Then the integral Eq. (2) has a solution in X.

Proof. Let <sup>X</sup> <sup>¼</sup> C b½ � ;<sup>c</sup> ;IR<sup>n</sup> ð Þ and <sup>k</sup>zk ¼ maxr∈½ � <sup>b</sup>;<sup>c</sup> <sup>∣</sup>z rð Þ∣, for <sup>z</sup>∈C a ð Þ ½ � ; <sup>b</sup> . Consider a partial order defined on X by

y, z∈C b½ � ;<sup>c</sup> ;IR<sup>n</sup> ð Þ, y <sup>≼</sup><sup>z</sup> if and only if y rð Þ <sup>≪</sup> z rð Þ, for r∈½ � <sup>b</sup>;<sup>c</sup> : (58)

Then ð Þ X; k:k; ≼ is a complete partial ordered metric space and for any increasing sequence f g zn in X converging to z∈ X, we have znð Þr ≪ z rð Þ for any r∈½ � b;c (see [36]). By using Eq. (56), conditions (2, 3) and taking <sup>ϑ</sup>ð Þ¼ <sup>r</sup>; <sup>s</sup> <sup>1</sup> <sup>2</sup> s � r for all y, z∈ X with y≼z, we obtain

$$\begin{split} |T\mathbf{y}(r) - Tz(r)| &= \left| \int\_{b}^{c} \mathcal{B}(r, s, \mathbf{y}(s)) \mathbf{ds} - \int\_{b}^{c} \mathcal{B}(r, s, z(s)) \mathbf{ds} \right| \\ &\leq \int\_{b}^{c} |\mathcal{B}(r, s, \mathbf{y}(s)) - \mathcal{B}(r, s, z(s))| \mathbf{ds} \\ &\leq \int\_{b}^{c} p(r, s)|\mathbf{y}(s) - z(s)| \mathbf{ds} \\ &\leq \frac{1}{4} ||\mathbf{y} - z||. \end{split}$$

This implies that

$$\frac{1}{2}||y-z||-||Ty-Tz||\quad \geq \frac{1}{2}||y-z||-\frac{1}{4}||y-z|| = \frac{1}{4}||y-z||.$$

So ϑð Þ dð Þ T y; T z ; d yð Þ ; z ≥0 for all y, z∈ X with y≼z. Hence all the conditions of Theorem 3.18 are satisfied. Therefore T has a fixed point, consequently, integral Eq. (2) has a solution in <sup>X</sup>. □

### 4.2 Solution of non-linear matrix equation

Theorem 4.2 Let γ : H nð Þ! H nð Þ be an order-preserving mapping which maps P nð Þ into P nð Þ and Q ∈P nð Þ. Assume that there exists a positive number N for which ∑<sup>m</sup> <sup>i</sup>¼<sup>1</sup>AiA<sup>∗</sup> <sup>i</sup> <sup>≺</sup>NIn and <sup>∑</sup><sup>m</sup> <sup>i</sup>¼<sup>1</sup>A<sup>∗</sup> <sup>i</sup> γð Þ Q Ai≻0 such that for all X ≼ Y we have

$$d(\gamma(X), \gamma(Y)) \le \frac{1}{N} m(Y, X) e^{-\left(\frac{2 + d(X, Y)}{2d(X, Y)}\right)},\tag{59}$$

where

$$m(X,Y) = \max\left\{d(X,Y), d(X,TX), d(Y,TY), \frac{d(Y,TY) + d(X,TX)}{2}, 0\right\}$$

$$\frac{d(Y,TY)[1+d(X,TX)]}{1+d(X,Y)}, \frac{d(Y,TX)[1+d(X,TY)]}{1+d(X,Y)}$$

Then (1) has a solution in P nð Þ.

