For square nonsingular matrices P and Q, the following conditions hold:
- i.
If P ≥ Q > 0 then P−1 ≤ Q−1.
- ii.
The spectral norm is monotonic norm, i.e., if 0 < P ≤ Q, then ‖ P ‖ ≤ ‖ Q ‖.
- iii.
The mathematical expression P ≥ Q (P > Q) means that P − Q is a Hermitian positive semi-definite (definite) matrix and [P, Q ] = { Y : P ≤ Y ≤ Q }.
Definition 2.1: [33] Let H = (hij)r × s , then differentiating of the matrix H is defined by dH = (dhij)r × s . For example,
$$ let\ H=\left(\begin{array}{ccc}{v}^2& 3v-1& w+2v\\ {}{w}^2-3v& 2v& w+2\\ {}w-v& -5w& {w}^3+{v}^3\end{array}\right) $$
(2)
$$ Then, dH=\left(\begin{array}{ccc}2v\; dv& 3\; dv& dw+2\; dv\\ {}2w\; dw-3\; dv& 2\; dv& dw\\ {} dw- dv& -5 dw& 3{w}^2\; dw+3{v}^2\; dv\end{array}\right) $$
(3)
Lemma 2.1: [33]. The matrix differentiation has the following properties:
-
1)
d (H1 ± H2) = d H1 ± d H2 ;
-
2)
d (c H) = c (d H),where c is a complex number ;
-
3)
d (H∗) = (d H)∗;
-
4)
d (H1H2H3) = (d H1)H2H3 + H1(d H2)H3 + H1H2(dH3);
-
5)
d (H−1) = − H−1(d H) H−1;
-
6)
d H = 0, where H is a constant matrix and 0 is the zero matrix of the same dimension of H.
Theorem 2.1: [34]
Let (Y, ≤) be a partially ordered set granted with a metric space d such that (Y, d) is complete. Let G : Y × Y → Y be a continuous mapping with the mixed monotone property on Y. If there exists ε ∈ [0, 1), where \( d\;\left(\;G\left(y,z\right),G\left(v,w\right)\;\right)\le \frac{\varepsilon }{2}\;\left[\;d\;\left(y,v\right)+d\;\left(z,w\right)\;\right] \) for all (y, z), (v, w) ∈ Y × Y where y ≥ v and z ≤ w. Moreover, there exist y0, z0 ∈ Y such that y0 ≤ G (y0, z0) and z0 ≥ G (z0, y0). Then,
- (a)
G has a coupled fixed point \( \left(\tilde{y},\tilde{z}\right)\in Y\times Y \);
- (b)
The sequences { yk} and { zk} defined by yk + 1 = G ( yk, zk) and zk + 1 = G ( zk, yk) converge to \( \tilde{y} \) and \( \tilde{z} \), respectively ;
In addition, suppose that every pair of elements has a lower bound and an upper bound, then
- (c)
G has a unique coupled fixed point \( \left(\tilde{y},\tilde{z}\right)\in Y\times Y \);
- (d)
\( \tilde{y}=\tilde{z} \); and
- (e)
We have the following estimate:
$$ \max\;\left\{\;d\;\left({y}_k,\tilde{y}\right),d\;\left({z}_k,\tilde{y}\right)\;\right\}\le \frac{\varepsilon^k}{2\;\left(1-\varepsilon \right)}\;\left[\;d\;\left(G\left({y}_0,{z}_0\right),{y}_0\right)+d\;\right(\;G\left({z}_0,{y}_0\right),{z}_0\;\Big]. $$
Theorem 2.2 (Theorem of Schauder Fixed Point) [35]
Every continuous function g : T → T mapping T into itself has a fixed point, whereT be a nonempty compact convex subset of a normed vector space.
Suppose that the set of matrices Ψ defined by \( \varPsi =\left\{\;X\in H(M):X\ge \frac{1}{2}I\;\right\} \).
let the mapping G : Ψ × Ψ → Ψ associated with Eq. (1) is defined by
$$ G\left(X,Y\right)=I-\sum \limits_{j=1}^n{B}_j^{\ast }{X}^{-1}{B}_j+\sum \limits_{i=1}^m{A}_i^{\ast }{Y}^{-1}{A}_i $$
(4)
Theorem 2.3: [21] Suppose that the following assumptions hold
$$ \sum \limits_{j=1}^n{\left\Vert {B}_j\right\Vert}^2<\frac{1}{4^2},\sum \limits_{i=1}^m{\left\Vert\;{A}_i\right\Vert}^2<\frac{1}{4^2} $$
(5)
$$ 6\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j-2\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\le \frac{3}{2}I $$
(6)
$$ 6\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i-2\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j\le \frac{3}{2}I $$
(7)
Then Eq. (1) has a unique maximal solution XL with
$$ {X}_L\in \left[I-2\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j+\frac{2}{3}\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i,I-\frac{2}{3}\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j+2\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\right]. $$
Proof: [21] We demand that there exists (X, Y ) ∈ H(M) × H(M) a solution to the system
$$ X=I-\sum \limits_{j=1}^n{B}_j^{\ast }{X}^{-1}{B}_j+\sum \limits_{i=1}^m{A}_i^{\ast }{Y}^{-1}{A}_i, $$
$$ Y=I-\sum \limits_{j=1}^n{B}_j^{\ast }{Y}^{-1}{B}_j+\sum \limits_{i=1}^m{A}_i^{\ast }{X}^{-1}{A}_i $$
(8)
Now, taking \( {X}_0=\frac{1}{2}I \) and \( {Y}_0=\frac{3}{2}I \).
