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Common zeros of the solutions of two differential equations with transcendental coefficients

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Research

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.

Keywords

Nevanlinna theory differential equations 

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
w + P w = 0 , Open image in new window
(1)

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 non-trivial solution of (1) and P is a transcendental entire function, then we have:
  1. (i)

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

     
  2. (ii)
    If a is a non-zero complex number, then f takes the value a infinitely often, and in fact
    N r , 1 f - a = T ( r , f ) + S ( r , f ) . Open image in new window
     

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, 8, 9, 10, 11, 12, 13, 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
w ( j ) = Q j w + R j w . Open image in new window
(2)
In particular,
Q 0 = 1 , Q 1 = 0 , Q 2 = - P , Q 3 = - P , Q 4 = P 2 - P , R 0 = 0 , R 1 = 1 , R 2 = 0 , R 3 = - P , R 4 = - 2 P . Open image in new window
(3)

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.
λ ( w ) = limsup r log + N r , 1 w log r = . Open image in new window
(4)
Let v ≢ 0 be entire solution of the differential equation
v ( k ) + 1 j k - 2 B j v ( j ) + A v = 0 , k 2 , Open image in new window
(5)
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 v w Open image in new window. Let

Then, one of the following two possibilities holds.

(a) L is constant, and
A = - Q k - j = 1 k - 2 B j Q j , Open image in new window
(7)

where Q k and Q j are deffined by Lemma 1.1.

(b) L is not constant, but L satisfies
m = 0 k k m L ( m ) R k - m + j = 1 k - 2 j m B j L ( m ) R j - m = 0 , Open image in new window
(8)
and A satisfies
A = - m = 0 k k m L ( m ) L Q k - m + j = 1 k - 2 j m B j L ( m ) L Q j - m , Open image in new window
(9)

where Rk - m, Rj - m, Qk - 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 Bk - 1in (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
w + P w = 0 , v ( k ) + Q v = 0 , Open image in new window

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
v + A v = 0 , Open image in new window
(10)

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 v w Open image in new window. Then, L = v w Open image in new window 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
- A = v v = w w + 2 g w w + g + g 2 = - P + g + g 2 + 2 g w w . Open image in new window
Now, by putting g' = w, we obtain
- A = - P + w + w 2 + 2 w . Open image in new window

Thus, A is entire, v w Open image in new window 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
v + B v + A v = 0 , Open image in new window
(11)

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 v w Open image in new window. 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 = 1 3 P L - 1 3 B L Open image in new window and
A = 8 3 P L L + 2 3 P + 1 3 B - 2 3 B L L . Open image in new window

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
P 3 = L L = Q 2 + Q . Open image in new window
If w solves (1), then v = Lw satisffies
v = L ( - P w - P w ) + 3 L ( - P w ) + 3 P 3 L w + P 3 L + P 3 L w Open image in new window
and so v solves v‴ + Av = 0 with
A = 2 3 P + 8 3 P Q . Open image in new window
It should be noted that P is transcendental since Q is transcendental and in fact writing
Q = P 3 Q - Q Q Open image in new window
shows that
m ( r , Q ) m ( r , P ) + S ( r , Q ) , Open image in new window
therefore,
ρ ( P ) = ρ ( Q ) = ρ ( Q ) . Open image in new window

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
v ( 4 ) + A v = 0 , Open image in new window
(12)
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 v w Open image in new window. Then, v = Lw where L is non-constatnt,
L = L P + 1 2 L P , A = 5 P L L + 5 2 P L L + 1 2 P - P 2 Open image in new window
(13)
and L = y1 y2 where yl, y2 are solutions of
y - P 4 y = 0 . Open image in new window
In particular, if v and w have the same zeros with the same multiplicities, then L is entire with no zeros and so are y1 and y2. In addition, when v and w have the same zeros:
  1. (i)

    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';

     
  2. (ii)

    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'.

