Further results on common zeros of the solutions of two differential equations

R E S E A R C H

Open Access

Further results on common zeros of the

solutions of two differential equations

Asim Asiri

*_{Correspondence:}

[email protected] Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia

Abstract

Purpose:Two problems are discussed in this paper. In the first problem, we consider one homogeneous and one non-homogeneous differential equations and study when the solutions of these differential equations can have (nearly) the same zeros. In the second problem, we consider two linear second-order differential equations and investigate when the solutions of these differential equations can take the value 0 and a non-zero value at (nearly) the same points.

Method:We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.

Conclusion:In the first problem, the results determine all pairs of such equations having solutions with the same zeros or nearly the same zeros. Regarding the second problem, the results also show all pairs of such equations having solutions taking the value 0 and a non-zero value at (nearly) the same points.

Keywords: Nevanlinna theory; differential equations

1 Introduction

There has been much research [–] on zeros of solutions of linear differential equations with entire coefficients. The principal paper [] that was published in  by Bank and Laine has stimulated many studies on this kind of problems. The reader is referred to [– ] for background on some applications of the Nevanlinna theory. We use the standard notation of the Nevanlinna theory from [].

In , Wittich [] proved the following theorem.

Theorem . If f is a non-trivial solution of w+Aw= ,i.e.,f ≡and A≡is entire,

then we have:

(i) T(r,A) =S(r,f).

(ii) Iff has finite order,thenAis a polynomial.

(iii) Ifais a non-zero complex number,thenf takes the valueainfinitely often,and in fact,outside a set of finite measure,

N

r, 

f–a

T(r,f).

The following facts follow from the asymptotic representation for solutions of the equa-tion

w+Pw= . ()

Theorem .[, ] Let P be a polynomial of degree n,and let w be a non-trivial solution of the equation().Then,w has order of growth equal ton+

 .Moreover,if w is a solution of

()which has infinitely many zeros,then we have

lim inf

r→∞

N(r,_w)

r(n+)/ > . ()

Our previous paper [] studied homogeneous linear differential equations having so-lutions with nearly the same zeros and proved several results, including the following.

Theorem .[] Let P≡be a polynomial of degree n.Let w≡be a solution of().

Assume that w has infinitely many zeros.Suppose that we have a solution v≡of the differential equation

v+Av=  ()

such that A is an entire function and N(r)counts zeros of v which are not zeros of w and zeros of w which are not zeros of v.Assume that

N(r) +T(r,A) =or(n+)/.

Then _wv is a constant and A=P.

The paper [] includes further results for homogeneous linear differential equations, and the corresponding problem wherePis a transcendental entire function of finite order is studied in [].

Recently, the same problem, but with non-homogeneous first-order differential equa-tions, has been studied in [], including the following result.

Theorem .[] Assume that v=Av+B and w=Cw+D,where A,B,C and D are entire functions of order less thanand v,w are transcendental functions.Assume that v=Lw,

where L has finitely many zeros and poles,and

T(r,A) +T(r,B) =S(r,v), T(r,C) +T(r,D) =S(r,w). ()

Then the following conclusions hold.

(I) IfLis a rational function,thenA≡C,Lis a constant andB=LD. (II) IfLis a transcendental function,then one of the following cases holds:

(i) B≡D≡andv,whave no zeros.

(ii) A= –CandB/A,D/Care non-zero constants,and

If,in addition,Lhas finite order incase (ii),thenA,B,C,Dare polynomials and so isA.

In this paper, our first result (Theorem .) looks at the same problem but with one ho-mogeneous and one non-hoho-mogeneous differential equations. In particular, we consider the first equation to be homogeneous of the second-order with a polynomial coefficient and the second equation to be non-homogeneous of the first-order with entire coefficients. A further result (Theorem .) studies the case where the solutions of two second-order homogeneous differential equations can take the value  and a non-zero value at (nearly) the same points.

2 Our results

Our first result is the following theorem.

Theorem . Suppose that P≡is a polynomial of degree n,and w solves(),and w≡

has infinitely many zeros.Suppose that v≡solves

v=Av+B, ()

where A,B are entire and AB≡,and

T(r,A) +T(r,B) =or(n+)/. ()

Suppose that L=w_v has finitely many zeros and poles(i.e.,w and v have the same zeros with finitely many exceptions).

Then A is a polynomial and there exists a polynomial Q such that

w=ceQ

e–Qdz and v=ceA

e–Qdz,

and

P= –Q+Q and B=ceA–Q,

where A_=A and c,care constants.

Example . TakeQto be a polynomial. Let

w=eQ

z



e–Qdz.

Then

w=Qw+e–Q, and

So, we haveP= –(Q+Q_).

Now, letAbe another polynomial, and let

v=eA

z



e–Qdz.

