Selçuk J. Appl. Math. Selçuk Journal of Vol. 10. No. 1. pp. 3-18, 2009 Applied Mathematics
Existence and Uniquness Theorems for A Certain Class of Non Linear Singular Integral Equations
Binali Musayev1, Murat Düz2
1Kırıkkale University, Faculty of Sciences and Arts, Department of Mathematics,
71450, Yah¸sihan - Kırıkkale, Turkey
2Uludag University, Faculty of Sciences and Arts, Department of Mathematics, Gorukle
-Bursa, Turkey
e-mail: m duz7837@ yaho o.com
Received: October 4, 2007
Summary. In this study, the existence of solution of the non-linear singular integral equation system
() = 1( () () 1( () ())) () () = 2( () () Q2( () ())) () which is more general than the problem
= ( ) |= () ∈ () (0) = 0 0∈
which has been studied by Ishak Altun , Kerin Koca and Binali Musayev, has been investigated by the same method. Here, () andQ () are Vekua integral operators defined by
() = −1 RR () − Q () = − 1 RR () (−)2
Key words: Singular integral equations,Holder continuity, fixed point theorem, contraction mapping.
1. Introduction
Let ⊂ be a simply connected region with smooth boundary .As known, the system of real partial differential equations of the form
− = 1( ) + = 2( ) is equivalent to the complex partial differential equation
(1.1) = ( ) where = + = + = 1 2 µ + ¶ = 1 2 µ − ¶
The existence of the solution of the equation (1.1) satisfying the Dirichlet bound-ary conditions
|= () ∈ () (0) = 0 0∈
in Holder space ¡¢, under suitable conditions, had been investigated by W. Tutschke(see[1]).
Let the function in (1.1) be a complex valued scalar function defined on the region
=©( ) : ∈ ∈ ª= 2 and let us consider the operators
() = −1RR () − Q () = −1 RR () (−)2 = +
for ∈ ¡¢. In this case, the solutions w of the equation(1.1) satisfy the system of nonlinear singular integral equations
() = () + ( () ()) () () = 0() +Q ( () ()) ()
where = and ()’s are arbitrary holomorphic functions defined on . In this paper, the more general nonlinear singular integral system
(1.2) () = 1( () () 1( () ())) () () = 2( () () Q2( () ())) ()
will be discussed for given fonctions 1 2 1 2under some conditions.
2. The existence, uniqueness and determination of the solution of the system of singular integral equations
In this section, we will present some theorems related to the solutions of the system (1.2) under suitable conditions.
Definition 1.If for every 1 2∈ there are constants 0 and satisfying the inequality
| (2) − (1)| ≤ |2− 1| 0 1
then the function : → is said to satisfy the Holder condition in the region or to be Holder continuous.
Let us denote the class of Holder continuous functions defined on by ¡¢. This class is a vector space. On the other hand, (0)¡¢≡ ¡¢is the class of the continuous functions on
and for w∈ ¡¢in this class, the norm is defined to be kk∞= kk() = max
©
| ()| : ∈ ª On the other hand, if the norm for ∈ ¡¢is defined as
kk≡ kk()() = kk∞+ ( )
where
( ) = supn| (1) − (2)| |1− 2|−: 16= 2 1 2∈ o
then the class ¡¢becomes a Banach space with this norm.
Let us denote the Holder continuous functions defined on and having partial derivatives of first order with respect to variables z and by (1)¡¢. This class constitutes a Banach space with norm
kk1 ≡ kk(1)() = max {kk kk kk} for ∈ (1)¡¢.
Moreover the vector spaces
2¡¢= ¡¢× ¡¢=©( ) : ∈ ¡¢ª
(2)¡¢= ()¡¢× ()¡¢=©( ) : ∈ ()¡¢ª having following norms;
k( )k∞2 ≡ k( )k2() = max {kk∞ kk∞}
k( )k2 ≡ k( )k(2)() = max {kk kk}
become Banach spaces. We denote these spaces by ³ 2¡¢; k( )k∞2´ and ³(2)¡¢; k( )k2´ respectively. Let ¡¢ = ½ :RR | ()| ∞ ¾ 1 ≤ ∞.Then for ∈ ¡¢ consider the norm
k( )k2≡ k( )k2
() = max
n
kk kk o
Defined for ( ) ∈ 2() where 2 ¡ ¢= ¡¢× ¡¢ and kk≡ kk() = ⎛ ⎝ZZ | ()| ⎞ ⎠ 1 Let = max 12∈ |1− 2|.
