Consider the following one-dimensional Sturm–Liouville equation

− u " ( t ) + q ( t ) u ( t ) = λ u ( t ) , t ∈ [ 0 , 1 ]

where the real-valued function q ∈ C [ 0 , 1 ] ; λ = s 2 is a complex spectral parameter and s = x + ι y ; x , y ∈ ℝ . We will use the notation Q ( t ) = 1 2 ∫ 0 t q ( τ ) d τ .

In this article S ∈ ℂ s : = ℝ s U ℂ s + U ℂ s − , where ℝ s : = ℝ s − U ℝ s + , U ℝ s 0 ℝ s − : = { s = x + ι y ∈ ℂ : x = 0 , y > 0 } , ℝ s + : = { s = x + ι y ∈ ℂ : x > 0 , y = 0 } , ℝ s 0 : = { s = 0 } , ℂ s + : = { s = x + ι y ∈ ℂ : x > 0 , y > 0 } and ℂ s − : = { s = x + ι y ∈ ℂ : x > 0 , y < 0 } .

Then a map λ = S 2 is the bijection between ℂ s and ℂ λ : = ℂ .

We shall investigate Sturm–Liouville Problem (SLP) which consist of equation (1) on [ 0 , 1 ] with one classical (local) Robin type Boundary Condition (BC)

cos α u ( 0 ) + sin α u ' ( 0 ) = 0 , α ∈ ( 0 , π ) ,

and another two-point Nonlocal Boundary Condition (NBC)

( C a s e 1 ) u ' ( 1 ) = γ u ( ξ ) , ξ ∈ [ 0 , 1 ] ,

( C a s e 2 ) u ' ( 1 ) = γ u ' ( ξ ) , ξ ∈ [ 0 , 1 ) ,

( C a s e 2 ) u ( 1 ) = γ u ( ξ ) , ξ ∈ [ 0 , 1 ) ,

where γ ∈ ℝ . We consider the Dirichlet and the Neumann BC:

( C a s e d ) u ( 0 ) = 0 ,

( C a s e n ) u ' ( 0 ) = 0 ,

too. The Sturm–Liouville problem (1), (4_d), (3₃) was investigated in [2], the Sturm–Liouville problem (1), (4_n), (3) was investigated in [4].

In this section we present some statements about solution of IVP. These statements were proved in [3]. We will use them for investigation asymptotic expansions for SLP (1)–(3). Additionally, we introduce some notation related to our asymptotical analysis of this problem.

Let λ = s 2 , s ∈ ℂ s and w α s ( t ) be a solution of equation (1) satisfying the initial conditions

w α s ( 0 ) = sin α , w α s ' ( 0 ) = − cos α .

The function w ( t , s , α ) = w α s ( t ) is an analytic (holomorphic) function of s and this function satisfies boudary condition (2). We denote φ s ( t ) = φ o s ( t ) : = w ( t , s , 0 ) and ψ s ( t ) = φ − 1 , s ( t ) = w ( t , s , π / 2 ) .

Under the condition that q ∈ ℂ r [ 0 , 1 ] , r ∈ ℕ 0 : = ℕ U { 0 } , asymptotic expansions may be obtained for φ s ( t ) [3] and ψ s ( t ) [4]. We will use recursive formula

p i + 1 0 ( t ) = − 1 2 ∫ 0 t q ( τ ) p i 0 ( τ ) d τ − ∑ j = 2 + ∂ i ( q p j - 1 0 ) ( i − j ) ( t ) + ( − 1 ) i ( q p j − 1 0 ) ( i − j ) ( 0 ) 2 i − j + 2

Lemma 1. (See [3, Lemma 7]) Let s ∈ ℂ s and q ∈ C r [ 0 , 1 ] . Then for | S | ≥ q 0 we have the asymptotic expansions

