Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 1. pp. 53-70, 2011 Applied Mathematics
On Solution of A Certain Class of Non Linear Singular Integral Equa-tions
Murat Düz
Karabuk University, Faculty of Sciences and Arts, Department of Mathematics, 78050 Karabuk, Turkiye
e-mail:m duz@ karabuk.edu.tr
Received Date: February 22, 2010 Accepted Date: September 2, 2010
Abstract. In this study, the existence of solution of the non-linear singular integral equation system
() = 1 Ã () () 1( () ()) Y 1( () ()) ! () () = 2 Ã () () 2( () ()) Y 2( () ()) ! () has been investigated.
Here,
()
and Y
() are Vekua integral operators defined by
() = − 1 ZZ () − Y () = − 1 ZZ () ( − )2
Key words: Singular integral equations; Holder continuity; fixed point theo-rem; contraction mapping.
1. Introduction
Let ⊂ be a simple 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) = ( ) where = + = + = 1 2 µ + ¶ = 1 2 µ − ¶ The existence of the solution of the equation (1) satisfying the Dirichlet bound-ary conditions
Re |= () ∈ () Im (0) = 0 0∈
in Holder space ¡¢, under suitable conditions, had been investigated by [1],[4].Let the function in (1) be a complex valued scalar function defined on the region
=©( ) : ∈ ∈ ª= x2
and let ∈ ¡¢ In this case, the solutions of the equation (1) satisfy the system of nonlinear singular integral equations
(2) () = () + ( () ()) () () = 0() +Q ( () ()) ()
where = and ()’s are arbitrary holomorphic functions defined on G. In [3], we studied the more general nonlinear singular integral system
(3) () = 1( () () 1( () ()) ()) () = _2( () ()Q2( () ())())
In [5], the more general than (3) nonlinear singular integral system (4) () = 1( () () 1( () ())
Q
1( () ())) () () = 2( () () 2( () ()) Q2( () ())) () was studied. In this paper the more general than (4) nonlinear singular integral equation system
(5) () = 1( () () 1( () ()) Q
1( () ())) () () = 2( () () 2( () ()) Q2( () ())) () will be discussed for given functions 1 2 1 2 1 2 under some conditions. 2. The Existence, Uniqueness and Determination of The Solution The System of Singular Integral Equations
In this section, some theorems related to the solutions of the system (5 ) under suitable conditions will be presented.
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 ∈ ¡¢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 ( ) = sup n | (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
for ∈ (1)¡¢.Moreover the vector spaces 2¡¢= ¡¢x¡¢=©( ) : ∈ ¡¢ª ()2¡¢= ()¡¢x()¡¢=n( ) : ∈ ()¡¢o having 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 ¡¢= ⎧ ⎨ ⎩ : ZZ | ()| ∞ ⎫ ⎬ ⎭ 1 ≤ ∞ Define for ( ) ∈ 2 ¡ ¢ k( )k2≡ k( )k2 () = max n kk kk o where 2¡¢= ¡¢x¡¢ and kk≡ kk() = ⎛ ⎝ZZ | ()| ⎞ ⎠ 1 Let = max 12∈ |1− 2| Lemma 1.([1]). If ( ) ∈ ()2¡¢ 0 1
then for 1 ∞ and 0 ≤ we have the following inequality: k( )k∞2 ≤ 2k( )k2+
1
(2)1 k( )k2
Theorem 1. ([1]). For
( ) ∈ ()2¡¢ 0 1 and 1 ∞ the following inequality holds:
k( )k∞2≤ ( ) k( )k 2 2+ 2 k( )k 2+ 2 Here ( ) = max {1( ) 2( )} where ( ) =¡√ ¢− 2+ 1( ) = 2( ) + ¡ 2( )¢− 1 2( ) = 2√ 4 √ 4 − 1 ( )
Definition 2. Let : → where = × 4be given. If for every (1 1 1 1 1) (2 2 2 2 2) ∈
there are positive numbers 1 2 3 4satisfying
(6) | (1 1 1 1 1) − (2 2 2 2 2)| ≤ 1|1− 2|
+ 2|1− 2| +3|1− 2| + 4|1− 2| + 5|1− 2|
then the function is said to be of class 111¡1 2 3 4; ¢over and we write ∈ 1111¡1 2 3 4 5; ¢
Definition 3. Let ∗ :
1 → where 1 = × 2If for every (1 1 1) (2 2 2) ∈ 1 there are positive numbers 1 2 3 satisfying
(7) |∗(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 ¢ over 1 and we write ∗∈ 11 ¡ 1 2 3; 1 ¢ Let for the bounded operators
kk= sup n kk: ∈ () ¡ ¢ kk 1o ° ° ° ° ° Y ° ° ° ° ° = supnkk: ∈ ()¡¢ kk 1 o .
