⊆E(f)where jois even. By Lemma 2.5,there exists(α,β)⊆Sjo(P2,θ)such that

for all z∈Ω(α,β)with|z|=r/∈E1,mesE1<∞,where K,M are positive constants. Substituting(3.2)–(3.4)into the first inequality of(3.1),we obtain that for θ∈(α,β)and enough large r/∈E1,

It is impossible since δ(P2,θ)>o.

(2)Assume that k1=k2and a1/a2=b<1,so δ(P1,θ)=bδ(P2,θ).Set c=max{b,o}and take o<ε<(1−c)/(1+c),so c(1+ε)<1−ε.Applying Lemma 2.6 again to

we obtain that for all z=reiθwith θ∈Sj(P2,θ)(H2∪H3)and even j,there is Ro(θ)such that

holds for r≥Ro(θ),where H3is a set of linear measure zero,and depends onCombining (3.5),(3.6)with(3.1),it leads that for θ∈Sj(P2,θ)(H1∪H2∪H3)with even j,

Since[(1−ε)−(1+ε)c]δ(P2,θ)>o,it impliesµ(f)=∞.Similarly as in the proof of(1),if Sj(P2,θ)/⊆E(f)with even j,we also have(3.4).Then taking(3.4),(3.5)and(3.6)into(3.1) gives a contradiction.Thus,we have mesE(f)≥π.

(3)Assume that k1=k2and a1/a2=b/∈ℝ,so argb∈(o,π)∪(π,2π).Without loss of generality,we can assume a2=1.Thus,δ(P2,θ)=cos(k2θ)and δ(P1,θ)=|b|cos(k2θ+argb). There are the following two basic facts.

Case(i)Suppose that argb∈(o,π).By Remark 2.2,we see that

are k2open intervals,each interval is of linear measure(argb)/k2.Such intervals Sj(θ)satisfy that if θ∈Sj(θ),then

Then,by Lemma 2.6,for all z=reiθand θ∈Sj(θ)(H1∪H2∪H3)(j=o,2,···,2k2−2)and enough large r,we have(3.3)and

Substituting(3.3)and(3.8)into(3.1),we obtainwe obtainµ(f)=∞.

If So(θ)is not contained in E(f),then by Lemma 2.5,there exists(α,β)⊆So(θ)such that (3.4)holds for z∈Ω(α,β)with|z|=r/∈E1,mesE1<∞.Thus,taking(3.3),(3.4)and(3.8) into the first inequality of(3.1)yields

This is a contradiction since

Case(ii)Suppose that argb∈(π,2π).Now Sj(θ)in(3.7)is the interval of linear measure (2π−argb)/k2.Using the same method in Case(i),we obtainµ(f)=∞,and all Sj(θ)in(3.7) are contained in E(f),so mesE(f)≥2π−argb.

Combining Cases(i)and(ii),we haveµ(f)=∞and

This completes the proof of Theorem 1.1.

4　Proof of Theorem 1.2

For the case k1

For the case k1=k2and b∈(o,1),we know Sj(P2,θ)=Sj(P1,θ).

For the case k1=k2and b/∈ℝ,so argb∈(o,π)∪(π,2π).From the definition of Sj(P,θ) in Remark 2.2,Sj(P1,θ)can be obtained by rotating Sj(P2,θ)with the angle argb for argb∈(π,2π)and 2π−argb for argb∈(π,2π).Thus,there exists odd numbers j1and j2such that the open interval Sj1(P1,θ)∩Sj2(P2,θ)is not nonempty.

Thus,there exist θ1,θ2and A satisfying θ1<θ2,max{k1,k2}≤A and π/(θ2−θ1)