( φ s ) s ( l ) ( t , s ) = − ∑ p j ( t ) cos l j = 1 r + 1 ( s t + π 2 ( j − l ) ) s − j + O ( s − ( r + 2 ) e ( r + 2 ) | y | t ) ,

( φ s ' ) s ( l ) ( t , s ) = − ∑ j = 0 r p j − l cos ( s t + π 2 ( j − l ) ) s − j + O ( s − ( r + 1 ) e ( r + 2 ) | y | t )

for l ∈ ℕ 0 , where p 1 k ( t ) = − t p 1 k − 1 ( t ) , p i k ( t ) = ( 1 − i ) p i − 1 k − 1 ( t ) − t p i k − 1 ( t ) , i = 2 , r + 1 , ‾ p 0 - k ( t ) = − t p 0 − k − 1 ( t ) , p i − k ( t ) = ( 1 − i ) p i − 1 − k − 1 ( t ) − t p i − k − 1 ( t ) , = 1 , r k ∈ ℕ , ‾ p i − 0 ( t ) = p i 0 ' ( t ) − p i + 1 0 ( t ) , i = 1 , r , ‾ p 0 − 0 ( t ) = 1 , and p j 0 ( t ) is calculated by (6) i = 1 , r ‾ ,with p 1 0 ( t ) = − 1 .

Lemma 2. (See [4, Lemma 9].) Let S ∈ ℂ s and q ∈ ℂ r [ 0 , 1 ] . Then for | s | ≥ q 0 we have the asymptotic expansions

( ψ s ) s ( l ) ( t , s ) = − ∑ p j l ( t ) cos ( s t + π 2 ( j − l ) s − j + O ( s − ( r + 1 ) e ( r + 2 ) | y | t ) j = 0 r

( ψ s ' ) s ( l ) ( t , s ) = − ∑ j = − 1 r − 1 p j − l cos ( s t + π 2 ( j − l ) s − j + O ( s − r e ( r + 2 ) | y | t )

for l ∈ ℕ 0 . Where p 0 k ( t ) = − t p 0 k − 1 ( t ) , p i k ( t ) = ( 1 − i ) p i − 1 k − 1 ( t ) − t p i k − 1 ( t ) , i = 1 , r , ‾ p − 1 − k ( t ) = − t p − 1 − k − 1 ( t ) , p i − k ( t ) = ( 1 − i ) p i − 1 − k − 1 ( t ) − t p i − k − 1 ( t ) , i = 0 , r − 1 , ‾ k ∈ ℕ , p i − 0 ( t ) = p i 0 ' ( t ) − p i + 1 0 ( t ) , i = 0 , r − 1 , ‾ p − 1 − 0 ( t ) = 1 and p j 0 ( t ) is calculated by (6) for i = 0 , r − 1 ‾ with p 0 0 ( t ) = − 1 .

Remark 1. (See [3, Lemma 7], [4, Lemma 7].) In the case q ∈ C [ 0 , 1 ] and l = 0 we have the asymptotic expansions:

φ s ( t ) = − sin ( s t ) s − 1 + O ( s − 2 e | y | t ) , ψ s ( t ) = cos ( s t ) + O ( s − 1 e | y | t ) ,

φ s ' ( t ) = − cos ( s t ) + O ( s − 1 e | y | t ) , ψ s ' ( t ) = − s sin ( s t ) + O ( e | y | t ) .

Remark 2. In the case q ∈ ℂ 1 [ 0 , 1 ] and l = 0 we have the asymptotic expansions

φ s ( t ) = − sin ( s t ) s − 1 + Q ( t ) cos ( s t ) s 2 + O ( s − 3 e 3 | y | t ) ,

ψ s ( t ) = cos ( s t ) + Q ( t ) sin ( s t ) s − 1 + O ( s − 2 e 3 | y | t ) ,

φ s ' ( t ) = − cos ( s t ) − Q ( t ) sin ( s t ) s − 1 + O ( s − 2 e 3 | y | t ) ,

ψ s ' ( t ) = − s sin ( s t ) + Q ( t ) cos ( s t ) + O ( s − 1 e 3 | y | t ) ,

where Q ( t ) = 1 2 ∫ 0 t q ( τ ) d τ .