Lemma 2. Let ∈ 1111 ¡ 1 2 3 4 5; ¢ ∈ 11 ¡ 1 2 3; 1 ¢ ∈ 11¡1 2 3; 1¢ ( = 1 2) = (0 0) and ( ) = n ( ) : k( )k2≤ o If 0 = max © |( 0 0 0 0)| : ∈ ª = max©|( 0 0)| : ∈ ª = max © |( 0 0)| : ∈ ª 1= 01+ 11+ 2 (12+ 13) + [201+ 411+ 4 (12+ 13) ] 14kk+ [201+ 411+ 4 (12+ 13) ] 15 ° ° ° ° ° Y ° ° ° ° ° 2= 02+ 21+ 2 (22+ 23) + [202+ 421+ 4 (22+ 23) ] 24kk+ [202+ 421+ 4 (22+ 23) ] 25 ° ° ° ° ° Y ° ° ° ° ° , max {1 2} ≤ then for ˜ () = 1 à () () 1( () ()) () Y 1( () ()) () ! ˜ () = 2 à () () 2( () ()) () Y 2( () ()) () ! the operator : ()2¡¢→ ()2¡¢ 0 1 ( ) → ( ) =³ ˜˜ ´ transforms the ball
(; ) into itself.
Proof. From the definition of ˜ (), | ˜ ()| = ¯ ¯ ¯ ¯ ¯1 à () () 1( () ()) () Y 1( () ()) () !¯¯¯ ¯ ¯ ≤ ¯ ¯ ¯ ¯ ¯1 à () () 1( () ()) () Y 1( () ()) () ! −1 à 0 0 1( 0 0) () Y 1( 0 0) () !¯¯¯ ¯ ¯ + ¯ ¯ ¯ ¯ ¯1 à 0 0 1( 0 0) () Y 1( 0 0) () ! − 1( 0 0 0 0) ¯ ¯ ¯ ¯ ¯ + |1( 0 0 0 0)| .
From the inequality (6) and (7) we can write
(8)
| ˜ ()| ≤ 12| ()| + 13| ()| +14|[1( () ()) () − 1( 0 0) ()]| +15|Q[1( () ()) () − 1( 0 0) ()]| +14|1( 0 0) ()| + 15|Q1( 0 0) ()| + 01 Now let us obtain a bound for
k1( () ()) − 1( 0 0)k()(1) For every 1 2∈ (9) |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| (10) ≤ [211+ (12+ 13) ] |1− 2| from the inequality (9) and (10), we can write
(11) k1( () ()) − 1( 0 0)k()(
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
(12) k1( 0 0)k()(1) ≤ 11+ 01 Similarly we can obtain
k1( () ()) − 1( 0 0)k()(
1) ≤ 2 [11+ (12+ 13) ]
(13) k1( 0 0)k()(
1) ≤ 11+ 01
Using the inequalities (11), (12), (13) for every ∈ , we obtain
| ˜ ()| ≤ (12+ 13) + 14 kk[311+ 2 (12+ 13) + 01] +15 ° ° ° ° ° Y ° ° ° ° ° [311+ 2 (12+ 13) + 01] + 01
