C om mun.Fac.Sci.U niv.A nk.Series A 1 Volum e 66, N umb er 2, Pages 175–194 (2017) D O I: 10.1501/C om mua1_ 0000000810 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
SPECTRAL ANALYSIS OF BOUNDARY VALUE PROBLEMS WITH RETARDED ARGUMENT
ERDO ¼GAN ¸SEN AND AZAD BAYRAMOV
Abstract. In this paper, by modifying some techniques of [S.B. Norkin, Dif-ferential equations of the second order with retarded argument, Translations of Mathematical Monographs, Vol. 31, AMS, Providence, RI, 1972]and sug-gesting own approaches we …nd asymptotic formulas for the eigenvalues and eigenfunctions of boundary value problems of Sturm-Liouville type for the second order di¤erential equation with retarded argument.
1. Formulation of the problem
In this study we shall investigate discontinuous eigenvalue problems which consist of Sturm-Liouville equation
a(x)y00(x) + M (x)y(x (x)) + y(x) = 0 (1)
on 0;2 [ 2; ; with boundary conditions
y(0) cos + y0(0) sin = 0; (2)
y( ) cos + y0( ) sin = 0; (3)
and transmission conditions y( 2 0) 1y(2 + 0) = 0; (4) y0( 2 0) 2y 0( 2 + 0) = 0; (5) where a(x) = a2
1 for x 2 0;2 and a(x) = a 2
2 for x 2 2; ; the real-valued
function M (x) is continuous in 0;2 [ 2; and has a …nite limit M (2 0) = limx!2 0M (x); the real valued function (x) 0 continuous in 0;2 [ 2; and
has a …nite limit (2 0) = limx!2 0 (x), x (x) 0; if x 2 0;2 ; x (x) Received by the editors: May 18, 2016, Accepted: January 02, 2017.
2010 Mathematics Subject Classi…cation. 34L20, 35R10.
Key words and phrases. Di¤erential equation with retarded argument; transmission conditions; asymptotics of eigenvalues and eigenfunctions.
c 2 0 1 7 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis tic s .
2; if x 2 2; ; is a real spectral parameter; i’s (i = 1; 2) are arbitrary real
numbers.
In the book [1] and papers [2-9] the asymptotic formulas for the eigenvalues and eigenfunctions of boundary value problems with retarded argument and a spectral parameter in the di¤erential equation and/or boundary conditions and/or trans-mission conditions were obtained.
The articles [10-18] are devoted to study of the spectral properties of eigenvalues and eigenfunctions of the classical Sturm-Liouville problems.
If we take (x) 0 and/or 1 = 2 = 1 and/or a(x) 1 then the asymptotic
formulas for eigenvalues and eigenfunctions correspond to those for the classical Sturm-Liouville problem [10, 14, 16-18].
Let 1(x; ) be a solution of Eq. (1) on 0; 2 ; satisfying the initial conditions
1(0; ) = sin ; 01(0; ) = cos : (6)
The conditions (6) de…ne a unique solution of Eq. (1) on 0;2 [1].
After de…ning the above solution we shall de…ne the solution 2(x; ) of Eq. (1)
on 2; by means of the solution 1(x; ) using the initial conditions
2 2; = 1 1 1 2; ; 0 2 2; = 1 2 01 2; : (7)
The conditions (7) are de…ned as a unique solution of Eq. (1) on 2; :
Consequently, the function (x; ) is de…ned on 0;2 [ 2; by the equality (x; ) = 1(x; ); x 2 [0;2)
2(x; ); x 2 (2; ]
is a solution of the Eq. (1) on 0;2 [ 2; ; which satis…es one of the boundary conditions and both transmission conditions.
Lemma 1.1. Let (x; ) be a solution of Eq.(1) and > 0: Then the following integral equations hold:
1(x; ) = sin cos s a1 x a1cos s sin s a1 x 1 a1s x Z 0 M ( ) sin s a1 (x ) 1( ( ) ; ) d s =p ; > 0 ; (8) 2(x; ) = 1 2; 1 cos s a2 x 2 + a2 01 2; s 2 sin s a2 x 2 1 a2s x Z =2 M ( ) sin s a2 (x ) 2( ( ) ; ) d s =p ; > 0 : (9)
Proof. To prove this, it is enough to substitute s2 a2 1 1( ; ) @2 1( ; ) @ 2 and s 2 a2 2 2( ; ) @2 2( ; ) @ 2 instead of M ( ) a2 1 1( ( ); ) and M ( ) a2 2 2( ( ); ) in the integrals
in (8) and (9) respectively and integrate by parts twice.