We can calculate the first functions p j l and p j − l in Case d (for function φ ):

p 1 0 = − 1 , p 2 0 = Q ( t ) , p 3 0 = − 1 2 ( Q ( t ) ) 2 1 4 q ( t ) + 1 4 q ( 0 ) ,

p 1 1 = t , p 2 1 = 1 − t Q ( t ) , p 1 2 = − t 2 ,

p ‾ 0 0 = 1 , p ‾ 1 0 = − Q ( t ) , p ‾ 2 0 = 1 2 ( Q ( t ) ) 2 + 1 4 q ( t ) − 1 4 q ( 0 ) ,

p ‾ 0 1 = − t , p ‾ 1 1 = t Q ( t ) , p ‾ 0 2 = t 2 ;

and in Case n (for function ψ ):

p 0 0 = − 1 , p 1 0 = Q ( t ) , p 2 0 = − 1 2 ( Q ( t ) ) 2 + 1 4 q ( t ) − 1 4 q ( 0 ) ,

p 0 1 = t , p 1 1 = − t Q ( t ) , p 0 2 = − t 2 ,

p ‾ − 1 0 = 1 , p ‾ 0 0 = − Q ( t ) , p ‾ 1 0 = 1 2 ( Q ( t ) ) 2 + 1 4 q ( t ) + 1 4 q ( 0 ) ,

p ‾ − 1 1 = − t , p ‾ 0 1 = 1 + t Q ( t ) , p ‾ − 1 2 = t 2 .

We will use an additional index to distinguish cases: p j d , l ( t ) , p j − d , l ( t ) (Case d), p j n , l ( t ) , p j − n , l ( t ) (in Case n).

The following integral equation holds [1,5]:

w α s ( t ) − 1 s ∫ 0 t q ( τ ) sin ( s ( t − τ ) ) w α s ( τ ) d τ = sin α cos ( s t ) − cos α sin ( s t ) s .

If α = 0 , then the solution of this integral equation is φ s , if α = π / 2 , then the solution of this integral equation is ψ s . By superposition principle we have

w α s ( t ) = cos α . φ s ( t ) + sin α . ψ s ( t ) .

So, we get asymptotic expansions for function w α s .

Lemma 3. Let s ∈ ℂ s and q ∈ C r [ 0 , 1 ] . Then for | s | ≥ q 0 we have the asymptotic expansions

( w α s ) s ( l ) ( t , s ) = − ∑ j = 0 r p j l ( t ) cos ( s t + π 2 ( j − l ) ) s − j + O ( s − ( r + 1 ) e ( r + 2 ) | y | t )

( w α s ' ) s ( l ) ( t , s ) = − ∑ p ‾ j − l ( t ) cos ( s t + π 2 ( j − l ) ) s − j j = − 1 r − 1 + O ( s − r e ( r + 2 ) | y | t ) )

for l ∈ ℕ 0 , where

Remark 3. Formulas (15)–(16) are valid for α = 0 , but in (7)–(8) we have more accurate the asymptotic expansions with

p j d , l ( t ) cos ( s t + π 2 ( r + 1 − l ) ) s − ( r + 1 ) + O ( s − ( r + 2 ) e ( r + 2 ) | y | t ) ,

p j − d , l ( t ) cos ( s t + π 2 ( r − l ) ) s − r + O ( s − ( r + 1 ) e ( r + 1 ) | y | t ) ,

instead O ( s − ( r + 1 ) e ( r + 2 ) | y | t ) and O ( s − r e r + 2 ) | y | t ) in (13)–(14).

