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.
We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.
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
There has been much research [1-8] on zeros of solutions of linear differential equations with entire coefficients. The principal paper  that was published in 1982 by Bank and Laine has stimulated many studies on this kind of problems. The reader is referred to [10-12] for background on some applications of the Nevanlinna theory. We use the standard notation of the Nevanlinna theory from .
In 1955, Wittich  proved the following theorem.
Theorem 1.1Iffis a non-trivial solution of , i.e., and is entire, then we have:
(ii) Iffhas finite order, thenAis a polynomial.
(iii) Ifais a non-zero 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 non-trivial 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  studied homogeneous linear differential equations having solutions with nearly the same zeros and proved several results, including the following.
Let be a polynomial of degreen. Let be a solution of (1). Assume thatwhas infinitely many zeros. Suppose that we have a solution of the differential equation
such thatAis an entire function and counts zeros ofvwhich are not zeros ofwand zeros ofwwhich are not zeros ofv. Assume that
Then is a constant and .
Recently, the same problem, but with non-homogeneous first-order differential equations, has been studied in , including the following result.
Assume that and , 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 .
(i) andv, whave no zeros.
(ii) and , are non-zero constants, and
where , and .
In this paper, our first result (Theorem 2.1) looks at the same problem but with one homogeneous and one non-homogeneous 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 2.2) studies the case where the solutions of two second-order homogeneous differential equations can take the value 0 and a non-zero value at (nearly) the same points.
2 Our results
Our first result is the following theorem.
Theorem 2.1Suppose that is a polynomial of degreen, andwsolves (1), and has infinitely many zeros. Suppose that solves
whereA, Bare entire and , and
Suppose that has finitely many zeros and poles (i.e., wandvhave the same zeros with finitely many exceptions).
ThenAis a polynomial and there exists a polynomialQsuch that
where and , are constants.
Example 2.1 Take Q to be a polynomial. Let
So, we have .
Now, let be another polynomial, and let
Note that v has the same zeros as w. Now, we have
We choose so that
For example, let .
We now state our second result.
Theorem 2.2Suppose that is a polynomial of degreen, andAis an entire function, and suppose thatwsolves (1) andvsolves (3), and . Let andwhave, 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 non-zero constants and is non-constant and
whereσ, , , are non-zero constants.
Example 2.2 If and , then
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
Then , where .
Also , where .
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
We also have, using (1), (5), (8), (9) and (10),
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 . Therefore,
Claim 3.1We claim that .
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
Therefore, we get
We use Lemma 2.3 in [, p.38] with to get
Now, we have . So
We also have .
This completes the proof of Claim 3.1.
Using Claim 3.1 and (14), we get
Also, by Theorem 1.2, we get
Therefore, we must have
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
where , are polynomials.
But then, we can write
where is rational.
Then we also can write
where is rational and .
Substitute (18) in (16), we obtain
So, we get
Substituting (21) in (19), we obtain
Thus, , H is entire, and B has no zeros.
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
where is a constant and Q is a polynomial.
Now, we have
So, we can write
Therefore, we have
Hence, using (22), we obtain
where is a constant and .
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
Moreover, solves , and so
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 4.1Let be distinct, and let be rational functions such that
Then there exists such that and , and for .
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)
and and . □
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 Wiman-Valiron theory . Also, since has infinitely many zeros.
Now, two cases have to be considered.
Case (I). Assume that P is a non-zero constant; then and A is constant. Therefore, we can write
where , and ( ).
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
where , and so
Now, by using Lemma 4.1, is constant and so we can write (28) as
where δ is constant.
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:
I(c): and .
I(d): and .
I(e): and .
I(f): and .
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 non-constant. Also, we have (7) and case (B) of the conclusion.
Case (II). Suppose that P is non-constant. 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
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 non-constant, then . Therefore,
where is rational because is rational and is rational, and so is rational.
Then we can write (33) as
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
because if not, w has finitely many zeros, a contradiction. Also,
where is a rational function.
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 non-zero constant. So,
By integrating, we get
Also, using (31), (38) and (39),
where is a rational function.
Now, we can assume that because if , we can multiply w by .
We differentiate (40) to get
is a rational function.
So, from (3) and (41), we get
where and .
Since v is transcendental and , are rational functions, we must have
Claim 5.1We claim that .
To show this, let .
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
By (45), v has finitely many zeros, so we can write
where , are polynomials, , and is non-constant because v is transcendental.
Now, we can write this as
where , are rational functions and , are polynomials.
Here, and are linearly independent because is non-constant. 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 , which contradicts our assumption in Case (II) that P is non-constant. □
The author declares that he has no competing interests.
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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