In this paper, we examine the asymptotic form of two linearly independant solutions of the general second-order differential equation.

( p y ′ ) ′ + q y ′ + r y = 0 , (1)

as x → ∞ , where x is the independant variable and the prime denotes d d x .

The coefficients p,q and r are nowhere zero in some interval [ a , ∞ ) . We shall consider the situation where p and r are small compared to q see (15) to identify the following case:

q ′ q = o ( r q ) ,             ( x → ∞ ) (2)

and under (2) we shall obtain the forms of the asymptotic solutions for (1) as x → ∞ which is given in Theorem 1.

If p = 1 , then (1) reduces to the differential equation considered by Walker [1] . We do not investigate the case where q 2 = o ( p r ) , the analysis for this case is already known for the Sturm-Liouville equation

( p y ′ ) ′ + r y = 0,

see Eastham [2] and Atkinson [3] .

We shall use the asymptotic Theorem of Eastham ( [3] , Section 2), [4] to obtain our main result of (1) in Section 4. The general feature of our method are given in Sections (2) and (3), with some examples in Section (5).

We write (1) in a standard way [5] as a first-order system:

Y ′ = A Y (3)

where

Y = ( y p y ′ ) (4)

and the matrix is given by

A = ( 0 p − 1 − r − q p − 1 ) . (5)

As in [6] we express the matrix A in the diagonal form:

T − 1 A T = Λ = d i a g ( λ 1 , λ 2 ) (6)

and we therefore require the eigenvalues λ j and the eigenvectors v j of A, j = 1 , 2 .

The characteristic equation of is given by:

p λ 2 + q λ + r = 0. (7)

An eigenvector v j corresponding to λ j is

v j = ( 1               p λ j ) * (8)

where the superscript * denote the transpose.

Now by (7)

λ j = − q 2 p ± ( q 2 − 4 p r ) 1 / 2 2 p                     ( j = 1 , 2 ) (9)

Now we define the matrix T in (6) by

T = [ 1 1 p λ 1 p λ 2 ] (10)

Hence by (6), the transformation

Y = T Z , (11)

takes (3) into

Z ′ = ( Λ − T − 1 T ′ ) Z (12)

Now if we write

T − 1 T ′ = ( t j k ) , (13)

then by (7) and (10)

t 1 j = ( λ 1 − λ 2 ) − 1 [ ( p ′ λ j 2 + q ′ λ j + r ′ ) ( 2 p λ j + q ) − 1 − p ′ p λ j ] t 2 j = − t 1 j                                                             ( j = 1 , 2 ) . (14)

Now we need to work (14) in terms of r , p and q in order to determine (12) and then make progress for (1).

At this stage we require the following conditions in the coefficients r , p and q as x → ∞ .

Condition I. r , p and q are nowhere zero in some interval [ a , ∞ ) , and

r p = o ( q 2 ) ,               ( x → ∞ ) (15)

we write

δ = r p q 2 → 0               ( x → ∞ ) (16)

Condition II.

δ r ′ r , δ p ′ p , δ q ′ q     are   all   L ( a , ∞ ) . (17)

Now if we let

D = ( q 2 − 4 p r ) 1 / 2 2 p (18)

then (9) gives

λ j = − q 2 p ± D                 ( j = 1 , 2 ) (19)

where by(18) and (16)

D = q 2 p ( 1 − 4 δ ) 1 / 2 ~ q 2 p               ( x → ∞ ) . (20)

Now by (19) and (20)

λ 1 = − r q [ 1 + δ + O ( δ 2 ) ] , (21)

and

λ 2 = − q p [ 1 − δ + O ( δ 2 ) ] (22)

Now using (14), (21) and (22) we obtain

t 11 = t 21 = O ( Δ ) , (23)

t 12 = − t 22 = − q ′ q + O ( Δ ) , (24)

where

Δ = ( | r ′ r δ | + | p ′ p δ | + | q ′ q δ | ) (25)

Hence by (17),

Δ ∈ L ( a , ∞ ) . (26)

Therefore, by (23), (24) and (26), we can write (12) as:

Z ′ = ( Λ + R + S ) Z , (27)

where

R = [ 0 q ′ q 0 − q ′ q ] , (28)

and S is L ( a , ∞ ) by (26).

Theorem 1. Let the coefficients r and p in (1) be C 1 [ a , ∞ ) while q to be C 2 [ a , ∞ ) .