Corollary 1. If q ∈ C [ 0 , 1 ] and α ∈ ( 0 , π ) , then we have asymptotic expansions:

w α s ( t ) = sin α · cos ( s t ) + O ( s − 1 e | y | t ) , w α s ' ( t ) = − sin α · sin ( s t ) s + O ( e | y | t ) .

Corollary 2. If q ∈ C 1 [ 0 , 1 ] and α ∈ ( 0 , π ) , then we have asymptotic expansions:

w α s ( t ) = sin α . cos ( s t ) + ( − cos α + sin α Q ( t ) ) sin ( s t ) s − 1 + O ( s − 2 e 3 | y | t ) ,

w α s ( t ) = − sin α . sin ( s t ) s + ( − cos α + sin α Q ( t ) ) cos ( s t ) + O ( s − 1 e 3 | y | t ) .

Lemma 4. Let x ∈ ℝ s + , δ ∈ ℝ , q ∈ C s [ 0 , 1 ] , Q j ( x ) , j = 1 , r ‾ , are bounded functions.

If s = x + δ ,

δ = ∑ j = 1 r Q j ( x ) x − j + O ( x − ( r + 1 ) ) ,

then we have the following equations

w α s ( t ) = ∑ R j ( t , x ) x − j j = 0 r + O ( x − ( r + 1 ) ) , w α s ' ( t ) = ∑ R ‾ j ( t , x ) x − j j = − 1 r − 1 + O ( x − r ) ,

where

R 0 ( t , x ) = sin α · R 0 n ( t , x ) , R j ( t , x ) = cos α · R j d ( t , x ) + sin α · R j n ( t , x ) , j = 1 , r , ‾

R ‾ − 1 ( t , x ) = sin α · R ‾ − 1 n ( t , x ) , R ‾ j ( t , x ) = cos α · R ‾ j d ( t , x ) + sin α · R ‾ j n ( t , x ) , j = 0 , r − 1 , ‾

and functions

m = 0 , r . ‾

Proof. The proof follows from (12) and asymptotic expansions for φ s ( t ) [3,Corollary 2] and ψ s ( t ) [4,Corollary2].

Remark 4. In the case α = 0 we have more accurate the asymptotic expansions

φ s ( t ) = ∑ R j d ( t , x ) x − j j = 1 r + 1 + O ( x − ( r + 2 ) ) , φ s ' ( t ) = ∑ R ‾ j d j = 0 r ( t , x ) x − j + O ( x − ( r + 1 ) ) .

Corollary 3. If q ∈ C [ 0 , 1 ] and α ∈ ( 0 , π ) , the n we have asymptotic expansions:

w α s ( t ) = sin α · cos ( x t ) + O ( x − 1 ) , w α s ' ( t ) = − sin α · sin ( x t ) x + O ( 1 ) .

Corollary 4. If q ∈ C 1 [ 0 , 1 ] and α ∈ ( 0 , π ) , then we have asymptotic expansions:

w α s ( t ) = sin α . cos ( x t ) + ( − cos α + sin α ( Q ( t ) − t Q 1 ( x ) ) ) sin ( x t ) x − 1 + O ( x − 2 ) ,

w α s ' ( t ) = − sin α . sin ( x t ) x + ( − cos α + sin α ( Q ( t ) − t Q 1 ( x ) ) ) cos ( x t ) + O ( x − 1 ) .

Substituting w α s ( t ) into (3) we get the characteristic equation

h α ( s ) : = w α s ' ( 1 ) − γ w α s ( ξ ) = 0 ,

h α ( s ) : = w α s ' ( 1 ) − γ w α s ' ( ξ ) = 0 ,

h α ( s ) : = w α s ( 1 ) − γ w α s ( ξ ) = 0 .