Theorem 1.2. The problem (1) (5) can have only simple eigenvalues. Proof. It is similar to the proof of Theorem 1 in [6].
2. Existence of solutions of the problem
The function (x; )de…ned in introduction is a nontrivial solution of Eq. (1) sat-isfying conditions (2), (4) and (5). Putting (x; ) into (3), we get the characteristic equation
R( ) ( ; ) cos + 0( ; ) sin = 0: (10)
By Theorem 1.2 the set of eigenvalues of boundary-value problem (1)-(5) co-incides with the set of real roots of Eq. (10). Let M1 = a11
=2 R 0 jM( )jd and M2 = a21 R =2
jM( )j d . Also let us assume that max k2M2
1; k2M22 ; k > 1
(k 2 R). Then for the solution 1(x; ) of Eq. (8), the following inequality holds:
j 1(x; )j q k2M2 1sin2 + a21cos2 (k 1) jM1j ; x 2 h 0; 2 i : (11)
Di¤erentiating (8) with respect to x, we have
0 1(x; ) = s a1 sin sin s a1 x cos cos s a1 x 1 a2 1 x Z 0 M ( ) cos s a1 (x ) 1( ( ) ; )d : (12) Then from (11) and (12) for the solution 2(x; ) of Eq. (9), the following inequality holds: j 2(x; )j q k2M2 1sin 2 + a2 1cos2 (k 1)2ja1M1 1 2j ; x 2h 2; i : (13)
Theorem 2.1. The problem (1) (5) has an in…nite set of positive eigenvalues. Proof. Di¤erentiating (9) with respect to x, we get
0 2(x; ) = s 1 2; a2 1 sin s a2 x 2 + 0 1 2; 2 cos s a2 x 2 1 a2 2 x Z =2 M ( ) cos s a2 (x ) 2( ( ) ; )d (s = p ; > 0): (14)
From (8)-(10), (12) and (14), we get 1 1 sin cos s 2a1 a1cos s sin s 2a1 1 a1s Z 2 0 M ( ) sin s a1 ( 2 ) 1( ( ); )d ! cos s 2a2 + a2 s 2 s a1 sin sin s 2a1 cos cos s 2a1 1 a2 1 Z 2 0 M ( ) cos s a1 ( 2 ) 1( ( ); )d ! sin s 2a2 1 a2s Z =2 M ( ) sin s a2 ( ) 2( ( ); )d # cos + s a2 1 sin cos s 2a1 a1cos s sin s 2a1 1 a1s Z 2 0 M ( ) sin s a1 ( 2 ) 1( ( ); )d ! sin s 2a2 + 1 2 s a1 sin sin s 2a1 cos cos s 2a1 1 a2 1 Z 2 0 M ( ) cos s a1 ( 2 ) 1( ( ); )d ! cos s 2a2 1 a2 2 Z 2 M ( ) cos s a2 ( ) 2( ( ); )d # sin = 0 (15) There are four possible cases:
1: sin 6= 0; sin 6= 0; 2: sin 6= 0; sin = 0; 3: sin = 0; sin 6= 0; 4: sin = 0; sin = 0:
Let be su¢ ciently large and 1a2= 2a1.
Cases 1 and 4. Then, by (11) and (13), Eq. (15) may be rewritten in the form s sins (a1+ a2)
2a1a2
+ O(1) = 0: (16)
For large s Eq. (16) has an in…nite set of roots.
Cases 2 and 3. In these cases, Eq. (15) assumes the form s coss (a1+ a2)
2a1a2
Obviously, for large s Eq. (17) has, evidently, an in…nite set of roots. The proof is complete.
Thus, by Theorem 1.2 we conclude that the problem (1)-(5) has in…nitely many nontrivial solutions.
3. Asymptotic formulas for eigenvalues and eigenfunctions The function (x; )de…ned in introduction is a nontrivial solution of Eq. (1) sat-isfying conditions (2), (4) and (5). Putting (x; ) into (3), we get the characteristic equation
R( ) ( ; ) cos + 0( ; ) sin = 0: (10)
By Theorem 1 the set of eigenvalues of boundary-value problem (1)-(5) coincides with the set of real roots of Eq. (10). Let M1 = a11
=2 R 0 jM( )jd and M 2 = a21 R =2
jM( )j d . Also let us assume that max k2M12; k2M22 ; k > 1 (k 2 R).