Now, let us obtain the Holder constant ( ˜ ) For every 1 2∈ | ˜ (1) − ˜ (2)| ≤ ¯ ¯ ¯ ¯ ¯1 à 1 (1) (1) 1( () ()) (1) Y 1( () ()) ! (1) − 1 à 2 (2) (2) 1( () ()) (2) Y 1( () ()) (2) !¯¯ ¯ ¯ ¯ ≤ 11|1− 2|+ 12| (1) − (2)| + 13| (1) − (2)| +14|[1( () ()) (1) − 1( () ()) (2)] | +15 ¯ ¯ ¯ ¯ ¯ Y [1( () ()) (1) − 1( () ()) (2)] ¯ ¯ ¯ ¯ ¯ ≤ [11+ (12+ 13) ] |1− 2|+ 14kkk1( () ())k|1− 2| +15 ° ° ° ° ° Y ° ° ° ° ° k1( () ())k|1− 2|
Moreover, for every 1 2∈
|1( () ())| ≤ |1( () ()) − 1( 0 0)| + |1( 0 0)| ≤ (12+ 13) + 01
From the definition of (7)
|1(1 (1) (1)) − 1(2 (2) (2))| ≤ [11+ (12+ 13) ] |1− 2| Thus k1( () ())k≤ [01+ 11+ 2 (12+ 13) ] Similarly |1( () ())| ≤ |1( () ()) − 1( 0 0)| + |1( 0 0)| ≤ (12+ 13) + 01 and thus k1( () ())k≤ [01+ 11+ 2 (12+ 13) ] Hence, for any 1 2∈
| ˜ (2) − ˜ (1)| ≤][11+ (12+ 13) + 14kk(01+ 11+ 2 (12+ 13) ) +15 ° ° ° ° ° Y ° ° ° ° ° (01+ 11+ 2 (12+ 13) ) # |1− 2|. or we obtain ( ˜ ) = [11+ (12+ 13) + 14kk(01+ 11+ 2 (12+ 13) ) +15 ° ° ° ° ° Y ° ° ° ° ° (01+ 11+ 2 (12+ 13) ) # Thus, for 1 = 01+ 11+ 2 (12+ 13) + 14kk[201+ 411+ 4 (12+ 13) ] +15 ° ° ° ° ° Y ° ° ° ° ° [201+ 411+ 4 (12+ 13) ] we get k ˜k≤ 1 In a similar way, for
2 = 02+ 21+ 2 (22+ 23) + 24kk[202+ 421+ 4 (22+ 23) ] +25 ° ° ° ° ° Y ° ° ° ° ° [202+ 421+ 4 (22+ 23) ]
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. ([1]) If ∈ 1¡¢, then the ball
( ) is compact in 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 ¡ ¢x()2¡¢→ [0 ∞) define metrics on ()2¡¢ Thus,¡()2¡¢ ;
∞2 ¢
and¡()2¡¢ ; 2
¢ become metric spaces.
Lemma 4. ([1]) Let 0 1 and 1 ≤ ∞. Then the convergence on the ball ( ) with respect to the metrics ∞2 and 2 are equivalent.
Lemma 5. Let ∈ 111 ¡ 1 2 3 4 5; ¢ and ∈ 11 ¡ 1 2 3; 1 ¢ ∈ 11 ¡ 1 2 3; 1 ¢ , ( = 1 2) (0 1)
and 1 ∞. In this case, for the operator A defined in Lemma 2, the inequality
(14) 2[ (1 1) (2 2)] ≤ 3() ∞2[(1 1) (2 2)] is satisfied for all (1 1) (2 2) ∈ ( ), where
3() = () 1 max ⎧ ⎨ ⎩12+ 13+ ⎛ ⎝14kk+ 15 ° ° ° ° ° Y ° ° ° ° ° ⎞ ⎠ (12+ 13) 22+ 23+ ⎛ ⎝24kk+ 25 ° ° ° ° ° Y ° ° ° ° ° ⎞ ⎠ (22+ 23) ⎫ ⎬ ⎭ .