From condition (6) we have
\( 6\overset{n}{\sum \limits_{j=1}}{B}_j^{\ast }{B}_j-2\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\le \frac{3}{2}I \) then \( 2\overset{n}{\sum \limits_{j=1}}{B}_j^{\ast }{B}_j-\frac{2}{3}\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\le \frac{1}{2}I \) so we have \( G\left(\frac{1}{2}I,\frac{3}{2}I\right)=I-2\overset{n}{\sum \limits_{j=1}}{B}_j^{\ast }{B}_j+\frac{2}{3}\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\ge \frac{1}{2}I \) , that is, \( {X}_0=\frac{1}{2}I\le G\;\left(\frac{1}{2}I,\frac{3}{2}I\right) \).
Moreover from condition (7) we get
\( 6\overset{m}{\sum \limits_{i=1}}{A}_i^{\ast }{A}_i-2\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j\le \frac{3}{2}I \) then \( 2\overset{m}{\sum \limits_{i=1}}{A}_i^{\ast }{A}_i-\frac{2}{3}\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j\le \frac{1}{2}I \) so we have
\( G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)=I-\frac{2}{3}\overset{n}{\sum \limits_{j=1}}{B}_j^{\ast }{B}_j+2\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\le \frac{3}{2}I \) and \( {Y}_0=\frac{3}{2}I\ge G\;\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right). \)
From Theorem 2.1 (a), there exists (X, Y ) ∈ H(M) × H(M)where G(X, Y) = X and G(Y, X) = Y that is, (X, Y) is a solution to (8). On the other hand, for every X, Y ∈ H(M) there is a greatest lower bound and a least upper bound. Note also that the partial order G is a continuous mapping, by Theorem 2.1, (X, Y) is the unique coupled fixed point of G that is X = Y = XL.
Thus, the unique solution of Eq. (1) is XL.
Now, using the Theorem of Schauder Fixed Point, we state the mapping \( F:\left[G\;\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\;\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right]\to \varPsi \) by
$$ F\left({X}_L\right)=G\left({X}_L,{X}_L\right)=I-\sum \limits_{j=1}^n{B}_j^{\ast }{X}_L^{-1}{B}_j+\sum \limits_{i=1}^m{A}_i^{\ast }{X}_L^{-1}{A}_i, $$
For all \( {X}_L\in \left[G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right]. \)
We want to prove that
$$ F\kern0.36em \left(\left[G\;\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\;\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right]\right)\subseteq \left[G\;\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\;\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right]. $$
Let \( {X}_L\in \left[G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right] \) , that is \( G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right)\le {X}_L\le G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right) \).
Applying the property of mixed monotone of G yields that
$$ G\;\left(G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right)\le F\left({X}_L\right)=G\left({X}_L,{X}_L\right)\le G\;\left(G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right),G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right)\right), $$
since \( G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right)\ge \frac{1}{2}I \) and \( G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\le \frac{3}{2}I \).
Applying the property of the mixed monotone of G again implies that
$$ G\;\left(G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right)\ge G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right), $$
(9)
$$ G\;\left(G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right),G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right)\right)\le G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right). $$
(10)
From (9) and (10), it follows that
$$ G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right)\le F\left({X}_L\right)\le G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right). $$
Thus, our claim that
$$ F\kern0.36em \left(\left[G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right]\right)\subseteq \left[G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right].\mathrm{holds} $$
Now, we have a continuous mapping F that maps the compact convex set \( \left[G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right] \) into itself, from Schuader fixed point theorem we get that F has at least one fixed point in this set, but a fixed point of F is a solution of Eq. (1), and we proved already that Eq. (1) has a unique solution in Ψ. Thus, this solution must be in the set \( \left[G\left(\frac{1}{2}I,\kern0.36em \frac{3}{2}I\right),G\left(\frac{3}{2}I,\kern0.36em \frac{1}{2}I\right)\right] \). That is,
$$ {X}_L\in \left[I-2\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j+\frac{2}{3}\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i,I-\frac{2}{3}\sum \limits_{j=1}^n{B}_j^{\ast }{B}_j+2\sum \limits_{i=1}^m{A}_i^{\ast }{A}_i\right]. $$
Which completes the proof of the theorem.