Then for the solution 1(x; ) of Eq. (8), the following inequality holds:
j 1(x; )j q k2M2 1sin 2 + a2 1cos2 (k 1) jM1j ; x 2h0; 2 i : (11)
Di¤erentiating (8) with respect to x, we have
0 1(x; ) = s a1 sin sin s a1 x cos cos s a1 x 1 a2 1 x Z 0 M ( ) cos s a1 (x ) 1( ( ) ; )d : (12) Then from (11) and (12) for the solution 2(x; ) of Eq. (9), the following inequality holds: j 2(x; )j q k2M2 1sin2 + a21cos2 (k 1)2ja1M1 1 2j ; x 2h 2; i : (13)
Theorem 3.1. The problem (1) (5) has an in…nite set of positive eigenvalues. Proof. Di¤erentiating (9) with respect to x, we get
0 2(x; ) = s 1 2; a2 1 sin s a2 x 2 + 0 1 2; 2 cos s a2 x 2 1 a2 2 x Z =2 M ( ) cos s a2 (x ) 2( ( ) ; )d (s = p ; > 0): (14)
From (8)-(10), (12) and (14), we get 1 1 sin cos s 2a1 a1cos s sin s 2a1 1 a1s Z 2 0 M ( ) sin s a1 ( 2 ) 1( ( ); )d ! cos s 2a2 + a2 s 2 s a1 sin sin s 2a1 cos cos s 2a1 1 a2 1 Z 2 0 M ( ) cos s a1 ( 2 ) 1( ( ); )d ! sin s 2a2 1 a2s Z =2 M ( ) sin s a2 ( ) 2( ( ); )d 3 7 5 cos + s a2 1 sin cos s 2a1 a1cos s sin s 2a1 1 a1s Z 2 0 M ( ) sin s a1 ( 2 ) 1( ( ); )d ! sin s 2a2 + 1 2 s a1 sin sin s 2a1 cos cos s 2a1 1 a2 1 Z 2 0 M ( ) cos s a1 ( 2 ) 1( ( ); )d ! cos s 2a2 1 a2 2 Z 2 M ( ) cos s a2 ( ) 2( ( ); )d 3 7 5 sin = 0: (15) There are four possible cases:
1: sin 6= 0; sin 6= 0; 2: sin 6= 0; sin = 0; 3: sin = 0; sin 6= 0; 4: sin = 0; sin = 0:
Let be su¢ ciently large and 1a2= 2a1.
Cases 1 and 4. Then, by (11) and (13), Eq. (15) may be rewritten in the form s sins (a1+ a2)
2a1a2
+ O(1) = 0: (16)
Cases 2 and 3. In these cases, Eq. (15) assumes the form s coss (a1+ a2)
2a1a2
+ O(1) = 0: (17)
Obviously, for large s Eq. (17) has, evidently, an in…nite set of roots. The proof is complete.
Thus, by Theorem 1 we conclude that the problem (1)-(5) has in…nitely many nontrivial solutions. We shall now study the asymptotic properties of eigenvalues and eigenfunctions. In the following we shall assume that s is su¢ ciently large. From (8) and (11), we get
1(x; ) = O(1) on [0;
2] (18)
and from (9) and (13), we get
2(x; ) = O(1) on [2; ] (19)
in the cases 1 and 2. In cases 3 and 4 we have
1(x; ) = O( 1 s) on [0;2] (20) and 2(x; ) = O( 1 s) on [2; ]: (21)
The existence and continuity of the derivatives 01s(x; ) for 0 x 2; j j < 1, and 02s(x; ) for 2 x ; j j < 1, follows from Theorem 1.4.1 in [1]. Using the same technique in the proof of Lemma 3.1 in [1], we have the following equalities:
In cases 1 and 2 0 1s(x; ) = O(1); x 2 [0;2]; (22) 0 2s(x; ) = O(1); x 2 [ 2; ] (23)
and in cases 3 and 4
0 1s(x; ) = O( 1 s); x 2 [0;2]; (24) 0 2s(x; ) = O( 1 s); x 2 [2; ] (25) hold.