Lemma 1. 1 2 ∈ ([7]). If ( ) ∈ ()2¡¢ 0 1 then for 1 ∞and 0 ≤ we have the following inequality:
k( )k∞2≤ 2k( )k2+ 1
(2)1k( )k2
Theorem 1([7]). For ( ) ∈ ()2¡¢ 0 1 and 1 ∞ the following inequality holds:
(2.1) k( )k∞2≤ ( ) k( )k 2 2+ 2 k( )k 2+ Here ( ) = max {1( ) 2( )} where ( ) = (√)− 2+ 1( ) = 2( ) +¡2( )¢− 1 2( ) = 2 √ 4 √ 4−1 ( )
Definition2. Let : → where = 3 be given. If, for every (1 1 1 1) (2 2 2 2) ∈ , there are positive numbers 1 2 3 4 satisfy-ing (2.2) ¯ ¯ (1 1 1 1) − ¡ 2 2 2 2¢¯¯≤ 1|1−2| +2 ¯ ¯1−2 ¯ ¯ +3 ¯ ¯1−2 ¯ ¯ +4|1−2|
then the function is said to be of class 111 ¡ 1 2 3 4; ¢ over and we write ∈ 111 ¡ 1 2 3 4; ¢ .
Definition3. Let ∗ : 1 → where 1 = 2. If, for every (
1 1 1) (2 2 2) ∈ 1, there are positive numbers 1 2 3 satisfying
(2.3) |∗(1 1 1) − ∗(2 2 2)| ≤ 1|1−2|+2|1−2| +3|1−2| then the function ∗ is said to be of class 11¡1 2 3 4; 1¢ over1 and we write ∈ 11¡1 2 3; 1¢.
Let the Vekua integral operators
() () = − 1 ZZ () − Y () () = −1 ZZ () ( − )2 where = + and = + be given for ∈ ¡¢ (0 1) Let for the bounded operators
kk= sup©kk: ∈ ()¡¢ kk 1ª kQk= sup © kk: ∈ () ¡ ¢ kk 1ª Lemma 2:Let ∈ 111 ¡ 1 2 3 4; ¢ ∈ 11 ¡ 1 2 3; 1 ¢ ( = 1 2) = (0 0) and ( ) = n ( ) : k( )k2≤ o .If 0= max©|( 0 0 0)| : ∈ ª = max © |( 0 0)| : ∈ ª 1= 01+ 11+ 2 (12+ 13) + 214[01+ 211+ 2 (12+ 13) ] kk 2= 02+ 21+ 2 (22+ 23) + 224[02+ 221+ 2 (22+ 23) ] kQk and max {1 2} ≤ then for
˜ () = 1( () () 1( () ()) ()) ˜ () = 2( () () Q2( () ()) ()) the operator : (2)¡¢→ (2)¡¢ 0 1 ( ) → ( ) =³ ˜˜ ´ transforms the ball (; ) into itself.
