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Common zeros of the solutions of two differential equations with transcendental coefficients
Journal of Inequalities and Applications volume 2011, Article number: 134 (2011)
Abstract
Purpose
We consider a pair of homogeneous linear differential equations with transcendental entire coefficients of finite order, and the question of when solutions of these equations can have the same zeros or nearly the same zeros.
Method
We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.
Conclusion
The results determine all pairs of such equations having solutions with the same zeros, or nearly the same zeros.
1 Introduction
This paper continues our study from [1] of the question of when two linear differential equations in the complex domain can have solutions with (nearly) the same zeros. Our starting point is the equation
and [1] considered the case where P is a polynomial. In this paper, P will be a transcendental entire function, but we will retain the same notation since we will sometimes refer the reader to [1].
The following theorem was proved in 1955 by Wittich [2].
Theorem 1.1 If f ≢ 0 is a nontrivial solution of (1) and P is a transcendental entire function, then we have:

(i)
f has infinite order and T (r, P) = S (r, f).

(ii)
If a is a nonzero complex number, then f takes the value a infinitely often, and in fact
N\phantom{\rule{2.77695pt}{0ex}}\left(r,\phantom{\rule{2.77695pt}{0ex}}\frac{1}{fa}\right)=T\left(r,\phantom{\rule{2.77695pt}{0ex}}f\right)+S\left(r,\phantom{\rule{2.77695pt}{0ex}}f\right).
In this paper, we use standard notation of Nevanlinna theory from [3]. The reader is referred to the books of Hille [4] and Laine [5], the keynote paper [6], and to [7–14] for comprehensive results on the zeros of solutions of linear differential equations with entire coefficients.
The following lemma from [1] is needed in order to state our main results: the proof is by induction as in [1].
Lemma 1.1 Suppose w" = Pw where P is an entire function with order of growth ρ (P) < ∞. Then for j ≥ 0, there exist entire functions Q_{ j } and R_{ j } of finite order such that
In particular,
Now, our first main result can be stated.
Theorem 1.2 Let P be a transcendental entire function with ρ (P) < ∞. Let w ≢ 0 be a solution of (1). Assume that the zeros of w have infinite exponent of convergence, i.e.
Let v ≢ 0 be entire solution of the differential equation
such that A and B_{ j } are entire functions and ρ (A) < ∞ and ρ (B_{ j }) < ∞. Assume that N (r) has finite order, where N (r) counts both zeros and poles of \frac{v}{w}. Let
Then, one of the following two possibilities holds.
(a) L is constant, and
where Q_{ k } and Q_{ j } are deffined by Lemma 1.1.
(b) L is not constant, but L satisfies
and A satisfies
where R_{k  m}, R_{j  m}, Q_{k  m}, and Q_{ j  m }are also defined by Lemma 1.1.
As in [1], we note that there is no loss of generality in assuming that there is no term in w' in (1) and that there is no term B_{k  1}in (5), since we are considering zeros of solutions.
We also note that the hypothesis (4) is not redundant: to see this, let w = e^{B} and v = e^{C} where B and C are any entire functions of finite order. Then, w and v solve
where P =  (B" + B'^{2}) and Q =  v^{(k)}/v =  (C')^{k} + ... are entire of finite order, but since B and C are arbitrary, there is no relationship between P and Q.
The following results for the cases k = 2, k = 3 and k = 4 will be deduced from Theorem 1.2.
Theorem 1.3 Let P be a transcendental entire function and ρ (P) < ∞. Let w ≢ 0 be a solution of (1). Assume that the zeros of w have infinite exponent of convergence, i.e. (4) holds. Let v ≢ 0 be entire solution of the differential equation
such that A is an entire function and ρ (A) < ∞. Assume that N (r) has finite order, where N (r) counts zeros and poles of \frac{v}{w}. Then, L=\frac{v}{w} is a constant and A = P.
Example 1.1 This example shows that ρ (A) < ∞ is vital in Theorem 1.3. To show this, let v = we^{g} where g is an entire function. Then, we get
Now, by putting g' = w, we obtain
Thus, A is entire, \frac{v}{w} is non constant and v has the same zeros as w.
Theorem 1.4 Let P be a transcendental entire function and ρ(P) < ∞. Let w ≢ 0 be a solution of (1). Assume that the zeros of w have infinite exponent of convergence, i.e. (4) holds. Let v ≢ 0 be entire solution of the differential equation
such that A and B are entire functions with ρ (A) < ∞ and ρ (B) < ∞. Assume that N (r) has finite order, where N (r) counts zeros and poles of \frac{v}{w}. Then, v = Lw and one of the following holds.
(a) L is constant and A = P', B = P.
(b) L is non constant and {L}^{\u2033}=\frac{1}{3}PL\frac{1}{3}BL and
Remark 1.1 If B ≡ 0 in Theorem 1.4 then case (a) cannot hold since P is transcendental.
Example 1.2 In this example, we show that case (b) can occur in Theorem 1.4. To show this, we can use Example 2.3 from [1] with Q a transcendental entire function of finite order.
Let L = e^{Q} and set
If w solves (1), then v = Lw satisffies
and so v solves v‴ + Av = 0 with
It should be noted that P is transcendental since Q is transcendental and in fact writing
shows that
therefore,
Also, P and A have finite order.
Theorem 1.5 Let P be a transcendental entire function and ρ (P) < ∞. Let w ≢ 0 be a solution of (1). Assume that the zeros of w have infinite exponent of convergence, i.e. (4) holds. Let v ≢ 0 be entire solution of the differential equation
such that A is an entire function with ρ (A) < ∞. Assume that N (r) has finite order, where N (r) counts zeros and poles of \frac{v}{w}. Then, v = Lw where L is nonconstatnt,
and L = y_{1} y_{2} where y_{l}, y_{2} are solutions of
In particular, if v and w have the same zeros with the same multiplicities, then L is entire with no zeros and so are y_{1} and y_{2}. In addition, when v and w have the same zeros:

(i)
if y_{l}, y_{2} are linearly dependent then L = e^{2C}with C an entire function, P = 4 (C" + C'^{2}) and A is a differential polynomial in C';

(ii)
if y_{l}, y_{2} are linearly independent then L = e^{C} with C an entire function, P = 2C" + C'^{2} + k^{2}e^{2C}where k is a constant and A is a differential polynomial in e^{C}and C'.
Example 1.3 To show that (13) can occur, let L = Y^{2} = e^{Q} where Q is a transcendental entire function of finite order and set
so that P is an entire function of finite order, and the same argument as in Example 1.2 shows that P is transcendental. Then, as in Example 2.6 of [1],
If w solves (1) then v = Lw satisfies
and so v solves (12) with
which is also entire of finite order.
2 Proof of Theorem 1.2
In this proof, we use M_{1}, M_{2}, ... to denote positive constants.
Claim 1: We claim that w has simple zeros and
This holds by the existenceuniqueness theorem [4].
Lemma 2.1 There exists N > 0 such that m\left(r,\frac{{w}^{\prime}}{w}\right)<{r}^{N} as r→ ∞.
Proof: We can get this as follows. Use N_{1}, N_{2}, ... to denote positive constants. Since A has finite order,
So for r outside a set E_{0} of finite logarithmic measure, we get from WimanValiron theory [5]
where v(r, w) denotes the central index, and so
Hence, by [5, Lemma 1.1.2] and the fact that v(r, w) is nondecreasing,
So the maximum term μ(r, w) satisfies
and so
Now, we can use Lemma 2.3 in [[3], p. 36] with R = 2r to get
This completes the proof of this lemma.
From (1) we have (2). From (2), (6) and by using Leibniz' rule, we get, for 1 ≤ j ≤ k,
From (5), (14) and the fact that \left(\begin{array}{c}\hfill j\hfill \\ \hfill m\hfill \end{array}\right)=0 when j < m ≤ k, we find that
Then, we have
Now, there are three cases which should be considered.
Case (I): If L is a constant, then w solves (5) and, by using Lemma 1.1, we obtain the following equations
By adding these two equations and using (2) again, we get
Therefore, we can write
Now, if {R}_{k}+{\sum}_{1\le j\le k2}{B}_{j}{R}_{j}\equiv 0, then A+{Q}_{k}+{\sum}_{1\le j\le k2}{B}_{j}{Q}_{j}\equiv 0 and so we have (7) and conclusion (a).
Suppose next that {R}_{k}+{\sum}_{1\le j\le k2}{B}_{j}{R}_{j}\not\equiv 0; then
Recall that all zeros of w are simple. We deduce that
But this contradicts (4).
Case (II): If L is not constant and (8) holds, then from (15), we get (9) and conclusion (b) of the theorem.
It remains only to show that the following case cannot occur.
Case (III): Suppose that L is not constant and (8) does not hold and let S = L'/L.
We first compare N(r, S) with N(r). Recall that all zeros of w are simple. On the other hand, v solves a differential equation of order k. So, zeros of v have multiplicities less than or equal to k  1.
So, L=\frac{v}{w} has zeros with multiplicities at most k  1 and has simple poles. Then, we get
Claim 2: We claim that
for r outside a set E of finite linear measure.
To prove this, we use the fact that Q_{0} = 1 and R_{0} = 0 in Lemma 1.1 to write (15) in the form
We can write, for 1 ≤ m ≤ k,
where U_{m1}(S) is a polynomial in S, S', S″,..., S^{(k)}with constant coefficients and total degree at most m  1. This follows immediately from Lemma 3.5 in [3] and can be easily proved by induction.
This gives us an integer q > 0 such that (17) may be written as
where i_{μ, j} ≥ 0 are integers and
for each j. Lemma 2.1 gives m\left(r,\frac{{w}^{\prime}}{w}\right)<{r}^{{M}_{3}}. Also, a_{ j } and b_{ j } are polynomials in A, B_{ μ }, Q_{ μ } and R_{ μ }, and so satisfy
By Clunie's lemma [[5], p. 39], we should have
for r outside a set E of finite linear measure.
Now, we use (16) and (19) to obtain
and so
for r outside a set E of finite linear measure. This proves Claim 2.
Claim 3: We claim that
This follows from Claim 1, [5, Lemma 1.1.1] and the fact that T(r, S) is nondecreasing.
Now, dividing (15) by L shows that if at z the function \frac{{w}^{\prime}}{w} has a pole, then either
or
Using Claim 3, we can write (15) as
Where T\left(r,{A}_{j}\right)=O\left({r}^{{M}_{7}}\right), j = 1, 2 and A_{2} ≢ 0 by the assumption of Case (III).
Now, by using Claim 1 and (20), we get
So, N\left(r,\frac{1}{w}\right) has finite order. But this contradicts (4). Hence, Case (III) cannot occur.
3 Proof of Theorem 1.3
Assume the hypotheses of Theorem 1.3. Taking k = 2 in Theorem 1.2, two cases have to be considered as in [1].
In case (a): L is a constant and A = P by (7) and Lemma 1.1.
In case (b): L is not constant, but (8) and Lemma 1.1 give
But this requires that L should be constant, a contradiction.
4 Proof of Theorem 1.4
Assume the hypotheses of Theorem 1.4. Taking k = 3 and B_{1} = B in Theorem 1.2, we have two cases to consider as in [1].
In case (a), L is a constant, and (7) and Lemma 1.1 give A = P'. But, since w solves (1) and (11), we have
which gives P = B.
In case (b), L is not constant and, using (8) and Lemma 1.1,
and therefore
Differentiating (21), we obtain
Also, we have, using (22),
5 Proof of Theorem 1.5
We will need the following lemma, which is due to Bank and Laine [6, 8].
Lemma 5.1 Let B an entire function. Then, every solution of the equation
is of the form u = y_{1}y_{2}, where y_{1}, y_{2} are solutions (possibly linearly dependent) of