Note thatvhas the same zeros asw. Now, we have

v=Av+eA–Q_, whereA=A_.

We chooseAso that

deg(A– Q) <

deg(P) +   .

For example, letA= Q.

We now state our second result.

Theorem . Suppose that P≡is a polynomial of degree n,and A is an entire function,

and suppose that w solves()and v solves(),and vw≡.Let v– and w have,with finitely many exceptions,the same zeros and the same multiplicities.Then one of the following holds.

(A) whas finitely many zeros andvis a polynomial andA= .

(B) whas infinitely many zeros andP,Aare non-zero constants and v–

w is non-constant

and

w=λeσz+λe–σz,

v=λeσz,

()

whereσ,λ,λ,λare non-zero constants.

Example . Ifw=ez_–e–z_andv=ez_{, then}

v–  =ez–  =ezez–e–z=ezw.

Hence,v–  has the same zeros asw. HereP= – andA= –.

Example . We give an example to show that the zeros ofv–  andwmust necessarily have the same multiplicities. To show this, let

w=sinz

, v=cosz.

Thenw= ⇔_z=kπ, wherek∈Z.

Alsov= ⇔z=kπ, wherek∈Z.

Therefore,wandv–  have the same zeros but the zeros are simple forw, double for

3 Proof of Theorem 2.1

Proof We have

w=Lv. ()

So,

w=Lv+ Lv+Lv. () but

v=Av+Av+B=Av+A(Av+B) +B=vA+A+AB+B. () We also have, using (), (), (), () and (),

 =w+Pw=Lv+ Lv+Lv+PLv

=Lv+ L(Av+B) +LvA+A+AB+B+PLv

=L+ LA+LA+A+PLv+ LB+LAB+B. () Let

M=L

L. ()

Then

L L =M

_+M_{. ()}

We divide () byL, and by using () and (), we get

 = L

L +  L LA+A

_+A+P

v+ L

LB+AB+B

=M+M+ MA+A+A+Pv+ MB+AB+B. () The next step is to estimate the growth ofM.

We know thatρ(w) =n+_ from []. Therefore,

m

r,w

w

=O(logr) =or(n+)/. ()

Claim . We claim that T(r,M) =o(r(n+)/).

To show this, we know thatN(r,m) =O(logr) sinceMhas finitely many poles. Write () as

whereA_=A.

Then, there exists a constantcsuch that

v=eA

c+

z



e–A(t)_B(t)dt

.

Also, using (), we can write

logM(r,A)≤T(r,A) =or(n+)/, M(r,A)≤expor(n+)/. Also,

A(z) =A() +

z



A(t)dt.

So,

M(r,A)≤A()+rM(r,A)≤O() +rexp

or(n+)/≤expor(n+)/. Therefore, we get

M(r,v)≤exp expor(n+)/, T(r,v)≤logM(r,v)≤expor(n+)/. We use Lemma . in [, p.] withR= rto get

m

r,v

v

=O(logr) +Olog+T(r,v)≤or(n+)/.

Now, we haveM=w_w –v_v. So

m(r,M) =or(n+)/.

We also haveN(r,M) =O(logr). Hence,

T(r,M) =or(n+)/+O(logr) =or(n+)/. This completes the proof of Claim .. Using Claim . and (), we get

Tr,M+M+ MA+A+A+P=or(n+)/

and

Tr, MB+AB+B=or(n+)/. Also, by Theorem ., we get

T(r,v)≥N

r,

v

Therefore, we must have

M=M+M+ MA+A+A+P≡ ()

and

M= MB+AB+B≡ ()

because otherwise we can writev= –M/Mto get a contradiction.

We now divide () byBto get

M+A+B

B ≡.

So,B/Bhas finitely many poles, and soBhas finitely many zeros. Then we can writeB

in the form

B=PeP,

whereP,Pare polynomials.

But then, we can write

B/B=R,

whereRis rational.

Then we also can write

M= –A

 +R, ()

whereRis rational andR= –_R.

Substitute () in (), we obtain

≡

–A

 +R  + A

 –AR+R

 

+–A+ AR

+A+A+P

≡

A

 +R  + A

 +AR+R

  +P ≡ A

 +R

+

A

 +R



+P. ()

Now, let

N=A  +R=

A

 –

B

B.

Also, let

So, we get

H H =

A

 –

B

B=N. ()

Substituting () in (), we obtain

 =N+N+P=H

H +P, H

_{+PH= .}

Thus,ρ(H) <∞,His entire, andBhas no zeros. Then,

m

r,H

H

=O(logr)

Therefore,Ais a polynomial.

SinceBhas no zeros, from () we can write

H=eQ, ()

whereQis a polynomial.

SincewandH solve the same equation and are linearly independent (becausewhas zeros butHdoes not), we can write

w H

=(constant)

H .