Proof: From the definition of ˜ (),
| ˜ ()| = |1( () () 1( () ()) ())|
≤ |1( () () 1( () ()) ()) − 1( 0 0 1( 0 0) ())| + |1( 0 0 1( 0 0) ())| From the inequality (2.2) we can write
| ˜ ()| ≤ 12| ()| +13| ()| +14|[1( () ()) () −1( 0 0) ()]| + 14|1( 0 0) ()| + |1( 0 0 0)| Thus (2.4) ≤ 12| ()| +13| ()| +14kkk1( () ()) − 1( 0 0)k()( 1) +14kkk1( 0 0)k()(1) +01 Now let us obtain a bound for k1( () ()) − 1( 0 0)k()(
1).For every 1 2∈ (2.5) |1( () ()) − 1( 0 0)| ≤ 12| ()| +13| ()| ≤ (12+ 13) and |[1( () ()) − 1( 0 0)] (1) − [1( () ()) − 1( 0 0)] (2)| = |1(1 (1) (1)) − 1(2 (2) (2)) − [1(1 0 0) − 1(2 0 0)]| ≤ 11|1− 2|+ 12| (1) − (2)| + 13| (1) − (2)| + 11|1− 2| ≤ 211|1− 2|+ 12kk()() |1− 2|+ 13kk()() |1− 2| ≤ [211+ (12+ 13) ] |1− 2|
from the inequality(2.5), we can write
(2.6) k1( () ()) − 1( 0 0)k()(1) ≤ 2 [11+ (12+ 13) ] Now let us obtain a bound for k1( 0 0)k()(1). For any 1 2∈ , since
|1( 0 0) (2) − 1( 0 0) (1)| = |1(2 0 0) − 1(1 0 0)| ≤ 11|1− 2| we can write
(2.7) k1( 0 0)k()(1) ≤ 11+ 01
Using the inequalities (2.6) and (2.7) in (2.4), for every ∈ , we obtain
| ˜ ()| ≤ 01+ (12+ 13) + 14[01+ 311+ 2 (12+ 13) ] kk Now, let us obtain the Holder constant ( ˜ ) . For every 1 2∈
| ˜ (1) − ˜ (2)| ≤ |1(1 (1) (1) 1( () ()) (1))
−1(2 (2) (2) 1( () ()) (2))| ≤ 11|1− 2|+12| (1) − (2)| +13| (1) − (2)| + 14|[1( () ()) (1) − 1( () ()) (2)]|
≤ [11+ (12+ 13) ] |1− 2|+ 14kkk1( () ())k()(1) |1− 2| Moreover, for every 1 2∈
|1(1 () ())| ≤ |1(1 () ()) − 1( 0 0)| + |1( 0 0)| ≤ (12+ 13) + 01
From the definition of (2.3)
|1(1 (1) (1)) − 1(2 (2) (2))| ≤ [11+ (12+ 13) ] |1− 2| Hence, for any 1 2∈
| ˜ (2) − ˜ (1)| ≤ [11+(12+13) + 14kk(01+11+2(12+13))] |1−2| or we obtain
Thus, for 1= 01+ 2 (12+ 13) + 11+ 214(01+ 211+ 2 (12+ 13) ) kk we get k ˜k≤ 1 in a similarway, for 2= 02+ 2 (22+ 23) + 11+ 224(02+ 221+ 2 (22+ 23) ) ° ° ° ° ° Y ° ° ° ° °
it can be shown that
° ° °˜°°° ≤ 2 Therefore, ° ° °³ ˜˜ ´°°° 2 = max n k ˜k ° ° °˜°°° o ≤ max {1 2} If max {1 2} ≤ , then ° ° °³ ˜˜ ´°°° 2 ≤ , i.e., ( ) = ³ ˜ ˜´∈ ( ) Lemma 3: If Φ ∈ (1)¡¢, then the ball ()2[(Φ Φ0); ] is compact in ³
2¡¢; k( )k∞2´
Proof. For ( ) ∈ ()2[(Φ Φ0); ] we have k( ) − (Φ Φ0)k ≤ .Therefore, if for any 0 we choose =¡
¢1
, then for all 1 2∈ we have ¯
¯( (1) (1)) −(Φ (1) Φ0(1))− ( (2) (2)) +(Φ (2) Φ0(2)) ¯
¯ ≤ |1−2|
Whenever |1− 2| ; whence it follows that ()2[(Φ Φ0); ] is uniformly bounded, and its elements are continuous of the same order. Therefore, by Arzela-Ascoli Theorem, the ball ()2[(Φ Φ0); ] is a compact subset of ³
2¡¢; k( )k∞2´.
Corollary 1. The sphere ( ) is a complete subspace of ³ 2¡¢; k( )k ∞2 ´ . For ( ) ³ ˜˜ ´∈ ()2¡¢ (0 1) let ∞2h( ) ³ ˜˜ ´i=°°°( ) −³ ˜˜ ´°°° ∞2
and for 1 ≤ ∞ 2 h ( ) ³ ˜˜ ´i=°°°( ) −³ ˜˜ ´°°° 2 2 h ( ) ³ ˜˜ ´i=°°°( ) −³ ˜˜ ´°°° 2 The transformations ∞2 2: ()2¡¢()2¡¢→ [0 ∞) Define metrics on ()2¡¢. Thus,¡()2¡¢;
∞2¢ and ¡()2¡¢; 2¢ become metric spaces.