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

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

$$T(X) = Q + \sum\_{i=1}^{m} A\_i^\* \boldsymbol{\gamma}(X) A\_i \tag{60}$$

and F rð Þ¼ ln r respectively. Then a fixed point of T is a solution of (1). Let 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> <sup>t</sup> <sup>þ</sup> <sup>1</sup> 2 , we have

$$\begin{split}d(\mathbf{TX}, \mathbf{TY}) &= \|(\mathbf{T}\mathbf{Y} - \mathbf{T}\mathbf{X})\|\_{1} \\ &= \text{tr}(\mathbf{T}\mathbf{Y} - \mathbf{T}\mathbf{X}) \\ &= \sum\_{i=1}^{m} \text{tr}\big(A\_{i}A\_{i}^{\*}(\boldsymbol{\gamma}(\mathbf{Y}) - \boldsymbol{\gamma}(\mathbf{X}))\big) \\ &= \text{tr}\big(\left(\sum\_{i=1}^{m} A\_{i}A\_{i}^{\*}\right)(\boldsymbol{\gamma}(\mathbf{Y}) - \boldsymbol{\gamma}(\mathbf{X}))\big) \\ &\leq \left|\big|\sum\_{i=1}^{m} A\_{i}A\_{i}^{\*}\right| \left||\boldsymbol{\gamma}(\mathbf{Y}) - \boldsymbol{\gamma}(\mathbf{X})\|\_{1} \\ &\leq \frac{\left|\sum\_{i=1}^{m} A\_{i}A\_{i}^{\*}\right|}{N} \text{m}(\boldsymbol{\gamma}, \mathbf{X})\big{e^{-\left(\frac{2+\|\mathbf{Y}-\mathbf{X}\|\_{1}}{2\|\mathbf{Y}-\mathbf{X}\|\_{1}}\right)}}\text{f} \\ &\leq m(\mathbf{Y}, \mathbf{X})e^{-\left(\frac{2+\|\mathbf{Y}-\mathbf{X}\|\_{1}}{2\|\mathbf{Y}-\mathbf{X}\|\_{1}}\right)}\text{,\end{split}$$

and so,

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

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

Proof. Let <sup>X</sup> <sup>¼</sup> C b½ � ;<sup>c</sup> ;IR<sup>n</sup> ð Þ and <sup>k</sup>zk ¼ maxr∈½ � <sup>b</sup>;<sup>c</sup> <sup>∣</sup>z rð Þ∣, for <sup>z</sup>∈C a ð Þ ½ � ; <sup>b</sup> . Consider a

y, z∈C b½ � ;<sup>c</sup> ;IR<sup>n</sup> ð Þ, y <sup>≼</sup><sup>z</sup> if and only if y rð Þ <sup>≪</sup> z rð Þ, for r∈½ � <sup>b</sup>;<sup>c</sup> : (58)

Then ð Þ X; k:k; ≼ is a complete partial ordered metric space and for any increasing sequence f g zn in X converging to z∈ X, we have znð Þr ≪ z rð Þ for any r∈½ � b;c

> <sup>b</sup> <sup>B</sup>ðr; <sup>s</sup>; z sð ÞÞd<sup>s</sup> � � �

<sup>b</sup> ∣Bð Þ� r; s; y sð Þ Bð Þ r; s; z sð Þ ∣ds

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;

(see [36]). By using Eq. (56), conditions (2, 3) and taking <sup>ϑ</sup>ð Þ¼ <sup>r</sup>; <sup>s</sup> <sup>1</sup>

<sup>b</sup> <sup>B</sup>ðr; <sup>s</sup>; y sð ÞÞd<sup>s</sup> � <sup>Ð</sup> <sup>c</sup>

<sup>b</sup> p rð Þ ; s ∣y sðÞ� z sð Þ∣ds

1 2 <sup>k</sup><sup>y</sup> � <sup>z</sup>k � <sup>1</sup>

So ϑð Þ dð Þ T y; T z ; d yð Þ ; z ≥0 for all y, z∈ X with y≼z. Hence all the conditions of Theorem 3.18 are satisfied. Therefore T has a fixed point, consequently, integral Eq. (2) has a solution in <sup>X</sup>. □