Let n be a natural number. We shall say that the number is situated near the number 4n2a21a 2 2 (a1+a2)2 or (2n+1)2a21a 2 2 (a1+a2)2 if, respectively 2na1a2 a1+a2 p <k12 or (2n+1)a1a2 a1+a2 p < 1 k2 (k > 1; k 2 R) :
Theorem 3.2. Let n be a natural number. For each su¢ ciently large n; in cases 1 and 4, there is exactly one eigenvalue of the problem (1) (5) near 4n2a21a22
(a1+a2)2 and
in cases 2 and 3, there is exactly one eigenvalue of this problem near (2n+1)2a21a 2 2
(a1+a2)2 :
Proof. It is similar to proof of Theorem 3.3.1 in [1].
With the helps of Eqs. (16) and (17) we can …nd asymptotic formulas for eigen-values of the problem (1)-(5). Let n be su¢ ciently large. In what follows we shall denote by n = s2n the eigenvalue of the problem (1) (5) situated near
4n2a2 1a 2 2 (a1+a2)2 (or near (2n+1)2a21a 2 2 (a1+a2)2 ).
Cases 1 and 4. We set sn = 2naa1+a1a22 + n. From Eq. (16) it follows that
n = O n1 . Consequently sn= 2na1a2 a1+ a2 + O 1 n : (26)
Cases 2 and 3. We set sn = (2n+1)aa1+a21a2 + n. From Eq. (17) it follows that n = O n1 . Consequently sn= (2n + 1) a1a2 a1+ a2 + O 1 n : (27)
Formula (26) and (27) make it possible to obtain asymptotic expressions for eigen-function of the problem (1) (5). In cases 1 and 2, from (8), (12) and (18), we get 1(x; ) = sin cos s a1 x + O 1 s ; (28)
From (9), (19) and (28), we get
2(x; ) = sin 1 cos s (a2 a1) 2a1a2 + x a2 + O 1 s : (29)
In cases 3 and 4, from (8), (12) and (20), we obtain
1(x; ) = a1cos s sin s a1 x + O 1 s2 ; (30)
From (9), (21) and (30), we get
2(x; ) = a1cos s 1 sin s (a2 a1) 2a1a2 + x a2 + O 1 s2 : (31)
Now we can write the asymptotic representations of eigenfunctions yn(x) = 1
(x; n) for x 2 [0;2); 2(x; n) for x 2 (2; ]:
Case 1. By substituting (26) into (28) and (29), we …nd that 1(x; n) = sin cos 2na2x a1+ a2 + O 1 n ; 2(x; n) = sin 1 cos n (a2 a1) a1+ a2 + 2na1x a1+ a2 + O 1 n : Case 2. By substituting (27) into (28) and (29), we …nd that
1(x; n) = sin cos (2n + 1) a2x a1+ a2 + O 1 n ; 2(x; n) = sin 1 cos (2n + 1) (a2 a1) 2 (a1+ a2) +(2n + 1) a1x a1+ a2 + O 1 n : Case 3. By substituting (27) into (30) and (31), we …nd that
1(x; n) = (a1+ a2) cos (2n + 1) a2 sin(2n + 1) a2x a1+ a2 + O 1 n2 ; 2(x; n) = (a1+ a2) cos (2n + 1) a2 1 sin (2n + 1) (a2 a1) 2 (a1+ a2) +(2n + 1) a1x a1+ a2 + O 1 n2 :
Case 4. By substituting (26) into (30) and (31), we …nd that
1(x; n) = (a1+ a2) cos 2na2 sin 2na2x a1+ a2 + O 1 n2 ; 2(x; n) = (a1+ a2) cos 2na2 1 sin n (a2 a1) a1+ a2 + 2na1x a1+ a2 + O 1 n2 :
