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

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.

There has been much research [1-8] 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 [10-12] 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 non-trivial solution of, i.e., andis entire, then we have:

(i) .

(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

Theorem 1.2[9,11]

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 [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

Thenis a constant and.

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 non-homogeneous first-order 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.

(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.

Our first result is the following theorem.

Theorem 2.1Suppose thatis a polynomial of degreen, andwsolves (1), andhas infinitely many zeros. Suppose thatsolves

whereA, Bare entire and, and

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

whereand, are constants.

Example 2.1 Take Q to be a polynomial. Let

Then

and

So, we have .

Now, let be another polynomial, and let

Note that v has the same zeros as w. Now, we have

where .

We choose so that

For example, let .

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 non-zero constants andis 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 .

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,

Claim 3.1We claim that.

To show this, we know that since M has finitely many poles.

Write (5) as

where .

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

Now, we have . So

We also have .

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

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

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

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.  □

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)

and and . □

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 [17]. 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.

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:

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

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 non-constant, 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

where is a rational function.

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 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

where

is a rational function.

So, from (3) and (41), we get

and so

where and .

Since v is transcendental and , are rational functions, we must have

and

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

and so

By (45), v has finitely many zeros, so we can write

where , are polynomials, , and is non-constant 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 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 [9], 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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