Let’s define functions:

h − 1 l ( s ) : = − p ‾ − 1 l ( 1 ) sin ( s − π 2 l ) = ( − 1 ) l − 1 sin α sin ( s − π 2 l ) ,

h j l ( s ) : = γ p j l ( ξ ) cos ( ξ s + π 2 ( j − l ) ) − p ‾ j l ( 1 ) cos ( s + π 2 ( j − l ) ) , j = 0 , r − 1 , ‾

h j l ( s ) : = γ p ‾ j l ( ξ ) cos ( ξ s + π 2 ( j − l ) ) − p ‾ j l ( 1 ) cos ( s + π 2 ( j − l ) ) , j = − 1 , r − 1 , ‾

h j l ( s ) : = γ p j l ( ξ ) cos ( ξ s + π 2 ( j − l ) ) − p j l ( 1 ) cos ( s + π 2 ( j − ) ) , j = 0 , r , ‾

where functions p j l and p ‾ j l are defined by formulas (15) – (16). For example,

h − 1 0 ( s ) = − sin α sin s , h − 1 1 ( s ) = − sin α cos s ,

h 0 0 ( s ) = ( sin α Q ( 1 ) − cos α ) cos s − sin α γ cos ( ξ s ) ,

h − 1 0 ( s ) = sin α ( sin s − γ sin ( ξ s ) ) , h − 1 1 ( s ) = − sin α ( cos s − γ ξ cos ( ξ s ) ) ,

h 0 0 ( s ) = ( sin α Q ( 1 ) − cos α ) cos s − ( sin α Q ( ξ ) − cos α ) γ cos ( ξ s ) ,

h 0 0 ( s ) = sin α ( cos s − γ cos ( ξ s ) ) , h 0 1 ( s ) = − sin α ( sin s − γ ξ sin ( ξ s ) ) ,

h − 1 0 ( s ) = ( sin α Q ( 1 ) − cos α ) sin s − ( sin α Q ( ξ ) − cos α ) γ sin ( ξ s ) .

We will use the notation: ρ = − 1 , a k : = ( k − 1 / 2 ) π in Cases 1 , 2 ; ρ = 0 , a k : = ( k − 1 ) π in Cases 3 , k ∈ ℕ .

Lemma 5. Suppose | γ | < 1 in Cases 2 and 3. Then | h ρ 0 ( a k + ι y ) | ≥ κe | y | , κ > 0 .

Proof. In Case 1 we have

| h − 1 0 ( a k + ι y ) | = sin α | sin ( a k + ι y ) | = sin α | sin a k cos h y + ι cos a k sin h y ) | = sin α cos h y ≥ sin α e | y | / 2 .

From inequalities

| sin s − γ sin ( ξ s ) | ≥ ( | sin x | − | γ | | sin ( ξ s ) | ) cos h y ≥ ( | sin x | − | γ | ) cos h y ,

| cos s − γ cos ( ξ s ) | ≥ ( | cos x | − | γ | | cos ( ξ s ) | ) cos h y ≥ ( | cos x | − | γ | ) cos h y

we get (in Cases 2 and 3)

| h ρ 0 ( a k + ι y ) | ≥ sin α ( 1 − | γ | ) cos h y ≥ sin α ( 1 − | γ | ) e | y | / 2 .

Lemma 6. Suppose | γ | < 1 in Cases 2 and 3. There exists B > 0 such that | h ρ 0 ( s ) | ≥ κe | y | , k > 0 for | y | ≥ B .

Proof. We estimate

| sin s − γ sin ( ξ s ) | ≥ | sin s | − | γ | | sin ( ξ s ) | ≥ sin h | y | − | γ | cos h ( ξ y ) ,

| cos h s − γ cos ( ξ s ) | ≥ | cos s | − | γ | | cos ( ξ s ) | ≥ sin h | y | − | γ | cos h ( ξ y ) .

For | γ | < 1 we have

lim y → + ∞ ( sin h y − | γ | cos h ( ξ y ) ) e − y = 1 2 ( 1 − | γ | . ⌊ ξ ⌋ ≥ 1 2 ( 1 − | γ | ) > 0 .