Therefore,

w=ceQ

e–Qdz, ()

wherecis a constant andQis a polynomial.

Now, we have

L

L =M= – A

 +R=N–A=

H H –A.

So, we can write

L= (constant)·He–A_. Therefore, we have

v=w

L =

(constant)·HH–_dz

He–A . Hence, using (), we obtain

v=ceA

wherecis a constant andA=A.

Now, from () and (), we notice thatwandvhave the same zeros. Also, differentiating (), using (), we have

v=cAeA

H–dz+ceAH–=Av+ceAH–=Av+ceA–Q.

Comparing this with (), we get

B=ceA–Q.

Moreover,H=eQ_{solvesH+PH= , and so}

–P=H

H =Q

_+Q_.

This completes the proof of Theorem ..

4 A lemma needed to prove Theorem 2.2

In order to prove Theorem ., we must state and prove the following lemma. We include a proof for completeness.

Lemma . Let P, . . . ,Pn∈Cbe distinct,and let A, . . . ,Anbe rational functions such that

≡A(z)ePz+· · ·+An(z)ePnz. ()

Then there exists k∈ {, . . . ,n}such that Pk= and Ak= ,and Aj= for j=k.

Proof The proof is by induction. It is obvious that the lemma is true whenn= . Assume that the lemma is true form≤n– . Differentiating (), we get

≡B(z)ePz+· · ·+Bn(z)ePnz, Bj=Aj+PjAj.

Now, we have two cases to consider.

Case (): Suppose there existsksuch thatBk≡. Without loss of generality, letk= ,

then we can write

 =  +B

B

e(P–P)z₊· · ·₊Bn

B

e(Pn–P)z_.

Since we assumed the lemma is true for m≤n– , there existsj∈ {, . . . ,n}such that

Pj–P= . But this contradicts our assumption thatP, . . . ,Pnare distinct.

Case (): Suppose thatBj=  for eachj,i.e.,

A_j+PjAj≡.

IfPj= , thenAj≡ because otherwise we have

A_j Aj

But this contradicts the fact thatPj= .

So, we haveAj≡ forPj= . Thus, () becomes (for somek)

 =Ak(z)ez=Ak(z),

andPk=  andAk= .

5 Proof of Theorem 2.2

We first note that, outside a set of finite measure, by Theorem .,

T(r,v)∼N

r, 

v– 

≤N

r, 

w

+O(logr)≤Or(n+)/+O(logr). ()

In particular, ifwhas finitely many zeros, thenvis a polynomial, which givesA= . This completes the proof of part (A) in the conclusion.

Assume henceforth thatwhas infinitely many zeros. Then () implies thatρ(v)≤(n+ )/, and soAis a polynomial of degree at mostnby the Wiman-Valiron theory []. Also,

A≡ sincev–  has infinitely many zeros. Now, two cases have to be considered.

Case (I). Assume thatPis a non-zero constant; thenn=  andAis constant. Therefore, we can write

w=ceαz+ce–αz, v=deβz+de–βz, ()

whereα,β∈C\ {},cj,dj∈Candcj=  (j= , ).

Sincewandv–  have the same zeros with finitely many exceptions, we can write

v–  =RePw, ()

whereRis a rational function andPis a polynomial. We know thatdeg(P)≤ because ρ(w),ρ(v)≤. We can now write

deβz+de–βz–  =Reγz

ceαz+ce–αz

,

whereγ ∈C, and so

 =deβz+de–βz–cRe(γ+α)z–cRe(γ–α)z.

Now, by using Lemma .,Ris constant and so we can write () as

v– 

w =e

γz+δ

, ()

whereδis constant. Therefore,

deβz+de–βz–  =

eγz+δceαz+ce–αz

Now, by using Lemma ., we get

α+γ=β, –βor , –α+γ =β, –βor ,

andα+γ, –α+γ,β, –β,  cannot all be different. We must now try six cases:

I(a): Ifα+γ =βand –α+γ = –β, thenγ =  andα=β. But this contradicts (). Thus, this case cannot happen.

I(b): Ifα+γ =βand –α+γ= , thenβ= γ andα=γ. Substituting these in () gives

deγz+de–γz–  =heγz+h,

whereh,hare constants, which yieldsd= . Putting this in () gives () withσ=γ.

There are four more cases: I(c):α+γ = –βand –α+γ=β. I(d):α+γ= –βand –α+γ = . I(e):α+γ =  and –α+γ =β. I(f ):α+γ =  and –α+γ= –β.

It is easy to check that case I(c) is impossible and cases I(d), I(e), I(f ) all lead to () with

σ=γ.

From these cases, we find thatγ = , and so v–_w =eγz+δ_{is non-constant. Also, we have}

() and case (B) of the conclusion.