     
Example 1.3 To show that (13) can occur, let L = Y2 = e Q where Q is a transcendental entire function of finite order and set
Q = S = 2 y = 2 Y Y , P = S 2 + 2 S = 4 ( y 2 + y ) Open image in new window
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],
L = 2 Y Y , L = 2 Y 2 + 2 Y Y = 2 Y 2 + P 2 L , L = P L + P 2 L , Open image in new window
If w solves (1) then v = Lw satisfies
v ( 4 ) = L ( ( P 2 - P ) w - 2 P w ) + 4 L ( - P w - P w ) + 6 L ( - P w ) + 4 P L + P 2 L w + P L + 3 2 P L + P 2 L w Open image in new window
and so v solves (12) with
A = 5 P ( Q 2 + Q ) + 5 2 P Q + 1 2 P - P 2 , Open image in new window

which is also entire of finite order.

2 Proof of Theorem 1.2

In this proof, we use M1, M2, ... to denote positive constants.

Claim 1: We claim that w has simple zeros and
N r , 1 w = N r , w w . Open image in new window

This holds by the existence-uniqueness theorem [4].

Lemma 2.1 There exists N > 0 such that m ( r , w w ) < r N Open image in new window as r→ ∞.

Proof: We can get this as follows. Use N1, N2, ... to denote positive constants. Since A has finite order,
M ( r , A ) exp ( r N 1 ) as  r . Open image in new window
So for r outside a set E0 of finite logarithmic measure, we get from Wiman-Valiron theory [5]
ν ( r , w ) r 2 ( 1 + o ( 1 ) ) M ( r , A ) , Open image in new window
where v(r, w) denotes the central index, and so
ν ( r , w ) exp ( r N 2 ) ( r , r E 0 ) Open image in new window
Hence, by [5, Lemma 1.1.2] and the fact that v(r, w) is non-decreasing,
ν ( r , w ) exp ( r N 3 ) as  r , for all  r . Open image in new window
So the maximum term μ(r, w) satisfies
log μ ( r , w ) exp ( r N 4 ) Open image in new window
and so
T ( r , w ) log M ( r , w ) exp ( r N 5 ) . Open image in new window
Now, we can use Lemma 2.3 in [[3], p. 36] with R = 2r to get
m r , w w O ( log + T ( R , w ) + log r ) r N 6 . Open image in new window

This completes the proof of this lemma.

From (1) we have (2). From (2), (6) and by using Leibniz' rule, we get, for 1 ≤ jk,
v ( j ) = m = 0 j j m L ( m ) ( Q j - m w + R j - m w ) = m = 0 j j m L ( m ) Q j - m w + m = 0 j j m L ( m ) R j - m w . Open image in new window
(14)
From (5), (14) and the fact that j m = 0 Open image in new window when j < mk, we find that
- A L w = - A v = v ( k ) + 1 j k - 2 B j v ( j ) = m = 0 k k m L ( m ) Q k - m w + m = 0 k k m L ( m ) R k - m w + j = 1 k - 2 B j m = 0 j j m L ( m ) Q j - m w + m = 0 j j m L ( m ) R j - m w . = m = 0 k k m L ( m ) Q k - m + j = 1 k - 2 B j m = 0 j j m L ( m ) Q j - m w + m = 0 k k m L ( m ) R k - m + j = 1 k - 2 B j m = 0 j j m L ( m ) R j - m w = m = 0 k k m L ( m ) Q k - m + j = 1 k - 2 j m B j L ( m ) Q j - m w + m = 0 k k m L ( m ) R k - m + j = 1 k - 2 j m B j L ( m ) R j - m w . Open image in new window
Then, we have
0 = m = 0 k k m L ( m ) Q k - m + j = 1 k - 2 j m B j L ( m ) Q j - m + A L + m = 0 k k m L ( m ) R k - m + j = 1 k - 2 j m B j L ( m ) R j - m w w . Open image in new window
(15)

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
{ w ( k ) + 1 j k 2 B j w ( j ) + A w = 0 , w ( k ) + Q k w + R k w ' = 0. Open image in new window
By adding these two equations and using (2) again, we get
0 = 1 j k - 2 B j ( Q j w + R j w ) + A w + Q k w + R k w . Open image in new window
Therefore, we can write
R k + 1 j k - 2 B j R j w w + A + Q k + 1 j k - 2 B j Q j = 0 . Open image in new window

Now, if R k + 1 j k - 2 B j R j 0 Open image in new window, then A + Q k + 1 j k - 2 B j Q j 0 Open image in new window and so we have (7) and conclusion (a).