So, in Cases 2 and 3 there exists B > 0 such that

| h ρ 0 ( s ) | ≥ 1 4 sin α ( 1 − | γ | ) e | y | .

for | y | ≥ B . The proof in Case 1 repeats the proof in Case 2 with γ = 0 .

Lemma 7. Let s ∈ ℂ s and q ∈ C r [ 0 , 1 ] . Then for | s | ≥ q 0 the asymptotic expansion

h α ( l ) ( s ) = ∑ h j l ( s ) s − j + O ( s − ( r + 1 + ρ ) e ( r + 2 ) | y | ) j = ρ r + ρ

is valid, l ℕ ∈ ℕ 0 .

Proof. The proof for α ∈ ( 0 , π ) is the same as in case α = π / 2 [4,see Lemma11].

Remark 5. In the case q ∈ C [ 0 , 1 ] we have (see Corollary 1) the asymptotic expansion

h α ( s ) = h ρ 0 ( s ) s − ρ + O ( s − ( 1 + ρ ) e | y | ) .

Remark 6. If α = 0 (the problem (1), (4_d), (3)) formula (21) is valid with ρ = 0 in Cases 1, 2; ρ = 1 in Case 3, and

h 0 0 ( s ) = − cos s , h 0 1 ( s ) = − sin s ,

h 0 0 ( s ) = γ cos ( ξ s ) − cos s , h 0 1 ( s ) = sin s − γ ξ sin ( ξ s ) ,

h 1 0 ( s ) = γ sin ( ξ s ) − sin s , h 1 1 ( s ) = γ ξ cos ( ξ s ) − cos s .

So, if α = 0 , then Lemma 5 and Lemma 6 are valid for the problem (1), (4_d), (3) with ρ = 0 , α k = k π , k ∈ ℕ , in Cases 1 and 2, ρ = 0 , α k = ( k − 1 / 2 ) π , k ∈ ℕ , in Case 3.

Let us consider positive s = x > 0 , q ∈ C r [ 0 , 1 ] . We investigate equation h α ( x + δ ) = 0 , δ ∈ ℝ , with additional condition

| h ρ 1 ( x ) | ≥ κ > 0

Lemma 8. Suppose | γ | < 1 in Cases 2 and 3. If h ρ 0 ( x ) = 0 , then (24) is valid. The constant κ is the same for all such x.

Proof. In Case 1 if h − 1 0 ( x ) = − sin α sin x = 0 , then x = x k = k π , k ∈ ℕ and | h − 1 1 ( x k ) | = | − sin α cos x k | = sin α > 0 .

In Case 2 we have equation h − 1 0 ( x ) = − sin α ( sin x − γ sin ( ξ x ) ) . If x k are the root of equation x − γ sin ( ξ x ) = 0 , then we get [2, see Lemma 4 and Corollary 3] that | h − 1 1 ( x k ) | = | − sin α ( cos x k − γ ξ cos ( ξ x ) ) | ≥ sin α | cos x k − γ ξ cos ( ξ x k ) | ≥ sin α . k = κ ≥ 0 .

In Case 3 we have equation h 0 0 ( x ) = sin α ( cos x − γ cos ( ξ x ) ) . If x k are the root of equation cos x − γ cos ( ξ x ) = 0 , then we get [2, see Lemma 5] that | h 0 1 ( x k ) | ≥ sin α | sin x k − γ ξ sin ( ξ x k ) | ≥ sin α ( | sin x k | − | γ | . | sin ( ξ x k ) | ) ≥ sin α . k = κ > 0 .

Remark 7. Lemma 4 and Lemma 5 in [2] were proved for ξ ∈ ( 0 , 1 ) , but it is easy to see, that they are valid for ξ = 0 too. So, Lemma 8 is proved for all ξ (see (3)).

Remark 8. If α = 0 then Lemma 8 is valid with ρ = 0 Cases 1 and 2, ρ = 1 in Case 3. The proof is the same.