Proof. For all (1 1) (2 2) ∈ ( ) and ∈ using
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− ˜2kand ° ° °˜1− ˜2 ° ° °
For all ∈ since | ˜1() − ˜2()| = ¯ ¯ ¯ ¯ ¯1 à 1() 1() 1( 1() 1()) () Y 1( 1() 1()) () ! −1 à 2() 2() 1( 2() 2()) () Y 1( 2() 2()) () !¯¯¯ ¯ ¯ ≤ 12|1() − 2()| + 13|1() − 2()| +14|(1( 1() 1()) − 1( 2() 2())) ()| +15 ¯ ¯ ¯ ¯ ¯ Y (1( 1() 1()) − 1( 2() 2())) () ¯ ¯ ¯ ¯ ¯
from the Minkowski inequality and 1 1 ∈ 11 ¡ 11 12 13; 1 ¢ we ob-tain ⎛ ⎝ZZ | ˜1() − ˜2()| ⎞ ⎠ 1 ≤ ⎧ ⎨ ⎩ ZZ [12|1() − 2()| + +13|1() − 2()| + 14|[1( 1() 1()) − 1( 2() 2())]| +15 ¯ ¯ ¯ ¯ ¯ Y [1( 1() 1()) − 1( 2() 2())] ¯ ¯ ¯ ¯ ¯ )1 ≤ 12k1− 2k+ 13k1− 2k+ +14kkk1( 1() 1()) − 1( 2() 2())k +15 ° ° ° ° ° Y ° ° ° ° ° k1( 1() 1()) − 1( 2() 2())k ⎞ ⎠ ()1 ≤ ³(12+ 13) max n k1− 2k k1− 2k o +14kk ³ 12k1− 2k+ 13k1− 2k ´ +15 ° ° ° ° ° Y ° ° ° ° ° ³ 12k1− 2k+ 13k1− 2k ´⎞ ⎠ ()1 ≤ ⎛ ⎝ ⎡ ⎣(12+ 13) + ⎛ ⎝14kk(12+ 13) + 15 ° ° ° ° ° Y ° ° ° ° ° (12+ 13) ⎞ ⎠ × ⎤ ⎦ × maxnk1− 2k k1− 2k o´ ()1 ≤ ()1 ⎡ ⎣12+ 13+ ⎛ ⎝14kk(12+ 13) + 15 ° ° ° ° ° Y ° ° ° ° ° (12+ 13) ⎞ ⎠ ⎤ ⎦ × ×∞2[(1 1) (2 2)] . Thus we get (15) k ˜1− ˜2k≤ () 1 h 12+ 13+ ³ 14kk(12+ 13) +15kQk(12+ 13) ´i ˆ ∞2
and similarly (16) ° ° °˜1− ˜2 ° ° ° ≤ () 1 h 22+ 23+ ³ 24kk(22+ 23) ´i +25kQk(22+ 23) ´ ˆ ∞2i
where ˆ∞2 = ∞2[(1 1) (2 2)] With the help of the inequalities (15) and (16), the required inequality (14) is obtained.
Lemma 6. ([1]) Assume the conditions of the Lemma 2 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. Theorem 2. ([1]) Let ∈ 111¡1 2 3 4; ¢,
∈ 11¡1 2 3; 1¢, ∈ 11¡1 2 3; 1¢( = 1 2) and max {1 2} ≤
The nonlinear singular integral system (5) has at least one solution on the sphere ( ) Now, let us study the uniqueness of the solution of the system (5) and how to find it. For this, we use a variant of Banach fixed point theorem: Theorem 3.([2]).Assume that the following hypothesis 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 in 2.
(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 ∈ 1111 ¡ 1 2 3 4 5; ¢ ∈ 11 ¡ 1 2 3; 1 ¢ ∈ 11 ¡ 1 2 3; 1 ¢ , ( = 1 2) (0 1), max {1 2} ≤
and = max ⎧ ⎨ ⎩01+ 13+ ⎛ ⎝14kk+ 15 ° ° ° ° ° Y ° ° ° ° ° ⎞ ⎠ (12+ 13) 22+ 23+ ⎛ ⎝24kk+ 25 ° ° ° ° ° Y ° ° ° ° ° ⎞ ⎠ (22+ 23) ⎫ ⎬ ⎭ 1 hold. Then the system (5) of nonlinear singular integral equations has a unique solution (∗ ∗) ∈ ( ) This solution is the limit of the sequence ( ) defined by (17) () = 1 Ã −1() −1() 1( −1() −1()) Q 1( −1() −1()) ! () () = 2 Ã −1() −1() 2( −1() −1()) Q 2( −1() −1()) ! () = 1 2 where (0 0) ∈ ( ) is any initial element. Moreover, the inequality
2[( ) (∗ ∗)] ≤
1 − 2[(1 1) (0 0)] hold.
Proof. Let = ( ), 1= 2and 2= 2in Theorem 3. Let be an operator defined in the Lemma 6. Since
max {1 2} ≤
, the operator transform the space ( 2) into itself. Now let us show that when 1 the operator is a contraction operator on the sphere ( ) with respect to the metric 2 .For
( ) ∈ ( ) ( ) () = (1( ) () 2( ) ()) where 1( ) () = 1 Ã () () 1( () ()) Y 1( () ()) () ! 2( ) () = 2 Ã () () 2( () ()) Y 2( () ()) () !
