Common zeros of the solutions of two differential equations with transcendental coefficients
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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.
We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.
The results determine all pairs of such equations having solutions with the same zeros, or nearly the same zeros.
KeywordsNevanlinna theory differential equations
The following theorem was proved in 1955 by Wittich .
f has infinite order and T (r, P) = S (r, f).
- (ii)If a is a non-zero complex number, then f takes the value a infinitely often, and in fact
In this paper, we use standard notation of Nevanlinna theory from . The reader is referred to the books of Hille  and Laine , the keynote paper , and to [7, 8, 9, 10, 11, 12, 13, 14] for comprehensive results on the zeros of solutions of linear differential equations with entire coefficients.
Now, our first main result can be stated.
Then, one of the following two possibilities holds.
where Q k and Q j are deffined by Lemma 1.1.
where Rk - m, Rj - m, Qk - m, and Q j - m are also defined by Lemma 1.1.
As in , 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 Bk - 1in (5), since we are considering zeros of solutions.
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.
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 . Then, is a constant and A = P.
Thus, A is entire, is non constant and v has the same zeros as w.
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 . Then, v = Lw and one of the following holds.
(a) L is constant and A = P', B = P.
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  with Q a transcendental entire function of finite order.
Also, P and A have finite order.
if yl, y2 are linearly dependent then L = e2Cwith C an entire function, P = 4 (C" + C'2) and A is a differential polynomial in C';
if yl, y2 are linearly independent then L = e C with C an entire function, P = 2C" + C'2 + k2e-2Cwhere k is a constant and A is a differential polynomial in e-Cand C'.
which is also entire of finite order.
2 Proof of Theorem 1.2
In this proof, we use M1, M2, ... to denote positive constants.
This holds by the existence-uniqueness theorem .
Lemma 2.1 There exists N > 0 such that as r→ ∞.
This completes the proof of this lemma.
Now, there are three cases which should be considered.
Now, if , then and so we have (7) and conclusion (a).
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.
for r outside a set E of finite linear measure.
where Um-1(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  and can be easily proved by induction.
for r outside a set E of finite linear measure.
for r outside a set E of finite linear measure. This proves Claim 2.
This follows from Claim 1, [5, Lemma 1.1.1] and the fact that T(r, S) is non-decreasing.
Where , j = 1, 2 and A2 ≢ 0 by the assumption of Case (III).
So, 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 .
In case (a): L is a constant and A = P by (7) and Lemma 1.1.
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 B1 = B in Theorem 1.2, we have two cases to consider as in .
which gives P = B.
5 Proof of Theorem 1.5
- If y1, y2are linearly dependent, then u = y2with y a solution of (24) and
- If y1, y2are linearly independent, then(25)
where k = W(y1, y2).
We remark that (25) is the well known Bank-Laine product formula .
Now, assume the hypotheses of Theorem 1.5. Taking k = 4 and B1 = B2 = 0 in Theorem 1.2, there are two cases have to be considered.
So P must be constant, but this contradicts our assumption that P is transcendental.
Hence, Case (a) cannot occur.
Since this is a linear differential equation and P is an entire function, it follows that L is an entire function.
Therefore, (26) and (28) prove (13).
Suppose that w, v have the same zeros. Then, L has no zeros and poles.
Now, we set in (26) and apply Lemma 5.1. Then, L = y1y2 where y1, y2 are solutions of (24).
Then, we have the following two cases:
Substituting these in (28) shows that A is a differential polynomial in C'.
where k = W(y1, y2). Also, L is not a polynomial, since P (∞) ≠ 0.
Substituting these in (28) shows that A is a differential polynomial in e-Cand C'.
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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