Let’s denote the function

Q 1 ( x ) = − h 1 + ρ 0 ( x ) ( h ρ 1 ( x ) ) - 1

If functions Q 1 , . . . , Q k − 1 are defined, then we can find functions

l = 1 , k − 1 ‾ and function

If q ∈ C 1 [ 0 , 1 ] , then

Q 1 ( x ) = ( sin α Q ( 1 ) − cos α ) cos x − sin α γ cos ( ξ x ) sin α cos x

Q 1 ( x ) = ( sin α Q ( 1 ) − cos α ) cos x − ( sin α Q ( ξ ) − cos α ) γ cos ( ξ x ) sin α ( cos x − γ ξ cos ( ξ x ) )

Q 1 ( x ) = ( sin α Q ( 1 ) − cos α ) sin x − ( sin α Q ( ξ ) − cos α ) γ sin ( ξ x ) sin α ( sin x − γ ξ sin ( ξ x ) )

Lemma 9. If q ∈ C r [ 0 , 1 ] and δ = o ( 1 ) , h ρ 0 ( x ) = 0 , then asymptotic expansion

δ = ∑ Q j ( x ) x − j j = 1 r + O ( x − ( r + 1 ) )

is valid, where Q j ( x ) , j = 1 , r , ‾ , are bounded functions.

Proof. The proof can be found in [4, see proof of Lemma 14].

In this section we assume, that | γ | < 1 in Cases 2, 3. Let us denote domains D k = { s ∈ ℂ : | x | ≤ a k , | y | ≤ a k } , D s k = ℂ s ∩ D k , k ∈ ℕ ( k > 1 in Case 3), contours T s k = ℂ s ∩ ∂ D k , and intervals I k : = ( a k , a k + 1 ) ⊂ D s , k + 1 ∖ D s k , k ∈ ℕ .

Lemma 10. Suppose | γ | < 1 in Cases 2 and 3. If q ∈ C [ 0 , 1 ] , then it follows that the number of zeros of functions h α ( s ) and h ρ 0 ( s ) s − ρ is the same inside T s k for sufficiently large k.

Proof. We have (see (22)) h α ( s ) = h ρ 0 ( s ) s − ρ + O ( s − 1 − ρ e | y | ) . Using Lemma 5 and Lemma 6we estimate | O ( s − 1 − ρ e | y | ) | ≤ c 1 | s | − 1 − ρ e | y | < m i n { k , κ } | s | − ρ e | y | ≤ | h ρ 0 ( s ) s − ρ | on the contours T s k for sufficiently large k. Therefore, by Rouché theorem it follows that the number of zeros of h α ( s ) and h ρ 0 ( s ) s − ρ are the same inside T s k for sufficiently large k.

Corollary 5. If q ∈ C [ 0 , 1 ] , then it follows that the number of zeros of functions h α ( s ) and h ρ 0 ( s ) is the same between T s k and T s , k + 1 for sufficiently large k.

Remark 9. If α = 0 , then Lemma 10 and Corollary 5 are valid. The proof is the same. In Case 3 s = 0 isn’t zero of the function h 1 0 ( s ) s − 1 = γ s − 1 sin ( ξ s ) − s − 1 sin s for | γ | < 1 .

From (19) (and (23) for α = 0 ) we have that function h ρ 0 has only one positive root x k ∈ I k , k ∈ ℕ . For example, x k = π k (and x k = π ( k + 1 / 2 ) for α = 0 ) in Case 1. In Cases 2 and 3 existence of such root follows from [4, Lemma 4 and Lemma 5]. Thus, function h α ( s ) has only one root sk between T s k and T s , k + 1 for sufficiently large k.

In Case 2 ( sin α ≠ 0 ) a k = ( k − 1 / 2 ) π and

h α ( a k ) = − sin α ( sin a k − γ sin ( ξ a k ) ) a k + O ( 1 ) = a k sin α ( ( − 1 ) k + γ sin ( ξ a k ) + O ( k − 2 ) ) .