Abstract
Purpose
Two problems are discussed in this paper. In the first problem, we consider one homogeneous and one nonhomogeneous 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 secondorder differential equations and investigate when the solutions of these differential equations can take the value 0 and a nonzero 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 nonzero value at (nearly) the same points.
Keywords:
Nevanlinna theory; differential equations1 Introduction
There has been much research [18] on zeros of solutions of linear differential equations with entire coefficients. The principal paper [9] that was published in 1982 by Bank and Laine has stimulated many studies on this kind of problems. The reader is referred to [1012] for background on some applications of the Nevanlinna theory. We use the standard notation of the Nevanlinna theory from [13].
In 1955, Wittich [12] proved the following theorem.
Theorem 1.1Iffis a nontrivial solution of, i.e., andis entire, then we have:
(ii) Iffhas finite order, thenAis a polynomial.
(iii) Ifais a nonzero complex number, thenftakes the valueainfinitely often, and in fact, outside a set of finite measure,
The following facts follow from the asymptotic representation for solutions of the equation
LetPbe a polynomial of degreen, and letwbe a nontrivial solution of the equation (1). Then, whas order of growth equal to. Moreover, ifwis a solution of (1) which has infinitely many zeros, then we have
Our previous paper [14] studied homogeneous linear differential equations having solutions with nearly the same zeros and proved several results, including the following.
Theorem 1.3[14]
Letbe a polynomial of degreen. Letbe a solution of (1). Assume thatwhas infinitely many zeros. Suppose that we have a solutionof the differential equation
such thatAis an entire function andcounts zeros ofvwhich are not zeros ofwand zeros ofwwhich are not zeros ofv. Assume that
The paper [14] includes further results for homogeneous linear differential equations, and the corresponding problem where P is a transcendental entire function of finite order is studied in [15].
Recently, the same problem, but with nonhomogeneous firstorder differential equations, has been studied in [16], including the following result.
Theorem 1.4[16]
Assume thatand, whereA, B, CandDare entire functions of order less than 1 andv, ware transcendental functions. Assume that, whereLhas finitely many zeros and poles, and
Then the following conclusions hold.
(I) IfLis a rational function, then, Lis a constant and.
(II) IfLis a transcendental function, then one of the following cases holds: If, in addition, Lhas finite order in case (ii), thenA, B, C, Dare polynomials and so is.
(ii) and, are nonzero constants, and
In this paper, our first result (Theorem 2.1) looks at the same problem but with one homogeneous and one nonhomogeneous differential equations. In particular, we consider the first equation to be homogeneous of the secondorder with a polynomial coefficient and the second equation to be nonhomogeneous of the firstorder with entire coefficients.
A further result (Theorem 2.2) studies the case where the solutions of two secondorder homogeneous differential equations can take the value 0 and a nonzero value at (nearly) the same points.
2 Our results
Our first result is the following theorem.
Theorem 2.1Suppose thatis a polynomial of degreen, andwsolves (1), andhas infinitely many zeros. Suppose thatsolves
Suppose thathas finitely many zeros and poles (i.e., wandvhave the same zeros with finitely many exceptions).
ThenAis a polynomial and there exists a polynomialQsuch that
and
Example 2.1 Take Q to be a polynomial. Let
Then
and
Now, let be another polynomial, and let
Note that v has the same zeros as w. Now, we have
We now state our second result.
Theorem 2.2Suppose thatis a polynomial of degreen, andAis an entire function, and suppose thatwsolves (1) andvsolves (3), and. Letandwhave, 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 and.
(B) whas infinitely many zeros andP, Aare nonzero constants andis nonconstant and
whereσ, , , are nonzero constants.
Hence, has the same zeros as w. Here and .
Example 2.3 We give an example to show that the zeros of and w must necessarily have the same multiplicities. To show this, let
Therefore, w and have the same zeros but the zeros are simple for w, double for . Here, and .
3 Proof of Theorem 2.1
Proof We have
So,
but
We also have, using (1), (5), (8), (9) and (10),
Let
Then
We divide (11) by L, and by using (12) and (13), we get
The next step is to estimate the growth of M.
We know that from [9]. Therefore,
To show this, we know that since M has finitely many poles.
Write (5) as
Then, there exists a constant c such that
Also, using (6), we can write
Also,
So,
Therefore, we get
We use Lemma 2.3 in [[13], p.38] with to get
Hence,
This completes the proof of Claim 3.1.
Using Claim 3.1 and (14), we get
and
Also, by Theorem 1.2, we get
Therefore, we must have
and
because otherwise we can write to get a contradiction.
We now divide (17) by B to get
So, has finitely many poles, and so B has finitely many zeros. Then we can write B in the form
But then, we can write
Then we also can write
Substitute (18) in (16), we obtain
Now, let
Also, let
So, we get
Substituting (21) in (19), we obtain
Thus, , H is entire, and B has no zeros.
