4. Existence of solution

In this section, by using the fixed point results proved in the previous section, we obtain the existence of the solution of integral Eq. (2) and matrix Eq. (1).

### 4.1 Solution of Fredholm integral equation of second kind

Let <sup>≪</sup> be a partial order relation on IR<sup>n</sup>. Define <sup>T</sup> : <sup>X</sup> ! <sup>X</sup> by

$$\mathcal{T}\mathbf{z}(r) = \int\_{b}^{c} \mathcal{B}(r, s, \mathbf{z}(s)) \mathbf{ds} + \mathbf{g}(r), \qquad r \in [a, b]. \tag{56}$$

Theorem 4.1 Let <sup>X</sup> <sup>¼</sup> C b½ � ;<sup>c</sup> ;IR<sup>n</sup> ð Þ with the usual spermium norm. Suppose that 1. <sup>B</sup> : ½ �� <sup>b</sup>;<sup>c</sup> ½ �� <sup>b</sup>;<sup>c</sup> IR<sup>n</sup> ! IR<sup>n</sup> and <sup>g</sup> : IR<sup>n</sup> ! IR<sup>n</sup> are continuous;

2. there exists a continuous function p : ½ �� b;c ½ �! b;c ½ � b;c such that

$$|\mathcal{B}(r,s,u) - \mathcal{B}(r,s,v)| \le p(r,s)|u - v|,\tag{57}$$

Proof. Define T : H nð Þ! H nð Þ and F : IRþ ! IR by

Contraction Mappings and Applications DOI: http://dx.doi.org/10.5772/intechopen.81571

d TX ð Þ¼ ; TY k k T Y � T X <sup>1</sup>

¼ ∑ m i¼1

¼ tr ∑ m i¼1 AiA<sup>∗</sup> i � �

> ∑ m i¼1 AiA<sup>∗</sup> i

, m Yð Þ ;X e

≤ ∑<sup>m</sup> i¼1 AiA<sup>∗</sup> i

� � � �

� � � �

� � � �

� � � �

≤

have

and so,

ln k k T Y � T X <sup>1</sup>

This implies that

Consequently,

Also, from ∑<sup>m</sup>

23

� � , ln m Yð Þ ;<sup>X</sup> <sup>e</sup>

1 k k Y � X <sup>1</sup>

<sup>i</sup>¼<sup>1</sup>A<sup>∗</sup>

þ 1

T ð Þ¼ X Q þ ∑

X, Y <sup>∈</sup> H nð Þ with <sup>X</sup> <sup>≼</sup>Y, then <sup>γ</sup>ð Þ <sup>X</sup> <sup>≼</sup>γð Þ<sup>y</sup> . So, for d Xð Þ ; <sup>Y</sup> . 0 and <sup>τ</sup>ðÞ¼ <sup>t</sup> <sup>1</sup>

¼ trð Þ T Y � T X

tr AiA<sup>∗</sup>

� � � �

� � � �

� � � �

� � � � <sup>N</sup> m Yð Þ ;<sup>X</sup> <sup>e</sup>

� <sup>2</sup>þk k <sup>Y</sup>�<sup>X</sup> <sup>1</sup> <sup>2</sup>k k <sup>Y</sup>�<sup>X</sup> <sup>1</sup> � �

� <sup>2</sup>þk k <sup>Y</sup>�<sup>X</sup> <sup>1</sup> <sup>2</sup>k k <sup>Y</sup>�<sup>X</sup> <sup>1</sup> ! � �

<sup>2</sup> <sup>þ</sup> ln k k <sup>T</sup> <sup>Y</sup> � <sup>T</sup> <sup>X</sup> <sup>1</sup>

τð Þþ d Xð Þ ; Y Fð Þ d TX ð Þ ; TY , Fð Þ m Xð Þ ; Y :

we conclude that <sup>T</sup> has a fixed point and hence (1) has a solution in P nð Þ. □

m i¼1 A∗

and F rð Þ¼ ln r respectively. Then a fixed point of T is a solution of (1). Let

<sup>i</sup> ð Þ <sup>γ</sup>ð Þ� <sup>Y</sup> <sup>γ</sup>ð Þ <sup>X</sup> � �

� �

k k γð Þ� Y γð Þ X <sup>1</sup>

,

ð Þ γð Þ� Y γð Þ X

� <sup>2</sup>þk k <sup>Y</sup>�<sup>X</sup> <sup>1</sup> <sup>2</sup>k k <sup>Y</sup>�<sup>X</sup> <sup>1</sup> � �

� � , ln ð Þ m Xð Þ ; <sup>Y</sup> :

<sup>i</sup> γð Þ Q Ai≻0, we have Q ≼T ð Þ Q . Thus, by using Theorem 3.24,

<sup>i</sup> γð Þ X Ai (60)

<sup>¼</sup> ln ð Þ� m Xð Þ ; <sup>Y</sup> <sup>2</sup> <sup>þ</sup> k k <sup>Y</sup> � <sup>X</sup> <sup>1</sup>

2k k Y � X <sup>1</sup>

:

<sup>t</sup> <sup>þ</sup> <sup>1</sup> 2 , we

for each r, s∈½ � <sup>b</sup>;<sup>c</sup> and u, v<sup>∈</sup> IR<sup>n</sup> with <sup>u</sup> <sup>≪</sup> <sup>v</sup>.

3. sup<sup>r</sup> <sup>∈</sup>½ � <sup>b</sup>;<sup>c</sup> Ð c <sup>b</sup> p rð Þ ; <sup>s</sup> ds <sup>¼</sup> <sup>q</sup> <sup>≤</sup> <sup>1</sup> 4;

4. there exists z<sup>0</sup> ∈ X and z<sup>1</sup> ∈T z<sup>0</sup> such that z<sup>0</sup> ≼z1;

5. for a sequence f g zn ⊂ X, limn!<sup>∞</sup>f g zn ¼ z and zn ≼znþ<sup>1</sup> for all n∈ IN, we have zn ≼z for all n∈ IN.