4. Sharper asymptotic formulas for eigenvalues and eigenfunctions Under some additional conditions the more exact asymptotic formulas which depend upon the retardation may be obtained. Let us assume that the following conditions are ful…lled:
a)The derivatives M0(x) and 00(x) exist and are bounded in [0; 2)
S
(2; ] and have …nite limits M0(
2 0) = xlim !2 0 M0(x) and 00( 2 0) = xlim !2 0 00(x), respectively. b) 0(x) 1 in [0; 2)S(2; ], (0) = 0 and lim x!2+0 (x) = 0. By using b), we have x (x) 0; x 2 [0; 2); (32) x (x) 2; x 2 (2; ]: (33)
From (28), (29), (32) and (33); in the cases 1 and 2 we have
1( ( ) ; ) = sin cos
s a1
( ( )) + O 1
2( ( ) ; ) = sin 1 cos s a2 (a2 a1) 2a1 + ( ) + O 1 s : (35)
From (30), (31), (32) and (33); in the cases 3 and 4 we have
1( ( ) ; ) = a1cos s sin s a1 ( ( )) + O 1 s2 ; (36) 2( ( ) ; ) = a1cos s 1 sin s a2 (a2 a1) 2a1 + ( ) + O 1 s2 : (37) Let A (x; s; ( )) = 12 x R 0 M ( ) sin as 1 ( ) d ; B(x; s; ( )) = 1 2 x R 0 M ( ) cos s a1 ( ) d :
It is obvious that these functions are bounded for 0 x 2 and 0 < s < 1: Let C(x; s; ( )) = 1 2 x R 2 M ( ) sin s a2 ( ) d ; D(x; s; ( )) = 12 x R 2 M ( ) cos as 2 ( ) d :
It is obvious that these functions are bounded for 2 x and 0 < s < 1: Under the conditions a) and b) the following formulas
8 > > > > > > > > > > > < > > > > > > > > > > > : x R 0 M ( ) cosas 1(2 ( ))d x R 0 M ( ) sinas 1(2 ( ))d x R 2 M ( ) cos s a2(2 ( ))d x R 2 M ( ) sinas 2(2 ( ))d = O 1 s (38)
can be proved by the same technique in Lemma 3.3.3 in [1].
Case 1. Putting the expressions (34) and (35) into (15), and using the equalities in (15), after long operations we have
tans (a1+ a2) 2a1a2 =1 s cos sin 1 +sin cos 2 +B 2; s; ( ) sin a1a2 1 +D ( ; s; ( )) sin sin a2 2 1 # + O 1 s2 :
Again if we take sn= 2naa1+a1a22 + n, then
tan n + n(a1+ a2) 2a1a2 = tan n (a1+ a2) 2a1a2 = a1+ a2 2na1a2
2 4 cos sin 1 +sin cos 2 + B 2;2na1a2 a1+a2; ( ) sin a1a2 1 + D ;2na1a2 a1+a2; ( ) sin sin a2 2 1 3 5 + O 1 n2 :
Hence for large n,
n= 1 n 2 4 cos sin 1 +sin cos 2 + B 2;2na1a2 a1+a2; ( ) sin a1a2 1 + D ;2na1a2 a1+a2; ( ) sin sin a2 2 1 3 5 + O 1 n2 and …nally sn= 2na1a2 a1+ a2 + 1 n 2 4 cos sin 1 +sin cos 2 + B 2;2na1a2 a1+a2; ( ) sin a1a2 1 + D ;2na1a2 a1+a2; ( ) sin sin a2 2 1 3 5 + O 1 n2 : (39)
Case 2. Putting the expressions (34) and (35) into (15), and using the equalities in (15), after long operations we have
cots (a1+ a2) 2a1a2 =1 s " a1cos 1 +B 2; s; ( ) a1 1 +D ( ; s; ( )) a2 1 # + O 1 s2
Again if we take sn= (2n+1)aa1+a21a2 + n, then
cot 2n + 1 2 + n(a1+ a2) 2a1a2 = tan n (a1+ a2) 2a1a2 = a1+ a2 (2n + 1) a1a2 2 4a1cos 1 + B 2;(2n+1)a1a2 a1+a2 ; ( ) a1 1 + D ;(2n+1)a1a2 a1+a2 ; ( ) a2 1 3 5 + O 1 n2 :