Suppose next that R k + 1 j k - 2 B j R j 0 Open image in new window; then
w = 0 w w = R k + 1 j k - 2 B j R j = 0 . Open image in new window
Recall that all zeros of w are simple. We deduce that
N ( r , 1 w ) N r , 1 R k + 1 j k - 2 B j R j = O ( r M 1 ) . Open image in new window

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 = v w Open image in new window has zeros with multiplicities at most k - 1 and has simple poles. Then, we get
N ( r , S ) N r , 1 L + N ( r , L ) = N ( r ) Open image in new window
(16)
Claim 2: We claim that
T ( r , S ) O ( r M 2 ) Open image in new window

for r outside a set E of finite linear measure.

To prove this, we use the fact that Q0 = 1 and R0 = 0 in Lemma 1.1 to write (15) in the form
0 = L ( k ) L + A + m = 0 k - 1 L ( m ) L k m Q k - m + R k - m w w + j = 1 k - 2 j m B j Q j - m + R j - m w w . Open image in new window
(17)
We can write, for 1 ≤ mk,
L ( m ) L = S m + U m - 1 ( S ) , Open image in new window

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 [3] and can be easily proved by induction.

This gives us an integer q > 0 such that (17) may be written as
S k = j = 0 q a j + b j w w S i 0 , j ( S ) i 1 , j ( S ) i 2 , j ( S ( k ) ) i k , j , Open image in new window
(18)
where iμ, j ≥ 0 are integers and
μ = 0 k i μ , j k - 1 Open image in new window
for each j. Lemma 2.1 gives m ( r , w w ) < r M 3 Open image in new window. Also, a j and b j are polynomials in A, B μ , Q μ and R μ , and so satisfy
m ( r , a j ) + m ( r , b j ) = O ( r M 4 ) as  r , Open image in new window
By Clunie's lemma [[5], p. 39], we should have
m ( r , S ) O ( r M 4 ) + O ( log + T ( r , S ) ) Open image in new window
(19)

for r outside a set E of finite linear measure.

Now, we use (16) and (19) to obtain
T ( r , S ) = N ( r , S ) + m ( r , S ) N ( r ) + m ( r , S ) O ( r M 5 ) + O ( log + T ( r , S ) ) Open image in new window
and so
T ( r , S ) = O ( r M 6 ) Open image in new window

for r outside a set E of finite linear measure. This proves Claim 2.

Claim 3: We claim that
T ( r , S ) O ( r M 6 ) , for all large  r . Open image in new window

This follows from Claim 1, [5, Lemma 1.1.1] and the fact that T(r, S) is non-decreasing.

Now, dividing (15) by L shows that if at z the function w w Open image in new window has a pole, then either
A 2 = m = 0 k k m L ( m ) L R k - m + j = 1 k - 2 j m B j L ( m ) L R j - m = 0 Open image in new window
or
A 1 = m = 0 k k m L ( m ) L Q k - m + j = 1 k - 2 j m B j L ( m ) L Q j - m + A = . Open image in new window
Using Claim 3, we can write (15) as
A 1 + A 2 w w = 0 , Open image in new window
(20)

Where T ( r , A j ) = O ( r M 7 ) Open image in new window, j = 1, 2 and A2 ≢ 0 by the assumption of Case (III).

Now, by using Claim 1 and (20), we get
N r , 1 w = N r , w w N r , 1 A 2 + N ( r , A 1 ) = O ( r M 8 ) . Open image in new window

So, N ( r , 1 w ) Open image in new window 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
0 = m = 0 2 2 m L ( m ) R 2 - m = L R 2 + 2 L R 1 + L R 0 = 2 L . Open image in new window

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 [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
0 = w + B w + A w = w + B w + P w = w + P w + P w , Open image in new window

which gives P = B.