Then,
Therefore, A is a polynomial.
Since B has no zeros, from (20) we can write
where Q is a polynomial.
Since w and H solve the same equation and are linearly independent (because w has zeros but H does not), we can write
Therefore,
where is a constant and Q is a polynomial.
Now, we have
So, we can write
Therefore, we have
Hence, using (22), we obtain
Now, from (23) and (24), we notice that w and v have the same zeros.
Also, differentiating (24), using (22), we have
Comparing this with (5), we get
This completes the proof of Theorem 2.1. □
4 A lemma needed to prove Theorem 2.2
In order to prove Theorem 2.2, we must state and prove the following lemma. We include a proof for completeness.
Lemma 4.1Letbe distinct, and letbe rational functions such that
Then there existssuch thatand, andfor.
Proof The proof is by induction. It is obvious that the lemma is true when .
Assume that the lemma is true for . Differentiating (25), we get
Now, we have two cases to consider.
Case (1): Suppose there exists k such that . Without loss of generality, let , then we can write
Since we assumed the lemma is true for , there exists such that . But this contradicts our assumption that are distinct.
Case (2): Suppose that for each j, i.e.,
If , then because otherwise we have
But this contradicts the fact that .
So, we have for . Thus, (25) becomes (for some k)
5 Proof of Theorem 2.2
We first note that, outside a set of finite measure, by Theorem 1.1,
In particular, if w has finitely many zeros, then v is a polynomial, which gives . This completes the proof of part (A) in the conclusion.
Assume henceforth that w has infinitely many zeros. Then (26) implies that , and so A is a polynomial of degree at most n by the WimanValiron theory [17]. Also, since has infinitely many zeros.
Now, two cases have to be considered.
Case (I). Assume that P is a nonzero constant; then and A is constant. Therefore, we can write
Since w and have the same zeros with finitely many exceptions, we can write
where is a rational function and is a polynomial. We know that because . We can now write
Now, by using Lemma 4.1, is constant and so we can write (28) as
where δ is constant.
Therefore,
Now, by using Lemma 4.1, we get
and , , β, −β, 0 cannot all be different.
We must now try six cases:
I(a): If and , then and . But this contradicts (30). Thus, this case cannot happen.
I(b): If and , then and . Substituting these in (30) gives
where , are constants, which yields . Putting this in (27) gives (7) with .
There are four more cases:
It is easy to check that case I(c) is impossible and cases I(d), I(e), I(f) all lead to (7) with .
From these cases, we find that , and so is nonconstant. Also, we have (7) and case (B) of the conclusion.
Case (II). Suppose that P is nonconstant. We will show that this leads to a contradiction. Let
where L is a rational function and Q is an entire function.
From (26), we have , and so Q is a polynomial.
Also, from (31), we have
So,
Now, we have two cases to consider.
Case (i): If M is constant, then either and , so that , which is a contradiction, or
is a rational function, which is a contradiction since w has infinitely many zeros.
Case (ii): If M is nonconstant, then . Therefore,
where is rational because is rational and is rational, and so is rational.
Also,
Then we can write (33) as
where
are rational functions and Q is a polynomial.
Let , then we can write (34) as
Now, we have two cases to consider:
Case ii(a): If in (35), then (34) gives
which is a contradiction since w has infinitely many zeros.
Case ii(b): Assume that in (35); then (35) gives
Now, (1) and (35) give
and so
Therefore,
because if not, w has finitely many zeros, a contradiction. Also,
Put
Then,
From (36), we get
So, G solves (1), and since and is a polynomial of degree n, we see that G is a transcendental entire function with finitely many zeros and has order .
Since w and G solve the same equation but w has infinitely many zeros and G has finitely many zeros, w and G are linearly independent, and we can write
where c is a nonzero constant. So,
By integrating, we get
Also, using (31), (38) and (39),
Now, we can assume that because if , we can multiply w by .
We differentiate (40) to get
where
is a rational function.
So, from (3) and (41), we get
and so
Since v is transcendental and , are rational functions, we must have
and
From (44), we have
From (42), we get
We integrate to get
where a is a constant. But this contradicts the fact that H and K are rational functions and G is a transcendental function. This completes the proof of Claim 5.1.
Once we have Claim 5.1, (41) gives
and so
By (45), v has finitely many zeros, so we can write
where , are polynomials, , and is nonconstant because v is transcendental.
Therefore,
Now, we can write this as
where , are rational functions and , are polynomials.
Here, and are linearly independent because is nonconstant. Now, we get
where , are rational and satisfy
Therefore, because otherwise or . Thus, , both solve and have finitely many zeros, and they are linearly independent.
Hence, P is constant by [9], which contradicts our assumption in Case (II) that P is nonconstant. □
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.
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