Hence for large n,
n = 2 (2n + 1) 2 4a1cos 1 + B 2;(2n+1)a1a2 a1+a2 ; ( ) a1 1 + D ;(2n+1)a1a2 a1+a2 ; ( ) a2 1 3 5 + O 1 n2
and …nally sn= (2n + 1) a1a2 a1+ a2 2 (2n + 1) 2 4a1cos 1 + B 2;(2n+1)a1a2 a1+a2 ; ( ) a1 1 + D ;(2n+1)a1a2 a1+a2 ; ( ) a2 1 3 5 + O 1 n2 : (40) Case 3. Putting the expressions (36) and (37) into (15), and using the equalities in (15), after long operations we have
cots (a1+ a2) 2a1a2 = 1 s " a1cos 1 A 2; s; ( ) sin a2 1 a1D ( ; s; ( )) sin a2 2 1 + O 1 s2
Again if we take sn= (2n+1)aa1+a21a2 + n, then
cot 2n + 1 2 + n(a1+ a2) 2a1a2 = tan n (a1+ a2) 2a1a2 = a1+ a2 (2n + 1) a1a2 2 4a1cos 1 A 2;(2n+1)a1a2 a1+a2 ; ( ) sin a2 1 a1D ;(2n+1)aa1+a21a2; ( ) sin a2 2 1 3 5 + O 1 n2 :
Hence for large n,
n = 2 (2n + 1) 2 4a1cos 1 A 2;(2n+1)a1a2 a1+a2 ; ( ) sin a2 1 a1D ;(2n+1)aa1+a21a2; ( ) sin a2 2 1 3 5 + O 1 n2 and …nally sn= (2n + 1) a1a2 a1+ a2 2 (2n + 1) 2 4a1cos 1 A 2;(2n+1)a1a2 a1+a2 ; ( ) sin a2 1 a1D ;(2n+1)aa1+a21a2; ( ) sin a2 2 1 3 5 + O 1 n2 : (41)
Case 4. Putting the expressions (36) and (37) into (15), and using the equalities in (15), after long operations we have
tans (a1+ a2) 2a1a2 = 1 s " a1D ( ; s; ( )) a2 1 B 2; s; ( ) 1 +a1C ( ; s; ( )) a2 1 # +O 1 s2 :
Again if we take sn= 2naa1+a1a22 + n, then
tan n + n(a1+ a2) 2a1a2 = tan n (a1+ a2) 2a1a2 = a1+ a2 2na1a2 2 4a1D ; 2na1a2 a1+a2; ( ) a2 1 B 2;2na1a2 a1+a2; ( ) 1 + a1C ;2naa1+a1a22; ( ) a2 1 3 5 + O 1 n2 :
Hence for large n,
n= 1 n 2 4a1D ; 2na1a2 a1+a2; ( ) a2 1 B 2;2na1a2 a1+a2; ( ) 1 + a1C ;2naa1+a1a22; ( ) a2 1 3 5 + O 1 n2 and …nally sn= 2na1a2 a1+ a2 + 1 n 2 4a1D ; 2na1a2 a1+a2; ( ) a2 1 B 2;2na1a2 a1+a2; ( ) 1 + a1C ;2naa1+a1a22; ( ) a2 1 3 5+O 1 n2 : (42) Now, we are ready to obtain a sharper asymptotic formula for the eigenfunctions.
Case 1. From (8) and (34)
1(x; ) = sin cos s a1 x a1cos s sin s a1 x sin a1s x Z 0 M ( ) sin s a1 (x ) cos s a1 ( ( )) d : (43)
Thus, using (38) and making necessary arrangements we have 1(x; ) = sin cos s a1 x 1 +A (x; s; ( )) a1s sinas 1x s a1cos + sin B (x; s; ( )) a1 + O 1 s2 : (44)
Now replacing s by sn and using (39) we get
1n(x) = sin cos 2na2x a1+ a2 2 41 + (a1+ a2) A x; 2na1a2 a1+a2; ( ) 2na2 1a2 3 5
sin sin 2na2x a1+ a2 8 < : x n 2 4 cos sin 1 +sin cos 2 + B 2;2na1a2 a1+a2; ( ) sin a1a2 1 + D ;2na1a2 a1+a2; ( ) sin sin a2 2 1 3 5 9 = ; a1+ a2 2na1a2 sin 2na2x a1+ a2 2 4a1cos + sin B x;2na1a2 a1+a2; ( ) a1 3 5 + O 1 n2 :
From (12), (34) and (38), we have