If y_{1}, y_{2}are linearly dependent, then u = y^{2}with y a solution of (24) and
B=\frac{{y}^{\u2033}}{y}=\frac{1}{4}{\left(\frac{{u}^{\prime}}{u}\right)}^{2}\frac{1}{2}\frac{{u}^{\u2033}}{u}. 
If y_{1}, y_{2}are linearly independent, then
4B={\left(\frac{{u}^{\prime}}{u}\right)}^{2}2\frac{{u}^{\u2033}}{u}\frac{{k}^{2}}{{u}^{2}},(25)
where k = W(y_{1}, y_{2}).
We remark that (25) is the well known BankLaine product formula [6].
Now, assume the hypotheses of Theorem 1.5. Taking k = 4 and B_{1} = B_{2} = 0 in Theorem 1.2, there are two cases have to be considered.
Case (a): L is a constant and, by using (3), we have
But, differentiating (1) two times gives
.
Since we also have w^{(4)} + Aw = 0, this gives
So P must be constant, but this contradicts our assumption that P is transcendental.
Hence, Case (a) cannot occur.
Case (b): L is nonconstant and L satisfies, using (3),
and therefore
Since this is a linear differential equation and P is an entire function, it follows that L is an entire function.
By using (3) and (9), we obtain
Differentiating (26) and dividing by L gives
By substituting this in (27), we get
Therefore, (26) and (28) prove (13).
Suppose that w, v have the same zeros. Then, L has no zeros and poles.
Now, we set B=\frac{P}{4} in (26) and apply Lemma 5.1. Then, L = y_{1}y_{2} where y_{1}, y_{2} are solutions of (24).
Then, we have the following two cases:
Case 1: If y_{1}, y_{2} are linearly dependent, then L = y^{2} with y a solution of (24) and
Let L = e^{2C}with C an entire function. Then,
Substituting these in (28) shows that A is a differential polynomial in C'.
Case 2: If y_{1}, y_{2} are linearly independent, then
where k = W(y_{1}, y_{2}). Also, L is not a polynomial, since P (∞) ≠ 0.
Let L = e^{C} with C an entire function. Then,
Substituting these in (28) shows that A is a differential polynomial in e^{C}and C'.
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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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Asiri, A. Common zeros of the solutions of two differential equations with transcendental coefficients. J Inequal Appl 2011, 134 (2011). https://doi.org/10.1186/1029242X2011134
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DOI: https://doi.org/10.1186/1029242X2011134