Theorem 4.2 Let γ : H nð Þ! H nð Þ be an order-preserving mapping which maps P nð Þ into P nð Þ and Q ∈P nð Þ. Assume that there exists a positive number N for which

1

<sup>i</sup> γð Þ Q Ai≻0 such that for all X ≼ Y we have

� <sup>2</sup>þd Xð Þ ;<sup>Y</sup> 2d Xð Þ ;Y � �

d Yð Þ ; T X ½ � 1 þ d Xð Þ ; T Y 1 þ d Xð Þ ; Y

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

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>.

<sup>b</sup> p rð Þ ; <sup>s</sup> ds <sup>¼</sup> <sup>q</sup> <sup>≤</sup> <sup>1</sup>

Then the integral Eq. (2) has a solution in X.

≤ Ð <sup>c</sup>

≤ Ð <sup>c</sup>

<sup>4</sup> <sup>k</sup><sup>y</sup> � <sup>z</sup>k:

≤ 1

ky � zk�kT y � T zk ≥

4.2 Solution of non-linear matrix equation

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

�

Then (1) has a solution in P nð Þ.

dð Þ γð Þ X ; γð Þ Y ≤

m Xð Þ¼ ; Y max d Xð Þ ; Y ; dðX; T XÞ; dðY; T YÞ;

d Yð Þ ; T Y ½ � 1 þ d Xð Þ ; T X <sup>1</sup> <sup>þ</sup> d Xð Þ ; <sup>Y</sup> ;

<sup>i</sup> <sup>≺</sup>NIn and <sup>∑</sup><sup>m</sup>

3. sup<sup>r</sup> <sup>∈</sup>½ � <sup>b</sup>;<sup>c</sup>

Ð c

Recent Advances in Integral Equations

zn ≼z for all n∈ IN.

partial order defined on X by

y, z∈ X with y≼z, we obtain

This implies that

1 2

∑<sup>m</sup> <sup>i</sup>¼<sup>1</sup>AiA<sup>∗</sup>

22

where

<sup>∣</sup><sup>T</sup> y rð Þ� <sup>T</sup> z rð Þ<sup>∣</sup> <sup>¼</sup> <sup>Ð</sup> <sup>c</sup>

j j Bð Þ� r; s; u Bð Þ r; s; v ≤p rð Þ ; s ∣u � v∣, (57)

<sup>2</sup> s � r for all

�

<sup>4</sup> <sup>k</sup><sup>y</sup> � <sup>z</sup>k ¼ <sup>1</sup>

<sup>4</sup> <sup>k</sup><sup>y</sup> � <sup>z</sup>k:

, (59)

<sup>2</sup> ;

� :

d Yð Þþ ; T Y d Xð Þ ; T X

$$\ln\left(\|\mathcal{T}Y - \mathcal{T}X\|\_1\right) < \ln\left(m(Y,X)e^{-\left(\frac{2+\|Y-X\|\_1}{2\|Y-X\|\_1}\right)}\right) = \ln\left(m(X,Y)\right) - \frac{2+\|Y-X\|\_1}{2\|Y-X\|\_1}.$$

This implies that

$$\frac{1}{||Y - X||\_1} + \frac{1}{2} + \ln\left(||TY - TX||\_1\right) \le \ln\left(m(X, Y)\right).$$

Consequently,

$$
\sigma(d(X, Y)) + \mathcal{F}(d(TX, TY)) \le \mathcal{F}(m(X, Y)).
$$

Also, from ∑<sup>m</sup> <sup>i</sup>¼<sup>1</sup>A<sup>∗</sup> <sup>i</sup> γð Þ Q Ai≻0, we have Q ≼T ð Þ Q . Thus, by using Theorem 3.24, we conclude that <sup>T</sup> has a fixed point and hence (1) has a solution in P nð Þ. □ Recent Advances in Integral Equations