0 1(x; ) s = sin sin s a1x a1 1 +A (x; s; ( )) a1s cos s a1x s a1cos + sin B (x; s; ( )) a1 + O 1 s2 : (45) From (9), (35), (38), (44) and (45) 2(x; ) = 1 1 ( sin cos s 2a1 " 1 + A 2; s; ( ) a1s # sin2as 1 s " a1cos + B 2; s; ( ) sin a1 # + O 1 s2 ) cos s a2 x 2 +a2 2 ( sin sin2as 1 a1 " 1 +A 2; s; ( ) a1s # cos s 2a1 a1s " a1cos + B 2; s; ( ) sin a1 # + O 1 s2 ) sin s a2 x 2 sin a2 1s Z x =2 M ( ) sin s a2 (x ) cos s a2 (a2 a1) 2a1 + ( ) + O 1 s2
= sin 1 cos s a2 (a2 a1) 2a1 + x " 1 + A 2; s; ( ) a1s # 1 1s sin s a2 (a2 a1) 2a1 + x " a1cos + B 2; s; ( ) sin a1 # D (x; s; ( )) sin sin s a2 (a2 a1) 2a1 + x + a2 1s C (x; s; ( )) sin cosas2 (a2 a1) 2a1 + x a2 1s + O 1 s2 : (46)
Now replacing s by sn and using (39) we get
2n(x) = sin 1 cos 2na1 a1+ a2 (a2 a1) 2a1 + x 2 41 + (a1+ a2) A 2; 2na1a2 a1+a2; ( ) 2na2 1a2 3 5 sin 1 sin 2na1 a1+ a2 (a2 a1) 2a1 + x x n cos sin 1 +sin cos 2 + B 2;2na1a2 a1+a2; ( ) sin a1a2 1 + D ;2na1a2 a1+a2; ( ) sin sin a2 2 1 3 5 9 = ; a1+ a2 2na1a2 1 sin 2na1 a1+ a2 (a2 a1) 2a1 + x 2 4a1cos + B 2;2na1a2 a1+a2; ( ) sin a1 3 5 (a1+ a2) D x;2naa1+a1a22; ( ) sin 2na1a22 1 sin 2na1 a1+ a2 (a2 a1) 2a1 + x + (a1+ a2) C x;2naa1+a1a22; ( ) sin 2na1a22 1 cos 2na1 a1+ a2 (a2 a1) 2a1 + x +O 1 n2 :
Case 2. Replacing s by sn in (44), we …nd, by use of (40), that 1n(x) = sin cos (2n + 1) a2x a1+ a2 2 41 +(a1+ a2) A x; (2n+1)a1a2 a1+a2 ; ( ) (2n + 1) a2 1a2 3 5 sin sin(2n + 1) a2x a1+ a2 8 < : 2x (2n + 1) 2 4a1cos 1 + B 2;(2n+1)a1a2 a1+a2 ; ( ) a1 1 + D ;(2n+1)a1a2 a1+a2 ; ( ) a2 1 3 5 9 = ; a1+ a2 (2n + 1) a1a2 sin(2n + 1) a2x a1+ a2 2 4a1cos + sin B x;(2n+1)a1a2 a1+a2 ; ( ) a1 3 5+O 1 n2 :
Now replacing s by sn in (46), we …nd, by use of (40), that
2n(x) = sin 1 cos (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x 2 41 + (a1+ a2) A 2; (2n+1)a1a2 a1+a2 ; ( ) (2n + 1) a2 1a2 3 5 sin 1 sin (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x 2x (2n + 1) 2 4a1cos 1 + B 2;(2n+1)a1a2 a1+a2 ; ( ) a1 1 + D ;(2n+1)a1a2 a1+a2 ; ( ) a2 1 3 5 9 = ; a1+ a2 (2n + 1) a1a2 1 sin (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x 2 4a1cos + B 2;(2n+1)a1a2 a1+a2 ; ( ) sin a1 3 5 (a1+ a2) D x;(2n+1)aa1+a21a2; ( ) sin (2n + 1) a1a22 1 sin (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x + (a1+ a2) C x;(2n+1)aa1+a21a2; ( ) sin (2n + 1) a1a22 1 cos (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x + O 1 n2 :
Case 3. By use of (8), (36) and (38) and making necessary arrangements we have 1(x; ) = B (x; s; ( )) cosas 1x s2 sinas 1x s a1cos + A (x; s; ( )) s +O 1 s3 : (47) Now replacing s by sn and using (41) we get
1n(x) =
(a1+ a2)2B x;(2n+1)aa1+a21a2; ( ) cos(2n+1)aa1+a22x