Thus for any¡(1) (1)¢¡(2) (2)¢∈ (; ) we can write
2 ³ ³(1) (1)´ ³(2) (2)´´= max ½ ° ° °1 ³ (1) (1)´− 1 ³ (2) (2)´°°°
° ° °2 ³ (1) (1)´− 2 ³ (2) (2)´°°° ¾ Since 1∈ 1111¡11 12 13 14 15; ¢ 1∈ 11¡11 12 13; 1¢ 1∈ 11 ¡ 11 12 13; 1 ¢ ° ° °1 ³ (1) (1)´− 1 ³ (2) (2)´°°° = ° ° ° ° °1 à (1) (1) 1 ³ (1) (1)´Y 1 ³ (1) (1)´ ! − −1 à (2) (2) 1 ³ (2) (2)´Y 1 ³ (2) (2) ´!°° ° ° ° = ⎧ ⎨ ⎩ ZZ [12 ¯ ¯ ¯(1) () − (2)() ¯ ¯ ¯ + 13 ¯ ¯ ¯(1) () − (2)() ¯ ¯ ¯ + +14 ¯ ¯ ¯ ³ 1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´()¯¯¯ +15 ¯ ¯ ¯ ¯ ¯ Y ³ 1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´() ¯ ¯ ¯ ¯ ¯ # )1 ≤ 12 ° ° °(1)− (2)°°° + 13 ° ° °(1)− (2)°°° +14 ⎛ ⎝ZZ ¯ ¯ ¯ ³ 1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´()¯¯¯ ⎞ ⎠ 1 +15 ⎛ ⎝ZZ ¯ ¯ ¯ ¯ ¯ Y ³ 1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´() ¯ ¯ ¯ ¯ ¯ ⎞ ⎠ 1 ≤ 12 ° ° °(1)− (2)°°°+ 13 ° ° °(1)− (2)°°° +14kk ⎛ ⎝ZZ ¯ ¯ ¯³1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´¯¯¯ ⎞ ⎠ 1
+15 ° ° ° ° ° Y ° ° ° ° ° ⎛ ⎝ZZ ¯ ¯ ¯³1 ³ (1)() (1)()´− 1 ³ (2)() (2)()´´¯¯¯ ⎞ ⎠ 1 ≤ 12 ³³ (1) (1)´³(2) (2)´´ where 1= 12+ 13+ (12+ 13) 14kk+ (12+ 13) 15 ° ° ° ° ° Y ° ° ° ° ° Similarly ° ° °2 ³ (1) (1)´− 2 ³ (2) (2)´°°° ≤ 22 ³³ (1) (1)´³(2) (2)´´ where 2= 22+ 23+ (22+ 23) 24kk+ (22+ 23) 25 ° ° ° ° ° 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 Theorem 2, the system (17) has at least one solution in the ball ( ) Now let us show this fact: Since
( ) = (−1 −1) = 1 2 we obtain
2[(+1 +1) ( )] = 2[ ( ) (−1 −1)] ≤ 2[( ) (−1 −1)] Repeating this process consecutively, it follows that
2[(+1 +1) (0 0)] ≤ 2[(1 1) (0 0)] .
Thus, for any two natural numbers and we can write 2[(+ +) ( )] ≤ ¡ 1 + + 2+ + −1¢2[(1 1) (0 0)] . (18) = 1 − 1 − 2[(1 1) (0 0)]
Since lim →∞
= 0, the sequence {( )}∞1 is Cauchy by (18). 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 solution of the system (17). In this case, we can write
2[(∗ ∗) (∗∗ ∗∗)] = 2[ (∗ ∗) (∗∗ ∗∗)] ≤ 2[(∗ ∗) (∗∗ ∗∗)] However, this is possible only if 2[(∗ ∗) (∗∗ ∗∗)] = 0
Remark 1. Since, by (17), the sequence {( )}∞1 , whose terms are defined by
( ) = (−1 −1)
is convergent to the solution (∗ ∗) in the ball (; ) with respect to the metric 2 , it is also convergent with respect to the metric ∞2. Thus the metrics ∞2 and 2are equivalent on X.
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