Lemma 4. Let 0 1 and 1 ≤ ∞. Then the convergence on the ball ( ) with respect to the metrics∞2 and 2 are equivalent.
Proof.Let (0 0), ( ) ∈ ( ) and lim
→∞∞2[( ) (0 0)] = 0, ( = 1 2 ). Then from the inequality
2[( ) (0 0)] = max n k− 0k k− 0k o ≤ ()1∞2[( ) (0 0)] we obtain lim →∞2[( ) (0 0)] = 0 where mG is the area in .
Let us prove the converse statement: From Theorem 1 (inequality(2.1)) we have
k( )− (0 0)k∞2 ≤ ( ) k( ) − (0 0)k 2 2+ 2 k( ) − (0 0)k 2+ 2 or ∞2[( ) (0 0)] ≤ (2) 2 2+ ( ) 2+ 2 [( ) (0 0)] Thus, we obtain lim →∞2[( ) (0 0)] = 0 and consequently lim →∞∞2[( ) (0 0)] = 0
Lemma 5. Let ∈ 111 ¡ 1 2 3 4; ¢ and ∈ 11 ¡ 1 2 3;1 ¢ , ( = 1 2) (0 1) and 1 ∞. In this case, for the operator A defined in Lemma 2, the inequality
(2.8) 2[ (1 1) (2 2)] ≤ 3() ∞2[(1 1) (2 2)] is satisfied for all
(1 1) (2 2) ∈ ( ) where 3() = ()1max n 12+13+14(12+ 13) kk 22+23 +24(22+ 23) kQk o
Proof. For all (1 1) (2 2) ∈ ( ) and ∈ using
(2.9) k (1 1) − (2 2)k2= ° ° °³˜1 ˜1 ´ −³˜2 ˜2´°°° 2 = max ½ k ˜1− ˜2k ° ° °˜1− ˜2 ° ° ° ¾
let us turn to account the norms k ˜1− ˜2k and ° ° °˜1− ˜2 ° ° ° . For all ∈ since | ˜1() − ˜2()| = |1( 1() 1() 1( 1() 1()) ()) − 1( 2() 2() 1( 2() 2()) ())| ≤ 12|1() − 2()| + 13|1() − 2()| +14|(1( 1() 1()) − 1( 2() 2())) ()| from the Minkowski inequality and 1∈ 11
¡ 11 12 13; 1 ¢ we obtain à RR ¨ | ˜1() − ˜2()| !1 ≤ ⎧ ⎨ ⎩ Z Z [12|1() − 2()| + 13|1() − 2()| +14|(1( 1() 1()) − 1( 2() 2()) ())|]} 1 ≤ 12k1− 2k + 13k1− 2k+ 14kkk1( 1() 1()) − 1( 2() 2())k ≤ 12k1− 2k+ 13k1− 2k+ 14kk ³ 12k1− 2k+ 13k1− 2k ´ ≤h12+ 13+ 14(12+ 13) kk i maxnk1− 2k k1− 2k o ≤ ()1h12+ 13+ 14(12+ 13) kk i ∞2[(1 1) (2 2)] Thus we get
(2.10) k ˜1− ˜2k≤ () 1h 12+ 13+ 14(12+ 13) kkˆ∞2 i and similarly (2.11) ° ° °˜1− ˜2 ° ° ° ≤ () 1 ⎡ ⎣22+ 23+ 24(22+ 23) ° ° ° ° ° Y ° ° ° ° ° ˆ ∞2 ⎤ ⎦
where ˆ∞2 = ∞2[(1 1) (2 2)] With the help of the inequalities (2.10) and (2.11), the required inequality (2.8) is obtained from (2.9).
Lemma 6. Assume the conditions of the Lemma 5 are satisfied. Let max {1 2} ≤ . In this case, the operator : (; ) → (; ) defined in the Lemma 2, is a continuous operator according to the metric ∞2.