Case (II). Suppose thatPis non-constant. We will show that this leads to a contradiction. Let

v– 

w =M=Le

Q_{, ()}

whereLis a rational function andQis an entire function. From (), we haveρ(v) <∞, and soQis a polynomial. Also, from (), we have

v=Mw+ , v=Mw+Mw,

v=Mw+ Mw+Mw= Mw+wM–PM= –Av= –A(Mw+ ).

So,

Mw+M+AM–PMw= –A. ()

Now, we have two cases to consider.

Case (i): IfMis constant, then eitherA=PandA= , so thatP= , which is a contra-diction, or

w= –A

AM–PM

Case (ii): IfMis non-constant, thenM≡. Therefore,

w+

M

M+

(A–P)M

M

w= –A

M, ()

where _M_M +(A_–_MP)M is rational because M

M = L

L +Qis rational and M

M is rational, and so M

M = M

M / M

M is rational.

Also,

–A

M =

–A

(L+QL)eQ.

Then we can write () as

w=Rw+Se–Q, ()

where

R=PM–AM–M

M , S=

–A

(L+QL) are rational functions andQis a polynomial.

LetU=Se–Q_{, then we can write () as}

w=Rw+U. ()

Now, we have two cases to consider: Case ii(a): IfR≡ in (), then () gives

w=Se–Q, w=–w

P =

–(S–QS)

P e

–Q_,

which is a contradiction sincewhas infinitely many zeros. Case ii(b): Assume thatR≡ in (); then () gives

w=Rw+Rw+U.

Now, () and () give

–Pw=R(Rw+U) +Rw+U,

and so

R+R+Pw+RU+U= .

Therefore,

because if not,whas finitely many zeros, a contradiction. Also,

RU+U≡. ()

Put

G= 

U=Te

Q_{, ()}

whereT= /Sis a rational function. Then,

R=–U

U = G

G.

From (), we get

 =R+R+P=G

G +P, G

_{+PG= .}

So,Gsolves (), and sinceP≡ and is a polynomial of degree n, we see thatGis a transcendental entire function with finitely many zeros and has order (n+ )/.

SincewandGsolve the same equation butwhas infinitely many zeros andGhas finitely many zeros,wandGare linearly independent, and we can write

wG–wG=c,

wherecis a non-zero constant. So,

w G

= c

G.

By integrating, we get

w=G

z _c

Gdζ. ()

Also, using (), () and (),

v=  +Mw=  +LeQTeQ

z _c

Gdζ=  +HG 

z _c

Gdζ, ()

whereH=L/T is a rational function.

Now, we can assume thatc=  because ifc= , we can multiplywby /c. We differentiate () to get

v=HG

z _

Gdζ+ HGG

z 

Gdζ+H=H+K(v– ), ()

where

K=H

H +  G

G ()

So, from () and (), we get

–Av=v=H+K(v– ) +KH+K(v– ), and so

 =vK+K+A+H–K+KH–K=vU+U,

whereU=K+K+AandU=H–K+KH–K.

Sincevis transcendental andU,Uare rational functions, we must have

U=K+K+A=  ()

and

U=H–K+KH–K= . ()

Claim . We claim that H≡K. To show this, letH≡K. From (), we have

H–K

H–K +K= .

From (), we get

H–K H–K = –K=

–H H – 

G G.

We integrate to get

H–K= a

HG, G

₌ a

H(H–K),

whereais a constant. But this contradicts the fact thatHandKare rational functions and

Gis a transcendental function. This completes the proof of Claim .. Once we have Claim ., () gives

v=H+H(v– ) =Hv, and so

H=v

v. ()

By (),vhas finitely many zeros, so we can write

v=PeQ,

Therefore,

w=v– 

LeQ =

PeQ– 

LeQ .

Now, we can write this as

w=ReS+ReS, ()

whereR=P/L≡,R= –/L≡ are rational functions andS=Q–Q,S= –Qare

polynomials.

Here,ReSandReS are linearly independent becauseQis non-constant. Now, we get

 =w+Pw=JeS+JeS=JeQ–Q+Je–Q,

whereJ,Jare rational and satisfy

Jk=Rk+ RkSk+Rk

S_k+S_k+P.

Therefore,J=J=  because otherwiseeQ= –J/Jore–Q=J/J. Thus,ReS,ReS

both solvey+Py=  and have finitely many zeros, and they are linearly independent. Hence,Pis constant by [], which contradicts our assumption in Case (II) thatPis

non-constant.

Competing interests

The author declares that he has no competing interests.

Acknowledgements

The author would like to thank his supervisor Prof. Jim Langley for his support and guidance. Also, he would like to thank King Abdulaziz University for financial support for his PhD study.

Received: 8 May 2012 Accepted: 17 September 2012 Published: 4 October 2012 References

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doi:10.1186/1029-242X-2012-222