(2n + 1)2a2 1a22 (a1+ a2) sin(2n+1)aa1+a22x (2n + 1) a1a2 2 4a1cos + (a1+ a2) A x;(2n+1)aa1+a21a2; ( ) sin (2n + 1) a1a2 3 5 +2 (a1+ a2) x cos cos (2n+1)a2x a1+a2 (2n + 1)2a2 2 4a1cos 1 A 2;(2n+1)a1a2 a1+a2 ; ( ) sin a2 1 a1D ;(2n+1)aa1+a21a2; ( ) sin a2 2 1 3 5 + O 1 n3 :
From (12), (37) and (38), we have
0 1(x; ) s = a1 h D (x; s; ( )) cosas 2 (a2 a1) 2a1 + x i s2a 2 1 + a1 h C (x; s; ( )) sinas 2 (a2 a1) 2a1 + x i s2a 2 1 + O 1 s3 : (48) From (9), (37), (38), (47) and (48) 2(x; ) = B 2; s; ( ) cosas2 (a2 a1) 2a1 + x s2 1 sin s a2 (a2 a1) 2a1 + x a1cos + A(2;s; ( )) s s 1 + a1 h D (x; s; ( )) cosas 2 (a2 a1) 2a1 + x i s2a 2 1 + a1 h C (x; s; ( )) sinas 2 (a2 a1) 2a1 + x i s2a 2 1 + O 1 s3 : (49)
Now replacing s by sn and using (41) we get 2n(x) = (a1+ a2)2B 2;(2n+1)aa1+a21a2; ( ) (2n + 1)2a2 1a22 1 cos (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x (a1+ a2) (2n + 1) a1a2 1 sin (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x 2 4a1cos + (a1+ a2) A 2;(2n+1)aa1+a21a2; ( ) sin (2n + 1) a1a2 3 5 +2 (a1+ a2) x cos (2n + 1)2a2 1 cos (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x 2 4a1cos 1 A 2;(2n+1)a1a2 a1+a2 ; ( ) sin a2 1 a1D ;(2n+1)aa1+a21a2; ( ) sin a2 2 1 3 5 + (a1+ a2) 2 (2n + 1)2a1a32 1 D x;(2n + 1) a1a2 a1+ a2 ; ( ) cos (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x +C x;(2n + 1) a1a2 a1+ a2 ; ( ) sin (2n + 1) a1 a1+ a2 (a2 a1) 2a1 + x + O 1 n3 :
Case 4. Replacing s by sn in (47), we …nd, by use of (42), that
1n(x) =
(a1+ a2)2B x;2naa1+a1a22; ( ) cos2naa1+a2x2
4n2a2 1a22 (a1+ a2) sin2naa1+a2x2 2na1a2 2 4a1cos + (a1+ a2) A x;2naa1+a1a22; ( ) 2na1a2 3 5 +(a1+ a2) x cos cos 2na2x a1+a2 2n2a 2 2 4a1 h D ;2na1a2 a1+a2; ( ) + C ; 2na1a2 a1+a2; ( ) i a2 1 B 2;2na1a2 a1+a2; ( ) 1 3 5+O 1 n3 :
Now replacing s by sn in (49), we …nd, by use of (42), that
2n(x) = (a1+ a2)2B 2;2naa1+a1a22; ( ) 4n2a2 1a22 1 cos 2na1 a1+ a2 (a2 a1) 2a1 + x (a1+ a2) 2na1a2 1 sin 2na1 a1+ a2 (a2 a1) 2a1 + x
2 4a1cos + (a1+ a2) A 2;2naa1+a1a22; ( ) sin 2na1a2 3 5 (a1+ a2) x cos 2n2a 2 1 cos 2na1x a1+ a2 (a2 a1) 2a1 + x 2 4a1 h D ;2na1a2 a1+a2; ( ) + C ; 2na1a2 a1+a2; ( ) i a2 1 B 2;2na1a2 a1+a2; ( ) 1 3 5 +(a1+ a2) 2 4n2a 1a32 1 D x;2na1a2 a1+ a2 ; ( ) cos 2na1 a1+ a2 (a2 a1) 2a1 + x +C x;2na1a2 a1+ a2 ; ( ) sin 2na1 a1+ a2 (a2 a1) 2a1 + x + O 1 n3 : References
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Current address : Erdo¼gan ¸Sen: Department of Mathematics, Nam¬k Kemal University, 59030, Tekirda¼g, Turkey
E-mail address : erdogan.math@gmail.com
Current address : Azad Bayramov: Azerbaijan State Pedagogical University, 1000, Baku, Azer-baijan