In case (b), L is not constant and, using (8) and Lemma 1.1,
0 = m = 0 3 3 m L ( m ) R 3 - m + 1 m B 1 L ( m ) R 1 - m = - P L + B L + 3 L Open image in new window
and therefore
L = 1 3 P L - 1 3 B L . Open image in new window
(21)
Differentiating (21), we obtain
L = 1 3 P L + 1 3 P L - 1 3 B L - 1 3 B L . Open image in new window
(22)
Also, we have, using (22),
A = - m = 0 3 3 m L ( m ) L Q 3 - m + 1 m B 1 L ( m ) L Q 1 - m = P + 3 P L L - B L L - L L = P + 3 P L L - B L L - 1 3 P - 1 3 P L L + 1 3 B + 1 3 B L L = 8 3 P L L + 2 3 P + 1 3 B - 2 3 B L L . Open image in new window

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
u + 4 B u + 2 B u = 0 Open image in new window
(23)
is of the form u = y1y2, where y1, y2 are solutions (possibly linearly dependent) of
y + B y = 0 . Open image in new window
(24)
  • If y1, y2are linearly dependent, then u = y2with y a solution of (24) and
    B = - y y = 1 4 u u 2 - 1 2 u u . Open image in new window
  • If y1, y2are linearly independent, then
    4 B = u u 2 - 2 u u - k 2 u 2 , Open image in new window
    (25)

where k = W(y1, y2).

We remark that (25) is the well known Bank-Laine product formula [6].

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.

Case (a): L is a constant and, by using (3), we have
A = - Q 4 = - P 2 + P . Open image in new window
But, differentiating (1) two times gives
0 = w ( 4 ) + ( P - P 2 ) w + 2 P w Open image in new window
.
Since we also have w(4) + Aw = 0, this gives
0 = 2 P w . Open image in new window

So P must be constant, but this contradicts our assumption that P is transcendental.

Hence, Case (a) cannot occur.

Case (b): L is non-constant and L satisfies, using (3),
0 = m = 0 4 4 m L ( m ) R 4 - m = 4 L - 4 L P - 2 L P Open image in new window
and therefore
L = L P + 1 2 L P . Open image in new window
(26)

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
A = - m = 0 4 4 m L ( m ) L Q 4 - m = - P 2 + P + 4 P L L + 6 P L L - L ( 4 ) L . Open image in new window
(27)
Differentiating (26) and dividing by L gives
L ( 4 ) L = P L L + 3 2 P L L + 1 2 P . Open image in new window
By substituting this in (27), we get
A = - P 2 + P + 4 P L L + 6 P L L - P L L - 3 2 P L L - 1 2 P = 5 P L L + 5 2 P L L + 1 2 P - P 2 . Open image in new window
(28)

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 = - P 4 Open image in new window in (26) and apply Lemma 5.1. Then, L = y1y2 where y1, y2 are solutions of (24).

Then, we have the following two cases:

Case 1: If y1, y2 are linearly dependent, then L = y2 with y a solution of (24) and
- P 4 = 1 4 L L 2 - 1 2 L L , P = - L L 2 + 2 L L . Open image in new window
Let L = e2Cwith C an entire function. Then,
P = - ( 2 C ) 2 + 2 ( 2 C + 4 C 2 ) = 4 ( C 2 + C ) . Open image in new window

Substituting these in (28) shows that A is a differential polynomial in C'.

Case 2: If y1, y2 are linearly independent, then
P = - L L 2 + 2 L L + k 2 L 2 , Open image in new window

where k = W(y1, y2). Also, L is not a polynomial, since P (∞) ≠ 0.

Let L = e C with C an entire function. Then,
P = - C 2 + 2 C + 2 C 2 + k 2 e - 2 C = 2 C + C 2 + k 2 e - 2 C . Open image in new window

Substituting these in (28) shows that A is a differential polynomial in e-Cand C'.

Notes

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; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of Mathematics, Faculty of EducationKing Abdulaziz UniversityJeddahSaudi Arabia
  2. 2.School of Mathematical SciencesUniversity of NottinghamNottinghamUK

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