Proof. Let (0 0) ( ) ∈ ()2[(Φ Φ0) ; ] and let lim
→∞∞2[( ) (0 0)] = 0 Now, we have to show that
lim
→∞∞2[ ( ) (0 0)] = 0 from the Lemma 1 we can write
k[ ( ) (0 0)]k∞2≤ ( ) k ( ) − (0 0)k 2 2+ 2 k ( ) − (0 0)k 2+ 2
Since k[ ( ) (0 0)]k2≤ 2, making use of (2.8), we obtain
k[ ( ) (0 0)]k∞2≤ ( ) [3()] 2+k( ) − (0 0)k 2+ ∞2 k ( ) − (0 0)k 2 2+ 2 ≤ (2)2+2 ( ) [ 3()] 2+k( ) − (0 0)k 2+ ∞2 Thus, the proof is complete.
Theorem 2. Let ∈ 111¡1 2 3 4; ¢ ∈ 11¡1 2 3; 1¢, ( = 1 2) and max {1 2} ≤ . The nonlinear singular integral system (1.2) has at least one solution on the sphere ( ).
Proof.Since max {1 2} ≤ the operator A defined in Lemma 2 transforms the convex, (closed) and compact subset ( ) of ()2
¡
itself and in addition, since the operator A is continuous on this set with respect to the metric ∞2 , it has a at least one solution in the set ( )(or in the space ()2¡¢of the system (1.2)). Thus, the theorem is proved.
Now, let us study the uniqueness of the solution of the system (1.2) and how to find it. For this, we use a variant of Banach fixed point theorem:
Theorem 3.([6]).Assume that the following hypothses hold: 1. Let ( 1) be a compact metric space.
2. Let 2 be another metric on X such that any sequence converging with respect to 1 is also convergent in2.
3. Let the operator : → be a contraction mapping with respect to 2, i.e. let for any ∈ there exist a number 0 ≤ 1 such that 2( ) ≤ 2( ).
Then the equation = has a unique solution ∗ and 0 ∈ being any initial element, the sequence () defined by = −1, = 1 2
converges to ∗ with speed
2( ∗) ≤
1 − 2(1 0)
Theorem 4. Let the conditions ∈ 111 ¡ 1 2 3 4; ¢ ∈ 11 ¡ 1 2 3; 1 ¢ ,( = 1 2) (0 1), max {1 2} ≤ and = max ( 12+13+ (12+ 13) 14kk 22+23+ (22+ 23) 24 ° ° ° °Q ° ° ° ° ) 1 hold. Then the system (1.2) of nonlinear singular integral equations has a unique solution (∗ ∗) ∈ ( ). This solution is the limit of the sequence ( ) defined by
(2.12) () = 1( −1() −1() 1( −1() −1())) () () = 2( −1() −1() Q2( −1() −1())) () = 1 2 where (0 0) ∈ ( ) is any initial element. Moreover, the inequality
(2.13) 2[( ) (∗ ∗)] ≤
1 − 2[(1 1) (0 0)] hold.
Proof. Let = ( ) 1 = 2 and 1 = 2 in Theorem 3. Let A be an operator defined in the Lemma 2. Since max {1 2} ≤ , the operator A transform the space ( 2) into itself.