Let (15) and (17) hold.

Let

q ′ q = o ( r q )                 ( x → ∞ ) (29)

( q ′ p q 2 ) ′ , r 2 p q 3       are       L ( a , ∞ ) (30)

Let

R e [ q p − 2 r q + q ′ q ]     be     of   one   sign   in   [ a , ∞ ) . (31)

Then (1) has solutions y 1 and y 2 such that

y 1 ~ e x p ( − ∫ a x r q d t ) , (32)

y ′ 1 = o [ q p − 1 e x p ( − ∫ a x r q d t ) ] (33)

while

y 2 ~ q − 1 e x p ( ∫ a x [ − q p + r q ] d t ) , (34)

y ′ 2 ~ p − 1 e x p ( ∫ a x [ − q p + r q ] d t ) . (35)

Proof. As in [6] , we apply the Eastham theorem ( [3] , section 2) to the system (27) provided only that Λ and R , satisfy the required conditions.

We shall use (15), (17), (29), and (31).

We first require that

q ′ q = o ( λ 1 − λ 2 ) , (36)

this being [2] for our system,

λ 1 − λ 2 = q p ( 1 − 4 δ ) 1 / 2 , (37)

Thus (36) holds by (15) and (29).

Second, we need

[ ( λ 1 − λ 2 ) − 1 q ′ q ] ′ ∈ L ( a , ∞ ) . (38)

this being [2] for our system. By (38), this requirement is implied by (17) and (30).

Finally we show that the eigenvalues μ k of Λ + R satisfy the dichotomy condition [2] .

As in [6] and [7] , the dichotomy condition holds if

R e ( μ 1 − μ 2 ) = f + g , (39)

where f has one sign in [ a , ∞ ) and g is L ( a , ∞ ) [2] .

Now by (6) and (28):

μ 1 ( x ) = λ 1 ( x ) ,             μ 2 ( x ) = λ 2 ( x ) − q ′ q , (40)

then by (21), (22) and (40)

R e ( μ 1 − μ 2 ) = R e ( q p − 2 r p + q ′ q ) + O ( r 2 p q 3 ) , (41)

Thus, by (31) and (30), (39) holds. Since (27) satisfies all the conditions for the asymptotic result [3, section 2], it follows that as x → ∞ , (27) has two linearly independant solutions.

Z k ( x ) = [ e k + o ( 1 ) ] e x p ( ∫ a x μ k ( t ) d t ) (42)

with e k the coordinate vector with k-th coponment unity and other coponments zero.

Finally, on transforming back to y via (10), (11), (4) and making use of (40), (21), (22) and (30), we obtain (33), also (32) after adjusing y 1 by a constant multiple, and similary for y 2 and y ′ 2 .□

Example 1. We consider the cofficients in (1) given by

r ( x ) = c 1 x α 1 ,             q ( x ) = c 2 x α 2 ,             p ( x ) = c 3 x α 3 .

α i and c i ( 1 ≤ i ≤ 3 ) are real constants with c i ≠ 0 . Then (15) and (17) of Theorem 4.1 hold under the conditions

2 α 2 − α 1 − α 3 > 0. (43)

Also (29) true if

α 1 − α 2 + 1 > 0 (44)

Now in (30) ( q ′ p q 2 ) ′ is L ( a , ∞ ) if

α 2 − α 3 + 1 > 0 (45)

wich is true by (43) and (44).

Also, in (30), r 2 p q 3 is L ( a , ∞ ) if

3 α 2 − 2 α 1 − α 3 > 1. (46)

So all conditions of theorem 4.1 are true under (43), (44) and (46). For example if we take α 1 = α 2 .

Then all condition are true if

α 2 − α 3 > 1. (47)

Example 2. Let r ( x ) = c 1 x α 1 exp ( x a ) , p ( x ) = c 2 x α 2 exp ( − 4 x b ) , q ( x ) = c 3 x α 3 exp ( − x b )

where b ≥ a > 0 , α i and c i ( 1 ≤ i ≤ 3 ) are real constants with c i ≠ 0 .

Again it is easy to check that all conditions of Theorem 4.1 are satisfied.

Al-Hammadi, A.S.A. (2017) Asymptotic Theory for a General Second-Order Differential Equation. Advances in Pure Mathematics, 7, 407-412. https://doi.org/10.4236/apm.2017.78026