Now let us show that when 1 the operator A is a contraction operator on the sphere ( )with respect to the metric 2
For ( ) ∈ ( )
( ) () = (1( ) () 2( ) ()) where
1( ) () = 1( () () 1( () ()) ()) 2( ) () = 2( () () 2( () ()) ()) Thus for any¡(1) (1)¢¡(2) (2)¢∈ (; ) we can write
2 ³ ³(1) (1)´ ³(2) (2)´´ = max½°°°1 ³ (1) (1)´− 1 ³ (2) (2)´°°° ° ° °2 ³ (1) (1)´− 2 ³ (2) (2)´°°° ¾ Since 1∈ 111¡11 12 13 14; ¢ 1∈ 11¡11 12 13; 1¢ ° ° °1 ³ (1) (1)´− 1 ³ (2) (2)´°°° = ° ° °1 ³ (1)() (1)() 1 ³ (1)() (1)()´´ −1 ³ (2)() (2)() 1 ³ (2)() (2)()´´°°° ≤ ⎧ ⎨ ⎩ Z Z h 12 ¯ ¯ ¯(1)() − (2)()¯¯¯ + 13 ¯ ¯ ¯(1)() − (2)()¯¯¯ +14 ¯ ¯ ¯ ³ 1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´()¯¯¯io 1 ≤ 12 ° ° °(1)− (2)°°° +13 ° ° °(1)− (2)°°° +14 ⎛ ⎝Z Z ¯ ¯ ¯ ³ 1 ³ (1)() (1)()´ −1 ³ (2)() (2)()´´()¯¯¯´ 1 ≤ 12 ° ° °(1)− (2)°°° + 13 ° ° °(1)− (2)°°° + +14kk ⎛ ⎝Z Z ¯ ¯ ¯³1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´¯¯¯ ⎞ ⎠ 1
≤ 12 ° ° °(1)− (2)°°° + 13 ° ° °(1)− (2)°°° + 14kk µ 12 ° ° °(1)− (2)°°° +13 ° ° °(1)− (2)°°° ¶ =³12+ 1214kk´ °°° (1) − (2)°°° + ³ 13+ 1413kk´ °°° (1) − (2)°°° ≤ 12 ³³ (1) (1)´³(2) (2)´´ where 1= 12+ 13+ (12+ 13) 14kk Similarly ° ° °2 ³ (1) (1)´− 2 ³ (2) (2)´°°° ≤ 22 ³³ (1) (1)´³(2) (2)´´ where 2= 22+ 23+ (22+ 23) 24 ° ° ° ° ° Y ° ° ° ° ° Thus when = max {1 2} for every
¡ (1) (1)¢¡(2) (2)¢∈ ( ) we can write 2 h ³(1) (1)´ ³(2) (2)´i≤ 2 h³ (1) (1)´³(2) (2)´i Thus when 1, the operator is contraction operator on the sphere ( ) with respect to the metric 2.
By Lemma 2,the system (1.2) has at least one solution in the ball ()2[(Φ Φ0) ; ]. Now let us show this fact: Since
( ) = (−1 −1) = 1 2 by considering (2.13) we obtain
2[(+1 +1) ( )] = 2[ ( ) (−1 −1)] ≤ 2[( ) (−1 −1)] Repeating this process consecutively, it follows that
Thus, for any two natural numbers m and n we can write (2.14) 2[(+ +) ( )] ≤ ¡ 1 + + 2+ + −1¢2[(1 1) (0 0)] = 1−1−2[(1 1) (0 0)] Since lim →∞ = 0, the sequence {( )}∞1 is Cauchy by (2.14). Since ( 2) is complete, there is an element (∗ ∗) ∈ such that lim
→∞( ) = (∗ ∗). On the other hand,
2[(+1 +1) (∗ ∗)] = 2[ ( ) (∗ ∗)] ≤ 2[( ) (∗ ∗)] and lim →∞2[( ) (∗ ∗)] = 0 imply that lim →∞2[(+1 +1) (∗ ∗)] = 0 and thus lim →∞2(+1 +1) = (∗ ∗)
So we get (∗ ∗) = (∗ ∗), and this shows that (∗ ∗) is a solution to the equation ( ) = ( ).
Now let us prove the uniqueness of this solution: Let(∗∗ ∗∗) be another so-lution of the system (2.12). In this case, by (2.13), we can write
2[(∗ ∗) (∗∗ ∗∗)] = 2[ (∗ ∗) (∗∗ ∗∗)] ≤ 2[(∗ ∗) (∗∗ ∗∗)] However, this is possible only if 2[(∗ ∗) (∗∗ ∗∗)] = 0.
Remark 1.Since, by (2.14), the sequence {( )}∞1 ,whose terms are de-fined by ( ) = (−1 −1)(or by (2.12)), is convergent to the solution (∗ ∗) in the ball ()2[(Φ Φ0) ; ] with respect to the metric
2 , it is also convergent with respect to the metric ∞2. Thus the metrics ∞2 and 2are equivalent on .
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